INTRODUCTION TO STRUCTURAL MOTION CONTROL J. J. Connor
Contents i
Contents Preface
x
Chapter 1
Introduction 1.1 Mo...

Author:
Jerome J. Connor

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INTRODUCTION TO STRUCTURAL MOTION CONTROL J. J. Connor

Contents i

Contents Preface

x

Chapter 1

Introduction 1.1 Motivation for structural motion control 1 • Limitations of conventional structural design 1 • Motion based structural design and motion control 3 1.2 Motion versus strength issues for building type structures 4 • Example 1.1: Cantilever shear beam 6 • Example 1.2: Cantilever bending beam 7 • Example 1.3: Quasi-shear beam frame 9 1.3 Design of a single-degree-of freedom system for dynamic loading 11 • Response for periodic excitation 11 • Design criteria 16 • Methodology for acceleration controlled design 17 • Example 1.4: An illustration of acceleration controlled design 20 • Methodology for displacement controlled design 21 • Example 1.5: An illustration of displacement controlled design 21 • Methodology for force controlled design 23 • Example 1.6: Force reduction 25 1.4 Design of a single-degree-of-freedom system for support motion 26 • Response for periodic support motion 26 • Design scenarios 27 • Example 1.7: Controlling acceleration due to ground motion 27 • Example 1.8: Controlling relative motion due to ground motion 28 1.5 Stiffness distribution for a two degree-of-freedom system 30 • Example 1.9: 2DOF system - with equal masses 33 1.6 Control strategies for motion based design 33 1.7 Scope of text 42 Problems 45

ii Contents

Part I: Passive Control

Chapter 2

Optimal stiffness distribution 2.1 Introduction 51 2.2 Governing equations - planar beam 53 • Planar deformation-displacement relations 54 • Optimal deformation and displacement profiles 55 • Equilibrium equations 57 • Force-deformation relations 58 • Example 2.1: Composite sandwich beam 59 • Example 2.2: Equivalent rigidities for a discrete truss beam 60 • Governing equations for buildings modelled as pseudo shear beams 64 2.3 Stiffness distribution for a continuous cantilever beam under static loading 69 • Example 2.3: Cantilever beam - quasi static seismic loading 71 • Example 2.4: Truss-beam revisited 72 2.4 Stiffness distribution for a discrete cantilever shear beam - static loading 75 • Example 2.5: 3DOF shear beam 76 2.5 Stiffness distribution - truss under static loading 77 • Example 2.6: Application of least square approaches 81 • Example 2.7: Comparison of strength vs displacement based design 89 2.6 Stiffness distribution for a cantilever beam - dynamic response 93 • Rigidity distribution - undamped response 93 2.7 Stiffness distribution for a discrete shear beam - dynamic response 98 • Example 2.8: 3DOF shear beam 99 2.8 Stiffness calibration 100 • Governing equation - fundamental mode response of a discrete shear beam 100 • Example 2.9: 3DOF shear beam revisited 101 • Governing equations - fundamental mode response of a continuous beam 105 • Example 2.10: One-dimensional parameters - continuous beam108 • Stiffness calibration - periodic excitation 109 • Example 2.11: 3DOF shear beam revisited again 111 • Example 2.12: Stiffness calibration - continuous beam 113 • Example 2.13: Example 2.12 revisited 114

Contents iii • Stiffness calibration - seismic excitation • Seismic response spectra • Example 2.14: Example 2.12 revisited for seismic excitation • Example 2.15: 5DOF shear beam 2.9 Examples of stiffness distribution for buildings 2.10 Stiffness modification • Iterative procedure - seismic excitation • Multiple mode response • Building examples - iterated stiffness distribution Problems

116 119 128 129 132 139 139 140 143 149

Chapter 3

Optimal passive damping distribution 3.1 Introduction 167 3.2 Viscous, frictional, and hysteretic damping 171 • Viscous damping 171 • Example 3.1: Viscous damper 172 • Friction damping 174 • Hysteretic damping 176 • Example 3.2: Stiffness of a rod hysteretic damper 177 • Example 3.3: Stiffness of two hysteretic dampers in series 178 3.3 Viscoelastic material damping 179 • Example 3.4: Viscoelastic damper 182 3.4 Equivalent viscous damping 184 • Example 3.5: Structural and hysteretic damping comparison seismic excitation 186 • Example 3.6: Determining α d for 3M ISD110 damping material 193 3.5 Damping parameters - discrete shear beam 193 • Damping systems 193 • Rigid structural members - Linear viscous behavior 197 • Example 3.7: Example 2.15 revisited 198 • Rigid structural members - Linear viscoelastic behavior 200 • Example 3.8: 201 • Example 3.9: Example 3.7 revisited 202 • Example 3.10: Viscoelastic damper design 203 • Example 3.11: Hysteretic damper design - diagonal element 205 • Flexible structural members - linear viscoelastic behavior 206 • Example 3.12: Coupled spring-damper model 209 3.6 Damping parameters - truss beam 211 • Linear viscous behavior 213 • Linear viscoelastic behavior 214 3.7 Damping distribution for MDOF systems 215

iv Contents • Multi mode free vibration response • Example 3.13: Eigenvalue problem - 2DOF • Example 3.14: Modal response for nonproportional damping • Stiffness proportional viscous damping • Examples - low rise buildings • Examples - building #4 Problems

216 219 226 229 232 239 245

Chapter 4

Tuned mass damper systems 4.1 Introduction 261 4.2 An introductory example 262 • Example 4.1: Preliminary design of a TMD for a SDOF system 264 4.3 Examples of existing tuned mass damper systems 266 • Translational tuned mass dampers 266 • Pendulum tuned mass damper 272 4.4 Tuned mass damper theory for SDOF systems 277 • Undamped structure - undamped TMD 277 • Undamped structure - damped TMD 280 • Example 4.2: Design of a TMD for an undamped SDOF system293 • Damped structure - damped TMD 293 • Example 4.3: Design of a TMD for a damped SDOF system 299 4.5 Case studies - SDOF systems 300 4.6 Tuned mass damper theory for MDOF systems 307 • Example 4.4: Design of a TMD for a damped MDOF system 313 • Example 4.5: Design of TMD’s for a simply supported beam 315 4.7 Case studies - MDOF systems 322 Problems 331 Chapter 5

Base isolation systems 5.1 Introduction 339 5.2 Isolation for SDOF systems 340 • SDOF examples 340 • Bearing terminology 344 • Modified SDOF model 347 • Periodic excitation - modified SDOF model 349 • Example 5.1: Stiffness factors for prescribed structure and base motion 351 • Seismic excitation - modified SDOF model 352 • Example 5.2: Stiffness parameters - modified SDOF model of

Contents v Building example #2 356 5.3 Design issues for structural isolation systems 357 • Flexibility 357 • Rigidity under low level lateral loads 358 • Energy dissipation/absorption 359 • Modeling of a natural rubber bearing (NRB) 360 • Modeling of a lead rubber bearing (LRB) 364 • Applicability of base isolation systems 367 5.4 Examples of existing base isolation systems 368 • USC University Hospital 369 • Fire Department Command and Control Facility 369 • Evans and Sutherland Manufacturing Facility 370 • Salt Lake City Building 371 • The Toushin 24 Ohmori Building 371 • Bridgestone Toranomon Building 374 • San Francisco City Hall 374 • Long Beach V.A. Hospital 375 5.5 Optimal stiffness distribution - discrete shear beam 376 • Scaled stiffness distribution 376 • Example 5.3: Scaled stiffness for a 4 DOF beam with base isolation 379 • Fundamental mode response 380 • Example 5.4: Example 5.3 revisited 380 • Stiffness calibration for seismic isolation 382 • Example 5.5: Stiffness calibration for Example 5.4 382 5.6 Optimal stiffness distribution - continuous cantilever beam 386 • Stiffness distribution - undamped response 386 • Fundamental mode equilibrium equation 393 • Rigidity calibration - seismic excitation 395 • Example 5.6: Stiffness calibration - Example Building #2 396 5.7 Building design examples 397 • Stiffness distribution on fundamental mode response 397 • Stiffness distribution including the contribution of the higher modes 404 Problems 413

vi Contents

Part II: Active Control

Chapter 6

Introduction to active structural motion control 6.1 The nature of active control 423 • Active control versus passive control 423 • The role of feedback 427 • Computational requirements and models for active control 427 6.2 An introductory example of quasi-static feedback control 428 • Example 6.1: Shape control for uniform loading 431 • Example 6.2: Discrete displacement data 432 6.3 An introductory example of dynamic feedback control 433 • Example 6.3: Illustrative example - influence of velocity feedback 436 6.4 Actuator technologies 441 • Introduction 441 • Force application schemes 442 • Large scale linear actuators 446 • Large scale adaptive configuration based actuators 450 • Small scale adaptive material based actuators 451 6.5 Examples of existing large scale active structural control systems 462 • AMD in Kyobashi Seiwa Building 463 • AVS Control at Kajima Technical Research Institute 465 • DUOX Active-Passive TMD in Ando Nishikicho Building 468 • ABS - 600 Ton Full Scale Test Structure 472 Problems 474

Chapter 7

Quasi-static control algorithms 7.1 Introduction to control algorithms 7.2 Active prestressing of a simply supported beam • Passive prestressing • Active prestressing • Active prestressing with concentrated forces • Example 7.1: A single force actuator • Example 7.2: Two force actuators • A general active prestressing methodology

479 480 480 482 488 488 492 499

Contents vii • Example 7.3: Multiple actuators 503 7.3 Quasi-static displacement control of beams 506 • Continuous least square formulation 506 • Discrete least square formulation 508 • Example 7.4: Cantilever beam - Least square algorithm 510 • Extended least square formulation 516 • Example 7.5: Example 7.4 revisited 518 7.4 Quasi-static control of MDOF systems 521 • Introduction 521 • Selection of measures 522 • Example 7.6: Illustrative examples of observability 523 • Example 7.7: Illustrative examples of controllability 525 • Least square control algorithms 527 • Example 7.8: Example 7.7 revisited 528 • Example 7.9: Example 7.7 revisited with an extended least square algorithm 531 • Problems 534

Chapter 8

Dynamic Control Algorithms 8.1 Introduction 545 8.2 State-space representation - time invariant SDOF system 546 • Governing equations 546 • Free vibration uncontrolled response 548 • General solution - time invariant systems 550 • Example 8.1: Equivalence of equations (8.18) and (8.24) 551 • Stability criterion 552 • Linear negative feedback 553 • Effect of time delay on feedback control 555 • Stability analysis for time delay 560 8.3 Discrete time formulation - SDOF systems 566 • Governing equation 566 • Linear negative feedback control 568 • Stability analysis time invariant linear feedback control 569 • Example 8.2: Stability analysis-SDOF system with no time delay 575 • Example 8.3: Stability analysis - SDOF system with time delay 584 8.4 Optimal linear feedback - time invariant SDOF systems 586 • Quadratic performance index 586 • An example - linear quadratic regulator control algorithm 588 • The continuous time algebraic Riccati equation 593

viii Contents • The discrete time algebraic Riccati equation 598 • Example 8.4: Expanded form of discrete algebraic Riccati equation for an undamped SDOF system 600 • Finite interval discrete time algebraic Riccati equation 609 • Example 8.5: Example revised 610 • Continuous time Riccati different equation 611 • Variational formulation of the continuous time Riccati equation 612 • Example 8.6: Application to scalar case 617 8.5 State-space of formulation for MDOF systems 623 • Free vibration reponse-time invariant uncontrolled system 625 • Example 8.7: Free vibration solution for proportional damping 629 • Example 8.8: General uncoupled damping 630 • Orthogonality properties of the state eigenvectors 631 • Example 8.9: Initial conditions-free vibration response 632 • Determination of W and f j 633 • General solution-time invariant system 634 • Modal state space formulation-uncoupled damping 635 • Modal state space formulation-arbitrary damping 639 • Example 8.10: Modal formulation-undamped case 642 • Example 8.11: Modal formulation-uncoupled damping 643 • Example 8.12: Modal parameters-4DOF system 644 • Example 8.13: Modal response for example 8.12 651 • Example 8.14: Modal response response with feedback for example 8.12 654 • Stability analysis-discrete modal formulation 659 • Example 8.15: Stability analysis for example 8.14 662 • Controllability of a particular modal response 679 • Example 8.16: Controllability analysis for a 20 DOF Modal 680 • Observability of a particular modal response 682 • Example 8.17: Observability analysis for a 20 DOD Modal 684 8.6 Optimal linear feedback-MDOF time invariant systems 686 • Continuous time modal formulation 686 • Discrete time modal formulation 688 • Application studies - LQR control 690 • Example 8.18: Control force design studies for a 20 DOF shear beam 698 • Example 8.19: Alternate choice of response measures 711 Problems 714

Contents ix

References

725

Bibliography

735

Index

737

x

Introduction to Structural Motion Control

1

Chapter 1

Introduction 1.1 Motivation for structural motion control Limitations of conventional structural design The word, design, has two meanings. When used as a verb it is defined as the act of creating a description of an artifact. It is also used as a noun, and in this case, is defined as the output of the activity, i.e., the description. In this text, structural design is considered to be the activity involved in defining the physical makeup of the structural system. In general, the “designed” structure has to satisfy a set of requirements pertaining to safety and serviceability. Safety relates to extreme loadings which are likely to occur no more than once during a structure’s life. The concerns here are the collapse of the structure, major damage to the structure and its contents, and loss of life. Serviceability pertains to moderate loadings which may occur several times during the structure’s lifetime. For service loadings, the structure should remain fully operational, i.e. the structure should suffer negligible damage, and furthermore, the motion experienced by the structure should not exceed specified comfort limits for humans and motion sensitive equipment mounted on the structure. An example of a human comfort limit is the restriction on the acceleration; humans begin to feel uncomfortable when the acceleration reaches about 0.02g . A comprehensive discussion of human comfort

2 Chapter 1: Introduction criteria is given by Bachmann and Ammann (1987). Safety concerns are satisfied by requiring the resistance (i.e. strength) of the individual structural elements to be greater than the demand associated with the extreme loading. The conventional structural design process proportions the structure based on strength requirements, establishes the corresponding stiffness properties, and then checks the various serviceability constraints such as elastic behavior. Iteration is usually necessary for convergence to an acceptable structural design. This approach is referred to as strength based design since the elements are proportioned according to strength requirements. Applying a strength based approach for preliminary design is appropriate when strength is the dominant design requirement. In the past, most structural design problems have fallen in this category. However, a number of developments have occurred recently which have limited the effectiveness of the strength based approach. Firstly, the trend toward more flexible structures such as tall buildings and longer span horizontal structures has resulted in more structural motion under service loading, thus shifting the emphasis from safety toward serviceability. For instance, the wind induced lateral deflection of the Empire State Building in New York City, one of the earliest tall buildings in the United States, is several inches whereas the wind induced lateral deflection of the World Trade Center tower is several feet, an order of magnitude increase. This difference is due mainly to the increased height and slenderness of the World Trade Center in comparison to the Empire State tower. Furthermore, satisfying the limitation on acceleration is a difficult design problem for tall slender buildings. Secondly, some of the new types of facilities such as space platforms and semi-conductor manufacturing centers have more severe design constraints on motion than the typical civil structure. In the case of microdevice manufacturing, the environment has to be essentially motion free. Space platforms used to support mirrors have to maintain a certain shape to a small tolerance in order for the mirror to properly function. The design strategy for motion sensitive structures is to proportion the members based on the stiffness needed to satisfy the motion constraints, and then check if the strength requirements are satisfied. Thirdly, recent advances in material science and engineering have resulted in significant increases in the strength of traditional civil engineering materials

1.1 Motivation for Structural Motion Control

3

such as steel and concrete, as well as a new generation of composite materials. Although the strength of structural steel has essentially doubled, its elastic modulus has remained constant. Also, there has been some increase in the elastic modulus for concrete, but this improvement is still small in comparison to the increment in strength. The lag in material stiffness versus material strength has led to a problem with satisfying the serviceability requirements on the various motion parameters. Indeed, for very high strength materials, it is possible for the serviceability requirements to be dominant. Some examples presented in the following sections illustrate this point. Motion based structural design and motion control Motion based structural design is an alternate design process which is more effective for the structural design problem described above. This approach takes as its primary objective the satisfaction of motion related design requirements, and views strength as a constraint, not as a primary requirement. Motion based structural design employs structural motion control methods to deal with motion issues. Structural motion control is an emerging engineering discipline concerned with the broad range of issues associated with the motion of structural systems such as the specification of motion requirements governed by human and equipment comfort, and the use of energy storage, dissipation, and absorption devices to control the motion generated by design loadings. Structural motion control provides the conceptional framework for the design of structural systems where motion is the dominant design consideration. Generally, one seeks the optimal deployment of material and motion control mechanisms to achieve the design targets on motion as well as satisfy the constraints on strength. In what follows, a series of examples which reinforce the need for an alternate design paradigm having motion rather than strength as its primary focus, and illustrate the application of structural motion control methods to simple structures is presented. The first three examples deal with the issue of strength versus serviceability from a static perspective for building type structures. The discussion then shifts to the dynamic regime. A single-degree-offreedom (SDOF) system is used to introduce the strategy for handling motion constraints for dynamic excitation. The last example extends the discussion further to multi-degree-of-freedom (MDOF) systems, and illustrates how to deal with one of the key issues of structural motion control, determining the optimal stiffness distribution. Following the examples, an overview of structural motion control methodology is presented.

4 Chapter 1: Introduction

1.2 Motion versus strength issues for building type structures Building configurations have to simultaneously satisfy the requirements of site (location and geometry), building functionality (occupancy needs), appearance, and economics. These requirements significantly influence the choice of the structural system and the corresponding design loads. Buildings are subjected to two types of loadings: gravity loads consisting of the actual weight of the structural system and the material, equipment, and people contained in the building, and lateral loads consisting mainly of wind and earthquake loads. Both wind and earthquake loadings are dynamic in nature and produce significant amplification over their static counterpart. The relative importance of wind versus earthquake depends on the site location, building height, and structural makeup. For steel buildings, the transition from earthquake dominant to wind dominant loading for a seismically active region occurs when the building height reaches approximately 100m . Concrete buildings, because of their larger mass, are controlled by earthquake loading up to at least a height of 250m , since the additional gravity load increases the seismic forces. In regions where the earthquake action is low (e.g. Chicago in the USA), the transition occurs at a much lower height, and the design is governed primarily by wind loading. When a low rise building is designed for gravity loads, it is very likely that the underlying structure can carry most of the lateral loads. As the building height increases, the overturning moment and lateral deflection resulting from the lateral loads increase rapidly, requiring additional material over and above that needed for the gravity loads alone. Figure 1.1 (Taranath, 1988) illustrates how the unit weight of the structural steel required for the different loadings varies with the number of floors. There is a substantial weight cost associated with lateral loading.

1.2 Motion Versus Strength Issues for Building Type Structure 5

140

120

Gravity loads

Lateral loads

Number of floors

100

Floor

Columns

80

60

40

20

0 0

50 500

100 1000

150 1500

Structural steel - N/m

2

200 2000

250 2500

Fig. 1.1: Structural steel quantities for gravity and wind systems To illustrate the dominance of motion over strength as the slenderness of the structure increases, the uniform cantilever beam shown in Fig. 1.2 is considered. The lateral load is taken as a concentrated force p applied to the tip of the beam, and is assumed to be static. The limiting cases of a pure shear beam and a pure bending beam are examined. u

p

d w

H a

a d section a-a

Fig. 1.2: Building modeled as a uniform cantilever beam

6 Chapter 1: Introduction

Example 1.1: Cantilever shear beam The shear stress τ is given by p τ = -----As

(1.1)

where A s is the cross sectional area over which the shear stress can be considered to be constant. When the bending rigidity is very large, the displacement, u , at the tip of the beam is due mainly to shear deformation, and can be estimated as pH u = ----------GA s

(1.2)

where G is the shear modulus and H is the height of the beam. This model is called a shear beam. The shear area needed to satisfy the strength requirement follows from eqn (1.1): As

p ≥ ----strength τ∗

(1.3)

where τ∗ is the allowable stress. Noting eqn (1.2), the shear area needed to satisfy the serviceability requirement on displacement is As

p H ≥ ---- ⋅ -----serviceability G u∗

(1.4)

where u∗ denotes the allowable displacement. The ratio of the area required to satisfy serviceability to the area required to satisfy strength provides an estimate of the relative importance of the motion design constraints versus the strength design constraints As τ∗ H serviceability r = -------------------------------------- = ----- ⋅ -----G u∗ As

(1.5)

strength

Figure 1.3 shows the variation of r with H ⁄ u∗ . Increasing H ⁄ u∗ places

1.2 Motion Versus Strength Issues for Building Type Structure 7 more emphasis on the motion constraint since it corresponds to a decrease in the allowable displacement, u∗ . Furthermore, an increase in the allowable shear stress, τ∗ , also increases the dominance of the displacement constraint. r

*

*

τ2 > τ1

*

τ1

100

200

300

400

H -----u∗

Fig. 1.3: Plot of r versus H ⁄ u∗ for a pure shear beam

Example 1.2: Cantilever bending beam When the shear rigidity is very large, shear deformation is negligible, and the beam is called a “bending” beam. The maximum bending moment M in the structure occurs at the base and equals M = pH

(1.6)

The resulting maximum stress σ is M Md pHd σ = ----- = --------- = ----------S 2I 2I

(1.7)

where S is the section modulus, I is the moment of inertia of the cross-section about the bending axis, and d is the depth of the cross-section (see Fig. 1.2). The corresponding displacement at the tip of the beam u becomes

8 Chapter 1: Introduction 3

pH u = ---------3EI

(1.8)

The moment of inertia needed to satisfy the strength requirement is given by pHd I strength ≥ ----------2σ∗

(1.9)

Using eqn (1.8), the moment of inertia needed to satisfy the serviceability requirement is 3

pH I serviceability ≥ ------------3Eu∗

(1.10)

Here, u∗ and σ∗ denote the allowable displacement and stress respectively. The ratio of the moment of inertia required to satisfy serviceability to the moment of inertia required to satisfy strength has the form 3 I serviceability pH 2σ∗ 2H σ∗ H ------------ ⋅ ----------- = -------- ⋅ ------ ⋅ -----r = ------------------------------- = 3d E u∗ I strength 3Eu∗ pHd

(1.11)

Figure 1.4 shows the variation of r with H ⁄ u∗ for a constant value of the aspect ratio H ⁄ d ( H ⁄ d ≈ 7 for tall buildings). Similar to the case of the shear beam, an increase in H ⁄ u∗ places more emphasis on the displacement since it corresponds to a decrease in the allowable displacement, u∗ , for a constant H . Also, an increase in the allowable stress, σ∗ , increases the importance of the displacement constraint. For example, consider a standard strength steel beam with an allowable stress of σ∗ = 200MPa , a modulus of elasticity of E = 200,000MPa , and an aspect ratio of H ⁄ d = 7 . The value of H ⁄ u∗ at which a transition from strength to serviceability occurs is 3 E d H = --- ⋅ ------ ⋅ ---- ≈ 200 -----2 σ∗ H ∗ u r=1

(1.12)

For H ⁄ u∗ > 200 , r > 1 and motion controls the design. On the other hand, if high

1.2 Motion Versus Strength Issues for Building Type Structure 9 strength steel is utilized ( σ∗ = 400MPa and E = 200,000MPa ) H -----u∗

≈ 100

(1.13)

r=1

and motion essentially controls the design for the full range of allowable displacement.

r

*

*

σ2 > σ1

*

σ1

100

200

300

400

H -----u∗

Fig. 1.4: Plot of r versus H ⁄ u∗ for a pure bending beam

Example 1.3: Quasi-shear beam frame This example compares strength vs. motion based design for a single bay frame of height H and load p (see Fig. 1.5). For simplicity, a very stiff girder is assumed, resulting in a frame that displays quasi-shear beam behavior. Furthermore, the columns are considered to be identical, each characterized by a modulus of elasticity E c , and a moment of inertia about the bending axis I c . The maximum moment, M , in each column is equal to pH M = -------4

(1.14)

10 Chapter 1: Introduction The lateral displacement of the frame under the load is expressed as

u

Ig = ∞ p --2

p --2 Ec , I c

H

Ec , I c

Fig. 1.5: Quasi-shear beam example pH u = -------DT

(1.15)

where D T denotes the equivalent shear rigidity which, for this structure, is given by 24E c I c D T = ----------------2 H

(1.16)

The strength constraint requires that the maximum stress in the column be less than the allowable stress σ∗ pHd Md --------- = ----------- ≤ σ∗ 8I c 2I c

(1.17)

where d represents the depth of the column in the bending plane. Equation (1.17) is written as Ic

pHd ≥ ----------strength 8σ∗

(1.18)

The serviceability requirement constrains the maximum displacement to be less than the allowable displacement u∗

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 11 3

pH ----------------- ≤ u∗ 24E c I c

(1.19)

The corresponding requirement for I c is 3

pH ≥ -----------------Ic serviceability 24E u∗

(1.20)

c

Forming the ratio of the moment of inertia required to satisfy the serviceability requirement to the moment of inertia required to satisfy the strength requirement, Ic σ∗ H H serviceability r = ----------------------------------- = --------- ⋅ ---- ⋅ -----3E c d u∗ Ic strength

(1.21)

leads to the value of H ⁄ u∗ for which motion dominates the design H 3E c d ------ ≥ --------- ⋅ ---u∗ σ∗ H

(1.22)

1.3 Design of a single-degree-of freedom system for dynamic loading The previous examples dealt with motion based design for static loading. A similar approach applies for dynamic loading once the relationship between the excitation and the response is established. The procedure is illustrated for the single-degree-of-freedom (SDOF) system shown in Fig. 1.6. Response for periodic excitation The governing equation of motion of the system has the form mu˙˙( t ) + cu˙ ( t ) + ku ( t ) = p ( t )

(1.23)

12 Chapter 1: Introduction

k R m

p

c u

Fig. 1.6: Single-degree-of-freedom system where m , k , c are the mass, stiffness, and viscous damping parameters of the system respectively, p is the applied loading, u is the displacement, and t is the independent time variable. The dot operator denotes differentiation with respect to time. Of interest is the case where p is a periodic function of time. Taking p to be sinusoidal in time with frequency Ω , p ( t ) = pˆ sin Ωt

(1.24)

the corresponding forced vibration response is given by u ( t ) = uˆ sin ( Ωt – δ )

(1.25)

where uˆ and δ characterize the response. They are related to the system and loading parameters as follows: pˆ uˆ = --- H 1 k

(1.26)

1 H 1 = -------------------------------------------------2 2

[ 1 – ρ ] + [ 2ξρ ]

ω =

k ---m

c c ξ = ------------ = --------------2ωm 2 km

(1.27)

2

(1.28)

(1.29)

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 13 Ω m ρ = ---- = Ω ---ω k

(1.30)

2ξρ tan δ = --------------2 1–ρ

(1.31)

The term pˆ ⁄ k is the displacement response that would occur if the loading were applied statically; H 1 represents the effect of the time varying nature of the response. Figure 1.7 shows the variation of H 1 with the frequency ratio, ρ , for various levels of damping. The maximum value of H 1 and corresponding frequency ratio are related to the damping ratio ξ by H1

1 = ------------------------2 max 2ξ 1 – ξ

(1.32)

5 4.5

ξ = 0.0

4 3.5

H1

3

ξ = 0.2

2.5 2

ξ = 0.4

1.5 1 0.5 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Ω m ρ = ---- = Ω ---ω k Fig. 1.7: Plot of H 1 versus ρ and ξ ρ max =

1 – 2ξ

2

(1.33)

14 Chapter 1: Introduction 2

When ξ << 1 , ρ max ≈ 1 H1

(1.34)

1 ≈ -----max 2ξ

(1.35)

Since ξ is usually small, the maximum dynamic response is significantly greater than the static response and ρ max is close to 1. For example, for ξ = 0.2 which corresponds to a high level of damping, the peak response is H1

max

≈ 2.55

(1.36)

ρ max = 0.96

(1.37)

When the forcing frequency, Ω , is close to the natural frequency, ω , the response is controlled by adjusting the damping. Outside of this region, damping has less influence, and has essentially no effect for ρ < 0.4 and ρ > 1.6 . Differentiating u twice with respect to time leads to the acceleration, a , 2

a ( t ) = u˙˙( t ) = – Ω uˆ sin ( Ωt – δ ) = – aˆ sin ( Ωt – δ )

(1.38)

Noting eqn (1.26), the magnitude of a can be written as pˆ pˆ 2 aˆ = --- Ω H 1 = ---- H 2 m k

(1.39)

where 2

H2 = ρ H1 =

4

ρ ---------------------------------------------2 2 2 [ 1 – ρ ] + [ 2ξρ ]

(1.40)

The variation of H 2 with ρ for different damping ratios is shown in Fig. 1.8. Note that the behavior of H 2 for small and large ρ is opposite to H 1 . The maximum value of H 2 is the same as the maximum value for H 1 , but the location (i.e. the corresponding value of ρ ) is different. They are related to ξ by

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 15 1 ρ max = ---------------------2 1 – 2ξ H2

(1.41)

1 = ------------------------2 max 2ξ 1 – ξ

(1.42)

The ratio pˆ ⁄ m is the acceleration the mass would experience if it were unrestrained and subjected to a constant force of magnitude pˆ . One can interpret H 2 as a modification factor which takes into account the time varying nature of the loading and the system restraints associated with stiffness and damping.

5 4.5

ξ = 0.0

4 3.5

H2

3 2.5

ξ = 0.2

2 1.5 1

1 ξ = ------2

0.5 0 0

0.2

0.4

0.6

0.8

1

1.2

Ω ρ = ---- = Ω ω Fig. 1.8: Plot of H 2 versus

1.4

1.6

1.8

2

m ---k ρ and ξ

Once the system and loading are defined (i.e. m , k , c , pˆ , and Ω are specified), one determines H 1 and computes the peak amplitudes using the following relations pˆ uˆ = --- H 1 k 2

aˆ = Ω uˆ

(1.43)

(1.44)

16 Chapter 1: Introduction H 1 = H 1 ( Ω, m, k, c ) = H 1 ( ρ, ξ )

(1.45)

Note that for periodic response, the acceleration is related to the displacement by the square of the forcing frequency. One can also work with H 2 instead of H 1. Design criteria The design problem differs from analysis in that one starts with the mass of the system, m , and the loading characteristics, pˆ and Ω , and determines k and c such that the motion parameters, uˆ and aˆ , satisfy the specified criteria. In general, one has limits on both displacement and acceleration uˆ ≤ u∗

(1.46)

aˆ ≤ a∗

(1.47)

where u∗ and a∗ are the target design values. In this case, since uˆ and aˆ are related by 2

aˆ = Ω uˆ

(1.48)

2 one needs to determine which constraint controls. If a∗ ≤ Ω u∗ , the acceleration limit controls and the optimal solution will be

a∗ uˆ = ------- < u∗ 2 Ω

(1.49)

aˆ = a∗

(1.50)

2 If, on the other hand, a∗ ≥ Ω u∗ , the displacement limit controls. For this case, the optimal solution satisfies

uˆ = u∗

(1.51)

2 aˆ = Ω u∗ < a∗

(1.52)

In what follows, both cases are illustrated.

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 17 Methodology for acceleration controlled design One works with eqn (1.39). Expressing the target design acceleration as a function of the gravitational acceleration, a∗ = fg

(1.53)

and defining H 2∗ as a∗ H 2∗ = ----------pˆ ⁄ m

(1.54)

the design constraint takes the form W H 2 ≤ H 2∗ = ----- f pˆ

(1.55)

where W is the weight of the system. The totality of possible solutions is contained in the region below H 2 = H 2∗ . Figure 1.9 illustrates the region for H 2∗ = 2 . For low damping, the intersection of H 2 = H 2∗ and the H 2 curve for a particular value of ξ , ξ∗ , establishes two limiting ρ values, ρ 1 [ H 2∗, ξ∗ ] and ρ 2 [ H 2∗, ξ∗ ] . Permissible values of ρ for the damping ratio ξ∗ , are

18 Chapter 1: Introduction

5 4.5

ξ = 0.0

4 3.5

H2

3 2.5 2

ξ = ξ∗ H 2∗ possible solutions

1.5 1 0.5 0 0

ρ1 0.2

0.4

0.6

0.8

ρ2 1

Ω ρ = ---- = Ω ω Fig. 1.9: Possible values of

1.2

1 ξ = ------2

1.4

1.6

1.8

2

m ---k H 2 ≤ H 2∗

0 < ρ ≤ ρ 1 [ H 2∗, ξ∗ ] ρ ≥ ρ 2 [ H 2∗, ξ∗ ]

(1.56)

The second region does not exist when H 2∗ < 1 . Noting eqn (1.40), the expressions for ρ 1 and ρ 2 are

ρ 1, 2 =

2 2 2 1 2 1 – 2ξ∗ − + [ 1 – 2ξ∗ ] – 1 + ---------H 2∗ ---------------------------------------------------------------------------------------------1 2 1 – ---------H 2∗

These functions are plotted in Fig. 1.10 for representative values of ξ∗ . The limiting values of ρ for ξ∗ = 0 reduce to

(1.57)

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 19

1 -------------------1 1 ± ---------H 2∗

ρ 1, 2 =

(1.58)

2

ρ2

1.8

ξ∗ = 0.2

1.6

ξ∗ = 0.1

1.4

ξ∗ = 0

1.2 1

m ρ = Ω ---k

0.6

ρ1

0.4 0.2 0 0

1

H 2∗ = 2

0.8

2

3

4

5

6

H2 Fig. 1.10: Plot of ρ 1 and ρ 2 versus ξ∗ and H 2 Noting eqn (1.29), one can express eqn (1.56) in terms of limiting values of stiffness. By definition, 2

Ω m k = -----------2 ρ

(1.59)

Then, letting 2

Ω m k j = -----------2 ρj

(1.60)

and noting that k 2 < k 1 , the allowable ranges for k are given by: H 2∗ < 1

k1 < k < ∞

(1.61)

20 Chapter 1: Introduction H 2∗ > 1

0 < k < k 2 and k 1 < k < ∞

(1.62)

Given H 2∗ , one specifies a value of ξ , computes ρ j with eqn (1.57), and selects a value for k which satisfies the above constraints on stiffness. The damping parameter is determined from (1.63)

c = 2ξωm = 2ξ km

Example 1.4: An illustration of acceleration controlled design Suppose pˆ = 0.1W and a∗ = 0.05g . Applying eqn (1.54) leads to H 2∗ = 0.5 . Figure 1.10 shows that damping has a negligible effect for this value of H 2∗ ; the design is essentially controlled by stiffness. Taking ξ = 0 , and using eqn (1.58) results in 2 1 ρ 1 = --3

2

k > 3Ω m To illustrate the other extreme, pˆ = 0.1W and a∗ = 0.2g is considered. Here, H 2∗ = 2.0 . The two allowable regions for k corresponding to different values of ξ are obtained by applying equations (1.57), (1.60), and (1.62). k ≥ k1

k ≤ k2 ξ

k1/Ω2m

k2/Ω2m

c1/Ω m

c2/Ω m

0

1.5

0.5

0

0

0.1

1.439

0.521

0.24

0.144

0.2

1.231

0.610

0.444

0.312

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 21 Methodology for displacement controlled design The starting point is eqn. (1.26). Noting equations (1.40) and (1.59), eqn (1.26) can be written as pˆ pˆ pˆ 2 uˆ = --- H 1 = ------------ ρ H 1 = ------------ H 2 2 2 k Ω m Ω m

(1.64)

**

Then, defining H 2 as ** H2

2

*

Ω mu = -----------------pˆ

(1.65)

*

where u is the target displacement, the design constraint is given by **

H2 < H2

(1.66)

The remaining steps are the same as for the previous case; the only difference is **

**

the definition of H 2 . One applies equations (1.57) thru (1.63), using H 2 instead of H 2∗ .

Example 1.5: An illustration of displacement controlled design Suppose pˆ = 10kN *

u = 10cm m = 1000kg Then 2 ( 1000 ) ( 0.1 ) 2 H 2 ** = ----------------------------- Ω = ( 0.01 )Ω 10000

22 Chapter 1: Introduction Various values for Ω are considered. 1. Ω = 2π r ⁄ s H 2 ** = 0.394 **

For this value of H 2 , the design is controlled by stiffness. Taking ξ = 0 , and using eqn (1.58), 1 1 ------ = 1 + -------- = 3.538 2 ** ρ1 H2 2

Ω m k 1 = ------------ = 139.7kN ⁄ m 2 ρ1 The value of k is constrained by k ≥ k 1 = 139.7kN ⁄ m which corresponds to ρ ≤ ρ 1 = 0.532 2. Ω = 4π r/s H 2 ** = 1.576 **

Since H 2 is greater than 1, there are 2 allowable regions for k . Also these regions depend on the damping ratio, ξ . Results for different values of ξ are listed below. They are generated using equations (1.57), (1.60), and (1.62). ρ < ρ1

k ≥ k1

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 23 ρ > ρ2

k ≤ k2

ξ

ρ1

k1(kN/m)

c1 (kN*s/m)

ρ2

k2(kN/m)

c2 (kN*s/m)

0

0.782

258.1

0

1.654

57.72

0

0.1

0.795

249.6

3.16

1.627

59.63

1.54

0.2

0.839

224.1

5.98

1.541

66.5

3.26

The solution for k ≤ k 2 is π radians out of phase with the forcing function. Decreasing k reduces the natural frequency, ω , and moves the system away from the resonant zone, ρ = 1 . For small k , H 2 approaches unity, and the response measures tend toward the following limits pˆ aˆ → ---m

pˆ uˆ → ---------------2 Ω ⋅m

(1.67)

Methodology for force controlled design The previous examples dealt with limiting the displacement or acceleration of the single degree of freedom mass. Another design scenario is associated with the concept of isolation, i.e., one wants to limit the internal force that is generated by the applied force and transmitted to the support. The reaction force, R , shown in Fig (1.6), is given by R = p – ma = ku + cu˙

(1.68)

Expressing R as R = Rˆ sin ( Ωt – δ + δ 1 )

(1.69)

and using equations (1.24) thru (1.31), the magnitude and phase shift are given by

24 Chapter 1: Introduction ˆ = H pˆ R 3

(1.70) 2

1 + [ 2ξρ ] ---------------------------------------------2 2 2 [ 1 – ρ ] + [ 2ξρ ]

H3 =

(1.71)

tan δ 1 = 2ξρ

(1.72)

5

*

ξ = 0

4.5 4 3.5

H3

3 2.5

*

ξ = 0.2

2

*

ξ = 0.4

1.5 1 0.5 0 0

2 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

m ρ = Ω ---k Fig. 1.11: Plot of H 3 versus ρ and ξ Figure 1.11 shows the variation of H 3 with ρ and ξ . At ρ = 2 , H 3 = 1 for all values of ξ . When ρ > 2 , the minimum value of H 3 corresponds to ξ = 0 , which implies that damping magnifies rather than decreases the response in this region. The strategy for reducing the reaction is to take the stiffness as 2

k<Ω m⁄2

(1.73)

Decreasing k “softens” the system and reduces the internal force. However, the displacement and acceleration “increase”, and approach the limiting values given by eqn (1.67)

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 25

Example 1.6: Force reduction Suppose pˆ = 10kN m = 1000kg Ω = 4π r ⁄ s and the desired magnitude of the reaction is 10% of the applied force. Then, H 3 = 0.10 Taking ξ = 0 in eqn (1.71) 1 H 3 = ------------------ = 0.1 1 – ρ2 results in ρ =

11 = 3.317

Finally, the corresponding stiffness is 2

Ω m k = ------------ = 14.36kN ⁄ m 2 ρ

26 Chapter 1: Introduction

1.4 Design of a single-degree-of-freedom system for support motion Response for periodic support motion

k m c ug

u + ug

Fig. 1.12: Single-degree-of-freedom system The equation of motion for a SDOF system subjected to support motion is written as mu˙˙( t ) + cu˙ ( t ) + ku ( t ) = – mu˙˙g ( t )

(1.74)

where u is now considered to be the relative displacement with respect to the support, and u g denotes the motion of the support. Assuming periodic excitation u g ( t ) = uˆ g sin Ωt

(1.75)

the relative and total displacements are given by u ( t ) = uˆ sin ( Ωt – δ ) u t ( t ) = u ( t ) + u g ( t ) = uˆ t sin ( Ωt + δ 1 – δ )

(1.76) (1.77)

where uˆ = H 2 uˆ g

(1.78)

uˆ t = H 3 uˆ g

(1.79)

The total acceleration is related to the support acceleration by a similar expression u˙˙t ( t ) = a t ( t ) = aˆ t sin ( Ωt + δ 1 – δ )

(1.80)

1.4 Design of a Single-degree-of Freedom System for Support Motion 27 2

aˆ t = H 3 [ – Ω uˆ g ] = H 3 aˆ g

(1.81)

Design scenarios There are a number of design scenarios for support motion. If the system is sensitive to the total acceleration, one can limit the response by requiring H 3 to be small with respect to unity. In effect, one isolates the mass from the support motion. From Fig. 1.11, ρ must be greater than 2 for isolation to be feasible. The 2 corresponding constraint on stiffness is k < mΩ ⁄ 2 . If the support motion is interpreted as seismic excitation, and the system represents a structure, one wants to control the relative motion of the structure and the total acceleration of the equipment attached to the structure. One approach is to isolate the structure so that the acceleration constraint is satisfied. The relative motion will be essentially equal to the ground motion, so an additional mechanism is needed to localize the motion. An example of this approach is a rigid building supported by low stiffness springs; the motion in this approach is confined to the support springs. Another approach would be to work with H 2 so as to satisfy the displacement constraint with stiffness and adjust the damping to satisfy as close as possible the constraint on acceleration. The examples presented below illustrate both approaches. A detailed discussion of isolation of building structures for seismic excitation is contained in Chapter 5.

Example 1.7: Controlling acceleration due to ground motion This problem concerns selecting ρ and ξ such that the total acceleration is a fraction of the support acceleration, say aˆ t = νaˆ g

ν << 1

(1.82)

Then from eqn (1.81) aˆ t H 3 = ----- = ν aˆ g

(1.83)

As pointed out earlier, ξ = 0 yields the minimum value of H 3 . Specializing eqn

28 Chapter 1: Introduction (1.71), one obtains 1 ρ ≥ 1 + --ν

(1.84)

as the lower limit on ρ . The stiffness limit follows from the definition of ρ 2 Ω 2 mΩ k = m ---- ≤ -----------1 ρ 1 + --ν

(1.85) 2

mΩ For example, suppose ν = 1 ⁄ 3 . Then, ρ ≥ 2 and k ≤ ------------ . 4

Example 1.8: Controlling relative motion due to ground motion This example illustrates how to constrain the relative motion due to support movement. A typical scenario is a structure subjected to seismic excitation. Defining u∗ as the desired relative displacement, the appropriate equation follows from eqn (1.78). u∗ H 2 ≤ ------ = H 2∗ uˆ g

(1.86)

For this type of problem, H 2∗ is usually greater than unity. The procedure for solving eqn (1.86) was discussed in the previous section. Considering the solution zone corresponding to ρ > ρ 2 (See Fig. 1.9) and starting with ξ = 0 leads to ρ2 ≈ ρ2

ξ=0

=

1 ------------------1 1 – ---------H 2∗

(1.87)

2

mΩ k ξ = 0 = --------------------------2 ρ2

(1.88)

ξ=0

as the first estimate. The change in ρ 2 and k due to including damping can be evaluated using eqn (1.57).

1.4 Design of a Single-degree-of Freedom System for Support Motion 29 To illustrate the process, the following numerical values are considered. Ω = 4π rd/s

2π T = ------ = 0.5s Ω

u∗ = 0.1m

uˆ g = 0.033m

Applying eqns (1.86) through (1.88), the values of the various parameters are 0.1 H 2∗ = ------------- = 3 0.033

ρ2

ξ=0

=

1 ------------ = 1.22 1 1 – --3

T ξ = 0 = ρ2 T = 0.61s ξ=0

2 4π 2 mΩ k ξ = 0 = --------------------------- = m ---------- = 105.27m 2 1.22 ρ2 ξ=0

For no damping, one has to take k ≤ k ξ = 0 in order to satisfy the constraint on amplification. To assess the sensitivity to damping, ξ∗ is taken as 0.1 . Using eqn (1.57), the corresponding value of ρ 2 is ρ2

ξ = 0.1

= 1.18

The other parameters are determined with T ξ = 0.1 = ρ 2 T = 0.59s ξ = 0.1

30 Chapter 1: Introduction 2

k

mΩ = ------------------------------- = 112.53m ξ = 0.1 2 ρ2 ξ = 0.1

The solution is fairly insensitive to damping. However, if one uses the larger stiffness value, and the damping assumed in the calculation is not realized, the amplification will be greater than the design value, H 2∗ .

1.5 Stiffness distribution for a two degree-of-freedom system The last two examples were concerned with how to determine the stiffness for a SDOF system subjected to dynamic loading when motion is the controlling design criterion. The procedure for a multi-degree-of-freedom (MDOF) system is more involved since one has now to determine a set of stiffnesses rather than just a single stiffness. A two-degree-of-freedom system is considered here to illustrate the approach. The next chapter extends the approach to deal with continuous beam type structures and discrete mass systems having both stiffness and damping. Consider the two degree-of-freedom system shown in Fig. 1.13. The SDOF notation is extended to the MDOF case by including a subscript to indicate the corresponding object. Enforcing equilibrium for the individual masses leads to the following governing equations m 1 u˙˙1 ( t ) + ( k 1 + k 2 )u 1 ( t ) – k 2 u 2 ( t ) = p 1 ( t )

(1.89)

m 2 u˙˙2 ( t ) + k 2 u 2 ( t ) – k 2 u ( t ) = p 2 ( t ) 1

(1.90)

Specializing these equations for free vibration, the solution is expressed as u 1 ( t ) = Φ 1 sin ωt

(1.91)

u 2 ( t ) = Φ 2 sin ωt

(1.92)

1.5 Stiffness Distribution for a Two Degree-of -freedom System 31 p1 ( t ) = p2 ( t ) = 0

(1.93)

where Φ j , are the amplitudes and ω is a frequency measure. Requiring eqns (1.89) and (1.90) to be satisfied for Φ j ≠ 0 yields the following equation for the frequency, k1 + k2 k2 k1 k2 2 λ + – ----------------- – ------- λ + --------------- = 0 m1 m2 m1 m2

(1.94)

where λ = ω 2 .

p2

p1 k1

m1

k2

m2

u1

u2

Fig. 1.13: Two-degree-of-freedom system The displacement amplitudes are related by k 1 + k 2 – λm 1 Φ 2 = ---------------------------------- Φ 1 k2

(1.95)

There are 2 values of λ , and 2 corresponding sets of Φ ’s. Each set of Φ ’s defines a displacement pattern which is called a mode shape. A detailed description of free vibration analysis for multi-degree-of-freedom systems is presented later in chapter 2. When k 1 and k 2 are specified, one solves eqn (1.94) for λ and then determines the displacement amplitude ratio with eqn (1.95). The actual mode shape is defined only in a relative sense since Φ 1 is an arbitrary constant. An inverse approach is followed in motion based design. One specifies the relative mode shape based on the desired deformation, and then determines the relative stiffness distribution. The details are presented below.

32 Chapter 1: Introduction The differences in displacement amplitudes can be interpreted as the relative displacements between the masses ∆u 1 = Φ 1

(1.96)

∆u 2 = Φ 2 – Φ 1

(1.97)

For optimal motion behavior, the relative displacements should be equal. This constraint results in a linear mode shape. ∆u 1 = ∆u 2

(1.98)

Φ 2 = 2Φ 1

(1.99)

Setting Φ 2 = 2Φ 1 in eqn (1.95) and solving for λ yields k 1∗ – k 2∗ λ 1∗ = ---------------------m1

(1.100)

where an asterisk is used to indicate the value associated with a prescribed relative mode shape. Introducing λ 1∗ in eqn (1.94) leads to a relation between k 1∗ and k 2∗ m1 k 1∗ = 1 + ---------- k 2∗ 2m 2

(1.101)

Finally, substituting for k 1∗ in eqn (1.100), results in a relation between the fundamental frequency (lowest) and k 2∗ : k 2∗ 2 λ 1∗ = ω 1 = ---------2m 2

(1.102)

Once k 2∗ is specified, the fundamental frequency and stiffness parameters are uniquely determined. One can also specify the desired fundamental frequency and then use eqn (1.102) to determine the necessary value of k 2∗ . Given k 1 and k 2 , eqn (1.94) can be solved for λ 2 , and using this value, the second mode shape can be established with eqn (1.95).

1.6 Control Strategies for Motion Based Design

33

Example 1.9: 2DOF system - with equal masses The case where m 1 = m 2 = m is considered to illustrate the procedure. Applying eqn (1.101), k 1∗ is related to k 2∗ by 3 k 1∗ = --- k 2∗ 2

(1.103)

Suppose the design objective is to have T 1 = 1s (i.e. ω 1 = 2π rd/s ). Then one sets ω 1 = 2π in eqn (1.102) and solves for k 2∗ k 2∗ = 2m ( 2π )

2

(1.104)

Once m is specified, the stiffness parameters can be evaluated. Substituting for these parameters in eqn (1.94) leads to λ 2 = 3k 2∗ ⁄ m . The corresponding mode shape is Φ 2 = – 0.5Φ 1.

1.6 Control strategies for motion based design The previous sections dealt with the problem of limiting the response of a single degree of freedom system subjected to periodic excitation. Although the problem considered was rather simple, the strategies used to generate appropriate values for the system stiffness and damping are basic paradigms that are also applicable to more complex structures. In what follows, the SDOF strategies are reviewed to provide the basis for establishing a general motion control methodology for structural systems. Consider the linear elastic SDOF system shown in Fig (1.14). Suppose the design loading is static, and the motion criteria is u < u * . Enforcing equilibrium leads to an expression for the required stiffness. p k ≥ ----- ≡ k * u*

(1.105)

34 Chapter 1: Introduction From a stiffness perspective, control for static loading is realized by adjusting the stiffness such that k is greater than a certain limiting value which depends on the design criteria for displacement.

k p m

F

c ug

u + u g = ut

Fig. 1.14: Single degree of freedom model for passive and active control Another way of expressing the motion control requirement is in terms of energy. In general, the work done by the external forces acting on a system is equal to the sum of the mechanical energy stored in the system and the energy transformed to another form, through either energy dissipation or absorption mechanisms. This system is elastic, and the loading is static. It follows that the stored energy is equal to the strain energy, and since no energy is dissipated, the work done by the external loading must be balanced by the strain energy. Expanding these quantities leads to eqn (1.106). External Work ≤ Strain Energy 1 * 1 --- pu ≤ --- k ( u * ) 2 ⇒ 2 2

p ----- ≤ k u*

(1.106)

From an energy perspective, static control is achieved by providing sufficient energy storage capacity to satisfy the energy demand. The design value of stiffness defined by eqn (1.105) is based on the hypothesis that the system properties (k,m) remain constant as the load is being applied, and there is no other agent which assists in resisting the load. Control is achieved solely by the action of the stiffness embedded within the system. This type of control action is called “passive control” since the system responds in a passive manner. A different strategy for limiting the displacement is based on using an external energy source to satisfy some of the energy demand. In this approach, the

1.6 Control Strategies for Motion Based Design

35

energy requirement is taken as: External work ≤ Strain energy + Supplied energy

(1.107)

Energy can be supplied in different ways. For the SDOF system, a force actuator can be used to generate the force, F, shown in Fig (1.14). Force actuators are devices which convert energy into a mechanical force, and provide the ability to implement this control strategy. The work done by F is negative, and its magnitude represents the external energy that needs to be supplied. Work

F

1 = – --- Fu * 2

(1.108)

Equation (1.107) expands to 1 * 1 1 --- pu ≤ --- k ( u * ) 2 + --- Fu * 2 2 2

(1.109)

and is follows that p ≤ ku * + F

(1.110)

One can also arrive at this result by enforcing equilibrium. Controlling motion by supplying external energy is referred to as “active control”. One of the key decisions in motion based design is the selection of the control strategy. Fully passive static control is simple to implement, since it involves only providing stiffness. Active control employs actuator technology which is costly and less reliable. However, active control has considerable potential, particularly for structural applications where weight is a critical issue, such as aerospace vehicles and long span bridges. The concepts discussed above are also applicable for dynamic loading. When the loading is periodic, the response amplitudes are given by: pˆ uˆ = ------------ H 2 Ω2m pˆ aˆ = ----H 2 m

(1.111)

36 Chapter 1: Introduction where H 2 is plotted in Fig (1.15). Since civil structures are generally lightly damped, a realistic upper limit for ξ is 0.3. Assuming pˆ , Ω , and m are specified, one needs to select values for stiffness and damping that satisfy the design motion criteria. H 2 < H 2*

(1.112)

where the value of H 2* depends on whether displacement or acceleration is the limiting motion constraint. mΩ 2 u * H 2* = -----------------pˆ ma * H 2* = --------p

( displacement ) (1.113) ( acceleration )

4

ξ = 0 3.5

3

Stiffness and damping are coupled

Stiffness dominates response

2.5

Stiffness dominates response

H 2*

2

H2

H 2*

1.5

1

H 2*

ξ = 0.3

0.5

0

0

0.5

1

1.5

2

Ω m ρ = ---- = Ω ---ω k

2.5

3

Fig. 1.15: Response function for periodic excitation

3.5

4

1.6 Control Strategies for Motion Based Design

37

Figure (1.15) shows that H 2* is essentially independent of damping for ρ < ≈ 0.5 and ρ> ≈ 3.0 . If H 2* < ≈ 0.5 , stiffness is the only parameter that can be “adjusted” to satisfy the design constraint, and the design process is essentially quasi-static. For 0.5 < H 2* < 1.0 , both stiffness and damping are available as control variables. One can generate a spectrum of designs by specifying stiffness ( ρ ) , and determining the corresponding damping ( ξ ) that satisfies H 2 = H 2* . When H 2* > 1 , there are 2 possible design zones, one for ρ < 1 and the other for ρ > 1. As H 2* increases, these zones merge into a single zone centered at ρ = 1, the resonant state of the system. Damping dominates the response in the neighborhood of ρ = 1. The strategy employed for 0.5 < H 2* < 1.0 can also be applied for H 2* > 1.0 . Given H 2* , one selects a stiffness value within the stiffness range defined by the specified value of H 2* , and the boundary curves ξ = 0 and ξ = 0.3 , and then determines the required value of ξ . For 1.0 ≤ H 2* ≤ 1.5 , the second stiffness range is outside ρ = 3.0 . Designs for this region are controlled by stiffness, and are characterized by their low stiffness. Design for the first region are controlled by both stiffness and damping, and have a higher stiffness. The response for periodic ground excitation is given by uˆ t = H 3 uˆ g

(1.114)

where H 3 is plotted in Fig (1.16). This function also relates the applied periodic force and the corresponding reaction, ˆ = H pˆ R 3

(1.115)

These equations are used to establish a control strategy for isolating a system from an external action, either an applied loading or a support motion. In general, one wants H 3 < 1 for isolation. Defining H 3* as the design requirement, the allowable stiffness range is bounded by the curves for ξ = 0 and ξ = 0.3 , as indicated in Fig (1.16). The absolute upper limit on stiffness is ρ =

Ω2m 2 ⇒ k max = -----------2

(1.116)

Designs are generated by specifying stiffness, and then determining the corresponding value of ξ . Isolation is achieved by reducing the stiffness below the critical level, k max . If k > k max , isolation is not possible even for the case where maximum damping ( ξ = 0.3 ) is introduced.

38 Chapter 1: Introduction

4

3.5

ξ = 0 3

2.5

H3 2

ξ = 0.3

1.5

1

H 3* 0.5

ξ = 0.3

ξ = 0 0

0

0.5

1

1.5

2

2.5

3

3.5

Ω m ρ = ---- = Ω ---ω k Fig. 1.16: Response function for periodic ground excitation

4

Active control of a SDOF system for dynamic loading involves applying a control force, F(t ), and adjusting its magnitude over time in such a manner that the resulting motion is constrained within the desired limits. One needs a force actuator that responds essentially in real time. Specifying the magnitude and sense of F(t ) is the key issue for active control. Various approaches are discussed later in the text. In general, F is selected to oppose the motion and does negative work on the mass. This work has to be supplied by an external energy source. Taking F to be proportional to the displacement and velocity, F ( t ) = +k d u ( t ) + k v u˙ ( t )

(1.117)

corresponds to increasing the stiffness and damping of the original system. This observation follows by substituting for F in the equilibrium equation. Then, active control based on eqn (1.117) can be interpreted as introducing virtual stiffness and

1.6 Control Strategies for Motion Based Design

39

damping. From an energy perspective, stable active control reduces the energy demand for the system. The energy balance equation for a linear SDOF system initially at rest is given by t

∫0

pu˙ dt +

∫

t

t 1 1 ( – Fu˙ ) dt = --- mu˙ 2 + --- ku 2 + cu˙ 2 dt 2 2 0 0

∫

(1.118)

The last term is the energy dissipated through viscous action. Passive control provides energy storage and energy dissipation mechanisms to resist the demand. When F(t ) is taken to have the same sense as u˙ , active control decreases the energy input to the system since the integral is always negative. The effect of this action is a lower energy demand that has to be met by passive control mechanisms. Selecting a motion control strategy for dynamic excitation involves a number of decisions. Firstly, should the strategy be purely passive or a combination of passive and active control? Secondly, what percentage of the energy demand should be met by energy storage (stiffness) vs. energy dissipation (damping)? Thirdly, the type and properties of passive energy dissipation device needs to be specified. There are a number of passive energy dissipation devices that are appropriate for structural systems. Fourthly, if active control is used to supplement passive control, the type and capacity of the force actuator needs to be established. Although the discussion has been focussed on a SDOF system, the concepts introduced here are also applicable for a general multi degree of freedom structural system. The only difference is that now one is dealing with sets of stiffness and damping parameters, and motion criteria involving the displacement variables associated with the degrees of freedom. Using vector notation, the problem can be stated as: given a desired displacement response vector, u * ( t ) , determine k, c, and F, the vectors containing the stiffness, damping, and active control force variables. These vectors are functions when the system is continuous. The desired response for a structure is related to the nature of the loading and the critical performance measures chosen for the structure. For service

40 Chapter 1: Introduction loading, the damage that non-structural elements, such as wall panels of buildings, can experience constrains the magnitude and distribution of displacements, while human and equipment comfort limits the peak acceleration. The controlling criterion for wind dominant design tends to be the peak velocity and acceleration. Displacement is the controlling criterion for earthquake dominant designs. In general, the structure is required to remain elastic under service loading. Under extreme loading, structural performance and stability requirements constrain the magnitude and distribution of inelastic deformation that the structural components can experience. Structural deformation is the key measure for earthquake dominant design. Although design codes allow a structure to experience inelastic deformation under an extreme earthquake with no collapse or loss of life, the current trend is to reduce the allowable inelastic deformation (Fajfar & Krawinkler 1997). This shift is driven by the need to lower the cost of repair. Furthermore, recent studies have shown that the initial cost of a structure designed to remain elastic under extreme seismic load may be less than the cost corresponding to the conventional design approach which allows for residual deformation (Connor et al, 1997). Combining these savings with the decrease of damage repair and increased revenue due to maintaining operability can result in a substantial reduction in life cycle cost. In order to satisfy the complete set of motion requirements, a multi-level control strategy is required. Figure (1.17) shows the strategy adapted for this text. The first step for passive control is to establish the distribution of stiffness, which involves the choice of material stiffness and cross-sectional properties, by enforcing the requirements on the spatial distribution of displacements corresponding to the design loading condition. Example 1.9 illustrates this computation for a 2-DOF system. The ideal state is uniform deformation throughout the structure. However, this state is possible only for a statically determinate structure. The requirement on the magnitude of the response is met with a combination of scaling the stiffness distribution and incorporating passive energy dissipation devices such as viscous dampers, tuned mass and liquid sloshing dampers, and base isolation systems. Depending on the level of damping chosen, damping and stiffness may be coupled. In this case, one cannot independently scale stiffness and damping, and iteration on the scaling process and possibly also on the stiffness distribution may be required. Typical civil structures are usually lightly damped, with ξ < = 0.05 , and coupling is not a

1.6 Control Strategies for Motion Based Design

41

problem. However, the current trend is toward more heavily damped structures with ξ ≈ 0.2 to 0.3. Seismic design for a heavily damped structure requires iteration between stiffness and damping. The response for extreme loading is controlled by supplementing stiffness and damping with mechanisms that absorb energy. An example of an absorption mechanism is a hysteretic damper. These devices experience residual inelastic deformation, which is a measure of the energy absorbed. The first step for active control is to decide on a location for force actuators which provides the capability for controlling the displacement response in the desired manner. Controllability is a critical issue for active control. It relates to the sensitivity of the response to the control force magnitude. Given the distribution, the desired force magnitudes are established using a control algorithm, and instructions then are sent to the actuators to deliver these forces. Examples of force actuators are active mass drivers and active tendon systems. Active control can be utilized to supplement passive control for service loading. Since the level of power, energy, and force required for active control of large scale civil structures under extreme loading cannot be economically achieved with the present technology, passive control is the predominant choice for this loading.

Load

Functional

Level

requirements

Service

Extreme

Control

Action

Passive

Active

-Control the spatial distribution of displacements

-Establish the spatial distribution of stiffness

-Select an appropriate spatial distribution of force actuators

-Control the magnitude of the response

-Specify a distribution of damping (energy dissipation) devices -Scale the stiffness and damping magnitudes

-Establish the magnitude of the control forces -Activate the force actuators

-Control the distribution and magnitude of the response

-Include energy absorption mechanisms

Fig. 1.17: Multilevel control strategy

42 Chapter 1: Introduction

1.7 Scope of text This text presents a systematic treatment of the concepts and computational procedures for motion control of civil structures. The material is organized according to the nature of the control process. Part I concerns passive control, and includes Chapters 2 thru 5. Methodologies for stiffness and damping based control are presented, and applied to typical building type structures. Part II consists of Chapters 6 thru 9. These chapters are intended to provide an introduction to active control concepts and computational algorithms.Quasi static control is discussed first, since it is easier to deal with analytically. Classical dynamic control algorithms are treated next.Their application to civil structures is illustrated with a series of computer simulation studies. A brief description of the individual chapters is provided below. Chapter 2 deals with optimizing the distribution of stiffness for beam and frame type structures. The chapter begins with an overview of the governing equations for beam type structures and then addresses the static problem. A method for distributing rigidity to attain the desired static deformation profiles is presented. The method is extended to the dynamic case and an analytic formulation is developed for structural systems characterized by a dominant fundamental mode response. How to distribute stiffness in structures so as to generate a fundamental mode shape with uniform shear and bending deformation profiles is the key issue. An iterative numerical scheme based on the response spectrum method is developed to incorporate the effect of the higher modes which get amplified as the slenderness of the structure increases. Results for a set of representative buildings subjected to seismic excitation are presented. They provide an assessment of the effectiveness of controlling displacement with stiffness. Chapter 3 examines the role of damping for motion control of dynamically excited structures. The chapter starts by examining the response characteristics of single-degree-of-freedom (SDOF) systems with various types of damping mechanisms. The concept of equivalent viscous damping is introduced and is used to express both structural damping and hysteretic damping in terms of their viscous counterpart. Numerical results are presented to evaluate the validity of this concept for SDOF systems. Multi-degree-of-freedom (MDOF) systems are discussed next. A formulation is presented for the case where the damping distribution is taken proportional to the stiffness distribution generated in Chapter 2. The effectiveness of damping under seismic excitation is assessed by

1.7 Scope of Text 43 numerical simulation. Examples illustrating the design of viscoelastic and hysteretic dampers are also included. Chapter 4 assesses the effectiveness of tuned mass dampers (TMD) for controlling the motion of dynamically excited structures. The theory of TMDs for SDOF systems is developed and applied to a range of SDOF systems subjected to harmonic and seismic excitations. The theory is then extended to MDOF systems, and a procedure for designing a TMD to control a specific modal response is developed and illustrated. Descriptions of existing implementations of tuned mass dampers in building structures are also presented. Chapter 5 considers base isolated structures (BIS). Base isolation is a control strategy that modifies the stiffness distribution. A base isolation control strategy for SDOF systems is developed, and then extended to the MDOF case. The procedure is similar to the approach followed in Chapter 2. One distributes stiffness along the height of the building so that the fundamental mode experiences uniform deformation, and selects the stiffness of the base isolation system such that the deformation in the isolator is bounded. The iterative procedure developed in Chapter 2 to account for the effect of the higher modes, is modified to incorporate the effect of the base isolation system. Chapter 6 provides an overview of active control. The architecture of an active control system is viewed as having 3 main components: a monitoring system which acquires data; a cognitive unit called a controller which processes the acquired date and decides on a course of action; and an actuator system composed of a set of physical devices that execute the instructions from the controller. Various types of controllers are described, and examples illustrating the application of particular type of control action called linear feedback control are presented. Existing established actuator technologies are discussed in considerable detail. An overview of emerging actuator technologies based on adaptive materials is also included. Lastly, examples of existing structural control systems for large scale structures are presented. Chapter 7 addresses quasi-static control, i.e., where the structural response to applied loading can be approximated as static response. Since time dependent effects are neglected, stiffness is the only quantity available for passive control. Active control combines stiffness with a set of pseudo-static control forces. Fundamental concepts such as observability, controllability, optimal control, and computational procedures are introduced in this chapter. Both continuous and

44 Chapter 1: Introduction discrete physical systems are treated. Chapter 8 considers dynamic feedback control of time-invariant multidegree-of freedom structural systems. A combination of stiffness, damping, and time dependent forces is used for motion control of dynamic systems. The statespace formulations of the governing equations for SDOF and MDOF systems are used to discuss stability, controllability, and observability aspects of dynamically controlled systems. Continuous and discrete forms of the linear quadratic regulator (LQR) control algorithm are derived, and examples illustrating their application to a set of shear beam type buildings are presented. The effect of time delay in the stability of LQR control, and several other linear control algorithms are also discussed.

Problems

45

Problems Problem 1.1 Refer to example 1.2 and equation 1.11. Construct plots of ( H * ⁄ u * ) γ = 1 , as a function of the aspect ratio ( H ⁄ d ) for the following ranges: •

E = 200, 000MPa

•

σ from 200MPa to 600MPa

•

*

( H ⁄ d ) from 3 to 8

Recommended design values of H ⁄ u∗ for a building are in the region 400 to 500. Comment on when “motion” rather than strength controls the design.

Problem 1.2 Refer to example 1.3. A typical value for the ratio of column depth to story height is 0.10. Using eqn (1.22), determine the value of ( H ⁄ u * ) for which the c constraint on displacement controls the design taking the following ranges for *

and σ : c

•

E = 200, 000MPa

•

σ from 200MPa to 600MPa

*

Problem 1.3 Refer to example 1.3. Suppose a mass, m , is attached to the “infinitely stiff” girder. Assuming the columns have negligible mass, determine the expression for the natural frequency, ω , for lateral vibration in terms of the material and geometric properties of the frame.

46 Chapter 1: Introduction

Problem 1.4 Consider a SDOF system having m = 1000kg and subjected to a sinusoidal force with amplitude pˆ = 500N and frequency 2πr ⁄ s . Recommend design values for stiffness k , and damping, c , corresponding to the following limiting values for peak acceleration: *

•

a = 0.1m ⁄ s

•

a = 1m ⁄ s

•

a = 3m ⁄ s

*

2

*

2

2

Problem 1.5 Consider a SDOF system with mass of 5000kg . The system is to be subjected to a periodic loading having a magnitude of 4kN and frequency 2 hertz. Recommend design values for stiffness and damping corresponding to the following limiting values of peak displacement: *

•

u = 1mm

•

u = 10mm

•

u = 20mm

* *

Problem 1.6 Suppose the mass of a SDOF system is known, but the stiffness, k , and damping, c , are unknown. Discuss how you would determine the stiffness by applying a periodic loading for which both the frequency and magnitude can be varied, and monitoring the response. How would you determine the damping parameter, c?

Problem 1.7 Consider a SDOF system having m = 1000kg . The system is to be subjected to a periodic force with a magnitude of 5kN and frequency 3Hz .

Problems

47

Recommend design values for k , and c such that the magnitude of the reaction force is less than: •

10kN

•

3kN

Problem 1.8 A SDOF system having m = 1000kg is to be subjected to the following ground motion: u g = 0.3 sin 4πt meters Recommend values for k and c for the following design values of peak relative displacement: *

•

u = 0.2m

•

u = 0.4m

*

Problem 1.9 Recommend values for k and c for a SDOF system with mass of 2000kg subjected to the following ground motion: u g = 0.5 sin 3πt meters The motion constraint is uˆ t ≤ 0.1m . What is the corresponding value for the peak relative motion, uˆ ? Discuss how the response measures, uˆ t and uˆ , vary as the spring stiffness is reduced from the value you select.

Problem 1.10 Consider a SDOF system subjected to ground excitation. Take m = 3000kg and the ground acceleration as

48 Chapter 1: Introduction a g = 4.0 sin 4πt Recommend values for k and c such that the peak total acceleration is less than 2

1.0m ⁄ s .

Problem 1.11 A machine represented by the mass m is to be supported by the spring and dashpot shown in the following figure. The machine is sensitive to total acceleration and therefore needs to be “isolated” form the ground motion. m=2000 kg

u+ug m

k

c ug

Consider the ground acceleration to consist of two dominant components ag ----- = 0.1 sin ( 2πt + δ 1 ) + 0.2 sin ( 4πt + δ 2 ) g where 2

g is the gravitational acceleration ( 9.81m ⁄ s ) and δ 1, δ 2 are random phase angles that can range from 0 to 2π A reasonable approximation for the peak acceleration of the combined response is at =

2

( a t, 1 ) + ( a t, 2 )

2

where a t, 1 and a t, 2 are the total accelerations due to the individual harmonic

Problems

49

excitations with random phasing. Suppose the desired maximum total acceleration is 0.05g . Describe how you would establish design values for k and c .

Problem 1.12 Consider the 2 DOF system discussed in Section 1.5. Take m 2 = 0.5m 1 . Determine the stiffness parameters k and k corresponding to m 1 = 1000kg and the following range of ω 1 : ω 1 = π, 2π, 4π Also determine ω 2 and the shape of the second mode corresponding to each stiffness distribution.

Problem 1.13 The derivation in Section 1.5 was based on specifying the fundamental mode to be linear. Suppose one takes ϕ 2 = αϕ 1 where α is a scale factor.

Generalize eqn (1.101) to incorporate α . What is the significance of taking α ≠ 2 ? What happens when α is taken to be negative?

50 Chapter 1: Introduction

51

Part I: Passive Control Chapter 2

Optimal stiffness distribution 2.1 Introduction This chapter is concerned with the first step in passive motion control, establishing a distribution of structural stiffness which produces the desired displacement profile. When the design loading is quasi-static, the stiffness parameters are determined by solving the equilibrium equations in an inverse way. Dynamic loading is handled by selecting the stiffness parameters such that the fundamental mode shape has the desired displacement profile. The implicit assumptions here are that one can incorporate sufficient damping to minimize the contributions of the higher modes, and the fundamental mode shape is independent of damping. The latter assumption is reasonable for lightly damped structures. Discrete systems are governed by algebraic equations, and the problem reduces to finding the elements of the system stiffness matrix. The static case involves solving *

KU = P

*

*

(2.1) *

for K , where U and P are the prescribed displacement and loading vectors. Some novel numerical procedures for solving eqn (2.1) are presented in a later section.

52 Chapter 2: Optimal Stiffness Distribution In the dynamic case, the equilibrium equation for undamped periodic excitation of the fundamental mode is used: *

2

KΦ = ω 1 MΦ

*

(2.2) *

where M is the discrete mass matrix, Φ is a scaled version of the desired displacement profile, and ω 1 is the fundamental frequency. Taking 1 K' = ------K 2 ω1 P' = M Φ

*

(2.3)

(2.4)

reduces eqn (2.2) to *

K'Φ = P'

(2.5)

The solution technique for eqn (2.1) also applies for eqn (2.5). Once K' is known, the stiffness can be scaled by specifying the frequency, ω 1 . An appropriate value for ω 1 is established by converting the system to an equivalent one degree-of-freedom system and using the SDOF design approaches discussed in the introduction. Continuous systems such as beams are governed by partial differential equations, and the degree of complexity that can be dealt with analytically is limited. The general strategy of working with equilibrium equations is the same, but now one has to determine analytic functions rather than discrete values for the stiffness. Analytical solutions are useful since they allow the key dimensionless parameters to be identified and contain generic information concerning the behavior. In what follows, the first topic discussed concerns establishing the stiffness distribution for static loading applied to a set of structures consisting of continuous cantilever beams, building type structures modeled as equivalent discontinuous beams with lumped masses, and truss-type structures. Closed form solutions are generated for the continuous cantilever beam example. The next topic concerns establishing the stiffness distribution for the case of dynamic loading applied to beam-type structures. The process of calibration of the

2.2 Governing Equations - Transverse Bending of Planar Beams

53

fundamental frequency is described and illustrated for both periodic and seismic excitation. The last section of the chapter deals with the situation where the higher modes cannot be ignored. An iterative numerical scheme is presented and applied to a representative range of beam-type structures.

2.2 Governing equations - transverse bending of planar beams In this section, the governing equations for a specialized form of a beam are developed. The beam is considered to have a straight centroidal axis and a crosssection that is symmetrical with respect to a plane containing the centroidal axis. Figure 2.1 shows the notation for the coordinate axes and the displacement measures (translations and rotations) that define the motion of the member. The beam cross-section is assumed to remain a plane under loading. This restriction is the basis for the technical theory of beams, and reduces the number of displacement variables down to three translations and three rotations which are functions of x and time. When the loading is constrained to act in the symmetry plane for the crosssection, the behavior involves only those motion measures associated with this plane. In this discussion, the x-y plane is taken as the plane of symmetry, and u x , u y , and β z are the relevant displacement variables. If the loading is further restricted to act only in the y direction, the axial displacement measure, u x , is identically equal to zero. The behavior for this case is referred to as transverse bending. In what follows, the governing equations for transverse bending of a continuous planar beam are derived. The derivation is then extended to deal with discontinuous structures such as trusses and frames that are modelled as equivalent beams.

54 Chapter 2: Optimal Stiffness Distribution

Note:

x-y plane is a plane of symmetry for the cross-section y y βy uy

ux

βx x

uz z

z

βz

Fig. 2.1: Notation - planar beam.

Planar deformation-displacement relations Figure 2.2 shows the initial and deformed configurations of a differential beam element. The cross-sectional rotation, β z , is assumed to be sufficiently small such 2 that β z << 1 . In this case, linear strain-displacement relations are acceptable. Letting γ denote the transverse shearing strain and ε the extensional strain at an arbitrary location y from the reference axis, and taking β ≡ β z and u ≡ u y , the deformation relations take the form ε = – yχ

(2.6)

∂u γ = ------ – β ∂x

(2.7)

∂β χ = -----∂x

(2.8)

where χ denotes the bending deformation parameter.

2.2 Governing Equations - Transverse Bending of Planar Beams

Y'

X'

Y

B

A y

A y

B

O

55

γ β

O

X

X

Fig. 2.2: Initial and deformed elements.

Optimal deformation and displacement profiles Optimal design from a motion perspective corresponds to a state of uniform shear and bending deformation under the design loading. This goal is expressed as γ = γ∗

(2.9)

χ = χ∗

(2.10)

Uniform deformation states are possible only for statically determinate structures. Building type structures can be modeled as cantilever beams, and therefore the goal of uniform deformation can be achieved for these structures. Consider the vertical cantilever beam shown in Fig. 2.3. Integrating eqns (2.7) and (2.8) and enforcing the boundary conditions at x = 0 leads to β = χ∗ x

(2.11) 2

χ∗ x u = γ ∗ x + -----------2

(2.12)

The deflection at the end of the beam is given by 2

χ∗ H u ( H ) = γ ∗ H + -------------2

(2.13)

56 Chapter 2: Optimal Stiffness Distribution 2

where γ ∗ H is the contribution from shear deformation and χ∗ H ⁄ 2 is the contribution from bending deformation. For actual buildings, the ratio of height to width (i.e. aspect ratio) provides an indication of the relative contribution of shear versus bending deformation. Buildings with aspect ratios on the order of unity tend to display shear beam behavior and χ ≈ 0 . On the other hand, buildings with aspect ratios greater than about 7 display bending beam behavior and γ ≈ 0 . x

b(x) H

y,u

Fig. 2.3: Simple cantilever beam. One establishes the values of γ ∗ , χ∗ based on the performance constraints imposed on the motion, and selects the stiffness such that these target deformations are reached. Introducing a dimensionless factor s , which is equal to the ratio of the displacement due to bending and the displacement due to shear at x=H, 2 H χ∗ Hχ* s = -------------- ⁄ ( γ *H ) = ---------- 2 2γ *

(2.14)

transforms eqn (2.13) to a form that is more convenient for low rise buildings. u ( H ) = ( 1 + s )γ ∗ H A shear beam is defined by s = 0 . Tall buildings tend to have s ≈ 1 .

(2.15)

2.2 Governing Equations - Transverse Bending of Planar Beams

57

Equilibrium equations Figure 2.4 shows a differential beam element subjected to an external transverse loading, b , and restrained by the internal transverse shear, V , and bending moment, M . By definition, V =

∫

τ dA

(2.16)

∫

(2.17)

M = – yσ dA

where τ and σ are the stresses acting on the cross-section. Summing forces and moments leads to 2

∂ u ∂V ------- + b = ρ m ∂x ∂t2

(2.18)

2

∂ β ∂M -------- + V = J ∂x ∂t2

(2.19)

where ρ m , J are the mass and rotatory inertia per unit length. When the member is supported only at x = 0 (see Fig. 2.3), the equilibrium equations can be expressed in the following integral form H

V(x) =

∫

2

∂ u b – ρ m 2 dx ∂t

(2.20)

x

H

M(x) =

∫ x

2

∂ β V – J 2 dx ∂t

(2.21)

58 Chapter 2: Optimal Stiffness Distribution

y b

∂M M + --------dx ∂x

∂τ τ + ------ dx ∂x

M

∂σ σ + -------dx

σ

∂x

x

τ V

dx

∂V V + -------dx ∂x

Fig. 2.4: Forces acting on a differential element. In the case of static loading, the acceleration terms are equal to 0, and V and M can be determined by integrating eqns (2.20) and (2.21). Force-deformation relations The force-deformation relations, also referred to as the constitutive relations, depend on the characteristics of the materials that make up the beam. For the case of static loading and linear elastic behavior, the expressions relating the shear force and bending moment to the shear deformation and bending deformation respectively are expressed as V ( x ) = D T ( x )γ ( x )

(2.22)

M ( x ) = D B ( x )χ ( x )

(2.23)

where D T and D B are defined as the transverse shear and bending rigidities. These equations have to be modified when the deformation varies with time. This aspect is addressed in Section 2.4. Examples which illustrate how to determine the rigidity coefficients for a range of beam cross-sections are presented below.

2.2 Governing Equations - Transverse Bending of Planar Beams

59

Example 2.1. Composite sandwich beam Figure 2.5 shows a sandwich beam composed of 2 face plates and a core. b d

tf Fig. 2.5: Composite beam cross section.

The face material is usually much stiffer than the core material, and therefore the core is assumed to carry only shear stress. Noting eqn (2.6), the strains in the face and core are d ε f = ± --- χ 2

(2.24)

γc = γ

(2.25)

The face thickness is also assumed to be small in comparison to the depth. Considering the material to be linear elastic, the expressions for shear and moment are: V = ( bd )τ c = ( bdG c )γ

(2.26) 2

bt f d M = ( bt f d )σ f = -------------- E f χ 2

(2.27)

where G c is the shear modulus for the core and E f is the Young’s modulus for the face plate. The corresponding rigidity coefficients are: D T = bdG c

(2.28)

60 Chapter 2: Optimal Stiffness Distribution 2

bt f d D B = -------------- E f 2

(2.29)

Example 2.2. Equivalent rigidities for a discrete Truss-beam The term, truss-beam, refers to a beam type structure composed of a pair of chord members and a diagonal bracing system. Figure 2.6 illustrates an x-bracing scheme. Truss beams are used as girders for long span horizontal systems. Truss beams are also deployed to form rectangular space structures which are the primary lateral load carrying mechanisms for very tall buildings. The typical “mega-truss” has large columns located at the 4 corners of a rectangular crosssection, and diagonal bracing systems placed on the outside force planes. These structures are usually symmetrical, and the behavior in one of the symmetry directions can be modelled using an equivalent truss beam. When the spacing, h, is small in comparison to the overall length, one can approximate the discrete structure as a continuous beam having equivalent properties. In this example, approximate expressions for these equivalent properties are derived for the case of x-bracing.

M V c

Ec A

d

E d A

θ

h

F

c

F

d

F

d

F

c

B

Fig. 2.6: Parameters and internal forces - Truss-beam. The key assumption is that the members carry only axial force. This approximation is reasonable when the members are slender, and diagonal or chevron bracing is used. Noting Fig. 2.6, the cross section force quantities are related to the member forces by

2.2 Governing Equations - Transverse Bending of Planar Beams

M = BF

c

61

(2.30)

d

V = 2F cos θ

(2.31)

Assuming linear elastic behavior, the member forces are also related to the extensional strains by c

c c c

d

= A E ε

F = A E ε F

(2.32)

d d d

(2.33)

It remains to express the extensional strains in terms of the bending and shear deformation measures. Figure 2.7 shows the deformed shapes of a panel of the truss beam. The d extensional strain in the diagonals, ε , due to the relative motion between adjacent nodes, ∆ h , is a function of ∆ h and θ . ∆ h cos θ sin θ d ε ∆ = ------------------------------h h

(2.34)

Neglecting the extensional strain in the diagonal due to ∆β , and approximating γ as ∆h ------ ≈ γ h

(2.35)

one obtains the following approximation for the total extensional strain: d d γ sin 2θ ε ≈ε ≈ γ cos θ sin θ ≈ -----------------∆h 2

(2.36) c

Similarly, the extensional strain in the chord, ε , is related to the change in angle, ∆β , between adjacent sections by ∆v c B∆β ε = ------ = ----------h 2h Noting that ∆β ⁄ h is related to the bending deformation χ ,

(2.37)

62 Chapter 2: Optimal Stiffness Distribution ∆β χ = ------h

(2.38)

the strain can be expressed as c Bχ ε = ------2

(2.39)

+u

+β

∆h

∆β

∆v

h

θ

θ

x

B

y

B

Fig. 2.7: Deformed truss-beam section. c

d

Substituting for ε and ε and combining eqns (2.30)-(2.33) results in c c 2

A E B M = -------------------- χ 2 d d

V = [ A E sin 2θ cos θ ]γ

(2.40)

(2.41)

Comparing these expressions with the definition equations for the rigidity parameters leads to the following relations for the equivalent continuous beam properties: c c 2

A E B D B = -------------------2 d d

D T = A E sin 2θ cos θ

(2.42)

(2.43)

2.2 Governing Equations - Transverse Bending of Planar Beams

63

When the truss beam model is used to represent a tall building, the chords correspond to the columns of the building. These elements are required to carry both gravity and lateral loading whereas the diagonals carry only lateral loading. Since the column force required by the gravity loading can be of the same order as the force generated by the lateral loading, the allowable incremental deformation in the column due to lateral loading should be less than the corresponding value for the diagonal. To allow for this reduction, a factor, f ∗ , which is defined as the ratio of the diagonal strain to the chord strain for lateral loading is introduced. d

ε f ∗ = ----c ε

(2.44)

This factor is greater than 1. Substituting for the strain measures from eqns (2.36) and (2.39) results in γ sin 2θ χ = -----------------f ∗B

(2.45)

Once the shear deformation level γ ∗ is specified, the bending deformation is determined with γ ∗ sin 2θ χ∗ = --------------------f ∗B

(2.46)

Substituting for χ∗ in eqn (2.14), the ratio of the contributions from bending and shear deformation expands to H sin 2θ s = -------------------2 f ∗B

(2.47)

Typical values of f ∗ for buildings range from about 3 for elastic behavior to 6 for inelastic behavior. Equation (2.47) shows that the bending contribution becomes more important as the aspect ratio, H ⁄ B , increases. The shear and bending contributions to the elastic displacement at the top of the building are essentially equal when H ≈ 6B .

64 Chapter 2: Optimal Stiffness Distribution

Governing equations for buildings modelled as pseudo shear beams This section considers a class of planar rectangular building frames having aspect ratios of order O(1) and moment resisting connections. Figure 2.8 shows a typical case. This type of structure is the exact opposite to the truss beam with respect to the way the lateral loading is carried. In the case of the truss beam, the transverse shear is provided by the axial forces in the braces. Here, the shear is produced by bending of the columns. The axial deformation of the columns is usually small for low rise frames, so it is reasonable to assume the “floors” experience only lateral displacement and slide with respect to each other. Considering the structure as a pseudo-beam, there is no rotation of the cross-section, i.e., β = 0 ; there is only one displacement variable per floor; and the transverse shearing strain at a story location is equal to the interstory displacement divided by the story height. In what follows, the formulation of the governing equations is illustrated using a simple structure and then generalized for more complex structures.

h H

B Fig. 2.8: Low rise rigid frame. The 2-story frame shown in Fig 2.9 is modeled as a 2 DOF system having masses concentrated at the floor locations and shear beam segments which represent the action of the columns and beams in resisting lateral displacement. The shear forces in the equivalent beam segments (see Fig. 2.10) are expressed in terms of shear stiffness factors:

2.2 Governing Equations - Transverse Bending of Planar Beams V 1 = k1 u1

V 2 = k2 ( u2 – u1 )

65

(2.48)

Noting the definition of transverse shear strain, γ 1 = u1 ⁄ h1

u2 – u1 γ 2 = -----------------h2

(2.49)

one can relate the k ‘s to the equivalent transverse shear rigidity factors: V 1 = D T, 1 γ 1 , V 2 = D T, 2 γ 2

⇒ D T, i = h i k i , i = 1, 2

Lb h2

m2

interior

u2, p2 k2

exterior

u1, p1

m1 h1

(2.50)

k1

Fig. 2.9: A discrete shear beam model. m2

u2, p2 V2 V2

m1

u1, p1 V1 V1

Fig. 2.10: External and internal forces for discrete shear beam model. The equivalent shear stiffness factors are determined by displacing the floors of the actual frame, determining the shear forces in the columns, summing these forces for each story, and equating the total shear forces to V 1 and V 2 as

66 Chapter 2: Optimal Stiffness Distribution defined by eqn (2.48). The shear force in the j ’th column of story i is expressed as: V i, j = k i, j ( u i – u i – 1 )

(2.51)

where k i, j depends on the frame geometry and member properties. Then, summing the column shears for story i and generalizing eqn (2.48) leads to ki =

∑j ki, j

(2.52)

For this example, i = 1, 2 and j ranges from 1 to 5 . An approximate expression for the column shear stiffness factors can be obtained by assuming the location of the inflection points in the columns and beams. Taking these points at the mid-points, as indicated in Fig. 2.9, leads to the following estimates for interior and exterior columns: 12EI c k ( interior column ) = ---------------------3 h (1 + r)

(2.53)

12EI c k ( exterior column ) = -------------------------3 h ( 1 + 2r )

(2.54)

where r is a dimensionless parameter, I c Lb r = ---- ⋅ ----h Ib

(2.55)

and the subscripts denote column and beam properties. A typical frame has r = O(1) . The equilibrium equations for the “discrete” beam are established by enforcing equilibrium for the lumped masses shown in Fig. 2.10. p 1 = V 1 – V 2 + m 1 u˙˙1

p 2 = V 2 + m 2 u˙˙2

Substituting for V 1 and V 2 , eqn (2.56) expands to

(2.56)

2.2 Governing Equations - Transverse Bending of Planar Beams

p 1 = k 1 u 1 + k 2 ( u 1 – u 2 ) + m 1 u˙˙1 p 2 = k 2 ( u 2 – u 1 ) + m 2 u˙˙2

67

(2.57)

It is convenient to express eqn (2.57) in matrix form. The various matrices are defined as:

U =

P =

u1

(2.58)

u2 p1

(2.59)

p2

M =

m1 0

K =

k1 + k2

–k2

–k2

k2

(2.60)

0 m2

(2.61)

With these definitions, eqn (2.57) has the form ˙˙ P = KU + M U

(2.62)

Equation 2.20 expresses the shear force in a continuous beam as an integral of the applied lateral loading. The corresponding equations for this discrete system are obtained from eqn (2.56) by combining the individual equations: V 2 = p 2 – m 2 u˙˙2 V 1 = p 1 + p – m 1 u˙˙1 – m 2 u˙˙2

(2.63)

2

In general, the shear force in a particular story is determined by summing the forces acting on the stories above this story. A building having n stories is considered next. The building is modeled as

68 Chapter 2: Optimal Stiffness Distribution an n DOF system with lumped mass and equivalent shear springs, as shown in Fig. 2.11. The strategy for determining the equivalent shear stiffness factors is the same as discussed above. The only difference for this case is the form of the system matrices. They are now of order n . mn kn

Vi+1 pi

mi

m2

u2, p2

Vi

k2 m1 k1 Fig. 2.11: General shear beam model The equilibrium equation for mass i is given by p i = m i u˙˙i + V i – V i + 1

(2.64)

Expressing the shear forces in terms of the nodal displacements V j = k j(u j – u j – 1)

(2.65)

and substituting in eqn (2.64) results in p i = m i u˙˙i – k i u i – 1 + ( k i + k i + 1 )u i – k i + 1 u i + 1

(2.66)

This equation defines the entries in the i ‘th row of M and K . The expanded forms are listed below. m1 m2 M =

. . . mn

2.3 Stiffness Distribution for a Continuous Cantilever Beam under Static Loading

k1 + k2

–k2

0 . ..

0

0

0

–k2

k2 + k3

–k3 . . .

0

0

0

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

0

0

kn – 1 + kn

–kn

0

0

0 . . . –kn – 1 0 . .. 0

–kn

kn

K =

. . .

69

(2.67)

2.3 Stiffness distribution for a continuous cantilever beam under static loading Once the shear and bending moment distributions are specified, the rigidity distributions required to produce a specific deformation profile can be evaluated using eqns (2.22) and (2.23). The equations corresponding to uniform deformation reduce to V D T = ----γ∗

(2.68)

M D B = -----χ∗

(2.69)

For example, taking a uniform loading b ( x ) = b as shown in Fig. 2.12, which is a reasonable assumption for the wind action on a tall building, results in V (x) = b[H – x]

(2.70) 2

b[H – x] M ( x ) = -----------------------2 and

(2.71)

70 Chapter 2: Optimal Stiffness Distribution bH b[H – x] x D T ( x ) = --------------------- = ------- 1 – ---H γ∗ γ∗

(2.72)

3 2 bH x 2 b[H – x] D B ( x ) = ------------------------ = ----------- 1 – ---H 4sγ ∗ 2χ∗

(2.73)

b

H

x

B

Fig. 2.12: Continuous cantilever beam. A cantilever beam having a linear shear rigidity distribution and a quadratic bending rigidity distribution will be in a state of uniform deformation under uniform transverse loading. Taking typical values for γ ∗ , f ∗, and the aspect ratio B ⁄ H for a tall building modelled as a truss beam, 1 γ ∗ = --------400

f∗ = 3

B 1 ---- = --H 6

(2.74)

and evaluating s , H s = ------------- = 1 2 f ∗B

(2.75)

one obtains 2

χ∗ H H u ( H ) = γ ∗ H + -------------- = γ ∗ H ( 1 + s ) = --------2 200

(2.76)

This result corresponds to the extreme load value for displacement. One would

2.3 Stiffness Distribution for a Continuous Cantilever Beam under Static Loading

71

use these typical values together with b and H to establish an appropriate value for D T at x = 0 . As will be seen later, the rigidity distributions need to be modified near x = H in order to avoid excessive deformation under dynamic load.

Example 2.3. Cantilever beam - quasi static seismic loading The cantilever beam loading shown in Fig. 2.13 is used to simulate, in a quasi-static way, seismic excitation for low rise buildings. The triangular loading is related to the inertia forces associated with the fundamental mode response, and the concentrated force is included to represent the effect of the higher modes. Evaluating V and applying eqn (2.68) leads to b0 H x 2 1 D T = ----- P + ---------- 1 – ---- H 2 γ∗

(2.77)

A combination of constant and quadratic terms is a reasonable starting point for the transverse shear rigidity distribution.

P

b0

H

Fig. 2.13: Cantilever beam - quasi-static seismic loading

72 Chapter 2: Optimal Stiffness Distribution Example 2.4. Truss-beam revisited This example extends the treatment of the truss beam discussed in Example 2.2 and focuses on comparing the cross-sectional parameters required to satisfy the strength based versus the stiffness based performance criteria. Considering elastic behavior and given the desired design deformations d∗ c∗ γ ∗ and χ∗ , the corresponding extensional strains ε and ε must be less than d c the yield strains for the element materials, ε y and ε y respectively. That is d

d∗

γ ∗ sin 2θ = γ ∗ cos θ sin θ = --------------------2

(2.78)

c

c∗

Bχ∗ = ---------2

(2.79)

εy ≥ ε εy ≥ ε

Once the dimensions and the design deformations are specified, the structural material can be chosen to satisfy the motion design constraints defined by eqns (2.78) and (2.79). When the column strain is constrained to be related to the diagonal strain by ε

c∗

d∗ ε = -------f∗

(2.80)

eqn (2.79) can be written as d

c εy

εy ≥ -----f∗

(2.81)

To provide more options in satisfying the design requirements, different materials may be used. One must also insure that the stresses due to the design forces, V and M , are less than the yield stresses. c

d

The axial forces in the columns, F , and diagonals, F , are related to the transverse shear and moment by d

V = 2F cos θ M = BF

c

(2.82) (2.83)

2.3 Stiffness Distribution for a Continuous Cantilever Beam under Static Loading

73

The cross-sectional areas required to provide the strength capacity follow from eqns (2.82) and (2.83) d V A strength ≥ -------------------------d∗ 2σ cos θ

(2.84)

c M A strength ≥ ------------c∗ Bσ

(2.85)

where σ∗ denotes the allowable stresses based on strength considerations. The rigidity terms for this model are d d

2

D T = 2A E sin θ cos θ

(2.86)

c c

A E B2 D B = -------------------2

(2.87)

Substituting eqns (2.86) and (2.87) in the motion based design criteria, V D T = ----γ∗

(2.88)

M D B = -----χ∗

(2.89)

one obtains the following expressions for the cross-sectional areas required to satisfy the stiffness requirement. d V A stiffness ≥ --------------------------------------------d 2 2E γ ∗ sin θ cos θ

(2.90)

c 2M f ∗M A stiffness ≥ ------------------- = ----------------------------------------c 2 c E B χ∗ E Bγ ∗ sin θ cos θ

(2.91)

The ratio of areas provides a measure of the relative importance of strength versus stiffness

74 Chapter 2: Optimal Stiffness Distribution d d d∗ d A strength ε E γ ∗ E sin θ cos θ ----------------------- = -------------------------------------- = ---------------d d∗ d∗ A stiffness σ σ

(2.92)

c c c∗ c A strength ε E χ∗ E B ----------------------- = ---------------- = -------------c c∗ c∗ A stiffness σ 2σ

(2.93)

Stiffness controls when the ratios are less than unity. The limit on γ ∗ follows from eqn (2.92) d∗ σ ∗ γ = -------------------------------d E sin θ cos θ

(2.94)

For γ ∗ < γ ∗ , the cross-sectional area is governed by the deformation constraint, and eqns (2.90) and (2.91) apply. When γ ∗ > γ ∗ , the allowable stress is the controlling factor, and eqns (2.84) and (2.85) apply. Inelastic behavior occurs in this case. Values of γ ∗ for a range of allowable stress levels for steel calculated using an angle of 45° are listed below in Table 2.1. With high-strength steel, the structure can experience substantial transverse shear deformation and still remain elastic. Table 2.1: γ ∗ values for various steel strengths. σ∗ ( MPa )

γ∗

250

1 ⁄ 400

500

1 ⁄ 200

1000

1 ⁄ 100

2.4 Stiffness Distribution for a Discrete Cantilever Shear Beam - Static Loading

75

2.4 Stiffness distribution for a discrete cantilever shear beam - static loading Consider the set of equilibrium equations relating the nodal forces and story displacements for an n ‘th order discrete shear beam: p1

k1 + k2

p2

–k2

. = . . pn

. . . 0

–k2

. . . 0

u1

k2 + k3 . . . 0

u2

. . . 0

.

. . . . . . . . . . . kn un

(2.95)

P = KU These equations are derived in section 2.2. In the normal analysis problem, one specifies P and K , and solves for U . The problem is statically determinate since there are n equations for the n unknown displacements. In this problem, one specifies U and P , and attempts to determine the n stiffness factors. Since there are n linear algebraic equations, it should be possible to solve for the n stiffness coefficients by rearranging the equations such that the k ‘s are the unknowns. The vector containing the stiffness coefficients is denoted by k . k1 k2 k =

. . . kn

(2.96)

With this definition, eqn (2.95) is written as Sk = P

*

(2.97)

where the elements of S are linear combinations of the prescribed displacement components, u* i , and P * contains the prescribed loads. The entries in the i ‘th

76 Chapter 2: Optimal Stiffness Distribution row of S follow from eqn (2.66). S ( i, i ) = u* i – u* i – 1 S ( i, i + 1 ) = u* i – u* i + 1

(2.98)

S ( i, j ) = 0 for j ≠ i, i + 1 These entries define S to be an “upper triangular” matrix. The diagonal entries for S are interstory displacements. Normally, one would not specify an interstory displacement to be 0 since it would require an infinite shear stiffness. It follows that S will be non-singular, and there will be a unique solution for k .

Example 2.5. 3 DOF shear beam Consider a 3 DOF cantilever beam subjected to uniform nodal loading and required to have a linear displacement profile. The design values are: 19.6 P = 19.6 kN 19.6

(2.99)

0.025 U = 0.050 m 0.075

(2.100)

*

*

Applying eqn (2.98) results in k1 0.025 – 0.025 0 19.6 = 0 0.025 – 0.025 k 2 19.6 0 0 0.025 k 19.6 3

(2.101)

Since the coefficient matrix is upper triangular, one solves for k 3 , k 2 , and then k 1 by backsubstitution. The solution is

2.5 Stiffness Distribution - Truss under Static Loading 77

2352 k = 1558 kN/m 784

(2.102)

As illustrated with the above example, solving eqn (2.97) is relatively simple, and one can easily handle arbitrary loading and permissible (non-zero interstory drift) displacement profiles.

2.5 Stiffness distribution - truss under static loading The approach discussed in the previous sections can be extended to deal with other types of structures such as trusses and frames. Since trusses involve simple equations, the approach is illustrated using a planar truss. A more general discussion is contained in Gallegos (1998). P2, u2 P1, u1

1

3 2

45o

60o

Fig. 2.14: Planar truss. The truss shown in Fig 2.14 has 3 members and 2 displacement variables; the supports are assumed to be rigid so there are no support movements. Each member has an extension, e , and corresponding force, F . Assuming linear elastic behavior, the force-displacement relation for member i is Fi = ki ei

(2.103)

78 Chapter 2: Optimal Stiffness Distribution where k is the member stiffness factor, AE k i = -------- L member i

(2.104)

Using a geometric analysis, the member extensions are expressed in terms of the nodal displacements: 2 e 1 = ------- ( u 1 + u 2 ) 2 e2 = u2

(2.105)

1 3 e 3 = – --- u 1 + ------- u 2 2 2 The nodal and member forces are related by the nodal force equilibrium equations 2 1 P 1 = ------- F 1 – --- F 3 2 2

(2.106)

2 3 P 2 = ------- F 1 + F 2 + ------- F 3 2 2 Since there are 3 extensions and only 2 displacements, one cannot arbitrarily specify values for all 3 extensions. They are constrained by the geometric compatibility equation that follows by combining the first and third rows of eqn (2.105) and substituting for u 2 in terms of e 2 2 1 ------- e 1 + e 3 = --- ( 1 + 3 )e 2 2 2

(2.107)

In general, one cannot have a state of uniform deformation in a statically indeterminate system. One possible strategy is to specify Once the extensions are known, one equilibrium equations, and establish a stiffness factors. The result is

u 1 and u 2 , and evaluate the extensions. can substitute for the forces in the set of equations relating the member

2.5 Stiffness Distribution - Truss under Static Loading 79 Sk = P* 2 ------- e 1* 2 = * 2 P2 ------- e 1* 2 P1

*

0 e 2*

1 k1 – --- e 3* 2 k2 3 * ------- e 3 k 3 2

(2.108)

Since there are 3 unknown stiffness factors and only 2 equations, eqn (2.108) represents an undetermined system of equations and one cannot obtain a unique solution for k . This situation is typical of indeterminate structures. A solution of eqn (2.108) can be generated by selecting 2 stiffness factors as the primary variables, such as k 1 and k 2 , and solving for these variables in terms of P 1* , P 2* , and the third stiffness factor, k 3 . The result is written as k1 = k1 + a1 k3

(2.109)

k2 = k2 + a2 k3 where k 1 , k 2 depend on the forces, P 1 and P 2 . Equation (2.109) represent a constraint on the stiffness parameters. An additional condition must be introduced in order to determine k 3 . This additional condition is usually established by formulating an optimization problem. One possibility is to work with a weighted volume measure that is related to the member stiffness factor, and seek the solution that corresponds to the minimum value of the sum of the weighted volumes. For example, one can take V i = ( AEL ) i

(2.110)

as the member volume variable. Expressing k i in terms of V i, AE 1 k i = -------- = ------ V i L i L i2

(2.111)

and substituting for k i in the constraint equations transforms these equations to

80 Chapter 2: Optimal Stiffness Distribution 2

V 1 = L 12 k 1 + a 1 ( L 1 ⁄ L 3 ) V 3 2

V 2 = L 22 k 2 + a 2 ( L 2 ⁄ L 3 ) V 3

(2.112)

The optimization problem can be stated as: minimize f ( V ) = V 1 + V 2 + V 3

(2.113)

Using eqn (2.112), the objective function reduces to (2.114)

f = bo + b1 V 3

where b o and b 1 are constants. Since f is a linear function, there is no minimum value and it follows that the approach does not introduce an additional independent constraint. Another possibility is to use a least square approach, i.e., to select V 3 such that the sum of the squares of the weighted member volumes is a minimum. The objective function for this optimization problem is 2 2 1 2 f = --- ( V 1 + V 2 + V 3 ) 2

(2.115)

Substituting for V 1 and V 2 transforms f to f = co + c1 V 3 + c2 V 3

2

(2.116)

Requiring f to be stationary with respect to a change in V 3 , ∂f = c1 + 2 c2 V 3 = 0 ∂V3

(2.117)

leads to c1 V 3 = – -------2c 2

(2.118)

The problem with this approach is that it tends to eliminate redundant members,

2.5 Stiffness Distribution - Truss under Static Loading 81 such as member 3 for this example. This tendency is evidenced by “optimal” volume values which are low, and may even be negative. A negative value for the volume indicates the member should be deleted. A third approach partially overcomes the limitation of the least square approach by working with the deviation from the mean value as the member variable. Let V m denote the mean value, 1 V m = --- ( V 1 + V 2 + V 3 ) 3

(2.119)

and V i represent the deviation from the mean value for member i, (2.120)

Vi = Vi – Vm The optimization problem is stated as 2 2 1 2 minimize f = --- ( V 1 + V 2 + V 3 ) 2

(2.121)

subject to the constraints on V 1 and V 2 . Substituting for V 1 and V 2 transforms f to 2

f = do + d1 V 3 + d2 V 3

(2.122)

The remaining steps are the same as for the classical least square approach. This approach generates a solution which tends toward the same value for each member. However, as will be illustrated in the following example, it does not eliminate the possibility of negative volume members.

Example 2.6. Application of least square approaches The procedures described above are applied to Fig 2.14 specialized for the data contained in (1)

82 Chapter 2: Optimal Stiffness Distribution

P2, u2 P1, u1

4m

1

3 2

4m

P 1 = 100kN

P 2 = 300kN

u 1 = 0.01m

u 2 = 0.01m

2.31m

(1)

Using eqn (2.105), the extensions are 2 e 1 = ------- ( 0.01 + 0.01 ) = 0.01414m 2 e 2 = 0.01m

(2)

1 3 e 3 = – --- ( 0.01 ) + ------- ( 0.01 ) = 0.00366m 2 2 Substituting for the e’s in eqn (2.108) results in 0.01k 1 – 0.00183k 3 = 100, 000 0.01k 1 + 0.01k 2 + 0.00317k 3 = 300, 000 Solving for k 1 and k 2 in terms of k 3 leads to

(3)

2.5 Stiffness Distribution - Truss under Static Loading 83

k 1 = 0.183k 3 + 10 7 k 2 = – 0.500k 3 + 2 ⋅ 10 7

(4)

Equation (4) represents the constraint on k 1 and k 2 . Substituting for k in terms of weighted volume measure, V=AEL, using eqn (2.111) transforms eqn (4) to V 1 = 0.275V 3 + 32 ⋅ 10 7 V 2 = – 0.375V 3 + 32

⋅ 10 7

(5)

Least square procedure The objective function is 2 2 2 1 f = --- ( V 1 + V 2 + V 3 ) = f ( V 3 ) 2

(6)

Differentiating with respect to V 3 , ∂V 1 ∂V 2 ∂f = V1 + V2 + V3 ∂V3 ∂V3 ∂V3

(7)

substituting for V 1 and V 2 , and then setting the resulting expression equal to zero results in 7

1.216V 3 – 3.222 ⋅ 10 = 0

(8)

The “optimal” least square values are V 3 = 2.65 ⋅ 10 7 V 1 = 32.71 ⋅ 10 7 V 2 = 31.00

⋅ 10 7

( units are newtons ⋅ meters )

(9)

84 Chapter 2: Optimal Stiffness Distribution The low value for V 3 indicates that this member is “redundant”, and could be removed. Mean value least square procedure One starts by establishing the mean value, 1 V m = --- ( V 1 + V 2 + V 3 ) = 0.3V 3 + 21.33 ⋅ 10 7 3

(10)

and the deviations V 1 = V 1 – V m = – 0.025V 3 + 10.71 ⋅ 10 7 V 2 = V 2 – V m = – 0.675V 3 + 10.67 ⋅ 10 7

(11)

V 3 = V 3 – V m = 0.700V 3 – 21.33 ⋅ 10 7 The objective function for this case is

1 f = --- ( V 12 + V 22 + V 32 ) = f ( V 3 ) 2

(12)

Setting the derivative of f with respect to V 3 equal to zero, ∂V 1 ∂V 2 ∂V 3 ∂f = V1 + V2 + V3 = 0 ∂V3 ∂V3 ∂V3 ∂V3

(13)

and noting eqn (11), reduces eqn (13) to 0.946V 3 – 22.40 ⋅ 10 7 = 0 The final mean value least square results are

(14)

2.5 Stiffness Distribution - Truss under Static Loading 85 V 3 = 23.67 ⋅ 10 7 V 1 = 38.50 ⋅ 10 7 V 2 = 23.10

(15)

⋅ 10 7

( units are newtons ⋅ meters ) Note that now V 3 is the same order of magnitude as V 1 and V 2 . The solution is sensitive to the prescribed loading and displacement quantities. To illustrate this point, the computation is repeated taking the same imposed displacements but different sets of nodal forces. The results are listed below: Case 1 P 1 = 100kN

P 2 = 200kN

Least square: V 1 = 31.37 ⋅ 10 7 V 2 = 16.86 ⋅ 10 7 V 3 = – 2.29 ⋅ 10 7 Mean value least square: V 1 = 35.37 ⋅ 10 7 V 2 = 11.40 ⋅ 10 7 V 3 = 12.26 ⋅ 10 7 Case 2 P 1 = 100kN Least square: V 1 = 30.02 ⋅ 10 7 V 2 = 2.71 ⋅ 10 7 V 3 = – 7.25 ⋅ 10 7

P 2 = 100kN

86 Chapter 2: Optimal Stiffness Distribution Mean value least square: V 1 = 32.23 ⋅ 10 7 V 2 = – 0.32 ⋅ 10 7 V 3 = 0.86 ⋅ 10 7 The case 2 result indicates that members 2 and 3 should be deleted. This is the correct solution for the specified loading and displacement. Taking P 1 = P 2 corresponds to a load at 45o, which coincides with the direction of member 1. Furthermore, an extension of member 1 corresponds to u 1 = u 2 , which is the specified displacement condition.

The procedures described above can be generalized to deal with an arbitrary truss. Suppose there are n degrees of freedom and m members. It follows that there are n equilibrium equations relating the m stiffness factors and the n prescribed nodal forces. These equations are written as Sk = P

(2.123)

where S is of order n × m . When m > n , the structure is indeterminate, and eqn (2.123) does not have a unique solution for k . In this case the problem is undetermined, and an optimization statement has to be formulated. One can work with either stiffness or weighted volume measures as the variables. To allow for different choices, the form of eqn (2.123) is generalized to AX = P

(2.124)

where X is an m th order vector containing the selected variables, and A is of order n × m . If one selects V defined by eqn (2.111) as the measure, A is obtained 2 by multiplying the i’th column of S by 1 ⁄ L i , and taking i from 1 to m . Background material on computational techniques for solving linear algebraic equations is contained in Strang,1993. In what follows, the use of one of these techniques to solve eqn (2.124) is described in general terms. The same procedure is also applied in quasi-static active control problems in Chapter 7.

2.5 Stiffness Distribution - Truss under Static Loading 87 The least square approach works with a scalar quantity, f, that is equal to the sum of the squares of the elements of X . Using matrix notation, f is given by 1 T f = --- X X 2

(2.125)

The elements of X are constrained by eqn (2.124). It follows that n elements can be expressed in terms of m – n elements, which can be interpreted as the unknowns that need to be selected such that f is a minimal value. Assuming the first n columns of A are independent, eqn (2.124) can be expressed in partitioned form X1 A1 A2 = P X2

(2.126)

where A 1 is of order n × n . Solving eqn (2.126) for X 1 leads to X 1 = X 1 + BX 2

(2.127)

where X 2 represents the unknown elements. Using eqn (2.127), X is expressed in terms of X 2 , X = X 1 + B X 2 = X + CX 2 I 0

(2.128)

Substituting for X , the objective function expands to 1 f = --- ( X + CX 2 ) T ( X + CX 2 ) 2

(2.129)

Requiring f to be stationary with respect to X 2 leads to the following set of m – n equations C T CX 2 = – C T X

(2.130)

One solves eqn (2.130) for X 2 , and then determines X 1 with eqn (2.127). This

88 Chapter 2: Optimal Stiffness Distribution solution corresponds to an absolute minimum value of f (see Strang, 1993). The MATLAB statement, X = pinv ( A )*P , generates this least square solution. The mean value least square approach works with the deviation from the mean, e i = x i – x mean

(2.131)

For this case, there are m members, and x mean is given by 1 x mean = ---- ( x 1 + x 2 + … + x m ) m

(2.132)

Using matrix notation, the deviation vector, e , can be expressed as, e = DX = D ( X + CX 2 )

(2.133)

where the entries in row i of D are 1 D ( i, i ) = 1 – ---m 1 D ( i, j ) = – ---m

(2.134) for j ≠ i

The objective function is a quadratic form in e . 1 f = --- e T e 2

(2.135)

Substituting for e using eqn (2.133), and enforcing stationarity with respect to X 2 leads to the equation for X 2 . ( D T C T CD )X 2 = – D T C T X

(2.136)

An alternative way of generating the solution is based on using Lagrangian multipliers to incorporate the constraint equations in the objective function (Strang, 1993). The “generalized” function is defined as

2.5 Stiffness Distribution - Truss under Static Loading 89 1 J = --- e T e + λ T ( AX – P ) 2 ˜

(2.137)

1 = --- X T ( D T D )X + λ T ( AX – P ) 2 ˜

where λ is a vector containing the n Lagrangian multipliers. Requiring J to be stationary with respect to both X and λ leads to the following set of equations: D T DX + A T λ = 0 ˜ AX = P

(2.138)

These equations can be combined, DT D A

AT X = 0 0 λ˜ P

(2.139)

and solved in a single step using a linear equation solver such as one of the MATLAB functions Example 2.7. Comparison of strength vs. displacement based design P 2, u 2 u1 θ θ 1 2

3

Consider the 3 member truss shown in the figure. Suppose members 1 and 3 to have the same properties. It follows that u 1 = 0 and e 3 ≡ e 1 . The member elongations are:

90 Chapter 2: Optimal Stiffness Distribution e 1 = u 2 cos θ e2 = u2

(1)

Noting eqn (1), the elongations are constrained by e 1 = e 2 cos θ

(2)

The member flexibility factor, f , is defined as 1 L f = --- = -------k AE

(3)

Then, the member force-deformation relation can be expressed as e = fF

(4)

Using (4), the geometric compatibility equation expressed in terms of member forces has the form f 1 F 1 = cos θ f 2 F 2

(5)

The solution for F 1 and F 2 is f1 F 1 = cos θ ------ F 2 f2 P2 F 2 = ---------------------------------f1 1 + 2 cos2 θ -----f2 •

(6)

Strength based design

Ideally, one would want the member stresses to be equal to an allowable stress, σ a . However, a state of uniform stress is not possible because of the indeterminate nature of the structure. Noting that e = Lε for a member, eqn (1) can be written as

(7)

2.5 Stiffness Distribution - Truss under Static Loading 91 L 1 ε 1 = cos θ L 2 ε 2

(8)

Then assuming the same material for members 1 and 2, and substituting L 2 = L 1 cos θ results in the constraint equation for the member stresses: σ 1 = ( cos2 θ )σ 2

(9)

According to eqn (9), σ 1 ≤ σ 2 . Let σ2 = σa σ 1 = σ a cos2 θ

(10)

represent the design values for the stresses. The member cross-sectional areas are: F2 A 2 = -----σa

(11)

F1 F1 A 1 = ------ = --------------------σ1 σ a cos2 θ Substituting for F 1 and F 2 in the equilibrium equation, F 2 + 2F 1 cos θ = P 2

(12)

leads to a relationship between the cross-sectional areas: P2 A 2 + [ 2 cos3 θ ]A 1 ≥ -----σa

(13)

92 Chapter 2: Optimal Stiffness Distribution The allowable solutions lie outside the triangle shaded in Figure 1. A2 P2 -----σa

Acceptable zone

0

A1 P2 – 1 3 ------ ( 2 cos θ ) σa Figure 1

The constraint equation is expressed as A 2 + αA 1 = β

(14)

There is no unique solution for the areas. One approach is to apply a least square approach to the member areas. Minimizing the function 2

J = ( A 1 + A 22 + A 32 )

(15)

subject to the constraint, eqn (14), results in the following design: αβ A 1 = --------------2 + α2 2β A 2 = --------------2 + α2 •

(16)

Motion based design

The member forces are expressed in terms of the displacement, u 2 , by using equations (1) and (4)

2.6 Stiffness Distribution for a Cantilever Beam - Dynamic Response 93

A1 E u2 F 1 = ----------- e 1 = A 1 E ------ cos θ L1 L1 A2 E u2 F 2 = ----------- e 2 = A 2 E -----L2 L2

(17)

Substituting for the forces in eqn (12) leads to P2 L2 A 2 + [ 2 cos3 θ ]A 1 ≥ -----------Eu 2

(18)

This equation is similar to the strength based constraint equation. They differ only with respect to the right hand sides. Comparing these terms, it follows that motion based design controls when P2 L2 P2 ------------ > -----Eu 2 σ a

(19)

which translates to σa u 2* < L 2 -----E

(20)

2.6 Stiffness distribution for a cantilever beam - dynamic response Rigidity distribution - undamped response The distribution of rigidity is established by requiring the fundamental mode for a dynamically loaded cantilever beam to have a certain profile. For undamped free vibration, the distributed loading consists only of the inertia terms

94 Chapter 2: Optimal Stiffness Distribution 2

∂ u ( x, t ) b ( x, t ) = – ρ m ( x ) ---------------------- = – ρ m ( x )u˙˙( x, t ) 2 ∂t m ( x, t ) = – J ( x )

2

(2.140)

∂ ˙˙ ( x, t ) β ( x, t ) = – J ( x )β 2 ∂t

For simplicity, a uniform mass distribution is assumed. The equilibrium equations specialized for uniform mass, elastic behavior, and no rotatory inertia loading take the form H

∫

(2.141)

V ( x, t ) dx = D B ( x )χ ( x, t )

(2.142)

V ( x, t ) = – ρ m u˙˙( x, t ) dx = D T ( x )γ ( x, t ) x

H

M ( x, t ) =

∫ x

The desired behavior is periodic vibration with uniform shear and bending deformation throughout the length. Requiring periodic vibration with uniform deformation modes is achieved by specifying the deformations as follows: γ ( x, t ) = γ ∗ cos ( ω 1 t + δ )

(2.143)

χ ( x, t ) = χ∗ cos ( ω 1 t + δ )

(2.144)

where ω 1 is the fundamental circular frequency of the beam and δ is an arbitrary scalar. The displacement expressions are generated using eqns (2.11) and (2.12) 2 χ∗ x u = γ ∗ x + ------------ cos ( ω 1 t + δ ) 2

(2.145)

β = [ χ∗ x ] cos ( ω 1 t + δ )

(2.146)

2s χ∗ = -----γ ∗ H

(2.147)

2.6 Stiffness Distribution for a Cantilever Beam - Dynamic Response 95 Substituting for u , γ and χ in eqns (2.141) and (2.142), the cosine terms cancel out, and one obtains the following expressions for the rigidity distributions, 2

2

ρm ω1 H x x 2 2s x 3 D T ---- = ---------------------- 1 – ---- + ----- 1 – ---- H 2 H 3 H 2

(2.148)

4

ρm ω1 H 1 1 x x 3 x 4x 1 x 4 D B ---- = ---------------------- --- 1 – -------- + --- ---- + ----- 2 – 3 ---- + ---- H H H 2 4 3H 3 H 6s

(2.149)

The magnitude of D T at x = 0 can be determined by specifying representative values of base shear and transverse shear deformation [ V ( 0 ) ] max D T ( 0 ) = --------------------------γ∗

(2.150)

Once D T ( 0 ) is defined, the fundamental frequency can be obtained with eqn (2.148) specialized for x = 0 . The final expressions are 2

2

ρm ω1 H 2s D T ( 0 ) = ---------------------- 1 + ----2 3

ω1 =

2 [ V ( 0 ) ] max -----------------------------------------2 2s γ ∗ ρ m H 1 + ----3

(2.151)

(2.152)

Figures 2.15 and 2.16 show the normalized distribution of D T and D B for typical low and high rise buildings. Figure 2.17 contains plots of the first five mode shapes corresponding to the above rigidity distributions, and Fig. 2.18 shows the shear deformation profiles of those modes. The results are relatively insensitive to the parameter s , which ranges from 0 for a pure shear beam to about 1 for a tall building. It should be noted that the mode shape and frequency expressions are the exact solution for the fundamental modal response of the beam having the rigidity distributions defined by eqns (2.148) and (2.149). This approach leads to the proper distribution of rigidities, but the actual magnitude is not defined since one has to specify D T ( 0 ) using a representative value of base shear. In effect, one has to calibrate the rigidities for a particular

96 Chapter 2: Optimal Stiffness Distribution loading. The calibration procedure is discussed in detail in later sections.

1

x Normalized height ---H

0.9 0.8 0.7

T = 5.35s H = 200m ρ m = 20000kg/m s = 0.63

0.6 0.5 0.4 0.3

T = 0.49s H = 25m ρ m = 20000kg/m s = 0.15

0.2 0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Shear rigidity distribution D T ⁄ D T ( 0 )

1

Fig. 2.15: Normalized shear rigidity distribution.

1

T = 5.35s H = 200m ρ m = 20000kg/m s = 0.63

x Normalized height ---H

0.9 0.8 0.7 0.6

T = 0.49s H = 25m ρ m = 20000kg/m s = 0.15

0.5 0.4 0.3 0.2 0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

Bending rigidity distribution

0.7

0.8

0.9

DB ⁄ DB ( 0 )

Fig. 2.16: Normalized bending rigidity distribution.

1

2.6 Stiffness Distribution for a Cantilever Beam - Dynamic Response 97

1

x Normalized height ---H

0.9

T = 5.35s H = 200m ρ m = 20000kg/m s = 0.63

0.8 0.7 0.6 0.5 0.4

T = 0.49s H = 25m ρ m = 20000kg/m s = 0.15

0.3 0.2 0.1 0

1

2

3

4

5

Mode number Fig. 2.17: Mode shapes.

1 0.9

T = 5.35s H = 200m ρ m = 20000kg/m s = 0.63

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

T = 0.49s H = 25m ρ m = 20000kg/m s = 0.15

0.3 0.2 0.1 0

1

2

3

4

5

Mode number Fig. 2.18: Mode shear deformation profiles.

98 Chapter 2: Optimal Stiffness Distribution

2.7 Stiffness distribution for a discrete shear beam - dynamic response Our starting point is eqn (2.62), the equilibrium equation for an undamped discrete system. ˙˙ + KU = P MU

(2.153)

The forms of M and K specialized for an n ‘th order shear beam are defined by eqn (2.67). Setting P = 0 and expressing U as U = Φ cos ( ωt + δ )

(2.154)

transforms eqn (2.153) to 2

KΦ = ω MΦ

(2.155)

Solutions of eqn (2.155) define the frequencies ( ω j ) and mode shapes ( Φ j ) corresponding to a particular beam, i.e., when M and K are specified. In the inverse approach, one specifies a mode shape and determines K . Let Φ = Φ

*

(2.156)

denote the specified modal profile. Substituting in eqn (2.155) leads to *

2

KΦ = ω MΦ

*

(2.157)

Defining a scaled version of K , 1 K' = ------K 2 ω

(2.158)

reduces eqn (2.157) to *

*

K' Φ = M Φ = P'

(2.159)

The remaining steps are the same as for the static case discussed in section 2.4. Eqn (2.159) is written as

2.7 Stiffness Distribution for a Discrete Shear Beam - Dynamic Response 99 (2.160)

S'k' = P'

where k' contains the n scaled stiffness factors, and S' is an upper triangular * matrix containing the elements of Φ . *

*

S' ( i, i ) = Φ i – Φ i – 1 *

*

S' ( i, i + 1 ) = Φ i – Φ i + 1 S' ( i, j ) = 0 for j ≠ i, i + 1

(2.161)

One solves eqn (2.160) for k' . Once ω is specified, the actual values of the stiffness parameters can be determined with 2

k j = ω k' j

(2.162)

Example 2.8. 3 DOF shear beam This example illustrates the determination of the stiffness distribution that produces the following linear profile for the fundamental mode. 1⁄3 * Φ = 2⁄3 1

(2.163)

*

It is convenient to scale Φ so that the magnitude of the maximum element is equal to 1 . The corresponding forms of P' and S' are

*

m1 ⁄ 3

P' = M Φ = 2m 2 ⁄ 3 m3

(2.164)

100 Chapter 2: Optimal Stiffness Distribution

S' =

1 1 --- – --3 3 1 0 --3

0

1 – --3 1 0 0 --3

(2.165)

Solving eqn (2.160) results in k 3' = 3m 3 k 2' = 2m 2 + 3m 3

(2.166)

k 1' = m 1 + 2 m 2 + 3m 3 *

*

When all the elements of Φ are positive, one can show that Φ is actually the fundamental mode shape for the system having the stiffness distribution defined by eqn (2.160), and ω is the corresponding fundamental frequency.

2.8 Stiffness calibration Governing equations- fundamental mode response of a discrete shear beam The shear beam is assumed to have some damping and this energy dissipation mechanism is represented as linear viscous damping. The corresponding governing equations have the form ˙˙ + CU˙ + KU = P MU

(2.167)

where CU˙ defines the nodal forces due to viscous damping. How one establishes C is treated in the next chapter. Our objective here is to reduce eqn (2.167) to a single scalar equation which governs the response of the fundamental mode. The displacement vector is expressed as U = q ( t )Φ

*

(2.168)

2.7 Stiffness Calibration 101 *

where Φ is the dimensionless prescribed mode shape used to establish the relative stiffness distribution in the previous section, and q ( t ) can be interpreted as the displacement parameter that defines the response of the fundamental mode. Substituting for U in eqn (2.167) *

*

*

(2.169)

MΦ q˙˙ + CΦ q˙ + KΦ q = P * T

and then premultiplying by ( Φ ) results in the following scalar equation for q : ˜ q˙˙ + c˜ q˙ + k˜ q = p˜ m

(2.170)

where the quantities with superscript tilde are equivalent one degree of freedom mass, damping, stiffness, and force quantities. Their definition equations are: T

˜ = ( Φ* ) M Φ* m

(2.171)

* T * ˜ c˜ = ( Φ ) CΦ = 2ξ˜ ωm

(2.172)

2 ˜ k˜ = ω m

(2.173) T

* p˜ = ( Φ ) P

(2.174)

The form of eqn (2.173) follows from eqn (2.157)

Example 2.9. 3 DOF shear beam revisited *

Consider the 3 DOF system defined in example 2.8. Taking Φ according to eqn (2.163), the one dimensional mass and force measures are

˜ = 1--- 2--- 1 m 3 3

m1 ⁄ 3

m1 4 2m 2 ⁄ 3 = ------- + --- m 2 + m 3 9 9 m3

(2.175)

102 Chapter 2: Optimal Stiffness Distribution

p˜ = 1--3

2 --3

p1 1

1 2 p 2 = --- p 1 + --- p 2 + p 3 3 3 p3

(2.176)

The nodal forces due to seismic excitation are proportional to the nodal masses. A typical force is pi = –mi a g

(2.177)

where a g is the ground acceleration. Substituting for p i in eqn (2.176) results in m 1 2m 2 p˜ = – ------- + ---------- + m 3 a g 3 3

(2.178)

It is convenient to express p˜ due to seismic acceleration as ˜a p˜ = – Γm g

(2.179)

where Γ is a dimensionless factor. For this example, Γ is given by m 1 2m 2 ------- + ---------- + m 3 3 3 Γ = -------------------------------------m 1 4m 2 ------- + ---------- + m 3 9 9

(2.180)

When the masses are equal, Γ = 9 ⁄ 7 . In order to evaluate the one dimensional damping coefficient, c˜ , the system damping matrix, C must be specified. The form of C depends on how the viscous damping elements are located throughout the structure. Figure 2.19 shows an arrangement consisting of viscous dampers inserted between adjacent masses. In this case, the shear force in an element depends on both the shear deformation and the time rate of change of the shear deformation. Assuming linear behavior, the typical equations have the form V i = k i ( u i – u i – 1 ) + c i ( u˙ i – u˙ i – 1 ) p i = m i u˙˙i + V i – V i + 1

(2.181)

2.7 Stiffness Calibration 103 Equation (2.181) shows that the forms of K and C are similar; one replaces k i with c i . Taking C as c1 + c2 C =

–c2

0

–c2

c2 + c3

–c3

0

–c3

c3

(2.182)

leads to 1 c˜ = --- ( c 1 + c 2 + c 3 ) 9

(2.183)

The contribution of the individual dampers to c˜ is the same since the assumed displacement profile corresponds to constant shear deformation. m3 k3, c3

Vi+1 m2

u2, p2 k2, c2

ui, pi

mi Vi

m1 k1, c1

Fig. 2.19: Dampers included in shear beam elements.

104 Chapter 2: Optimal Stiffness Distribution

c3 ’

k3

Vi+1

c2’ u2, p2

. ci’ ui

k2

pi Vi

c1 ’

k1

Fig. 2.20: Dampers attached to nodal masses. The damper arrangement shown in Fig. 2.20 produces nodal forces that depend on the nodal velocities. Assuming linear behavior, the typical equations are V i = ki ( ui – ui – 1 ) p i = m i u˙˙i + c' i u˙ + V i – V i – 1

(2.184)

With this arrangement, C is similar to M . Then c' 1 0 0 C =

0 c' 2 0 0 0 c' 3

(2.185)

and c' 1 4 c˜ = ------ + --- c' 2 + c' 3 9 9

(2.186)

Equation 2.186 shows that the most effective damper location is node 3. This

2.7 Stiffness Calibration 105 result can also be deduced by examining the assumed displacement profile. Node 3 corresponds to the maximum element in Φ * , and therefore has the largest velocity.

Governing equations - fundamental mode response of a continuous beam The displacement variables for a continuous beam are the translation, u , and rotation, β , of the cross-section. Noting equations (2.145) and (2.146), the displacement expressions corresponding to uniform shear and bending deformation are expressed as u = q ( t )φ ( x )

(2.187)

β = q ( t )ψ ( x )

(2.188)

where φ , ψ are the fundamental mode shape functions, x 2 x φ = ---- + s ---- H H

(2.189)

2s x ψ = ----- ---- H H

(2.190)

q is the modal amplitude parameter, and 2s ⁄ H is the ratio of prescribed bending deformation to prescribed shear deformation. A pure shear beam has s = 0 . The deformations and maximum displacement are related to q by 1 γ = u, x – β = q ---- H

(2.191)

2s χ = β, x = q ------- 2 H

(2.192)

u(H ) = q(1 + s)

(2.193)

The equilibrium equation expressed in terms of q(t) can be generated using the Principle of Virtual Displacements (Bathe, 1996). This Principle is used to

106 Chapter 2: Optimal Stiffness Distribution transform continuous equilibrium equations into a set of discrete equations, and can be interpreted as a variational statement. It provides the basis for the Finite Element Displacement Method. One starts by introducing an expansion for the displacement variables in terms of generalized coordinates. For this case, φ ( x ) and ψ ( x ) are the prescribed displacement fields, and q(t) is the generalized coordinate. A virtual displacement is defined as a displacement distribution generated by perturbing the generalized coordinate by a small amount, δq . According to this principle, the first order work done by the external loads during this virtual displacement is equal to the first order work done by the internal forces during the corresponding virtual deformation. The transverse shear force and bending moment are the internal forces for a beam; the corresponding deformations are γ and χ . Assuming the external loading consists only of the transverse load, b, the mathematical form of the principle is H

∫

H

( Mδχ + Vδγ ) dx =

0

∫

bδu dx

(2.194)

0

Taking u according to eqn (2.187), the virtual terms are related to δq by: δu = δqφ 1 δγ = δq ---- H

(2.195)

2s δχ = δq ------- 2 H Using eqn (2.195), and requiring the equality to be satisfied for arbitrary δq results in

∫

H

2s 1 ------ M + ----V dx = 2 H 0 H

H

∫

bφ dx

(2.196)

0

To proceed further, M , V , and b must be related to q . Assuming linear behavior and including viscous damping in both the

2.7 Stiffness Calibration 107 material and external loading terms, the expressions for the internal shear force and moment reduce to 1 V = D T γ + C T γ˙ = ---- ( D T q + C T q˙ ) H

(2.197)

2s M = D B χ + C B χ˙ = ------- ( D B q + C B q˙ ) 2 H

(2.198)

where C T , C B are material damping parameters. Including inertia and damping contributions, the external distributed loading expands to b = – ρ m u˙˙ – c'u˙ + b = – ρ m φq˙˙ – c'φq˙ + b

(2.199)

where c' is a damping parameter and b is the prescribed loading. Substituting in eqn (2.196), and writing the resulting equation in the “same” form as eqn (2.170), ˜ q˙˙ + c˜ q˙ + k˜ q = p˜ , leads to the following definition equations for the onem dimensional parameters:

˜ = m

p˜ =

∫ ∫

H

2

ρ m φ dx

(2.200)

φb dx

(2.201)

0

H

0

k˜ =

c˜ =

H

2 1 2 4s ˜ ----------- 2 D T + 4- D B dx = ω 1 m H 0 H

∫

H

2 1 2 4s ------2- C T + -------4- C B + c'φ dx H 0 H

∫

(2.202)

(2.203)

Equation (2.202)follows from the definition of the rigidity distributions (see eqns (2.148) and (2.149)). The other terms depend on the mass density distribution and prescribed loading. The following example considers various cases.

108 Chapter 2: Optimal Stiffness Distribution

Example 2.10. One dimensional parameters - continuous beam 1. Uniform mass density * ρ m = cons tan t = ρ m 2

s ˜ = ρ * H 1--- + --s- + ---m m 3 2 5

(2.204)

2. Uniform loading *

b = b (t) *

b H 2s p˜ = ---------- 1 + ----- 2 3

(2.205)

3. Linear loading x b = b 0 ( t ) ---H b0 H 3s p˜ = ---------- 1 + ----- 3 4

(2.206)

4. Seismic loading b = –ρm a g p˜ = – a g

∫

H

ρ m φ dx 0

A more convenient form of p˜ is

(2.207)

2.7 Stiffness Calibration 109

˜Γ p˜ = – a g m

Γ=

∫ ∫

H

ρ m φ dx 0 ----------------------------H 2 ρ m φ dx 0

(2.208)

When ρ m is constant, Γ depends only on φ .

∫ ∫

H

φ dx

2s 1 + ----* 3 0 Γ ρ = const = Γ = --------------------- = --------------------------H 2 m 2 2s 2 --- + s + -------φ dx 5 3

(2.209)

0

*

Table 2.2 shows the variation of Γ with s . *

Table 2.2: Variation of Γ with s . s

*

0.00

Γ 1.50

0.125

1.36

0.25

1.24

0.50

1.05

0.75

0.91

1.00

0.81

Stiffness calibration - periodic excitation Equation 2.170 is the governing 1 DOF equation for both discrete and continuous beams. One just has to use the appropriate mass, damping stiffness, and loading terms. The solution for a 1 DOF system subjected to periodic excitation is developed in section 1.3. That solution is applicable here with a change in notation to account for the difference between eqns (1.23) and (2.170).

110 Chapter 2: Optimal Stiffness Distribution The response due to the loading, p˜ = pˆ sin ( Ωt )

(2.210)

is written as q = qˆ sin ( Ωt – δ )

(2.211)

where qˆ is related to pˆ by pˆ qˆ = ------------ H 2 ˜ Ω2 m

(2.212)

2 2 ˜ 2 Ω m Ω ρ = ------------ = ------2 k˜ ω

(2.213)

c˜ ξ˜ = -----------˜ 2ωm

(2.214) 4

H2 =

ρ ---------------------------------------------2 2 2 ( 1 – ρ ) + ( 2ξ˜ ρ )

(2.215)

˜ are assumed to be prescribed. The constraint on The terms p˜ , Ω , and m displacement, say u ( x ) = u * ( x ) , is converted to a constraint on q˜ using the assumed displacement expansion, u = qφ . Taking qˆ ≤ q *

(2.216)

in eqn (2.212), the calibration problem reduces to determining ρ and ξ which satisfy the following constraint: **

H2 ≤ H2 ** H2

2

˜ q* Ω m = ----------------pˆ

Once ρ is known, the frequency is determined with

(2.217)

2.7 Stiffness Calibration 111 2

2 Ω ω = ------2 ρ

(2.218) 2

Lastly, one scales the stiffness parameters with ω .

Example 2.11. 3 DOF shear beam revisited again Consider the case of equal nodal masses and loads for the system defined in example 2.9. m 1 = m 2 = m 3 = 1000 kg pˆ 1 = pˆ 2 = pˆ 3 = 10kN Applying equations (2.175) and (2.176) leads to ˜ = 14 ------ ( 1000 ) = 1555kg m 9 p˜ = 2 ( 10 ) = 20kN *

The displacement constraint is taken as u 3 = 0.1m . Since Φ ( 3 ) ≡ 1 , it * ** follows that q = 0.1m . Evaluating H 2 results in **

H 2 = 0.007775Ω

2

Various values for Ω are considered below. 1. Ω = 2π **

H 2 = 0.307 **

Since H 2 is less than 1 , there is only one value for ρ . Stiffness dominates the response and one can neglect ξ in the expression for H 2 . Then, using eqn. (2.215),

112 Chapter 2: Optimal Stiffness Distribution 1 1 ------ = 1 + -------- = 4.257 2 ** ρ1 H2 and ω = ( 2π ) ( 4.257 )

1⁄2

= 12.96 rad/s

The stiffness parameters are derived in example 2.8. For equal nodal masses, eqn (2.166) reduces to 3000 k' = 5000 6000

(kg)

Finally, the scaled stiffnesses are 504 k = ω k' = 840 1067 2

(kN/m)

2. Ω = 4π **

H 2 = 1.228 **

Since H 2 is greater than 1 , there are two values for ρ . Results for different values of ξ˜ are listed below. Selecting ω 2 generates the lowest stiffness and shifts the response away from the resonance zone. ξ˜

ρ1

ω 1 (rad/s)

ρ2

ω 2 (rad/s)

0

0.633

19.85

2.383

5.27

0.1

0.656

19.16

2.300

5.46

0.2

0.742

16.93

2.033

6.18

2.7 Stiffness Calibration 113 Example 2.12. Stiffness calibration - continuous beam Consider a continuous beam model of a moderately tall building having the following properties H = 100 m ρ m = 20000 kg/m Let s = 0.4 , which corresponds to allocating 40% of the deflection at the top of the building to bending deformation. The equivalent mass follows from eqn (2.204). ˜ = ( 20000 ) ( 100 ) ( 0.541 ) = 1.082 × 10 6 kg m Assuming the displacement at the top is prescribed and noting eqn (2.193), the nodal amplitude is determined with 1 q * = -----------u * ( H ) 1+s 1 100 A representative value for u * ( H ) is H ⁄ 400 . Then, q * = ------- --------- = 0.178 m . 1.4 400 Lastly, the loading is assumed to be uniform over the building height. Applying eqn (2.205), b * ( 100 ) 0.8 p˜ = ------------------- 1 + ------- = 63.33 b * 2 3 where b * has units of N ⁄ m . Combining these terms, the constraint condition is given by **

H2 ≤ H2

2

** Ω H 2 = ------- ( 3043 ) b*

The remaining steps are the same as for the previous example. For example, ** assuming Ω = 1 rad/s and b * = 6086 N ⁄ m , H 2 = 0.5 and the appropriate value of ω is 3 rad/s . Substituting for the various parameters in eqns (2.148)

114 Chapter 2: Optimal Stiffness Distribution and (2.149), the corresponding rigidity distributions are 5 x 2 2s x 3 D T = 3 × 10 1 – ---- + ----- 1 – ---- H 3 H

(kN)

9 1 1 x x 3 4x 1 x 4 D B = 3 × 10 --- 1 – -------- + --- ---- + ----- 2 – 3 ---- + ---- 4 H H 3H 3 H 6s

(kNm)

The previous examples dealt with the case where the displacement controls the design. When acceleration is the limiting condition, one works with the differentiated form of eqn (2.211) 2

q˙˙ = – Ω qˆ sin ( Ωt – δ )

(2.219)

and establishes a relation between the maximum allowable acceleration, ( ˙˙ u ) max , and ( q˙˙) max . Letting ( q˙˙) max = a *

(2.220)

the constraint condition for acceleration controlled design is *

H2 < H2 * H2

˜ a* m = --------p˜

(2.221)

The remaining steps are the same as before. One selects a value for ξ , determines the allowable values of ρ and then converts to values for ω .

Example 2.13. Example 2.12 revisited Consider the continuous beam model introduced in example 2.12. Suppose the acceleration at the top of the building is the critical motion parameter. This acceleration is expressed as a fraction of the gravitational acceleration,

2.7 Stiffness Calibration 115 u˙˙( H ) ≤ fg

(2.222)

Noting eqn (2.193), u˙˙( H ) is related to q˙˙ by u˙˙( H ) = ( 1 + s )q˙˙ Then, the limiting value for q˙˙ is f ( q˙˙) max = a * = ----------- g 1+s

(2.223)

Substituting for a * , eqn (2.221) takes the form ˜g f * m H 2 = -------- ----------- p˜ 1 + s

(2.224)

The limiting acceleration for human comfort is about 1.5% of g. Taking ˜ , p˜ , and s for example 2.12, the design f = 0.015 and using the values of m constraint is * 1805 H 2 = -----------b*

where b * has units of N/m. To illustrate the computational aspects, the case where b * = 3610 N ⁄ m is considered. The corresponding value of H 2* is 0.5. Taking ξ = 0 and using eqn (2.215), one obtains 2 1 ρ 1 = --3 2

2 2 Ω ω 1 = -------- = 3Ω 2 ρ1

The acceleration constraint requires ω ≥ ω 1 .

116 Chapter 2: Optimal Stiffness Distribution Stiffness calibration - seismic excitation The equivalent loading for seismic excitation has been expressed as ˜ Γa p˜ = – m g

(2.225)

˜ and Γ depend on the model, i.e., discrete or continuous beam. where m Substituting for p˜ in eqn (2.170), one obtains the following equation for the modal amplitude factor, q . 2 q˙˙ + 2ξ˜ ωq˙ + ω q = – Γa g

(2.226)

˜ is the equivalent modal damping ratio. The corresponding where ξ˜ = c˜ ⁄ 2ωm equation for a 1 DOF system follows from eqn (1.23) 2

u˙˙ + 2ξωu˙ + ω u = – a g

(2.227)

Taking ξ = ξ˜ , the displacement responses for the same accelerogram are related by q ( t ) = Γu ( t )

(2.228)

Therefore, one can work with a 1 DOF system and then scale the response using eqn (2.228). Assuming the “design” ground motion time history is specified, one can theoretically solve eqn (2.227) and determine the relationship between the peak response u max , and the system parameters ω and ξ . It was possible to find an analytic solution for the case of periodic loading. However, seismic accelograms are considerably more complex than periodic forcing. Typical accelograms are shown in figures 2.21 and 2.22. One has to resort to using a numerical scheme, and generate a set of solutions corresponding to a representative range for ω and ξ . A particular solution for u max is expressed as 1 u max ( ω i, ξ i ) = ---------------------------------- S v ( ω i, ξ i ) ω i ( 1 – ξ i2 ) 1 ⁄ 2

(2.229)

where S v is defined as the pseudo spectral velocity. The complete solution is represented by a set of plots of S v vs. ω (or period T ) for different values of ξ .

2.7 Stiffness Calibration 117 Figure 2.23 contains pseudo spectral velocity profiles for the accelerogram plotted in Fig. 2.21.

4

El Centro (component S00E) 2 Peak acceleration @ 2.12s = 3.41 m/s Peak velocity @ 2.18s = 0.33 m/s Peak displacement @ 8.58s = 0.10m

Acceleration a g - m/s

2

3

2

1

0

−1

−2

−3

−4 0

10

20

30

40

50

60

Time - s Fig. 2.21: Time history of El Centro accelerogram (component S00E). 2

1.5

Acceleration a g - m/s

2

Taft (component N21E) 2 Peak acceleration @ 9.10s = 1.52 m/s Peak velocity @ 3.40s = 0.16 m/s Peak displacement @ 44.14s = 0.07m

1

0.5

0

−0.5

−1

−1.5

−2 0

10

20

30

40

50

60

Time - s Fig. 2.22: Time history of Taft accelerogram (component N21E).

118 Chapter 2: Optimal Stiffness Distribution

1

Pseudo spectral velocity S v - m/s

10

ξ = 2% ξ = 5% ξ = 10% ξ = 20%

0

10

−1

10

El Centro (component S00E)

−2

10

−1

10

0

10

1

10

Period T - s Fig. 2.23: Pseudo spectral velocity profiles. Although there has been substantial progress in earthquake engineering, numerous uncertainties still exist, especially in defining the magnitude and frequency content of the earthquakes that a particular site is expected to encounter. Earthquake records differ significantly as illustrated by the accelerograms for El Centro (Component S00E) and Taft (Component N21E). Given the difficulty in estimating the ground motion that a structure may experience over its useful life, it is prudent to base seismic design on a range of possible earthquake motions. One approach to dealing with this uncertainty is based on using a seismic response spectrum which is an envelop of upper bound responses corresponding to a set of ground motions which are representative for the site. Assuming S v ( ω, ξ ) represents the design response spectrum, the stiffness calibration proceeds as follows. Given the design values for the motion parameters, one establishes the desired value for q max . Let q *denote this value. For the cantilever beam case, q * = γ * H where γ * is the targeted transverse shear deformation. Noting equations (2.228) and (2.229), the design constraint takes the form

2.7 Stiffness Calibration 119 ΓS v ( ω, ξ ) q * ≡ -------------------------------ω( 1 – ξ2 )1 ⁄ 2

(2.230)

Equation (2.230) is written as ΓS v ( ω, ξ ) ω = --------------------------------q*( 1 – ξ2 )1 ⁄ 2

(2.231)

One determines ω by iterating on S v ( ω, ξ ) . Construction of the design response spectrum for S v and the application of eqn (2.231) are discussed in the following section. Seismic response spectra A seismic response spectrum is a plot of the maximum value of a response variable such as displacement, velocity, acceleration, or any other quantity of interest versus the natural frequency or period for a broad range of single-degreeof-freedom (SDOF) systems subjected to the same loading, in this case, an earthquake excitation. In what follows, the generation of response spectra for a SDOF linear elastic system is described and approaches for establishing design spectra are suggested. The solution of eqn (2.227) for a specified ground acceleration a g ( t ) can be expressed as a Duhamel integral (Chopra, 1995) t

∫

1 1 u ( t ) = – ----- a g ( τ )e – ξω ( t – τ ) sin [ ω' ( t – τ ) ]dτ ≡ ----- Θ ω' ω'

(2.232)

0 t

∫

ω u˙ ( t ) = – ----- a g ( τ )e – ξω ( t – τ ) sin [ ω' ( t – τ ) + δ ]dτ ω' 0

(2.233)

120 Chapter 2: Optimal Stiffness Distribution t

ω2

∫

u˙˙( t ) = – a g ( t ) – ------ a g ( τ )e – ξω ( t – τ ) sin [ ω' ( t – τ ) + δ 1 ] dτ ω' 0 ω2

(2.234)

= – a g ( t ) + ------Θ' ω' where ω' = ω 1 – ξ 2 is the damped frequency and δ and δ 1 are defined as ( 1 – ξ2 )1 / 2 tan δ = – --------------------------ξ

(2.235)

2ξ ( 1 – ξ 2 ) 1 / 2 tan δ 1 = ---------------------------------1 – 2ξ 2

(2.236)

For lightly damped structures, ξ < 0.10 , and therefore the difference between the damped and the undamped frequencies can be neglected. The response quantities for this case reduce to t

∫

Θ 1 u ( t ) ≈ – ---- a g ( τ )e – ξω ( t – τ ) sin [ ω ( t – τ ) ] dτ ≡ ---ω ω

(2.237)

0

t

∫

u˙ ( t ) ≈ – a g ( τ )e – ξω ( t – τ ) { cos [ ω ( t – τ ) ] – ξ sin [ ω ( t – τ ) ] } dτ

(2.238)

0

t

∫

u˙˙( t ) ≈ – a g ( t ) – ω a g ( τ )e – ξω ( t – τ ) sin [ ω ( t – τ ) + 2ξ ] dτ ≡ – a g ( t ) – ωΘ' (2.239) 0

The spectral displacement, S d , is defined as the maximum value of the response relative to the ground. Noting eqn (2.232), S d is given by 1 S d = ----- Θ max ω'

(2.240)

2.7 Stiffness Calibration 121 The maximum value of Θ has units of velocity. It is referred to as the pseudo spectral velocity and is denoted as S v . S v ( ω, ξ ) = Θ max

(2.241)

The spectral acceleration, S a , is defined as the maximum value of the absolute acceleration. ω2 S a = ------Θ' max ω'

(2.242)

1 When ξ is small, the measures can be approximated by S d ≈ ---- S v and S a ≈ ωS v. ω The energy input E I to a SDOF system initially at rest and subjected to ground excitation is given by t

∫

E I = – m a g ( τ )u˙ ( τ ) dτ

(2.243)

0

Part of this energy is stored temporarily in the structure in the form of kinetic and strain energy; the remaining portion is dissipated by damping. The various energy terms are Kinetic energy t

∫

2 m E K = m u˙˙( τ )u˙ ( τ ) dτ = ---- [ u˙ ( t ) ] 2

(2.244)

0

Strain energy t

∫

2 k E S = k u ( τ )u˙ ( τ ) dτ = --- [ u ( t ) ] 2 0

(2.245)

122 Chapter 2: Optimal Stiffness Distribution Dissipated energy t

∫

2

E D = c [ u˙ ( τ ) ] dτ

(2.246)

0

With this notation, the energy balance equation takes the form EK + ED + ES = EI

(2.247)

Equation (2.247) applies for arbitrary t . When t is large in comparison to the duration of the earthquake, E K = E S = 0 and E I is equal to the energy dissipated by damping. An alternate way of expressing the net input energy is in terms of the equivalent velocity, V E , defined as 2E I ( t e ) -----------------m

VE =

(2.248)

where E I ( t e ) is the input energy evaluated at t e , the duration of the earthquake. 1

Equivalent velocity V E - m/s

10

ξ = 20% ξ = 10% 0

10

ξ = 5% ξ = 2%

−1

10

−2

10

El Centro (component S00E) −1

10

0

10

Period T - s Fig. 2.24: Equivalent velocity profiles.

1

10

2.7 Stiffness Calibration 123 Figures 2.23 and 2.24 illustrate the variation of the pseudo spectral velocity and the equivalent velocity with period and damping ratio. Whereas the pseudo spectral velocity profiles are smoothened and lowered with increasing damping, the equivalent velocity profiles tend to converge toward a single curve. This result indicates that the work done by an earthquake is not very sensitive to damping. Furthermore, for low damping, the equivalent velocity plot is close to the pseudo spectral velocity curve. Therefore, it is reasonable to assume that the pseudo spectral velocity provides a measure of the energy input at low damping ratios. An ensemble of design earthquakes is used to produce the design spectral velocity function S v ( ω, ξ ) . The earthquakes that compose the ensemble are selected such that frequency content is representative of the site where the structure is to be located. They are then scaled such that their maximum spectral velocity is equal to a reference value S v R for a damping ratio of 0.02 . This value represents the magnitude of excitation for which a lightly damped structure would be designed for. Two methods for generating the design spectral velocity function are discussed here. Both methods are based on the ensemble average of the design earthquake spectral velocity functions. One method develops an analytical expression for a design spectral velocity function and involves only simple calculations. The second method generates a numerical approximation and is a computer based approach. The analytical spectral velocity function is generated for each damping ratio by assuming a bilinear log-log relationship between 0.1 and 0.6 seconds, and a constant value for T > 0.6 seconds. The maximum value of S v is obtained by averaging the peak values of the individual spectra. Similarly, the minimum value is the average of the spectral values at T = 0.1 seconds. Figure 2.25 shows spectral velocity functions normalized such that the maximum value of S v = 1.2 ( m ⁄ s ) for ξ = 0.02 - a representative value for a major seismic event. The accelerograms corresponding to the earthquakes are also normalized based on these values. Once accelerograms are normalized, one can compute how the spectral velocity function varies with ξ by increasing the damping ratio and generating new spectral velocity functions. Spectral velocity functions for these “scaled” accelerograms and different damping ratios are shown in Figure 2.26. Figure 2.27 illustrates how the averaged maximum value of S v varies with ξ .

124 Chapter 2: Optimal Stiffness Distribution

Spectral Velocity Plots Scaled Such That Sv = 1.2 for ξ = 0.02

1

10

S = 1.2 m/s v

ξ = 0.02

0

Spectral Velocity − m/s

10

−1

10

−2

10

−1

0

10

1

10 Period − Seconds

10

Fig. 2.25: Normalized spectral velocity plots

Spectral Velocity Plots for ξ = 0.15

1

Spectral Velocity Plots for ξ = 0.30

1

10

10

S = 0.60 m/s ξ = 0.15

S = 0.47 m/s ξ = 0.30

v

v

0

0

Spectral Velocity − m/s

10

Spectral Velocity − m/s

10

−1

−1

10

10

−2

10

−2

−1

10

0

10 Period − Seconds

1

10

10

−1

10

0

10 Period − Seconds

Fig. 2.26: Spectral velocity plots for ξ = 0.15 and ξ = 0.30

1

10

2.7 Stiffness Calibration 125

1.2

1

Svmax [m/s]

0.8

0.6

0.4

0.2

0

0

0.05

0.1

0.15 ξ

0.2

0.25

Fig. 2.27: S v vs. ξ max The spectral velocity function based on the bilinear log-log relationship can be used to solve equation 2.231. Since there is no unique solution, a family of solutions for T is generated by specifying different values of ξ . For each value of ξ , the computation reduces to iterating on the following equations: T ≥ 0.6 sec *

2πq T = -----------------ΓS v

(2.249)

max

T < 0.6 sec=T max T

a+1

2πq * a = ------------------ ( T max ) ΓS v max

log ( S v ⁄ Sv ) max min a = --------------------------------------------log ( T max ⁄ T min )

(2.250)

(2.251)

126 Chapter 2: Optimal Stiffness Distribution The analytical expressions developed for the spectral velocity tend to be conservative, since the piecewise linear log-log approximation for the average design spectrum is over-simplified. An improvement in the process of establishing ω can be obtained if an average spectral velocity function is generated by taking the ensemble average of the spectral velocity functions, instead of attempting to create an analytical expression. Equation 2.231 can be solved by iterating through the average spectral velocity function and finding the values of ω and S v that satisfy the equation for a given value of ξ . Figure 2.28 shows average spectral velocity functions for various values of ξ . For a given value of T , numerical functions similar to that of Fig 2.27 can be used to determine S v for a given value of ξ .

2.7 Stiffness Calibration 127

1

10

ξ = 0.1

0

Sv [m/s]

10

−1

10

0

10 Period T − seconds

1

10

1

10

ξ = 0.02, 0.05, 0.1, 0.15, 0.2, 0.25, 0.3

0

Sv [m/s]

10

−1

10

0

10 Period T − seconds

Fig. 2.28: Average spectral velocity functions

1

10

128 Chapter 2: Optimal Stiffness Distribution

Example 2.14. Example 2.12 revisited for seismic excitation The properties of the beam considered in example 2.12 are H = 100 m ρ m = 20000 kg ⁄ m s = 0.4 ˜ = 1.082 × 10 6 kg m Using eqn (2.209), the corresponding value of Γ is 1.12 . The design value for q is established by specifying the desired peak transverse shear deformation. Taking γ * = 1 ⁄ 200 and noting that q * = γH (see eqn (2.191) results in q * = γ * H = 0.5 m An estimate for T is determined by first evaluating the right hand side of eqn (2.249). If the value is greater than 0.6 , that estimate is the actual value. If less than 0.6 , eqn (2.250) applies. For this example, the right hand term is 2πq * 2.80 ------------------ = -------------ΓS v Sv max max Taking ξ = 0.02 and S v = 1.2 , max 2.80 T = ---------- = 2.34 sec 1.2 2π ω = ------ = 2.68 rad/s T The modal damping parameter is determined with ˜ = ( 2 ) ( 0.02 ) ( 2.68 ) ( 1.082 ) × 10 6 = 116 (kN)(sec/meter) c˜ = 2ξωm

2.7 Stiffness Calibration 129 Increasing damping to 0.1 and taking the corresponding maximum value of S v according to Fig 2.27, ξ = 0.1

→

Sv

max

= 0.7

leads to 2.80 T = ---------- = 4.0 sec 0.7 ω = 1.57 rad/s c˜ = 340 (kN)(sec/meter) 2

One evaluates the rigidity parameters D T and D B by substituting for ω in eqns (2.148) and (2.149).

Example 2.15. 5 DOF shear beam Consider a 5 DOF shear beam having equal nodal masses. The fundamental mode vector is specified such that the relative nodal displacements for the 5 elements are equal. Its form is U = qΦ * 1 2 1 Φ * = --- 3 5 4 5 The equivalent modal mass follows from eqn (2.171): T

˜ = ( Φ * ) M Φ * = 2.2 m m where m is the “constant” nodal mass. Substituting for Φ * and M in eqn (2.160), leads to the following set of stiffness coefficients

130 Chapter 2: Optimal Stiffness Distribution k' 1 = 15 m

k' 2 = 14 m

k' 3 = 12 m

k' 4 = 9 m

k' 5 = 5 m

2

The actual stiffness values depend on ω , i.e., k = ω k' Specializing eqn (2.174) for seismic excitation and equal nodal masses, one obtains p˜ = – ma g

∑ Φ*i = –3ma g

Then, noting the definition of Γ , – p˜ 3m Γ = ---------- = ------------ = 1.36 ˜a 2.2m m g The maximum displacement at node 5 (the top level) is considered to be the controlling motion measure. u5

= q max ≡ q * max

Assuming the beam is a model of a building, u 5 is related to the height of the max building and the story shear deformation. u5

max

= γ *H

To illustrate the remaining steps, suppose H and γ * have the following values H = 20 (meters) γ * = 1 ⁄ 200 Then, q * = 0.1 (meters ) and the right hand term in eqn (2.249) becomes 2πq * 0.462 ------------------- = --------------ΓS vmax S vmax Case 1: ξ=0.02 Taking ξ = 0.02 and S vmax = 1.2 , the value of the above term is less than

2.7 Stiffness Calibration 131 0.6 and eqn (2.250) applies. Noting Fig 2.25, T min = 0.1 , and S v, min = 0.085 . Using these values leads to: T = 0.49 sec ω = 12.82 r/s ˜ = 1.128 m (newtons-sec/meters) c˜ = 2ξωm Case 2: ξ=0.1 Taking S vmax = 0.7 for ξ = 0.1 , 0.462 --------------- = 0.66 S vmax and eqn (2.249) applies. T = 0.66 sec ω = 9.52 r/s c˜ = 4.19 m Increasing the damping reduces S vmax and the required stiffness. In this case, the 2 reduction factor is ( 9.52 ⁄ 12.82 ) = 0.55 , a 45 % decrease in stiffness. The modal damping parameter is related to the system damping matrix by eqn (2.172): T

c˜ = ( Φ * ) C ( Φ * ) Given c˜ , one needs to establish C . When linear viscous dampers are connected to adjacent nodes, C is similar in form to the stiffness matrix K . Using Φ * defined above, the triple matrix product evaluates to 1 c˜ = ------ ( c 1 + c 2 + c 3 + c 4 + c 5 ) 25 where c i is the damping coefficient for element i . Since there is only one equation relating the five coefficients, there is no unique solution for the c ‘s. Additional

132 Chapter 2: Optimal Stiffness Distribution equations can be derived by imposing conditions on the damping ratios for the higher modes and optimizing “cost” and “performance”. Establishing the optimal damping distribution is addressed in the next chapter.

2.9 Examples of stiffness distribution for buildings In this section, a series of simulation studies for a representative range of buildings are presented. The primary focus here is on assessing the adequacy of the rigidity distribution derived with the single mode expression to control the deformation profiles. How the rigidity distribution can be corrected to account for the influence of the higher modes is discussed in the following section. Only shear rigidity distributions and shear deformation profiles are presented here. The corresponding bending rigidity and deformation plots can be found in AbboudKlink (1995). The motion based design data are the uniformly distributed mass ρ m , the building height H , the s factor relating the shear and bending deformations, and the values of the maximum allowable shear deformation γ ∗ corresponding to a specified level of ground excitation as measured by the pseudo spectral velocity S v . Table 2.3 lists the levels of excitation considered here and their corresponding allowable deformations. For preliminary design, the extreme earthquake level is used. Equation (2.249) gives the fundamental period corresponding to the rigidity distributions based on the fundamental mode response. These distributions are generated using eqns (2.148) and (2.149). The relevant design parameters for the buildings modelled as a continuous beam are listed in Table 2.4.

Table 2.3: Design loads and allowable deformation values. Earthquake level

S v (m/s)

γ∗

Service load

0.6

1/400

Extreme load

1.2

1/200

Ultimate load

2.4

1/100

2.9 Examples of Stiffness Distribution for Building 133 Table 2.4: Design parameters for the continuous models. Building #

ρm (kg/m)

(m)

H⁄B

f∗

s

γ∗

Sv (m/s)

ξ1 (%)

T1

1

20000

25

2

7

0.15

1/200

0.9

2

0.49

2

20000

50

3

6

0.25

1/200

1.2

2

1.06

3

20000

100

4

5

0.40

1/200

1.2

2

2.34

4

20000

200

5

4

0.63

1/200

1.2

2

5.35

H

(s)

The continuous building models are discretized and the corresponding lumped parameter MDOF models are then subjected to earthquake excitation. The time history response due to seismic excitation is generated by numerically integrating the MDOF matrix equilibrium equations. Two accelerograms, El Centro (component S00E) and Taft (component N21E), both scaled to a maximum pseudo spectral velocity of 1.2m/s corresponding to the extreme loading condition, are used. Table 2.5 contains data for the first 3 modes of the MDOF models. Stiffness proportional damping is assumed to establish the damping ratios for the higher modes. This data is included here to provide background information needed to interpret the response. An alternative solution procedure based on working with the equilibrium equations expressed in terms of modal coordinates is described in the next section. Table 2.5: Modal parameters for the discrete models. T1(s)

T2(s)

T3(s)

ξ1 ( % )

ξ2 ( % )

ξ3 ( % ) Γ2 ⁄ Γ1 Γ3 ⁄ Γ1

Bldg #1 Bldg #2

0.52 1.09

0.21 0.43

0.13 0.26

2.00 2.00

5.04 5.09

8.17 8.36

0.40 0.43

0.23 0.25

Bldg #3 Bldg #4

2.39 5.48

0.92 2.06

0.55 1.21

2.00 2.00

5.18 5.32

8.64 9.06

0.45 0.48

0.26 0.27

Figures 2.29 through 2.36 show the generated shear rigidity distributions and the resulting maximum shear deformations for the different cases. The vertical dashed line in the deformation plots indicates the target maximum deformation; the solid line shows the maximum deformation due to El Centro excitation, and the dashed line shows the maximum deformation due to Taft excitation. The mean ( γ m and χ m ) and standard deviation ( γ sd and χ sd ) of a

134 Chapter 2: Optimal Stiffness Distribution deformation profile provide measures of the adequacy of the rigidity distribution. These values are summarized in Table 2.6. Table 2.6: Deformation results for the example buildings. El

Centro

γm

γ sd

χm

χ sd

γm

γ sd

χm

(10 )

(10 )

(10 )

(10 )

(10 )

(10 )

(10 )

(10-6)

3.42 4.40 6.13 3.50

1.25 4.42 9.82 19.42

3.84 4.44 4.61 2.20

1.14 4.10 10.06 14.22

1.87 3.42 4.42 4.54

1.27 6.37 16.52 7.66

2.18 3.61 3.61 2.74

0.80 6.06 15.88 5.55

-3

Bldg #1 Bldg #2 Bldg #3 Bldg #4

-4

-5

Taft

-6

-3

-4

-5

χ sd

The results show that the deviation from the state of uniform deformation occurs in the upper portion of the building, and is far more pronounced in tall buildings. This trend is attributed to the contributions of the higher modes which have the greatest deformation near the top (see Fig 2.18). For low rise buildings, the deformation pattern is essentially dominated by the fundamental mode response. A procedure for incorporating the effect of the higher modes on the rigidity distribution is presented in the next section. An alternate strategy is to increase the damping for the fundamental mode. These results are for a low level of damping ( ξ 1 = 2%). Damping is discussed in the next chapter. The results also show that the magnitude of the maximum deformation is sensitive to the excitation. For example, the difference in the behavior of Building 4 under the two excitations is mainly due to the difference in the response spectra. For El Centro, the actual value of S v at a period of about 5s , shown in Fig. 2.37, is much less than the value used for the initial rigidity estimate, S v = 1.2m/s , and as a result, the actual contribution of the first mode is overestimated. The Taft spectrum on the other hand, shows a closer agreement with the design value of S v and consequently, the actual and design deformations are closer together.

2.9 Examples of Stiffness Distribution for Building 135

1 0.9

Building #1 Quadratic based Initial

x Normalized height ---H

0.8 0.7

H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.29: Shear rigidity distribution for Building 1.

1 0.9

Building #1 Quadratic based Initial

x Normalized height ---H

0.8

H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.30: Maximum shear deformation for Building 1.

0.01

136 Chapter 2: Optimal Stiffness Distribution

1

x Normalized height ---H

0.9

Building #2 Quadratic based Initial

0.8 0.7

H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.31: Shear rigidity distribution for Building 2.

1

x Normalized height ---H

0.9

Building #2 Quadratic based Initial

0.8

H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.32: Maximum shear deformation for Building 2.

0.01

2.9 Examples of Stiffness Distribution for Building 137

1 0.9

x Normalized height ---H

Building #3 Quadratic based Initial

0.8 0.7

H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.33: Shear rigidity distribution for Building 3.

1

x Normalized height ---H

0.9 0.8

Building #3 Quadratic based Initial

0.7 0.6

H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2%

0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.34: Maximum shear deformation for Building 3.

0.01

138 Chapter 2: Optimal Stiffness Distribution

1 0.9

x Normalized height ---H

Building #4 Quadratic based Initial

0.8 0.7

H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.35: Shear rigidity distribution for Building 4.

1

x Normalized height ---H

0.9

Building #4 Quadratic based Initial

0.8 0.7

H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.36: Maximum shear deformation for Building 4.

0.01

2.10 Stiffness Modification for Seismic Excitation 139

1

Pseudo spectral velocity S v - m/s

10

ξ = 2% 0

10

-1

10

El Centro S00E (Scale = 1.66) Taft N21E (Scale = 3.00) -2

10 -1 10

0

10

1

10

Period T - s Fig. 2.37: Modified response spectrum for Building 4.

2.10 Stiffness modification for seismic excitation Iterative procedure The numerical studies presented in the previous section show that the method for establishing the shear and bending rigidities based on the fundamental mode is adequate for low rise buildings but needs to be modified for moderate rise buildings. This section presents a procedure for adjusting the stiffness which is based on updating the shear and bending deformation profiles by including the contribution of the higher modes, and then determining improved estimates for the rigidity measures with (i + 1) (x) DT

(i + 1) (x) DB

=

(i) ( i ) [ γ ( x ) ] max D T ------------------------------∗

(2.252)

=

(i) ( i ) [ χ ( x ) ] max D B -------------------------------∗

(2.253)

γ

χ

The iteration is continued until the change in the rigidity is within the acceptable range. Details of the procedure for seismic excitation applied to beam type

140 Chapter 2: Optimal Stiffness Distribution structures are presented below. Multiple mode response Given the mass distribution ρ m , one specifies: γ ∗ , the desired deformation; S v ( ω 1, ξ 1 ) , the level of earthquake to be designed for; and the parameter s relating the shear and bending deformations. An initial estimate of (0) (0) D T ( x ) and D B ( x ) is generated using the fundamental mode response approach presented in the previous sections. The structure is then discretized as an nth order MDOF system, leading to the following governing equations ˙˙( t ) + CU˙ ( t ) + KU ( t ) = P ( t ) = – MEa ( t ) MU g

(2.254)

where M , C , and K represent the mass, damping, and stiffness matrices of the discrete system, U is the displacement vector relative to the base, and E is a vector representing the rigid body motion of the system due to a unit translation at the base. Equation (2.254) can be transformed into a set of uncoupled equations by expressing U in terms of a “special” set of vectors, φ j , and assuming a particular form for C. n

U(t) =

∑

Φ j q j ( t ) = Φq ( t )

(2.255)

j=1

These vectors are solutions of the following eigenvalue problem: KΦ j = ω 2j MΦ j j = 1, 2, 3, . . . , n

(2.256)

and satisfy the orthogonality relation: Φ iT KΦ j = ω 2j Φ iT MΦ j δ ij

(2.257)

where δ ij is the dirac delta function. Background material on the eigenvalue problem is presented in Appendix I. Assuming C to be proportional to K (the basis for this assumption is discussed in Section 3.7 which deals with arbitrary damping),

2.10 Stiffness Modification for Seismic Excitation 141 C = αK

(2.258)

and noting eqn (2.257) leads to the following set of uncoupled scalar equations q˙˙j + 2ξ j ω j q˙ j + ω 2j q j = – Γ j a g

j = 1, 2, 3, . . . , n

(2.259)

The damping ratio and modal participation factor for the j’th mode are αω j ξ j = ---------2

(2.260)

ΦT j ME Γ j = --------------------ΦT j MΦ j

(2.261)

The individual modal responses are determined following the procedure established for the SDOF system. Noting eqn (2.232), the solution for the j’th modal response is Γj q j ( t ) = ------- Θ j ( t ) ω' j

(2.262)

t

Θ j(t) =

∫

– a g ( τ )e

–ξ j ω j ( t – τ )

sin [ ω' j ( t – τ ) ] dτ

(2.263)

0

U j(t) = Φ jqj(t)

(2.264)

Finally, the relative displacement vector due to all modal responses is obtained by superposition n

U ( t ) = Φq ( t ) =

∑ j=1

Γ jΦ jΘ j(t) -------------------------ω' j

(2.265)

To determine an estimate of the earthquake response of a lumped MDOF system, one can make use of the response spectra method. The maximum response for a mode is obtained from the response spectrum for the SDOF system

142 Chapter 2: Optimal Stiffness Distribution using the corresponding values of frequency, damping, and participation factor. The expressions for the jth mode are

qj Uj

Γj = ------- S v ( ω j, ξ j ) ω' j max

(2.266)

= Φ jq j max

(2.267)

max

where S v ( ω j, ξ j ) is the pseudo spectral velocity corresponding to the damping and frequency of the jth mode. The shear and bending deformations in the i’th element due to the maximum response of the j’th mode are determined with the following differential approximations: γ (i)

1 ---- u ( i + 1 ) – u ( i ) jmax = jmax hi

(2.268)

1 – --- β ( i + 1 ) – β ( i ) jmax 2 1 χ ( i ) jmax = ---- β ( i + 1 ) – β ( i ) jmax hi

(2.269)

where the displacement measures are for the nodes at the base and upper end of element i. The individual modal time histories are generally not in phase so one needs to introduce an assumption as to the relative phasing. Taking the algebraic sum assumes they are all in phase. Using a square root sum of squares (SRSS) procedure assumes the phase lag is a uniformly distributed random variable, and tends to place more emphasis on the dominant terms. The latter procedure is discussed here. The maximum value of a variable, say v, is determined with 1⁄2

2 2 2 v max = ( v 1max ) + ( v 2max ) + … + ( v Nmax )

(2.270)

where N is the number of modes retained for the stiffness modification. This

2.10 Stiffness Modification for Seismic Excitation 143 computational algorithm is used to determine γ max and χ max for each segment. (0)

One starts with D , computes the peak modal responses using eqn (2.266), and then evaluates the peak deformations with equations (2.268) and (2.269). The rigidity coefficients are then updated with equations (2.252) and (2.253). (1)

(1)

The process is now repeated. Given D T ( x ) and D B ( x ) , the system stiffness matrix is generated, and updated mode shapes, frequencies, and participation factors are evaluated, leading to revised peak deformation distributions and ultimately to new estimates for rigidities. Building examples - iterated stiffness distribution The iterative procedure is applied to the four building examples defined in Table 2.4. Table 2.7 lists the periods for the first three modes of the discretized system corresponding to the iterated rigidity distributions. The mean and standard deviation of the shear and bending deformations resulting from the El Centro and Taft excitations are summarized for different iterations in Table 2.8. Results in the form of converged shear rigidity distributions and maximum shear deformation profiles for El Centro and Taft excitations are provided in Figs 2.33 through 2.38. These profiles illustrate the convergence of the iterative procedure over the spectrum of building examples starting with the quasi-quadratic distribution as an initial estimate for the rigidity. Significant improvement in the maximum deformation profiles are obtained after just one iteration. The contribution of the higher modes becomes more significant as the period of the structure increases, leading to a decreasing rate of convergence with increasing period. Another important feature is the substantial difference in response for the two earthquakes over the spectrum of periods. El Centro dominates for Building 3 ( T 1 = 2.15s ) while Taft produces the most deformation for Building 4 ( T 1 = 4.86s ). This result stresses the need to use a set of representative earthquakes, and to establish bounds on the deformation profiles. In general, the rigidity is increased; the increment ranges from 20% at the base to about 100% at the top for the high period range. There is only a slight correction for the low period range since the response for small period is essentially governed by the fundamental mode. The increase in rigidity results in a decrease in period from the initial estimate based on eqn (2.160). An improved estimate for the period which accounts for the increase in rigidity is

144 Chapter 2: Optimal Stiffness Distribution 2πγ ∗ H = 0.9 -----------------T 1 = 0.9T 1 Γ1 Sv initial

for

T 1 > 0.6

(2.271)

Table 2.7: Modal parameters - building examples. T1(s)

T2(s)

T3(s)

ξ1 ( % )

ξ2 ( % )

ξ3 ( % ) Γ2 ⁄ Γ1 Γ3 ⁄ Γ1

Bldg #1 Initial (Q) Iteration 1 Iteration 2

0.52 0.52 0.52

0.21 0.21 0.21

0.13 0.13 0.13

2.00 2.00 2.00

5.04 5.04 5.04

8.17 8.12 8.12

0.40 0.40 0.40

0.23 0.23 0.23

Bldg #2 Initial (Q) Iteration 1 Iteration 2 Iteration 3

1.09 1.07 1.05 1.05

0.43 0.40 0.40 0.40

0.26 0.24 0.24 0.24

2.00 2.00 2.00 2.00

5.09 5.31 5.27 5.27

8.36 8.80 8.68 8.69

0.43 0.42 0.42 0.42

0.25 0.24 0.24 0.24

Bldg #3 Initial (Q) Iteration 1 Iteration 2 Iteration 3 Iteration 4

2.39 2.27 2.20 2.16 2.15

0.92 0.82 0.79 0.78 0.77

0.55 0.48 0.46 0.46 0.45

2.00 2.00 2.00 2.00 2.00

5.18 5.54 5.55 5.55 5.55

8.64 9.52 9.47 9.46 9.45

0.45 0.45 0.45 0.45 0.45

0.26 0.25 0.25 0.25 0.25

Bldg #4 Initial (Q) Iteration 1 Iteration 2 Iteration 3 Iteration 4

5.48 5.18 4.98 4.89 4.86

2.06 1.81 1.73 1.70 1.69

1.21 1.03 0.99 0.97 0.96

2.00 2.00 2.00 2.00 2.00

5.32 5.72 5.75 5.74 5.75

9.06 10.04 10.04 10.05 10.05

0.48 0.48 0.48 0.48 0.48

0.27 0.26 0.26 0.26 0.26

2.10 Stiffness Modification for Seismic Excitation 145 Table 2.8: Mean and standard deviation deformations. El

Centro

Taft

γm

γ sd

χm

χ sd

γm

γ sd

χm

χ sd

(10-3)

(10-4)

(10-5)

(10-6)

(10-3)

(10-4)

(10-5)

(10-6)

Bldg #1 Initial (Q) Iteration 1 Iteration 2

3.42 3.45 3.45

1.25 1.45 1.45

3.84 3.01 3.01

1.14 4.37 4.36

1.87 1.91 1.91

1.27 1.52 1.51

2.18 1.72 1.72

0.80 2.43 2.42

Bldg #2 Initial (Q) Iteration 1 Iteration 2 Iteration 3

4.40 4.32 4.39 4.41

4.42 3.77 3.47 3.76

4.44 3.50 3.54 3.52

4.10 7.71 7.45 7.56

3.42 3.35 3.21 3.13

6.37 3.69 3.29 2.77

3.61 2.92 2.75 2.66

6.06 4.04 3.28 3.18

Bldg #3 Initial (Q) Iteration 1 Iteration 2 Iteration 3 Iteration 4

6.13 5.51 4.97 4.67 4.54

9.82 5.72 4.44 4.79 4.89

4.61 3.78 3.35 3.10 2.99

10.06 7.56 6.07 5.83 5.69

4.42 3.84 3.60 3.50 3.51

16.52 5.09 3.39 3.01 3.41

3.61 2.78 2.58 2.46 2.45

15.88 4.61 2.92 2.63 2.86

Bldg #4 Initial (Q) Iteration 1 Iteration 2 Iteration 3 Iteration 4

3.50 2.62 2.56 2.52 2.49

19.42 3.81 3.75 3.55 3.42

2.20 1.43 1.42 1.39 1.36

14.22 2.86 2.17 2.08 2.03

4.54 4.36 4.58 4.31 4.10

7.66 5.47 5.55 5.22 4.92

2.74 2.41 2.60 2.49 2.39

5.55 5.98 6.48 5.98 5.51

146 Chapter 2: Optimal Stiffness Distribution

1 0.9

x Normalized height ---H

Building #2 Quadratic based Iteration 3

0.8 0.7

H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2

mode 2 mode 3

mode 1

0.1 0 0

0.2

SRSS

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.38: Converged shear rigidity distribution for Building 2.

1 0.9

x Normalized height ---H

Building #2 Quadratic based Iteration 3

0.8

H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.39: Maximum shear deformation for Building 2.

0.01

2.10 Stiffness Modification for Seismic Excitation 147

1

x Normalized height ---H

0.9

Building #3 Quadratic based Iteration 4

0.8 0.7

H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2%

0.6

mode 1

0.5 0.4 0.3 0.2

mode 3

mode 2

SRSS

0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.40: Converged shear rigidity distribution for Building 3.

1

x Normalized height ---H

0.9

Building #3 Quadratic based Iteration 4

0.8

H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.41: Maximum shear deformation for Building 3.

0.01

148 Chapter 2: Optimal Stiffness Distribution

1

x Normalized height ---H

0.9

Building #4 Quadratic based Iteration 4

0.8 0.7

H = 200m ρ m = 20000kg/m s = 0.625 S v = 1.2m/s ξ 1 = 2%

0.6

mode 1

0.5 0.4

mode 2

0.3 0.2

mode 3

SRSS

0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.42: Converged shear rigidity distribution for Building 4.

1

x Normalized height ---H

0.9

Building #4 Quadratic based Iteration 4

0.8

H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.43: Maximum shear deformation for Building 4.

0.01

Problems 149

Problems Problem 2.1 Determine the transverse shear and bending deformations corresponding to the following displacement distributions a) nπx u = sin ---------2L β = 0 n = 1, 3, … b) u = x 2 + a1 x 3 + a2 x 4 β = 2x + 3a 1 x 2 + 4a 2 x 3 c) u = a1 x + a2 x 2 + a3 x 3 β = 2a 2 x + 3a 3 x 2 d) x2 x2 x4 u = a 1 x – ------ + a 2 ----- – ------------ 2 12L 2 2L x3 β = a 2 x – ---------- 3H 2

Problem 2.2 Determine the shear and bending rigidity coefficients for the cross section shown below. The dimensions are in centimeters.

150 Chapter 2: Optimal Stiffness Distribution

1

face

50

E f = 210, 000 MPa

core

G c = 2, 000 MPa

1

25

Problem 2.3 Consider the chevron bracing scheme shown below. Determine an expression for D T . Assume the members carry only axial force.

h

Ad

Ad

α

α

B

Problem 2.4 The force-displacement relationship for the structures shown below is written as P = ku .

Problems 151

Ib h

P, u

Ib

P, u h

Ic

Ic

P, u h

B

B

a)

b)

Ad

B c)

Establish the expression for k for each structure. Comment on the relative efficiency of the structures for resisting lateral loading.

Problem 2.5 Consider a five story five bay rigid frame modeled as a shear beam. Assume all the columns in a story have the same properties, but allow for variation in column size over the stories. Establish an approximate expression for the equivalent shear beam stiffness for a typical story.

i H

h

B

Problem 2.6 Diagonal bracing is added to a rigid frame as indicated in the figure below.

152 Chapter 2: Optimal Stiffness Distribution a) Establish an approximate expression for the equivalent shear beam stiffness for a typical story. Assume the column properties are constant in a story, but vary from story to story.

i H

h

B b) Suppose the diagonal bracing system is expected to carry a specified fraction of the total shear in a story. Discuss how you would select the crosssectional area of the diagonal braces and moment of inertia for the columns in a typical story.

Problem 2.7 Determine the expressions for the distribution of shear and bending rigidities corresponding to constant deformations γ ∗ and χ∗ .

b0

H x

πx b = b 0 sin -----H

Problems 153

Problem 2.8 Determine the shear and bending rigidity distributions. Take: 1 γ ∗ = --------400 1 χ∗ = -----------------20, 000 Note: these values correspond to s = 0.1 (see eqn 2.14).

10kN

1kN/m

10m

Problem 2.9 A particular steel having an allowable stress of 600 MPa and Young’s modulus equal to 210,000 MPa has been selected as the “design” material. Assuming the material is to be used in 45° bracing members, for what range of transverse shear strains will the design be controlled by motion constraints rather than strengths?

Problem 2.10 a) Refer to problem 2.5. Consider a uniform lateral loading of 30 kN/m, a story height of 4m, and a design shear strain, γ ∗ , equal to 1/300 for each story. Estimate the value of I c for each story assuming steel is selected as the material. b) Refer to problem 2.6. Use the loading defined in part a. Select the

154 Chapter 2: Optimal Stiffness Distribution bracing properties such that the bracing carries 25% of the total story shear.

Problem 2.11

P 2, u 2 P 1, u 1 4m

2 1

3m Determine the member stiffness factors for the following prescribed loading and displacement quantities. P 1 = 100 kN

P 2 = 50 kN

u 1 = 0.01 m

u 2 = 0.005 m

Problem 2.12

P 2, u 2 P 1, u 1 4m

3

2 1

4m

3m

The design objective is to determine the member stiffness factors for the

Problems 155 above truss such that the nodal displacements corresponding to P 1 = 25 kN , P 2 = 50 kN are u 1 = 0.01 m , u 2 = 0.01 m . Generate solutions using the least square and mean value least square approaches.

Problem 2.13

P 2, u 2 P 1, u 1 4m

3

2

1 P 3, u 3

4

5 4m

4m

Recommend stiffness factors for the truss shown above based in the following requirements u 3 = – 0.0019m u 1 = – 0.0087m u 2 = 0.0012m P 1 = – 100kN

P 2 = 50kN

P 3 = – 100kN

Problem 2.14 u∗

u∗

P Beam 1 is a shear beam. Beam 2 is a bending beam. H

1

αP

2

( 1 – α )P

Consider two cantilever beams coupled with a rigid link at the top.

156 Chapter 2: Optimal Stiffness Distribution Assume beam 1 is a shear beam and beam 2 is a bending beam. Determine the optimal rigidity distributions that satisfy the displacement requirement u∗ = H ⁄ β and divide the lateral load between the two beams as indicated.

Problem 2.15 The bending beam model is based on the assumption of no transverse shear strain, i.e. γ = 0 . Theoretically, one sets D T = ∞ , which results in s = ∞ . The displacement distributions for periodic vibration with constant curvature, χ∗ , and γ = 0 are x 2 u = q B ---- H 2x β = q B ------H2 H2 q B = -------χ∗ cos ( ω 1 t + δ ) 2 Consider uniform mass density. Determine D B ( x ) for this prescribed mode shape.

Problem 2.16 m 1 = aρ m H

x

H

bx ρ m = ρ m 1 – ------ H

Consider a cantilever beam having the mass distribution indicated in the figure. Assuming the beam acts as a shear beam, determine the transverse shear

Problems 157 rigidity distribution required in order that the fundamental mode shape have the following form: x φ ( x ) = ---H Investigate how shear rigidity varies with a , taking b = 0 .

Problem 2.17 The structure shown below consists of a cantilever beam and a rigid weightless link connecting 2 masses to the top end of the beam. Assume the cantilever beam has uniform mass density ρ m and negligible transverse shear deformation.

b

b m 1 = ( ρ m H )a

m1

m1

H

a) Determine the bending rigidity distribution required to produce the fundamental mode shape corresponding to constant curvature. The displacement expressions for this case are: x u = q B ---- H

2

2x β = q B ------H2 H2 q B = -------χ∗ cos ( ω 1 t + δ ) 2 b) Investigate the sensitivity of the solution to a variation of a .

158 Chapter 2: Optimal Stiffness Distribution c) Suppose outriggers are attached to the rigid links and the beam is extended as shown below. Determine the bending rigidity distribution for the case where a uniform static loading is applied, and the design objective is constant curvature. H/2 m1 A

m1 A

d) Repeat part a using the structure defined in part c.

Problem 2.18 Consider a cantilever shear beam having uniform mass density and constant transverse shear rigidity. Allowing for material damping and external damping, the equilibrium equation for free vibration has the following form (see eqns (2.197) and (2.199)): D T u, xx + C T u, xxt – ρ m u, tt – C′u, t = 0

(1)

The boundary conditions are u(x = 0) = 0 V(x = L) = 0

(2) (3)

Substituting for V in eqn (3) leads to D T u, x ( L ) + C T u, xt ( L ) = 0

(4)

The general solution of eqn (1) which satisfies the boundary conditions at x = 0, L is u = e αt sin λx

(5)

Problems 159 2n – 1 λ = ---------------π 2L

n = 1, 2, …

Let DT 1 / 2 ω = λ -------ρm

(6)

λ 2 C T + C′ -------------------------- = 2ξω ρm

(7)

a) Derive the expression for α in terms of ω and ξ . Assume ξ < 1 . The solution corresponding to n = 1 is the fundamental solution. Comment on the nature of this solution. b) Compare the mode shapes with the profiles based on constant shear deformation for the fundamental mode. Consider n from 1 to 3. c) Let ξ n be the value of ξ corresponding to λ = λ n , i.e., the n ‘th mode. Comment on how ξ n varies with n .

Problem 2.19 Refer to the figure in problem 2.5. Assume the structure is a discrete shear beam. Determine the distribution of equivalent shear beam stiffness based on a linear profile for the fundamental mode and the following floor masses: m 1 = m 2 = m 3 = m 4 = 10, 000 kg m 5 = 5, 000 kg Relate the shear beam stiffness to the column size for each story.

Problem 2.20 Repeat problem 2.19 using the structure defined in problem 2.6. Assume the bracing contributes 25% of the story stiffness.

160 Chapter 2: Optimal Stiffness Distribution

Problem 2.21

u + ug m k

c

ug

A single degree of freedom system is to be designed to displace a given amount, u max , under a specified seismic excitation. Use the response spectrum for the pseudo spectral velocity shown in the figure, and take m = 10, 000 kg. Determine design values for k and c for a range of damping coefficients ( ξ = 0.05, 0.10, 0.20 ) and two values of u max : u max = 0.1 m, 0.2 m . Also determine the maximum values of the stiffness and damper forces. Spectral Velocity S v (m/s) S v = 1.4, ξ = 0.05 S v = 1.0, ξ = 0.1

1.0

S v = 0.8, ξ = 0.2

0.1 0.1

0.6 1.0

10

Period T (sec)

Problems 161

Problem 2.22 m⁄2 h h

p 3, u 3

m

p 2, u 2

m

p 1, u 1

h

The structure consists of 3 shear beam segments and the lumped masses indicated in the figure. Take m = 1, 000 kg .

h = 4 meters

a) Determine the shear rigidity distribution required for the fundamental mode vector to have the form 1 2 Φ = ---, ---, 1 3 3 b) Assume a periodic force, p = pˆ sin ( Ωt ) is applied at node 3. Calibrate the rigidity for the following conditions: pˆ = 10 kN

Ω = 4π

h q max = --------150

ξ = 0.05

c) Calibrate the shear rigidity factors for the following conditions: h q max = --------100

Sv

max

= 1.2 m/s

ξ = 0.02

Use equations (2.249) - (2.251). d) Assume linear viscous dampers having the same property, c 1 = c 2 = c 3 = c , are installed. Determine c based on the results of part c.

162 Chapter 2: Optimal Stiffness Distribution

Problem 2.23 Refer to problem 2.19. a) Calibrate the stiffness for the following seismic excitation criteria: u ( H ) max = 0.125 m Sv

max

= 1.0 m/s

ξ = 0.05

b) Assume viscous dampers are installed in the third, fourth, and fifth stories. Take the damper coefficient to be the same for each story. Determine the coefficient required by the calibration specified in part a.

Problem 2.24 Refer to problem 2.20. a) Calibrate the structure for the following seismic criteria: u ( H ) max = 0.125 m Sv

max

= 1.2 m/s

ξ = 0.02

b) Assume a uniform distribution of damping over the height. Determine the damping coefficient required by part a. c) Suppose a single damper is connected to the top node (node 5). Determine the value of c required by part a.

Problem 2.25 Refer to problem 2.15. Take H = 200 m and ρ m = 10, 000 kg/m . ˜ and Γ . a) Determine m b) Assume the following loading x sin Ωt b = b 0 ------------------H

where b 0 = 6 kN/m

Problems 163 is applied. Calibrate the stiffness based on limiting u ( H ) to 0.5 m for a frequency of 1 radian per second. c) Suppose the design is constrained by the peak acceleration at x = H under periodic loading. Calibrate the stiffness for u˙˙( H ) max = 0.015 g where g = 9.86 m/s 2 . Consider ξ to be equal to 0.05 and use the loading defined in part b. d) Assume the stiffness calibration is dictated by seismic excitation. Take u ( H ) max = 1 m , S v = 1 m/s , ξ = 0.05 . Determine D B ( x ) . Assume max C T ( x ) = αD B ( x ) where α is a constant. Determine α corresponding to ξ = 0.05 .

Problem 2.26 Refer to problem 2.16. ˜ and Γ in terms of ρ , a , and b . a) Determine m m b) Assume a uniform periodic loading is applied to the beam. Calibrate the shear rigidity for the following specifications. H = 10 m

u ( H ) max = H ⁄ 300

ρ m = 100 kg/m a = 1 b = 0.5 Ω = 2π rad/s c) Calibrate the rigidity for the case of seismic loading. Take u ( H ) max = H ⁄ 300 Sv

max

= 1.2 m/s

ξ = 0.02 Use eqn (2.249).

164 Chapter 2: Optimal Stiffness Distribution

Problem 2.27 ˜ and Γ . Refer to problem 2.17. Determine m

Problem 2.28 Consider eqn (2.249). Let q∗ = H ⁄ α where α is a design parameter. Express H as H = nh where h is the typical story height and n is the number of stories. This leads to T = fn . Estimate a typical value of f .

Problem 2.29

10 m

5m

This problem concerns the preliminary design of the lateral stiffness system for a 10 story rectangular rigid frame. The frame properties and design criteria are as follows: • • • •

Height = 5 m/story Width = 10 m/bay Mass/floor = 10,000 kg Seismic loading: S v = 1.2 m/s, ξ = 0.02

•

Max. Deflection at top = 0.25 m

Problems 165 •

γ max = 1/200 Model the frame as a 10 DOF shear beam.

a) Determine an initial estimate for the lateral stiffness at each story level considering a single mode response and ξ = 0.02 . Use the spectrum shown below. b) Consider linear viscous dampers to be installed at each story level. Take c i = αk i for story i . Using the stiffness distribution established in Part a (frequencies and mode shapes), establish the modal equations for a modal decomposition in terms of the first 3 modes. Determine α such that ξ 1 = 0.02 . Determine ξ 2 and ξ 3 corresponding to this value of α . Estimate the maximum ‘‘mean square’’ deformation response using the response spectrum shown below, and modify the initial stiffness distribution using eqn (2.252). Carry out this computation for 2 iterations. c) Suppose the lateral stiffness is maintained constant over 2 story heights, i.e. k2 = k1 k 4 = k 3 , etc. Discuss how you would determine initial estimates for k 1, k 3, … considering a single mode response and ξ = 0.02 . Also discuss how you would modify the stiffness distribution.

Spectral Velocity S v (m/s) S v = 1.2, ξ = 0.02 S v = 0.92, ξ = 0.05 S v = 0.8, ξ = 0.1

1.0

S v = 0.6, ξ = 0.2

0.1 0.1

0.6 1.0

10

Period T (sec)

166 Chapter 2: Optimal Stiffness Distribution

167

Chapter 3

Optimal passive damping distribution 3.1 Introduction Damping is the process by which physical systems such as structures dissipate and absorb the energy input from external excitations. Damping reduces the build-up of strain energy and the system response, especially near resonance conditions where damping governs the response. Figure 3.1 illustrates the influence of damping on the time history of the strain energy for a system with mass of 1 kg and a period of 1 second subjected to the unscaled El Centro accelerogram shown in Fig. 2.21. The symbols in Fig. 3.1 refer to the energy input ( E I ), the energy stored ( E s ), and the energy dissipated ( E D ). During the early stage of the response, there is a rapid build-up of the input energy, similar to an impulsive loading. Damping dissipates energy over a response cycle, in this case, 1 second. For low damping ratio, the energy dissipated per cycle is small, and many cycles are required before the input energy is eventually dissipated. As ξ is increased, the energy dissipated per cycle increases, and the stored energy build up is reduced. Shifting from ξ = 0.02 to ξ = 0.1 reduces the peak stored energy demand by a factor of 3.7 for this particular system and earthquake excitation. It should be noted that seismic accelerograms differ with respect to their frequency content and intensity, and therefore one needs to carry out energy time history studies for individual excitations applied to a specific system. For example, Fig 3.2 shows the response of the same system for a typical Northridge accelerogram. The input energy build up for the Northridge loading is quite different than for the El Centro loading.

168 Chapter 3: Optimal Passive Damping Distribution

1.0 ξ = 2%

Energy (J)

0.8 0.6 0.4 0.2 0

0

2

4

6

8

10

12

14

16

18

20

0.8

Energy (J)

ξ = 5% 0.6 0.4 0.2 0

0

2

4

6

8

10

12

14

16

18

20

0.7 Energy (J)

0.6

ξ = 10%

0.4 EI ED

0.2

0

ES

0

2

4

6

8

10 Time (s)

12

14

16

18

Fig. 3.1: Energy Build Up, El Centro (S00E), Imperial Valley 1940

20

3.1 Introduction 169

1.0

Energy (J)

0.8 0.6 EI ED

0.4

ES

0.2 0

0

2

4

6

8

10 Time (s)

12

14

16

18

20

Fig. 3.2: Energy Build Up, Arleta Station (90 DEG), Northridge 1994, ξ = 2% Dissipation and absorption are attributed to a number of external and internal mechanisms, including the following: • Energy dissipation due to the viscosity of the material. This process depends on the time rate of change of the deformations, and is referred to as material damping. Viscoelastic materials belong to this category. • Energy dissipation and absorption caused by the material undergoing cyclic inelastic deformation and ending up with some residual deformation. The cyclic inelastic deformation path forms a hysteresis loop which correspond to energy dissipation; the residual deformation is a measure of the energy absorption. This process is generally termed hysteretic damping. • Energy dissipation associated with overcoming the friction between moving bodies in contact, such as flexible connections. Coulomb damping refers to the case where the magnitude of the friction force is constant. Structural damping is a more general friction damping mechanism which allows for a variable magnitude of the friction force. • Energy dissipation resulting from the interaction of the structure with its surrounding environment. Relative motion of the structure

170 Chapter 3: Optimal Passive Damping Distribution generates forces which oppose the motion and extract energy from the structure. Fluid-structure interaction is a typical case. The fluid exerts a drag force which depends on the relative velocity and functions as an equivalent viscous damping force. • Damping devices installed at discrete locations in structures to supplement their natural energy dissipation/absorption capabilities. These mechanisms may be passive or active. Passive mechanisms require no external energy, whereas active mechanisms cannot function without an external source of energy. Passive devices include viscous, friction, tuned mass, and liquid sloshing dampers. Active damping is achieved by applying external control forces to the structure over discrete time intervals. The magnitudes of the control forces are adjusted at each time point according to a control algorithm. • Passive damping removes energy from the response, and therefore can not cause the response to become unstable. Since active control involves an external source of energy, there is the potential for introducing an instability in the system. The term “semi-active” refers to a particular class of active devices that require a relatively small amount of external energy and apply the control force in such a way that the resulting motion is always stable. Chapter 6 discusses active control devices. In this chapter, the response characteristics for material, hysteretic, and friction damping mechanisms are examined for a single degree-of-freedom (SDOF) system. The concept of equivalent viscous damping is introduced and is used to express viscoelastic, structural, and hysteretic damping in terms of their equivalent viscous counterpart. Numerical simulations are presented to demonstrate the validity of this concept for SDOF systems subjected to seismic excitation. This introductory material is followed by a discussion of the influence of distributed viscous damping on the deformation profiles of multi-degree-offreedom (MDOF) systems. The damping distribution is initially taken to be proportional to the converged stiffness distribution generated in the previous chapter, and then modified to allow for non-proportional damping. Numerical results and deformation profiles for a range of structures subjected to seismic loading are presented, and the adequacy of this approach for distributing damping is assessed. Distributed passive damping can be supplemented with one or more discrete damping devices to improve the response profile. Discrete viscous

3.2 Viscous,Frictional, and Hysteretic Damping 171 dampers inserted in discrete shear beam type structures are considered in this chapter; the basic theory for tuned mass dampers is presented in the next chapter. Subsequent chapters deal with base isolation, a form of passive stiffness/ damping control, and active control.

3.2 Viscous, frictional, and hysteretic damping Viscous damping Viscous damping is defined as the energy dissipation mechanism where the damping force is a function of the time rate of change of the corresponding displacement measure: F = f ( u˙ )

(3.1)

where F is the damping force and u˙ is the velocity in the direction of F . The linearized form is written as: F = cu˙

(3.2)

where c , the damping coefficient, is a property of the damping device. Linear viscous damping is convenient to deal with mathematically and therefore is the preferred way of representing energy dissipation. In general, the work W done on the device during the time interval u ( t 2 ) by t2 [ t 1, t 2 ] , is given W =

∫

F du =

∫

Fu˙ dt

(3.3)

u ( t1 ) t1 Considering periodic excitation

u = uˆ sin Ωt

(3.4)

and evaluating eqn (3.3) for one full cycle under linear viscous damping leads to W viscous = cπΩuˆ

2

(3.5)

This term represents the energy dissipated per cycle by the damping device, as the system to which it is attached undergoes a periodic motion of amplitude uˆ and frequency Ω . Figure 3.3 shows the force-displacement relationship for periodic excitation; the enclosed area represents W .

172 Chapter 3: Optimal Passive Damping Distribution

t = 0

F u = uˆ sin Ωt

cΩuˆ

uˆ

– uˆ

u

π t = ---Ω

Fig. 3.3: Viscous response - periodic excitation

Example 3.1: Viscous damper Figure 3.4 shows a possible design for a viscous damping device. The gap between the plunger and external plates is filled with a linear viscous fluid characterized by τ = G v γ˙

(3.6)

where τ and γ are the shearing stress and strain measures respectively and G v is the viscosity coefficient. Assuming no slip between the fluid and plunger, the shear strain is related to the plunger motion by u γ = ----td

(3.7)

plunger

td

fluid

F,u td

fluid

L

w

Side View

End View

Fig. 3.4: Viscous damping device

3.2 Viscous,Frictional, and Hysteretic Damping 173 where t d is the thickness of the viscous layer. Letting L and w represent the initial wetted length and width of the plunger respectively, the damping force is equal to (3.8)

F = 2wLτ Substituting for τ results in 2wLG v F = ------------------- u˙ td

(3.9)

Finally, eqn (3.9) is written as (3.10)

F = cu˙ where c represents the viscous coefficient of the device, 2wL c = ----------- G v td

(3.11)

The design parameters are the geometric measures w, L, t d and the fluid viscosity, Gv . A schematic diagram of a typical viscous damper employed for structural applications is contained in Fig 3.5; an actual damper is shown in Fig 3.6. Fluid is forced through orifices located in the piston head as the piston rod position is changed, creating a resisting force which depends on the velocity of the rod. The damping coefficient can be varied by adjusting the control valve. Variable damping devices are useful for active control. Section 6.4 contains a description of a particular variable damping device that is used as a semi-active force actuator. This chapter considers only passive damping, i.e. a fixed damping coefficient.

Fig. 3.5: Schematic diagram - viscous damper

174 Chapter 3: Optimal Passive Damping Distribution

Fig. 3.6: Viscous damper - 450 kN capacity (Taylor Devices Inc. http://www.taylordevices.com)

Equation (3.5) shows that the energy loss per cycle for viscous damping depends on the frequency of the excitation. This dependency is at variance with observations for real structural systems which indicate that the energy loss per cycle tends to be independent of the frequency. In what follows, a number of damping models which exhibit the latter property are presented. Friction damping Coulomb damping is characterized by a damping force which is in phase with the deformation rate and has constant magnitude. Mathematically, the force can be expressed as F = F sgn ( u˙ )

(3.12)

where sgn ( u˙ ) denotes the sign of u˙ . Figure 3.7 shows the variation of F with u for periodic excitation. The work per cycle is the area enclosed by the response curve

3.2 Viscous,Frictional, and Hysteretic Damping 175

F F

–u

u = uˆ sin Ωt u = uˆ

u

u

Fig. 3.7: Coulomb damping force versus displacement W coulomb = 4Fu

(3.13)

Figure 3.8 shows a coulomb friction damper used with diagonal X bracing in structures. Friction pads are inserted at the bolt-plate connections. Interstory displacement results in relative rotation at the connections, and the energy dissipated is equal to the work done by the frictional moments during this relative rotation.

Fig. 3.8: Friction brace damper Structural damping removes the restriction on the magnitude of the damping force, and considers the force to be proportional to the displacement amplitude. The definition equation for this friction model has the form F = k s u sgn ( u˙ )

(3.14)

where k s is a pseudo-stiffness factor. Figure 3.9 shows the corresponding cyclic response path. The energy dissipated per cycle is equal to 2 ks u 2 (3.15) W structural = 4 ----------- = 2k s u 2

176 Chapter 3: Optimal Passive Damping Distribution

u = uˆ sin Ωt u = uˆ

–u

F

π t = ------2Ω ks

u

u

Fig. 3.9: Structural damping force versus displacement Hysteretic damping Hysteretic damping is due to the inelastic deformation of the material composing the device. The form of the damping force-deformation relationship depends on the stress-strain relationship for the material and the make-up of the device. Figure 3.10 illustrates the response path for the case where the material forcedeformation relationship is elastic-perfectly plastic. The limiting values are F y , the yield force, and u y , the displacement at which the material starts to yield; k h is the elastic damper stiffness. The ratio of the maximum displacement to the yield displacement is referred to as the ductility ratio and is denoted by µ . With these definitions, the work per cycle for hysteretic damping has the form 2 µ–1 (3.16) W hysteretic = 4k h u y [ µ – 1 ] = 4F y u -----------µ

3.2 Viscous,Frictional, and Hysteretic Damping 177

F

u = uˆ sin Ωt u = uˆ u µ = ----uy

Fy

–u uy

u

u

kh

Fig. 3.10: Hysteretic damping force versus displacement Figure 3.11 shows a bracing element that functions as an hysteretic damper (Watanabe, 1998). The element is composed of a core member fabricated with highly ductile low strength steel (yield strength of 200 MPa., maximum percent strain of 60%), a cylindrical jacket, and mortar placed between the core member and the jacket. The jacket functions as an additional bending element, and its cross sectional moment of inertia is selected such that the buckling load is equal to the yield force. This design feature allows the brace to be used for both tensile and compressive loading. Triangular plate hysteretic dampers that dissipate energy through bending action have also been used for buildings (Tsai, 1993 and 1998).

Fig. 3.11: Hysteretic damper brace element (Watanabe, 1998)

Example 3.2: Stiffness of a rod hysteretic damper Consider a damping device consisting of a cylindrical rod of length L and area A .

178 Chapter 3: Optimal Passive Damping Distribution Suppose the material is elastic-perfectly plastic, as shown in Fig. 3.12. The relevant terms are u ε = --L

(3.17)

F = Aσ

(3.18)

σ

σy

F,u

F E

L εy

ε

Fig. 3.12: Elastic-perfectly plastic damper device. Then, u y = Lε y

(3.19)

F y = Aσ y = k h u y AE k h = -------L

(3.20) (3.21)

Example 3.3: Stiffness of two hysteretic dampers in series The device treated in Example 3.2 is modified by adding a second rod in series, as shown in Fig. 3.13. The yield force for the second rod is assumed to be greater than the yield force for the first rod, A 2 σ y, 2 > A 1 σ y, 1 ≡ F y, 1

(3.22)

Since the force is the same for both devices, the total elastic displacement is the sum of the individual contributions.

3.3 Viscoelastic Material Damping 179

A1 , E1

A2 , E2

F

F,u L1

L2

Fig. 3.13: Two rod hysteretic damping device. L1 L2 1 (3.23) u = ------------- + ------------- F = ----- F kh A1 E1 A2 E2 Specializing eqn (3.23) for the onset of yielding, one obtains A1 L2 E1 F y, 1 (3.24) u y = ----------- = L 1 ε y, 1 1 + -------------------kh A2 L1 E2 A1 E1 1 (3.25) k h = ------------- ----------------------------L1 A1 L2 E1 1 + -------------------When two elements areAused, 2 L 1 E 2one can vary both the yield force, F y , and the elastic yield deformation u y . The energy dissipation increases with decreasing u y , for a given deformation amplitude u .

3.3 Viscoelastic material damping A material is considered to be elastic when the stresses due to an excitation are unique functions of the associated deformation. Similarly, a material is said to be viscous when the stress state depends only on the deformation rates. For simple shear, these definitions translate to Elastic τ = Ge γ

(3.26)

Viscous τ = G v γ˙

(3.27)

The stress-deformation paths for periodic strain are illustrated in Figs 3.14(a) and

180 Chapter 3: Optimal Passive Damping Distribution 3.14(b). There is no time lag between stress and strain for elastic behavior, whereas the stress is π ⁄ 2 radians out of phase with the strain for viscous behavior. If these relations are linearly combined, one obtains the path shown in Fig. 3.14(c). τ

τ

γ = γˆ sin Ωt

τ G v Ωγˆ

G e γˆ

γˆ

γ

γˆ

G v Ωγˆ

(a) Elastic

γ

(b) Viscous

γ

(c) Viscoelastic

Fig. 3.14: Stress-deformation relations. Materials that behave similar to Fig. 3.14(c) are called viscoelastic. The properties of a linear viscoelastic material are determined by applying a periodic excitation and observing the response, which involves both an amplification and a phase shift. The basic relations are expressed as γ = γˆ sin Ωt

(3.28)

τ = γˆ [ G s sin Ωt + G l cos Ωt ]

(3.29)

where G s is the storage modulus and G l is the loss modulus. The ratio of the loss modulus to the storage modulus is defined as the loss factor, η Gl (3.30) η = ------ = tan δ Gs An alternate form for eqn (3.29) is ˆ sin ( Ωt + δ ) τ = γˆ G ˆ = G

2

2

Gs + Gl = Gs 1 + η

(3.31) 2

(3.32)

The angle δ is the phase shift between stress and strain. Delta ranges from 0 for elastic behavior to π ⁄ 2 for pure viscous behavior. Experimental observations show that the material properties G s and η

3.3 Viscoelastic Material Damping 181 vary with temperature and the excitation frequency. Figure 3.15 illustrates these trends for ISD110, a 3M product. The dependency on frequency makes it difficult to generalize the stress-strain relationships based on periodic excitation to allow for an arbitrary time varying loading such as seismic excitation. This problem is addressed in the next section.

To determine the damping properties at the desired temperature and frequency from the data graph shown above, proceed as follows: • Locate the desired frequency on the RIGHT vertical scale. • Follow the chosen frequency line to the desired temperature isotherm. • From this intersect, go vertically up and/or down until crossing both the shear (storage) modulus G s and loss factor curves η . • Read the storage modulus and loss factor values from the appropriate LEFT hand scale.

Fig. 3.15: Variation of 3M viscoelastic material, ISD110, with frequency and temperature. The energy dissipated per unit volume of material for one cycle of deformation is determined from 2π ⁄ Ω W viscoelastic =

∫

τγ˙ dt

(3.33)

0 Substituting for τ and γ using eqns (3.28) and (3.29) results in

W viscoelastic = πG l γˆ

2

(3.34)

182 Chapter 3: Optimal Passive Damping Distribution The corresponding expression for a pure viscous material is generated using eqn (3.27) W viscous = πG v Ωγˆ

2

(3.35)

This expression involves the frequency explicitly whereas the effect of frequency is embedded in G l for the viscoelastic case.

Example 3.4: Viscoelastic damper A damper device is fabricated by bonding thin sheets of a viscoelastic material to steel plates, as illustrated in Fig. 3.16. Since the elastic modulus for steel is considerably greater than the shear modulus G s for the sheet material, one can consider all the motion to be due to shear deformation of the sheets. Defining u as the relative displacement of the ends of the damper device, the shearing strain is u γ = ----td

(3.36)

Given γ , one evaluates τ with the stress-strain relation and then F using the equilibrium equation for the system (3.37)

F = 2wLτ Applying a periodic excitation u = uˆ sin Ωt

(3.38)

td

F,u

td

w

L

L

Side View

Top View

Fig. 3.16: Viscoelastic damper device. and taking τ according to eqn (3.29), one obtains

F,u

3.3 Viscoelastic Material Damping 183 F = f d G s uˆ [ sin Ωt + η cos Ωt ]

(3.39)

2wL f d = ----------td Equation (3.39) can also be written as

(3.40)

ˆ uˆ sin ( Ωt + δ ) F = f dG

(3.41)

ˆ = G 1 + η2 G s

(3.42)

Finally, the energy dissipated per cycle is given by W = πη f d G s uˆ

2

(3.43)

Typical polymer materials, such as Scotchdamp ISD110 (3M Company, 1993) have G s in the range of 1.5MPa and η ≈ 1 .

Based on the result of the previous example, the expressions defining the response of a viscoelastic damper due to periodic excitation can be written in a generalized form u = uˆ sin Ωt

(3.44)

F = f d G s uˆ ( sin Ωt + η cos Ωt ) W viscoelastic = πη f d G s uˆ

2

(3.45) (3.46)

where f d depends on the geometric configuration of the device, G s is the storage modulus, and η is the material loss factor. Figure 3.17 shows the variation of F with u over the loading cycle.

184 Chapter 3: Optimal Passive Damping Distribution

π t = ------2Ω

F

u = uˆ sin Ωt

f d Gs η f d G s uˆ

t = 0 uˆ

u

π t = ---Ω Fig. 3.17: Variation of F d with u d for viscoelastic material.

3.4 Equivalent viscous damping The expression for the damping force corresponding to linear viscous damping is the most convenient mathematical form, in comparison to the other damping force expressions, for deriving approximate analytical solutions to the force equilibrium equations. Therefore, one way of handling the different damping models is to convert them to equivalent viscous damping models. In what follows, a conversion strategy based on equating the energy dissipated per cycle of periodic excitation to the corresponding value for linear viscous damping is described. Linear viscous damping is defined by eqn (3.2): (3.47)

F = cu˙ Specializing eqn (3.47) for periodic excitation u = uˆ sin Ωt

(3.48)

leads to F = cΩuˆ cos Ωt

(3.49)

Also, noting eqn (3.5), the energy loss per cycle is W = cπΩuˆ

2

(3.50)

The force and energy loss for the other models are expressed in terms of an equivalent damping coefficient, c eq

3.4 Equivalent Viscous Damping 185 F = c eq Ωuˆ cos Ωt W = c eq πΩuˆ

2

(3.51) (3.52)

Substituting W for a particular damping model in eqn (3.52), and taking u = uˆ , one obtains the equivalent damping coefficient. The coefficients for the various models are listed below: Coulomb 4F c eq = ----------πΩuˆ

(3.53)

Structural 2k s c eq = -------πΩ

(3.54)

Hysteretic 4F y µ – 1 c eq = ----------- -----------πΩuˆ µ

(3.55)

Viscoelastic

f dG f d ηG s l c eq = ----------------- = ------------Ω Ω

(3.56)

These expressions are valid for periodic excitation of amplitude uˆ and frequency Ω . They can be used to approximate structural and hysteretic damping as pseudo-viscous damping but require specifying a representative frequency, Ω r , and amplitude, u r . In this case, eqns (3.54) and (3.55) are written as Structural 2k s c eq = ---------πΩ r

(3.57)

Hysteretic 4F y µ – 1 c eq = ---------------- -----------πΩ r u r µ

(3.58)

where Fy u y = -----k u rh µ = -----uy

(3.59) (3.60)

Numerical simulations illustrating the accuracy of this approximation are

186 Chapter 3: Optimal Passive Damping Distribution provided by the following examples.

Example 3.5: Structural and hysteretic damping comparison - seismic excitation A 1 DOF shear beam having the following properties is considered: 6

m = 4 ×10 kg

ω 1 = 1.17rd ⁄ s

k = 5517kN ⁄ m

T 1 = 5.35s

c = 187.9kNs ⁄ m

ξ 1 = 2.0%

The equivalent structural stiffness is generated using eqn (3.57), taking c = c e and Ω r equal to the fundamental frequency ω 1 . The corresponding structural stiffness is πω 1 c e π ( 1.17 ) ( 187.9 ) (3.61) k s = --------------- = ------------------------------------- = 346.6kN ⁄ m 2 2 Results for this model subjected to Taft excitation are compared with the corresponding results for the linear viscous model in Figs 3.18 and 3.19. Close agreement is observed.

3.4 Equivalent Viscous Damping 187

−3

6

x 10

Structural damping versus Viscous damping T 1 = 5.35 s

4

ξ 1 = 2%

γ - m/m

2

0

−2

Viscous damping Structural damping

−4

−6 0

10

20

30

40

50

60

Time t - s Fig. 3.18: Response of SDOF with structural damping.

5

4

x 10

Structural damping T 1 = 5.35 s

3

ξ 1 = 2%

F - N

2

Taft excitation

1

0

−1

−2

−3

−4 −6

−4

−2

0

2

4

6 −3

γ - m/m Fig. 3.19: Structural damping force versus deformation.

x 10

188 Chapter 3: Optimal Passive Damping Distribution

The hysteretic model calibration defined by eqn (3.58) is not as straight forward since both the yield force and the ductility are involved. For periodic motion, the maximum displacement u d = uˆ d is known. Then, one can specify the desired ductility µ∗ and compute the required force level and initial stiffness with πω 1 uc eq µ∗ (3.62) F y = ---------------------- --------------4 µ∗ – 1 u (3.63) u y = -----µ∗ Fy (3.64) k h = -----uy For non-periodic motion, one needs to specify the limiting elastic displacement u y , and estimate the maximum amplitude u . This leads to estimates for the ductility ratio µ∗ and the peak force F y . Figures 3.20 and 3.21 show the results based on taking u equal to the peak amplitude observed for pure viscous damping, and a ductility ratio µ∗ = 7.5 . One can adjust µ∗ and F y to obtain closer agreement. Since the energy is dissipated only during this inelastic phase, hysteretic damping is generally less effective than either viscous or structural damping for low intensity loading.

−3

6

x 10

4

Hysteretic damping versus Viscous damping T 1 = 5.35 s

ξ 1 = 2%

γ - m/m

2

0

−2

−4

−6 0

Viscous damping Hysteretic d amping 10

20

30

40

50

Time t - s Fig. 3.20: Response of SDOF with hysteretic damping.

60

3.4 Equivalent Viscous Damping 189

5

F - N

4

x 10

3

Hysteretic damping T 1 = 5.35 s

2

Taft excitation

ξ 1 = 2%

1

0

−1

−2

−3 −6

−4

−2

0

2

4

6 −3

γ - m/m

x 10

Fig. 3.21: Hysteretic damping force versus deformation. The calibration of the equivalent viscous damping coefficient was based on assuming a periodic excitation. As discussed above, non-periodic excitation requires some assumptions as to the response. An improved estimate of the equivalent damping coefficient can be obtained by evaluating the actual work done by the damping force. Starting with t ED

actual

=

∫ Fu˙ dt 0

and writing ED leads to

(3.65) t

eq. viscous t

∫

= c eq u˙ 2 dt

∫ Fu˙ dt

0

0 c eq ( t ) = --------------- = 2ξ eq ωm t

∫

(3.66)

(3.67)

2 dt Equation (3.67) canu˙ be used to evaluate the variation over time of the equivalent damping ratio. Taking t = t end , the total duration of the response, provides an 0 estimate of the effective damping ratio. Figure 3.22 shows results generated for a range of seismic excitations and hysteretic damper yield force levels. As expected, the effective damping increases with increasing excitation and decreases with

190 Chapter 3: Optimal Passive Damping Distribution

Equivalent Viscous Damping Ratio

increasing yield force.

0.2 T = 4.91sec

ξH

ζN = 2 %

0.1

S

v

= 2.4 m/s

S S

0

0

0.01

0.02

0.03

v

v

= 1.2 m/s = 0.6 m/s

0.04

F y ⁄ mg 0.05

Fig. 3.22: Equivalent viscous damping ratio vs. yield force

The viscoelastic model calibration is more involved since the material properties are also frequency dependent. Referring back to eqn (3.45), the damping force for periodic excitation u = uˆ sin Ωt

(3.68)

was expressed as F = uˆ f d G s ( sin Ωt + η cos Ωt )

(3.69)

where f d is a geometric factor defined by the geometry of the device. Our objective is to express F as F = k eq u + c eq u˙

(3.70)

where k eq and c eq are equivalent stiffness and damping terms. Considering periodic excitation, eqn (3.70) takes the form F = uˆ [ k eq sin Ωt + Ωc eq cos Ωt ]

(3.71)

One can obtain estimates for k eq and c eq with a least square approach. Assuming

3.4 Equivalent Viscous Damping 191 there are N material property data sets, and summing the squares of the errors for k eq and c eq over N the ensemble results in

∑ ∑

2

[ k eq – f d G s ( Ω i ) ] f d Gs ( Ωi ) ⋅ η ( Ωi ) 2 J c = i = 1 c e – -------------------------------------------Ωi Jk =

(3.72)

N

(3.73)

i=1

∑

)

Minimizing eqn (3.72) with respect to k eq yields N 1 Gs ( Ωi ) = f d G s k eq = f d ---N

(3.74)

i=1

Similarly, minimizing eqn N (3.73) with respect to c eq results in G s ( Ω i )η ( Ω i ) 1 c eq = f d -----------------------------------N Ωi

∑

(3.75)

i=1

The form of eqn (3.75) suggests that c eq be expressed as c eq = α dN k eq Gs η Substituting for k eq --------and- c eq leads to the definition equation for α d Ω i

(3.76)

∑

i=1 α d = ---------------------------N

(3.77)

∑

Gs ( Ω Note that α d depends only i ) on the material, i.e. it is independent of the geometry of the device. i = 1 With this notation, the equivalent viscous force-deformation relation for a linear viscoelastic damper is written as F = k eq u + α d k eq u˙

(3.78)

Example 3.6: Determining α d for 3M ISD110 damping material This example illustrates how the procedure discussed above can be applied to compute the parameters for the 3M Scotchdamp ISD110 material. Using Fig. 3.15, data corresponding to five frequencies is generated. Table 3.1 contains this data. Applying eqns (3.74), (3.75), and (3.77), one obtains k eq = 5.7 f d

α d = 0.104

c eq = 0.593 f d

192 Chapter 3: Optimal Passive Damping Distribution Table 3.1: Data for ISD110 Scotchdamp material (from Fig. 3.15). Ω (r/s)

G s (MPa)

η

0.628

1.0

1.0

3.14

2.5

1.0

6.28

3.7

0.93

12.56

5.0

0.85

31.4

9.0

0.65

62.8

13.0

0.55

3.5 Damping parameters - discrete shear beam Damping systems This section extends the treatment of discrete shear beams to include damping devices located between the floors. Figure 3.23 illustrates 2 different placement schemes of viscous type dampers for a typical panel. Scheme a combines the damper with a structural element and deploys the composite element on the diagonal between floors. Scheme b places the damper on a roller support at the floor level, and connects the device to the adjacent floor with structural elements. An actual installation of a scheme b system is shown in Fig. 3.24. The structural elements are modelled as linear springs and the representation defined in Fig 3.25 is used.

3.5 Damping Parameters - Discrete Shear Beam 193

ui

θ

ui – 1

(a) ui

ui – 1

(b) Fig. 3.23: Damper placement schemes.

194 Chapter 3: Optimal Passive Damping Distribution

Fig. 3.24: Viscous dampers coupled with chevron bracing.

3.5 Damping Parameters - Discrete Shear Beam 195

u i, V d, i

c k'

V d, i

ui – 1 (a) u i, V d, i k' c

ui – 1 V d, i

(b) Fig. 3.25: Idealized models of structures with viscous dampers. A differential story displacement generates a deformation of the damper, resulting in a damper force which produces the story shear, V d, i . A subscript d is used to denote quantities associated with the damper. The total story shear is the sum of the “elastic” shear force due to elastic frame/brace action and the “damper” shear force. The former was considered in Chapter 2. This contribution is written as V e, i = k e, i ( u i – u i – 1 )

(3.79)

where subscript e refers to “elastic” frame/brace action. The damper shear force is a function of both the relative displacement and the relative velocity. This term

196 Chapter 3: Optimal Passive Damping Distribution is expressed in a form similar to eqn (3.78): V d, i = k d, i ( u i – u i – 1 ) + c d, i ( u˙ i – u˙ i – 1 )

(3.80)

where k d and c d are “equivalent” properties that depend on the makeup of the damping system. Various cases are considered in the following sections. Rigid structural members - linear viscous behavior Consider first the case where the stiffness k' , of the structural members contained in the damping system is sufficiently large so that the extension of the member is negligible in comparison to the extension of the damper. Defining e d as the extension of the damper, and considering scheme a, the damper force for linear viscous behavior is given by F d = c d e˙d = c d ( u˙ i – u˙ i – 1 ) cos θ

(3.81)

The corresponding shear force is V d, i = F d cos θ = c d ( u˙ i – u˙ i – 1 ) cos2 θ

(3.82)

The equivalent damping coefficient for story i is obtained by summing the contributions of the dampers present in story i . c d, i =

∑

( c d cos2 θ )

(3.83)

story i

Equation (3.83) also applies for scheme b; θ = 0 for this arrangement of structural members and dampers. Scheme b is more effective than scheme a (a factor of 2 for 45° bracing), and is more frequently adopted. The general spring-dashpot model shown in Fig 3.26 is useful for representing the different contributions to the story shear force. For this case, the damper ( C ) acts in parallel with the elastic shear stiffness of the frame/bracing system ( K ) and e is equal to the interstory displacement. An extended version of this model is used to study other damping systems.

3.5 Damping Parameters - Discrete Shear Beam 197

K V C

e

Fig. 3.26: Spring and dashpot in parallel model.

Example 3.7: Example 2.15 revisited Consider the 5 DOF shear beam defined in example 2.15. Taking the constant nodal mass as 10,000 kg, and using the stiffness calibration based on S vmax = 0.7 m/s , ξ = 0.1 results in the following values for the element shear stiffness factors, fundamental frequency, nodal mass, and damping: k 1 = 13.59 MN/m , k 2 = 12.68 MN/m , k 3 = 10.87 MN/m k 4 = 8.15 MN/m , k 5 = 4.53 MN/m ω 1 = 9.52 rad/sec

(1)

˜ = 22, 000 kg m c˜ = 41.9 kN-s/m The element damping coefficients are related to c˜ by 1 c˜ = ------ ( c 1 + c 2 + c 3 + c 4 + c 5 ) 25

(2)

• Taking c i constant leads to c 1 = c 2 = … = c 5 = 5c˜ = 210 kN-s/m

(3)

• Assuming damping is proportioned to the element stiffness, c i = αk i , and selecting α according to α = 2ξ 1 ⁄ ω 1 , (the basis for this equation is established in Section 3.7) one obtains the following: α = 0.0210

198 Chapter 3: Optimal Passive Damping Distribution c 1 = 285 kN-s/m

c 2 = 266 kN-s/m

c 4 = 171 kN-s/m

c 5 = 95 kN-s/m

c 3 = 228 kN-s/m (4)

Suppose the chevron brace scheme (scheme b) is used, and 2 dampers are deployed per floor. The “design” values for the dampers are obtained by dividing the above results by a factor 2. For the uniform case, c = 105 kN-s/m . In order to design the damper, one also needs to specify the peak value of the damper force. This quantity is determined with F max = cv max where v max is the maximum relative velocity of the damper piston. For this damper deployment scheme, the relative damper displacement is equal to the interstory displacement. It follows that v max for level i is equal to v max

level i

= ( u˙ i – u˙ i – 1 ) max = h i ( γ˙ i ) max

(5)

The nodal displacements for this 5 DOF model are considered to vary linearly with height: 1 u i – u i – 1 = --- q 5 where q is the modal amplitude. Then 1 v max = --- ( q˙ ) max 5 The peak amplitude is determined with 1 q max = ---- ΓS v ( ξ 1, ω 1 ) ω

(7)

One can estimate q˙max by assuming the response is periodic, with frequency ω 1 . q˙max ≈ ΓS v ( ξ 1, ω 1 )

(8)

Using the problem data, q˙max ≈ ( 1.36 ) ( 0.7 ) = 0.95 m/s and the peak damper force is estimated as

(9)

3.5 Damping Parameters - Discrete Shear Beam 199 0.95 F max ≈ ( 105 ) ---------- = 20 kN 5

(10)

Rigid structural members - linear viscoelastic behavior

K Ve

V

K1

Vd C

e

Fig. 3.27: Spring-dashpot model for viscoelastic damping. The case where the damping mechanism is viscoelastic is represented by the model shown in Fig. 3.27. Here, the damping force has an elastic component as well as a viscous component. Noting eqn (3.78), the damping force is expressed as V d = K 1 e + Ce˙ = K 1 e + α d K 1 e˙

(3.84)

where K 1 and C (or α d ) are the equivalent stiffness and damping parameters for the visco-elastic device, and e is the interstory displacement. This formulation assumes the viscous device is attached to a rigid element so that all the deformation occurs in the device. The more general case is treated later. Using eqn (3.84), one obtains V = ( K + K 1 )e + Ce˙ = ( K + K 1 )e + α d K 1 e˙

(3.85)

When K 1 ⁄ K is small with respect to unity, the contribution of the visco-elastic damper to the stiffness can be neglected.

200 Chapter 3: Optimal Passive Damping Distribution

Example 3.8: Consider a SDOF system having an elastic spring and a visco-elastic damper modeled as shown in the figure. Suppose m , ω , and ξ are specified, and the objective is to establish values for the spring stiffness and damper properties.

m

k k1 c

p u

The governing equation has the form mu˙˙ + cu˙ + ( k + k 1 )u = p

(1)

By definition, k+k ω 2 = -------------1m

(2)

c = 2ξωm

(3)

Given ξ and ω , c is determined with eqn (3). The stiffness factors are related by k 1 + k = mω 2

(4)

Our strategy for dealing with a visco-elastic device is based on expressing the equivalent damper coefficient as (see eqn (3.78)): c = αd k1

(5)

where α d is a “material” property. Example 3.6 illustrates how to evaluate α d for a typical visco-elastic material. The procedure followed here is to first determine k 1 , using eqns (3) and (5),

3.5 Damping Parameters - Discrete Shear Beam 201 2ξωm c k 1 = --------------- ≡ -----αd αd

(6)

and then substitute for k 1 in eqn (4). This operation results in an equation for k. 2 2ξωm 2ξ k = mω – --------------- = mω ω – ------ α d αd

(7)

Suppose m =10,000 kg, ω = 2 π rad/s, and ξ = 0.1 . Then, c = 2 ( 0.1 ) ( 2π ) ( 10 4 ) = 12.56 kN-s/m Using a typical value for α d , α d = 0.15 leads to k 1 = 84 kN/m k = 394 – 84 = 310 kN/m For these parameters, the visco-elastic element contributes approximately 20% of the stiffness.

Example 3.9: Example 3.7 revisited Suppose visco-elastic dampers are used for the 5 DOF system considered in example 3.7. The damper force is taken as F d = k d u d + c d u˙ d = k d u d + α d k d u˙ d

(1)

where α d and k d depend on the device, and u d is the displacement of the damper. Consider the case where a chevron brace with 2 dampers is installed in each floor and the damping distribution defined by eqn 4 in example 3.7 is used. The damper coefficients are determined by dividing the values listed in eqn 4 by 2 (2 dampers per floor): c d, 1 = 142.5

c d, 2 = 133

c d, 3 = 114

202 Chapter 3: Optimal Passive Damping Distribution c d, 4 = 85.5

c d, 5 = 47.5

(2)

(units of kN-s/m) The damper stiffness is determined with 1 k d = ------ c d αd

(3)

Assuming α d = 0.15 , the corresponding values of damper stiffness are: k d, 1 = 950

k d, 2 = 887

k d, 4 = 570

k d, 5 = 317

k d, 3 = 760 (4)

(units are kN/m) The total story shear stiffness distribution is given by eqn (1) in example 3.7. This value is the sum of the elastic stiffness due to frame/brace action and the stiffness due to the 2 dampers k = ke + 2 ⋅ kd

(5)

Using (4) and the data from example 3.7, the frame/brace story shear stiffness factors for this choice of α d are k e, 1 = 11.69 k e, 4 = 7.01

k e, 2 = 10.93

k e, 3 = 9.35

k e, 5 = 3.89

(6)

(units are MN/m) The contribution of the damper stiffness is about 14% of the total stiffness for this example.

Example 3.10: Viscoelastic damper design Referring back to eqn (3.74), the elastic stiffness of the damper depends on the average storage modulus of the viscoelastic material and a geometric parameter fd.

)

3.5 Damping Parameters - Discrete Shear Beam 203

K1 = f d G s )

(1)

Given K 1 and G s , one solves for f d K1 f d = -------Gs )

(2)

To proceed further, one needs to specify the geometry of the device. The figure listed below shows a system consisting of 2n layers of a viscoelastic material located between metal plates. Considering the metal elements to be rigid with respect to the viscoelastic elements, the shape factor is given by wL d f d = 2n ---------td

(3)

The layer thickness is usually fixed by the material manufacturer, and therefore the design variables are the number, length, and width of the viscous plates.

n

w 2 1

td

Ld

Ld As an illustration, suppose (4)

)

k d, 1 = 10, 000 kN/m

Taking G s = 2.5 MPa as the “average” modulus for 3M ISD110, the corresponding shape factor is 3

10, 000 ×10 f d = ----------------------------= 4.0 m 6 2.5 ×10

(5)

204 Chapter 3: Optimal Passive Damping Distribution Substituting for f d in eqn (3), the variables are related by 2n ( wL d ) = t d ⋅ f d = 4.0t d (meters)

(6)

Suppose n = 2 and t d = 10 –2 m = 1 cm . –2

wL d = 1.0 ×10 m 2

(7)

Taking w = L d results in w = L d = 0.1 m = 10 cm

(8)

Example 3.11: Hysteretic damper design - diagonal element Equation (3.58) defines the equivalent viscous damping parameter for hysteretic damping. Substituting the extension, e , for the displacement measure u , and solving for the yield force, F y , results in πΩ r e r µ F y = --------------- ------------ C 4 µ–1

(1)

where e r and Ω r are representative extension and frequency values, and µ is given by er e r sin θ er µ = ----- = --------- = ---------------ey Lε y hε y

(2)

where e y is the extension at which the diagonal material yields. The representative extension is a function of the representative transverse shear deformation γ r . Taking γ r equal to γ ∗ , the design level for γ , leads to e r = γ ∗ h cos θ

(3)

sin 2θ γ ∗ µ = --------------- ----2 εy

(4)

and

3.5 Damping Parameters - Discrete Shear Beam 205 A typical design value for γ ∗ is 1 ⁄ 200 . Ideally, one should use a low strength material so that the response is essentially inelastic throughout the loading duration, thus maximizing the energy dissipation. One potential candidate material is the 200MPa yield strength steel developed by Nippon Steel (Nippon Steel Corp., 1990); the corresponding yield strain is 1 ⁄ 2000 . Using these values and taking θ = 45° provides an upper bound estimate for the ductility ratio 2000 1 µ ≈ ------------ --------- = 5 2 200

(5)

Flexible structural members - linear viscoelastic behavior For completeness, the analysis for the refined viscoelastic model shown in Fig. 3.28 is presented. The component attached to the damper device is modeled as a spring in series with the damper which is considered to be linear viscoelastic with frequency dependent properties G s and η . It is convenient to deal first with a periodic excitation, and then average the properties over the appropriate frequency range.

Primary elastic element

K

Fe Fd

K1 K2

C

F e

Secondary elastic element

Fig. 3.28: General spring-dashpot model. Letting e d represent the displacement of the damper and considering e d to be periodic, the corresponding damper force, F d , follows from eqn (3.69) e d = eˆ d sin Ωt

(3.86)

F d = G s f d eˆ d ( sin Ωt + η cos Ωt ) = G s f d eˆ d sin ( Ωt + δ )

(3.87)

206 Chapter 3: Optimal Passive Damping Distribution where f d is a characteristic geometric parameter for the device and tan δ = η . Since the force in the secondary element must be equal to the damping force, the extensions are related by Fd e – e d = -----(3.88) K2 Substituting for e d and F d leads to the expression for the total displacement e Fd G s f d G s f d (3.89) e = e d + ------ = eˆ d 1 + ------------- sin Ωt + eˆ d ------------- η cos Ωt K2 K2 K2 Equation (3.89) can also be written as e = eˆ sin ( Ωt + δ 1 ) where

ηG s f d 2 Gs f d 2 eˆ = eˆ d 1 + ------------- + ----------------- = ℵeˆ d K2 K2 1 tan δ 1 = η ----------------------K2 1 + ------------The force in theGprimary elastic element depends only on e sfd F e = Ke = Keˆ sin ( Ωt + δ 1 )

(3.90)

(3.91) (3.92)

(3.93)

Combining F e and F d , the total force is given by F = F e + F d = Keˆ sin ( Ωt + δ 1 ) + G s f d eˆ d sin ( Ωt + δ )

(3.94)

A more compact form for F is F = Kˆ eˆ sin ( Ωt + δ 1 + δ 2 )

(3.95)

where δ 2 represents the phase shift between the excitation and the force response, and Kˆ is the total stiffness measure. The definition equations are 2 2 Gs f d Gs f d (3.96) Kˆ = K cos δ 1 + ------------- cosGδ f + K sin δ 1 + ------------- sin δ s d ℵδ + -----------ℵ sin δ K sin 1 ℵ (3.97) tan ( δ 1 + δ 2 )ℜ= ---------------------------------------------------- = ℜ f G --------------- – 1 s d tan δ 1 K cos δ + ------------ cos δ ℵ (3.98) tan δ 2 = -------------------------- 1 1 ℜ + --------------δ 1be expressed in conventional form by shifting the time The equationstan can reference point. Defining t' as

3.5 Damping Parameters - Discrete Shear Beam 207 δ1 t = t' – ----Ω

(3.99)

transforms eqns (3.90) and (3.95) to e = eˆ sin Ωt'

(3.100)

F = Kˆ eˆ sin ( Ωt' + δ 2 )

(3.101)

Finally, expanding eqn (3.101), the result can be expressed in a form similar to the conventional viscoelastic form F = K s eˆ sin Ωt' + ηK s eˆ cos Ωt'

(3.102)

K s = Kˆ cos δ 2

(3.103)

η = tan δ 2

(3.104)

where

One can interpret K s as the storage stiffness and η as the loss factor for the assemblage. Equivalent parameters can be generated following the procedure described in Section 3.4. One writes (3.105)

F = K eq e + C eq e˙

and equates eqn (3.105) with (3.102). The error terms for a periodic excitation are (3.106)

E 1 = K eq – K s ηK s E 2 = C eq – ---------Ω

(3.107)

Minimizing the sum of the square of these terms over the frequency range with respect to K eq and C eq produces N the following expressions 1 (3.108) K s ( Ωi ) K eq = K s ≡ ---N N average η i K s, i ( Ωi i=) 1 1 (3.109) C eq = -----------------------------N Ωi

∑

∑

i=1

where N is the number of frequencies composing the data set. If one considers K 2 to be very large with respect to the visco-elastic damper stiffness, f d G s, the various terms simplify to ℵ≈1

(3.110)

208 Chapter 3: Optimal Passive Damping Distribution δ1 ≈ 0

(3.111)

Gs f d 2 2G s f d Kˆ ≈ K 1 + ----------------- cos δ + ------------K K 1 η = tan δ 2 = ℜ = η -----------------------------------K 1 + -------------------------G s f d cos δ

(3.112) (3.113)

Example 3.12: Coupled spring-damper model This example considers the case where the damper is viscous, and the structural element connecting the damper to the floors is flexible. This is a simplified version of the model considered above; the stiffness component of the visco-elastic device is deleted.

K

Fe

F

Fd K2

C

e

The steps are similar to the previous steps. We take e d = eˆ d sin Ωt

(1)

Then, F d = CΩeˆ d cos Ωt e = eˆ sin ( Ωt + δ 1 ) CΩ 2 eˆ = eˆ d 1 + -------- K2

1/2

CΩ tan δ 1 = -------K2 The total force is expressed as

(2)

3.5 Damping Parameters - Discrete Shear Beam 209 F = Kˆ eˆ sin ( Ωt + δ 1 + δ 2 )

(3)

where CΩ Kˆ = K cos δ 1 1 + tan δ 1 + -------K

2 1/2

CΩ 1 tan δ 2 = -------- ⋅ ------------------------------------------------------------K 1 1 1 1 + ( CΩ ) 2 ------ ------ + ---- K 2 K 2 K

(4)

(5)

The remaining steps are the same. One expresses F as F = K eq e + C eq e˙

(6)

and determines K eq and C eq using eqns (3.108) and (3.109). When K 2 = ∞ , δ 1 = 0 and CΩ tan δ 2 = -------K

K cos δ 2 = ---Kˆ

(7)

Then, Ks = K

nK s = CΩ

(8)

K eq ≡ K

C eq ≡ C

(9)

and

3.6 Damping parameters - truss-beam This section extends the treatment of the truss-beam discussed in Example 2.2 to include damping. The typical panel shown in Fig. 3.29 is considered to be composed of two sets of elements: an elastic system which provides the stiffness (shear and bending) and a second system which functions as a distributed energy dissipation/absorption mechanism.

210 Chapter 3: Optimal Passive Damping Distribution

M V c

E c A

E

d

A

d

h

F

c

F

d

F

d

F

c

θ

B

Fig. 3.29: Truss-beam structure - geometry and forces. The various force quantities are determined with F

d c

d

d

c

c

= Fe + Fd

F = Fe + Fd d

V = 2F cos θ M = BF

c

(3.114) (3.115) (3.116) (3.117)

where superscripts ‘c’ and ‘d’ refer to the column and diagonal elements, and subscripts ‘e’ and ‘d’ denote the elastic and damping force components. Defining e as the extension, i.e. the total change in length of a structural element, and noting eqns (2.36) and (2.39), the extensions for the diagonal and chord elements are related to the transverse shear and bending deformations by e

d

= γh cos θ

c Bh e = ------- χ 2

The elastic force is a function of e and the element properties AE F e = -------- e = K ⋅ e L

(3.118) (3.119)

(3.120)

where L denotes the length, A the cross-sectional area, and E the Young’s Modulus for the elastic element.

3.6 Damping Parameters - Truss Beam 211 Linear viscous behavior The dissipative force-deformation relation depends on the nature of the damping device. For viscous behavior, F d is a function of e˙ . The linear viscous force is expressed as (3.121)

F d = Ce˙

where C is a property of the device, which may be an actual viscous damper, or an equivalent viscous coefficient. Summing the elastic and damping forces leads to the element force-extension relations c

c c

c c

(3.122)

F = K e + C e˙ F

d

d d

d d

(3.123)

= K e + C e˙

One can associate these relations with the parallel spring/dashpot model shown in Fig. 3.26. The member force-deformation relations are obtained by introducing these expressions into the definition equations for V and M . They are written as = D T γ + C T γ˙

(3.124)

M = D B χ + C B χ˙

(3.125)

where 2

d

2

d

D T = [ 2hcos θ ]K C T = [ 2hcos θ ]C

(3.126) (3.127)

2

B h c D B = --------- K 2 2 B h c C B = --------- C 2

(3.128) (3.129)

It is convenient to express the damping coefficients in terms of the stiffness parameters, CT = αT DT

(3.130)

CB = αB DB

(3.131)

where α T and α B follow from the previous equations d C α T = ------d K

(3.132)

212 Chapter 3: Optimal Passive Damping Distribution c

C α B = -----(3.133) c K Note that α T and α B depend on the damping device configuration and stiffness of the diagonal and chord elements. Expressing C in terms of D and taking α constant over the height simplifies the analysis of the beam as a continuum using modal shape functions by uncoupling the equations for the modal coordinates. Linear viscoelastic behavior The case where the damping mechanism is viscoelastic is represented by the model shown in Fig. 3.27. Here, the damping force has an elastic component as well as a viscous component, and is expressed as F d = K 1 e + Ce˙ = K 1 e + α d K 1 e˙

(3.134)

where K 1 and α d are the equivalent stiffness and damping parameters for the device, and e is the total extension of the member. This formulation assumes the viscous device is attached to a rigid element so that all the deformation occurs in the device. Using eqn (3.134), one obtains c

c

c

c c

c

F = ( K + K 1 )e + α dc K 1 e˙ F

d

d

d

d d

d

= ( K + K 1 )e + α dd K 1 e˙

(3.135) (3.136)

Substituting for the forces and deformations leads to 2

d

d

D T = [ 2hcos θ ] ( K + K 1 ) 2

d

C T = [ 2hcos θ ]α dd K 1 and

(3.137) (3.138)

2

c B h c D B = --------- ( K + K 1 ) 2 2 B h c c C B = ---------α d K 1 2

The proportionality factors in this case are given by c c K1 ⁄ K CB α B = ------- = α dc -------------------------DB dc ⁄K dc 1 +K K ⁄ K CT 1 1 α T = ------- = α dd ---------------------------d d DT 1 + K1 ⁄ K

(3.139) (3.140)

(3.141) (3.142)

3.6 Damping Parameters - Truss Beam 213

3.7 Damping distribution for MDOF systems Chapter 2 dealt with the problem of specifying stiffness distributions for beamtype structures. Numerical simulations for seismic excitation showed that one can establish stiffness distributions with an iterative procedure that accounted for the contribution of the higher modes. However, the amplitude of the deformation profile cannot be totally controlled with just stiffness, and other mechanisms need to be incorporated. An examination of the effectiveness of distributed linear viscous damping to decrease the deformation amplitude and therefore provide more control on the response is presented in this section. The structure is assumed to be discretized as an n ’th order MDOF system governed by the following equation. ˙˙ + CU˙ + KU = P MU

(3.143)

where U contains the nodal translation and rotation measures, M is the system mass matrix, C is the system damping matrix, K represents the system stiffness, and P represents the load vector. A similar set of equations was considered in Section 2.10, where a restriction was placed on C and a procedure for transferring the coupled matrix equations to an uncoupled set of equations in terms of modal coordinates was described. The procedure is extended here to allow for an arbitrary form of C . Multi mode free vibration response The case where P = 0 is considered first. The general homogeneous solution can be expressed as U = Ae λt Φ

(3.144)

where A and λ are arbitrary scalars, and Φ is an unknown vector of order n. Substituting for U in eqn (3.143) leads to an equation relating λ and Φ . ( λ 2 M + λC + K )Φ = 0

(3.145)

Solving eqn (3.145) for λ and Φ is referred to as the quadratic eigenvalue problem. A detailed discussion of the mathematical aspects of this problem is contained in Strang,1993.The discussion presented here is intended to provide an introduction to this topic which is also revisited in Chapter 8.

214 Chapter 3: Optimal Passive Damping Distribution When C = 0 , eqn (3.145) reduces to ( K + λ 2 M )Φ = 0

(3.146)

Since K and M are positive definite, λ 2 must be negative. Then, λ can be expressed as λ = ± iω

ω>0

(3.147)

The solution corresponding to the pair of values for λ is U = A 1 e iωt Φ + A 2 e –iωt Φ

(3.148)

Since U must 1be real, the A’s must be complex conjugates, A 1 = --- ( A R + iA I ) 2 1 A 2 = --- ( A R – iA I ) 2 A and A transforms the solution to Substituting for 1 2 U = ( A R cos ωt – A I sin ωt )Φ

(3.149)

(3.150)

Equation (3.146) has n distinct solutions, i.e., n Φ ‘s and corresponding ω ‘s. Each solution satisfies 2

KΦ i = ω i MΦ i

i = 1, 2, …, n

(3.151)

and the following orthogonality relations T

T

Φ j KΦ i = Φ j MΦ i = 0

i≠j

(3.152)

˜i Φ i KΦ i = ω i2 Φ i MΦ i = ω i2 m

(3.153)

T

T

Suppose C is a scaled version of K . C = αK

(3.154)

where α is a positive scalar. This condition is called “stiffness proportional damping”. Substituting for C in eqn (3.145), and rearranging terms leads to λ2 K + ----------------M Φ = 0 (3.155) 1 + αλ Setting λ2 ---------------- = – ω 2 1 + αλ

3.7 Damping Distribution for MDOF Systems 215 converts eqn (3.155) to the “undamped” form. It follows that the eigenvectors (mode shapes) are the “same”, and the eigenvalues ( λ ) depend on ω and α . λ 2j + αω 2j λ j + ω 2j = 0

(3.156)

The eigenvalues are expressed as λ j = – ξ j ω j ± iω j ( 1 – ξ 2j ) 1 / 2 = – ξ j ω j ± iω' j where ξ j is the damping ratio for the j ‘th mode. αω ξ j = ----------j 2

(3.157)

(3.158)

The modal damping ratio increases with increasing mode number. Substituting for λ , the typical free vibration modal response is still periodic, but with a time decaying amplitude. U = e –ξi wi t Φ i [ A R, i cos ω' i t – A I, i sin ω' i t ]

(3.159)

The case where C is an arbitrary symmetric positive definite matrix involves an additional complication; the eigenvectors as well as the eigenvalues are complex quantities. These quantities occur as complex conjugates and are expressed as λ j = λ R, j + iλ I, j λ˜ j = λ R, j – iλ I, j Φ j = Φ R, j + iΦ I, j

(3.160)

˜ = Φ – iΦ Φ j R, j I, j The j ‘th solution is generated by combining the corresponding complex solutions. ˜ e λ˜ j t Φ ˜ U = A j eλ jt Φ j + A j j

(3.161)

Expanding the complex products leads to the following “real” solution: = e λR, j t { ( A R, j Φ R, j – A I, j Φ I, j ) cos λ I t + ( – A I, j Φ R, j – A R, j Φ I, j ) sin λ I t } (3.162)

216 Chapter 3: Optimal Passive Damping Distribution Equation (3.162) is expressed in a form similar to eqn (3.159) U = e –ξ j ω j t { ( A R, j Φ R, j ) cos ω' j t + ( – A I, j Φ R, j ) sin ω' j t } + e –ξ j ω j t { ( – A I, j Φ I, j ) cos ω' j t + ( – A R, j Φ I, j ) sin ω' j t }

(3.163)

Combining the time varying terms leads to a simpler form: U = f j Φ R, j + g j Φ I , j

(3.164)

The second term vanishes for stiffness proportional damping. Our strategy for generating an initial estimate for the stiffness distribution is based on using Φ R as the modal shape function. The actual modal response involves an additional shape function when C is an arbitrary matrix. The importance of this second term depends on the amount of damping. Various computational issues related to stiffness are addressed by the following examples. This topic is discussed in more detail later in Chapter 8, where the general solution for arbitrary loading is presented.

Example 3.13: Eigenvalue problem - 2DOF

m2

u2 k 2, c 2

m1

u1 k 1, c 1

The governing equations for free vibration of the 2DOF system shown above have the form: m 1 u˙˙1 + ( c 1 + c 2 )u˙ 1 + ( k 1 + k 2 )u 1 – c 2 u˙ 2 – k 2 u 2 = 0 m 2 u˙˙2 + c 2 u˙ 2 + k 2 u 2 – c 2 u˙ 1 – k 2 u 1 = 0

(1)

3.7 Damping Distribution for MDOF Systems 217 Expressing the displacements as u 1 = φ 1 e λt

u 2 = φ 2 e λt

(2)

and substituting in eqn (1) leads to a set of equations for the φ terms. ( m 1 λ 2 + λ ( c 1 + c 2 ) + k 1 + k 2 )φ 1 – ( λc 2 + k 2 )φ 2 = 0 – ( λc 2 + k 2 ) φ 1 + ( m 2 λ 2 + λc 2 + k 2 )φ 2 = 0

(3)

For a non-trivial solution to exist, the determinant of the coefficient matrix must vanish. This requirement results in the following equation for λ . ( m 1 λ 2 + λ ( c 1 + c 2 ) + k 1 + k 2 ) ( m 2 λ 2 + λc 2 + k 2 ) – ( λc 2 + k 2 ) 2 = 0

(4)

The elements of Φ are not uniquely defined since eqn. 3 is a homogeneous equation. Solving the first equation leads to a relation between φ 1 and φ 2 . m1 λ 2 + λ ( c1 + c2 ) + k1 + k2 φ 2 = φ 1 ⋅ ------------------------------------------------------------------λc 2 + k 2

(5)

The undamped solution is obtained by setting c 1 = c 2 = 0 . ( m 1 λ 2 + k 1 + k 2 ) ( m 2 λ 2 + k 2 ) – k 22 = 0

(6)

m1 λ 2 + k1 + k2 φ 2 = φ 1 ⋅ -----------------------------------k2

(7)

Expanding eqn (6) results in ( m 1 m 2 )λ 4 + ( m 1 k 2 + m 2 ( k 1 + k 2 ) )λ 2 + k 1 k 2 = 0

(8)

Noting that the coefficients are all positive, it follows that λ 2 must be a negative real number. Expressing λ as λ = ±i ω

(9)

and solving eqn (8) for ω 2 leads to ω 2 = a(1 ± b) where

(10)

218 Chapter 3: Optimal Passive Damping Distribution m1 k2 + m2 ( k1 + k2 ) a = ----------------------------------------------2m 1 m 2

(11)

1/2 4m 1 m 2 k 1 k 2 b = 1 – ------------------------------------------------------2- ( m1 k2 + m2 ( k1 + k2 ) )

(12)

Equation (10) has 2 real positive roots. The elements of φ are also real quantities in this case. Each eigenvalue/eigenvector produces a solution of the form u 1 = Φ ( j )1 ( A R, j cos ω j t – A I, j sin ω j t ) u 2 = Φ ( j )2 ( A R, j cos ω j t – A I, j sin ω j t )

(13)

where A R and A I are integration constants that are evaluated using the initial conditions at time t = 0 . To illustrate the computations, the system examined in example 1.9 is considered. The mass and stiffness are: m 1 = m 2 = 1, 000 kg = 1 kN-s 2 /m k 2 = ( 2 ) ( 2π ) 2 = 78.88 kN/m

(14)

k 1 = ( 3 ) ( 2π ) 2 = 118.3 kN/m These stiffness values produce a fundamental mode which is linear and correspond to a fundamental frequency of 2π rad/s. Evaluating a and b for this set of properties results in a = 138.0 (1/sec 2 ) 1/2

24 b = 1 – ------ 49

5 = --7

(15)

and ω 1, 2 = 2π , 2π ( 2.45 ) = 6.28, 15.29 The mode shapes are determined with eqn (7). Fundamental mode

(16)

3.7 Damping Distribution for MDOF Systems 219

λ 2 = – ω 2 = – ( 2π ) 2 φ 2 = 2φ 1

(17)

Second mode λ 2 = – 6 ( 2π ) 2 φ2 = –φ1 ⁄ 2

(18)

It is convenient to normalize the modal vectors such that the magnitude of the maximum element is unity. For this example, the normalized vectors are Φ1 = 1 ⁄ 2 1

Φ2 =

1 –1 ⁄ 2

(19)

The case where c is proportional to k is considered next. Setting c i = αk i and defining λ′ as λ2 ( λ′ ) 2 = ---------------1 + αλ

(20)

reduces eqns (4) and (5) to the “same” form as eqns (6) and (7) with λ replaced with λ′ . ( m 1 ( λ′ ) 2 + k 1 + k 2 ) ( m 2 ( λ′ ) 2 + k 2 ) – k 22 = 0

(21)

m 1 ( λ′ ) 2 + k 1 + k 2 φ 2 = φ 1 ------------------------------------------k2

(22)

One sets λ′ = – ω 2

(23)

and determines ω 2 with eqn (10). The eigenvector is the same as the undamped vector, and the eigenvalues are determined by combining eqns (20) and (23). λ2 ---------------- = – ω 2 1 + αλ The solution is written in the following general form:

(24)

220 Chapter 3: Optimal Passive Damping Distribution

λ j = – ξ j ω j ± iω j ′ ξ j = αω j ⁄ 2 ω j ′ = ω j [ 1 – ξ 2j ] 1 / 2

(25)

u i = Φ ( j )i e –ξ j ω j t ( A R, j cos ω j ′t – A I, j sin ω j ′t ) j = 1, 2 Continuing with the numerical example, suppose the design objective is ξ 1 = 0.1 . The required value of α is 2ξ 2 ( 0.1 ) α = --------1 = --------------- = 0.0318 sec ω1 2π

(26)

Then, c 1 = αk 1 = 3.762 kN-s/m c 2 = αk 2 = 2.508 kN-s/m

(27)

The damping ratio for the second mode is proportional to the frequency ratio: ω ξ ----2- = -----2- = 2.45 ξ1 ω1

(28)

ξ 2 = 2.45 ( 0.10 ) = 0.245 Lastly, the modified frequencies are: ω 1 ′ = ω 1 [ 1 – ξ 12 ] 1 / 2 = ω 1 ( 0.99 ) 1 / 2 = 0.995ω 1 = 6.25 ω 2 ′ = ω 2 ( 0.94 ) 1 / 2 = 0.97ω 2 = 14.83

(29)

The case where C is not proportional to either the mass or stiffness matrices involves solving a complete 4’th degree equation for λ . Equation (4) is written as λ4 + f 3λ3 + f 2λ2 + f 1λ + f 0 = 0 where the coefficients are

(30)

3.7 Damping Distribution for MDOF Systems 221

f 3 = ( m1 m2 ) –1 ( m1 c 2 + m2 ( c 1 + c 2 ) ) f 2 = ( m1 m2 ) –1 ( m1 k 2 + m2 ( k 1 + k 2 ) + c 1 c 2 ) f 1 = ( m1 m2

) –1 ( k

2 c1

+ k1 c2 )

(31)

f 0 = ( m1 m2 ) –1 ( k 1 k 2 ) Since all the coefficients are positive, the real part of λ must be negative. The factored form of eqn (30) is expressed as ( λ 2 + a1 λ + a2 ) ( λ 2 + b1 λ + b2 ) = 0

(32)

where a1 + b1 = f 3 a1 b1 + a2 + b2 = f 2 a1 b2 + a2 b1 = f 1

(33)

a2 b2 = f 0 With this notation, the 2 pairs of complex conjugate roots are 1 ( λ 1, λ˜ 1 ) = --- [ – a 1 ± i ( 4a 2 – a 12 ) 1 / 2 ] = – ξ 1 ω 1 ± iω 1 ′ 2 1 ( λ 2, λ˜ 2 ) = --- [ – b 1 ± i ( 4b 2 – b 12 ) 1 / 2 ] = – ξ 2 ω 2 ± iω 2 ′ 2

(34)

The eigenvectors are determined with eqn (5) which is written as φ 2 = hφ 1

(35)

where h is a function of λ . Each pair of complex conjugate roots generates a corresponding pair of complex conjugates for h and φ . Considering the first pair of eigenvalues, ( λ 1, λ˜ 1 ) ⇒ ( h 1, h˜ 1 )

(36)

and φ 1, 2 = h 1 φ 1, 1 φ˜ 1, 2 = h˜ 1 φ˜ 1, 1

(37)

222 Chapter 3: Optimal Passive Damping Distribution The modes defined by eqn (37) are represented as: ˜ ) = Φ ( Φ 1, Φ 1 R, 1 ± iΦ I, 1

(38)

Since Φ is not uniquely defined, one can set the magnitude of one element, and scale the other elements using eqn (35). Adopting the normalized form defined by eqn (19), the definition equations for the elements are:

Mode 1 φ2 ≡ 1 1 1 φ 1, φ˜ 1 = -----, ----h 1 h˜ 1

(39)

Mode 2 φ1 ≡ 1 φ 2, φ˜ 2 = h 2, h˜ 2

(40)

Results based on the mass and stiffness properties defined by eqn (14) and 5 different distributions of the damping coefficients are listed below in Table 3.2. The first case is stiffness proportional damping with ξ 1 = 0.1 . The next 3 cases correspond to uniform, linear, and concentrated distributions with the sum of the damping coefficients held constant. The last case is a repeat of case 2 with the damping doubled. Concentrating all the damping in the second element produces the maximum value of ξ 2 , and does not result in a significant change in the mode shape. However, it increase the damping force, and requires a “larger” damper. Selecting a uniform distribution has the advantage that the damper size is less than the sizes for the other choices. To arrive at an optimal distribution, one needs to consider the total damper costs and other constraints on placement of dampers. There are situations where dampers cannot be placed in certain floors. Computer based simulation is used in this case to “optimize” damper placement.

3.7 Damping Distribution for MDOF Systems 223

Table 3.2: Frequencies, damping ratios, and mode shapes for 2DOF system c 1 = 3.762

c 1 = 3.135

c 1 = 2.09

c 1 = 0.0

c 1 = 6.27

c 2 = 2.513

c 2 = 3.135

c 2 = 4.18

c 2 = 6.27

c 2 = 6.27

ω1

6.28

6.286

6.325

6.504

6.305

ξ1

0.1

0.1

0.0987

0.0926

0.1987

φ R, 1

0.5

0.502

0.509

0.528

0.511

φ R, 2

1.0

1.0

1.0

1.0

1.0

φ I, 1

0

-0.025

-0.066

-0.151

-0.0486

φ I, 2

0

0

0

0

0

ω2

15.29

15.37

15.276

14.854

15.32

ξ2

0.245

0.265

0.301

0.385

0.532

φ R, 1

1.0

1.0

1.0

1.0

1.0

φ R, 2

-0.5

-0.515

-0.544

-0.620

-0.564

φ I, 1

0

0

0

0

0

φ I, 2

0

0.059

0.161

0.37

0.107

Note: m 1 = m 2 = 1, 000 kg , k 1 = 118.3 kN/m , k 2 = 78.9 kN/m Units of c are kN s/m

Example 3.14: Modal response for nonproportional damping Equation (3.163) defines the general homogeneous solution for non-

224 Chapter 3: Optimal Passive Damping Distribution proportional damping. A typical solution is actually the sum of 2 solutions (the subscript j is dropped for convenience): U = f Φ R + gΦ I

(1)

f = A R ( e –ξωt cos ω′t ) + A I ( – e –ξωt sin ω′t )

(2)

g = A R ( – e –ξωt sin ω ′t ) + A I ( – e –ξωt cos ω ′t )

(3)

where

The responses associated with A R and A I are 90° out of phase with each other. This formulation is applied to the 2 DOF system treated in example (3.13). Table 3.2 shows that the mode shapes have the following general form: Mode 1 ΦR =

aR 1

ΦI =

aI

ΦI =

0 bI

(4)

0

Mode 2 ΦR =

1 bR

(5)

where a and b depend on the level of damping. Substituting for the Φ ‘s in eqn (1) leads to: Mode 1 u1 = aR f + aI g u2 = f

(6)

Mode 2 u1 = f u2 = bR f + bI g

(7)

3.7 Damping Distribution for MDOF Systems 225 Expanding the terms in eqn (6) shows the phasing effect on the fundamental mode. u 1 = e –ξ1 ω1 t [ A R ( a R cos ω 1 ′t – a I sin ω 1 ′t ) + A I ( – a R sin ω 1 ′t – a I cos ω 1 ′t ) ] (8)

u 2 = e –ξ1 ω1 t [ A R ( cos ω 1 ′t ) + A I ( – sin ω 1 ′t ) ]

(9)

Comparing there terms shows that the solution for u 1 is out of phase with respect to u 2 . Defining β and aˆ as a tan β 1 = -----I aR

(10)

aˆ 1 = [ a R2 + a I2 ] 1 / 2

(11)

transforms eqn (8) to u 1 = e –ξ1 ω1 t [ A R aˆ 1 cos ( ω 1 ′t + β 1 ) – A I aˆ 1 sin ( ω 1 ′t + β 1 ) ]

(12)

The shear strains in the element are related to the difference in displacements. 1 γ 1 = ----- u 1 h1 1 γ 2 = ----- ( u 2 – u 1 ) h2

(13)

Using eqns (8) and (9), u2 – u1 = e –ξ1 ω1 t [ A R ( ( 1 – a R ) cos ω 1 ′t + a I sin ω 1 ′t ) – A I ( ( 1 – a R ) sin ω 1 ′t – a I cos ω 1 ′t ) ] (14)

Equation (14) is written in a form similar to eqn (12). The amplitude and phase terms are:

226 Chapter 3: Optimal Passive Damping Distribution aI tan β 2 = – -------------1 – aR

(15)

aˆ 2 = [ ( 1 – a R ) 2 + a I2 ] 1 / 2

(16)

With this notation, eqn (14) takes the form –ξ ω t u 2 – u 1 = e 1 1 [ A R aˆ 2 cos ( ω 1 ′t + β 2 ) – A I aˆ 2 sin ( ω 1 ′t + β 2 ) ]

(17)

The peak differential displacements are: u 1 max = aˆ 1 ( A R2 + A I2 ) 1 / 2 u 2 – u 1 max = aˆ 2 ( A R2 + A I2 ) 1 / 2

(18)

Table 3.2 lists results for a R and a I corresponding to various distributions of c 1 and c 2 . The largest deviation from the proportional damping case ( a R = 0.5 ; a I = 0 ) is case 4. a R = 0.528

a I = – 0.151

(19)

Using the above values, the various parameters are β 1 = – 15.96°

β 2 = – 17.74°

aˆ 1 = 0.549

aˆ 2 = 0.495

(20)

For this example, the peak shear strains in elements 1 and 2 differ by only 10%

Stiffness proportional viscous damping An introductory treatment of the case where C = αK was presented in section 2.10. This section expands upon that analytical treatment and also contains some examples which illustrate the influence of damping on the response of the higher modes. Expressing U as an expansion in terms of the undamped mode shape vectors,

3.7 Damping Distribution for MDOF Systems 227 n

U =

∑

qi Φi

(3.165)

i=1

and noting the orthogonality relations leads to the following set of equations for the q’s T Φi P p˜ i 2 (3.166) q˙˙i + 2ξ i ω i q˙i + ω i q i = ----------- = -----i = 1, 2, 3, . . . . ˜i ˜i m m where ξ i , the damping ratio for the ith mode, is related to α and ω i by eqn (3.158). The magnitude of α is determined by specifying the damping ratio for the first mode 2ξ 1 (3.167) α = --------ω1 Given α , one calculates the other damping ratios with ωi ξ i = ξ 1 ------ ω 1 Suppose the initial conditions on U are U ( 0 ) = U∗ U˙ ( 0 ) = U˙ ∗

(3.168)

(3.169)

Expressing U as U = Φq

(3.170)

and noting the orthogonality relation for M which follows from eqn (3.151), ˜ i δ ij ] = m ˜ Φ T MΦ = [ m

(3.171)

one can transform eqn (3.170) to ˜ q = Φ T MU m

(3.172)

by premultiplying by Φ T M . The i ‘th term has the form ˜ i q i = φ iT MU m

(3.173)

Applying this transformation to the initial conditions on U leads to the initial conditions for q i . 1 (3.174) q i ( 0 ) = ------ φ iT MU∗ = q i∗ ˜i m 1 (3.175) q˙i ( 0 ) = ------ φ iT MU˙ ∗ = q˙i∗ ˜i m i = 1, 2, …, n

228 Chapter 3: Optimal Passive Damping Distribution The orthogonality relationships for the coefficient matrices reduces the problem of solving an n ‘th order coupled equation to one of solving n uncoupled equations for the modal coordinates. A similar result can be obtained for nonproportional damping, but that formulation is considerably more complicated than this formulation. The nonproportional case is discussed again in Chapter 8 where a general approach for dealing with arbitrary damping based on the “statespace” formulation is presented. The total response for the i’th modal coordinate 1 ξi ωi t * q i ( t ) = e –as: q i cos ω′t + ----- ( q˙* + ξωq * ) sin ω′t + can be expressed ω′ ξi ωi (3.176) q˙i ( t ) = e –ξi ωi t tq˙i* cos ω′t + – q˙* ---------- + ( – ω i ′ 2 – ξ i2 ω i2 )q * sin ω′t – 1 p˜ i –ξi ωi ( t – τ ) ω i ′ + ------- ------ e sin [ ω i ′ ( t – τ ) ] dτ ˜it ωi ′ m (3.177) ˜i 20 ω 1 / i2 p ( 1 – ξ–i )------ ------ e –ξi ωi ( t – τ ) sin [ ω i ′ ( t – τ ) – δ i ] dτ ˜ ω' - m (3.178) tan δ i = ------------------------ξi i 0 i

∫

∫

i = 1, 2, …, n The procedure for generating the stiffness distribution presented in Chapter 2 assumes the fundamental mode ( i = 1 ) is the dominate term and bases the stiffness distribution on the displacement profile defined by q 1 Φ 1 . Other modes may contribute, depending on the frequency of excitation and modal damping. Ideally, one wants high modal damping for i = 2, 3, … to minimize their contribution. With stiffness proportional damping, the modal damping ratios are fixed by the frequency ratios. They increase with mode number, which is desirable, but one cannot arbitrarily increase a particular modal damping ratio to further reduce the modal response. The only recourse is to increase the fundamental mode damping ratio, ξ 1 . This action increases α since α = 2ξ 1 ⁄ ω 1 , and requires additional damping. A set of examples illustrating the effect of increasing ξ 1 on the response are presented in the following section. Example 3.15: Low rise buildings. Results for the shear beam type buildings defined in Table 3.3 are presented in this section. The building sites are assumed to be in California, and the reference spectral velocity is taken as S v = 1.2 m/s for ξ = 0.02 . Three earthquakes representative of the sites, namely, El Centro, Northridge Station01, and Northridge Station03, are used to generate ensemble averaged response spectra. Their scaled accelerograms are shown in Figure 3.30 through 3.32.

3.7 Damping Distribution for MDOF Systems 229

Table 3.3: Example shear buildings Building

Number of stories

Story height (meters)

Mass/story (103kg)

γ obj

1

3

4

10

0.005

2

6

5

10

0.005

3

9

5

10

0.005

230 Chapter 3: Optimal Passive Damping Distribution

4

acceleration m/s2

3 2 1 0 −1 −2 −3

0

10

20

30 time s

40

50

60

Spectrum

150

100

50

0

0

20

40

60

80 ω (rad/s)

100

120

140

160

Fig. 3.30: Scaled time history and frequency content of El Centro ( S v = 1.2 & ξ = 0.02 )

3.7 Damping Distribution for MDOF Systems 231

4

acceleration m/s

2

3 2 1 0 −1 −2 −3

0

10

20

30 time s

40

50

60

120

Spectrum

100 80 60 40 20 0

0

20

40

60

80 ω (rad/s)

100

120

140

160

Fig. 3.31: Scaled time history and frequency content of Northridge: Station 1 component 090 ( S v = 1.2 & ξ = 0.02 )

232 Chapter 3: Optimal Passive Damping Distribution

2

acceleration m/s

2

3

1 0 −1 −2

0

10

20

30 time s

40

50

60

140 120

Spectrum

100 80 60 40 20 0

0

20

40

60

80 ω (rad/s)

100

120

140

160

Fig. 3.32: Scaled time history and frequency content of Northridge: Station 3 component 090 ( S v = 1.2 & ξ = 0.02 )

3.7 Damping Distribution for MDOF Systems 233 Stiffness and damping distributions are generated for values of ξ 1 ranging from 0.02 to 0.3. For each ξ 1 value, the following quantities are determined: ω 1 ; the story stiffness coefficients, k i ; the damping parameter α = 2ξ 1 ⁄ ω 1 ; and lastly the story damping coefficients, c i = αk i . Figure 3.33 shows the variation of the frequency and period for the first three modes with increasing damping of the fundamental mode.

0.8

Mode 1 Mode 2 Mode 3

40 20 0

0

0

0.4

0

0.1 0.2 0.3 ξξ1 − 3 Story Building

0.4

0

0.1 0.2 0.3 ξ 1ξ − 6 Story Building

0.4

0

0.1 0.2 0.3 ξξ1 − 9 Story Building

0.4

2 1.5 T (s)

ω (rad/s)

0.1 0.2 0.3 ξ 1ξ − 3 Story Building

30 20 10 0

0.4 0.2

40

1 0.5

0

30

0.1 0.2 0.3 ξξ1 − 6 Story Building

0

0.4

3

20

2 T (s)

ω (rad/s)

0.6 T (s)

ω (rad/s)

60

10 0

1

0

0.1 0.2 0.3 ξξ1 − 9 Story Building

0.4

0

Fig. 3.33: Frequencies and periods of first three modes

234 Chapter 3: Optimal Passive Damping Distribution As ξ 1 is increased, the required stiffness decreases, and the corresponding damping increases. The total stiffness is defined as the sum of the story stiffnesses. A similar definition is used for the story damping coefficients. These two quantities provide a measure of the “design” cost. The data for the three example structures are plotted in Fig 3.34. Total damping varies essentially linearly with ξ , whereas total stiffness behaves nonlinearly, and the degree of nonlinearly increases with the number of stories. Given costs for stiffness elements and dampers, one can generate data on cost as a function of ξ 1 and establish an “optimal” damping system for stiffness proportional damping.

7

3

x 10

5

Total Stiffness 8

N/(m s)

N/m

1

8

0 7 x 10

0.1 0.2 0.3 ξ − 3 Story Building

0

0.4

3

N/(m s)

N/m

4 2

8

0 7 x 10

0.1 0.2 0.3 ξ − 6 Story Building

0.4

0 6 x 10

0.1 0.2 0.3 ξ − 6 Story Building

0.4

4 2 0

0

0.1 0.2 0.3 ξ − 9 Story Building

0.4

1

6

N/(m s)

N/m

0.1 0.2 0.3 ξ − 3 Story Building

2

0

0.4

6

0

0 6 x 10

4 2

6

0

Total Damping

6

2

0

x 10

0.1 0.2 0.3 ξ − 9 Story Building

0.4

4 2 0

Fig. 3.34: Total stiffness (N/m) and total damping (N-s/m) for the three example buildings

3.7 Damping Distribution for MDOF Systems 235

Using time-history analysis, the peak shear deformation was generated for each story of each building, and each earthquake. Figure 3.35 shows the variation of the mean deformation (the average over the building height) and standard deviation with ξ 1 . As expected, the response depends on the earthquake even though the earthquakes are scaled to have the same S v . The ensemble average provides a better estimate of the response. In general, the deformations are lower than the target value and the gap widens as ξ 1 is increased. Taking ξ 1 = 0.1 as the typical value, the results are within 80% of the target. Considering the uncertainty associated with specifying the seismic excitation and modeling the structure, one needs to apply a conservative design procedure.

−3

−3

El Centro

x 10 6

6

4

4

2

2

Northridge03−090

Northridge01−090

0

Ensemble Average

x 10

0 −3 x 10

0.1

0.2

0.3

0.4

0

6

6

4

4

2

2

0

0 −3 x 10

0.1

0.2

0.3

0.4

0

6

6

4

4

2

2

0

0 −3 x 10

0.1

0.2

0.3

0.4

0

6

6

4

4

2

2

0

0

0.1 0.2 0.3 ξ − 3 Story Building

(a) 3 story building

0.4

0

0 −3 x 10

0.1

0.2

0.3

0.4

0 −3 x 10

0.1

0.2

0.3

0.4

0 −3 x 10

0.1

0.2

0.3

0.4

0.1

0.2

0.3

0.4

0

6 Story Building (b) 6ξ −story building

Fig. 3.35: Mean story deformation and standard deviation for design earthquakes

236 Chapter 3: Optimal Passive Damping Distribution

−3

El Centro

x 10 6 4 2

Ensemble Average

Northridge03−090

Northridge01−090

0

0 −3 x 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 −3 x 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 −3 x 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0

0.05

0.1

0.15

0.2 0.25 ξ − 9 Story Building (c) 9 story building

0.3

0.35

0.4

6 4 2 0 6 4 2 0 6 4 2 0

Fig. 3.35: Mean story deformation and standard deviation for design earthquakes Example 3.16: Building #4 The building designated as example #4 in section 2.9, Table 2.4 is considered here. This building has a height of 200m, which corresponds to about 50 stories. The relevant properties are listed below for convenience.

3.7 Damping Distribution for MDOF Systems 237 Table 3.4: Properties for Building #4 H = 200m

H/B = 5

ρ m = 20,000 kg/m Design earthquake:

Sv = 1.2 m/s for ξ = 0.02

Design deformation:

γ * = 1/200

f

*

s = 0.63

=4

Three values of the damping ratio for the fundamental mode are considered. Figure 2.27 is used to obtain the value of Sv corresponding to each value of ξ . Given Sv, one finds T with eqn (2.220), and the rigidity distributions with eqns (2.119) and (2.220). These rigidity distributions are modified using the approach described in sections 2.10, and are scaled during the iteration process such that the final values of T are equal to the values predicted by eqn (2.220). Table 3.5 contains the design data for Sv and the corresponding periods. The “converged” rigidity distributions for each damping ratio are plotted in Figure 3.36, 3.37, and 3.38. Table 3.5: Design data for Building #4 Case

Design ξ1

values

Sv (m/s)

Period γ*

T1 (s)

1

0.02

1.2

1/200

5.36

2

0.05

0.94

1/200

6.85

3

0.10

0.70

1/200

9.25

238 Chapter 3: Optimal Passive Damping Distribution

1

x Normalized height ---H

0.9

0.8

Building #4 Quadratic Based T=5.36 s

0.7

0.6

mode 1

0.5

0.4

mode 2

0.3

SRSS

mode 3

0.2

0.1

0

0

2

4

6

8

10

12

14 8

x 10

Shear rigidity distribution DT - N Fig. 3.36: Converged shear rigidity distribution for Building 4, ξ = 0.02 1 0.9

x Normalized height ---H

Building #4 Quadratic based T = 6.85 s

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 3.37: Converged shear rigidity distribution for Building 4, ξ = 0.05

3.7 Damping Distribution for MDOF Systems 239

1

x Normalized height ---H

0.9

Building #4 Quadratic based T = 9.2 s

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 3.38: Converged shear rigidity distribution for Building 4, ξ = 0.10 Profiles of the maximum transverse shear deformation due to seismic excitation scaled such that Sv = 1.2 m/s for ξ = 0.02 are plotted in Figures 3.39, 3.40, and 3.41. The trend is similar to what was observed for the shear building examples examined in the previous section. As the damping is increased, the mean deformation trends to decrease, even though the design value of the spectral velocity is decreased with increasing ξ . This behavior indicates that the design procedure is conservative, at least for this example building. It should be noted that the periods are in the range where the dynamic amplification is not as significant as for low period buildings. Also, one should consider a number of excitations in order to generate a more representative “ensemble” average.

240 Chapter 3: Optimal Passive Damping Distribution

1

0.9

Building #4 T = 5.36 s

0.8

x Normalized height ---H

0.7

0.6

0.5

0.4

0.3

_________ El Centro

0.2

- - - - - - - Taft

0.1

0

(Sv=1.2 m/s, ξ=0.02)

γ∗ 0

0.002

0.004

0.006

0.008

0.01

Maximum shear deformation γ - m/m

0.012

Fig. 3.39: Maximum shear deformation for Building 4, ξ = 0.02 1 0.9

x Normalized height ---H

Building #4 T = 6.85 s

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

0.01

Fig. 3.40: Maximum shear deformation for Building 4, ξ 1 = 0.05

3.7 Damping Distribution for MDOF Systems 241

1 0.9

x Normalized height ---H

Building #4 T = 9.2 s

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

0.01

Fig. 3.41: Maximum shear deformation for building 4, ξ 1 = 0.10

242 Chapter 3: Optimal Passive Damping Distribution

Problems 243

Problems Problem 3.1 Consider a SDOF system. The total stored energy is the sum of the kinetic energy and the strain energy. Stored Energy = E S + E K 1 1 = --- ku 2 + --- mu˙ 2 2 2 Note that E S and E K are out of phase for periodic excitation. Suppose there is a rapid build-up of strain energy during the initial phase of a seismic excitation, and the design objective is to limit u max to a target value, u∗ . What strategy would you apply? Assume the mass cannot be varied.

Problem 3.2 Give examples of energy dissipation and absorption devices that are used in structures such as vehicles, crash barriers, machine supports, buildings, and bridges.

Problem 3.3 Consider a SDOF system having a linear spring and linear viscous damper. Suppose the initial conditions at t = 0 are: u(0) = 0

u˙ ( 0 ) = u˙∗

(1)

and there is no external loading.

k

c

u

244 Chapter 3: Optimal Passive Damping Distribution a) Determine the expression for the free vibration response corresponding to these initial conditions. b) Construct a plot of u˙ vs. u for t ranging from 0 to 4π ⁄ ω . Comment on the influence of damping on the time history response. Illustrate for m = 1, 000 kg , k = 9, 000 kN/m .

Problem 3.4 Refer to problem 3.3. Suppose the device is a Coulomb function damper. a) Determine the free vibration response. Since the sense of the friction force is determined by the sense of the velocity, one needs to generate solutions for different time intervals. During the first interval, 0 ≤ t ≤ t 1 , the velocity is positive. b) Construct a plot of u˙ vs. u for t sufficient to include one full cycle.

Problem 3.5 The SDOF system shown below is subjected to a periodic force. Determine the expressions for the spring and damper forces. Let F denote the total internal force. F = Fs + Fd

(1)

Fs

k

F

pˆ sin Ωt

m Fd

c

u

Determine the expression for F . Compare the magnitude of F with the magnitudes of F s and F d .

Problems 245 Problem 3.6 Refer to example 3.3. Consider material “one” to be low strength steel with σ y = 200 MPa , and material “two” to have σ y = 500 MPa steel. Design a hysteretic damper for the following criteria: F y = 100 kN L1 + L2 = 5 m k h = 6, 000 kN/m

Problem 3.7 Design a viscoelastic damper for the following conditions: • • • • •

Temperature = 20° Frequency excitation = 1 Hz Material 3M-ISD110 Maximum displacement = 0.025 m Maximum force = 10 kN

Problem 3.8 k p

m F

u

The SDOF system shown above has a linear viscoelastic damper which produces the force F. Assume u = uˆ sin Ωt and use F given by eqn (3.45). Determine p .

246 Chapter 3: Optimal Passive Damping Distribution

Problem 3.9 A convenient way of dealing with periodic excitation is to introduce complex notation. The basic identity is e iθ = cos θ + i sin θ

(1)

where θ is an arbitrary scalar. Using this notation, an arbitrary periodic forcing can be expressed in terms of a complex amplitude, p p = pe iΩt = ( pˆ e iδ )e iΩt = pˆ e i ( Ωt + δ ) = pˆ ( cos ( Ωt + δ ) + i sin ( Ωt + δ ) )

(2)

One can work with the complex form, and then retain either the real term (for cosine forcing) or the imaginary term (for sine forcing). a) Consider a 1 DOF system having a spring and linear viscous damper in parallel. Suppose p is a general periodic excitation, p = pe iΩt . Let u = ue iΩt represent the response. Express the relationship between p and u as p = Ku where K is the complex system stiffness. Determine K .

k p, u

m c

b) Consider a linear spring and viscous damper in series. Determine K .

k

c m

p, u

c) Refer to part a. Express the relationship between p and u as u = Hp . ˆ e iα . Suppose the loading consists of a set of periodic Determine H = H

Problems 247 excitations, p = p1 e

iΩ 1 t

+ p2 e

iΩ 2 t

+ … + pn e

iΩ n t

(3)

Express u in terms of p j and H j . Discuss how you would evaluate u at some time t for a cosine forcing, i.e., the loading corresponding to the real part of eqn (3). Problem 3.10 Equations (3.28) and (3.29) define the stress-strain response of a linear viscoelastic material to periodic sinusoidal excitation. Using complex notation, one can generalize these equations for arbitrary periodic excitation, γ = γ e iΩt , where γ is the complex amplitude. The complex shear modulus is defined as ˆ e iδ G = G

(1)

where tan δ = η, the loss factor ˆ = [ G 2 + G 2 ]1 / 2 G s l

(2)

With these definitions, the shear stress can be written as τ = τe iΩt τ = Gγ

(3)

a) Consider a linear viscoelastic damper subjected to periodic excitation, u = ue iΩt . Express the force as F = Fe iΩt and let F = k v u . Determine k v by generalizing eqn (3.41).

b) Consider a SDOF system having a spring and linear viscoelastic device in parallel. Assume p is a general periodic excitation and let p = Ku . Determine K . Equate this result to the result for a linear spring and linear viscous damper in parallel (see problem 3.9a), and determine the equivalent stiffness and damping factors, k' and c' .

248 Chapter 3: Optimal Passive Damping Distribution

k p, u

m (Actual)

k' m c’ (Equivalent Viscous) c) Consider a linear spring and linear viscoelastic device in series. Determine K .

k

p, u

m

d) Consider a SDOF system having a linear spring in parallel with a linear spring/linear viscoelastic device. Determine K

k m

p, u

k1

Problem 3.11 Consider a single degree of freedom system having m = 1000 kg and k = 36 kN/m. This problem concerns designing various types of damping devices.

Problems 249 Assume the system is subjected to seismic excitation, and the design objective is a damping ratio of 0.1. Take S v = 1 m/s as the response for pure viscous damping with ξ = 0.1. Propose damper properties for the following cases: 1.Viscous 2.Structural 3.Coulomb 4.Hysteretic with µ = 5 Discuss the basis for your recommendations.

Problem 3.12 Refer to problem 3.10, part b. Take m = 2000 kg and k = 25 kN/m. Determine the equivalent stiffness and viscous damping coefficients for the case where the material is 3M ISD110, the excitation frequency ranges from 0.5 Hz to 5 –2 Hz, and f d = 10 m .

Problem 3.13 Refer to problem 3.10, part b. Suppose m = 4000 kg and k = 50 kN/m. Design a visco-elastic damper to produce an equivalent viscous damping of 3 kNs/m at an excitation frequency of 2 Hz. Use 3M ISD110 material. Also determine the corresponding equivalent stiffness. Problem 3.14 Refer to problem 3.10, part b. The equivalent stiffness and viscous damping coefficients corresponding to a periodic excitation of frequency Ω are: k' = k + f d G s

(1)

η f d Gs c' = ----------------Ω

(2)

where G s and η are functions of Ω . Suppose one wants to approximate these coefficients with “constant” values over the frequency range of interest.

250 Chapter 3: Optimal Passive Damping Distribution

k' ( Ω ) ≈ k

*

c' ( Ω ) ≈ c

(3)

*

(4)

Show that a least square approximation leads to: *

k = k + k eq

(5)

*

c = c eq

(6)

where k eq and c eq are defined by equations (3.74) and (3.75). Problem 3.15 Consider a SDOF system having a spring and a viscoelastic damper modelled as shown below. Take c = α d k 1 . a) Determine expressions for the fundamental frequency, ω , and damping ratio, ξ. b) Suppose m = 1000 kg and k = 9 kN/m. Determine the damper properties c and k1 that correspond to a damping ratio of 0.1. Take α d = 0.15 . Also determine the fundamental frequency. c) Suppose m = 1000 kg and the total stiffness, k + k 1 , is equal to 9kN/m. Determine the stiffness and damping parameters required for ξ = 0.1 . Take α d = 0.15 . k k1

c Problem 3.16 Refer to example 3.12

m

u

Problems 251 a) Take C = 600 N s/m. Determine K eq and C eq corresponding to a periodic excitation, Ω = 2π r ⁄ s , and the following values for stiffness: 1. K = 9 kN ⁄ m

K 2 = 4.5 kN ⁄ m

2. K = 9 kN ⁄ m

K 2 = 18 kN ⁄ m

3. K = 9 kN ⁄ m

K2 = ∞

b) Suppose m = 1000kg. Determine ω and ξ corresponding to the values of K eq and C eq obtained in part a. Problem 3.17 Consider the truss beam segment shown below. Assume the diagonal bracing system consists of elastic elements and linear viscous dampers attached to rigid links connecting the end nodes. Dimension the elastic element and damper for the following requirements: 3

D T = 260 × 10 kN 3

T

= 5 × 10 kN s/m

Use steel for the elastic element.

4m 45

0

Problem 3.18 Refer to problem 3.17. Assume the diagonal bracing system consists of

252 Chapter 3: Optimal Passive Damping Distribution elastic elements and linear viscoelastic elements. Dimension the components for the requirements specified in problem 3.17. Use equivalent stiffness and damping parameters for the viscoelastic component, and take α d = 0.15 . Problem 3.19 Consider a structural system composed of 2 subsystems: the first is linear elastic and provides only stiffness; the second system is linear viscoelastic and provides both stiffness and damping. Let K e denote the stiffness matrix for the elastic component and K 1, C the stiffness and damping matrices for the viscoelastic component. The equations of motion are: ˙˙ + CU˙ + ( K + K )U = P MU e 1

(1)

where C is related to K 1 . Specializing eqn(1) for free vibration and expressing the solution as: λt

U = Ae Φ

(2)

leads to the following equation: 2

[ λ M + λC + ( K e + K 1 ) ]Φ = 0

(3)

The solution for the primary elastic system is expressed as: λ = ± iω

ω >0

( K e – ω 2 M )Φ e = 0

(4) (5)

a) Consider the pure elastic case, C = 0 and K 1 ≠ 0 . Determine the corresponding λ and Φ for the case where K 1 as a scalar multiple of K e , say K 1 = α1 K e . b) Consider the pure viscous case C ≠ 0 and K 1 = 0 . Take C = α 2 K e . Determine λ and Φ . Express λ as: λ = – ξω ± iω'

(6)

c) Consider the linear viscoelastic case. Take C = α d K 1 where α d is a material parameter. Assume stiffness proportional damping and consider 2 cases:

Problems 253 1. C = α 2 K e 2. C = α 3 ( K e + K 1 ) Determine expressions for the corresponding λ and Φ . d) Consider the case where C = α 3 ( K e + K 1 ) and C = α d K 1 . Suppose the solution is expressed as: N

U =

∑ q jΦe, j

(7)

j=1

where Φ e, j are the undamped modal vectors, i.e., the solution of eqn (5). Assume Φ e are normalized with respect to M . T ˜j Φ e, d M Φ e, d = m

(8)

Derive the governing equation and initial conditions for q j . Express it in a form similar to eqn (3.166), using “equivalent” terms for ω j and ξ j . Utilize the results generated in part a to establish the natural frequency for the case where K = K e + K 1 . Discuss how damping and frequency vary with α 3 , for a given viscoelastic material. Note that α d = ∞ for pure viscous damping. Problem 3.20 The governing equation for an n'th order system is given by eqn (3.143) ˙˙ + CU˙ + KU = P MU

(1)

Expressing the homogeneous solution as λt

U = Ae Φ

(2)

and specializing eqn (1) for P = 0 leads to: 2

( λ M + Cλ + K )Φ = 0

(3)

Various cases ( C = 0 , C = αK , and C arbitrary) are discussed in section 3.7.

254 Chapter 3: Optimal Passive Damping Distribution Suppose one defines a new variable, X ,

X =

U U˙

(4)

˙˙ using eqn (1), one obtains: Differentiating X and substituting for U 0 ˙ X˙ = U = – 1 ˙˙ U –M K

I –M

–1

0 U + – 1 C U˙ M P

(5)

Denoting the coefficient matrix as A , eqn (5) simplifies to X˙ = AX + B

(6)

The homogeneous solution of eqn (6) has the general form: αt

X = e Ψ a) Determine the equation relating α and Ψ . b) Show that α = λ, i.e., α is an eigenvalue of eqn (3). c) What is the significance of the real part of α ? d) Consider a 2 DOF discrete shear beam having the following properties: 2

m 1 = m 2 = 1000 kg = 1 kNs ⁄ m k 1 = 118.3 kN ⁄ m

k 2 = 78.9 kN/m

c 1 = 2.09 kNs/m

c 2 = 4.18 kNs/m

Form A and determine the eigenvalues and eigenvectors using the eigen routine in MATLAB. e) Repeat part d using the MOTIONLAB program.

Problems 255 Problem 3.21 m 3 = 2m

u3 k 3, c 3

m2 = m

u2 k 2, c 2

m1 = m

u1 k 1, c 1

Consider the 3 DOF system shown above. a) Determine the magnitudes of k 1 , k 2 , and k 3 such that the first mode has the form 1 2 Φ 1 = ---, ---, 1 3 3 and the fundamental frequency is 2π rad/sec. Take m = 2000 kg. b) Determine the viscous damping coefficients c 1 , c 2 , and c 3 such that the damping ratio, ξ 1 , for the first mode is 0.1. Consider both stiffness proportional damping and uniform damping. Use MOTIONLAB to determine the modal properties for the case of non-proportional damping. c) Suppose viscoelastic dampers are used. Let k d and c d denote the equivalent stiffness and damping coefficients, and take c d = α d k d where α d is a material property. Determine c d, i and the elastic stiffness k e, i for each element. Assume α d = 0.15 . Consider stiffness proportional damping and uniform damping. Compare the modal properties for these 2 cases. Problem 3.22 Refer to example 3.7. Use MOTIONLAB to determine the modal properties for the first 3 modes. Take k i according to eqn (1) and c i based on:

256 Chapter 3: Optimal Passive Damping Distribution a) damping proportional to stiffness b) uniform damping c) c 1 = c 2 = 0 and c 3 = c 4 = c 5 d) other combinations of the c' that you believe may be more optimal in the sense of resulting in higher damping ratios for the higher modes. Problem 3.23 Refer to example 3.7. Suppose viscoelastic dampers are used for elements 3, 4, 5 and there is no damping in elements 1 and 2. Assume uniform viscoelastic damper properties and take α d = 0.15 . Starting with the data contained in eqn (1) of example 3.7, determine the damper properties and modified elastic stiffness which correspond to a damping ratio of 0.1 for the fundamental mode. Using MOTIONLAB, assess the effect of non-proportional damping on the mode shape for this design value of ξ 1 . What would be the effect if ξ 1 is increased to 0.2 and the stiffness is maintained at the level defined by eqn (1)? Problem 3.24

10 m

5m

This problem concerns the preliminary design of a 10 story rectangular rigid frame for seismic excitation. The frame properties and design criteria are: • Height = 5 m/story

Problems 257 • Width = 10 m/bay • Mass/floor = 10,000 kg • Max. deflection at top = 0.25 m • Max. story shear deformation = 1/200 • Response spectrum shown below.

Spectral Velocity S v (m/s) S v = 1.2, ξ = 0.02 S v = 0.8, ξ = 0.1

1.0

S v = 0.6, ξ = 0.2

0.1 0.1

0.6 1.0

10

Period T (sec)

The chevron bracing system is similar to the scheme shown in Fig. 3.23(b); it allows 2 viscous dampers to be placed on each floor. a) Determine the stiffness distributions based on a linear fundamental mode k i , for profile and ξ equal to 0.02, 0.10, and 0.20. Evaluate the stiffness cost, each distribution.

∑

b) Assume stiffness proportional damping. Determine the corresponding ci . distributions for the story damping coefficients, and the damping cost,

∑

c) Repeat b) for “uniform” damping. Using MOTIONLAB, determine the properties for the fundamental mode. Compare the displacement profiles. *

d) Repeat b) for c 1 = c 2 = c 3 = c 4 = c 5 = 0 and c 6 = c 7 = c 8 = c 9 = c 10 = c . Determine the actual damping ratio and profile for the fundamental mode. e) The damper cost increases nonlinearly with the damper coefficient, i.e., the cost

258 Chapter 3: Optimal Passive Damping Distribution for 2c is more than twice the cost for c. Also, the damper force increases with c and places more loading on the brace-floor connection. With these limitations, discuss how you would select a damper placement that satisfies the performance requirement on the maximum transverse shear deformation for each story and minimizes a “cost” function. Problem 3.25 Consider a 5 DOF shear beam having equal masses and equal nodal forces. Suppose the force consists of a combination of a static component and a periodic excitation, p = p s + p d sin ωt where ω is a random quantity. Take m i = 10, 000kg p s = 100kN p d = 1kN a) Determine the stiffness distribution such that the interstory displacement under the static loading is 0.02m for each story. b) Assume ω coincides with the frequency for the fundamental mode corresponding to the stiffness distribution generated in part (a). Suppose the design objective is to have the peak acceleration less than 0.02g where g is the acceleration due to gravity and is equal to 9.87m/s2. Suggest various schemes for generating the required energy dissipation. Comment on what you consider to be the optimal solution.

259

Chapter 4

Tuned mass damper systems 4.1 Introduction A tuned mass damper (TMD) is a device consisting of a mass, a spring, and a damper that is attached to a structure in order to reduce the dynamic response of the structure. The frequency of the damper is tuned to a particular structural frequency so that when that frequency is excited, the damper will resonate out of phase with the structural motion. Energy is dissipated by the damper inertia force acting on the structure. The TMD concept was first applied by Frahm in 1909 (Frahm, 1909) to reduce the rolling motion of ships as well as ship hull vibrations. A theory for the TMD was presented later in the paper by Ormondroyd & Den Hartog (1928), followed by a detailed discussion of optimal tuning and damping parameters in Den Hartog’s book on Mechanical Vibrations (1940). The initial theory was applicable for an undamped SDOF system subjected to a sinusoidal force excitation. Extension of the theory to damped SDOF systems has been investigated by numerous researchers. Significant contributions were made by Randall et al. (1981), Warburton (1980,1981,1982), and Tsai & Lin (1993). This chapter starts with an introductory example of a TMD design and a brief description of some of the implementations of tuned mass dampers in building structures. A rigorous theory of tuned mass dampers for SDOF systems subjected to harmonic force excitation and harmonic ground motion is discussed next. Various cases including an undamped TMD attached to an undamped

260 Chapter 4: Tuned Mass Damper Systems SDOF system, a damped TMD attached to an undamped SDOF system, and a damped TMD attached to a damped SDOF system are considered. Time history responses for a range of SDOF systems connected to optimally tuned TMD and subjected to harmonic and seismic excitations are presented. The theory is then extended to MDOF systems where the TMD is used to dampen out the vibrations of a specific mode. An assessment of the optimal placement locations of TMDs in building structures is included. Numerous examples are provided to illustrate the level of control that can be achieved with such passive devices for both harmonic and seismic excitations.

4.2 An introductory example In this section, the concept of the tuned mass damper is illustrated using the twomass system shown in Fig. 4.1. Here, the subscript d refers to the tuned mass damper; the structure is idealized as a single degree of freedom system. Introducing the following notation 2 k ω = ---m

(4.1)

c = 2ξωm

(4.2)

kd 2 ω d = ------md

(4.3)

c d = 2ξ d ω d m d

(4.4)

and defining m as the mass ratio, md m = ------m p

(4.5)

kd

k

md

m c

cd u Fig. 4.1: SDOF - TMD system.

u + ud

4.2 An Introductory Example 261 the governing equations of motion are given by Primary mass

2 p ( 1 + m )u˙˙ + 2ξωu˙ + ω u = ---- – mu˙˙d m

Tuned mass

u˙˙d + 2ξ d ω d u˙ d + ω d u d = – u˙˙

2

(4.6) (4.7)

The purpose of adding the mass damper is to limit the motion of the structure when it is subjected to a particular excitation. The design of the mass damper involves specifying the mass m d , stiffness k d , and damping coefficient c d . The optimal choice of these quantities is discussed in Section 4.4. In this example, the near-optimal approximation for the frequency of the damper, (4.8) ωd = ω is used to illustrate the design procedure. The stiffnesses for this frequency combination are related by k d = mk

(4.9)

Equation (4.8) corresponds to tuning the damper to the fundamental period of the structure. Considering a periodic excitation, p = pˆ sin Ωt

(4.10)

the response is given by u = uˆ sin ( Ωt + δ 1 )

(4.11)

u d = uˆ d sin ( Ωt + δ 1 + δ 2 )

(4.12)

where uˆ and δ denote the displacement amplitude and phase shift respectively. The critical loading scenario is the resonant condition, Ω = ω . The solution for this case has the following form pˆ 1 uˆ = ------- ---------------------------------------km 2ξ 1 2 1 + ------ + --------- m 2ξ d 1 uˆ d = --------- uˆ 2ξ d

(4.13)

(4.14)

262 Chapter 4: Tuned Mass Damper Systems 1 2ξ tan δ 1 = – ------ + --------m 2ξ d

(4.15)

π tan δ 2 = – --2

(4.16)

Note that the response of the tuned mass is 900 out of phase with the response of the primary mass. This difference in phase produces the energy dissipation contributed by the damper inertia force. The response for no damper is given by pˆ uˆ = --k

1 ---- 2ξ

π δ 1 = – --2

(4.17) (4.18)

To compare these two cases, one can express eqn (4.13) in terms of an equivalent damping ratio pˆ uˆ = --k

1 ------- 2ξ

(4.19)

e

where m 2ξ 1 2 ξ e = ---- 1 + ------ + --------- m 2ξ 2 d

(4.20)

Equation (4.20) shows the relative contribution of the damper parameters to the total damping. Increasing the mass ratio magnifies the damping. However, since the added mass also increases, there is a practical limit on m . Decreasing the damping coefficient for the damper also increases the damping. Noting eqn (4.14), the relative displacement also increases in this case, and just as for the mass, there is a practical limit on the relative motion of the damper. Selecting the final design requires a compromise between these two constraints.

Example 4.1: Preliminary design of a TMD for a SDOF system Suppose ξ = 0 and one wants to add a tuned mass damper such that the equivalent damping ratio is 10% . Using eqn (4.20), and setting ξ e = 0.1 , the following relation between m and ξ d is obtained.

4.2 An Introductory Example 263 m 2ξ 1 2 ---- 1 + ------ + --------- = 0.1 m 2ξ 2 d

(4.21)

The relative displacement constraint is given by eqn (4.14) 1 uˆ d = --------- uˆ 2ξ d

(4.22)

Combining eqn (4.21) and eqn (4.22), and setting ξ = 0 leads to uˆ d 2 m ---- 1 + ------ = 0.1 uˆ 2

(4.23)

Usually, uˆ d is taken to be an order of magnitude greater than uˆ . In this case eqn (4.23) can be approximated as m uˆ d ---- ------ ≈ 0.1 2 uˆ

(4.24)

The generalized form of eqn (4.24) follows from eqn (4.20): 1 m ≈ 2ξ e ------------- uˆ ⁄ uˆ

(4.25)

d

Finally, taking uˆ d = 10uˆ yields an estimate for m 2 ( 0.1 ) m = --------------- = 0.02 10

(4.26)

This magnitude is typical for m . The other parameters are 1 uˆ ξ d = --- ------ = 0.05 2 uˆ d

(4.27)

and from eqn (4.9) k d = mk = 0.02k

(4.28)

It is important to note that with the addition of only 2% of the primary mass, one obtains an effective damping ratio of 10% . The negative aspect is the large relative motion of the damper mass; in this case, 10 times the displacement of the primary mass. How to accommodate this motion in an actual structure is an important design consideration.

A description of some applications of tuned mass dampers to building

264 Chapter 4: Tuned Mass Damper Systems structures is presented in the following section to provide additional background on this type of device prior to entering into a detailed discussion of the underlying theory.

4.3 Examples of existing tuned mass damper systems Although the majority of applications have been for mechanical systems, tuned mass dampers have been used to improve the response of building structures under wind excitation. A short description of the various types of dampers and several building structures that contain tuned mass dampers follows. Translational tuned mass dampers Figure 4.2 illustrates the typical configuration of a unidirectional translational tuned mass damper. The mass rests on bearings that function as rollers and allow the mass to translate laterally relative to the floor. Springs and dampers are inserted between the mass and the adjacent vertical support members which transmit the lateral “out-of-phase” force to the floor level, and then into the structural frame. Bidirectional translational dampers are configured with springs/dampers in 2 orthogonal directions and provide the capability for controlling structural motion in 2 orthogonal planes. Some examples of early versions of this type of damper are described below.

Support md Floor Beam

Direction of motion Fig. 4.2: Schematic diagram of a translational tuned mass damper.

4.3 Examples of Existing Tuned Mass Damper Systems 265

• John Hancock Tower (Engineering News Record, Oct. 1975) Two dampers were added to the 60-story John Hancock Tower in Boston to reduce the response to wind gust loading. The dampers are placed at opposite ends of the 58th story, 67m apart, and move to counteract sway as well as twisting due to the shape of the building. Each damper weighs 2700 kN and consists of a lead-filled steel box about 5.2m square and 1m deep that rides on a 9m long steel plate. The lead-filled weight, laterally restrained by stiff springs anchored to the interior columns of the building and controlled by servo-hydraulic cylinders, slides back and forth on a hydrostatic bearing consisting of a thin layer of oil forced through holes in the steel plate. Whenever the horizontal acceleration exceeds 0.003g for two consecutive cycles, the system is automatically activated. This system was designed and manufactured by LeMessurier Associates/SCI in association with MTS System Corp., at a cost of around 3 million dollars, and is expected to reduce the sway of the building by 40% to 50%. • Citicorp Center (Engineering News Record Aug. 1975, McNamara 1977, Petersen 1980) The Citicorp (Manhattan) TMD was also designed and manufactured by LeMessurier Associates/SCI in association with MTS System Corp. This building is 279m high, has a fundamental period of around 6.5s with an inherent damping ratio of 1% along each axis. The Citicorp TMD, located on the 63rd floor in the crown of the structure, has a mass of 366 Mg, about 2% of the effective modal mass of the first mode, and was 250 times larger than any existing tuned mass damper at the time of installation. Designed to be biaxially resonant on the building structure with a variable operating period of 6.25s ± 20% , adjustable linear damping from 8% to 14%, and a peak relative displacement of ± 1.4m , the damper is expected to reduce the building sway amplitude by about 50%. This reduction corresponds to increasing the basic structural damping by 4%. The concrete mass block is about 2.6m high with a plan cross-section of 9.1m by 9.1m and is supported on a series of twelve 60cm diameter hydraulic pressurebalanced bearings. During operation, the bearings are supplied oil from a separate hydraulic pump which is capable of raising the mass block about 2cm to its operating position in about 3 minutes. The damper system is activated automatically whenever the horizontal acceleration exceeds 0.003g for two consecutive cycles, and will automatically shut itself down when the building

266 Chapter 4: Tuned Mass Damper Systems acceleration does not exceed 0.00075g in either axis over a 30 minute interval. LeMessurier estimates Citicorp’s TMD, which cost about 1.5 million dollars, saved 3.5 to 4 million dollars. This sum represents the cost of some 2,800 tons of structural steel that would have been required to satisfy the deflection constraints. • Canadian National Tower (Engineering News Record, 1976) The 102m steel antenna mast on top of the Canadian National Tower in Toronto (553m high including the antenna) required two lead dampers to prevent the antenna from deflecting excessively when subjected to wind excitation. The damper system consists of two doughnut-shaped steel rings, 35cm wide, 30cm deep, and 2.4m and 3m in diameter, located at elevations 488m and 503m. Each ring holds about 9 metric tons of lead and is supported by three steel beams attached to the sides of the antenna mast. Four bearing universal joints that pivot in all directions connect the rings to the beams. In addition, four separate hydraulically activated fluid dampers mounted on the side of the mast and attached to the center of each universal joint dissipate energy. As the leadweighted rings move back and forth, the hydraulic damper system dissipates the input energy and reduces the tower’s response. The damper system was designed by Nicolet, Carrier, Dressel, and Associates, Ltd, in collaboration with Vibron Acoustics, Ltd. The dampers are tuned to the second and fourth modes of vibration in order to minimize antenna bending loads; the first and third modes have the same characteristics as the prestressed concrete structure supporting the antenna and did not require additional damping. • Chiba Port Tower (Kitamura et al. 1988) Chiba Port Tower (completed in 1986) was the first tower in Japan to be equipped with a TMD. Chiba Port Tower is a steel structure 125m high weighing 1950 metric tons and having a rhombus shaped plan with a side length of 15m. The first and second mode periods are 2.25s and 0.51s respectively for the X direction and 2.7s and 0.57s for the Y direction. Damping for the fundamental mode is estimated at 0.5%. Damping ratios proportional to frequencies were assumed for the higher modes in the analysis. The purpose of the TMD is to increase damping of the first mode for both the X and Y directions. Figure 4.3 shows the damper system. Manufactured by Mitsubishi Steel Manufacturing Co., Ltd, the damper has: mass ratios with respect to the modal mass of the first mode of about 1/120 in the X direction and 1/80 in the Y direction; periods in the X and Y directions of 2.24s and 2.72s respectively; and a damper damping ratio of 15%. The maximum

4.3 Examples of Existing Tuned Mass Damper Systems 267 relative displacement of the damper with respect to the tower is about ± 1m in each direction. Reductions of around 30% to 40% in the displacement of the top floor and 30% in the peak bending moments are expected.

Fig. 4.3: Tuned mass damper for Chiba-Port Tower. The early versions of TMD’s employ complex mechanisms for the bearing and damping elements, have relatively large masses, occupy considerably space, and are quite expensive. Recent versions, such as the scheme shown in Fig 4.4, have been designed to minimize these limitations. This scheme employs a multiassemblage of elastomeric rubber bearings, which function as shear springs, and bitumen rubber compound (BRC) elements, which provide viscoelastic damping capability. The device is compact in size, requires unsophisticated controls, is multidirectional, and is easily assembled and modified. Figure 4.5 shows a full scale damper being subjected to dynamic excitation by a shaking table. An actual installation is contained in Fig. 4.6.

268 Chapter 4: Tuned Mass Damper Systems

Fig. 4.4: Tuned mass damper with spring and damper assemblage.

Fig. 4.5: Deformed position - tuned mass damper.

4.3 Examples of Existing Tuned Mass Damper Systems 269

Fig. 4.6: Tuned mass damper - Huis Ten Bosch Tower, Nagasaki. The effectiveness of a tuned mass damper can be increased by attaching an auxiliary mass and an actuator to the tuned mass and driving the auxiliary mass with the actuator such that its response is out of phase with the response of the tuned mass. Fig 4.7 illustrates this scheme. The effect of driving the auxiliary mass is to produce an additional force which complements the force generated by the tuned mass, and therefore increases the equivalent damping of the TMD (one can obtain the same behavior by attaching the actuator directly to the tuned mass, thereby eliminating the need for an auxiliary mass). Since the actuator requires an external energy source, this system is referred to as an active tuned mass damper. The scope of this chapter is restricted to passive TMD’s. Active TMD’s are discussed in Chapter 6.

270 Chapter 4: Tuned Mass Damper Systems

Auxiliary mass Actuator Support

Floor Beam

Direction of motion Fig. 4.7: An active tuned mass damper configuration. Pendulum tuned mass damper The problems associated with the bearings can be eliminated by supporting the mass with cables which allow the system to behave as a pendulum. Fig 4.8a shows a simple pendulum attached to a floor. Movement of the floor excites the pendulum. The relative motion of the pendulum produces a horizontal force which opposes the floor motion. This action can be represented by an equivalent SDOF system which is attached to the floor as indicated in Fig 4.8b. u

L

keq

θ md

t=0

t ud

u

(a) actual system

md u+ud (b) equivalent system

Fig. 4.8: A simple pendulum tuned mass damper.

4.3 Examples of Existing Tuned Mass Damper Systems 271 The equation of motion for the horizontal direction is Wd T sin θ + -------- ( u˙˙ + u˙˙d ) = 0 g where T is the tension in the cable. When θ approximations apply

(4.29)

is small, the following

u d = L sin θ ≈ Lθ T ≈ Wd Introducing these approximations transforms eqn (4.29) to Wd m d u˙˙d + --------u d = – m d u˙˙ L and it follows that the equivalent shear spring stiffness is Wd k eq = -------L The natural frequency of the pendulum is related to keq by k eq g ω d2 = ------- = --md L Noting eqn (4.33), the natural period of the pendulum is L T d = 2π --g

(4.30)

(4.31)

(4.32)

(4.33)

(4.34)

The simple pendulum tuned mass damper concept has a serious limitation. Since the period depends on L, the required length for large Td may be greater than the typical story height. For instance, the length for Td=5 secs is 6.2 meters whereas the story height is between 4 to 5 meters. This problem can be eliminated by resorting to the scheme illustrated in Fig 4.9. The interior rigid link magnifies the support motion for the pendulum, and results in the following equilibrium equation Wd (4.35) m d ( u˙˙ + u˙˙1 + u˙˙d ) + --------u d = 0 L The rigid link moves in phase with the damper, and has the same displacement amplitude. Then, taking u1= ud in eqn (4.35) results in Wd md m d u˙˙d + --------u d = – ------- u˙˙ (4.36) 2L 2

272 Chapter 4: Tuned Mass Damper Systems The equivalent stiffness is Wd/2L , and it follows that the effective length is equal to 2L. Each additional link increases the effective length by L. An example of a pendulum type damper is described below.

u

L

md u+u1

u+u1+ud

Fig. 4.9: Compound pendulum. • Crystal Tower (Nagase & Hisatoku 1990) The tower, located in Osaka Japan, is 157m high and 28m by 67m in plan, weighs 44,000 metric tons, and has a fundamental period of approximately 4s in the north-south direction and 3s in the east-west direction. A tuned pendulum mass damper was included in the early phase of the design to decrease the windinduced motion of the building by about 50%. Six of the nine air cooling and heating ice thermal storage tanks (each weighing 90 tons) are hung from the top roof girders and used as a pendulum mass. Four tanks have a pendulum length of 4m and slide in the north-south direction; the other two tanks have a pendulum length of about 3m and slide in the east-west direction. Oil dampers connected to the pendulums dissipate the pendulum energy. Fig 4.10 shows the layout of the ice storage tanks that were used as damper masses. Views of the actual building and one of the tanks are presented in Fig 4.11. The cost of this tuned mass damper system was around $350,000, less than 0.2% of the construction cost.

4.3 Examples of Existing Tuned Mass Damper Systems 273

Fig. 4.10: Pendulum damper layout - Crystal Tower.

274 Chapter 4: Tuned Mass Damper Systems

Fig. 4.11: Ice storage tank - Crystal Tower. A modified version of the pendulum damper is shown in Fig 4.12. The restoring force provided by the cables is generated by introducing curvature in the support surface and allowing the mass to roll on this surface. The vertical motion of the weight requires an energy input. Assuming θ is small, the equations for the case where the surface is circular are the same as for the conventional pendulum with the cable length L, replaced with the surface radius R.

4.4 Tuned Mass Damper Theory for SDOF Systems 275

θ

R

md

u

(a)

Floor md keq

(b)

Fig. 4.12: Rocker pendulum.

4.4 Tuned mass damper theory for SDOF systems In what follows, various cases ranging from fully undamped to fully damped conditions are analyzed and design procedures are presented. Undamped structure - undamped TMD Figure 4.13 shows a SDOF system having mass m and stiffness k , subjected to both external forcing and ground motion. A tuned mass damper with mass m d and stiffness k d is attached to the primary mass. The various displacement measures are: u g , the absolute ground motion; u , the relative motion between the

276 Chapter 4: Tuned Mass Damper Systems primary mass and the ground; and u d , the relative displacement between the damper and the primary mass. With this notation, the governing equations take the form m d [ u˙˙d + u˙˙] + k d u d = – m d a g

(4.37)

mu˙˙ + ku – k d u d = – ma g + p

(4.38)

where a g is the absolute ground acceleration and p is the force loading applied to the primary mass.

p

k

m

ug

kd

md

u + ug

ud + u + ug

Fig. 4.13: SDOF system coupled with a TMD. The excitation is considered to be periodic of frequency Ω , a g = aˆ g sin Ωt

(4.39)

p = pˆ sin Ωt

(4.40)

Expressing the response as u = uˆ sin Ωt

(4.41)

u d = uˆ d sin Ωt

(4.42)

and substituting for these variables, the equilibrium equations are transformed to [ – m d Ω 2 + k d ]uˆ d – m d Ω 2 uˆ = – m d aˆ g

(4.43)

– k d uˆ d + [ – mΩ 2 + k ]uˆ = – maˆ g + pˆ

(4.44)

4.4 Tuned Mass Damper Theory for SDOF Systems 277 The solutions for uˆ and uˆ d are given by 2

2

pˆ 1 – ρ d maˆ g 1 + m – ρ d uˆ = --- --------------- – ---------- -------------------------- k D1 k D1

(4.45)

2 pˆ mρ maˆ g m uˆ d = ----- ----------- – ---------- ------- kd D1 k d D 1

(4.46)

where 2

2

D 1 = [ 1 – ρ ] [ 1 – ρ d ] – mρ

2

and the ρ terms are dimensionless frequency ratios, Ω Ω ρ = ---- = --------------ω k⁄m Ω Ω ρ d = ------- = ---------------------ωd kd ⁄ md

(4.47)

(4.48) (4.49)

Selecting the mass ratio and damper frequency ratio such that 2

1 – ρd + m = 0

(4.50)

reduces the solution to pˆ uˆ = --k

(4.51)

pˆ 2 maˆ g uˆ d = – -----ρ + ---------kd kd

(4.52)

This choice isolates the primary mass from ground motion and reduces the response due to external force to the pseudo-static value, pˆ ⁄ k . A typical range for m is 0.01 to 0.1 . Then, the optimal damper frequency is very close to the forcing frequency. The exact relationship follows from eqn (4.50). ωd

Ω = -----------------opt 1+m

(4.53)

One determines the corresponding damper stiffness with 2

Ω mm = ωd kd m d = ----------------1+m opt opt 2

(4.54)

278 Chapter 4: Tuned Mass Damper Systems Finally, substituting for k d , eqn (4.52) takes the following form aˆ g 1 + m uˆ d = -------------- --pˆ- + ------ m k 2 Ω

(4.55)

One specifies the amount of relative displacement for the damper and determines m with eqn (4.55). Given m and Ω , the stiffness is found using eqn (4.54). It should be noted that this stiffness applies for a particular forcing frequency. Once the mass damper properties are defined, eqns (4.45) and (4.46) can be used to determine the response for a different forcing frequency. The primary mass will move under ground motion excitation in this case. Undamped structure - damped TMD The next level of complexity has damping included in the mass damper, as shown in Fig. 4.14. The equations of motion for this case are m d u˙˙d + c d u˙ d + k d u d + m d u˙˙ = – m d a g

(4.56)

mu˙˙ + ku – c d u˙ d – k d u d = – ma g + p

(4.57)

The inclusion of the damping terms in eqns (4.56) and (4.57) produces a phase shift between the periodic excitation and the response. It is convenient to work initially with the solution expressed in terms of complex quantities. One expresses the excitation as a g = aˆ g e p = pˆ e

iΩt

iΩt

(4.58) (4.59)

where aˆ g and pˆ are real quantities. The response is taken as u = ue

iΩt

ud = ud e

iΩt

(4.60) (4.61)

4.4 Tuned Mass Damper Theory for SDOF Systems 279

p kd k

md

m cd ug

ud + u + ug

u + ug

Fig. 4.14: Undamped SDOF system coupled with a damped TMD system.

where the response amplitudes, u and u d are considered to be complex quantities. The real and imaginary parts of a g correspond to cosine and sinusoidal input. Then, the corresponding solution is given by either the real (for cosine) or imaginary (for sine) parts of u and u d . Substituting eqns (4.60) and iΩt (4.61) in the set of governing equations and cancelling e from both sides results in (4.62) [ – m d Ω 2 + ic d Ω + k d ]u d – m d Ω 2 u = – m d aˆ g – [ ic d Ω + k d ]u d + [ – mΩ 2 + k ]u = – maˆ g + pˆ

(4.63)

The solution of the governing equations is aˆ g m pˆ 2 2 2 2 u = ---------- [ f – ρ + i2ξ d ρf ] – ---------- [ ( 1 + m ) f – ρ + i2ξ d ρf ( 1 + m ) ] kD 2 kD 2

(4.64)

2 aˆ m pˆ ρ g u d = ---------- – ---------kD 2 kD 2

(4.65)

where 2

2

2

2 2

2

D 2 = [ 1 – ρ ] [ f – ρ ] – mρ f + i2ξ d ρf [ 1 – ρ ( 1 + m ) ]

(4.66)

ωd f = ------ω

(4.67)

and ρ was defined earlier as the ratio of Ω to ω (see eqn (4.48)). Converting the complex solutions to polar form leads to the following

280 Chapter 4: Tuned Mass Damper Systems expressions iδ 1 aˆ g m iδ 2 pˆ u = --- H 1 e – ----------H 2 e k k

(4.68)

– i δ 3 aˆ g m –i δ3 pˆ u d = --- H 3 e – ----------H 4 e k k

(4.69)

where the H factors define the amplification of the pseudo-static responses, and the δ ‘s are the phase angles between the response and the excitation. The various H and δ terms are listed below 2

2 2

2

[ f – ρ ] + [ 2ξ d ρf ] H 1 = -----------------------------------------------------------D2

(4.70)

2 2

2

2

[ ( 1 + m ) f – ρ ] + [ 2ξ d ρf ( 1 + m ) ] H 2 = -------------------------------------------------------------------------------------------------D2

(4.71)

2

ρ H 3 = ---------D2

(4.72)

1 H 4 = ---------D2

(4.73)

D2 =

2

2

2

2 2 2

2

( [ 1 – ρ ] [ f – ρ ] – mρ f ) + ( 2ξ d ρf [ 1 – ρ ( 1 + m ) ] )

2

(4.74)

Also δ1 = α1 – δ3

(4.75)

δ2 = α2 – δ3

(4.76) 2

2ξ d ρf [ 1 – ρ ( 1 + m ) ] tan δ 3 = -----------------------------------------------------------------2 2 2 2 2 [ 1 – ρ ] [ f – ρ ] – mρ f

(4.77)

2ξ d ρf tan α 1 = -----------------2 2 f –ρ

(4.78)

2ξ d ρf ( 1 + m ) tan α 2 = ------------------------------------2 2 (1 + m) f – ρ

(4.79)

4.4 Tuned Mass Damper Theory for SDOF Systems 281 For most applications, the mass ratio is less than about 0.05 . Then, the amplification factors for external loading ( H 1 ) and ground motion ( H 2 ) are essentially equal. A similar conclusion applies for the phase shift. In what follows, the solution corresponding to ground motion is examined and the optimal values of the damper properties for this loading condition are established. An in-depth treatment of the external forcing case is contained in Den Hartog’s text (Den Hartog, 1940). Figure 4.15 shows the variation of H 2 with forcing frequency for specific values of damper mass m and frequency ratio f , and various values of the damper damping ratio, ξ d . When ξ d = 0 , there are two peaks with infinite amplitude located on each side of ρ = 1 . As ξ d is increased, the peaks approach each other and then merge into a single peak located at ρ ≈ 1 . The behavior of the amplitudes suggests that there is an optimal value of ξ d for a given damper configuration ( m d and k d , or equivalently, m and f ). Another key observation is that all the curves pass through two common points, P and Q . Since these curves correspond to different values of ξ d , the location of P and Q must depend only on m and f . Proceeding with this line of reasoning, the expression for H 2 can be written as 2

H2 =

2 2

2

2

2

a2 a1 ⁄ a2 + ξd a1 + ξd a2 ----------------------- = ----- --------------------------2 2 2 a4 a2 ⁄ a2 + ξ2 a3 + ξd a4 3 4 d

(4.80)

where the ‘a’ terms are functions of m , ρ , and f . Then, for H 2 to be independent of ξ d , the following condition must be satisfied a1 a3 ----- = ----a2 a4

(4.81)

The corresponding values for H 2 are H2

P, Q

=

a2 ----a4

(4.82)

282 Chapter 4: Tuned Mass Damper Systems

30

ξd = 1

m = 0.01 f = 1 25

ξd = 0 20

H2

P 15

0 < ξd < 1

Q 10

5

0 0.8

ρ1 0.85

0.9

0.95

ρ2 1

Ω ρ = ---ω

1.05

1.1

1.15

Fig. 4.15: Plot of H 2 versus ρ . Substituting for the ‘a’ terms in eqn (4.81), one obtains a quadratic equation for ρ 4 2 1 + 0.5m 2 2 ρ – ( 1 + m ) f + --------------------- ρ + f = 0 1+m

2

(4.83)

The two positive roots ρ 1 and ρ 2 are the frequency ratios corresponding to points P and Q . Similarly, eqn (4.82) expands to H2

1+m = -----------------------------------------2 P, Q 1 – ρ 1, 2 ( 1 + m )

(4.84)

Figure 4.15 shows different values for H 2 at points P and Q . For optimal behavior, one wants to minimize the maximum amplitude. As a first step, one requires the values of H 2 for ρ 1 and ρ 2 to be equal. This produces a distribution which is symmetrical about ρ 2 = 1 ⁄ ( 1 + m ) , as illustrated in Fig. 4.16. Then, by increasing the damping ratio, ξ d , one can lower the peak amplitudes until the peaks coincide with points P and Q . This state represents the optimal performance of the TMD system. A further increase in ξ d causes the peaks to merge and the amplitude to increase beyond the optimal value.

4.4 Tuned Mass Damper Theory for SDOF Systems 283

30

25

H2

20

o

ξd > ξd o

ξd < ξd

Q

P

15

ξd 10

opt

5

ρ1 0

0.85

0.9

0.95

opt 1

Ω ρ = ---ω

ρ2

opt

1.05

1.1

1.15

Fig. 4.16: Plot of H 2 versus ρ for f opt .

Requiring the amplitudes to be equal at P and Q is equivalent to the following condition on the roots 2

2

1 – ρ1 ( 1 + m ) = 1 – ρ2 ( 1 + m )

(4.85)

Then, substituting for ρ 1 and ρ 2 using eqn (4.83), one obtains a relation between the optimal tuning frequency and the mass ratio 1 – 0.5m f opt = ------------------------1+m

(4.86)

ωd

(4.87)

opt

= f opt ω

The corresponding roots and optimal amplification factors are ρ 1, 2 H2

opt

=

1 ± 0.5m -------------------------1+m

1+m = ---------------opt 0.5m

(4.88) (4.89)

284 Chapter 4: Tuned Mass Damper Systems The expression for the optimal damping at the optimal tuning frequency is ξd

opt

=

m ( 3 – 0.5m ) ------------------------------------------------8 ( 1 + m ) ( 1 – 0.5m )

(4.90)

Figures 4.17 through 4.20 show the variation of the optimal parameters with the mass ratio, m .

1

0.98

f opt

0.96

0.94

0.92

0.9

0.88 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

m

Fig. 4.17: Optimum tuning frequency ratio, f opt.

0.09

0.1

4.4 Tuned Mass Damper Theory for SDOF Systems 285

1.1

f opt 1.05

ρ2

ρ 1, 2

opt

1

opt

0.95

ρ1

opt

0.9

0.85

0.8 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.18: Input frequency ratios at which the response is independent of damping.

0.2 0.18 0.16 0.14

ξd

opt

0.12 0.1 0.08 0.06 0.04 0.02 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

m

Fig. 4.19: Optimal damping ratio for TMD.

0.09

0.1

286 Chapter 4: Tuned Mass Damper Systems

25

15

H2

opt

20

10

5

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.20: Maximum dynamic amplification factor for SDOF system (optimal tuning and damping). The response of the damper is defined by eqn (4.69). Specializing this equation for the optimal conditions leads to the plot of amplification versus mass ratio contained in Fig. 4.21. A comparison of the damper motion with respect to the motion of the primary mass for optimal conditions is shown in Fig. 4.22.

4.4 Tuned Mass Damper Theory for SDOF Systems 287

300

250

H4

opt

200

150

100

50

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.21: Maximum dynamic amplification factor for TMD.

20

f opt, ρ opt, ξ d

18

opt

16

H4 uˆ d ------ = -----H2 uˆ

14 12 10 8 6 4 2 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.22: Ratio of maximum TMD amplitude to maximum system amplitude.

288 Chapter 4: Tuned Mass Damper Systems Lastly, response curves for a typical mass ratio, m = 0.01 , and optimal tuning are plotted in Figs 4.23 and 4.24. The response for no damper is also plotted in Fig. 4.23. One observes that the effect of the damper is to limit the motion in a frequency range centered on the natural frequency of the primary mass and extending about 0.15ω . Outside of this range, the motion is not significantly influenced by the damper. 30

25

ξ d = 0.0 ξ d = 0.03 ξ d = 0.061 (optimal) ξ d = 0.1 No damper

m = 0.01 f opt = 0.9876

H2

20

P

15

Q

10

5

ρ1 0 0.8

0.85

0.9

0.95

ρ2

opt 1

Ω ρ = ---ω

opt

1.05

1.1

1.15

1.2

Fig. 4.23: Response curves for amplitude of system with optimally tuned TMD.

4.4 Tuned Mass Damper Theory for SDOF Systems 289

150

ξd ξd ξd ξd ξd

m = 0.01 f opt = 0.9876

0.0 0.03 0.061 (optimal) 0.1 0.2

H4

100

= = = = =

50

0

0.85

0.9

0.95

1

Ω ρ = ---ω

1.05

1.1

1.15

Fig. 4.24: Response curves for amplitude of optimally tuned TMD. The maximum amplification for a damped SDOF system without a TMD, undergoing harmonic excitation is given by eqn (1.32) 1 H = ------------------------2 2ξ 1 – ξ

(4.91)

Since ξ is small, a reasonable approximation is 1 H ≈ -----2ξ

(4.92)

Expressing the optimal H 2 in a similar form provides a measure of the equivalent damping ratio ξ e for the primary mass 1 ξ e = -------------------2H 2

(4.93)

opt

Figure 4.25 shows the variation of ξ e with the mass ratio. A mass ratio of 0.02 is equivalent to about 5% damping in the primary system.

290 Chapter 4: Tuned Mass Damper Systems

0.12

f opt, ρ opt, ξ d 0.1

opt

ξe

0.08

0.06

0.04

0.02

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.25: Equivalent damping ratio for optimally tuned TMD.

The design of a TMD involves the following steps: • Establish the allowable values of displacement of the primary mass and the TMD for the design loading. This data provides the design values for H 2 and H 4 . opt

opt

• Determine the mass ratios required to satisfy these motion constraints from Figs 4.20 and 4.21. Select the largest value of m . • Determine f opt form Fig. 4.17. • Compute ω d ω d = f opt ω

(4.94)

• Compute k d 2

2

k d = m d ω d = mk f opt • Determine ξ d from Fig. 4.19. opt

(4.95)

4.4 Tuned Mass Damper Theory for SDOF Systems 291 • Compute c d (4.96)

c d = 2ξ d ω d m d = m f opt 2ξ d ωm opt opt

Example 4.2: Design of a TMD for an undamped SDOF system Consider the following motion constraints < 7

(4.97)

H4 ----------------- < 6 H2

(4.98)

H2

opt

opt

Constraint eqn(4.97) requires m ≥ 0.05 . For constraint eqn(4.98), one needs to take m ≥ 0.02 . Therefore, m ≥ 0.05 controls the design. The relevant parameters are: m = 0.05

f opt = 0.94

ξd

m d = 0.05m

ω d = 0.94ω

k d = m f opt k = 0.044k

opt

= 0.135

Then 2

Damped structure - damped TMD All real systems contain some damping. Although an absorber is likely to be added only to a lightly damped system, assessing the effect of damping in the real system on the optimal tuning of the absorber is an important design consideration. The main system in Fig. 4.26 consists of the mass m , spring stiffness k , and viscous damping c . The TMD system has mass m d , stiffness k d , and viscous damping c d . Considering the system to be subjected to both external forcing and ground excitation, the equations of motion are m d u˙˙d + c d u˙ d + k d u d + m d u˙˙ = – m d a g

(4.99)

292 Chapter 4: Tuned Mass Damper Systems mu˙˙ + cu˙ + ku – c d u˙ d – k d u d = – ma g + p

(4.100)

p k

kd md

m c

cd

ug

ud + u + ug

u + ug

Fig. 4.26: Damped SDOF system coupled with a damped TMD system. Proceeding in the same way as for the undamped case, the solution due to periodic excitation (both p and ug) is expressed in polar form: aˆ g m iδ iδ pˆ u = --- H 5 e 5 – ----------H 6 e 6 k k aˆ g m –i δ iδ pˆ u d = --- H 7 e 7 – ----------H 8 e 8 k k

(4.101) (4.102)

The various H and δ terms are defined below 2

2 2

2

[ f – ρ ] + [ 2ξ d ρf ] H 5 = -----------------------------------------------------------D3 2

2 2

(4.103) 2

[ ( 1 + m ) f – ρ ] + [ 2ξ d ρf ( 1 + m ) ] H 6 = -------------------------------------------------------------------------------------------------D3

(4.104)

ρ2 H 7 = ---------D3

(4.105)

1 + [ 2ξρ ] 2 H 8 = ------------------------------D3

(4.106)

4.4 Tuned Mass Damper Theory for SDOF Systems 293 D 3 = { [ – f 2 ρ 2 m + ( 1 – ρ 2 ) ( f 2 – ρ 2 ) – 4ξξ d f ρ 2 ] 2

(4.107)

+ 4 [ ξρ ( f 2 – ρ 2 ) + ξ d fρ ( 1 – ρ 2 ( 1 + m ) ) 2 ] } δ5 = α1 – δ7

(4.108)

δ6 = α2 – δ7

(4.109)

δ8 = α3 – δ7

(4.110)

ξρ ( f 2 – ρ 2 ) + ξ d fρ ( 1 – ρ 2 ( 1 + m ) ) tan δ 7 = 2 ---------------------------------------------------------------------------------------------------– f 2 ρ 2 m + ( 1 – ρ 2 ) ( f 2 – ρ 2 ) – 4ξξ d f ρ 2

(4.111)

tan α 3 = 2ξρ

(4.112)

The α 1 and α 2 terms are defined by eqns 4.78 and 4.79. In what follows, the case of an external force applied to the primary mass is considered. Since D 3 involves ξ, one cannot establish analytical expressions for the optimal tuning frequency and optimal damping ratio in terms of the mass ratio. In this case, these parameters also depend on ξ . Numerical simulations can be applied to evaluate H 5 and H 7 for a range of ρ , given the values for m , ξ , f , and ξ d . Starting with specific values for m and ξ , plots of H 5 versus ρ can be generated for a range of f and ξ d . Each H 5 – ρ plot has a peak value of H 5 . The particular combination of f and ξ d that correspond to the lowest peak value of H 5 is taken as the optimal state. Repeating this process for different values of m and ξ produces the behavioral data needed to design the damper system. Figure 4.27 shows the variation of the maximum value of H 5 for the optimal state. The corresponding response of the damper is plotted in Fig. 4.28. Adding damping to the primary mass has an appreciable effect for small m . Noting eqns (4.101) and (4.102), the ratio of damper displacement to primary mass displacement is given by H7 uˆ d ρ2 --------- = ------- = -----------------------------------------------------------(4.113) H5 uˆ 2 2 2 2 [ f – ρ ] + [ 2ξ d ρf ] Since ξ is small, this ratio is essentially independent of ξ . Figure 4.29 confirms this statement. The optimal values of the frequency and damping ratios generated through simulation are plotted in Figs 4.30 and 4.31. Lastly, using eqn (4.93), can be converted to an equivalent damping ratio for the primary system. H5 opt

294 Chapter 4: Tuned Mass Damper Systems 1 ξ e = -------------------2H 5

(4.114)

opt

Figure 4.32 shows the variation of ξ e with m and ξ . Tsai & Lin (1993) suggest equations for the optimal tuning parameters f and ξ d determined by curve fitting schemes. The equations are listed below for completeness 2 1 – 0.5m f = ------------------------- + 1 – 2ξ – 1 1+m

(4.115)

– [ 2.375 – 1.034 m – 0.426m ]ξ m – ( 3.730 – 16.903 m + 20.496m )ξ 2 m 2 3m ------------------------------------------------- + ( 0.151ξ – 0.170ξ ) 8 ( 1 + m ) ( 1 – 0.5m )

ξd =

(4.116) 2

+ ( 0.163ξ + 4.980ξ ) m

40

ξ ξ ξ ξ ξ

35

30

= = = = =

0.0 0.01 0.02 0.05 0.1

H5

opt

25

20

15

10

5

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.27: Maximum dynamic amplification factor for damped SDOF system.

4.4 Tuned Mass Damper Theory for SDOF Systems 295

300

ξ ξ ξ ξ ξ

250

0.0 0.01 0.02 0.05 0.1

150

H7

opt

200

= = = = =

100

50

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.28: Maximum dynamic amplification factor for TMD.

20

ξ ξ ξ ξ ξ

18 16

H7 uˆ d ------ = -----H5 uˆ

14

= = = = =

0.0 0.01 0.02 0.05 0.1

12 10 8 6 4 2 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.29: Ratio of maximum TMD amplitude to maximum system amplitude.

296 Chapter 4: Tuned Mass Damper Systems

1

ξ ξ ξ ξ ξ

0.98 0.96 0.94

= = = = =

0.0 0.01 0.02 0.05 0.1

f opt

0.92 0.9 0.88 0.86 0.84 0.82 0.8 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.30: Optimum tuning frequency ratio for TMD, f opt.

0.2 0.18 0.16 0.14

0.1

ξd

opt

0.12

0.08

ξ ξ ξ ξ ξ

0.06 0.04 0.02 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

= = = = =

0.08

m

Fig. 4.31: Optimal damping ratio for TMD.

0.0 0.01 0.02 0.05 0.1 0.09

0.1

4.4 Tuned Mass Damper Theory for SDOF Systems 297

0.2 0.18 0.16 0.14

ξe

0.12 0.1 0.08

ξ ξ ξ ξ ξ

0.06 0.04 0.02 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

= = = = =

0.0 0.01 0.02 0.05 0.1 0.09

0.1

m

Fig. 4.32: Equivalent damping ratio for optimally tuned TMD.

Example 4.3: Design of a TMD for a damped SDOF system Example 4.2 is reworked here, allowing for 2% damping in the primary system. The same design motion constraints are considered: < 7

(4.117)

H7 ----------------- < 6 H5

(4.118)

H5

opt

opt

Using Fig. 4.27, the required mass ratio for ξ = 0.02 is m ≈ 0.03 . The other optimal values are f opt = 0.965 and ξ d = 0.105 . opt

Then m d = 0.03m

ω d = 0.955ω

2

k d = m f opt k = 0.027k

In this case, there is a significant reduction in the damper mass required for this

298 Chapter 4: Tuned Mass Damper Systems set of motion constraints. The choice between including damping in the primary system versus incorporating a tuned mass damper depends on the relative costs and reliability of the two alternatives, and the nature of the structural problem. A TMD system is generally more appropriate for upgrading an existing structure where access to the structural elements is difficult.

4.5 Case studies - SDOF systems Figures 4.33 to 4.44 show the time history responses for two SDOF systems with periods of 0.49s and 5.35s respectively under harmonic (at resonance conditions), El Centro, and Taft ground excitations. All examples have a system damping ratio of 2% and an optimally tuned TMD with a mass ratio of 1%. The excitation magnitudes have been scaled so that the peak amplitude of the response of the system without the TMD is unity. The plots show the response of the system without the TMD (the dotted line) as well as the response of the system with the TMD (the solid line). Figures showing the time history of the relative displacement of the TMD with respect to the system are also presented. Significant reduction in the response of the primary system under harmonic excitation is observed. However, optimally tuned mass dampers are relatively ineffective under seismic excitation, and in some cases produce a negative effect, i.e. they amplify the response slightly. This poor performance is attributed to the ineffectiveness of tuned mass dampers for impulsive loadings as well as their inability to reach a resonant condition and therefore dissipate energy under random excitation. These results are in close agreement with the data presented by Kaynia et al. (1981).

4.5 Cases Studies - SDOF Systems 299

1 0.8

T = 0.49s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.33: Response of SDOF to harmonic excitation.

6

4

T = 0.49s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

60

Time - s

Fig. 4.34: Relative displacement of TMD under harmonic excitation.

300 Chapter 4: Tuned Mass Damper Systems

1 0.8

T = 0.49s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.35: Response of SDOF to El Centro excitation.

6

4

T = 0.49s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

60

Time - s

Fig. 4.36: Relative displacement of TMD under El Centro excitation.

4.5 Cases Studies - SDOF Systems 301

1 0.8

T = 0.49s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.37: Response of SDOF to Taft excitation.

6

4

T = 0.49s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

Time - s

Fig. 4.38: Relative displacement of TMD under Taft excitation.

60

302 Chapter 4: Tuned Mass Damper Systems

1 0.8

T = 5.35s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.39: Response of SDOF to harmonic excitation.

6

4

T = 5.35s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

60

Time - s

Fig. 4.40: Relative displacement of TMD under harmonic excitation.

4.5 Cases Studies - SDOF Systems 303

1 0.8

T = 5.35s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.41: Response of SDOF to El Centro excitation.

6

4

T = 5.35s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

60

Time - s

Fig. 4.42: Relative displacement of TMD under El Centro excitation.

304 Chapter 4: Tuned Mass Damper Systems

1 0.8

T = 5.35s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.43: Response of SDOF to Taft excitation.

6

4

T = 5.35s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

Time - s

Fig. 4.44: Relative displacement of TMD under Taft excitation.

60

4.6 Tuned Mass Damper Theory for MDOF Systems 305

4.6 Tuned mass damper theory for MDOF systems The theory of a SDOF system presented earlier is extended here to deal with a MDOF system having a number of tuned mass dampers located throughout the structure. Numerical simulations, which illustrate the application of this theory to the set of example building structures used as the basis for comparison of the different schemes throughout the text are presented in the next section.

p2

p1

ug

md

m2

m1 c1

kd

k2

k1

c2 u1 + ug

cd u2 + ug

u2 + ug + ud

Fig. 4.45: 2DOF system with TMD.

A 2DOF system having a damper attached to mass 2 is considered first to introduce the key ideas. The governing equations for the system shown in Fig. 4.45 are: m 1 u˙˙1 + c 1 u˙ 1 + k 1 u 1 – k 2 ( u 2 – u 1 ) – c 2 ( u˙ 2 – u˙ 1 ) = p 1 – m 1 u˙˙g

(4.119)

m 2 u˙˙2 + c 2 ( u˙ 2 – u˙ 1 ) + k 2 ( u 2 – u 1 ) – k d u d – c d u˙ d = p 2 – m 2 u˙˙g

(4.120)

m d u˙˙d + k d u d + c d u˙ d = – m d ( u˙˙2 + u˙˙g )

(4.121)

The key step is to combine eqns (4.119) and (4.120) and express the resulting equation in a form similar to the SDOF case defined by eqn (4.100). This operation reduces the problem to an equivalent SDOF system, for which the theory of Section 4.4 is applicable. The approach followed here is based on transforming the original matrix equation to scalar modal equations. Introducing matrix notation, eqns (4.119) and (4.120) are written as

306 Chapter 4: Tuned Mass Damper Systems

˙˙ + CU˙ + KU = MU

p1 – m1 ag p2 – m2 ag

+

0 k d u d + c d u˙ d

(4.122)

where the various matrices are

U =

u1

M =

m1

(4.123)

u2

(4.124)

m2

K =

k1 + k2 –k2 –k2 k2

(4.125)

C =

c1 + c2 –c2 –c2 c2

(4.126)

One substitutes for U in terms of the modal vectors and coordinates U = Φ1 q1 + Φ2 q2

(4.127)

The modal vectors satisfy the following orthogonality relations (see eqn (2.257)) T

2

T

Φ j KΦ i = δ ij ω j Φ j MΦ i

(4.128)

Defining modal mass, stiffness, and damping terms, ˜ j = Φ T MΦ m j j

(4.129)

T 2 ˜j k˜ j = Φ j KΦ j = ω j m

(4.130)

T c˜ j = Φ j CΦ j

(4.131)

expressing the elements of Φ j as

4.6 Tuned Mass Damper Theory for MDOF Systems 307 Φj =

Φ j1

(4.132)

Φ j2

and assuming damping is proportional to stiffness C = αK

(4.133)

one obtains a set of uncoupled equations for q 1 and q 2 ˜ j q˙˙ + c˜ j q˙ + k˜ j q = Φ ( p – m a ) m j = 1,2 j j j j1 1 1 g

(4.134)

+ Φ j2 ( p 2 – m 2 a g + k d u d + c d u˙ d ) With this assumption, the modal damping ratio is given by αω j c˜ j ξ j = ---------------- = ---------˜j 2 2ω j m

(4.135)

Equation (4.134) represents two equations. Each equation defines a ˜ , k˜ , and particular SDOF system having mass, stiffness, and damping equal to m ξ . Since a TMD is effective for a narrow frequency range, one has to decide on which modal resonant response is to be controlled with the TMD. Once this decision is made, the analysis can proceed using the selected modal equation and the initial equation for the TMD, i.e. eqn (4.121). Suppose the first modal response is to be controlled. Taking j = 1 in eqn (4.133) leads to ˜ 1 q˙˙ + c˜ 1 q˙ + k˜ 1 q = Φ p + Φ p m 1 1 1 11 1 12 2

(4.136)

– [ m 1 Φ 11 + m 2 Φ 12 ]a g + Φ 12 [ k d u d + c d u˙ d ] In general, u 2 is obtained by superposing the modal contributions u 2 = Φ 12 q 1 + Φ 22 q 2

(4.137)

However, when the external forcing frequency is close to ω 1 , the first mode response will dominate, and it is reasonable to assume u 2 ≈ Φ 12 q 1 Solving for q 1

(4.138)

308 Chapter 4: Tuned Mass Damper Systems 1 q 1 = --------- u 2 Φ 12

(4.139)

and then substituting in eqn (4.136), one obtains ˜ 1e u˙˙2 + c˜ 1e u˙ 2 + k˜ 1e u = k u + c u˙ d m 2 d d d

(4.140)

˜ 1e a + p˜ 1e – Γ 1e m g ˜ 1e , c˜ 1e , k˜ 1e , p˜ 1e , and Γ represent the equivalent SDOF parameters where m 1e for the combination of mode 1 and node 2, the node at which the TMD is attached. Their definition equations are 1 ˜ ˜ 1e = --------m1 m 2 Φ 12

(4.141)

1 k˜ 1e = --------- k˜ 1 2 Φ 12

(4.142)

c˜ 1e = αk˜ 1e

(4.143)

Φ 11 p 1 + Φ 12 p 2 p˜ 1e = -------------------------------------Φ 12 Φ 12 Γ 1e = --------- ( m 1 Φ 11 + m 2 Φ 22 ) ˜1 m

(4.144) (4.145)

Equations 4.121 and 4.140 are similar in form to the SDOF equations treated in the previous section. Both set of equations are compared below. TMD equation m d u˙˙d + c d u˙ d + k d u d = – m d ( u˙˙ – a g ) vs m d u˙˙d + c d u˙ d + k d u d = – m d ( u˙˙2 – a g )

(4.146)

4.6 Tuned Mass Damper Theory for MDOF Systems 309 Primary mass equation mu˙˙ + cu˙ + ku = c d u˙ d + k d u d + p – ma g vs ˜ ˜ 1e u˙˙2 + c˜ 1e u˙ 2 + k˜ 1e u = c u˙ d + k u + p˜ – Γ m m 1e 2 d d d 1e 1e a g Taking

u2 ≡ u

˜ 1e ≡ m m

c˜ 1e ≡ c

p˜ 1e ≡ p

Γ 1e ≡ Γ

k˜ 1e ≡ k

(4.147)

(4.148)

transforms the primary mass equation for the MDOF case to mu˙˙ + cu˙ + ku = c d u˙ d + k d u d + p – Γma g

(4.149)

which differs from the corresponding SDOF equation by the factor Γ. Therefore, the solution for ground excitation generated earlier has to be modified to account for the presence of Γ. The “generalized” solution is written in the same form as the SDOF case. One needs only to modify the terms associated with a g , i.e., H6, H8 and δ6 , δ8 . Their expanded form is listed below. 2 2

2

2

[ ( Γ + m ) f – Γρ ] + [ 2ξ d ρf ( Γ + m ) ] H 6 = ------------------------------------------------------------------------------------------------------D3 2

2

(4.150)

2

[ 1 + ρ ( Γ – 1 ) ] + [ 2ξρ ] H 8 = --------------------------------------------------------------------D3

(4.151)

2ξ d fρ ( Γ + m ) tan a 2 = -----------------------------------------f 2 ( Γ + m ) – Γρ 2

(4.152)

2ξρ tan a 3 = --------------------------------1 + ( Γ – 1 )ρ 2

(4.153)

δ6 = a2 – δ7

(4.154)

δ8 = a3 – δ7

(4.155)

310 Chapter 4: Tuned Mass Damper Systems where D 3 is defined by eqn (4.107), and δ 7 is given by eqn (4.111). From this point on, one proceeds as described in Section 4.4. The mass ratio is defined in terms of the equivalent SDOF mass. md m = --------˜ 1e m

(4.156)

Given m and ξ 1 , one finds the tuning frequency and damper damping ratio using Figs 4.30 and 4.31. The damper parameters are determined with ˜ 1e md = m m

(4.157)

ω d = f opt ω 1

(4.158)

c d = 2ξ d ωd md opt

(4.159)

Expanding the expression for the damper mass, T

m [ Φ 1 MΦ 1 ] ˜ 1e = -----------------------------md = m m 2 Φ 12

(4.160)

shows that one should select the TMD location to coincide with the maximum amplitude of the mode shape that is being controlled. In this case, the first mode is the target mode, and Φ 12 is the maximum amplitude for Φ 1 . This derivation can be readily generalized to allow for tuning on the i th modal frequency. One writes eqn (4.139) as 1 q i ≈ -------- u 2 Φ i2

(4.161)

where i is either 1 or 2 . The equivalent parameters are 1 ˜ ˜ ie = -------mi m 2 Φ i2 2 ˜ ie k˜ ie = ω i m

˜ ie and ξ , one specifies m and finds the optimal tuning with Given m i

(4.162)

(4.163)

4.6 Tuned Mass Damper Theory for MDOF Systems 311 ω d = f opt ω i

(4.164)

Example 4.4: Design of a TMD for a damped MDOF system To illustrate the above procedure, a two DOF system having m 1 = m 2 = 1 is considered. Designing the system for a fundamental period of T 1 = 1s and a uniform deformation fundamental mode profile yields the following stiffnesses (refer to Example 1.6) 2

k 1 = 12π = 118.44 2

k 2 = 8π = 78.96 Requiring a fundamental mode damping ratio of 2% , and taking damping proportional to stiffness ( C = αK ), the corresponding α is 2ξ 1 0.02 α = --------- = ---------- = 0.0064 ω1 π The mass, stiffness, and damping matrices for these design conditions are: M = 1 0 0 1 K = 197.39 – 78.96 – 78.96 78.96 C =

1.26 – 0.51 – 0.51 0.51

Performing an eigenvalue analysis yields the following frequencies and mode shapes ω 1 = 6.28r/s Φ 1 = 0.5 1.0

ω 2 = 15.39r/s Φ2 =

1.0 – 0.5

The corresponding modal mass, stiffness, and damping terms are: ˜ 1 = Φ T MΦ = 1.25 m 1 1

˜ 2 = Φ T MΦ = 1.25 m 2 2

312 Chapter 4: Tuned Mass Damper Systems T k˜ 1 = Φ 1 KΦ 1 = 39.48

T k˜ 2 = Φ 2 KΦ 2 = 236.87

T c˜ 1 = Φ 1 CΦ 1 = 0.25 c˜ 1 ξ 1 = ----------------- = 0.02 ˜1 2ω 1 m

T c˜ 2 = Φ 2 CΦ 2 = 1.51 c˜ 2 ξ 2 = ----------------- = 0.049 ˜2 2ω 2 m

The optimal parameters for a TMD having a mass ratio of 0.01 and tuned to a specific mode are listed below. Mode 1 - optimum location is node 2 f opt = 0.982

ξd

m d = 0.0125

k d = 0.4754

opt

= 0.062 c d = 0.0096

Mode 2 - optimum location is node 1 f opt = 0.972

ξd

m d = 0.05

k d = 11.1894

opt

= 0.068 c d = 0.1017

The general case of a MDOF system with a tuned mass damper connected to the nth degree of freedom is treated in a similar manner. Using the notation defined above, the jth modal equation can be expressed as ˜ j q˙˙ + c˜ j q˙ + k˜ j q = p˜ + Φ [ k u + c u˙ d ] m j j j j jn d d d

j = 1,2,...

(4.165)

where p˜ j denotes the modal force due to ground motion and external forcing, and Φ jn is the element of Φ j corresponding to the nth displacement variable. To control the ith modal response, one sets j = i in eqn (4.165), and introduces the approximation 1 q i ≈ --------- u n Φ in This leads to the following equation for u n ˜ ie u˙˙n + c˜ ie u˙ n + k˜ ie u = p˜ + k u + c u˙ d m ie n d d d where

(4.166)

(4.167)

4.6 Tuned Mass Damper Theory for MDOF Systems 313 (4.168)

T 1 ˜ 1 ˜ ie = -------- M i = --------- Φ i MΦ i m 2 2 Φ in Φ in 2 ˜ ie k˜ ie = ω i m

(4.169)

c˜ ie = αk˜ ie

(4.170)

1 p˜ ie = --------- p˜ i Φ in

(4.171)

The remaining steps are the same as described above. One specifies m and ξ i , determines the optimal tuning and damping values with Figs 4.30 and 4.31, and then computes m d and ω d . T m ˜ ie = -------- Φ i MΦ i md = m m 2 Φ in

(4.172)

ω d = f opt ω i

(4.173)

The optimal mass damper for mode i is obtained by selecting n such that Φ in is the maximum element in Φ i .

Example 4.5: Design of TMD’s for a simply supported beam y,u

P*

x*

EI constant

x L Figure 1 Consider the simply supported beam shown in Figure 1. The modal shapes

314 Chapter 4: Tuned Mass Damper Systems and frequencies for the case where the cross sectional properties are constant and the transverse shear deformation is negligible are: nπx Φ n ( x ) = sin ---------L

(1)

EI nπ 4 ω n2 = ------- ------ ρm L

(2)

n = 1, 2, … One obtains a set of N equations in terms of N modal coordinates by expressing the transverse displacement, u(x,t), as N

u =

∑

q i ( t )Φ j ( x )

(3)

j=1

and substituting for u in the Principle of Virtual Displacements specialized for negligible transverse shear deformation (see eqn (2.194)), L

∫

Mδχ dx =

0

∫

bδu dx

(4)

Substituting for δχ , 2

d δχ = – ( δu ) dx2

(5)

and taking δu = δq j Φ j leads to the following equations

(6)

4.6 Tuned Mass Damper Theory for MDOF Systems 315

∫

∫

– MΦ j, xx dx =

bΦ j dx

(7)

j = 1, 2, …, N

Lastly, one substitutes for M and b in terms of Φ and q, and evaluates the integrals. The expressions for M and b are N

M = EIχ = – EI

∑

q l Φ l, xx

(8)

l=1

N

b = – ρ m u˙˙ + b ( x, t ) = – ρ m

∑

Φ l q˙˙l + b ( x, t )

(9)

l=1

Noting the orthogonality properties of the modal shape functions, L

∫

L Φ j Φ k dx = δ jk --2

(10)

0

L

∫

jπ 4 L Φ j, xx Φ k, xx dx = ----- δ jk -- L 2

(11)

0

the modal equations uncouple and reduce to ˜ j q˙˙ + k˜ j q = p˜ m j j j

(12)

Lρ m ˜ j = ---------m 2

(13)

jπ 4 L k˜ j = EI ----- -- L 2

(14)

where

316 Chapter 4: Tuned Mass Damper Systems L

p˜ j =

∫

jπx b sin --------- dx L

(15)

0

When the external loading consists of a concentrated force applied at the location x = x * (see Fig 1), the corresponding modal load for the j’th mode is jπx * p˜ j = P * sin ----------L

(16)

In this example, the force is considered to be due to a mass attached to the beam as indicated in Fig 2. The equations for the tuned mass and the force are m d ( u˙˙* + u˙˙d ) + k d u d + c d u˙ d = 0

(17)

m d ( u˙˙* + u˙˙d ) = – P *

(18)

u*+ud

md kd

cd

u*

x* Figure 2 Suppose one wants to control the i’th modal response with a tuned mass damper attached at x = x * . Taking j equal to i in eqns (12) and (13), the i’th modal equation has the form iπx * ˜ i q˙˙i + k˜ i q = ( k u + c u˙ d ) sin ---------m i d d d L

(19)

4.6 Tuned Mass Damper Theory for MDOF Systems 317 Assuming the response is dominated by the i’th mode, u * ( x *, t ) is approximated by iπx * u * ( x *, t ) ≈ q i sin ---------L

(20)

and eqn (19) is transformed to an equation relating u * and u d . ˜ ie u˙˙* + k˜ ie u * = k u + c u˙ d m d d d

(21)

where 1 ˜i ˜ ie = ---------------------------m m 2 * iπx sin --------- L

(22)

The remaining steps utilize the results generated for the SDOF undamped ˜ ie and k˜ ie structure - damped TMD system considered in section 4.3. One uses m as the mass and stiffness parameters for the primary system. To illustrate the procedure, consider the damper to be located at mid-span, and the first mode is to be controlled. Taking i=1 and x * = L ⁄ 2 , the corresponding parameters are iπx * sin ---------- = 1 L

(23)

Lρ m ˜ 1 = ---------˜ ie = m m 2

(24)

EIL π 4 k˜ ie = k˜ 1 = --------- --- 2 L

(25)

One specifies the equivalent damping ratio, ξ e , and determines the required mass ratio from Fig 4.32. For example, taking ξ e = 0.06 requires m = 0.03 . The other parameters corresponding to m = 0.03 follow from Figs 4.29, 4.30, and 4.31. ωd f opt = ------ = 0.965 ω1

(26)

318 Chapter 4: Tuned Mass Damper Systems ξd

opt

= 0.105

uˆ d ------ = 5 uˆ *

(27)

(28)

Using these parameters, the corresponding expression for the damper properties are ˜1 m d = 0.03m

(29)

ω d = 0.965ω 1

(30)

k d = ω d2 m d

(31)

c d = 2ξ d ω d m d

(32)

˜ 1 and ω are specified, the damper properties can be evaluated. Once m 1 For example, consider the beam to be a steel beam having the following properties L = 20m ρ m = 1000k g ⁄ m I = 8 × 10 – 4 m 4

(33)

E = 2 × 10 11 N ⁄ m 2 The beam parameters are ˜ 1 = 10, 000kg m ω 1 = 9.87r ⁄ s Applying eqns (29)-(32), results in

(34)

4.6 Tuned Mass Damper Theory for MDOF Systems 319 m d = 300kg ω d = 9.52r ⁄ s k d = 27, 215N ⁄ m

(35)

c d = 599.8Ns ⁄ m The total mass of the girder is 20,000kg. Adding 300kg, which is just 1.5% of the total mass, produces an effective damping ratio of 0.06 for the first mode response. The mode shape for the second mode has a null point at x=L/2, and therefore locating a tuned mass at this point would have no effect on the second modal response. The optimal locations are x * = L ⁄ 4 and x * = 3L ⁄ 4 . Taking x * = L ⁄ 4 and i=2, one obtains iπx * sin ---------- = 1 L

(36)

Lρ m ˜ 2e = m ˜ 2 = ---------m 2

(37)

π 4 k˜ 2e = k˜ 2 = 8EIL --- L

(38)

16EI π 4 ω 22 = ------------ --- ρ m L

(39)

The procedure from here on is the same as before. One specifies ξ e , determines the required mass ratio, and then the frequency and damping parameters. It is of interest to compare the damper properties corresponding to the same equivalent damping ratio. Taking ξ e = 0.06 , the damper properties for the example steel beam are m d = 300kg

(40)

k d = 435, 440N ⁄ m

(41)

c d = 2400Ns ⁄ m

(42)

320 Chapter 4: Tuned Mass Damper Systems The required damper stiffness is an order of magnitude greater than the corresponding value for the first mode response.

4.7 Case studies - MDOF systems This section presents shear deformation profiles for the standard set of building examples defined in Table 2.4. A single TMD is placed at the top floor and tuned to either the first or second mode. The structures are subjected to harmonic ground acceleration with a frequency equal to the fundamental frequency of the buildings, as well as scaled versions of El Centro and Taft ground accelerations. As expected, significant reduction in the response is observed for the harmonic excitations (see Figs 4.46 through 4.49). The damper is generally less effective for seismic excitation versus harmonic excitation (see Figs 4.50 through 4.61). Results for the low period structures show more influence of the damper which is to be expected since the response is primarily due to the first mode. This data indicates that a TMD is not the optimal solution for controlling the motion due to seismic excitation.

4.6 Case Studies - MDOF Systems 321

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% Harmonic TMD - Mode 1

x Normalized height ---H

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.46: Maximum shear deformation for Building 1.

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% Harmonic TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.47: Maximum shear deformation for Building 2.

0.01

322 Chapter 4: Tuned Mass Damper Systems

1

Building #3 Quadratic based Initial H = 100m ρ = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% Harmonic TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.48: Maximum shear deformation for Building 3.

1

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% Harmonic TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.49: Maximum shear deformation for Building 4.

0.01

4.6 Case Studies - MDOF Systems 323

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 1

x Normalized height ---H

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5%

0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.50: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.51: Maximum shear deformation for Building 1.

0.01

324 Chapter 4: Tuned Mass Damper Systems

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.52: Maximum shear deformation for Building 2.

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.53: Maximum shear deformation for Building 2.

0.01

4.6 Case Studies - MDOF Systems 325

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

Building #3 Quadratic based Initial H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 1

0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.54: Maximum shear deformation for Building 3.

1 0.9

Building #3 Quadratic based Initial H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 1

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.55: Maximum shear deformation for Building 3.

0.01

326 Chapter 4: Tuned Mass Damper Systems

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

Building #3 Quadratic based Initial H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 2

0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.56: Maximum shear deformation for Building 3.

1 0.9

Building #3 Quadratic based Initial H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 2

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.57: Maximum shear deformation for Building 3.

0.01

4.6 Case Studies - MDOF Systems 327

1 0.9

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 1

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.58: Maximum shear deformation for Building 4.

1

x Normalized height ---H

0.9

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 1

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.59: Maximum shear deformation for Building 4.

0.01

328 Chapter 4: Tuned Mass Damper Systems

1 0.9

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 2

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.60: Maximum shear deformation for Building 4.

1

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 2

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.61: Maximum shear deformation for Building 4.

0.01

Problems 329

Problems Problem 4.1 Verify eqns (4.13) through (4.17). Hint: express p, u, and ud in complex form p = pˆ e iΩt u = ue iΩt u d = u d e iΩt and solve eqns (4.6) and (4.7) for u and u d . Then take u = uˆ e

iδ 1

u d = uˆ d e

i ( δ1 + δ2 )

ω = ωd = Ω

Problem 4.2 Refer to eqns (4.14) and (4.20). Express ξ e as a function of m , ξ , and ˆu ⁄ uˆ d . Take ξ = 0.05 , and plot ξ e vs. m for a representative range of the magnitude of the displacement ratio, uˆ ⁄ uˆ d .

Problem 4.3 Figure 4.7 illustrates an active tuned mass damper configuration. The damper can be modelled with the 2 DOF system shown below. The various terms are: u s is the total displacement of the support attached to the floor beam; F a is the self equilibrating force provided by the actuator; m d, k d, c d are parameters for the damper mass; k a and m a are parameters for the auxillary mass.

330 Chapter 4: Tuned Mass Damper Systems

ka

kd

R

Fa cd us

md

ma us + ud

us + ud + ua

a) Derive the governing equation for m d and m a . Also determine an expression for the resultant force, R, that the system applies to the floor beam. b) Consider m a to be several orders of magnitude smaller than m d , eg, m a = 0.01m d . Also take the actuator force to be a linear function of the relative velocity of the damper mass. F a = c a u˙ d Specialize the equations for this case. How would you interpret the contribution of the actuator force to the governing equation for the damper mass?

Problem 4.4 Design a pendulum damper system having a natural period of 6 seconds and requiring less than 4 meters of vertical space.

Problems 331 Problem 4.5

L1 m1

L2

m2

The pendulum shown above is connected to a second mass which is free to move horizontally. The connection between mass 1 and mass 2 carries only shear. Derive an equation for the period of the compound pendulum and the length of an equivalent simple pendulum. Assume the links are rigid.

Problem 4.6 Refer to Fig 4.12. Establish the equations of motion for the mass, m d , considering θ to be small. Verify that the equivalent stiffness is equal to W d ⁄ R .

Problem 4.7 Refer to Fig 4.15 and eqn (4.84). Derive the corresponding expression for P, Q starting with eqn (4.70) and using the same reasoning strategy. Considering the mass ratio, m , to be less than 0.03, estimate the difference in the “optimal” values for the various parameters. H1

Problem 4.8 Generate plots of H 1 vs. ρ for ξ d ranging from 0 to 0.2, m = 0.01 , and f = 0.9876 . Compare the results with the plots shown in Fig 4.23.

332 Chapter 4: Tuned Mass Damper Systems Problem 4.9 Consider a system composed of an undamped primary mass and a tuned mass damper. The solution for periodic force excitation is given by (see eqns (4.60) to (4.79)) u = ue iΩt

(1)

u d = u d e iΩt

(2)

iδ p u = --- H 1 e 1 k

(3)

iδ p u d = --- H 3 e 3 k

(4)

2

2 2

2

[ f – ρ ] + [ 2ξ d ρf ] H 1 = -----------------------------------------------------------D2

(5)

2

ρ H 3 = ---------D2 D2 =

(6)

2

2

2

2 2 2

2

( [ 1 – ρ ] [ f – ρ ] – mρ f ) + ( 2ξ d ρf [ 1 – ρ ( 1 + m ) ] )

2

(7)

The formulation for the optimal damper properties carried out in Section 4.3 was based on minimizing the peak value of H1 (actually H2 but H1 behaves in a similar way), i.e., on controlling the displacement of the primary mass. Suppose the design objective is to control the acceleration of the primary mass. Noting eqns (1) and (3), the acceleration is given by u˙˙ = a = ae iΩt

(8)

i(δ + π) pΩ 2 a = ----------H 1 e 1 k

(9)

Substituting for k transforms eqn (9) to

Problems 333 i(δ + π) p a = ----H1′ e 1 m

(10)

H1′ = ρ 2 H 1

(11)

where

Investigate the behavior of H1′ with ρ, f , m and ξ d . If it behaves similar to H 2 as shown in Fig 4.15, describe how you would establish the optimal values for the various parameters, and also how you would design a tuned mass system when H1′ is specified.

Problem 4.10 Design a TMD for a damped SDOF system having ξ = 0.02 . The design motion constraints are: a) H5

opt

< 10

H7 ---------------- < 5 H5 opt

b) H5

opt

<5

H7 ---------------- < 5 H5 opt

c) Repeat part b considering ξ to be equal to 0.05.

Problem 4.11 This problem concerns the design of a tuned-mass damper for a damped

334 Chapter 4: Tuned Mass Damper Systems single degree of freedom system. The performance criteria are: ξ eq = 0.1

uˆ d ⁄ uˆ = 5

a) Determine the damper properties for a system having m = 10,000 kg and k = 395 kN/m for the following values of ξ : • ξ = 0.02 • ξ = 0.05 b) Will the damper be effective for an excitation with frequency 2.5π rad/sec ? Discuss the basis for you conclusion.

Problem 4.12 Refer to example 3.7. Suppose a tuned mass damper is installed at the top level (at mass 5). a) Determine the damper properties such that the equivalent damping ratio for the fundamental mode is 0.16. Use the values of m , k , c from example 3.7. Assume stiffness proportional damping for c. b) u5

m5 L md

k 5, c 5

Consider the tuned mass damper to be a pendulum attached to m 5 . Determine m d and L for the damper properties established in part a. c) Repeat part a for the case where the mass damper is tuned for the second mode rather than for the first mode. Use the same values of m , k , c and assume stiffness proportional damping.

Problems 335 Problem 4.13 Consider a cantilever shear beam with the following properties: • H = 50 m • ρ m = 20, 000 kg/m 5 0.6x • D T = 8 ×10 1 – ---------- kN H

a) Model the beam as a 10 DOF discrete shear beam having 5m segments. Determine the first 3 mode shapes and frequencies. Normalize the mode shapes such that the peak amplitude is unity for each mode. b) Design tuned mass dampers to provide an effective modal damping ratio of 0.10 for the first and third modes. Take ξ 1 = 0.02 and assume modal damping is proportional to stiffness. Note: you need to first establish the ‘‘optimal location’’ of the tuned mass dampers for the different modes.

Problem 4.14 Consider a simply supported steel beam having the following properties L = 30m ρ m = 1500k g ⁄ m –2

I = 1 ×10 m 4 a) Design tuned mass damper systems that provide an equivalent damping of 0.05 for each of the first 3 modes. b) Repeat part a) with the constraint that an individual damper mass cannot exceed 300kg. Hint: utilize symmetry of a particular mode shape to locate a pair of dampers whose function is to control that mode.

336 Chapter 4: Tuned Mass Damper Systems Problem 4.15 15 m

15 m constant EI W u1

Consider the simply supported beam shown above. The beam has a uniform weight of 15 kN/m and a concentrated weight at mid span of 100 kN. The flexural rigidity is constant and equal to 200,000 kN-m2. a) Assume the first mode can be approximated by: π u = u 1 sin --- x L Determine the governing equation for u 1 using the Principle of Virtual Displacements. b) Design a tuned mass damper to provide an equivalent damping ratio of 0.05 for the first mode. Assume no damping for the beam itself. c) Will the damper designed in part b be effective for the second mode? Explain your answer.

Problem 4.16 Refer to problem 3.25 part (b). Suggest a tuned mass damper for generating the required energy dissipation.

337

Chapter 5

Base isolation systems 5.1 Introduction The term isolation refers to the degree of interaction between objects. An object is said to be isolated if it has little interaction with other objects. The act of isolating an object involves providing an interface between the object and its neighbors which minimizes interaction. These definitions apply directly to various physical systems. For example, one speaks of isolating a piece of equipment from its support by mounting the equipment on an isolation system which acts as a buffer between the equipment and the support. The design of isolation systems for vibrating machinery is a typical application. The objective here is to minimize the effect of the machine induced loading on the support. Another application is concerned with minimizing the effect of support motion on the structure. This issue is becoming increasingly more important for structures containing motion sensitive equipment and also for structures located adjacent to railroad tracks or other sources of ground disturbance. Although isolation as a design strategy for mounting mechanical equipment has been employed for over seventy years, only recently has the concept been seriously considered for civil structures, such as buildings and bridges, subjected to ground motion. This type of excitation interacts with the structure at the foundation level, and is transmitted up through the structure. Therefore, it is logical to isolate the structure at its base, and prevent the ground

338 Chapter 5: Base Isolation Systems motion from acting on the structure. The idea of seismic isolation dates back to the late nineteenth century, but the application was delayed by the lack of suitable commercial isolation components. Substantial development has occurred since the mid 1980’s (Naeim and Kelly, 1999), and base isolation for certain types of civil structures is now considered to be a highly viable design option by the seismic engineering community, particularly in Japan (Wada, 1998), for moderate to extreme seismic excitation. A set of simple examples are presented in the next section to identify the key parameters and illustrate the quantitative aspects of base isolation. This material is followed by a discussion of practical aspects of seismic base isolation and a description of some seismically isolated buildings. The remaining sections deal with the behavioral and design issues for base isolated MDOF structural systems. Numerical results illustrating the level of performance feasible with seismic base isolation are included to provide a basis of comparison with the other motion control schemes considered in this text.

5.2 Isolation for SDOF systems The application of base isolation to control the motion of a SDOF system subjected to ground motion was discussed earlier in Section 1.3 as part of a general treatment of design for dynamic excitation. The analytical formulation developed in that section provides the basis for designing an isolation system for simple structures that can be accurately represented with a SDOF model. Examples illustrating the reasoning process one follows are presented below. The formulation is also extended to deal with a modified version of a SDOF model that is appropriate for a low-rise building isolated at its base. This model is useful for preliminary design. SDOF examples The first example considers external periodic forcing of the SDOF system shown in Fig. 5.1. The solution of this problem is contained in Section 1.3. For convenience, the relevant equations are listed below:

5.2 Isolation for SDOF Systems 339

k R

m

p

c u Fig. 5.1: SDOF system. p = pˆ sin Ωt

(5.1)

u = uˆ sin ( Ωt – δ )

(5.2)

H1 uˆ = ------- pˆ k

(5.3)

1 H 1 = -------------------------------------------------2 2

[ 1 – ρ ] + [ 2ξρ ]

(5.4)

2

Ω ρ = ---ω

(5.5)

2ξρ tan δ = --------------2 1–ρ

(5.6)

Given pˆ and Ω , one can determine uˆ for a specific system having mass m , stiffness k , and damping c . With uˆ known, the forces in the spring and damper can be evaluated. The reaction can be found by either summing the internal forces, or combining p with the inertia force. With the latter approach, one writes R = p – mu˙˙

(5.7)

and expands the various terms using eqns (5.1) through (5.6). The result is expressed as ˆ sin ( Ωt – δ ) R = R r

(5.8)

ˆ = H pˆ R 3

(5.9)

340 Chapter 5: Base Isolation Systems 2

1 + [ 2ξρ ] ---------------------------------------------2 2 2 [ 1 – ρ ] + [ 2ξρ ]

H3 =

(5.10)

2

ρ H 1 sin δ tan δ r = – -----------------------------------2 1 + ρ H 1 cos δ

(5.11)

The function, H3, is referred to as the transmissibility of the system. It is a measure of how much of the load p is transmitted to the support. When ξ = 0 , δ = 0 and H 3 reduces to H 1. Figure 5.2 shows the variation of H 3 with ρ and ξ .

5 4.5

ξ = 0 4 3.5

H3

3 2.5

ξ = 0.2

2 1.5

ξ = 0.4

1 0.5 0 0

2 0.2

0.4

0.6

0.8

1

Ω ρ = ---ω

1.2

1.4

1.6

1.8

2

Fig. 5.2: Plot of H 3 versus ρ . The model presented above can be applied to the problem of designing a support system for a machine with an eccentric rotating mass. Here, one wants to minimize the reaction force for a given pˆ , i.e. one takes H 3 < 1 . Noting Fig. 5.2, this constraint requires the frequency ratio, ρ , to be greater than 2 , and it follows that

5.2 Isolation for SDOF Systems 341 Ω ω < ------2

(5.12)

The corresponding periods are related by T > Tf 2 =

2π 2 ------ Ω

(5.13)

where T f is the forcing period. For example, taking T = 3T f results in ˆ = 0.125pˆ , a reduction of 87.5% from the static value. R The second example illustrates the strategy for isolating a system from support motion. Applying the formulation derived in Section 1.4 to the system shown in Fig. 5.3, the amplitudes of the relative and total displacement of the mass, uˆ and uˆ t , are related to the support displacement by uˆ = ρ 2 H 1 uˆ g = H 2 uˆ g

(5.14)

uˆ t = H 3 uˆ g

(5.15)

Taking H 3 small with respect to unity reduces the effect of support motion on the position of the mass. The frequency and period criteria are the same as those of the previous example. One takes ρ > 2 to reduce uˆ t . However, since H2 approaches unity as ρ increases, the magnitude of the relative motion increases and approaches the ground motion, uˆ g . Therefore, this relative motion needs to be accomodated. k m c ug

ut = u g + u

Fig. 5.3: SDOF system subjected to support motion.

These examples show that isolation is obtained by taking the period of the SDOF system to be large in comparison to the forcing (either external or support)

342 Chapter 5: Base Isolation Systems period. Expressing this requirement as T ≥ ρ∗ T f

Ω ω < ----ρ*

⇒

(5.16)

where ρ∗ depends on the desired reduction in amplitude, the constraint on the stiffness of the spring is given by Ω 2 2π 2 k < m ------ = m -----------ρ∗ ρ∗ T f

(5.17)

It should be noted that this derivation assumes that a single periodic excitation is applied. The result is applicable for narrow band excitations which are characterized by a dominant frequency. A more complex analysis involving iteration on the stiffness is required to deal with broad band excitations. One has to ensure that the forcing near the fundamental frequency is adequately controlled by damping in this case. Bearing terminology The spring and damper elements connecting the mass to the support are idealizations of physical objects called bearings. They provide a constraint against motion relative to a support plane, as illustrated in Fig. 5.4. The bearing in Fig. 5.4(a) functions as an axial element and resists the displacement normal to the plane with normal stresses (tension and compression). The bearing shown in Fig. 5.4(b) constrains relative tangential motion through shearing action over the height of the bearing. These elements are usually combined into a single compound bearing, but it is more convenient to view them as being uncoupled when modeling the system.

Fn , un axial bearing (a)

Fig. 5.4: Axial and shear bearings.

Ft , ut

shear bearing (b)

5.2 Isolation for SDOF Systems 343 When applying the formulation developed above, one distinguishes between normal and tangential support motion. For normal motion, axial type bearings such as springs and rubber cushions are used; the k defined by eqn (5.17) is the axial stiffness of the bearing F n ⁄ u n . Shear bearings such as laminated rubber cushions and inverted pendulum type sliding devices are used when the induced motion is parallel to the ground surface. In this case, k represents the required shearing stiffness of the bearing, F t ⁄ u t . Figure 5.5 shows an air spring/damper scheme used for vertical support. Single and multiple stage laminated rubber bearings are illustrated in Fig 5.6. Rubber bearings used for seismic isolation can range up to 1 m in diameter and are usually inserted between the foundation footings and the base of the structure. A particular installation for a building is shown in Fig 5.7.

Fig. 5.5: Air spring bearing.

344 Chapter 5: Base Isolation Systems

a) Single stage

b) multiple stage Fig. 5.6: Laminated rubber bearings.

5.2 Isolation for SDOF Systems 345

Fig. 5.7: Rubber bearing seismic isolation system. Modified SDOF Model In what follows, the support motion is considered to be due to seismic excitation. Although both normal (vertical) and tangential (horizontal) motions occur during a seismic event, the horizontal ground motion is generally more significant for structural systems since it leads to lateral loading. Typical structural systems are designed for vertical loading and then modified for lateral loading. Since the vertical motion is equivalent to additional vertical loading, it is not as critical as the horizontal motion. The model shown in Fig. 5.3 represents a rigid structure supported on flexible shear bearings. To allow for the flexibility of the structure, the structure can be modeled as a MDOF system. Figure 5.8 illustrates a SDOF beam type idealization. One can estimate the equivalent SDOF properties of the structure by assuming that the structural response is dominated by the fundamental mode. The data provided in earlier chapters shows that this assumption is reasonable for low-rise buildings subjected to seismic excitation. An in-depth analysis of low rise buildings modeled as MDOF beams is presented later in this chapter. The objective here is to derive a simple relationship showing the effect of the bearing stiffness on the relative displacement of the

346 Chapter 5: Base Isolation Systems structure, u , with respect to the base displacement, u b + u g . The governing equations for the lumped mass model consist of an equilibrium equation for the mass, and an equation relating the shear forces in the spring and the bearing. mu˙˙ + cu˙ + ku = – m ( u˙˙b + u˙˙g )

(5.18)

k b u b + c b u˙ b = ku + cu˙

(5.19)

u + u b + ug m

k,c kb , cb

u b + ug ug

(a) Actual structure

(b) Beam idealization

(c) Lumped mass model

Fig. 5.8: Base isolation models. Neglecting damping, eqn (5.19) can be solved for ub in terms of u. k u b = ----- u kb

(5.20)

Then, substituting for ub in eqn (5.18) leads to k m 1 + ----- u˙˙ + ku = – mu˙˙g kb

(5.21)

Equation (5.21) is written in the conventional form for a SDOF system 2

u˙˙ + ω eq u = – Γu˙˙g

(5.22)

where Γ is a participation factor, kb kb kb Γ = -------------- = ----- ⁄ 1 + ----- k k + kb k

(5.23)

5.2 Isolation for SDOF Systems 347 and ωeq is an equivalent frequency measure 2 2 Γk ω eq = ------ = Γω m

(5.24)

In this case, ω eq is the fundamental frequency of the system consisting of the structure plus bearing. Taking k b small with respect to k decreases the inertia loading on the structure as well as the effective frequency. Consequently, the structural response is reduced. Periodic excitation - modified SDOF model To illustrate the effect of base stiffness on the response, the case of periodic ground motion, u g = uˆ g sin Ωt , is considered. The various response amplitudes are given by 2

Γρ eq uˆ = --------------------uˆ g 2 1 – ρ eq

(5.25)

k uˆ b = ----- uˆ kb

(5.26)

1 uˆ t = uˆ + uˆ b + uˆ g = --------------------uˆ g 2 1 – ρ eq

(5.27)

where the brackets indicate absolute values, and ρeq is the frequency ratio Ω ρ eq = --------ω eq

(5.28)

Comparing eqn (5.27) with eqn (5.15) shows that the results are similar. One replaces ω with ω eq in the expression for H 3 . The limiting cases are k b = 0 and k b = ∞ . The former is the fully isolated case where u b ≈ – uˆ g and u t ≈ 0 ; the latter corresponds to a fixed support where u b ≈ 0 and u t ≈ u + u g . Suppose the structure is defined, and the problem concerns selecting a bearing stiffness such that the total response satisfies uˆ t ≤ νuˆ g

ν<1

348 Chapter 5: Base Isolation Systems One needs to take ρ eq > 2 . Noting eqn (5.27), the required value of ρ eq is 1 2 = 1 + -ρ eq ν

(5.29)

Substituting for ρ eq in eqn (5.28) leads to Ω2 2 = -----------ω eq 1 1 + --ν

(5.30)

Finally, using eqn (5.23) and (5.24), the required bearing stiffness is given by 1 k k b = k ---------------------------- = ----------------------------------------k k ( 1 + ( 1 ⁄ ν)) -------------- – 1 -------------------------------- – 1 ω 2 m 2 mΩ eq

(5.31)

The more general problem is the case where both structural stiffness and the bearing stiffness need to be established subject to the following constraints on the magnitudes of uˆ and uˆ b. uˆ = ν s uˆ g uˆ b = ν b uˆ g

(5.32)

The typical design scenario has ν b larger than ν s . Noting eqn (5.26), the stiffness factors are related by νs k b = -----k νb

(5.33)

Equation (5.25) provides the second equation relating the stiffness factors. It reduces to 1 Γ – 1 + -------- = ----2 νs ρ eq where

(5.34)

5.2 Isolation for SDOF Systems 349

νs kb ⁄ k Γ = --------------------- = ----------------νs + νb 1 + kb ⁄ k

(5.35)

2 leads to ω 2 , and then k. Solving eqn (5.34) for ρ eq eq

k Ω2 2 = ------ω eq - = Γ ---2 m ρ eq

(5.36)

The following example illustrates the computational steps.

Example 5.1: Stiffness factors for prescribed structure and base motion. Suppose ν s = 0.1 and ν b = 1.0 . The relative motion of the base with respect to the ground is allowed to be 10 times greater than the relative motion of the structure with respect to the base. (1)

uˆ b = 10uˆ The stiffness factors are related by νs k b = -----k = 0.1k νb

(2)

Evaluating Γ and ρ eq , using eqns (5.34) and (5.35), νs 0.1 ----------------- = ------- = 0.0909 νs + νb 1.1 1 Γ 1 1 – -------- = ----- = ------- = 0.909 2 νs 1.1 ρ eq 2 ρ eq

(3)

(4)

= 11.0011

leads to 2 = 0.0909Ω 2 ω eq

and finally to k

(5)

350 Chapter 5: Base Isolation Systems m 2 k = ----ω eq = mΩ 2 Γ

(6)

Seismic excitation - modified SDOF model An estimate of the stiffness parameters required to satisfy the motion constraints under seismic excitation can be obtained with the response spectra approach described in Chapter 2. Taking u˙˙g to be the seismic excitation, the solution of eqn (5.22) is related to the spectral velocity by ΓS v u max = ---------ω eq

(5.37)

where S v is a function of the equivalent frequency, ω eq , and the equivalent damping ratio for the structure/bearing system, ξ eq . Substituting for Γ and ω eq , eqn (5.37) expands to mk b u max = S v ---------------------k(k + kb)

(5.38)

The relation between the maximum relative displacement of the bearing and the maximum structural motion follows from eqn (5.20) ub

k = ----- u max max kb

(5.39)

In this development, the criteria for motion based design of a base isolated structure are expressed as limits on the relative motion terms u max = u∗ ub

max

= u b∗

(5.40) (5.41)

The values of k and k b required to satisfy these constraints follow by solving eqns (5.38) and (5.39).

5.2 Isolation for SDOF Systems 351 ku∗ k b = --------u b∗

(5.42)

2

2

mS v mS v 1 k = ------------- ------------------- = --------------------------------2 u b∗ u∗ ( u∗ + u b∗ ) [ u∗ ] 1 + -------u∗

(5.43)

One assumes S v is constant, evaluates k and k b , determines the frequency ω eq with eqn (5.24), and then updates S v if necessary. It is of interest to compare the stiffness required by the base isolated structure with the stiffness of the corresponding fixed base structure. Taking k b = ∞ reduces eqn (5.38) to m u max = S v ---k

(5.44)

The fixed base structural stiffness k f follows from eqn (5.44) 2

mS v k f = k k = ∞ = ------------2 b [ u∗ ]

(5.45)

Using eqn (5.45) and assuming the value of S v is the same for both cases, the stiffness ratios reduce to,

1 k ----- = ------------------kf u b∗ 1 + --------u∗ u∗ -------u b∗ kb ----- = ------------------kf u b∗ 1 + --------u∗

(5.46)

(5.47)

The ratio of the isolated period to the fixed base period can be generated with eqn (5.24)

352 Chapter 5: Base Isolation Systems T eq ωf u b∗ --------- = --------- = 1 + --------Tf ω eq u∗

(5.48)

Figures 5.9 and 5.10 show the variation of k ⁄ k f and k b ⁄ k f with u b∗ ⁄ u∗ for a given constant S v . The increase in the period is plotted in Fig. 5.11. There is a significant reduction in the structural stiffness required by the seismic excitation when the base is allowed to move. For example, taking u b∗ = 2u∗ decreases the design stiffness by a factor of 3 . However, one has to ensure that a potential resonant condition is not created by shifting the period. There may be a problem with wind gust loading as the period is increased beyond 3 seconds. This problem can be avoided by providing additional stiffness that functions under wind loading but not under seismic loading. Section 5.3 deals with this problem.

1 0.9 0.8 0.7

k ----kf

0.6 0.5 0.4 0.3 0.2 0.1 0 0

1

2

3

4

5

u b∗ --------u∗

6

7

8

Fig. 5.9: Variation of k ⁄ k f with u b∗ ⁄ u∗ .

9

10

5.2 Isolation for SDOF Systems 353

1 0.9 0.8 0.7

kb ----kf

0.6 0.5 0.4 0.3 0.2 0.1 0 0

1

2

3

4

5

u b∗ --------u∗

6

7

8

9

10

9

10

Fig. 5.10: Variation of k b ⁄ k f with u b∗ ⁄ u∗ .

12

10

T eq --------Tf

8

6

4

2

0 0

1

2

3

4

5

u b∗ --------u∗

6

7

8

Fig. 5.11: Variation of T eq ⁄ T f with u b∗ ⁄ u∗ .

354 Chapter 5: Base Isolation Systems

Example 5.2:Stiffness parameters - modified SDOF model of Building example #2. The procedure for establishing the appropriate values for u∗ and u b∗ is illustrated using building Example 2 as the reference structure. Table 2.4 lists the relevant design information. The period for the fixed base case is 1.06 sec. Since bad isolation increases the period, the assumption that S v is constant is valid. The relative displacement at the top of the building is estimated as Hγ ∗ where H is the height of the structure and γ ∗ is the prescribed shear deformation. Taking H = 50m and γ ∗ = 1 ⁄ 200 leads to u∗ = 0.25m . The allowable bearing displacement depends on the bearing configuration and response characteristics, as well as the seismic excitation. For the totally soft case, u b is equal to the ground excitation. Hardening the bearing reduces u b somewhat, so a reasonable upper limit is the peak ground displacement corresponding to the design value of S v for representative earthquakes. A typical design value for u b∗ is 0.3m. Using u∗ = 0.25m and u b∗ = 0.3m corresponds to the following stiffness factors k = 0.455k f k b = 0.833k T = 2.2T f The required structural stiffness is reduced by 55% for this degree of base isolation.

These scenarios provide an indication of the potential benefit of base isolation for seismic excitation. However, one should note that the isolated structure is less stiff than the fixed base structure, and therefore will experience larger displacement under other types of loading such as wind. Also, the simplified model considered here is based on linear undamped behavior, whereas the actual bearings have some damping and may behave in a nonlinear manner. More complex models are considered in a later section.

5.3 Design Issues for Structural Isolation Systems 355

5.3 Design issues for structural isolation systems The most important requirements for an isolation system concern flexibility, energy dissipation, and rigidity under low level loading. A number of solutions have been proposed for civil type structures over the past thirty years. The most significant aspects of these designs is discussed below. Flexibility A structural isolation system generally consists of a set of flexible support elements that are proportioned such that the period of vibration of the isolated structure is considerably greater than the dominant period of the excitation. Systems proposed to date employ plates sliding on a curved surface (eg., an inverted pendulum), sleeved piles, and various types of rubber bearings. The most popular choice at this point in time is the rubber bearing, with about 90% of the applications. Rubber bearings consist of layers of natural rubber sheets bonded to steel plates, as shown in Fig. 5.12. The steel plates constrain the lateral deformation of the rubber under vertical loading, resulting in a vertical stiffness several orders of magnitude greater than the horizontal stiffness. The lateral stiffness depends on the number and thickness of the rubber sheets. Increasing either quantity decreases the stiffness; usually one works with a constant sheet thickness and increases the number of layers. As the height increases, buckling becomes the controlling failure mechanism, and therefore, the height is usually limited to about half the diameter. Natural rubber is a nonlinear viscoelastic material, and is capable of deforming up to about 300% without permanent damage. Shear strain on the order of 100% is a common design criterion. Bearing diameters up to 1m and load capacities up to 5 MN are commercially available.

356 Chapter 5: Base Isolation Systems

Mounting plate Rubber h Steel shims

D

Fig. 5.12: Typical natural rubber bearing (NRB). Rigidity under low level lateral loads Increasing the lateral flexibility by incorporating a base isolation system provides an effective solution for high level seismic excitation. Although the relative motion between the structure and the support may be large, the absolute structural motion is generally small, so that the structure does not feel the earthquake. The effect of other types of lateral loading such as wind is quite different. In this case, the loading is applied directly to the structure, and the low lateral stiffness can result in substantial lateral displacement of the structure relative to the fixed support. To control the motion under service loading, one can incorporate an additional stiffness system that functions for service loading but is not operational for high level loading. Systems composed of rods and/or springs that are designed to behave elastically up to a certain level of service loading and then yield have been developed and are commercially available. There are a variety of steel dampers having the above characteristics that can be combined with the rubber bearings. Figure 5.13 illustrate a particular scheme. The steel rod is dimensioned (length and area) such that it provides the initial stiffness and yields at the intended force level. The earliest solution and still the most popular approach is to incorporate a lead rod in the rubber bearing, as illustrated in Fig. 5.14. The lead plug is dimensioned according to the force level at which the system is intended to yield.

5.3 Design Issues for Structural Isolation Systems 357

Fig. 5.13: Steel rod damper combined with a NRB. Mounting plate Rubber h

Lead plug Steel shims

D

Fig. 5.14: Typical lead rubber bearing (LRB). Energy dissipation/absorption Rubber bearings behave in a viscoelastic manner and have some energy dissipation capacity. Additional damping can be provided by separate devices such as viscous, hysteretic, and friction dampers acting in parallel with the rubber bearings. The lead rubber bearing (LRB) is representative of this design approach; the lead plug provides both initial stiffness and hysteretic damping. Since hysteretic damping action occurs only at high level loading, hysteretic-type systems require additional viscous damping to control the response for low level loading. High damping natural rubber with a dissipation capacity about 4 times the conventional value is used together with other devices to improve the energy dissipation capacity of the isolation system. Figure 5.15 illustrates the deployment of a combination of NRB’s, steel dampers, and viscous dampers. This scheme allows one to adjust both stiffness and damping for each load level, i.e., for both low and high level loading.

358 Chapter 5: Base Isolation Systems

Fig. 5.15: Isolation devices of Bridgestone Toranomon Building. Modeling of a natural rubber bearing (NRB) For the purpose of preliminary design, a NRB can be modeled as a simple shear element having a cylindrical shape and composed of a viscoelastic material. Figure 5.16 defines the notation and shows the mode of deformation. The relevant equations are u γ = --h

(5.49)

F = τA

(5.50)

h = nt b

(5.51)

5.3 Design Issues for Structural Isolation Systems 359 where A is the cross-sectional area, t b is the thickness of an individual rubber sheet, and n is the total number of sheets. Each sheet is assumed to be in simple shear. Applying the viscoelastic constitutive relations developed in Section 3.3, the behavior for harmonic shear strain is given by γ = γˆ sin Ωt

(5.52)

τ = G s γˆ sin Ωt + ηG s γˆ cos Ωt

(5.53)

u

F τ

tb h

Fig. 5.16: Natural rubber bearing under horizontal loading. where G s is the storage modulus and η is the loss factor. In general, G s and η are functions of the forcing frequency and temperature. They are also functions of the strain amplitude in the case of high damping rubbers which exhibit nonlinear viscoelastic behavior. Combining the above equations leads to u = uˆ sin Ωt

(5.54)

F = f d G s uˆ [ sin Ωt + η cos Ωt ]

(5.55)

uˆ = γˆ h = γˆ nt b

(5.56)

A A f d = ---- = -------h nt b

(5.57)

where

Note that f d depends on the bearing geometry whereas η and G s are material properties. The standard form of the linearized force-displacement relation is defined

360 Chapter 5: Base Isolation Systems by eqn (3.70) F = k eq u + c eq u˙

(5.58)

where k eq and c eq are the equivalent linear stiffness and viscous damping terms. Estimates for k eq and c eq can be obtained with a least squares approach. Assuming there are N material property data sets covering the expected range of strain amplitude and frequency, the resulting approximate expressions are eqns (3.74), (3.76), and (3.77) which are listed below for convenience. 1 k eq = f d ---N

∑

)

N

Gs ( Ωi ) = f d G s

(5.59)

i=1

c eq = αk eq N

∑

Gs η --------- Ω i

i=1 α = ---------------------------N

∑

(5.60)

(5.61)

Gs ( Ωi )

i=1

Equation (5.58) is used in the MDOF analysis presented in a later section. Figures 5.17 and 5.18 show that the material properties for natural and filled rubber are essentially constant for the frequency range of interest. Assuming G s and η are constant, the equivalent properties reduce to k eq = f d G s

(5.62)

η α = ------T av 2π

(5.63)

where Tav is the average period for the excitation and G s , η are the “constant” values.

5.3 Design Issues for Structural Isolation Systems 361

G s ( Pa )

η

Fig. 5.17: Storage modulus and loss factor for natural rubber vs. frequency (Snowden, 1979)

362 Chapter 5: Base Isolation Systems

G s ( Pa )

η

Fig. 5.18: Storage modulus and loss factor for filled natural rubber vs. frequency (Snowden, 1979) Modeling of a lead rubber bearing (LRB) As a first approximation, the LRB can be considered to consist of two elements: i) a linear viscoelastic element representing the rubber component, and ii) a linear elastic-perfectly plastic element simulating the lead plug. This model assumes that the static force response relationship is bilinear, as indicated in Fig. 5.19. The stiffness defined by eqn (5.62) can be used for the rubber bearing, i.e. for k 1 . k ( rubber ) ≡ k 1 = f d G s

(5.64)

Considering lead to behave in a linear elastic manner, the plug stiffness can be expressed as

5.3 Design Issues for Structural Isolation Systems 363 Ap Gp k ( lead ) ≡ k 2 = --------------hp

(5.65)

where A p , h p , and G p denote the cross-sectional area, height, and shear modulus for the plug. Lastly, the displacement corresponding to the onset of yielding is related to the yield strain for lead by uy = hp γ y

(5.66)

F k1

k 1 ( rubber )

Fy k1 + k2

F

ks k 2 ( lead )

u uy

uˆ

u

Fig. 5.19: Lead rubber bearing model - quasi static response. Interpreting the behavior of the lead rubber bearing for large deformation as viscoelastic, the response due to harmonic motion is expressed in terms of a secant stiffness, k s , and equivalent loss factor, η˜ , u = uˆ sin Ωt

(5.67)

F = k s uˆ sin Ωt + η˜ k s uˆ cos Ωt

(5.68)

where k s is related to the elastic energy storage capacity and η˜ is a measure of the energy dissipated through hysteretic damping of the rubber and lead components. Defining µ as the ductility ratio γˆ uˆ µ = ------ = ------uy γy the secant stiffness is related to the individual stiffness terms by

(5.69)

364 Chapter 5: Base Isolation Systems k2 k s = k 1 + ----µ

(5.70)

The equivalent loss factor is defined as 1 W η˜ = ------ -----2π E S

(5.71)

where W is the hysteretic work per cycle and E S is the maximum strain energy. Evaluating the energy terms 2

W = 4 ( µ – 1 )k 2 u y + πηk 1 µ 2 u y2

(5.72)

2 1 E S = --- k s [ µu y ] 2

(5.73)

and substituting in eqn (5.71) leads to k1 4 ( µ – 1 )k 2 η˜ = -------------------------- + η ----2 ks πk s µ

(5.74)

–3

Noting that γ y is about 5 ×10 about 0.5 , one can estimate µ as

and the typical peak response strain is

γˆ µ = ------- ≈ 100 γy

(5.75)

A typical value for the ratio of k 1 to k 2 is k 1 ≈ 0.1k 2

(5.76)

Then, reasonable estimates for k s and η˜ are k s = 1.1k 1

(5.77)

0.1 4 η˜ = --------- + ---------- η = 0.12 + 0.909η 0.11 11π

(5.78)

The loss coefficient for high damping rubber can be as high as 0.15 . Combining a

5.3 Design Issues for Structural Isolation Systems 365 high damping rubber bearing with a lead plug provides an effective solution for both initial stiffness and damping over the range from low to high excitation. The last step involves transforming eqn (5.68) to the standard form, eqn (5.58). Applying a least square approach and treating k s and η˜ as functions of both the strain amplitude and frequency leads to N

1 k eq = ---N

∑

k s ( µ i, Ω i )

(5.79)

i=1 N

1 c eq = ---N

∑ i=1

k s ( µ i, Ω i )η˜ ( µ i, Ω i ) -----------------------------------------------Ωi

(5.80)

where N is the number of data sets, i.e., values of µ and Ω . It is reasonable to assume G s and η are constant, and evaluate these parameters for a representative range of the ductility parameter, µ . Applicability of base isolation systems The feasibility of base isolation depends on whether it is needed, whether the proposed structure is suitable for base isolation, and whether it is cost effective compared with alternative solutions (Mayes et al. 1990). The need for base isolation may arise if the location is an area of high seismicity, if increased building safety and post earthquake operability are required, if reduced lateral design forces are desired, or if an existing structure needs upgrading to satisfy current safety requirements. A structure is considered suitable if: i) the subsoil conditions do not produce long period input motions to the structure, ii) the structure is less than about 10 to 15 stories and has a height-to-width ratio that prevents overturning, iii) the site permits the required level of motion of the base with respect to ground, and iv) the non-seismic lateral loads (such as wind) are less than approximately 10% of the weight of the structure. The cost effectiveness of a base isolated structure can be assessed by assigning values to both the initial and life cycle costs and benefits. Examples of cost items are: the bearings, changes to accommodate the isolation system, maintenance and inspection of the isolation system, and the cost of maintaining

366 Chapter 5: Base Isolation Systems operability after earthquakes. Examples of savings are: lower initial cost of the structural system, less construction time, lower insurance premium, reduction in earthquake structural and nonstructural damage, and the reduction in injuries, deaths, and lawsuits from related damages. When disruption costs and the value of the building contents are important, seismic isolation has a substantial economic advantage over other systems provided that such an isolation scheme is technically feasible. Under such conditions, initial cost savings of up to 5% of the building cost have been noticed. For conventional buildings where disruption of operation is not important, there may not be sufficient cost savings in the structural system to offset the cost of the isolators (Mayes et al. 1990). The greatest advantage of base isolation is achieved when it is considered in the early planning stages of the project, since it is possible to take advantage of the reduced response due to the isolation system. If the Base Isolation System is selected and added after completion of the structural design, many complications may arise since the construction techniques may have to be altered. For bridge construction on the other hand, the economic issues are very different from those for buildings. In bridges, the implementation of seismic isolation simply requires the use of a seismic isolation bearing rather than a conventional bearing. Since bearings are only one or two percent of the cost of a bridge, an increase in the cost of isolation bearings will have very little impact on the overall construction cost and consequently, the use of a seismic isolation system is expected to reduce the overall construction cost (Billings et al. 1985).

5.4 Examples of existing base isolation systems The past few years, especially since the Kobe earthquake in Japan, have seen a significant increase in the number of base isolated structures which suggests that the technology is gaining acceptance. A short description of some of the first implementations of base isolation systems is presented here to provide an indication of the type of buildings that are being isolated and the cost savings, if any, achieved by employing this technology. More comprehensive descriptions are contained in Kelly (1993), the Architectural Institute of Japan Guide to Base Isolated Buildings in Japan (1993), and various company web sites listed in the Electronic Reference Section of the text.

5.4 Examples of Existing Base Isolation Systems 367 USC University Hospital (Myers 1989, Asher & Van Volkingburg 1989) This eight-story structure, shown in Fig 5.20, is used as a teaching hospital by the University of Southern California. It resists seismic forces with a steel braced frame located on the perimeter, and is supported on 68 LRB and 81 NRB isolators. The seismic design was based on a 0.4g response spectrum increased by 20% to account for near-fault effects. The decision to incorporate seismic isolation was made in the preliminary design phase of the project. Structural cost comparisons for conventional and isolated structures were developed and the benefits of seismic isolation were assessed. It was determined that the cost savings in the structural frame would be sufficient to pay for the new structural ground floor slab and the isolation system. The additional cost of mechanical and architectural details was 1.3% and there was a 1.4% cost savings in the soil nailed retaining wall used in the isolation design versus the conventional retaining wall. Consequently, there was no net additional cost for incorporating seismic isolation on this hospital project.

Fig. 5.20: USC University Hospital Fire Department Command and Control Facility (Mayes et al. 1990) This is a two-story, steel perimeter braced frame structure that utilizes 36 highdamping elastomeric isolation bearings. The decision to utilize seismic isolation on this project was based on a comparison of two designs (conventional and isolation) that required maintaining the functionality of the structure after the extreme design event. This project reflects the first such detailed comparison for two designs to meet a performance specification. In the case of this two-story

368 Chapter 5: Base Isolation Systems structure, the isolated structure was found to be 6% less expensive than conventional design. A reduction in losses by a factor of 40 is expected with the seismic isolation. Evans and Sutherland Manufacturing Facility (Reaveley et al. 1989) The building, (see Fig 5.21), is a four-story manufacturing site for flight simulators located near the Warm Springs and East faults in Salt Lake City. The building measures 280ft x 160ft in plan and rests on 40 LRB and 58 NRB isolators. Preliminary costs for conventional and isolated designs were developed and the benefits of seismic isolation assessed at the conceptual design phase. The structural engineers decided to design the structural framing system for the UBC code forces for conventional design and, consequently, there were no structural framing cost savings. The additional structural cost was the basement structural floor (versus a slab-on-grade) and the heavy fail safe system used. Based on cost data developed by the contractors, the cost premium for incorporating seismic isolation was 5% or $400,000 on an $8 million project. Important in the decision to employ seismic isolation was protecting the building contents, including work in progress, the value of which exceeds $100 million (approximately 12 times the cost of the structure).

Fig. 5.21: Evans and Sutherland Facility

5.4 Examples of Existing Base Isolation Systems 369 Salt Lake City Building (Mayes et al. 1987, Walters et al. 1986) This facility, shown in Fig 5.22, is a five-story, Richardson Romanesque Revival structure constructed between 1892 and 1894, 265ft x 130ft in plan, and built of unreinforced brick and sandstone. Its 12 story tower is centrally located and is also constructed of unreinforced masonry. The building was restored and a combination of 208 LRB and 239 NRB isolators were installed, separating the building from its foundation. The structure is now protected against damage for the 0.2g design earthquake event. This project was the subject of a detailed study of several retrofit schemes among which were base isolation and UBC strengthening. The schemes were developed in sufficient detail to permit cost estimates and an evaluation of performance. Although the cost of these two alternatives was comparable, the decision to use seismic isolation was made based on the considerably better performance that results from the implementation of such a scheme. The complete architectural and historic restoration, and seismic rehabilitation work was estimated to be $24 million. The approximate value of the seismic isolation work reported by the contractor was $4,414,000 including the cost of the 447 seismic isolators.

Fig. 5.22: Salt Lake City Building The Toushin 24 Ohmori Building (Kajima, 1989) This building has 1 underground story which is used as a parking garage, and 9 stories above ground. It is located adjacent to 2 of the busiest railway lines in Tokyo, and the isolation system was required to reduce the traffic induced vibration as well as seismic motion. Figure 5.23 shows a view of the building, a

370 Chapter 5: Base Isolation Systems sectional plan, and the isolation scheme. A combination of laminated natural rubber bearings and steel rod dampers were deployed. Thick layers of rubber were used to decrease the vertical stiffness and thus filter out vertical microtremors.

a)View of building

5.4 Examples of Existing Base Isolation Systems 371

b) Section

c) Devices Fig. 5.23: The Toushin 24 Ohmori Building

372 Chapter 5: Base Isolation Systems Bridgestone Toranomon Building (Shimizu, 1987) The Bridgestone Toranomon Building (see Fig 5.24) is an office building of the Bridgestone Corporation, a major supplier of rubber products such as bearings. The base isolation system consists of 12 laminated rubber bearings, 25 steel dampers, and 8 viscous (oil) dampers. Figure 5.15 shows the layout of the devices. The viscous dampers are intended to dissipate the energy associated with wind and low intensity excitations. At this load level, the steel dampers are designed to behave elastically and provide stiffness. Energy associated with a large seismic excitation is dissipated/absorbed primarily by the steel dampers.

Fig. 5.24: Bridgestone Toranomon Building San Francisco City Hall (1994) San Francisco City Hall is an historic structure that is currently being

5.4 Examples of Existing Base Isolation Systems 373 retrofitted with a seismic isolation system consisting of 530 lead rubber isolators. The design basis earthquake is 0.50g. Cost of retrofitting the structure is estimated at $105 million.

Fig. 5.25: San Francisco City Hall Long Beach V.A. Hospital The hospital is 12 story concrete structure with shear walls. A combination of 110 LRB, 18 NRB and 18 sliding bearings were installed in the mechanical crawl spaces below the building to improve the building’s ability to survive earthquakes up to magnitude 0.32g.

Fig. 5.26: Long Beach V.A. Hospital

374 Chapter 5: Base Isolation Systems

5.5 Optimal stiffness distribution - discrete shear beam The theory developed earlier in this chapter for the SDOF case is extended here to deal with the more general case of a deformable beam-type structure supported by a base isolation system. Linear behavior is assumed since the objective is to generate results which are suitable for preliminary design. The approach followed to establish the stiffness distribution for the structure is similar to what was presented in Chapter 2. The only modification required is to include the effect of the stiffness and damping associated with the base isolation system. Most of the notation and relevant equations have been introduced in Chapter 2. In what follows, the stiffness distribution corresponding to uniform deformation for the fundamental mode of the composite system consisting of a discrete shear beam and isolation system is derived. The theory is extended to deal with continuous beams in the next section. Scaled stiffness distribution Figure 5.27 defines the notation used for the base isolated shear beam. The bearing system is represented by an equivalent linear spring, k1, and linear viscous damper, c1; m1 represents the mass lumped at the foundation level above the bearings; ui is the displacement of the mass mi with respect to the ground; ki and ci are the story stiffness and viscous damping coefficients for the actual structure. The governing equations for free undamped vibration are expressed as ˙˙ + KU = 0 MU where the various matrices are the same as defined in chapter 2.

(5.81)

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 375

un

mn kn,cn

un-1

mn-1 kn-1,cn-1

u2

m2 k2,c2

m1

u1

k1,c1

Fig. 5.27: Notations for base isolated discrete shear beam. In the previous development, the modal displacement profile was selected such that the interstory displacement was constant over the beam. That strategy is modified here to allow for a different interstory displacement for the first story, which, in this model, represents the relative displacement of the bearing.

4

ub

us

4

3

3

2

2

ν +1

H

1

ub

a) actual

1

ν

b) scaled

Fig. 5.28: Example displacement profile.

376 Chapter 5: Base Isolation Systems Figure 5.28a illustrates the choice of displacement profile; us is the displacement at the top node due to deformation of the beam, and ub is the bearing displacement. For equal story height, the bilinear profile corresponds to uniform shear in the beam, γ = u s ⁄ H . The bearing displacement is expressed as a multiple of the maximum structural displacement, ub = ν us

(5.82)

and the profile is scaled by taking u s as the independent displacement parameter. Fig 5.28b shows the scaled profile. With this choice of displacement parameter, the displacement vector takes the form U = qΦ = u s ν,

1 ν + ---, 3

2 ν + ---, 3

ν + 1

(5.83)

Note that the choice of q as the maximum structural displacement due to deformation of the structure is consistent with the approach followed for the fixed base case. The modified displacement profile introduced here allows for an additional story at the bottom of the beam and distinguishes between the deformation at the base and within the beam. Generalizing this approach for an n’th order system, the fundamental mode profile is taken as U = qΦ Φ = ν,

1 ν + ------------, n–1

2 ν + ------------, …, n–1

ν + 1

(5.84)

The remaining steps are the same as followed in section 2.7. One writes U = e ± iωt Φ and substitutes for U in eqn (5.81). This leads to KΦ = ω 2 MΦ

(5.85)

Scaling K and rearranging the equations results in Sk' = MΦ

(5.86)

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 377 where k' i = k i ⁄ ω 2 and S is defined by eqn (2.161), listed below for convenience. S' ( i, i ) = Φ i – Φ i – 1

(5.87)

S' ( i, i + 1 ) = Φ i – Φ i + 1

S' ( i, j ) = 0 for j ≠ i, i + 1 Given M and Φ, one solves eqn (5.86) for k' . This procedure is illustrated with the following example.

Example 5.3:Scaled stiffness for a 4DOF beam with base isolation. Consider the beam shown in Fig 5.28. The various matrices are 1 2 Φ = ν, ν + ---, ν + ---, ν + 1 (1) 3 3 MΦ = m 1 ν, ν S =

1 m 2 ν + --- , 3

–1 ⁄ 3 1⁄3

–1 ⁄ 3 1⁄3

k' = { k' 1 k' 2 k' 4 = 3m 4 ( ν + 1 )

2 m 3 ν + --- , 3

(2)

(3)

–1 ⁄ 3 1⁄3 k' 3

m4 ( ν + 1 )

k' 4 }

(4)

2 k' 3 = 3m 3 ν + --- + k' 4 3 1 k' 2 = 3m 2 ν + --- + k' 3 3 k' 2 k' 1 = m 1 + -----3ν When the masses are equal, eqn (5) reduces to

(5)

378 Chapter 5: Base Isolation Systems 2 k' 1 = m * 4 + --- ν k' 2 = m * ( 6 + 9ν )

(6)

k' 3 = m * ( 5 + 6ν ) k' 4 = m * ( 3 + 3ν )

Fundamental mode response Taking U according to eqn (5.84), the response of the fundamental mode is governed by ˜ q˙˙ + c˜ q˙ + k˜ q = p˜ – m ˜ Γa m g

(5.88)

where the modal parameters are defined as ˜ = Φ T MΦ m p˜ = Φ T P

c˜ = Φ T CΦ Φ T ME Γ = -----------------˜ m

k˜ = Φ T KΦ c˜ ξ˜ = -----------˜ 2ωm

(5.89)

Since Φ now involves the relative displacement factor, ν , these terms will also depend on ν .

Example 5.4:Example 5.3 revisited. Modal parameters for the 4DOF shear beam considered in example 5.3 are listed below

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 379

˜ = m ν 2 + m ν 2 + 2--- ν + 1--- m 1 2 3 9 4 4 + m 3 ν 2 + --- ν + --- + m 4 ( ν 2 + 2ν + 1 ) 3 9 1 2 Φ T ME = m 1 ν + m 2 ν + --- + m 3 ν + --- + m 4 ( ν + 1 ) 3 3

(1)

1 c˜ = ν 2 c 1 + --- ( c 2 + c 3 + c 4 ) 9 1 2 p˜ = νp 1 + ν + --- p 2 + ν + --- p 3 + ( ν + 1 )p 4 3 3 ˜ and Γ simplify to When the masses are equal, m ˜ = m * 4ν 2 + 2ν + 14 m ------ 9 (2)

1 + 2ν 9 2ν + 1 Γ = ---------------------------- = --- ------------------------------------7 9 7 2ν 2 + ν + --1 + --- ν ( 1 + 2ν ) 7 9

˜ ⁄ m * and Γ for a range of values of ν are listed below in Table 5.1. Values of m There is a significant reduction in Γ with increasing ν , and this results in a reduced response to seismic excitation. Table 5.1: Modal mass and participation factors for 4DOF shear beam with equal modal masses. ν 0 1 2 3 4 5

˜ ⁄ m* m 1.556 7.556 21.556 43.556 73.556 111.556

Γ 1.286 0.794 0.464 0.321 0.245 0.197

The modal damping parameter, c˜ , depends on both the bearing damping c 1 and the structural damping ( c 2, c 3, c 4 ) properties. Incorporating damping in the bearing is more effective than distributing damping over the structure for the

380 Chapter 5: Base Isolation Systems fundamental mode response. Structural damping is needed mainly to control the higher modes.

Stiffness calibration for seismic isolation The peak fundamental mode response due to seismic excitation is given by 1 q max = ---- ΓS v ( ω, ξ ) ω

(5.90)

One specifies q max , ξ , and S v ( ω, ξ ) , and determines ω by iterating on eqn (5.90). By definition, q max is the maximum structural displacement relative to the base motion due to deformation of the structure. It is evaluated using the design value for the maximum transverse shear strain and the structural height, q max = γ * H

(5.91)

The peak amplitude of the bearing displacement relative to the ground follows from eqn (5.82) ub

max

≡ u1

max

= νq max

(5.92)

Given q max and u b , ν is determined with eqn (5.92). This approach has to max be modified when the structure is taken to be rigid, i.e., when q max ≈ 0 . In this case, the system reduces to a SDOF model, and the formulation presented in section 5.2 is applicable.

Example 5.5:Stiffness calibration for Example 5.4 Returning to the 4DOF example structure, the following data is assumed. H = 15m γ * = 1 ⁄ 200 S v ( ω, ξ ) defined by Fig 5.29

(1)

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 381 Using (1), q max = ( 15 ) ( 1 ⁄ 200 ) = 0.075m

(2)

To proceed further, one needs to specify ν . Various cases are considered below.

Spectral Velocity S v (m/s)

S v = 1.2, ξ = 0.02 S v = 0.92, ξ = 0.05

1.0

S v = 0.8, ξ = 0.1 S v = 0.6, ξ = 0.2

0.1

Period 0.1

0.6 1.0

10

T (sec)

Fig. 5.29: Spectral velocity. Case 1

ub

max

= 0.3m

The parameters corresponding to this bearing displacement are ν = 0.3 ⁄ 0.075 = 4 Γ = 0.245

(3)

Substituting in eqn (5.90) leads to an expression for the period, T. 2πq max 1.922 T = ------------------------- = --------------------ΓS v ( T, ξ ) S v ( T, ξ )

(4)

Suppose ξ = 0.05. From Fig 5.29, S v = 0.92 for T>0.6 sec. No iteration is required here, since eqn (4) predicts T>0.6.

382 Chapter 5: Base Isolation Systems

1.922 T = ------------- = 2.13s 0.92 2π ω = ------ = 2.94r ⁄ s T

(5)

The stiffness coefficients are generated using the results contained in example 5.3. For the case of uniform mass, eqns (7) of example 5.3 apply. Taking ν = 4 and ω according to eqn (5) above leads to k 4 = 129.6m *

k 3 = 244.6m *

k 2 = 354.3m *

k 1 = 37.9m *

(6)

Damping is determined with eqn (3) of example 5.4. For ν = 4 and ξ = 0.05 , ˜ = 111.6m * m

(7)

˜ = 2 ( 0.05 ) ( 2.94 ) ( 111.6m * ) = 32.8m * c˜ = 2ξωm

(8)

The individual damping coefficients are related to c˜ by 1 c˜ = 16c 1 + --- ( c 2 + c 3 + c 4 ) = 32.8m * 9

(9)

One has to decide how to allocate damping to the various components. For example, assuming 75% of c˜ is contributed by the bearing requires c 1 = 1.54m * c 2 + c 3 + c 4 = 73.8m *

(10)

Placing damping at the base is an order of magnitude more effective than distributing the damping throughout the structure for this degree of isolation. Case 2 u b = 0.15m max For this case, ν = 2 . The various parameters for ξ = 0.05 , ν = 2 , and uniform mass are as follows.

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 383 Γ = 0.464 T = 1.025s ω = 6.16r ⁄ s ˜ = 21.556m * m c˜ = 13.28m * 1 4c 1 + --- ( c 2 + c 3 + c 4 ) = 13.28m * 9 k 4 = 341m *

k 3 = 644m *

k 2 = 909m *

k 1 = 189m *

(11)

Case 3 Fixed base The fixed base case is treated in examples 2.9 and 2.10. Specializing these results for uniform mass and ξ = 0.05 results in the following parameters and properties. Γ = 1.286 T = 0.5s ω = 12.56r ⁄ s ˜ = 1.556m * m 1 --- ( c 2 + c 3 + c 4 ) = 1.95m * 9 k 4 = 473m * k 3 = 788m * k 2 = 946m * k1 = ∞

(12)

384 Chapter 5: Base Isolation Systems

5.6 Optimal stiffness distribution - continuous cantilever beam Stiffness distribution - undamped response The equilibrium equations for undamped motion of the base isolated continuous beam shown in Fig. 5.30 are H

∫

V ( x, t ) = – ρ m u˙˙( x, t ) dx

(5.93)

x

H

M ( x, t ) =

∫

V ( x, t ) dx

(5.94)

x

uT

uB

u

M + dM D T, D B

H

V + dV dx

x

V

ub

M kb

Fig. 5.30: Base isolated continuous beam. The transverse shear and bending deformation measures for the beam are related to the translation and rotation quantities by

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 385 ∂u = γ+β ∂x

(5.95)

∂β ∂x

(5.96)

χ =

Considering γ and χ to be functions only of time, integrating the resulting equations with respect to x, and imposing the boundary conditions at x = 0 , one obtains expressions for u and β in terms of γ ( t ) , χ ( t ) , and u b ( t ) 1 2 u = γx + --- χx + u b 2

(5.97)

β = χx

(5.98)

where u b denotes the displacement of the base of the structure with respect to ground. Taking γ = γ∗e

iω 1 t

(5.99)

iω 1 t

(5.100)

χ = χ∗ e

u b = u b∗ e

iω 1 t

(5.101)

produces a periodic motion of the beam. Noting that the deformation measures γ ∗ and χ∗ are related by (see eqn (2.14) and Fig 5.30) uB χ∗ H s = ------ = ----------uT 2γ ∗

(5.102)

and expressing u b∗ in terms of the displacement at x = H due to shear deformation (see Fig. 5.30) u b∗ = νu T∗ = νγ ∗ H

(5.103)

transforms eqn (5.97) into iω t iω t x sx 2 u = ---- + -------- + ν γ * He 1 = Φ ( x ) (γ *He 1 ) = q ( t )Φ ( x ) H H2

(5.104)

The function Φ ( x ) defines the fundamental mode. The corresponding expression

386 Chapter 5: Base Isolation Systems for the fixed base case is eqn (2.189). Differentiating u with respect to time, 2

iω t sx u˙˙ = – x + -------- + νH γ ∗ ω 12 e 1 H

(5.105)

and substituting for u˙˙ in eqn (5.93) leads to iω t sx 3 νx 1 s x2 V = ρ m ω 12 H 2 --- + --- + ν – ---------- – ---------- – ------ γ ∗ e 1 2 3 2 3 H 2H 3H

(5.106)

The corresponding relation for the bending moment is M = ρ m ω 12 H 4 χ∗ e

iω 1 t

1 1 ν 1 1 ν x ----- + --- + ----- – ----- + --- + ----- ---6s 8 4s 4s 6 2s H

(5.107)

νx 2 x3 x4 + ------------- + ---------------- + ------------4sH 2 12sH 3 24H 4 Lastly, the shear and bending rigidity distributions are determined with the definition equations V 1 s x2 sx 3 νx D T = ----- = ρ m ω 12 H 2 --- + --- + ν – ---------- – ---------- – -----2 3 γ* 2H 2 3H 3 H

(5.108)

νx 2 M 1 1 ν 1 1 ν x x3 x4 D B = ----- = ρ m ω 12 H 4 ----- + --- + ----- – ----- + --- + ----- ---- + ------------- + ---------------- + ------------6s 8 4s 4s 6 2s H 4sH 2 12sH 3 24H 4 χ* (5.109) Equation (5.108) is written as DT ( 0 ) 1 s x 2 sx 3 νx D T = --------------------------- --- + --- + ν – ---------- – -------- – ------ 2 3 H 1 s 2H 2 3 --- + --- + ν 2 3

(5.110)

where D T ( 0 ) is the shear rigidity at the base. 1 s D T ( 0 ) = ρ m ω 12 H 2 --- + --- + ν ≡ k s H 2 3

(5.111)

The parameter, k s , can be interpreted as an equivalent shear stiffness

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 387 measure.The shear force at the base of the beam must equal the shear force in the bearing to satisfy the force equilibrium condition for undamped motion. Equating these forces D T ( 0 )γ ∗ = k b u b∗ = k b νγ ∗ H

(5.112)

and solving for k b results in DT ( 0 ) 1 k b = ---------------- = --- k s ν νH

(5.113)

The fundamental frequency follows from eqn (5.111) DT ( 0 ) ω 12 = ------------------------------------------1 s ρ m H 2 --- + --- + ν 2 3

(5.114)

Figures 5.31 and 5.32 shows the mode shapes and shear deformation profiles for the first five modes of a typical low rise building. The variation in the mode shape profiles with the ratio of the stiffness of the isolator, k b , to the shear beam stiffness k s are illustrated by Figures 5.33, 5.34 and 5.35. This ratio is equal to 1 ⁄ ν . Figure 5.36 displays the variation in the periods of the highest three fundamental modes. The primary influence is on the period of the fundamental mode which is significantly increased when the stiffness of the isolator is several orders of magnitude lower than the beam stiffness. The effect on the periods of the second and third modes is relatively insignificant. Figure 5.37 shows the variation of the participation of the second and third modes relative to the first. The plot shows that the contribution of the second and third modes is also significantly reduced by decreasing the stiffness of the isolator with respect to the beam stiffness.

388 Chapter 5: Base Isolation Systems

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

1

2

3

4

5

Mode number

Fig. 5.31: Mode shapes for a typical base isolated structure.

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

1

2

3

4

5

Mode number

Fig. 5.32: Mode deformation profiles for a typical base isolated structure.

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 389

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

k b ⁄ k s = 0.001 k b ⁄ k s = 0.1 k b ⁄ k s = 10.0

0.3 0.2 0.1 0

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Mode 1

Fig. 5.33: Variation of mode 1 shape with relative stiffness of isolator.

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

k b ⁄ k s = 0.001 k b ⁄ k s = 0.1 k b ⁄ k s = 10.0

0.3 0.2 0.1 0

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Mode 2

Fig. 5.34: Variation of mode 2 shape with relative stiffness of isolator.

390 Chapter 5: Base Isolation Systems

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

k b ⁄ k s = 0.001 k b ⁄ k s = 0.1 k b ⁄ k s = 10.0

0.3 0.2 0.1 0

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Mode 3

Fig. 5.35: Variation of mode 3 shape with relative stiffness of isolator.

8

T1 T2 T3

7

6

Period

5

4

3

2

1

0 −3 10

−2

10

−1

10

0

10

1

10

kb ⁄ ks

Fig. 5.36: Variation of periods with relative stiffness of isolator.

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 391

Relative participation factors

1

Γ2 ⁄ Γ1 Γ3 ⁄ Γ1

0.8

0.6

0.4

0.2

0 −3 10

−2

−1

10

10

0

10

1

10

kb ⁄ ks

Fig. 5.37: Variation of relative participation factors with relative stiffness of isolator.

Fundamental mode equilibrium equation Incorporating the contribution of the base isolation system, the principle of virtual displacements has the form H

∫

H

( M ⋅ δχ + V ⋅ δγ )dx + F b ⋅ δu b =

0

∫

( b ⋅ δu )dx

(5.115)

0

where F b is the shear force in the bearing. The equations relating internal forces to deformations and deformation rates are taken as V = D T γ + C T γ˙

(5.116)

M = D B χ + C B χ˙

(5.117)

F b = k b u b + c b u˙ b

(5.118)

392 Chapter 5: Base Isolation Systems The form of the modal expansion follows from eqn (5.104). x sx 2 u = qΦ ( x ) = q ---- + -------- + ν H H2

(5.119)

2sx β = qΨ ( x ) = q --------- H2

(5.120)

Assuming external loading and seismic excitation, the loading term is b = – ρ m a – ρ m u˙˙ + b ( x, t ) g

(5.121)

Finally, introducing the various terms in the principle of virtual displacements leads to the equilibrium equation for q ˜ q˙˙ + c˜ q˙ + k˜ q = p˜ m

(5.122)

where H

˜ = m

∫

2 x sx 2 1 s 2sν s 2 ρ m ---- + -------- + ν dx = ρ m H --- + ν + ν 2 + --- + --------- + ----H H2 2 3 5 3

(5.123)

C T 4s 2 C B ------- + ---------------- dx + c b ν 2 H2 H4

(5.124)

D T 4s 2 D B ------- + ---------------- dx + k b ν 2 H2 H4

(5.125)

0

H

c˜ =

∫ 0

H

k˜ =

∫ 0

H

p˜ =

∫

x sx 2 1 s ( b – ρ m a g ) ---- + -------- + ν dx = – ρ m H --- + ν + --- a g – p˜ e H H2 2 3

0

Expressing c˜ and k˜ as

(5.126)

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 393 ˜ c˜ = 2ξωm

(5.127)

˜ k˜ = ω 2 m

(5.128)

transforms eqn (5.122) to 2 1 q˙˙ + 2ξωq˙ + ω q = – Γa g + ----p˜ e ˜ m

(5.129)

1 s ν + --- + --2 3 Γ = -------------------------------------------------------------1 s 2 s 2sν --- + ν 2 + ν + ----- + --- + --------3 5 2 3

(5.130)

where

For a pure shear beam, s = 0 and the participation factor for the fundamental mode reduces to 1 ν + --2 Γ = ------------------------1 --- + ν 2 + ν 3

(5.131)

The expression for the modal damping ratio depends on how one specifies the damping over the beam. Rigidity calibration - seismic excitation The calibration procedure presented in Chapter 2 is applied to the base isolated model. Starting with ΓS v ( ω, ξ ) q max = ------------------------ω

(5.132)

and substituting for q max results in ΓS v νΓS v ω = ----------- = ------------γ∗H u b∗

(5.133)

394 Chapter 5: Base Isolation Systems One specifies ν in addition to the other parameters ( γ *, ξ, S v ) , and solves for ω . This value is then used to determine D T ( 0 ) and k b with eqns (5.111) and (5.113).

Example 5.6:Stiffness calibration - Example Building #2. The stiffness calibration for Building example #2 was considered in Chapter 2. In what follows, the calibration procedure is extended to include stiffness and damping components located at the base. The design data are: H = 50m

ρ m = 20, 000k g ⁄ m

γ * = 1 ⁄ 200 ξ 1 = 0.05

s = 0.25 S v = 0.92m ⁄ s

(1)

Using (1), the peak relative structural displacement is q max = γ * H = 0.25m

(2)

To proceed further, one needs to specify the base displacement and then establish the value of ν with eqn (5.103). We take u b = 0.25m

(3)

Then ub ν = ------------ = 1.0 q max

(4)

Given ν , the participation factor follows from eqn (5.130) Γ = 0.600

(5)

The modal mass is determined with eqn (5.123) ˜ = 2.638 × 10 6 kg m

(6)

Assuming CT = constant and CB =0 in eqn (5.124), the modal damping

5.6 Building Design Examples 395 coefficient reduces to CT c˜ = ------- + ν 2 c b = 0.02C T + c b H

(7)

Equation (5.127) relates c˜ to ω . ˜ = 0.2638 × 10 6 ω Ns ⁄ m c˜ = 2ξωm

(8)

Lastly, the frequency is found using eqn (5.133) and the design data for S v ( ξ = 0.05 and S v = 0.92m ⁄ s ) ΓS v ω = ---------- = 2.4S v = 2.61r ⁄ s γ *H T = 2.41sec

(9)

With ω known, the modal damping coefficient follows from eqn (8), c˜ = 0.688MNs ⁄ m

(10)

and the transverse shear rigidity at the base of the beam is determined with eqn (5.111). D T ( 0 ) = 538.6MN

(11)

Finally, given D T ( 0 ) , the isolation stiffness is estimated using eqn (5.113), which is based on neglecting the contribution of the damping force in the bearing. k b = 10.77MN

(12)

5.7 Building design examples Stiffness distribution based on fundamental mode response Since base isolation is a potential solution for buildings with less than about 10 stories and having an aspect ratio that prevents overturning, only Buildings 1 and

396 Chapter 5: Base Isolation Systems 2 of Chapter 2 are considered in the simulation. Tables 2.4 and 2.5 lists the design data. The damping in the structure is obtained by specifying ξ 1 and computing a stiffness proportional damping matrix considering a fixed based structure, as carried out in Chapter 2. The damping in the bearing for specified bearing damping ratio ξ b is obtained by assuming the structure to be a SDOF system having a mass equal to the total mass of the structure and a stiffness equal to the stiffness of the bearing as obtained from eqn (5.113). Figures 5.38 and 5.39 show the shear rigidity distributions for Buildings 1 and 2 obtained with eqns (5.108) and (5.109), taking S v = 1.2m ⁄ s . The corresponding shear deformation profiles for different combinations of building damping ratios and isolator damping ratios when the structures are subjected to scaled versions (to S v = 1.2m ⁄ s for ξ = 0.02 ) of the El Centro and Taft accelerograms are plotted in Figures 5.40 through 5.47. The maximum deformation in the bearing is also indicated on the plots. The plots show that increasing the damping in the bearing tends to reduce the bearing deformation without significantly altering the shape of the deformation profile along the structure’s height. Furthermore, just as for the fixed base structures, the contribution of the higher modes becomes more significant as the structure becomes more slender. The structural parameters, as well as the mean and standard deviation results for the deformations of the above buildings are tabulated in the following section.

5.6 Building Design Examples 397

1 0.9

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% 6 k b = 6.35 ×10 N/m

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N

Fig. 5.38: Initial shear rigidity distribution for Building 1.

1 0.9

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% 6 k b = 9.93 ×10 N/m

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N

Fig. 5.39: Initial shear rigidity distribution for Building 2.

398 Chapter 5: Base Isolation Systems

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ b = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.18m 5%, u b = 0.18m 10%, u b = 0.17m 20%, u b = 0.17m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.40: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.18m 5%, u b = 0.16m 10%, u b = 0.13m 20%, u b = 0.11m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.41: Maximum shear deformation for Building 1.

0.01

5.6 Building Design Examples 399

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ b = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.09m 5%, u b = 0.08m 10%, u b = 0.07m 20%, u b = 0.07m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.42: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.09m 5%, u b = 0.06m 10%, u b = 0.06m 20%, u b = 0.05m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.43: Maximum shear deformation for Building 1.

0.01

400 Chapter 5: Base Isolation Systems

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ b = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.24m 5%, u b = 0.23m 10%, u b = 0.21m 20%, u b = 0.20m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.44: Maximum shear deformation for Building 2.

1

Building #2 Quadratic Based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.24m 5%, u b = 0.22m 10%, u b = 0.19m 20%, u b = 0.14m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.45: Maximum shear deformation for Building 2.

0.01

5.6 Building Design Examples 401

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ b = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.14m 5%, u b = 0.13m 10%, u b = 0.11m 20%, u b = 0.09m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.46: Maximum shear deformation for Building 2.

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.14m 5%, u b = 0.11m 10%, u b = 0.08m 20%, u b = 0.07m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.47: Maximum shear deformation for Building 2.

0.01

402 Chapter 5: Base Isolation Systems Stiffness distribution including the contribution of the higher modes This section extends the iterative procedure developed in Section 2.10 to incorporate iterating over the stiffness for the beam and isolator. For simplicity, only shear deformation in the isolator is considered. The method consists of including the contribution of the higher modes to the transverse shear, the bending deformation, and the base shear, and then updating the shear and bending rigidities and isolation stiffness using (i + 1) DT (x)

(i + 1) DB (x)

(i + 1) kb

=

(i) [ γ ( x ) ] max ( i ) D T ------------------------------∗

(5.134)

=

(i) [ χ ( x ) ] max ( i ) D B -------------------------------∗

(5.135)

γ

χ

(i)

[ V b ] max = ------------------------u b∗

(5.136)

The peak values are found with eqn (2.270). Rigidity iterations are performed on building examples 1 and 2. Table 5.2 lists the parameters of the buildings for the initial rigidity distributions as well as for a single iteration. For both building examples, one iteration was sufficient to achieve convergence.Tables 5.3 and 5.4 contain the deformations averaged over the height of the structure and the corresponding standard deviations for Buildings 1 and 2 subjected to scaled versions of El Centro and Taft excitations.

5.6 Building Design Examples 403 Table 5.2: Modal parameters - building examples. T1(s)

T2(s)

T3(s)

ξ1 ( % ) ξ2 ( % ) ξ3 ( % ) Γ2 ⁄ Γ1 Γ3 ⁄ Γ1

Bldg #1 Initial (Q) Iteration 1

2.09 2.01

0.62 0.60

0.33 0.32

1.56 1.54

3.55 3.64

6.26 6.37

0.11 0.12

0.03 0.03

Bldg #2 Initial (Q) Iteration 1

2.51 2.45

0.90 0.85

0.49 0.46

1.48 1.48

3.70 3.83

6.60 6.89

0.20 0.19

0.06 0.06

Figures 5.48 through 5.57 show the shear and bending rigidity distributions resulting from the first iteration, as well as the shear deformation profiles corresponding to different combinations of building damping ratios and isolator damping ratios under El Centro and Taft excitation.Results from one iteration provide sufficient convergence accuracy. For Building 2, the iterative scheme tends to pull the top back in, resulting in a more uniform deformation profile. The effect of damping is similar to that noticed in the examples of the previous section.

404 Chapter 5: Base Isolation Systems Table 5.3: Mean and standard deviation deformation results for Building 1. Bldg #1 El Centro Initial (Q)

Iteration 1

Taft Initial (Q)

Iteration 1

γm

γ sd

χm

χ sd

ξ1 (%)

ξb (%)

ub (m)

(10-3)

(10-4)

(10-5)

(10-6)

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.18 0.18 0.17 0.17

2.80 2.74 2.61 2.43

0.70 0.46 0.11 0.22

3.18 3.11 2.93 2.72

0.56 0.31 0.07 0.24

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.18 0.16 0.13 0.11

2.80 2.55 2.36 2.19

0.70 1.09 1.92 2.35

3.18 2.93 2.79 2.62

0.56 0.95 1.31 1.57

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.17 0.17 0.15 0.15

3.28 3.09 2.79 2.37

1.36 1.05 0.66 0.62

2.86 2.68 2.41 1.98

4.40 4.31 4.10 3.63

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.17 0.14 0.12 0.10

3.28 2.55 2.26 2.16

1.36 1.23 0.92 1.38

2.86 2.22 1.96 1.90

4.40 3.08 3.21 2.98

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.09 0.08 0.07 0.07

1.61 1.42 1.25 1.08

3.42 2.22 1.32 0.49

2.04 1.74 1.49 1.25

2.76 1.86 0.96 0.31

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.09 0.06 0.06 0.05

1.61 1.34 1.21 1.17

3.42 2.78 1.80 1.68

2.04 1.67 1.45 1.41

2.76 2.60 1.79 1.54

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.08 0.08 0.08 0.07

1.80 1.57 1.38 1.17

2.26 1.00 0.39 0.27

1.64 1.38 1.19 0.99

2.00 1.97 1.97 1.84

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.08 0.07 0.06 0.05

1.80 1.41 1.25 1.22

2.26 1.73 0.63 0.93

1.64 1.27 1.07 1.07

2.00 1.45 1.43 1.29

5.6 Building Design Examples 405 Table 5.4: Mean and standard deviation deformation results for Building 2. Bldg #2 El Centro Initial (Q)

Iteration 1

Taft Initial (Q)

Iteration 1

γm

γ sd

χm

χ sd

ξ1 (%)

ξb (%)

ub (m)

(10-3)

(10-4)

(10-5)

(10-6)

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.24 0.23 0.21 0.20

4.61 4.13 3.77 3.26

7.49 4.17 2.49 1.22

4.90 4.25 3.81 3.25

5.96 3.28 1.83 0.81

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.24 0.22 0.19 0.14

4.61 4.15 3.61 2.99

7.49 6.25 4.98 4.10

4.90 4.39 3.79 3.15

5.96 4.88 3.72 2.78

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.23 0.22 0.21 0.20

4.67 4.11 3.65 3.12

4.36 1.84 1.68 2.34

4.04 3.44 2.98 2.50

5.92 5.67 5.54 5.14

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.23 0.21 0.18 0.14

4.67 4.21 3.65 2.98

4.36 3.61 2.87 2.43

4.04 3.62 3.12 2.55

5.92 5.43 4.82 4.15

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.14 0.13 0.11 0.09

2.77 2.30 1.89 1.49

6.24 3.49 1.74 1.33

3.00 2.41 1.91 1.52

5.56 3.19 1.74 1.17

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.14 0.11 0.08 0.07

2.77 2.34 1.99 1.87

6.24 6.53 6.03 5.69

3.00 2.56 2.23 2.16

5.56 6.05 4.98 4.48

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.11 0.10 0.09 0.08

2.45 2.05 1.69 1.38

4.92 2.68 1.22 0.75

2.22 1.80 1.43 1.16

2.84 2.38 2.03 2.00

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.11 0.09 0.08 0.07

2.45 2.16 1.87 1.74

4.92 4.57 3.92 2.93

2.22 1.97 1.69 1.57

2.84 2.73 2.47 1.72

406 Chapter 5: Base Isolation Systems

1 0.9

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% 6 k b = 6.94 ×10 N/m

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N

Fig. 5.48: Converged shear rigidity distribution for Building 1.

1 0.9

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% 7 k b = 1.05 ×10 N/m

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N

Fig. 5.49: Converged shear rigidity distribution for Building 2.

5.6 Building Design Examples 407

1

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ b = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.17m 5%, u b = 0.16m 10%, u b = 0.15m 20%, u b = 0.15m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.50: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.17m 5%, u b = 0.14m 10%, u b = 0.12m 20%, u b = 0.10m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.51: Maximum shear deformation for Building 1.

0.01

408 Chapter 5: Base Isolation Systems

1

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ b = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.08m 5%, u b = 0.08m 10%, u b = 0.08m 20%, u b = 0.07m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.52: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.08m 5%, u b = 0.07m 10%, u b = 0.06m 20%, u b = 0.03m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.53: Maximum shear deformation for Building 1.

0.01

5.6 Building Design Examples 409

1

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ b = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.23m 5%, u b = 0.22m 10%, u b = 0.21m 20%, u b = 0.20m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.54: Maximum shear deformation for Building 2.

1

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.23m 5%, u b = 0.21m 10%, u b = 0.18m 20%, u b = 0.14m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.55: Maximum shear deformation for Building 2.

0.01

410 Chapter 5: Base Isolation Systems

1

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ b = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.11m 5%, u b = 0.10m 10%, u b = 0.09m 20%, u b = 0.08m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.56: Maximum shear deformation for Building 2.

1

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.11m 5%, u b = 0.09m 10%, u b = 0.08m 20%, u b = 0.07m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.57: Maximum shear deformation for Building 2.

0.01

Problems 411

Problems Problem 5.1

k p

m c ug

ut = u g + u

Consider a SDOF system with m=1000kg. The design requirements for the system involve 2 nonconcurent loading conditions, a periodic external forcing and a periodic ground motion.Under the external forcing, the relative motion is prescribed; the total motion is prescribed for the case of ground excitation. These design conditions are summarized below. The asterisk denotes a specified quantity. 1) p = p * sin Ω P* t

(1)

u < u*

(2)

2) u g = u g* sin Ω g* t

(3)

u t < u t* ≡ H 3* u g*

(4)

The design problem involves determining k and c such that the above performance requirements are satisfied. a) Recommand values for k and c corresponding to the following design data: 1)

Ω g* = 2πr ⁄ s p * = 400N

2)

Ω g* = 2πr ⁄ s p * = 800N

u g* = 0.2m π Ω P* = --- r ⁄ s 2 * H 3 = 0.125 π Ω P* = --- r ⁄ s 2

H 3* = 0.125 u * = 0.2m

u * = 0.2m

412 Chapter 5: Base Isolation Systems

3)

Ω g* = 2πr ⁄ s

H 3* = 0.125

p * = 1600N

π Ω P* = --- r ⁄ s 2

u * = 0.2m

b) It may not be possible to satisfy both constraints with the same value of stiffness. When this problem arises, one can determine the stiffness corresponding to each design condition and design a spring mechanism that allows the stiffness to be varied. Suggest design concepts for such a mechanism. This situation may occur for lateral loading applied to a seismically isolated structure. Discuss how you would implement a variable stiffness scheme for non simultaneous wind and seismic loading.

Problem 5.2 Refer to the modified SDOF model defined by Fig 5.8. Take m=10,000 kg and k=400 kN/m. Suppose the system is to be subjected to a ground motion, u g ( meters ) = 0.5 sin 4πt , and the total displacement is required to be less than 0.2m. Determine the appropriate bearing stiffness.

Problem 5.3 Refer to the modified SDOF model defined by Fig 5.8. Take m=10,000 kg and the ground motion to be u g ( meters ) = 0.5 sin 4πt . Determine k and k b such that a) u ≤ 0.3m and u b ≤ 0.3m . b) u ≤ 0.1m and u b ≤ 0.3m

Problem 5.4 Refer to the modified SDOF model defined by Fig 5.8. Take m=10,000 kg and consider the system to be subjected to seismic excitation of intensity S v = 1.2m ⁄ s . Determine k , k b and T eq for the following design conditions:

Problems 413 a) u * = 0.1m

u b* = 0.1m

b) u * = 0.1m

u b* = 0.2m

c) u * = 0.1m

u b* = 0.3m

Problem 5.5 Consider a cylindrical bearing having a diameter of 0.6m, a height of 0.3m and composed of filled natural rubber. Take the rubber properties according to Fig 5.18. Assume the frequency range is from 1 to 5 Hz, and the temperature is 20oC. a) Estimate the equivalent linear stiffness and linear viscous damping parameters, k eq and c eq . b) Determine the diameter of a lead plug for the case where the plug stiffness is 10 times the stiffness of the rubber bearing. Assume the lead plug and 3 rubber cylinder have the same height. Take G P = 4 ×10 MPa . c) Assume the bearing experiences periodic excitation resulting in a shear strain amplitude of 50%. Determine the secant stiffness, k s , and loss factor, η˜ , using eqns (5.70) and (5.74). d) Instead of lead, consider using low strength steel as the material for the 3 initial stiffness element, k 2 . Take τ y = 150MPa and G = 80 ×10 MPa for the steel plug. Repeat parts b and c.

Problem 5.6

Consider an isolation system composed of a NRB, a steel hysteretic

414 Chapter 5: Base Isolation Systems k 1 ( rubber ) k 2 ( damper )

F, u

c F

k1

Fy k1+k2

uˆ y uˆ * uˆ damper, and a viscous damper. Neglect the damping provided by the NRB. a) For low level loading, one specifies the initial stiffness, damping, and yield force level. Describe how you would design the steel damper. b) For high level loading, one specifies the secant stiffness and equivalent viscous damping based on a seismic analysis. Describe how you would design the NRB. How would you select the viscous damper? c) Suppose the isolation system is composed of spring and damper elements whose properties can be varied instantaneously. Assuming the elements behave linearly, the force is given by F = k ( t )u + c ( t )u˙

k F, u c

Problems 415 where k(t) and c(t) are properties that can be “adjusted”. Describe how you would utilize this system for a building subjected (nonsimultaneously) to both wind and seismic excitation. How would you design these devices? Note: a system which has the ability to change its properties is said to be adaptive. Adaptive systems are discussed in Chapter 6.

Problem 5.7 m7

m1

k 7, c 7

u7

u1 k 1, c 1

Consider a 6 story building with base isolation modelled as a 7 DOF system. Take the floor masses as m 1 = m 2 = … = m 6 = 10, 000kg and m 7 = 20, 000kg . a) Find the scaled stiffness for the profile based on eqn (5.84). Take ν = 2.0 . ˜ , c˜ , p˜ , and Γ . b) Evaluate the expressions for m c) Calibrate the stiffness distribution for seismic excitation. Use the spectral = 0.15m , and velocity plot contained in Fig 5.29. Take ξ = 0.05 , u s max ub = 0.3m . Allocate 75% of the damping to the bearing. Use MOTIONLAB max

to determine the modal shapes, frequencies, and damping ratios for the 2nd and 3rd modes. d) Repeat part c assuming the base is fixed. Take the sum of the stiffness factors as

416 Chapter 5: Base Isolation Systems a measure of the cost of stiffness. Compare the costs of the fixed base and base isolation solutions. Also compare the modal properties. e) Suppose filled rubber bearings having a diameter of 0.5m and a height of 0.25m are to be used for the isolation system. Assume G = 4MPa and η = 0.15 for the rubber. How many bearings are required for the design conditions specified in part c? Also discuss how you would provide the damping required for the isolation system. f) Discuss how you would deal with lateral wind loading. Assume the dominant wind gust frequency is 0.2Hz.

Problem 5.8

30m

ρ m constant x

Consider a base isolated continuous cantilever beam having a uniform mass density of 2000 kg/m and a height of 30m. a) Generate shear and bending rigidity distributions and isolation stiffness which correspond to a fundamental frequency of 0.33 Hz and various values for s and ν . Take 0 ≤ s ≤ 0.25 and 1 ≤ ν ≤ 3 . Comment on the sensitivity of the rigidity parameters to variation in s and ν . b) Approximate the continuous beam with 6 shear beam segments plus an additional segment to simulate the bearing. Determine the first 3 mode shapes and frequencies corresponding to s=0 and ν = 0 , 1.5, and 3. Consider the lumped masses to be equal. Comment on the sensitivity of the modal properties to the ratio of isolation stiffness, k b , to the shear beam stiffness measure, k s .

Problems 417

Problem 5.9 ub

q

H

D T and ρ x

m

constant

k b, c b ug

Consider a base isolated continuous shear beam having a uniform mass density and constant transverse shear rigidity. Assume the lateral displacement is approximated by x x u = u g + u b + q ---- = u g + q ν + ---- H H a) Establish an equation for q(t) using the Principle of Virtual Displacements. Allow for linear viscous damping in the bearing and uniform material damping in the beam. Write the result in the same form as eqn (5.122) and determine the ˜ , c˜ , k˜ , p˜ , ω, ξ , and Γ . expressions for m b) Suppose H=30m, ρ m = 2000k g ⁄ m , and v=2. Calibrate the stiffness distribution ( D T and k b ) for ξ = 0.05 , q max = 0.15, and S υ defined by Fig 5.29. Determine the damping parameters assuming the bearing contributes 75% to c˜ . c) Approximate the continuous beam with 6 shear beam segments plus an additional segment to simulate the bearing. Determine the first 2 mode shapes and frequencies, using the design data generated in part b.

418 Chapter 5: Base Isolation Systems

Problem 5.10

m1

ub + u1 + u g p

ρ m, D T ( x )

H

kb cb ub + u g

ug

Consider the base isolated shear beam shown above. Assume uniform mass density, m 1 = a ( ρ m H ) , and the shear rigidity to be defined as x D T ( x ) = D∗ 1 – -------- 2H a) Derive the equilibrium corresponding to the following approximate displacement expansion x u = u 1 ( t ) ---- + u b ( t ) H Discuss how the undamped free vibration response behaves as D∗ , k b , and a are varied. b) ub + u1 m1

md ub + u1 + ud

Suppose the force, p , applied at the top of the structure is generated by a

Problems 419 tuned mass damper as shown in the figure. How would you estimate the properties of the tuned mass damper to obtain an effective damping ratio of 0.05 for the mode shape approximation considered in part a? c) Suppose p has the form p = – b 1 u 1 – b 2 u˙ 1 + p ( t 1 ) where b 1, b 2 are constants and p is prescribed. Assume c b = 0 and determine the equation for u 1. What effect do b 1 and b 2 have on the response? Illustrate for a periodic excitation.

420 Chapter 5: Base Isolation Systems

421

Part II: Active Control Chapter 6

Introduction to active structural motion control 6.1 The nature of active structural control Active versus passive control The design methodologies presented in the previous chapters provide systematic procedures for distributing passive motion control resources which, by definition, have fixed properties and do not require an external source of energy. Once installed, a passive system cannot be modified instantaneously, and therefore one needs a reliable estimate of the design loading and an accurate numerical model of the physical system for any passive control scheme to be effective. The inability to change a passive control system dynamically to compensate for an unexpected loading tends to result in an over-conservative design. When self-weight is an important design constraint, one cannot afford to be too conservative. Also, simulation studies on the example building structures show that passive control is not very effective in fine tuning the response in a local region. Considering these limitations, the potential for improving the performance by dynamically modifying the loading and system properties exists. An active structural control system is one which has the ability to determine the present state of the structure, decide on a set of actions that will change this state to a more desirable one, and carry out these actions in a controlled manner and in a short period of time. Such control systems can theoretically accommodate unpredictable environmental

422 Chapter 6: Introduction to Active Structural Motion Control changes, meet exacting performance requirements over a wide range of operating conditions, and compensate for the failure of a limited number of structural components. In addition, they may be able to offer more efficient solutions for a wide range of applications, from both technical and financial points of view. Active motion control is obtained by integrating within the structure a control system consisting of three main components: a) monitor, a data acquisition system, b) controller, a cognitive module which decides on a course of action in an intelligent manner, and c) actuator, a set of physical devices which execute the instructions from the controller. Fig 6.1 shows the interaction and function of these components; the information processing elements for active control are illustrated in Fig 6.2. This control strategy is now possible due to significant recent advances in materials that react to external stimuli in a nonconventional manner, sensor and actuator technologies, real-time information processing, and intelligent decision systems.

EXCITATION

STRUCTURE

RESPONSE

ACTUATOR AGENTS TO CARRY OUT INSTRUCTIONS

MONITOR SENSORS to measure external loading

MONITOR DEVELOP THE ACTION PLAN (SET OF INSTRUCTIONS TO BE COMMUNICATED TO THE ACTUATORS)

DECIDE ON COURSE OF ACTION IDENTIFY THE STATE OF SYSTEM

CONTROLLER Fig. 6.1: Components of an active control system

SENSORS to measure response

6.1 The Nature of Active Structural Control 423

Physical System

Sensor

Sensor Sensor

Decision Making

Action

Data Fusion Modeling & Analysis Processing

Transmission channel

Visualization

Processing

Archival and Access

Fig. 6.2: Information Processing Elements for an Active Control System

h( p)

p

∆p f

u

decide on ∆p f (a) Passive decide on changing h to h'

u + εu observe u

+ p

h' ( p + ∆p e + ∆p f )

+ observe p p + εp

(b) Active decide on ∆p e

∆p e

Fig. 6.3: Passive and active feedback diagrams

u

424 Chapter 6: Introduction to Active Structural Motion Control The simple system shown in Fig. 6.3 is useful for comparing active and passive control. Figure 6.3(a) corresponds to passive control. The input, p , is transformed to an output, u , by the operation h ( p ) u = h( p)

(6.1)

One can interpret this system as a structure with p denoting the loading, u the displacement, and h the flexibility of the structure. The strategy for passive motion control is to determine h ( p ) such that the estimated output due to the expected loading is contained within the design limits, and then design the structure for this specific flexibility. Active control involves monitoring the input and output, and adjusting the input and possibly also the system itself, to bring the response closer to the desired response. Figure 6.3(b) illustrates the full range of possible actions. Assuming the input corrections and system modifications are introduced instantaneously, the input-output relation for the actively controlled system is given by u = h' ( p + ∆p e + ∆p f )

(6.2)

Monitoring the input, and adjusting the loading is referred to as open-loop control. Observing the response, and using the information to apply a correction to the loading is called feedback control. The terminology closed-loop control is synonymous with feedback control. In addition to applying a correction to the input, the control system may also adjust certain properties of the actual system represented by the transformation h ( p ) . For example, one can envision changing the geometry, the connectivity, and the properties of structural elements in real time. One can also envision modifying the decision system. A system that can adjust its properties and cognitive processes is said to be “adaptive”. The distinguishing characteristic of an adaptive system is the self-adjustment feature. Non-adaptive active structural control involves monitoring and applying external forces using an invariant decision system. The make up of the structure is not changed. Adaptive control is the highest level of active control. The role of feedback Feedback is a key element of the active control process. The importance of feedback can be easily demonstrated by considering a linear static system and

6.1 The Nature of Active Structural Control 425 taking the input correction to be a linear function of the output. For this case, u = hp

(6.3)

∆p f = k f [ u + ε u ]

(6.4)

where h and k f are constants. Substituting in eqn (6.2) specialized for h’=h and solving for u results in hk f h u = ---------------- [ p + ∆p e ] + ---------------- ε u 1 – hk f 1 – hk f

(6.5)

When k f is positive, the sensitivity of the system to loading is increased by feedback, i.e. the response is amplified. Taking k f negative has the opposite effect on the response. Specializing eqn (6.5) for negative feedback ( k f < 0 ), the response becomes hk f h (6.6) u = -------------------- [ p + ∆p e ] + -------------------- ε u 1 + hk f 1 + hk f Increasing k f decreases the effect of external loading. However, the influence of ε u , the noise in the response observation, increases with k f and, for sufficiently large k f , is essentially independent of the feedback parameter. This result indicates that the accuracy of the monitoring system employed to observe the response is an important design issue for a control system. Computational requirements and models for active control The monitor component identified in Figs 6.1 and 6.2 employs sensors to measure a combination of variables relevant to motion such as strain, acceleration, velocity, displacement, and other physical quantities such as pressure, temperature, and ground motion. This data is usually in the form of analog signals which are converted to discrete time sequences, fused with other data, and transmitted to the controller module. Data compression is an important issue for large scale remote sensing systems. Wavelet based data compression (Amaratunga, 1997) is a promising approach for solving the data processing problem. The functional requirements of the controller are to compare the observed response with the desired response, establish the control action such as the level of feedback force, and communicate the appropriate commands to the actuator

426 Chapter 6: Introduction to Active Structural Motion Control which then carries out the actual control actions such as apply force or modify a structural property. The controller unit is composed of a digital computer and software designed to evaluate the input and generate the instructions for the actuators. There are 2 information processing tasks: state identification and decision making. Given a limited amount of data on the response, one needs to generate a more complete description of the state of the system. Some form of model characterizing the spatial distributions of the response and data analysis are required. Once the state has been identified, the corrective actions which bring the present state closer to the desired state can be established. In this phase, a model which defines the input - output relationship for the structure is used together with an optimization method to decide upon an appropriate set of actions. For algorithmic non-adaptive systems, the decision process is based on a numerical procedure that is invariant during the period when the structure is being controlled. Time invariant linear feedback is a typical non-adaptive control algorithm. An adaptive controller may have, in addition to a numerical control algorithm, other symbolic computational models in the form of rule-based systems and neural networks which provide the capability of modifying the structure and control algorithm in an intelligent manner when there is a change in the environmental conditions. Examples illustrating time invariant linear feedback control algorithms are presented in the following sections; a detailed treatment of the algorithms is contained in Chapters 7 and 8.

6.2 An introductory example of quasi-static feedback control Consider the cantilever beam shown in Fig 6.4. Suppose the beam acts like a bending beam, and the design objective is to control the deflected shape such that it has constant curvature. The target displacement distribution corresponding to this constraint has the form 1 u * ( x ) = --- χ * x 2 2

(6.7)

where χ * is the desired curvature. One option is to select the bending rigidity according to M(x) D B ( x ) = ---------------χ*

(6.8)

6.2 An Introductory Example of Quasi-static Feedback Conrol 427 where M(x) is the moment at location x due to the design loading. This strategy is a stiffness based passive control approach. A second option is to select a representative bending rigidity distribution, and apply a set of control forces which produce a displacement distribution that, when combined with the displacement due to the design loading, results is a displacement profile that is close to the desired distribution. In what follows, the latter option is discussed.

u(x) x F L Fig. 6.4: Cantilever beam with control force Suppose the control force system consists of a single force applied at x=L. Assuming linear elastic behavior, and using the linear technical theory of beams as the model for the structure, the displacement due to F is estimated as F F x3 (6.9) u c ( x ) = ----------- Lx 2 – ----- = -------h ( x ) DB 2D B 3 where D B is the bending rigidity, considered constant in this example. The displacement due to the design loading is also determined with the technical beam theory. This term is denoted as u o ( x ) , and expressed as 1 u o ( x ) = ------- g ( x ) DB

(6.10)

Combining the 2 displacement patterns results in the total displacement, u(x). u ( x ) = uo ( x ) + uc ( x )

(6.11)

The expanded form of eqn (6.11) corresponding to the particular choice of control force location for this example is 1 u ( x ) = ------- [ g ( x ) + Fh ( x ) ] DB

(6.12)

The difference between u ( x ) and u * ( x ) is defined as e ( x ) and interpreted

428 Chapter 6: Introduction to Active Structural Motion Control as the displacement error. e ( x ) = uo + uc – u *

(6.13)

For this example, D B is considered to be fixed, and therefore e ( x ) is a function only of the single control force magnitude F. 1 e ( x ) = ------- [ g ( x ) + Fh ( x ) ] – u * ( x ) DB

(6.14)

Ideally, one wants e = 0 for 0 ≤ x ≤ L . This goal cannot be achieved, and it is necessary to work with a relaxed condition. The simplest choice is collocation, which involves setting e equal to zero at a set of prescribed locations. For example, setting e = 0 at x=L leads to DB u * ( L ) – g ( L ) F = -----------------------------------------(6.15) h(L) A more demanding condition is a least square requirement, which involves first forming the sum of e 2 evaluated at a set of prescribed points, and then selecting F such that the sum is a minimum. The continuous least square sum is given by the following integral 1 J = --2

L

∫

e 2 dx = J ( F )

(6.16)

0

Taking J(F) as the measure of the square error sum, F is determined with the stationary condition ∂J = 0 ∂F

(6.17)

Differentiating the integral expression for J, ∂J = ∂F

L

∫

e

0

∂e dx = 0 ∂F

(6.18)

and using eqn (6.14), which defines e ( x ) for this particular example, results in L

∫

h ( x ) [ D B u * ( x ) – g ( x ) ] dx

0

F = ----------------------------------------------------------------------L

∫

(6.19)

( h ( x ) ) 2 dx

0

The value of F defined by eqn (6.19) produces the absolute minimum value of J. A proof of this statement is presented in Section 7.2 which treats in more detail the

6.2 An Introductory Example of Quasi-static Feedback Conrol 429 least square procedure for quasi-static loading.

Example 6.1: Shape control for uniform loading This example illustrates the application of the approach described above to the case where the design loading is a uniform distributed load extending over the entire length of the beam. The corresponding deflected shape is 1 w 1 u o ( x ) = ------- g ( x ) = ------- ------ ( x 4 – 4x 3 L + 6x 2 L 2 ) D B 24 DB

(1)

Applying collocation at x=L leads to * 3 D B χ 3 F e ( L ) = 0 = --- ------------- – --- wL 2 L 2

(2)

The least square solution is * 91 D B χ 2065 F ls = ------ ------------- – ------------ wL 66 L 5280

DB χ * = ( 1.379 ) ------------- – ( 0.391 )wL L

(3)

Both solutions are approximate since they do not satisfy e ( x ) = 0 . One can improve the performance by taking additional control forces. Selecting the spatial distribution of the control forces is a key decision for the design of a control system.

The above discussion assumes that there is some initial loading, and one can determine the corresponding displacement field with the physical model of the structural system. This control strategy is similar to the concept of prestressing. A more general scenario is the case where one is observing the response at a set of “observation” points and the loading is being applied gradually so that there is negligible dynamic amplification and sufficient time to

430 Chapter 6: Introduction to Active Structural Motion Control adjust the control forces. Here, one needs to establish u o ( x ) using the observed displacement data. Suppose there are s observation points located at x j ( j= 1, 2, ..., s), and at time t the monitoring system produces the data set u o ( x j, t ) . This data can be used together with an interpolation scheme to generate an estimate of u o for the region adjacent to the observation points. A typical spatial interpolation model has the form s

u o ( x, t ) =

∑

u o ( x j, t )Ψ j ( x )

(6.20)

j=1

where Ψ j ( x ) are interpolation functions. Given u o ( x, t ) , one forms the displacement error, 1 e ( x, t ) = u o ( x, t ) – u * ( x ) + -------h ( x )F ( t ) DB

(6.21)

and determines F(t) with either collocation or a least square method. The continuous least square estimate for F(t) is given by L

∫

h ( x ) ( u * ( x ) – u o ( x, t ) ) dx 0

F ( t ) = D B --------------------------------------------------------------------L ( h ( x ) ) 2 dx

(6.22)

∫

0

Example 6.2: Discrete displacement data

Suppose the displacement observation points are located at x=L/2 and x=L. Given these 2 values of displacement, one needs to employ an interpolation scheme in order to estimate u o ( x ) . Taking a quadratic expansion, uo ( x ) = ao + a1 x + a2 x 2

(1)

6.3 An Introductory Example of Dynamic Feedback Control 431

u1 L/2

u2 L/2

x and specializing eqn(1) for points 1 and 2 results in the following approximation, x x 2 x x 2 u o ( x ) ≈ u 1 4 --- – 4 --- + u 2 – --- + 2 --- L L L L

(2)

The expression for the control force is obtained using eqn (6.22) and eqn (2). Evaluating the integrals leads to * 133 91 D B χ D B 98 ----------------F≈ – ------- ------ u 1 + --------- u 2 66 L 3 33 66 L

(3)

as an estimate for F.

6.3 An introductory example of dynamic feedback control To gain further insight on the nature of feedback control, the simple SDOF system shown in Fig. 6.5 is considered. The system is assumed to be subjected to both an external force and ground motion, and controlled with the force F . Starting with the governing equation, mu˙˙ + cu˙ + ku = – ma g + F + p

(6.23)

and introducing the definitions for frequency and damping ratio leads to the

432 Chapter 6: Introduction to Active Structural Motion Control

k m

F

p

c ug

u + ug

Fig. 6.5: Single-degree-of-freedom system.

standardized form of the governing equation 2 F p u˙˙ + 2ξωu˙ + ω u = – a g + ---- + ---m m

(6.24)

The free vibration response of the uncontrolled system has the general form u = Ae

λt

(6.25)

Substituting for u in eqn (6.24), one obtains two possible solutions u = A1 e

λ1 t

+ A2 e

λ2 t

(6.26) 2

λ 1, 2 = – ξω ± iω 1 – ξ = – ξω ± iω′

(6.27)

Considering A 1 and A 2 to be complex conjugates, 1 A 1, 2 = --- [ A R ± iA I ] 2

(6.28)

where A R and A I are real numbers representing the real and imaginary parts of A , the solution takes the form u = e

– ξωt

2 2 A R cos ωt 1 – ξ + A I sin ωt 1 – ξ

(6.29)

6.3 An Introductory Example of Dynamic Feedback Control 433 One determines A R and A I resulting expressions are

with the initial conditions for u and u˙ . The

AR = u ( 0 ) 1 A I = – ----- ( u˙ ( 0 ) + ξωu ( 0 ) ) ω′

(6.30)

Considering negative linear feedback, the control force is expressed as a linear combination of velocity and displacement F = – k v u˙ – k d u

(6.31)

where the subscripts ‘v’ and ‘d’ refer to the nature of the feedback, i.e. velocity or displacement. Feedback is implemented in the actual physical system by: • observing the response • determining u and u˙ • calculating F with eqn (6.31) • applying F with an actuator Mathematically, one incorporates feedback by substituting for F in eqn (6.24). The result is kv 2 kd p u˙˙ + 2ξω + ----- u˙ + ω + ----- u = – a g + --- m m m

(6.32)

Equation (6.32) can be transformed to the standardized form by defining equivalent damping and frequency parameters as follows: 2 2 kd ω eq = ω + ----m

(6.33)

kv 2ξ eq ω eq = 2ξω + ----m

(6.34)

With this notation, the solution for the free vibration response of the linear feedback controlled case has the same general form as for no control; one just replaces ξ and ω with ξ eq and ω eq respectively in eqn (6.29). It follows that the effect of linear feedback is to change the fundamental frequency and damping

434 Chapter 6: Introduction to Active Structural Motion Control ratio. Solving eqns (6.33) and (6.34) for ω eq and ξ eq , results in kd ω eq = ω 1 + ----k

(6.35)

ξ eq = ξ + ξ a

(6.36)

where ξ a is the increment in damping ratio due to active control kv kd 1 ⁄ 2 1 ξ a = ----------------------------- ------------ – ξ 1 + ----– 1 k d 1 ⁄ 2 2ωm k 1 + ----k

(6.37)

Critical damping corresponds to ξ eq = 1 kv -----------= 2mω ξ = 1 eq

kd 1 + ----- – ξ k

(6.38)

Equation (6.35) shows that negative displacement feedback increases the frequency. According to eqn (6.37), the damping ratio is increased by velocity feedback and decreased by displacement feedback. If the objective of including active control is to limit the response amplitude, velocity feedback is the appropriate mechanism. Displacement feedback is destabilizing in the sense that it reduces the effect of damping. Stability and other issues associated with linear feedback are discussed in Chapter 8.

Example 6.3: Illustrative example - influence of velocity feedback This example demonstrates the influence of velocity feedback on the response of 2 SDOF systems subjected to seismic excitation. The properties of the systems are System 1: m = 10, 000kg

ω 1 = 6.32rd ⁄ s

k = 400, 000N ⁄ m

T 1 = 0.99s

c = 2, 500Ns ⁄ m

6.3 An Introductory Example of Dynamic Feedback Control 435 System 2: m = 10, 000kg

ω 1 = 2rd ⁄ s

k = 40, 000N ⁄ m

T 1 = 3.14s

c = 830Ns ⁄ m

ξ = 0.0208

The models are excited with the El Centro and Taft accelerograms scaled to a max = 0.5g . Figures 6.6 through 6.8 contain plots of the maximum relative displacement, maximum control force magnitude, and maximum power requirement. The power requirement is computed using the following expression Power = force × velocity = Fu˙

(1)

which assumes the control force is a set of self-equilibrating forces applied as shown in Fig 6.5. Ground motion has no effect on the work done by F with this force scheme.

436 Chapter 6: Introduction to Active Structural Motion Control

0.16

Maximum displacement - m

System 1

T1=0.99s El Centro Taft

0.14

0.12

0.1

0.08

0.06

0.04 0.05

0.1

0.15

ξa

0.2

0.25

0.3

0.35

System 2 T1=3.14s El Centro Maximum displacement - m

0.3

Taft

0.25

0.2

0.15

0.1 0.05

0.1

0.15

ξa

0.2

0.25

0.3

Fig. 6.6: Variation of maximum displacement with active damping.

6.3 An Introductory Example of Dynamic Feedback Control 437

4

2.4

x 10

System 1 T1=0.99s 2.2

Maximum force - N

El Centro 2

Taft

1.8

1.6

1.4

1.2

1

0.8

0.6

0.4 0.05

0.1

0.15

ξa

0.2

0.25

0.3

7000

T1=3.14s

System 2 6000

El Centro

Maximum force - N

Taft 5000

4000

3000

2000

1000 0.05

0.1

0.15

ξa

0.2

0.25

0.3

Fig. 6.7: Variation of maximum control force level with active damping.

438 Chapter 6: Introduction to Active Structural Motion Control

16000

System 1 T1=0.99s Maximum power - N.m/s

14000

El Centro Taft

12000

10000

8000

6000

4000

2000 0.05

0.1

0.15

ξa

0.2

0.25

0.3

4500

System 2 T1=3.14s Maximum power - N.m/s

4000

El Centro 3500

Taft

3000

2500

2000

1500

1000

500 0.05

0.1

0.15

0.2

0.25

0.3

ξa

Fig. 6.8: Variation of maximum power requirement with active damping.

6.4 Actuator Technologies 439

6.4 Actuator technologies Introduction The actuator component of the control system generates and applies the control forces at specific locations on the structure according to instructions from the controller. Over the past several decades, a number of force generation devices have been developed for a broad range of motion control applications. These devices can be described in terms of performance parameters such as response time, peak force, and operating requirements such as peak power and total energy demand. The ideal device is one that can deliver a large force in a short period of time for a small energy input. Civil structures generally require large control forces, on the order of a meganewton and, for seismic excitation, response times on the order of milliseconds. The requirement on peak force coupled with the constraint on energy demand is very difficult to achieve with a fully active force actuator system. There are force actuator systems that are capable of delivering large force, but they also have a high energy demand. Included in this group are hydraulic, electromechanical, and electromagnetic devices. All these types are based on very mature technology. There is considerable on-going research and development of new force actuators that have a low energy demand. One approach is based on modifying the physical makeup of the device in such a way that only a small amount of energy is required to produce a significant increase in force. Varying the orifice area of a viscous type damper is an example of this strategy. A typical viscous damper can deliver a force on the order of a meganewton. By adjusting the damping parameter, the damper can be transformed into a large scale force actuator. The second approach employs adaptive materials as the force generating mechanism. These materials respond to a low energy input by changing their properties and their state in a nonconventional manner which results in a force. Although these technologies are promising, the current devices can produce only low forces, on the order of a kilonewton, and therefore their applicability for civil structures is limited at this time. There are 2 issues that need to be addressed: i) how the force generation mechanism works, and ii) how the forces are applied to the structure. The first

440 Chapter 6: Introduction to Active Structural Motion Control issue is related to the physical make up and underlying physics of the device. The second question is concerned with how the device is attached to the structure so as to produce the “desired” control force. In what follows, the attachment issue is discussed first, then the state of the art for linear actuator technologies is reviewed, and lastly some adaptive material based actuators are described. Force application schemes The schematic drawing contained in Fig 6.9 shows the typical makeup of hydraulic, electro-mechanical, and electro-magnetic linear actuators. There are 2 basic elements, a piston and a mechanism that translates the piston linearly either by applying a force to one end or by moving the end with a gear mechanism. The interaction of the piston with an adjacent body produces a pair of contact forces F at the contact point and a corresponding reaction force at the actuator support. If the body moves under the action of F, the mechanism usually compensates for this motion such that the force remains constant until instructed by the controller to change the force magnitude.

u

Mechanism

Piston

F

Structure

Actuator Support F Fig. 6.9: Linear actuator

Consider the structural frame shown in Fig 6.10a. Suppose the objective is to apply a horizontal force at point A and there is no adjacent structure which could support the actuator. One option (Fig 6.10b) is to fasten a tendon to point A, pass it over a pulley attached to the base, and then connect it to a linear actuator which can generate a tensile force in the tendon. In this scheme, the actuator reaction force is transmitted directly to the base. A second option would be to place the actuator directly on the structure. The actuator reaction force is now

6.4 Actuator Technologies 441 transmitted to the structure; however, the other end of the piston needs to be restrained in order to generate a control force. If the restraining body is rigidly connected to the structure as shown in Fig 6.10c, the force system is selfequilibrating and the structure “feels” no lateral force. Member AC is in tension. In order to have a “non zero” lateral force acting on the structure, the restraining body must be allowed to move laterally. This objective can be achieved by attaching an auxiliary mass, ma, to the piston and supporting the mass on rollers. The mass moves with respect to the structure with an absolute acceleration equal to F/ma. One specifies the peak force and magnitude of the auxiliary mass, and designs the actuator so that it can provide the required force at that level of acceleration. Since the force is generated by driving the mass, this scheme is referred to as an active mass driver.

A

A

F Tendon Pulley F (b) Active tendon

(a) Active force

a

C F

A

(c) Self-equilibrating forces

ma

F

(d) Active mass driver

Fig. 6.10: Control force schemes The extension of these schemes to a multi-story structure is shown in Fig. 6.11. A linear actuator placed on a diagonal produces a set of self-equilibrating forces which impose a shearing action on the particular story to which it is attached. The other stories experience no deformation since the story shear due to this actuator is zero. It follows that one needs to incorporate active braces

442 Chapter 6: Introduction to Active Structural Motion Control throughout the structure in order to achieve global displacement control. Forces generated with active mass drivers are not self-equilibrating and consequently have more influence on the global displacement response.

F3 F3 T3

F3

F2

F2 T2

T1

(a) Active brace

F2

F1

F1 F1 (b) Active mass driver

Fig. 6.11: Control force schemes for a multi-story structure The previous examples relate to shear beam structures which require forces that act in the transverse direction. For bending beam problems, control force systems that produce bending moments are required. This action can be obtained with linear actuators placed on the upper and lower surfaces, as illustrated in Fig 6.12. The region between A and B is subjected to a constant moment equal to F d. Another scheme is shown in Fig 6.13. The actuator is attached to the beam with rods that provide the resistance to the piston motion, resulting in the selfequilibrating system that produces a triangular moment distribution over the region A-B-C. By combing a number of these actuator-rod configurations, one can generate a piecewise linear bending moment distribution.

6.4 Actuator Technologies 443

F d

F

a) A Fd

B Fd

b)

B

A (-) Fd

c)

Fig. 6.12: Constant moment field L

F/2

F/2 F

F a)

b) C

A

B

(-)

c)

FL/4

Fig. 6.13: Triangular moment field

444 Chapter 6: Introduction to Active Structural Motion Control Linear actuators generate control force systems composed of concentrated forces. For discrete structures such as frames, this type of force distribution is appropriate. However, for continuous structures such as beams and plates, a continuous force distribution is more desirable. One strategy that has been examined is based on using a adaptive material in the form of a thin plate. Fig 6.14 illustrates this approach for a continuous beam. Plates are attached by epoxy to the upper and lower surfaces. Applying a voltage to the plate generates a longitudinal strain. Since the plate is attached to the surface, motion of the plate is restrained and an interfacial shear stress τ ( t ) is generated. This stress produces a distributed control moment m c ( x, t ) equal to m c ( x, t ) = τb f d

(6.39)

where b f is the width of the plate. Spatial and temporal variation of the control force system is achieved by varying the voltage applied to the plate.

film τ beam

mo dx d

+ τ dx

Fig. 6.14: Moment generated by strain actuators Large scale linear actuators Referring back to Fig 6.9, a linear actuator can be considered to consist of a piston and a mechanism which applies a force to the piston and also controls the motion of the piston. This actuator type is the most widely available and extensively used, particularly for applications requiring a large force and short response time. The descriptors hydraulic, electro-mechanical, and electromagnetic refer to the nature of the force generation mechanisms. These devices generally have a high energy demand.

6.4 Actuator Technologies 445 Hydraulic systems generate the force by applying a pressure on the face of a piston head contained within a cylinder. Fig 6.15 illustrates this concept. Fluid is forced in or out of the cylinder through the orifice at C1 to compensate for the piston displacement and maintain a certain pressure. These systems have the highest force capacity of the linear actuator group, on the order of meganewtons (Dorey et al., 1996). Precise control movement and force can be achieved with a suitable control system. Protection against overload is provided by a pressure relief value. The disadvantages of this type of system are: the requirements for fluid storage systems, complex valves and pumps to regulate the flow and pressure, seals, and continuous maintenance. Durability of the seals and the potential for fluid spills are critical issues.

Fig. 6.15: Schematic cross section view - a hydraulic cylinder (Dorey et al, 1996) Electromechanical linear actuators generate the force by moving the piston with a gear mechanism that is driven by an electric motor. The motion, and therefore the force, is controlled by adjusting the power input to the motor. These devices are compact in size, environmentally safe, and economical. Figure 6.16 illustrates the various components of a linear electric actuator system (Raco, www.raco.de). The largest electric actuator that can be ordered “off the shelf” is rated for 600kN force.

446 Chapter 6: Introduction to Active Structural Motion Control

Positioning Control

Amplifier unit for motor

Servomotor

Electric Actuator

Fig. 6.16: Components of linear electric actuator systems (Raco, http//:www.raco.de) Hydraulic and electromechanical actuators are composed of many parts that are in contact with each other, and therefore have a high risk of breakdown. Since electromagnetic actuators are driven by magnetic forces, which do not require mechanical contact, they are theoretically more reliable. Small scale electromagnetic actuators with a force capacity ranging from tens of newtons to several kilonewtons are commercially available. In addition to their compact nature and low voltage and amperage requirements, their response time is low, on the order of milliseconds. These features are ideal for active force generation, and electromagnetic actuators are a popular choice for small scale structures. Large scale electromagnetic actuator technology is still in the research and development phase, and there is no commercial product with a force capacity in the meganewton range available at this time. However, since the force generation mechanism is so simple, it is reasonable to expect that economical designs will eventually emerge, and be viable candidates for controlling large scale structures. Figure 6.17 shows a schematic cross-sectional view of a design concept for a large scale electromagnetic actuator developed by Chaniotakis et al (99) at the Plasma Science and Fusion Center, Massachusetts Institute of Technology.

6.4 Actuator Technologies 447

driving coil

Fc

piston

field coil

Fem

R=Fem Fig. 6.17: Schematic view - MIT Electromagnetic actuator design The unit consists of a cylindrical shell housing and a piston. Two sets of axisymmetric electromagnets are used. The field coil is embedded in the cylinder, and generates a stationary magnetic field. A driving coil is attached to the piston which translates with respect to the housing. The force mechanism is based on the interaction between the magnetic field generated by the stationary field magnet and the current in the driving coil. For this particular design, the electromagnetic force is a linear function of the coil currents and is independent of the position of the piston. F em = cI f I d

(6.40)

where I f , I d are the currents in the field and driving coils, and c is a design parameter. The equation of motion for the piston relates the electromagnetic force, the inertia force for the piston mass, and the external contact force, F c . m p u˙˙ = F em – F c

(6.41)

When the device is used as an active mass driver, an auxiliary mass, m a , is

448 Chapter 6: Introduction to Active Structural Motion Control attached to the end of the piston, and F c is set equal to m a u˙˙ . The reaction at the base of the housing is equal to the electromagnetic force which acts on the field coil and is transmitted to the housing. There are several advantages to this concept: i) the response time is on the order of milliseconds, ii) there is minimal mechanical contact, and iii) the technology for controlling the current is mature. The disadvantages are: i) the current and voltage requirements for a force on the order of meganewton cannot be satisfied with conventional electrical power supply technology, and ii) there is minimal experience related to the design, fabrication, and operating performance of large scale electromagnetic actuators. Large scale adaptive configuration based actuators This category includes mechanical devices such as dampers, friction elements, and stiffness elements which have the ability to generate a large force by changing their physical makeup. Their distinguishing features are their low ratio of energy demand to force output, and their large scale force capacity. These features are very desirable for applications to large scale civil structures. Adaptive devices which have a low energy demand and also function as energy dissipation mechanisms are referred to as semi-active actuators since they behave like passive devices in the sense that they increase the stability of a structure and require no external energy. The key point is stability. A semi-active actuator will never destabilize a system whereas an active actuator may destabilize a system even though it has a low energy demand. The use of a variable orifice damper as a force actuator was suggested by Feng et al. (1990), and developed further by Shinozuka et al (1992). Kurata et al. (1994) implemented variable dampers in a large scale 3 story frame structure and Sack and Patten (1994) installed a hydraulic actuator with a controllable orifice on a bridge on interstate highway I-35 in Oklahoma. Figure 6.18 illustrates the concept of a variable orifice damper. The flow control valve is operated by a controller which in turn responds to the demand for a particular magnitude of force specified by the feedback control algorithm; the controller adjusts the valve opening according to the force demand. Since the valve motion is perpendicular to the flow, the force required to adjust the valve position is small, and therefore the energy demand is low. Kurata’s experiment required only 30 Watts to operate a valve. Letting v denote the velocity of the piston, the damping force, F, can be

6.4 Actuator Technologies 449 expressed as F= c v

(6.42)

where c is a function of v and the geometry of the valve. In the case of a passive damper, c is a prescribed function of v, and F is determined uniquely by specifying v. In the case of a variable damper, c is a function of the valve position as well as v. When used as an actuator, F is specified, v is observed, c is calculated with eqn (6.42), and the valve position is determined using the values of c and v. The limitation of this device is the dependency of the sense of F on the sense of v; F is always opposite in sense to v. Therefore, if the force required by the control algorithm at a particular time has the same sense as the observed velocity, the force demand cannot be met, and the actuator needs to be inactivated until a later time when the phase is reversed.

Fig. 6.18: Variable damping mechanism (Kajima Corporation, http://www.kajima.com) Small scale adaptive material based actuators Low force capacity electromechanical and electromagnetic linear actuators are standard “off-the-shelf” products offered by a number of suppliers. Our interest here is not with these devices but rather with a new generation of small

450 Chapter 6: Introduction to Active Structural Motion Control scale force actuators that utilize the unique properties of adaptive materials to generate the force. Research and development in this area was initiated by the aerospace industry as a potential solution for shape control of satellite arms and airplane control surfaces. As the technology evolved, other applications related to motion control of small scale structures such as robot arms and biomedical devices have occurred. Although the technology continues to evolve, and reliability is still a major concern, these devices are being seriously considered as candidates for force control where the required force level is on the order of a kilonewton. Brief descriptions of the various adaptive material based actuators are presented below. • Piezoelectric actuators Piezoelectric materials belong to the electrostrictive material category. When subjected to a voltage, they undergo a molecular transformation which results in the material extending (or contracting) in a manner similar to the Poisson effect for applied stress (Ikeda, 1997). Figure 6.19 illustrates this behavioral mode; a voltage V z in the Z direction produces extensional strains ε x and ε y in the X and Y directions. The opposite behavioral mode occurs when the material is stressed in the X-Y plane; a voltage V z' in the Z direction is generated by σ x and σ y . This behavior was first reported by Pierre and Jacques Curie in 1880. Pierre later won a Nobel price in Physics with his wife Marie for their work on radioactivity. Historically, piezoelectric materials have been used as strain sensors. Their use as actuators is more recent and stimulated primarily by the Aerospace Industry. (Crawley, 1987)

6.4 Actuator Technologies 451

VZ

Z

V' Z

εy

σy Y

X

εx

σx

Fig. 6.19: Piezoelectric electrical-mechanical interaction Piezoelectric actuators are fabricated with piezoceramic block type elements or piezopolymer films. Lead zirconate titanate (PZT) is the dominant piezoceramic composite used for sensors and actuators in the frequency range up to 10 6 Hz. Polyvinyldene fluoride (PVDF) is the most common piezoelectric film. Since it has a relatively low strength, PVDF is used mainly as a sensor, particularly for the high frequency range up to 10 9 Hz. The underlying principle is the same for both materials. The piezoelectric object is attached to a surface which restrains its motion. When a voltage is applied, the object tends to expand immediately, and contact forces between the object and the restraining medium are produced. Two actuator configurations have been developed. The first model is a conventional linear actuator, such as shown in Fig 6.20. Piezoceramic wafers are stacked vertically, bonded, enclosed in a protective housing, and fitted with electrical connectors. These devices can deliver large forces, on the order of 20 kN, with a response time of several milliseconds (Kinetic Ceramics Inc., www.kineticceramics.com).

452 Chapter 6: Introduction to Active Structural Motion Control

Fig. 6.20: Cylindrical piezoceramic linear actuator (Kinetic Ceramics Inc) The second configuration has the form of a thin plate as illustrated in Fig 6.21. Piezoceramic wafers are distributed over the area in a regular pattern. They may also be stacked through the thickness. This type of device is bonded to a surface, and applies a pair of self equilibrating tangential forces to the surface. The peak force depends on the applied voltage and degree of restraint. A force level of 500 N at 200 volts, and millisecond response, is the typical upper limit for off-the-shelf plate type piezoceramic actuators (Active Control Experts, www.acx.com).

impressed voltage

bond

wafer

bond

Fig. 6.21: Plate type piezoelectric actuator • Shape memory alloys Shape memory alloys are metal alloys which, if deformed inelastically at

6.4 Actuator Technologies 453 room temperature, return to their original shape when heated above a certain temperature. Figure 6.22 illustrates this behavior. The initial straight form is deformed inelastically at room temperature to the curved form. When the temperature is elevated, the curved form shifts back to the straight form, and remains in that form when the temperature is lowered to room temperature. If no further deformation is introduced, the straight form remains invariant during any subsequent thermal cycling.

a) Initial shape at room temperature

b) Deformed shape at room temperature

c) Shape at elevated temperature

d) Shape after cooling to room temperature Fig. 6.22: One Way Shape Memory Behavior The ability to return to its initial shape when heated is due to a phase transformation from martensite at room temperature to austenite at elevated temperature (Wayman et al., 1972). Inelastic deformation introduced during the martensite phase is eliminated when the state passes over to the austenite phase. The phase transitions are illustrated in Fig 6.23: As and Af define the temperatures for the start and finish of the transition from martensite to austenite for the case when the material is being heated; the corresponding temperatures for the cooling case are Ms and Mf. When T is greater than Af, the phase is austenite but it is possible to convert it back to martensite by applying stress. The quantity Md is the temperature beyond which austenite cannot be converted to martensite by

454 Chapter 6: Introduction to Active Structural Motion Control stress, i.e., the phase remains austenite for arbitrary applied stress.

As

Volume fraction martensite

Mf 1

0.5

0 Ms

Af

Md

Temperature, T Fig. 6.23: Martensitic transformation on cooling and heating The stress-strain behavior is strongly dependent on temperature. Figure 6.24 shows the limiting stress-strain curves for Nitinol, a nickel-titanium alloy (Jackson et al, 1972). For T<Mf, the lowest temperature for the fully martensitic phase, the material behaves like a typical ductile metal. Yielding occurs at about 150 MPa, and inelastic deformation is introduced. The behavior for T>Md is elastic up to 650 MPa, and the modulus, E1, is about 4 times larger than the initial value, E0. Between these limiting temperatures, there is a graduation in behavior between fully ductile to fully elastic.

6.4 Actuator Technologies 455

σ (MPa)

700 c’ 600

T > Md

500 T < Mf c’’

400 σ∗

b’

b

300 E1 200

100 E0

E0 a 0

2

4

c

6

8

10

ε (%)

Fig. 6.24: Effect of temperature on stress strain behavior of Nitinol (Jackson et al., 1972) Referring to Fig 6.24 the one way effect can be explained by tracking the response as a stress, σ * , is applied and then removed at T<Mf. This action produces the path a-b-c. Increasing the temperature at this time to T>Md shifts c

456 Chapter 6: Introduction to Active Structural Motion Control back to a since the behavior for T>Md is elastic. Further cycling of the temperature with no stress applied will not cause the point to shift from a. A different type of response is obtained when the stress is held constant and the temperature is cycled a number of times between Mf and Md. In this case, the path is b-b’-b. The effect of thermal cycling is to introduce 2 way shape memory. Repeating the scenario described above, starting at b and removing the stress shifts the b position to c. Then, increasing the temperature causes the position to shift to a. However, when the temperature is now lowered to Mf, the position shifts back to point c instead of remaining at a. Further thermal cycling at no stress results in the deformation switching between a and c. At high temperature, the material remembers the initial state; at low temperature it remembers the deformed state. Figure 6.25 is a modified version of Fig 6.22 which shows this behavioral mode. Applying the stress and subsequent thermal cycling trains the material to remember 2 shapes (Fremond et al, 1996).

a) Deformed shape at low temperature

b) Initial shape at elevated temperature Fig. 6.25: Two-way shape memory behavior The two-way shape memory behavior provides the basis for force actuation. If a trained shape memory alloy is restrained at low temperature so that it cannot deform, a force is generated when the alloy is heated since it wants to return to its initial undeformed shape. Referring back to the stress-strain plots in Fig 6.24, suppose the position at low temperature is point c. When heat is applied, the behavior is governed by the curve for T>Md. The material reacts as if it were subjected to the positive strain a-c, and the position jumps to c’. This behavior is “ideal”; the actual position is lower, such as point c’’, but the induced stress is still a significant value, on the order of 300 to 400 MPa.

6.4 Actuator Technologies 457 Nitinol alloys in the form of small diameter wires ( ≈ 0.4mm ) are used to assemble a force actuator. Heating is applied by passing an electric current through the wire. This process limits the response time to seconds vs. milliseconds for piezoelectric materials. Another limitation is the material cost; a typical price is on the order of $300/kg. Most of the applications of shape memory actuators are for products requiring low capacity, and where cost and response time are not critical issues. • Controllable fluids Controllable fluids are characterized by their ability to change from a fluid to a semi-solid in milliseconds when subjected to an electric or a magnetic field. This effect was identified by Winslow in 1949. Two materials belonging to this category are electrorheological (ER) and magnetorheological (MR) fluids. In their initial state, they behave as viscous fluids. Application of the field (electric or magnetic) introduces an additional plastic solid type behavior mode, and the response is now a combination of plastic and viscous actions. Figure 6.26 illustrates this transformation for the case where the material is idealized as a Bingham solid, which is defined as an ideal plastic solid in parallel with a linear viscous fluid. The stress-strain relation for a Bingham solid subjected to shear deformation has the form: τ = τ y sgn ( γ˙ ) + ηγ˙

(6.43)

where τ y denotes the yield stress and η is the viscosity. Experimental results show that the yield stress increases significantly with the field strength, f, up to a limiting value τ max . However, the viscosity is essentially constant (Carlson, J.D. et al, 1995). Therefore, it is reasonable to take η equal to a constant, η o , and interpret the material behavior as a combination of variable coulomb hysteretic damping and constant linear viscous damping. The equivalent linear material viscosity, η eq , is dependent on the amplitude and frequency of the shear deformation. For the case of periodic excitation, η eq is given by 4τ y (6.44) η eq = η o + ----------πΩγˆ where γˆ is the shear strain amplitude, and Ω is the excitation frequency.

458 Chapter 6: Introduction to Active Structural Motion Control

Saturation limit τ

f 2 > f1 η

f1

τy,max τy

f=0

γ˙

f= field strength measure

Fig. 6.26: Effect of electric and magnetic fields on the stress-strain relationship Devices containing controllable fluids can be used as either variable dampers or as semi-active force actuators. Figure 6.27 shows a schematic view of a prototype developed by the Lord Corporation (Lord Corp., www.lordcorp.com). The main cylinder houses the piston, the MR fluid, and the magnetic circuit. A small electromagnet is embedded in the piston head and supplied with current that generates the magnetic field across the annular orifice. Typical small scale versions have a force capacity of about 3 kN, millisecond response, and draw about 10 watts of power. A large scale version with capacity of 200 kN and millisecond response has also been produced (Spencer, B.F. et al, 1998).

6.4 Actuator Technologies 459

Fig. 6.27: Rheonetics SD-1000-2 MR damper (Pang, L. et al, 1998) Experimental results for quasi-static loading applied to the small scale version are plotted in Figure 6.28. These results show that the Bingham solid idealization is a reasonable approximation for the actual behavior, and also that the plastic yield force is limited by saturation of the fluid. For this device, saturation occurs in the vicinity of 1.5 amps. When used as a force actuator, the amperage is adjusted such that, for the observed value of velocity, the desired

460 Chapter 6: Introduction to Active Structural Motion Control force magnitude is generated. A controllable fluid type device is more effective than a variable orifice damper since the yield force is the primary component, and this force is independent of velocity. The low external power requirement, rapid response, and the potential for large capacity are attractive features. With further development, MR based actuators should be competitive solutions for large scale civil structures.

Fig. 6.28: Quasi-steady force velocity response for Rheonetics SD-1000-2 MR damper (Pang, L. et al., 1998)

6.5 Examples of existing large scale active structural control systems Kajima Corporation has pioneered the research, design, experimentation, and implementation of active control of large scale building structures (Sakamoto et al. 1994). Their work has been concerned with the following active motion control schemes: active mass driver (AMD), active variable stiffness (AVS), and hybrid mass damper (HMD). These schemes have been implemented in the set of buildings listed in Table 6.1. Other Japanese organizations such as Shimizu Corp., Takenaka Corp., and Mitsubishi have also carried out substantial research and

6.5 Examples of Existing Large Scale Active Structural Control Systems 461 implementation in the field of active structural control. In the USA, various government organizations such as NSF, ARO, AFOSR, ONR are funding research on active control of structures. A description of some of the implementations is given below. A detailed assessment of the performance of these structures during the earthquakes and strong winds that they have already encountered is found in Sakamoto et al. (1994). Table 6.1: Implementations of active and hybrid control systems in buildings Kyobashi Seiwa

KaTRI No.21

Completion date

Aug. 1989

Nov. 1990

July 1993

April 1994

Jan. 1995

Number of floors

11+1(BG)

3+2(BG)

14+2(BG)

52+5(BG)

29+3(BG)

Building height (m)

33.10

16.30

68.00

232.60

144.45

Total floor area (m2)

423.37

465.00

4,928.30

264,140.91

30,369.66

37.32

150.00

324.15

4,523.54

1,072.36

Control system

AMD

AVS

HMD

HMD

HMD

Type of disturbance

ME,SW

LE

ME,SW

ME,SW

ME,SW

2)

Typical floor area (m

Ando Shinjuku Nishikicho Park

Dowa Kasai Phoenix

Name of Building

(by Kajima Corporation) Note:

BG: Below ground level. ME: Moderate earthquake. LE: Large earthquake. SW: Strong wind.

AMD in Kyobashi Seiwa Building (Kajima 91-63E, 1991a) The Kyobashi Seiwa Building (see Fig 6.29) is a very slender building with a width of only 4m, a length of 12m, and a height of 33m (11 floors). It is constructed of rigidly connected steel frames consisting of box columns and H-shaped beams. The total structural weight is about 400 metric tons (1 metric ton equals 104 newtons). An active mass driver (AMD) system is installed on the top floor. The objective of the AMD is to reduce the maximum lateral response associated with 2 frequent earthquakes (i.e. peak ground acceleration of 10cm ⁄ s ) and strong winds (i.e. maximum speed of 20m ⁄ s ) to about one third of the uncontrolled value.

462 Chapter 6: Introduction to Active Structural Motion Control

a. AMD System installation

b.Schematic of AMD Fig. 6.29: AMD for Kyobashi Seiwa Building

6.5 Examples of Existing Large Scale Active Structural Control Systems 463 The AMD system is illustrated in Fig 6.29(b). The basic components are: • Sensors installed at several locations including: ground level, midheight, and the roof level to detect seismic motions and tremors at the ground level and in the building. • The control computer which analyzes each signal and issues a drive order. The control algorithm of the system is of the closed-loop control type where the active control force is determined through linear velocity feedback of the structure’s response. • Actuators which execute the control order and drive the masses. The hydraulic pressure source for the actuator consists of two pumps and an accumulator. One pump is small in comparison to the other; its function is to provide a minimum level of pressure continuously. The larger pump is activated when the earthquake occurs. • Two added masses driven by the actuators. Lateral vibration in the width direction is controlled with one mass (weighing about 4 metric tons, about 1% of the building weight) located at the center of the building. Torsional vibration is reduced with a second mass, weighing about 1 metric ton, located at one end of the structure. These masses are suspended by steel cables to reduce the frictional effects. The lag in the response is about 0.01s . AVS Control at Kajima Technical Research Institute (Kajima 91-65E, 1991b) An active variable stiffness (AVS) device is a non-resonant control type system which is designed to reduce the energy input to buildings from external excitation by continuously altering the building’s stiffness, based on the nature of the excitation, to avoid a resonant state. The major advantage of an AVS is that it can be activated and driven with a small amount of power, partially solving the energy requirement problem of active control. An AVS system was implemented in a steel structure, three stories high, weighing about 400 metric tons, located at the Kajima Technical Research Institute (see Fig 6.30). The lateral bracing system consists of steel braces placed in the transverse direction (gable side), and variable stiffness devices (VSD) installed between the brace tops and the lateral beams. These devices alter the building’s stiffness by shifting from the locked mode (engage: brace is effective) to the

464 Chapter 6: Introduction to Active Structural Motion Control unlocked mode (release: brace is ineffective). The variable stiffness range of the building was designed such that the natural resonant frequency is about 2.5Hz for the locked condition and about 1Hz for the unlocked condition. Auxiliary reinforcing braces are installed in the longitudinal direction to increase the building’s stiffness so that the control for the transverse direction is executed with minimum torsion.

a. View of building

6.5 Examples of Existing Large Scale Active Structural Control Systems 465

b.Schematic of locking device

c. Close up view of locking device Fig. 6.30: Kajima Technical Research Institute AVS System The AVS system consists of: • A measurement and control device consisting of an accelerometer placed on the first floor of the building which feeds the earthquake input into a motion analyzer. The analyzer consists of several special band-pass filters which approximate the response transfer characteristics to each stiffness type. Three stiffness types (eight stiffness types are possible) with mutually different resonant frequencies were selected for the building: all braces locked, all braces

466 Chapter 6: Introduction to Active Structural Motion Control unlocked, and only the braces in the bottom floor locked. Based on the filtered output, the control computer selects the instantaneous stiffness that yields the minimum building response. The control interval required to judge stiffness selection is 0.004s. To maximize the reliability and redundancy of the system, the control computer system is made up of a host computer which determines the operation of the entire system, and three controllers. As the host computer requires extremely fast data acquisition and processing, the real-time UNIX operating system was used. Personal computers utilizing MS-DOS were selected for the controllers which are placed between the host computer and each VSD. • The VSD switches over the connection condition between the brace and the beam in accordance with the signals from the control computer. A VSD consists of a two-ended-type enclosed hydraulic cylinder with a regulator valve inserted in the tube connecting the two cylinder chambers. The open/close function is controlled by oil movements, thus locking or unlocking the connection between the beam and the braces. Twenty watts of electric power per device is required for the valve function. The time needed to alter the lock/unlock condition is about 0.03s. • The emergency power source counteracts power blackout and enables the entire control system to continue to operate for at least 30 minutes even after sudden termination of the regular power supply. Moreover, in case of sudden power termination, the mechanism will cause the devices to automatically adopt a locked condition, thus increasing the building’s strength capacity. Under normal conditions, all the braces are kept in the locked condition. DUOX Active-Passive TMD in Ando Nishikicho Building (Kajima 93-82E, 1993) A combination of an AMD and a TMD, the DUOX system is installed in the Ando Nishikicho Building show in Fig 6.31. This structure consists of four main steel columns located at the corners, 14 stories above ground, and two basements. The building is located in an area of Tokyo that has mainly small, low-rise buildings, and consequently is susceptible to strong winds. The DUOX system was installed near the center of the top floor (i.e. at the building’s center of gravity) to control vibration in both horizontal directions.

6.5 Examples of Existing Large Scale Active Structural Control Systems 467

a. View of building

468 Chapter 6: Introduction to Active Structural Motion Control

b. Schematic of control system

c. Conceptual diagram of DUOX

6.5 Examples of Existing Large Scale Active Structural Control Systems 469

d. Bi-directional control Fig. 6.31: Nishikicho building motion control system

The DUOX system consists of: • Sensors installed in the basement and on the top floor of the building to monitor the ground acceleration and building motion. Sensors are also installed on the AMD and TMD to measure the motion of these devices. • A digital control computer which receives the feedback signals from the sensors, analyzes them, and determines the optimal control forces that will achieve the required control effect and also maintain the stroke of the AMD within the allowable range. • A passive TMD weighing 18 metric tons, about 0.8% of the weight of the above ground portion of the building (about 2,600 metric tons). The TMD is supported by laminated rubber bearings which provide the required stiffness. Oil dampers provide additional viscous damping in the system.

470 Chapter 6: Introduction to Active Structural Motion Control • Two AMD units, driven by AC servo motors and ball screws and weighing about 10% of the weight of the TMD (about 0.08% of the building’s weight). The units are mounted one on top of the other in a criss-cross manner to provide control in the two horizontal directions. The mass of the TMD moves out of phase with the building so that the building motion is always being retarded. The active mass is driven in the direction opposite to that of the TMD so as to magnify the motion of the TMD. When the building response falls within the allowable zone, the AMD operates to dampen out the motion of the TMD. ABS - 600 Ton Full-Scale Test Structure (Soong et al. 1991, Reinhorn et al. 1993) This 600 metric ton test structure, located in Tokyo, Japan, consists of a symmetric two-bay six-story building, 10m x 10m x 21.7m, constructed of rigidly connected steel frames composed of box columns and W-shaped beams with reinforced concrete slabs at each of the floors. The structure was intentionally designed to be a relatively flexible structure in order to simulate a typical high-rise building. The fundamental periods are 1s and 1.5s. The spectral properties of the structure obtained both analytically (with a shear beam model) and experimentally, were within 3% for all modes. A peak ground acceleration of 0.1g was taken as the design loading. The control forces were applied with an Active Bracing System (ABS). The ABS system consists of: • Velocity sensors, with an output range of ± 100cm ⁄ s , installed at the base, the first, the third, and the sixth floor, and an accelerometer, with 2 an output range of ± 1000cm ⁄ s , placed at the base of the structure. This data is fed into the controller and used as the state variables for feedback. • A PC-Limited 386 microcomputer facilitated with Analog-to-Digital (A/D) and Digital-to-Analog (D/A) boards is used for on-line computation. The control logic is implemented on an Intel 80386 25MHz processor equipped with an Intel 80387 25-MHz coprocessor. • Two algorithms based on classical linear optimal control theory were used. One control model works with the complete state vector and requires about 0.014s to process the data at each time point. The other

6.5 Examples of Existing Large Scale Active Structural Control Systems 471 control law is specialized for a reduced state vector corresponding to a 3DOF approximation which considers the translations of the first, third, and sixth floors as variables. The computation time for this algorithm is estimated at around 0.005s. • A biaxial active bracing system (ABS) consisting of a pair of hydraulic actuators (length 73.5cm, piston diameter 152.4mm, rod diameter 63.5mm) aligned in each of the principal directions and located at the first floor level. The actuators, monitored by two hydraulic servovalve controllers with auxiliary hydraulic accumulators, are capable of generating 685kN of control force in each direction with a time delay between the command signal and the actual actuator response of about 0.012s. The primary design parameters for the ABS are the required control force, the maximum displacement of the actuator ( ± 5cm ), and the actuator velocity (6.6cm/s). These parameters influence the actuator capacity, cylinder stroke, and flow rate requirement of the hydraulic servovalve. The servo-controlled system was designed to pump the hydraulic oil at a constant speed during control action, and the difference between the required and supplied oil flow is adjusted through the hydraulic accumulators. Each actuator is equipped with a linear variable displacement transducer (LVDT) having a range of ± 12mm , which is used to adjust the length of the brace via the servovalve loop. • Fail-safe features are incorporated in the hardware and software to protect the system in the event of a failure of one or more components of the system.

472

Introduction to Structural Motion Control

Problems 473

Problems Problem 6.1

b 0 = 20 kN/m

x L = 10 m

Consider the cantilever beam shown above. Assume transverse shear deformation is negligible. The desired state is constant curvature and u ( L ) ≤ L ⁄ 400 . a) Determine the bending rigidity distribution based on the equilibrium model, M(x) D B = ------------χ∗

(1)

b) Determine a “constant” value of D B using the least square procedure described below. The displacement function for the triangular loading is given by: b0 L 4 1 1 u 0 = ------------ x 2 1 – --- x + ------ x 3 = D B– 1 g ( x ) 6D B 2 20

(2)

where x = x ⁄ L . One forms the “displacement” error e ( x ) = u 0 ( x ) – u∗ ( x )

(3)

and the error functional, L

J =

∫0 e2dx

(4)

474 Chapter 6: Introduction to Active Structural Motion Control Let D B– 1 = f B . The value of f B is found by minimizing J with respect to f B . ∂J --------- = 0 ∂fB

(5)

Establish the expression for f B . c) Assume D B is specified. The displacement u 0 is now fixed. Suppose a vertical force, F , is now applied at x = L , and the magnitude of F can be adjusted. Determine a value for F by minimizing J with respect to F . The appropriate form of the error is e = u 0 + u c – u∗ , where F FL 3 x3 u c = ----------- x 2 – ----- = -------h ( x ) DB 2D B 3

(6)

Evaluate F for a representative range of D B . d) Let e i represent the displacement error at point x i . e i = u 0 ( x i ) + u c ( x i ) – u∗ ( x i )

(7)

Suppose one uses a “discrete” error functional consisting of N terms. 1 J = --2

N

∑ ei2

(8)

i=1

Derive the general expression for F corresponding to requiring J to be stationary. Evaluate F taking D B constant = 1/2 the value obtained in part b, N = 3 , and x i = 1/3, 2/3, and 1. Note: it is convenient to introduce matrix notation and express the error as: e =

e1 e2 = e + f F e3

(9)

1 J = --- e T e 2 e) Refer to part d. Consider 2 control forces, F 1 at x = L ⁄ 2 and F 2 at

Problems 475 x = L . Evaluate F 1 , F 2 for the conditions specified in part d. Compare the displacement solutions for part d and part e. f) Suppose one introduces weights for the error at the discrete points, x i , and also introduces a cost associated with the control force F applied at x = L. The expanded functional has the form, 1 J = --2

N

∑

1 q i e i2 + --- rF 2 2

(10)

i=1

where q i is the weight for e i and r is the cost weight for F . Increasing r places more weight on F vs. e . Similarly, increasing q i places more weight on e i . Derive the expression for F corresponding to requiring J to be stationary. Rework part d using this formulation, taking q 1 = q 2 = q 3 = 1 and r = r∗ ⁄ D B2 . Investigate the variation of F with r∗ .

Problem 6.2

b0

u

x L Consider a cantilever beam subjected to a uniform loading. Take D B = EI as constant and neglect transverse shear deformation. The deflected shape is given by: b0 L 4 u = ------------ [ x 2 ( 6 – 4x + x 2 ) ] 24EI

476 Chapter 6: Introduction to Active Structural Motion Control where x = x ⁄ L . Suppose the ‘‘desired’’ deflected shape is u = aLx 2 where a is a specified value. a)

M

x L Consider a concentrated moment, M , applied at x = L to ‘‘correct’’ the difference between the actual and desired deflected shape. How would you determine M ? b) m0

x L Suppose a uniform distributed moment, m 0 , is used as the control force system. How would you determine m 0 ?

Problems 477 Problem 6.3

v P

M

L

L

The structure show above consists of two single spans interconnected at the interior support with a force actuator system that can instantaneously apply a moment, M(t). The structure also has the capability to observe the transverse displacement at a limited number of points along both spans. Consider a force, P, which moves with velocity v along the span. Assume the velocity is sufficiently small such that the actual response of the structure can be approximated by the corresponding static response. Describe how you would establish the value of the actuator moment, M , such that the maximum transverse displacement at any time is less than a specified value, u*.

478

Introduction to Structural Motion Control

479

Chapter 7

Quasi-static Control Algorithms 7.1 Introduction to control algorithms Referring back to Fig 6.1, an active structural control system has 3 main components: i) a data acquisition system that collects observations on the excitation and response, ii) a controller that identifies the state of the structure and decides on a course of action and iii) a set of actuators that apply the actions specified by the controller. The decision process utilizes both information about how the structure responds to different inputs and optimization techniques to arrive at an “optimal” course of action. When this decision process is based on a specific procedure involving a set of prespecified operations, the process is said to be algorithmic, and the procedure is called a “control algorithm”. A non-adaptive control algorithm is time invariant, i.e., the procedure is not changed over the time period during which the structure is being controlled. Adaptive control algorithms have the ability to modify their decision making process over the time period, and can deal more effectively with unanticipated loadings. They also can upgrade their capabilities by incorporating a learning mechanism. This text is concerned primarily with time invariant control algorithms which are well established in the control literature. Adaptive control is an on-going research area which holds considerable promise but is not well defined at this time. A brief discussion is included here to provide an introduction to the topic. The topic addressed in this chapter is quasi-static control, i.e., where the

480 Chapter 7: Quasi-static Control Algorithms structural response to applied loading can be approximated as static response. Since time dependent effects are neglected, stiffness is the only quantity available for passive control. Active control combines stiffness with a set of pseudo-static control forces. The quasi-static case is useful for introducing fundamental concepts such as observability, controllability, and optimal control. Both continuous and discrete physical systems are treated. The next chapter considers time- invariant dynamic feedback control of multi-degree-of freedom structural systems. A combination of stiffness, damping, and time dependent forces is used for motion control of dynamic systems. The state-space formulations of the governing equations for SDOF and MDOF systems are used to discuss stability, controllability, and observability aspects of dynamically controlled systems. Continuous and discrete forms of the linear quadratic regulator (LQR) control algorithm are derived, and examples illustrating their application to a set of shear beam type buildings are presented. The effect of time delay in the stability of LQR control, and several other linear control algorithms are also discussed.

7.2 Active prestressing of a simply supported beam Passive prestressing The concept of introducing an initial stress in a structure to offset the stress produced by the design loading is known as prestressing. This strategy has been used for over 60 years to improve the performance of concrete structures, particularly beams. The approach is actually a form of quasi-static control, where the variables being controlled are the stresses. Figure 7.1 illustrates prestressing of a single span beam with a single cable. When the cable shape is parabolic, the tension introduced in the cable creates an “upward” uniform loading, wo, that is related to the tension by wo L 2 (7.1) ------------- = T 8d The initial moment distribution is parabolic, and the moment is negative according to the conventional notation.

7.2 Active Prestressing of a Simple Supported Beam 481

d cable L

d = distance between top and bottom location of the cable

wo

(-)

woL2/8

M+

P

PL/4

(+) M+ Fig. 7.1: Passive prestressing scheme

Suppose the design loading is a concentrated force that can act at any point on the span. The maximum positive moment due to the force occurs when the force acts at mid-span, and the resultant positive moment at mid-span is given by 2 L PL w o L (7.2) M --- = ------- – ------------ 2 8 4 The initial mid-span moment is negative and equal to wo L 2 L M --- = – ------------ 2 8

(7.3)

If the prestress level is selected such that wo L 2 1 PL ------------- = --- ------- = M * 2 4 8 which requires M* T = -------- ≡ T * d

(7.4)

(7.5)

482 Chapter 7: Quasi-static Control Algorithms then the maximum positive and negative moments are equal. The cross section can now be proportioned for M * , which is 1/2 the design moment corresponding to the case of no prestress. This reduction is the optimal value; taking T > T * will increase the initial moment beyond M * and result in the cross-section being controlled by the initial prestress. The limitation of this approach is the need to apply the total prestress loading prior to the application of the actual loading. Since the tension is not adjusted while the loading is being applied, the scheme can be viewed as a form of passive control. The best result that can be obtained with prestressing for this example is equal design moment values for the unloaded and loaded states. Active prestressing Suppose the cable tension can be adjusted at any time. The equivalent uniform upward loading due to the cable action can now be considered to be an active loading. Deforming w a as the equivalent active loading and noting eqn (7.1), the loading is related to the “active” tension force, T(t ), by 8d w a = ------ T ( t ) L 2

(7.6)

The time history of T(t ) can be established using simple static equilibrium relations. Given the spatial distribution and time history of the loading, T(t ) is determined such that the maximum moment at any time is less than the design moment M * for the cross-section. When the applied loading is uniformly distributed, the moment distributions are similar in form. The expression for the net positive moment has the form 1 M ( x, t ) = ( w ( t ) – w a ( t ) ) --- ( Lx – x 2 ) 2

(7.7)

Enforcing the constraint on the maximum moment, which occurs at mid-span, L2 ( w ( t ) – w a ( t ) ) ------ ≤ M * 8

(7.8)

results in the following control algorithm, 8 1. w a ( t ) = 0 for w ( t ) ≤ ------ M* ≡ w s L2 2. w a ( t ) = w ( t ) – w s for w ( t ) > w s

(7.9)

7.2 Active Prestressing of a Simple Supported Beam 483 No action needs to be taken until w reaches w s , since the maximum moment is less than M * . Above this load level, the active loading counteracts the difference between w ( t ) and w s . With active prestressing, the constraint imposed on the initial prestressing is eliminated. Theoretically, the total applied load can be carried by the active system for this example. This result is due to the fact that the moment distributions for the actual and active loadings have the same form. When these distributions are different, the effectiveness of active prestressing depends on the difference between the distributions. The following discussion addresses this point. Consider the case where the loading is a concentrated force that can act at any point on the span, and the prestressing action is provided by a single cable. The moment diagrams for the individual loadings are shown in Fig 7.2. When these distributions are combined, there is a local positive maximum at point B, the point of application of the load, and possibly also at another point, say C. Whether the second local negative maximum occurs depends on the level of prestressing. As w a is increased, the positive moment at B decreases, and the negative moment at C increases. For a given position of the loading, the control problem involves establishing whether w a can be selected such that the magnitudes of both local moment maxima are less than the prescribed target design value, M * , indicated in Fig 7.2. With passive prestressing, the optimal prestressing scheme produced a 50% reduction in the required design moment, i.e., it resulted in M * =0.5(PL/4). Whether an additional reduction can be achieved with active prestressing remains to be determined.

484 Chapter 7: Quasi-static Control Algorithms

P a

L –a

x ab P ----L

(+)

M+

M+

(-)

L2 – w a -----8

parabola + M*

M net

low prestress

A

B

–M*

C

D

M+

high prestress xc

Fig. 7.2: Active prestressing scheme for a concentrated load The net moment is given by Region A-B Lx – x 2 Px ( L – a ) M ( x, t ) = ------------------------ – w a ------------------ 2 L

(7.10)

7.2 Active Prestressing of a Simple Supported Beam 485 Region B-C-D Lx – x 2 Pa ( L – x ) M ( x, t ) = ------------------------ – w a ------------------ 2 L

(7.11)

Specializing eqn (7.10) for x = a leads to P 1 M + ≡ M ( a, t ) = a ( L – a ) --- – --- w a L 2

(7.12)

The location of the second maxima is established by differentiating eqn (7.11) with respect to x and setting the resulting expression equal to 0. This operation yields L Pa x c = --- + ---------2 Lw a The value for M at x = x c has the following form: Pa w a x c M - ≡ M ( x c, t ) = ( L – x c ) ------ – ------------ for x c < L L 2

(7.13)

(7.14)

When x c > L , the maximum negative moment occurs outside the span, and M – is taken as 0. The control algorithm is established by requiring M+ ≤ M *

(7.15)

M- ≤ M * for all a . Starting at a = 0 , no action is required as a increases until the moment due to the force P is equal to M * . The limiting value is denoted as a s , and determined with P ---a s ( L – a s ) = M * L

(7.16)

When a is greater than a s , the maximum positive moment, M + , is set equal to M* , P 1 a ( L – a ) --- – --- w a = M * L 2 Solving eqn (7.17) for w a leads to P M* w a = 2 --- – ------------------- L a ( L – a )

(7.17)

(7.18)

The last step involves checking whether for this value of w a , the maximum negative moment, M - , exceeds – M * .

486 Chapter 7: Quasi-static Control Algorithms It is convenient to work with dimensionless variables for x and M. x x = --L

M M = -------M*

(7.19)

where PL M * = f ------- 4

(7.20)

The factor, f, can be interpreted as the “reduction” due to prestressing. No prestress corresponds to f=1; passive prestress for this loading and prestressing scheme corresponds to f=0.5. Using this notation, a s is given by 1 a s = --- [ 1 – ( 1 – f ) 1 ⁄ 2 ] 2 The dimensionless form of eqn (7.18) is written as wa L 2 8d 8 2 ------------- ≡ -------- T = --- – ------------------- = g ( a, f ) * * f a ( 1 – a) M M Lastly, the dimensionless peak negative moment is expressed as M4a gx c -------- = M - = ( 1 – x c ) ------ – -------f 2 M*

(7.21)

(7.22)

(7.23)

where 1 4a x c = --- + -----2 fg

(7.24)

The peak negative moment is a function of the position coordinate, a , and the design moment reduction factor, f. Since M - must be less than 1 for all values of a between 0 and 0.5, the magnitude of f is constrained to be above a limiting value, fmin. Figure 7.3 shows plots of M - vs. a for a range of values f. For this case, the limiting value of f is equal to 0.26. Therefore, with active prestressing, the design moment can be reduced to 50% of the corresponding value for passive prestressing. The influence line for the cable tension required for the “optimal” active prestressing algorithm is plotted in Fig 7.4. Also plotted is the required tension corresponding to f=0.5, the optimal passive value. As expected, lowering the cross-sectional design moment results in an increase in the required cable tension. In order to arrive at an optimal design, the costs associated with the material (cross-section) and prestressing need to be considered.

7.2 Active Prestressing of a Simple Supported Beam 487

0

f=0.5

−0.2

f=0.4

−0.4

f=0.3

−0.6

M–

−0.8

−1

−1.2

f=0.2

−1.4

−1.6

−1.8

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.25

0.3

0.35

0.4

0.45

0.5

a⁄L

0

−0.2

−0.4

−0.6

−0.8

0.27 0.26 0.25

M– −1

−1.2

−1.4

−1.6

−1.8

0

0.05

0.1

0.15

0.2

a⁄L

Fig. 7.3: Influence lines for the peak negative moment

488 Chapter 7: Quasi-static Control Algorithms

6

5

f=0.26 4

8d ------------- T PL ⁄ 4

3

f=0.5 2

1

0

0

0.05

0.1

0.15

0.2

a⁄L

0.25

0.3

0.35

0.4

0.45

0.5

Fig. 7.4: Influence lines for the optimal cable tension Active prestressing with concentrated forces In this section, the use of concentrated forces to generate prestress moment fields is examined. The design loading is assumed to be a single concentrated force that can act anywhere along the span.

Example 7.1: A single force actuator Consider the structure shown in Fig (1). The active prestressing is provided by a single force, F, acting at mid-span. This loading produces 2 local moment maxima, M1 and M2. The moment at mid-span may be negative for certain combinations of a and F, and therefore it is necessary to check both M1 and M2 when selecting a value for the control force. Adopting the strategy discussed earlier, the control algorithm is based on the following requirements M1 ≤ M * M2 ≤ M *

(1)

7.2 Active Prestressing of a Simple Supported Beam 489 for all a where Pa ( L – a ) Fa M 1 = ----------------------- – -----L 2

(2)

Pa FL M 2 = ------ – ------2 4

(3)

and M * is the design moment for the cross-section.

a

P

L –a L/2

F a(L – a) P ------------------L

(+) M+

(-)

M+

FL – ------4 M1 M2

Figure 1

M+

490 Chapter 7: Quasi-static Control Algorithms Shifting to dimensionless variables, a a = --L PL M * = f ------4 Mi M i = -------M* F F = --P

(4)

transforms the equations to 4 aF M 1 = --- a ( 1 – a ) – -----f 2

(5)

1 M 2 = --- [ 2a – F ] f

(6)

No action is required for a ≤ a s where a s is defined by eqn (7.21). 4 a s ( 1 – a s ) --- ≡ 1 f

(7)

When a > a s , the force is selected such that M 1 ≡ 1 . f F = 2 1 – a – ------ 4a

(8)

The corresponding expression for M 2 is 4a – 2 1 M 2 = --------------- + -----f 2a

(9)

Figures 2 and 3 show the variation of M 2 and F with a and f. The limiting value of f for active prestressing is 0.345; when f>0.345, the negative moment at midspan is greater than the design moment, M * . For passive prestressing, the optimal solution is f=0.5. Shifting from passive to active control results in an additional 30 percent reduction in the allowable design moment.

7.2 Active Prestressing of a Simple Supported Beam 491

1

0.5

0.5

0

0.45 0.4

−0.5

0.35 −1 0.3 −1.5

M2

0.25

−2

−2.5

0.2

−3

−3.5

−4

0

0.05

0.1

0.15

0.2

0.25

a Figure 2

0.3

0.35

0.4

0.45

0.5

1.4

1.2 0.2 0.25

1

0.3 0.35

0.8

0.4

F

0.45

0.6

0.5 0.4

0.2

0

0

0.05

0.1

0.15

0.2

0.25

a Figure 3

0.3

0.35

0.4

0.45

0.5

492 Chapter 7: Quasi-static Control Algorithms

Example 7.2: Two force actuators The previous example showed that the effectiveness of a prestressing scheme depends on the difference between the moment distributions for the applied loading and the prestressing forces. In the case of a single control force applied at mid-span, the limiting condition occurs when the applied load is near the end support, where the difference in the moment distributions is a maximum. Increasing the number of control forces provides the capability to modify the “shape” of the “prestress” moment distribution to conform better with the applied moment field, and therefore increase the amount of prestressing that can be applied. This example illustrates the use of self-equilibrating control force systems which provide the maximum flexibility for adjusting the moment field. Consider the self-equilibrating force system shown in Fig. 1. This force system produces a bilinear moment field which is local, i.e., confined to the loaded region. Therefore, perturbing the control force magnitude, F, has no effect outside of this region.

F/2

F l

F/2 l

M+ Fl – ----2 Figure 1 Applying a set of these self-equilibrating systems results in a piecewize linear moment distribution. Figure 2 illustrates the case of 2 force systems located immediately adjacent to each other. The corresponding moment field is defined in terms of 2 force parameters, F1 and F2.

7.2 Active Prestressing of a Simple Supported Beam 493 Region A-B M1 M = – -------- ( x – x A ) l

(1)

Region B-C M1 M2 M = – -------- ( l + x B – x ) – -------- ( x – x B ) l l

(2)

Region C-D M2 M = – -------- ( l + x c – x ) l

(3)

F1 l M 1 = -------2

(4)

where F2 l M 2 = -------2

F1

F1 -----2 A

l

F2 -----2 B

F1 -----2 l

F2 C

F2 -----2 l

D M+

M1 x

(-)

M2

Figure 2 Example 7.1 treated the case of a single actuator deployed on a simply supported beam subjected to a single concentrated force that can act at any point on the beam. Suppose the control force system now consists of 2 local moment fields centered at the third points of the span. Figure 3 shows the 2 loading scenarios for this example. The moments at B,C, and D corresponding to the different loading scenarios are:

494 Chapter 7: Quasi-static Control Algorithms Region A-B (Fig 3a): M p = PL ( a ( 1 – a ) ) – 3aM 1

(5)

2 M B = --- PLa – M 1 3

(6)

1 M C = --- PLa – M 2 3

(7)

Region B-C (Fig 3b): M p = PL ( a ( 1 – a ) ) – M 1 ( 2 – 3a ) – M 2 ( 3a – 1 )

(8)

1 M B = --- PL ( 1 – a ) – M 1 3

(9)

1 M C = --- PLa – M 2 3

(10)

a = aL

P (+) B (-)

A

C M2

M1 MP

D

MB

MC M+

L/3

L/3

Figure 3a

L/3

7.2 Active Prestressing of a Simple Supported Beam 495

P

a = aL

(+) B (-)

A M1 MB

C

D

M2 MP

MC M+

L/3

L/3

L/3

Figure 3b Let M * represent the design moment. Expressing M * as a fraction of the maximum moment for the case where there is no prestressing, PL M * = f ------4

(11)

and working with dimensionless moments, M( ) M ( ) = -----------M*

(12)

transforms eqns (5) thru (10) to the following: Region A-B 4 M p = --- ( a ( 1 – a ) ) – 3aM 1 f

(13)

8 M B = ------a – M 1 3f

(14)

4 M C = ------a – M 2 3f

(15)

496 Chapter 7: Quasi-static Control Algorithms Region B-C 4 M p = --- a ( 1 – a ) – M 1 ( 2 – 3a ) – M 2 ( 3a – 1 ) f

(16)

4 M B = ------ ( 1 – a ) – M 1 3f

(17)

4 M C = ------a – M 2 3f

(18)

The control objective for this example is to limit the peak value of each of the dimensionless moment variables to be less than unity. MP ≤ 1

MB ≤ 1

MC ≤ 1

1 0 ≤ a ≤ --2

(19)

Since there are 3 constraints and only 2 control parameters, the problem is overconstrained. The strategy followed here is based on determining M 1 and M 2 using 2 of the constraints in eqn (19), and then adjusting f such that the third constraint is also satisfied. Figure 4 shows the variation in the moment measures with a , the coordinate defining the position of the load, corresponding to the following choice of constraints: Region A-B M P = +1 M C = +1

(20)

MB < 1 Region B-C M P = +1 M C = +1 – 4 ( 3a – 1 )

(21)

MB < 1 The minimum value of f is equal to 0.228. This value is controlled by the constraint on M B in the region A-B. For the case of a single actuator applied at

7.2 Active Prestressing of a Simple Supported Beam 497 mid-span, the optimum value for f was found to be 0.345. Applying 2 force actuators leads to an additional reduction in the “permissible” design moment. 4

3.5

3

f = 0.228 2.5

M1

2

1.5

1

0.5

0

0

0.05

0.1

0.15

0.2

a 0.25 Figure 4a

0.3

0.35

0.4

0.45

0.5

498 Chapter 7: Quasi-static Control Algorithms

4

3.5

3

2.5

M2

2

1.5

f = 0.228 1

0.5

0

0

0.05

0.1

0.15

0.2

0.25

a Figure 4b

0.3

0.35

0.4

0.45

0.5

0.3

0.35

0.4

0.45

0.5

1

0.5

0

MB −0.5

f = 0.228 −1

−1.5

0

0.05

0.1

0.15

0.2

a 0.25 Figure 4c

7.2 Active Prestressing of a Simple Supported Beam 499

1

0.8

0.6

f = 0.228

0.4

0.2

MC

0

−0.2

−0.4

−0.6

−0.8

−1

0

0.05

0.1

0.15

0.2

0.25

a Figure 4d

0.3

0.35

0.4

0.45

0.5

A general active prestressing methodology The discussion to this point has been concerned with a specific loading and a specific prestressing scheme. In what follows, a general methodology for dealing with the combination of an arbitrary design loading and prestressing scheme applied to a simply supported beam is described, and the control algorithm corresponding to a particular choice of error measure is formulated. This methodology is also applicable for displacement control which is discussed in the following section. Let M d ( x ) denote the moment due to the design loading, M c ( x ) the moment generated by the prestress system, and M(x) the net moment. By definition, M ( x ) = Md ( x ) + Mc ( x )

(7.25)

The design objective is to limit the magnitude of M ( x ) to be less than M * , the design moment for the cross-section.

500 Chapter 7: Quasi-static Control Algorithms Md + Mc ≤ M *

0≤x≤L

(7.26)

Equation (7.26) imposes a constraint on the magnitude of M c . – Md – M * ≤ Mc ≤ M * – Md

0≤x≤L

(7.27)

These limits establish the lower and upper bounds for M c ( x ) . Given M d ( x ) , one generates the limiting boundaries and then decides on a “target” distribution for Mc ( x ) . 3 Md 1 -------- 0 M* -1 -2

(a)

upper bound Mc 1 -------- 0 M* -1 -2

a

b

c

e

d

(b)

lower bound

Permissible region -4 1 M c* -------- 0 M * -1 -2

a

b

c

d

e

(c)

Optimum solution

Fig. 7.5: Limiting prestress moment fields Figure (7.5) illustrates the process of establishing the desired distribution for the control moment. The curves shown in Fig (7.5b) correspond to M c = – M d ± M * ; allowable values of M c are defined by the shaded area. The distribution corresponding to selecting the minimum allowable value of M c ( x ) at each x is plotted in Fig (7.5c). Assuming magnitude is the dominant measure,

7.2 Active Prestressing of a Simple Supported Beam 501 this distribution represents the optimal “target” prestress moment field. Let M c* ( x ) denote the desired “target” prestress moment field. Suppose the actual prestress moment field is a linear combination of r individual fields, r

Mc ( x ) =

∑

m jψ j(x)

(7.28)

j=1

where m j are moment amplitude parameters and ψ j ( x ) are dimensionless functions. The error associated with a specific choice of moment parameters is represented by the difference, e ( x ) , e ( x ) = M c* ( x ) – M c ( x )

(7.29)

Ideally, one wants e ( x ) = 0 for 0 ≤ x ≤ L . However this goal cannot be achieved when M c* is an arbitrary function, and it is necessary to work with an approximate error condition established using collocation, the least square method, or some other weighted residual scheme. The least square method is based on taking the integral of ( e ( x ) ) 2 as a measure of the accuracy of the approximation represented by eqn (7.28). This integral is denoted as J. 1 J = --2

L

∫

e 2 dx

(7.30)

0

In general, J is a function of the r moment parameters. Equations for these parameters are generated by requiring J to be stationary. ∂J = 0 ∂ mi

i = 1, 2, …, r

(7.31)

Expanding eqn (7.31) results in the following linear matrix equation, (7.32)

am = b where a, b, m are r’th order matrices, and the elements of a and b are: a ij =

∫

L

Ψ i ( x )Ψ j ( x ) dx

0

bi =

i, j = 1, 2…, r

L

∫

Ψ i ( x )M c* ( x ) dx 0

(7.33)

(7.34)

Given M c* ( x ) ,one determines b and then solves for m . This solution produces the least value for J, for a particular set of Ψ ‘s. A sense of convergence can be

502 Chapter 7: Quasi-static Control Algorithms obtained by expanding the set of basis functions, and comparing the corresponding values of J. It should be noted that the exact condition, e(x)=0 for 0<x<1, is generally not satisfied by eqn (7.32). This formulation works with continuous functions, and requires the evaluation of a set of integrals. It is more convenient to work with vectors rather than functions, since the computation reduces to matrix operations. Suppose the moment is monitored at n points within the interval 0 ≤ x ≤ L . The desired prestress moment vector is of order n × 1. M c* = { M c*, 1 , M c*, 2 , …, M c*, n }

(7.35)

Evaluating eqn (7.28) at these observation points leads to M c = Ψm

(7.36)

where Ψ is of order n × r . The error vector is taken to be the difference between M c* and M c , e = M c* – M c = M c* – Ψm

(7.37)

When r = n and the individual prestress moment fields are linearly independent, Ψ is non-singular and it is possible to determine an m that exactly satisfies e = 0 . m = Ψ – 1 M c*

(7.38)

When r ≠ n , a least square procedure can be used to establish an approximate solution for e = 0 . The error measure is taken as the norm of e . 1 J = --- e T e = J ( m ) 2

(7.39)

Requiring J to be stationary with respect to m leads to an equation having the same form as eqn (7.32), with a and b now given by a = ΨT Ψ

(7.40)

b = Ψ T M c*

(7.41)

The following example illustrates the application of the discrete formulation.

7.2 Active Prestressing of a Simple Supported Beam 503

Example 7.3: Multiple actuators Consider the design moment field shown in Fig (1). The longitudinal axis is discretized with 10 equal segments, resulting in 11 (n=11) observation points. Applying the criteria defined by eqn (7.27), and taking M * = 1 leads to the bounding curves for M c* plotted in Fig (2). The problem now consists of generating a prestress moment distribution which lies between these bounds. Suppose 4 self-equilibrating force system (r=4) that produce bilinear moment fields are applied at equally spaced interior points. The corresponding Ψ functions are shown in Fig (3a) and the typical field is plotted in Fig (3b). Since the prestress moment field is defined in terms of the moments at only 4 fixed points (3,5,7,9), and there are 9 interior points, the solution for the case of an arbitrary target distribution will be approximate. Various solutions for a particular target distribution are plotted in Fig (4). Curve (1) corresponds to taking m 1 = m 2 = – 1, m 3 = 0 , and m 4 = 1. Curve (2) is the least square solution for the 4 actuator system defined in Fig (3a). Curve (3) is based on the 7 actuator system defined in Fig (3c). The 3 additional actuators applied at points 2,4, and 6 eliminate the error up to point 6. Incorporating 2 more actuators at points 8 and 10 would completely eliminate the error associated with the 4 actuator system. It produces a moment field that is fully contained within the allowable zone and has the lowest cost as measured by the sum of the actuator moment magnitudes, Σm i . Md

3 2 1 1

2

3

4

5

6

7

8

9

-1 -2 Figure 1: Moment due to design loading

10

11

504 Chapter 7: Quasi-static Control Algorithms

M c* 3

upper bound

M* = 1

2 1 1

2

3

4

5

6

7

8

9

10

11

-1 -2

lower bound

-3 -4 Figure 2: Upper and lower bounds on the prestress moment field

7.2 Active Prestressing of a Simple Supported Beam 505 1

1

2

3

4

5

6

7

8

10

9

11

Ψ1

-1 1

1

2

3

4

5

6

7

8

9

10

11

1

2

3

4

5

6

7

8

9

10

11

2

3

4

5

6

7

8

9

10

11

Ψ2

-1 1 Ψ3

-1 1

1 -1

(a) r=4

1

1

5

3

m5 m1 2

3

m3

m2

m1

m6

4

m4

7 (b) r=4

m3 m2 m 7 5 7 6 (c) r=7

11

9

m4 8

9

10

11

Figure 3: Prestress moment fields for 4 and 7 actuators

Ψ4

506 Chapter 7: Quasi-static Control Algorithms

1

0.5

0

3

−0.5

2 −1

Target distribution

1

4 actuators (1)

−1.5

4 actuators (2) o...o... 7 actuators (3)

−2

−2.5

1

2

3

4

5

6

7

8

9

10

11

Figure 4: Prestress moment fields

7.3 Quasi-static displacement control of beams The previous section discussed how active prestressing can be applied to limit the magnitude of the bending moment in a beam subjected to a quasi-static loading. In this section, a procedure for controlling the quasi-static displacement profile of a beam is described. Both prestress and displacement control are based on the same least square error minimization algorithm which is generalized later in the next section. Consider a beam subjected to a loading that produces the displacement field u o ( x ) .Suppose the desired displacement profile is u * ( x ) , and the difference

7.3 Quasi-static Displacement Control of Beams 507 between u o ( x ) and u * ( x ) is to be reduced by applying a set of control force systems. Let u c ( x ) denote the displacement due to the control forces. The resultant error in displacement is denoted by e ( x ) , and is defined as e ( x ) = uo ( x ) + uc – u * ( x )

(7.42)

Assuming u c ( x ) is a function of r control force parameters, u c ( x ) = u c ( x, F 1, F 2, …, F r )

(7.43)

the issue now is how to establish approximate values for the force parameters. The least square procedure works with the square of the displacement error summed up over the domain, in this case the length of the beam. Both continuous and discrete forms of the least square error function are considered here. Continuous least square formulation In the continuous case, J is taken as 1 J = --2

L

∫

e 2 dx

(7.44)

0

and u c ( x ) is expressed as a linear combination of prescribed functions and unknown scalar force parameters, r

uc ( x ) =

∑

Fi φi ( x )

(7.45)

i=1

Substituting for u c ( x ) and e transforms J to a quadratic form in F, 1 J = --2

r

∫

L

( uo –

u * ) 2 dx

+

0

∑∫

L

i=1

r

1 + --2

r

∑∑ ∫ j = 1i = 1

( u o – u * )φ i dx F i +

0

(7.46)

L 0

Introducing the following notation,

φ i φ j dx F i F j

508 Chapter 7: Quasi-static Control Algorithms F = { Fi } b = { bi } a = [ a ij ] L

∫

a ij =

φ i φ j dx

(7.47)

0

L

bi =

∫

( – u o + u * )φ i dx 0

L

c =

∫

( u o – u * ) 2 dx 0

reduces eqn (7.46) to a matrix expression, 1 1 J = --- c – b T F + --- F T aF 2 2

(7.48)

The first and second differentials of J are r

dJ =

∑ i=1

∂J dF = dF T ( – b + aF ) ∂ Fi i

r

d2 J

=

r

∑ ∑ j=1

∂ ∂F j

i=1

∂J dF dF j = dF T adF ∂ Fi i

(7.49)

(7.50)

Requiring J to be stationary is enforced by setting the coefficient of dF T equal to 0 in eqn (7.49), and leads to a set of r equations for the r force parameters. dJ = 0 for arbitrary dF

⇒

aF = b

(7.51)

The stationary point corresponds to an absolute minimum value when a is a positive definite matrix. This property exists when the displacement functions, Ψ i ( x ) , are linearly independent. Discrete least square formulation The discrete least square formulation is based on selecting a displacement

7.3 Quasi-static Displacement Control of Beams 509 error vector, e , and taking J as 1 J = --- e T e 2

(7.52)

One forms e by specifying n points along the longitudinal axis and evaluating the difference between the various displacement components at these points. Using matrix notation, e is expressed as e = U o – U * + ΦF

(7.53)

where U o and U * are displacement vectors of order n × 1, and Φ is of order n × r . Substituting for e in eqn (7.52) and expressing J in the same form as for the continuous case 1 1 J = --- e T e = c – b T F + --- F T aF 2 2

(7.54)

leads to the definition equations for a , b and c . b = φ T ( – Uo + U * )

(7.55)

a = ΦT Φ

(7.56)

c = ( Uo – U * ) T ( Uo – U * )

(7.57)

The stationary requirement for J is still aF = b. Since a is now generated by pre-multiplying Φ by its transpose, the properties of a depend on the relative size of n and r. If n > r, a will be nonsingular provided that the columns of Φ are linearly independent. This condition is equivalent to requiring the rank of Φ to be equal to the number of its columns. If the rank of Φ is less than r, say r - s, then s control forces do not contribute to the error and need to be deleted. If n < r, a will be singular and a special solution procedure needs to be applied. Strang(1993) contains a detailed discussion of the pseudo-inverse procedure for solving aF = b when a is singular. This procedure is executed by MATLAB with the statement, F = pinv ( a )*b. The following example illustrates different scenarios for n vs. r.

510 Chapter 7: Quasi-static Control Algorithms

Example 7.4: Cantilever beam - Least square algorithm Consider the cantilever beam shown in Fig (1). Introducing the dimensionless variable x x = --L

(1)

and assuming the bending rigidity, D B , is constant, the expression for the initial and desired displacement fields are written as: wL 4 u o = ---------- g ( x ) DB

(2)

1 g ( x ) = ------ [ x 2 ( 6 – 4x + x 2 ) ] 24

(3)

L u * = ---x 2 α

(4)

where α is a dimensionless parameter.

u

F1

F2

L/2

w x L

Figure 1 Suppose a single control force F 1 is applied at x = L. The corresponding displacement distribution is

7.3 Quasi-static Displacement Control of Beams 511 L3 x3 u c = φ 1 F 1 = ----------- x 2 – ----- F 1 3 2D B

(5)

Substituting for the various functions in eqn (7.47) results in a 11 F 1 = b 1

(6)

L7 a 11 = -------- ( 0.0262 ) D B2

(7)

L5 wL 8 b 1 = – ---------- ( 0.0102 ) + ------------ ( 0.0722 ) αD B D B2

(8)

where

Solving for F 1 leads to: DB F 1 = ---------- ( 2.7576 ) – wL ( 0.3911 ) L2α

(9)

The error function associated with the solution is: wL 4 L e ( x ) = ---------- f ( x ) + --- g ( x ) DB α wL 4 = ---------- { 0.0545x 2 – 0.1015x 3 + 0.417x 4 } + DB L + --- { 0.3789x 2 – 0.4596x 3 } α Table 1 contains the errors at locations x = 0.5 and x = 1.0 .

(10)

512 Chapter 7: Quasi-static Control Algorithms

x = 0.5

x = 0.5

x = 1.0

x = 1.0

Case

f

g

f

g

1. Continuous least square F 1

0.0035

0.0373

-0.0053

-0.0808

2. Discrete least square F 1 and 2 obs. points

0.0048

0.0570

-0.0015

-0.0178

3. Discrete least square F 2 and 2 obs. points

-0.0049

0.1292

0.0020

-0.0521

4. Discrete least square F 1, F 2 and 2 obs. points

0

0

0

0

5. Discrete least square F 1, F 2 and 1 obs. point

0.0043

0.0703

0

0

Table 1 Comparison of displacement errors To illustrate the discrete formulation, observation points at x = 0.5 and x = 1.0 are selected. For this case, n = 2 and r = 1. Using eqns (1), (3), and (5), the terms defining e are: 17 --------

Uo –

U*

1 L --4- = ---------- – --D B 1 α --1 8 wL 4 384

(11)

5 -----

L 3 48

Φ = ------- DB 1 --3 Substituting in eqns (7.55) and (7.56) results in

(12)

7.3 Quasi-static Displacement Control of Beams 513 L6 a = a 11 = -------- ( 0.1220 ) D B2

(13)

wL 7 L4 b = b 1 = ---------- ( – 0.0463 ) + ------------ ( 0.3594 ) αD B D B2

(14)

DB F 1 = ---------- ( 2.9467 ) – wL ( 0.3795 ) αL 2

(15)

and

This solution approaches the continuous solution, eqn (9), as the number of observation points is increased. For example, the solution corresponding to 4 observation points located at x = 0.25, 0.50, 0.75, and 1.0 is DB F 1 = ---------- ( 2.8685 ) – wL ( 0.3842 ) αL 2

(16)

The location of the point of application of the control force also has an influence on the solution. Considering a single force, F 2 , applied at x = 0.5 , and working with 2 observation points at x = 0.5 and 1.0 generates the following least square solution, DB F 2 = ---------- ( 9.0997 ) – wL ( 1.1807 ) αL 2

(17)

Comparison of the errors listed in Table 1 indicates that applying the force at x = 1.0 is more effective for this structure/loading combination. Applying both F 1 and F 2 , and taking 2 observation points results in Φ being of order 2x2. In this case, one can set e = 0 and solve ΦF = U * – U o

(18)

for F . The least square formulation produces the same solution for F when Φ is square. Evaluating eqn (18) results in

514 Chapter 7: Quasi-static Control Algorithms F1 – 0.1964 D B 5.1429 = wL + ---------2- F – 0.5714 2 αL – 6.8571

(19)

Suppose control forces are applied at x = 0.5 and 1.0, and only one observation point is selected. In this case, n = 1 and r = 2. Considering the point to be at x = 1.0 , the corresponding error term defined by eqn (7.53) has the following form: L3 wL 4 L e ( x = 1 ) = ----------- – --- + ----------3D B 8D B α

5 L 3 F1 ------ ------- 48 D B F 2

(20)

e = U o – U c + ΦF Expanding a and b , 1 L3 2 a = ----------5 1 3D ----B 16

1 5 L3 ------ = ---------- 3D 5 16 B -----16

5 -----16 5 5 ------ ------ 16 16

1 L3 wL 4 L b = ----------- 5 – ----------- + --- 3D ------ 8D B B α 16

(21)

(22)

shows that a is singular; the second row is 5/16 times the first row. The same relationship holds for the elements of b . Therefore, in this case, there is only one independent equation in the set, aF = b, generated by requiring e T e to be stationary. Another equation has to be established. The pseudo inverse solution procedure starts by setting e equal to 0. Noting eqn (20), e(x = 1) = 0

⇒

DB 3 5 F 1 + ------ F 2 = 3 ---------- – --- wL 16 αL 2 8

(23)

An additional equation relating F 1 and F 2 is generated by requiring the mean square sum of the force magnitudes to be a minimum, considering eqn (23) to be a constraint between F 1 and F 2 . The steps are:

7.3 Quasi-static Displacement Control of Beams 515 a) Form the objective function 1 I = --- ( F 12 + F 22 ) 2

(24)

b) Form the first differential dI = F 1 dF 1 + F 2 dF 2 = 0

(25)

c) Establish the constraint between the differentials Noting eqn (23), 5 dF 1 + ------ dF 2 = 0 16

(26)

d) Establish the additional equation Substituting for dF 2 in terms of dF 1 in eqn (25), F + – 16 ------ F dF = 0 1 5 2 1

(27)

and requiring eqn (27) to be satisfied for arbitrary dF 1 yields 16 F 1 = ------ F 2 5

(28)

e) Solve for the primary variable Using eqn (28), the solution follows form eqn (23). DB 3D B 3 1 F 1 = ----------------------------- ----------- – --- wL = ---------- ( 2.7331 ) – wL ( 0.3416 ) 5 2 αL 2 8 αL 2 1 + ---- 16

(29)

5 F 2 = ------ F 1 16 Table 1 lists the error at x = 0.5 . The error at x = 1.0 was taken to be zero at the start of the solution process. One can obtain the solution defined by eqn (29) with the MATLAB statement, F = pinv ( a )*b.

516 Chapter 7: Quasi-static Control Algorithms The previous example illustrated the use of concentrated forces to adjust the deflected shape of a cantilever beam subjected to a uniform transverse loading. The approach involved selecting a set of points at which the displacements are monitored, and a set of control forces. Considering the displacement field to be continuous corresponds to taking an infinite number of observation points. In order to apply the procedure for an arbitrary loading, a finite number of points needs to be specified. The control algorithm is established using the discrete least square formulation. A square nonsingular set of equations is obtained when the number of observation points, n, is taken to be greater or equal to the number of control forces, r. If n is less than r, the problem is underspecified in the sense that additional equations have to be established. In this case, a pseudo inverse procedure can be applied. Extended least square formulation The standard least square formulation works with a weighted displacement error, e ( x ) , which represents the difference between the actual and target displacement fields. No priority is assigned to a specific location, say e ( x i ) , and also, no limit is placed on the magnitude of the control forces. This approach can be generalized to include priorities by introducing weighting functions for e ( x ) and F i , and employing an extended form of the error functional, J. Considering first the continuous case, the error at x is weighted with a function, q(x), and the error measure is taken as the integral of qe 2 , 1 J ⇒ J' = --2

∫

L

q ( x )e 2 dx

(7.58)

0

Setting q(x) = constant assigns equal priority in the interval 0 ≤ x ≤ L . Taking q(x) as a set of n Dirac delta functions centered at x = x i , i=1, 2, ..., n converts eqn (7.58) to the discrete form and allows for different weighting factors, 1 J' = --2

n

∑

q i e i2 ( x i )

(7.59)

i=1

The magnitudes of the control forces influence J’ indirectly through e ( x ) . Selecting the force magnitudes such that J’ is stationary corresponds to forcing the weighted displacement error to be a minimum. There is a “cost” associated with a

7.3 Quasi-static Displacement Control of Beams 517 particular choice of the force magnitudes, and this cost is not reflected by the stationary solution for J’. Consequently, as the displacement error is driven toward zero, the magnitude and therefore the “cost” of the forces may become excessive. This situation can be avoided by combining J’ with a force cost function, and requiring the sum to be stationary. A convenient choice for the force cost function is the weighted quadratic form, 1 Force cos t fuction = --2

r

∑

r j F 2j

(7.60)

j=1

where r j is a weight associated with F j . In general, r j > 0 . Combining this term with J’ leads to the extended functional, J 1 . 1 J 1 = --2

∫

L o

1 qe 2 dx + --2

r

∑

r j F 2j

(7.61)

j=1

The discrete form of J 1 is taken as 1 1 J 1 = --- e T Qe + --- F T RF 2 2

(7.62)

where e is the displacement error vector, and Q, R are diagonal matrices with nonnegative elements, Q = [ q i δ ij ]

i, j = 1, 2, …, n

(7.63)

R = [ r k δ kl ]

k, l = 1, 2, …, r

(7.64)

Substituting for e using eqn (7.53), J 1 expands to 1 1 J 1 = --- c 1 – b 1T F + --- F T a 1 F 2 2

(7.65)

where b1 = Φ T Q ( – Uo + U * )

(7.66)

a 1 = Φ T QΦ + R

(7.67)

c1 = ( Uo –U * ) T Q ( Uo – U * )

(7.68)

518 Chapter 7: Quasi-static Control Algorithms Requiring J 1 to be stationary with respect to F leads to a set of r equations for the r force variables, (7.69)

a1 F = b1

Although the order is the same as for the conventional formulation that considered only the displacement error, a 1 differs from a in 2 ways. Firstly, the quadratic term, Φ T Φ , is replaced with a weighted form, Φ T QQ . Secondly, positive quantities are added to the diagonal elements. The latter operation ensures that a 1 is non-singular for all values of n vs. r. Holding Q fixed and increasing the magnitude of R places more emphasis on lowering the cost of the control forces. Consequently, the displacement error will increase. Decreasing R with respect to Q has the opposite effect on the displacement error. Ideally, one wants minimum displacement error and minimum force cost, but that goal is impossible. Generating solutions corresponding to a range of Q and R values provides information concerning the sensitivity of the 2 weighting measures which is useful for selecting design values. The following example illustrates this computation.

Example 7.5: Example 7.4 revisited Example 7.4 considered observation points at x = 0.5 and 1.0, and a single control force applied at x = 1.0 . The example is reworked here using the extended least square algorithm. Starting with the basic data contained in eqns (11) and (12) of example 7.4, 17 --------

1 – --- L + --- 4 U o – U * = ---------- DB 1 α -- –1 8 wL 4 384

5 - L 3 ----Φ = ----------- 16 3D B 1 and applying eqns (7.66), (7.67), and (7.69) results in

(1)

7.3 Quasi-static Displacement Control of Beams 519

L4 85 5 wL 7 1 b 1 = ---------- ------ – --------- q 1 – q 2 + --------------- ------ q 1 + q 2 3D α 64 D B2 24 768 B

(2)

L6 5 2 a 1 = ----------- ------ q 1 + q 2 + r 1 9D B2 16

(3)

F 1 = a 1– 1 b 1

(4)

Suppose the q’s are equal to a common value, q . In this case, the coefficients simplify to L 4 23 wL 7 853 b 1 = ---------- – --------------- q + ------------ ------ q D B α 64 D B2 18432

(5)

L 6 281 a 1 = -------- ------------ q + r 1 D B2 2304

(6)

The first term on the right hand side of eqn (6) is a measure of the priority for the displacement error; the second term reflects the importance of the force cost. Since q is present in both b 1 and a 1 the force depends only on the ratio of r 1 to q. To establish the appropriate magnitude for r 1 ⁄ q , r 1 and q can be scaled such that the 2 terms on the right hand side of eqn (6) have the same order of magnitude. One possible approach is to take r 1 = 1 and scale q. r1 = 1 D B2 2304 ------- ------------ q * q = 6 L 281

(7)

where q * is a measure of the relative importance of the force cost vs. the displacement error. Using eqn (7) the expression for F 1 takes the form DB wl ( – 0.3794 ) + ---------- ( 2.9466 ) q * L2α F 1 = -------------------------------------------------------------------------------1 + q*

(8)

Equal weighting for displacement and force as assigned when q * = 1. Increasing

520 Chapter 7: Quasi-static Control Algorithms q * beyond 1 places more emphasis on the displacement error, and the solution approaches the solution corresponding to no constraint on the control force magnitude. Taking q * ≈ 0 ignores the displacement error, and the procedure will minimize the functional corresponding to only the force cost. Consequently, the solution will be F = 0.

The scaling operation on r and q described in example 7.5 can be applied to the general formulation represented by eqns (7.66) thru (7.69) in the following manner. Firstly, R is taken such that all its elements are between 0 and 1. This matrix is denoted as R * . R = [ r *j δ ij ] ≡ R *

(7.70)

where r *j are individual force cost “relative” weighting factors that are now of order 1. Secondly, Q is scaled with a factor, q , and expressed as Q = qQ *

(7.71)

The magnitude of q can be chosen such that both of the terms comprising a 1 have essentially the same magnitude. Using these definitions, eqn (7.69) is transformed to ( q ( Φ T Q * Φ ) + R * )F = qΦ T Q * ( – U o + U * )

(7.72)

Taking q to be equal to the inverse of the maximum element in the matrix product, Φ T Φ , is a convenient choice since now the elements of Q * can be considered to be of order 1. Sensitivity can be assessed by varying the elements of Q * and R * . Obviously Q * cannot be a null matrix, whereas R * = 0 is a limiting case. Setting R * = I and perturbing the elements of Q * is a reasonable strategy for assessing how the solution varies as the priorities with respect to the displacement error and force cost are shifted. Increasing Q * places more emphasis on the displacement error.

7.4 Quasi-static Control of MDOF Systems 521

7.4 Quasi-static control of MDOF systems Introduction The governing equations for the linear static behavior of an n’th order system controlled with r forces are written as: (7.73)

KU = P + E f F

where U contains the n displacement variables, K is the system stiffness matrix, P is the prescribed load vector, F contains the r control forces, and E f defines the distribution of the r control forces in the load vector. For n DOF and r control forces, E f is of order n x r. Figure 7.6 illustrates the form of E f for a particular set of control forces applied to a discrete shear beam model. u4 u3 u2 u1

F1 F2 F2

0 Ef = 0 1 0

–1 1 0 0

Fig. 7.6: Control force distribution matrix Selecting a control force system involves 2 decisions: firstly, the nature and location of the control forces which establishes E f and secondly, the control algorithm which determines the magnitude of the control force parameters corresponding to the choice of E f . The decision on E f depends in turn on the objective of adding control which usually involves limiting the response of certain variables. Therefore, the first step is to identify the performance measures. Given the performance measures, a set of response quantities which can be “observed” and then used to establish the magnitudes of the performance measures are identified. Also, once the performance measures are defined, the equilibrium equations are applied to establish the relationship between the “performance”

522 Chapter 7: Quasi-static Control Algorithms and the control force measures. Lastly, a mathematical procedure such as the least square method is employed to generate a control algorithm which is used to establish a quantitative estimate of the control forces. Selection of measures By definition performance measures are those quantities whose magnitudes are to be limited by the application of control forces. In the previous sections dealing with beam type structures, bending moments and displacements were selected. For a low rise building modelled as a shear beam, limiting the transverse shear deformation at certain story locations is the objective for motion based design. Assuming there are np performance measures, the individual measures are denoted by v i ( i = 1, 2, …, n p ) and included in the vector v .

u3 h

3 u2

h

2

h

u1

γ 1 v = γ 2

1 θ

Fig. 7.7: Deformations as performance measures Figure 7.7 illustrates the selection process for a single truss modelled as a 3 DOF shear beam. The performance measures are taken to be the transverse shear strains in the first and second stories. They are related to the displacements by u1 u2 – u1 (7.74) v 2 = -----------------v 1 = -----h h The general matrix form of eqn (7.74) is written as: v = BU

(7.75)

7.4 Quasi-static Control of MDOF Systems 523 where B can be interpreted as a transformation matrix which converts displacements measures into performance measures. Observation measures are those quantities which are monitored and used to establish the magnitude of the performance measures. They may be either deformations or displacements. Let w i denote the i’th observation measure, n ob the total number of observations, and w the observation vector. The key decisions here are selecting the number, the nature (i.e., displacement, deformation,...), and the location of the observations such that the performance measures are uniquely defined. Suppose w * is a particular choice of observation variables. Assuming the behavior is linear, v associated with this choice of w can be expressed as v

w*

= B1 w *

(7.76)

One evaluates the i’th row of B 1 using the definition equation for v i . If all the elements in row i are equal to 0, v i cannot be determined by this choice of w. Similarly, if all entries in column j are 0, w j contributes nothing to v and must be deleted from w * . Observability is the term used to characterize the relationship between v and w. A monitoring scheme is said to be observable when the rank of B 1 is equal to n p , the number of performance measures. In general, a necessary condition is n ob ≥ n p ; the sufficient condition is for B 1 to have at least n p linearly independent columns. Various scenarios are illustrated by the following example.

Example 7.6: Illustrative examples of observability Consider the physical model defined by Fig 7.7. There are 2 performance measures, the shear strains in the first 2 stories. Various cases are considered below. Observability for this example requires the rank of B to be 2. Case 1.

w * = { u 2, u 3 }

Applying eqns (7.74) leads to 1 B 1 = --- 0 0 h 1 0

(1)

524 Chapter 7: Quasi-static Control Algorithms The second column contains all zero’s since u 3 has no effect on either v 1 or v 2 . Case 2. w * = { u 1 } 1 B 1 = --- 1 h –1

(2)

The rank is 1. The problem here is the incomplete determination of γ 2 , which requires u 2 as well as u 1 , Case 3. w * = { u 1, u 2 } 1 B 1 = --- 1 0 h –1 1

(3)

Here, r(B1)=2. Observability is satisfied. Case 4. w * = { ε 1, ε 2 } where ε 1 and ε 2 are the extensional strains in the diagonal elements located in the first and second stories. Noting the definition equations for the extensions, h e 1 = u 1 cos θ = -----------ε 1 sin θ h e 2 = ( u 2 – u 1 ) cos θ = -----------ε 2 sin θ

(4)

and substituting for u 1 and u 2 in eqn (7.74) results in 1 B 1 = ------------------------ 1 0 sin θ cos θ 0 1

(5)

In this case, r(B1)=2 for arbitrary finite θ > 0 .

The last group of variables to be defined is the set of control forces represented by the vector F. Once the location of the forces is chosen, E f is known, and the equilibrium equations can be solved to determine the displacements generated by applying F. Using eqn (7.73), U due to F is given by

7.4 Quasi-static Control of MDOF Systems 525 U

F

= K –1 E f F

(7.77)

Then, substituting for U in eqn (7.75) leads to a relation between v and F. v F = BU F = ( BK – 1 E f )F = C f F

(7.78)

The coefficient matrix, Cf, is of order n p × r , and defines the effect of the control forces on the performance measures. If the elements in the i’th row of Cf are all zero, the choice of control forces has no influence on the i’th performance measure, i.e., v i cannot be controlled by F. Similarly, if all the elements in the j’th column of Cf are zero, F j contributes nothing and should be deleted from F. In the usual control scenario, the number of force parameters is less than the number of performance parameters. Requiring r(Cf)=r for n p > r ensures that the columns of Cf are linearly independent and eliminates the situation where Cf has a zero column. Controllability is defined for an individual performance measure, and requires the corresponding row of Cf to have at least one non-zero element. The following example illustrates some choices of control force systems for example 7.6

Example 7.7: Illustrative examples of controllability Consider the shear beam model defined in Fig (7.7) and discussed in example 7.6. There are 2 performance measures, and therefore one would not normally take more than 2 control forces. The forms of Cf corresponding to various force systems are listed below. Case 3 has an excess of control forces. Case 4 has a deficiency with respect to controllability; v 1 cannot be controlled with this scheme.

526 Chapter 7: Quasi-static Control Algorithms Case 1. F1 0 Ef = 0 1

Cf =

1 ---------D T1 1 ---------D T2

Case 2. F1

F2

0 1 Ef = 0 0 1 0

Cf =

1 1 ---------- ---------D T1 D T1 1 ---------D T2

0

Case 3. F1 F3 F2

010 Ef = 0 0 1 100

1 1 1 ---------- ---------- ---------D T1 D T1 D T1 Cf = 1 1 ---------- 0 ---------D T2 D T2

7.4 Quasi-static Control of MDOF Systems 527 Case 4. F1 F1

F2

F2

0 –1 E f = –1 1 1 0

0

0 Cf = 1 0 ---------D T2

Least square control algorithms Suppose the system is subjected to its design loading, and the values of the performance parameters resulting from this loading are observed with a monitoring scheme. Let v o denote the “observed” response, and v * the desired response. The difference between these vectors is the target response for the force control system. Taking v c as the response due to the control forces, the net performance error vector is e = vo + vc – v *

(7.79)

The goal is to select F such that e is less than the allowable error. Considering the behavior to be linear, and noting eqn (7.78), e is related to F by e = vo + C f F – v *

(7.80)

Equation (7.80) represents n p equations in r unknowns; Cf is of order n p × r . When n p = r , there is a unique F which satisfies e = 0. For all other combinations of n p and r, the solution for F is not unique and also may not satisfy e = 0. The least square solution for the case where n p > r is generated by forming the error norm, 1 J = --- e T e 2

(7.81)

and requiring J to be stationary with respect to F. This operation leads to a set of r equations, which are written as:

528 Chapter 7: Quasi-static Control Algorithms

aF = b a = C Tf C f

(7.82)

b = C Tf ( v * – v o ) The coefficient matrix will be non-singular when Cf satisfies the controllability requirement, r(Cf)=r. Since the square error norm is used, the solution of eqn (7.82) does not result in e = 0.

Example 7.8: Example 7.7 revisited The least square control algorithm is applied to the model considered in example 7.7. The following values are specified for the performance measures, 0.007 vo = 0.004

0.005 v* = 0.005

(1)

and the transverse shear rigidity is assumed to be constant. Case 1. Using eqn (1) of example 7.7, eqn (7.82) reduces to 2 1 -------- F 1 = -------- [ ( v 1* – v 1, 0 ) + ( v 2* – v 2, 0 ) ] 2 D DT T

(2)

The solution for F 1 and the corresponding error vector are F 1 ≡ 0.0005D T 0.0015 e = – 0.0015

(3)

Case 2 Since the number of control forces is equal to the number of performance measures, solving eqn (7.82) is equivalent to solving eqn (7.80) with e set equal to

7.4 Quasi-static Control of MDOF Systems 529 0. This result follows by writing eqn (7.82) as C Tf ( C f F – v * + v o ) = 0

(4)

and noting that the force vector which satisfies C f F = v * – vo

(5)

also satisfies (4). The solution is F 1 = 0.001D T F 2 = – 0.003D T

(6)

e = 0 Case 3 In this case, there are 3 force parameters and only 2 performance variables. The problem is undetermined. Setting e = 0 in eqn (7.80) provides 2 equations relating F 1 , F 2 , and F 3

C f F = v * – vo

⇒

1 -------- ( F 1 + F 2 + F 3 ) = – 0.002 DT 1 -------- ( F 1 + F 3 ) = 0.001 DT

(7)

Taking F 3 as the “excess” force parameter, and solving for F 1 and F 2 leads to F 1 = – F 3 + 0.001D T F 2 = – 0.003D T

(8)

The additional equation is obtained by requiring the norm of F to be stationary with respect to F 3 . For this example, the norm is taken as the Euclidean norm, I. 1 1 I = --- F T F = --- ( F 12 + F 22 + F 32 ) = I ( F 3 ) 2 2 Then

(9)

530 Chapter 7: Quasi-static Control Algorithms ∂I = 0 ∂ F3

⇒

F1

∂F 1

∂ F3

+ F3 = 0

⇒

F1 ( –1 ) + F3 = 0

(10)

Finally, the control force magnitudes are F 1 = ( 0.0005 )D T F 2 = ( – 0.003 )D T

(11)

F 3 = ( 0.0005 )D T Other possible solutions that satisfy e = 0 are 1 0 F = 10 – 3 D T – 3 and F = 10 – 3 D T – 3 0 1

(12)

The pseudo inverse solution given by (11) has the lowest norm, i.e., the smallest value of I.

The generalized version of the least square algorithm is based on the extended form of the error norm, 1 1 J 1 = --- qe T Q * e + --- F T R * F = J 1 ( F ) 2 2

(7.83)

where Q * and R * are diagonal weighting matrices with elements of order 1, and q is a scaling factor defined as q = 1 ⁄ ( maximum element in C Tf C f )

(7.84)

Requiring J 1 to be stationary with respect to F leads to r equations for F, a1 F = b1 a 1 = q C Tf Q * C f + R *

(7.85)

b 1 = q C Tf Q * ( v * – v o ) Different priorities are assigned by perturbing the individual elements of Q * and

7.4 Quasi-static Control of MDOF Systems 531 R * . If all the elements of R * are finite, a 1 will be non-singular for r > n p as well as for r ≤ n p , and it is not necessary to generate additional equations for F. However, since this property is due to including R * , the equations may be illconditioned if r > n p and R * is small. The preferred strategy is to take R * = I and perturb the elements of Q * . The following example illustrates different scenarios.

Example 7.9: Example 7.7 revisited with an extended least square algorithm Certain cases considered in example (7.8) are reworked here using the extended algorithm defined by eqn (7.85) and the same performance specifications as for example (7.8). Case 1 2 C Tf C f = -------D T2

(1)

D T2 1 q = --------------------------------------------------------------------------- = -------2 maximum element in C Tf C f

(2)

q 1* + q 2* a 1 = ------------------ + r 1* 2

(3)

DT b 1 = [ q 1* ( v 1* – v 1, 0 ) + q 2* ( v 2* – v 2, 0 ) ] -------2 DT = [ ( – 0.002 )q 1* + 0.001q 2* ] -------2 D T ( – 0.002 )q 1* + 0.001q 2* F 1 = -------- ----------------------------------------------------2 1 * --- ( q 1 + q 2* ) + r 1* 2

(4)

(5)

Taking equal weights for the displacement errors and the force cost ( q 1* = q 2* = r 1* = 1 ) reduces F 1 by 50%.

532 Chapter 7: Quasi-static Control Algorithms Case 3 212 1 C Tf C f = -------- 1 1 1 D T2 212

(6)

D T2 q = -------2

(7)

1 a 1 = --2

q 1* + q 2* + 2r 1*

q 1*

q 1* + q 2*

q 1*

q 1* + 2r 2*

q 1*

q 1* + q 2*

q 1*

q 1* + q 2* + 2r 3*

– 2q * + q * 1 2 10 – 3 D T b 1 = ------------------- – 2q 1* 2 – 2q 1* + q 2*

(8)

(9)

Note that the elements of a 1 are of the same order. When r 1* = r 3* = 0 , the third row of a 1 is equal to the first row, and a 1 is singular. The third row of b 1 is also equal to the first row, and a 1 F = b 1 represents only 2 independent equations in this case. The third equation relating the 3 control forces can be generated by requiring the norm, ( 1 ⁄ 2 )F T F , to be stationary. As mentioned earlier, the MATLAB function, F = pinv ( a 1 )*b 1 , generates this solution. The standard MATLAB statement, F = a 1 \b 1, will also generate a solution for this case, but the force norm will be greater than the value corresponding to the pseudo inverse procedure. Results for F and e based on taking q 1* = q 2* = q * , r 1* = r 2* = r 3* = 1, and letting q * range from 1 to ∞ are listed in Table 1. The last row contains the error for the case where no control forces are applied. Taking q * = ∞ places all the emphasis on the error reduction, and results in the maximum control force magnitude. Decreasing q * shifts some of the emphasis to reducing the control force.

7.4 Quasi-static Control of MDOF Systems 533

q*

10 3 F 1 ⁄ D T

10 3 F 2 ⁄ D T

10 3 F 3 ⁄ D T

10 3 e 1

10 3 e 2

∞

0.5

-3.0

0.5

0

0

∞ ( a1 \ b1 )

0

-3.0

1.0

0

0

16

0.3314

-2.3670

0.3314

0.2954

-0.3373

8

0.2264

-1.9623

0.2264

0.4906

-0.5472

4

0.1053

-1.4737

0.1053

0.7363

-0.7895

2

0

-1.0

0

1.0

-1.0

1

-0.0625

-0.625

-0.0625

1.25

-1.125

No control

0

0

0

2.0

-1.0

( pinv ( a 1 )*b 1 )

Table 1

534

Introduction to Structural Motion Control

535 Chapter 7: Quasi-static Control Algorithms

Problems Problem 7.1 A simply supported beam is to be subjected to a concentrated force which can be applied to any point along the longitudinal axis of the beam. The maximum moment that the beam can resist is 250kNm. Since the maximum moment due to the force is 500kNm, it is necessary to apply a force system which controls the magnitude of the maximum moment. Consider the force control scheme shown below. The control forces produce a piecewize - linear moment distribution defined by the nodal moments m 1 , m 2 , and m 3 . Take m 1 , m 2 , and m 3 as the “force” parameters a) Assuming the force is translated along the beam, determine the values of m 1 , m 2 , and m 3 corresponding to the 7 load points located at an interval of L/8 along the beam. b) Suppose the maximum value of m 2 that the force actuator can deliver is 200kNm. Assuming there are no restrictions on m 1 and m 3 , can the control force system prevent the moment in the beam from exceeding 250kNm? How would you modify the control scheme if this requirement cannot be satisfied?

x

P

L m2

m1

L = 20meters m3

L⁄4

L⁄4

L⁄4

L⁄4

Problems 536 Problem 7.2

b 0 = 20 kN/m

x L = 10 m

Consider the cantilever beam shown above. Assume transverse shear deformation is negligible. The desired state is constant curvature and u ( L ) ≤ L ⁄ 400 . a) Determine the bending rigidity distribution based on the equilibrium model, M(x) D B = ------------χ∗

(1)

b) Determine a “constant” value of D B using the least square procedure described below. The displacement function for the triangular loading is given by: 4

b0 L 1 1 u 0 = ------------ x 2 1 – --- x + ------ x 3 = D B– 1 g ( x ) 6D B 2 20

(2)

where x = x ⁄ L . One forms the “displacement” error e ( x ) = u 0 ( x ) – u∗ ( x )

(3)

and the error functional, L

J =

∫0 e2dx

(4)

Let D B– 1 = f B . The value of f B is found by minimizing J with respect to f B .

537 Chapter 7: Quasi-static Control Algorithms ∂J --------- = 0 ∂fB

(5)

Establish the expression for f B . c) Assume D B is specified. The displacement u 0 is now fixed. Suppose a vertical force, F , is now applied at x = L and the magnitude of F can be adjusted. Determine a value for F by minimizing J with respect to F . The appropriate form of the error is e = u 0 + u c – u∗ where FL 3 F x3 u c = ----------- x 2 – ----- = -------h ( x ) 2D B DB 3

(6)

Evaluate F for a representative range of D B . d) Let e i represent the displacement error at point x i . e i = u 0 ( x i ) + u c ( x i ) – u∗ ( x i )

(7)

Suppose one uses a “discrete” error functional consisting of N terms. 1 J = --2

N

∑ ei2

(8)

i=1

Derive the general expression for F corresponding to requiring J to be stationary. Evaluate F taking D B constant = 1/2 the value obtained in part b, N = 3 , and x i = 1/3, 2/3, and 1. Note: it is convenient to introduce matrix notation and express the error as: e =

e1 e2 = e + f F e3

(9)

1 J = --- e T e 2 e) Refer to part d. Consider 2 control forces, F 1 at x = L ⁄ 2 and F 2 at x = L . Evaluate F 1 , F 2 for the conditions specified in part d. Compare the displacement solutions for part d and part e.

Problems 538 Problem 7.3

u

b0

x L Consider a cantilever beam subjected to a uniform loading. Take D B = EI as constant and neglect transverse shear deformation. The deflected shape is given by: b0 L 4 u = ------------ [ x 2 ( 6 – 4x + x 2 ) ] 24EI where x = x ⁄ L . Suppose the ‘‘desired’’ deflected shape is u = aLx 2 where a is a specified value, and a concentrated moment, M , is applied at x = L to ‘‘correct’’ the difference between the actual and desired deflected shape. a) M

x L Determine an expression for M using the continuous least square error functional. b) Determine M using the discrete least square error functional. Take 10 equally spaced points in the interval 0 < x ≤ L . Compare the error norms for parts (a) and (b).

539 Chapter 7: Quasi-static Control Algorithms c) M

αL L Suppose M is applied at x = αL . Repeat part (b), taking α = 0.5 and 0.75. Compare the error norms corresponding to α = 0.5, 0.75 , and 1.0.

Problem 7.4 a) Consider the bending beam - outrigger system shown below. Assume the outriggers are infinitely stiff, the beam bending rigidity is constant, and the cables are initially tensioned to a level of T o . Suppose the cable tensions can be adjusted to counteract the effect of a lateral load. Consider a force, P, applied at x = H . Suppose the desired deflected shape is u = x 2 ⁄ ( 300H ) . Determine an estimate for ∆T . P, u C b

b B

T o + ∆T

H DB

A

T o – ∆T

2H -------3 x

Problems 540 b) Extend the analysis procedure of part (a) to deal with the case of 2 outriggers shown below. Assume ∆T 1 and ∆T 2 can be independently adjusted. P, u

C T o + ∆T 1

DB b

T o – ∆T 1 b H

B T o + ∆T 2

DB

T o – ∆T 2

H ---2 x

A

c) Refer to part (a). Suppose the outrigger is located a distance αH from the base. Determine an estimate for ∆T in terms of α . Consider α to vary from 0.25 to 0.75.

Problem 7.5

v P

Ma L

L

541 Chapter 7: Quasi-static Control Algorithms The structure shown above consists of two single spans interconnected at the interior support with a force actuator system that can instantaneously apply a moment, M a ( t ) . The structure also has the capability to observe the transverse displacement at a limited number of points along both spans. Consider a force, P , which moves with velocity v along the span. Assume the velocity is sufficiently small such that the actual response of the structure can be approximated by the corresponding static response. a) Describe how you would establish the time history of the actuator moment, M a , such that the maximum transverse displacement at any time in both spans is less than a specified value, u∗ . Take PL 3 u∗ = -----------96EI Note: PL 3 ⁄ ( 48EI ) is the maximum value of u due to P moving across the span with M a = 0 . b) Discuss how you would implement this control scheme.

Problem 7.6 Solve problem 7.2, part (d), using the extended error function. Take r 1* = 1 and q 1* = q 2* = q 3* = q * . Investigate the variation of F and the displacement error norm with q * .

Problem 7.7 a) Solve problem 7.4, part (b), considering a “force” cost as well as a displacement error norm. Generate solutions for a range of the weighting parameters. Assume equal relative weights for the displacements ( q 1 = q 2 = q* ) and the control forces ( r 1 = r 2 = r * ) . b) Problem 7.4 considered a single force applied at x = 1. Describe how you

Problems 542 would extend the solution scheme to handle an arbitrary lateral loading applied to member ABC.

Problem 7.8

12kN D T3 = 2400kN 8kN D T2 = 4000kN 4kN D T1 = 4800kN

Consider the 3 DOF shear beam shown above. Suppose the performance measures are the transverse shear strains and the desired value of strain is 1/300 for each segment. The 3 control force schemes shown below are proposed. Establish values for the control forces associated with each scheme using the extended error norm defined by eqn (7.83) and the following uniform weighting factors: r i* = 1

q *j = q *

q * = 1, 2, 5, 100

Also investigate how the norm of the strain error vector varies with q * for each scheme. Which scheme would you recommend? What is the basis for your decision?

543 Chapter 7: Quasi-static Control Algorithms

F3 F3

F1

F1 F2

F2

F2

F1

F1 (1)

(2)

(3)

Problem 7.9 Consider the discrete shear beam shown below. Assume the transverse shear rigidity is constant for all the elements. a) Suppose the goal is to have uniform transverse shear strain over the height. Formulate the control algorithms for the 2 control force schemes shown. Which scheme would you employ? b) Suppose the goal is to have a prescribed nodal displacement vector U * . Let U o denote the initial displacement vector. Formulate the control algorithms for the 2 schemes. Does scheme 1 still correspond to fully decentralized control? Elaborate on your answer.

Problems 544

F5 (5) h (4) h (3) h (2) h

F4

F1

F5

F4

F3

F2

F2

F3

F2

F1

(1) h (1)

(2)

545

Chapter 8

Dynamic Control Algorithms - Time invariant systems 8.1 Introduction This chapter extends the active control strategy introduced in the previous chapter to deal with time dependent loading. The discussion is restricted to time invariant systems, i.e., the case where the system properties and force feedback algorithm are constant over the duration of the time response. The material presented here is organized as follows. Firstly, the state-space formulation for a SDOF system is developed, and used to generate the free vibration response. This solution provides the basis for establishing a criterion for dynamic stability of a SDOF system. Secondly, linear negative feedback is introduced, and the topic of stability is revisited. The primary focus is on assessing the effect of time delay in applying the control force on the stability. Thirdly, the SDOF state-space formulation is specialized to deal with discrete time control, where the feedback forces are computed at discrete time points and held constant over time intervals. Stability for discrete time feedback with time delay is examined in detail, and a numerical procedure for determining the time increment corresponding to a stability transition is presented and illustrated with examples. Fourthly, the choice of the optimal magnitudes of the feedback

546 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems parameters is considered. Optimality is related to the magnitude of a quadratic performance index (LQR) which is taken as a time integral involving weighted response and control force terms. This approach is referred to as the “linear quadratic regulator problem” and leads to a time invariant linear relationship between the control forces and state variables. Fifthly, the state-space formulation is extended to MDOF systems. The modal properties for an arbitrary damping scheme are derived, and used to generate the governing equations expressed in terms of the modal coordinates. The last section deals with optimal feedback based on the LQR performance index generalized for MDOF systems. Examples which illustrate the sensitivity of the response and cost parameters to variations in the location and nature of the control forces, and weighting coefficients are presented.

8.2 State-space formulation - time invariant SDOF systems Governing equations

k

F m p

c ug

u + ug

Fig. 8.1: SDOF system. The dynamic response of the SDOF linear system shown in Fig. 8.1 is governed by the second order equation, mu˙˙ + cu˙ + ku = – ma g + p + F

(8.1)

where p is the applied external loading, F is the active force, and m, k, c are system parameters. Integrating eqn (8.1) in time, and enforcing the initial conditions on u and u˙ at t = 0 , one obtains the velocity and displacement as functions of time. These quantities characterize the state of the system in the sense that once u and u˙ are specified, the acceleration and internal forces can be determined by back substitution.

8.2 State-space Formulation-Time Invariant SDOF System 547 Rather than working with a second order equation, it is more convenient to transform eqn (8.1) to a set of first order equations involving the state variables u and u˙ . The new form is

du ------ = u˙ dt du˙ 1 1 c k ------ = – ---- u˙ + – ---- u + ( – 1 )a g + ---- p + ---- F m m m m dt

(8.2)

This form is called the state-space representation. The motivation for the statespace representation is mainly the reduced complexity in generating both analytical and numerical solutions. Matrix notation is convenient for expressing the state-space equations in a compact form. Defining X as the state vector, X = u = X (t) u˙

(8.3)

the matrix equilibrium equation is written as dX ------- = X˙ = AX + B f F + B g a g + B p p dt

(8.4)

where the various coefficient matrices are defined below. A =

0 1 k c – ---- – ---m m

Bf = Bp =

Bg =

0 1 ---m

0 –1

The initial conditions at t=0 are denoted by X o .

(8.5)

(8.6)

(8.7)

548 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems u(0) ≡ Xo u˙ ( 0 )

X(0) =

(8.8)

With this representation, the problem is reduced to solving a first order equation involving X . Free vibration uncontrolled response The free vibration uncontrolled response is governed by a reduced form of eqn (8.4) X˙ = AX

(8.9)

When A is constant, the general solution has the form X = Ve

λt

(8.10)

where V is an unspecified vector of order 2 and λ is a scalar. Substituting for X results in ( A – λI )V = 0

(8.11)

where I is the identity matrix. According to eqn (8.11), the eigenvalues of A define the frequency and damping characteristics of the free vibration response. Expanding A – λI = 0 , –λ 1 k c – ---- – ---- – λ m m

= 0

(8.12)

leads to the characteristic equation 2 c k λ + ----λ + ---- = 0 m m

(8.13)

and two eigenvalues 1 c k c 2 λ 1, 2 = --- – ---- ± i 4 ---- – ---- = λ R ± iλ I m m 2 m 2

(8.14)

Noting that k ⁄ m = ω and c ⁄ m = 2ξω , eqn (8.14) is identical to eqn (6.27) which was obtained from the second order equation. Given λ , eqn (8.11) can be solved for the eigenvectors which define the state-space modes. Since λ is complex, the eigenvectors occur as complex conjugates.

8.2 State-space Formulation-Time Invariant SDOF System 549

0 1 ˜ ±i = V R ± iV I = V 1, V 1 λI λR

V 1, 2 =

(8.15)

The total free vibration response is obtained by combining the 2 complex solutions such that the resulting expression is real. Starting with X = A1 e

λ1 t

V 1 + A2 e

λ˜ 1 t

V˜ 1

(8.16)

and taking 1 A 1 = --- ( A R + iA I ) 2 ˜ A = A

(8.17)

1

2

where A R and A I are real scalars, results in X (t) = e

λR t

{ ( A R V R – A I V I ) cos λ I t + ( – A R V I – A I V R ) sin λ I t }

(8.18)

The constants A R and A I are determined by enforcing the initial conditions on X at t=0. ˜ V˜ = A V – A V X ( 0 ) = A1 V 1 + A 1 1 R R I I ⇒ uo

=

u˙ o

AR AR λR – AI λI

(8.19)

⇒ AR = uo 1 A I = – ----- ( u˙ o + λ R u o ) λI Lastly, the solution for u(t) is given by the first scalar equation in eqn (8.18). u(t) = e

λR t

( A R cos λ I t – A I sin λ I t )

(8.20)

550 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems General solution - time invariant systems The general solution for an arbitrary loading can be expressed as a Duhamel integral involving a specialized form of the free vibration response. Considering first a first order scalar equation, (8.21)

y˙ = ay + g

where a is constant, and g is a function of t, the complete solution has the form y(t) = e

a ( t – to )

t

yo +

∫

e a ( t – τ ) g ( τ ) dτ

(8.22)

to

A similar form can be generated for the first order matrix equation, X˙ ( t ) = AX + G

(8.23)

The free vibration solution defined by eqn (8.18), can be expressed as X ( t ) = e At X o

(8.24)

where e At is defined by the following series: 1 1 e At = I + At + --- AAt 2 + … + ----- A n t n + … 2 n!

(8.25)

This matrix exponential function has the “same” property as the corresponding scalar function. d At ( e ) = Ae At dt Using eqn (8.24), the Duhamel integral matrix form of the total solution for eqn (8.4) is X (t) = e

A ( t – to )

t

Xo +

∫

e A ( t – τ ) G ( τ ) dτ

(8.26)

to

where G ( t ) = B f F + B g a g + Bp p The corresponding scalar form of the solution for u(t) is

(8.27)

8.2 State-space Formulation-Time Invariant SDOF System 551

u(t) = e

λR t

1 u cos λ I t + ----- ( u˙ o + λ R u o ) sin λ I t + o λ I

t

+

∫

0

(8.28)

1 λ (t – τ) p(τ) F(τ) ----- e R sin λ I ( t – τ ) – a g ( τ ) + ---------- + ----------- dτ λI m m

Equation (8.26) applies for an arbitrary linear time invariant system. It is convenient for establishing a discrete formulation of the governing equations. This topic is addressed in the next section.

Example 8.1: Equivalence of equations (8.18) and (8.24) Consider eqn (8.16). The total free vibration response is given by ˜ e λ˜ t V˜ X ( t ) = A 1 e λt V 1 + A 1 1 ˜ V˜ X (0) = A V + A 1

1

1

(1)

1

Noting eqn (8.11), the λ and V terms are related by AV 1 = λV 1 AV˜ = λ˜ V˜ 1

(2)

1

Expanding the product, e λt V 1 , and using eqn (2), leads to t2 e λt V 1 = V 1 + ( λV 1 ) t + λ ( λV 1 ) ---- + … 2 t2 = I + At + AA ---- + … V 1 2

(3)

= e At V 1 It follows that eqn (1) can be written as ˜ V˜ ) = e At X X ( t ) = e At ( A 1 V 1 + A 1 1 o

(4)

552 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Stability criterion Another advantage of the state-space representation is the ability to relate the stability of the physical system to the eigenvalues of A . A system is said to be stable when the motion resulting from some initial disturbance is bounded. Assuming the system state is X o at time t = 0 , stability requires X ( t ) – Xo ≤ ε

for all t

(8.29)

where ε defines the bound on the perturbation from X o . Equation (8.18) defines the general homogeneous solution for a SDOF time invariant system. The terms contained inside the brackets depend on the initial conditions and are bounded since the time dependency is harmonic. Therefore, it follows that the exponential term must be bounded. This requirement is satisfied when the exponent is negative, λR ≤ 0

(8.30)

In words, the real part of the eigenvalues of A must be equal to or less than zero. When λ R = 0 , the response is pure harmonic oscillation. A negative λ R produces a damped harmonic response. Plotting λ in the complex plane provides a geometric interpretation of the stability. For the SDOF case, there are two eigenvalues, λ = λ R ± iλ I c λ R = – -------- = – ξω 2m λI =

(8.31)

2 c 2 k ---- – -------- = ω 1 – ξ m 2m

Figure 8.2 shows the corresponding points in the complex plane. These points are referred to as poles. Undamped motion has poles on the imaginary axis. Holding stiffness constant and increasing c causes the poles to move along the circle of radius ω toward the critical damping point, ξ = 1 . With further increase in damping, the curves bifurcate with one branch heading in the negative (real axis)

8.2 State-space Formulation-Time Invariant SDOF System 553 direction, and the other toward the origin. Increasing the stiffness with c held constant moves the poles in the imaginary direction. With this terminology, the stability criterion requires all the poles corresponding to the eigenvalues of A to be on or to the left of the imaginary axis, as shown in Fig. 8.3. The uncontrolled SDOF system is, according to this definition, always stable since ξ ≤ 0 . k increases λ I ξ = 0 ω c increases ξ = 1

θ

λR

ω ξ = 0 Fig. 8.2: Poles for SDOF system.

λI

stable

unstable

λR

Fig. 8.3: Stability condition for SDOF system. Linear negative feedback The response of a SDOF time invariant system with negative linear feedback is governed by eqn (8.4) with F taken as a linear function of the state variables

554 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems F = –Kf X = – kd kv u u˙

(8.32)

Substituting for F , the governing equation is transformed to X˙ = A c X + B g a g + B p p

(8.33)

where 0 Ac = A – Bf Kf =

1

c kv k kd – ---- – ----- – ---- – ----m m m m

(8.34)

The general form of the free vibration solution of eqn (8.33) is X = Ve

λt

(8.35)

where λ and V are the eigenvalues and eigenvectors of A c , the modified coefficient matrix. They are related by [ A – B f K f – λI ]V = O

(8.36)

The eigenvalues of A c are written as λ = λR ±i λI c + kν λ R = – -------------- = – ξ eq ω eq 2m 2

λ I = ω eq 1 – ξ eq kd 1 ⁄ 2 ω eq = ω 1 + ----------ω2m ξ eq = ξ + ξ a

(8.37)

kν ω ξ a = ----------------- – ξ 1 – -------- 2mω eq ω eq Since k v is positive for negative feedback (note that the minus sign is incorporated in the definition equation, eqn (8.32)), the system is stable for arbitrary k v . Displacement feedback moves the poles in the imaginary direction, and consequently has no effect on the stability. It follows that increasing the negative

8.2 State-space Formulation-Time Invariant SDOF System 555 velocity feedback is a more appropriate mechanism for improving the stability of a SDOF system. Noting eqns (8.23) and (8.26), the total solution for time invariant linear feedback can be expressed in a form similar to eqn (8.26). X (t) = e

Ac ( t – to )

t

Xo +

∫

e

Ac ( t – τ )

to

( B g a g ( τ ) + B p p ( τ ) ) dτ

(8.38)

The identity established in example 8.1 is also applicable here. e

Ac ( t – to )

X o ≡ Ae

λ ( t – to )

˜

˜ e λ ( t – t o ) V˜ V+A

(8.39)

where A is a complex scalar, and ( λ, V ) are the solution of eqn (8.36). Requiring A ( t – to ) λ R ≤ 0 ensures that e c is bounded. The above discussion assumes there is no delay between observing the state and generating the force. In general, there is some delay and the force at time t is computed using data observed at an earlier time, t – t d . The force for linear negative feedback is taken as F ( t ) = –Kf X ( t – td )

(8.40)

where t d represents the delay time. Delay introduces additional terms in λ R and under certain conditions, can cause λ R to become positive and consequently, the system becomes unstable. Therefore, although ideal linear negative feedback is unconditionally stable, one needs to examine the potential destabilizing effect of delay for the actual control system. One procedure for investigating the effect of time delay on the stability of a controlled time invariant SDOF system is described in the following section. Additional approaches are discussed in later sections. Effect of time delay on feedback control Time delay in feedback control systems in the sum of the times required to execute the following actions: • acquire the data from sensors placed at different locations in the structure; • process the sensor data and calculate the control force;

556 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems • transmit the control force signal to the actuator; • ramp up the actuator to the desired force level. The resultant time delay affects the synchronization between the control force and the response of the system and may, under certain conditions, cause a significant degradation in the performance of the control system which could result to instability. In what follows, an analytical procedure for assessing the stability of a SDOF system with time delay is presented. This approach follows the method proposed by Agrawal et al. (1993). Results based on numerical simulations are included here to illustrate the effect of time delay on the response. Assuming the feedback control algorithm consists of a linear combination of displacement and velocity terms, and uses data associated with the time, t – t d , the feedback force at time t is written as kd kv 1 ----F ( t ) = – -----u ( t – t d ) – -----u˙ ( t – t d ) = – g d u ( t – t d ) – g v u˙ ( t – t d ) (8.41) m m m Substituting for F in eqn (8.1), the governing equation for this case is 2 p(t) u˙˙( t ) + 2ξωu˙ ( t ) + ω u ( t ) + g d u ( t – t d ) + g v u˙ ( t – t d ) = ---------m

(8.42)

The general form of the homogeneous solution of eqn (8.42) is u = Ae

λt

(8.43)

where A is an arbitrary constant and λ satisfies 2

λ + 2ξω + g v e

– λt d

2

λ + ω + gd e

– λt d

= 0

(8.44)

Letting λ = λ R ± iλ I represent the roots, the stability requirement is λ R ≤ 0 . Expressing λ R and λ I in a convenient form is complicated by the presence of the exponential terms. A first order approximation can be obtained by introducing the following expansion e

– λt d

1 1 = 1 – λt d + --- ( λt d ) 2 – --- ( λt d ) 3 + … 6 2

(8.45)

and retaining only the first 2 terms. The result is expressed in the same form as eqn (8.37) with ξ eq and ω eq replaced with modified terms.

8.2 State-space Formulation-Time Invariant SDOF System 557 λ R = – ξ' eq ω' eq (8.46)

2

λ I = ω' eq 1 – ξ' eq

The modified equivalent frequency and damping are related to the time delay by 2

1 + gd ⁄ ω ω' eq = ω -------------------------1 – gv td

1 ξ + ------- ( g v – g d t d ) 2ω ξ' eq = ---------------------------------------------------gd [ 1 – g v t d ] 1 + -----2 ω

(8.47)

(8.48)

Equation (8.46) is convenient for identifying behavioral trends. For no initial damping and no velocity feedback ( g v = 0 ), the approximation for λ R reduces to gd td k d td λ R = ------------ = -----------2ω 2ωm

(8.49)

Since λ R is positive, it follows that according to this first order approximation, displacement feedback with time delay produces unstable behavior for an initially undamped system ( ξ = 0 ) . The real part of λ corresponding to pure velocity feedback ( g d = 0 ) is estimated as kv gv ξ + -----------ξ + ------2ω 2ωm λ R = – ω -------------------- = – ω ---------------------1 – gv td kv td 1 – ---------m

(8.50)

This result suggests that the response is stable when ( k v t d ⁄ m ) < 1 . An improved approximation can be obtained by substituting the Pade

558 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems expansion e

– λt d

1 – λt d ⁄ 2 3 ≈ ------------------------- + O [ ( λt d ) ] 1 + λt d ⁄ 2

(8.51)

in eqn (8.44). The resulting expression is now a 3rd degree polynomial in λ td 3 td 2 g 2 2 ----- λ + 1 + t d ξω – -----v- λ + 2ξω + g v + ----- ( ω – g d ) λ + ω + g d = 0 2 2 2 (8.52) Introducing the notation for the equivalent frequency and damping defined earlier for the instantaneous feedback case (see eqn (8.37)), 2

2

ω eq = ω + g d 1 ξ eq ω eq = ξω + --- g ν 2 ξ eq = ξ + ξ a

(8.53)

transforms eqn (8.52) to td 3 2 ----- λ + [ 1 + ( 2ξω – ξ eq ω eq )t d ]λ + 2

(8.54) 2

2 ω eq 2 2ξ eq ω eq + t d ω – --------- λ + ω eq = 0 2 For pure displacement feedback, ξ eq ω eq = ξω . For pure velocity feedback, ω eq = ω and ξ a = k ν ⁄ ( 2ωm ) . Equation (8.54) is solved numerically for a specific SDOF system with a period of 5s , and the two limiting cases of pure displacement and pure velocity feedback with no initial damping. Figure 8.4 shows the movement of the poles for pure displacement feedback as a function of ω eq and t d . When t d = 0 , the poles move on the imaginary axis with increasing ω eq . As t d increases, the path shifts to the right, and when t d is about 1s , the direction is essentially along the positive real axis.

8.2 State-space Formulation-Time Invariant SDOF System 559

2.5 2

Imaginary axis, λ I

1.5

T = 5s ξ = 0% kv = 0

1 0.5 0

ω eq ω eq ω eq

td = 0 = 2.0ω = 1.5ω = 1.0ω

t d = 0.25

t d = 0.5 t d = 1.0

k+k 2 = -------------dω eq m

−0.5 −1

ω eq = 1.0ω ω eq = 1.5ω ω eq = 2.0ω td = 0

−1.5 −2 −2.5 −0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

Real axis, λ R

t d = 1.0 t d = 0.25 0.2

0.3

t d = 0.5 0.4

0.5

Fig. 8.4: Variation of pole profile with t d under pure displacement feedback. The effect of time delay for pure velocity feedback is illustrated by Fig. 8.5. For no time delay, increasing k ν moves the poles further back in the negative real half plane until the state of critical damping is reached. As t d increases, the paths tend to bend toward the positive real half plane, and eventually intersect the imaginary axis. For a given value of k ν , there may be a limiting time delay beyond which the system is unstable. These observations are based on an approximation and apply for a particular system, i.e. specific values of ω and ξ . An exact analysis of the instability problem is presented in the next section.

560 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

2.5 2

t d = 0.5

T = 5s ξ = 0%

t d = 1.0

Imaginary axis, λ I

1.5

t d = 0.25

1

0.5

td = 0

ξ a = 0.4 0.3

0.2

0.1 kν ξ a = -----------2ωm kd = 0

0 −0.5

td = 0

−1

t d = 0.25

ξ a = 0.4 0.3

0.2

0.1

−1.5

t d = 1.0

−2 −2.5 −1

t d = 0.5 −0.8

−0.6

−0.4

−0.2

0

Real axis, λ R

0.2

0.4

Fig. 8.5: Variation of pole profile with t d under pure velocity feedback.

Stability analysis for time delay Figure 8.5 shows that, for a given system, there is a particular value of t d which corresponds to a transition in the behavior of λ R . For t d greater than this value, λ R increases and eventually becomes positive. When λ R = 0 , there is a transition from stable to unstable behavior, and the corresponding value of t d defines the stability limit for the system, i.e., for the particular combination of k, c, k ν , and k d . Equation (8.44) is the definition equation for λ . Noting that the critical time delay corresponds to λ R = 0 , by expressing λ as λ = iΩ

(8.55)

where Ω is a real scalar, one can specialize eqn (8.44) for this case and determine the critical value of t d . Substituting for λ leads to 2

2

– Ω + i2ξωΩ + ω + g d e

– it d Ω

+ ig v Ωe

– it d Ω

= 0

(8.56)

8.2 State-space Formulation-Time Invariant SDOF System 561 Replacing the exponential term by e

– it d Ω

= cos ( t d Ω ) – i sin ( t d Ω )

(8.57)

yields 2

2

– Ω + ω + g d cos ( t d Ω ) + g v Ω sin ( t d Ω )

(8.58)

+ i [ 2ξωΩ – g d sin ( t d Ω ) + g v Ω cos ( t d Ω ) ] = 0 Equation (8.58) is satisfied when the real and imaginary terms vanish. Then, g d sin ( t d Ω ) – g v Ω cos ( t d Ω ) = 2ξωΩ 2

g d cos ( t d Ω ) + g v Ω sin ( t d Ω ) = Ω – ω

(8.59) 2

(8.60)

It remains to solve these equations for t d in terms of g d , g v , ω , and ξ . Squaring both sides and adding the equations eliminates the trigonometric terms and results in a quartic equation for Ω . 4

2 2

2

2

2

4

2

Ω + ( 4ξ ω – 2ω – g v )Ω + ω – g d = 0

(8.61)

The roots of eqn (8.61) are given by 2 2

2

2

2 2

2

2 2

( 4ξ ω – 2ω – g v ) ( 4ξ ω – 2ω – g v ) 4 2 Ω 1, 2 = − + – ------------------------------------------------- − + ---------------------------------------------------- – ( ω – g d ) 2 4 (8.62) Since the poles correspond to ± iΩ , only the positive value of Ω needs to be considered, resulting in two values of Ω . The next step is to determine t d Ω . Noting the trigonometric identities, td Ω 2 tan ---------2 sin ( t d Ω ) = -----------------------------------2 td Ω 1 + tan ---------2

(8.63)

562 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems 2 td Ω 1 – tan ---------2 cos ( t d Ω ) = -----------------------------------2 td Ω 1 + tan ---------2

(8.64)

eqn (8.59) can be expressed as td Ω 2 td Ω ( g v Ω – 2ξωΩ )tan ---------- + 2g d tan ---------- – ( g v Ω + 2ξωΩ ) = 0 2 2

(8.65)

The two roots of eqn (8.65) are td Ω – gd − + g d + g v Ω – 4ξ ω Ω tan ---------- = -----------------------------------------------------------------------------2 g v Ω – 2ξωΩ 2

2

2

2 2

2

(8.66)

Finally, the maximum time delay t d can be determined from max 2 2 2 2 2 2 − – 1 – g d + g d + g v Ω – 4ξ ω Ω 2 = ----tan -----------------------------------------------------------------------------td Ω g v Ω – 2ξωΩ max

(8.67)

The minimum positive value of t d is the maximum allowable time delay. max For the limiting case of pure velocity feedback control of an undamped system (i.e. g d = 0 and ξ = 0 ), eqn (8.59) reduces to g v cos ( t d Ω ) = 0

(8.68)

and it follows that td

π = ------2Ω max

(8.69)

The expression for Ω can be obtained from either eqn (8.60) or eqn (8.62), 2 Ω ---- = ξ a + 1 + ξ a ω

(8.70)

gν kν ξ a = ------- = -----------2ω 2ωm

(8.71)

where

Finally, the maximum delay can be expressed in terms of the fundamental period

8.2 State-space Formulation-Time Invariant SDOF System 563 of the uncontrolled system. td 1 max ----------------- = ---------------------------------------T 2 4 ξa + 1 + ξa

(8.72)

Figures 8.6 through 8.8 illustrate the effect of varying the displacement feedback, velocity feedback, system damping, and system period on the maximum allowable time delay. Figure 8.6 shows plots of the maximum allowable time delay for an undamped 5s SDOF system as a function of the active damping ratio ξ a (i.e. velocity feedback) for three values of displacement feedback k d . The central curve corresponds to k d = 0 . The lower curve corresponds to k d > 0 , i.e. leading to an increase in the frequency of the controlled system, and shows that t d = 0 for no active damping. Furthermore, Fig. 8.6 illustrates the effect of underestimating/overestimating the stiffness of the system on the maximum allowable time delay. If the model used to establish the system properties underestimates the stiffness, the actual limit on the time delay will be less than the predicted limit and stability may be a problem. Figure 8.7 shows the effect of the damping in the system on the maximum allowable time delay. In general, damping increases the allowable time delay. The effect of the fundamental period of the system on the maximum allowable time delay is illustrated by Fig. 8.8. As expected from eqn (8.20), the maximum allowable time delay increases with period and decreases with active damping. Finally, Fig. 8.9 illustrates the degradation in performance, with increasing time delay, of the SDOF system subjected to seismic excitation. Instability occurs beyond a time delay of 1.1s which, according to Fig. 8.7, corresponds to the maximum allowable time delay for this level of damping. The instability is due to the unbounded growth of the homogeneous solution, and will occur for any arbitrary external excitation.

564 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

2.5

T = 5s ξ = 0% 2

k 2 = ω 2 + ----dω eq m

1.5

ω

td

max

ω eq = 0.8ω

1

ω eq = 1.2ω 0.5

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

kv ξ a = ----------2ωm Fig. 8.6: Maximum allowable time delay t d as a function of k d and k v . max 2

ξ = 5% 10%15% 20%25%

T = 5s kd = 0

1.8 1.6 1.4

td

max

1.2 1

ξ = 0%

0.8 0.6 0.4 0.2 0 0

0.1

0.2

0.3 k v 0.4 ξ a = ----------2ωm Fig. 8.7: Maximum allowable time delay t d

max

0.5

0.6

as a function of k v and ξ .

8.2 State-space Formulation-Time Invariant SDOF System 565

1.5

ξ a = 0%

ξ = 0 kd = 0

10% 20% 30% 40% 50% 60% 70%

td

max

1

0.5

0 0

1

2

3

4

5

6

T - s Fig. 8.8: Maximum allowable time delay t d as a function of k v and T . max 10 9 8

T = 5s ξ = 2% kd = 0 ξ a = 20%

u ( td ) ⁄ u ( td = 0 )

7 6 5

El Centro Taft

4 3 2 1 0 0

0.5

td - s

1

1.5

Fig. 8.9: Degradation in performance of a SDOF system with time delay.

566 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

8.3 Discrete time formulation - SDOF system Governing equation The continuous state-space linear formulation considers X and F to be continuous functions of time that satisfy the following ordinary linear differential equation and initial conditions, X˙ = AX + B f F + B g a g + B p p X ( t = 0 ) = X o*

(8.73)

In the case of a SDOF system, the coefficients are second order matrices involving the system properties,

A =

0 1 k c – ---- – ---m m

B f = Bp =

Bg =

0 1 ---m

(8.74)

0 –1

The formulation presented in the previous section was based on the assumption that m, k, and c are constant. This restriction allows one to obtain eqn (8.26), the exact analytical solution expressed in terms of a convolution integral. A numerical integration procedure is required to evaluate the convolution integral when the loading is a complex function such as a ground acceleration time history. If the system parameters and/or the feedback parameters are time dependent, an exact analytic solution of the equilibrium equation can not be established, and one must resort to generating an approximate solution with a numerical procedure such as a finite difference method which works with values of the variables at discrete points in time. A discrete time approach is also necessary for real time feedback control, since the control force is computed using “observed” values for the response at discrete time points.

8.3 Discrete Time Formulation- SDOF System 567 One generates a discrete time formulation by subdividing the time domain into intervals, say t o – t 1 , t 1 – t 2 , . . . , t n – t f , and taking as unknowns the values of X ( t ) and F ( t ) at the discrete time points. The notation, X(t j) ≡ X j

(8.75)

F(t j) ≡ F j

is convenient for representing these discrete variables. Equation (8.73) is approximated at each time point by an algebraic equation relating the values of X and F at that point and neighboring points, and is used to estimate the value of X at a later time. In what follows, the procedure is illustrated using a simple approximation for eqn (8.73). When the system parameters are constant, the solution is given by eqn (8.26) which is listed below for convenience. t

X(t) = e

A ( t – to )

X ( to ) +

∫

t

e

A(t – τ)

to

B f F ( τ ) dτ +

∫

e

A(t – τ)

( B p p + B a g ( τ ) ) dτ (a) g

to

One can use this result to obtain an approximate solution between two time points, say t j and t j + 1 = t j + ∆t , by introducing assumptions for the variation of the force terms during the interval ∆t . The simplest model is based on using the values at the initial time t j . Taking the time limits as t = t j + 1 and t o = t j , assuming the force terms are equal to their value for t = t j , F(τ) = F(t j) = F j p(τ) = p(t j) = p j a g ( τ ) = a g ( t j ) = a g, j

(8.76)

At

and noting the expansion for e transforms the convolution integrals contained in eqn (a) to the following algebraic form, Xj + 1 = e

A∆t

X j + A – 1 ( e A∆t – I ) [ B g a g, j + B f F j + B p p j ]

(8.77)

568 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Equation (8.77) provides an estimate for X j + 1 based on data associated with the time point, t j . The first term on the right hand side is the exact free vibration response at t j + 1 , considering X j to be the initial state at t j . The remaining terms represent the contribution of the “constant” loading terms over the time interval, ∆t . Starting at t = 0 which corresponds to j = 0, one specifies X o and computes X 1 . This process is repeated until the desired time is reached. The computation will be bounded when eqn (8.75) is numerically stable. Stability is discussed in more detail later in the section. This approach can also be applied to an adaptive system. The mass is considered to be constant, and therefore B f and B g are constant. Stiffness and damping are assumed to be constant over a time interval, and to vary from one interval to another. Discrete values of stiffness and damping are defined for the time interval, t j ≤ t < t j + 1, as follows: k(t) = k(t j) ≡ k j c(t) = c(t j) ≡ c j

(8.78)

A(t) = A(t j) ≡ A j Since A is considered constant over an interval, eqn (a) is still applicable for the interval. The resulting form of the discrete equilibrium equation is obtained by replacing A with A j in eqn (8.77). Xj + 1 = e

A j ∆t

X j + A –j 1 ( e

A j ∆t

– I ) [ B g a g, j + B f F j + B p p j ]

(8.79)

Starting at j = 0, one forms A o and determines X 1 . Then, A is updated to A 1 and used to compute X 2 . This process is continued for successive time points. Linear negative feedback control The discrete formulation represented by eqn (8.79) assumes properties and forces are constant over a time interval and are updated at the starting point of the interval. A feedback law consistent with this assumption is tj ≤ t < tj + 1 F j = –K f , j X j where K f , j is the feedback gain matrix at t = t j .

(8.80)

8.3 Discrete Time Formulation- SDOF System 569 K f , j = k d ( t j)

kv ( t j )

(8.81)

Substituting for F transforms eqn (8.79) to Xj + 1 = e

A j ∆t

+ A j –1 ( e

– A –j 1 ( e

A j ∆t

A j ∆t

– I )B f K f X j

(8.82)

– I ) [ B g a g, j + B p p j ]

One can also derive this expression by specializing eqn (8.38). The time history response is generated by starting with X o , and computing X 1 , X 2 , ... etc. Key issues here are the specification of: the time interval; the magnitude and temporal distribution of the system stiffness and damping; and the feedback parameters, k d and k v . Stability analysis for time invariant linear feedback control The numerical stability of the discrete feedback control algorithm is determined by examining the nature of the homogeneous solution. The analysis presented here assumes the coefficient matrices are constant, i.e., the system is time invariant. Specializing eqn (8.79) for this case, and considering no external loading other than feedback leads to the governing equation X j + 1 = e A∆t X j + A – 1 ( e A∆t – I )B f F j

(8.83)

For no time delay, F j is taken as a linear function of X j . The effect of time delay is to shift the value of X used to compute F j back to X ( t j – ν∆t ) ≡ X j – ν . For j < ν , the control force is taken as 0. Fj = 0

j<ν

F j = –K f X j – ν

j≥ν

(8.84)

Substituting for F j , the discrete equilibrium equation allowing for time delay takes the following form X j + 1 = e A∆t X j X j + 1 = e A∆t X j – A – 1 ( e A∆t – I )B f K f X j – ν

0≤j<ν ν≤j

(8.85)

570 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Considering first the case of no time delay, the free vibration response is determined with X j + 1 = ( e A∆t – A – 1 ( e A∆t – I )B f K f )X j

j = 0, 1, 2, …

(8.86)

This equation applies when the control force is updated at the discrete time points, and assumed constant during the time intervals. The corresponding form for the case where the force is assumed to be continuously updated can be obtained by specializing eqn (8.38). Xj + 1 = e

( A – B f K f )∆t

(8.87)

Xj

The solution generated with eqn (8.87) is the “exact” damped solution. According to eqn (8.39), this solution is “bounded” for negative feedback applied to an initially stable system. The error introduced by approximating the control force is represented by the difference between the coefficient matrices. It remains to determine whether this “error” causes eqn (8.86) to became numerically unstable for particular combinations of K f and ∆t , and therefore to generate homogeneous solutions which increase rather than decrease with time. In what follows, a strategy for determining the numerical stability of first order matrix difference equations such as eqns (8.86) and (8.87) is presented. The approach is introduced using a simple scalar equation and is then generalized to handle an n’th order equation. Consider the first order differential equation, (8.88)

x˙ = ax + bf

where a and b are constants and f = f(t). The exact solution is determined using eqn (8.22). x(t) = e

a ( t – to )

x ( to ) +

∫

t

e a ( t – τ ) bf ( τ ) dτ

(8.89)

to

This solution is bounded when a<0. Specializing the limits for t leads to the difference form of (8.89)

8.3 Discrete Time Formulation- SDOF System 571

xj + 1 =

e a∆t

xj +

∫

tj + 1

e

a(t j + 1 – τ)

bf ( τ ) dτ

(8.90)

tj

The homogeneous solution is “exact”, and therefore is bounded when the system is stable, i.e., when a<0. A similar solution procedure can be followed for the case where the forcing term, f, is a continuous function of x. Assuming linear negative feedback, f ( t ) = – kx ( t )

k>0

(8.91)

transforms eqn (8.88) to x˙ = ( a – bk )x

(8.92)

The exact solution has the form x j + 1 = e ( a – bk )∆t x j

(8.93)

This solution is bounded when bk>a. Normally one starts with an initially stable system, (a<0), and adds negative feedback so that “boundness” is always ensured. Equation (8.93) is based on continuous feedback. The solution corresponding to the following discrete feedback f ( τ ) = – kx j tj ≤ τ ≤ tj + 1

(8.94)

is given by x j + 1 = cx j

(8.95)

bk c = e a∆t + ----- ( 1 – e a∆t ) a

(8.96)

where

Taking j=0, 1, 2, ... leads to x j = ( c ) j xo

(8.97)

572 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems For x j to be bounded, the absolute magnitude of c must be less than 1. c <1 ⇒

(8.98)

bk e a∆t + ----- ( 1 – e a∆t ) < 1 a Equation (8.98) represents a constraint on k and ∆t . When the system is initially stable, a is negative, a = –a

(8.99)

Specializing eqn (8.98) for this case, and noting that the negative value controls, results in kb ----- ( 1 – e – a ∆t ) – e – a ∆t < 1 a

(8.100)

which is written as: kb 1 + e – a ∆t k max b ----- < ------------------------ ≡ --------------a 1 – e – a ∆t a

(8.101)

Equation (8.101) defines the maximum value of k for a particular value of ∆t . The variation of k max b ⁄ a with a ∆t is shown in Fig 8.10. For small a ∆t , the curve tends to ∞ as ∆t → 0 ; for large a ∆t , the curve is bounded by k max b ⁄ a = 1 . If kb ⁄ a is less than 1, there is no restriction on the magnitude of ∆t . When, kb ⁄ a is greater than 1, the value of ∆t has to satisfy eqn (8.101). Using Fig (8.10), one determines the limiting value of a ∆t corresponding to the specified value of kb ⁄ a . The range of allowable values for a ∆t is between zero and this “limiting” value. For example, suppose kb ⁄ a is equal to 1.6. According to Fig (8.10), the maximum allowable value of a ∆t is 1.4, and the range is 0 < a ∆t < 1.4 .

8.3 Discrete Time Formulation- SDOF System 573

10

9

8

7

6

k max b ---------------5 a 4

3

eqn ( 8.101 ) 2

1

0

0

0.5

1

1.5

2

2.5

a ∆t

3

3.5

4

4.5

5

Fig. 8.10: Stability limits for x j + 1 = cx j According to the analysis presented above, the scalar equation, x j + 1 = cx j , is numerically stable when c < 1 . The matrix equation, X j + 1 = cX j , has a similar constraint involving the matrix, c. Observing that X j can be expressed as j

X j = c Xo

(8.102)

it follows that the elements of c j must be bounded as j → ∞ . A measure for c j is established by noting that the expression, j

Xj = ρ Ψ

(8.103)

where ρ and Ψ satisfy the characteristic equation for c, ( c – ρI )Ψ = 0

(8.104)

is a solution of X j + 1 = cX j . Suppose c is of order 2. There are 2 eigensolutions which may be real or complex. When the solution is complex, the individual solutions are complex conjugates, and the total real solution is a linear combination of these j solutions,

574 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems ˜ ρ˜ j Ψ ˜ X j = Aρ j Ψ + A

(8.105)

˜ are complex constants. These constants are evaluated using the where A , A initial conditions for X o . ˜Ψ ˜ X o = AΨ + A

(8.106)

Substituting for A and Ψ , A = A R + iA I Ψ = Ψ R + iΨ I

(8.107)

results in 2 scalar equations for A R and A I . 2 ( AR ΨR – AI ΨI ) = X o

(8.108)

The complex eigenvalue can be expressed as ρ = ρ R + iρ I = ρe iθ

(8.109)

where ρ is the magnitude of ρ and θ is the polar angle. 2 + ρ 2 ]1 ⁄ 2 ρ = [ ρR I ρI tan θ = -----ρR

(8.110)

Using the polar form for ρ , the j’th power is given by ρ j = ( ρ ) j e i ( jθ )

(8.111)

Introducing the expressions for A , Ψ , and ρ , the solution for X j expands to X j = ( ρ ) j [ 2 ( A R Ψ R – A I Ψ I ) cos ( jθ ) – 2 ( A R Ψ I + A I Ψ R ) sin ( jθ ) ]

(8.112)

j = 1, 2, … Since the terms within the brackets are bounded, it follows that X j is finite as j → ∞ when the modulus of ρ is less than unity, ρ<1

(8.113)

8.3 Discrete Time Formulation- SDOF System 575 Extending this analysis to the case where c is of order 2n, there may be n pairs of complex conjugates solutions having the same form as eqn (8.105), and the total solution is given by n

Xj =

∑

j ˜ ρ˜ lΨ ˜ ( Al ρl Ψl + A l l l)

l=1 n

=

∑

(8.114) ( ρ l ) j [ 2 ( A R, l Ψ R, l – A I, l Ψ I, l ) cos ( jθ l )

l=1

– 2 ( A R, l Ψ I, l + A I, l Ψ R, l ) sin ( jθ l ) ] Defining ρ max as the maximum modulus for the set of eigenvalues of c, the stability requirement is ρ max < 1

(8.115)

This requirement also applies when some of the eigenvalues are real. For this case, the maximum absolute value must be less than unity. In general, c depends on the nature of the finite difference approximation, the system parameters, and the magnitude of the time step, ∆t . For the case of time invariant negative linear discrete feedback with no time delay, the form of c follows from eqn (8.82) specialized for an external loading: c = e A∆t – A – 1 ( e A∆t – I )B f K f

(8.116)

Given A , B f , and K f , one selects a value for ∆t , determines the corresponding eigenvalues of c, and then adjusts the value of ∆t if ρ max is greater than 1. The complexity of the equation for c necessitates the use of a numerical eigenvalue routine such as the eig( ) function in MATLAB. The following example illustrates computational details and presents the stability limit for a SDOF system.

Example 8.2: Stability analysis - SDOF system with no time delay The various matrices for the SDOF case with negative velocity feedback

576 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems are:

A =

0

1

–ω 2

– 2ξω

– ωω 0 BfKf =

0

2ξ – -----ω 1

1 – -----ω2 0

ω ---ω – 2ξω

0

A∆t =

A –1 =

0 0 kν = 0 ----m

(1)

0 2ξ a ω

where ω is a dimensionless parameter. ∆t ω = ω∆t = 2π -----T

(2)

When ξ = 0 , the expansion of e A∆t can be expressed as 1 e A∆t = cos ωI + ---- sin ω A ω

(3)

Using eqn (3), one can obtain analytical expressions for the elements of c. The resulting form of c is

c =

cos ω – ω sin ω

1 ---- ( sin ω + 2ξ a ( cos ω – 1 ) ) ω cos ω – 2ξ a sin ω

(4)

Evaluating the eigenvalues of c leads to ρ 1, 2 = b 1 ± ( b 12 – b 2 ) 1 ⁄ 2 b 1 = cos ω – ξ a sin ω

(5)

b 2 = 1 – 2ξ a sin ω Equation (5) shows that the eigenvalues depend on the dimensionless parameter,

8.3 Discrete Time Formulation- SDOF System 577 ω = 2π ( ∆t ⁄ T ) , and the active damping ratio, ξ a . Figure 1 contains plots of the absolute values of the eigenvalues vs. ∆t ⁄ T for a representative range of ξ a . The curves intersect at ∆t ⁄ T = 0.5 , which corresponds to a transition in behavior. When ∆t ⁄ T ≥ 0.5 , the eigenvalues are complex and ρ > 1 . For the range 0 < ∆t ⁄ T < 0.5 , the eigenvalues start out as complex quantities, and then shift over to real quantities at a particular value of ∆t ⁄ T which depends on ξ a . The curve bifurcates at this value. When ξ a = 0 , the bifurcation occurs at ∆t ⁄ T = 0.5 . The intersection of the upper branch with ρ = 1 defines the limiting value of ∆t ⁄ T for the corresponding value of ξ a . This value corresponds to ρ = – 1 and is related to ξ a by cos ω – 2ξ a sin ω = – 1 . Although ρ = 1 for ξ = 0 and arbitrary ∆t ⁄ T , the modulus is greater than 1 for ξ a > 0 and ∆t ⁄ T > 0.5 . Therefore, ∆t ⁄ T = 0.5 represents the “limit” for an undamped system. 2

1.8

ξ = 0 kd = 0

1.6

ξ a = 0.4 0.3 0.2 0.1 0.0

1.4

1.2

ρ

ξ a = 0.0

1

0.1 0.2 0.3 0.4

0.8

0.6

0.4

0.2

0

0

0.1

0.2

0.3

0.4

0.5

∆t ⁄ T

0.6

0.7

0.8

0.9

1

Figure 1 The effect of including damping in the system is to offset the destabilizing effect of feedback. Figure 2 illustrates this trend. Increasing ξ lowers ρ and

578 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems decreases the range of ω for which ρ is greater than 1. When ξ a = 0 , c reduces to e A∆t . The eigenvalues of e A∆t are related to the eigenvalues of A by ρ = e λ∆t

(7)

Noting that λ = – ξω ± i ( 1 – ξ 2 )ω , eqn (7) expands to 2 1⁄2 ρ = e – ξω [ e ± i ( 1 – ξ ) ω ]

(8)

and it follows that ρ = e – ξω

(9)

Equation (9) explains the trend for ρ to decrease with increasing ω . The second term in eqn (8) is the source of the transition from real to complex eigenvalues. This shift occurs when ( 1 – ξ 2 ) 1 ⁄ 2 ω = π . 2

1.8

ξ = 0.05 kd = 0

1.6

1.4

ξ a = 0.4 0.3 0.2 0.1 0.0

1.2

ρ

ξ a = 0.0

1

0.1 0.2 0.3 0.4

0.8

0.6

0.4

0.2

0

0

0.1

0.2

0.3

0.4

0.5

∆t ⁄ T

a) ξ = 0.05

0.6

0.7

0.8

0.9

1

8.3 Discrete Time Formulation- SDOF System 579

2

1.8

ξ = 0.1 kd = 0

1.6

1.4

1.2

ρ

ξ a = 0.0

1

ξ a = 0.4 0.3 0.2 0.1 0.0

0.1 0.2 0.3

0.8

0.6

0.4

0.4

0.2

0

0

0.1

0.2

0.3

0.4

0.5

∆t ⁄ T

0.6

0.7

0.8

0.9

1

b) ξ = 0.1 Figure 2 The limits on ∆t ⁄ T depend on both ξ and ξ a . As Fig 2 shows, ξ has a significant effect on ρ , particularly in the region, ∆t ⁄ T > 0.5. When ξ = 0.1 , this region is stable even for ξ a = 0.4 , which represents significant active damping. A conservative strategy would be to require ∆t ⁄ T to be less than the value for which the upper branch first intersects ρ = 1 . These values are listed in Table 1. Table 1: Stability limit for ∆t ⁄ T ξ

ξa = 0

ξ a = 0.1

ξ a = 0.2

ξ a = 0.3

ξ a = 0.4

0

0.50

0.44

0.38

0.33

0.29

0.05

∞

0.48

0.38

0.33

0.29

0.10

∞

∞

0.39

0.34

0.29

Figure 3 illustrates the onset of instability for a SDOF system having

580 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems X ( 0 ) = { 1, 0 } . The discrete time history response is obtained by evaluating eqn (8.86) for j ranging from 0 to 100. Increasing ∆t from 0.33T to 0.34T causes the free vibration response of the initially displaced system to shift from stable to unstable behavior. 1

0.8

ξ = 0.1

Displacement, u

0.6

ξ a = 0.3

∆t ------ = 0.33 T kd = 0

0.4

0.2

0

−0.2

−0.4

0

10

20

30

40

50

60

Discrete time point a)

70

80

90

100

8.3 Discrete Time Formulation- SDOF System 581

5

4

ξ = 0.1 ξ a = 0.3 ∆t ------ = 0.34 T

3

Displacement, u

2

1

0

−1

−2

−3

−4

−5

0

10

20

30

40

50

60

70

80

90

100

Discrete time point b) Figure 3

The stability analysis allowing for time delay starts with eqn (8.86) which is written here in a simpler form. X j + 1 = c1 X J + c3 X j – ν

(8.117)

c 1 = e A∆t

(8.118)

c 3 = – A – 1 ( e A∆t – I )B f K f

(8.119)

where

The previous analysis corresponds to ν = 0 and c 1 + c 3 = c . Given ν and the damping parameters, ξ and ξ a , one needs to establish the limiting value of the

582 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems ratio of the time increment to the fundamental period, ∆t ⁄ T . One option is to employ numerical simulation and solve eqn (8.117) for a range of values of ∆t ⁄ T . Figure 8.11 illustrates the stability transition for ν = 1 and a specific set of parameters. The corresponding results for ν = 0 are shown in Figure 3 of example 8.2. In this case, the maximum time step is reduced by about 50%. The advantage of numerical simulation is that it allows arbitrary values of ξ , ξ a , and ν to be considered. The disadvantage is that the search can be inefficient if the starting point is not reasonably close to the solution. One can start with the solution for ν = 0 and decrease ∆t ⁄ T . Another option is to utilize a simpler formulation based on ξ = 0 to generate an estimate of the starting value. The latter option is described below. 1

ν = 1

0.8

ξ = 0.1 ξ a = 0.3 ∆t ------ = 0.15 T

Displacement, u

0.6

0.4

0.2

0

−0.2

−0.4

−0.6

0

50

100

150

200

a)

250

300

350

400

Discrete Time Points

8.3 Discrete Time Formulation- SDOF System 583

1500

ν = 1 ξ = 0.1 ξ a = 0.3 ∆t ------ = 0.16 T

Displacement, u

1000

500

0

−500

−1000

−1500

0

50

100

150

200

250

b)

300

350

400

Discrete Time Points

Fig. 8.11: Illustration of instability due to time delay The solution of eqn (8.117) can be expressed as X J = ρ jΨ

(8.120)

where ρ is a complex scalar, and Ψ is an unknown vector. In order for the solution to be stable, the absolute magnitude of ρ must be less that 1. Substituting for X j , eqn (8.117) takes the following form ( c 3 + ρ ν c 1 – ρ ν + 1 I )Ψ = 0

(8.121)

When ξ = 0 , the coefficient matrix reduces to (see example 8.2) ρ ν cos ω – ρ ν + 1 – ρ ν ω sin ω

sin ω cos ω – 1 ρ ν ------------ + 2ξ a ---------------------ω ω ρ ν cos ω – 2ξ

a

sin ω – ρ ν + 1

(8.122)

584 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems For Ψ to have a non-trivial value, the determinant of the coefficient matrix must vanish. Enforcing this requirement leads to an equation of degree ν + 2 in ρ . p ( ρ ) = ρ ν + 2 + aρ ν + 1 + ρ ν + bρ – b = 0

(8.123)

a = – 2 cos ω b = 2ξ a sin ω

(8.124)

where

Given ν and ξ a , one evaluates the roots of p ( ρ ) =0 for a representative range of ω = 2π ( ∆t ⁄ T ) . The following example illustrates this computation.

Example 8.3: Stability analysis -SDOF system with time delay Figure 1 contains plots of ρ vs. ∆t ⁄ T for a range of ξ a and ν = 1, 2 . As expected, increasing the time delay lowers the magnitude of the time interval at which instability occurs. The effect of time delay becomes more significant as the active damping is increased. Table 1 contains the limiting values of ∆t ⁄ T for ξ = 0 and ξ a ranging up to 0.4. Both active damping and time delay result in a decrease in the allowable time step. Including passive damping has the opposite effect on the time increment, i.e., it increases the allowable time increment. For example, the critical time step ratio for ν = 1 , ξ a = 0.3 is 0.15 for ξ = 0.1 vs. 0.124 for ξ = 0.0 .

8.3 Discrete Time Formulation- SDOF System 585

1.4

ξ a = 0.5

1.3

1.2

0.4

ξ = 0 ν = 1

0.3 0.2

1.1

0.1 ρ

1

0.0

0.9

0.8

0.7

0.6 0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

a) ν = 1

∆t ⁄ T

0.2

1.4

ξ a = 0.5 0.4 0.3

1.3

1.2

ξ = 0 ν = 2

0.2 0.1

1.1

ρ

1

0.9

0.8

0.7

0.6 0.04

0.06

0.08

0.1

0.12

b) ν = 2 Figure 1

0.14

0.16

0.18

0.2

∆t ⁄ T

586 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Table 1: Limiting values of ∆t ⁄ T for ξ = 0 Stability limit for ∆t ⁄ T ξa

ν = 0

ν = 1

ν = 2

0

0.50

0.500

0.5000

0.158

0.0955

0.151

0.0908

0.144

0.0860

0.137

0.0822

0.131

0.0780

0.124

0.0745

0.118

0.07085

0.113

0.0677

0.05 0.10

0.44

0.15 0.20

0.38

0.25 0.30

0.33

0.35 0.40

0.29

The actual time increment is governed by the responses of the monitoring and actuator systems. Given data on these response times, one can evaluate the expected ratio, ∆t ⁄ T , for a particular system and determine, using Table 1 as a starting point for numerical simulation, whether stability will be a problem.

8.4 Optimal linear feedback - time invariant SDOF systems Quadratic performance index The previous sections were concerned with presenting the state-space formulation and numerical procedures for evaluating the time history response of SDOF systems having linear time invariant feedback. Negative velocity feedback was shown to be equivalent to damping, and to result in stable behavior when it is applied in a continuous manner with no time delay. The effect of time delay can be destabilizing under certain conditions that depend on the magnitude of the feedback force and the delay time. Discrete time control works with response data at discrete time points and considers the feedback force to be constant over a time interval. Instability can also occur if the time interval exceeds a critical value that depends on the level of “feedback” induced damping and the time delay. Given

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 587 the feedback parameters, k d and k v , and the delay time, one can determine the limiting time interval using the procedure described in the previous section. The remaining issue that needs to be addressed here is the selection of the magnitude of the feedback parameters. A strategy similar to the approach followed for quasistatic control is introduced in this section. The key step is the formulation of a performance measure which provides a rational basis for comparing different choices for the feedback parameters and establishing the “optimal” solution. The quasi-static formulation presented in Chapter 7 was concerned with controlling a displacement profile, and worked with a quadratic performance measure formed by integrating the square of the displacement error over the structural spatial domain. This measure was extended to reflect the cost of the control forces by incorporating an additional quadratic force term, and introducing weights for the displacement and force terms. By adjusting the weights, one can assess the effect of assigning more emphasis to either the displacement error or the force magnitude. Dynamic control is concerned with controlling the response of the system over a specified time period. Starting at some initial state, X(0), the objective is to minimize the deviation between the actual response, X(t), and the desired response, X * ( t ) , by applying a control force F(t) over the time interval. Considering the integral of the square of the difference between the actual and desired responses to be the measure of the “closeness” of the solution, and including a cost associated with the control force leads to the following quadratic performance index for a SDOF system, 1 J = --2

∫

t*

{ q d ( u – u * ) 2 + q v ( u˙ – u˙ * ) 2 + rF 2 } dt

(8.125)

0

where t * is the time period over which the performance is being measured, and q d, q v, r are the weighting functions for the displacement, velocity and force terms. Specializing F for linear time invariant feedback transforms J to: 1 J = --2

∫

t*

{ q d ( u – u * ) 2 + q v ( u˙ – u˙ * ) 2 + r ( k d u + k v u˙ ) 2 } dt

(8.126)

0

where k d and k v are constants. When the system and weighting parameters are specified, J reduces to a

588 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems function of the feedback parameters, the external excitation, the desired response, and the time period. The optimal feedback for a specific combination of excitation, desired response, and time interval, is defined as the set of values for k d and k v which correspond to a stationary value of J specialized for these conditions. They are determined with the following equations, ∂ * J = 0 ∂ kd

∂ * J = 0 ∂ kv

(8.127)

where J * is J evaluated for specific u, u * , and t * . In general the response depends on k d and k v , as well as the system parameters and external excitation. One option is to evaluate J * numerically and use simulation to locate the “optimal” values for k d and k v . This approach is not very efficient since it requires ranging over both k d and k v . Another option is to apply variational calculus methodology and transform the stationary requirement to a differential equation. In the following section, an example illustrating the first option is presented. This example deals with the special case where u * is taken as 0, the response is due to an initial velocity at t=0, and the time interval extends out to infinity. These conditions are typical for the regulator problem which is concerned with applying control forces to take a system from an arbitrary non-zero initial state X o , to a final state X f at time t f . The regulator problem is encountered when a system is perturbed from its steady-state equilibrium position and there is a need to eliminate the unwanted perturbation as quickly as possible. Since the required force increases as the time interval decreases, it is necessary to include the force cost as well as the response magnitude in the performance index in order to obtain a realistic solution. Subsequent sections describe the variational calculus formulation of the stationary requirement for J as a differential equation, and illustrate its application. This derivation is extended to deal with multi-degree of freedom systems later in the Chapter. An example - linear quadratic regulator control algorithm Suppose a single degree of freedom system is subjected to an impulse at t=0. Considering only velocity feedback, the response is given by u˙ o λ t (8.128) u = -----e R sin λ I t λI

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 589 where u˙ o is the initial velocity, and the lambda terms depend on the system properties and the velocity feedback parameter. λ R = – ξ eq ω 2 ]1 ⁄ 2 λ I = ω [ 1 – ξ eq

(8.129)

kv ξ eq = ξ + ------------ = ξ + ξ a 2ωm Interpreting u as a disturbance from the steady-state position, u * = 0 , the objective is to reduce u to a small quantity, i.e., to regulate the response, in a specified time interval. The linear quadratic regulator control algorithm specialized for only velocity feedback is based on the following form of J, 1 J = --2

∫

∞

{ q d u 2 + q v u˙ 2 + r ( k v u˙ ) 2 } dt

(8.130)

0

and determines the optimal value of k v by enforcing the stationary requirement with respect to k v , ∂J = 0 ∂ kv

(8.131)

Substituting for u in eqn (8.130), considering the weighting functions to be constant, and expressing k v in terms of ξ a leads to qd 1 8ω ------- J = ------------------- -----(8.132) + q v + 4m 2 ω 2 rξ a2 u˙ 2 ( ξ + ξ ) ω2 o

a

The form of the terms within the square bracket suggest the following scale factors for the weights: qd = ω 2 qd qv = qv r r = -----------------4m 2 ω 2

(8.133)

With these definitions, all the superscripted weights have the units of (1/ velocity)2 and the performance index simplifies to ξ a2 qd + qv 8ω ------- J = ---------------- + r -------------(8.134) u˙ 2 ξ+ξ ξ+ξ o

a

a

Finally, enforcing the stationary requirement on J with respect to ξ a results in the

590 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems following equation for the optimal active damping ratio, ξa

qd + qv 1 / 2 = ξ 2 + ----------------–ξ r optimal

(8.135)

Figure 8.12 shows the variation of ξ a with the weighting parameters. Increasing r places more emphasis on minimizing the control force, and ξ a decreases. Increasing the displacement and velocity weights places more emphasis on decreasing the response, and the result is an increase in ξ a . 1

0.9

0.8

ξ = 0

0.7

ξ = 0.1 0.6

ξa

ξ = 0.2

0.5

opt 0.4

0.3

0.2

0.1

0

0

0.1

0.2

0.3

0.4

0.5

0.6

qd + qv qd + ω 2 qv ----------------- = -----------------------r 4ω 4 m 2 r

0.7

0.8

0.9

1

Fig. 8.12: Variation of the optimal velocity feedback with the weighting parameters The previous analysis considered the weighting functions to be constant over the time period during which the performance index is being evaluated. This restriction fixes the relative priorities on the various terms that contribute to the index. If the weights are allowed to vary with time, the priorities can be adjusted

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 591 as to place more emphasis on a certain term at a particular point in time. For example, one may want to suppress the response more aggressively later in the period. This shift in emphasis can be achieved by taking the weights as t q d = ω 2 ------ q d T* t q v = ------ q v T* r r = -----------------4m 2 ω 2

(8.136)

where T * is a time constant that defines the transition for the priority shift. It is convenient to express T * in terms of the fundamental period for the system and a scaling factor, α . 2πα T * = αT = ---------ω The performance index for this case has the following form, ξ a2 qd + qv 8ω 1 ------ J = ---------- 2q d + ---------------------- + r ------------- u˙ 2 4πα ( ξ + ξa ) 2 ξ + ξa o

(8.137)

(8.138)

Requiring J to be stationary with respect to ξ a results in a cubic polynomial for the optimal value of ξ a . 1 qd + qv ξ a ( ξ a + ξ ) ( ξ a + 2ξ ) – ---------- ----------------- = 0 2πα r

(8.139)

Figure 8.13 contains plots corresponding to ξ = 0 and various values of α . Also shown is the plot for the uniform (time) weighting case defined by eqn (8.135). Assuming q d , q v , and r are fixed, the effect of increasing α is to reduce the optimal value of ξ a . This trend is due to the decrease in magnitude of the weighted response during the initial phase of the response where the actual response magnitude is a maximum. Lowering this term decreases the demand on the control force, and consequently the required value of ξ a decreases.

592 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.18

1 α = --4

0.16

0.14

1 α = --2

eqn ( 8.135 )

0.12

ξa

α = 2

α = 1

ξ = 0

0.1

opt 0.08

0.06

0.04

0.02

0

0

0.005

0.01

0.015

0.02

0.025

qd + qv ----------------r a)

0.03

0.035

0.04

0.35

1 α = --4

0.3

1 α = --2

0.25

ξa

eqn ( 8.135 )

α = 1

0.2

α = 2

opt

ξ = 0

0.15

0.1

0.05

0

0

0.02

0.04

0.06

0.08

0.1

qd + qv ----------------r b)

0.12

0.14

0.16

0.18

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 593

0.5

1 α = --4

0.45

1 α = --2 α = 1

0.4

0.35

α = 2

ξ = 0

0.3

ξa

0.25

opt 0.2

eqn ( 8.135 ) 0.15

0.1

0.05

0

0

0.2

0.4

0.6

qd + qv ----------------r

0.8

1

1.2

1.4

c) Fig. 8.13: Variation of the optimal velocity feedback with the time scale factor, α The continuous-time algebraic Riccati equation The previous example illustrated the application of the linear quadratic regulator algorithm for a particular initial state and velocity feedback. In what follows, the performance index is expressed in matrix form, and the analysis is extended to be applicable for both displacement and velocity feedback, and arbitrary initial conditions. The scalar form of the performance index for the linear quadratic regular problem is obtained by setting u * = u˙ * = 0 in eqn (8.126) and taking t * = ∞ . 1 J = --2

∞

∫

0

[ q d u 2 + q v u˙ 2 + r ( k d u + k v u˙ ) 2 ] dt

(8.140)

594 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Introducing the state vector notation, J is expressed as 1 J = --2

∞

∫

X T ( Q + K Tf RK f )X T dt

(8.141)

0

where Q and R are diagonal weighting matrices, and K f is the linear state feedback matrix.

Q =

qd qv

(8.142)

R = r[1] K f = kd kv

The governing differential equation for free vibration response follows from eqn (8.33). X˙ = ( A – B f K f )X = A c X

(8.143)

Solutions of eqn (8.143) satisfy X ( t ) → 0 as t → ∞

(8.144)

when the feedback parameters ( k v, k d ) are selected such that the eigenvalues of A c have a negative real part. When the weighting matrices are taken to be constant, an explicit expression for J can be obtained by expressing the integrand as X T ( Q + K Tf RK f )X = –

d ( X T HX ) dt

(8.145)

where H is a symmetric positive-definite second order matrix. Noting eqn (8.143), the right hand side expands to d ( X T HX ) = X˙ T HX + X T H X˙ = X T ( A cT H + H A c )X dt

(8.146)

and it follows that A cT H + H A c = – ( Q + K Tf RK f ) With this change in variables, the integral reduces to

(8.147)

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 595 1 J = --- X T ( 0 )HX ( 0 ) 2

(8.148)

Requiring J to be stationary with respect to K f and noting that X ( 0 ) is independent of K f leads to δH = 0 . The differential of H is generated by operating on eqn (8.147). A cT δH + δH A c + δA cT H + HδA c = – δK Tf RK f – K Tf RδK f

(8.149)

Substituting for δA c , δA c = – B f δK f

(8.150)

and setting δH = 0 , the stationary condition takes the form δK Tf ( B Tf H – RK f ) + ( H B f – K Tf R )δK f = 0

(8.151)

For eqn (8.151) to be satisfied for arbitrary δK f , the coefficients must vanish. Then the optimal feedback is related to H by R – 1 B Tf H = K f optimal

(8.152)

The last step involves substituting for K f in eqn (8.147). This operation leads to the following definition equation for H, which is called the continuous time algebraic Riccati equation A T H + HA – H B f R – 1 B Tf H = – Q

(8.153)

Given the system parameters ( A, B f ) , one specifies the weights (Q and R), and solves for the elements of H. Note that by definition, H is symmetric and positive definite. With H known, the feedback matrix is determined with eqn (8.152). Substituting for A and B f , A =

0

1

–ω 2

– 2ξω

0 Bf = 1 ---m and expressing H as

(8.154)

596 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

H =

H 11 H 12 H 12 H 22

(8.155)

in eqn (8.153) yields the following 3 equations for the elements of H 1 2 --------- H + 2ω 2 H 12 – q d = 0 rm 2 12 1 --------- H H – H 11 + 2ξωH 12 + ω 2 H 22 = 0 rm 2 12 22

(8.156)

1 2 --------- H + 4ξωH 22 – 2H 12 – q v = 0 rm 2 22 The elements of H are also constrained by the requirement that H must be positive definite. H 11 > 0

H 22 > 0

2 >0 H 11 H 22 – H 12

(8.157)

Given H, the elements of K f are determined with eqn (8.152). H 12 k d = ---------rm H 22 k v = ---------rm

(8.158)

Solving for H 12 and H 22 , and enforcing the positive definite requirement leads to 1/2 qd H 12 = rω 2 m 2 – 1 + 1 + ---------------- rω 4 m 2

qv H 12 1 / 2 2 2 H 22 = 2rωm – ξ + ξ + -------------------- + -------------------- 4rω 2 m 2 2rω 2 m 2

(8.159)

The equivalent frequency and damping for the SDOF system with feedback corresponding to eqn (8.158) are determined with eqn (8.37) and (8.158).

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 597 1/4 kd 1 / 2 qd ω eq = ω 1 + ----------⇒ ω 1 + ----------------ω2m rω 4 m 2

(8.160) 1 / 2 1 / 2 kv qd qv ω 2 ω 1 ξ eq = ξ -------- + ----------------- ⇒ -------- ξ + -------------------- + --- – 1 + 1 + ---------------- ω eq ω eq 2mω eq 4rω 2 m 2 2 rω 4 m 2

Weighting the displacement introduces additional stiffness and increases the frequency. Weighting the velocity generates additional damping, but has no effect on the frequency. It follows that pure velocity feedback corresponds to q d = 0, q v ≠ 0 . The active damping ratio for this case is given by: H 22 q v 1 / 2 2 ξ a = ----------------- = – ξ + ξ + -------------------- 2rm 2 ω 4rω 2 m 2

(8.161)

This result coincides with eqn (8.135) which was generated by integrating the analytic solution for the impulse generated response. These solutions must coincide since the approach followed earlier is a simplified version of this procedure. The matrix formulation presented above is also applicable to multi-degree of freedom systems. MATLAB has a function called CARE (Continuous -time Algebraic Riccati Equation) which generates H, K f , and the eigenvalues of A – B f K f for specified values of A, B f , Q and R. One option for scaling Q and R is to use the factors defined by eqn (8.133). The corresponding forms are

Q =

qd qv

=

ω 2 qd 0 0

qv

(8.162)

r R = [ r ] = -----------------4m 2 ω 2 where q d , q v , and r range from 0 to ≈ 1. MATLAB’s default option is to assume R ≡ I . In this case, taking Q as:

598 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Q

= R≡I

4ω 4 m 2 q d

0

0

4m 2 ω 2 q v

(8.163)

maintains the same range ( 0 → 1 ) for q d and q v . The form of the continuous algebraic Riccati solution corresponding to r = 1 and q d , q v defined by eqn (8.163) is kd ----- = – 1 + [ 1 + 4q d ] 1 / 2 k kv kd 1 / 2 ------------ = ξ a = – ξ + ξ 2 + q v + -----2ωm 2k

(8.164)

A similar scaling strategy can be employed for a multi-degree-of-freedom system. In this case the magnitude of the weighting factor for each of the displacement and velocity variables can be independently assigned. The discrete time algebraic Riccati equation The discrete time formulation of the algebraic Riccati equation is based on a performance index that involves a summation of weighted response terms evaluated at discrete times. The discrete form of the performance index corresponding to the same assumptions introduced for the continuous time formulation is ∞

1 J = --2

∑

T XT j ( Q + K f RK f )X j

(8.165)

j=0

where X j is determined with eqn (8.86) which is written here as X j + 1 = c1 X j + c2 F j c 1 = e A∆t

c 2 = A – 1 ( e A∆t – I )B f

(8.166)

Taking F j = – K f X j , eqn (8.166) reduces to X j + 1 = ( c 1 – c 2 K f )X j = cX j

(8.167)

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 599 Finally, eqn (8.165) is expressed as ∞

1 T 1 T j j T ( c ) ( Q + K f RK f )c X o = --- X oT H X o J = --- X o 2 2 j = 0

∑

(8.168)

where H is a symmetrical positive definite matrix. Stability requires that c j approach 0 as j → ∞ . Noting this constraint, the expression for H reduces to (see problem 8.11) H – c T Hc = Q + K Tf RK f

(8.169)

Imposing the stationary requirement on J leads to 1 δJ = --- X oT δH X o = 0 for arbitrary δK f 2 ⇒

(8.170)

δH = 0 Operating on eqns (8.169) and (8.166), δH – c T δHc – δc T Hc – c T Hδc = δK Tf RK f + K Tf RδK f

(8.171)

δc = – c 2 δK f

(8.172)

and setting δH = 0 results in the following expression for the optimal feedback matrix Kf

optimal

= ( R + c 2T H c 2 ) – 1 c 2T H c 1

(8.173)

The final form of the discrete algebraic Riccati equation is H – c 1T H c 1 + ( c 1T H c 2 ) ( R + c 2T H c 2 ) – 1 ( c 1T H c 2 ) T = Q

(8.174)

Given m, k, c, and ∆t, c 1 and c 2 can be evaluated. One specifies the weighting matrices R and Q and then determines H. MATLAB has a function called DARE which computes H, K f , and the eigenvalues of c specialized for opt K f set equal to the optimal value. The scaling strategies discussed earlier for the continuous time case are also applicable for the discrete time formulation. DARE assumes the default value of R to be I and therefore eqn (8.163) can be used to determine the corresponding value of Q. The following example describes the

600 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems solution procedure in considerable detail.

Example 8.4: Solution of the discrete time algebraic Riccati equation for a SDOF system The undamped case is considered first. Noting the results presented in Example 8.2, the various coefficient matrices are:

c1 =

cos ω – ω sin ω

sin ω -----------ω cos ω

(1)

1 c 2 = ----------- 1 – cos ω mω 2 ω sin ω

(2)

1 f c 1T H c 2 = ----------- 1 mω 2 f 2

(3)

1 c 2T H c 2 = -------------- f 3 2 m ω4

(4)

∆t ω = ω∆t = 2π -----T

(5)

where

f 1 = H 11 ( cos ω ) ( 1 – cos ω ) + H 12 ( ω sin ω ) ( 2 cos ω – 1 ) – H 22 ( ω sin ω ) 2 sin ω ( 1 – cos ω ) f 2 = H 11 ---------------------------------------- + H 12 [ 2 ( sin ω ) 2 – 1 + cos ω ] ω + H 22 ( ω cos ω sin ω)

(6)

(7)

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 601 f 3 = H 11 [ 1 – cos ω ] 2 + H 12 [ 2ω sin ω ( 1 – cos ω ) ] + H 22 [ ω sin ω ] 2

(8)

The optimal feedback matrix is determined using eqn (8.173). K f = kd

1 k v = ----------------------------------------- f 1 f3 mω 2 r + -------------- m 2 ω 4

f2

(9)

Finally, the elements of H are obtained by expanding eqn (8.174). This step leads to the following 3 scalar equations: ( sin ω ) 2 H 11 + [ 2ω sin ω cos ω ]H 12 – [ ω 2 sin2 ω ]H 22 f 12 = q d – -----------------------------f 3 + ω4m2r sin ω cos ω – -------------------------- H 11 + [ 2 sin2 ω ]H 12 + [ ω sin ω cos ω ]H 22 ω –f1f2 = -----------------------------f 3 + ω4m2r

(10)

2 sin ω cos ω sin2 ω – --------------- H 11 – ------------------------------ H 12 + [ – cos2 ω ]H 22 ω ω2 f 22 = q v – -----------------------------f + ω4m2r 3

The above equations are reduced to a dimensionless form by scaling the various terms according to the following laws

602 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems f 1 = ω4m2r f 1

q d = ω 4 m 2 rq d

f 2 = ω3m2r f 2

q v = ω 2 m 2 rq v

f 3 = ω4m2r f 3

(11)

H 11 = ω 4 m 2 rH 11 H 12 = ω 3 m 2 rH 12 H 22 = ω 2 m 2 rH 22

Using eqn (11), the expression for the optimal feedback matrix takes the form ω2m K f = --------------- f 1 1 + f3

1 ---- f 2 ω

(12)

Equations (10) reduce to 2

) sin2 ω +

f1 2 sin ω cos ω H 12 = 4q d – --------------1 + f3

) sin2 ω +

f2 2 sin ω cos ω H 12 = – 4 q v + --------------1 + f3

( H 11 – H 22

2

( H 11 – H 22

(13)

f1f2 ( H 11 – H 22 ) sin ( ω cos ω – 2 sin2 ωH 12 ) = --------------1 + f3 Lastly, the f terms are determined with f 1 = H 11 cos ω ( 1 – cos ω ) + H 12 sin ω ( 2 cos ω – 1 ) – H 22 sin2 ω f 2 = H 11 sin ω ( 1 – cos ω ) + H 12 [ 2 sin2 ω – 1 + cos ω ] + H 22 sin ω cos ω

(14)

f 3 = H 11 ( 1 – cos ω ) 2 + H 12 ( 2 sin ω ( 1 – cos ω ) ) + H 22 sin2 ω One specifies the relative weights ( q d, q v ) , and the relative time step ratio, ω = 2π ( ∆t ⁄ T ) , solves eqn (13) for the elements of H , computes the optimal feedback with eqn (12), and lastly checks for stability by evaluating the

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 603 eigenvalues of c 1 – c 2 K f . As ∆t → 0, the solution for the discrete formulation approaches the solution for the continuous case. This can be shown by introducing second order approximations for cos ω , sin ω and letting ω approach 0. Assuming ω 2 << 1 , the trigonometric terms are replaced with cos ω ≈ 1 sin ω ≈ ω

(15)

The corresponding equations are f 1 ≈ ωH 12 + O ( ω 2 ) f 2 ≈ ωH 22 + O ( ω 2 )

(16)

f 3 ≈ ω 2 H 22 + O ( ω 3 ) and 2

ω 2 ( H 11 – H 22 ) + 2ωH 12 ≈ 4q d – ω 2 H 12 2

ω 2 ( H 11 – H 22 ) + 2ωH 12 ≈ – 4q v + ω 2 H 22

(17)

ω ( H 11 – H 22 ) – 2ω 2 H 12 ≈ ω 2 H 11 H 22 Based on eqn (17), the limiting solution is 2ωH 12 ≈ 4q d 2

2 ω 2 H 22 ≈ 4q v + 4q d – ω 2 H 12

(18)

≈ 4q v + 4q d ( 1 – q d ) Noting eqn (16), the limiting forms for f 1, f 2 are f 1 ≈ 2q d f 2 ≈ 2 { qv + qd ( 1 – qd ) } 1 / 2 Finally, the feedback terms are

(19)

604 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

K f = 2ω 2 m q d

2ωm { q v + q d ( 1 – q ) } 1 / 2 d

(20)

Setting q d = 0 produces pure velocity feedback with v

= 2ωm ( q v ) 1 / 2

(21)

The corresponding damping ratio is 1/2 qv kv ξ a = ------------ = ( q v ) 1 / 2 = -------------------- 4ω 2 m 2 r 2ωm

(22)

This result coincides with eqn (8.161). In the discrete case, setting q d = 0 does not produce pure velocity feedback for finite ω . However, when ω is small, f 1 is small and k d is usually neglected. To determine the behavior of the feedback parameters, DARE is used to solve the discrete algebraic Riccati equation for a set of damped SDOF systems, taking a range of values for q v and ∆t ⁄ T , with q d set to 0, q v scaled according to eqn (11), and r set to 1. The results for k d ⁄ k are plotted in Fig 1. Figure 2 shows the variation of k v ⁄ ( 2ωm ) .

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 605

0

ξ = 0.02 qd = 0

−0.05

q v = 4m 2 ω 2 q v

−0.1

kd ⁄ k

−0.15

q v = 0.01 −0.2

q v = 0.03 −0.25

q v = 0.05 q v = 0.07

−0.3

−0.35

0

0.1

0.2

0

0.3

∆t ⁄ T a)

0.4

0.5

0.7

q v = 4m 2 ω 2 q v

ξ = 0.05 qd = 0

−0.05

0.6

−0.1

q v = 0.01

kd ⁄ k

−0.15

q v = 0.03 −0.2

q v = 0.05 q v = 0.07

−0.25

−0.3

−0.35

0

0.1

0.2

0.3

∆t ⁄ T b)

0.4

0.5

0.6

0.7

606 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0

q v = 4m 2 ω 2 q v

ξ = 0.10 qd = 0

−0.05

q v = 0.01 q v = 0.03

kd ⁄ k

−0.1

q v = 0.05 −0.15

q v = 0.07

−0.2

−0.25

0

0.1

0.2

0.3

∆t ⁄ T c)

0.4

0.5

0.6

Figure 1: Displacement feedback predicted by the discrete algebraic Riccati equation for a SDOF system

0.7

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 607

k v ⁄ ( 2mω )

0.3

0.25

q v = 0.07

0.2

q v = 0.05

0.15

q v = 0.03

0.1

q v = 0.01

ξ = 0.02 qd = 0

q v = 4m 2 ω 2 q v

0.05

0

−0.05

−0.1

−0.15

−0.2

0

0.1

0.2

0.25

k v ⁄ ( 2mω )

q v = 0.05

0.15

q v = 0.03

0.1

q v = 0.01

∆t ⁄ T a)

0.4

0.5

0.6

0.7

q v = 4m 2 ω 2 q v

ξ = 0.05 qd = 0

q v = 0.07 0.2

0.3

0.05

0

−0.05

−0.1

−0.15

0

0.1

0.2

0.3

∆t ⁄ T b)

0.4

0.5

0.6

0.7

608 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.25

0.2

q v = 0.07 q v = 0.05

0.15

k v ⁄ ( 2mω )

q v = 4m 2 ω 2 q v

ξ = 0.10 qd = 0

q v = 0.03

0.1

q v = 0.01

0.05

0

−0.05

−0.1

0

0.1

0.2

0.3

∆t ⁄ T c)

0.4

0.5

0.6

0.7

Figure 2: Velocity feedback predicted by the discrete algebraic Riccati equation In general, the feedback parameters decrease with increasing damping in the “uncontrolled” system. As expected, increasing q v with r held constant places more emphasis on lowering the magnitude of the response and consequently the magnitudes of the feedback terms increase. As ∆t ⁄ T approaches 0, the solution tends toward the “continuous” solution given by eqn (8.160). As ∆t ⁄ T increases, k v decreases, and eventually becomes negative in the region of ∆t ⁄ T ≈ 0.35 . The feedback stiffness term is also negative. All these solutions are stable, i.e., the maximum modulus of the eigenvalues of c 1 – c 2 K f is less than unity. The pattern of behavior near ∆t ⁄ T = 0.5 was also observed in the stability analysis for the discrete state-space formulation presented in Section 8.3 It should be noted that the typical value for ∆t ⁄ T is small, on the order of 0.02. For this range, it is reasonable to take ( k d ⁄ k ) ≈ 0 . For seismic excitation, the minimum value of ∆t is controlled by the ground motion data. The typical time interval for acceleration

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 609 time history records is 0.02 sec.

Finite interval discrete time algebraic Riccati equation The algebraic Riccati equations presented in the previous sections are based on performance indices which involve integration over an infinite time interval. By considering the time interval to be finite, one can generate a spectrum of Riccati type equations corresponding to specific values of the time interval. The minimum value is equal to the time step used to generate the discrete time solution; the maximum value is equal to infinity. Eqn (8.174) corresponds to the latter choice. An equation corresponding to the first choice is derived below. The discrete quadratic cost function associated with the time interval between t j and t j + 1 is a convenient choice. 1 J j, j + 1 = --- [ X T QX + FT j R jF j] 2 j+1 j j+1

(8.175)

It allows for updating the system properties and weighting parameters at each discrete time and consequently is applicable for adaptive as well as time invariant systems. The equilibrium equation follows from eqn (8.166). X j + 1 = c 1, j X j + c 2, j F j

(8.176)

Requiring J j, j + 1 to be stationary with respect to F j , T δJ j, j + 1 = δX T j + 1 Q j X j + 1 + δF j RF j = 0

(8.177)

and noting that δX j + 1 = c 2, j δF j

(8.178)

results in R j F j + c 2T, j Q j X j + 1 = 0

(8.179)

Finally, substituting for X j + 1 yields the following linear feedback law for F j F j = – ( R j + c 2T, j Q j c 2, j ) – 1 ( c 2T, j Q j c 1, j )X j = – K f , j X j

(8.180)

This expression is similar to eqn (8.173) which is based on an infinite time

610 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems interval. Replacing H with Q reduces eqn (8.173) to eqn (8.180). Working with the latter equations is more convenient since it avoids solving the nonlinear Riccati equation. The following example illustrates the expanded form of K f for the undamped case.

Example 8.5: Example 8.4 revisited The various matrices derived in ex 8.4 are applicable for this case when H is replaced by Q. The resulting forms are (the subscript j is dropped here to simplify the notation): f1 1 K f = ----------------------- ----------f 3 mω 2 r + -------------m2ω4

f2 ----------mω 2

(1)

q d = ω 4 m 2 rq d

(2)

q v = ω 2 m 2 rq v

(3)

f1 ----------- = rmω 2 ( q d cos ω ( 1 – cos ω ) – q v sin2 ω ) mω 2

(4)

f2 ----------- = rmω ( q d sin ω ( 1 – cos ω ) + q v sin ω cos ω) mω 2

(5)

f3 r + -------------- = r ( 1 + q d ( 1 – cos ω ) 2 + q v sin2 ω ) m2ω4

(6)

Expressing K f as K f = kd kv

(7)

and using the above equations leads to the dimensionless form of the feedback parameters

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 611 q d cos ω ( 1 – cos ω ) – q v sin2 ω kd ----------- = -------------------------------------------------------------------------mω 2 1 + q d ( 1 – cos ω ) 2 + q v sin2 ω

(8)

q d sin ω ( 1 – cos ω ) + q v sin ω cos ω kv ------------ = ------------------------------------------------------------------------------------2mω 2 ( 1 + q d ( 1 – cos ω ) 2 + q v sin2 ω )

(9)

These equations show that k d depends mainly on q d , and k v on q v. The limiting forms for q d = 0 are: – sin2 ω q v kd ----------- = -----------------------------mω 2 1 + sin2 ω q v

(10)

sin ω cos ω q v kv ------------ = -------------------------------------2mω 2 ( 1 + sin2 ω q v)

(11)

Suppose ∆t ⁄ T = 0.02 . The corresponding values of the feedback parameters are: – 0.015775 q v kd ----------- = ------------------------------------1 + 0.015775 q v mω 2

(12)

0.1243q v kv ------------ = ---------------------------------2mω 2 + 0.03155q v

(13)

Taking q v = 1 results in an active damping ratio of ≈ 0.06 . The change in stiffness is on the order of 1.5%, and can be neglected.

Continuous time Riccati differential equation The algebraic Riccati equation is based on eqn (8.141), which involves an integral extending from t = 0 to t = ∞ . A more general formulation considers the time interval to be finite, and allows for a non-zero final state. Suppose the initial state at t = t o is X o . Let t f denote the final time, and X *f the desired final state. The generalized form of the performance index is written as

612 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

∫

tf

1 1 J = --( X T QX + F T RF ) dt + --- ( X f – X *f ) T S ( X f – X *f ) 2 t 2 o

(8.181)

where S is a symmetric positive semi-definite matrix which provides a measure of the error associated with the boundary conditions at t f . Given some initial state the optimal control force F(t) is determined by requiring J to be stationary. Although the solution procedure for t f ≠ ∞ is more complicated that for t f = ∞ , the optimal feedback law has the same form, F ( t ) = – ( R – 1 B f H )X

(8.182)

However, H must now satisfy a differential equation rather than an algebraic equation, H˙ + A T H + H A T – H B f R – 1 B Tf H = – Q

(8.183)

and boundary conditions at t = t f . H (t f ) = S

(8.184)

These equations are derived in the following section. In addition to eqn (8.183), the state vector and control force are constrained by X˙ = AX + B f F X ( to ) = X o

to ≤ t ≤ t f

(8.185)

Since the boundary condition on H is at t = t f , one has to solve for H backward in time, i.e., from t f toward t o . Given H, one solves for X and F by moving forward in time from t o . The nature of the solution near t f is similar to an exponential decay to a steady state solution. Since solving eqn (8.183) is not convenient, H(t) is approximated by the steady state solution obtained with the algebraic Riccati equation. This simplification is possible only when A, Q, and R are constant. Variational formulation of the continuous time Riccati equation For completeness, the derivation of the Riccati equation is presented here. The proof is based on the requirement that A , B f , Q , and R are bounded and piecewise-continuous in the interval t o ≤ t ≤ t f . Assuming an optimal control, F ,

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 613 exists, then an optimal trajectory, X , which satisfies ˙ X = AX + B f F

(8.186)

X ( to ) = X o also exists. The corresponding optimal cost functional is given by t

f T T T 1 J ( F ) = --- [ X f – X ∗ ] S [ X f – X ∗ ] + [ X QX + F RF ]dt 2 to

∫

(8.187)

The proof proceeds as follows. Let F = F + δF = F + εF˜

(8.188)

where ε is an arbitrary positive or negative scalar and F˜ is an arbitrary valued time function representing a perturbation of the control force from the optimal value. Substituting for F , eqn (8.185) expands to X˙ = AX + B f F + εB f F˜ X ( to ) = X o

(8.189)

The solution for X can be expressed in terms of the optimal trajectory and a perturbation due to the variation in the control force, ˜ X = X + δX = X + εX

(8.190)

˜ is related to the control perturbation F˜ by where the state perturbation X ˜˙ = AX ˜ + B F˜ X f ˜ (t ) = O X o

(8.191)

The constraint equation can be incorporated in the cost function by introducing a Lagrangian multiplier, β ( t ) , and treating β as a variable as well as X and F (Strang, 1993). The modified cost function is

614 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems T 1 J ( F ) = --- ( [ X – X f ] S [ X – X f ] ) 2 tf

+

∫

1--- X T QX + 1--- F T RF + β T [ AX + B F – X˙ ] dt f 2 2

(8.192)

to

Noting that tf

tf

∫

T T T β X˙ dt = β ( t f )X ( t f ) – β ( t o )X ( t o ) –

to

∫

T β˙ Xdt

(8.193)

to

one can write eqn (8.192) as T T T 1 J ( F ) = --- [ X ( t f ) – X ∗ ] S [ X ( t f ) – X ∗ ] – β ( t f )X ( t f ) + β ( t o )X ( t o ) 2 tf

+

∫

T T T 1 T 1 T --- X QX + β AX + β˙ X + --- F RF + β B f F dt 2 2

(8.194)

to

The cost for optimal control is T T T 1 J ( F ) = --- [ X ( t f ) – X ∗ ] S [ X ( t f ) – X ∗ ] – β ( t f )X ( t f ) + β ( t o )X ( t o ) 2 tf

+

∫

T T T 1 T 1 T --- X QX + β AX + β˙ X + --- F RF + β B f F dt 2 2

(8.195)

to

Since by definition, F and X are optimal, the cost difference must be positive, J(F ) – J(F ) ≥ 0

(8.196)

for all F , t o ≤ t ≤ t f . Substituting for F and X in terms of the perturbations, one obtains an expansion for the cost difference in terms of ε .

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 615 2

T T ε ˜T ˜ ( t ) + ---˜ (t ) - X ( t f )SX J ( F ) – J ( F ) = ε ( [ X ( t f ) – X ∗ ] S – β ( t f ) )X f f 2 tf

∫

T ˜ T T T T + ε ( [ X Q + β A + β˙ ]X + [ F R + β B f ]F˜ ) dt to

(8.197)

tf

∫

T ε2 ˜ T QX ˜ ]dt + ----- [ F˜ RF˜ + X 2 to

A necessary condition for eqn (8.196) to be satisfied is that the first order terms in ˜ , and F˜ . This requirement leads to a first order differential ε vanish for all ε , X equation for β T T T X Q + β A + β˙ = O T

(8.198)

T

[ X ( t f ) – X∗ ] S – β ( t f ) = O and the control law T

T

F R + β Bf = O ⇒

(8.199) –1

F = – R B Tf β The sufficient condition for eqn (8.196) is related to the second order terms. tf 2 ε ˜T ˜ ( t ) + [ F˜ T RF˜ + X ˜ T QX ˜ ]dt J ( F ) – J ( F ) = ----- X ( t f )SX f 2 to

∫

(8.200)

Since S and Q are positive semi-definite and R is positive definite, this term is always positive, and it follows that X and F represent the solution that corresponds to the minimum value of J(F). Equations (8.186) and (8.198) can be combined into a set of coupled linear vector differential equations involving X and β .

616 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

˙ X = β˙

–1 T Bf X T β

–B f R

A –Q

–A

(8.201)

The solution can be expressed as X ( t2 ) β ( t2 )

= Φ ( t 2, t 1 )

X ( t1 ) β ( t1 )

(8.202)

where t 1 and t 2 are arbitrary times, and Φ is a 2n x 2n state transition matrix of the form Φ 11 ( t 2, t 1 ) Φ 12 ( t 2, t 1 ) Φ ( t 2, t 1 ) = (8.203) Φ 21 ( t 2, t 1 ) Φ 22 ( t 2, t 1 ) Taking t 1 = t and t 2 = t f one obtains X ( t f ) = Φ 11 ( t f , t )X ( t ) + Φ 12 ( t f , t )β ( t ) β ( t f ) = Φ 21 ( t f , t )X ( t ) + Φ 22 ( t f , t )β ( t )

(8.204)

Using the boundary condition, β( t f ) = S[ X ( t f ) – X∗ ]

(8.205)

one can solve for β ( t ) in terms of X ( t ) [ SΦ 12 ( t f , t ) – Φ 22 ( t f , t ) ]β ( t ) = [ Φ 21 ( t f , t ) – SΦ 11 ( t f , t ) ]X ( t ) + SX * (8.206) The above relation is expressed as β ( t ) = H ( t )X ( t ) + G ( t )

(8.207)

Finally, substituting for β ( t ) in eqn (8.199) results in a linear state feedback control law –1 T

F ( t ) = –R B f [ H X + G ]

(8.208)

The equation for H is derived by operating on the governing equations. Differentiating eqn (8.207), one obtains

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 617 β˙ = H X˙ + H˙ X + G˙

(8.209)

Substituting for X˙ , β˙ expands to –1 T –1 T β˙ = [ H˙ + HA – H B f R B f H ]X + G˙ – H B f R B f G

(8.210)

Finally, introducing β and β˙ in eqn (8.198), –1 T

T

[ H˙ + HA + A H – H B f R B f H + Q ]X T

+ [A –

–1 T H B f R B f ]G

(8.211)

+ G˙ = O

and setting the coefficient of X to zero results in equations for H and G T –1 T H˙ + HA + A H – H B f R B f H + Q = O

(8.212)

T –1 T G˙ + [ A – H B f R B f ]G = O

(8.213)

The boundary conditions follow from the condition on β at t f and have the form H (t f ) = S

(8.214)

G ( t f ) = – SX ∗

For the regulator problem, X ∗ is taken as zero. Then, G = O and β = HX . As a final point, one can show that the minimum remaining cost at time t is equal to 1 T J ( F ) = --- [ X ( t )HX ( t ) ] 2

(8.215)

Example 8.6: Application to scalar case To illustrate the solution procedure, the scalar form of the LQR formulation is considered here. The equations for t o < t < t f are : X˙ – AX – B f F = O

(1)

618 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems β˙ + Aβ + QX = O

(2)

1 F = – ---- B f β R

(3)

At t = t o X ( to ) = X o

(4)

At t = t f β ( t f ) = S [ X ( t f ) – X∗ ]

(5)

Substituting for F transforms eqn (1) to 2

Bf X˙ – AX + ------ β = 0 R

(6)

For stability, A < 0 . The homogeneous solution has the form X = Ce β = De

λt

(7)

λt

(8)

where λ satisfies 2

QB f λ = A + ----------R 2

2

(9)

and the constants are related by λ+A C = – -------------- D Q

(10)

λ 1, 2 = ± λ

(11)

Then

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 619

2

λ =

QB f A + ----------R 2

(12)

A + λi C i = – --------------- D i = ν i D i Q

(13)

and the full solution is β = D1 e

λt

+ D2 e

X = ν1 D e 1

λt

– λt

(14)

+ ν2 D e

– λt

(15)

2

Imposing the end conditions results in two equations for D 1 and D 2 . ν1 D1 + ν2 D2 = X o

(16)

[ 1 – Sν 1 ]D 1 + [ 1 – Sν 2 ]D 2 e

– 2λt f

= –e

– λt f

SX∗

(17)

The case of finite t f is handled by expressing D 1 as D1 = D1 e

– λt f

(18)

This substitution transforms the solution to β = D1 e

–λ ( tf – t )

X = ν1 D1 e

+ D2 e

–λ ( tf – t )

– λt

(19)

+ ν2 D e 2

– λt

(20)

The first term applies near t f and the second term near t o . Enforcing the boundary conditions leads to – λt f 1 D 2 = ----- X o – ν 1 D 1 e ν2

(21)

620 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Xo – λt f – SX∗ – ------- [ 1 – Sν 2 ]e ν2 D 1 = -------------------------------------------------------------------------------ν1 – 2λt f [ 1 – ν 1 S ] – ----- [ 1 – ν 2 S ]e ν2

(22)

When λt f > 3 , the solutions uncouple as follows: 3 For 0 < t ≤ --λ β = D2 e X = ν2 β

– λt

Bf Bf F = – ----- β = – ---------- X R ν2 R

(23)

Xo D 2 ≈ ------ν2 2

QB f A–λ A A ν 2 = – ------------- = – ---- + ------- 1 + ----------2 Q Q Q A R 3 3 For --- < t < t f – --λ λ β = 0 X = 0 F = 0 3 For t f – --- < t ≤ t f λ

(24)

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 621 –λ ( tf – t )

β = D1 e X = ν1 β

Bf Bf F = – ----- β = – ---------- X R ν1 R

(25)

– SX∗ D 1 = -----------------1 – ν1 S 2

QB f A+λ A A ν 1 = – -------------- = – ---- – ------- 1 + ----------2 Q Q Q A R Stability of the controlled system requires A < 0 . Then, for the stable case ν2 > 0

(26)

ν1 < 0

The feedback is negative near t = 0 and positive near t = t f . Letting X∗ → 0 eliminates the solution in the region of t f . The algebraic Riccati equation approach for this set of equations starts with the scalar form of eqn (8.212) 2

–H

2

Bf ------ + 2AH + Q = 0 R

(27)

The roots are R H 1, 2 = ------ A ± A 2 Bf

2

QB f 1 + ----------2 RA

(28)

Enforcing the requirement that the uncontrolled system be stable, A<0 transforms eqn (28) to

(29)

622 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

2

QB f RA H 1, 2 = ----------- – 1 ± 1 + ----------2 2 Bf RA

(30)

Finally, the requirement that H > 0 eliminates the negative root, and one obtains RA H opt = ----------2 Bf

2

QB f 1 + ----------- – 1 2 RA

(31)

Referring back to eqn (23), the solution near t = 0 for the case where t f is large has the form – λt

X = Xoe X o – λt β = ------- e ν2

(32) 2

QB f A A ν 2 = – ---- + ------- 1 + ----------2 Q Q RA Taking A < 0 , the expression for ν 2 becomes 2

QB f A ν 2 = ------- 1 + 1 + ----------2 Q RA

(33)

Forming the ratio of β to X , β 1 ---- = ----X ν2

(34)

and substituting for ν 2 leads to RA 1 Q ----- = ----------------------------------------------- = ----------2 ν2 2 Bf QB f A 1 + 1 + ----------2 RA

2

QB f 1 + ----------- – 1 2 RA

(35)

8.5 State-space Formulation for MDOF Systems 623 Comparing eqn (35) with eqn (31) shows that H opt = 1 ⁄ ν 2 . The negative root of eqn (30) is equal to 1 ⁄ ν 1 , and corresponds to positive feedback in the region of t f .

8.5 State-space formulation for MDOF systems Notation and governing equations The material presented in the previous sections can be readily extended to the case of a multi-degree-of-freedom system. One has only to generalize the definition equations for the various matrices involved in the state-space representation. In what follows, the essential steps for an nth order linear system are discussed. Using the notation defined in Chapter 2, the equations for an nth order linear system subjected to seismic excitation and a set of r control forces applied at various locations on the system are written as ˙˙ + CU˙ + KU = – MEa + E F + P MU g f

(8.216)

where E f is an n x r matrix that defines the location of the control forces with respect to the degrees of freedom. Figure 8.14 illustrates the notation for the case of a lumped mass model of a shear beam having four degrees of freedom and two control forces. The initial conditions involve constraints on the displacements and velocities at time t = 0 . U ( 0 ) = Uo U˙ ( 0 ) = U˙ o

(8.217)

Following the approach of section 8.2, the state-space form for eqn (8.216) is taken as X˙ = AX + B f F + B g a g + B p P

(8.218)

624 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

u4

m4

F1 u3

m3

u1 u2

m2

U

F2 u1

=

u2 u3 u4

1 1 E = 1 1

0 0 Ef = 0 1

0 1 0 0

m1

Fig. 8.14: 4DOF system with two control forces where X is now a vector of order 2n, X =

U U˙

(8.219)

and the coefficient matrices are given by A =

O –1

I –1

(2n x 2n)

(8.220)

(2n x r)

(8.221)

(2n x 1)

(8.222)

(2n x 1)

(8.223)

–M K –M C

Bf =

Bg =

Bp =

O –1

M Ef

O –E

O M –1

8.5 State-space Formulation for MDOF Systems 625 Free vibration response - time invariant uncontrolled system Specializing eqn (8.218) for no external forcing leads to a set of 2n homogeneous first order differential equations. X˙ = AX

(8.224)

Considering A to be constant, the general solution has the form X = e λt V

(8.225)

where λ and V satisfy the following set of 2n homogeneous algebraic equations ( A – λI )V = 0

(8.226)

Substituting for A, eqn (8.226) expands to – λI

I

– M –1 K

– M – 1 C – λI

V = 0

The solution of eqn (8.227) has the form Φ V = λΦ

(8.227)

(8.228)

where λ and Φ are constrained by n conditions, λ 2 Φ + λM – 1 CΦ + M – 1 KΦ = 0

(8.229)

For the SDOF case, the complete solution of eqn (8.229) consists of 2 solutions involving a pair of complex roots for λ λ = λ1 λ = λ˜ 1

Φ = 1 Φ = 1

(8.230)

Example 3.13 dealt with a 2 DOF system. In this case, there are 4 solutions involving 2 pairs of complex roots:

626 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems λ = λ 1, λ˜ 1 and λ = λ , λ˜ 2

˜ Φ = Φ 1, Φ 1 (8.231) ˜ Φ = Φ 2, Φ 2

2

Then, for an n’th order system, it follows that there are 2n complex solutions involving n pairs of complex roots. λ = λ 1, λ˜ 1, λ 2, λ˜ 2, …, λ n, λ˜ n ˜ , Φ ,Φ ˜ , …, Φ , Φ ˜ Φ = Φ 1, Φ 2 n 2 n 1 ˜ ˜ Φ1 Φ1 Φn Φn V = , , …, , ˜ ˜ λ 1 Φ 1 λ˜ 1 Φ λ n Φ n λ˜ n Φ 1 n

(8.232)

Each term in eqn (8.232) is complex λ j = λ R, j + iλ I, j Φ j = Φ R, j + iΦ I, j

(8.233)

V j = V R, j + iV I, j The complete solution is obtained by combining the 2n solutions using ˜ . complex constants, A j and A j n

X =

∑

( A je

λ jt

˜ e λ˜ j t V˜ ) Vj + A j j

(8.234)

j=1

Substituting for A j and e

λ jt

,

A j = A R, j + iA I, j e

λ jt

= e

λ R, j t

( cos λ I t + i sin λ I t )

transforms eqn (8.234) to

(8.235)

8.5 State-space Formulation for MDOF Systems 627 n

∑

X (t) =

e

λ R, j t

[ ( A R, j V R, j – A I, j V I, j ) cos λ I, j t

j=1

(8.236)

– ( A R, j V I, j – A I, j V R, j ) sin λ I, j t ] One determines A R, j and A I, j using the initial conditions on X at time t=0. This computation is discussed in the next section. Stability requires the homogeneous solutions to be bounded. Noting eqn (8.236), a system having n DOF is stable when λ R, j < 0

j = 1, 2, …, n

(8.237)

The free vibration solution given by eqn (8.234) can also be expressed as X ( t ) = e At X ( 0 )

(8.238)

where A is defined by eqn (8.220) and e At is generated with the following series: 1 1 e At = I + At + --- AAt 2 + … + ----- A n t n + … 2 n!

(8.239)

Equation (8.238) is a generalized form of the SDOF solution. One can establish this result by specializing eqn (8.234) for t = 0, n

X (0) =

∑

˜ V˜ ) ( A jV j + A j j

(8.240)

j=1

and noting that e At V j = e e At V˜

j

= e

λ jt λ˜ j t

Vj V˜ j

Example 8.7: Free vibration solution for proportional damping Consider eqn (8.229) premultiplied by M.

(8.241)

628 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems λ 2 MΦ + λCΦ + KΦ = 0

(1)

Suppose C is a linear function of K and M C = αK + βM

(2)

Substituting for C results in 2+βλ λ ------------------- MΦ + KΦ = 0 1 + αλ

(3)

When the coefficient of MΦ is expressed as λ 2 + βλ ------------------- = – ω 2 1 + αλ

(4)

eqn (3) takes the form of the classical eigenvalue equation. KΦ = ω 2 MΦ

(5)

There are n eigensolutions to eqn (5) ω 1 2, ω 2 2, …, ω n 2 Φ 1, Φ 2, …, Φ n

(6)

All the eigenvectors are real vectors. Also, all the eigenvalues are positive real quantities. ωj 2 > 0

(7)

Given ω j 2 , one solves eqn (4) for λ j . The pair of complex roots are written as λ j, λ˜ j = – ξ j ω j ± i ω j { 1 – ξ 2j } 1 / 2

(8)

where the damping ratio is related to α and β by β + α ωj 2 ξ j = ------------------------2 ωj

(9)

8.5 State-space Formulation for MDOF Systems 629 In what follows, ω j is considered to be positive, and ω j is set equal to ω j . The eigenvectors, Φ j , are real and satisfy the following orthogonality relationships ΦT j M Φ k = δ jk m k

(10)

2 ΦT j KΦ k = δ jk ω k m k

(11)

Since Φ is real, the state eigenvector, V, has the following form Φj Φj 0 = Vj = + i λ jΦ j λ R, j Φ j λ I, j Φ j

(12)

Undamped free vibration response is the special case where α = β = 0 . The corresponding solution is λ j = ± iω j

(13)

Φ j 0 V j = + i 0 ω jΦ j

(14)

Example 8.8: General uncoupled damping The previous example dealt with the case where the damping matrix is a linear combination of the mass and stiffness matrices. This approach can be extended to deal with a more general form of C. Suppose C satisfies the following orthogonality conditions, ΦT j CΦ k = 2δ jk m k ω k ξ k

j = 1, 2, …, r

(1)

where ξ k are now considered to be independent parameters. The free vibration solution is the same as presented in example 8.7.

630 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems λ j = – ξ j ω j ± iω j [ 1 – ξ 2j ] 1 / 2 Φj Vj = λ jΦ j

(2)

KΦ j = ω 2j M Φ j The ability to independently specify the modal damping ratio for the first r modes by adjusting C is very convenient since it allows one to deal more effectively with the problem of suppressing the response of a particular subset of modes contained within this group. The form of C that satisfies eqn (1) is established by noting the following identities: n 2n –1 ΦT j M ( M K ) Φ k = ω j m j δ jk

(3)

n 2n + 2 m δ –1 ΦT j K ( M K ) Φk = ω j j jk

A linear combination of these matrices is a candidate solution. s

C =

∑

n

( an [ M ( M –1 K ) n ] + bn [ K ( M –1 K ) ] )

(4)

n=0

Taking s = 0 corresponds to Rayleigh damping. Taking s ≤ r allows one to specify r modal damping ratios. The coefficients for this case are determined with s

∑

– 1 + b ω 2n + 1 ) = 2ξ ( a n ω 2n j n j j

j = 1, 2, …, r

(5)

n=0

If both a and b are used, s = r/2. The basic problem with this approach is the form of C generated with eqn (3). The coupling between the off-diagonal elements extends beyond adjacent nodes when n ≥ 1. This pattern requires a deployment of dampers that involves connecting dampers to nodes that are not adjacent, e.g., a damper between floor 1 and floor 3. Realistically, the required pattern cannot be achieved.

8.5 State-space Formulation for MDOF Systems 631

Orthogonality properties of the state eigenvectors The eigenvector V j satisfies eqn (8.226). AV j = λ j V j

(8.242)

Since A is not symmetrical, V j is not orthogonal to V k . VT j V k ≠ δ jk

(8.243)

A set of vectors which are orthogonal to V can be established by considering the eigenvalue problem for the transpose of A: A T W j = λ *j W j

j = 1, 2, …, n

(8.244)

where λ * , W represent the eigenvalue and corresponding eigenvector for A T . Eqn (8.244) requires the determinant of the coefficient matrix to vanish. AT – λ*I = 0

(8.245)

Noting that the determinant of the transpose of a matrix is equal to the determinant of the original matrix, b ≡ bT

(8.246)

it follows that λ *j ≡ λ j

(8.247)

and eqn (8.244) is equivalent to AT W j = λ jW j

(8.248)

Starting with AV j = λ j V j AT W k = λk W k

k≠j

(8.249)

and premultiplying the first equation with W k and the second with V j , one obtains

632 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems W kT AV j = λ j W kT V j

(8.250)

T T VT j A W k = λk V j W k

(8.251)

Eqn (8.251) can be written as W kT AV j = λ k W kT V j

(8.252)

Then subtracting eqn (8.250) from (8.252) results in 0 = ( λ k – λ j )W kT V j

(8.253)

According to eqn (8.253), W k and V j are orthogonal W kT V j = 0

j≠k

for

(8.254)

The result for j = k is written as WT j Vj = f j

(8.255)

One can show that ˜ TV = 0 W k j W kT V˜ j = 0

for all j, k

(8.256)

and ˜ T V˜ = ˜f W j j j

(8.257)

Example 8.9: Initial conditions - free vibration response Consider the general free vibration solution defined by eqn (8.234). n

X (t) =

∑

( A je

λ jt

˜ e λ˜ j t V˜ ) Vj + A j j

(1)

j=1

The integration constants are determined using the initial conditions on X at t = 0.

8.5 State-space Formulation for MDOF Systems 633 n

X (0) =

∑

˜ V˜ ) ( A jV j + A j j

(2)

j=1

Noting the orthogonality relations, the expressions for the constants are: 1 A j = ----- W T X (0) fj j

(3)

1 ˜ T ˜ = ----W j X (0) A j ˜f j

Determination of W and f j One establishes W by solving the following eigenvalue problem A T W = λW

(8.258)

Substituting for A T , the coefficient matrix expands to – λI

– K M –1

I

– CM – 1 – λI

W = 0

(8.259)

Expressing W in partitioned form, W 1 W = W 2

(8.260)

and expanding eqn (8.259) leads to – λW 1 – K M – 1 W 2 = 0 W 1 – CM – 1 W 2 – λW 2 = 0

(8.261)

Solving the first equation for W 2 , W 2 = – λM K – 1 W 1

(8.262)

634 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems and substituting into the second equation results in W 1 = – λCK – 1 W 1 – λ 2 M K – 1 W 1

(8.263)

This equation can be transformed to a form similar to the eigen equation for Φ by expressing W 1 as (8.264)

W 1 = KW 1 The result is λ 2 M W 1 + λCW 1 + KW 1 = 0

(8.265)

Comparing this form with eqn (8.229) shows that W1 ≡ Φ

(8.266)

It follows that the forms of V j and W j are related by KΦ j Wj = –λ j M Φ j

Φj Vj = λ jΦ j

(8.267)

Using these expressions, the definition equation for f expands to T 2 T f j = WT j V j = Φ j KΦ j – λ j Φ j M Φ j T = – 2λ 2j Φ T j M Φ j – λ j Φ j CΦ j

(8.268)

When C is proportional to either K or M, Φ is real and f j is given by f j = m j ( ω 2j – λ 2j )

(8.269)

The value for no damping is f j = 2m j ω 2j . General Solution - time invariant system Following the approach employed for the SDOF case, the general solution for an arbitrary loading applied to a time invariant MDOF system can be expressed as a Duhamel integral involving the free vibration response. Noting eqns (8.238) and (8.239), the complete solution has the form

8.5 State-space Formulation for MDOF Systems 635

X (t) = e

A ( t – to )

Xo +

∫

t

to

e A ( t – τ ) ( B f F + B p P + B g a g ) dτ

(8.270)

where the coefficient matrices ( A, B f , B p, B g ) are defined by eqns (8.220) to (8.223). This equation is similar to the SDOF solution given by eqn (8.26); one has only to introduce the appropriate forms of A, B f , B p , and B g . The discrete time equilibrium equation for the MDOF case follows from eqn (8.77). X j + 1 = e A∆t X j + A – 1 ( e A∆t – I ) [ B g a g, j + B f F j + B p P j ]

(8.271)

Working with the full set of equations for an n’th order system requires dealing with matrices of order 2n. As n increases, the computational costs for e A∆t and A – 1 become excessive, and one needs to consider an alternative approach. In conventional dynamic analysis, this reduction in computational effort is achieved by expressing the state vector as a linear combination of prescribed modal vectors multiplied by unknown coordinate variables which are functions of time. Depending upon the nature of the loading, one can obtain a reasonably accurate solution by suitably selecting the participating modes so as to minimize the required number of modal coordinate functions, and consequently reduce the computational cost. Interpreting the total response as a superposition of modal responses is also useful from the perspective of active control. A key issue of active control is the optimal location of “active” forces so as to minimize the response of a particular mode. In what follows, the formulation of the governing equations in terms of modal coordinates is presented. This formulation is used later to establish the form of the LQR algorithm in terms of modal coordinates. Modal state space formulation - uncoupled damping The displacement vector is expressed as a linear combination of a subset of the eigenvectors for the undamped system scaled with functions of time. s

U(t) =

∑

q j ( t )Φ j

(8.272)

j=1

Assuming the system has n DOF, this expression will produce the exact solution when s is taken to be n. Substituting for U ( t ) in the force equilibrium equation,

636 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems premultiplying the result with Φ k , and noting the orthogonality conditions for Φ k , one obtains s equations having the following form, p

m k q˙˙k +

∑

( Φ kT CΦ j )q˙ j + m k ω k2 q k

j=1

= – Φ kT MEa g + Φ kT P + Φ kT E f F

k = 1, 2, …, s

(8.273)

These equations uncouple when C is orthogonal to Φ j . Assuming this condition is satisfied, and taking Φ kT CΦ j = 2δ jk m k ω k ξ k

(8.274)

results in q˙˙k + 2ξ k ω k q˙k + ω k2 q k 1 1 1 = – ------- Φ kT MEa g + ------- Φ kT P + ------- Φ kT E f F mk mk mk

k = 1, 2, …, s

(8.275)

The initial conditions for q k are 1 q k ( 0 ) = ------- Φ kT MU ( 0 ) mk 1 q˙k ( 0 ) = ------- Φ kT M U˙ ( 0 ) mk

(8.276)

When F = 0, these equations can be solved separately. The effect of feedback is to introduce coupling between the modal equations. In order to deal with feedback, the equations are transformed to the standard state-space form by defining q as the modal coordinate vector, q1 q2 q = … qs

s×1

(8.277)

8.5 State-space Formulation for MDOF Systems 637 and X m as the modal coordinate state vector, q Xm = q˙

2s × 1

(8.278)

Note that now X m contains only 2s modal coordinate terms. The equations are expressed in the same form as eqn (8.218), X˙ m = A m X m + B fm F + B pm P + B gm a g

(8.279)

where the “modal” forms of the coefficient matrices along with their sizes are: m = [ m j δ ij ]

(s × s)

(8.280)

Λ = [ ω 2j δ ij ]

(s × s)

(8.281)

Λ 1 = [ 2ξ j ω j δ ij ]

(s × s)

(8.282)

Φ = Φ1 Φ2 … Φs

(n × s)

(8.283)

( 2s × 2s )

(8.284)

( 2s × r )

(8.285)

( 2s × r )

(8.286)

Am =

B fm =

B pm =

B gm =

0 I –Λ –Λ1 0 m –1 Φ T E f

0 m –1 Φ T

0 – m – 1 Φ T ME

( 2s × 1 )

Lastly, the initial conditions follow from eqn (8.276)

(8.287)

638 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

q(0) m – 1 Φ T MU ( 0 ) X m(0) = = q˙ ( 0 ) m – 1 Φ T M U˙ ( 0 )

( 2s × 1 )

(8.288)

Feedback is introduced by expressing F as a linear function of the modal state vector (8.289)

F = – K fm X m = – k d q – k v q˙

The individual feedback matrices are of order r x s. For pure velocity feedback, k d is set equal to 0. Substituting for F in eqn (8.279) leads to the state equation specialized for continuous negative linear feedback. X˙ m = ( A m – B fm K fm )X m + B pm P + B gm a g

(8.290)

where

A m – B fm K fm =

0

I

– Λ – m –1 Φ T E f kd

– Λ1 – m –1 Φ T E f kv

(8.291)

The discrete form is similar to eqn (8.271). X m, j + 1 = e

A m ∆t

–1 ( e X m, j + A m

A m ∆t

– I ) [ B fm F j + B pm P j + B gm a g ]

(8.292)

F j = – K fm X m, j For the continuous case, the free vibration solution with feedback is stable when A m is stable and k v is positive. This condition is always satisfied. The discrete solution is stable provided that the modulus of the largest eigenvalue of c m is less than unity. ρ max ( c m ) < 1 cm = e

A m ∆t

–1 ( e – Am

(8.293) A m ∆t

– I )B fm K fm

(8.294)

Given the selection of the modes and the feedback matrix, K fm , one can establish the limit on ∆t by computing the eigenvalues of c m for a range of values of ∆t .

8.5 State-space Formulation for MDOF Systems 639 This computation is illustrated in example 8.14 Modal state space formulation - arbitrary damping When C is an arbitrary symmetric positive definite matrix, the expansion in terms of the eigenvectors for the undamped system and real modal coordinates does not lead to uncoupled modal equations. In this case, one has to work with an expansion involving complex modal coordinates and complex state eigenvectors. The state space vector is approximated as a linear combination of s modal vectors and coordinates, U(t) 1 X (t) = = --2 U˙ ( t )

s

∑

( q j V j + q˜ j V˜ j )

(8.295)

j=1

where q and V are complex quantities defined by q j ( t ) = q R, j ( t ) + iq I, j ( t )

Vj =

Φj

(8.296)

(8.297)

λ jΦ j

The expanded real forms for U ( t ) and U˙ ( t ) are s

U(t) =

∑

( q R, j Φ R, j – q I, j Φ I, j )

(8.298)

j=1 s

U˙ ( t ) =

∑

[ q R, j ( λ R, j Φ R, j – λ I, j Φ I, j ) – q I, j ( λ R, j Φ I, j + λ I, j Φ R, j ) ] (8.299)

j=1

When damping is uncoupled, Φ I, j = 0 . The fully undamped case has λ R, j = 0 , λ I, j = ω j , and Φ I, j = 0. The initial conditions for the modal coordinates are determined by specializing X ( t ) for t = 0.

640 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems s

1 X ( 0 ) = --2

∑

( q j ( 0 )V j + q˜ j ( 0 )V˜ j )

(8.300)

j=1

Premultiplying by W k , and noting the orthogonality relation between W k and V j , one obtains 2 q k ( 0 ) = -----W kT X ( 0 ) fk

(8.301)

Substituting for W k , the expression for q k ( 0 ) expands to 2λ k 2 q k ( 0 ) = -----Φ kT K U ( 0 ) – --------- Φ kT M U˙ ( 0 ) f fk k

k = 1, 2, …, s

(8.302)

Introducing the expression for X ( t ) in the state equilibrium equation, and premultiplying for W k results in a set of s complex scalar equations, 2 2 2 q˙k = λ k q k + -----W kT B f F + -----W kT B g a g + -----W kT B p P f f f k

k

k

(8.303)

k = 1, 2, …, s The terms involving F, a g , and P can be interpreted as complex modal forces associated with the k’th mode. Expanding the matrix products and substituting the notation, 2λ k 2 b f , k = -----W kT B f = – --------- Φ kT E f fk fk 2λ k 2 b g, k = -----W kT B g = --------- Φ kT ME fk fk

(8.304)

2λ k 2 b p, k = -----W kT B p = – --------- Φ kT fk fk reduces the governing modal equations to a simpler form, q˙k = λ k q k + b f , k F + b p, k P + b g, k a g

k = 1, 2, …, s

(8.305)

8.5 State-space Formulation for MDOF Systems 641 Since q k is complex, there are a total of 2s equations.The corresponding set of “real” equations for the k’th mode are: q˙R, k = λ R, k q R, k – λ I, k q I, k + b f R, k F + b pR, k P + b a gR, k g q˙I, k = λ I, k q R, k + λ R, k q I, k + b fI, k F + b pI, k P + b a gI, k g

(8.306)

k=1,2,...,s One establishes the initial conditions for q R and q I by expanding eqn (8.302). The resulting expressions are: 2 T + f T q R, k ( 0 ) = ---------------------------- [ ( f R, k Φ R ,k I, k Φ I, k )KU ( 0 ) 2 2 f R, k + f I, k

(8.307)

T – λ ΦT ) + f T T ˙ – { f R, k ( λ R, k Φ R ,k I, k I, k I, k ( λ R, k Φ I, k + λ I, k Φ R, k ) }MU ( 0 ) ]

f I, k q I, k ( 0 ) = – -----------q R, k + 2Φ IT, k KU ( 0 ) f R, k 2 T )M U ˙ (0) – ----------- ( λ R, k Φ IT, k + λ I, k Φ R ,k f R, k

(8.308)

Various specialized forms of these equations are presented in the examples listed below.

Example 8.10: Modal formulation - undamped case It is of interest to specialize the general formulation presented in this section for the case of no damping and compare this result with the solution obtained with the conventional formulation derived in the previous section. Setting C = 0 reduces the various terms to:

642 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Φ I, k = 0 λ R, k = 0 λ I, k = ω k f k = 2m k ω k2 i b f , k = – -------------- Φ kT E f mk ωk

(1)

i b p, k = – -------------- Φ kT mk ωk i b g, k = -------------- Φ kT ME mk ωk Substituting in eqns (8.307) and (8.308), leads to q˙R, k = – ω k q I, k

(2)

1 q˙I, k = ω k q R, k – -------------- Φ kT ( E f F + P – MEa g ) m ω

(3)

k k

Equation (2) shows that 1 q I, k = – ------ q˙R, k ωk

(4)

Then, eqn (3) can be written as 1 q˙˙R, k + ω k2 q R, k = ------- Φ kT ( E f F + P – MEa g ) mk

(5)

This result is identical to eqn (8.225).

Example 8.11: Modal formulation - uncoupled damping When the damping matrix is proportional to K and M, the eigenvector, Φ , is real. However, the eigenvalue is complex. The various terms specialized for this case are as follows:

8.5 State-space Formulation for MDOF Systems 643 Φ I, k = 0 λ R, k = – ξ k ω k λ I, k = ω k [ 1 – ξ k2 ] 1 / 2 f k = 2m k λ I, k ( λ I, k – iλ R, k ) = – ( 2m k λ I λ ) i λk i ----- = -------------------fk 2m k λ I, k i b f , k = – ----------------- Φ kT E f m k λ I, k

(1)

i b g, k = – ----------------- Φ kT ME m k λ I, k i b p, k = – ----------------- Φ kT m k λ I, k Since the forcing terms are pure imaginary, the equations simplify to q˙R, k = λ R, k q R, k – λ I, k q I, k 1 q˙I, k = λ I, k q R, k + λ R, k q I, k – ----------------- Φ kT ( E f F + P – MEa g ) m λ

(2)

k I, k

The initial conditions are 1 q R, k ( 0 ) = ------- Φ kT MU ( 0 ) mk 1 1 q I, k ( 0 ) = ---------- λ R, k q R, k ( 0 ) – ------- Φ T M U˙ ( 0 ) λ I, k mk

(3)

This example shows that q I, k involves a combination of q R, k and q˙R, k . The actual displacements are given by

644 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems s

U(t) =

∑

q R, j Φ j

j=1

U˙ ( t ) =

∑

(4)

s

s

( λ R, j q R, j – λ I, j q I, j )Φ j =

j=1

∑

q˙R, j Φ j

j=1

Example 8.12: Modal parameters - 4DOF system This example presents modal data for the 4 DOF system shown in Fig. 1. The element stiffness and damping values are listed in Table 1. The element stiffness factors are selected so that the displacement profile for the first mode is essentially linear. Two damping distributions are considered. Case 1 represents a low damping level for the first mode. The damping coefficients for case 2 are adjusted so that the sum, Σc j , is 3 times the corresponding sum for the first case. Since the internodal displacement for the first mode is essentially constant, this adjustment increases the modal damping ratio for the first mode by a factor of 3. More emphasis is placed on element 1 to exaggerate the non-proportional distribution and therefore increase the magnitude of the imaginary part of the modal vector.

1000kg

m4

u4

k 4, c 4 1000kg

m3

u3

k 3, c 3 1000kg

m2

u2

k 2, c 2 1000kg

m1 k 1, c 1

Figure 1

u1

8.5 State-space Formulation for MDOF Systems 645 Table 1 Element number

Element stiffness, k

Element damping, c (kN.s/m)

(kN/m)

case 1

case 2

1

1700

4

20

2

1400

3

7

3

1000

2

2

4

700

1

1

Σ = 10

Σ = 30

Scaled versions of the modal displacement profiles for the 2 damping distributions are plotted in Figures 2 and 3. The real parts are essentially identical. Since case 1 represents a low level of damping, the amplitude of the corresponding imaginary part is negligible in comparison to the real part which is of order 1. Case 2 results show a significant increase in the amplitude for the imaginary part. This shift is due to the large increment in c assigned to element 1. The modal periods and damping ratios are shown in Figures 4 and 5. There is essentially no change in the periods. As expected, all the modal damping ratios are higher for case 2. Lastly, to illustrate the effect of damping, time history responses were generated for the El-Centro ground excitation. The maximum values of the inter-nodal displacements for the elements are plotted in Figure 6. Increasing the damping reduces the absolute maximum value from 0.0258 to 0.0217, and also decreases the average value.

646 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Node number

4

3

Mode 2

Mode 3 2

Mode 1 Mode 4

1 −1

−0.8

−0.6

−0.4

−0.2

0

Amplitude case 1

0.2

0.4

0.6

0.8

1

Node number

4

3

Mode 2

Mode 3 2

Mode 1 Mode 4

1 −1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

Amplitude case 2 Figure 2: Modal displacement profile - real part

0.8

1

8.5 State-space Formulation for MDOF Systems 647

Node number

4

3

Mode 3 Mode 2

Mode 1 2

Mode 4

1 −0.02

−0.015

−0.01

−0.005

0

0.005

Amplitude case 1

0.01

0.015

0.02

0.025

Node number

4

Mode 2

3

Mode 1

Mode 3

2

Mode 4

1 −0.4

−0.3

−0.2

−0.1

0

Amplitude case 2

0.1

0.2

Figure 3: Modal displacement profile - imaginary part

0.3

648 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.5

0.45

Modal period, ( sec )

0.4

0.35

0.3

0.25

Damping case 1 0.2

Damping case 2 0.15

0.1

0.05

1

2

3

4

Mode number Figure 4: Modal period - No feedback

0.2

0.18

Modal damping ratio

0.16

0.14

0.12

0.1

Damping case 2

0.08

0.06

0.04

Damping case 1

0.02

0

1

2

Mode number

3

Figure 5: Modal damping ratio - No feedback

4

8.5 State-space Formulation for MDOF Systems 649

4

Element number

El Centro ground motion 3

Damping case 2

2

1 0.017

0.018

0.019

0.02

0.021

Damping case 1

0.022

0.023

0.024

0.025

0.026

Maximum element internodal displacement, (m) Figure 6: Element internodal displacement profile - No feedback

When F = 0, the problem reduces to solving a set of 2 equations for each mode. Including F couples the modal equations, and now one has to solve 2s simultaneous equations. In this case, it is convenient to shift over to a state-space type formulation. The state vector is defined in a similar way as for the previous formulations. Firstly, the real and imaginary modal coordinates are grouped separately, q R1 q R2 qR = … q Rs

q I1 q I2 qI = … q Is

Secondly, the “modal” state vector is taken as

(8.309)

650 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems qR Xm = qI

(8.310)

Lastly, the state equilibrium equation is expressed in the same form as for uncoupled damping X˙ m = A m X m + B fm F + B pm P + B gm a

(8.311)

where the corresponding form of the coefficient matrices are λ R = [ λ R, k δ jk ]

λ I = [ λ I, k δ jk ]

b fR, 1 b fR =

b pR =

b gR =

b fR, 2 …

b fI =

b fI, 2

b fR, s b pR, 1

b pI, 1

…

b pI =

b pI, 2 …

b pR, s

b pI, s

b gR, 1

b gI, 1

b gR, 2 … b gR, s

Am =

b fI, 1 … b fI, s

b pR, 2

λR

–λI

λI

λR

(8.312)

b gI =

b gI, 2 …

(8.313)

(8.314)

(8.315)

b gI, s

(8.316)

8.5 State-space Formulation for MDOF Systems 651

B fm =

b fR b fI

B pm =

b pR b pI

B gm =

b gR b gI

(8.317)

The initial conditions for X m are obtained with eqns (8.307) and (8.308). Negative linear feedback is taken as q F = – K fm X m = – k d k v R qI

(8.318)

Example 8.10 showed that q I is a linear function of q˙ R for no damping. Including damping results in q I being a function of both q˙ R and q R . For a lightly damped system, it is reasonable to assume q I depends only on q˙ R , and therefore to approximate pure velocity feedback by setting k d = 0 . The feedback term for this case reduces to:

B fm F =

0 0

– b fR k v q R –b k qI

(8.319)

fI v

Example 8.13: Modal response for example 8.12 The state space modal formulation is applied to the 4 DOF system with Case 1 damping defined in example 8.12. Results for the real and imaginary parts of the modal coordinates for the first 2 modes are plotted below. Comparison of the plots in Figure (1) shows that the response is dominated by the first mode. The ratio of the amplitudes for the first and second modes is approximately 30, which indicates that the second mode contributes about 3% to the total response for this combination of structure and excitation.

652 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.1

Modal coordinate, q R1

0.08

El Centro Damping Case 1

0.06

0.04

0.02

0

−0.02

−0.04

−0.06

−0.08

0

5

10

15

20

25

Time (sec) a) Mode 1

30

35

40

45

50

−3

4

x 10

Modal coordinate, q R2

3

El Centro Damping Case 1

2

1

0

−1

−2

−3

−4

0

5

10

15

20

25

Time (sec) b) Mode 2

30

35

40

45

Figure 1: Time history response of the real part of the modal coordinate

50

8.5 State-space Formulation for MDOF Systems 653

0.08

0.06

El Centro Damping Case 1

Modal coordinate, q I1

0.04

0.02

0

−0.02

−0.04

−0.06

−0.08

−0.1

0

5

10

15

20

25

30

35

40

45

50

Time (sec) a) Mode 1 −3

3

x 10

Modal coordinate, q I2

2

El Centro Damping Case 1

1

0

−1

−2

−3

0

5

10

15

20

25

Time (sec)

30

35

40

45

50

b) Mode 2 Figure 2: Time history response of the imaginary part of the modal coordinate

654 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Example 8.14: Modal response with feedback for example 8.12 This example illustrates the effect of linear negative velocity feedback on the modal damping ratios. The modal state space formulation defined by eqns (8.309) through (8.319) is applied to the 4 DOF system with Case 1 damping described in example 8.12. Two control force systems are considered. The first system is a single force applied at node 4; the second system consists of selfequilibrating sets of nodal forces applied on each element. Figure (1) shows the spatial distribution of the control forces. m4

u4

k 4, c 4 m3

F1 F2

u3

k 3, c 3 m2

F2 F3

u2

k 2, c 2 m1

F1

F1

F3 F4

u1

k 1, c 1 Scheme 1

Scheme 2

Figure 1 Given the feedback matrix, K fm , one determines F using eqn (8.318), F = – K fm X m

(1)

and then solves for X m with eqn (8.311). X˙ m = ( A m – B fm K fm )X m + B pm P + B gm a g

(2)

The frequency and damping parameters for the system with feedback are related to the eigenvalues of A m – B fm K fm , and depend on the control force scheme as

8.5 State-space Formulation for MDOF Systems 655 well as K fm . Particular forms of K fm obtained with the algebraic Riccati equation and specialized for pure velocity feedback (see eqn (8.319)) are used here to generate results for the 2 force schemes. This formulation is discussed in detail in the next section. Our objective here is to illustrate the variation in behavior associated with feedback systems. Figure (2) shows the modal damping ratios for the 4 DOF model with Case 1 damping, force scheme 1, and a particular choice for K fm . Feedback results in a significant increase (a factor of 10) in the damping ratio for the first mode. Since this response is dominated by the first mode, there also is a reduction in the displacement response. The relevant displacement quantities are plotted in Figs (2b) and (2c). The last plot shows the variation of the control force magnitude with time. Results for the control force scheme 2 are plotted in Figure (3). This combination of control forces and weighting produces essentially equal nodal damping ratios whereas scheme 1 places the priority on the first mode. Since the damping ratio is less, the displacement response is greater. There also is a reduction in the peak magnitudes of the 4 control forces. Modal damping ratio with feedback

0.24

0.22

Case 1 damping Model defined in example 8.12

0.2

0.18

0.16

0.14

0.12

0.1

0.08

0.06

0.04

1

2

3

Mode number a) Modal damping ratio with continuous feedback - no saturation limit

4

656 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

4

Element number

El Centro excitation

3

2

1 0.0092

0.0094

0.0096

0.0098

0.01

0.0102

0.0104

0.0106

0.0108

Maximum element internodal displacement b) Element internodal displacement profile

0.011

0.0112

0.04

Nodal displacement (m)

0.03

El Centro excitation 0.02

0.01

0

−0.01

−0.02

−0.03

−0.04

−0.05

0

5

10

15

Time (sec) c) Displacement of node 4

20

25

30

8.5 State-space Formulation for MDOF Systems 657

3000

El Centro excitation

Control force magnitude

2000

1000

0

−1000

−2000

−3000

−4000

Max Force = 4860 N −5000

0

500

Time step number

1000

1500

d) Control force time history Figure 2: Modal response for control force scheme 1

Modal damping ratio with feedback

0.135

Case 1 damping Model defined in example 8.12

0.13

0.125

0.12

0.115

0.11

1

2

3

Mode number a) Modal damping ratio with continuous feedback - no saturation limit

4

658 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Element number

4

3

2

1 0.0122

0.0124

0.0126

0.0128

0.013

0.0132

0.0134

0.0136

0.0138

0.014

0.0142

Maximum element internodal displacement b) Element internodal displacement profile 0.06

El Centro excitation

Nodal displacement (m)

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

Time (sec) c) Displacement of node 4

20

25

30

8.5 State-space Formulation for MDOF Systems 659

Maximum control force magnitude (Newtons)

3800

El Centro excitation 3700

3600

3500

3400

3300

3200

3100

3000

2900

1

2

3

4

Control force number d) Control force time history Figure 3: Modal response for control force scheme 2

Stability analysis - discrete modal formulation The discrete form of the state equilibrium equation expressed in terms of modal coordinates is used here to evaluate the effect of the time interval and the delay time on the stability. Allowing for linear feedback, the governing equation for free vibration response is X m, j + 1 = e

A m ∆t

–1 ( e X m, j + A m

A m ∆t

– I )B fm F j

(8.320)

Time delay is introduced by taking F j as Fj = 0 F j = – K fm X m, j – ν

0≤j<ν j≥ν

(8.321)

660 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems With this specification for F j , the response equations take the following form X m, j + 1 = c m1 X m, j X m, j + 1 = c m1 X m, j + c m3 X m, j – ν

0≤j<ν j≥ν

(8.322)

where c m1 = e c m3 =

A m ∆t

– 1 ( e A m ∆t – Am

(8.323) – I )B fm K fm

The stability analysis for the SDOF system showed that time delay lowered the limiting value of the time interval. Therefore, one can obtain an upper bound estimate for ∆t by considering no time delay. For this case, eqn (8.322) simplifies to X m, j + 1 = ( c m1 + c m3 )X m, j = c m X m, j

(8.324)

Stability requires the modulus of the largest eigenvalue of c m to be less than 1. ρ max ( c m ) < 1

(8.325)

The complexity of c m necessitates the use of a numerical computational procedure. Example 8.2 illustrates the computational details for a SDOF system with negative velocity feedback. In general, one ranges over ∆t to determine the value corresponding to ρ max ( c m ) = 1. When there is time delay, the eigenvalue problem is of order ν + 1 , ( c m3 + ρ ν c m1 – ρ ν + 1 I )Ψ = 0

(8.326)

For the SDOF case, it was possible to expand the determinant of the coefficient matrix and obtain a polynomial expansion for ρ . However, a numerical procedure is still required to evaluate the roots. For MDOF systems, this hybrid approach is not feasible and one must resort to a fully numerical approach. Given ∆t and ν , eqn (8.322) can be used to generate the corresponding free vibration response. By decreasing ∆t in increments from the critical value for no time delay, one can converge on the limiting value for the specified time delay. This solution strategy is efficient when the number of modal DOF is small, on the order of 5.

8.5 State-space Formulation for MDOF Systems 661 The computational cost for the exponential matrix increases nonlinearly with the number of modes and eventually becomes excessive. In this case, one can resort to an approximate formulation which assumes the modal equations are fully uncoupled. This assumption reduces the stability analysis for a problem involving s modes to s individual SDOF stability analyses. The approximate stability formulation proceeds as follows. Firstly the feedback force is considered to involve only q I . s

F =

∑

k v, j q I, j

(8.327)

j=1

Secondly, b fR, k is taken as a null matrix in eqn (8.306). These conditions correspond to pure velocity feedback for a system with uncoupled damping. Lastly, the feedback terms in the k’th equation are assumed to involve only q I, k . With these assumptions, eqn (8.306) takes the form q˙R, k = λ R, k q R, k – λ I, k q I, k q˙I, k = λ I, k q R, k + λ R, k q I, k – b k q I, k

(8.328)

where b k = b fI, k k v, k

(8.329)

The remaining steps are the same as for the SDOF formulation presented in section 8.3. Introducing matrix notation, eqn (8.328) is expressed as q˙ k = A k q k – B k q k

(8.330)

q R, k qk = q I, k

(8.331)

where

662 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Ak =

Bk =

λ R, k – λ I, k λ I, k λ R, k

(8.332)

0

(8.333)

bk

The discrete form allowing for time delay is written as q k, j + 1 = c k1 q k, j + c k3 q k, j – ν

(8.334)

where the c matrices are now 2 x 2. c k1 = e

A k ∆t

c k3 = – A k– 1 ( e

(8.335) A k ∆t

– I )B k

(8.336)

One determines the limiting value of ∆t for each k ranging from 1 to s, the total number of modes selected to represent the solution. The lowest value controls the change of the time interval.

Example 8.15: Stability analysis for example 8.14 This example examines the effect of time delay on the response of the 4 DOF system considered in Example 8.14. Modal properties corresponding to the 2 control force schemes are listed below in Table 1. The time increment is considered to be equal to 0.02 secs. An estimate of the limiting value of ν for each force scheme can be obtained using the data contained in Table 1 of Example 8.3. These results are for ξ = 0 , and provide a lower bound estimate, ν o . The limiting value of ν for ξ > 0 is higher than ν o . For convenience, the table is listed below as Table 2.

8.5 State-space Formulation for MDOF Systems 663 Table 1: Force scheme 1

Force scheme 2

ξa

ξa

0.013

0.212

0.100

0.102

0.029

0.045

0.096

0.129

0.155

0.046

0.005

0.078

0.095

0.211

0.071

0.000

0.060

Mode

T

∆t -----T

ξ

1

0.498

0.040

2

0.197

3 4

Table 2: Limiting values of ∆t ⁄ T for ξ = 0 (Example 8.3, Table 1) Stability limit for ∆t ⁄ T ξa

ν = 0

ν = 1

ν = 2

0

0.50

0.500

0.5000

0.158

0.0955

0.151

0.0908

0.144

0.0860

0.137

0.0822

0.131

0.0780

0.124

0.0745

0.118

0.07085

0.113

0.0677

0.05 0.10

0.44

0.15 0.20

0.38

0.25 0.30

0.33

0.35 0.40

0.29

For force scheme 1, the first and second modes have the larger values of ξ a and therefore control the delay. Assuming ξ = 0 , Table 2 indicates that mode 2 is close to instability for ν = 2 (∆t ⁄ T = 0.102 vs. 0.096). The value for mode 1 is closer to 3. Since ξ a is ≈ 0 for modes 3 and 4, they are not critical even though the corresponding time ratios are higher. Force scheme 2 produces some damping for modes 3 and 4, and since these time ratios are the highest, they control the stability limit. Allowing for some increase due to initial damping, the critical value for ν is between 1 and 2 for modes 3, 4, and >2 for modes 1, 2. The least critical mode is mode 1 since it has the lowest time ratio. Figure 1 contains results generated for control force scheme 1, with ν

664 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems taken equal to 2. They are in close agreement with the solution for ν = 0 presented in Example 8.14. Increasing ν to 3 produces the responses plotted in Figure 2. Mode 2 has gone critical, and dominates the response whereas mode 1 remains at essentially the same peak level as for ν = 2 . These results are based on limiting the peak magnitude of the control force to 15,000N, which is 3 times the peak value for ν = 2 . Limiting F simulates saturation of the force actuators. Comparison of the power input plots shows that shifting ν from 2 to 3 transforms the energy dissipation process, resulting in energy being supplied to the system and a periodic forced vibration response following the seismic induced response. Reducing the peak magnitude down to 5,000N, the value required for ν = 0 (and also ν = 2 ), leads to the responses plotted in Figure 3. The magnitudes have been reduced, but the system is still responding as if it were subjected to a periodic forced vibration as well as a seismic excitation. Decreasing F max further will not eliminate this additional motion, as illustrated by the plot for F = 2,500 and 1,000N. Results for force scheme 2 are plotted in Figs 5 and 6. Mode 4 controls the allowable delay. For ν = 1 , there is a negligible effect on the response. For ν = 2 , mode 1 is essentially unchanged, while mode 2 shows some additional motion due to delay. Two out of 4 of the control forces reach the prescribed peak value of 15,000N. The response corresponding to F max = 4,000N shown in Fig 7 exhibits similar features.

8.5 State-space Formulation for MDOF Systems 665

0.04

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

Nodal displacement (m)

0.03

0.02

0.01

0

−0.01

−0.02

−0.03

−0.04

0

5

10

15

20

25

30

35

40

45

50

Time (sec) a) Displacement of node 4 −3

Modal coordinate response

4

x 10

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

3

2

1

0

−1

−2

−3

0

5

10

15

20

25

30

35

40

45

Time (sec) b) Time history response of the real part of the coordinate for mode 2

50

666 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

−4

2.5

x 10

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

Modal coordinate response

2

1.5

1

0.5

0

−0.5

−1

−1.5

−2

0

5

10

15

20

25

Time (sec)

30

35

40

45

50

c) Time history response of the real part of the coordinate for mode 4 5000

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

Control force magnitude

4000

3000

2000

1000

0

−1000

−2000

−3000

−4000

−5000

0

500

1000

1500

2000

Time step number d) Time history response of the magnitude of the control force

2500

8.5 State-space Formulation for MDOF Systems 667

Power supplied by the control force

500

0

−500

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

−1000

−1500

−2000

−2500

−3000

−3500

0

500

1000

1500

Time step number e) Time history response of the input power Figure 1

2000

2500

668 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.15

Nodal displacement (m)

0.1

0.05

0

−0.05

−0.1

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 15000N

−0.15

−0.2

0

5

10

15

20

25

30

35

40

45

50

35

40

45

50

Time (sec) a) Displacement of node 4 0.1

Modal coordinate response

0.08

0.06

0.04

0.02

0

−0.02

−0.04

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 15000N

−0.06

−0.08

−0.1

0

5

10

15

20

25

30

Time (sec) b) Time history response of the real part of the coordinate for mode 2

8.5 State-space Formulation for MDOF Systems 669

−4

2.5

x 10

Modal coordinate response

2

1.5

1

0.5

0

−0.5

−1

−1.5

−2

−2.5

0

5

10

15

20

25

30

35

40

45

Time (sec) c) Time history response of the real part of the coordinate for mode 4

50

4

Control force magnitude

1.5

x 10

1

0.5

0

−0.5

−1

−1.5

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 15000N 0

500

1000

1500

Time step number

d) Control force magnitude

2000

2500

670 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

4

5

Power supplied by the control force

4

3

x 10

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 15000N

2

1

0

−1

−2

−3

0

500

1000

1500

Time step number

e) Input power Figure 2

2000

2500

8.5 State-space Formulation for MDOF Systems 671

0.06

Nodal displacement (m)

Force scheme 1 Damping case 1 El Centro 0.04 ν = 3 Fmax = 5000N 0.02

0

−0.02

−0.04

−0.06

0.03

Modal coordinate response

0.02

0

5

10

15

20

25

30

35

40

45

50

15

20

25

30

35

40

45

50

Time (sec) a) Displacement of node 4

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 5000N

0.01

0

−0.01

−0.02

−0.03

0

5

10

Time (sec) b) Time history response of the real part of the coordinate for mode 2

672 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

−4

2.5

x 10

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 5000N

Modal coordinate response

2

1.5

1

0.5

0

−0.5

−1

−1.5

−2

0

5

10

15

20

25

30

35

40

45

50

Time (sec) c) Time history response of the real part of the coordinate for mode 4

Power supplied by the control force

6000

4000

2000

0

−2000

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 5000N

−4000

−6000

0

500

1000

1500

Time step number d) Time history response of the input power Figure 3

2000

2500

8.5 State-space Formulation for MDOF Systems 673

0.08

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 1500N

Nodal displacement (m)

0.06

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

30

Time (sec) a) Displacement of node 4

0.08

40

45

50

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 1000N

0.06

Nodal displacement (m)

35

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

30

Time (sec) b) Displacement of node 4 Figure 4

35

40

45

50

674 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.05

Force scheme 2 Damping case 1 El Centro ν = 1 Fmax = 15000N

0.04

Nodal displacement (m)

0.03

0.02

0.01

0

−0.01

−0.02

−0.03

−0.04

−0.05

0

5

10

15

20

25

30

Time (sec) a) Displacement of node 4

35

40

45

50

−4

2.5

x 10

Force scheme 2 Damping case 1 El Centro ν = 1 Fmax = 15000N

Modal coordinate response

2

1.5

1

0.5

0

−0.5

−1

−1.5

−2

0

5

10

15

20

25

30

35

40

45

Time (sec) b) Time history response of the real part of the coordinate for mode 4

50

8.5 State-space Formulation for MDOF Systems 675

Maximum control force magnitude

3800

Force scheme 2 Damping case 1 El Centro ν = 1 Fmax = 15000N

3700

3600

3500

3400

3300

3200

3100

3000

2900

1

2

3

Control force number c) Peak values of the control forces Figure 5

4

676 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.06

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

Nodal displacement (m)

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

30

35

40

35

40

45

50

Time (sec) a) Displacement of node 4 −3

Modal coordinate response

3

x 10

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

2

1

0

−1

−2

−3

0

5

10

15

20

25

30

45

Time (sec) b) Time history response of the real part of the coordinate for mode 4

50

8.5 State-space Formulation for MDOF Systems 677

−3

4

x 10

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

Modal coordinate response

3

2

1

0

−1

−2

−3

0

5

10

15

20

25

Time (sec)

30

35

40

45

50

c) Time history response of the real part of the coordinate for mode 2 0.06

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

Modal coordinate response

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

30

35

40

45

Time (sec) d) Time history response of the real part of the coordinate for mode 1

50

678 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

4

Maximum control force magnitude

1.5

x 10

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

1.4

1.3

1.2

1.1

1

0.9

0.8

0.7

0.6

0.5

1

2

3

Control force number e) Peak values of the control forces Figure 6

4

8.5 State-space Formulation for MDOF Systems 679

0.06

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 4000N

Nodal displacement (m)

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

Time (sec)

30

35

40

45

50

Figure 7: Displacement of node 4

Controllability of a particular modal response The governing equation for the response of the k’th mode is given by eqn (8.305). 2λ k q˙k = λ k q k + b f , k F = λ k q k – --------- Φ kT E f F f

(8.337)

k

By definition, the j’th column of E f defines the nodal force pattern corresponding to F j . Premultiplication by Φ kT generates the equivalent modal force for mode k. If Φ kT is orthogonal to the j’th column of E f , F j will have no effect on the response of the k’th mode. Therefore, for mode k to be controllable, the nodal location of the active control forces must satisfy the following constraint Φ kT E f ≠ 0

(8.338)

680 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems When damping is uncoupled, Φ k real, and this constraint is obvious. One cannot apply a control force at a null point of a mode.

Example 8.16: Controllability analysis for a 20 DOF Model Controllability is illustrated using a 20 DOF shear beam model having constant mass and stiffness. Figure 1 shows the first 4 modal displacement profiles which are real in this case since there is no damping. The corresponding internodal displacement profiles are plotted in Fig 2. The latter curves are scaled versions of the element shear deformation profiles. 20

18

Node number

16

14

12

10

8

6

4

No feedback m = 1000kg k = 1700000N ⁄ m

2

1 0 −1

−0.8

−0.6

−0.4

−0.2

0

2 0.2

3

4 0.4

Amplitude Figure 1: Modal displacement profiles

0.6

0.8

1

8.5 State-space Formulation for MDOF Systems 681

20

18

16

Element number

14

12

10

8

6

4

2

0 −0.8

1 −0.6

−0.4

−0.2

Amplitude

0

2 0.2

3

4 0.4

0.6

Figure 2: Scaled modal element shear deformation profiles Suppose the j’th control force system consists of a single force applied at one of the 20 nodes, say node l. If node l is a null point for a particular mode, the j’th force has no effect on the response of that mode. Considering Fig 1, the first mode has no null points, and it follows that this mode can be controlled by applying the force at any of the 20 nodes. The second mode, has a null point between nodes 13 and 14. Therefore, these nodes are ineffective. The optimum locations for controlling mode 2 are nodes 7 and 20 which correspond to the maximum amplitudes. Mode 3 has null points adjacent to nodes 8 and 16. Nodes 4,12, and 20 are the optimal force locations. Lastly, mode 4 has null points adjacent to nodes 6, 12, and 18. The corresponding optimal locations are nodes 3, 9, 15, and 20. These results are summarized below in Table 1.

682 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Table 1 Nodal location - Single Control Force Mode Optimal

Not allowed

1

20

1

2

7, 20

13, 14

3

4, 12, 20

8, 16

4

3, 9, 15, 20

6, 12, 18

Another option is for the j’th control force system to consist of a set of selfequilibrating force applied at adjacent nodes, say l and l+1. The product of the k’th modal vector, Φ k , and the j’th column of E f is proportional to the shear deformation for the element located between nodes l and l+1. Therefore, if this element corresponds to a null point for the modal element shear deformation profile, control force system j has no effect on the response of mode k. It follows that null points are to be avoided, and the elements corresponding to the peak values are the optimal locations for the pair of self-equilibrating control forces. Considering element j to be between nodes j and j-1, and the forces are applied at the ends of an element, the constraints and optimal locations for this case follow from Fig 2, and are listed in Table 2. Table 2

Mode

Element location - Self-equilibrating Force Optimal

Not allowed

1

1

20

2

1, 14

7

3

1, 9, 17

5, 13

4

1, 6, 12, 18

3, 4, 9, 10, 15

Observability of a particular modal response Feedback is implemented by observing the response, computing the

8.5 State-space Formulation for MDOF Systems 683 control forces with an algorithm and then applying these forces with actuators. When the general modal formulation is used, F is taken as a linear function of the modal coordinates, (8.339)

F = – kd qR – kv qI

Assuming s modes are considered, each q is of order s x 1. If both terms are retained, 2s observations are required to uniquely define the modal response. Let y 1, y 2, …, y N denote the observations. These observations may be displacements, velocities, strains, strain rates, or some other measured that are linearly related to the state variables. The k’th observation is expressed as U yk = Dk = Dk X U˙

(8.340)

where D k depends on the nature and location of the observation represented by y k . One can interpret D k as defining the nodal variables that y k is monitoring. Substituting for the state vector in terms of the full set of n coordinates leads to the following exact expression for y k n

1 y k = --2

∑

n

D k ( q j V j + q˜ j V˜ j ) =

j=1

∑

D k ( V R, j q R, j – V I, j q I, j )

(8.341)

j=1

If D k is orthogonal to V R, j , q R, j does not contribute to y k and it follows that y k does not observe q R, j . A similar statement applies for q I, j . Therefore, for the j’th mode to be observable, at least one of the set of k D’s cannot be orthogonal to the modal vectors for those modes which are included in the modal approximation for the solution. Assuming s modes are retained and N observations are made at a particular instant in time, eqn (8.341) is approximated by y ≈ DV R q R – DV I q I

(8.342)

where y is of order N x 1; q R, q I are of order s x 1; the j’th row of D contains D j ; and the k’th column of V contains V k . Observability for the modal coordinates selected to approximate the solution requires each column of the products, DV R

684 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems and DV I , to have at least one non-zero term. This constraint can be expressed in terms of the rank of the products. rank ( DV R ) = s rank ( DV I ) = s

(8.343)

Since DV is of order N x s, it follows that N ≥ s . If N = 2s and the rank constraint is satisfied, the solution for q R and q I is unique. Otherwise, a least square type solution has to be generated. When damping is uncoupled, the linear velocity feedback law expressed in terms of modal coordinates simplifies to F = – k v q˙

(8.344)

where q˙ and U˙ are related by U˙ ≈ Φq˙

(8.345)

Suppose the observations involve modal velocities. Expressing y as ˙ y = D' U

(8.346)

and substituting for U˙ results in y = ( D'Φ )q˙

(8.347)

Assuming s modes are retained, observability requires D'Φ j ≠ 0 for j = 1, 2, …, s . This constraint is equivalent to the condition that the observation points are not located at null points for the j’th modal vector. Requiring that rank of D'Φ to be equal to s satisfies the observability condition. Since it also requires N ≥ s , the solution is either unique (N = s), or over-determined.

Example 8.17: Observability analysis for a 20 DOF model Suppose 4 nodes are retained, and the observed variables are nodal velocities. At least 4 observations are necessary when damping is uncoupled. For arbitrary damping, 8 observations are required. If damping is uncoupled, the

8.6 Optimal Linear Feedback - MDOF Time Invariant System 685 constraint on the location of the observation points is that they do not coincide with null of the modal displacement profiles. The most desirable locations are those which coincide with the maximum modal amplitudes. This choice eliminates the possibility of an ill-conditioned set of equations for the modal velocities defined by eqn (8.347). The unacceptable and optimal locations for the velocity observations are listed below. Table 1 Modal response quantity

Nodal location - Single Control Force Optimal

Not allowed

q˙1

20

1

q˙2

7, 20

13, 14

q˙3

4, 12, 20

8, 16

q˙4

3, 9, 15, 20

6, 12, 18

One needs to select at least 4 observation nodes. Taking 7, 9, 12, and 20, and applying eqn (8.347) leads to the following set of equations for the first 4 modal velocities, u˙ 7

0.5114 u˙ 9 = 0.6367 0.7959 u˙ 12 1.0000 u˙ 20

1.0000 0.8792 0.3741 – 0.9941

0.4440 – 0.3019 – 0.9941 0.9824

q˙ – 0.5757 1 – 0.9941 q˙2 0.1528 q˙3 – 0.9650 q˙ 4

(1)

u˙ 7 0.3973 – 0.3375 u˙ 9 0.1761 u˙ 12 – 0.0975 u˙ 20

(2)

Inverting eqn (1) results in q˙1

0.6124 q˙2 = 0.7397 0.8881 q˙3 0.7767 q˙4

– 0.6095 – 0.0967 – 0.7186 – 1.2636

0.8513 0.0262 – 0.2170 0.6343

686 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

8.6 Optimal linear feedback - MDOF time invariant systems The formulation of the linear quadratic regulator problem developed for a time-invariant SDOF system in Section 8.4 is extended here to MDOF systems. Since matrix notation was used for the SDOF case, the only difference between the SDOF and MDOF formulations is the form of the weighting matrices, Q and R, contained in the quadratic performance index. In what follows, the LQR control algorithm corresponding to the modal state-space formulation is derived for both the continuous and discrete time feedback scenarios. The role of Q in this case is to assign relative weights to the different modal responses. Weighting the different control forces is achieved by adjusting the elements of R. Examples illustrating the sensitivity of the modal damping due to feedback as a function of Q and R are presented. Lastly, various issues involved in “selecting” a control force scheme are also discussed. Continuous time modal formulation The governing equation for this formulation is eqn (8.311). For convenience, the relevant equations are listed below. X˙ m = A m X m + B fm F + B pm P + B gm a g

(a)

where X m contains the real and imaginary parts of the modal coordinates, qR Xm = qI

(b)

and the coefficient matrices depend on the modal properties of the MDOF system and the nodal distribution of the control forces defined by E f . Negative linear feedback is taken as q F = – K fm X m = – k d k v R qI

(c)

An expression for K fm is established by requiring the following performance index to be stationary, 1 J = --2

∫

∞

T QX + F T QF ) dt (Xm m 0

(8.348)

8.6 Optimal Linear Feedback - MDOF Time Invariant System 687 where Q and R are taken as diagonal weighting matrices,

Q =

Qd 0 (8.349)

0 Qv

R = [ r i δ ij ] When Q and R are constant, the solution is K fm = R – 1 B Tfm H = k d

kv

(8.350)

where H is determined by T H + H A – H B R –1 B T H = Q Am m fm fm

(8.351)

Just as for the SDOF case, k d ≡ 0 when Q d is taken as a null matrix. This choice avoids the potential instability associated with displacement feedback. Given A m , B fm , R, and Q, one can determine K fm and the eigenvalues of A m – B fm K fm with the MATLAB function CARE. These eigenvalues define the modal damping ratios and frequencies for the system corresponding to the particular choice of control force locations, ( E f ) , and weighting matrices. Given E f , one can generate a range of modal damping ratio distributions by varying the individual weighting factors contained in Q and R. To avoid a potential instability due to displacement feedback, Q d is taken as a null matrix. The form of Q v is a generalized version of eqn (8.163) which applies for the SDOF case. Q v = [ 4m i2 ω i2 w i δ ij ]

(8.352)

where m i is the modal mass, ω i is the modal frequency, and w i is the relative weighting for the i’th mode. With this scaling law, w i and r j are if O(1). Increasing w n places more emphasis on reducing the response of mode n. Increasing r m places more emphasis on reducing the “cost” for the m’th control force. Both of these perturbations result in changes in the magnitudes of the modal damping ratios. In addition to these parameters, the modal damping distribution is also influenced by the nature of the control force scheme. A pair of self-equilibrating control forces is equivalent to “material” damping which tends to produce

688 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems damping that increases with the modal frequency. A single control force applied at a node is similar to mass proportional damping, and results in damping that decreases with the modal frequency. The force schemes considered in Example 8.14 exhibited these behavioral trends. Continuous time LQR control is useful for establishing an initial estimate of the control force system properties required to achieve certain performance objectives such as limiting the peak deformations and accelerations in a structure subject to constraints on the external power requirement and the peak value of the control forces. This “estimate” is then evaluated using the discrete time formulation specialized for the particular time step selected, and the design is modified if necessary. The effect of time delay is also considered at this stage. Discrete time modal formulation The discrete time free vibration formulation allowing for feedback is based on eqn (8.320) which is expressed here as X m, j + 1 = c m1 X m, j + c m2 F j

(8.353)

where c m1 = e c m2 =

A m ∆t

– 1 ( e A m ∆t – Am

(8.354) I )B fm

For the discrete time case, the quadratic performance index is taken as ∞

1 J = --2

∑

T QX T (Xm ,j m, j + F j RF j )

(8.355)

j=0

with Q, R defined by eqn (8.349). Expressing F j as F j = – K fm X m, j

(8.356)

and requiring J to be stationary with respect to K fm leads to the discrete algebraic Riccati equation, –1 T T T Hc T T H – c m1 m1 + ( c m1 H c m2 ) ( R + c m2 H c m2 ) ( c m1 H c m2 ) = Q

(8.357)

8.6 Optimal Linear Feedback - MDOF Time Invariant System 689 and the following expression for the optimal feedback, K fm

optimal

T H c ) –1 ( c T H c ) T = ( R + c m2 m2 m1 m2

(8.358)

This solution can be generated with the MATLAB function DARE. Substituting for F j , the free vibration response with optimal feedback is governed by X m, j + 1 = c m1 – c m2 K fm X j = c m X j opt

(8.359)

The frequency and damping properties of the discrete model with feedback are related to the eigenvalues of c m . Suppose s modes are considered. There are s pairs of complex conjugates, ρ = ρ 1, ρ˜ 1, ρ 2, ρ˜ 2, …, ρ s, ρ˜ s

(8.360)

The j’th pair is expressed in polar form. ρ j ρ˜ j = ρ j e

± iθ j

(8.361)

With this notation, the modal periods and corresponding damping ratios are given by δj ξ j = ------------------------------[ δ 2j + θ 2j ] 1 / 2 2π∆t T j = ------------θj

(8.362)

δ j = – ln ( ρ j ) These equations are useful for comparing discrete vs. continuous feedback. Although the actual control system is based on discrete time feedback, it is convenient to work with the continuous time formulation during the preliminary design phase which is concerned with selecting the location and nature of the control forces, and estimating the relative weighting factors in order to satisfy the specified performance requirements. Shifting from continuous to discrete time feedback with time delay changes the response characteristics, such as the modal damping, and can be potentially destabilizing. For this formulation, the free vibration response becomes unstable when ρ > 1, which corresponds to a

690 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems negative value for ξ . Examples illustrating various behavioral aspects of MDOF systems controlled with the LQR algorithm are presented in the following section. Application studies - LQR control The 4 DOF model defined in Fig 8.15 is utilized to illustrate the design of LQR based control force systems. Versions of this model were used in the preceding examples to demonstrate instability. The focus here is on modal damping introduced by feedback. Both global and self-equilibrating force schemes (Fig 8.15 b, c) are considered. m4

u4

k 4, c 4 m3

F1

F1

F1 F2

u3

k 3, c 3 m2

F2 F3

u2

F2

k 2, c 2 m1

F3 F4

u1

k 1, c 1 a)

b)

c)

Node/Element

m (kg)

k (kN/m)

c (kNs/m)

1

1000

1700

4

2

1000

1400

3

3

1000

1000

2

4

1000

700

1

Fig. 8.15: Definition of model and control force schemes Case 1: Global forcing The scheme shown in (b) involves specifying 2 force weighting parameters ( r 1 and r 2 ) in R and 4 modal velocity weighting parameters ( w 1 → w 4 ) in Q v . Suppose the design objective is a uniform distribution of the peak internodal displacements for a specified dynamic excitation, such as an earthquake. Starting

8.6 Optimal Linear Feedback - MDOF Time Invariant System 691 with uniform weighting, one can adjust the r’s and w’s separately until the desired uniform displacement profile is obtained. The strategy followed here is based on first perturbing the force weights to obtain a reasonable level of damping, then adjusting the modal weights to obtain essentially uniform modal damping, and finally scaling the r’s to shift the average value of the peak internodal displacement to the desired value. Results for the first step are shown in Figure 8.16 a). Starting with w and r set to unity, the r’s are reduced to 0.5 and 0.25. The modal damping for the first mode (period ≈ 0.5 sec) is essentially doubled, while the change in the corresponding damping ratios for the third and fourth modes is small. There is close agreement between the continuous and discrete feedback results for all but the second mode.

Modal damping ratio with feedback

0.09

0.085

0.08

0.075

0.07

0.065

0.06

w = 1 r1 = r2 = 1

0.055

0.05

0.045 0.05

Continuous feedback Discrete feedback 0.1

0.15

0.2

0.25

0.3

Modal period (sec) a)

0.35

0.4

0.45

0.5

692 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.13

Modal damping ratio with feedback

0.12

0.11

0.1

0.09

0.08

0.07

w = 1 r 1 = r 2 = 0.5

0.06

0.05

0.04 0.05

Continuous feedback Discrete feedback 0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

Modal period (sec) b) Modal damping ratio with feedback

0.18

0.16

0.14

0.12

0.1

0.08

w = 1 r 1 = r 2 = 0.25

0.06

Continuous feedback Discrete feedback 0.04 0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

Modal period (sec) c) Fig. 8.16: Sensitivity of the modal damping ratio to the control force weighting factor, r

8.6 Optimal Linear Feedback - MDOF Time Invariant System 693 Taking r = 0.25 as a first trial value for r, the modal weightings are adjusted to increase the damping ratios for the higher modes. Fig 8.17 shows results for a particular w distribution which places the primary emphasis on modes 3 and 4. The modal damping is nearly uniform, but note that there is a significant difference between the continuous and discrete damping ratios. This weighting scheme produces the peak internodal displacement profile plotted in Fig 8.17 b. 0.21

Modal damping ratio with feedback

0.2

0.19

0.18

0.17

w1 = 1 0.16

w2 = 4

0.15

w 3 = 70

0.14

w 4 = 50 r 1 = r 2 = 0.25

0.13

0.12 0.05

Continuous feedback Discrete feedback 0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Modal period (sec) a) Modal damping ratio with continuous and discrete feedback - no saturation limit

694 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

4

Element number

3

El Centro excitation w1 = 1 w2 = 4 2

w 3 = 70 w 4 = 50 r 1 = r 2 = 0.25 Maximum displacement Average displacement

1 0.01

0.0105

0.011

0.0115

0.012

0.0125

0.013

Maximum element internodal displacement (m)

0.0135

b) Element internodal profile Fig. 8.17: Modal damping and peak response for initial weighting scheme Holding the w values constant and decreasing r generates the results shown in Fig 8.18. The average damping in increased, and the average peak displacement is decreased. One continues this process until the “design” displacement value is obtained. In addition to motion based design requirements, there are also constraints on the peak values of the control forces and the peak power required. For this design the peak quantities are: F 1, max = 4.234kN F 2, max = 1.783kN Peak Power = 3.772kNm ⁄ s Average internodal displacement = 0.011m Optimal design involves a consideration of all of these requirements, and assigning priorities for the multiple objectives.

8.6 Optimal Linear Feedback - MDOF Time Invariant System 695

Modal damping ratio with feedback

0.28

0.26

0.24

0.22

w1 = 1 w2 = 4

0.2

w 3 = 70 0.18

w 4 = 50 r 1 = r 2 = 0.125

0.16

Continuous feedback Discrete feedback 0.14 0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Modal period (sec) a) Modal damping ratio with continuous and discrete feedback - no saturation limit

Element number

4

3

El Centro excitation w1 = 1 w2 = 4 2

w 3 = 70 w 4 = 50 r 1 = r 2 = 0.125 Maximum displacement Average displacement

1 0.0085

0.009

0.0095

0.01

0.0105

0.011

0.0115

Maximum element internodal displacement (m) b) Element internodal displacement profile Fig. 8.18:

0.012

696 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Case 2: Self-equilibrating control force The force scheme defined by Fig 8.15c involves 4 control forces. Starting with uniform weightings, the scale factors are adjusted until the modal damping ratio distribution is essentially the same as obtained with the final weighting scheme of Case 1. Figure 8.19 contains the final results for Case 2. Since the modal damping distributions are nearly identical, internodal displacement profiles are also in close agreement. The peak values of the control forces and power for this control force scheme are F 1, max = 3.818kN F 2, max = 5.074kN F 3, max = 4.849kN F 4, max = 4.449kN Peak Power = 3.599kNm ⁄ s Average internodal displacement = 0.0108m This scheme requires a larger force, 5.074kN vs. 4.234kN. The peak power required depends on the equivalent damping, and since this quantity is essentially the same, it follows that the power requirements will also be close.

Modal damping ratio with feedback

8.6 Optimal Linear Feedback - MDOF Time Invariant System 697

0.34

w1 = 8

0.32

w2 = 5

0.3

w 3 = 12

0.28

w 4 = 25

0.26

r 1 = r 2 = r 3 = r 4 = 0.35

0.24

0.22

0.2

0.18

0.16

Continuous feedback Discrete feedback

0.14 0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

Modal period (sec) a) Modal damping ratio with continuous and discrete feedback - no saturation limit

Element number

4

3

El Centro excitation w1 = 8 w2 = 5 2

w 3 = 12 w 4 = 25 r 1 = r 2 = r 3 = r 4 = 0.35 Maximum displacement Average displacement

1 0.0085

0.009

0.0095

0.01

0.0105

0.011

0.0115

0.012

Maximum element internodal displacement (m) b) Element internodal displacement profile Fig. 8.19: Modal damping and peak response - Self-equilibrating control force system

698 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Example 8.18: Control force design studies for a 20 DOF shear beam A 20 DOF shear beam with constant mass (1000kg) and constant element damping (10,000Nsec/m) is considered. An estimate for the element stiffness distribution is selected such that the element shear deformation profile for the first mode is essentially uniform, and the average element relative displacement response due to the El-Centro seismic excitation is approximately 0.0125m. Figure 1 shows the modal properties and response for this choice of stiffness and no feedback control. There is close agreement between the actual and desired deformation for nodes 1 through 13. Beyond this point, the difference increases rapidly and exhibits an exponential type growth pattern, similar to the internodal displacement profiles for modes 3 and 4. This result indictates that the contribution of the higher modes is dominating the response in the upper region.

20

18

Element number

16

14

12

10

8

6

4

2

0

0

0.5

1

1.5

2

2.5

3

3.5 6

x 10

Element stiffness-newtons per meter a) Element shear stiffness distribution Figure 1 Properties for no iteration and no feedback

8.6 Optimal Linear Feedback - MDOF Time Invariant System 699

20

18

16

Node number

14

12

10

8

6

4

2

0 −1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Amplitude b) Modal displacement profile-real part 20

18

16

Element number

14

12

10

8

6

4

2

0 −0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Amplitude c) Element internodal modal displacement profile-real part Figure 1 Properties for no iteration and no feedback

1.2

1.4

700 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.14

Modal damping ratio

0.12

0.1

0.08

0.06

0.04

0.02

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period d) Modal damping ratio without feedback 20 maximum average design value

18

16

Element number

14

12

10

8

6

4

2

0 0.012

0.014

0.016

0.018

0.02

0.022

0.024

Maximum element internodal displacement e) Element internodal displacement profile Figure 1 Properties for no iteration and no feedback

0.026

0.028

8.6 Optimal Linear Feedback - MDOF Time Invariant System 701

Various options are possible. One can alter the deformation profiles for modes 3 and 4 by modifying the stiffness in the upper region so that the gradients are decreased. Another option is to work with the initial stiffness and incorporate addtional damping with feedback control . These approaches generate the maximum values for stiffness (option 1) and damping (option 2). Combining these approaches results in intermediate values for these parameters. Figure 2 contains the results generated by iterating on the element stiffness according to the following algorithm maximum element displacement k new = k old ------------------------------------------------------------------------------------------- desired element displacement

(1)

The computation proceeds as follows. Using the initial stiffness, the time history reponse due to El-Centro is generated, and the peak value of inter-nodal displacement is determined for each element.Equation (1) is applied to update each element stiffness, and the complete analysis is then repeated. Covergence is quite rapid for this example. After 2 cycles, the correction process has essentially reached the final state. Figure 2c shows the modified element displacement profiles after 2 iterations. The peak value for element 20, the most critical location, has been reduced by approximately 33%. This correcton results in a significant improvement in the element reponse profiles plotted in Fig 2d. The “bulge” in the upper region has been eliminated.

702 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

20 initial iterated 18

16

Element number

14

12

10

8

6

4

2

0

0

0.5

1

1.5

2

2.5

3

3.5

4 6

x 10

Iterated element stiffness-newtons per meter (a) Two iterations 20 initial iterated 18

16

Element number

14

12

10

8

6

4

2

0

0

0.5

1

1.5

2

2.5

3

3.5

4 6

x 10

Iterated element stiffness-newtons per meter (b) Four iterations Figure 2 Iterated stiffness with no feedback

8.6 Optimal Linear Feedback - MDOF Time Invariant System 703

20

18

16

Element number

14

12

10

8

6

4

2

0 −0.4

−0.2

0

0.2

0.4

0.6

0.8

Amplitude-internodal modal displacement profile (real part) (c) Two iterations Figure 2 Iterated stiffness with no feedback

1

704 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

20 maximum average design value

18

16

12

10

8

6

4

2

0 0.0115

0.012

0.0125

0.013

Maximum element internodal displacement (d) Two iterations 20 maximum average design value

18

16

Element number

Element number

14

14

12

10

8

6

4

2

0 0.0118

0.012

0.0122

0.0124

0.0126

0.0128

Maximum element internodal displacement ( e) Four iterations Figure 2 Iterated stiffness with no feedback

0.013

8.6 Optimal Linear Feedback - MDOF Time Invariant System 705 Following the second approach, 4 self-equilibrating pairs of nodal forces are applied on elements 17,18,19 ,and 20. Figure 3 defines the notation for these forces. Results based on the initial stiffness and the following set of weighting coeffcients are plotted in Fig 4. w 1 = 0 w 2 =1 w 3 = 3 w 4 = 5 r 1 = r 2 = r 3 = r 4 = 0.125 These coefficients were selected to focus the control mainly on modes 3 and 4. Even though the modal damping for these modes is increased significantly, these still is a substantial difference between the actual and desired response in the upper zone.

20

F1

19

F1

F2

F2

F3

18 17 16

1

Fig 3

F3 F4

F4

706 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.5 continuous discrete−Riccati no feedback

0.45

Modal damping ratio

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period a) Modal damping ration with and without continuous or discrete feedback-no Saturation limit 20 maximum average design value

18

16

Element number

14

12

10

8

6

4

2

0 0.011

0.012

0.013

0.014

0.015

0.016

0.017

Maximum element internodal displacement b) Element internodal displacement profile Figure 4 Initial stiffness and feedback

0.018

0.019

8.6 Optimal Linear Feedback - MDOF Time Invariant System 707

Maximum control force magnitude

3800

3700

3600

3500

3400

3300

3200

3100

3000

1

2

3

4

Control force number c) Maximum value of the control force Figure 4 Initial stiffness and feedback Results based on iterating once on the stiffness, and then applying feedback to the “modifed ” system are plotted in Fig. 5. The weighting coefficients for this case are: w 1 = 0 w 2 =1 w 3 = 3 r 1 = r 2 = r 3 = r 4 = 1.0

w4 = 5

(3)

Increasing the r values reduces the equvialent damping, as shown in Fig 5a. However since the stiffness was corrected, the net effect is a significantly improved element displacement profile and lower magnitudes for the control forces.

708 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.14 continuous discrete−Riccati no feedback

Modal damping ration

0.12

0.1

0.08

0.06

0.04

0.02

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period a) Modal damping ration with and without continuous or discrete feeback-no saturation limit 20 maximum average design value

18

Element number

16

14

12

10

8

6

4

2

0 0.0114

0.0116

0.0118

0.012

0.0122

0.0124

0.0126

0.0128

Maximum element internodal displacement Element internodal displacement profile Figure 5 Single iteration on stiffness and then feedback

0.013

8.6 Optimal Linear Feedback - MDOF Time Invariant System 709

1500

Maximum control force magnitude

1490

1480

1470

1460

1450

1440

1430

1420

1

2

3

4

Control force number c) Maximum value of the control forces 20 initial iterated 18

16

Elememt number

14

12

10

8

6

4

2

0

0

0.5

1

1.5

2

2.5

3

3.5

4 6

x 10

Iterated element stiffness-newtons per meter d) Iterated element shear stiffness distribution Figure 5 Single iteration on stiffness and then feedback

710 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Applying a single control force at node 20 produces the results shown in Fig 6. The corresponding weights and peak value of the control force are w 1 = 0 w 2 =1 w 3 = 3 r1 = 3 F 1 max = 1719N

w4 = 5

(4)

0.1 continuous discrete−Riccati no feedback

0.09

0.08

Element number

0.07

0.06

0.05

0.04

0.03

0.02

0.01

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period a) Modal damping ratio with and without continuous or discrete feedback-no saturation limit Figure 6 Single iteration on stiffness and a single control force

8.6 Optimal Linear Feedback - MDOF Time Invariant System 711

20 maximum average design value

18

16

Element number

14

12

10

8

6

4

2

0 0.0114

0.0116

0.0118

0.012

0.0122

0.0124

0.0126

Maximum element internodal displacement b) Element internodal displacement profile Figure 6 Single iteration on stiffness and a single control force This example illustrates that there is no unique solution. One can vary the stiffness, damping, and active feedback control scheme to adjust the response. In order to determine the “optional”solution, cost measures need to be assigned to each of the paremeters. Simulation studies, as illustrated here, provide the data on sensitivities which allows for a more informed decision as to the final design.

Example 8.19: Alternate choice of response measures The previous examples are based on the specification of either nodal velocities or the first derivative of the modal coordinates as the response measures included in the performance index for the LQR control algorithm. Other response measures, such as element shear deformation rate, can also be selected. One needs only to specify the relationship between the alternate measures and the state vector, and specify weights for the alternate measures. Defining Y as the (1)

712 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems vector containing the alternate measures, the relationship is expressed as either: Y = DX (2) or Y = DX m (3) depending on whether the nodal or modal formulation is used. Taking the index T as Y Q 0 Y and substituting for Y leads to T

Q = D Q0 D The remaining steps are the same as for the standard formulation. This formulation is applied to the system considered in example 8.18. The reponse measures are taken as the shear deformation rates for elements 18,19, and 20. Uniform weighting ( w 1 = w 2 = w 3 = 1 ) is used for the element deformations. The location of the control force is the same as for example 8.18. Figure 1 shows some of the results for this case. These plots correspond to Fig 6 of example 8.18. The peak value of the control force is 1200 N for Fig1b vs. 1719 N for Fig 6. 0.1 continuous discrete−Riccati no feedback

0.09

Modal damping ratio

0.08

0.07

0.06

0.05

0.04

0.03

0.02

0.01

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period a) Modal damping ratio with and without continuous or discrete feedback-

8.6 Optimal Linear Feedback - MDOF Time Invariant System 713 no saturation limit Figure 1 Solution based on using deformation rates for elements 18,19,20 and a single control force at node 20 . 20 maximum average design value

18

16

Element number

14

12

10

8

6

4

2

0 0.01

0.0105

0.011

0.0115

0.012

0.0125

0.013

0.0135

Maximum element internodal displacement b) Element internodal displacement profile Fig 1 Solution based on using deformation rates for elements 18,19,20 and a single control force at node 20.

714

Introduction to Structural Motion Control

Problems 715

Problems Problem 8.1 Refer to eqn (8.28). Consider a SDOF system which is initially at rest ( u 0 = u˙ 0 = 0 ). a) Determine u ( t ) for the case of a sinusoidal ground acceleration, a g = aˆ g sin Ωt . b) Discuss how you would develop a numerical procedure for evaluating at various times such as t1 , t2 , ... etc. u Problem 8.2 Show that the error in the Pade approximation, (eqn 8.51), is of order ( ( λ td ) . 3

e

– λt d

1 1 – --- ( λt d ) 2 = -------------------------- + Error 1 1 + --- ( λt d ) 2

Problem 8.3 Consider a single degree-of-freedom system with continuous pure velocity feedback. Suppose the natural period is 1 second. Use eqn(8.54) to determine the maximum allowable time delay corresponding to the following values for ξ and ξa : ξ = 0.05 ξ a = 0.05, 0.10, 0.20, 0.30 Problem 8.4 Verify eqn(8.77). Problem 8.5 Consider eqn(8.73). Integrating this equation between

t j and t j + 1 leads

716 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems to tj + 1

X j + 1 –X j =

∫

( AX + B f F + B g a g + B p p) dt

tj

Suppose the integrand is assumed to vary linearly over the time interval, and the coefficient matrices A, B f, … ,are constant. a) Derive the expression for X j + 1 corresponding to these conditions. Compare this result with eqn(8.77). Comment on the nature of the error. b) Specialize a) for negative linear feedback, and compare with eqn(8.82) . c) Specialize b) for no time delay and free vibration response. Compare with eqn(8.86). Define the stability requirement for this approximation. Problem 8.6 A SDOF system has the following properties: m = 1000 kg k = 70, 000 N ⁄ m c = 320 N ⋅ s ⁄ m Using the data contained in example 8.2, estimate the limiting time step for each of the following values of the feedback parameters: kd = 0 k v = 0, 800, 1600, 3200, 6400 N⋅ s ⁄ m Problem 8.7 Apply numeral simulation to eqn(8.117) and establish an estimate for the stability limit of the following systems: m = 1000 kg k = 60, 000N ⁄ m c = 750 N⋅ s ⁄ m kd = 0 k v = 3000 N ⋅ s ⁄ m υ = 0, 1, 2

Problems 717

Example 8.3 contains data for c=0. The STABILITY option of MOTIONLAB solves eqn (8.117) for given values of the system properties and the time ratio, ∆t ⁄ T . Problem 8.8 Refer to Example 8.3. Extend Table 1 to include ξ = 0.05 . Consider υ = 0, 1, 2 and ξ a = 0, 0.1, 0.2, 0.3 . Problem 8.9 Verify eqn(8.158) using the CARE function of MATLAB. Take: m = 1000 kg c = 1

k = 60, 000N ⁄ m

c = 750N ˙ s ⁄ m

q v = 4ω 2 m 2 q˙˙v

qd = 0

and consider q v to have the following values: 0,0.5,1.0. Problem 8.10 Refer to eqn (8.160).Will the LQR algorithm ever produce an unstable system? Problem 8.11 Consider eqn(8.168). Let D = Q + KTf RK f n

Sn =

∑C

j, T

DC

j

j=0

Noting the identity, T

C Sn C – Sn = – D + C

n + 1, T

DC

and the limit condition, j

C →0

as

j→∞

n+1

718 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems derive eqn(8.169) Problem 8.12 Refer to Figs 1 and 2 of example 8.4. Suppose the time ratio ∆t ⁄ T is determined by the external loading, and is equal to 0.1. Suggest a value for q v such that ξ a is close to 0.2 when ξ = 0.05 . Problem 8.13 Consider the following system m = 1000 kg k = 60, 000 N ⁄ m c=1000 N.s ⁄ m Suppose ∆t = 0.02s . Select the parameters for discrete time feedback control such that the effective damping ratio is equal to 0.2, Use Figs 1 and 2 of example 8.4 to obtain an initial estimate, and the discrete time Riccati equation option of MOTIONLAB to refine the estimate. Note that the solution tends toward the continuous time feedback case as ∆t ⁄ T approaches 0. Problem 8.14 Refer to example 8.5. Compare the expressions for k d and k v corresponding to q d = 0 with the continuous time Riccati solution defined by eqn(8.161). Use the discrete time Riccati solution for ξ = 0.02 listed in example 8.4 to compare the values of q v required to produce ( k v ⁄ 2ωm ) = 0.2 for 2 time increments, ∆t ⁄ T = 0.02 and 0.1. Problem 8.15 Rework Problem 8.13 using the finite interval discrete time algebraic Riccati equation. Note that the weighting factors for the finite interval formulation are different than the corresponding weighing factors for the discrete time algebraic Riccati equation. Problem 8.16 Verify eqn(8.236).

Problems 719 Problem 8.17 Refer to eqn(5) of example 8.7. Show that the eigenvetors are real when K and M are symmetric and the eigenvalues are positive real quantities when K and M are positive definite. Problem 8.18 T

Show that W j V k = 0 for j ≠ k by substituting for W and V using eqn (8.267) and noting the definition equation for φ j , λ j . 2

λ j Mφ j + λ j Cφ j + Kφ j = 0 Problem 8.19 Refer to eqn(2) of example 8.11. Reduce these equations to a single equation in terms of q R and compare with the equation obtained with the conventional modal formulation(e.g, see eqn(8.275). Is there any advantage to working with the state-space formulation vs the conventional modal formulation? Problem 8.20 Consider a 20 DOF system having the following constant properties: mass = 1000 kg element stiffness = 1800 kN ⁄ m element damping = 25 kN ⋅ s ⁄ m a) Determine the modal properties (period, damping, modal displacement profile, modal intermodal displacement profile) for the first 4 modes. b) Determine the response for node 20 due to the Kobe ground excitation using the complete state-space formulation. Take ∆t = 0.02 sec. c) Repeat part b) using the modal state-space formulation and only the first 4 modes. Compare the time history and Fourier components for node 20 with the corresponding results obtained in b). Also compare the peak values of the modal coordinates.

720 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Problem 8.21 Consider a SDOF system having the following properties: m = 10, 000 kg k = 400, 000 N ⁄ m c=2500 N ⋅ s ⁄ m a) Using the LQR control algorithm, establish values for the weighting parameters such that the effective damping for continuous velocity feedback has the following values: ξ eq = 0.05, 0.1, 0.2 . b) Take ∆t = 0.02 seconds. Evaluate the discrete time damping ratios for a). c) Using the finite time interval control algorithm, establish values for the weighting parameters such that the discrete time damping ratios are the same as found in b). d) Using the model properties corresponding to ξ eq = 0.1 established in a), determine the maximum values of the displacement, control force, and power associated with the El-Centro accelerogram. Use ∆t = 0.02 . e) Repeat d) for the Kobe accelerogram. f) Repeat d) for the Mexico City 1 accelerogram. Problem 8.22 Consider a 4 DOF system having the following properties: Node/Element

m (kg)

k (kN/m)

c (kNs/m)

1

1000

1700

4

2

1000

1400

3

3

1000

1000

2

4

1000

700

1

Suppose a single control force is applied at the top node. Using the LQR algorithm, select the weighting parameters which result in a value of the damping ratio for discrete feedback equal to 0.2 for the first mode. Take ∆t = 0.02 seconds, and apply the following strategies:

Problems 721 a) Use the conventional state-space formulation and weight the nodal velocities uniformly. b) Use the modal state-space formulation and weight the first derivative of the modal coordinates uniformly. c) Use the conventional state-space formulation and weight the element deformation rates uniformly. d) Repeat c) using the modal state-space formulation. Problem 8.23 Consider a 5 DOF shear beam with the following constant mass and stiffness properties: m = 10, 000 kg k = 350, 000 N ⁄ m

a) Assuming uniform element viscous damping, determine the magnitude of element damping such that the first mode damping ratio is 0.02. b) Apply a single control force at mode 5. Assuming all 5 modes are retained, and they are weighted equally, determine the weighting parameters such that the equivalent damping for continuous feedback is 0.15 for the first mode. c) Determine the corresponding damping ratio for discrete time feedback. Take ∆t = 0.02 seconds. d) Investigate the effect of delay on the free vibration reponse of the modal coordinates due to an initial displacement. Use ∆t = 0.02 seconds and the parameters established in part b. Problem 8.24 Consider the following 5 DOF systems: Node/Element

m (kg)

k (kN/m)

c (kNs/m)

1

1000

2000

5

2

1000

1700

5

722 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Node/Element

m (kg)

k (kN/m)

c (kNs/m)

3

1000

1400

5

4

1000

1000

5

5

1000

700

5

a) Suppose control forces are applied at all 5 nodes. Determine the modal coordinate weighting parameters such that the equivalent damping ratio corresponding to continuous feedback is equal to 0.15 for the first mode. Assume uniform weighting. b) Suppose self-equilibrating sets of control forces are applied on all 5 elements and the weighting is applied to the element deformation time rates. Determine the weights such that the first mode damping ratio is 0.15. Assume uniform weighting. c) Apply the Northridge earthquake to each system and compare 1. the internode displacement profiles 2. the peak power 3. the peak value of the control forces Problem 8.25 Consider a 10 DOF shear beam with constant mass, element stiffness, and element damping. Take m=10000 kg . a) Determine the stiffness and damping constants such that the properties for the first mode are Period = 1 sec ond Damping ratio = 0.02

b) Select an active control force scheme which provides a damping ratio of 0.2 for the first mode. c) Apply the Kobe ground acceleration to the system defined in b). Examine the responses of the first 3 modes. Generate both the time histories and the Fourier components.

Problems 723 d) If the design objective is to have uniform peak element shear deformation throughout the system, what design modifications would you suggest? Illustrate your strategy for the case where the target value of the relative inter-nodal displacement is 0.0125m. Problem 8.26 Consider a 10 DOF shear beam with constant nodal mass equal to 10000kg. a) Select a parabolic distribution of element stiffness and a constant element viscous damping so that the period for the first mode is 1 second, and the modal damping ratio is 0.02. b) Carry out 2 cycles of iteration on the element stiffness using the ElCentro ground excitation and 0.0125m as the desired value of internodal displacement. c) Incorporate active control in the system obtained in part b. Select the weighting parameters such that the modal damping ratios ( for continuous feedback ) for the first 3 modes are approximately equal to 0.15. Consider a global forcing scheme and weight the modal coordinate velocities. d) Repeat c) using self-equilibrating control force schemes and weight the modal coordinate velocities. e) Repeat d) using the internodal element displacements performance measures. (See Example 8.19).

as the

Problem 8.27 Consider the bending beam-outrigger system shown below. Assume the outriggers are inifintely stiff, the beam bending rigidity is constant, and the cables are initially tensioned to a level of T 0 . Suppose the cable tensions can be continuously adjusted to counteract the effect of lateral load.

724 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

C

m1=a( ρ m H) m1

b

b B

T o + ∆T

m1 3H/2

D B, ρ m T o – ∆T u(x ,t)

A

H x

a) Take the lateral displacement and rotation at points B, C as the degreesof-freedom and establish the corresponding matrix form of the equilibrium equations. Work with "lumped" masses, rotatory inertias, and loads. b) Develop the state-space formulation for a) c) Describe how you would implement linear velocity feedback control. d) Suppose the bending rigidity is specified, and the critical dynamic loading is a uniform periodic excitation. Discuss how you would "calibrate" the feedback parameters for the case where the design objective is to limit the maximum acceleration. Illustrate your strategy.

Contents i

Contents Preface

x

Chapter 1

Introduction 1.1 Motivation for structural motion control 1 • Limitations of conventional structural design 1 • Motion based structural design and motion control 3 1.2 Motion versus strength issues for building type structures 4 • Example 1.1: Cantilever shear beam 6 • Example 1.2: Cantilever bending beam 7 • Example 1.3: Quasi-shear beam frame 9 1.3 Design of a single-degree-of freedom system for dynamic loading 11 • Response for periodic excitation 11 • Design criteria 16 • Methodology for acceleration controlled design 17 • Example 1.4: An illustration of acceleration controlled design 20 • Methodology for displacement controlled design 21 • Example 1.5: An illustration of displacement controlled design 21 • Methodology for force controlled design 23 • Example 1.6: Force reduction 25 1.4 Design of a single-degree-of-freedom system for support motion 26 • Response for periodic support motion 26 • Design scenarios 27 • Example 1.7: Controlling acceleration due to ground motion 27 • Example 1.8: Controlling relative motion due to ground motion 28 1.5 Stiffness distribution for a two degree-of-freedom system 30 • Example 1.9: 2DOF system - with equal masses 33 1.6 Control strategies for motion based design 33 1.7 Scope of text 42 Problems 45

ii Contents

Part I: Passive Control

Chapter 2

Optimal stiffness distribution 2.1 Introduction 51 2.2 Governing equations - planar beam 53 • Planar deformation-displacement relations 54 • Optimal deformation and displacement profiles 55 • Equilibrium equations 57 • Force-deformation relations 58 • Example 2.1: Composite sandwich beam 59 • Example 2.2: Equivalent rigidities for a discrete truss beam 60 • Governing equations for buildings modelled as pseudo shear beams 64 2.3 Stiffness distribution for a continuous cantilever beam under static loading 69 • Example 2.3: Cantilever beam - quasi static seismic loading 71 • Example 2.4: Truss-beam revisited 72 2.4 Stiffness distribution for a discrete cantilever shear beam - static loading 75 • Example 2.5: 3DOF shear beam 76 2.5 Stiffness distribution - truss under static loading 77 • Example 2.6: Application of least square approaches 81 • Example 2.7: Comparison of strength vs displacement based design 89 2.6 Stiffness distribution for a cantilever beam - dynamic response 93 • Rigidity distribution - undamped response 93 2.7 Stiffness distribution for a discrete shear beam - dynamic response 98 • Example 2.8: 3DOF shear beam 99 2.8 Stiffness calibration 100 • Governing equation - fundamental mode response of a discrete shear beam 100 • Example 2.9: 3DOF shear beam revisited 101 • Governing equations - fundamental mode response of a continuous beam 105 • Example 2.10: One-dimensional parameters - continuous beam108 • Stiffness calibration - periodic excitation 109 • Example 2.11: 3DOF shear beam revisited again 111 • Example 2.12: Stiffness calibration - continuous beam 113 • Example 2.13: Example 2.12 revisited 114

Contents iii • Stiffness calibration - seismic excitation • Seismic response spectra • Example 2.14: Example 2.12 revisited for seismic excitation • Example 2.15: 5DOF shear beam 2.9 Examples of stiffness distribution for buildings 2.10 Stiffness modification • Iterative procedure - seismic excitation • Multiple mode response • Building examples - iterated stiffness distribution Problems

116 119 128 129 132 139 139 140 143 149

Chapter 3

Optimal passive damping distribution 3.1 Introduction 167 3.2 Viscous, frictional, and hysteretic damping 171 • Viscous damping 171 • Example 3.1: Viscous damper 172 • Friction damping 174 • Hysteretic damping 176 • Example 3.2: Stiffness of a rod hysteretic damper 177 • Example 3.3: Stiffness of two hysteretic dampers in series 178 3.3 Viscoelastic material damping 179 • Example 3.4: Viscoelastic damper 182 3.4 Equivalent viscous damping 184 • Example 3.5: Structural and hysteretic damping comparison seismic excitation 186 • Example 3.6: Determining α d for 3M ISD110 damping material 193 3.5 Damping parameters - discrete shear beam 193 • Damping systems 193 • Rigid structural members - Linear viscous behavior 197 • Example 3.7: Example 2.15 revisited 198 • Rigid structural members - Linear viscoelastic behavior 200 • Example 3.8: 201 • Example 3.9: Example 3.7 revisited 202 • Example 3.10: Viscoelastic damper design 203 • Example 3.11: Hysteretic damper design - diagonal element 205 • Flexible structural members - linear viscoelastic behavior 206 • Example 3.12: Coupled spring-damper model 209 3.6 Damping parameters - truss beam 211 • Linear viscous behavior 213 • Linear viscoelastic behavior 214 3.7 Damping distribution for MDOF systems 215

iv Contents • Multi mode free vibration response • Example 3.13: Eigenvalue problem - 2DOF • Example 3.14: Modal response for nonproportional damping • Stiffness proportional viscous damping • Examples - low rise buildings • Examples - building #4 Problems

216 219 226 229 232 239 245

Chapter 4

Tuned mass damper systems 4.1 Introduction 261 4.2 An introductory example 262 • Example 4.1: Preliminary design of a TMD for a SDOF system 264 4.3 Examples of existing tuned mass damper systems 266 • Translational tuned mass dampers 266 • Pendulum tuned mass damper 272 4.4 Tuned mass damper theory for SDOF systems 277 • Undamped structure - undamped TMD 277 • Undamped structure - damped TMD 280 • Example 4.2: Design of a TMD for an undamped SDOF system293 • Damped structure - damped TMD 293 • Example 4.3: Design of a TMD for a damped SDOF system 299 4.5 Case studies - SDOF systems 300 4.6 Tuned mass damper theory for MDOF systems 307 • Example 4.4: Design of a TMD for a damped MDOF system 313 • Example 4.5: Design of TMD’s for a simply supported beam 315 4.7 Case studies - MDOF systems 322 Problems 331 Chapter 5

Base isolation systems 5.1 Introduction 339 5.2 Isolation for SDOF systems 340 • SDOF examples 340 • Bearing terminology 344 • Modified SDOF model 347 • Periodic excitation - modified SDOF model 349 • Example 5.1: Stiffness factors for prescribed structure and base motion 351 • Seismic excitation - modified SDOF model 352 • Example 5.2: Stiffness parameters - modified SDOF model of

Contents v Building example #2 356 5.3 Design issues for structural isolation systems 357 • Flexibility 357 • Rigidity under low level lateral loads 358 • Energy dissipation/absorption 359 • Modeling of a natural rubber bearing (NRB) 360 • Modeling of a lead rubber bearing (LRB) 364 • Applicability of base isolation systems 367 5.4 Examples of existing base isolation systems 368 • USC University Hospital 369 • Fire Department Command and Control Facility 369 • Evans and Sutherland Manufacturing Facility 370 • Salt Lake City Building 371 • The Toushin 24 Ohmori Building 371 • Bridgestone Toranomon Building 374 • San Francisco City Hall 374 • Long Beach V.A. Hospital 375 5.5 Optimal stiffness distribution - discrete shear beam 376 • Scaled stiffness distribution 376 • Example 5.3: Scaled stiffness for a 4 DOF beam with base isolation 379 • Fundamental mode response 380 • Example 5.4: Example 5.3 revisited 380 • Stiffness calibration for seismic isolation 382 • Example 5.5: Stiffness calibration for Example 5.4 382 5.6 Optimal stiffness distribution - continuous cantilever beam 386 • Stiffness distribution - undamped response 386 • Fundamental mode equilibrium equation 393 • Rigidity calibration - seismic excitation 395 • Example 5.6: Stiffness calibration - Example Building #2 396 5.7 Building design examples 397 • Stiffness distribution on fundamental mode response 397 • Stiffness distribution including the contribution of the higher modes 404 Problems 413

vi Contents

Part II: Active Control

Chapter 6

Introduction to active structural motion control 6.1 The nature of active control 423 • Active control versus passive control 423 • The role of feedback 427 • Computational requirements and models for active control 427 6.2 An introductory example of quasi-static feedback control 428 • Example 6.1: Shape control for uniform loading 431 • Example 6.2: Discrete displacement data 432 6.3 An introductory example of dynamic feedback control 433 • Example 6.3: Illustrative example - influence of velocity feedback 436 6.4 Actuator technologies 441 • Introduction 441 • Force application schemes 442 • Large scale linear actuators 446 • Large scale adaptive configuration based actuators 450 • Small scale adaptive material based actuators 451 6.5 Examples of existing large scale active structural control systems 462 • AMD in Kyobashi Seiwa Building 463 • AVS Control at Kajima Technical Research Institute 465 • DUOX Active-Passive TMD in Ando Nishikicho Building 468 • ABS - 600 Ton Full Scale Test Structure 472 Problems 474

Chapter 7

Quasi-static control algorithms 7.1 Introduction to control algorithms 7.2 Active prestressing of a simply supported beam • Passive prestressing • Active prestressing • Active prestressing with concentrated forces • Example 7.1: A single force actuator • Example 7.2: Two force actuators • A general active prestressing methodology

479 480 480 482 488 488 492 499

Contents vii • Example 7.3: Multiple actuators 503 7.3 Quasi-static displacement control of beams 506 • Continuous least square formulation 506 • Discrete least square formulation 508 • Example 7.4: Cantilever beam - Least square algorithm 510 • Extended least square formulation 516 • Example 7.5: Example 7.4 revisited 518 7.4 Quasi-static control of MDOF systems 521 • Introduction 521 • Selection of measures 522 • Example 7.6: Illustrative examples of observability 523 • Example 7.7: Illustrative examples of controllability 525 • Least square control algorithms 527 • Example 7.8: Example 7.7 revisited 528 • Example 7.9: Example 7.7 revisited with an extended least square algorithm 531 • Problems 534

Chapter 8

Dynamic Control Algorithms 8.1 Introduction 545 8.2 State-space representation - time invariant SDOF system 546 • Governing equations 546 • Free vibration uncontrolled response 548 • General solution - time invariant systems 550 • Example 8.1: Equivalence of equations (8.18) and (8.24) 551 • Stability criterion 552 • Linear negative feedback 553 • Effect of time delay on feedback control 555 • Stability analysis for time delay 560 8.3 Discrete time formulation - SDOF systems 566 • Governing equation 566 • Linear negative feedback control 568 • Stability analysis time invariant linear feedback control 569 • Example 8.2: Stability analysis-SDOF system with no time delay 575 • Example 8.3: Stability analysis - SDOF system with time delay 584 8.4 Optimal linear feedback - time invariant SDOF systems 586 • Quadratic performance index 586 • An example - linear quadratic regulator control algorithm 588 • The continuous time algebraic Riccati equation 593

viii Contents • The discrete time algebraic Riccati equation 598 • Example 8.4: Expanded form of discrete algebraic Riccati equation for an undamped SDOF system 600 • Finite interval discrete time algebraic Riccati equation 609 • Example 8.5: Example revised 610 • Continuous time Riccati different equation 611 • Variational formulation of the continuous time Riccati equation 612 • Example 8.6: Application to scalar case 617 8.5 State-space of formulation for MDOF systems 623 • Free vibration reponse-time invariant uncontrolled system 625 • Example 8.7: Free vibration solution for proportional damping 629 • Example 8.8: General uncoupled damping 630 • Orthogonality properties of the state eigenvectors 631 • Example 8.9: Initial conditions-free vibration response 632 • Determination of W and f j 633 • General solution-time invariant system 634 • Modal state space formulation-uncoupled damping 635 • Modal state space formulation-arbitrary damping 639 • Example 8.10: Modal formulation-undamped case 642 • Example 8.11: Modal formulation-uncoupled damping 643 • Example 8.12: Modal parameters-4DOF system 644 • Example 8.13: Modal response for example 8.12 651 • Example 8.14: Modal response response with feedback for example 8.12 654 • Stability analysis-discrete modal formulation 659 • Example 8.15: Stability analysis for example 8.14 662 • Controllability of a particular modal response 679 • Example 8.16: Controllability analysis for a 20 DOF Modal 680 • Observability of a particular modal response 682 • Example 8.17: Observability analysis for a 20 DOD Modal 684 8.6 Optimal linear feedback-MDOF time invariant systems 686 • Continuous time modal formulation 686 • Discrete time modal formulation 688 • Application studies - LQR control 690 • Example 8.18: Control force design studies for a 20 DOF shear beam 698 • Example 8.19: Alternate choice of response measures 711 Problems 714

Contents ix

References

725

Bibliography

735

Index

737

x

Introduction to Structural Motion Control

1

Chapter 1

Introduction 1.1 Motivation for structural motion control Limitations of conventional structural design The word, design, has two meanings. When used as a verb it is defined as the act of creating a description of an artifact. It is also used as a noun, and in this case, is defined as the output of the activity, i.e., the description. In this text, structural design is considered to be the activity involved in defining the physical makeup of the structural system. In general, the “designed” structure has to satisfy a set of requirements pertaining to safety and serviceability. Safety relates to extreme loadings which are likely to occur no more than once during a structure’s life. The concerns here are the collapse of the structure, major damage to the structure and its contents, and loss of life. Serviceability pertains to moderate loadings which may occur several times during the structure’s lifetime. For service loadings, the structure should remain fully operational, i.e. the structure should suffer negligible damage, and furthermore, the motion experienced by the structure should not exceed specified comfort limits for humans and motion sensitive equipment mounted on the structure. An example of a human comfort limit is the restriction on the acceleration; humans begin to feel uncomfortable when the acceleration reaches about 0.02g . A comprehensive discussion of human comfort

2 Chapter 1: Introduction criteria is given by Bachmann and Ammann (1987). Safety concerns are satisfied by requiring the resistance (i.e. strength) of the individual structural elements to be greater than the demand associated with the extreme loading. The conventional structural design process proportions the structure based on strength requirements, establishes the corresponding stiffness properties, and then checks the various serviceability constraints such as elastic behavior. Iteration is usually necessary for convergence to an acceptable structural design. This approach is referred to as strength based design since the elements are proportioned according to strength requirements. Applying a strength based approach for preliminary design is appropriate when strength is the dominant design requirement. In the past, most structural design problems have fallen in this category. However, a number of developments have occurred recently which have limited the effectiveness of the strength based approach. Firstly, the trend toward more flexible structures such as tall buildings and longer span horizontal structures has resulted in more structural motion under service loading, thus shifting the emphasis from safety toward serviceability. For instance, the wind induced lateral deflection of the Empire State Building in New York City, one of the earliest tall buildings in the United States, is several inches whereas the wind induced lateral deflection of the World Trade Center tower is several feet, an order of magnitude increase. This difference is due mainly to the increased height and slenderness of the World Trade Center in comparison to the Empire State tower. Furthermore, satisfying the limitation on acceleration is a difficult design problem for tall slender buildings. Secondly, some of the new types of facilities such as space platforms and semi-conductor manufacturing centers have more severe design constraints on motion than the typical civil structure. In the case of microdevice manufacturing, the environment has to be essentially motion free. Space platforms used to support mirrors have to maintain a certain shape to a small tolerance in order for the mirror to properly function. The design strategy for motion sensitive structures is to proportion the members based on the stiffness needed to satisfy the motion constraints, and then check if the strength requirements are satisfied. Thirdly, recent advances in material science and engineering have resulted in significant increases in the strength of traditional civil engineering materials

1.1 Motivation for Structural Motion Control

3

such as steel and concrete, as well as a new generation of composite materials. Although the strength of structural steel has essentially doubled, its elastic modulus has remained constant. Also, there has been some increase in the elastic modulus for concrete, but this improvement is still small in comparison to the increment in strength. The lag in material stiffness versus material strength has led to a problem with satisfying the serviceability requirements on the various motion parameters. Indeed, for very high strength materials, it is possible for the serviceability requirements to be dominant. Some examples presented in the following sections illustrate this point. Motion based structural design and motion control Motion based structural design is an alternate design process which is more effective for the structural design problem described above. This approach takes as its primary objective the satisfaction of motion related design requirements, and views strength as a constraint, not as a primary requirement. Motion based structural design employs structural motion control methods to deal with motion issues. Structural motion control is an emerging engineering discipline concerned with the broad range of issues associated with the motion of structural systems such as the specification of motion requirements governed by human and equipment comfort, and the use of energy storage, dissipation, and absorption devices to control the motion generated by design loadings. Structural motion control provides the conceptional framework for the design of structural systems where motion is the dominant design consideration. Generally, one seeks the optimal deployment of material and motion control mechanisms to achieve the design targets on motion as well as satisfy the constraints on strength. In what follows, a series of examples which reinforce the need for an alternate design paradigm having motion rather than strength as its primary focus, and illustrate the application of structural motion control methods to simple structures is presented. The first three examples deal with the issue of strength versus serviceability from a static perspective for building type structures. The discussion then shifts to the dynamic regime. A single-degree-offreedom (SDOF) system is used to introduce the strategy for handling motion constraints for dynamic excitation. The last example extends the discussion further to multi-degree-of-freedom (MDOF) systems, and illustrates how to deal with one of the key issues of structural motion control, determining the optimal stiffness distribution. Following the examples, an overview of structural motion control methodology is presented.

4 Chapter 1: Introduction

1.2 Motion versus strength issues for building type structures Building configurations have to simultaneously satisfy the requirements of site (location and geometry), building functionality (occupancy needs), appearance, and economics. These requirements significantly influence the choice of the structural system and the corresponding design loads. Buildings are subjected to two types of loadings: gravity loads consisting of the actual weight of the structural system and the material, equipment, and people contained in the building, and lateral loads consisting mainly of wind and earthquake loads. Both wind and earthquake loadings are dynamic in nature and produce significant amplification over their static counterpart. The relative importance of wind versus earthquake depends on the site location, building height, and structural makeup. For steel buildings, the transition from earthquake dominant to wind dominant loading for a seismically active region occurs when the building height reaches approximately 100m . Concrete buildings, because of their larger mass, are controlled by earthquake loading up to at least a height of 250m , since the additional gravity load increases the seismic forces. In regions where the earthquake action is low (e.g. Chicago in the USA), the transition occurs at a much lower height, and the design is governed primarily by wind loading. When a low rise building is designed for gravity loads, it is very likely that the underlying structure can carry most of the lateral loads. As the building height increases, the overturning moment and lateral deflection resulting from the lateral loads increase rapidly, requiring additional material over and above that needed for the gravity loads alone. Figure 1.1 (Taranath, 1988) illustrates how the unit weight of the structural steel required for the different loadings varies with the number of floors. There is a substantial weight cost associated with lateral loading.

1.2 Motion Versus Strength Issues for Building Type Structure 5

140

120

Gravity loads

Lateral loads

Number of floors

100

Floor

Columns

80

60

40

20

0 0

50 500

100 1000

150 1500

Structural steel - N/m

2

200 2000

250 2500

Fig. 1.1: Structural steel quantities for gravity and wind systems To illustrate the dominance of motion over strength as the slenderness of the structure increases, the uniform cantilever beam shown in Fig. 1.2 is considered. The lateral load is taken as a concentrated force p applied to the tip of the beam, and is assumed to be static. The limiting cases of a pure shear beam and a pure bending beam are examined. u

p

d w

H a

a d section a-a

Fig. 1.2: Building modeled as a uniform cantilever beam

6 Chapter 1: Introduction

Example 1.1: Cantilever shear beam The shear stress τ is given by p τ = -----As

(1.1)

where A s is the cross sectional area over which the shear stress can be considered to be constant. When the bending rigidity is very large, the displacement, u , at the tip of the beam is due mainly to shear deformation, and can be estimated as pH u = ----------GA s

(1.2)

where G is the shear modulus and H is the height of the beam. This model is called a shear beam. The shear area needed to satisfy the strength requirement follows from eqn (1.1): As

p ≥ ----strength τ∗

(1.3)

where τ∗ is the allowable stress. Noting eqn (1.2), the shear area needed to satisfy the serviceability requirement on displacement is As

p H ≥ ---- ⋅ -----serviceability G u∗

(1.4)

where u∗ denotes the allowable displacement. The ratio of the area required to satisfy serviceability to the area required to satisfy strength provides an estimate of the relative importance of the motion design constraints versus the strength design constraints As τ∗ H serviceability r = -------------------------------------- = ----- ⋅ -----G u∗ As

(1.5)

strength

Figure 1.3 shows the variation of r with H ⁄ u∗ . Increasing H ⁄ u∗ places

1.2 Motion Versus Strength Issues for Building Type Structure 7 more emphasis on the motion constraint since it corresponds to a decrease in the allowable displacement, u∗ . Furthermore, an increase in the allowable shear stress, τ∗ , also increases the dominance of the displacement constraint. r

*

*

τ2 > τ1

*

τ1

100

200

300

400

H -----u∗

Fig. 1.3: Plot of r versus H ⁄ u∗ for a pure shear beam

Example 1.2: Cantilever bending beam When the shear rigidity is very large, shear deformation is negligible, and the beam is called a “bending” beam. The maximum bending moment M in the structure occurs at the base and equals M = pH

(1.6)

The resulting maximum stress σ is M Md pHd σ = ----- = --------- = ----------S 2I 2I

(1.7)

where S is the section modulus, I is the moment of inertia of the cross-section about the bending axis, and d is the depth of the cross-section (see Fig. 1.2). The corresponding displacement at the tip of the beam u becomes

8 Chapter 1: Introduction 3

pH u = ---------3EI

(1.8)

The moment of inertia needed to satisfy the strength requirement is given by pHd I strength ≥ ----------2σ∗

(1.9)

Using eqn (1.8), the moment of inertia needed to satisfy the serviceability requirement is 3

pH I serviceability ≥ ------------3Eu∗

(1.10)

Here, u∗ and σ∗ denote the allowable displacement and stress respectively. The ratio of the moment of inertia required to satisfy serviceability to the moment of inertia required to satisfy strength has the form 3 I serviceability pH 2σ∗ 2H σ∗ H ------------ ⋅ ----------- = -------- ⋅ ------ ⋅ -----r = ------------------------------- = 3d E u∗ I strength 3Eu∗ pHd

(1.11)

Figure 1.4 shows the variation of r with H ⁄ u∗ for a constant value of the aspect ratio H ⁄ d ( H ⁄ d ≈ 7 for tall buildings). Similar to the case of the shear beam, an increase in H ⁄ u∗ places more emphasis on the displacement since it corresponds to a decrease in the allowable displacement, u∗ , for a constant H . Also, an increase in the allowable stress, σ∗ , increases the importance of the displacement constraint. For example, consider a standard strength steel beam with an allowable stress of σ∗ = 200MPa , a modulus of elasticity of E = 200,000MPa , and an aspect ratio of H ⁄ d = 7 . The value of H ⁄ u∗ at which a transition from strength to serviceability occurs is 3 E d H = --- ⋅ ------ ⋅ ---- ≈ 200 -----2 σ∗ H ∗ u r=1

(1.12)

For H ⁄ u∗ > 200 , r > 1 and motion controls the design. On the other hand, if high

1.2 Motion Versus Strength Issues for Building Type Structure 9 strength steel is utilized ( σ∗ = 400MPa and E = 200,000MPa ) H -----u∗

≈ 100

(1.13)

r=1

and motion essentially controls the design for the full range of allowable displacement.

r

*

*

σ2 > σ1

*

σ1

100

200

300

400

H -----u∗

Fig. 1.4: Plot of r versus H ⁄ u∗ for a pure bending beam

Example 1.3: Quasi-shear beam frame This example compares strength vs. motion based design for a single bay frame of height H and load p (see Fig. 1.5). For simplicity, a very stiff girder is assumed, resulting in a frame that displays quasi-shear beam behavior. Furthermore, the columns are considered to be identical, each characterized by a modulus of elasticity E c , and a moment of inertia about the bending axis I c . The maximum moment, M , in each column is equal to pH M = -------4

(1.14)

10 Chapter 1: Introduction The lateral displacement of the frame under the load is expressed as

u

Ig = ∞ p --2

p --2 Ec , I c

H

Ec , I c

Fig. 1.5: Quasi-shear beam example pH u = -------DT

(1.15)

where D T denotes the equivalent shear rigidity which, for this structure, is given by 24E c I c D T = ----------------2 H

(1.16)

The strength constraint requires that the maximum stress in the column be less than the allowable stress σ∗ pHd Md --------- = ----------- ≤ σ∗ 8I c 2I c

(1.17)

where d represents the depth of the column in the bending plane. Equation (1.17) is written as Ic

pHd ≥ ----------strength 8σ∗

(1.18)

The serviceability requirement constrains the maximum displacement to be less than the allowable displacement u∗

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 11 3

pH ----------------- ≤ u∗ 24E c I c

(1.19)

The corresponding requirement for I c is 3

pH ≥ -----------------Ic serviceability 24E u∗

(1.20)

c

Forming the ratio of the moment of inertia required to satisfy the serviceability requirement to the moment of inertia required to satisfy the strength requirement, Ic σ∗ H H serviceability r = ----------------------------------- = --------- ⋅ ---- ⋅ -----3E c d u∗ Ic strength

(1.21)

leads to the value of H ⁄ u∗ for which motion dominates the design H 3E c d ------ ≥ --------- ⋅ ---u∗ σ∗ H

(1.22)

1.3 Design of a single-degree-of freedom system for dynamic loading The previous examples dealt with motion based design for static loading. A similar approach applies for dynamic loading once the relationship between the excitation and the response is established. The procedure is illustrated for the single-degree-of-freedom (SDOF) system shown in Fig. 1.6. Response for periodic excitation The governing equation of motion of the system has the form mu˙˙( t ) + cu˙ ( t ) + ku ( t ) = p ( t )

(1.23)

12 Chapter 1: Introduction

k R m

p

c u

Fig. 1.6: Single-degree-of-freedom system where m , k , c are the mass, stiffness, and viscous damping parameters of the system respectively, p is the applied loading, u is the displacement, and t is the independent time variable. The dot operator denotes differentiation with respect to time. Of interest is the case where p is a periodic function of time. Taking p to be sinusoidal in time with frequency Ω , p ( t ) = pˆ sin Ωt

(1.24)

the corresponding forced vibration response is given by u ( t ) = uˆ sin ( Ωt – δ )

(1.25)

where uˆ and δ characterize the response. They are related to the system and loading parameters as follows: pˆ uˆ = --- H 1 k

(1.26)

1 H 1 = -------------------------------------------------2 2

[ 1 – ρ ] + [ 2ξρ ]

ω =

k ---m

c c ξ = ------------ = --------------2ωm 2 km

(1.27)

2

(1.28)

(1.29)

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 13 Ω m ρ = ---- = Ω ---ω k

(1.30)

2ξρ tan δ = --------------2 1–ρ

(1.31)

The term pˆ ⁄ k is the displacement response that would occur if the loading were applied statically; H 1 represents the effect of the time varying nature of the response. Figure 1.7 shows the variation of H 1 with the frequency ratio, ρ , for various levels of damping. The maximum value of H 1 and corresponding frequency ratio are related to the damping ratio ξ by H1

1 = ------------------------2 max 2ξ 1 – ξ

(1.32)

5 4.5

ξ = 0.0

4 3.5

H1

3

ξ = 0.2

2.5 2

ξ = 0.4

1.5 1 0.5 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Ω m ρ = ---- = Ω ---ω k Fig. 1.7: Plot of H 1 versus ρ and ξ ρ max =

1 – 2ξ

2

(1.33)

14 Chapter 1: Introduction 2

When ξ << 1 , ρ max ≈ 1 H1

(1.34)

1 ≈ -----max 2ξ

(1.35)

Since ξ is usually small, the maximum dynamic response is significantly greater than the static response and ρ max is close to 1. For example, for ξ = 0.2 which corresponds to a high level of damping, the peak response is H1

max

≈ 2.55

(1.36)

ρ max = 0.96

(1.37)

When the forcing frequency, Ω , is close to the natural frequency, ω , the response is controlled by adjusting the damping. Outside of this region, damping has less influence, and has essentially no effect for ρ < 0.4 and ρ > 1.6 . Differentiating u twice with respect to time leads to the acceleration, a , 2

a ( t ) = u˙˙( t ) = – Ω uˆ sin ( Ωt – δ ) = – aˆ sin ( Ωt – δ )

(1.38)

Noting eqn (1.26), the magnitude of a can be written as pˆ pˆ 2 aˆ = --- Ω H 1 = ---- H 2 m k

(1.39)

where 2

H2 = ρ H1 =

4

ρ ---------------------------------------------2 2 2 [ 1 – ρ ] + [ 2ξρ ]

(1.40)

The variation of H 2 with ρ for different damping ratios is shown in Fig. 1.8. Note that the behavior of H 2 for small and large ρ is opposite to H 1 . The maximum value of H 2 is the same as the maximum value for H 1 , but the location (i.e. the corresponding value of ρ ) is different. They are related to ξ by

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 15 1 ρ max = ---------------------2 1 – 2ξ H2

(1.41)

1 = ------------------------2 max 2ξ 1 – ξ

(1.42)

The ratio pˆ ⁄ m is the acceleration the mass would experience if it were unrestrained and subjected to a constant force of magnitude pˆ . One can interpret H 2 as a modification factor which takes into account the time varying nature of the loading and the system restraints associated with stiffness and damping.

5 4.5

ξ = 0.0

4 3.5

H2

3 2.5

ξ = 0.2

2 1.5 1

1 ξ = ------2

0.5 0 0

0.2

0.4

0.6

0.8

1

1.2

Ω ρ = ---- = Ω ω Fig. 1.8: Plot of H 2 versus

1.4

1.6

1.8

2

m ---k ρ and ξ

Once the system and loading are defined (i.e. m , k , c , pˆ , and Ω are specified), one determines H 1 and computes the peak amplitudes using the following relations pˆ uˆ = --- H 1 k 2

aˆ = Ω uˆ

(1.43)

(1.44)

16 Chapter 1: Introduction H 1 = H 1 ( Ω, m, k, c ) = H 1 ( ρ, ξ )

(1.45)

Note that for periodic response, the acceleration is related to the displacement by the square of the forcing frequency. One can also work with H 2 instead of H 1. Design criteria The design problem differs from analysis in that one starts with the mass of the system, m , and the loading characteristics, pˆ and Ω , and determines k and c such that the motion parameters, uˆ and aˆ , satisfy the specified criteria. In general, one has limits on both displacement and acceleration uˆ ≤ u∗

(1.46)

aˆ ≤ a∗

(1.47)

where u∗ and a∗ are the target design values. In this case, since uˆ and aˆ are related by 2

aˆ = Ω uˆ

(1.48)

2 one needs to determine which constraint controls. If a∗ ≤ Ω u∗ , the acceleration limit controls and the optimal solution will be

a∗ uˆ = ------- < u∗ 2 Ω

(1.49)

aˆ = a∗

(1.50)

2 If, on the other hand, a∗ ≥ Ω u∗ , the displacement limit controls. For this case, the optimal solution satisfies

uˆ = u∗

(1.51)

2 aˆ = Ω u∗ < a∗

(1.52)

In what follows, both cases are illustrated.

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 17 Methodology for acceleration controlled design One works with eqn (1.39). Expressing the target design acceleration as a function of the gravitational acceleration, a∗ = fg

(1.53)

and defining H 2∗ as a∗ H 2∗ = ----------pˆ ⁄ m

(1.54)

the design constraint takes the form W H 2 ≤ H 2∗ = ----- f pˆ

(1.55)

where W is the weight of the system. The totality of possible solutions is contained in the region below H 2 = H 2∗ . Figure 1.9 illustrates the region for H 2∗ = 2 . For low damping, the intersection of H 2 = H 2∗ and the H 2 curve for a particular value of ξ , ξ∗ , establishes two limiting ρ values, ρ 1 [ H 2∗, ξ∗ ] and ρ 2 [ H 2∗, ξ∗ ] . Permissible values of ρ for the damping ratio ξ∗ , are

18 Chapter 1: Introduction

5 4.5

ξ = 0.0

4 3.5

H2

3 2.5 2

ξ = ξ∗ H 2∗ possible solutions

1.5 1 0.5 0 0

ρ1 0.2

0.4

0.6

0.8

ρ2 1

Ω ρ = ---- = Ω ω Fig. 1.9: Possible values of

1.2

1 ξ = ------2

1.4

1.6

1.8

2

m ---k H 2 ≤ H 2∗

0 < ρ ≤ ρ 1 [ H 2∗, ξ∗ ] ρ ≥ ρ 2 [ H 2∗, ξ∗ ]

(1.56)

The second region does not exist when H 2∗ < 1 . Noting eqn (1.40), the expressions for ρ 1 and ρ 2 are

ρ 1, 2 =

2 2 2 1 2 1 – 2ξ∗ − + [ 1 – 2ξ∗ ] – 1 + ---------H 2∗ ---------------------------------------------------------------------------------------------1 2 1 – ---------H 2∗

These functions are plotted in Fig. 1.10 for representative values of ξ∗ . The limiting values of ρ for ξ∗ = 0 reduce to

(1.57)

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 19

1 -------------------1 1 ± ---------H 2∗

ρ 1, 2 =

(1.58)

2

ρ2

1.8

ξ∗ = 0.2

1.6

ξ∗ = 0.1

1.4

ξ∗ = 0

1.2 1

m ρ = Ω ---k

0.6

ρ1

0.4 0.2 0 0

1

H 2∗ = 2

0.8

2

3

4

5

6

H2 Fig. 1.10: Plot of ρ 1 and ρ 2 versus ξ∗ and H 2 Noting eqn (1.29), one can express eqn (1.56) in terms of limiting values of stiffness. By definition, 2

Ω m k = -----------2 ρ

(1.59)

Then, letting 2

Ω m k j = -----------2 ρj

(1.60)

and noting that k 2 < k 1 , the allowable ranges for k are given by: H 2∗ < 1

k1 < k < ∞

(1.61)

20 Chapter 1: Introduction H 2∗ > 1

0 < k < k 2 and k 1 < k < ∞

(1.62)

Given H 2∗ , one specifies a value of ξ , computes ρ j with eqn (1.57), and selects a value for k which satisfies the above constraints on stiffness. The damping parameter is determined from (1.63)

c = 2ξωm = 2ξ km

Example 1.4: An illustration of acceleration controlled design Suppose pˆ = 0.1W and a∗ = 0.05g . Applying eqn (1.54) leads to H 2∗ = 0.5 . Figure 1.10 shows that damping has a negligible effect for this value of H 2∗ ; the design is essentially controlled by stiffness. Taking ξ = 0 , and using eqn (1.58) results in 2 1 ρ 1 = --3

2

k > 3Ω m To illustrate the other extreme, pˆ = 0.1W and a∗ = 0.2g is considered. Here, H 2∗ = 2.0 . The two allowable regions for k corresponding to different values of ξ are obtained by applying equations (1.57), (1.60), and (1.62). k ≥ k1

k ≤ k2 ξ

k1/Ω2m

k2/Ω2m

c1/Ω m

c2/Ω m

0

1.5

0.5

0

0

0.1

1.439

0.521

0.24

0.144

0.2

1.231

0.610

0.444

0.312

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 21 Methodology for displacement controlled design The starting point is eqn. (1.26). Noting equations (1.40) and (1.59), eqn (1.26) can be written as pˆ pˆ pˆ 2 uˆ = --- H 1 = ------------ ρ H 1 = ------------ H 2 2 2 k Ω m Ω m

(1.64)

**

Then, defining H 2 as ** H2

2

*

Ω mu = -----------------pˆ

(1.65)

*

where u is the target displacement, the design constraint is given by **

H2 < H2

(1.66)

The remaining steps are the same as for the previous case; the only difference is **

**

the definition of H 2 . One applies equations (1.57) thru (1.63), using H 2 instead of H 2∗ .

Example 1.5: An illustration of displacement controlled design Suppose pˆ = 10kN *

u = 10cm m = 1000kg Then 2 ( 1000 ) ( 0.1 ) 2 H 2 ** = ----------------------------- Ω = ( 0.01 )Ω 10000

22 Chapter 1: Introduction Various values for Ω are considered. 1. Ω = 2π r ⁄ s H 2 ** = 0.394 **

For this value of H 2 , the design is controlled by stiffness. Taking ξ = 0 , and using eqn (1.58), 1 1 ------ = 1 + -------- = 3.538 2 ** ρ1 H2 2

Ω m k 1 = ------------ = 139.7kN ⁄ m 2 ρ1 The value of k is constrained by k ≥ k 1 = 139.7kN ⁄ m which corresponds to ρ ≤ ρ 1 = 0.532 2. Ω = 4π r/s H 2 ** = 1.576 **

Since H 2 is greater than 1, there are 2 allowable regions for k . Also these regions depend on the damping ratio, ξ . Results for different values of ξ are listed below. They are generated using equations (1.57), (1.60), and (1.62). ρ < ρ1

k ≥ k1

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 23 ρ > ρ2

k ≤ k2

ξ

ρ1

k1(kN/m)

c1 (kN*s/m)

ρ2

k2(kN/m)

c2 (kN*s/m)

0

0.782

258.1

0

1.654

57.72

0

0.1

0.795

249.6

3.16

1.627

59.63

1.54

0.2

0.839

224.1

5.98

1.541

66.5

3.26

The solution for k ≤ k 2 is π radians out of phase with the forcing function. Decreasing k reduces the natural frequency, ω , and moves the system away from the resonant zone, ρ = 1 . For small k , H 2 approaches unity, and the response measures tend toward the following limits pˆ aˆ → ---m

pˆ uˆ → ---------------2 Ω ⋅m

(1.67)

Methodology for force controlled design The previous examples dealt with limiting the displacement or acceleration of the single degree of freedom mass. Another design scenario is associated with the concept of isolation, i.e., one wants to limit the internal force that is generated by the applied force and transmitted to the support. The reaction force, R , shown in Fig (1.6), is given by R = p – ma = ku + cu˙

(1.68)

Expressing R as R = Rˆ sin ( Ωt – δ + δ 1 )

(1.69)

and using equations (1.24) thru (1.31), the magnitude and phase shift are given by

24 Chapter 1: Introduction ˆ = H pˆ R 3

(1.70) 2

1 + [ 2ξρ ] ---------------------------------------------2 2 2 [ 1 – ρ ] + [ 2ξρ ]

H3 =

(1.71)

tan δ 1 = 2ξρ

(1.72)

5

*

ξ = 0

4.5 4 3.5

H3

3 2.5

*

ξ = 0.2

2

*

ξ = 0.4

1.5 1 0.5 0 0

2 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

m ρ = Ω ---k Fig. 1.11: Plot of H 3 versus ρ and ξ Figure 1.11 shows the variation of H 3 with ρ and ξ . At ρ = 2 , H 3 = 1 for all values of ξ . When ρ > 2 , the minimum value of H 3 corresponds to ξ = 0 , which implies that damping magnifies rather than decreases the response in this region. The strategy for reducing the reaction is to take the stiffness as 2

k<Ω m⁄2

(1.73)

Decreasing k “softens” the system and reduces the internal force. However, the displacement and acceleration “increase”, and approach the limiting values given by eqn (1.67)

1.3 Design of a Single-degree-of Freedom System for Dynamic Loading 25

Example 1.6: Force reduction Suppose pˆ = 10kN m = 1000kg Ω = 4π r ⁄ s and the desired magnitude of the reaction is 10% of the applied force. Then, H 3 = 0.10 Taking ξ = 0 in eqn (1.71) 1 H 3 = ------------------ = 0.1 1 – ρ2 results in ρ =

11 = 3.317

Finally, the corresponding stiffness is 2

Ω m k = ------------ = 14.36kN ⁄ m 2 ρ

26 Chapter 1: Introduction

1.4 Design of a single-degree-of-freedom system for support motion Response for periodic support motion

k m c ug

u + ug

Fig. 1.12: Single-degree-of-freedom system The equation of motion for a SDOF system subjected to support motion is written as mu˙˙( t ) + cu˙ ( t ) + ku ( t ) = – mu˙˙g ( t )

(1.74)

where u is now considered to be the relative displacement with respect to the support, and u g denotes the motion of the support. Assuming periodic excitation u g ( t ) = uˆ g sin Ωt

(1.75)

the relative and total displacements are given by u ( t ) = uˆ sin ( Ωt – δ ) u t ( t ) = u ( t ) + u g ( t ) = uˆ t sin ( Ωt + δ 1 – δ )

(1.76) (1.77)

where uˆ = H 2 uˆ g

(1.78)

uˆ t = H 3 uˆ g

(1.79)

The total acceleration is related to the support acceleration by a similar expression u˙˙t ( t ) = a t ( t ) = aˆ t sin ( Ωt + δ 1 – δ )

(1.80)

1.4 Design of a Single-degree-of Freedom System for Support Motion 27 2

aˆ t = H 3 [ – Ω uˆ g ] = H 3 aˆ g

(1.81)

Design scenarios There are a number of design scenarios for support motion. If the system is sensitive to the total acceleration, one can limit the response by requiring H 3 to be small with respect to unity. In effect, one isolates the mass from the support motion. From Fig. 1.11, ρ must be greater than 2 for isolation to be feasible. The 2 corresponding constraint on stiffness is k < mΩ ⁄ 2 . If the support motion is interpreted as seismic excitation, and the system represents a structure, one wants to control the relative motion of the structure and the total acceleration of the equipment attached to the structure. One approach is to isolate the structure so that the acceleration constraint is satisfied. The relative motion will be essentially equal to the ground motion, so an additional mechanism is needed to localize the motion. An example of this approach is a rigid building supported by low stiffness springs; the motion in this approach is confined to the support springs. Another approach would be to work with H 2 so as to satisfy the displacement constraint with stiffness and adjust the damping to satisfy as close as possible the constraint on acceleration. The examples presented below illustrate both approaches. A detailed discussion of isolation of building structures for seismic excitation is contained in Chapter 5.

Example 1.7: Controlling acceleration due to ground motion This problem concerns selecting ρ and ξ such that the total acceleration is a fraction of the support acceleration, say aˆ t = νaˆ g

ν << 1

(1.82)

Then from eqn (1.81) aˆ t H 3 = ----- = ν aˆ g

(1.83)

As pointed out earlier, ξ = 0 yields the minimum value of H 3 . Specializing eqn

28 Chapter 1: Introduction (1.71), one obtains 1 ρ ≥ 1 + --ν

(1.84)

as the lower limit on ρ . The stiffness limit follows from the definition of ρ 2 Ω 2 mΩ k = m ---- ≤ -----------1 ρ 1 + --ν

(1.85) 2

mΩ For example, suppose ν = 1 ⁄ 3 . Then, ρ ≥ 2 and k ≤ ------------ . 4

Example 1.8: Controlling relative motion due to ground motion This example illustrates how to constrain the relative motion due to support movement. A typical scenario is a structure subjected to seismic excitation. Defining u∗ as the desired relative displacement, the appropriate equation follows from eqn (1.78). u∗ H 2 ≤ ------ = H 2∗ uˆ g

(1.86)

For this type of problem, H 2∗ is usually greater than unity. The procedure for solving eqn (1.86) was discussed in the previous section. Considering the solution zone corresponding to ρ > ρ 2 (See Fig. 1.9) and starting with ξ = 0 leads to ρ2 ≈ ρ2

ξ=0

=

1 ------------------1 1 – ---------H 2∗

(1.87)

2

mΩ k ξ = 0 = --------------------------2 ρ2

(1.88)

ξ=0

as the first estimate. The change in ρ 2 and k due to including damping can be evaluated using eqn (1.57).

1.4 Design of a Single-degree-of Freedom System for Support Motion 29 To illustrate the process, the following numerical values are considered. Ω = 4π rd/s

2π T = ------ = 0.5s Ω

u∗ = 0.1m

uˆ g = 0.033m

Applying eqns (1.86) through (1.88), the values of the various parameters are 0.1 H 2∗ = ------------- = 3 0.033

ρ2

ξ=0

=

1 ------------ = 1.22 1 1 – --3

T ξ = 0 = ρ2 T = 0.61s ξ=0

2 4π 2 mΩ k ξ = 0 = --------------------------- = m ---------- = 105.27m 2 1.22 ρ2 ξ=0

For no damping, one has to take k ≤ k ξ = 0 in order to satisfy the constraint on amplification. To assess the sensitivity to damping, ξ∗ is taken as 0.1 . Using eqn (1.57), the corresponding value of ρ 2 is ρ2

ξ = 0.1

= 1.18

The other parameters are determined with T ξ = 0.1 = ρ 2 T = 0.59s ξ = 0.1

30 Chapter 1: Introduction 2

k

mΩ = ------------------------------- = 112.53m ξ = 0.1 2 ρ2 ξ = 0.1

The solution is fairly insensitive to damping. However, if one uses the larger stiffness value, and the damping assumed in the calculation is not realized, the amplification will be greater than the design value, H 2∗ .

1.5 Stiffness distribution for a two degree-of-freedom system The last two examples were concerned with how to determine the stiffness for a SDOF system subjected to dynamic loading when motion is the controlling design criterion. The procedure for a multi-degree-of-freedom (MDOF) system is more involved since one has now to determine a set of stiffnesses rather than just a single stiffness. A two-degree-of-freedom system is considered here to illustrate the approach. The next chapter extends the approach to deal with continuous beam type structures and discrete mass systems having both stiffness and damping. Consider the two degree-of-freedom system shown in Fig. 1.13. The SDOF notation is extended to the MDOF case by including a subscript to indicate the corresponding object. Enforcing equilibrium for the individual masses leads to the following governing equations m 1 u˙˙1 ( t ) + ( k 1 + k 2 )u 1 ( t ) – k 2 u 2 ( t ) = p 1 ( t )

(1.89)

m 2 u˙˙2 ( t ) + k 2 u 2 ( t ) – k 2 u ( t ) = p 2 ( t ) 1

(1.90)

Specializing these equations for free vibration, the solution is expressed as u 1 ( t ) = Φ 1 sin ωt

(1.91)

u 2 ( t ) = Φ 2 sin ωt

(1.92)

1.5 Stiffness Distribution for a Two Degree-of -freedom System 31 p1 ( t ) = p2 ( t ) = 0

(1.93)

where Φ j , are the amplitudes and ω is a frequency measure. Requiring eqns (1.89) and (1.90) to be satisfied for Φ j ≠ 0 yields the following equation for the frequency, k1 + k2 k2 k1 k2 2 λ + – ----------------- – ------- λ + --------------- = 0 m1 m2 m1 m2

(1.94)

where λ = ω 2 .

p2

p1 k1

m1

k2

m2

u1

u2

Fig. 1.13: Two-degree-of-freedom system The displacement amplitudes are related by k 1 + k 2 – λm 1 Φ 2 = ---------------------------------- Φ 1 k2

(1.95)

There are 2 values of λ , and 2 corresponding sets of Φ ’s. Each set of Φ ’s defines a displacement pattern which is called a mode shape. A detailed description of free vibration analysis for multi-degree-of-freedom systems is presented later in chapter 2. When k 1 and k 2 are specified, one solves eqn (1.94) for λ and then determines the displacement amplitude ratio with eqn (1.95). The actual mode shape is defined only in a relative sense since Φ 1 is an arbitrary constant. An inverse approach is followed in motion based design. One specifies the relative mode shape based on the desired deformation, and then determines the relative stiffness distribution. The details are presented below.

32 Chapter 1: Introduction The differences in displacement amplitudes can be interpreted as the relative displacements between the masses ∆u 1 = Φ 1

(1.96)

∆u 2 = Φ 2 – Φ 1

(1.97)

For optimal motion behavior, the relative displacements should be equal. This constraint results in a linear mode shape. ∆u 1 = ∆u 2

(1.98)

Φ 2 = 2Φ 1

(1.99)

Setting Φ 2 = 2Φ 1 in eqn (1.95) and solving for λ yields k 1∗ – k 2∗ λ 1∗ = ---------------------m1

(1.100)

where an asterisk is used to indicate the value associated with a prescribed relative mode shape. Introducing λ 1∗ in eqn (1.94) leads to a relation between k 1∗ and k 2∗ m1 k 1∗ = 1 + ---------- k 2∗ 2m 2

(1.101)

Finally, substituting for k 1∗ in eqn (1.100), results in a relation between the fundamental frequency (lowest) and k 2∗ : k 2∗ 2 λ 1∗ = ω 1 = ---------2m 2

(1.102)

Once k 2∗ is specified, the fundamental frequency and stiffness parameters are uniquely determined. One can also specify the desired fundamental frequency and then use eqn (1.102) to determine the necessary value of k 2∗ . Given k 1 and k 2 , eqn (1.94) can be solved for λ 2 , and using this value, the second mode shape can be established with eqn (1.95).

1.6 Control Strategies for Motion Based Design

33

Example 1.9: 2DOF system - with equal masses The case where m 1 = m 2 = m is considered to illustrate the procedure. Applying eqn (1.101), k 1∗ is related to k 2∗ by 3 k 1∗ = --- k 2∗ 2

(1.103)

Suppose the design objective is to have T 1 = 1s (i.e. ω 1 = 2π rd/s ). Then one sets ω 1 = 2π in eqn (1.102) and solves for k 2∗ k 2∗ = 2m ( 2π )

2

(1.104)

Once m is specified, the stiffness parameters can be evaluated. Substituting for these parameters in eqn (1.94) leads to λ 2 = 3k 2∗ ⁄ m . The corresponding mode shape is Φ 2 = – 0.5Φ 1.

1.6 Control strategies for motion based design The previous sections dealt with the problem of limiting the response of a single degree of freedom system subjected to periodic excitation. Although the problem considered was rather simple, the strategies used to generate appropriate values for the system stiffness and damping are basic paradigms that are also applicable to more complex structures. In what follows, the SDOF strategies are reviewed to provide the basis for establishing a general motion control methodology for structural systems. Consider the linear elastic SDOF system shown in Fig (1.14). Suppose the design loading is static, and the motion criteria is u < u * . Enforcing equilibrium leads to an expression for the required stiffness. p k ≥ ----- ≡ k * u*

(1.105)

34 Chapter 1: Introduction From a stiffness perspective, control for static loading is realized by adjusting the stiffness such that k is greater than a certain limiting value which depends on the design criteria for displacement.

k p m

F

c ug

u + u g = ut

Fig. 1.14: Single degree of freedom model for passive and active control Another way of expressing the motion control requirement is in terms of energy. In general, the work done by the external forces acting on a system is equal to the sum of the mechanical energy stored in the system and the energy transformed to another form, through either energy dissipation or absorption mechanisms. This system is elastic, and the loading is static. It follows that the stored energy is equal to the strain energy, and since no energy is dissipated, the work done by the external loading must be balanced by the strain energy. Expanding these quantities leads to eqn (1.106). External Work ≤ Strain Energy 1 * 1 --- pu ≤ --- k ( u * ) 2 ⇒ 2 2

p ----- ≤ k u*

(1.106)

From an energy perspective, static control is achieved by providing sufficient energy storage capacity to satisfy the energy demand. The design value of stiffness defined by eqn (1.105) is based on the hypothesis that the system properties (k,m) remain constant as the load is being applied, and there is no other agent which assists in resisting the load. Control is achieved solely by the action of the stiffness embedded within the system. This type of control action is called “passive control” since the system responds in a passive manner. A different strategy for limiting the displacement is based on using an external energy source to satisfy some of the energy demand. In this approach, the

1.6 Control Strategies for Motion Based Design

35

energy requirement is taken as: External work ≤ Strain energy + Supplied energy

(1.107)

Energy can be supplied in different ways. For the SDOF system, a force actuator can be used to generate the force, F, shown in Fig (1.14). Force actuators are devices which convert energy into a mechanical force, and provide the ability to implement this control strategy. The work done by F is negative, and its magnitude represents the external energy that needs to be supplied. Work

F

1 = – --- Fu * 2

(1.108)

Equation (1.107) expands to 1 * 1 1 --- pu ≤ --- k ( u * ) 2 + --- Fu * 2 2 2

(1.109)

and is follows that p ≤ ku * + F

(1.110)

One can also arrive at this result by enforcing equilibrium. Controlling motion by supplying external energy is referred to as “active control”. One of the key decisions in motion based design is the selection of the control strategy. Fully passive static control is simple to implement, since it involves only providing stiffness. Active control employs actuator technology which is costly and less reliable. However, active control has considerable potential, particularly for structural applications where weight is a critical issue, such as aerospace vehicles and long span bridges. The concepts discussed above are also applicable for dynamic loading. When the loading is periodic, the response amplitudes are given by: pˆ uˆ = ------------ H 2 Ω2m pˆ aˆ = ----H 2 m

(1.111)

36 Chapter 1: Introduction where H 2 is plotted in Fig (1.15). Since civil structures are generally lightly damped, a realistic upper limit for ξ is 0.3. Assuming pˆ , Ω , and m are specified, one needs to select values for stiffness and damping that satisfy the design motion criteria. H 2 < H 2*

(1.112)

where the value of H 2* depends on whether displacement or acceleration is the limiting motion constraint. mΩ 2 u * H 2* = -----------------pˆ ma * H 2* = --------p

( displacement ) (1.113) ( acceleration )

4

ξ = 0 3.5

3

Stiffness and damping are coupled

Stiffness dominates response

2.5

Stiffness dominates response

H 2*

2

H2

H 2*

1.5

1

H 2*

ξ = 0.3

0.5

0

0

0.5

1

1.5

2

Ω m ρ = ---- = Ω ---ω k

2.5

3

Fig. 1.15: Response function for periodic excitation

3.5

4

1.6 Control Strategies for Motion Based Design

37

Figure (1.15) shows that H 2* is essentially independent of damping for ρ < ≈ 0.5 and ρ> ≈ 3.0 . If H 2* < ≈ 0.5 , stiffness is the only parameter that can be “adjusted” to satisfy the design constraint, and the design process is essentially quasi-static. For 0.5 < H 2* < 1.0 , both stiffness and damping are available as control variables. One can generate a spectrum of designs by specifying stiffness ( ρ ) , and determining the corresponding damping ( ξ ) that satisfies H 2 = H 2* . When H 2* > 1 , there are 2 possible design zones, one for ρ < 1 and the other for ρ > 1. As H 2* increases, these zones merge into a single zone centered at ρ = 1, the resonant state of the system. Damping dominates the response in the neighborhood of ρ = 1. The strategy employed for 0.5 < H 2* < 1.0 can also be applied for H 2* > 1.0 . Given H 2* , one selects a stiffness value within the stiffness range defined by the specified value of H 2* , and the boundary curves ξ = 0 and ξ = 0.3 , and then determines the required value of ξ . For 1.0 ≤ H 2* ≤ 1.5 , the second stiffness range is outside ρ = 3.0 . Designs for this region are controlled by stiffness, and are characterized by their low stiffness. Design for the first region are controlled by both stiffness and damping, and have a higher stiffness. The response for periodic ground excitation is given by uˆ t = H 3 uˆ g

(1.114)

where H 3 is plotted in Fig (1.16). This function also relates the applied periodic force and the corresponding reaction, ˆ = H pˆ R 3

(1.115)

These equations are used to establish a control strategy for isolating a system from an external action, either an applied loading or a support motion. In general, one wants H 3 < 1 for isolation. Defining H 3* as the design requirement, the allowable stiffness range is bounded by the curves for ξ = 0 and ξ = 0.3 , as indicated in Fig (1.16). The absolute upper limit on stiffness is ρ =

Ω2m 2 ⇒ k max = -----------2

(1.116)

Designs are generated by specifying stiffness, and then determining the corresponding value of ξ . Isolation is achieved by reducing the stiffness below the critical level, k max . If k > k max , isolation is not possible even for the case where maximum damping ( ξ = 0.3 ) is introduced.

38 Chapter 1: Introduction

4

3.5

ξ = 0 3

2.5

H3 2

ξ = 0.3

1.5

1

H 3* 0.5

ξ = 0.3

ξ = 0 0

0

0.5

1

1.5

2

2.5

3

3.5

Ω m ρ = ---- = Ω ---ω k Fig. 1.16: Response function for periodic ground excitation

4

Active control of a SDOF system for dynamic loading involves applying a control force, F(t ), and adjusting its magnitude over time in such a manner that the resulting motion is constrained within the desired limits. One needs a force actuator that responds essentially in real time. Specifying the magnitude and sense of F(t ) is the key issue for active control. Various approaches are discussed later in the text. In general, F is selected to oppose the motion and does negative work on the mass. This work has to be supplied by an external energy source. Taking F to be proportional to the displacement and velocity, F ( t ) = +k d u ( t ) + k v u˙ ( t )

(1.117)

corresponds to increasing the stiffness and damping of the original system. This observation follows by substituting for F in the equilibrium equation. Then, active control based on eqn (1.117) can be interpreted as introducing virtual stiffness and

1.6 Control Strategies for Motion Based Design

39

damping. From an energy perspective, stable active control reduces the energy demand for the system. The energy balance equation for a linear SDOF system initially at rest is given by t

∫0

pu˙ dt +

∫

t

t 1 1 ( – Fu˙ ) dt = --- mu˙ 2 + --- ku 2 + cu˙ 2 dt 2 2 0 0

∫

(1.118)

The last term is the energy dissipated through viscous action. Passive control provides energy storage and energy dissipation mechanisms to resist the demand. When F(t ) is taken to have the same sense as u˙ , active control decreases the energy input to the system since the integral is always negative. The effect of this action is a lower energy demand that has to be met by passive control mechanisms. Selecting a motion control strategy for dynamic excitation involves a number of decisions. Firstly, should the strategy be purely passive or a combination of passive and active control? Secondly, what percentage of the energy demand should be met by energy storage (stiffness) vs. energy dissipation (damping)? Thirdly, the type and properties of passive energy dissipation device needs to be specified. There are a number of passive energy dissipation devices that are appropriate for structural systems. Fourthly, if active control is used to supplement passive control, the type and capacity of the force actuator needs to be established. Although the discussion has been focussed on a SDOF system, the concepts introduced here are also applicable for a general multi degree of freedom structural system. The only difference is that now one is dealing with sets of stiffness and damping parameters, and motion criteria involving the displacement variables associated with the degrees of freedom. Using vector notation, the problem can be stated as: given a desired displacement response vector, u * ( t ) , determine k, c, and F, the vectors containing the stiffness, damping, and active control force variables. These vectors are functions when the system is continuous. The desired response for a structure is related to the nature of the loading and the critical performance measures chosen for the structure. For service

40 Chapter 1: Introduction loading, the damage that non-structural elements, such as wall panels of buildings, can experience constrains the magnitude and distribution of displacements, while human and equipment comfort limits the peak acceleration. The controlling criterion for wind dominant design tends to be the peak velocity and acceleration. Displacement is the controlling criterion for earthquake dominant designs. In general, the structure is required to remain elastic under service loading. Under extreme loading, structural performance and stability requirements constrain the magnitude and distribution of inelastic deformation that the structural components can experience. Structural deformation is the key measure for earthquake dominant design. Although design codes allow a structure to experience inelastic deformation under an extreme earthquake with no collapse or loss of life, the current trend is to reduce the allowable inelastic deformation (Fajfar & Krawinkler 1997). This shift is driven by the need to lower the cost of repair. Furthermore, recent studies have shown that the initial cost of a structure designed to remain elastic under extreme seismic load may be less than the cost corresponding to the conventional design approach which allows for residual deformation (Connor et al, 1997). Combining these savings with the decrease of damage repair and increased revenue due to maintaining operability can result in a substantial reduction in life cycle cost. In order to satisfy the complete set of motion requirements, a multi-level control strategy is required. Figure (1.17) shows the strategy adapted for this text. The first step for passive control is to establish the distribution of stiffness, which involves the choice of material stiffness and cross-sectional properties, by enforcing the requirements on the spatial distribution of displacements corresponding to the design loading condition. Example 1.9 illustrates this computation for a 2-DOF system. The ideal state is uniform deformation throughout the structure. However, this state is possible only for a statically determinate structure. The requirement on the magnitude of the response is met with a combination of scaling the stiffness distribution and incorporating passive energy dissipation devices such as viscous dampers, tuned mass and liquid sloshing dampers, and base isolation systems. Depending on the level of damping chosen, damping and stiffness may be coupled. In this case, one cannot independently scale stiffness and damping, and iteration on the scaling process and possibly also on the stiffness distribution may be required. Typical civil structures are usually lightly damped, with ξ < = 0.05 , and coupling is not a

1.6 Control Strategies for Motion Based Design

41

problem. However, the current trend is toward more heavily damped structures with ξ ≈ 0.2 to 0.3. Seismic design for a heavily damped structure requires iteration between stiffness and damping. The response for extreme loading is controlled by supplementing stiffness and damping with mechanisms that absorb energy. An example of an absorption mechanism is a hysteretic damper. These devices experience residual inelastic deformation, which is a measure of the energy absorbed. The first step for active control is to decide on a location for force actuators which provides the capability for controlling the displacement response in the desired manner. Controllability is a critical issue for active control. It relates to the sensitivity of the response to the control force magnitude. Given the distribution, the desired force magnitudes are established using a control algorithm, and instructions then are sent to the actuators to deliver these forces. Examples of force actuators are active mass drivers and active tendon systems. Active control can be utilized to supplement passive control for service loading. Since the level of power, energy, and force required for active control of large scale civil structures under extreme loading cannot be economically achieved with the present technology, passive control is the predominant choice for this loading.

Load

Functional

Level

requirements

Service

Extreme

Control

Action

Passive

Active

-Control the spatial distribution of displacements

-Establish the spatial distribution of stiffness

-Select an appropriate spatial distribution of force actuators

-Control the magnitude of the response

-Specify a distribution of damping (energy dissipation) devices -Scale the stiffness and damping magnitudes

-Establish the magnitude of the control forces -Activate the force actuators

-Control the distribution and magnitude of the response

-Include energy absorption mechanisms

Fig. 1.17: Multilevel control strategy

42 Chapter 1: Introduction

1.7 Scope of text This text presents a systematic treatment of the concepts and computational procedures for motion control of civil structures. The material is organized according to the nature of the control process. Part I concerns passive control, and includes Chapters 2 thru 5. Methodologies for stiffness and damping based control are presented, and applied to typical building type structures. Part II consists of Chapters 6 thru 9. These chapters are intended to provide an introduction to active control concepts and computational algorithms.Quasi static control is discussed first, since it is easier to deal with analytically. Classical dynamic control algorithms are treated next.Their application to civil structures is illustrated with a series of computer simulation studies. A brief description of the individual chapters is provided below. Chapter 2 deals with optimizing the distribution of stiffness for beam and frame type structures. The chapter begins with an overview of the governing equations for beam type structures and then addresses the static problem. A method for distributing rigidity to attain the desired static deformation profiles is presented. The method is extended to the dynamic case and an analytic formulation is developed for structural systems characterized by a dominant fundamental mode response. How to distribute stiffness in structures so as to generate a fundamental mode shape with uniform shear and bending deformation profiles is the key issue. An iterative numerical scheme based on the response spectrum method is developed to incorporate the effect of the higher modes which get amplified as the slenderness of the structure increases. Results for a set of representative buildings subjected to seismic excitation are presented. They provide an assessment of the effectiveness of controlling displacement with stiffness. Chapter 3 examines the role of damping for motion control of dynamically excited structures. The chapter starts by examining the response characteristics of single-degree-of-freedom (SDOF) systems with various types of damping mechanisms. The concept of equivalent viscous damping is introduced and is used to express both structural damping and hysteretic damping in terms of their viscous counterpart. Numerical results are presented to evaluate the validity of this concept for SDOF systems. Multi-degree-of-freedom (MDOF) systems are discussed next. A formulation is presented for the case where the damping distribution is taken proportional to the stiffness distribution generated in Chapter 2. The effectiveness of damping under seismic excitation is assessed by

1.7 Scope of Text 43 numerical simulation. Examples illustrating the design of viscoelastic and hysteretic dampers are also included. Chapter 4 assesses the effectiveness of tuned mass dampers (TMD) for controlling the motion of dynamically excited structures. The theory of TMDs for SDOF systems is developed and applied to a range of SDOF systems subjected to harmonic and seismic excitations. The theory is then extended to MDOF systems, and a procedure for designing a TMD to control a specific modal response is developed and illustrated. Descriptions of existing implementations of tuned mass dampers in building structures are also presented. Chapter 5 considers base isolated structures (BIS). Base isolation is a control strategy that modifies the stiffness distribution. A base isolation control strategy for SDOF systems is developed, and then extended to the MDOF case. The procedure is similar to the approach followed in Chapter 2. One distributes stiffness along the height of the building so that the fundamental mode experiences uniform deformation, and selects the stiffness of the base isolation system such that the deformation in the isolator is bounded. The iterative procedure developed in Chapter 2 to account for the effect of the higher modes, is modified to incorporate the effect of the base isolation system. Chapter 6 provides an overview of active control. The architecture of an active control system is viewed as having 3 main components: a monitoring system which acquires data; a cognitive unit called a controller which processes the acquired date and decides on a course of action; and an actuator system composed of a set of physical devices that execute the instructions from the controller. Various types of controllers are described, and examples illustrating the application of particular type of control action called linear feedback control are presented. Existing established actuator technologies are discussed in considerable detail. An overview of emerging actuator technologies based on adaptive materials is also included. Lastly, examples of existing structural control systems for large scale structures are presented. Chapter 7 addresses quasi-static control, i.e., where the structural response to applied loading can be approximated as static response. Since time dependent effects are neglected, stiffness is the only quantity available for passive control. Active control combines stiffness with a set of pseudo-static control forces. Fundamental concepts such as observability, controllability, optimal control, and computational procedures are introduced in this chapter. Both continuous and

44 Chapter 1: Introduction discrete physical systems are treated. Chapter 8 considers dynamic feedback control of time-invariant multidegree-of freedom structural systems. A combination of stiffness, damping, and time dependent forces is used for motion control of dynamic systems. The statespace formulations of the governing equations for SDOF and MDOF systems are used to discuss stability, controllability, and observability aspects of dynamically controlled systems. Continuous and discrete forms of the linear quadratic regulator (LQR) control algorithm are derived, and examples illustrating their application to a set of shear beam type buildings are presented. The effect of time delay in the stability of LQR control, and several other linear control algorithms are also discussed.

Problems

45

Problems Problem 1.1 Refer to example 1.2 and equation 1.11. Construct plots of ( H * ⁄ u * ) γ = 1 , as a function of the aspect ratio ( H ⁄ d ) for the following ranges: •

E = 200, 000MPa

•

σ from 200MPa to 600MPa

•

*

( H ⁄ d ) from 3 to 8

Recommended design values of H ⁄ u∗ for a building are in the region 400 to 500. Comment on when “motion” rather than strength controls the design.

Problem 1.2 Refer to example 1.3. A typical value for the ratio of column depth to story height is 0.10. Using eqn (1.22), determine the value of ( H ⁄ u * ) for which the c constraint on displacement controls the design taking the following ranges for *

and σ : c

•

E = 200, 000MPa

•

σ from 200MPa to 600MPa

*

Problem 1.3 Refer to example 1.3. Suppose a mass, m , is attached to the “infinitely stiff” girder. Assuming the columns have negligible mass, determine the expression for the natural frequency, ω , for lateral vibration in terms of the material and geometric properties of the frame.

46 Chapter 1: Introduction

Problem 1.4 Consider a SDOF system having m = 1000kg and subjected to a sinusoidal force with amplitude pˆ = 500N and frequency 2πr ⁄ s . Recommend design values for stiffness k , and damping, c , corresponding to the following limiting values for peak acceleration: *

•

a = 0.1m ⁄ s

•

a = 1m ⁄ s

•

a = 3m ⁄ s

*

2

*

2

2

Problem 1.5 Consider a SDOF system with mass of 5000kg . The system is to be subjected to a periodic loading having a magnitude of 4kN and frequency 2 hertz. Recommend design values for stiffness and damping corresponding to the following limiting values of peak displacement: *

•

u = 1mm

•

u = 10mm

•

u = 20mm

* *

Problem 1.6 Suppose the mass of a SDOF system is known, but the stiffness, k , and damping, c , are unknown. Discuss how you would determine the stiffness by applying a periodic loading for which both the frequency and magnitude can be varied, and monitoring the response. How would you determine the damping parameter, c?

Problem 1.7 Consider a SDOF system having m = 1000kg . The system is to be subjected to a periodic force with a magnitude of 5kN and frequency 3Hz .

Problems

47

Recommend design values for k , and c such that the magnitude of the reaction force is less than: •

10kN

•

3kN

Problem 1.8 A SDOF system having m = 1000kg is to be subjected to the following ground motion: u g = 0.3 sin 4πt meters Recommend values for k and c for the following design values of peak relative displacement: *

•

u = 0.2m

•

u = 0.4m

*

Problem 1.9 Recommend values for k and c for a SDOF system with mass of 2000kg subjected to the following ground motion: u g = 0.5 sin 3πt meters The motion constraint is uˆ t ≤ 0.1m . What is the corresponding value for the peak relative motion, uˆ ? Discuss how the response measures, uˆ t and uˆ , vary as the spring stiffness is reduced from the value you select.

Problem 1.10 Consider a SDOF system subjected to ground excitation. Take m = 3000kg and the ground acceleration as

48 Chapter 1: Introduction a g = 4.0 sin 4πt Recommend values for k and c such that the peak total acceleration is less than 2

1.0m ⁄ s .

Problem 1.11 A machine represented by the mass m is to be supported by the spring and dashpot shown in the following figure. The machine is sensitive to total acceleration and therefore needs to be “isolated” form the ground motion. m=2000 kg

u+ug m

k

c ug

Consider the ground acceleration to consist of two dominant components ag ----- = 0.1 sin ( 2πt + δ 1 ) + 0.2 sin ( 4πt + δ 2 ) g where 2

g is the gravitational acceleration ( 9.81m ⁄ s ) and δ 1, δ 2 are random phase angles that can range from 0 to 2π A reasonable approximation for the peak acceleration of the combined response is at =

2

( a t, 1 ) + ( a t, 2 )

2

where a t, 1 and a t, 2 are the total accelerations due to the individual harmonic

Problems

49

excitations with random phasing. Suppose the desired maximum total acceleration is 0.05g . Describe how you would establish design values for k and c .

Problem 1.12 Consider the 2 DOF system discussed in Section 1.5. Take m 2 = 0.5m 1 . Determine the stiffness parameters k and k corresponding to m 1 = 1000kg and the following range of ω 1 : ω 1 = π, 2π, 4π Also determine ω 2 and the shape of the second mode corresponding to each stiffness distribution.

Problem 1.13 The derivation in Section 1.5 was based on specifying the fundamental mode to be linear. Suppose one takes ϕ 2 = αϕ 1 where α is a scale factor.

Generalize eqn (1.101) to incorporate α . What is the significance of taking α ≠ 2 ? What happens when α is taken to be negative?

50 Chapter 1: Introduction

51

Part I: Passive Control Chapter 2

Optimal stiffness distribution 2.1 Introduction This chapter is concerned with the first step in passive motion control, establishing a distribution of structural stiffness which produces the desired displacement profile. When the design loading is quasi-static, the stiffness parameters are determined by solving the equilibrium equations in an inverse way. Dynamic loading is handled by selecting the stiffness parameters such that the fundamental mode shape has the desired displacement profile. The implicit assumptions here are that one can incorporate sufficient damping to minimize the contributions of the higher modes, and the fundamental mode shape is independent of damping. The latter assumption is reasonable for lightly damped structures. Discrete systems are governed by algebraic equations, and the problem reduces to finding the elements of the system stiffness matrix. The static case involves solving *

KU = P

*

*

(2.1) *

for K , where U and P are the prescribed displacement and loading vectors. Some novel numerical procedures for solving eqn (2.1) are presented in a later section.

52 Chapter 2: Optimal Stiffness Distribution In the dynamic case, the equilibrium equation for undamped periodic excitation of the fundamental mode is used: *

2

KΦ = ω 1 MΦ

*

(2.2) *

where M is the discrete mass matrix, Φ is a scaled version of the desired displacement profile, and ω 1 is the fundamental frequency. Taking 1 K' = ------K 2 ω1 P' = M Φ

*

(2.3)

(2.4)

reduces eqn (2.2) to *

K'Φ = P'

(2.5)

The solution technique for eqn (2.1) also applies for eqn (2.5). Once K' is known, the stiffness can be scaled by specifying the frequency, ω 1 . An appropriate value for ω 1 is established by converting the system to an equivalent one degree-of-freedom system and using the SDOF design approaches discussed in the introduction. Continuous systems such as beams are governed by partial differential equations, and the degree of complexity that can be dealt with analytically is limited. The general strategy of working with equilibrium equations is the same, but now one has to determine analytic functions rather than discrete values for the stiffness. Analytical solutions are useful since they allow the key dimensionless parameters to be identified and contain generic information concerning the behavior. In what follows, the first topic discussed concerns establishing the stiffness distribution for static loading applied to a set of structures consisting of continuous cantilever beams, building type structures modeled as equivalent discontinuous beams with lumped masses, and truss-type structures. Closed form solutions are generated for the continuous cantilever beam example. The next topic concerns establishing the stiffness distribution for the case of dynamic loading applied to beam-type structures. The process of calibration of the

2.2 Governing Equations - Transverse Bending of Planar Beams

53

fundamental frequency is described and illustrated for both periodic and seismic excitation. The last section of the chapter deals with the situation where the higher modes cannot be ignored. An iterative numerical scheme is presented and applied to a representative range of beam-type structures.

2.2 Governing equations - transverse bending of planar beams In this section, the governing equations for a specialized form of a beam are developed. The beam is considered to have a straight centroidal axis and a crosssection that is symmetrical with respect to a plane containing the centroidal axis. Figure 2.1 shows the notation for the coordinate axes and the displacement measures (translations and rotations) that define the motion of the member. The beam cross-section is assumed to remain a plane under loading. This restriction is the basis for the technical theory of beams, and reduces the number of displacement variables down to three translations and three rotations which are functions of x and time. When the loading is constrained to act in the symmetry plane for the crosssection, the behavior involves only those motion measures associated with this plane. In this discussion, the x-y plane is taken as the plane of symmetry, and u x , u y , and β z are the relevant displacement variables. If the loading is further restricted to act only in the y direction, the axial displacement measure, u x , is identically equal to zero. The behavior for this case is referred to as transverse bending. In what follows, the governing equations for transverse bending of a continuous planar beam are derived. The derivation is then extended to deal with discontinuous structures such as trusses and frames that are modelled as equivalent beams.

54 Chapter 2: Optimal Stiffness Distribution

Note:

x-y plane is a plane of symmetry for the cross-section y y βy uy

ux

βx x

uz z

z

βz

Fig. 2.1: Notation - planar beam.

Planar deformation-displacement relations Figure 2.2 shows the initial and deformed configurations of a differential beam element. The cross-sectional rotation, β z , is assumed to be sufficiently small such 2 that β z << 1 . In this case, linear strain-displacement relations are acceptable. Letting γ denote the transverse shearing strain and ε the extensional strain at an arbitrary location y from the reference axis, and taking β ≡ β z and u ≡ u y , the deformation relations take the form ε = – yχ

(2.6)

∂u γ = ------ – β ∂x

(2.7)

∂β χ = -----∂x

(2.8)

where χ denotes the bending deformation parameter.

2.2 Governing Equations - Transverse Bending of Planar Beams

Y'

X'

Y

B

A y

A y

B

O

55

γ β

O

X

X

Fig. 2.2: Initial and deformed elements.

Optimal deformation and displacement profiles Optimal design from a motion perspective corresponds to a state of uniform shear and bending deformation under the design loading. This goal is expressed as γ = γ∗

(2.9)

χ = χ∗

(2.10)

Uniform deformation states are possible only for statically determinate structures. Building type structures can be modeled as cantilever beams, and therefore the goal of uniform deformation can be achieved for these structures. Consider the vertical cantilever beam shown in Fig. 2.3. Integrating eqns (2.7) and (2.8) and enforcing the boundary conditions at x = 0 leads to β = χ∗ x

(2.11) 2

χ∗ x u = γ ∗ x + -----------2

(2.12)

The deflection at the end of the beam is given by 2

χ∗ H u ( H ) = γ ∗ H + -------------2

(2.13)

56 Chapter 2: Optimal Stiffness Distribution 2

where γ ∗ H is the contribution from shear deformation and χ∗ H ⁄ 2 is the contribution from bending deformation. For actual buildings, the ratio of height to width (i.e. aspect ratio) provides an indication of the relative contribution of shear versus bending deformation. Buildings with aspect ratios on the order of unity tend to display shear beam behavior and χ ≈ 0 . On the other hand, buildings with aspect ratios greater than about 7 display bending beam behavior and γ ≈ 0 . x

b(x) H

y,u

Fig. 2.3: Simple cantilever beam. One establishes the values of γ ∗ , χ∗ based on the performance constraints imposed on the motion, and selects the stiffness such that these target deformations are reached. Introducing a dimensionless factor s , which is equal to the ratio of the displacement due to bending and the displacement due to shear at x=H, 2 H χ∗ Hχ* s = -------------- ⁄ ( γ *H ) = ---------- 2 2γ *

(2.14)

transforms eqn (2.13) to a form that is more convenient for low rise buildings. u ( H ) = ( 1 + s )γ ∗ H A shear beam is defined by s = 0 . Tall buildings tend to have s ≈ 1 .

(2.15)

2.2 Governing Equations - Transverse Bending of Planar Beams

57

Equilibrium equations Figure 2.4 shows a differential beam element subjected to an external transverse loading, b , and restrained by the internal transverse shear, V , and bending moment, M . By definition, V =

∫

τ dA

(2.16)

∫

(2.17)

M = – yσ dA

where τ and σ are the stresses acting on the cross-section. Summing forces and moments leads to 2

∂ u ∂V ------- + b = ρ m ∂x ∂t2

(2.18)

2

∂ β ∂M -------- + V = J ∂x ∂t2

(2.19)

where ρ m , J are the mass and rotatory inertia per unit length. When the member is supported only at x = 0 (see Fig. 2.3), the equilibrium equations can be expressed in the following integral form H

V(x) =

∫

2

∂ u b – ρ m 2 dx ∂t

(2.20)

x

H

M(x) =

∫ x

2

∂ β V – J 2 dx ∂t

(2.21)

58 Chapter 2: Optimal Stiffness Distribution

y b

∂M M + --------dx ∂x

∂τ τ + ------ dx ∂x

M

∂σ σ + -------dx

σ

∂x

x

τ V

dx

∂V V + -------dx ∂x

Fig. 2.4: Forces acting on a differential element. In the case of static loading, the acceleration terms are equal to 0, and V and M can be determined by integrating eqns (2.20) and (2.21). Force-deformation relations The force-deformation relations, also referred to as the constitutive relations, depend on the characteristics of the materials that make up the beam. For the case of static loading and linear elastic behavior, the expressions relating the shear force and bending moment to the shear deformation and bending deformation respectively are expressed as V ( x ) = D T ( x )γ ( x )

(2.22)

M ( x ) = D B ( x )χ ( x )

(2.23)

where D T and D B are defined as the transverse shear and bending rigidities. These equations have to be modified when the deformation varies with time. This aspect is addressed in Section 2.4. Examples which illustrate how to determine the rigidity coefficients for a range of beam cross-sections are presented below.

2.2 Governing Equations - Transverse Bending of Planar Beams

59

Example 2.1. Composite sandwich beam Figure 2.5 shows a sandwich beam composed of 2 face plates and a core. b d

tf Fig. 2.5: Composite beam cross section.

The face material is usually much stiffer than the core material, and therefore the core is assumed to carry only shear stress. Noting eqn (2.6), the strains in the face and core are d ε f = ± --- χ 2

(2.24)

γc = γ

(2.25)

The face thickness is also assumed to be small in comparison to the depth. Considering the material to be linear elastic, the expressions for shear and moment are: V = ( bd )τ c = ( bdG c )γ

(2.26) 2

bt f d M = ( bt f d )σ f = -------------- E f χ 2

(2.27)

where G c is the shear modulus for the core and E f is the Young’s modulus for the face plate. The corresponding rigidity coefficients are: D T = bdG c

(2.28)

60 Chapter 2: Optimal Stiffness Distribution 2

bt f d D B = -------------- E f 2

(2.29)

Example 2.2. Equivalent rigidities for a discrete Truss-beam The term, truss-beam, refers to a beam type structure composed of a pair of chord members and a diagonal bracing system. Figure 2.6 illustrates an x-bracing scheme. Truss beams are used as girders for long span horizontal systems. Truss beams are also deployed to form rectangular space structures which are the primary lateral load carrying mechanisms for very tall buildings. The typical “mega-truss” has large columns located at the 4 corners of a rectangular crosssection, and diagonal bracing systems placed on the outside force planes. These structures are usually symmetrical, and the behavior in one of the symmetry directions can be modelled using an equivalent truss beam. When the spacing, h, is small in comparison to the overall length, one can approximate the discrete structure as a continuous beam having equivalent properties. In this example, approximate expressions for these equivalent properties are derived for the case of x-bracing.

M V c

Ec A

d

E d A

θ

h

F

c

F

d

F

d

F

c

B

Fig. 2.6: Parameters and internal forces - Truss-beam. The key assumption is that the members carry only axial force. This approximation is reasonable when the members are slender, and diagonal or chevron bracing is used. Noting Fig. 2.6, the cross section force quantities are related to the member forces by

2.2 Governing Equations - Transverse Bending of Planar Beams

M = BF

c

61

(2.30)

d

V = 2F cos θ

(2.31)

Assuming linear elastic behavior, the member forces are also related to the extensional strains by c

c c c

d

= A E ε

F = A E ε F

(2.32)

d d d

(2.33)

It remains to express the extensional strains in terms of the bending and shear deformation measures. Figure 2.7 shows the deformed shapes of a panel of the truss beam. The d extensional strain in the diagonals, ε , due to the relative motion between adjacent nodes, ∆ h , is a function of ∆ h and θ . ∆ h cos θ sin θ d ε ∆ = ------------------------------h h

(2.34)

Neglecting the extensional strain in the diagonal due to ∆β , and approximating γ as ∆h ------ ≈ γ h

(2.35)

one obtains the following approximation for the total extensional strain: d d γ sin 2θ ε ≈ε ≈ γ cos θ sin θ ≈ -----------------∆h 2

(2.36) c

Similarly, the extensional strain in the chord, ε , is related to the change in angle, ∆β , between adjacent sections by ∆v c B∆β ε = ------ = ----------h 2h Noting that ∆β ⁄ h is related to the bending deformation χ ,

(2.37)

62 Chapter 2: Optimal Stiffness Distribution ∆β χ = ------h

(2.38)

the strain can be expressed as c Bχ ε = ------2

(2.39)

+u

+β

∆h

∆β

∆v

h

θ

θ

x

B

y

B

Fig. 2.7: Deformed truss-beam section. c

d

Substituting for ε and ε and combining eqns (2.30)-(2.33) results in c c 2

A E B M = -------------------- χ 2 d d

V = [ A E sin 2θ cos θ ]γ

(2.40)

(2.41)

Comparing these expressions with the definition equations for the rigidity parameters leads to the following relations for the equivalent continuous beam properties: c c 2

A E B D B = -------------------2 d d

D T = A E sin 2θ cos θ

(2.42)

(2.43)

2.2 Governing Equations - Transverse Bending of Planar Beams

63

When the truss beam model is used to represent a tall building, the chords correspond to the columns of the building. These elements are required to carry both gravity and lateral loading whereas the diagonals carry only lateral loading. Since the column force required by the gravity loading can be of the same order as the force generated by the lateral loading, the allowable incremental deformation in the column due to lateral loading should be less than the corresponding value for the diagonal. To allow for this reduction, a factor, f ∗ , which is defined as the ratio of the diagonal strain to the chord strain for lateral loading is introduced. d

ε f ∗ = ----c ε

(2.44)

This factor is greater than 1. Substituting for the strain measures from eqns (2.36) and (2.39) results in γ sin 2θ χ = -----------------f ∗B

(2.45)

Once the shear deformation level γ ∗ is specified, the bending deformation is determined with γ ∗ sin 2θ χ∗ = --------------------f ∗B

(2.46)

Substituting for χ∗ in eqn (2.14), the ratio of the contributions from bending and shear deformation expands to H sin 2θ s = -------------------2 f ∗B

(2.47)

Typical values of f ∗ for buildings range from about 3 for elastic behavior to 6 for inelastic behavior. Equation (2.47) shows that the bending contribution becomes more important as the aspect ratio, H ⁄ B , increases. The shear and bending contributions to the elastic displacement at the top of the building are essentially equal when H ≈ 6B .

64 Chapter 2: Optimal Stiffness Distribution

Governing equations for buildings modelled as pseudo shear beams This section considers a class of planar rectangular building frames having aspect ratios of order O(1) and moment resisting connections. Figure 2.8 shows a typical case. This type of structure is the exact opposite to the truss beam with respect to the way the lateral loading is carried. In the case of the truss beam, the transverse shear is provided by the axial forces in the braces. Here, the shear is produced by bending of the columns. The axial deformation of the columns is usually small for low rise frames, so it is reasonable to assume the “floors” experience only lateral displacement and slide with respect to each other. Considering the structure as a pseudo-beam, there is no rotation of the cross-section, i.e., β = 0 ; there is only one displacement variable per floor; and the transverse shearing strain at a story location is equal to the interstory displacement divided by the story height. In what follows, the formulation of the governing equations is illustrated using a simple structure and then generalized for more complex structures.

h H

B Fig. 2.8: Low rise rigid frame. The 2-story frame shown in Fig 2.9 is modeled as a 2 DOF system having masses concentrated at the floor locations and shear beam segments which represent the action of the columns and beams in resisting lateral displacement. The shear forces in the equivalent beam segments (see Fig. 2.10) are expressed in terms of shear stiffness factors:

2.2 Governing Equations - Transverse Bending of Planar Beams V 1 = k1 u1

V 2 = k2 ( u2 – u1 )

65

(2.48)

Noting the definition of transverse shear strain, γ 1 = u1 ⁄ h1

u2 – u1 γ 2 = -----------------h2

(2.49)

one can relate the k ‘s to the equivalent transverse shear rigidity factors: V 1 = D T, 1 γ 1 , V 2 = D T, 2 γ 2

⇒ D T, i = h i k i , i = 1, 2

Lb h2

m2

interior

u2, p2 k2

exterior

u1, p1

m1 h1

(2.50)

k1

Fig. 2.9: A discrete shear beam model. m2

u2, p2 V2 V2

m1

u1, p1 V1 V1

Fig. 2.10: External and internal forces for discrete shear beam model. The equivalent shear stiffness factors are determined by displacing the floors of the actual frame, determining the shear forces in the columns, summing these forces for each story, and equating the total shear forces to V 1 and V 2 as

66 Chapter 2: Optimal Stiffness Distribution defined by eqn (2.48). The shear force in the j ’th column of story i is expressed as: V i, j = k i, j ( u i – u i – 1 )

(2.51)

where k i, j depends on the frame geometry and member properties. Then, summing the column shears for story i and generalizing eqn (2.48) leads to ki =

∑j ki, j

(2.52)

For this example, i = 1, 2 and j ranges from 1 to 5 . An approximate expression for the column shear stiffness factors can be obtained by assuming the location of the inflection points in the columns and beams. Taking these points at the mid-points, as indicated in Fig. 2.9, leads to the following estimates for interior and exterior columns: 12EI c k ( interior column ) = ---------------------3 h (1 + r)

(2.53)

12EI c k ( exterior column ) = -------------------------3 h ( 1 + 2r )

(2.54)

where r is a dimensionless parameter, I c Lb r = ---- ⋅ ----h Ib

(2.55)

and the subscripts denote column and beam properties. A typical frame has r = O(1) . The equilibrium equations for the “discrete” beam are established by enforcing equilibrium for the lumped masses shown in Fig. 2.10. p 1 = V 1 – V 2 + m 1 u˙˙1

p 2 = V 2 + m 2 u˙˙2

Substituting for V 1 and V 2 , eqn (2.56) expands to

(2.56)

2.2 Governing Equations - Transverse Bending of Planar Beams

p 1 = k 1 u 1 + k 2 ( u 1 – u 2 ) + m 1 u˙˙1 p 2 = k 2 ( u 2 – u 1 ) + m 2 u˙˙2

67

(2.57)

It is convenient to express eqn (2.57) in matrix form. The various matrices are defined as:

U =

P =

u1

(2.58)

u2 p1

(2.59)

p2

M =

m1 0

K =

k1 + k2

–k2

–k2

k2

(2.60)

0 m2

(2.61)

With these definitions, eqn (2.57) has the form ˙˙ P = KU + M U

(2.62)

Equation 2.20 expresses the shear force in a continuous beam as an integral of the applied lateral loading. The corresponding equations for this discrete system are obtained from eqn (2.56) by combining the individual equations: V 2 = p 2 – m 2 u˙˙2 V 1 = p 1 + p – m 1 u˙˙1 – m 2 u˙˙2

(2.63)

2

In general, the shear force in a particular story is determined by summing the forces acting on the stories above this story. A building having n stories is considered next. The building is modeled as

68 Chapter 2: Optimal Stiffness Distribution an n DOF system with lumped mass and equivalent shear springs, as shown in Fig. 2.11. The strategy for determining the equivalent shear stiffness factors is the same as discussed above. The only difference for this case is the form of the system matrices. They are now of order n . mn kn

Vi+1 pi

mi

m2

u2, p2

Vi

k2 m1 k1 Fig. 2.11: General shear beam model The equilibrium equation for mass i is given by p i = m i u˙˙i + V i – V i + 1

(2.64)

Expressing the shear forces in terms of the nodal displacements V j = k j(u j – u j – 1)

(2.65)

and substituting in eqn (2.64) results in p i = m i u˙˙i – k i u i – 1 + ( k i + k i + 1 )u i – k i + 1 u i + 1

(2.66)

This equation defines the entries in the i ‘th row of M and K . The expanded forms are listed below. m1 m2 M =

. . . mn

2.3 Stiffness Distribution for a Continuous Cantilever Beam under Static Loading

k1 + k2

–k2

0 . ..

0

0

0

–k2

k2 + k3

–k3 . . .

0

0

0

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

0

0

kn – 1 + kn

–kn

0

0

0 . . . –kn – 1 0 . .. 0

–kn

kn

K =

. . .

69

(2.67)

2.3 Stiffness distribution for a continuous cantilever beam under static loading Once the shear and bending moment distributions are specified, the rigidity distributions required to produce a specific deformation profile can be evaluated using eqns (2.22) and (2.23). The equations corresponding to uniform deformation reduce to V D T = ----γ∗

(2.68)

M D B = -----χ∗

(2.69)

For example, taking a uniform loading b ( x ) = b as shown in Fig. 2.12, which is a reasonable assumption for the wind action on a tall building, results in V (x) = b[H – x]

(2.70) 2

b[H – x] M ( x ) = -----------------------2 and

(2.71)

70 Chapter 2: Optimal Stiffness Distribution bH b[H – x] x D T ( x ) = --------------------- = ------- 1 – ---H γ∗ γ∗

(2.72)

3 2 bH x 2 b[H – x] D B ( x ) = ------------------------ = ----------- 1 – ---H 4sγ ∗ 2χ∗

(2.73)

b

H

x

B

Fig. 2.12: Continuous cantilever beam. A cantilever beam having a linear shear rigidity distribution and a quadratic bending rigidity distribution will be in a state of uniform deformation under uniform transverse loading. Taking typical values for γ ∗ , f ∗, and the aspect ratio B ⁄ H for a tall building modelled as a truss beam, 1 γ ∗ = --------400

f∗ = 3

B 1 ---- = --H 6

(2.74)

and evaluating s , H s = ------------- = 1 2 f ∗B

(2.75)

one obtains 2

χ∗ H H u ( H ) = γ ∗ H + -------------- = γ ∗ H ( 1 + s ) = --------2 200

(2.76)

This result corresponds to the extreme load value for displacement. One would

2.3 Stiffness Distribution for a Continuous Cantilever Beam under Static Loading

71

use these typical values together with b and H to establish an appropriate value for D T at x = 0 . As will be seen later, the rigidity distributions need to be modified near x = H in order to avoid excessive deformation under dynamic load.

Example 2.3. Cantilever beam - quasi static seismic loading The cantilever beam loading shown in Fig. 2.13 is used to simulate, in a quasi-static way, seismic excitation for low rise buildings. The triangular loading is related to the inertia forces associated with the fundamental mode response, and the concentrated force is included to represent the effect of the higher modes. Evaluating V and applying eqn (2.68) leads to b0 H x 2 1 D T = ----- P + ---------- 1 – ---- H 2 γ∗

(2.77)

A combination of constant and quadratic terms is a reasonable starting point for the transverse shear rigidity distribution.

P

b0

H

Fig. 2.13: Cantilever beam - quasi-static seismic loading

72 Chapter 2: Optimal Stiffness Distribution Example 2.4. Truss-beam revisited This example extends the treatment of the truss beam discussed in Example 2.2 and focuses on comparing the cross-sectional parameters required to satisfy the strength based versus the stiffness based performance criteria. Considering elastic behavior and given the desired design deformations d∗ c∗ γ ∗ and χ∗ , the corresponding extensional strains ε and ε must be less than d c the yield strains for the element materials, ε y and ε y respectively. That is d

d∗

γ ∗ sin 2θ = γ ∗ cos θ sin θ = --------------------2

(2.78)

c

c∗

Bχ∗ = ---------2

(2.79)

εy ≥ ε εy ≥ ε

Once the dimensions and the design deformations are specified, the structural material can be chosen to satisfy the motion design constraints defined by eqns (2.78) and (2.79). When the column strain is constrained to be related to the diagonal strain by ε

c∗

d∗ ε = -------f∗

(2.80)

eqn (2.79) can be written as d

c εy

εy ≥ -----f∗

(2.81)

To provide more options in satisfying the design requirements, different materials may be used. One must also insure that the stresses due to the design forces, V and M , are less than the yield stresses. c

d

The axial forces in the columns, F , and diagonals, F , are related to the transverse shear and moment by d

V = 2F cos θ M = BF

c

(2.82) (2.83)

2.3 Stiffness Distribution for a Continuous Cantilever Beam under Static Loading

73

The cross-sectional areas required to provide the strength capacity follow from eqns (2.82) and (2.83) d V A strength ≥ -------------------------d∗ 2σ cos θ

(2.84)

c M A strength ≥ ------------c∗ Bσ

(2.85)

where σ∗ denotes the allowable stresses based on strength considerations. The rigidity terms for this model are d d

2

D T = 2A E sin θ cos θ

(2.86)

c c

A E B2 D B = -------------------2

(2.87)

Substituting eqns (2.86) and (2.87) in the motion based design criteria, V D T = ----γ∗

(2.88)

M D B = -----χ∗

(2.89)

one obtains the following expressions for the cross-sectional areas required to satisfy the stiffness requirement. d V A stiffness ≥ --------------------------------------------d 2 2E γ ∗ sin θ cos θ

(2.90)

c 2M f ∗M A stiffness ≥ ------------------- = ----------------------------------------c 2 c E B χ∗ E Bγ ∗ sin θ cos θ

(2.91)

The ratio of areas provides a measure of the relative importance of strength versus stiffness

74 Chapter 2: Optimal Stiffness Distribution d d d∗ d A strength ε E γ ∗ E sin θ cos θ ----------------------- = -------------------------------------- = ---------------d d∗ d∗ A stiffness σ σ

(2.92)

c c c∗ c A strength ε E χ∗ E B ----------------------- = ---------------- = -------------c c∗ c∗ A stiffness σ 2σ

(2.93)

Stiffness controls when the ratios are less than unity. The limit on γ ∗ follows from eqn (2.92) d∗ σ ∗ γ = -------------------------------d E sin θ cos θ

(2.94)

For γ ∗ < γ ∗ , the cross-sectional area is governed by the deformation constraint, and eqns (2.90) and (2.91) apply. When γ ∗ > γ ∗ , the allowable stress is the controlling factor, and eqns (2.84) and (2.85) apply. Inelastic behavior occurs in this case. Values of γ ∗ for a range of allowable stress levels for steel calculated using an angle of 45° are listed below in Table 2.1. With high-strength steel, the structure can experience substantial transverse shear deformation and still remain elastic. Table 2.1: γ ∗ values for various steel strengths. σ∗ ( MPa )

γ∗

250

1 ⁄ 400

500

1 ⁄ 200

1000

1 ⁄ 100

2.4 Stiffness Distribution for a Discrete Cantilever Shear Beam - Static Loading

75

2.4 Stiffness distribution for a discrete cantilever shear beam - static loading Consider the set of equilibrium equations relating the nodal forces and story displacements for an n ‘th order discrete shear beam: p1

k1 + k2

p2

–k2

. = . . pn

. . . 0

–k2

. . . 0

u1

k2 + k3 . . . 0

u2

. . . 0

.

. . . . . . . . . . . kn un

(2.95)

P = KU These equations are derived in section 2.2. In the normal analysis problem, one specifies P and K , and solves for U . The problem is statically determinate since there are n equations for the n unknown displacements. In this problem, one specifies U and P , and attempts to determine the n stiffness factors. Since there are n linear algebraic equations, it should be possible to solve for the n stiffness coefficients by rearranging the equations such that the k ‘s are the unknowns. The vector containing the stiffness coefficients is denoted by k . k1 k2 k =

. . . kn

(2.96)

With this definition, eqn (2.95) is written as Sk = P

*

(2.97)

where the elements of S are linear combinations of the prescribed displacement components, u* i , and P * contains the prescribed loads. The entries in the i ‘th

76 Chapter 2: Optimal Stiffness Distribution row of S follow from eqn (2.66). S ( i, i ) = u* i – u* i – 1 S ( i, i + 1 ) = u* i – u* i + 1

(2.98)

S ( i, j ) = 0 for j ≠ i, i + 1 These entries define S to be an “upper triangular” matrix. The diagonal entries for S are interstory displacements. Normally, one would not specify an interstory displacement to be 0 since it would require an infinite shear stiffness. It follows that S will be non-singular, and there will be a unique solution for k .

Example 2.5. 3 DOF shear beam Consider a 3 DOF cantilever beam subjected to uniform nodal loading and required to have a linear displacement profile. The design values are: 19.6 P = 19.6 kN 19.6

(2.99)

0.025 U = 0.050 m 0.075

(2.100)

*

*

Applying eqn (2.98) results in k1 0.025 – 0.025 0 19.6 = 0 0.025 – 0.025 k 2 19.6 0 0 0.025 k 19.6 3

(2.101)

Since the coefficient matrix is upper triangular, one solves for k 3 , k 2 , and then k 1 by backsubstitution. The solution is

2.5 Stiffness Distribution - Truss under Static Loading 77

2352 k = 1558 kN/m 784

(2.102)

As illustrated with the above example, solving eqn (2.97) is relatively simple, and one can easily handle arbitrary loading and permissible (non-zero interstory drift) displacement profiles.

2.5 Stiffness distribution - truss under static loading The approach discussed in the previous sections can be extended to deal with other types of structures such as trusses and frames. Since trusses involve simple equations, the approach is illustrated using a planar truss. A more general discussion is contained in Gallegos (1998). P2, u2 P1, u1

1

3 2

45o

60o

Fig. 2.14: Planar truss. The truss shown in Fig 2.14 has 3 members and 2 displacement variables; the supports are assumed to be rigid so there are no support movements. Each member has an extension, e , and corresponding force, F . Assuming linear elastic behavior, the force-displacement relation for member i is Fi = ki ei

(2.103)

78 Chapter 2: Optimal Stiffness Distribution where k is the member stiffness factor, AE k i = -------- L member i

(2.104)

Using a geometric analysis, the member extensions are expressed in terms of the nodal displacements: 2 e 1 = ------- ( u 1 + u 2 ) 2 e2 = u2

(2.105)

1 3 e 3 = – --- u 1 + ------- u 2 2 2 The nodal and member forces are related by the nodal force equilibrium equations 2 1 P 1 = ------- F 1 – --- F 3 2 2

(2.106)

2 3 P 2 = ------- F 1 + F 2 + ------- F 3 2 2 Since there are 3 extensions and only 2 displacements, one cannot arbitrarily specify values for all 3 extensions. They are constrained by the geometric compatibility equation that follows by combining the first and third rows of eqn (2.105) and substituting for u 2 in terms of e 2 2 1 ------- e 1 + e 3 = --- ( 1 + 3 )e 2 2 2

(2.107)

In general, one cannot have a state of uniform deformation in a statically indeterminate system. One possible strategy is to specify Once the extensions are known, one equilibrium equations, and establish a stiffness factors. The result is

u 1 and u 2 , and evaluate the extensions. can substitute for the forces in the set of equations relating the member

2.5 Stiffness Distribution - Truss under Static Loading 79 Sk = P* 2 ------- e 1* 2 = * 2 P2 ------- e 1* 2 P1

*

0 e 2*

1 k1 – --- e 3* 2 k2 3 * ------- e 3 k 3 2

(2.108)

Since there are 3 unknown stiffness factors and only 2 equations, eqn (2.108) represents an undetermined system of equations and one cannot obtain a unique solution for k . This situation is typical of indeterminate structures. A solution of eqn (2.108) can be generated by selecting 2 stiffness factors as the primary variables, such as k 1 and k 2 , and solving for these variables in terms of P 1* , P 2* , and the third stiffness factor, k 3 . The result is written as k1 = k1 + a1 k3

(2.109)

k2 = k2 + a2 k3 where k 1 , k 2 depend on the forces, P 1 and P 2 . Equation (2.109) represent a constraint on the stiffness parameters. An additional condition must be introduced in order to determine k 3 . This additional condition is usually established by formulating an optimization problem. One possibility is to work with a weighted volume measure that is related to the member stiffness factor, and seek the solution that corresponds to the minimum value of the sum of the weighted volumes. For example, one can take V i = ( AEL ) i

(2.110)

as the member volume variable. Expressing k i in terms of V i, AE 1 k i = -------- = ------ V i L i L i2

(2.111)

and substituting for k i in the constraint equations transforms these equations to

80 Chapter 2: Optimal Stiffness Distribution 2

V 1 = L 12 k 1 + a 1 ( L 1 ⁄ L 3 ) V 3 2

V 2 = L 22 k 2 + a 2 ( L 2 ⁄ L 3 ) V 3

(2.112)

The optimization problem can be stated as: minimize f ( V ) = V 1 + V 2 + V 3

(2.113)

Using eqn (2.112), the objective function reduces to (2.114)

f = bo + b1 V 3

where b o and b 1 are constants. Since f is a linear function, there is no minimum value and it follows that the approach does not introduce an additional independent constraint. Another possibility is to use a least square approach, i.e., to select V 3 such that the sum of the squares of the weighted member volumes is a minimum. The objective function for this optimization problem is 2 2 1 2 f = --- ( V 1 + V 2 + V 3 ) 2

(2.115)

Substituting for V 1 and V 2 transforms f to f = co + c1 V 3 + c2 V 3

2

(2.116)

Requiring f to be stationary with respect to a change in V 3 , ∂f = c1 + 2 c2 V 3 = 0 ∂V3

(2.117)

leads to c1 V 3 = – -------2c 2

(2.118)

The problem with this approach is that it tends to eliminate redundant members,

2.5 Stiffness Distribution - Truss under Static Loading 81 such as member 3 for this example. This tendency is evidenced by “optimal” volume values which are low, and may even be negative. A negative value for the volume indicates the member should be deleted. A third approach partially overcomes the limitation of the least square approach by working with the deviation from the mean value as the member variable. Let V m denote the mean value, 1 V m = --- ( V 1 + V 2 + V 3 ) 3

(2.119)

and V i represent the deviation from the mean value for member i, (2.120)

Vi = Vi – Vm The optimization problem is stated as 2 2 1 2 minimize f = --- ( V 1 + V 2 + V 3 ) 2

(2.121)

subject to the constraints on V 1 and V 2 . Substituting for V 1 and V 2 transforms f to 2

f = do + d1 V 3 + d2 V 3

(2.122)

The remaining steps are the same as for the classical least square approach. This approach generates a solution which tends toward the same value for each member. However, as will be illustrated in the following example, it does not eliminate the possibility of negative volume members.

Example 2.6. Application of least square approaches The procedures described above are applied to Fig 2.14 specialized for the data contained in (1)

82 Chapter 2: Optimal Stiffness Distribution

P2, u2 P1, u1

4m

1

3 2

4m

P 1 = 100kN

P 2 = 300kN

u 1 = 0.01m

u 2 = 0.01m

2.31m

(1)

Using eqn (2.105), the extensions are 2 e 1 = ------- ( 0.01 + 0.01 ) = 0.01414m 2 e 2 = 0.01m

(2)

1 3 e 3 = – --- ( 0.01 ) + ------- ( 0.01 ) = 0.00366m 2 2 Substituting for the e’s in eqn (2.108) results in 0.01k 1 – 0.00183k 3 = 100, 000 0.01k 1 + 0.01k 2 + 0.00317k 3 = 300, 000 Solving for k 1 and k 2 in terms of k 3 leads to

(3)

2.5 Stiffness Distribution - Truss under Static Loading 83

k 1 = 0.183k 3 + 10 7 k 2 = – 0.500k 3 + 2 ⋅ 10 7

(4)

Equation (4) represents the constraint on k 1 and k 2 . Substituting for k in terms of weighted volume measure, V=AEL, using eqn (2.111) transforms eqn (4) to V 1 = 0.275V 3 + 32 ⋅ 10 7 V 2 = – 0.375V 3 + 32

⋅ 10 7

(5)

Least square procedure The objective function is 2 2 2 1 f = --- ( V 1 + V 2 + V 3 ) = f ( V 3 ) 2

(6)

Differentiating with respect to V 3 , ∂V 1 ∂V 2 ∂f = V1 + V2 + V3 ∂V3 ∂V3 ∂V3

(7)

substituting for V 1 and V 2 , and then setting the resulting expression equal to zero results in 7

1.216V 3 – 3.222 ⋅ 10 = 0

(8)

The “optimal” least square values are V 3 = 2.65 ⋅ 10 7 V 1 = 32.71 ⋅ 10 7 V 2 = 31.00

⋅ 10 7

( units are newtons ⋅ meters )

(9)

84 Chapter 2: Optimal Stiffness Distribution The low value for V 3 indicates that this member is “redundant”, and could be removed. Mean value least square procedure One starts by establishing the mean value, 1 V m = --- ( V 1 + V 2 + V 3 ) = 0.3V 3 + 21.33 ⋅ 10 7 3

(10)

and the deviations V 1 = V 1 – V m = – 0.025V 3 + 10.71 ⋅ 10 7 V 2 = V 2 – V m = – 0.675V 3 + 10.67 ⋅ 10 7

(11)

V 3 = V 3 – V m = 0.700V 3 – 21.33 ⋅ 10 7 The objective function for this case is

1 f = --- ( V 12 + V 22 + V 32 ) = f ( V 3 ) 2

(12)

Setting the derivative of f with respect to V 3 equal to zero, ∂V 1 ∂V 2 ∂V 3 ∂f = V1 + V2 + V3 = 0 ∂V3 ∂V3 ∂V3 ∂V3

(13)

and noting eqn (11), reduces eqn (13) to 0.946V 3 – 22.40 ⋅ 10 7 = 0 The final mean value least square results are

(14)

2.5 Stiffness Distribution - Truss under Static Loading 85 V 3 = 23.67 ⋅ 10 7 V 1 = 38.50 ⋅ 10 7 V 2 = 23.10

(15)

⋅ 10 7

( units are newtons ⋅ meters ) Note that now V 3 is the same order of magnitude as V 1 and V 2 . The solution is sensitive to the prescribed loading and displacement quantities. To illustrate this point, the computation is repeated taking the same imposed displacements but different sets of nodal forces. The results are listed below: Case 1 P 1 = 100kN

P 2 = 200kN

Least square: V 1 = 31.37 ⋅ 10 7 V 2 = 16.86 ⋅ 10 7 V 3 = – 2.29 ⋅ 10 7 Mean value least square: V 1 = 35.37 ⋅ 10 7 V 2 = 11.40 ⋅ 10 7 V 3 = 12.26 ⋅ 10 7 Case 2 P 1 = 100kN Least square: V 1 = 30.02 ⋅ 10 7 V 2 = 2.71 ⋅ 10 7 V 3 = – 7.25 ⋅ 10 7

P 2 = 100kN

86 Chapter 2: Optimal Stiffness Distribution Mean value least square: V 1 = 32.23 ⋅ 10 7 V 2 = – 0.32 ⋅ 10 7 V 3 = 0.86 ⋅ 10 7 The case 2 result indicates that members 2 and 3 should be deleted. This is the correct solution for the specified loading and displacement. Taking P 1 = P 2 corresponds to a load at 45o, which coincides with the direction of member 1. Furthermore, an extension of member 1 corresponds to u 1 = u 2 , which is the specified displacement condition.

The procedures described above can be generalized to deal with an arbitrary truss. Suppose there are n degrees of freedom and m members. It follows that there are n equilibrium equations relating the m stiffness factors and the n prescribed nodal forces. These equations are written as Sk = P

(2.123)

where S is of order n × m . When m > n , the structure is indeterminate, and eqn (2.123) does not have a unique solution for k . In this case the problem is undetermined, and an optimization statement has to be formulated. One can work with either stiffness or weighted volume measures as the variables. To allow for different choices, the form of eqn (2.123) is generalized to AX = P

(2.124)

where X is an m th order vector containing the selected variables, and A is of order n × m . If one selects V defined by eqn (2.111) as the measure, A is obtained 2 by multiplying the i’th column of S by 1 ⁄ L i , and taking i from 1 to m . Background material on computational techniques for solving linear algebraic equations is contained in Strang,1993. In what follows, the use of one of these techniques to solve eqn (2.124) is described in general terms. The same procedure is also applied in quasi-static active control problems in Chapter 7.

2.5 Stiffness Distribution - Truss under Static Loading 87 The least square approach works with a scalar quantity, f, that is equal to the sum of the squares of the elements of X . Using matrix notation, f is given by 1 T f = --- X X 2

(2.125)

The elements of X are constrained by eqn (2.124). It follows that n elements can be expressed in terms of m – n elements, which can be interpreted as the unknowns that need to be selected such that f is a minimal value. Assuming the first n columns of A are independent, eqn (2.124) can be expressed in partitioned form X1 A1 A2 = P X2

(2.126)

where A 1 is of order n × n . Solving eqn (2.126) for X 1 leads to X 1 = X 1 + BX 2

(2.127)

where X 2 represents the unknown elements. Using eqn (2.127), X is expressed in terms of X 2 , X = X 1 + B X 2 = X + CX 2 I 0

(2.128)

Substituting for X , the objective function expands to 1 f = --- ( X + CX 2 ) T ( X + CX 2 ) 2

(2.129)

Requiring f to be stationary with respect to X 2 leads to the following set of m – n equations C T CX 2 = – C T X

(2.130)

One solves eqn (2.130) for X 2 , and then determines X 1 with eqn (2.127). This

88 Chapter 2: Optimal Stiffness Distribution solution corresponds to an absolute minimum value of f (see Strang, 1993). The MATLAB statement, X = pinv ( A )*P , generates this least square solution. The mean value least square approach works with the deviation from the mean, e i = x i – x mean

(2.131)

For this case, there are m members, and x mean is given by 1 x mean = ---- ( x 1 + x 2 + … + x m ) m

(2.132)

Using matrix notation, the deviation vector, e , can be expressed as, e = DX = D ( X + CX 2 )

(2.133)

where the entries in row i of D are 1 D ( i, i ) = 1 – ---m 1 D ( i, j ) = – ---m

(2.134) for j ≠ i

The objective function is a quadratic form in e . 1 f = --- e T e 2

(2.135)

Substituting for e using eqn (2.133), and enforcing stationarity with respect to X 2 leads to the equation for X 2 . ( D T C T CD )X 2 = – D T C T X

(2.136)

An alternative way of generating the solution is based on using Lagrangian multipliers to incorporate the constraint equations in the objective function (Strang, 1993). The “generalized” function is defined as

2.5 Stiffness Distribution - Truss under Static Loading 89 1 J = --- e T e + λ T ( AX – P ) 2 ˜

(2.137)

1 = --- X T ( D T D )X + λ T ( AX – P ) 2 ˜

where λ is a vector containing the n Lagrangian multipliers. Requiring J to be stationary with respect to both X and λ leads to the following set of equations: D T DX + A T λ = 0 ˜ AX = P

(2.138)

These equations can be combined, DT D A

AT X = 0 0 λ˜ P

(2.139)

and solved in a single step using a linear equation solver such as one of the MATLAB functions Example 2.7. Comparison of strength vs. displacement based design P 2, u 2 u1 θ θ 1 2

3

Consider the 3 member truss shown in the figure. Suppose members 1 and 3 to have the same properties. It follows that u 1 = 0 and e 3 ≡ e 1 . The member elongations are:

90 Chapter 2: Optimal Stiffness Distribution e 1 = u 2 cos θ e2 = u2

(1)

Noting eqn (1), the elongations are constrained by e 1 = e 2 cos θ

(2)

The member flexibility factor, f , is defined as 1 L f = --- = -------k AE

(3)

Then, the member force-deformation relation can be expressed as e = fF

(4)

Using (4), the geometric compatibility equation expressed in terms of member forces has the form f 1 F 1 = cos θ f 2 F 2

(5)

The solution for F 1 and F 2 is f1 F 1 = cos θ ------ F 2 f2 P2 F 2 = ---------------------------------f1 1 + 2 cos2 θ -----f2 •

(6)

Strength based design

Ideally, one would want the member stresses to be equal to an allowable stress, σ a . However, a state of uniform stress is not possible because of the indeterminate nature of the structure. Noting that e = Lε for a member, eqn (1) can be written as

(7)

2.5 Stiffness Distribution - Truss under Static Loading 91 L 1 ε 1 = cos θ L 2 ε 2

(8)

Then assuming the same material for members 1 and 2, and substituting L 2 = L 1 cos θ results in the constraint equation for the member stresses: σ 1 = ( cos2 θ )σ 2

(9)

According to eqn (9), σ 1 ≤ σ 2 . Let σ2 = σa σ 1 = σ a cos2 θ

(10)

represent the design values for the stresses. The member cross-sectional areas are: F2 A 2 = -----σa

(11)

F1 F1 A 1 = ------ = --------------------σ1 σ a cos2 θ Substituting for F 1 and F 2 in the equilibrium equation, F 2 + 2F 1 cos θ = P 2

(12)

leads to a relationship between the cross-sectional areas: P2 A 2 + [ 2 cos3 θ ]A 1 ≥ -----σa

(13)

92 Chapter 2: Optimal Stiffness Distribution The allowable solutions lie outside the triangle shaded in Figure 1. A2 P2 -----σa

Acceptable zone

0

A1 P2 – 1 3 ------ ( 2 cos θ ) σa Figure 1

The constraint equation is expressed as A 2 + αA 1 = β

(14)

There is no unique solution for the areas. One approach is to apply a least square approach to the member areas. Minimizing the function 2

J = ( A 1 + A 22 + A 32 )

(15)

subject to the constraint, eqn (14), results in the following design: αβ A 1 = --------------2 + α2 2β A 2 = --------------2 + α2 •

(16)

Motion based design

The member forces are expressed in terms of the displacement, u 2 , by using equations (1) and (4)

2.6 Stiffness Distribution for a Cantilever Beam - Dynamic Response 93

A1 E u2 F 1 = ----------- e 1 = A 1 E ------ cos θ L1 L1 A2 E u2 F 2 = ----------- e 2 = A 2 E -----L2 L2

(17)

Substituting for the forces in eqn (12) leads to P2 L2 A 2 + [ 2 cos3 θ ]A 1 ≥ -----------Eu 2

(18)

This equation is similar to the strength based constraint equation. They differ only with respect to the right hand sides. Comparing these terms, it follows that motion based design controls when P2 L2 P2 ------------ > -----Eu 2 σ a

(19)

which translates to σa u 2* < L 2 -----E

(20)

2.6 Stiffness distribution for a cantilever beam - dynamic response Rigidity distribution - undamped response The distribution of rigidity is established by requiring the fundamental mode for a dynamically loaded cantilever beam to have a certain profile. For undamped free vibration, the distributed loading consists only of the inertia terms

94 Chapter 2: Optimal Stiffness Distribution 2

∂ u ( x, t ) b ( x, t ) = – ρ m ( x ) ---------------------- = – ρ m ( x )u˙˙( x, t ) 2 ∂t m ( x, t ) = – J ( x )

2

(2.140)

∂ ˙˙ ( x, t ) β ( x, t ) = – J ( x )β 2 ∂t

For simplicity, a uniform mass distribution is assumed. The equilibrium equations specialized for uniform mass, elastic behavior, and no rotatory inertia loading take the form H

∫

(2.141)

V ( x, t ) dx = D B ( x )χ ( x, t )

(2.142)

V ( x, t ) = – ρ m u˙˙( x, t ) dx = D T ( x )γ ( x, t ) x

H

M ( x, t ) =

∫ x

The desired behavior is periodic vibration with uniform shear and bending deformation throughout the length. Requiring periodic vibration with uniform deformation modes is achieved by specifying the deformations as follows: γ ( x, t ) = γ ∗ cos ( ω 1 t + δ )

(2.143)

χ ( x, t ) = χ∗ cos ( ω 1 t + δ )

(2.144)

where ω 1 is the fundamental circular frequency of the beam and δ is an arbitrary scalar. The displacement expressions are generated using eqns (2.11) and (2.12) 2 χ∗ x u = γ ∗ x + ------------ cos ( ω 1 t + δ ) 2

(2.145)

β = [ χ∗ x ] cos ( ω 1 t + δ )

(2.146)

2s χ∗ = -----γ ∗ H

(2.147)

2.6 Stiffness Distribution for a Cantilever Beam - Dynamic Response 95 Substituting for u , γ and χ in eqns (2.141) and (2.142), the cosine terms cancel out, and one obtains the following expressions for the rigidity distributions, 2

2

ρm ω1 H x x 2 2s x 3 D T ---- = ---------------------- 1 – ---- + ----- 1 – ---- H 2 H 3 H 2

(2.148)

4

ρm ω1 H 1 1 x x 3 x 4x 1 x 4 D B ---- = ---------------------- --- 1 – -------- + --- ---- + ----- 2 – 3 ---- + ---- H H H 2 4 3H 3 H 6s

(2.149)

The magnitude of D T at x = 0 can be determined by specifying representative values of base shear and transverse shear deformation [ V ( 0 ) ] max D T ( 0 ) = --------------------------γ∗

(2.150)

Once D T ( 0 ) is defined, the fundamental frequency can be obtained with eqn (2.148) specialized for x = 0 . The final expressions are 2

2

ρm ω1 H 2s D T ( 0 ) = ---------------------- 1 + ----2 3

ω1 =

2 [ V ( 0 ) ] max -----------------------------------------2 2s γ ∗ ρ m H 1 + ----3

(2.151)

(2.152)

Figures 2.15 and 2.16 show the normalized distribution of D T and D B for typical low and high rise buildings. Figure 2.17 contains plots of the first five mode shapes corresponding to the above rigidity distributions, and Fig. 2.18 shows the shear deformation profiles of those modes. The results are relatively insensitive to the parameter s , which ranges from 0 for a pure shear beam to about 1 for a tall building. It should be noted that the mode shape and frequency expressions are the exact solution for the fundamental modal response of the beam having the rigidity distributions defined by eqns (2.148) and (2.149). This approach leads to the proper distribution of rigidities, but the actual magnitude is not defined since one has to specify D T ( 0 ) using a representative value of base shear. In effect, one has to calibrate the rigidities for a particular

96 Chapter 2: Optimal Stiffness Distribution loading. The calibration procedure is discussed in detail in later sections.

1

x Normalized height ---H

0.9 0.8 0.7

T = 5.35s H = 200m ρ m = 20000kg/m s = 0.63

0.6 0.5 0.4 0.3

T = 0.49s H = 25m ρ m = 20000kg/m s = 0.15

0.2 0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

Shear rigidity distribution D T ⁄ D T ( 0 )

1

Fig. 2.15: Normalized shear rigidity distribution.

1

T = 5.35s H = 200m ρ m = 20000kg/m s = 0.63

x Normalized height ---H

0.9 0.8 0.7 0.6

T = 0.49s H = 25m ρ m = 20000kg/m s = 0.15

0.5 0.4 0.3 0.2 0.1 0 0

0.1

0.2

0.3

0.4

0.5

0.6

Bending rigidity distribution

0.7

0.8

0.9

DB ⁄ DB ( 0 )

Fig. 2.16: Normalized bending rigidity distribution.

1

2.6 Stiffness Distribution for a Cantilever Beam - Dynamic Response 97

1

x Normalized height ---H

0.9

T = 5.35s H = 200m ρ m = 20000kg/m s = 0.63

0.8 0.7 0.6 0.5 0.4

T = 0.49s H = 25m ρ m = 20000kg/m s = 0.15

0.3 0.2 0.1 0

1

2

3

4

5

Mode number Fig. 2.17: Mode shapes.

1 0.9

T = 5.35s H = 200m ρ m = 20000kg/m s = 0.63

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

T = 0.49s H = 25m ρ m = 20000kg/m s = 0.15

0.3 0.2 0.1 0

1

2

3

4

5

Mode number Fig. 2.18: Mode shear deformation profiles.

98 Chapter 2: Optimal Stiffness Distribution

2.7 Stiffness distribution for a discrete shear beam - dynamic response Our starting point is eqn (2.62), the equilibrium equation for an undamped discrete system. ˙˙ + KU = P MU

(2.153)

The forms of M and K specialized for an n ‘th order shear beam are defined by eqn (2.67). Setting P = 0 and expressing U as U = Φ cos ( ωt + δ )

(2.154)

transforms eqn (2.153) to 2

KΦ = ω MΦ

(2.155)

Solutions of eqn (2.155) define the frequencies ( ω j ) and mode shapes ( Φ j ) corresponding to a particular beam, i.e., when M and K are specified. In the inverse approach, one specifies a mode shape and determines K . Let Φ = Φ

*

(2.156)

denote the specified modal profile. Substituting in eqn (2.155) leads to *

2

KΦ = ω MΦ

*

(2.157)

Defining a scaled version of K , 1 K' = ------K 2 ω

(2.158)

reduces eqn (2.157) to *

*

K' Φ = M Φ = P'

(2.159)

The remaining steps are the same as for the static case discussed in section 2.4. Eqn (2.159) is written as

2.7 Stiffness Distribution for a Discrete Shear Beam - Dynamic Response 99 (2.160)

S'k' = P'

where k' contains the n scaled stiffness factors, and S' is an upper triangular * matrix containing the elements of Φ . *

*

S' ( i, i ) = Φ i – Φ i – 1 *

*

S' ( i, i + 1 ) = Φ i – Φ i + 1 S' ( i, j ) = 0 for j ≠ i, i + 1

(2.161)

One solves eqn (2.160) for k' . Once ω is specified, the actual values of the stiffness parameters can be determined with 2

k j = ω k' j

(2.162)

Example 2.8. 3 DOF shear beam This example illustrates the determination of the stiffness distribution that produces the following linear profile for the fundamental mode. 1⁄3 * Φ = 2⁄3 1

(2.163)

*

It is convenient to scale Φ so that the magnitude of the maximum element is equal to 1 . The corresponding forms of P' and S' are

*

m1 ⁄ 3

P' = M Φ = 2m 2 ⁄ 3 m3

(2.164)

100 Chapter 2: Optimal Stiffness Distribution

S' =

1 1 --- – --3 3 1 0 --3

0

1 – --3 1 0 0 --3

(2.165)

Solving eqn (2.160) results in k 3' = 3m 3 k 2' = 2m 2 + 3m 3

(2.166)

k 1' = m 1 + 2 m 2 + 3m 3 *

*

When all the elements of Φ are positive, one can show that Φ is actually the fundamental mode shape for the system having the stiffness distribution defined by eqn (2.160), and ω is the corresponding fundamental frequency.

2.8 Stiffness calibration Governing equations- fundamental mode response of a discrete shear beam The shear beam is assumed to have some damping and this energy dissipation mechanism is represented as linear viscous damping. The corresponding governing equations have the form ˙˙ + CU˙ + KU = P MU

(2.167)

where CU˙ defines the nodal forces due to viscous damping. How one establishes C is treated in the next chapter. Our objective here is to reduce eqn (2.167) to a single scalar equation which governs the response of the fundamental mode. The displacement vector is expressed as U = q ( t )Φ

*

(2.168)

2.7 Stiffness Calibration 101 *

where Φ is the dimensionless prescribed mode shape used to establish the relative stiffness distribution in the previous section, and q ( t ) can be interpreted as the displacement parameter that defines the response of the fundamental mode. Substituting for U in eqn (2.167) *

*

*

(2.169)

MΦ q˙˙ + CΦ q˙ + KΦ q = P * T

and then premultiplying by ( Φ ) results in the following scalar equation for q : ˜ q˙˙ + c˜ q˙ + k˜ q = p˜ m

(2.170)

where the quantities with superscript tilde are equivalent one degree of freedom mass, damping, stiffness, and force quantities. Their definition equations are: T

˜ = ( Φ* ) M Φ* m

(2.171)

* T * ˜ c˜ = ( Φ ) CΦ = 2ξ˜ ωm

(2.172)

2 ˜ k˜ = ω m

(2.173) T

* p˜ = ( Φ ) P

(2.174)

The form of eqn (2.173) follows from eqn (2.157)

Example 2.9. 3 DOF shear beam revisited *

Consider the 3 DOF system defined in example 2.8. Taking Φ according to eqn (2.163), the one dimensional mass and force measures are

˜ = 1--- 2--- 1 m 3 3

m1 ⁄ 3

m1 4 2m 2 ⁄ 3 = ------- + --- m 2 + m 3 9 9 m3

(2.175)

102 Chapter 2: Optimal Stiffness Distribution

p˜ = 1--3

2 --3

p1 1

1 2 p 2 = --- p 1 + --- p 2 + p 3 3 3 p3

(2.176)

The nodal forces due to seismic excitation are proportional to the nodal masses. A typical force is pi = –mi a g

(2.177)

where a g is the ground acceleration. Substituting for p i in eqn (2.176) results in m 1 2m 2 p˜ = – ------- + ---------- + m 3 a g 3 3

(2.178)

It is convenient to express p˜ due to seismic acceleration as ˜a p˜ = – Γm g

(2.179)

where Γ is a dimensionless factor. For this example, Γ is given by m 1 2m 2 ------- + ---------- + m 3 3 3 Γ = -------------------------------------m 1 4m 2 ------- + ---------- + m 3 9 9

(2.180)

When the masses are equal, Γ = 9 ⁄ 7 . In order to evaluate the one dimensional damping coefficient, c˜ , the system damping matrix, C must be specified. The form of C depends on how the viscous damping elements are located throughout the structure. Figure 2.19 shows an arrangement consisting of viscous dampers inserted between adjacent masses. In this case, the shear force in an element depends on both the shear deformation and the time rate of change of the shear deformation. Assuming linear behavior, the typical equations have the form V i = k i ( u i – u i – 1 ) + c i ( u˙ i – u˙ i – 1 ) p i = m i u˙˙i + V i – V i + 1

(2.181)

2.7 Stiffness Calibration 103 Equation (2.181) shows that the forms of K and C are similar; one replaces k i with c i . Taking C as c1 + c2 C =

–c2

0

–c2

c2 + c3

–c3

0

–c3

c3

(2.182)

leads to 1 c˜ = --- ( c 1 + c 2 + c 3 ) 9

(2.183)

The contribution of the individual dampers to c˜ is the same since the assumed displacement profile corresponds to constant shear deformation. m3 k3, c3

Vi+1 m2

u2, p2 k2, c2

ui, pi

mi Vi

m1 k1, c1

Fig. 2.19: Dampers included in shear beam elements.

104 Chapter 2: Optimal Stiffness Distribution

c3 ’

k3

Vi+1

c2’ u2, p2

. ci’ ui

k2

pi Vi

c1 ’

k1

Fig. 2.20: Dampers attached to nodal masses. The damper arrangement shown in Fig. 2.20 produces nodal forces that depend on the nodal velocities. Assuming linear behavior, the typical equations are V i = ki ( ui – ui – 1 ) p i = m i u˙˙i + c' i u˙ + V i – V i – 1

(2.184)

With this arrangement, C is similar to M . Then c' 1 0 0 C =

0 c' 2 0 0 0 c' 3

(2.185)

and c' 1 4 c˜ = ------ + --- c' 2 + c' 3 9 9

(2.186)

Equation 2.186 shows that the most effective damper location is node 3. This

2.7 Stiffness Calibration 105 result can also be deduced by examining the assumed displacement profile. Node 3 corresponds to the maximum element in Φ * , and therefore has the largest velocity.

Governing equations - fundamental mode response of a continuous beam The displacement variables for a continuous beam are the translation, u , and rotation, β , of the cross-section. Noting equations (2.145) and (2.146), the displacement expressions corresponding to uniform shear and bending deformation are expressed as u = q ( t )φ ( x )

(2.187)

β = q ( t )ψ ( x )

(2.188)

where φ , ψ are the fundamental mode shape functions, x 2 x φ = ---- + s ---- H H

(2.189)

2s x ψ = ----- ---- H H

(2.190)

q is the modal amplitude parameter, and 2s ⁄ H is the ratio of prescribed bending deformation to prescribed shear deformation. A pure shear beam has s = 0 . The deformations and maximum displacement are related to q by 1 γ = u, x – β = q ---- H

(2.191)

2s χ = β, x = q ------- 2 H

(2.192)

u(H ) = q(1 + s)

(2.193)

The equilibrium equation expressed in terms of q(t) can be generated using the Principle of Virtual Displacements (Bathe, 1996). This Principle is used to

106 Chapter 2: Optimal Stiffness Distribution transform continuous equilibrium equations into a set of discrete equations, and can be interpreted as a variational statement. It provides the basis for the Finite Element Displacement Method. One starts by introducing an expansion for the displacement variables in terms of generalized coordinates. For this case, φ ( x ) and ψ ( x ) are the prescribed displacement fields, and q(t) is the generalized coordinate. A virtual displacement is defined as a displacement distribution generated by perturbing the generalized coordinate by a small amount, δq . According to this principle, the first order work done by the external loads during this virtual displacement is equal to the first order work done by the internal forces during the corresponding virtual deformation. The transverse shear force and bending moment are the internal forces for a beam; the corresponding deformations are γ and χ . Assuming the external loading consists only of the transverse load, b, the mathematical form of the principle is H

∫

H

( Mδχ + Vδγ ) dx =

0

∫

bδu dx

(2.194)

0

Taking u according to eqn (2.187), the virtual terms are related to δq by: δu = δqφ 1 δγ = δq ---- H

(2.195)

2s δχ = δq ------- 2 H Using eqn (2.195), and requiring the equality to be satisfied for arbitrary δq results in

∫

H

2s 1 ------ M + ----V dx = 2 H 0 H

H

∫

bφ dx

(2.196)

0

To proceed further, M , V , and b must be related to q . Assuming linear behavior and including viscous damping in both the

2.7 Stiffness Calibration 107 material and external loading terms, the expressions for the internal shear force and moment reduce to 1 V = D T γ + C T γ˙ = ---- ( D T q + C T q˙ ) H

(2.197)

2s M = D B χ + C B χ˙ = ------- ( D B q + C B q˙ ) 2 H

(2.198)

where C T , C B are material damping parameters. Including inertia and damping contributions, the external distributed loading expands to b = – ρ m u˙˙ – c'u˙ + b = – ρ m φq˙˙ – c'φq˙ + b

(2.199)

where c' is a damping parameter and b is the prescribed loading. Substituting in eqn (2.196), and writing the resulting equation in the “same” form as eqn (2.170), ˜ q˙˙ + c˜ q˙ + k˜ q = p˜ , leads to the following definition equations for the onem dimensional parameters:

˜ = m

p˜ =

∫ ∫

H

2

ρ m φ dx

(2.200)

φb dx

(2.201)

0

H

0

k˜ =

c˜ =

H

2 1 2 4s ˜ ----------- 2 D T + 4- D B dx = ω 1 m H 0 H

∫

H

2 1 2 4s ------2- C T + -------4- C B + c'φ dx H 0 H

∫

(2.202)

(2.203)

Equation (2.202)follows from the definition of the rigidity distributions (see eqns (2.148) and (2.149)). The other terms depend on the mass density distribution and prescribed loading. The following example considers various cases.

108 Chapter 2: Optimal Stiffness Distribution

Example 2.10. One dimensional parameters - continuous beam 1. Uniform mass density * ρ m = cons tan t = ρ m 2

s ˜ = ρ * H 1--- + --s- + ---m m 3 2 5

(2.204)

2. Uniform loading *

b = b (t) *

b H 2s p˜ = ---------- 1 + ----- 2 3

(2.205)

3. Linear loading x b = b 0 ( t ) ---H b0 H 3s p˜ = ---------- 1 + ----- 3 4

(2.206)

4. Seismic loading b = –ρm a g p˜ = – a g

∫

H

ρ m φ dx 0

A more convenient form of p˜ is

(2.207)

2.7 Stiffness Calibration 109

˜Γ p˜ = – a g m

Γ=

∫ ∫

H

ρ m φ dx 0 ----------------------------H 2 ρ m φ dx 0

(2.208)

When ρ m is constant, Γ depends only on φ .

∫ ∫

H

φ dx

2s 1 + ----* 3 0 Γ ρ = const = Γ = --------------------- = --------------------------H 2 m 2 2s 2 --- + s + -------φ dx 5 3

(2.209)

0

*

Table 2.2 shows the variation of Γ with s . *

Table 2.2: Variation of Γ with s . s

*

0.00

Γ 1.50

0.125

1.36

0.25

1.24

0.50

1.05

0.75

0.91

1.00

0.81

Stiffness calibration - periodic excitation Equation 2.170 is the governing 1 DOF equation for both discrete and continuous beams. One just has to use the appropriate mass, damping stiffness, and loading terms. The solution for a 1 DOF system subjected to periodic excitation is developed in section 1.3. That solution is applicable here with a change in notation to account for the difference between eqns (1.23) and (2.170).

110 Chapter 2: Optimal Stiffness Distribution The response due to the loading, p˜ = pˆ sin ( Ωt )

(2.210)

is written as q = qˆ sin ( Ωt – δ )

(2.211)

where qˆ is related to pˆ by pˆ qˆ = ------------ H 2 ˜ Ω2 m

(2.212)

2 2 ˜ 2 Ω m Ω ρ = ------------ = ------2 k˜ ω

(2.213)

c˜ ξ˜ = -----------˜ 2ωm

(2.214) 4

H2 =

ρ ---------------------------------------------2 2 2 ( 1 – ρ ) + ( 2ξ˜ ρ )

(2.215)

˜ are assumed to be prescribed. The constraint on The terms p˜ , Ω , and m displacement, say u ( x ) = u * ( x ) , is converted to a constraint on q˜ using the assumed displacement expansion, u = qφ . Taking qˆ ≤ q *

(2.216)

in eqn (2.212), the calibration problem reduces to determining ρ and ξ which satisfy the following constraint: **

H2 ≤ H2 ** H2

2

˜ q* Ω m = ----------------pˆ

Once ρ is known, the frequency is determined with

(2.217)

2.7 Stiffness Calibration 111 2

2 Ω ω = ------2 ρ

(2.218) 2

Lastly, one scales the stiffness parameters with ω .

Example 2.11. 3 DOF shear beam revisited again Consider the case of equal nodal masses and loads for the system defined in example 2.9. m 1 = m 2 = m 3 = 1000 kg pˆ 1 = pˆ 2 = pˆ 3 = 10kN Applying equations (2.175) and (2.176) leads to ˜ = 14 ------ ( 1000 ) = 1555kg m 9 p˜ = 2 ( 10 ) = 20kN *

The displacement constraint is taken as u 3 = 0.1m . Since Φ ( 3 ) ≡ 1 , it * ** follows that q = 0.1m . Evaluating H 2 results in **

H 2 = 0.007775Ω

2

Various values for Ω are considered below. 1. Ω = 2π **

H 2 = 0.307 **

Since H 2 is less than 1 , there is only one value for ρ . Stiffness dominates the response and one can neglect ξ in the expression for H 2 . Then, using eqn. (2.215),

112 Chapter 2: Optimal Stiffness Distribution 1 1 ------ = 1 + -------- = 4.257 2 ** ρ1 H2 and ω = ( 2π ) ( 4.257 )

1⁄2

= 12.96 rad/s

The stiffness parameters are derived in example 2.8. For equal nodal masses, eqn (2.166) reduces to 3000 k' = 5000 6000

(kg)

Finally, the scaled stiffnesses are 504 k = ω k' = 840 1067 2

(kN/m)

2. Ω = 4π **

H 2 = 1.228 **

Since H 2 is greater than 1 , there are two values for ρ . Results for different values of ξ˜ are listed below. Selecting ω 2 generates the lowest stiffness and shifts the response away from the resonance zone. ξ˜

ρ1

ω 1 (rad/s)

ρ2

ω 2 (rad/s)

0

0.633

19.85

2.383

5.27

0.1

0.656

19.16

2.300

5.46

0.2

0.742

16.93

2.033

6.18

2.7 Stiffness Calibration 113 Example 2.12. Stiffness calibration - continuous beam Consider a continuous beam model of a moderately tall building having the following properties H = 100 m ρ m = 20000 kg/m Let s = 0.4 , which corresponds to allocating 40% of the deflection at the top of the building to bending deformation. The equivalent mass follows from eqn (2.204). ˜ = ( 20000 ) ( 100 ) ( 0.541 ) = 1.082 × 10 6 kg m Assuming the displacement at the top is prescribed and noting eqn (2.193), the nodal amplitude is determined with 1 q * = -----------u * ( H ) 1+s 1 100 A representative value for u * ( H ) is H ⁄ 400 . Then, q * = ------- --------- = 0.178 m . 1.4 400 Lastly, the loading is assumed to be uniform over the building height. Applying eqn (2.205), b * ( 100 ) 0.8 p˜ = ------------------- 1 + ------- = 63.33 b * 2 3 where b * has units of N ⁄ m . Combining these terms, the constraint condition is given by **

H2 ≤ H2

2

** Ω H 2 = ------- ( 3043 ) b*

The remaining steps are the same as for the previous example. For example, ** assuming Ω = 1 rad/s and b * = 6086 N ⁄ m , H 2 = 0.5 and the appropriate value of ω is 3 rad/s . Substituting for the various parameters in eqns (2.148)

114 Chapter 2: Optimal Stiffness Distribution and (2.149), the corresponding rigidity distributions are 5 x 2 2s x 3 D T = 3 × 10 1 – ---- + ----- 1 – ---- H 3 H

(kN)

9 1 1 x x 3 4x 1 x 4 D B = 3 × 10 --- 1 – -------- + --- ---- + ----- 2 – 3 ---- + ---- 4 H H 3H 3 H 6s

(kNm)

The previous examples dealt with the case where the displacement controls the design. When acceleration is the limiting condition, one works with the differentiated form of eqn (2.211) 2

q˙˙ = – Ω qˆ sin ( Ωt – δ )

(2.219)

and establishes a relation between the maximum allowable acceleration, ( ˙˙ u ) max , and ( q˙˙) max . Letting ( q˙˙) max = a *

(2.220)

the constraint condition for acceleration controlled design is *

H2 < H2 * H2

˜ a* m = --------p˜

(2.221)

The remaining steps are the same as before. One selects a value for ξ , determines the allowable values of ρ and then converts to values for ω .

Example 2.13. Example 2.12 revisited Consider the continuous beam model introduced in example 2.12. Suppose the acceleration at the top of the building is the critical motion parameter. This acceleration is expressed as a fraction of the gravitational acceleration,

2.7 Stiffness Calibration 115 u˙˙( H ) ≤ fg

(2.222)

Noting eqn (2.193), u˙˙( H ) is related to q˙˙ by u˙˙( H ) = ( 1 + s )q˙˙ Then, the limiting value for q˙˙ is f ( q˙˙) max = a * = ----------- g 1+s

(2.223)

Substituting for a * , eqn (2.221) takes the form ˜g f * m H 2 = -------- ----------- p˜ 1 + s

(2.224)

The limiting acceleration for human comfort is about 1.5% of g. Taking ˜ , p˜ , and s for example 2.12, the design f = 0.015 and using the values of m constraint is * 1805 H 2 = -----------b*

where b * has units of N/m. To illustrate the computational aspects, the case where b * = 3610 N ⁄ m is considered. The corresponding value of H 2* is 0.5. Taking ξ = 0 and using eqn (2.215), one obtains 2 1 ρ 1 = --3 2

2 2 Ω ω 1 = -------- = 3Ω 2 ρ1

The acceleration constraint requires ω ≥ ω 1 .

116 Chapter 2: Optimal Stiffness Distribution Stiffness calibration - seismic excitation The equivalent loading for seismic excitation has been expressed as ˜ Γa p˜ = – m g

(2.225)

˜ and Γ depend on the model, i.e., discrete or continuous beam. where m Substituting for p˜ in eqn (2.170), one obtains the following equation for the modal amplitude factor, q . 2 q˙˙ + 2ξ˜ ωq˙ + ω q = – Γa g

(2.226)

˜ is the equivalent modal damping ratio. The corresponding where ξ˜ = c˜ ⁄ 2ωm equation for a 1 DOF system follows from eqn (1.23) 2

u˙˙ + 2ξωu˙ + ω u = – a g

(2.227)

Taking ξ = ξ˜ , the displacement responses for the same accelerogram are related by q ( t ) = Γu ( t )

(2.228)

Therefore, one can work with a 1 DOF system and then scale the response using eqn (2.228). Assuming the “design” ground motion time history is specified, one can theoretically solve eqn (2.227) and determine the relationship between the peak response u max , and the system parameters ω and ξ . It was possible to find an analytic solution for the case of periodic loading. However, seismic accelograms are considerably more complex than periodic forcing. Typical accelograms are shown in figures 2.21 and 2.22. One has to resort to using a numerical scheme, and generate a set of solutions corresponding to a representative range for ω and ξ . A particular solution for u max is expressed as 1 u max ( ω i, ξ i ) = ---------------------------------- S v ( ω i, ξ i ) ω i ( 1 – ξ i2 ) 1 ⁄ 2

(2.229)

where S v is defined as the pseudo spectral velocity. The complete solution is represented by a set of plots of S v vs. ω (or period T ) for different values of ξ .

2.7 Stiffness Calibration 117 Figure 2.23 contains pseudo spectral velocity profiles for the accelerogram plotted in Fig. 2.21.

4

El Centro (component S00E) 2 Peak acceleration @ 2.12s = 3.41 m/s Peak velocity @ 2.18s = 0.33 m/s Peak displacement @ 8.58s = 0.10m

Acceleration a g - m/s

2

3

2

1

0

−1

−2

−3

−4 0

10

20

30

40

50

60

Time - s Fig. 2.21: Time history of El Centro accelerogram (component S00E). 2

1.5

Acceleration a g - m/s

2

Taft (component N21E) 2 Peak acceleration @ 9.10s = 1.52 m/s Peak velocity @ 3.40s = 0.16 m/s Peak displacement @ 44.14s = 0.07m

1

0.5

0

−0.5

−1

−1.5

−2 0

10

20

30

40

50

60

Time - s Fig. 2.22: Time history of Taft accelerogram (component N21E).

118 Chapter 2: Optimal Stiffness Distribution

1

Pseudo spectral velocity S v - m/s

10

ξ = 2% ξ = 5% ξ = 10% ξ = 20%

0

10

−1

10

El Centro (component S00E)

−2

10

−1

10

0

10

1

10

Period T - s Fig. 2.23: Pseudo spectral velocity profiles. Although there has been substantial progress in earthquake engineering, numerous uncertainties still exist, especially in defining the magnitude and frequency content of the earthquakes that a particular site is expected to encounter. Earthquake records differ significantly as illustrated by the accelerograms for El Centro (Component S00E) and Taft (Component N21E). Given the difficulty in estimating the ground motion that a structure may experience over its useful life, it is prudent to base seismic design on a range of possible earthquake motions. One approach to dealing with this uncertainty is based on using a seismic response spectrum which is an envelop of upper bound responses corresponding to a set of ground motions which are representative for the site. Assuming S v ( ω, ξ ) represents the design response spectrum, the stiffness calibration proceeds as follows. Given the design values for the motion parameters, one establishes the desired value for q max . Let q *denote this value. For the cantilever beam case, q * = γ * H where γ * is the targeted transverse shear deformation. Noting equations (2.228) and (2.229), the design constraint takes the form

2.7 Stiffness Calibration 119 ΓS v ( ω, ξ ) q * ≡ -------------------------------ω( 1 – ξ2 )1 ⁄ 2

(2.230)

Equation (2.230) is written as ΓS v ( ω, ξ ) ω = --------------------------------q*( 1 – ξ2 )1 ⁄ 2

(2.231)

One determines ω by iterating on S v ( ω, ξ ) . Construction of the design response spectrum for S v and the application of eqn (2.231) are discussed in the following section. Seismic response spectra A seismic response spectrum is a plot of the maximum value of a response variable such as displacement, velocity, acceleration, or any other quantity of interest versus the natural frequency or period for a broad range of single-degreeof-freedom (SDOF) systems subjected to the same loading, in this case, an earthquake excitation. In what follows, the generation of response spectra for a SDOF linear elastic system is described and approaches for establishing design spectra are suggested. The solution of eqn (2.227) for a specified ground acceleration a g ( t ) can be expressed as a Duhamel integral (Chopra, 1995) t

∫

1 1 u ( t ) = – ----- a g ( τ )e – ξω ( t – τ ) sin [ ω' ( t – τ ) ]dτ ≡ ----- Θ ω' ω'

(2.232)

0 t

∫

ω u˙ ( t ) = – ----- a g ( τ )e – ξω ( t – τ ) sin [ ω' ( t – τ ) + δ ]dτ ω' 0

(2.233)

120 Chapter 2: Optimal Stiffness Distribution t

ω2

∫

u˙˙( t ) = – a g ( t ) – ------ a g ( τ )e – ξω ( t – τ ) sin [ ω' ( t – τ ) + δ 1 ] dτ ω' 0 ω2

(2.234)

= – a g ( t ) + ------Θ' ω' where ω' = ω 1 – ξ 2 is the damped frequency and δ and δ 1 are defined as ( 1 – ξ2 )1 / 2 tan δ = – --------------------------ξ

(2.235)

2ξ ( 1 – ξ 2 ) 1 / 2 tan δ 1 = ---------------------------------1 – 2ξ 2

(2.236)

For lightly damped structures, ξ < 0.10 , and therefore the difference between the damped and the undamped frequencies can be neglected. The response quantities for this case reduce to t

∫

Θ 1 u ( t ) ≈ – ---- a g ( τ )e – ξω ( t – τ ) sin [ ω ( t – τ ) ] dτ ≡ ---ω ω

(2.237)

0

t

∫

u˙ ( t ) ≈ – a g ( τ )e – ξω ( t – τ ) { cos [ ω ( t – τ ) ] – ξ sin [ ω ( t – τ ) ] } dτ

(2.238)

0

t

∫

u˙˙( t ) ≈ – a g ( t ) – ω a g ( τ )e – ξω ( t – τ ) sin [ ω ( t – τ ) + 2ξ ] dτ ≡ – a g ( t ) – ωΘ' (2.239) 0

The spectral displacement, S d , is defined as the maximum value of the response relative to the ground. Noting eqn (2.232), S d is given by 1 S d = ----- Θ max ω'

(2.240)

2.7 Stiffness Calibration 121 The maximum value of Θ has units of velocity. It is referred to as the pseudo spectral velocity and is denoted as S v . S v ( ω, ξ ) = Θ max

(2.241)

The spectral acceleration, S a , is defined as the maximum value of the absolute acceleration. ω2 S a = ------Θ' max ω'

(2.242)

1 When ξ is small, the measures can be approximated by S d ≈ ---- S v and S a ≈ ωS v. ω The energy input E I to a SDOF system initially at rest and subjected to ground excitation is given by t

∫

E I = – m a g ( τ )u˙ ( τ ) dτ

(2.243)

0

Part of this energy is stored temporarily in the structure in the form of kinetic and strain energy; the remaining portion is dissipated by damping. The various energy terms are Kinetic energy t

∫

2 m E K = m u˙˙( τ )u˙ ( τ ) dτ = ---- [ u˙ ( t ) ] 2

(2.244)

0

Strain energy t

∫

2 k E S = k u ( τ )u˙ ( τ ) dτ = --- [ u ( t ) ] 2 0

(2.245)

122 Chapter 2: Optimal Stiffness Distribution Dissipated energy t

∫

2

E D = c [ u˙ ( τ ) ] dτ

(2.246)

0

With this notation, the energy balance equation takes the form EK + ED + ES = EI

(2.247)

Equation (2.247) applies for arbitrary t . When t is large in comparison to the duration of the earthquake, E K = E S = 0 and E I is equal to the energy dissipated by damping. An alternate way of expressing the net input energy is in terms of the equivalent velocity, V E , defined as 2E I ( t e ) -----------------m

VE =

(2.248)

where E I ( t e ) is the input energy evaluated at t e , the duration of the earthquake. 1

Equivalent velocity V E - m/s

10

ξ = 20% ξ = 10% 0

10

ξ = 5% ξ = 2%

−1

10

−2

10

El Centro (component S00E) −1

10

0

10

Period T - s Fig. 2.24: Equivalent velocity profiles.

1

10

2.7 Stiffness Calibration 123 Figures 2.23 and 2.24 illustrate the variation of the pseudo spectral velocity and the equivalent velocity with period and damping ratio. Whereas the pseudo spectral velocity profiles are smoothened and lowered with increasing damping, the equivalent velocity profiles tend to converge toward a single curve. This result indicates that the work done by an earthquake is not very sensitive to damping. Furthermore, for low damping, the equivalent velocity plot is close to the pseudo spectral velocity curve. Therefore, it is reasonable to assume that the pseudo spectral velocity provides a measure of the energy input at low damping ratios. An ensemble of design earthquakes is used to produce the design spectral velocity function S v ( ω, ξ ) . The earthquakes that compose the ensemble are selected such that frequency content is representative of the site where the structure is to be located. They are then scaled such that their maximum spectral velocity is equal to a reference value S v R for a damping ratio of 0.02 . This value represents the magnitude of excitation for which a lightly damped structure would be designed for. Two methods for generating the design spectral velocity function are discussed here. Both methods are based on the ensemble average of the design earthquake spectral velocity functions. One method develops an analytical expression for a design spectral velocity function and involves only simple calculations. The second method generates a numerical approximation and is a computer based approach. The analytical spectral velocity function is generated for each damping ratio by assuming a bilinear log-log relationship between 0.1 and 0.6 seconds, and a constant value for T > 0.6 seconds. The maximum value of S v is obtained by averaging the peak values of the individual spectra. Similarly, the minimum value is the average of the spectral values at T = 0.1 seconds. Figure 2.25 shows spectral velocity functions normalized such that the maximum value of S v = 1.2 ( m ⁄ s ) for ξ = 0.02 - a representative value for a major seismic event. The accelerograms corresponding to the earthquakes are also normalized based on these values. Once accelerograms are normalized, one can compute how the spectral velocity function varies with ξ by increasing the damping ratio and generating new spectral velocity functions. Spectral velocity functions for these “scaled” accelerograms and different damping ratios are shown in Figure 2.26. Figure 2.27 illustrates how the averaged maximum value of S v varies with ξ .

124 Chapter 2: Optimal Stiffness Distribution

Spectral Velocity Plots Scaled Such That Sv = 1.2 for ξ = 0.02

1

10

S = 1.2 m/s v

ξ = 0.02

0

Spectral Velocity − m/s

10

−1

10

−2

10

−1

0

10

1

10 Period − Seconds

10

Fig. 2.25: Normalized spectral velocity plots

Spectral Velocity Plots for ξ = 0.15

1

Spectral Velocity Plots for ξ = 0.30

1

10

10

S = 0.60 m/s ξ = 0.15

S = 0.47 m/s ξ = 0.30

v

v

0

0

Spectral Velocity − m/s

10

Spectral Velocity − m/s

10

−1

−1

10

10

−2

10

−2

−1

10

0

10 Period − Seconds

1

10

10

−1

10

0

10 Period − Seconds

Fig. 2.26: Spectral velocity plots for ξ = 0.15 and ξ = 0.30

1

10

2.7 Stiffness Calibration 125

1.2

1

Svmax [m/s]

0.8

0.6

0.4

0.2

0

0

0.05

0.1

0.15 ξ

0.2

0.25

Fig. 2.27: S v vs. ξ max The spectral velocity function based on the bilinear log-log relationship can be used to solve equation 2.231. Since there is no unique solution, a family of solutions for T is generated by specifying different values of ξ . For each value of ξ , the computation reduces to iterating on the following equations: T ≥ 0.6 sec *

2πq T = -----------------ΓS v

(2.249)

max

T < 0.6 sec=T max T

a+1

2πq * a = ------------------ ( T max ) ΓS v max

log ( S v ⁄ Sv ) max min a = --------------------------------------------log ( T max ⁄ T min )

(2.250)

(2.251)

126 Chapter 2: Optimal Stiffness Distribution The analytical expressions developed for the spectral velocity tend to be conservative, since the piecewise linear log-log approximation for the average design spectrum is over-simplified. An improvement in the process of establishing ω can be obtained if an average spectral velocity function is generated by taking the ensemble average of the spectral velocity functions, instead of attempting to create an analytical expression. Equation 2.231 can be solved by iterating through the average spectral velocity function and finding the values of ω and S v that satisfy the equation for a given value of ξ . Figure 2.28 shows average spectral velocity functions for various values of ξ . For a given value of T , numerical functions similar to that of Fig 2.27 can be used to determine S v for a given value of ξ .

2.7 Stiffness Calibration 127

1

10

ξ = 0.1

0

Sv [m/s]

10

−1

10

0

10 Period T − seconds

1

10

1

10

ξ = 0.02, 0.05, 0.1, 0.15, 0.2, 0.25, 0.3

0

Sv [m/s]

10

−1

10

0

10 Period T − seconds

Fig. 2.28: Average spectral velocity functions

1

10

128 Chapter 2: Optimal Stiffness Distribution

Example 2.14. Example 2.12 revisited for seismic excitation The properties of the beam considered in example 2.12 are H = 100 m ρ m = 20000 kg ⁄ m s = 0.4 ˜ = 1.082 × 10 6 kg m Using eqn (2.209), the corresponding value of Γ is 1.12 . The design value for q is established by specifying the desired peak transverse shear deformation. Taking γ * = 1 ⁄ 200 and noting that q * = γH (see eqn (2.191) results in q * = γ * H = 0.5 m An estimate for T is determined by first evaluating the right hand side of eqn (2.249). If the value is greater than 0.6 , that estimate is the actual value. If less than 0.6 , eqn (2.250) applies. For this example, the right hand term is 2πq * 2.80 ------------------ = -------------ΓS v Sv max max Taking ξ = 0.02 and S v = 1.2 , max 2.80 T = ---------- = 2.34 sec 1.2 2π ω = ------ = 2.68 rad/s T The modal damping parameter is determined with ˜ = ( 2 ) ( 0.02 ) ( 2.68 ) ( 1.082 ) × 10 6 = 116 (kN)(sec/meter) c˜ = 2ξωm

2.7 Stiffness Calibration 129 Increasing damping to 0.1 and taking the corresponding maximum value of S v according to Fig 2.27, ξ = 0.1

→

Sv

max

= 0.7

leads to 2.80 T = ---------- = 4.0 sec 0.7 ω = 1.57 rad/s c˜ = 340 (kN)(sec/meter) 2

One evaluates the rigidity parameters D T and D B by substituting for ω in eqns (2.148) and (2.149).

Example 2.15. 5 DOF shear beam Consider a 5 DOF shear beam having equal nodal masses. The fundamental mode vector is specified such that the relative nodal displacements for the 5 elements are equal. Its form is U = qΦ * 1 2 1 Φ * = --- 3 5 4 5 The equivalent modal mass follows from eqn (2.171): T

˜ = ( Φ * ) M Φ * = 2.2 m m where m is the “constant” nodal mass. Substituting for Φ * and M in eqn (2.160), leads to the following set of stiffness coefficients

130 Chapter 2: Optimal Stiffness Distribution k' 1 = 15 m

k' 2 = 14 m

k' 3 = 12 m

k' 4 = 9 m

k' 5 = 5 m

2

The actual stiffness values depend on ω , i.e., k = ω k' Specializing eqn (2.174) for seismic excitation and equal nodal masses, one obtains p˜ = – ma g

∑ Φ*i = –3ma g

Then, noting the definition of Γ , – p˜ 3m Γ = ---------- = ------------ = 1.36 ˜a 2.2m m g The maximum displacement at node 5 (the top level) is considered to be the controlling motion measure. u5

= q max ≡ q * max

Assuming the beam is a model of a building, u 5 is related to the height of the max building and the story shear deformation. u5

max

= γ *H

To illustrate the remaining steps, suppose H and γ * have the following values H = 20 (meters) γ * = 1 ⁄ 200 Then, q * = 0.1 (meters ) and the right hand term in eqn (2.249) becomes 2πq * 0.462 ------------------- = --------------ΓS vmax S vmax Case 1: ξ=0.02 Taking ξ = 0.02 and S vmax = 1.2 , the value of the above term is less than

2.7 Stiffness Calibration 131 0.6 and eqn (2.250) applies. Noting Fig 2.25, T min = 0.1 , and S v, min = 0.085 . Using these values leads to: T = 0.49 sec ω = 12.82 r/s ˜ = 1.128 m (newtons-sec/meters) c˜ = 2ξωm Case 2: ξ=0.1 Taking S vmax = 0.7 for ξ = 0.1 , 0.462 --------------- = 0.66 S vmax and eqn (2.249) applies. T = 0.66 sec ω = 9.52 r/s c˜ = 4.19 m Increasing the damping reduces S vmax and the required stiffness. In this case, the 2 reduction factor is ( 9.52 ⁄ 12.82 ) = 0.55 , a 45 % decrease in stiffness. The modal damping parameter is related to the system damping matrix by eqn (2.172): T

c˜ = ( Φ * ) C ( Φ * ) Given c˜ , one needs to establish C . When linear viscous dampers are connected to adjacent nodes, C is similar in form to the stiffness matrix K . Using Φ * defined above, the triple matrix product evaluates to 1 c˜ = ------ ( c 1 + c 2 + c 3 + c 4 + c 5 ) 25 where c i is the damping coefficient for element i . Since there is only one equation relating the five coefficients, there is no unique solution for the c ‘s. Additional

132 Chapter 2: Optimal Stiffness Distribution equations can be derived by imposing conditions on the damping ratios for the higher modes and optimizing “cost” and “performance”. Establishing the optimal damping distribution is addressed in the next chapter.

2.9 Examples of stiffness distribution for buildings In this section, a series of simulation studies for a representative range of buildings are presented. The primary focus here is on assessing the adequacy of the rigidity distribution derived with the single mode expression to control the deformation profiles. How the rigidity distribution can be corrected to account for the influence of the higher modes is discussed in the following section. Only shear rigidity distributions and shear deformation profiles are presented here. The corresponding bending rigidity and deformation plots can be found in AbboudKlink (1995). The motion based design data are the uniformly distributed mass ρ m , the building height H , the s factor relating the shear and bending deformations, and the values of the maximum allowable shear deformation γ ∗ corresponding to a specified level of ground excitation as measured by the pseudo spectral velocity S v . Table 2.3 lists the levels of excitation considered here and their corresponding allowable deformations. For preliminary design, the extreme earthquake level is used. Equation (2.249) gives the fundamental period corresponding to the rigidity distributions based on the fundamental mode response. These distributions are generated using eqns (2.148) and (2.149). The relevant design parameters for the buildings modelled as a continuous beam are listed in Table 2.4.

Table 2.3: Design loads and allowable deformation values. Earthquake level

S v (m/s)

γ∗

Service load

0.6

1/400

Extreme load

1.2

1/200

Ultimate load

2.4

1/100

2.9 Examples of Stiffness Distribution for Building 133 Table 2.4: Design parameters for the continuous models. Building #

ρm (kg/m)

(m)

H⁄B

f∗

s

γ∗

Sv (m/s)

ξ1 (%)

T1

1

20000

25

2

7

0.15

1/200

0.9

2

0.49

2

20000

50

3

6

0.25

1/200

1.2

2

1.06

3

20000

100

4

5

0.40

1/200

1.2

2

2.34

4

20000

200

5

4

0.63

1/200

1.2

2

5.35

H

(s)

The continuous building models are discretized and the corresponding lumped parameter MDOF models are then subjected to earthquake excitation. The time history response due to seismic excitation is generated by numerically integrating the MDOF matrix equilibrium equations. Two accelerograms, El Centro (component S00E) and Taft (component N21E), both scaled to a maximum pseudo spectral velocity of 1.2m/s corresponding to the extreme loading condition, are used. Table 2.5 contains data for the first 3 modes of the MDOF models. Stiffness proportional damping is assumed to establish the damping ratios for the higher modes. This data is included here to provide background information needed to interpret the response. An alternative solution procedure based on working with the equilibrium equations expressed in terms of modal coordinates is described in the next section. Table 2.5: Modal parameters for the discrete models. T1(s)

T2(s)

T3(s)

ξ1 ( % )

ξ2 ( % )

ξ3 ( % ) Γ2 ⁄ Γ1 Γ3 ⁄ Γ1

Bldg #1 Bldg #2

0.52 1.09

0.21 0.43

0.13 0.26

2.00 2.00

5.04 5.09

8.17 8.36

0.40 0.43

0.23 0.25

Bldg #3 Bldg #4

2.39 5.48

0.92 2.06

0.55 1.21

2.00 2.00

5.18 5.32

8.64 9.06

0.45 0.48

0.26 0.27

Figures 2.29 through 2.36 show the generated shear rigidity distributions and the resulting maximum shear deformations for the different cases. The vertical dashed line in the deformation plots indicates the target maximum deformation; the solid line shows the maximum deformation due to El Centro excitation, and the dashed line shows the maximum deformation due to Taft excitation. The mean ( γ m and χ m ) and standard deviation ( γ sd and χ sd ) of a

134 Chapter 2: Optimal Stiffness Distribution deformation profile provide measures of the adequacy of the rigidity distribution. These values are summarized in Table 2.6. Table 2.6: Deformation results for the example buildings. El

Centro

γm

γ sd

χm

χ sd

γm

γ sd

χm

(10 )

(10 )

(10 )

(10 )

(10 )

(10 )

(10 )

(10-6)

3.42 4.40 6.13 3.50

1.25 4.42 9.82 19.42

3.84 4.44 4.61 2.20

1.14 4.10 10.06 14.22

1.87 3.42 4.42 4.54

1.27 6.37 16.52 7.66

2.18 3.61 3.61 2.74

0.80 6.06 15.88 5.55

-3

Bldg #1 Bldg #2 Bldg #3 Bldg #4

-4

-5

Taft

-6

-3

-4

-5

χ sd

The results show that the deviation from the state of uniform deformation occurs in the upper portion of the building, and is far more pronounced in tall buildings. This trend is attributed to the contributions of the higher modes which have the greatest deformation near the top (see Fig 2.18). For low rise buildings, the deformation pattern is essentially dominated by the fundamental mode response. A procedure for incorporating the effect of the higher modes on the rigidity distribution is presented in the next section. An alternate strategy is to increase the damping for the fundamental mode. These results are for a low level of damping ( ξ 1 = 2%). Damping is discussed in the next chapter. The results also show that the magnitude of the maximum deformation is sensitive to the excitation. For example, the difference in the behavior of Building 4 under the two excitations is mainly due to the difference in the response spectra. For El Centro, the actual value of S v at a period of about 5s , shown in Fig. 2.37, is much less than the value used for the initial rigidity estimate, S v = 1.2m/s , and as a result, the actual contribution of the first mode is overestimated. The Taft spectrum on the other hand, shows a closer agreement with the design value of S v and consequently, the actual and design deformations are closer together.

2.9 Examples of Stiffness Distribution for Building 135

1 0.9

Building #1 Quadratic based Initial

x Normalized height ---H

0.8 0.7

H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.29: Shear rigidity distribution for Building 1.

1 0.9

Building #1 Quadratic based Initial

x Normalized height ---H

0.8

H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.30: Maximum shear deformation for Building 1.

0.01

136 Chapter 2: Optimal Stiffness Distribution

1

x Normalized height ---H

0.9

Building #2 Quadratic based Initial

0.8 0.7

H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.31: Shear rigidity distribution for Building 2.

1

x Normalized height ---H

0.9

Building #2 Quadratic based Initial

0.8

H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.32: Maximum shear deformation for Building 2.

0.01

2.9 Examples of Stiffness Distribution for Building 137

1 0.9

x Normalized height ---H

Building #3 Quadratic based Initial

0.8 0.7

H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.33: Shear rigidity distribution for Building 3.

1

x Normalized height ---H

0.9 0.8

Building #3 Quadratic based Initial

0.7 0.6

H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2%

0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.34: Maximum shear deformation for Building 3.

0.01

138 Chapter 2: Optimal Stiffness Distribution

1 0.9

x Normalized height ---H

Building #4 Quadratic based Initial

0.8 0.7

H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.35: Shear rigidity distribution for Building 4.

1

x Normalized height ---H

0.9

Building #4 Quadratic based Initial

0.8 0.7

H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.36: Maximum shear deformation for Building 4.

0.01

2.10 Stiffness Modification for Seismic Excitation 139

1

Pseudo spectral velocity S v - m/s

10

ξ = 2% 0

10

-1

10

El Centro S00E (Scale = 1.66) Taft N21E (Scale = 3.00) -2

10 -1 10

0

10

1

10

Period T - s Fig. 2.37: Modified response spectrum for Building 4.

2.10 Stiffness modification for seismic excitation Iterative procedure The numerical studies presented in the previous section show that the method for establishing the shear and bending rigidities based on the fundamental mode is adequate for low rise buildings but needs to be modified for moderate rise buildings. This section presents a procedure for adjusting the stiffness which is based on updating the shear and bending deformation profiles by including the contribution of the higher modes, and then determining improved estimates for the rigidity measures with (i + 1) (x) DT

(i + 1) (x) DB

=

(i) ( i ) [ γ ( x ) ] max D T ------------------------------∗

(2.252)

=

(i) ( i ) [ χ ( x ) ] max D B -------------------------------∗

(2.253)

γ

χ

The iteration is continued until the change in the rigidity is within the acceptable range. Details of the procedure for seismic excitation applied to beam type

140 Chapter 2: Optimal Stiffness Distribution structures are presented below. Multiple mode response Given the mass distribution ρ m , one specifies: γ ∗ , the desired deformation; S v ( ω 1, ξ 1 ) , the level of earthquake to be designed for; and the parameter s relating the shear and bending deformations. An initial estimate of (0) (0) D T ( x ) and D B ( x ) is generated using the fundamental mode response approach presented in the previous sections. The structure is then discretized as an nth order MDOF system, leading to the following governing equations ˙˙( t ) + CU˙ ( t ) + KU ( t ) = P ( t ) = – MEa ( t ) MU g

(2.254)

where M , C , and K represent the mass, damping, and stiffness matrices of the discrete system, U is the displacement vector relative to the base, and E is a vector representing the rigid body motion of the system due to a unit translation at the base. Equation (2.254) can be transformed into a set of uncoupled equations by expressing U in terms of a “special” set of vectors, φ j , and assuming a particular form for C. n

U(t) =

∑

Φ j q j ( t ) = Φq ( t )

(2.255)

j=1

These vectors are solutions of the following eigenvalue problem: KΦ j = ω 2j MΦ j j = 1, 2, 3, . . . , n

(2.256)

and satisfy the orthogonality relation: Φ iT KΦ j = ω 2j Φ iT MΦ j δ ij

(2.257)

where δ ij is the dirac delta function. Background material on the eigenvalue problem is presented in Appendix I. Assuming C to be proportional to K (the basis for this assumption is discussed in Section 3.7 which deals with arbitrary damping),

2.10 Stiffness Modification for Seismic Excitation 141 C = αK

(2.258)

and noting eqn (2.257) leads to the following set of uncoupled scalar equations q˙˙j + 2ξ j ω j q˙ j + ω 2j q j = – Γ j a g

j = 1, 2, 3, . . . , n

(2.259)

The damping ratio and modal participation factor for the j’th mode are αω j ξ j = ---------2

(2.260)

ΦT j ME Γ j = --------------------ΦT j MΦ j

(2.261)

The individual modal responses are determined following the procedure established for the SDOF system. Noting eqn (2.232), the solution for the j’th modal response is Γj q j ( t ) = ------- Θ j ( t ) ω' j

(2.262)

t

Θ j(t) =

∫

– a g ( τ )e

–ξ j ω j ( t – τ )

sin [ ω' j ( t – τ ) ] dτ

(2.263)

0

U j(t) = Φ jqj(t)

(2.264)

Finally, the relative displacement vector due to all modal responses is obtained by superposition n

U ( t ) = Φq ( t ) =

∑ j=1

Γ jΦ jΘ j(t) -------------------------ω' j

(2.265)

To determine an estimate of the earthquake response of a lumped MDOF system, one can make use of the response spectra method. The maximum response for a mode is obtained from the response spectrum for the SDOF system

142 Chapter 2: Optimal Stiffness Distribution using the corresponding values of frequency, damping, and participation factor. The expressions for the jth mode are

qj Uj

Γj = ------- S v ( ω j, ξ j ) ω' j max

(2.266)

= Φ jq j max

(2.267)

max

where S v ( ω j, ξ j ) is the pseudo spectral velocity corresponding to the damping and frequency of the jth mode. The shear and bending deformations in the i’th element due to the maximum response of the j’th mode are determined with the following differential approximations: γ (i)

1 ---- u ( i + 1 ) – u ( i ) jmax = jmax hi

(2.268)

1 – --- β ( i + 1 ) – β ( i ) jmax 2 1 χ ( i ) jmax = ---- β ( i + 1 ) – β ( i ) jmax hi

(2.269)

where the displacement measures are for the nodes at the base and upper end of element i. The individual modal time histories are generally not in phase so one needs to introduce an assumption as to the relative phasing. Taking the algebraic sum assumes they are all in phase. Using a square root sum of squares (SRSS) procedure assumes the phase lag is a uniformly distributed random variable, and tends to place more emphasis on the dominant terms. The latter procedure is discussed here. The maximum value of a variable, say v, is determined with 1⁄2

2 2 2 v max = ( v 1max ) + ( v 2max ) + … + ( v Nmax )

(2.270)

where N is the number of modes retained for the stiffness modification. This

2.10 Stiffness Modification for Seismic Excitation 143 computational algorithm is used to determine γ max and χ max for each segment. (0)

One starts with D , computes the peak modal responses using eqn (2.266), and then evaluates the peak deformations with equations (2.268) and (2.269). The rigidity coefficients are then updated with equations (2.252) and (2.253). (1)

(1)

The process is now repeated. Given D T ( x ) and D B ( x ) , the system stiffness matrix is generated, and updated mode shapes, frequencies, and participation factors are evaluated, leading to revised peak deformation distributions and ultimately to new estimates for rigidities. Building examples - iterated stiffness distribution The iterative procedure is applied to the four building examples defined in Table 2.4. Table 2.7 lists the periods for the first three modes of the discretized system corresponding to the iterated rigidity distributions. The mean and standard deviation of the shear and bending deformations resulting from the El Centro and Taft excitations are summarized for different iterations in Table 2.8. Results in the form of converged shear rigidity distributions and maximum shear deformation profiles for El Centro and Taft excitations are provided in Figs 2.33 through 2.38. These profiles illustrate the convergence of the iterative procedure over the spectrum of building examples starting with the quasi-quadratic distribution as an initial estimate for the rigidity. Significant improvement in the maximum deformation profiles are obtained after just one iteration. The contribution of the higher modes becomes more significant as the period of the structure increases, leading to a decreasing rate of convergence with increasing period. Another important feature is the substantial difference in response for the two earthquakes over the spectrum of periods. El Centro dominates for Building 3 ( T 1 = 2.15s ) while Taft produces the most deformation for Building 4 ( T 1 = 4.86s ). This result stresses the need to use a set of representative earthquakes, and to establish bounds on the deformation profiles. In general, the rigidity is increased; the increment ranges from 20% at the base to about 100% at the top for the high period range. There is only a slight correction for the low period range since the response for small period is essentially governed by the fundamental mode. The increase in rigidity results in a decrease in period from the initial estimate based on eqn (2.160). An improved estimate for the period which accounts for the increase in rigidity is

144 Chapter 2: Optimal Stiffness Distribution 2πγ ∗ H = 0.9 -----------------T 1 = 0.9T 1 Γ1 Sv initial

for

T 1 > 0.6

(2.271)

Table 2.7: Modal parameters - building examples. T1(s)

T2(s)

T3(s)

ξ1 ( % )

ξ2 ( % )

ξ3 ( % ) Γ2 ⁄ Γ1 Γ3 ⁄ Γ1

Bldg #1 Initial (Q) Iteration 1 Iteration 2

0.52 0.52 0.52

0.21 0.21 0.21

0.13 0.13 0.13

2.00 2.00 2.00

5.04 5.04 5.04

8.17 8.12 8.12

0.40 0.40 0.40

0.23 0.23 0.23

Bldg #2 Initial (Q) Iteration 1 Iteration 2 Iteration 3

1.09 1.07 1.05 1.05

0.43 0.40 0.40 0.40

0.26 0.24 0.24 0.24

2.00 2.00 2.00 2.00

5.09 5.31 5.27 5.27

8.36 8.80 8.68 8.69

0.43 0.42 0.42 0.42

0.25 0.24 0.24 0.24

Bldg #3 Initial (Q) Iteration 1 Iteration 2 Iteration 3 Iteration 4

2.39 2.27 2.20 2.16 2.15

0.92 0.82 0.79 0.78 0.77

0.55 0.48 0.46 0.46 0.45

2.00 2.00 2.00 2.00 2.00

5.18 5.54 5.55 5.55 5.55

8.64 9.52 9.47 9.46 9.45

0.45 0.45 0.45 0.45 0.45

0.26 0.25 0.25 0.25 0.25

Bldg #4 Initial (Q) Iteration 1 Iteration 2 Iteration 3 Iteration 4

5.48 5.18 4.98 4.89 4.86

2.06 1.81 1.73 1.70 1.69

1.21 1.03 0.99 0.97 0.96

2.00 2.00 2.00 2.00 2.00

5.32 5.72 5.75 5.74 5.75

9.06 10.04 10.04 10.05 10.05

0.48 0.48 0.48 0.48 0.48

0.27 0.26 0.26 0.26 0.26

2.10 Stiffness Modification for Seismic Excitation 145 Table 2.8: Mean and standard deviation deformations. El

Centro

Taft

γm

γ sd

χm

χ sd

γm

γ sd

χm

χ sd

(10-3)

(10-4)

(10-5)

(10-6)

(10-3)

(10-4)

(10-5)

(10-6)

Bldg #1 Initial (Q) Iteration 1 Iteration 2

3.42 3.45 3.45

1.25 1.45 1.45

3.84 3.01 3.01

1.14 4.37 4.36

1.87 1.91 1.91

1.27 1.52 1.51

2.18 1.72 1.72

0.80 2.43 2.42

Bldg #2 Initial (Q) Iteration 1 Iteration 2 Iteration 3

4.40 4.32 4.39 4.41

4.42 3.77 3.47 3.76

4.44 3.50 3.54 3.52

4.10 7.71 7.45 7.56

3.42 3.35 3.21 3.13

6.37 3.69 3.29 2.77

3.61 2.92 2.75 2.66

6.06 4.04 3.28 3.18

Bldg #3 Initial (Q) Iteration 1 Iteration 2 Iteration 3 Iteration 4

6.13 5.51 4.97 4.67 4.54

9.82 5.72 4.44 4.79 4.89

4.61 3.78 3.35 3.10 2.99

10.06 7.56 6.07 5.83 5.69

4.42 3.84 3.60 3.50 3.51

16.52 5.09 3.39 3.01 3.41

3.61 2.78 2.58 2.46 2.45

15.88 4.61 2.92 2.63 2.86

Bldg #4 Initial (Q) Iteration 1 Iteration 2 Iteration 3 Iteration 4

3.50 2.62 2.56 2.52 2.49

19.42 3.81 3.75 3.55 3.42

2.20 1.43 1.42 1.39 1.36

14.22 2.86 2.17 2.08 2.03

4.54 4.36 4.58 4.31 4.10

7.66 5.47 5.55 5.22 4.92

2.74 2.41 2.60 2.49 2.39

5.55 5.98 6.48 5.98 5.51

146 Chapter 2: Optimal Stiffness Distribution

1 0.9

x Normalized height ---H

Building #2 Quadratic based Iteration 3

0.8 0.7

H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2%

0.6 0.5 0.4 0.3 0.2

mode 2 mode 3

mode 1

0.1 0 0

0.2

SRSS

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.38: Converged shear rigidity distribution for Building 2.

1 0.9

x Normalized height ---H

Building #2 Quadratic based Iteration 3

0.8

H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.39: Maximum shear deformation for Building 2.

0.01

2.10 Stiffness Modification for Seismic Excitation 147

1

x Normalized height ---H

0.9

Building #3 Quadratic based Iteration 4

0.8 0.7

H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2%

0.6

mode 1

0.5 0.4 0.3 0.2

mode 3

mode 2

SRSS

0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.40: Converged shear rigidity distribution for Building 3.

1

x Normalized height ---H

0.9

Building #3 Quadratic based Iteration 4

0.8

H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.41: Maximum shear deformation for Building 3.

0.01

148 Chapter 2: Optimal Stiffness Distribution

1

x Normalized height ---H

0.9

Building #4 Quadratic based Iteration 4

0.8 0.7

H = 200m ρ m = 20000kg/m s = 0.625 S v = 1.2m/s ξ 1 = 2%

0.6

mode 1

0.5 0.4

mode 2

0.3 0.2

mode 3

SRSS

0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 2.42: Converged shear rigidity distribution for Building 4.

1

x Normalized height ---H

0.9

Building #4 Quadratic based Iteration 4

0.8

H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2%

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

Fig. 2.43: Maximum shear deformation for Building 4.

0.01

Problems 149

Problems Problem 2.1 Determine the transverse shear and bending deformations corresponding to the following displacement distributions a) nπx u = sin ---------2L β = 0 n = 1, 3, … b) u = x 2 + a1 x 3 + a2 x 4 β = 2x + 3a 1 x 2 + 4a 2 x 3 c) u = a1 x + a2 x 2 + a3 x 3 β = 2a 2 x + 3a 3 x 2 d) x2 x2 x4 u = a 1 x – ------ + a 2 ----- – ------------ 2 12L 2 2L x3 β = a 2 x – ---------- 3H 2

Problem 2.2 Determine the shear and bending rigidity coefficients for the cross section shown below. The dimensions are in centimeters.

150 Chapter 2: Optimal Stiffness Distribution

1

face

50

E f = 210, 000 MPa

core

G c = 2, 000 MPa

1

25

Problem 2.3 Consider the chevron bracing scheme shown below. Determine an expression for D T . Assume the members carry only axial force.

h

Ad

Ad

α

α

B

Problem 2.4 The force-displacement relationship for the structures shown below is written as P = ku .

Problems 151

Ib h

P, u

Ib

P, u h

Ic

Ic

P, u h

B

B

a)

b)

Ad

B c)

Establish the expression for k for each structure. Comment on the relative efficiency of the structures for resisting lateral loading.

Problem 2.5 Consider a five story five bay rigid frame modeled as a shear beam. Assume all the columns in a story have the same properties, but allow for variation in column size over the stories. Establish an approximate expression for the equivalent shear beam stiffness for a typical story.

i H

h

B

Problem 2.6 Diagonal bracing is added to a rigid frame as indicated in the figure below.

152 Chapter 2: Optimal Stiffness Distribution a) Establish an approximate expression for the equivalent shear beam stiffness for a typical story. Assume the column properties are constant in a story, but vary from story to story.

i H

h

B b) Suppose the diagonal bracing system is expected to carry a specified fraction of the total shear in a story. Discuss how you would select the crosssectional area of the diagonal braces and moment of inertia for the columns in a typical story.

Problem 2.7 Determine the expressions for the distribution of shear and bending rigidities corresponding to constant deformations γ ∗ and χ∗ .

b0

H x

πx b = b 0 sin -----H

Problems 153

Problem 2.8 Determine the shear and bending rigidity distributions. Take: 1 γ ∗ = --------400 1 χ∗ = -----------------20, 000 Note: these values correspond to s = 0.1 (see eqn 2.14).

10kN

1kN/m

10m

Problem 2.9 A particular steel having an allowable stress of 600 MPa and Young’s modulus equal to 210,000 MPa has been selected as the “design” material. Assuming the material is to be used in 45° bracing members, for what range of transverse shear strains will the design be controlled by motion constraints rather than strengths?

Problem 2.10 a) Refer to problem 2.5. Consider a uniform lateral loading of 30 kN/m, a story height of 4m, and a design shear strain, γ ∗ , equal to 1/300 for each story. Estimate the value of I c for each story assuming steel is selected as the material. b) Refer to problem 2.6. Use the loading defined in part a. Select the

154 Chapter 2: Optimal Stiffness Distribution bracing properties such that the bracing carries 25% of the total story shear.

Problem 2.11

P 2, u 2 P 1, u 1 4m

2 1

3m Determine the member stiffness factors for the following prescribed loading and displacement quantities. P 1 = 100 kN

P 2 = 50 kN

u 1 = 0.01 m

u 2 = 0.005 m

Problem 2.12

P 2, u 2 P 1, u 1 4m

3

2 1

4m

3m

The design objective is to determine the member stiffness factors for the

Problems 155 above truss such that the nodal displacements corresponding to P 1 = 25 kN , P 2 = 50 kN are u 1 = 0.01 m , u 2 = 0.01 m . Generate solutions using the least square and mean value least square approaches.

Problem 2.13

P 2, u 2 P 1, u 1 4m

3

2

1 P 3, u 3

4

5 4m

4m

Recommend stiffness factors for the truss shown above based in the following requirements u 3 = – 0.0019m u 1 = – 0.0087m u 2 = 0.0012m P 1 = – 100kN

P 2 = 50kN

P 3 = – 100kN

Problem 2.14 u∗

u∗

P Beam 1 is a shear beam. Beam 2 is a bending beam. H

1

αP

2

( 1 – α )P

Consider two cantilever beams coupled with a rigid link at the top.

156 Chapter 2: Optimal Stiffness Distribution Assume beam 1 is a shear beam and beam 2 is a bending beam. Determine the optimal rigidity distributions that satisfy the displacement requirement u∗ = H ⁄ β and divide the lateral load between the two beams as indicated.

Problem 2.15 The bending beam model is based on the assumption of no transverse shear strain, i.e. γ = 0 . Theoretically, one sets D T = ∞ , which results in s = ∞ . The displacement distributions for periodic vibration with constant curvature, χ∗ , and γ = 0 are x 2 u = q B ---- H 2x β = q B ------H2 H2 q B = -------χ∗ cos ( ω 1 t + δ ) 2 Consider uniform mass density. Determine D B ( x ) for this prescribed mode shape.

Problem 2.16 m 1 = aρ m H

x

H

bx ρ m = ρ m 1 – ------ H

Consider a cantilever beam having the mass distribution indicated in the figure. Assuming the beam acts as a shear beam, determine the transverse shear

Problems 157 rigidity distribution required in order that the fundamental mode shape have the following form: x φ ( x ) = ---H Investigate how shear rigidity varies with a , taking b = 0 .

Problem 2.17 The structure shown below consists of a cantilever beam and a rigid weightless link connecting 2 masses to the top end of the beam. Assume the cantilever beam has uniform mass density ρ m and negligible transverse shear deformation.

b

b m 1 = ( ρ m H )a

m1

m1

H

a) Determine the bending rigidity distribution required to produce the fundamental mode shape corresponding to constant curvature. The displacement expressions for this case are: x u = q B ---- H

2

2x β = q B ------H2 H2 q B = -------χ∗ cos ( ω 1 t + δ ) 2 b) Investigate the sensitivity of the solution to a variation of a .

158 Chapter 2: Optimal Stiffness Distribution c) Suppose outriggers are attached to the rigid links and the beam is extended as shown below. Determine the bending rigidity distribution for the case where a uniform static loading is applied, and the design objective is constant curvature. H/2 m1 A

m1 A

d) Repeat part a using the structure defined in part c.

Problem 2.18 Consider a cantilever shear beam having uniform mass density and constant transverse shear rigidity. Allowing for material damping and external damping, the equilibrium equation for free vibration has the following form (see eqns (2.197) and (2.199)): D T u, xx + C T u, xxt – ρ m u, tt – C′u, t = 0

(1)

The boundary conditions are u(x = 0) = 0 V(x = L) = 0

(2) (3)

Substituting for V in eqn (3) leads to D T u, x ( L ) + C T u, xt ( L ) = 0

(4)

The general solution of eqn (1) which satisfies the boundary conditions at x = 0, L is u = e αt sin λx

(5)

Problems 159 2n – 1 λ = ---------------π 2L

n = 1, 2, …

Let DT 1 / 2 ω = λ -------ρm

(6)

λ 2 C T + C′ -------------------------- = 2ξω ρm

(7)

a) Derive the expression for α in terms of ω and ξ . Assume ξ < 1 . The solution corresponding to n = 1 is the fundamental solution. Comment on the nature of this solution. b) Compare the mode shapes with the profiles based on constant shear deformation for the fundamental mode. Consider n from 1 to 3. c) Let ξ n be the value of ξ corresponding to λ = λ n , i.e., the n ‘th mode. Comment on how ξ n varies with n .

Problem 2.19 Refer to the figure in problem 2.5. Assume the structure is a discrete shear beam. Determine the distribution of equivalent shear beam stiffness based on a linear profile for the fundamental mode and the following floor masses: m 1 = m 2 = m 3 = m 4 = 10, 000 kg m 5 = 5, 000 kg Relate the shear beam stiffness to the column size for each story.

Problem 2.20 Repeat problem 2.19 using the structure defined in problem 2.6. Assume the bracing contributes 25% of the story stiffness.

160 Chapter 2: Optimal Stiffness Distribution

Problem 2.21

u + ug m k

c

ug

A single degree of freedom system is to be designed to displace a given amount, u max , under a specified seismic excitation. Use the response spectrum for the pseudo spectral velocity shown in the figure, and take m = 10, 000 kg. Determine design values for k and c for a range of damping coefficients ( ξ = 0.05, 0.10, 0.20 ) and two values of u max : u max = 0.1 m, 0.2 m . Also determine the maximum values of the stiffness and damper forces. Spectral Velocity S v (m/s) S v = 1.4, ξ = 0.05 S v = 1.0, ξ = 0.1

1.0

S v = 0.8, ξ = 0.2

0.1 0.1

0.6 1.0

10

Period T (sec)

Problems 161

Problem 2.22 m⁄2 h h

p 3, u 3

m

p 2, u 2

m

p 1, u 1

h

The structure consists of 3 shear beam segments and the lumped masses indicated in the figure. Take m = 1, 000 kg .

h = 4 meters

a) Determine the shear rigidity distribution required for the fundamental mode vector to have the form 1 2 Φ = ---, ---, 1 3 3 b) Assume a periodic force, p = pˆ sin ( Ωt ) is applied at node 3. Calibrate the rigidity for the following conditions: pˆ = 10 kN

Ω = 4π

h q max = --------150

ξ = 0.05

c) Calibrate the shear rigidity factors for the following conditions: h q max = --------100

Sv

max

= 1.2 m/s

ξ = 0.02

Use equations (2.249) - (2.251). d) Assume linear viscous dampers having the same property, c 1 = c 2 = c 3 = c , are installed. Determine c based on the results of part c.

162 Chapter 2: Optimal Stiffness Distribution

Problem 2.23 Refer to problem 2.19. a) Calibrate the stiffness for the following seismic excitation criteria: u ( H ) max = 0.125 m Sv

max

= 1.0 m/s

ξ = 0.05

b) Assume viscous dampers are installed in the third, fourth, and fifth stories. Take the damper coefficient to be the same for each story. Determine the coefficient required by the calibration specified in part a.

Problem 2.24 Refer to problem 2.20. a) Calibrate the structure for the following seismic criteria: u ( H ) max = 0.125 m Sv

max

= 1.2 m/s

ξ = 0.02

b) Assume a uniform distribution of damping over the height. Determine the damping coefficient required by part a. c) Suppose a single damper is connected to the top node (node 5). Determine the value of c required by part a.

Problem 2.25 Refer to problem 2.15. Take H = 200 m and ρ m = 10, 000 kg/m . ˜ and Γ . a) Determine m b) Assume the following loading x sin Ωt b = b 0 ------------------H

where b 0 = 6 kN/m

Problems 163 is applied. Calibrate the stiffness based on limiting u ( H ) to 0.5 m for a frequency of 1 radian per second. c) Suppose the design is constrained by the peak acceleration at x = H under periodic loading. Calibrate the stiffness for u˙˙( H ) max = 0.015 g where g = 9.86 m/s 2 . Consider ξ to be equal to 0.05 and use the loading defined in part b. d) Assume the stiffness calibration is dictated by seismic excitation. Take u ( H ) max = 1 m , S v = 1 m/s , ξ = 0.05 . Determine D B ( x ) . Assume max C T ( x ) = αD B ( x ) where α is a constant. Determine α corresponding to ξ = 0.05 .

Problem 2.26 Refer to problem 2.16. ˜ and Γ in terms of ρ , a , and b . a) Determine m m b) Assume a uniform periodic loading is applied to the beam. Calibrate the shear rigidity for the following specifications. H = 10 m

u ( H ) max = H ⁄ 300

ρ m = 100 kg/m a = 1 b = 0.5 Ω = 2π rad/s c) Calibrate the rigidity for the case of seismic loading. Take u ( H ) max = H ⁄ 300 Sv

max

= 1.2 m/s

ξ = 0.02 Use eqn (2.249).

164 Chapter 2: Optimal Stiffness Distribution

Problem 2.27 ˜ and Γ . Refer to problem 2.17. Determine m

Problem 2.28 Consider eqn (2.249). Let q∗ = H ⁄ α where α is a design parameter. Express H as H = nh where h is the typical story height and n is the number of stories. This leads to T = fn . Estimate a typical value of f .

Problem 2.29

10 m

5m

This problem concerns the preliminary design of the lateral stiffness system for a 10 story rectangular rigid frame. The frame properties and design criteria are as follows: • • • •

Height = 5 m/story Width = 10 m/bay Mass/floor = 10,000 kg Seismic loading: S v = 1.2 m/s, ξ = 0.02

•

Max. Deflection at top = 0.25 m

Problems 165 •

γ max = 1/200 Model the frame as a 10 DOF shear beam.

a) Determine an initial estimate for the lateral stiffness at each story level considering a single mode response and ξ = 0.02 . Use the spectrum shown below. b) Consider linear viscous dampers to be installed at each story level. Take c i = αk i for story i . Using the stiffness distribution established in Part a (frequencies and mode shapes), establish the modal equations for a modal decomposition in terms of the first 3 modes. Determine α such that ξ 1 = 0.02 . Determine ξ 2 and ξ 3 corresponding to this value of α . Estimate the maximum ‘‘mean square’’ deformation response using the response spectrum shown below, and modify the initial stiffness distribution using eqn (2.252). Carry out this computation for 2 iterations. c) Suppose the lateral stiffness is maintained constant over 2 story heights, i.e. k2 = k1 k 4 = k 3 , etc. Discuss how you would determine initial estimates for k 1, k 3, … considering a single mode response and ξ = 0.02 . Also discuss how you would modify the stiffness distribution.

Spectral Velocity S v (m/s) S v = 1.2, ξ = 0.02 S v = 0.92, ξ = 0.05 S v = 0.8, ξ = 0.1

1.0

S v = 0.6, ξ = 0.2

0.1 0.1

0.6 1.0

10

Period T (sec)

166 Chapter 2: Optimal Stiffness Distribution

167

Chapter 3

Optimal passive damping distribution 3.1 Introduction Damping is the process by which physical systems such as structures dissipate and absorb the energy input from external excitations. Damping reduces the build-up of strain energy and the system response, especially near resonance conditions where damping governs the response. Figure 3.1 illustrates the influence of damping on the time history of the strain energy for a system with mass of 1 kg and a period of 1 second subjected to the unscaled El Centro accelerogram shown in Fig. 2.21. The symbols in Fig. 3.1 refer to the energy input ( E I ), the energy stored ( E s ), and the energy dissipated ( E D ). During the early stage of the response, there is a rapid build-up of the input energy, similar to an impulsive loading. Damping dissipates energy over a response cycle, in this case, 1 second. For low damping ratio, the energy dissipated per cycle is small, and many cycles are required before the input energy is eventually dissipated. As ξ is increased, the energy dissipated per cycle increases, and the stored energy build up is reduced. Shifting from ξ = 0.02 to ξ = 0.1 reduces the peak stored energy demand by a factor of 3.7 for this particular system and earthquake excitation. It should be noted that seismic accelerograms differ with respect to their frequency content and intensity, and therefore one needs to carry out energy time history studies for individual excitations applied to a specific system. For example, Fig 3.2 shows the response of the same system for a typical Northridge accelerogram. The input energy build up for the Northridge loading is quite different than for the El Centro loading.

168 Chapter 3: Optimal Passive Damping Distribution

1.0 ξ = 2%

Energy (J)

0.8 0.6 0.4 0.2 0

0

2

4

6

8

10

12

14

16

18

20

0.8

Energy (J)

ξ = 5% 0.6 0.4 0.2 0

0

2

4

6

8

10

12

14

16

18

20

0.7 Energy (J)

0.6

ξ = 10%

0.4 EI ED

0.2

0

ES

0

2

4

6

8

10 Time (s)

12

14

16

18

Fig. 3.1: Energy Build Up, El Centro (S00E), Imperial Valley 1940

20

3.1 Introduction 169

1.0

Energy (J)

0.8 0.6 EI ED

0.4

ES

0.2 0

0

2

4

6

8

10 Time (s)

12

14

16

18

20

Fig. 3.2: Energy Build Up, Arleta Station (90 DEG), Northridge 1994, ξ = 2% Dissipation and absorption are attributed to a number of external and internal mechanisms, including the following: • Energy dissipation due to the viscosity of the material. This process depends on the time rate of change of the deformations, and is referred to as material damping. Viscoelastic materials belong to this category. • Energy dissipation and absorption caused by the material undergoing cyclic inelastic deformation and ending up with some residual deformation. The cyclic inelastic deformation path forms a hysteresis loop which correspond to energy dissipation; the residual deformation is a measure of the energy absorption. This process is generally termed hysteretic damping. • Energy dissipation associated with overcoming the friction between moving bodies in contact, such as flexible connections. Coulomb damping refers to the case where the magnitude of the friction force is constant. Structural damping is a more general friction damping mechanism which allows for a variable magnitude of the friction force. • Energy dissipation resulting from the interaction of the structure with its surrounding environment. Relative motion of the structure

170 Chapter 3: Optimal Passive Damping Distribution generates forces which oppose the motion and extract energy from the structure. Fluid-structure interaction is a typical case. The fluid exerts a drag force which depends on the relative velocity and functions as an equivalent viscous damping force. • Damping devices installed at discrete locations in structures to supplement their natural energy dissipation/absorption capabilities. These mechanisms may be passive or active. Passive mechanisms require no external energy, whereas active mechanisms cannot function without an external source of energy. Passive devices include viscous, friction, tuned mass, and liquid sloshing dampers. Active damping is achieved by applying external control forces to the structure over discrete time intervals. The magnitudes of the control forces are adjusted at each time point according to a control algorithm. • Passive damping removes energy from the response, and therefore can not cause the response to become unstable. Since active control involves an external source of energy, there is the potential for introducing an instability in the system. The term “semi-active” refers to a particular class of active devices that require a relatively small amount of external energy and apply the control force in such a way that the resulting motion is always stable. Chapter 6 discusses active control devices. In this chapter, the response characteristics for material, hysteretic, and friction damping mechanisms are examined for a single degree-of-freedom (SDOF) system. The concept of equivalent viscous damping is introduced and is used to express viscoelastic, structural, and hysteretic damping in terms of their equivalent viscous counterpart. Numerical simulations are presented to demonstrate the validity of this concept for SDOF systems subjected to seismic excitation. This introductory material is followed by a discussion of the influence of distributed viscous damping on the deformation profiles of multi-degree-offreedom (MDOF) systems. The damping distribution is initially taken to be proportional to the converged stiffness distribution generated in the previous chapter, and then modified to allow for non-proportional damping. Numerical results and deformation profiles for a range of structures subjected to seismic loading are presented, and the adequacy of this approach for distributing damping is assessed. Distributed passive damping can be supplemented with one or more discrete damping devices to improve the response profile. Discrete viscous

3.2 Viscous,Frictional, and Hysteretic Damping 171 dampers inserted in discrete shear beam type structures are considered in this chapter; the basic theory for tuned mass dampers is presented in the next chapter. Subsequent chapters deal with base isolation, a form of passive stiffness/ damping control, and active control.

3.2 Viscous, frictional, and hysteretic damping Viscous damping Viscous damping is defined as the energy dissipation mechanism where the damping force is a function of the time rate of change of the corresponding displacement measure: F = f ( u˙ )

(3.1)

where F is the damping force and u˙ is the velocity in the direction of F . The linearized form is written as: F = cu˙

(3.2)

where c , the damping coefficient, is a property of the damping device. Linear viscous damping is convenient to deal with mathematically and therefore is the preferred way of representing energy dissipation. In general, the work W done on the device during the time interval u ( t 2 ) by t2 [ t 1, t 2 ] , is given W =

∫

F du =

∫

Fu˙ dt

(3.3)

u ( t1 ) t1 Considering periodic excitation

u = uˆ sin Ωt

(3.4)

and evaluating eqn (3.3) for one full cycle under linear viscous damping leads to W viscous = cπΩuˆ

2

(3.5)

This term represents the energy dissipated per cycle by the damping device, as the system to which it is attached undergoes a periodic motion of amplitude uˆ and frequency Ω . Figure 3.3 shows the force-displacement relationship for periodic excitation; the enclosed area represents W .

172 Chapter 3: Optimal Passive Damping Distribution

t = 0

F u = uˆ sin Ωt

cΩuˆ

uˆ

– uˆ

u

π t = ---Ω

Fig. 3.3: Viscous response - periodic excitation

Example 3.1: Viscous damper Figure 3.4 shows a possible design for a viscous damping device. The gap between the plunger and external plates is filled with a linear viscous fluid characterized by τ = G v γ˙

(3.6)

where τ and γ are the shearing stress and strain measures respectively and G v is the viscosity coefficient. Assuming no slip between the fluid and plunger, the shear strain is related to the plunger motion by u γ = ----td

(3.7)

plunger

td

fluid

F,u td

fluid

L

w

Side View

End View

Fig. 3.4: Viscous damping device

3.2 Viscous,Frictional, and Hysteretic Damping 173 where t d is the thickness of the viscous layer. Letting L and w represent the initial wetted length and width of the plunger respectively, the damping force is equal to (3.8)

F = 2wLτ Substituting for τ results in 2wLG v F = ------------------- u˙ td

(3.9)

Finally, eqn (3.9) is written as (3.10)

F = cu˙ where c represents the viscous coefficient of the device, 2wL c = ----------- G v td

(3.11)

The design parameters are the geometric measures w, L, t d and the fluid viscosity, Gv . A schematic diagram of a typical viscous damper employed for structural applications is contained in Fig 3.5; an actual damper is shown in Fig 3.6. Fluid is forced through orifices located in the piston head as the piston rod position is changed, creating a resisting force which depends on the velocity of the rod. The damping coefficient can be varied by adjusting the control valve. Variable damping devices are useful for active control. Section 6.4 contains a description of a particular variable damping device that is used as a semi-active force actuator. This chapter considers only passive damping, i.e. a fixed damping coefficient.

Fig. 3.5: Schematic diagram - viscous damper

174 Chapter 3: Optimal Passive Damping Distribution

Fig. 3.6: Viscous damper - 450 kN capacity (Taylor Devices Inc. http://www.taylordevices.com)

Equation (3.5) shows that the energy loss per cycle for viscous damping depends on the frequency of the excitation. This dependency is at variance with observations for real structural systems which indicate that the energy loss per cycle tends to be independent of the frequency. In what follows, a number of damping models which exhibit the latter property are presented. Friction damping Coulomb damping is characterized by a damping force which is in phase with the deformation rate and has constant magnitude. Mathematically, the force can be expressed as F = F sgn ( u˙ )

(3.12)

where sgn ( u˙ ) denotes the sign of u˙ . Figure 3.7 shows the variation of F with u for periodic excitation. The work per cycle is the area enclosed by the response curve

3.2 Viscous,Frictional, and Hysteretic Damping 175

F F

–u

u = uˆ sin Ωt u = uˆ

u

u

Fig. 3.7: Coulomb damping force versus displacement W coulomb = 4Fu

(3.13)

Figure 3.8 shows a coulomb friction damper used with diagonal X bracing in structures. Friction pads are inserted at the bolt-plate connections. Interstory displacement results in relative rotation at the connections, and the energy dissipated is equal to the work done by the frictional moments during this relative rotation.

Fig. 3.8: Friction brace damper Structural damping removes the restriction on the magnitude of the damping force, and considers the force to be proportional to the displacement amplitude. The definition equation for this friction model has the form F = k s u sgn ( u˙ )

(3.14)

where k s is a pseudo-stiffness factor. Figure 3.9 shows the corresponding cyclic response path. The energy dissipated per cycle is equal to 2 ks u 2 (3.15) W structural = 4 ----------- = 2k s u 2

176 Chapter 3: Optimal Passive Damping Distribution

u = uˆ sin Ωt u = uˆ

–u

F

π t = ------2Ω ks

u

u

Fig. 3.9: Structural damping force versus displacement Hysteretic damping Hysteretic damping is due to the inelastic deformation of the material composing the device. The form of the damping force-deformation relationship depends on the stress-strain relationship for the material and the make-up of the device. Figure 3.10 illustrates the response path for the case where the material forcedeformation relationship is elastic-perfectly plastic. The limiting values are F y , the yield force, and u y , the displacement at which the material starts to yield; k h is the elastic damper stiffness. The ratio of the maximum displacement to the yield displacement is referred to as the ductility ratio and is denoted by µ . With these definitions, the work per cycle for hysteretic damping has the form 2 µ–1 (3.16) W hysteretic = 4k h u y [ µ – 1 ] = 4F y u -----------µ

3.2 Viscous,Frictional, and Hysteretic Damping 177

F

u = uˆ sin Ωt u = uˆ u µ = ----uy

Fy

–u uy

u

u

kh

Fig. 3.10: Hysteretic damping force versus displacement Figure 3.11 shows a bracing element that functions as an hysteretic damper (Watanabe, 1998). The element is composed of a core member fabricated with highly ductile low strength steel (yield strength of 200 MPa., maximum percent strain of 60%), a cylindrical jacket, and mortar placed between the core member and the jacket. The jacket functions as an additional bending element, and its cross sectional moment of inertia is selected such that the buckling load is equal to the yield force. This design feature allows the brace to be used for both tensile and compressive loading. Triangular plate hysteretic dampers that dissipate energy through bending action have also been used for buildings (Tsai, 1993 and 1998).

Fig. 3.11: Hysteretic damper brace element (Watanabe, 1998)

Example 3.2: Stiffness of a rod hysteretic damper Consider a damping device consisting of a cylindrical rod of length L and area A .

178 Chapter 3: Optimal Passive Damping Distribution Suppose the material is elastic-perfectly plastic, as shown in Fig. 3.12. The relevant terms are u ε = --L

(3.17)

F = Aσ

(3.18)

σ

σy

F,u

F E

L εy

ε

Fig. 3.12: Elastic-perfectly plastic damper device. Then, u y = Lε y

(3.19)

F y = Aσ y = k h u y AE k h = -------L

(3.20) (3.21)

Example 3.3: Stiffness of two hysteretic dampers in series The device treated in Example 3.2 is modified by adding a second rod in series, as shown in Fig. 3.13. The yield force for the second rod is assumed to be greater than the yield force for the first rod, A 2 σ y, 2 > A 1 σ y, 1 ≡ F y, 1

(3.22)

Since the force is the same for both devices, the total elastic displacement is the sum of the individual contributions.

3.3 Viscoelastic Material Damping 179

A1 , E1

A2 , E2

F

F,u L1

L2

Fig. 3.13: Two rod hysteretic damping device. L1 L2 1 (3.23) u = ------------- + ------------- F = ----- F kh A1 E1 A2 E2 Specializing eqn (3.23) for the onset of yielding, one obtains A1 L2 E1 F y, 1 (3.24) u y = ----------- = L 1 ε y, 1 1 + -------------------kh A2 L1 E2 A1 E1 1 (3.25) k h = ------------- ----------------------------L1 A1 L2 E1 1 + -------------------When two elements areAused, 2 L 1 E 2one can vary both the yield force, F y , and the elastic yield deformation u y . The energy dissipation increases with decreasing u y , for a given deformation amplitude u .

3.3 Viscoelastic material damping A material is considered to be elastic when the stresses due to an excitation are unique functions of the associated deformation. Similarly, a material is said to be viscous when the stress state depends only on the deformation rates. For simple shear, these definitions translate to Elastic τ = Ge γ

(3.26)

Viscous τ = G v γ˙

(3.27)

The stress-deformation paths for periodic strain are illustrated in Figs 3.14(a) and

180 Chapter 3: Optimal Passive Damping Distribution 3.14(b). There is no time lag between stress and strain for elastic behavior, whereas the stress is π ⁄ 2 radians out of phase with the strain for viscous behavior. If these relations are linearly combined, one obtains the path shown in Fig. 3.14(c). τ

τ

γ = γˆ sin Ωt

τ G v Ωγˆ

G e γˆ

γˆ

γ

γˆ

G v Ωγˆ

(a) Elastic

γ

(b) Viscous

γ

(c) Viscoelastic

Fig. 3.14: Stress-deformation relations. Materials that behave similar to Fig. 3.14(c) are called viscoelastic. The properties of a linear viscoelastic material are determined by applying a periodic excitation and observing the response, which involves both an amplification and a phase shift. The basic relations are expressed as γ = γˆ sin Ωt

(3.28)

τ = γˆ [ G s sin Ωt + G l cos Ωt ]

(3.29)

where G s is the storage modulus and G l is the loss modulus. The ratio of the loss modulus to the storage modulus is defined as the loss factor, η Gl (3.30) η = ------ = tan δ Gs An alternate form for eqn (3.29) is ˆ sin ( Ωt + δ ) τ = γˆ G ˆ = G

2

2

Gs + Gl = Gs 1 + η

(3.31) 2

(3.32)

The angle δ is the phase shift between stress and strain. Delta ranges from 0 for elastic behavior to π ⁄ 2 for pure viscous behavior. Experimental observations show that the material properties G s and η

3.3 Viscoelastic Material Damping 181 vary with temperature and the excitation frequency. Figure 3.15 illustrates these trends for ISD110, a 3M product. The dependency on frequency makes it difficult to generalize the stress-strain relationships based on periodic excitation to allow for an arbitrary time varying loading such as seismic excitation. This problem is addressed in the next section.

To determine the damping properties at the desired temperature and frequency from the data graph shown above, proceed as follows: • Locate the desired frequency on the RIGHT vertical scale. • Follow the chosen frequency line to the desired temperature isotherm. • From this intersect, go vertically up and/or down until crossing both the shear (storage) modulus G s and loss factor curves η . • Read the storage modulus and loss factor values from the appropriate LEFT hand scale.

Fig. 3.15: Variation of 3M viscoelastic material, ISD110, with frequency and temperature. The energy dissipated per unit volume of material for one cycle of deformation is determined from 2π ⁄ Ω W viscoelastic =

∫

τγ˙ dt

(3.33)

0 Substituting for τ and γ using eqns (3.28) and (3.29) results in

W viscoelastic = πG l γˆ

2

(3.34)

182 Chapter 3: Optimal Passive Damping Distribution The corresponding expression for a pure viscous material is generated using eqn (3.27) W viscous = πG v Ωγˆ

2

(3.35)

This expression involves the frequency explicitly whereas the effect of frequency is embedded in G l for the viscoelastic case.

Example 3.4: Viscoelastic damper A damper device is fabricated by bonding thin sheets of a viscoelastic material to steel plates, as illustrated in Fig. 3.16. Since the elastic modulus for steel is considerably greater than the shear modulus G s for the sheet material, one can consider all the motion to be due to shear deformation of the sheets. Defining u as the relative displacement of the ends of the damper device, the shearing strain is u γ = ----td

(3.36)

Given γ , one evaluates τ with the stress-strain relation and then F using the equilibrium equation for the system (3.37)

F = 2wLτ Applying a periodic excitation u = uˆ sin Ωt

(3.38)

td

F,u

td

w

L

L

Side View

Top View

Fig. 3.16: Viscoelastic damper device. and taking τ according to eqn (3.29), one obtains

F,u

3.3 Viscoelastic Material Damping 183 F = f d G s uˆ [ sin Ωt + η cos Ωt ]

(3.39)

2wL f d = ----------td Equation (3.39) can also be written as

(3.40)

ˆ uˆ sin ( Ωt + δ ) F = f dG

(3.41)

ˆ = G 1 + η2 G s

(3.42)

Finally, the energy dissipated per cycle is given by W = πη f d G s uˆ

2

(3.43)

Typical polymer materials, such as Scotchdamp ISD110 (3M Company, 1993) have G s in the range of 1.5MPa and η ≈ 1 .

Based on the result of the previous example, the expressions defining the response of a viscoelastic damper due to periodic excitation can be written in a generalized form u = uˆ sin Ωt

(3.44)

F = f d G s uˆ ( sin Ωt + η cos Ωt ) W viscoelastic = πη f d G s uˆ

2

(3.45) (3.46)

where f d depends on the geometric configuration of the device, G s is the storage modulus, and η is the material loss factor. Figure 3.17 shows the variation of F with u over the loading cycle.

184 Chapter 3: Optimal Passive Damping Distribution

π t = ------2Ω

F

u = uˆ sin Ωt

f d Gs η f d G s uˆ

t = 0 uˆ

u

π t = ---Ω Fig. 3.17: Variation of F d with u d for viscoelastic material.

3.4 Equivalent viscous damping The expression for the damping force corresponding to linear viscous damping is the most convenient mathematical form, in comparison to the other damping force expressions, for deriving approximate analytical solutions to the force equilibrium equations. Therefore, one way of handling the different damping models is to convert them to equivalent viscous damping models. In what follows, a conversion strategy based on equating the energy dissipated per cycle of periodic excitation to the corresponding value for linear viscous damping is described. Linear viscous damping is defined by eqn (3.2): (3.47)

F = cu˙ Specializing eqn (3.47) for periodic excitation u = uˆ sin Ωt

(3.48)

leads to F = cΩuˆ cos Ωt

(3.49)

Also, noting eqn (3.5), the energy loss per cycle is W = cπΩuˆ

2

(3.50)

The force and energy loss for the other models are expressed in terms of an equivalent damping coefficient, c eq

3.4 Equivalent Viscous Damping 185 F = c eq Ωuˆ cos Ωt W = c eq πΩuˆ

2

(3.51) (3.52)

Substituting W for a particular damping model in eqn (3.52), and taking u = uˆ , one obtains the equivalent damping coefficient. The coefficients for the various models are listed below: Coulomb 4F c eq = ----------πΩuˆ

(3.53)

Structural 2k s c eq = -------πΩ

(3.54)

Hysteretic 4F y µ – 1 c eq = ----------- -----------πΩuˆ µ

(3.55)

Viscoelastic

f dG f d ηG s l c eq = ----------------- = ------------Ω Ω

(3.56)

These expressions are valid for periodic excitation of amplitude uˆ and frequency Ω . They can be used to approximate structural and hysteretic damping as pseudo-viscous damping but require specifying a representative frequency, Ω r , and amplitude, u r . In this case, eqns (3.54) and (3.55) are written as Structural 2k s c eq = ---------πΩ r

(3.57)

Hysteretic 4F y µ – 1 c eq = ---------------- -----------πΩ r u r µ

(3.58)

where Fy u y = -----k u rh µ = -----uy

(3.59) (3.60)

Numerical simulations illustrating the accuracy of this approximation are

186 Chapter 3: Optimal Passive Damping Distribution provided by the following examples.

Example 3.5: Structural and hysteretic damping comparison - seismic excitation A 1 DOF shear beam having the following properties is considered: 6

m = 4 ×10 kg

ω 1 = 1.17rd ⁄ s

k = 5517kN ⁄ m

T 1 = 5.35s

c = 187.9kNs ⁄ m

ξ 1 = 2.0%

The equivalent structural stiffness is generated using eqn (3.57), taking c = c e and Ω r equal to the fundamental frequency ω 1 . The corresponding structural stiffness is πω 1 c e π ( 1.17 ) ( 187.9 ) (3.61) k s = --------------- = ------------------------------------- = 346.6kN ⁄ m 2 2 Results for this model subjected to Taft excitation are compared with the corresponding results for the linear viscous model in Figs 3.18 and 3.19. Close agreement is observed.

3.4 Equivalent Viscous Damping 187

−3

6

x 10

Structural damping versus Viscous damping T 1 = 5.35 s

4

ξ 1 = 2%

γ - m/m

2

0

−2

Viscous damping Structural damping

−4

−6 0

10

20

30

40

50

60

Time t - s Fig. 3.18: Response of SDOF with structural damping.

5

4

x 10

Structural damping T 1 = 5.35 s

3

ξ 1 = 2%

F - N

2

Taft excitation

1

0

−1

−2

−3

−4 −6

−4

−2

0

2

4

6 −3

γ - m/m Fig. 3.19: Structural damping force versus deformation.

x 10

188 Chapter 3: Optimal Passive Damping Distribution

The hysteretic model calibration defined by eqn (3.58) is not as straight forward since both the yield force and the ductility are involved. For periodic motion, the maximum displacement u d = uˆ d is known. Then, one can specify the desired ductility µ∗ and compute the required force level and initial stiffness with πω 1 uc eq µ∗ (3.62) F y = ---------------------- --------------4 µ∗ – 1 u (3.63) u y = -----µ∗ Fy (3.64) k h = -----uy For non-periodic motion, one needs to specify the limiting elastic displacement u y , and estimate the maximum amplitude u . This leads to estimates for the ductility ratio µ∗ and the peak force F y . Figures 3.20 and 3.21 show the results based on taking u equal to the peak amplitude observed for pure viscous damping, and a ductility ratio µ∗ = 7.5 . One can adjust µ∗ and F y to obtain closer agreement. Since the energy is dissipated only during this inelastic phase, hysteretic damping is generally less effective than either viscous or structural damping for low intensity loading.

−3

6

x 10

4

Hysteretic damping versus Viscous damping T 1 = 5.35 s

ξ 1 = 2%

γ - m/m

2

0

−2

−4

−6 0

Viscous damping Hysteretic d amping 10

20

30

40

50

Time t - s Fig. 3.20: Response of SDOF with hysteretic damping.

60

3.4 Equivalent Viscous Damping 189

5

F - N

4

x 10

3

Hysteretic damping T 1 = 5.35 s

2

Taft excitation

ξ 1 = 2%

1

0

−1

−2

−3 −6

−4

−2

0

2

4

6 −3

γ - m/m

x 10

Fig. 3.21: Hysteretic damping force versus deformation. The calibration of the equivalent viscous damping coefficient was based on assuming a periodic excitation. As discussed above, non-periodic excitation requires some assumptions as to the response. An improved estimate of the equivalent damping coefficient can be obtained by evaluating the actual work done by the damping force. Starting with t ED

actual

=

∫ Fu˙ dt 0

and writing ED leads to

(3.65) t

eq. viscous t

∫

= c eq u˙ 2 dt

∫ Fu˙ dt

0

0 c eq ( t ) = --------------- = 2ξ eq ωm t

∫

(3.66)

(3.67)

2 dt Equation (3.67) canu˙ be used to evaluate the variation over time of the equivalent damping ratio. Taking t = t end , the total duration of the response, provides an 0 estimate of the effective damping ratio. Figure 3.22 shows results generated for a range of seismic excitations and hysteretic damper yield force levels. As expected, the effective damping increases with increasing excitation and decreases with

190 Chapter 3: Optimal Passive Damping Distribution

Equivalent Viscous Damping Ratio

increasing yield force.

0.2 T = 4.91sec

ξH

ζN = 2 %

0.1

S

v

= 2.4 m/s

S S

0

0

0.01

0.02

0.03

v

v

= 1.2 m/s = 0.6 m/s

0.04

F y ⁄ mg 0.05

Fig. 3.22: Equivalent viscous damping ratio vs. yield force

The viscoelastic model calibration is more involved since the material properties are also frequency dependent. Referring back to eqn (3.45), the damping force for periodic excitation u = uˆ sin Ωt

(3.68)

was expressed as F = uˆ f d G s ( sin Ωt + η cos Ωt )

(3.69)

where f d is a geometric factor defined by the geometry of the device. Our objective is to express F as F = k eq u + c eq u˙

(3.70)

where k eq and c eq are equivalent stiffness and damping terms. Considering periodic excitation, eqn (3.70) takes the form F = uˆ [ k eq sin Ωt + Ωc eq cos Ωt ]

(3.71)

One can obtain estimates for k eq and c eq with a least square approach. Assuming

3.4 Equivalent Viscous Damping 191 there are N material property data sets, and summing the squares of the errors for k eq and c eq over N the ensemble results in

∑ ∑

2

[ k eq – f d G s ( Ω i ) ] f d Gs ( Ωi ) ⋅ η ( Ωi ) 2 J c = i = 1 c e – -------------------------------------------Ωi Jk =

(3.72)

N

(3.73)

i=1

∑

)

Minimizing eqn (3.72) with respect to k eq yields N 1 Gs ( Ωi ) = f d G s k eq = f d ---N

(3.74)

i=1

Similarly, minimizing eqn N (3.73) with respect to c eq results in G s ( Ω i )η ( Ω i ) 1 c eq = f d -----------------------------------N Ωi

∑

(3.75)

i=1

The form of eqn (3.75) suggests that c eq be expressed as c eq = α dN k eq Gs η Substituting for k eq --------and- c eq leads to the definition equation for α d Ω i

(3.76)

∑

i=1 α d = ---------------------------N

(3.77)

∑

Gs ( Ω Note that α d depends only i ) on the material, i.e. it is independent of the geometry of the device. i = 1 With this notation, the equivalent viscous force-deformation relation for a linear viscoelastic damper is written as F = k eq u + α d k eq u˙

(3.78)

Example 3.6: Determining α d for 3M ISD110 damping material This example illustrates how the procedure discussed above can be applied to compute the parameters for the 3M Scotchdamp ISD110 material. Using Fig. 3.15, data corresponding to five frequencies is generated. Table 3.1 contains this data. Applying eqns (3.74), (3.75), and (3.77), one obtains k eq = 5.7 f d

α d = 0.104

c eq = 0.593 f d

192 Chapter 3: Optimal Passive Damping Distribution Table 3.1: Data for ISD110 Scotchdamp material (from Fig. 3.15). Ω (r/s)

G s (MPa)

η

0.628

1.0

1.0

3.14

2.5

1.0

6.28

3.7

0.93

12.56

5.0

0.85

31.4

9.0

0.65

62.8

13.0

0.55

3.5 Damping parameters - discrete shear beam Damping systems This section extends the treatment of discrete shear beams to include damping devices located between the floors. Figure 3.23 illustrates 2 different placement schemes of viscous type dampers for a typical panel. Scheme a combines the damper with a structural element and deploys the composite element on the diagonal between floors. Scheme b places the damper on a roller support at the floor level, and connects the device to the adjacent floor with structural elements. An actual installation of a scheme b system is shown in Fig. 3.24. The structural elements are modelled as linear springs and the representation defined in Fig 3.25 is used.

3.5 Damping Parameters - Discrete Shear Beam 193

ui

θ

ui – 1

(a) ui

ui – 1

(b) Fig. 3.23: Damper placement schemes.

194 Chapter 3: Optimal Passive Damping Distribution

Fig. 3.24: Viscous dampers coupled with chevron bracing.

3.5 Damping Parameters - Discrete Shear Beam 195

u i, V d, i

c k'

V d, i

ui – 1 (a) u i, V d, i k' c

ui – 1 V d, i

(b) Fig. 3.25: Idealized models of structures with viscous dampers. A differential story displacement generates a deformation of the damper, resulting in a damper force which produces the story shear, V d, i . A subscript d is used to denote quantities associated with the damper. The total story shear is the sum of the “elastic” shear force due to elastic frame/brace action and the “damper” shear force. The former was considered in Chapter 2. This contribution is written as V e, i = k e, i ( u i – u i – 1 )

(3.79)

where subscript e refers to “elastic” frame/brace action. The damper shear force is a function of both the relative displacement and the relative velocity. This term

196 Chapter 3: Optimal Passive Damping Distribution is expressed in a form similar to eqn (3.78): V d, i = k d, i ( u i – u i – 1 ) + c d, i ( u˙ i – u˙ i – 1 )

(3.80)

where k d and c d are “equivalent” properties that depend on the makeup of the damping system. Various cases are considered in the following sections. Rigid structural members - linear viscous behavior Consider first the case where the stiffness k' , of the structural members contained in the damping system is sufficiently large so that the extension of the member is negligible in comparison to the extension of the damper. Defining e d as the extension of the damper, and considering scheme a, the damper force for linear viscous behavior is given by F d = c d e˙d = c d ( u˙ i – u˙ i – 1 ) cos θ

(3.81)

The corresponding shear force is V d, i = F d cos θ = c d ( u˙ i – u˙ i – 1 ) cos2 θ

(3.82)

The equivalent damping coefficient for story i is obtained by summing the contributions of the dampers present in story i . c d, i =

∑

( c d cos2 θ )

(3.83)

story i

Equation (3.83) also applies for scheme b; θ = 0 for this arrangement of structural members and dampers. Scheme b is more effective than scheme a (a factor of 2 for 45° bracing), and is more frequently adopted. The general spring-dashpot model shown in Fig 3.26 is useful for representing the different contributions to the story shear force. For this case, the damper ( C ) acts in parallel with the elastic shear stiffness of the frame/bracing system ( K ) and e is equal to the interstory displacement. An extended version of this model is used to study other damping systems.

3.5 Damping Parameters - Discrete Shear Beam 197

K V C

e

Fig. 3.26: Spring and dashpot in parallel model.

Example 3.7: Example 2.15 revisited Consider the 5 DOF shear beam defined in example 2.15. Taking the constant nodal mass as 10,000 kg, and using the stiffness calibration based on S vmax = 0.7 m/s , ξ = 0.1 results in the following values for the element shear stiffness factors, fundamental frequency, nodal mass, and damping: k 1 = 13.59 MN/m , k 2 = 12.68 MN/m , k 3 = 10.87 MN/m k 4 = 8.15 MN/m , k 5 = 4.53 MN/m ω 1 = 9.52 rad/sec

(1)

˜ = 22, 000 kg m c˜ = 41.9 kN-s/m The element damping coefficients are related to c˜ by 1 c˜ = ------ ( c 1 + c 2 + c 3 + c 4 + c 5 ) 25

(2)

• Taking c i constant leads to c 1 = c 2 = … = c 5 = 5c˜ = 210 kN-s/m

(3)

• Assuming damping is proportioned to the element stiffness, c i = αk i , and selecting α according to α = 2ξ 1 ⁄ ω 1 , (the basis for this equation is established in Section 3.7) one obtains the following: α = 0.0210

198 Chapter 3: Optimal Passive Damping Distribution c 1 = 285 kN-s/m

c 2 = 266 kN-s/m

c 4 = 171 kN-s/m

c 5 = 95 kN-s/m

c 3 = 228 kN-s/m (4)

Suppose the chevron brace scheme (scheme b) is used, and 2 dampers are deployed per floor. The “design” values for the dampers are obtained by dividing the above results by a factor 2. For the uniform case, c = 105 kN-s/m . In order to design the damper, one also needs to specify the peak value of the damper force. This quantity is determined with F max = cv max where v max is the maximum relative velocity of the damper piston. For this damper deployment scheme, the relative damper displacement is equal to the interstory displacement. It follows that v max for level i is equal to v max

level i

= ( u˙ i – u˙ i – 1 ) max = h i ( γ˙ i ) max

(5)

The nodal displacements for this 5 DOF model are considered to vary linearly with height: 1 u i – u i – 1 = --- q 5 where q is the modal amplitude. Then 1 v max = --- ( q˙ ) max 5 The peak amplitude is determined with 1 q max = ---- ΓS v ( ξ 1, ω 1 ) ω

(7)

One can estimate q˙max by assuming the response is periodic, with frequency ω 1 . q˙max ≈ ΓS v ( ξ 1, ω 1 )

(8)

Using the problem data, q˙max ≈ ( 1.36 ) ( 0.7 ) = 0.95 m/s and the peak damper force is estimated as

(9)

3.5 Damping Parameters - Discrete Shear Beam 199 0.95 F max ≈ ( 105 ) ---------- = 20 kN 5

(10)

Rigid structural members - linear viscoelastic behavior

K Ve

V

K1

Vd C

e

Fig. 3.27: Spring-dashpot model for viscoelastic damping. The case where the damping mechanism is viscoelastic is represented by the model shown in Fig. 3.27. Here, the damping force has an elastic component as well as a viscous component. Noting eqn (3.78), the damping force is expressed as V d = K 1 e + Ce˙ = K 1 e + α d K 1 e˙

(3.84)

where K 1 and C (or α d ) are the equivalent stiffness and damping parameters for the visco-elastic device, and e is the interstory displacement. This formulation assumes the viscous device is attached to a rigid element so that all the deformation occurs in the device. The more general case is treated later. Using eqn (3.84), one obtains V = ( K + K 1 )e + Ce˙ = ( K + K 1 )e + α d K 1 e˙

(3.85)

When K 1 ⁄ K is small with respect to unity, the contribution of the visco-elastic damper to the stiffness can be neglected.

200 Chapter 3: Optimal Passive Damping Distribution

Example 3.8: Consider a SDOF system having an elastic spring and a visco-elastic damper modeled as shown in the figure. Suppose m , ω , and ξ are specified, and the objective is to establish values for the spring stiffness and damper properties.

m

k k1 c

p u

The governing equation has the form mu˙˙ + cu˙ + ( k + k 1 )u = p

(1)

By definition, k+k ω 2 = -------------1m

(2)

c = 2ξωm

(3)

Given ξ and ω , c is determined with eqn (3). The stiffness factors are related by k 1 + k = mω 2

(4)

Our strategy for dealing with a visco-elastic device is based on expressing the equivalent damper coefficient as (see eqn (3.78)): c = αd k1

(5)

where α d is a “material” property. Example 3.6 illustrates how to evaluate α d for a typical visco-elastic material. The procedure followed here is to first determine k 1 , using eqns (3) and (5),

3.5 Damping Parameters - Discrete Shear Beam 201 2ξωm c k 1 = --------------- ≡ -----αd αd

(6)

and then substitute for k 1 in eqn (4). This operation results in an equation for k. 2 2ξωm 2ξ k = mω – --------------- = mω ω – ------ α d αd

(7)

Suppose m =10,000 kg, ω = 2 π rad/s, and ξ = 0.1 . Then, c = 2 ( 0.1 ) ( 2π ) ( 10 4 ) = 12.56 kN-s/m Using a typical value for α d , α d = 0.15 leads to k 1 = 84 kN/m k = 394 – 84 = 310 kN/m For these parameters, the visco-elastic element contributes approximately 20% of the stiffness.

Example 3.9: Example 3.7 revisited Suppose visco-elastic dampers are used for the 5 DOF system considered in example 3.7. The damper force is taken as F d = k d u d + c d u˙ d = k d u d + α d k d u˙ d

(1)

where α d and k d depend on the device, and u d is the displacement of the damper. Consider the case where a chevron brace with 2 dampers is installed in each floor and the damping distribution defined by eqn 4 in example 3.7 is used. The damper coefficients are determined by dividing the values listed in eqn 4 by 2 (2 dampers per floor): c d, 1 = 142.5

c d, 2 = 133

c d, 3 = 114

202 Chapter 3: Optimal Passive Damping Distribution c d, 4 = 85.5

c d, 5 = 47.5

(2)

(units of kN-s/m) The damper stiffness is determined with 1 k d = ------ c d αd

(3)

Assuming α d = 0.15 , the corresponding values of damper stiffness are: k d, 1 = 950

k d, 2 = 887

k d, 4 = 570

k d, 5 = 317

k d, 3 = 760 (4)

(units are kN/m) The total story shear stiffness distribution is given by eqn (1) in example 3.7. This value is the sum of the elastic stiffness due to frame/brace action and the stiffness due to the 2 dampers k = ke + 2 ⋅ kd

(5)

Using (4) and the data from example 3.7, the frame/brace story shear stiffness factors for this choice of α d are k e, 1 = 11.69 k e, 4 = 7.01

k e, 2 = 10.93

k e, 3 = 9.35

k e, 5 = 3.89

(6)

(units are MN/m) The contribution of the damper stiffness is about 14% of the total stiffness for this example.

Example 3.10: Viscoelastic damper design Referring back to eqn (3.74), the elastic stiffness of the damper depends on the average storage modulus of the viscoelastic material and a geometric parameter fd.

)

3.5 Damping Parameters - Discrete Shear Beam 203

K1 = f d G s )

(1)

Given K 1 and G s , one solves for f d K1 f d = -------Gs )

(2)

To proceed further, one needs to specify the geometry of the device. The figure listed below shows a system consisting of 2n layers of a viscoelastic material located between metal plates. Considering the metal elements to be rigid with respect to the viscoelastic elements, the shape factor is given by wL d f d = 2n ---------td

(3)

The layer thickness is usually fixed by the material manufacturer, and therefore the design variables are the number, length, and width of the viscous plates.

n

w 2 1

td

Ld

Ld As an illustration, suppose (4)

)

k d, 1 = 10, 000 kN/m

Taking G s = 2.5 MPa as the “average” modulus for 3M ISD110, the corresponding shape factor is 3

10, 000 ×10 f d = ----------------------------= 4.0 m 6 2.5 ×10

(5)

204 Chapter 3: Optimal Passive Damping Distribution Substituting for f d in eqn (3), the variables are related by 2n ( wL d ) = t d ⋅ f d = 4.0t d (meters)

(6)

Suppose n = 2 and t d = 10 –2 m = 1 cm . –2

wL d = 1.0 ×10 m 2

(7)

Taking w = L d results in w = L d = 0.1 m = 10 cm

(8)

Example 3.11: Hysteretic damper design - diagonal element Equation (3.58) defines the equivalent viscous damping parameter for hysteretic damping. Substituting the extension, e , for the displacement measure u , and solving for the yield force, F y , results in πΩ r e r µ F y = --------------- ------------ C 4 µ–1

(1)

where e r and Ω r are representative extension and frequency values, and µ is given by er e r sin θ er µ = ----- = --------- = ---------------ey Lε y hε y

(2)

where e y is the extension at which the diagonal material yields. The representative extension is a function of the representative transverse shear deformation γ r . Taking γ r equal to γ ∗ , the design level for γ , leads to e r = γ ∗ h cos θ

(3)

sin 2θ γ ∗ µ = --------------- ----2 εy

(4)

and

3.5 Damping Parameters - Discrete Shear Beam 205 A typical design value for γ ∗ is 1 ⁄ 200 . Ideally, one should use a low strength material so that the response is essentially inelastic throughout the loading duration, thus maximizing the energy dissipation. One potential candidate material is the 200MPa yield strength steel developed by Nippon Steel (Nippon Steel Corp., 1990); the corresponding yield strain is 1 ⁄ 2000 . Using these values and taking θ = 45° provides an upper bound estimate for the ductility ratio 2000 1 µ ≈ ------------ --------- = 5 2 200

(5)

Flexible structural members - linear viscoelastic behavior For completeness, the analysis for the refined viscoelastic model shown in Fig. 3.28 is presented. The component attached to the damper device is modeled as a spring in series with the damper which is considered to be linear viscoelastic with frequency dependent properties G s and η . It is convenient to deal first with a periodic excitation, and then average the properties over the appropriate frequency range.

Primary elastic element

K

Fe Fd

K1 K2

C

F e

Secondary elastic element

Fig. 3.28: General spring-dashpot model. Letting e d represent the displacement of the damper and considering e d to be periodic, the corresponding damper force, F d , follows from eqn (3.69) e d = eˆ d sin Ωt

(3.86)

F d = G s f d eˆ d ( sin Ωt + η cos Ωt ) = G s f d eˆ d sin ( Ωt + δ )

(3.87)

206 Chapter 3: Optimal Passive Damping Distribution where f d is a characteristic geometric parameter for the device and tan δ = η . Since the force in the secondary element must be equal to the damping force, the extensions are related by Fd e – e d = -----(3.88) K2 Substituting for e d and F d leads to the expression for the total displacement e Fd G s f d G s f d (3.89) e = e d + ------ = eˆ d 1 + ------------- sin Ωt + eˆ d ------------- η cos Ωt K2 K2 K2 Equation (3.89) can also be written as e = eˆ sin ( Ωt + δ 1 ) where

ηG s f d 2 Gs f d 2 eˆ = eˆ d 1 + ------------- + ----------------- = ℵeˆ d K2 K2 1 tan δ 1 = η ----------------------K2 1 + ------------The force in theGprimary elastic element depends only on e sfd F e = Ke = Keˆ sin ( Ωt + δ 1 )

(3.90)

(3.91) (3.92)

(3.93)

Combining F e and F d , the total force is given by F = F e + F d = Keˆ sin ( Ωt + δ 1 ) + G s f d eˆ d sin ( Ωt + δ )

(3.94)

A more compact form for F is F = Kˆ eˆ sin ( Ωt + δ 1 + δ 2 )

(3.95)

where δ 2 represents the phase shift between the excitation and the force response, and Kˆ is the total stiffness measure. The definition equations are 2 2 Gs f d Gs f d (3.96) Kˆ = K cos δ 1 + ------------- cosGδ f + K sin δ 1 + ------------- sin δ s d ℵδ + -----------ℵ sin δ K sin 1 ℵ (3.97) tan ( δ 1 + δ 2 )ℜ= ---------------------------------------------------- = ℜ f G --------------- – 1 s d tan δ 1 K cos δ + ------------ cos δ ℵ (3.98) tan δ 2 = -------------------------- 1 1 ℜ + --------------δ 1be expressed in conventional form by shifting the time The equationstan can reference point. Defining t' as

3.5 Damping Parameters - Discrete Shear Beam 207 δ1 t = t' – ----Ω

(3.99)

transforms eqns (3.90) and (3.95) to e = eˆ sin Ωt'

(3.100)

F = Kˆ eˆ sin ( Ωt' + δ 2 )

(3.101)

Finally, expanding eqn (3.101), the result can be expressed in a form similar to the conventional viscoelastic form F = K s eˆ sin Ωt' + ηK s eˆ cos Ωt'

(3.102)

K s = Kˆ cos δ 2

(3.103)

η = tan δ 2

(3.104)

where

One can interpret K s as the storage stiffness and η as the loss factor for the assemblage. Equivalent parameters can be generated following the procedure described in Section 3.4. One writes (3.105)

F = K eq e + C eq e˙

and equates eqn (3.105) with (3.102). The error terms for a periodic excitation are (3.106)

E 1 = K eq – K s ηK s E 2 = C eq – ---------Ω

(3.107)

Minimizing the sum of the square of these terms over the frequency range with respect to K eq and C eq produces N the following expressions 1 (3.108) K s ( Ωi ) K eq = K s ≡ ---N N average η i K s, i ( Ωi i=) 1 1 (3.109) C eq = -----------------------------N Ωi

∑

∑

i=1

where N is the number of frequencies composing the data set. If one considers K 2 to be very large with respect to the visco-elastic damper stiffness, f d G s, the various terms simplify to ℵ≈1

(3.110)

208 Chapter 3: Optimal Passive Damping Distribution δ1 ≈ 0

(3.111)

Gs f d 2 2G s f d Kˆ ≈ K 1 + ----------------- cos δ + ------------K K 1 η = tan δ 2 = ℜ = η -----------------------------------K 1 + -------------------------G s f d cos δ

(3.112) (3.113)

Example 3.12: Coupled spring-damper model This example considers the case where the damper is viscous, and the structural element connecting the damper to the floors is flexible. This is a simplified version of the model considered above; the stiffness component of the visco-elastic device is deleted.

K

Fe

F

Fd K2

C

e

The steps are similar to the previous steps. We take e d = eˆ d sin Ωt

(1)

Then, F d = CΩeˆ d cos Ωt e = eˆ sin ( Ωt + δ 1 ) CΩ 2 eˆ = eˆ d 1 + -------- K2

1/2

CΩ tan δ 1 = -------K2 The total force is expressed as

(2)

3.5 Damping Parameters - Discrete Shear Beam 209 F = Kˆ eˆ sin ( Ωt + δ 1 + δ 2 )

(3)

where CΩ Kˆ = K cos δ 1 1 + tan δ 1 + -------K

2 1/2

CΩ 1 tan δ 2 = -------- ⋅ ------------------------------------------------------------K 1 1 1 1 + ( CΩ ) 2 ------ ------ + ---- K 2 K 2 K

(4)

(5)

The remaining steps are the same. One expresses F as F = K eq e + C eq e˙

(6)

and determines K eq and C eq using eqns (3.108) and (3.109). When K 2 = ∞ , δ 1 = 0 and CΩ tan δ 2 = -------K

K cos δ 2 = ---Kˆ

(7)

Then, Ks = K

nK s = CΩ

(8)

K eq ≡ K

C eq ≡ C

(9)

and

3.6 Damping parameters - truss-beam This section extends the treatment of the truss-beam discussed in Example 2.2 to include damping. The typical panel shown in Fig. 3.29 is considered to be composed of two sets of elements: an elastic system which provides the stiffness (shear and bending) and a second system which functions as a distributed energy dissipation/absorption mechanism.

210 Chapter 3: Optimal Passive Damping Distribution

M V c

E c A

E

d

A

d

h

F

c

F

d

F

d

F

c

θ

B

Fig. 3.29: Truss-beam structure - geometry and forces. The various force quantities are determined with F

d c

d

d

c

c

= Fe + Fd

F = Fe + Fd d

V = 2F cos θ M = BF

c

(3.114) (3.115) (3.116) (3.117)

where superscripts ‘c’ and ‘d’ refer to the column and diagonal elements, and subscripts ‘e’ and ‘d’ denote the elastic and damping force components. Defining e as the extension, i.e. the total change in length of a structural element, and noting eqns (2.36) and (2.39), the extensions for the diagonal and chord elements are related to the transverse shear and bending deformations by e

d

= γh cos θ

c Bh e = ------- χ 2

The elastic force is a function of e and the element properties AE F e = -------- e = K ⋅ e L

(3.118) (3.119)

(3.120)

where L denotes the length, A the cross-sectional area, and E the Young’s Modulus for the elastic element.

3.6 Damping Parameters - Truss Beam 211 Linear viscous behavior The dissipative force-deformation relation depends on the nature of the damping device. For viscous behavior, F d is a function of e˙ . The linear viscous force is expressed as (3.121)

F d = Ce˙

where C is a property of the device, which may be an actual viscous damper, or an equivalent viscous coefficient. Summing the elastic and damping forces leads to the element force-extension relations c

c c

c c

(3.122)

F = K e + C e˙ F

d

d d

d d

(3.123)

= K e + C e˙

One can associate these relations with the parallel spring/dashpot model shown in Fig. 3.26. The member force-deformation relations are obtained by introducing these expressions into the definition equations for V and M . They are written as = D T γ + C T γ˙

(3.124)

M = D B χ + C B χ˙

(3.125)

where 2

d

2

d

D T = [ 2hcos θ ]K C T = [ 2hcos θ ]C

(3.126) (3.127)

2

B h c D B = --------- K 2 2 B h c C B = --------- C 2

(3.128) (3.129)

It is convenient to express the damping coefficients in terms of the stiffness parameters, CT = αT DT

(3.130)

CB = αB DB

(3.131)

where α T and α B follow from the previous equations d C α T = ------d K

(3.132)

212 Chapter 3: Optimal Passive Damping Distribution c

C α B = -----(3.133) c K Note that α T and α B depend on the damping device configuration and stiffness of the diagonal and chord elements. Expressing C in terms of D and taking α constant over the height simplifies the analysis of the beam as a continuum using modal shape functions by uncoupling the equations for the modal coordinates. Linear viscoelastic behavior The case where the damping mechanism is viscoelastic is represented by the model shown in Fig. 3.27. Here, the damping force has an elastic component as well as a viscous component, and is expressed as F d = K 1 e + Ce˙ = K 1 e + α d K 1 e˙

(3.134)

where K 1 and α d are the equivalent stiffness and damping parameters for the device, and e is the total extension of the member. This formulation assumes the viscous device is attached to a rigid element so that all the deformation occurs in the device. Using eqn (3.134), one obtains c

c

c

c c

c

F = ( K + K 1 )e + α dc K 1 e˙ F

d

d

d

d d

d

= ( K + K 1 )e + α dd K 1 e˙

(3.135) (3.136)

Substituting for the forces and deformations leads to 2

d

d

D T = [ 2hcos θ ] ( K + K 1 ) 2

d

C T = [ 2hcos θ ]α dd K 1 and

(3.137) (3.138)

2

c B h c D B = --------- ( K + K 1 ) 2 2 B h c c C B = ---------α d K 1 2

The proportionality factors in this case are given by c c K1 ⁄ K CB α B = ------- = α dc -------------------------DB dc ⁄K dc 1 +K K ⁄ K CT 1 1 α T = ------- = α dd ---------------------------d d DT 1 + K1 ⁄ K

(3.139) (3.140)

(3.141) (3.142)

3.6 Damping Parameters - Truss Beam 213

3.7 Damping distribution for MDOF systems Chapter 2 dealt with the problem of specifying stiffness distributions for beamtype structures. Numerical simulations for seismic excitation showed that one can establish stiffness distributions with an iterative procedure that accounted for the contribution of the higher modes. However, the amplitude of the deformation profile cannot be totally controlled with just stiffness, and other mechanisms need to be incorporated. An examination of the effectiveness of distributed linear viscous damping to decrease the deformation amplitude and therefore provide more control on the response is presented in this section. The structure is assumed to be discretized as an n ’th order MDOF system governed by the following equation. ˙˙ + CU˙ + KU = P MU

(3.143)

where U contains the nodal translation and rotation measures, M is the system mass matrix, C is the system damping matrix, K represents the system stiffness, and P represents the load vector. A similar set of equations was considered in Section 2.10, where a restriction was placed on C and a procedure for transferring the coupled matrix equations to an uncoupled set of equations in terms of modal coordinates was described. The procedure is extended here to allow for an arbitrary form of C . Multi mode free vibration response The case where P = 0 is considered first. The general homogeneous solution can be expressed as U = Ae λt Φ

(3.144)

where A and λ are arbitrary scalars, and Φ is an unknown vector of order n. Substituting for U in eqn (3.143) leads to an equation relating λ and Φ . ( λ 2 M + λC + K )Φ = 0

(3.145)

Solving eqn (3.145) for λ and Φ is referred to as the quadratic eigenvalue problem. A detailed discussion of the mathematical aspects of this problem is contained in Strang,1993.The discussion presented here is intended to provide an introduction to this topic which is also revisited in Chapter 8.

214 Chapter 3: Optimal Passive Damping Distribution When C = 0 , eqn (3.145) reduces to ( K + λ 2 M )Φ = 0

(3.146)

Since K and M are positive definite, λ 2 must be negative. Then, λ can be expressed as λ = ± iω

ω>0

(3.147)

The solution corresponding to the pair of values for λ is U = A 1 e iωt Φ + A 2 e –iωt Φ

(3.148)

Since U must 1be real, the A’s must be complex conjugates, A 1 = --- ( A R + iA I ) 2 1 A 2 = --- ( A R – iA I ) 2 A and A transforms the solution to Substituting for 1 2 U = ( A R cos ωt – A I sin ωt )Φ

(3.149)

(3.150)

Equation (3.146) has n distinct solutions, i.e., n Φ ‘s and corresponding ω ‘s. Each solution satisfies 2

KΦ i = ω i MΦ i

i = 1, 2, …, n

(3.151)

and the following orthogonality relations T

T

Φ j KΦ i = Φ j MΦ i = 0

i≠j

(3.152)

˜i Φ i KΦ i = ω i2 Φ i MΦ i = ω i2 m

(3.153)

T

T

Suppose C is a scaled version of K . C = αK

(3.154)

where α is a positive scalar. This condition is called “stiffness proportional damping”. Substituting for C in eqn (3.145), and rearranging terms leads to λ2 K + ----------------M Φ = 0 (3.155) 1 + αλ Setting λ2 ---------------- = – ω 2 1 + αλ

3.7 Damping Distribution for MDOF Systems 215 converts eqn (3.155) to the “undamped” form. It follows that the eigenvectors (mode shapes) are the “same”, and the eigenvalues ( λ ) depend on ω and α . λ 2j + αω 2j λ j + ω 2j = 0

(3.156)

The eigenvalues are expressed as λ j = – ξ j ω j ± iω j ( 1 – ξ 2j ) 1 / 2 = – ξ j ω j ± iω' j where ξ j is the damping ratio for the j ‘th mode. αω ξ j = ----------j 2

(3.157)

(3.158)

The modal damping ratio increases with increasing mode number. Substituting for λ , the typical free vibration modal response is still periodic, but with a time decaying amplitude. U = e –ξi wi t Φ i [ A R, i cos ω' i t – A I, i sin ω' i t ]

(3.159)

The case where C is an arbitrary symmetric positive definite matrix involves an additional complication; the eigenvectors as well as the eigenvalues are complex quantities. These quantities occur as complex conjugates and are expressed as λ j = λ R, j + iλ I, j λ˜ j = λ R, j – iλ I, j Φ j = Φ R, j + iΦ I, j

(3.160)

˜ = Φ – iΦ Φ j R, j I, j The j ‘th solution is generated by combining the corresponding complex solutions. ˜ e λ˜ j t Φ ˜ U = A j eλ jt Φ j + A j j

(3.161)

Expanding the complex products leads to the following “real” solution: = e λR, j t { ( A R, j Φ R, j – A I, j Φ I, j ) cos λ I t + ( – A I, j Φ R, j – A R, j Φ I, j ) sin λ I t } (3.162)

216 Chapter 3: Optimal Passive Damping Distribution Equation (3.162) is expressed in a form similar to eqn (3.159) U = e –ξ j ω j t { ( A R, j Φ R, j ) cos ω' j t + ( – A I, j Φ R, j ) sin ω' j t } + e –ξ j ω j t { ( – A I, j Φ I, j ) cos ω' j t + ( – A R, j Φ I, j ) sin ω' j t }

(3.163)

Combining the time varying terms leads to a simpler form: U = f j Φ R, j + g j Φ I , j

(3.164)

The second term vanishes for stiffness proportional damping. Our strategy for generating an initial estimate for the stiffness distribution is based on using Φ R as the modal shape function. The actual modal response involves an additional shape function when C is an arbitrary matrix. The importance of this second term depends on the amount of damping. Various computational issues related to stiffness are addressed by the following examples. This topic is discussed in more detail later in Chapter 8, where the general solution for arbitrary loading is presented.

Example 3.13: Eigenvalue problem - 2DOF

m2

u2 k 2, c 2

m1

u1 k 1, c 1

The governing equations for free vibration of the 2DOF system shown above have the form: m 1 u˙˙1 + ( c 1 + c 2 )u˙ 1 + ( k 1 + k 2 )u 1 – c 2 u˙ 2 – k 2 u 2 = 0 m 2 u˙˙2 + c 2 u˙ 2 + k 2 u 2 – c 2 u˙ 1 – k 2 u 1 = 0

(1)

3.7 Damping Distribution for MDOF Systems 217 Expressing the displacements as u 1 = φ 1 e λt

u 2 = φ 2 e λt

(2)

and substituting in eqn (1) leads to a set of equations for the φ terms. ( m 1 λ 2 + λ ( c 1 + c 2 ) + k 1 + k 2 )φ 1 – ( λc 2 + k 2 )φ 2 = 0 – ( λc 2 + k 2 ) φ 1 + ( m 2 λ 2 + λc 2 + k 2 )φ 2 = 0

(3)

For a non-trivial solution to exist, the determinant of the coefficient matrix must vanish. This requirement results in the following equation for λ . ( m 1 λ 2 + λ ( c 1 + c 2 ) + k 1 + k 2 ) ( m 2 λ 2 + λc 2 + k 2 ) – ( λc 2 + k 2 ) 2 = 0

(4)

The elements of Φ are not uniquely defined since eqn. 3 is a homogeneous equation. Solving the first equation leads to a relation between φ 1 and φ 2 . m1 λ 2 + λ ( c1 + c2 ) + k1 + k2 φ 2 = φ 1 ⋅ ------------------------------------------------------------------λc 2 + k 2

(5)

The undamped solution is obtained by setting c 1 = c 2 = 0 . ( m 1 λ 2 + k 1 + k 2 ) ( m 2 λ 2 + k 2 ) – k 22 = 0

(6)

m1 λ 2 + k1 + k2 φ 2 = φ 1 ⋅ -----------------------------------k2

(7)

Expanding eqn (6) results in ( m 1 m 2 )λ 4 + ( m 1 k 2 + m 2 ( k 1 + k 2 ) )λ 2 + k 1 k 2 = 0

(8)

Noting that the coefficients are all positive, it follows that λ 2 must be a negative real number. Expressing λ as λ = ±i ω

(9)

and solving eqn (8) for ω 2 leads to ω 2 = a(1 ± b) where

(10)

218 Chapter 3: Optimal Passive Damping Distribution m1 k2 + m2 ( k1 + k2 ) a = ----------------------------------------------2m 1 m 2

(11)

1/2 4m 1 m 2 k 1 k 2 b = 1 – ------------------------------------------------------2- ( m1 k2 + m2 ( k1 + k2 ) )

(12)

Equation (10) has 2 real positive roots. The elements of φ are also real quantities in this case. Each eigenvalue/eigenvector produces a solution of the form u 1 = Φ ( j )1 ( A R, j cos ω j t – A I, j sin ω j t ) u 2 = Φ ( j )2 ( A R, j cos ω j t – A I, j sin ω j t )

(13)

where A R and A I are integration constants that are evaluated using the initial conditions at time t = 0 . To illustrate the computations, the system examined in example 1.9 is considered. The mass and stiffness are: m 1 = m 2 = 1, 000 kg = 1 kN-s 2 /m k 2 = ( 2 ) ( 2π ) 2 = 78.88 kN/m

(14)

k 1 = ( 3 ) ( 2π ) 2 = 118.3 kN/m These stiffness values produce a fundamental mode which is linear and correspond to a fundamental frequency of 2π rad/s. Evaluating a and b for this set of properties results in a = 138.0 (1/sec 2 ) 1/2

24 b = 1 – ------ 49

5 = --7

(15)

and ω 1, 2 = 2π , 2π ( 2.45 ) = 6.28, 15.29 The mode shapes are determined with eqn (7). Fundamental mode

(16)

3.7 Damping Distribution for MDOF Systems 219

λ 2 = – ω 2 = – ( 2π ) 2 φ 2 = 2φ 1

(17)

Second mode λ 2 = – 6 ( 2π ) 2 φ2 = –φ1 ⁄ 2

(18)

It is convenient to normalize the modal vectors such that the magnitude of the maximum element is unity. For this example, the normalized vectors are Φ1 = 1 ⁄ 2 1

Φ2 =

1 –1 ⁄ 2

(19)

The case where c is proportional to k is considered next. Setting c i = αk i and defining λ′ as λ2 ( λ′ ) 2 = ---------------1 + αλ

(20)

reduces eqns (4) and (5) to the “same” form as eqns (6) and (7) with λ replaced with λ′ . ( m 1 ( λ′ ) 2 + k 1 + k 2 ) ( m 2 ( λ′ ) 2 + k 2 ) – k 22 = 0

(21)

m 1 ( λ′ ) 2 + k 1 + k 2 φ 2 = φ 1 ------------------------------------------k2

(22)

One sets λ′ = – ω 2

(23)

and determines ω 2 with eqn (10). The eigenvector is the same as the undamped vector, and the eigenvalues are determined by combining eqns (20) and (23). λ2 ---------------- = – ω 2 1 + αλ The solution is written in the following general form:

(24)

220 Chapter 3: Optimal Passive Damping Distribution

λ j = – ξ j ω j ± iω j ′ ξ j = αω j ⁄ 2 ω j ′ = ω j [ 1 – ξ 2j ] 1 / 2

(25)

u i = Φ ( j )i e –ξ j ω j t ( A R, j cos ω j ′t – A I, j sin ω j ′t ) j = 1, 2 Continuing with the numerical example, suppose the design objective is ξ 1 = 0.1 . The required value of α is 2ξ 2 ( 0.1 ) α = --------1 = --------------- = 0.0318 sec ω1 2π

(26)

Then, c 1 = αk 1 = 3.762 kN-s/m c 2 = αk 2 = 2.508 kN-s/m

(27)

The damping ratio for the second mode is proportional to the frequency ratio: ω ξ ----2- = -----2- = 2.45 ξ1 ω1

(28)

ξ 2 = 2.45 ( 0.10 ) = 0.245 Lastly, the modified frequencies are: ω 1 ′ = ω 1 [ 1 – ξ 12 ] 1 / 2 = ω 1 ( 0.99 ) 1 / 2 = 0.995ω 1 = 6.25 ω 2 ′ = ω 2 ( 0.94 ) 1 / 2 = 0.97ω 2 = 14.83

(29)

The case where C is not proportional to either the mass or stiffness matrices involves solving a complete 4’th degree equation for λ . Equation (4) is written as λ4 + f 3λ3 + f 2λ2 + f 1λ + f 0 = 0 where the coefficients are

(30)

3.7 Damping Distribution for MDOF Systems 221

f 3 = ( m1 m2 ) –1 ( m1 c 2 + m2 ( c 1 + c 2 ) ) f 2 = ( m1 m2 ) –1 ( m1 k 2 + m2 ( k 1 + k 2 ) + c 1 c 2 ) f 1 = ( m1 m2

) –1 ( k

2 c1

+ k1 c2 )

(31)

f 0 = ( m1 m2 ) –1 ( k 1 k 2 ) Since all the coefficients are positive, the real part of λ must be negative. The factored form of eqn (30) is expressed as ( λ 2 + a1 λ + a2 ) ( λ 2 + b1 λ + b2 ) = 0

(32)

where a1 + b1 = f 3 a1 b1 + a2 + b2 = f 2 a1 b2 + a2 b1 = f 1

(33)

a2 b2 = f 0 With this notation, the 2 pairs of complex conjugate roots are 1 ( λ 1, λ˜ 1 ) = --- [ – a 1 ± i ( 4a 2 – a 12 ) 1 / 2 ] = – ξ 1 ω 1 ± iω 1 ′ 2 1 ( λ 2, λ˜ 2 ) = --- [ – b 1 ± i ( 4b 2 – b 12 ) 1 / 2 ] = – ξ 2 ω 2 ± iω 2 ′ 2

(34)

The eigenvectors are determined with eqn (5) which is written as φ 2 = hφ 1

(35)

where h is a function of λ . Each pair of complex conjugate roots generates a corresponding pair of complex conjugates for h and φ . Considering the first pair of eigenvalues, ( λ 1, λ˜ 1 ) ⇒ ( h 1, h˜ 1 )

(36)

and φ 1, 2 = h 1 φ 1, 1 φ˜ 1, 2 = h˜ 1 φ˜ 1, 1

(37)

222 Chapter 3: Optimal Passive Damping Distribution The modes defined by eqn (37) are represented as: ˜ ) = Φ ( Φ 1, Φ 1 R, 1 ± iΦ I, 1

(38)

Since Φ is not uniquely defined, one can set the magnitude of one element, and scale the other elements using eqn (35). Adopting the normalized form defined by eqn (19), the definition equations for the elements are:

Mode 1 φ2 ≡ 1 1 1 φ 1, φ˜ 1 = -----, ----h 1 h˜ 1

(39)

Mode 2 φ1 ≡ 1 φ 2, φ˜ 2 = h 2, h˜ 2

(40)

Results based on the mass and stiffness properties defined by eqn (14) and 5 different distributions of the damping coefficients are listed below in Table 3.2. The first case is stiffness proportional damping with ξ 1 = 0.1 . The next 3 cases correspond to uniform, linear, and concentrated distributions with the sum of the damping coefficients held constant. The last case is a repeat of case 2 with the damping doubled. Concentrating all the damping in the second element produces the maximum value of ξ 2 , and does not result in a significant change in the mode shape. However, it increase the damping force, and requires a “larger” damper. Selecting a uniform distribution has the advantage that the damper size is less than the sizes for the other choices. To arrive at an optimal distribution, one needs to consider the total damper costs and other constraints on placement of dampers. There are situations where dampers cannot be placed in certain floors. Computer based simulation is used in this case to “optimize” damper placement.

3.7 Damping Distribution for MDOF Systems 223

Table 3.2: Frequencies, damping ratios, and mode shapes for 2DOF system c 1 = 3.762

c 1 = 3.135

c 1 = 2.09

c 1 = 0.0

c 1 = 6.27

c 2 = 2.513

c 2 = 3.135

c 2 = 4.18

c 2 = 6.27

c 2 = 6.27

ω1

6.28

6.286

6.325

6.504

6.305

ξ1

0.1

0.1

0.0987

0.0926

0.1987

φ R, 1

0.5

0.502

0.509

0.528

0.511

φ R, 2

1.0

1.0

1.0

1.0

1.0

φ I, 1

0

-0.025

-0.066

-0.151

-0.0486

φ I, 2

0

0

0

0

0

ω2

15.29

15.37

15.276

14.854

15.32

ξ2

0.245

0.265

0.301

0.385

0.532

φ R, 1

1.0

1.0

1.0

1.0

1.0

φ R, 2

-0.5

-0.515

-0.544

-0.620

-0.564

φ I, 1

0

0

0

0

0

φ I, 2

0

0.059

0.161

0.37

0.107

Note: m 1 = m 2 = 1, 000 kg , k 1 = 118.3 kN/m , k 2 = 78.9 kN/m Units of c are kN s/m

Example 3.14: Modal response for nonproportional damping Equation (3.163) defines the general homogeneous solution for non-

224 Chapter 3: Optimal Passive Damping Distribution proportional damping. A typical solution is actually the sum of 2 solutions (the subscript j is dropped for convenience): U = f Φ R + gΦ I

(1)

f = A R ( e –ξωt cos ω′t ) + A I ( – e –ξωt sin ω′t )

(2)

g = A R ( – e –ξωt sin ω ′t ) + A I ( – e –ξωt cos ω ′t )

(3)

where

The responses associated with A R and A I are 90° out of phase with each other. This formulation is applied to the 2 DOF system treated in example (3.13). Table 3.2 shows that the mode shapes have the following general form: Mode 1 ΦR =

aR 1

ΦI =

aI

ΦI =

0 bI

(4)

0

Mode 2 ΦR =

1 bR

(5)

where a and b depend on the level of damping. Substituting for the Φ ‘s in eqn (1) leads to: Mode 1 u1 = aR f + aI g u2 = f

(6)

Mode 2 u1 = f u2 = bR f + bI g

(7)

3.7 Damping Distribution for MDOF Systems 225 Expanding the terms in eqn (6) shows the phasing effect on the fundamental mode. u 1 = e –ξ1 ω1 t [ A R ( a R cos ω 1 ′t – a I sin ω 1 ′t ) + A I ( – a R sin ω 1 ′t – a I cos ω 1 ′t ) ] (8)

u 2 = e –ξ1 ω1 t [ A R ( cos ω 1 ′t ) + A I ( – sin ω 1 ′t ) ]

(9)

Comparing there terms shows that the solution for u 1 is out of phase with respect to u 2 . Defining β and aˆ as a tan β 1 = -----I aR

(10)

aˆ 1 = [ a R2 + a I2 ] 1 / 2

(11)

transforms eqn (8) to u 1 = e –ξ1 ω1 t [ A R aˆ 1 cos ( ω 1 ′t + β 1 ) – A I aˆ 1 sin ( ω 1 ′t + β 1 ) ]

(12)

The shear strains in the element are related to the difference in displacements. 1 γ 1 = ----- u 1 h1 1 γ 2 = ----- ( u 2 – u 1 ) h2

(13)

Using eqns (8) and (9), u2 – u1 = e –ξ1 ω1 t [ A R ( ( 1 – a R ) cos ω 1 ′t + a I sin ω 1 ′t ) – A I ( ( 1 – a R ) sin ω 1 ′t – a I cos ω 1 ′t ) ] (14)

Equation (14) is written in a form similar to eqn (12). The amplitude and phase terms are:

226 Chapter 3: Optimal Passive Damping Distribution aI tan β 2 = – -------------1 – aR

(15)

aˆ 2 = [ ( 1 – a R ) 2 + a I2 ] 1 / 2

(16)

With this notation, eqn (14) takes the form –ξ ω t u 2 – u 1 = e 1 1 [ A R aˆ 2 cos ( ω 1 ′t + β 2 ) – A I aˆ 2 sin ( ω 1 ′t + β 2 ) ]

(17)

The peak differential displacements are: u 1 max = aˆ 1 ( A R2 + A I2 ) 1 / 2 u 2 – u 1 max = aˆ 2 ( A R2 + A I2 ) 1 / 2

(18)

Table 3.2 lists results for a R and a I corresponding to various distributions of c 1 and c 2 . The largest deviation from the proportional damping case ( a R = 0.5 ; a I = 0 ) is case 4. a R = 0.528

a I = – 0.151

(19)

Using the above values, the various parameters are β 1 = – 15.96°

β 2 = – 17.74°

aˆ 1 = 0.549

aˆ 2 = 0.495

(20)

For this example, the peak shear strains in elements 1 and 2 differ by only 10%

Stiffness proportional viscous damping An introductory treatment of the case where C = αK was presented in section 2.10. This section expands upon that analytical treatment and also contains some examples which illustrate the influence of damping on the response of the higher modes. Expressing U as an expansion in terms of the undamped mode shape vectors,

3.7 Damping Distribution for MDOF Systems 227 n

U =

∑

qi Φi

(3.165)

i=1

and noting the orthogonality relations leads to the following set of equations for the q’s T Φi P p˜ i 2 (3.166) q˙˙i + 2ξ i ω i q˙i + ω i q i = ----------- = -----i = 1, 2, 3, . . . . ˜i ˜i m m where ξ i , the damping ratio for the ith mode, is related to α and ω i by eqn (3.158). The magnitude of α is determined by specifying the damping ratio for the first mode 2ξ 1 (3.167) α = --------ω1 Given α , one calculates the other damping ratios with ωi ξ i = ξ 1 ------ ω 1 Suppose the initial conditions on U are U ( 0 ) = U∗ U˙ ( 0 ) = U˙ ∗

(3.168)

(3.169)

Expressing U as U = Φq

(3.170)

and noting the orthogonality relation for M which follows from eqn (3.151), ˜ i δ ij ] = m ˜ Φ T MΦ = [ m

(3.171)

one can transform eqn (3.170) to ˜ q = Φ T MU m

(3.172)

by premultiplying by Φ T M . The i ‘th term has the form ˜ i q i = φ iT MU m

(3.173)

Applying this transformation to the initial conditions on U leads to the initial conditions for q i . 1 (3.174) q i ( 0 ) = ------ φ iT MU∗ = q i∗ ˜i m 1 (3.175) q˙i ( 0 ) = ------ φ iT MU˙ ∗ = q˙i∗ ˜i m i = 1, 2, …, n

228 Chapter 3: Optimal Passive Damping Distribution The orthogonality relationships for the coefficient matrices reduces the problem of solving an n ‘th order coupled equation to one of solving n uncoupled equations for the modal coordinates. A similar result can be obtained for nonproportional damping, but that formulation is considerably more complicated than this formulation. The nonproportional case is discussed again in Chapter 8 where a general approach for dealing with arbitrary damping based on the “statespace” formulation is presented. The total response for the i’th modal coordinate 1 ξi ωi t * q i ( t ) = e –as: q i cos ω′t + ----- ( q˙* + ξωq * ) sin ω′t + can be expressed ω′ ξi ωi (3.176) q˙i ( t ) = e –ξi ωi t tq˙i* cos ω′t + – q˙* ---------- + ( – ω i ′ 2 – ξ i2 ω i2 )q * sin ω′t – 1 p˜ i –ξi ωi ( t – τ ) ω i ′ + ------- ------ e sin [ ω i ′ ( t – τ ) ] dτ ˜it ωi ′ m (3.177) ˜i 20 ω 1 / i2 p ( 1 – ξ–i )------ ------ e –ξi ωi ( t – τ ) sin [ ω i ′ ( t – τ ) – δ i ] dτ ˜ ω' - m (3.178) tan δ i = ------------------------ξi i 0 i

∫

∫

i = 1, 2, …, n The procedure for generating the stiffness distribution presented in Chapter 2 assumes the fundamental mode ( i = 1 ) is the dominate term and bases the stiffness distribution on the displacement profile defined by q 1 Φ 1 . Other modes may contribute, depending on the frequency of excitation and modal damping. Ideally, one wants high modal damping for i = 2, 3, … to minimize their contribution. With stiffness proportional damping, the modal damping ratios are fixed by the frequency ratios. They increase with mode number, which is desirable, but one cannot arbitrarily increase a particular modal damping ratio to further reduce the modal response. The only recourse is to increase the fundamental mode damping ratio, ξ 1 . This action increases α since α = 2ξ 1 ⁄ ω 1 , and requires additional damping. A set of examples illustrating the effect of increasing ξ 1 on the response are presented in the following section. Example 3.15: Low rise buildings. Results for the shear beam type buildings defined in Table 3.3 are presented in this section. The building sites are assumed to be in California, and the reference spectral velocity is taken as S v = 1.2 m/s for ξ = 0.02 . Three earthquakes representative of the sites, namely, El Centro, Northridge Station01, and Northridge Station03, are used to generate ensemble averaged response spectra. Their scaled accelerograms are shown in Figure 3.30 through 3.32.

3.7 Damping Distribution for MDOF Systems 229

Table 3.3: Example shear buildings Building

Number of stories

Story height (meters)

Mass/story (103kg)

γ obj

1

3

4

10

0.005

2

6

5

10

0.005

3

9

5

10

0.005

230 Chapter 3: Optimal Passive Damping Distribution

4

acceleration m/s2

3 2 1 0 −1 −2 −3

0

10

20

30 time s

40

50

60

Spectrum

150

100

50

0

0

20

40

60

80 ω (rad/s)

100

120

140

160

Fig. 3.30: Scaled time history and frequency content of El Centro ( S v = 1.2 & ξ = 0.02 )

3.7 Damping Distribution for MDOF Systems 231

4

acceleration m/s

2

3 2 1 0 −1 −2 −3

0

10

20

30 time s

40

50

60

120

Spectrum

100 80 60 40 20 0

0

20

40

60

80 ω (rad/s)

100

120

140

160

Fig. 3.31: Scaled time history and frequency content of Northridge: Station 1 component 090 ( S v = 1.2 & ξ = 0.02 )

232 Chapter 3: Optimal Passive Damping Distribution

2

acceleration m/s

2

3

1 0 −1 −2

0

10

20

30 time s

40

50

60

140 120

Spectrum

100 80 60 40 20 0

0

20

40

60

80 ω (rad/s)

100

120

140

160

Fig. 3.32: Scaled time history and frequency content of Northridge: Station 3 component 090 ( S v = 1.2 & ξ = 0.02 )

3.7 Damping Distribution for MDOF Systems 233 Stiffness and damping distributions are generated for values of ξ 1 ranging from 0.02 to 0.3. For each ξ 1 value, the following quantities are determined: ω 1 ; the story stiffness coefficients, k i ; the damping parameter α = 2ξ 1 ⁄ ω 1 ; and lastly the story damping coefficients, c i = αk i . Figure 3.33 shows the variation of the frequency and period for the first three modes with increasing damping of the fundamental mode.

0.8

Mode 1 Mode 2 Mode 3

40 20 0

0

0

0.4

0

0.1 0.2 0.3 ξξ1 − 3 Story Building

0.4

0

0.1 0.2 0.3 ξ 1ξ − 6 Story Building

0.4

0

0.1 0.2 0.3 ξξ1 − 9 Story Building

0.4

2 1.5 T (s)

ω (rad/s)

0.1 0.2 0.3 ξ 1ξ − 3 Story Building

30 20 10 0

0.4 0.2

40

1 0.5

0

30

0.1 0.2 0.3 ξξ1 − 6 Story Building

0

0.4

3

20

2 T (s)

ω (rad/s)

0.6 T (s)

ω (rad/s)

60

10 0

1

0

0.1 0.2 0.3 ξξ1 − 9 Story Building

0.4

0

Fig. 3.33: Frequencies and periods of first three modes

234 Chapter 3: Optimal Passive Damping Distribution As ξ 1 is increased, the required stiffness decreases, and the corresponding damping increases. The total stiffness is defined as the sum of the story stiffnesses. A similar definition is used for the story damping coefficients. These two quantities provide a measure of the “design” cost. The data for the three example structures are plotted in Fig 3.34. Total damping varies essentially linearly with ξ , whereas total stiffness behaves nonlinearly, and the degree of nonlinearly increases with the number of stories. Given costs for stiffness elements and dampers, one can generate data on cost as a function of ξ 1 and establish an “optimal” damping system for stiffness proportional damping.

7

3

x 10

5

Total Stiffness 8

N/(m s)

N/m

1

8

0 7 x 10

0.1 0.2 0.3 ξ − 3 Story Building

0

0.4

3

N/(m s)

N/m

4 2

8

0 7 x 10

0.1 0.2 0.3 ξ − 6 Story Building

0.4

0 6 x 10

0.1 0.2 0.3 ξ − 6 Story Building

0.4

4 2 0

0

0.1 0.2 0.3 ξ − 9 Story Building

0.4

1

6

N/(m s)

N/m

0.1 0.2 0.3 ξ − 3 Story Building

2

0

0.4

6

0

0 6 x 10

4 2

6

0

Total Damping

6

2

0

x 10

0.1 0.2 0.3 ξ − 9 Story Building

0.4

4 2 0

Fig. 3.34: Total stiffness (N/m) and total damping (N-s/m) for the three example buildings

3.7 Damping Distribution for MDOF Systems 235

Using time-history analysis, the peak shear deformation was generated for each story of each building, and each earthquake. Figure 3.35 shows the variation of the mean deformation (the average over the building height) and standard deviation with ξ 1 . As expected, the response depends on the earthquake even though the earthquakes are scaled to have the same S v . The ensemble average provides a better estimate of the response. In general, the deformations are lower than the target value and the gap widens as ξ 1 is increased. Taking ξ 1 = 0.1 as the typical value, the results are within 80% of the target. Considering the uncertainty associated with specifying the seismic excitation and modeling the structure, one needs to apply a conservative design procedure.

−3

−3

El Centro

x 10 6

6

4

4

2

2

Northridge03−090

Northridge01−090

0

Ensemble Average

x 10

0 −3 x 10

0.1

0.2

0.3

0.4

0

6

6

4

4

2

2

0

0 −3 x 10

0.1

0.2

0.3

0.4

0

6

6

4

4

2

2

0

0 −3 x 10

0.1

0.2

0.3

0.4

0

6

6

4

4

2

2

0

0

0.1 0.2 0.3 ξ − 3 Story Building

(a) 3 story building

0.4

0

0 −3 x 10

0.1

0.2

0.3

0.4

0 −3 x 10

0.1

0.2

0.3

0.4

0 −3 x 10

0.1

0.2

0.3

0.4

0.1

0.2

0.3

0.4

0

6 Story Building (b) 6ξ −story building

Fig. 3.35: Mean story deformation and standard deviation for design earthquakes

236 Chapter 3: Optimal Passive Damping Distribution

−3

El Centro

x 10 6 4 2

Ensemble Average

Northridge03−090

Northridge01−090

0

0 −3 x 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 −3 x 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0 −3 x 10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0

0.05

0.1

0.15

0.2 0.25 ξ − 9 Story Building (c) 9 story building

0.3

0.35

0.4

6 4 2 0 6 4 2 0 6 4 2 0

Fig. 3.35: Mean story deformation and standard deviation for design earthquakes Example 3.16: Building #4 The building designated as example #4 in section 2.9, Table 2.4 is considered here. This building has a height of 200m, which corresponds to about 50 stories. The relevant properties are listed below for convenience.

3.7 Damping Distribution for MDOF Systems 237 Table 3.4: Properties for Building #4 H = 200m

H/B = 5

ρ m = 20,000 kg/m Design earthquake:

Sv = 1.2 m/s for ξ = 0.02

Design deformation:

γ * = 1/200

f

*

s = 0.63

=4

Three values of the damping ratio for the fundamental mode are considered. Figure 2.27 is used to obtain the value of Sv corresponding to each value of ξ . Given Sv, one finds T with eqn (2.220), and the rigidity distributions with eqns (2.119) and (2.220). These rigidity distributions are modified using the approach described in sections 2.10, and are scaled during the iteration process such that the final values of T are equal to the values predicted by eqn (2.220). Table 3.5 contains the design data for Sv and the corresponding periods. The “converged” rigidity distributions for each damping ratio are plotted in Figure 3.36, 3.37, and 3.38. Table 3.5: Design data for Building #4 Case

Design ξ1

values

Sv (m/s)

Period γ*

T1 (s)

1

0.02

1.2

1/200

5.36

2

0.05

0.94

1/200

6.85

3

0.10

0.70

1/200

9.25

238 Chapter 3: Optimal Passive Damping Distribution

1

x Normalized height ---H

0.9

0.8

Building #4 Quadratic Based T=5.36 s

0.7

0.6

mode 1

0.5

0.4

mode 2

0.3

SRSS

mode 3

0.2

0.1

0

0

2

4

6

8

10

12

14 8

x 10

Shear rigidity distribution DT - N Fig. 3.36: Converged shear rigidity distribution for Building 4, ξ = 0.02 1 0.9

x Normalized height ---H

Building #4 Quadratic based T = 6.85 s

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 3.37: Converged shear rigidity distribution for Building 4, ξ = 0.05

3.7 Damping Distribution for MDOF Systems 239

1

x Normalized height ---H

0.9

Building #4 Quadratic based T = 9.2 s

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N Fig. 3.38: Converged shear rigidity distribution for Building 4, ξ = 0.10 Profiles of the maximum transverse shear deformation due to seismic excitation scaled such that Sv = 1.2 m/s for ξ = 0.02 are plotted in Figures 3.39, 3.40, and 3.41. The trend is similar to what was observed for the shear building examples examined in the previous section. As the damping is increased, the mean deformation trends to decrease, even though the design value of the spectral velocity is decreased with increasing ξ . This behavior indicates that the design procedure is conservative, at least for this example building. It should be noted that the periods are in the range where the dynamic amplification is not as significant as for low period buildings. Also, one should consider a number of excitations in order to generate a more representative “ensemble” average.

240 Chapter 3: Optimal Passive Damping Distribution

1

0.9

Building #4 T = 5.36 s

0.8

x Normalized height ---H

0.7

0.6

0.5

0.4

0.3

_________ El Centro

0.2

- - - - - - - Taft

0.1

0

(Sv=1.2 m/s, ξ=0.02)

γ∗ 0

0.002

0.004

0.006

0.008

0.01

Maximum shear deformation γ - m/m

0.012

Fig. 3.39: Maximum shear deformation for Building 4, ξ = 0.02 1 0.9

x Normalized height ---H

Building #4 T = 6.85 s

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

0.01

Fig. 3.40: Maximum shear deformation for Building 4, ξ 1 = 0.05

3.7 Damping Distribution for MDOF Systems 241

1 0.9

x Normalized height ---H

Building #4 T = 9.2 s

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

El Centro Taft

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

0.009

0.01

Fig. 3.41: Maximum shear deformation for building 4, ξ 1 = 0.10

242 Chapter 3: Optimal Passive Damping Distribution

Problems 243

Problems Problem 3.1 Consider a SDOF system. The total stored energy is the sum of the kinetic energy and the strain energy. Stored Energy = E S + E K 1 1 = --- ku 2 + --- mu˙ 2 2 2 Note that E S and E K are out of phase for periodic excitation. Suppose there is a rapid build-up of strain energy during the initial phase of a seismic excitation, and the design objective is to limit u max to a target value, u∗ . What strategy would you apply? Assume the mass cannot be varied.

Problem 3.2 Give examples of energy dissipation and absorption devices that are used in structures such as vehicles, crash barriers, machine supports, buildings, and bridges.

Problem 3.3 Consider a SDOF system having a linear spring and linear viscous damper. Suppose the initial conditions at t = 0 are: u(0) = 0

u˙ ( 0 ) = u˙∗

(1)

and there is no external loading.

k

c

u

244 Chapter 3: Optimal Passive Damping Distribution a) Determine the expression for the free vibration response corresponding to these initial conditions. b) Construct a plot of u˙ vs. u for t ranging from 0 to 4π ⁄ ω . Comment on the influence of damping on the time history response. Illustrate for m = 1, 000 kg , k = 9, 000 kN/m .

Problem 3.4 Refer to problem 3.3. Suppose the device is a Coulomb function damper. a) Determine the free vibration response. Since the sense of the friction force is determined by the sense of the velocity, one needs to generate solutions for different time intervals. During the first interval, 0 ≤ t ≤ t 1 , the velocity is positive. b) Construct a plot of u˙ vs. u for t sufficient to include one full cycle.

Problem 3.5 The SDOF system shown below is subjected to a periodic force. Determine the expressions for the spring and damper forces. Let F denote the total internal force. F = Fs + Fd

(1)

Fs

k

F

pˆ sin Ωt

m Fd

c

u

Determine the expression for F . Compare the magnitude of F with the magnitudes of F s and F d .

Problems 245 Problem 3.6 Refer to example 3.3. Consider material “one” to be low strength steel with σ y = 200 MPa , and material “two” to have σ y = 500 MPa steel. Design a hysteretic damper for the following criteria: F y = 100 kN L1 + L2 = 5 m k h = 6, 000 kN/m

Problem 3.7 Design a viscoelastic damper for the following conditions: • • • • •

Temperature = 20° Frequency excitation = 1 Hz Material 3M-ISD110 Maximum displacement = 0.025 m Maximum force = 10 kN

Problem 3.8 k p

m F

u

The SDOF system shown above has a linear viscoelastic damper which produces the force F. Assume u = uˆ sin Ωt and use F given by eqn (3.45). Determine p .

246 Chapter 3: Optimal Passive Damping Distribution

Problem 3.9 A convenient way of dealing with periodic excitation is to introduce complex notation. The basic identity is e iθ = cos θ + i sin θ

(1)

where θ is an arbitrary scalar. Using this notation, an arbitrary periodic forcing can be expressed in terms of a complex amplitude, p p = pe iΩt = ( pˆ e iδ )e iΩt = pˆ e i ( Ωt + δ ) = pˆ ( cos ( Ωt + δ ) + i sin ( Ωt + δ ) )

(2)

One can work with the complex form, and then retain either the real term (for cosine forcing) or the imaginary term (for sine forcing). a) Consider a 1 DOF system having a spring and linear viscous damper in parallel. Suppose p is a general periodic excitation, p = pe iΩt . Let u = ue iΩt represent the response. Express the relationship between p and u as p = Ku where K is the complex system stiffness. Determine K .

k p, u

m c

b) Consider a linear spring and viscous damper in series. Determine K .

k

c m

p, u

c) Refer to part a. Express the relationship between p and u as u = Hp . ˆ e iα . Suppose the loading consists of a set of periodic Determine H = H

Problems 247 excitations, p = p1 e

iΩ 1 t

+ p2 e

iΩ 2 t

+ … + pn e

iΩ n t

(3)

Express u in terms of p j and H j . Discuss how you would evaluate u at some time t for a cosine forcing, i.e., the loading corresponding to the real part of eqn (3). Problem 3.10 Equations (3.28) and (3.29) define the stress-strain response of a linear viscoelastic material to periodic sinusoidal excitation. Using complex notation, one can generalize these equations for arbitrary periodic excitation, γ = γ e iΩt , where γ is the complex amplitude. The complex shear modulus is defined as ˆ e iδ G = G

(1)

where tan δ = η, the loss factor ˆ = [ G 2 + G 2 ]1 / 2 G s l

(2)

With these definitions, the shear stress can be written as τ = τe iΩt τ = Gγ

(3)

a) Consider a linear viscoelastic damper subjected to periodic excitation, u = ue iΩt . Express the force as F = Fe iΩt and let F = k v u . Determine k v by generalizing eqn (3.41).

b) Consider a SDOF system having a spring and linear viscoelastic device in parallel. Assume p is a general periodic excitation and let p = Ku . Determine K . Equate this result to the result for a linear spring and linear viscous damper in parallel (see problem 3.9a), and determine the equivalent stiffness and damping factors, k' and c' .

248 Chapter 3: Optimal Passive Damping Distribution

k p, u

m (Actual)

k' m c’ (Equivalent Viscous) c) Consider a linear spring and linear viscoelastic device in series. Determine K .

k

p, u

m

d) Consider a SDOF system having a linear spring in parallel with a linear spring/linear viscoelastic device. Determine K

k m

p, u

k1

Problem 3.11 Consider a single degree of freedom system having m = 1000 kg and k = 36 kN/m. This problem concerns designing various types of damping devices.

Problems 249 Assume the system is subjected to seismic excitation, and the design objective is a damping ratio of 0.1. Take S v = 1 m/s as the response for pure viscous damping with ξ = 0.1. Propose damper properties for the following cases: 1.Viscous 2.Structural 3.Coulomb 4.Hysteretic with µ = 5 Discuss the basis for your recommendations.

Problem 3.12 Refer to problem 3.10, part b. Take m = 2000 kg and k = 25 kN/m. Determine the equivalent stiffness and viscous damping coefficients for the case where the material is 3M ISD110, the excitation frequency ranges from 0.5 Hz to 5 –2 Hz, and f d = 10 m .

Problem 3.13 Refer to problem 3.10, part b. Suppose m = 4000 kg and k = 50 kN/m. Design a visco-elastic damper to produce an equivalent viscous damping of 3 kNs/m at an excitation frequency of 2 Hz. Use 3M ISD110 material. Also determine the corresponding equivalent stiffness. Problem 3.14 Refer to problem 3.10, part b. The equivalent stiffness and viscous damping coefficients corresponding to a periodic excitation of frequency Ω are: k' = k + f d G s

(1)

η f d Gs c' = ----------------Ω

(2)

where G s and η are functions of Ω . Suppose one wants to approximate these coefficients with “constant” values over the frequency range of interest.

250 Chapter 3: Optimal Passive Damping Distribution

k' ( Ω ) ≈ k

*

c' ( Ω ) ≈ c

(3)

*

(4)

Show that a least square approximation leads to: *

k = k + k eq

(5)

*

c = c eq

(6)

where k eq and c eq are defined by equations (3.74) and (3.75). Problem 3.15 Consider a SDOF system having a spring and a viscoelastic damper modelled as shown below. Take c = α d k 1 . a) Determine expressions for the fundamental frequency, ω , and damping ratio, ξ. b) Suppose m = 1000 kg and k = 9 kN/m. Determine the damper properties c and k1 that correspond to a damping ratio of 0.1. Take α d = 0.15 . Also determine the fundamental frequency. c) Suppose m = 1000 kg and the total stiffness, k + k 1 , is equal to 9kN/m. Determine the stiffness and damping parameters required for ξ = 0.1 . Take α d = 0.15 . k k1

c Problem 3.16 Refer to example 3.12

m

u

Problems 251 a) Take C = 600 N s/m. Determine K eq and C eq corresponding to a periodic excitation, Ω = 2π r ⁄ s , and the following values for stiffness: 1. K = 9 kN ⁄ m

K 2 = 4.5 kN ⁄ m

2. K = 9 kN ⁄ m

K 2 = 18 kN ⁄ m

3. K = 9 kN ⁄ m

K2 = ∞

b) Suppose m = 1000kg. Determine ω and ξ corresponding to the values of K eq and C eq obtained in part a. Problem 3.17 Consider the truss beam segment shown below. Assume the diagonal bracing system consists of elastic elements and linear viscous dampers attached to rigid links connecting the end nodes. Dimension the elastic element and damper for the following requirements: 3

D T = 260 × 10 kN 3

T

= 5 × 10 kN s/m

Use steel for the elastic element.

4m 45

0

Problem 3.18 Refer to problem 3.17. Assume the diagonal bracing system consists of

252 Chapter 3: Optimal Passive Damping Distribution elastic elements and linear viscoelastic elements. Dimension the components for the requirements specified in problem 3.17. Use equivalent stiffness and damping parameters for the viscoelastic component, and take α d = 0.15 . Problem 3.19 Consider a structural system composed of 2 subsystems: the first is linear elastic and provides only stiffness; the second system is linear viscoelastic and provides both stiffness and damping. Let K e denote the stiffness matrix for the elastic component and K 1, C the stiffness and damping matrices for the viscoelastic component. The equations of motion are: ˙˙ + CU˙ + ( K + K )U = P MU e 1

(1)

where C is related to K 1 . Specializing eqn(1) for free vibration and expressing the solution as: λt

U = Ae Φ

(2)

leads to the following equation: 2

[ λ M + λC + ( K e + K 1 ) ]Φ = 0

(3)

The solution for the primary elastic system is expressed as: λ = ± iω

ω >0

( K e – ω 2 M )Φ e = 0

(4) (5)

a) Consider the pure elastic case, C = 0 and K 1 ≠ 0 . Determine the corresponding λ and Φ for the case where K 1 as a scalar multiple of K e , say K 1 = α1 K e . b) Consider the pure viscous case C ≠ 0 and K 1 = 0 . Take C = α 2 K e . Determine λ and Φ . Express λ as: λ = – ξω ± iω'

(6)

c) Consider the linear viscoelastic case. Take C = α d K 1 where α d is a material parameter. Assume stiffness proportional damping and consider 2 cases:

Problems 253 1. C = α 2 K e 2. C = α 3 ( K e + K 1 ) Determine expressions for the corresponding λ and Φ . d) Consider the case where C = α 3 ( K e + K 1 ) and C = α d K 1 . Suppose the solution is expressed as: N

U =

∑ q jΦe, j

(7)

j=1

where Φ e, j are the undamped modal vectors, i.e., the solution of eqn (5). Assume Φ e are normalized with respect to M . T ˜j Φ e, d M Φ e, d = m

(8)

Derive the governing equation and initial conditions for q j . Express it in a form similar to eqn (3.166), using “equivalent” terms for ω j and ξ j . Utilize the results generated in part a to establish the natural frequency for the case where K = K e + K 1 . Discuss how damping and frequency vary with α 3 , for a given viscoelastic material. Note that α d = ∞ for pure viscous damping. Problem 3.20 The governing equation for an n'th order system is given by eqn (3.143) ˙˙ + CU˙ + KU = P MU

(1)

Expressing the homogeneous solution as λt

U = Ae Φ

(2)

and specializing eqn (1) for P = 0 leads to: 2

( λ M + Cλ + K )Φ = 0

(3)

Various cases ( C = 0 , C = αK , and C arbitrary) are discussed in section 3.7.

254 Chapter 3: Optimal Passive Damping Distribution Suppose one defines a new variable, X ,

X =

U U˙

(4)

˙˙ using eqn (1), one obtains: Differentiating X and substituting for U 0 ˙ X˙ = U = – 1 ˙˙ U –M K

I –M

–1

0 U + – 1 C U˙ M P

(5)

Denoting the coefficient matrix as A , eqn (5) simplifies to X˙ = AX + B

(6)

The homogeneous solution of eqn (6) has the general form: αt

X = e Ψ a) Determine the equation relating α and Ψ . b) Show that α = λ, i.e., α is an eigenvalue of eqn (3). c) What is the significance of the real part of α ? d) Consider a 2 DOF discrete shear beam having the following properties: 2

m 1 = m 2 = 1000 kg = 1 kNs ⁄ m k 1 = 118.3 kN ⁄ m

k 2 = 78.9 kN/m

c 1 = 2.09 kNs/m

c 2 = 4.18 kNs/m

Form A and determine the eigenvalues and eigenvectors using the eigen routine in MATLAB. e) Repeat part d using the MOTIONLAB program.

Problems 255 Problem 3.21 m 3 = 2m

u3 k 3, c 3

m2 = m

u2 k 2, c 2

m1 = m

u1 k 1, c 1

Consider the 3 DOF system shown above. a) Determine the magnitudes of k 1 , k 2 , and k 3 such that the first mode has the form 1 2 Φ 1 = ---, ---, 1 3 3 and the fundamental frequency is 2π rad/sec. Take m = 2000 kg. b) Determine the viscous damping coefficients c 1 , c 2 , and c 3 such that the damping ratio, ξ 1 , for the first mode is 0.1. Consider both stiffness proportional damping and uniform damping. Use MOTIONLAB to determine the modal properties for the case of non-proportional damping. c) Suppose viscoelastic dampers are used. Let k d and c d denote the equivalent stiffness and damping coefficients, and take c d = α d k d where α d is a material property. Determine c d, i and the elastic stiffness k e, i for each element. Assume α d = 0.15 . Consider stiffness proportional damping and uniform damping. Compare the modal properties for these 2 cases. Problem 3.22 Refer to example 3.7. Use MOTIONLAB to determine the modal properties for the first 3 modes. Take k i according to eqn (1) and c i based on:

256 Chapter 3: Optimal Passive Damping Distribution a) damping proportional to stiffness b) uniform damping c) c 1 = c 2 = 0 and c 3 = c 4 = c 5 d) other combinations of the c' that you believe may be more optimal in the sense of resulting in higher damping ratios for the higher modes. Problem 3.23 Refer to example 3.7. Suppose viscoelastic dampers are used for elements 3, 4, 5 and there is no damping in elements 1 and 2. Assume uniform viscoelastic damper properties and take α d = 0.15 . Starting with the data contained in eqn (1) of example 3.7, determine the damper properties and modified elastic stiffness which correspond to a damping ratio of 0.1 for the fundamental mode. Using MOTIONLAB, assess the effect of non-proportional damping on the mode shape for this design value of ξ 1 . What would be the effect if ξ 1 is increased to 0.2 and the stiffness is maintained at the level defined by eqn (1)? Problem 3.24

10 m

5m

This problem concerns the preliminary design of a 10 story rectangular rigid frame for seismic excitation. The frame properties and design criteria are: • Height = 5 m/story

Problems 257 • Width = 10 m/bay • Mass/floor = 10,000 kg • Max. deflection at top = 0.25 m • Max. story shear deformation = 1/200 • Response spectrum shown below.

Spectral Velocity S v (m/s) S v = 1.2, ξ = 0.02 S v = 0.8, ξ = 0.1

1.0

S v = 0.6, ξ = 0.2

0.1 0.1

0.6 1.0

10

Period T (sec)

The chevron bracing system is similar to the scheme shown in Fig. 3.23(b); it allows 2 viscous dampers to be placed on each floor. a) Determine the stiffness distributions based on a linear fundamental mode k i , for profile and ξ equal to 0.02, 0.10, and 0.20. Evaluate the stiffness cost, each distribution.

∑

b) Assume stiffness proportional damping. Determine the corresponding ci . distributions for the story damping coefficients, and the damping cost,

∑

c) Repeat b) for “uniform” damping. Using MOTIONLAB, determine the properties for the fundamental mode. Compare the displacement profiles. *

d) Repeat b) for c 1 = c 2 = c 3 = c 4 = c 5 = 0 and c 6 = c 7 = c 8 = c 9 = c 10 = c . Determine the actual damping ratio and profile for the fundamental mode. e) The damper cost increases nonlinearly with the damper coefficient, i.e., the cost

258 Chapter 3: Optimal Passive Damping Distribution for 2c is more than twice the cost for c. Also, the damper force increases with c and places more loading on the brace-floor connection. With these limitations, discuss how you would select a damper placement that satisfies the performance requirement on the maximum transverse shear deformation for each story and minimizes a “cost” function. Problem 3.25 Consider a 5 DOF shear beam having equal masses and equal nodal forces. Suppose the force consists of a combination of a static component and a periodic excitation, p = p s + p d sin ωt where ω is a random quantity. Take m i = 10, 000kg p s = 100kN p d = 1kN a) Determine the stiffness distribution such that the interstory displacement under the static loading is 0.02m for each story. b) Assume ω coincides with the frequency for the fundamental mode corresponding to the stiffness distribution generated in part (a). Suppose the design objective is to have the peak acceleration less than 0.02g where g is the acceleration due to gravity and is equal to 9.87m/s2. Suggest various schemes for generating the required energy dissipation. Comment on what you consider to be the optimal solution.

259

Chapter 4

Tuned mass damper systems 4.1 Introduction A tuned mass damper (TMD) is a device consisting of a mass, a spring, and a damper that is attached to a structure in order to reduce the dynamic response of the structure. The frequency of the damper is tuned to a particular structural frequency so that when that frequency is excited, the damper will resonate out of phase with the structural motion. Energy is dissipated by the damper inertia force acting on the structure. The TMD concept was first applied by Frahm in 1909 (Frahm, 1909) to reduce the rolling motion of ships as well as ship hull vibrations. A theory for the TMD was presented later in the paper by Ormondroyd & Den Hartog (1928), followed by a detailed discussion of optimal tuning and damping parameters in Den Hartog’s book on Mechanical Vibrations (1940). The initial theory was applicable for an undamped SDOF system subjected to a sinusoidal force excitation. Extension of the theory to damped SDOF systems has been investigated by numerous researchers. Significant contributions were made by Randall et al. (1981), Warburton (1980,1981,1982), and Tsai & Lin (1993). This chapter starts with an introductory example of a TMD design and a brief description of some of the implementations of tuned mass dampers in building structures. A rigorous theory of tuned mass dampers for SDOF systems subjected to harmonic force excitation and harmonic ground motion is discussed next. Various cases including an undamped TMD attached to an undamped

260 Chapter 4: Tuned Mass Damper Systems SDOF system, a damped TMD attached to an undamped SDOF system, and a damped TMD attached to a damped SDOF system are considered. Time history responses for a range of SDOF systems connected to optimally tuned TMD and subjected to harmonic and seismic excitations are presented. The theory is then extended to MDOF systems where the TMD is used to dampen out the vibrations of a specific mode. An assessment of the optimal placement locations of TMDs in building structures is included. Numerous examples are provided to illustrate the level of control that can be achieved with such passive devices for both harmonic and seismic excitations.

4.2 An introductory example In this section, the concept of the tuned mass damper is illustrated using the twomass system shown in Fig. 4.1. Here, the subscript d refers to the tuned mass damper; the structure is idealized as a single degree of freedom system. Introducing the following notation 2 k ω = ---m

(4.1)

c = 2ξωm

(4.2)

kd 2 ω d = ------md

(4.3)

c d = 2ξ d ω d m d

(4.4)

and defining m as the mass ratio, md m = ------m p

(4.5)

kd

k

md

m c

cd u Fig. 4.1: SDOF - TMD system.

u + ud

4.2 An Introductory Example 261 the governing equations of motion are given by Primary mass

2 p ( 1 + m )u˙˙ + 2ξωu˙ + ω u = ---- – mu˙˙d m

Tuned mass

u˙˙d + 2ξ d ω d u˙ d + ω d u d = – u˙˙

2

(4.6) (4.7)

The purpose of adding the mass damper is to limit the motion of the structure when it is subjected to a particular excitation. The design of the mass damper involves specifying the mass m d , stiffness k d , and damping coefficient c d . The optimal choice of these quantities is discussed in Section 4.4. In this example, the near-optimal approximation for the frequency of the damper, (4.8) ωd = ω is used to illustrate the design procedure. The stiffnesses for this frequency combination are related by k d = mk

(4.9)

Equation (4.8) corresponds to tuning the damper to the fundamental period of the structure. Considering a periodic excitation, p = pˆ sin Ωt

(4.10)

the response is given by u = uˆ sin ( Ωt + δ 1 )

(4.11)

u d = uˆ d sin ( Ωt + δ 1 + δ 2 )

(4.12)

where uˆ and δ denote the displacement amplitude and phase shift respectively. The critical loading scenario is the resonant condition, Ω = ω . The solution for this case has the following form pˆ 1 uˆ = ------- ---------------------------------------km 2ξ 1 2 1 + ------ + --------- m 2ξ d 1 uˆ d = --------- uˆ 2ξ d

(4.13)

(4.14)

262 Chapter 4: Tuned Mass Damper Systems 1 2ξ tan δ 1 = – ------ + --------m 2ξ d

(4.15)

π tan δ 2 = – --2

(4.16)

Note that the response of the tuned mass is 900 out of phase with the response of the primary mass. This difference in phase produces the energy dissipation contributed by the damper inertia force. The response for no damper is given by pˆ uˆ = --k

1 ---- 2ξ

π δ 1 = – --2

(4.17) (4.18)

To compare these two cases, one can express eqn (4.13) in terms of an equivalent damping ratio pˆ uˆ = --k

1 ------- 2ξ

(4.19)

e

where m 2ξ 1 2 ξ e = ---- 1 + ------ + --------- m 2ξ 2 d

(4.20)

Equation (4.20) shows the relative contribution of the damper parameters to the total damping. Increasing the mass ratio magnifies the damping. However, since the added mass also increases, there is a practical limit on m . Decreasing the damping coefficient for the damper also increases the damping. Noting eqn (4.14), the relative displacement also increases in this case, and just as for the mass, there is a practical limit on the relative motion of the damper. Selecting the final design requires a compromise between these two constraints.

Example 4.1: Preliminary design of a TMD for a SDOF system Suppose ξ = 0 and one wants to add a tuned mass damper such that the equivalent damping ratio is 10% . Using eqn (4.20), and setting ξ e = 0.1 , the following relation between m and ξ d is obtained.

4.2 An Introductory Example 263 m 2ξ 1 2 ---- 1 + ------ + --------- = 0.1 m 2ξ 2 d

(4.21)

The relative displacement constraint is given by eqn (4.14) 1 uˆ d = --------- uˆ 2ξ d

(4.22)

Combining eqn (4.21) and eqn (4.22), and setting ξ = 0 leads to uˆ d 2 m ---- 1 + ------ = 0.1 uˆ 2

(4.23)

Usually, uˆ d is taken to be an order of magnitude greater than uˆ . In this case eqn (4.23) can be approximated as m uˆ d ---- ------ ≈ 0.1 2 uˆ

(4.24)

The generalized form of eqn (4.24) follows from eqn (4.20): 1 m ≈ 2ξ e ------------- uˆ ⁄ uˆ

(4.25)

d

Finally, taking uˆ d = 10uˆ yields an estimate for m 2 ( 0.1 ) m = --------------- = 0.02 10

(4.26)

This magnitude is typical for m . The other parameters are 1 uˆ ξ d = --- ------ = 0.05 2 uˆ d

(4.27)

and from eqn (4.9) k d = mk = 0.02k

(4.28)

It is important to note that with the addition of only 2% of the primary mass, one obtains an effective damping ratio of 10% . The negative aspect is the large relative motion of the damper mass; in this case, 10 times the displacement of the primary mass. How to accommodate this motion in an actual structure is an important design consideration.

A description of some applications of tuned mass dampers to building

264 Chapter 4: Tuned Mass Damper Systems structures is presented in the following section to provide additional background on this type of device prior to entering into a detailed discussion of the underlying theory.

4.3 Examples of existing tuned mass damper systems Although the majority of applications have been for mechanical systems, tuned mass dampers have been used to improve the response of building structures under wind excitation. A short description of the various types of dampers and several building structures that contain tuned mass dampers follows. Translational tuned mass dampers Figure 4.2 illustrates the typical configuration of a unidirectional translational tuned mass damper. The mass rests on bearings that function as rollers and allow the mass to translate laterally relative to the floor. Springs and dampers are inserted between the mass and the adjacent vertical support members which transmit the lateral “out-of-phase” force to the floor level, and then into the structural frame. Bidirectional translational dampers are configured with springs/dampers in 2 orthogonal directions and provide the capability for controlling structural motion in 2 orthogonal planes. Some examples of early versions of this type of damper are described below.

Support md Floor Beam

Direction of motion Fig. 4.2: Schematic diagram of a translational tuned mass damper.

4.3 Examples of Existing Tuned Mass Damper Systems 265

• John Hancock Tower (Engineering News Record, Oct. 1975) Two dampers were added to the 60-story John Hancock Tower in Boston to reduce the response to wind gust loading. The dampers are placed at opposite ends of the 58th story, 67m apart, and move to counteract sway as well as twisting due to the shape of the building. Each damper weighs 2700 kN and consists of a lead-filled steel box about 5.2m square and 1m deep that rides on a 9m long steel plate. The lead-filled weight, laterally restrained by stiff springs anchored to the interior columns of the building and controlled by servo-hydraulic cylinders, slides back and forth on a hydrostatic bearing consisting of a thin layer of oil forced through holes in the steel plate. Whenever the horizontal acceleration exceeds 0.003g for two consecutive cycles, the system is automatically activated. This system was designed and manufactured by LeMessurier Associates/SCI in association with MTS System Corp., at a cost of around 3 million dollars, and is expected to reduce the sway of the building by 40% to 50%. • Citicorp Center (Engineering News Record Aug. 1975, McNamara 1977, Petersen 1980) The Citicorp (Manhattan) TMD was also designed and manufactured by LeMessurier Associates/SCI in association with MTS System Corp. This building is 279m high, has a fundamental period of around 6.5s with an inherent damping ratio of 1% along each axis. The Citicorp TMD, located on the 63rd floor in the crown of the structure, has a mass of 366 Mg, about 2% of the effective modal mass of the first mode, and was 250 times larger than any existing tuned mass damper at the time of installation. Designed to be biaxially resonant on the building structure with a variable operating period of 6.25s ± 20% , adjustable linear damping from 8% to 14%, and a peak relative displacement of ± 1.4m , the damper is expected to reduce the building sway amplitude by about 50%. This reduction corresponds to increasing the basic structural damping by 4%. The concrete mass block is about 2.6m high with a plan cross-section of 9.1m by 9.1m and is supported on a series of twelve 60cm diameter hydraulic pressurebalanced bearings. During operation, the bearings are supplied oil from a separate hydraulic pump which is capable of raising the mass block about 2cm to its operating position in about 3 minutes. The damper system is activated automatically whenever the horizontal acceleration exceeds 0.003g for two consecutive cycles, and will automatically shut itself down when the building

266 Chapter 4: Tuned Mass Damper Systems acceleration does not exceed 0.00075g in either axis over a 30 minute interval. LeMessurier estimates Citicorp’s TMD, which cost about 1.5 million dollars, saved 3.5 to 4 million dollars. This sum represents the cost of some 2,800 tons of structural steel that would have been required to satisfy the deflection constraints. • Canadian National Tower (Engineering News Record, 1976) The 102m steel antenna mast on top of the Canadian National Tower in Toronto (553m high including the antenna) required two lead dampers to prevent the antenna from deflecting excessively when subjected to wind excitation. The damper system consists of two doughnut-shaped steel rings, 35cm wide, 30cm deep, and 2.4m and 3m in diameter, located at elevations 488m and 503m. Each ring holds about 9 metric tons of lead and is supported by three steel beams attached to the sides of the antenna mast. Four bearing universal joints that pivot in all directions connect the rings to the beams. In addition, four separate hydraulically activated fluid dampers mounted on the side of the mast and attached to the center of each universal joint dissipate energy. As the leadweighted rings move back and forth, the hydraulic damper system dissipates the input energy and reduces the tower’s response. The damper system was designed by Nicolet, Carrier, Dressel, and Associates, Ltd, in collaboration with Vibron Acoustics, Ltd. The dampers are tuned to the second and fourth modes of vibration in order to minimize antenna bending loads; the first and third modes have the same characteristics as the prestressed concrete structure supporting the antenna and did not require additional damping. • Chiba Port Tower (Kitamura et al. 1988) Chiba Port Tower (completed in 1986) was the first tower in Japan to be equipped with a TMD. Chiba Port Tower is a steel structure 125m high weighing 1950 metric tons and having a rhombus shaped plan with a side length of 15m. The first and second mode periods are 2.25s and 0.51s respectively for the X direction and 2.7s and 0.57s for the Y direction. Damping for the fundamental mode is estimated at 0.5%. Damping ratios proportional to frequencies were assumed for the higher modes in the analysis. The purpose of the TMD is to increase damping of the first mode for both the X and Y directions. Figure 4.3 shows the damper system. Manufactured by Mitsubishi Steel Manufacturing Co., Ltd, the damper has: mass ratios with respect to the modal mass of the first mode of about 1/120 in the X direction and 1/80 in the Y direction; periods in the X and Y directions of 2.24s and 2.72s respectively; and a damper damping ratio of 15%. The maximum

4.3 Examples of Existing Tuned Mass Damper Systems 267 relative displacement of the damper with respect to the tower is about ± 1m in each direction. Reductions of around 30% to 40% in the displacement of the top floor and 30% in the peak bending moments are expected.

Fig. 4.3: Tuned mass damper for Chiba-Port Tower. The early versions of TMD’s employ complex mechanisms for the bearing and damping elements, have relatively large masses, occupy considerably space, and are quite expensive. Recent versions, such as the scheme shown in Fig 4.4, have been designed to minimize these limitations. This scheme employs a multiassemblage of elastomeric rubber bearings, which function as shear springs, and bitumen rubber compound (BRC) elements, which provide viscoelastic damping capability. The device is compact in size, requires unsophisticated controls, is multidirectional, and is easily assembled and modified. Figure 4.5 shows a full scale damper being subjected to dynamic excitation by a shaking table. An actual installation is contained in Fig. 4.6.

268 Chapter 4: Tuned Mass Damper Systems

Fig. 4.4: Tuned mass damper with spring and damper assemblage.

Fig. 4.5: Deformed position - tuned mass damper.

4.3 Examples of Existing Tuned Mass Damper Systems 269

Fig. 4.6: Tuned mass damper - Huis Ten Bosch Tower, Nagasaki. The effectiveness of a tuned mass damper can be increased by attaching an auxiliary mass and an actuator to the tuned mass and driving the auxiliary mass with the actuator such that its response is out of phase with the response of the tuned mass. Fig 4.7 illustrates this scheme. The effect of driving the auxiliary mass is to produce an additional force which complements the force generated by the tuned mass, and therefore increases the equivalent damping of the TMD (one can obtain the same behavior by attaching the actuator directly to the tuned mass, thereby eliminating the need for an auxiliary mass). Since the actuator requires an external energy source, this system is referred to as an active tuned mass damper. The scope of this chapter is restricted to passive TMD’s. Active TMD’s are discussed in Chapter 6.

270 Chapter 4: Tuned Mass Damper Systems

Auxiliary mass Actuator Support

Floor Beam

Direction of motion Fig. 4.7: An active tuned mass damper configuration. Pendulum tuned mass damper The problems associated with the bearings can be eliminated by supporting the mass with cables which allow the system to behave as a pendulum. Fig 4.8a shows a simple pendulum attached to a floor. Movement of the floor excites the pendulum. The relative motion of the pendulum produces a horizontal force which opposes the floor motion. This action can be represented by an equivalent SDOF system which is attached to the floor as indicated in Fig 4.8b. u

L

keq

θ md

t=0

t ud

u

(a) actual system

md u+ud (b) equivalent system

Fig. 4.8: A simple pendulum tuned mass damper.

4.3 Examples of Existing Tuned Mass Damper Systems 271 The equation of motion for the horizontal direction is Wd T sin θ + -------- ( u˙˙ + u˙˙d ) = 0 g where T is the tension in the cable. When θ approximations apply

(4.29)

is small, the following

u d = L sin θ ≈ Lθ T ≈ Wd Introducing these approximations transforms eqn (4.29) to Wd m d u˙˙d + --------u d = – m d u˙˙ L and it follows that the equivalent shear spring stiffness is Wd k eq = -------L The natural frequency of the pendulum is related to keq by k eq g ω d2 = ------- = --md L Noting eqn (4.33), the natural period of the pendulum is L T d = 2π --g

(4.30)

(4.31)

(4.32)

(4.33)

(4.34)

The simple pendulum tuned mass damper concept has a serious limitation. Since the period depends on L, the required length for large Td may be greater than the typical story height. For instance, the length for Td=5 secs is 6.2 meters whereas the story height is between 4 to 5 meters. This problem can be eliminated by resorting to the scheme illustrated in Fig 4.9. The interior rigid link magnifies the support motion for the pendulum, and results in the following equilibrium equation Wd (4.35) m d ( u˙˙ + u˙˙1 + u˙˙d ) + --------u d = 0 L The rigid link moves in phase with the damper, and has the same displacement amplitude. Then, taking u1= ud in eqn (4.35) results in Wd md m d u˙˙d + --------u d = – ------- u˙˙ (4.36) 2L 2

272 Chapter 4: Tuned Mass Damper Systems The equivalent stiffness is Wd/2L , and it follows that the effective length is equal to 2L. Each additional link increases the effective length by L. An example of a pendulum type damper is described below.

u

L

md u+u1

u+u1+ud

Fig. 4.9: Compound pendulum. • Crystal Tower (Nagase & Hisatoku 1990) The tower, located in Osaka Japan, is 157m high and 28m by 67m in plan, weighs 44,000 metric tons, and has a fundamental period of approximately 4s in the north-south direction and 3s in the east-west direction. A tuned pendulum mass damper was included in the early phase of the design to decrease the windinduced motion of the building by about 50%. Six of the nine air cooling and heating ice thermal storage tanks (each weighing 90 tons) are hung from the top roof girders and used as a pendulum mass. Four tanks have a pendulum length of 4m and slide in the north-south direction; the other two tanks have a pendulum length of about 3m and slide in the east-west direction. Oil dampers connected to the pendulums dissipate the pendulum energy. Fig 4.10 shows the layout of the ice storage tanks that were used as damper masses. Views of the actual building and one of the tanks are presented in Fig 4.11. The cost of this tuned mass damper system was around $350,000, less than 0.2% of the construction cost.

4.3 Examples of Existing Tuned Mass Damper Systems 273

Fig. 4.10: Pendulum damper layout - Crystal Tower.

274 Chapter 4: Tuned Mass Damper Systems

Fig. 4.11: Ice storage tank - Crystal Tower. A modified version of the pendulum damper is shown in Fig 4.12. The restoring force provided by the cables is generated by introducing curvature in the support surface and allowing the mass to roll on this surface. The vertical motion of the weight requires an energy input. Assuming θ is small, the equations for the case where the surface is circular are the same as for the conventional pendulum with the cable length L, replaced with the surface radius R.

4.4 Tuned Mass Damper Theory for SDOF Systems 275

θ

R

md

u

(a)

Floor md keq

(b)

Fig. 4.12: Rocker pendulum.

4.4 Tuned mass damper theory for SDOF systems In what follows, various cases ranging from fully undamped to fully damped conditions are analyzed and design procedures are presented. Undamped structure - undamped TMD Figure 4.13 shows a SDOF system having mass m and stiffness k , subjected to both external forcing and ground motion. A tuned mass damper with mass m d and stiffness k d is attached to the primary mass. The various displacement measures are: u g , the absolute ground motion; u , the relative motion between the

276 Chapter 4: Tuned Mass Damper Systems primary mass and the ground; and u d , the relative displacement between the damper and the primary mass. With this notation, the governing equations take the form m d [ u˙˙d + u˙˙] + k d u d = – m d a g

(4.37)

mu˙˙ + ku – k d u d = – ma g + p

(4.38)

where a g is the absolute ground acceleration and p is the force loading applied to the primary mass.

p

k

m

ug

kd

md

u + ug

ud + u + ug

Fig. 4.13: SDOF system coupled with a TMD. The excitation is considered to be periodic of frequency Ω , a g = aˆ g sin Ωt

(4.39)

p = pˆ sin Ωt

(4.40)

Expressing the response as u = uˆ sin Ωt

(4.41)

u d = uˆ d sin Ωt

(4.42)

and substituting for these variables, the equilibrium equations are transformed to [ – m d Ω 2 + k d ]uˆ d – m d Ω 2 uˆ = – m d aˆ g

(4.43)

– k d uˆ d + [ – mΩ 2 + k ]uˆ = – maˆ g + pˆ

(4.44)

4.4 Tuned Mass Damper Theory for SDOF Systems 277 The solutions for uˆ and uˆ d are given by 2

2

pˆ 1 – ρ d maˆ g 1 + m – ρ d uˆ = --- --------------- – ---------- -------------------------- k D1 k D1

(4.45)

2 pˆ mρ maˆ g m uˆ d = ----- ----------- – ---------- ------- kd D1 k d D 1

(4.46)

where 2

2

D 1 = [ 1 – ρ ] [ 1 – ρ d ] – mρ

2

and the ρ terms are dimensionless frequency ratios, Ω Ω ρ = ---- = --------------ω k⁄m Ω Ω ρ d = ------- = ---------------------ωd kd ⁄ md

(4.47)

(4.48) (4.49)

Selecting the mass ratio and damper frequency ratio such that 2

1 – ρd + m = 0

(4.50)

reduces the solution to pˆ uˆ = --k

(4.51)

pˆ 2 maˆ g uˆ d = – -----ρ + ---------kd kd

(4.52)

This choice isolates the primary mass from ground motion and reduces the response due to external force to the pseudo-static value, pˆ ⁄ k . A typical range for m is 0.01 to 0.1 . Then, the optimal damper frequency is very close to the forcing frequency. The exact relationship follows from eqn (4.50). ωd

Ω = -----------------opt 1+m

(4.53)

One determines the corresponding damper stiffness with 2

Ω mm = ωd kd m d = ----------------1+m opt opt 2

(4.54)

278 Chapter 4: Tuned Mass Damper Systems Finally, substituting for k d , eqn (4.52) takes the following form aˆ g 1 + m uˆ d = -------------- --pˆ- + ------ m k 2 Ω

(4.55)

One specifies the amount of relative displacement for the damper and determines m with eqn (4.55). Given m and Ω , the stiffness is found using eqn (4.54). It should be noted that this stiffness applies for a particular forcing frequency. Once the mass damper properties are defined, eqns (4.45) and (4.46) can be used to determine the response for a different forcing frequency. The primary mass will move under ground motion excitation in this case. Undamped structure - damped TMD The next level of complexity has damping included in the mass damper, as shown in Fig. 4.14. The equations of motion for this case are m d u˙˙d + c d u˙ d + k d u d + m d u˙˙ = – m d a g

(4.56)

mu˙˙ + ku – c d u˙ d – k d u d = – ma g + p

(4.57)

The inclusion of the damping terms in eqns (4.56) and (4.57) produces a phase shift between the periodic excitation and the response. It is convenient to work initially with the solution expressed in terms of complex quantities. One expresses the excitation as a g = aˆ g e p = pˆ e

iΩt

iΩt

(4.58) (4.59)

where aˆ g and pˆ are real quantities. The response is taken as u = ue

iΩt

ud = ud e

iΩt

(4.60) (4.61)

4.4 Tuned Mass Damper Theory for SDOF Systems 279

p kd k

md

m cd ug

ud + u + ug

u + ug

Fig. 4.14: Undamped SDOF system coupled with a damped TMD system.

where the response amplitudes, u and u d are considered to be complex quantities. The real and imaginary parts of a g correspond to cosine and sinusoidal input. Then, the corresponding solution is given by either the real (for cosine) or imaginary (for sine) parts of u and u d . Substituting eqns (4.60) and iΩt (4.61) in the set of governing equations and cancelling e from both sides results in (4.62) [ – m d Ω 2 + ic d Ω + k d ]u d – m d Ω 2 u = – m d aˆ g – [ ic d Ω + k d ]u d + [ – mΩ 2 + k ]u = – maˆ g + pˆ

(4.63)

The solution of the governing equations is aˆ g m pˆ 2 2 2 2 u = ---------- [ f – ρ + i2ξ d ρf ] – ---------- [ ( 1 + m ) f – ρ + i2ξ d ρf ( 1 + m ) ] kD 2 kD 2

(4.64)

2 aˆ m pˆ ρ g u d = ---------- – ---------kD 2 kD 2

(4.65)

where 2

2

2

2 2

2

D 2 = [ 1 – ρ ] [ f – ρ ] – mρ f + i2ξ d ρf [ 1 – ρ ( 1 + m ) ]

(4.66)

ωd f = ------ω

(4.67)

and ρ was defined earlier as the ratio of Ω to ω (see eqn (4.48)). Converting the complex solutions to polar form leads to the following

280 Chapter 4: Tuned Mass Damper Systems expressions iδ 1 aˆ g m iδ 2 pˆ u = --- H 1 e – ----------H 2 e k k

(4.68)

– i δ 3 aˆ g m –i δ3 pˆ u d = --- H 3 e – ----------H 4 e k k

(4.69)

where the H factors define the amplification of the pseudo-static responses, and the δ ‘s are the phase angles between the response and the excitation. The various H and δ terms are listed below 2

2 2

2

[ f – ρ ] + [ 2ξ d ρf ] H 1 = -----------------------------------------------------------D2

(4.70)

2 2

2

2

[ ( 1 + m ) f – ρ ] + [ 2ξ d ρf ( 1 + m ) ] H 2 = -------------------------------------------------------------------------------------------------D2

(4.71)

2

ρ H 3 = ---------D2

(4.72)

1 H 4 = ---------D2

(4.73)

D2 =

2

2

2

2 2 2

2

( [ 1 – ρ ] [ f – ρ ] – mρ f ) + ( 2ξ d ρf [ 1 – ρ ( 1 + m ) ] )

2

(4.74)

Also δ1 = α1 – δ3

(4.75)

δ2 = α2 – δ3

(4.76) 2

2ξ d ρf [ 1 – ρ ( 1 + m ) ] tan δ 3 = -----------------------------------------------------------------2 2 2 2 2 [ 1 – ρ ] [ f – ρ ] – mρ f

(4.77)

2ξ d ρf tan α 1 = -----------------2 2 f –ρ

(4.78)

2ξ d ρf ( 1 + m ) tan α 2 = ------------------------------------2 2 (1 + m) f – ρ

(4.79)

4.4 Tuned Mass Damper Theory for SDOF Systems 281 For most applications, the mass ratio is less than about 0.05 . Then, the amplification factors for external loading ( H 1 ) and ground motion ( H 2 ) are essentially equal. A similar conclusion applies for the phase shift. In what follows, the solution corresponding to ground motion is examined and the optimal values of the damper properties for this loading condition are established. An in-depth treatment of the external forcing case is contained in Den Hartog’s text (Den Hartog, 1940). Figure 4.15 shows the variation of H 2 with forcing frequency for specific values of damper mass m and frequency ratio f , and various values of the damper damping ratio, ξ d . When ξ d = 0 , there are two peaks with infinite amplitude located on each side of ρ = 1 . As ξ d is increased, the peaks approach each other and then merge into a single peak located at ρ ≈ 1 . The behavior of the amplitudes suggests that there is an optimal value of ξ d for a given damper configuration ( m d and k d , or equivalently, m and f ). Another key observation is that all the curves pass through two common points, P and Q . Since these curves correspond to different values of ξ d , the location of P and Q must depend only on m and f . Proceeding with this line of reasoning, the expression for H 2 can be written as 2

H2 =

2 2

2

2

2

a2 a1 ⁄ a2 + ξd a1 + ξd a2 ----------------------- = ----- --------------------------2 2 2 a4 a2 ⁄ a2 + ξ2 a3 + ξd a4 3 4 d

(4.80)

where the ‘a’ terms are functions of m , ρ , and f . Then, for H 2 to be independent of ξ d , the following condition must be satisfied a1 a3 ----- = ----a2 a4

(4.81)

The corresponding values for H 2 are H2

P, Q

=

a2 ----a4

(4.82)

282 Chapter 4: Tuned Mass Damper Systems

30

ξd = 1

m = 0.01 f = 1 25

ξd = 0 20

H2

P 15

0 < ξd < 1

Q 10

5

0 0.8

ρ1 0.85

0.9

0.95

ρ2 1

Ω ρ = ---ω

1.05

1.1

1.15

Fig. 4.15: Plot of H 2 versus ρ . Substituting for the ‘a’ terms in eqn (4.81), one obtains a quadratic equation for ρ 4 2 1 + 0.5m 2 2 ρ – ( 1 + m ) f + --------------------- ρ + f = 0 1+m

2

(4.83)

The two positive roots ρ 1 and ρ 2 are the frequency ratios corresponding to points P and Q . Similarly, eqn (4.82) expands to H2

1+m = -----------------------------------------2 P, Q 1 – ρ 1, 2 ( 1 + m )

(4.84)

Figure 4.15 shows different values for H 2 at points P and Q . For optimal behavior, one wants to minimize the maximum amplitude. As a first step, one requires the values of H 2 for ρ 1 and ρ 2 to be equal. This produces a distribution which is symmetrical about ρ 2 = 1 ⁄ ( 1 + m ) , as illustrated in Fig. 4.16. Then, by increasing the damping ratio, ξ d , one can lower the peak amplitudes until the peaks coincide with points P and Q . This state represents the optimal performance of the TMD system. A further increase in ξ d causes the peaks to merge and the amplitude to increase beyond the optimal value.

4.4 Tuned Mass Damper Theory for SDOF Systems 283

30

25

H2

20

o

ξd > ξd o

ξd < ξd

Q

P

15

ξd 10

opt

5

ρ1 0

0.85

0.9

0.95

opt 1

Ω ρ = ---ω

ρ2

opt

1.05

1.1

1.15

Fig. 4.16: Plot of H 2 versus ρ for f opt .

Requiring the amplitudes to be equal at P and Q is equivalent to the following condition on the roots 2

2

1 – ρ1 ( 1 + m ) = 1 – ρ2 ( 1 + m )

(4.85)

Then, substituting for ρ 1 and ρ 2 using eqn (4.83), one obtains a relation between the optimal tuning frequency and the mass ratio 1 – 0.5m f opt = ------------------------1+m

(4.86)

ωd

(4.87)

opt

= f opt ω

The corresponding roots and optimal amplification factors are ρ 1, 2 H2

opt

=

1 ± 0.5m -------------------------1+m

1+m = ---------------opt 0.5m

(4.88) (4.89)

284 Chapter 4: Tuned Mass Damper Systems The expression for the optimal damping at the optimal tuning frequency is ξd

opt

=

m ( 3 – 0.5m ) ------------------------------------------------8 ( 1 + m ) ( 1 – 0.5m )

(4.90)

Figures 4.17 through 4.20 show the variation of the optimal parameters with the mass ratio, m .

1

0.98

f opt

0.96

0.94

0.92

0.9

0.88 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

m

Fig. 4.17: Optimum tuning frequency ratio, f opt.

0.09

0.1

4.4 Tuned Mass Damper Theory for SDOF Systems 285

1.1

f opt 1.05

ρ2

ρ 1, 2

opt

1

opt

0.95

ρ1

opt

0.9

0.85

0.8 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.18: Input frequency ratios at which the response is independent of damping.

0.2 0.18 0.16 0.14

ξd

opt

0.12 0.1 0.08 0.06 0.04 0.02 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

m

Fig. 4.19: Optimal damping ratio for TMD.

0.09

0.1

286 Chapter 4: Tuned Mass Damper Systems

25

15

H2

opt

20

10

5

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.20: Maximum dynamic amplification factor for SDOF system (optimal tuning and damping). The response of the damper is defined by eqn (4.69). Specializing this equation for the optimal conditions leads to the plot of amplification versus mass ratio contained in Fig. 4.21. A comparison of the damper motion with respect to the motion of the primary mass for optimal conditions is shown in Fig. 4.22.

4.4 Tuned Mass Damper Theory for SDOF Systems 287

300

250

H4

opt

200

150

100

50

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.21: Maximum dynamic amplification factor for TMD.

20

f opt, ρ opt, ξ d

18

opt

16

H4 uˆ d ------ = -----H2 uˆ

14 12 10 8 6 4 2 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.22: Ratio of maximum TMD amplitude to maximum system amplitude.

288 Chapter 4: Tuned Mass Damper Systems Lastly, response curves for a typical mass ratio, m = 0.01 , and optimal tuning are plotted in Figs 4.23 and 4.24. The response for no damper is also plotted in Fig. 4.23. One observes that the effect of the damper is to limit the motion in a frequency range centered on the natural frequency of the primary mass and extending about 0.15ω . Outside of this range, the motion is not significantly influenced by the damper. 30

25

ξ d = 0.0 ξ d = 0.03 ξ d = 0.061 (optimal) ξ d = 0.1 No damper

m = 0.01 f opt = 0.9876

H2

20

P

15

Q

10

5

ρ1 0 0.8

0.85

0.9

0.95

ρ2

opt 1

Ω ρ = ---ω

opt

1.05

1.1

1.15

1.2

Fig. 4.23: Response curves for amplitude of system with optimally tuned TMD.

4.4 Tuned Mass Damper Theory for SDOF Systems 289

150

ξd ξd ξd ξd ξd

m = 0.01 f opt = 0.9876

0.0 0.03 0.061 (optimal) 0.1 0.2

H4

100

= = = = =

50

0

0.85

0.9

0.95

1

Ω ρ = ---ω

1.05

1.1

1.15

Fig. 4.24: Response curves for amplitude of optimally tuned TMD. The maximum amplification for a damped SDOF system without a TMD, undergoing harmonic excitation is given by eqn (1.32) 1 H = ------------------------2 2ξ 1 – ξ

(4.91)

Since ξ is small, a reasonable approximation is 1 H ≈ -----2ξ

(4.92)

Expressing the optimal H 2 in a similar form provides a measure of the equivalent damping ratio ξ e for the primary mass 1 ξ e = -------------------2H 2

(4.93)

opt

Figure 4.25 shows the variation of ξ e with the mass ratio. A mass ratio of 0.02 is equivalent to about 5% damping in the primary system.

290 Chapter 4: Tuned Mass Damper Systems

0.12

f opt, ρ opt, ξ d 0.1

opt

ξe

0.08

0.06

0.04

0.02

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.25: Equivalent damping ratio for optimally tuned TMD.

The design of a TMD involves the following steps: • Establish the allowable values of displacement of the primary mass and the TMD for the design loading. This data provides the design values for H 2 and H 4 . opt

opt

• Determine the mass ratios required to satisfy these motion constraints from Figs 4.20 and 4.21. Select the largest value of m . • Determine f opt form Fig. 4.17. • Compute ω d ω d = f opt ω

(4.94)

• Compute k d 2

2

k d = m d ω d = mk f opt • Determine ξ d from Fig. 4.19. opt

(4.95)

4.4 Tuned Mass Damper Theory for SDOF Systems 291 • Compute c d (4.96)

c d = 2ξ d ω d m d = m f opt 2ξ d ωm opt opt

Example 4.2: Design of a TMD for an undamped SDOF system Consider the following motion constraints < 7

(4.97)

H4 ----------------- < 6 H2

(4.98)

H2

opt

opt

Constraint eqn(4.97) requires m ≥ 0.05 . For constraint eqn(4.98), one needs to take m ≥ 0.02 . Therefore, m ≥ 0.05 controls the design. The relevant parameters are: m = 0.05

f opt = 0.94

ξd

m d = 0.05m

ω d = 0.94ω

k d = m f opt k = 0.044k

opt

= 0.135

Then 2

Damped structure - damped TMD All real systems contain some damping. Although an absorber is likely to be added only to a lightly damped system, assessing the effect of damping in the real system on the optimal tuning of the absorber is an important design consideration. The main system in Fig. 4.26 consists of the mass m , spring stiffness k , and viscous damping c . The TMD system has mass m d , stiffness k d , and viscous damping c d . Considering the system to be subjected to both external forcing and ground excitation, the equations of motion are m d u˙˙d + c d u˙ d + k d u d + m d u˙˙ = – m d a g

(4.99)

292 Chapter 4: Tuned Mass Damper Systems mu˙˙ + cu˙ + ku – c d u˙ d – k d u d = – ma g + p

(4.100)

p k

kd md

m c

cd

ug

ud + u + ug

u + ug

Fig. 4.26: Damped SDOF system coupled with a damped TMD system. Proceeding in the same way as for the undamped case, the solution due to periodic excitation (both p and ug) is expressed in polar form: aˆ g m iδ iδ pˆ u = --- H 5 e 5 – ----------H 6 e 6 k k aˆ g m –i δ iδ pˆ u d = --- H 7 e 7 – ----------H 8 e 8 k k

(4.101) (4.102)

The various H and δ terms are defined below 2

2 2

2

[ f – ρ ] + [ 2ξ d ρf ] H 5 = -----------------------------------------------------------D3 2

2 2

(4.103) 2

[ ( 1 + m ) f – ρ ] + [ 2ξ d ρf ( 1 + m ) ] H 6 = -------------------------------------------------------------------------------------------------D3

(4.104)

ρ2 H 7 = ---------D3

(4.105)

1 + [ 2ξρ ] 2 H 8 = ------------------------------D3

(4.106)

4.4 Tuned Mass Damper Theory for SDOF Systems 293 D 3 = { [ – f 2 ρ 2 m + ( 1 – ρ 2 ) ( f 2 – ρ 2 ) – 4ξξ d f ρ 2 ] 2

(4.107)

+ 4 [ ξρ ( f 2 – ρ 2 ) + ξ d fρ ( 1 – ρ 2 ( 1 + m ) ) 2 ] } δ5 = α1 – δ7

(4.108)

δ6 = α2 – δ7

(4.109)

δ8 = α3 – δ7

(4.110)

ξρ ( f 2 – ρ 2 ) + ξ d fρ ( 1 – ρ 2 ( 1 + m ) ) tan δ 7 = 2 ---------------------------------------------------------------------------------------------------– f 2 ρ 2 m + ( 1 – ρ 2 ) ( f 2 – ρ 2 ) – 4ξξ d f ρ 2

(4.111)

tan α 3 = 2ξρ

(4.112)

The α 1 and α 2 terms are defined by eqns 4.78 and 4.79. In what follows, the case of an external force applied to the primary mass is considered. Since D 3 involves ξ, one cannot establish analytical expressions for the optimal tuning frequency and optimal damping ratio in terms of the mass ratio. In this case, these parameters also depend on ξ . Numerical simulations can be applied to evaluate H 5 and H 7 for a range of ρ , given the values for m , ξ , f , and ξ d . Starting with specific values for m and ξ , plots of H 5 versus ρ can be generated for a range of f and ξ d . Each H 5 – ρ plot has a peak value of H 5 . The particular combination of f and ξ d that correspond to the lowest peak value of H 5 is taken as the optimal state. Repeating this process for different values of m and ξ produces the behavioral data needed to design the damper system. Figure 4.27 shows the variation of the maximum value of H 5 for the optimal state. The corresponding response of the damper is plotted in Fig. 4.28. Adding damping to the primary mass has an appreciable effect for small m . Noting eqns (4.101) and (4.102), the ratio of damper displacement to primary mass displacement is given by H7 uˆ d ρ2 --------- = ------- = -----------------------------------------------------------(4.113) H5 uˆ 2 2 2 2 [ f – ρ ] + [ 2ξ d ρf ] Since ξ is small, this ratio is essentially independent of ξ . Figure 4.29 confirms this statement. The optimal values of the frequency and damping ratios generated through simulation are plotted in Figs 4.30 and 4.31. Lastly, using eqn (4.93), can be converted to an equivalent damping ratio for the primary system. H5 opt

294 Chapter 4: Tuned Mass Damper Systems 1 ξ e = -------------------2H 5

(4.114)

opt

Figure 4.32 shows the variation of ξ e with m and ξ . Tsai & Lin (1993) suggest equations for the optimal tuning parameters f and ξ d determined by curve fitting schemes. The equations are listed below for completeness 2 1 – 0.5m f = ------------------------- + 1 – 2ξ – 1 1+m

(4.115)

– [ 2.375 – 1.034 m – 0.426m ]ξ m – ( 3.730 – 16.903 m + 20.496m )ξ 2 m 2 3m ------------------------------------------------- + ( 0.151ξ – 0.170ξ ) 8 ( 1 + m ) ( 1 – 0.5m )

ξd =

(4.116) 2

+ ( 0.163ξ + 4.980ξ ) m

40

ξ ξ ξ ξ ξ

35

30

= = = = =

0.0 0.01 0.02 0.05 0.1

H5

opt

25

20

15

10

5

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.27: Maximum dynamic amplification factor for damped SDOF system.

4.4 Tuned Mass Damper Theory for SDOF Systems 295

300

ξ ξ ξ ξ ξ

250

0.0 0.01 0.02 0.05 0.1

150

H7

opt

200

= = = = =

100

50

0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.28: Maximum dynamic amplification factor for TMD.

20

ξ ξ ξ ξ ξ

18 16

H7 uˆ d ------ = -----H5 uˆ

14

= = = = =

0.0 0.01 0.02 0.05 0.1

12 10 8 6 4 2 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.29: Ratio of maximum TMD amplitude to maximum system amplitude.

296 Chapter 4: Tuned Mass Damper Systems

1

ξ ξ ξ ξ ξ

0.98 0.96 0.94

= = = = =

0.0 0.01 0.02 0.05 0.1

f opt

0.92 0.9 0.88 0.86 0.84 0.82 0.8 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

m

Fig. 4.30: Optimum tuning frequency ratio for TMD, f opt.

0.2 0.18 0.16 0.14

0.1

ξd

opt

0.12

0.08

ξ ξ ξ ξ ξ

0.06 0.04 0.02 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

= = = = =

0.08

m

Fig. 4.31: Optimal damping ratio for TMD.

0.0 0.01 0.02 0.05 0.1 0.09

0.1

4.4 Tuned Mass Damper Theory for SDOF Systems 297

0.2 0.18 0.16 0.14

ξe

0.12 0.1 0.08

ξ ξ ξ ξ ξ

0.06 0.04 0.02 0 0

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

= = = = =

0.0 0.01 0.02 0.05 0.1 0.09

0.1

m

Fig. 4.32: Equivalent damping ratio for optimally tuned TMD.

Example 4.3: Design of a TMD for a damped SDOF system Example 4.2 is reworked here, allowing for 2% damping in the primary system. The same design motion constraints are considered: < 7

(4.117)

H7 ----------------- < 6 H5

(4.118)

H5

opt

opt

Using Fig. 4.27, the required mass ratio for ξ = 0.02 is m ≈ 0.03 . The other optimal values are f opt = 0.965 and ξ d = 0.105 . opt

Then m d = 0.03m

ω d = 0.955ω

2

k d = m f opt k = 0.027k

In this case, there is a significant reduction in the damper mass required for this

298 Chapter 4: Tuned Mass Damper Systems set of motion constraints. The choice between including damping in the primary system versus incorporating a tuned mass damper depends on the relative costs and reliability of the two alternatives, and the nature of the structural problem. A TMD system is generally more appropriate for upgrading an existing structure where access to the structural elements is difficult.

4.5 Case studies - SDOF systems Figures 4.33 to 4.44 show the time history responses for two SDOF systems with periods of 0.49s and 5.35s respectively under harmonic (at resonance conditions), El Centro, and Taft ground excitations. All examples have a system damping ratio of 2% and an optimally tuned TMD with a mass ratio of 1%. The excitation magnitudes have been scaled so that the peak amplitude of the response of the system without the TMD is unity. The plots show the response of the system without the TMD (the dotted line) as well as the response of the system with the TMD (the solid line). Figures showing the time history of the relative displacement of the TMD with respect to the system are also presented. Significant reduction in the response of the primary system under harmonic excitation is observed. However, optimally tuned mass dampers are relatively ineffective under seismic excitation, and in some cases produce a negative effect, i.e. they amplify the response slightly. This poor performance is attributed to the ineffectiveness of tuned mass dampers for impulsive loadings as well as their inability to reach a resonant condition and therefore dissipate energy under random excitation. These results are in close agreement with the data presented by Kaynia et al. (1981).

4.5 Cases Studies - SDOF Systems 299

1 0.8

T = 0.49s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.33: Response of SDOF to harmonic excitation.

6

4

T = 0.49s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

60

Time - s

Fig. 4.34: Relative displacement of TMD under harmonic excitation.

300 Chapter 4: Tuned Mass Damper Systems

1 0.8

T = 0.49s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.35: Response of SDOF to El Centro excitation.

6

4

T = 0.49s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

60

Time - s

Fig. 4.36: Relative displacement of TMD under El Centro excitation.

4.5 Cases Studies - SDOF Systems 301

1 0.8

T = 0.49s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.37: Response of SDOF to Taft excitation.

6

4

T = 0.49s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

Time - s

Fig. 4.38: Relative displacement of TMD under Taft excitation.

60

302 Chapter 4: Tuned Mass Damper Systems

1 0.8

T = 5.35s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.39: Response of SDOF to harmonic excitation.

6

4

T = 5.35s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

60

Time - s

Fig. 4.40: Relative displacement of TMD under harmonic excitation.

4.5 Cases Studies - SDOF Systems 303

1 0.8

T = 5.35s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.41: Response of SDOF to El Centro excitation.

6

4

T = 5.35s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

60

Time - s

Fig. 4.42: Relative displacement of TMD under El Centro excitation.

304 Chapter 4: Tuned Mass Damper Systems

1 0.8

T = 5.35s ξ = 0.02 m = 0.01

0.6

u - m

0.4 0.2 0 -0.2 -0.4 -0.6 -0.8

without TMD with TMD

-1 0

10

20

30

40

50

60

Time - s

Fig. 4.43: Response of SDOF to Taft excitation.

6

4

T = 5.35s ξ = 0.02 m = 0.01

ud - m

2

0

-2

-4

-6 0

10

20

30

40

50

Time - s

Fig. 4.44: Relative displacement of TMD under Taft excitation.

60

4.6 Tuned Mass Damper Theory for MDOF Systems 305

4.6 Tuned mass damper theory for MDOF systems The theory of a SDOF system presented earlier is extended here to deal with a MDOF system having a number of tuned mass dampers located throughout the structure. Numerical simulations, which illustrate the application of this theory to the set of example building structures used as the basis for comparison of the different schemes throughout the text are presented in the next section.

p2

p1

ug

md

m2

m1 c1

kd

k2

k1

c2 u1 + ug

cd u2 + ug

u2 + ug + ud

Fig. 4.45: 2DOF system with TMD.

A 2DOF system having a damper attached to mass 2 is considered first to introduce the key ideas. The governing equations for the system shown in Fig. 4.45 are: m 1 u˙˙1 + c 1 u˙ 1 + k 1 u 1 – k 2 ( u 2 – u 1 ) – c 2 ( u˙ 2 – u˙ 1 ) = p 1 – m 1 u˙˙g

(4.119)

m 2 u˙˙2 + c 2 ( u˙ 2 – u˙ 1 ) + k 2 ( u 2 – u 1 ) – k d u d – c d u˙ d = p 2 – m 2 u˙˙g

(4.120)

m d u˙˙d + k d u d + c d u˙ d = – m d ( u˙˙2 + u˙˙g )

(4.121)

The key step is to combine eqns (4.119) and (4.120) and express the resulting equation in a form similar to the SDOF case defined by eqn (4.100). This operation reduces the problem to an equivalent SDOF system, for which the theory of Section 4.4 is applicable. The approach followed here is based on transforming the original matrix equation to scalar modal equations. Introducing matrix notation, eqns (4.119) and (4.120) are written as

306 Chapter 4: Tuned Mass Damper Systems

˙˙ + CU˙ + KU = MU

p1 – m1 ag p2 – m2 ag

+

0 k d u d + c d u˙ d

(4.122)

where the various matrices are

U =

u1

M =

m1

(4.123)

u2

(4.124)

m2

K =

k1 + k2 –k2 –k2 k2

(4.125)

C =

c1 + c2 –c2 –c2 c2

(4.126)

One substitutes for U in terms of the modal vectors and coordinates U = Φ1 q1 + Φ2 q2

(4.127)

The modal vectors satisfy the following orthogonality relations (see eqn (2.257)) T

2

T

Φ j KΦ i = δ ij ω j Φ j MΦ i

(4.128)

Defining modal mass, stiffness, and damping terms, ˜ j = Φ T MΦ m j j

(4.129)

T 2 ˜j k˜ j = Φ j KΦ j = ω j m

(4.130)

T c˜ j = Φ j CΦ j

(4.131)

expressing the elements of Φ j as

4.6 Tuned Mass Damper Theory for MDOF Systems 307 Φj =

Φ j1

(4.132)

Φ j2

and assuming damping is proportional to stiffness C = αK

(4.133)

one obtains a set of uncoupled equations for q 1 and q 2 ˜ j q˙˙ + c˜ j q˙ + k˜ j q = Φ ( p – m a ) m j = 1,2 j j j j1 1 1 g

(4.134)

+ Φ j2 ( p 2 – m 2 a g + k d u d + c d u˙ d ) With this assumption, the modal damping ratio is given by αω j c˜ j ξ j = ---------------- = ---------˜j 2 2ω j m

(4.135)

Equation (4.134) represents two equations. Each equation defines a ˜ , k˜ , and particular SDOF system having mass, stiffness, and damping equal to m ξ . Since a TMD is effective for a narrow frequency range, one has to decide on which modal resonant response is to be controlled with the TMD. Once this decision is made, the analysis can proceed using the selected modal equation and the initial equation for the TMD, i.e. eqn (4.121). Suppose the first modal response is to be controlled. Taking j = 1 in eqn (4.133) leads to ˜ 1 q˙˙ + c˜ 1 q˙ + k˜ 1 q = Φ p + Φ p m 1 1 1 11 1 12 2

(4.136)

– [ m 1 Φ 11 + m 2 Φ 12 ]a g + Φ 12 [ k d u d + c d u˙ d ] In general, u 2 is obtained by superposing the modal contributions u 2 = Φ 12 q 1 + Φ 22 q 2

(4.137)

However, when the external forcing frequency is close to ω 1 , the first mode response will dominate, and it is reasonable to assume u 2 ≈ Φ 12 q 1 Solving for q 1

(4.138)

308 Chapter 4: Tuned Mass Damper Systems 1 q 1 = --------- u 2 Φ 12

(4.139)

and then substituting in eqn (4.136), one obtains ˜ 1e u˙˙2 + c˜ 1e u˙ 2 + k˜ 1e u = k u + c u˙ d m 2 d d d

(4.140)

˜ 1e a + p˜ 1e – Γ 1e m g ˜ 1e , c˜ 1e , k˜ 1e , p˜ 1e , and Γ represent the equivalent SDOF parameters where m 1e for the combination of mode 1 and node 2, the node at which the TMD is attached. Their definition equations are 1 ˜ ˜ 1e = --------m1 m 2 Φ 12

(4.141)

1 k˜ 1e = --------- k˜ 1 2 Φ 12

(4.142)

c˜ 1e = αk˜ 1e

(4.143)

Φ 11 p 1 + Φ 12 p 2 p˜ 1e = -------------------------------------Φ 12 Φ 12 Γ 1e = --------- ( m 1 Φ 11 + m 2 Φ 22 ) ˜1 m

(4.144) (4.145)

Equations 4.121 and 4.140 are similar in form to the SDOF equations treated in the previous section. Both set of equations are compared below. TMD equation m d u˙˙d + c d u˙ d + k d u d = – m d ( u˙˙ – a g ) vs m d u˙˙d + c d u˙ d + k d u d = – m d ( u˙˙2 – a g )

(4.146)

4.6 Tuned Mass Damper Theory for MDOF Systems 309 Primary mass equation mu˙˙ + cu˙ + ku = c d u˙ d + k d u d + p – ma g vs ˜ ˜ 1e u˙˙2 + c˜ 1e u˙ 2 + k˜ 1e u = c u˙ d + k u + p˜ – Γ m m 1e 2 d d d 1e 1e a g Taking

u2 ≡ u

˜ 1e ≡ m m

c˜ 1e ≡ c

p˜ 1e ≡ p

Γ 1e ≡ Γ

k˜ 1e ≡ k

(4.147)

(4.148)

transforms the primary mass equation for the MDOF case to mu˙˙ + cu˙ + ku = c d u˙ d + k d u d + p – Γma g

(4.149)

which differs from the corresponding SDOF equation by the factor Γ. Therefore, the solution for ground excitation generated earlier has to be modified to account for the presence of Γ. The “generalized” solution is written in the same form as the SDOF case. One needs only to modify the terms associated with a g , i.e., H6, H8 and δ6 , δ8 . Their expanded form is listed below. 2 2

2

2

[ ( Γ + m ) f – Γρ ] + [ 2ξ d ρf ( Γ + m ) ] H 6 = ------------------------------------------------------------------------------------------------------D3 2

2

(4.150)

2

[ 1 + ρ ( Γ – 1 ) ] + [ 2ξρ ] H 8 = --------------------------------------------------------------------D3

(4.151)

2ξ d fρ ( Γ + m ) tan a 2 = -----------------------------------------f 2 ( Γ + m ) – Γρ 2

(4.152)

2ξρ tan a 3 = --------------------------------1 + ( Γ – 1 )ρ 2

(4.153)

δ6 = a2 – δ7

(4.154)

δ8 = a3 – δ7

(4.155)

310 Chapter 4: Tuned Mass Damper Systems where D 3 is defined by eqn (4.107), and δ 7 is given by eqn (4.111). From this point on, one proceeds as described in Section 4.4. The mass ratio is defined in terms of the equivalent SDOF mass. md m = --------˜ 1e m

(4.156)

Given m and ξ 1 , one finds the tuning frequency and damper damping ratio using Figs 4.30 and 4.31. The damper parameters are determined with ˜ 1e md = m m

(4.157)

ω d = f opt ω 1

(4.158)

c d = 2ξ d ωd md opt

(4.159)

Expanding the expression for the damper mass, T

m [ Φ 1 MΦ 1 ] ˜ 1e = -----------------------------md = m m 2 Φ 12

(4.160)

shows that one should select the TMD location to coincide with the maximum amplitude of the mode shape that is being controlled. In this case, the first mode is the target mode, and Φ 12 is the maximum amplitude for Φ 1 . This derivation can be readily generalized to allow for tuning on the i th modal frequency. One writes eqn (4.139) as 1 q i ≈ -------- u 2 Φ i2

(4.161)

where i is either 1 or 2 . The equivalent parameters are 1 ˜ ˜ ie = -------mi m 2 Φ i2 2 ˜ ie k˜ ie = ω i m

˜ ie and ξ , one specifies m and finds the optimal tuning with Given m i

(4.162)

(4.163)

4.6 Tuned Mass Damper Theory for MDOF Systems 311 ω d = f opt ω i

(4.164)

Example 4.4: Design of a TMD for a damped MDOF system To illustrate the above procedure, a two DOF system having m 1 = m 2 = 1 is considered. Designing the system for a fundamental period of T 1 = 1s and a uniform deformation fundamental mode profile yields the following stiffnesses (refer to Example 1.6) 2

k 1 = 12π = 118.44 2

k 2 = 8π = 78.96 Requiring a fundamental mode damping ratio of 2% , and taking damping proportional to stiffness ( C = αK ), the corresponding α is 2ξ 1 0.02 α = --------- = ---------- = 0.0064 ω1 π The mass, stiffness, and damping matrices for these design conditions are: M = 1 0 0 1 K = 197.39 – 78.96 – 78.96 78.96 C =

1.26 – 0.51 – 0.51 0.51

Performing an eigenvalue analysis yields the following frequencies and mode shapes ω 1 = 6.28r/s Φ 1 = 0.5 1.0

ω 2 = 15.39r/s Φ2 =

1.0 – 0.5

The corresponding modal mass, stiffness, and damping terms are: ˜ 1 = Φ T MΦ = 1.25 m 1 1

˜ 2 = Φ T MΦ = 1.25 m 2 2

312 Chapter 4: Tuned Mass Damper Systems T k˜ 1 = Φ 1 KΦ 1 = 39.48

T k˜ 2 = Φ 2 KΦ 2 = 236.87

T c˜ 1 = Φ 1 CΦ 1 = 0.25 c˜ 1 ξ 1 = ----------------- = 0.02 ˜1 2ω 1 m

T c˜ 2 = Φ 2 CΦ 2 = 1.51 c˜ 2 ξ 2 = ----------------- = 0.049 ˜2 2ω 2 m

The optimal parameters for a TMD having a mass ratio of 0.01 and tuned to a specific mode are listed below. Mode 1 - optimum location is node 2 f opt = 0.982

ξd

m d = 0.0125

k d = 0.4754

opt

= 0.062 c d = 0.0096

Mode 2 - optimum location is node 1 f opt = 0.972

ξd

m d = 0.05

k d = 11.1894

opt

= 0.068 c d = 0.1017

The general case of a MDOF system with a tuned mass damper connected to the nth degree of freedom is treated in a similar manner. Using the notation defined above, the jth modal equation can be expressed as ˜ j q˙˙ + c˜ j q˙ + k˜ j q = p˜ + Φ [ k u + c u˙ d ] m j j j j jn d d d

j = 1,2,...

(4.165)

where p˜ j denotes the modal force due to ground motion and external forcing, and Φ jn is the element of Φ j corresponding to the nth displacement variable. To control the ith modal response, one sets j = i in eqn (4.165), and introduces the approximation 1 q i ≈ --------- u n Φ in This leads to the following equation for u n ˜ ie u˙˙n + c˜ ie u˙ n + k˜ ie u = p˜ + k u + c u˙ d m ie n d d d where

(4.166)

(4.167)

4.6 Tuned Mass Damper Theory for MDOF Systems 313 (4.168)

T 1 ˜ 1 ˜ ie = -------- M i = --------- Φ i MΦ i m 2 2 Φ in Φ in 2 ˜ ie k˜ ie = ω i m

(4.169)

c˜ ie = αk˜ ie

(4.170)

1 p˜ ie = --------- p˜ i Φ in

(4.171)

The remaining steps are the same as described above. One specifies m and ξ i , determines the optimal tuning and damping values with Figs 4.30 and 4.31, and then computes m d and ω d . T m ˜ ie = -------- Φ i MΦ i md = m m 2 Φ in

(4.172)

ω d = f opt ω i

(4.173)

The optimal mass damper for mode i is obtained by selecting n such that Φ in is the maximum element in Φ i .

Example 4.5: Design of TMD’s for a simply supported beam y,u

P*

x*

EI constant

x L Figure 1 Consider the simply supported beam shown in Figure 1. The modal shapes

314 Chapter 4: Tuned Mass Damper Systems and frequencies for the case where the cross sectional properties are constant and the transverse shear deformation is negligible are: nπx Φ n ( x ) = sin ---------L

(1)

EI nπ 4 ω n2 = ------- ------ ρm L

(2)

n = 1, 2, … One obtains a set of N equations in terms of N modal coordinates by expressing the transverse displacement, u(x,t), as N

u =

∑

q i ( t )Φ j ( x )

(3)

j=1

and substituting for u in the Principle of Virtual Displacements specialized for negligible transverse shear deformation (see eqn (2.194)), L

∫

Mδχ dx =

0

∫

bδu dx

(4)

Substituting for δχ , 2

d δχ = – ( δu ) dx2

(5)

and taking δu = δq j Φ j leads to the following equations

(6)

4.6 Tuned Mass Damper Theory for MDOF Systems 315

∫

∫

– MΦ j, xx dx =

bΦ j dx

(7)

j = 1, 2, …, N

Lastly, one substitutes for M and b in terms of Φ and q, and evaluates the integrals. The expressions for M and b are N

M = EIχ = – EI

∑

q l Φ l, xx

(8)

l=1

N

b = – ρ m u˙˙ + b ( x, t ) = – ρ m

∑

Φ l q˙˙l + b ( x, t )

(9)

l=1

Noting the orthogonality properties of the modal shape functions, L

∫

L Φ j Φ k dx = δ jk --2

(10)

0

L

∫

jπ 4 L Φ j, xx Φ k, xx dx = ----- δ jk -- L 2

(11)

0

the modal equations uncouple and reduce to ˜ j q˙˙ + k˜ j q = p˜ m j j j

(12)

Lρ m ˜ j = ---------m 2

(13)

jπ 4 L k˜ j = EI ----- -- L 2

(14)

where

316 Chapter 4: Tuned Mass Damper Systems L

p˜ j =

∫

jπx b sin --------- dx L

(15)

0

When the external loading consists of a concentrated force applied at the location x = x * (see Fig 1), the corresponding modal load for the j’th mode is jπx * p˜ j = P * sin ----------L

(16)

In this example, the force is considered to be due to a mass attached to the beam as indicated in Fig 2. The equations for the tuned mass and the force are m d ( u˙˙* + u˙˙d ) + k d u d + c d u˙ d = 0

(17)

m d ( u˙˙* + u˙˙d ) = – P *

(18)

u*+ud

md kd

cd

u*

x* Figure 2 Suppose one wants to control the i’th modal response with a tuned mass damper attached at x = x * . Taking j equal to i in eqns (12) and (13), the i’th modal equation has the form iπx * ˜ i q˙˙i + k˜ i q = ( k u + c u˙ d ) sin ---------m i d d d L

(19)

4.6 Tuned Mass Damper Theory for MDOF Systems 317 Assuming the response is dominated by the i’th mode, u * ( x *, t ) is approximated by iπx * u * ( x *, t ) ≈ q i sin ---------L

(20)

and eqn (19) is transformed to an equation relating u * and u d . ˜ ie u˙˙* + k˜ ie u * = k u + c u˙ d m d d d

(21)

where 1 ˜i ˜ ie = ---------------------------m m 2 * iπx sin --------- L

(22)

The remaining steps utilize the results generated for the SDOF undamped ˜ ie and k˜ ie structure - damped TMD system considered in section 4.3. One uses m as the mass and stiffness parameters for the primary system. To illustrate the procedure, consider the damper to be located at mid-span, and the first mode is to be controlled. Taking i=1 and x * = L ⁄ 2 , the corresponding parameters are iπx * sin ---------- = 1 L

(23)

Lρ m ˜ 1 = ---------˜ ie = m m 2

(24)

EIL π 4 k˜ ie = k˜ 1 = --------- --- 2 L

(25)

One specifies the equivalent damping ratio, ξ e , and determines the required mass ratio from Fig 4.32. For example, taking ξ e = 0.06 requires m = 0.03 . The other parameters corresponding to m = 0.03 follow from Figs 4.29, 4.30, and 4.31. ωd f opt = ------ = 0.965 ω1

(26)

318 Chapter 4: Tuned Mass Damper Systems ξd

opt

= 0.105

uˆ d ------ = 5 uˆ *

(27)

(28)

Using these parameters, the corresponding expression for the damper properties are ˜1 m d = 0.03m

(29)

ω d = 0.965ω 1

(30)

k d = ω d2 m d

(31)

c d = 2ξ d ω d m d

(32)

˜ 1 and ω are specified, the damper properties can be evaluated. Once m 1 For example, consider the beam to be a steel beam having the following properties L = 20m ρ m = 1000k g ⁄ m I = 8 × 10 – 4 m 4

(33)

E = 2 × 10 11 N ⁄ m 2 The beam parameters are ˜ 1 = 10, 000kg m ω 1 = 9.87r ⁄ s Applying eqns (29)-(32), results in

(34)

4.6 Tuned Mass Damper Theory for MDOF Systems 319 m d = 300kg ω d = 9.52r ⁄ s k d = 27, 215N ⁄ m

(35)

c d = 599.8Ns ⁄ m The total mass of the girder is 20,000kg. Adding 300kg, which is just 1.5% of the total mass, produces an effective damping ratio of 0.06 for the first mode response. The mode shape for the second mode has a null point at x=L/2, and therefore locating a tuned mass at this point would have no effect on the second modal response. The optimal locations are x * = L ⁄ 4 and x * = 3L ⁄ 4 . Taking x * = L ⁄ 4 and i=2, one obtains iπx * sin ---------- = 1 L

(36)

Lρ m ˜ 2e = m ˜ 2 = ---------m 2

(37)

π 4 k˜ 2e = k˜ 2 = 8EIL --- L

(38)

16EI π 4 ω 22 = ------------ --- ρ m L

(39)

The procedure from here on is the same as before. One specifies ξ e , determines the required mass ratio, and then the frequency and damping parameters. It is of interest to compare the damper properties corresponding to the same equivalent damping ratio. Taking ξ e = 0.06 , the damper properties for the example steel beam are m d = 300kg

(40)

k d = 435, 440N ⁄ m

(41)

c d = 2400Ns ⁄ m

(42)

320 Chapter 4: Tuned Mass Damper Systems The required damper stiffness is an order of magnitude greater than the corresponding value for the first mode response.

4.7 Case studies - MDOF systems This section presents shear deformation profiles for the standard set of building examples defined in Table 2.4. A single TMD is placed at the top floor and tuned to either the first or second mode. The structures are subjected to harmonic ground acceleration with a frequency equal to the fundamental frequency of the buildings, as well as scaled versions of El Centro and Taft ground accelerations. As expected, significant reduction in the response is observed for the harmonic excitations (see Figs 4.46 through 4.49). The damper is generally less effective for seismic excitation versus harmonic excitation (see Figs 4.50 through 4.61). Results for the low period structures show more influence of the damper which is to be expected since the response is primarily due to the first mode. This data indicates that a TMD is not the optimal solution for controlling the motion due to seismic excitation.

4.6 Case Studies - MDOF Systems 321

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% Harmonic TMD - Mode 1

x Normalized height ---H

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.46: Maximum shear deformation for Building 1.

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% Harmonic TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.47: Maximum shear deformation for Building 2.

0.01

322 Chapter 4: Tuned Mass Damper Systems

1

Building #3 Quadratic based Initial H = 100m ρ = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% Harmonic TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.48: Maximum shear deformation for Building 3.

1

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% Harmonic TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.49: Maximum shear deformation for Building 4.

0.01

4.6 Case Studies - MDOF Systems 323

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 1

x Normalized height ---H

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5%

0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.50: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.51: Maximum shear deformation for Building 1.

0.01

324 Chapter 4: Tuned Mass Damper Systems

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.52: Maximum shear deformation for Building 2.

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 1

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.53: Maximum shear deformation for Building 2.

0.01

4.6 Case Studies - MDOF Systems 325

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

Building #3 Quadratic based Initial H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 1

0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.54: Maximum shear deformation for Building 3.

1 0.9

Building #3 Quadratic based Initial H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 1

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.55: Maximum shear deformation for Building 3.

0.01

326 Chapter 4: Tuned Mass Damper Systems

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

Building #3 Quadratic based Initial H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 2

0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.56: Maximum shear deformation for Building 3.

1 0.9

Building #3 Quadratic based Initial H = 100m ρ m = 20000kg/m s = 0.40 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 2

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.57: Maximum shear deformation for Building 3.

0.01

4.6 Case Studies - MDOF Systems 327

1 0.9

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 1

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.58: Maximum shear deformation for Building 4.

1

x Normalized height ---H

0.9

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 1

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.59: Maximum shear deformation for Building 4.

0.01

328 Chapter 4: Tuned Mass Damper Systems

1 0.9

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% El Centro TMD - Mode 2

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

= = = =

0.008

0% 1% 2% 5% 0.009

Maximum shear deformation γ - m/m

0.01

Fig. 4.60: Maximum shear deformation for Building 4.

1

Building #4 Quadratic based Initial H = 200m ρ m = 20000kg/m s = 0.63 S v = 1.2m/s ξ 1 = 2% Taft TMD - Mode 2

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

m m m m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

Maximum shear deformation γ - m/m

= = = =

0% 1% 2% 5% 0.009

Fig. 4.61: Maximum shear deformation for Building 4.

0.01

Problems 329

Problems Problem 4.1 Verify eqns (4.13) through (4.17). Hint: express p, u, and ud in complex form p = pˆ e iΩt u = ue iΩt u d = u d e iΩt and solve eqns (4.6) and (4.7) for u and u d . Then take u = uˆ e

iδ 1

u d = uˆ d e

i ( δ1 + δ2 )

ω = ωd = Ω

Problem 4.2 Refer to eqns (4.14) and (4.20). Express ξ e as a function of m , ξ , and ˆu ⁄ uˆ d . Take ξ = 0.05 , and plot ξ e vs. m for a representative range of the magnitude of the displacement ratio, uˆ ⁄ uˆ d .

Problem 4.3 Figure 4.7 illustrates an active tuned mass damper configuration. The damper can be modelled with the 2 DOF system shown below. The various terms are: u s is the total displacement of the support attached to the floor beam; F a is the self equilibrating force provided by the actuator; m d, k d, c d are parameters for the damper mass; k a and m a are parameters for the auxillary mass.

330 Chapter 4: Tuned Mass Damper Systems

ka

kd

R

Fa cd us

md

ma us + ud

us + ud + ua

a) Derive the governing equation for m d and m a . Also determine an expression for the resultant force, R, that the system applies to the floor beam. b) Consider m a to be several orders of magnitude smaller than m d , eg, m a = 0.01m d . Also take the actuator force to be a linear function of the relative velocity of the damper mass. F a = c a u˙ d Specialize the equations for this case. How would you interpret the contribution of the actuator force to the governing equation for the damper mass?

Problem 4.4 Design a pendulum damper system having a natural period of 6 seconds and requiring less than 4 meters of vertical space.

Problems 331 Problem 4.5

L1 m1

L2

m2

The pendulum shown above is connected to a second mass which is free to move horizontally. The connection between mass 1 and mass 2 carries only shear. Derive an equation for the period of the compound pendulum and the length of an equivalent simple pendulum. Assume the links are rigid.

Problem 4.6 Refer to Fig 4.12. Establish the equations of motion for the mass, m d , considering θ to be small. Verify that the equivalent stiffness is equal to W d ⁄ R .

Problem 4.7 Refer to Fig 4.15 and eqn (4.84). Derive the corresponding expression for P, Q starting with eqn (4.70) and using the same reasoning strategy. Considering the mass ratio, m , to be less than 0.03, estimate the difference in the “optimal” values for the various parameters. H1

Problem 4.8 Generate plots of H 1 vs. ρ for ξ d ranging from 0 to 0.2, m = 0.01 , and f = 0.9876 . Compare the results with the plots shown in Fig 4.23.

332 Chapter 4: Tuned Mass Damper Systems Problem 4.9 Consider a system composed of an undamped primary mass and a tuned mass damper. The solution for periodic force excitation is given by (see eqns (4.60) to (4.79)) u = ue iΩt

(1)

u d = u d e iΩt

(2)

iδ p u = --- H 1 e 1 k

(3)

iδ p u d = --- H 3 e 3 k

(4)

2

2 2

2

[ f – ρ ] + [ 2ξ d ρf ] H 1 = -----------------------------------------------------------D2

(5)

2

ρ H 3 = ---------D2 D2 =

(6)

2

2

2

2 2 2

2

( [ 1 – ρ ] [ f – ρ ] – mρ f ) + ( 2ξ d ρf [ 1 – ρ ( 1 + m ) ] )

2

(7)

The formulation for the optimal damper properties carried out in Section 4.3 was based on minimizing the peak value of H1 (actually H2 but H1 behaves in a similar way), i.e., on controlling the displacement of the primary mass. Suppose the design objective is to control the acceleration of the primary mass. Noting eqns (1) and (3), the acceleration is given by u˙˙ = a = ae iΩt

(8)

i(δ + π) pΩ 2 a = ----------H 1 e 1 k

(9)

Substituting for k transforms eqn (9) to

Problems 333 i(δ + π) p a = ----H1′ e 1 m

(10)

H1′ = ρ 2 H 1

(11)

where

Investigate the behavior of H1′ with ρ, f , m and ξ d . If it behaves similar to H 2 as shown in Fig 4.15, describe how you would establish the optimal values for the various parameters, and also how you would design a tuned mass system when H1′ is specified.

Problem 4.10 Design a TMD for a damped SDOF system having ξ = 0.02 . The design motion constraints are: a) H5

opt

< 10

H7 ---------------- < 5 H5 opt

b) H5

opt

<5

H7 ---------------- < 5 H5 opt

c) Repeat part b considering ξ to be equal to 0.05.

Problem 4.11 This problem concerns the design of a tuned-mass damper for a damped

334 Chapter 4: Tuned Mass Damper Systems single degree of freedom system. The performance criteria are: ξ eq = 0.1

uˆ d ⁄ uˆ = 5

a) Determine the damper properties for a system having m = 10,000 kg and k = 395 kN/m for the following values of ξ : • ξ = 0.02 • ξ = 0.05 b) Will the damper be effective for an excitation with frequency 2.5π rad/sec ? Discuss the basis for you conclusion.

Problem 4.12 Refer to example 3.7. Suppose a tuned mass damper is installed at the top level (at mass 5). a) Determine the damper properties such that the equivalent damping ratio for the fundamental mode is 0.16. Use the values of m , k , c from example 3.7. Assume stiffness proportional damping for c. b) u5

m5 L md

k 5, c 5

Consider the tuned mass damper to be a pendulum attached to m 5 . Determine m d and L for the damper properties established in part a. c) Repeat part a for the case where the mass damper is tuned for the second mode rather than for the first mode. Use the same values of m , k , c and assume stiffness proportional damping.

Problems 335 Problem 4.13 Consider a cantilever shear beam with the following properties: • H = 50 m • ρ m = 20, 000 kg/m 5 0.6x • D T = 8 ×10 1 – ---------- kN H

a) Model the beam as a 10 DOF discrete shear beam having 5m segments. Determine the first 3 mode shapes and frequencies. Normalize the mode shapes such that the peak amplitude is unity for each mode. b) Design tuned mass dampers to provide an effective modal damping ratio of 0.10 for the first and third modes. Take ξ 1 = 0.02 and assume modal damping is proportional to stiffness. Note: you need to first establish the ‘‘optimal location’’ of the tuned mass dampers for the different modes.

Problem 4.14 Consider a simply supported steel beam having the following properties L = 30m ρ m = 1500k g ⁄ m –2

I = 1 ×10 m 4 a) Design tuned mass damper systems that provide an equivalent damping of 0.05 for each of the first 3 modes. b) Repeat part a) with the constraint that an individual damper mass cannot exceed 300kg. Hint: utilize symmetry of a particular mode shape to locate a pair of dampers whose function is to control that mode.

336 Chapter 4: Tuned Mass Damper Systems Problem 4.15 15 m

15 m constant EI W u1

Consider the simply supported beam shown above. The beam has a uniform weight of 15 kN/m and a concentrated weight at mid span of 100 kN. The flexural rigidity is constant and equal to 200,000 kN-m2. a) Assume the first mode can be approximated by: π u = u 1 sin --- x L Determine the governing equation for u 1 using the Principle of Virtual Displacements. b) Design a tuned mass damper to provide an equivalent damping ratio of 0.05 for the first mode. Assume no damping for the beam itself. c) Will the damper designed in part b be effective for the second mode? Explain your answer.

Problem 4.16 Refer to problem 3.25 part (b). Suggest a tuned mass damper for generating the required energy dissipation.

337

Chapter 5

Base isolation systems 5.1 Introduction The term isolation refers to the degree of interaction between objects. An object is said to be isolated if it has little interaction with other objects. The act of isolating an object involves providing an interface between the object and its neighbors which minimizes interaction. These definitions apply directly to various physical systems. For example, one speaks of isolating a piece of equipment from its support by mounting the equipment on an isolation system which acts as a buffer between the equipment and the support. The design of isolation systems for vibrating machinery is a typical application. The objective here is to minimize the effect of the machine induced loading on the support. Another application is concerned with minimizing the effect of support motion on the structure. This issue is becoming increasingly more important for structures containing motion sensitive equipment and also for structures located adjacent to railroad tracks or other sources of ground disturbance. Although isolation as a design strategy for mounting mechanical equipment has been employed for over seventy years, only recently has the concept been seriously considered for civil structures, such as buildings and bridges, subjected to ground motion. This type of excitation interacts with the structure at the foundation level, and is transmitted up through the structure. Therefore, it is logical to isolate the structure at its base, and prevent the ground

338 Chapter 5: Base Isolation Systems motion from acting on the structure. The idea of seismic isolation dates back to the late nineteenth century, but the application was delayed by the lack of suitable commercial isolation components. Substantial development has occurred since the mid 1980’s (Naeim and Kelly, 1999), and base isolation for certain types of civil structures is now considered to be a highly viable design option by the seismic engineering community, particularly in Japan (Wada, 1998), for moderate to extreme seismic excitation. A set of simple examples are presented in the next section to identify the key parameters and illustrate the quantitative aspects of base isolation. This material is followed by a discussion of practical aspects of seismic base isolation and a description of some seismically isolated buildings. The remaining sections deal with the behavioral and design issues for base isolated MDOF structural systems. Numerical results illustrating the level of performance feasible with seismic base isolation are included to provide a basis of comparison with the other motion control schemes considered in this text.

5.2 Isolation for SDOF systems The application of base isolation to control the motion of a SDOF system subjected to ground motion was discussed earlier in Section 1.3 as part of a general treatment of design for dynamic excitation. The analytical formulation developed in that section provides the basis for designing an isolation system for simple structures that can be accurately represented with a SDOF model. Examples illustrating the reasoning process one follows are presented below. The formulation is also extended to deal with a modified version of a SDOF model that is appropriate for a low-rise building isolated at its base. This model is useful for preliminary design. SDOF examples The first example considers external periodic forcing of the SDOF system shown in Fig. 5.1. The solution of this problem is contained in Section 1.3. For convenience, the relevant equations are listed below:

5.2 Isolation for SDOF Systems 339

k R

m

p

c u Fig. 5.1: SDOF system. p = pˆ sin Ωt

(5.1)

u = uˆ sin ( Ωt – δ )

(5.2)

H1 uˆ = ------- pˆ k

(5.3)

1 H 1 = -------------------------------------------------2 2

[ 1 – ρ ] + [ 2ξρ ]

(5.4)

2

Ω ρ = ---ω

(5.5)

2ξρ tan δ = --------------2 1–ρ

(5.6)

Given pˆ and Ω , one can determine uˆ for a specific system having mass m , stiffness k , and damping c . With uˆ known, the forces in the spring and damper can be evaluated. The reaction can be found by either summing the internal forces, or combining p with the inertia force. With the latter approach, one writes R = p – mu˙˙

(5.7)

and expands the various terms using eqns (5.1) through (5.6). The result is expressed as ˆ sin ( Ωt – δ ) R = R r

(5.8)

ˆ = H pˆ R 3

(5.9)

340 Chapter 5: Base Isolation Systems 2

1 + [ 2ξρ ] ---------------------------------------------2 2 2 [ 1 – ρ ] + [ 2ξρ ]

H3 =

(5.10)

2

ρ H 1 sin δ tan δ r = – -----------------------------------2 1 + ρ H 1 cos δ

(5.11)

The function, H3, is referred to as the transmissibility of the system. It is a measure of how much of the load p is transmitted to the support. When ξ = 0 , δ = 0 and H 3 reduces to H 1. Figure 5.2 shows the variation of H 3 with ρ and ξ .

5 4.5

ξ = 0 4 3.5

H3

3 2.5

ξ = 0.2

2 1.5

ξ = 0.4

1 0.5 0 0

2 0.2

0.4

0.6

0.8

1

Ω ρ = ---ω

1.2

1.4

1.6

1.8

2

Fig. 5.2: Plot of H 3 versus ρ . The model presented above can be applied to the problem of designing a support system for a machine with an eccentric rotating mass. Here, one wants to minimize the reaction force for a given pˆ , i.e. one takes H 3 < 1 . Noting Fig. 5.2, this constraint requires the frequency ratio, ρ , to be greater than 2 , and it follows that

5.2 Isolation for SDOF Systems 341 Ω ω < ------2

(5.12)

The corresponding periods are related by T > Tf 2 =

2π 2 ------ Ω

(5.13)

where T f is the forcing period. For example, taking T = 3T f results in ˆ = 0.125pˆ , a reduction of 87.5% from the static value. R The second example illustrates the strategy for isolating a system from support motion. Applying the formulation derived in Section 1.4 to the system shown in Fig. 5.3, the amplitudes of the relative and total displacement of the mass, uˆ and uˆ t , are related to the support displacement by uˆ = ρ 2 H 1 uˆ g = H 2 uˆ g

(5.14)

uˆ t = H 3 uˆ g

(5.15)

Taking H 3 small with respect to unity reduces the effect of support motion on the position of the mass. The frequency and period criteria are the same as those of the previous example. One takes ρ > 2 to reduce uˆ t . However, since H2 approaches unity as ρ increases, the magnitude of the relative motion increases and approaches the ground motion, uˆ g . Therefore, this relative motion needs to be accomodated. k m c ug

ut = u g + u

Fig. 5.3: SDOF system subjected to support motion.

These examples show that isolation is obtained by taking the period of the SDOF system to be large in comparison to the forcing (either external or support)

342 Chapter 5: Base Isolation Systems period. Expressing this requirement as T ≥ ρ∗ T f

Ω ω < ----ρ*

⇒

(5.16)

where ρ∗ depends on the desired reduction in amplitude, the constraint on the stiffness of the spring is given by Ω 2 2π 2 k < m ------ = m -----------ρ∗ ρ∗ T f

(5.17)

It should be noted that this derivation assumes that a single periodic excitation is applied. The result is applicable for narrow band excitations which are characterized by a dominant frequency. A more complex analysis involving iteration on the stiffness is required to deal with broad band excitations. One has to ensure that the forcing near the fundamental frequency is adequately controlled by damping in this case. Bearing terminology The spring and damper elements connecting the mass to the support are idealizations of physical objects called bearings. They provide a constraint against motion relative to a support plane, as illustrated in Fig. 5.4. The bearing in Fig. 5.4(a) functions as an axial element and resists the displacement normal to the plane with normal stresses (tension and compression). The bearing shown in Fig. 5.4(b) constrains relative tangential motion through shearing action over the height of the bearing. These elements are usually combined into a single compound bearing, but it is more convenient to view them as being uncoupled when modeling the system.

Fn , un axial bearing (a)

Fig. 5.4: Axial and shear bearings.

Ft , ut

shear bearing (b)

5.2 Isolation for SDOF Systems 343 When applying the formulation developed above, one distinguishes between normal and tangential support motion. For normal motion, axial type bearings such as springs and rubber cushions are used; the k defined by eqn (5.17) is the axial stiffness of the bearing F n ⁄ u n . Shear bearings such as laminated rubber cushions and inverted pendulum type sliding devices are used when the induced motion is parallel to the ground surface. In this case, k represents the required shearing stiffness of the bearing, F t ⁄ u t . Figure 5.5 shows an air spring/damper scheme used for vertical support. Single and multiple stage laminated rubber bearings are illustrated in Fig 5.6. Rubber bearings used for seismic isolation can range up to 1 m in diameter and are usually inserted between the foundation footings and the base of the structure. A particular installation for a building is shown in Fig 5.7.

Fig. 5.5: Air spring bearing.

344 Chapter 5: Base Isolation Systems

a) Single stage

b) multiple stage Fig. 5.6: Laminated rubber bearings.

5.2 Isolation for SDOF Systems 345

Fig. 5.7: Rubber bearing seismic isolation system. Modified SDOF Model In what follows, the support motion is considered to be due to seismic excitation. Although both normal (vertical) and tangential (horizontal) motions occur during a seismic event, the horizontal ground motion is generally more significant for structural systems since it leads to lateral loading. Typical structural systems are designed for vertical loading and then modified for lateral loading. Since the vertical motion is equivalent to additional vertical loading, it is not as critical as the horizontal motion. The model shown in Fig. 5.3 represents a rigid structure supported on flexible shear bearings. To allow for the flexibility of the structure, the structure can be modeled as a MDOF system. Figure 5.8 illustrates a SDOF beam type idealization. One can estimate the equivalent SDOF properties of the structure by assuming that the structural response is dominated by the fundamental mode. The data provided in earlier chapters shows that this assumption is reasonable for low-rise buildings subjected to seismic excitation. An in-depth analysis of low rise buildings modeled as MDOF beams is presented later in this chapter. The objective here is to derive a simple relationship showing the effect of the bearing stiffness on the relative displacement of the

346 Chapter 5: Base Isolation Systems structure, u , with respect to the base displacement, u b + u g . The governing equations for the lumped mass model consist of an equilibrium equation for the mass, and an equation relating the shear forces in the spring and the bearing. mu˙˙ + cu˙ + ku = – m ( u˙˙b + u˙˙g )

(5.18)

k b u b + c b u˙ b = ku + cu˙

(5.19)

u + u b + ug m

k,c kb , cb

u b + ug ug

(a) Actual structure

(b) Beam idealization

(c) Lumped mass model

Fig. 5.8: Base isolation models. Neglecting damping, eqn (5.19) can be solved for ub in terms of u. k u b = ----- u kb

(5.20)

Then, substituting for ub in eqn (5.18) leads to k m 1 + ----- u˙˙ + ku = – mu˙˙g kb

(5.21)

Equation (5.21) is written in the conventional form for a SDOF system 2

u˙˙ + ω eq u = – Γu˙˙g

(5.22)

where Γ is a participation factor, kb kb kb Γ = -------------- = ----- ⁄ 1 + ----- k k + kb k

(5.23)

5.2 Isolation for SDOF Systems 347 and ωeq is an equivalent frequency measure 2 2 Γk ω eq = ------ = Γω m

(5.24)

In this case, ω eq is the fundamental frequency of the system consisting of the structure plus bearing. Taking k b small with respect to k decreases the inertia loading on the structure as well as the effective frequency. Consequently, the structural response is reduced. Periodic excitation - modified SDOF model To illustrate the effect of base stiffness on the response, the case of periodic ground motion, u g = uˆ g sin Ωt , is considered. The various response amplitudes are given by 2

Γρ eq uˆ = --------------------uˆ g 2 1 – ρ eq

(5.25)

k uˆ b = ----- uˆ kb

(5.26)

1 uˆ t = uˆ + uˆ b + uˆ g = --------------------uˆ g 2 1 – ρ eq

(5.27)

where the brackets indicate absolute values, and ρeq is the frequency ratio Ω ρ eq = --------ω eq

(5.28)

Comparing eqn (5.27) with eqn (5.15) shows that the results are similar. One replaces ω with ω eq in the expression for H 3 . The limiting cases are k b = 0 and k b = ∞ . The former is the fully isolated case where u b ≈ – uˆ g and u t ≈ 0 ; the latter corresponds to a fixed support where u b ≈ 0 and u t ≈ u + u g . Suppose the structure is defined, and the problem concerns selecting a bearing stiffness such that the total response satisfies uˆ t ≤ νuˆ g

ν<1

348 Chapter 5: Base Isolation Systems One needs to take ρ eq > 2 . Noting eqn (5.27), the required value of ρ eq is 1 2 = 1 + -ρ eq ν

(5.29)

Substituting for ρ eq in eqn (5.28) leads to Ω2 2 = -----------ω eq 1 1 + --ν

(5.30)

Finally, using eqn (5.23) and (5.24), the required bearing stiffness is given by 1 k k b = k ---------------------------- = ----------------------------------------k k ( 1 + ( 1 ⁄ ν)) -------------- – 1 -------------------------------- – 1 ω 2 m 2 mΩ eq

(5.31)

The more general problem is the case where both structural stiffness and the bearing stiffness need to be established subject to the following constraints on the magnitudes of uˆ and uˆ b. uˆ = ν s uˆ g uˆ b = ν b uˆ g

(5.32)

The typical design scenario has ν b larger than ν s . Noting eqn (5.26), the stiffness factors are related by νs k b = -----k νb

(5.33)

Equation (5.25) provides the second equation relating the stiffness factors. It reduces to 1 Γ – 1 + -------- = ----2 νs ρ eq where

(5.34)

5.2 Isolation for SDOF Systems 349

νs kb ⁄ k Γ = --------------------- = ----------------νs + νb 1 + kb ⁄ k

(5.35)

2 leads to ω 2 , and then k. Solving eqn (5.34) for ρ eq eq

k Ω2 2 = ------ω eq - = Γ ---2 m ρ eq

(5.36)

The following example illustrates the computational steps.

Example 5.1: Stiffness factors for prescribed structure and base motion. Suppose ν s = 0.1 and ν b = 1.0 . The relative motion of the base with respect to the ground is allowed to be 10 times greater than the relative motion of the structure with respect to the base. (1)

uˆ b = 10uˆ The stiffness factors are related by νs k b = -----k = 0.1k νb

(2)

Evaluating Γ and ρ eq , using eqns (5.34) and (5.35), νs 0.1 ----------------- = ------- = 0.0909 νs + νb 1.1 1 Γ 1 1 – -------- = ----- = ------- = 0.909 2 νs 1.1 ρ eq 2 ρ eq

(3)

(4)

= 11.0011

leads to 2 = 0.0909Ω 2 ω eq

and finally to k

(5)

350 Chapter 5: Base Isolation Systems m 2 k = ----ω eq = mΩ 2 Γ

(6)

Seismic excitation - modified SDOF model An estimate of the stiffness parameters required to satisfy the motion constraints under seismic excitation can be obtained with the response spectra approach described in Chapter 2. Taking u˙˙g to be the seismic excitation, the solution of eqn (5.22) is related to the spectral velocity by ΓS v u max = ---------ω eq

(5.37)

where S v is a function of the equivalent frequency, ω eq , and the equivalent damping ratio for the structure/bearing system, ξ eq . Substituting for Γ and ω eq , eqn (5.37) expands to mk b u max = S v ---------------------k(k + kb)

(5.38)

The relation between the maximum relative displacement of the bearing and the maximum structural motion follows from eqn (5.20) ub

k = ----- u max max kb

(5.39)

In this development, the criteria for motion based design of a base isolated structure are expressed as limits on the relative motion terms u max = u∗ ub

max

= u b∗

(5.40) (5.41)

The values of k and k b required to satisfy these constraints follow by solving eqns (5.38) and (5.39).

5.2 Isolation for SDOF Systems 351 ku∗ k b = --------u b∗

(5.42)

2

2

mS v mS v 1 k = ------------- ------------------- = --------------------------------2 u b∗ u∗ ( u∗ + u b∗ ) [ u∗ ] 1 + -------u∗

(5.43)

One assumes S v is constant, evaluates k and k b , determines the frequency ω eq with eqn (5.24), and then updates S v if necessary. It is of interest to compare the stiffness required by the base isolated structure with the stiffness of the corresponding fixed base structure. Taking k b = ∞ reduces eqn (5.38) to m u max = S v ---k

(5.44)

The fixed base structural stiffness k f follows from eqn (5.44) 2

mS v k f = k k = ∞ = ------------2 b [ u∗ ]

(5.45)

Using eqn (5.45) and assuming the value of S v is the same for both cases, the stiffness ratios reduce to,

1 k ----- = ------------------kf u b∗ 1 + --------u∗ u∗ -------u b∗ kb ----- = ------------------kf u b∗ 1 + --------u∗

(5.46)

(5.47)

The ratio of the isolated period to the fixed base period can be generated with eqn (5.24)

352 Chapter 5: Base Isolation Systems T eq ωf u b∗ --------- = --------- = 1 + --------Tf ω eq u∗

(5.48)

Figures 5.9 and 5.10 show the variation of k ⁄ k f and k b ⁄ k f with u b∗ ⁄ u∗ for a given constant S v . The increase in the period is plotted in Fig. 5.11. There is a significant reduction in the structural stiffness required by the seismic excitation when the base is allowed to move. For example, taking u b∗ = 2u∗ decreases the design stiffness by a factor of 3 . However, one has to ensure that a potential resonant condition is not created by shifting the period. There may be a problem with wind gust loading as the period is increased beyond 3 seconds. This problem can be avoided by providing additional stiffness that functions under wind loading but not under seismic loading. Section 5.3 deals with this problem.

1 0.9 0.8 0.7

k ----kf

0.6 0.5 0.4 0.3 0.2 0.1 0 0

1

2

3

4

5

u b∗ --------u∗

6

7

8

Fig. 5.9: Variation of k ⁄ k f with u b∗ ⁄ u∗ .

9

10

5.2 Isolation for SDOF Systems 353

1 0.9 0.8 0.7

kb ----kf

0.6 0.5 0.4 0.3 0.2 0.1 0 0

1

2

3

4

5

u b∗ --------u∗

6

7

8

9

10

9

10

Fig. 5.10: Variation of k b ⁄ k f with u b∗ ⁄ u∗ .

12

10

T eq --------Tf

8

6

4

2

0 0

1

2

3

4

5

u b∗ --------u∗

6

7

8

Fig. 5.11: Variation of T eq ⁄ T f with u b∗ ⁄ u∗ .

354 Chapter 5: Base Isolation Systems

Example 5.2:Stiffness parameters - modified SDOF model of Building example #2. The procedure for establishing the appropriate values for u∗ and u b∗ is illustrated using building Example 2 as the reference structure. Table 2.4 lists the relevant design information. The period for the fixed base case is 1.06 sec. Since bad isolation increases the period, the assumption that S v is constant is valid. The relative displacement at the top of the building is estimated as Hγ ∗ where H is the height of the structure and γ ∗ is the prescribed shear deformation. Taking H = 50m and γ ∗ = 1 ⁄ 200 leads to u∗ = 0.25m . The allowable bearing displacement depends on the bearing configuration and response characteristics, as well as the seismic excitation. For the totally soft case, u b is equal to the ground excitation. Hardening the bearing reduces u b somewhat, so a reasonable upper limit is the peak ground displacement corresponding to the design value of S v for representative earthquakes. A typical design value for u b∗ is 0.3m. Using u∗ = 0.25m and u b∗ = 0.3m corresponds to the following stiffness factors k = 0.455k f k b = 0.833k T = 2.2T f The required structural stiffness is reduced by 55% for this degree of base isolation.

These scenarios provide an indication of the potential benefit of base isolation for seismic excitation. However, one should note that the isolated structure is less stiff than the fixed base structure, and therefore will experience larger displacement under other types of loading such as wind. Also, the simplified model considered here is based on linear undamped behavior, whereas the actual bearings have some damping and may behave in a nonlinear manner. More complex models are considered in a later section.

5.3 Design Issues for Structural Isolation Systems 355

5.3 Design issues for structural isolation systems The most important requirements for an isolation system concern flexibility, energy dissipation, and rigidity under low level loading. A number of solutions have been proposed for civil type structures over the past thirty years. The most significant aspects of these designs is discussed below. Flexibility A structural isolation system generally consists of a set of flexible support elements that are proportioned such that the period of vibration of the isolated structure is considerably greater than the dominant period of the excitation. Systems proposed to date employ plates sliding on a curved surface (eg., an inverted pendulum), sleeved piles, and various types of rubber bearings. The most popular choice at this point in time is the rubber bearing, with about 90% of the applications. Rubber bearings consist of layers of natural rubber sheets bonded to steel plates, as shown in Fig. 5.12. The steel plates constrain the lateral deformation of the rubber under vertical loading, resulting in a vertical stiffness several orders of magnitude greater than the horizontal stiffness. The lateral stiffness depends on the number and thickness of the rubber sheets. Increasing either quantity decreases the stiffness; usually one works with a constant sheet thickness and increases the number of layers. As the height increases, buckling becomes the controlling failure mechanism, and therefore, the height is usually limited to about half the diameter. Natural rubber is a nonlinear viscoelastic material, and is capable of deforming up to about 300% without permanent damage. Shear strain on the order of 100% is a common design criterion. Bearing diameters up to 1m and load capacities up to 5 MN are commercially available.

356 Chapter 5: Base Isolation Systems

Mounting plate Rubber h Steel shims

D

Fig. 5.12: Typical natural rubber bearing (NRB). Rigidity under low level lateral loads Increasing the lateral flexibility by incorporating a base isolation system provides an effective solution for high level seismic excitation. Although the relative motion between the structure and the support may be large, the absolute structural motion is generally small, so that the structure does not feel the earthquake. The effect of other types of lateral loading such as wind is quite different. In this case, the loading is applied directly to the structure, and the low lateral stiffness can result in substantial lateral displacement of the structure relative to the fixed support. To control the motion under service loading, one can incorporate an additional stiffness system that functions for service loading but is not operational for high level loading. Systems composed of rods and/or springs that are designed to behave elastically up to a certain level of service loading and then yield have been developed and are commercially available. There are a variety of steel dampers having the above characteristics that can be combined with the rubber bearings. Figure 5.13 illustrate a particular scheme. The steel rod is dimensioned (length and area) such that it provides the initial stiffness and yields at the intended force level. The earliest solution and still the most popular approach is to incorporate a lead rod in the rubber bearing, as illustrated in Fig. 5.14. The lead plug is dimensioned according to the force level at which the system is intended to yield.

5.3 Design Issues for Structural Isolation Systems 357

Fig. 5.13: Steel rod damper combined with a NRB. Mounting plate Rubber h

Lead plug Steel shims

D

Fig. 5.14: Typical lead rubber bearing (LRB). Energy dissipation/absorption Rubber bearings behave in a viscoelastic manner and have some energy dissipation capacity. Additional damping can be provided by separate devices such as viscous, hysteretic, and friction dampers acting in parallel with the rubber bearings. The lead rubber bearing (LRB) is representative of this design approach; the lead plug provides both initial stiffness and hysteretic damping. Since hysteretic damping action occurs only at high level loading, hysteretic-type systems require additional viscous damping to control the response for low level loading. High damping natural rubber with a dissipation capacity about 4 times the conventional value is used together with other devices to improve the energy dissipation capacity of the isolation system. Figure 5.15 illustrates the deployment of a combination of NRB’s, steel dampers, and viscous dampers. This scheme allows one to adjust both stiffness and damping for each load level, i.e., for both low and high level loading.

358 Chapter 5: Base Isolation Systems

Fig. 5.15: Isolation devices of Bridgestone Toranomon Building. Modeling of a natural rubber bearing (NRB) For the purpose of preliminary design, a NRB can be modeled as a simple shear element having a cylindrical shape and composed of a viscoelastic material. Figure 5.16 defines the notation and shows the mode of deformation. The relevant equations are u γ = --h

(5.49)

F = τA

(5.50)

h = nt b

(5.51)

5.3 Design Issues for Structural Isolation Systems 359 where A is the cross-sectional area, t b is the thickness of an individual rubber sheet, and n is the total number of sheets. Each sheet is assumed to be in simple shear. Applying the viscoelastic constitutive relations developed in Section 3.3, the behavior for harmonic shear strain is given by γ = γˆ sin Ωt

(5.52)

τ = G s γˆ sin Ωt + ηG s γˆ cos Ωt

(5.53)

u

F τ

tb h

Fig. 5.16: Natural rubber bearing under horizontal loading. where G s is the storage modulus and η is the loss factor. In general, G s and η are functions of the forcing frequency and temperature. They are also functions of the strain amplitude in the case of high damping rubbers which exhibit nonlinear viscoelastic behavior. Combining the above equations leads to u = uˆ sin Ωt

(5.54)

F = f d G s uˆ [ sin Ωt + η cos Ωt ]

(5.55)

uˆ = γˆ h = γˆ nt b

(5.56)

A A f d = ---- = -------h nt b

(5.57)

where

Note that f d depends on the bearing geometry whereas η and G s are material properties. The standard form of the linearized force-displacement relation is defined

360 Chapter 5: Base Isolation Systems by eqn (3.70) F = k eq u + c eq u˙

(5.58)

where k eq and c eq are the equivalent linear stiffness and viscous damping terms. Estimates for k eq and c eq can be obtained with a least squares approach. Assuming there are N material property data sets covering the expected range of strain amplitude and frequency, the resulting approximate expressions are eqns (3.74), (3.76), and (3.77) which are listed below for convenience. 1 k eq = f d ---N

∑

)

N

Gs ( Ωi ) = f d G s

(5.59)

i=1

c eq = αk eq N

∑

Gs η --------- Ω i

i=1 α = ---------------------------N

∑

(5.60)

(5.61)

Gs ( Ωi )

i=1

Equation (5.58) is used in the MDOF analysis presented in a later section. Figures 5.17 and 5.18 show that the material properties for natural and filled rubber are essentially constant for the frequency range of interest. Assuming G s and η are constant, the equivalent properties reduce to k eq = f d G s

(5.62)

η α = ------T av 2π

(5.63)

where Tav is the average period for the excitation and G s , η are the “constant” values.

5.3 Design Issues for Structural Isolation Systems 361

G s ( Pa )

η

Fig. 5.17: Storage modulus and loss factor for natural rubber vs. frequency (Snowden, 1979)

362 Chapter 5: Base Isolation Systems

G s ( Pa )

η

Fig. 5.18: Storage modulus and loss factor for filled natural rubber vs. frequency (Snowden, 1979) Modeling of a lead rubber bearing (LRB) As a first approximation, the LRB can be considered to consist of two elements: i) a linear viscoelastic element representing the rubber component, and ii) a linear elastic-perfectly plastic element simulating the lead plug. This model assumes that the static force response relationship is bilinear, as indicated in Fig. 5.19. The stiffness defined by eqn (5.62) can be used for the rubber bearing, i.e. for k 1 . k ( rubber ) ≡ k 1 = f d G s

(5.64)

Considering lead to behave in a linear elastic manner, the plug stiffness can be expressed as

5.3 Design Issues for Structural Isolation Systems 363 Ap Gp k ( lead ) ≡ k 2 = --------------hp

(5.65)

where A p , h p , and G p denote the cross-sectional area, height, and shear modulus for the plug. Lastly, the displacement corresponding to the onset of yielding is related to the yield strain for lead by uy = hp γ y

(5.66)

F k1

k 1 ( rubber )

Fy k1 + k2

F

ks k 2 ( lead )

u uy

uˆ

u

Fig. 5.19: Lead rubber bearing model - quasi static response. Interpreting the behavior of the lead rubber bearing for large deformation as viscoelastic, the response due to harmonic motion is expressed in terms of a secant stiffness, k s , and equivalent loss factor, η˜ , u = uˆ sin Ωt

(5.67)

F = k s uˆ sin Ωt + η˜ k s uˆ cos Ωt

(5.68)

where k s is related to the elastic energy storage capacity and η˜ is a measure of the energy dissipated through hysteretic damping of the rubber and lead components. Defining µ as the ductility ratio γˆ uˆ µ = ------ = ------uy γy the secant stiffness is related to the individual stiffness terms by

(5.69)

364 Chapter 5: Base Isolation Systems k2 k s = k 1 + ----µ

(5.70)

The equivalent loss factor is defined as 1 W η˜ = ------ -----2π E S

(5.71)

where W is the hysteretic work per cycle and E S is the maximum strain energy. Evaluating the energy terms 2

W = 4 ( µ – 1 )k 2 u y + πηk 1 µ 2 u y2

(5.72)

2 1 E S = --- k s [ µu y ] 2

(5.73)

and substituting in eqn (5.71) leads to k1 4 ( µ – 1 )k 2 η˜ = -------------------------- + η ----2 ks πk s µ

(5.74)

–3

Noting that γ y is about 5 ×10 about 0.5 , one can estimate µ as

and the typical peak response strain is

γˆ µ = ------- ≈ 100 γy

(5.75)

A typical value for the ratio of k 1 to k 2 is k 1 ≈ 0.1k 2

(5.76)

Then, reasonable estimates for k s and η˜ are k s = 1.1k 1

(5.77)

0.1 4 η˜ = --------- + ---------- η = 0.12 + 0.909η 0.11 11π

(5.78)

The loss coefficient for high damping rubber can be as high as 0.15 . Combining a

5.3 Design Issues for Structural Isolation Systems 365 high damping rubber bearing with a lead plug provides an effective solution for both initial stiffness and damping over the range from low to high excitation. The last step involves transforming eqn (5.68) to the standard form, eqn (5.58). Applying a least square approach and treating k s and η˜ as functions of both the strain amplitude and frequency leads to N

1 k eq = ---N

∑

k s ( µ i, Ω i )

(5.79)

i=1 N

1 c eq = ---N

∑ i=1

k s ( µ i, Ω i )η˜ ( µ i, Ω i ) -----------------------------------------------Ωi

(5.80)

where N is the number of data sets, i.e., values of µ and Ω . It is reasonable to assume G s and η are constant, and evaluate these parameters for a representative range of the ductility parameter, µ . Applicability of base isolation systems The feasibility of base isolation depends on whether it is needed, whether the proposed structure is suitable for base isolation, and whether it is cost effective compared with alternative solutions (Mayes et al. 1990). The need for base isolation may arise if the location is an area of high seismicity, if increased building safety and post earthquake operability are required, if reduced lateral design forces are desired, or if an existing structure needs upgrading to satisfy current safety requirements. A structure is considered suitable if: i) the subsoil conditions do not produce long period input motions to the structure, ii) the structure is less than about 10 to 15 stories and has a height-to-width ratio that prevents overturning, iii) the site permits the required level of motion of the base with respect to ground, and iv) the non-seismic lateral loads (such as wind) are less than approximately 10% of the weight of the structure. The cost effectiveness of a base isolated structure can be assessed by assigning values to both the initial and life cycle costs and benefits. Examples of cost items are: the bearings, changes to accommodate the isolation system, maintenance and inspection of the isolation system, and the cost of maintaining

366 Chapter 5: Base Isolation Systems operability after earthquakes. Examples of savings are: lower initial cost of the structural system, less construction time, lower insurance premium, reduction in earthquake structural and nonstructural damage, and the reduction in injuries, deaths, and lawsuits from related damages. When disruption costs and the value of the building contents are important, seismic isolation has a substantial economic advantage over other systems provided that such an isolation scheme is technically feasible. Under such conditions, initial cost savings of up to 5% of the building cost have been noticed. For conventional buildings where disruption of operation is not important, there may not be sufficient cost savings in the structural system to offset the cost of the isolators (Mayes et al. 1990). The greatest advantage of base isolation is achieved when it is considered in the early planning stages of the project, since it is possible to take advantage of the reduced response due to the isolation system. If the Base Isolation System is selected and added after completion of the structural design, many complications may arise since the construction techniques may have to be altered. For bridge construction on the other hand, the economic issues are very different from those for buildings. In bridges, the implementation of seismic isolation simply requires the use of a seismic isolation bearing rather than a conventional bearing. Since bearings are only one or two percent of the cost of a bridge, an increase in the cost of isolation bearings will have very little impact on the overall construction cost and consequently, the use of a seismic isolation system is expected to reduce the overall construction cost (Billings et al. 1985).

5.4 Examples of existing base isolation systems The past few years, especially since the Kobe earthquake in Japan, have seen a significant increase in the number of base isolated structures which suggests that the technology is gaining acceptance. A short description of some of the first implementations of base isolation systems is presented here to provide an indication of the type of buildings that are being isolated and the cost savings, if any, achieved by employing this technology. More comprehensive descriptions are contained in Kelly (1993), the Architectural Institute of Japan Guide to Base Isolated Buildings in Japan (1993), and various company web sites listed in the Electronic Reference Section of the text.

5.4 Examples of Existing Base Isolation Systems 367 USC University Hospital (Myers 1989, Asher & Van Volkingburg 1989) This eight-story structure, shown in Fig 5.20, is used as a teaching hospital by the University of Southern California. It resists seismic forces with a steel braced frame located on the perimeter, and is supported on 68 LRB and 81 NRB isolators. The seismic design was based on a 0.4g response spectrum increased by 20% to account for near-fault effects. The decision to incorporate seismic isolation was made in the preliminary design phase of the project. Structural cost comparisons for conventional and isolated structures were developed and the benefits of seismic isolation were assessed. It was determined that the cost savings in the structural frame would be sufficient to pay for the new structural ground floor slab and the isolation system. The additional cost of mechanical and architectural details was 1.3% and there was a 1.4% cost savings in the soil nailed retaining wall used in the isolation design versus the conventional retaining wall. Consequently, there was no net additional cost for incorporating seismic isolation on this hospital project.

Fig. 5.20: USC University Hospital Fire Department Command and Control Facility (Mayes et al. 1990) This is a two-story, steel perimeter braced frame structure that utilizes 36 highdamping elastomeric isolation bearings. The decision to utilize seismic isolation on this project was based on a comparison of two designs (conventional and isolation) that required maintaining the functionality of the structure after the extreme design event. This project reflects the first such detailed comparison for two designs to meet a performance specification. In the case of this two-story

368 Chapter 5: Base Isolation Systems structure, the isolated structure was found to be 6% less expensive than conventional design. A reduction in losses by a factor of 40 is expected with the seismic isolation. Evans and Sutherland Manufacturing Facility (Reaveley et al. 1989) The building, (see Fig 5.21), is a four-story manufacturing site for flight simulators located near the Warm Springs and East faults in Salt Lake City. The building measures 280ft x 160ft in plan and rests on 40 LRB and 58 NRB isolators. Preliminary costs for conventional and isolated designs were developed and the benefits of seismic isolation assessed at the conceptual design phase. The structural engineers decided to design the structural framing system for the UBC code forces for conventional design and, consequently, there were no structural framing cost savings. The additional structural cost was the basement structural floor (versus a slab-on-grade) and the heavy fail safe system used. Based on cost data developed by the contractors, the cost premium for incorporating seismic isolation was 5% or $400,000 on an $8 million project. Important in the decision to employ seismic isolation was protecting the building contents, including work in progress, the value of which exceeds $100 million (approximately 12 times the cost of the structure).

Fig. 5.21: Evans and Sutherland Facility

5.4 Examples of Existing Base Isolation Systems 369 Salt Lake City Building (Mayes et al. 1987, Walters et al. 1986) This facility, shown in Fig 5.22, is a five-story, Richardson Romanesque Revival structure constructed between 1892 and 1894, 265ft x 130ft in plan, and built of unreinforced brick and sandstone. Its 12 story tower is centrally located and is also constructed of unreinforced masonry. The building was restored and a combination of 208 LRB and 239 NRB isolators were installed, separating the building from its foundation. The structure is now protected against damage for the 0.2g design earthquake event. This project was the subject of a detailed study of several retrofit schemes among which were base isolation and UBC strengthening. The schemes were developed in sufficient detail to permit cost estimates and an evaluation of performance. Although the cost of these two alternatives was comparable, the decision to use seismic isolation was made based on the considerably better performance that results from the implementation of such a scheme. The complete architectural and historic restoration, and seismic rehabilitation work was estimated to be $24 million. The approximate value of the seismic isolation work reported by the contractor was $4,414,000 including the cost of the 447 seismic isolators.

Fig. 5.22: Salt Lake City Building The Toushin 24 Ohmori Building (Kajima, 1989) This building has 1 underground story which is used as a parking garage, and 9 stories above ground. It is located adjacent to 2 of the busiest railway lines in Tokyo, and the isolation system was required to reduce the traffic induced vibration as well as seismic motion. Figure 5.23 shows a view of the building, a

370 Chapter 5: Base Isolation Systems sectional plan, and the isolation scheme. A combination of laminated natural rubber bearings and steel rod dampers were deployed. Thick layers of rubber were used to decrease the vertical stiffness and thus filter out vertical microtremors.

a)View of building

5.4 Examples of Existing Base Isolation Systems 371

b) Section

c) Devices Fig. 5.23: The Toushin 24 Ohmori Building

372 Chapter 5: Base Isolation Systems Bridgestone Toranomon Building (Shimizu, 1987) The Bridgestone Toranomon Building (see Fig 5.24) is an office building of the Bridgestone Corporation, a major supplier of rubber products such as bearings. The base isolation system consists of 12 laminated rubber bearings, 25 steel dampers, and 8 viscous (oil) dampers. Figure 5.15 shows the layout of the devices. The viscous dampers are intended to dissipate the energy associated with wind and low intensity excitations. At this load level, the steel dampers are designed to behave elastically and provide stiffness. Energy associated with a large seismic excitation is dissipated/absorbed primarily by the steel dampers.

Fig. 5.24: Bridgestone Toranomon Building San Francisco City Hall (1994) San Francisco City Hall is an historic structure that is currently being

5.4 Examples of Existing Base Isolation Systems 373 retrofitted with a seismic isolation system consisting of 530 lead rubber isolators. The design basis earthquake is 0.50g. Cost of retrofitting the structure is estimated at $105 million.

Fig. 5.25: San Francisco City Hall Long Beach V.A. Hospital The hospital is 12 story concrete structure with shear walls. A combination of 110 LRB, 18 NRB and 18 sliding bearings were installed in the mechanical crawl spaces below the building to improve the building’s ability to survive earthquakes up to magnitude 0.32g.

Fig. 5.26: Long Beach V.A. Hospital

374 Chapter 5: Base Isolation Systems

5.5 Optimal stiffness distribution - discrete shear beam The theory developed earlier in this chapter for the SDOF case is extended here to deal with the more general case of a deformable beam-type structure supported by a base isolation system. Linear behavior is assumed since the objective is to generate results which are suitable for preliminary design. The approach followed to establish the stiffness distribution for the structure is similar to what was presented in Chapter 2. The only modification required is to include the effect of the stiffness and damping associated with the base isolation system. Most of the notation and relevant equations have been introduced in Chapter 2. In what follows, the stiffness distribution corresponding to uniform deformation for the fundamental mode of the composite system consisting of a discrete shear beam and isolation system is derived. The theory is extended to deal with continuous beams in the next section. Scaled stiffness distribution Figure 5.27 defines the notation used for the base isolated shear beam. The bearing system is represented by an equivalent linear spring, k1, and linear viscous damper, c1; m1 represents the mass lumped at the foundation level above the bearings; ui is the displacement of the mass mi with respect to the ground; ki and ci are the story stiffness and viscous damping coefficients for the actual structure. The governing equations for free undamped vibration are expressed as ˙˙ + KU = 0 MU where the various matrices are the same as defined in chapter 2.

(5.81)

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 375

un

mn kn,cn

un-1

mn-1 kn-1,cn-1

u2

m2 k2,c2

m1

u1

k1,c1

Fig. 5.27: Notations for base isolated discrete shear beam. In the previous development, the modal displacement profile was selected such that the interstory displacement was constant over the beam. That strategy is modified here to allow for a different interstory displacement for the first story, which, in this model, represents the relative displacement of the bearing.

4

ub

us

4

3

3

2

2

ν +1

H

1

ub

a) actual

1

ν

b) scaled

Fig. 5.28: Example displacement profile.

376 Chapter 5: Base Isolation Systems Figure 5.28a illustrates the choice of displacement profile; us is the displacement at the top node due to deformation of the beam, and ub is the bearing displacement. For equal story height, the bilinear profile corresponds to uniform shear in the beam, γ = u s ⁄ H . The bearing displacement is expressed as a multiple of the maximum structural displacement, ub = ν us

(5.82)

and the profile is scaled by taking u s as the independent displacement parameter. Fig 5.28b shows the scaled profile. With this choice of displacement parameter, the displacement vector takes the form U = qΦ = u s ν,

1 ν + ---, 3

2 ν + ---, 3

ν + 1

(5.83)

Note that the choice of q as the maximum structural displacement due to deformation of the structure is consistent with the approach followed for the fixed base case. The modified displacement profile introduced here allows for an additional story at the bottom of the beam and distinguishes between the deformation at the base and within the beam. Generalizing this approach for an n’th order system, the fundamental mode profile is taken as U = qΦ Φ = ν,

1 ν + ------------, n–1

2 ν + ------------, …, n–1

ν + 1

(5.84)

The remaining steps are the same as followed in section 2.7. One writes U = e ± iωt Φ and substitutes for U in eqn (5.81). This leads to KΦ = ω 2 MΦ

(5.85)

Scaling K and rearranging the equations results in Sk' = MΦ

(5.86)

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 377 where k' i = k i ⁄ ω 2 and S is defined by eqn (2.161), listed below for convenience. S' ( i, i ) = Φ i – Φ i – 1

(5.87)

S' ( i, i + 1 ) = Φ i – Φ i + 1

S' ( i, j ) = 0 for j ≠ i, i + 1 Given M and Φ, one solves eqn (5.86) for k' . This procedure is illustrated with the following example.

Example 5.3:Scaled stiffness for a 4DOF beam with base isolation. Consider the beam shown in Fig 5.28. The various matrices are 1 2 Φ = ν, ν + ---, ν + ---, ν + 1 (1) 3 3 MΦ = m 1 ν, ν S =

1 m 2 ν + --- , 3

–1 ⁄ 3 1⁄3

–1 ⁄ 3 1⁄3

k' = { k' 1 k' 2 k' 4 = 3m 4 ( ν + 1 )

2 m 3 ν + --- , 3

(2)

(3)

–1 ⁄ 3 1⁄3 k' 3

m4 ( ν + 1 )

k' 4 }

(4)

2 k' 3 = 3m 3 ν + --- + k' 4 3 1 k' 2 = 3m 2 ν + --- + k' 3 3 k' 2 k' 1 = m 1 + -----3ν When the masses are equal, eqn (5) reduces to

(5)

378 Chapter 5: Base Isolation Systems 2 k' 1 = m * 4 + --- ν k' 2 = m * ( 6 + 9ν )

(6)

k' 3 = m * ( 5 + 6ν ) k' 4 = m * ( 3 + 3ν )

Fundamental mode response Taking U according to eqn (5.84), the response of the fundamental mode is governed by ˜ q˙˙ + c˜ q˙ + k˜ q = p˜ – m ˜ Γa m g

(5.88)

where the modal parameters are defined as ˜ = Φ T MΦ m p˜ = Φ T P

c˜ = Φ T CΦ Φ T ME Γ = -----------------˜ m

k˜ = Φ T KΦ c˜ ξ˜ = -----------˜ 2ωm

(5.89)

Since Φ now involves the relative displacement factor, ν , these terms will also depend on ν .

Example 5.4:Example 5.3 revisited. Modal parameters for the 4DOF shear beam considered in example 5.3 are listed below

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 379

˜ = m ν 2 + m ν 2 + 2--- ν + 1--- m 1 2 3 9 4 4 + m 3 ν 2 + --- ν + --- + m 4 ( ν 2 + 2ν + 1 ) 3 9 1 2 Φ T ME = m 1 ν + m 2 ν + --- + m 3 ν + --- + m 4 ( ν + 1 ) 3 3

(1)

1 c˜ = ν 2 c 1 + --- ( c 2 + c 3 + c 4 ) 9 1 2 p˜ = νp 1 + ν + --- p 2 + ν + --- p 3 + ( ν + 1 )p 4 3 3 ˜ and Γ simplify to When the masses are equal, m ˜ = m * 4ν 2 + 2ν + 14 m ------ 9 (2)

1 + 2ν 9 2ν + 1 Γ = ---------------------------- = --- ------------------------------------7 9 7 2ν 2 + ν + --1 + --- ν ( 1 + 2ν ) 7 9

˜ ⁄ m * and Γ for a range of values of ν are listed below in Table 5.1. Values of m There is a significant reduction in Γ with increasing ν , and this results in a reduced response to seismic excitation. Table 5.1: Modal mass and participation factors for 4DOF shear beam with equal modal masses. ν 0 1 2 3 4 5

˜ ⁄ m* m 1.556 7.556 21.556 43.556 73.556 111.556

Γ 1.286 0.794 0.464 0.321 0.245 0.197

The modal damping parameter, c˜ , depends on both the bearing damping c 1 and the structural damping ( c 2, c 3, c 4 ) properties. Incorporating damping in the bearing is more effective than distributing damping over the structure for the

380 Chapter 5: Base Isolation Systems fundamental mode response. Structural damping is needed mainly to control the higher modes.

Stiffness calibration for seismic isolation The peak fundamental mode response due to seismic excitation is given by 1 q max = ---- ΓS v ( ω, ξ ) ω

(5.90)

One specifies q max , ξ , and S v ( ω, ξ ) , and determines ω by iterating on eqn (5.90). By definition, q max is the maximum structural displacement relative to the base motion due to deformation of the structure. It is evaluated using the design value for the maximum transverse shear strain and the structural height, q max = γ * H

(5.91)

The peak amplitude of the bearing displacement relative to the ground follows from eqn (5.82) ub

max

≡ u1

max

= νq max

(5.92)

Given q max and u b , ν is determined with eqn (5.92). This approach has to max be modified when the structure is taken to be rigid, i.e., when q max ≈ 0 . In this case, the system reduces to a SDOF model, and the formulation presented in section 5.2 is applicable.

Example 5.5:Stiffness calibration for Example 5.4 Returning to the 4DOF example structure, the following data is assumed. H = 15m γ * = 1 ⁄ 200 S v ( ω, ξ ) defined by Fig 5.29

(1)

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 381 Using (1), q max = ( 15 ) ( 1 ⁄ 200 ) = 0.075m

(2)

To proceed further, one needs to specify ν . Various cases are considered below.

Spectral Velocity S v (m/s)

S v = 1.2, ξ = 0.02 S v = 0.92, ξ = 0.05

1.0

S v = 0.8, ξ = 0.1 S v = 0.6, ξ = 0.2

0.1

Period 0.1

0.6 1.0

10

T (sec)

Fig. 5.29: Spectral velocity. Case 1

ub

max

= 0.3m

The parameters corresponding to this bearing displacement are ν = 0.3 ⁄ 0.075 = 4 Γ = 0.245

(3)

Substituting in eqn (5.90) leads to an expression for the period, T. 2πq max 1.922 T = ------------------------- = --------------------ΓS v ( T, ξ ) S v ( T, ξ )

(4)

Suppose ξ = 0.05. From Fig 5.29, S v = 0.92 for T>0.6 sec. No iteration is required here, since eqn (4) predicts T>0.6.

382 Chapter 5: Base Isolation Systems

1.922 T = ------------- = 2.13s 0.92 2π ω = ------ = 2.94r ⁄ s T

(5)

The stiffness coefficients are generated using the results contained in example 5.3. For the case of uniform mass, eqns (7) of example 5.3 apply. Taking ν = 4 and ω according to eqn (5) above leads to k 4 = 129.6m *

k 3 = 244.6m *

k 2 = 354.3m *

k 1 = 37.9m *

(6)

Damping is determined with eqn (3) of example 5.4. For ν = 4 and ξ = 0.05 , ˜ = 111.6m * m

(7)

˜ = 2 ( 0.05 ) ( 2.94 ) ( 111.6m * ) = 32.8m * c˜ = 2ξωm

(8)

The individual damping coefficients are related to c˜ by 1 c˜ = 16c 1 + --- ( c 2 + c 3 + c 4 ) = 32.8m * 9

(9)

One has to decide how to allocate damping to the various components. For example, assuming 75% of c˜ is contributed by the bearing requires c 1 = 1.54m * c 2 + c 3 + c 4 = 73.8m *

(10)

Placing damping at the base is an order of magnitude more effective than distributing the damping throughout the structure for this degree of isolation. Case 2 u b = 0.15m max For this case, ν = 2 . The various parameters for ξ = 0.05 , ν = 2 , and uniform mass are as follows.

5.5 Optimal Stiffness Distribution - Discrete Shear Beam 383 Γ = 0.464 T = 1.025s ω = 6.16r ⁄ s ˜ = 21.556m * m c˜ = 13.28m * 1 4c 1 + --- ( c 2 + c 3 + c 4 ) = 13.28m * 9 k 4 = 341m *

k 3 = 644m *

k 2 = 909m *

k 1 = 189m *

(11)

Case 3 Fixed base The fixed base case is treated in examples 2.9 and 2.10. Specializing these results for uniform mass and ξ = 0.05 results in the following parameters and properties. Γ = 1.286 T = 0.5s ω = 12.56r ⁄ s ˜ = 1.556m * m 1 --- ( c 2 + c 3 + c 4 ) = 1.95m * 9 k 4 = 473m * k 3 = 788m * k 2 = 946m * k1 = ∞

(12)

384 Chapter 5: Base Isolation Systems

5.6 Optimal stiffness distribution - continuous cantilever beam Stiffness distribution - undamped response The equilibrium equations for undamped motion of the base isolated continuous beam shown in Fig. 5.30 are H

∫

V ( x, t ) = – ρ m u˙˙( x, t ) dx

(5.93)

x

H

M ( x, t ) =

∫

V ( x, t ) dx

(5.94)

x

uT

uB

u

M + dM D T, D B

H

V + dV dx

x

V

ub

M kb

Fig. 5.30: Base isolated continuous beam. The transverse shear and bending deformation measures for the beam are related to the translation and rotation quantities by

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 385 ∂u = γ+β ∂x

(5.95)

∂β ∂x

(5.96)

χ =

Considering γ and χ to be functions only of time, integrating the resulting equations with respect to x, and imposing the boundary conditions at x = 0 , one obtains expressions for u and β in terms of γ ( t ) , χ ( t ) , and u b ( t ) 1 2 u = γx + --- χx + u b 2

(5.97)

β = χx

(5.98)

where u b denotes the displacement of the base of the structure with respect to ground. Taking γ = γ∗e

iω 1 t

(5.99)

iω 1 t

(5.100)

χ = χ∗ e

u b = u b∗ e

iω 1 t

(5.101)

produces a periodic motion of the beam. Noting that the deformation measures γ ∗ and χ∗ are related by (see eqn (2.14) and Fig 5.30) uB χ∗ H s = ------ = ----------uT 2γ ∗

(5.102)

and expressing u b∗ in terms of the displacement at x = H due to shear deformation (see Fig. 5.30) u b∗ = νu T∗ = νγ ∗ H

(5.103)

transforms eqn (5.97) into iω t iω t x sx 2 u = ---- + -------- + ν γ * He 1 = Φ ( x ) (γ *He 1 ) = q ( t )Φ ( x ) H H2

(5.104)

The function Φ ( x ) defines the fundamental mode. The corresponding expression

386 Chapter 5: Base Isolation Systems for the fixed base case is eqn (2.189). Differentiating u with respect to time, 2

iω t sx u˙˙ = – x + -------- + νH γ ∗ ω 12 e 1 H

(5.105)

and substituting for u˙˙ in eqn (5.93) leads to iω t sx 3 νx 1 s x2 V = ρ m ω 12 H 2 --- + --- + ν – ---------- – ---------- – ------ γ ∗ e 1 2 3 2 3 H 2H 3H

(5.106)

The corresponding relation for the bending moment is M = ρ m ω 12 H 4 χ∗ e

iω 1 t

1 1 ν 1 1 ν x ----- + --- + ----- – ----- + --- + ----- ---6s 8 4s 4s 6 2s H

(5.107)

νx 2 x3 x4 + ------------- + ---------------- + ------------4sH 2 12sH 3 24H 4 Lastly, the shear and bending rigidity distributions are determined with the definition equations V 1 s x2 sx 3 νx D T = ----- = ρ m ω 12 H 2 --- + --- + ν – ---------- – ---------- – -----2 3 γ* 2H 2 3H 3 H

(5.108)

νx 2 M 1 1 ν 1 1 ν x x3 x4 D B = ----- = ρ m ω 12 H 4 ----- + --- + ----- – ----- + --- + ----- ---- + ------------- + ---------------- + ------------6s 8 4s 4s 6 2s H 4sH 2 12sH 3 24H 4 χ* (5.109) Equation (5.108) is written as DT ( 0 ) 1 s x 2 sx 3 νx D T = --------------------------- --- + --- + ν – ---------- – -------- – ------ 2 3 H 1 s 2H 2 3 --- + --- + ν 2 3

(5.110)

where D T ( 0 ) is the shear rigidity at the base. 1 s D T ( 0 ) = ρ m ω 12 H 2 --- + --- + ν ≡ k s H 2 3

(5.111)

The parameter, k s , can be interpreted as an equivalent shear stiffness

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 387 measure.The shear force at the base of the beam must equal the shear force in the bearing to satisfy the force equilibrium condition for undamped motion. Equating these forces D T ( 0 )γ ∗ = k b u b∗ = k b νγ ∗ H

(5.112)

and solving for k b results in DT ( 0 ) 1 k b = ---------------- = --- k s ν νH

(5.113)

The fundamental frequency follows from eqn (5.111) DT ( 0 ) ω 12 = ------------------------------------------1 s ρ m H 2 --- + --- + ν 2 3

(5.114)

Figures 5.31 and 5.32 shows the mode shapes and shear deformation profiles for the first five modes of a typical low rise building. The variation in the mode shape profiles with the ratio of the stiffness of the isolator, k b , to the shear beam stiffness k s are illustrated by Figures 5.33, 5.34 and 5.35. This ratio is equal to 1 ⁄ ν . Figure 5.36 displays the variation in the periods of the highest three fundamental modes. The primary influence is on the period of the fundamental mode which is significantly increased when the stiffness of the isolator is several orders of magnitude lower than the beam stiffness. The effect on the periods of the second and third modes is relatively insignificant. Figure 5.37 shows the variation of the participation of the second and third modes relative to the first. The plot shows that the contribution of the second and third modes is also significantly reduced by decreasing the stiffness of the isolator with respect to the beam stiffness.

388 Chapter 5: Base Isolation Systems

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

1

2

3

4

5

Mode number

Fig. 5.31: Mode shapes for a typical base isolated structure.

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0

1

2

3

4

5

Mode number

Fig. 5.32: Mode deformation profiles for a typical base isolated structure.

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 389

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

k b ⁄ k s = 0.001 k b ⁄ k s = 0.1 k b ⁄ k s = 10.0

0.3 0.2 0.1 0

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Mode 1

Fig. 5.33: Variation of mode 1 shape with relative stiffness of isolator.

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

k b ⁄ k s = 0.001 k b ⁄ k s = 0.1 k b ⁄ k s = 10.0

0.3 0.2 0.1 0

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Mode 2

Fig. 5.34: Variation of mode 2 shape with relative stiffness of isolator.

390 Chapter 5: Base Isolation Systems

1 0.9

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4

k b ⁄ k s = 0.001 k b ⁄ k s = 0.1 k b ⁄ k s = 10.0

0.3 0.2 0.1 0

−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Mode 3

Fig. 5.35: Variation of mode 3 shape with relative stiffness of isolator.

8

T1 T2 T3

7

6

Period

5

4

3

2

1

0 −3 10

−2

10

−1

10

0

10

1

10

kb ⁄ ks

Fig. 5.36: Variation of periods with relative stiffness of isolator.

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 391

Relative participation factors

1

Γ2 ⁄ Γ1 Γ3 ⁄ Γ1

0.8

0.6

0.4

0.2

0 −3 10

−2

−1

10

10

0

10

1

10

kb ⁄ ks

Fig. 5.37: Variation of relative participation factors with relative stiffness of isolator.

Fundamental mode equilibrium equation Incorporating the contribution of the base isolation system, the principle of virtual displacements has the form H

∫

H

( M ⋅ δχ + V ⋅ δγ )dx + F b ⋅ δu b =

0

∫

( b ⋅ δu )dx

(5.115)

0

where F b is the shear force in the bearing. The equations relating internal forces to deformations and deformation rates are taken as V = D T γ + C T γ˙

(5.116)

M = D B χ + C B χ˙

(5.117)

F b = k b u b + c b u˙ b

(5.118)

392 Chapter 5: Base Isolation Systems The form of the modal expansion follows from eqn (5.104). x sx 2 u = qΦ ( x ) = q ---- + -------- + ν H H2

(5.119)

2sx β = qΨ ( x ) = q --------- H2

(5.120)

Assuming external loading and seismic excitation, the loading term is b = – ρ m a – ρ m u˙˙ + b ( x, t ) g

(5.121)

Finally, introducing the various terms in the principle of virtual displacements leads to the equilibrium equation for q ˜ q˙˙ + c˜ q˙ + k˜ q = p˜ m

(5.122)

where H

˜ = m

∫

2 x sx 2 1 s 2sν s 2 ρ m ---- + -------- + ν dx = ρ m H --- + ν + ν 2 + --- + --------- + ----H H2 2 3 5 3

(5.123)

C T 4s 2 C B ------- + ---------------- dx + c b ν 2 H2 H4

(5.124)

D T 4s 2 D B ------- + ---------------- dx + k b ν 2 H2 H4

(5.125)

0

H

c˜ =

∫ 0

H

k˜ =

∫ 0

H

p˜ =

∫

x sx 2 1 s ( b – ρ m a g ) ---- + -------- + ν dx = – ρ m H --- + ν + --- a g – p˜ e H H2 2 3

0

Expressing c˜ and k˜ as

(5.126)

5.6 Optimal Stiffness Distribution - Continuous Cantilever Beam 393 ˜ c˜ = 2ξωm

(5.127)

˜ k˜ = ω 2 m

(5.128)

transforms eqn (5.122) to 2 1 q˙˙ + 2ξωq˙ + ω q = – Γa g + ----p˜ e ˜ m

(5.129)

1 s ν + --- + --2 3 Γ = -------------------------------------------------------------1 s 2 s 2sν --- + ν 2 + ν + ----- + --- + --------3 5 2 3

(5.130)

where

For a pure shear beam, s = 0 and the participation factor for the fundamental mode reduces to 1 ν + --2 Γ = ------------------------1 --- + ν 2 + ν 3

(5.131)

The expression for the modal damping ratio depends on how one specifies the damping over the beam. Rigidity calibration - seismic excitation The calibration procedure presented in Chapter 2 is applied to the base isolated model. Starting with ΓS v ( ω, ξ ) q max = ------------------------ω

(5.132)

and substituting for q max results in ΓS v νΓS v ω = ----------- = ------------γ∗H u b∗

(5.133)

394 Chapter 5: Base Isolation Systems One specifies ν in addition to the other parameters ( γ *, ξ, S v ) , and solves for ω . This value is then used to determine D T ( 0 ) and k b with eqns (5.111) and (5.113).

Example 5.6:Stiffness calibration - Example Building #2. The stiffness calibration for Building example #2 was considered in Chapter 2. In what follows, the calibration procedure is extended to include stiffness and damping components located at the base. The design data are: H = 50m

ρ m = 20, 000k g ⁄ m

γ * = 1 ⁄ 200 ξ 1 = 0.05

s = 0.25 S v = 0.92m ⁄ s

(1)

Using (1), the peak relative structural displacement is q max = γ * H = 0.25m

(2)

To proceed further, one needs to specify the base displacement and then establish the value of ν with eqn (5.103). We take u b = 0.25m

(3)

Then ub ν = ------------ = 1.0 q max

(4)

Given ν , the participation factor follows from eqn (5.130) Γ = 0.600

(5)

The modal mass is determined with eqn (5.123) ˜ = 2.638 × 10 6 kg m

(6)

Assuming CT = constant and CB =0 in eqn (5.124), the modal damping

5.6 Building Design Examples 395 coefficient reduces to CT c˜ = ------- + ν 2 c b = 0.02C T + c b H

(7)

Equation (5.127) relates c˜ to ω . ˜ = 0.2638 × 10 6 ω Ns ⁄ m c˜ = 2ξωm

(8)

Lastly, the frequency is found using eqn (5.133) and the design data for S v ( ξ = 0.05 and S v = 0.92m ⁄ s ) ΓS v ω = ---------- = 2.4S v = 2.61r ⁄ s γ *H T = 2.41sec

(9)

With ω known, the modal damping coefficient follows from eqn (8), c˜ = 0.688MNs ⁄ m

(10)

and the transverse shear rigidity at the base of the beam is determined with eqn (5.111). D T ( 0 ) = 538.6MN

(11)

Finally, given D T ( 0 ) , the isolation stiffness is estimated using eqn (5.113), which is based on neglecting the contribution of the damping force in the bearing. k b = 10.77MN

(12)

5.7 Building design examples Stiffness distribution based on fundamental mode response Since base isolation is a potential solution for buildings with less than about 10 stories and having an aspect ratio that prevents overturning, only Buildings 1 and

396 Chapter 5: Base Isolation Systems 2 of Chapter 2 are considered in the simulation. Tables 2.4 and 2.5 lists the design data. The damping in the structure is obtained by specifying ξ 1 and computing a stiffness proportional damping matrix considering a fixed based structure, as carried out in Chapter 2. The damping in the bearing for specified bearing damping ratio ξ b is obtained by assuming the structure to be a SDOF system having a mass equal to the total mass of the structure and a stiffness equal to the stiffness of the bearing as obtained from eqn (5.113). Figures 5.38 and 5.39 show the shear rigidity distributions for Buildings 1 and 2 obtained with eqns (5.108) and (5.109), taking S v = 1.2m ⁄ s . The corresponding shear deformation profiles for different combinations of building damping ratios and isolator damping ratios when the structures are subjected to scaled versions (to S v = 1.2m ⁄ s for ξ = 0.02 ) of the El Centro and Taft accelerograms are plotted in Figures 5.40 through 5.47. The maximum deformation in the bearing is also indicated on the plots. The plots show that increasing the damping in the bearing tends to reduce the bearing deformation without significantly altering the shape of the deformation profile along the structure’s height. Furthermore, just as for the fixed base structures, the contribution of the higher modes becomes more significant as the structure becomes more slender. The structural parameters, as well as the mean and standard deviation results for the deformations of the above buildings are tabulated in the following section.

5.6 Building Design Examples 397

1 0.9

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% 6 k b = 6.35 ×10 N/m

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N

Fig. 5.38: Initial shear rigidity distribution for Building 1.

1 0.9

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% 6 k b = 9.93 ×10 N/m

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N

Fig. 5.39: Initial shear rigidity distribution for Building 2.

398 Chapter 5: Base Isolation Systems

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ b = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.18m 5%, u b = 0.18m 10%, u b = 0.17m 20%, u b = 0.17m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.40: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.18m 5%, u b = 0.16m 10%, u b = 0.13m 20%, u b = 0.11m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.41: Maximum shear deformation for Building 1.

0.01

5.6 Building Design Examples 399

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ b = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.09m 5%, u b = 0.08m 10%, u b = 0.07m 20%, u b = 0.07m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.42: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Initial H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.09m 5%, u b = 0.06m 10%, u b = 0.06m 20%, u b = 0.05m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.43: Maximum shear deformation for Building 1.

0.01

400 Chapter 5: Base Isolation Systems

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ b = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.24m 5%, u b = 0.23m 10%, u b = 0.21m 20%, u b = 0.20m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.44: Maximum shear deformation for Building 2.

1

Building #2 Quadratic Based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.24m 5%, u b = 0.22m 10%, u b = 0.19m 20%, u b = 0.14m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.45: Maximum shear deformation for Building 2.

0.01

5.6 Building Design Examples 401

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ b = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.14m 5%, u b = 0.13m 10%, u b = 0.11m 20%, u b = 0.09m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.46: Maximum shear deformation for Building 2.

1

Building #2 Quadratic based Initial H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.14m 5%, u b = 0.11m 10%, u b = 0.08m 20%, u b = 0.07m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.47: Maximum shear deformation for Building 2.

0.01

402 Chapter 5: Base Isolation Systems Stiffness distribution including the contribution of the higher modes This section extends the iterative procedure developed in Section 2.10 to incorporate iterating over the stiffness for the beam and isolator. For simplicity, only shear deformation in the isolator is considered. The method consists of including the contribution of the higher modes to the transverse shear, the bending deformation, and the base shear, and then updating the shear and bending rigidities and isolation stiffness using (i + 1) DT (x)

(i + 1) DB (x)

(i + 1) kb

=

(i) [ γ ( x ) ] max ( i ) D T ------------------------------∗

(5.134)

=

(i) [ χ ( x ) ] max ( i ) D B -------------------------------∗

(5.135)

γ

χ

(i)

[ V b ] max = ------------------------u b∗

(5.136)

The peak values are found with eqn (2.270). Rigidity iterations are performed on building examples 1 and 2. Table 5.2 lists the parameters of the buildings for the initial rigidity distributions as well as for a single iteration. For both building examples, one iteration was sufficient to achieve convergence.Tables 5.3 and 5.4 contain the deformations averaged over the height of the structure and the corresponding standard deviations for Buildings 1 and 2 subjected to scaled versions of El Centro and Taft excitations.

5.6 Building Design Examples 403 Table 5.2: Modal parameters - building examples. T1(s)

T2(s)

T3(s)

ξ1 ( % ) ξ2 ( % ) ξ3 ( % ) Γ2 ⁄ Γ1 Γ3 ⁄ Γ1

Bldg #1 Initial (Q) Iteration 1

2.09 2.01

0.62 0.60

0.33 0.32

1.56 1.54

3.55 3.64

6.26 6.37

0.11 0.12

0.03 0.03

Bldg #2 Initial (Q) Iteration 1

2.51 2.45

0.90 0.85

0.49 0.46

1.48 1.48

3.70 3.83

6.60 6.89

0.20 0.19

0.06 0.06

Figures 5.48 through 5.57 show the shear and bending rigidity distributions resulting from the first iteration, as well as the shear deformation profiles corresponding to different combinations of building damping ratios and isolator damping ratios under El Centro and Taft excitation.Results from one iteration provide sufficient convergence accuracy. For Building 2, the iterative scheme tends to pull the top back in, resulting in a more uniform deformation profile. The effect of damping is similar to that noticed in the examples of the previous section.

404 Chapter 5: Base Isolation Systems Table 5.3: Mean and standard deviation deformation results for Building 1. Bldg #1 El Centro Initial (Q)

Iteration 1

Taft Initial (Q)

Iteration 1

γm

γ sd

χm

χ sd

ξ1 (%)

ξb (%)

ub (m)

(10-3)

(10-4)

(10-5)

(10-6)

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.18 0.18 0.17 0.17

2.80 2.74 2.61 2.43

0.70 0.46 0.11 0.22

3.18 3.11 2.93 2.72

0.56 0.31 0.07 0.24

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.18 0.16 0.13 0.11

2.80 2.55 2.36 2.19

0.70 1.09 1.92 2.35

3.18 2.93 2.79 2.62

0.56 0.95 1.31 1.57

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.17 0.17 0.15 0.15

3.28 3.09 2.79 2.37

1.36 1.05 0.66 0.62

2.86 2.68 2.41 1.98

4.40 4.31 4.10 3.63

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.17 0.14 0.12 0.10

3.28 2.55 2.26 2.16

1.36 1.23 0.92 1.38

2.86 2.22 1.96 1.90

4.40 3.08 3.21 2.98

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.09 0.08 0.07 0.07

1.61 1.42 1.25 1.08

3.42 2.22 1.32 0.49

2.04 1.74 1.49 1.25

2.76 1.86 0.96 0.31

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.09 0.06 0.06 0.05

1.61 1.34 1.21 1.17

3.42 2.78 1.80 1.68

2.04 1.67 1.45 1.41

2.76 2.60 1.79 1.54

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.08 0.08 0.08 0.07

1.80 1.57 1.38 1.17

2.26 1.00 0.39 0.27

1.64 1.38 1.19 0.99

2.00 1.97 1.97 1.84

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.08 0.07 0.06 0.05

1.80 1.41 1.25 1.22

2.26 1.73 0.63 0.93

1.64 1.27 1.07 1.07

2.00 1.45 1.43 1.29

5.6 Building Design Examples 405 Table 5.4: Mean and standard deviation deformation results for Building 2. Bldg #2 El Centro Initial (Q)

Iteration 1

Taft Initial (Q)

Iteration 1

γm

γ sd

χm

χ sd

ξ1 (%)

ξb (%)

ub (m)

(10-3)

(10-4)

(10-5)

(10-6)

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.24 0.23 0.21 0.20

4.61 4.13 3.77 3.26

7.49 4.17 2.49 1.22

4.90 4.25 3.81 3.25

5.96 3.28 1.83 0.81

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.24 0.22 0.19 0.14

4.61 4.15 3.61 2.99

7.49 6.25 4.98 4.10

4.90 4.39 3.79 3.15

5.96 4.88 3.72 2.78

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.23 0.22 0.21 0.20

4.67 4.11 3.65 3.12

4.36 1.84 1.68 2.34

4.04 3.44 2.98 2.50

5.92 5.67 5.54 5.14

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.23 0.21 0.18 0.14

4.67 4.21 3.65 2.98

4.36 3.61 2.87 2.43

4.04 3.62 3.12 2.55

5.92 5.43 4.82 4.15

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.14 0.13 0.11 0.09

2.77 2.30 1.89 1.49

6.24 3.49 1.74 1.33

3.00 2.41 1.91 1.52

5.56 3.19 1.74 1.17

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.14 0.11 0.08 0.07

2.77 2.34 1.99 1.87

6.24 6.53 6.03 5.69

3.00 2.56 2.23 2.16

5.56 6.05 4.98 4.48

2.00 5.00 10.00 20.00

2.00 2.00 2.00 2.00

0.11 0.10 0.09 0.08

2.45 2.05 1.69 1.38

4.92 2.68 1.22 0.75

2.22 1.80 1.43 1.16

2.84 2.38 2.03 2.00

2.00 2.00 2.00 2.00

2.00 5.00 10.00 20.00

0.11 0.09 0.08 0.07

2.45 2.16 1.87 1.74

4.92 4.57 3.92 2.93

2.22 1.97 1.69 1.57

2.84 2.73 2.47 1.72

406 Chapter 5: Base Isolation Systems

1 0.9

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% 6 k b = 6.94 ×10 N/m

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N

Fig. 5.48: Converged shear rigidity distribution for Building 1.

1 0.9

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% 7 k b = 1.05 ×10 N/m

x Normalized height ---H

0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2 9

x 10

Shear rigidity distribution D T - N

Fig. 5.49: Converged shear rigidity distribution for Building 2.

5.6 Building Design Examples 407

1

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ b = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.17m 5%, u b = 0.16m 10%, u b = 0.15m 20%, u b = 0.15m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.50: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.17m 5%, u b = 0.14m 10%, u b = 0.12m 20%, u b = 0.10m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.51: Maximum shear deformation for Building 1.

0.01

408 Chapter 5: Base Isolation Systems

1

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ b = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.08m 5%, u b = 0.08m 10%, u b = 0.08m 20%, u b = 0.07m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.52: Maximum shear deformation for Building 1.

1

Building #1 Quadratic based Iteration 1 H = 25m ρ m = 20000kg/m s = 0.15 S v = 1.2m/s ξ 1 = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.08m 5%, u b = 0.07m 10%, u b = 0.06m 20%, u b = 0.03m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.53: Maximum shear deformation for Building 1.

0.01

5.6 Building Design Examples 409

1

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ b = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.23m 5%, u b = 0.22m 10%, u b = 0.21m 20%, u b = 0.20m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.54: Maximum shear deformation for Building 2.

1

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% El Centro

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.23m 5%, u b = 0.21m 10%, u b = 0.18m 20%, u b = 0.14m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.55: Maximum shear deformation for Building 2.

0.01

410 Chapter 5: Base Isolation Systems

1

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ b = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξ1 ξ1 ξ1 ξ1

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.11m 5%, u b = 0.10m 10%, u b = 0.09m 20%, u b = 0.08m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

Maximum shear deformation γ - m/m

Fig. 5.56: Maximum shear deformation for Building 2.

1

Building #2 Quadratic based Iteration 1 H = 50m ρ m = 20000kg/m s = 0.25 S v = 1.2m/s ξ 1 = 2% Taft

0.9

x Normalized height ---H

0.8 0.7 0.6 0.5

ξb ξb ξb ξb

0.4 0.3 0.2 0.1 0 0

= = = =

2%, u b = 0.11m 5%, u b = 0.09m 10%, u b = 0.08m 20%, u b = 0.07m

γ∗ 0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

Maximum shear deformation γ - m/m

Fig. 5.57: Maximum shear deformation for Building 2.

0.01

Problems 411

Problems Problem 5.1

k p

m c ug

ut = u g + u

Consider a SDOF system with m=1000kg. The design requirements for the system involve 2 nonconcurent loading conditions, a periodic external forcing and a periodic ground motion.Under the external forcing, the relative motion is prescribed; the total motion is prescribed for the case of ground excitation. These design conditions are summarized below. The asterisk denotes a specified quantity. 1) p = p * sin Ω P* t

(1)

u < u*

(2)

2) u g = u g* sin Ω g* t

(3)

u t < u t* ≡ H 3* u g*

(4)

The design problem involves determining k and c such that the above performance requirements are satisfied. a) Recommand values for k and c corresponding to the following design data: 1)

Ω g* = 2πr ⁄ s p * = 400N

2)

Ω g* = 2πr ⁄ s p * = 800N

u g* = 0.2m π Ω P* = --- r ⁄ s 2 * H 3 = 0.125 π Ω P* = --- r ⁄ s 2

H 3* = 0.125 u * = 0.2m

u * = 0.2m

412 Chapter 5: Base Isolation Systems

3)

Ω g* = 2πr ⁄ s

H 3* = 0.125

p * = 1600N

π Ω P* = --- r ⁄ s 2

u * = 0.2m

b) It may not be possible to satisfy both constraints with the same value of stiffness. When this problem arises, one can determine the stiffness corresponding to each design condition and design a spring mechanism that allows the stiffness to be varied. Suggest design concepts for such a mechanism. This situation may occur for lateral loading applied to a seismically isolated structure. Discuss how you would implement a variable stiffness scheme for non simultaneous wind and seismic loading.

Problem 5.2 Refer to the modified SDOF model defined by Fig 5.8. Take m=10,000 kg and k=400 kN/m. Suppose the system is to be subjected to a ground motion, u g ( meters ) = 0.5 sin 4πt , and the total displacement is required to be less than 0.2m. Determine the appropriate bearing stiffness.

Problem 5.3 Refer to the modified SDOF model defined by Fig 5.8. Take m=10,000 kg and the ground motion to be u g ( meters ) = 0.5 sin 4πt . Determine k and k b such that a) u ≤ 0.3m and u b ≤ 0.3m . b) u ≤ 0.1m and u b ≤ 0.3m

Problem 5.4 Refer to the modified SDOF model defined by Fig 5.8. Take m=10,000 kg and consider the system to be subjected to seismic excitation of intensity S v = 1.2m ⁄ s . Determine k , k b and T eq for the following design conditions:

Problems 413 a) u * = 0.1m

u b* = 0.1m

b) u * = 0.1m

u b* = 0.2m

c) u * = 0.1m

u b* = 0.3m

Problem 5.5 Consider a cylindrical bearing having a diameter of 0.6m, a height of 0.3m and composed of filled natural rubber. Take the rubber properties according to Fig 5.18. Assume the frequency range is from 1 to 5 Hz, and the temperature is 20oC. a) Estimate the equivalent linear stiffness and linear viscous damping parameters, k eq and c eq . b) Determine the diameter of a lead plug for the case where the plug stiffness is 10 times the stiffness of the rubber bearing. Assume the lead plug and 3 rubber cylinder have the same height. Take G P = 4 ×10 MPa . c) Assume the bearing experiences periodic excitation resulting in a shear strain amplitude of 50%. Determine the secant stiffness, k s , and loss factor, η˜ , using eqns (5.70) and (5.74). d) Instead of lead, consider using low strength steel as the material for the 3 initial stiffness element, k 2 . Take τ y = 150MPa and G = 80 ×10 MPa for the steel plug. Repeat parts b and c.

Problem 5.6

Consider an isolation system composed of a NRB, a steel hysteretic

414 Chapter 5: Base Isolation Systems k 1 ( rubber ) k 2 ( damper )

F, u

c F

k1

Fy k1+k2

uˆ y uˆ * uˆ damper, and a viscous damper. Neglect the damping provided by the NRB. a) For low level loading, one specifies the initial stiffness, damping, and yield force level. Describe how you would design the steel damper. b) For high level loading, one specifies the secant stiffness and equivalent viscous damping based on a seismic analysis. Describe how you would design the NRB. How would you select the viscous damper? c) Suppose the isolation system is composed of spring and damper elements whose properties can be varied instantaneously. Assuming the elements behave linearly, the force is given by F = k ( t )u + c ( t )u˙

k F, u c

Problems 415 where k(t) and c(t) are properties that can be “adjusted”. Describe how you would utilize this system for a building subjected (nonsimultaneously) to both wind and seismic excitation. How would you design these devices? Note: a system which has the ability to change its properties is said to be adaptive. Adaptive systems are discussed in Chapter 6.

Problem 5.7 m7

m1

k 7, c 7

u7

u1 k 1, c 1

Consider a 6 story building with base isolation modelled as a 7 DOF system. Take the floor masses as m 1 = m 2 = … = m 6 = 10, 000kg and m 7 = 20, 000kg . a) Find the scaled stiffness for the profile based on eqn (5.84). Take ν = 2.0 . ˜ , c˜ , p˜ , and Γ . b) Evaluate the expressions for m c) Calibrate the stiffness distribution for seismic excitation. Use the spectral = 0.15m , and velocity plot contained in Fig 5.29. Take ξ = 0.05 , u s max ub = 0.3m . Allocate 75% of the damping to the bearing. Use MOTIONLAB max

to determine the modal shapes, frequencies, and damping ratios for the 2nd and 3rd modes. d) Repeat part c assuming the base is fixed. Take the sum of the stiffness factors as

416 Chapter 5: Base Isolation Systems a measure of the cost of stiffness. Compare the costs of the fixed base and base isolation solutions. Also compare the modal properties. e) Suppose filled rubber bearings having a diameter of 0.5m and a height of 0.25m are to be used for the isolation system. Assume G = 4MPa and η = 0.15 for the rubber. How many bearings are required for the design conditions specified in part c? Also discuss how you would provide the damping required for the isolation system. f) Discuss how you would deal with lateral wind loading. Assume the dominant wind gust frequency is 0.2Hz.

Problem 5.8

30m

ρ m constant x

Consider a base isolated continuous cantilever beam having a uniform mass density of 2000 kg/m and a height of 30m. a) Generate shear and bending rigidity distributions and isolation stiffness which correspond to a fundamental frequency of 0.33 Hz and various values for s and ν . Take 0 ≤ s ≤ 0.25 and 1 ≤ ν ≤ 3 . Comment on the sensitivity of the rigidity parameters to variation in s and ν . b) Approximate the continuous beam with 6 shear beam segments plus an additional segment to simulate the bearing. Determine the first 3 mode shapes and frequencies corresponding to s=0 and ν = 0 , 1.5, and 3. Consider the lumped masses to be equal. Comment on the sensitivity of the modal properties to the ratio of isolation stiffness, k b , to the shear beam stiffness measure, k s .

Problems 417

Problem 5.9 ub

q

H

D T and ρ x

m

constant

k b, c b ug

Consider a base isolated continuous shear beam having a uniform mass density and constant transverse shear rigidity. Assume the lateral displacement is approximated by x x u = u g + u b + q ---- = u g + q ν + ---- H H a) Establish an equation for q(t) using the Principle of Virtual Displacements. Allow for linear viscous damping in the bearing and uniform material damping in the beam. Write the result in the same form as eqn (5.122) and determine the ˜ , c˜ , k˜ , p˜ , ω, ξ , and Γ . expressions for m b) Suppose H=30m, ρ m = 2000k g ⁄ m , and v=2. Calibrate the stiffness distribution ( D T and k b ) for ξ = 0.05 , q max = 0.15, and S υ defined by Fig 5.29. Determine the damping parameters assuming the bearing contributes 75% to c˜ . c) Approximate the continuous beam with 6 shear beam segments plus an additional segment to simulate the bearing. Determine the first 2 mode shapes and frequencies, using the design data generated in part b.

418 Chapter 5: Base Isolation Systems

Problem 5.10

m1

ub + u1 + u g p

ρ m, D T ( x )

H

kb cb ub + u g

ug

Consider the base isolated shear beam shown above. Assume uniform mass density, m 1 = a ( ρ m H ) , and the shear rigidity to be defined as x D T ( x ) = D∗ 1 – -------- 2H a) Derive the equilibrium corresponding to the following approximate displacement expansion x u = u 1 ( t ) ---- + u b ( t ) H Discuss how the undamped free vibration response behaves as D∗ , k b , and a are varied. b) ub + u1 m1

md ub + u1 + ud

Suppose the force, p , applied at the top of the structure is generated by a

Problems 419 tuned mass damper as shown in the figure. How would you estimate the properties of the tuned mass damper to obtain an effective damping ratio of 0.05 for the mode shape approximation considered in part a? c) Suppose p has the form p = – b 1 u 1 – b 2 u˙ 1 + p ( t 1 ) where b 1, b 2 are constants and p is prescribed. Assume c b = 0 and determine the equation for u 1. What effect do b 1 and b 2 have on the response? Illustrate for a periodic excitation.

420 Chapter 5: Base Isolation Systems

421

Part II: Active Control Chapter 6

Introduction to active structural motion control 6.1 The nature of active structural control Active versus passive control The design methodologies presented in the previous chapters provide systematic procedures for distributing passive motion control resources which, by definition, have fixed properties and do not require an external source of energy. Once installed, a passive system cannot be modified instantaneously, and therefore one needs a reliable estimate of the design loading and an accurate numerical model of the physical system for any passive control scheme to be effective. The inability to change a passive control system dynamically to compensate for an unexpected loading tends to result in an over-conservative design. When self-weight is an important design constraint, one cannot afford to be too conservative. Also, simulation studies on the example building structures show that passive control is not very effective in fine tuning the response in a local region. Considering these limitations, the potential for improving the performance by dynamically modifying the loading and system properties exists. An active structural control system is one which has the ability to determine the present state of the structure, decide on a set of actions that will change this state to a more desirable one, and carry out these actions in a controlled manner and in a short period of time. Such control systems can theoretically accommodate unpredictable environmental

422 Chapter 6: Introduction to Active Structural Motion Control changes, meet exacting performance requirements over a wide range of operating conditions, and compensate for the failure of a limited number of structural components. In addition, they may be able to offer more efficient solutions for a wide range of applications, from both technical and financial points of view. Active motion control is obtained by integrating within the structure a control system consisting of three main components: a) monitor, a data acquisition system, b) controller, a cognitive module which decides on a course of action in an intelligent manner, and c) actuator, a set of physical devices which execute the instructions from the controller. Fig 6.1 shows the interaction and function of these components; the information processing elements for active control are illustrated in Fig 6.2. This control strategy is now possible due to significant recent advances in materials that react to external stimuli in a nonconventional manner, sensor and actuator technologies, real-time information processing, and intelligent decision systems.

EXCITATION

STRUCTURE

RESPONSE

ACTUATOR AGENTS TO CARRY OUT INSTRUCTIONS

MONITOR SENSORS to measure external loading

MONITOR DEVELOP THE ACTION PLAN (SET OF INSTRUCTIONS TO BE COMMUNICATED TO THE ACTUATORS)

DECIDE ON COURSE OF ACTION IDENTIFY THE STATE OF SYSTEM

CONTROLLER Fig. 6.1: Components of an active control system

SENSORS to measure response

6.1 The Nature of Active Structural Control 423

Physical System

Sensor

Sensor Sensor

Decision Making

Action

Data Fusion Modeling & Analysis Processing

Transmission channel

Visualization

Processing

Archival and Access

Fig. 6.2: Information Processing Elements for an Active Control System

h( p)

p

∆p f

u

decide on ∆p f (a) Passive decide on changing h to h'

u + εu observe u

+ p

h' ( p + ∆p e + ∆p f )

+ observe p p + εp

(b) Active decide on ∆p e

∆p e

Fig. 6.3: Passive and active feedback diagrams

u

424 Chapter 6: Introduction to Active Structural Motion Control The simple system shown in Fig. 6.3 is useful for comparing active and passive control. Figure 6.3(a) corresponds to passive control. The input, p , is transformed to an output, u , by the operation h ( p ) u = h( p)

(6.1)

One can interpret this system as a structure with p denoting the loading, u the displacement, and h the flexibility of the structure. The strategy for passive motion control is to determine h ( p ) such that the estimated output due to the expected loading is contained within the design limits, and then design the structure for this specific flexibility. Active control involves monitoring the input and output, and adjusting the input and possibly also the system itself, to bring the response closer to the desired response. Figure 6.3(b) illustrates the full range of possible actions. Assuming the input corrections and system modifications are introduced instantaneously, the input-output relation for the actively controlled system is given by u = h' ( p + ∆p e + ∆p f )

(6.2)

Monitoring the input, and adjusting the loading is referred to as open-loop control. Observing the response, and using the information to apply a correction to the loading is called feedback control. The terminology closed-loop control is synonymous with feedback control. In addition to applying a correction to the input, the control system may also adjust certain properties of the actual system represented by the transformation h ( p ) . For example, one can envision changing the geometry, the connectivity, and the properties of structural elements in real time. One can also envision modifying the decision system. A system that can adjust its properties and cognitive processes is said to be “adaptive”. The distinguishing characteristic of an adaptive system is the self-adjustment feature. Non-adaptive active structural control involves monitoring and applying external forces using an invariant decision system. The make up of the structure is not changed. Adaptive control is the highest level of active control. The role of feedback Feedback is a key element of the active control process. The importance of feedback can be easily demonstrated by considering a linear static system and

6.1 The Nature of Active Structural Control 425 taking the input correction to be a linear function of the output. For this case, u = hp

(6.3)

∆p f = k f [ u + ε u ]

(6.4)

where h and k f are constants. Substituting in eqn (6.2) specialized for h’=h and solving for u results in hk f h u = ---------------- [ p + ∆p e ] + ---------------- ε u 1 – hk f 1 – hk f

(6.5)

When k f is positive, the sensitivity of the system to loading is increased by feedback, i.e. the response is amplified. Taking k f negative has the opposite effect on the response. Specializing eqn (6.5) for negative feedback ( k f < 0 ), the response becomes hk f h (6.6) u = -------------------- [ p + ∆p e ] + -------------------- ε u 1 + hk f 1 + hk f Increasing k f decreases the effect of external loading. However, the influence of ε u , the noise in the response observation, increases with k f and, for sufficiently large k f , is essentially independent of the feedback parameter. This result indicates that the accuracy of the monitoring system employed to observe the response is an important design issue for a control system. Computational requirements and models for active control The monitor component identified in Figs 6.1 and 6.2 employs sensors to measure a combination of variables relevant to motion such as strain, acceleration, velocity, displacement, and other physical quantities such as pressure, temperature, and ground motion. This data is usually in the form of analog signals which are converted to discrete time sequences, fused with other data, and transmitted to the controller module. Data compression is an important issue for large scale remote sensing systems. Wavelet based data compression (Amaratunga, 1997) is a promising approach for solving the data processing problem. The functional requirements of the controller are to compare the observed response with the desired response, establish the control action such as the level of feedback force, and communicate the appropriate commands to the actuator

426 Chapter 6: Introduction to Active Structural Motion Control which then carries out the actual control actions such as apply force or modify a structural property. The controller unit is composed of a digital computer and software designed to evaluate the input and generate the instructions for the actuators. There are 2 information processing tasks: state identification and decision making. Given a limited amount of data on the response, one needs to generate a more complete description of the state of the system. Some form of model characterizing the spatial distributions of the response and data analysis are required. Once the state has been identified, the corrective actions which bring the present state closer to the desired state can be established. In this phase, a model which defines the input - output relationship for the structure is used together with an optimization method to decide upon an appropriate set of actions. For algorithmic non-adaptive systems, the decision process is based on a numerical procedure that is invariant during the period when the structure is being controlled. Time invariant linear feedback is a typical non-adaptive control algorithm. An adaptive controller may have, in addition to a numerical control algorithm, other symbolic computational models in the form of rule-based systems and neural networks which provide the capability of modifying the structure and control algorithm in an intelligent manner when there is a change in the environmental conditions. Examples illustrating time invariant linear feedback control algorithms are presented in the following sections; a detailed treatment of the algorithms is contained in Chapters 7 and 8.

6.2 An introductory example of quasi-static feedback control Consider the cantilever beam shown in Fig 6.4. Suppose the beam acts like a bending beam, and the design objective is to control the deflected shape such that it has constant curvature. The target displacement distribution corresponding to this constraint has the form 1 u * ( x ) = --- χ * x 2 2

(6.7)

where χ * is the desired curvature. One option is to select the bending rigidity according to M(x) D B ( x ) = ---------------χ*

(6.8)

6.2 An Introductory Example of Quasi-static Feedback Conrol 427 where M(x) is the moment at location x due to the design loading. This strategy is a stiffness based passive control approach. A second option is to select a representative bending rigidity distribution, and apply a set of control forces which produce a displacement distribution that, when combined with the displacement due to the design loading, results is a displacement profile that is close to the desired distribution. In what follows, the latter option is discussed.

u(x) x F L Fig. 6.4: Cantilever beam with control force Suppose the control force system consists of a single force applied at x=L. Assuming linear elastic behavior, and using the linear technical theory of beams as the model for the structure, the displacement due to F is estimated as F F x3 (6.9) u c ( x ) = ----------- Lx 2 – ----- = -------h ( x ) DB 2D B 3 where D B is the bending rigidity, considered constant in this example. The displacement due to the design loading is also determined with the technical beam theory. This term is denoted as u o ( x ) , and expressed as 1 u o ( x ) = ------- g ( x ) DB

(6.10)

Combining the 2 displacement patterns results in the total displacement, u(x). u ( x ) = uo ( x ) + uc ( x )

(6.11)

The expanded form of eqn (6.11) corresponding to the particular choice of control force location for this example is 1 u ( x ) = ------- [ g ( x ) + Fh ( x ) ] DB

(6.12)

The difference between u ( x ) and u * ( x ) is defined as e ( x ) and interpreted

428 Chapter 6: Introduction to Active Structural Motion Control as the displacement error. e ( x ) = uo + uc – u *

(6.13)

For this example, D B is considered to be fixed, and therefore e ( x ) is a function only of the single control force magnitude F. 1 e ( x ) = ------- [ g ( x ) + Fh ( x ) ] – u * ( x ) DB

(6.14)

Ideally, one wants e = 0 for 0 ≤ x ≤ L . This goal cannot be achieved, and it is necessary to work with a relaxed condition. The simplest choice is collocation, which involves setting e equal to zero at a set of prescribed locations. For example, setting e = 0 at x=L leads to DB u * ( L ) – g ( L ) F = -----------------------------------------(6.15) h(L) A more demanding condition is a least square requirement, which involves first forming the sum of e 2 evaluated at a set of prescribed points, and then selecting F such that the sum is a minimum. The continuous least square sum is given by the following integral 1 J = --2

L

∫

e 2 dx = J ( F )

(6.16)

0

Taking J(F) as the measure of the square error sum, F is determined with the stationary condition ∂J = 0 ∂F

(6.17)

Differentiating the integral expression for J, ∂J = ∂F

L

∫

e

0

∂e dx = 0 ∂F

(6.18)

and using eqn (6.14), which defines e ( x ) for this particular example, results in L

∫

h ( x ) [ D B u * ( x ) – g ( x ) ] dx

0

F = ----------------------------------------------------------------------L

∫

(6.19)

( h ( x ) ) 2 dx

0

The value of F defined by eqn (6.19) produces the absolute minimum value of J. A proof of this statement is presented in Section 7.2 which treats in more detail the

6.2 An Introductory Example of Quasi-static Feedback Conrol 429 least square procedure for quasi-static loading.

Example 6.1: Shape control for uniform loading This example illustrates the application of the approach described above to the case where the design loading is a uniform distributed load extending over the entire length of the beam. The corresponding deflected shape is 1 w 1 u o ( x ) = ------- g ( x ) = ------- ------ ( x 4 – 4x 3 L + 6x 2 L 2 ) D B 24 DB

(1)

Applying collocation at x=L leads to * 3 D B χ 3 F e ( L ) = 0 = --- ------------- – --- wL 2 L 2

(2)

The least square solution is * 91 D B χ 2065 F ls = ------ ------------- – ------------ wL 66 L 5280

DB χ * = ( 1.379 ) ------------- – ( 0.391 )wL L

(3)

Both solutions are approximate since they do not satisfy e ( x ) = 0 . One can improve the performance by taking additional control forces. Selecting the spatial distribution of the control forces is a key decision for the design of a control system.

The above discussion assumes that there is some initial loading, and one can determine the corresponding displacement field with the physical model of the structural system. This control strategy is similar to the concept of prestressing. A more general scenario is the case where one is observing the response at a set of “observation” points and the loading is being applied gradually so that there is negligible dynamic amplification and sufficient time to

430 Chapter 6: Introduction to Active Structural Motion Control adjust the control forces. Here, one needs to establish u o ( x ) using the observed displacement data. Suppose there are s observation points located at x j ( j= 1, 2, ..., s), and at time t the monitoring system produces the data set u o ( x j, t ) . This data can be used together with an interpolation scheme to generate an estimate of u o for the region adjacent to the observation points. A typical spatial interpolation model has the form s

u o ( x, t ) =

∑

u o ( x j, t )Ψ j ( x )

(6.20)

j=1

where Ψ j ( x ) are interpolation functions. Given u o ( x, t ) , one forms the displacement error, 1 e ( x, t ) = u o ( x, t ) – u * ( x ) + -------h ( x )F ( t ) DB

(6.21)

and determines F(t) with either collocation or a least square method. The continuous least square estimate for F(t) is given by L

∫

h ( x ) ( u * ( x ) – u o ( x, t ) ) dx 0

F ( t ) = D B --------------------------------------------------------------------L ( h ( x ) ) 2 dx

(6.22)

∫

0

Example 6.2: Discrete displacement data

Suppose the displacement observation points are located at x=L/2 and x=L. Given these 2 values of displacement, one needs to employ an interpolation scheme in order to estimate u o ( x ) . Taking a quadratic expansion, uo ( x ) = ao + a1 x + a2 x 2

(1)

6.3 An Introductory Example of Dynamic Feedback Control 431

u1 L/2

u2 L/2

x and specializing eqn(1) for points 1 and 2 results in the following approximation, x x 2 x x 2 u o ( x ) ≈ u 1 4 --- – 4 --- + u 2 – --- + 2 --- L L L L

(2)

The expression for the control force is obtained using eqn (6.22) and eqn (2). Evaluating the integrals leads to * 133 91 D B χ D B 98 ----------------F≈ – ------- ------ u 1 + --------- u 2 66 L 3 33 66 L

(3)

as an estimate for F.

6.3 An introductory example of dynamic feedback control To gain further insight on the nature of feedback control, the simple SDOF system shown in Fig. 6.5 is considered. The system is assumed to be subjected to both an external force and ground motion, and controlled with the force F . Starting with the governing equation, mu˙˙ + cu˙ + ku = – ma g + F + p

(6.23)

and introducing the definitions for frequency and damping ratio leads to the

432 Chapter 6: Introduction to Active Structural Motion Control

k m

F

p

c ug

u + ug

Fig. 6.5: Single-degree-of-freedom system.

standardized form of the governing equation 2 F p u˙˙ + 2ξωu˙ + ω u = – a g + ---- + ---m m

(6.24)

The free vibration response of the uncontrolled system has the general form u = Ae

λt

(6.25)

Substituting for u in eqn (6.24), one obtains two possible solutions u = A1 e

λ1 t

+ A2 e

λ2 t

(6.26) 2

λ 1, 2 = – ξω ± iω 1 – ξ = – ξω ± iω′

(6.27)

Considering A 1 and A 2 to be complex conjugates, 1 A 1, 2 = --- [ A R ± iA I ] 2

(6.28)

where A R and A I are real numbers representing the real and imaginary parts of A , the solution takes the form u = e

– ξωt

2 2 A R cos ωt 1 – ξ + A I sin ωt 1 – ξ

(6.29)

6.3 An Introductory Example of Dynamic Feedback Control 433 One determines A R and A I resulting expressions are

with the initial conditions for u and u˙ . The

AR = u ( 0 ) 1 A I = – ----- ( u˙ ( 0 ) + ξωu ( 0 ) ) ω′

(6.30)

Considering negative linear feedback, the control force is expressed as a linear combination of velocity and displacement F = – k v u˙ – k d u

(6.31)

where the subscripts ‘v’ and ‘d’ refer to the nature of the feedback, i.e. velocity or displacement. Feedback is implemented in the actual physical system by: • observing the response • determining u and u˙ • calculating F with eqn (6.31) • applying F with an actuator Mathematically, one incorporates feedback by substituting for F in eqn (6.24). The result is kv 2 kd p u˙˙ + 2ξω + ----- u˙ + ω + ----- u = – a g + --- m m m

(6.32)

Equation (6.32) can be transformed to the standardized form by defining equivalent damping and frequency parameters as follows: 2 2 kd ω eq = ω + ----m

(6.33)

kv 2ξ eq ω eq = 2ξω + ----m

(6.34)

With this notation, the solution for the free vibration response of the linear feedback controlled case has the same general form as for no control; one just replaces ξ and ω with ξ eq and ω eq respectively in eqn (6.29). It follows that the effect of linear feedback is to change the fundamental frequency and damping

434 Chapter 6: Introduction to Active Structural Motion Control ratio. Solving eqns (6.33) and (6.34) for ω eq and ξ eq , results in kd ω eq = ω 1 + ----k

(6.35)

ξ eq = ξ + ξ a

(6.36)

where ξ a is the increment in damping ratio due to active control kv kd 1 ⁄ 2 1 ξ a = ----------------------------- ------------ – ξ 1 + ----– 1 k d 1 ⁄ 2 2ωm k 1 + ----k

(6.37)

Critical damping corresponds to ξ eq = 1 kv -----------= 2mω ξ = 1 eq

kd 1 + ----- – ξ k

(6.38)

Equation (6.35) shows that negative displacement feedback increases the frequency. According to eqn (6.37), the damping ratio is increased by velocity feedback and decreased by displacement feedback. If the objective of including active control is to limit the response amplitude, velocity feedback is the appropriate mechanism. Displacement feedback is destabilizing in the sense that it reduces the effect of damping. Stability and other issues associated with linear feedback are discussed in Chapter 8.

Example 6.3: Illustrative example - influence of velocity feedback This example demonstrates the influence of velocity feedback on the response of 2 SDOF systems subjected to seismic excitation. The properties of the systems are System 1: m = 10, 000kg

ω 1 = 6.32rd ⁄ s

k = 400, 000N ⁄ m

T 1 = 0.99s

c = 2, 500Ns ⁄ m

6.3 An Introductory Example of Dynamic Feedback Control 435 System 2: m = 10, 000kg

ω 1 = 2rd ⁄ s

k = 40, 000N ⁄ m

T 1 = 3.14s

c = 830Ns ⁄ m

ξ = 0.0208

The models are excited with the El Centro and Taft accelerograms scaled to a max = 0.5g . Figures 6.6 through 6.8 contain plots of the maximum relative displacement, maximum control force magnitude, and maximum power requirement. The power requirement is computed using the following expression Power = force × velocity = Fu˙

(1)

which assumes the control force is a set of self-equilibrating forces applied as shown in Fig 6.5. Ground motion has no effect on the work done by F with this force scheme.

436 Chapter 6: Introduction to Active Structural Motion Control

0.16

Maximum displacement - m

System 1

T1=0.99s El Centro Taft

0.14

0.12

0.1

0.08

0.06

0.04 0.05

0.1

0.15

ξa

0.2

0.25

0.3

0.35

System 2 T1=3.14s El Centro Maximum displacement - m

0.3

Taft

0.25

0.2

0.15

0.1 0.05

0.1

0.15

ξa

0.2

0.25

0.3

Fig. 6.6: Variation of maximum displacement with active damping.

6.3 An Introductory Example of Dynamic Feedback Control 437

4

2.4

x 10

System 1 T1=0.99s 2.2

Maximum force - N

El Centro 2

Taft

1.8

1.6

1.4

1.2

1

0.8

0.6

0.4 0.05

0.1

0.15

ξa

0.2

0.25

0.3

7000

T1=3.14s

System 2 6000

El Centro

Maximum force - N

Taft 5000

4000

3000

2000

1000 0.05

0.1

0.15

ξa

0.2

0.25

0.3

Fig. 6.7: Variation of maximum control force level with active damping.

438 Chapter 6: Introduction to Active Structural Motion Control

16000

System 1 T1=0.99s Maximum power - N.m/s

14000

El Centro Taft

12000

10000

8000

6000

4000

2000 0.05

0.1

0.15

ξa

0.2

0.25

0.3

4500

System 2 T1=3.14s Maximum power - N.m/s

4000

El Centro 3500

Taft

3000

2500

2000

1500

1000

500 0.05

0.1

0.15

0.2

0.25

0.3

ξa

Fig. 6.8: Variation of maximum power requirement with active damping.

6.4 Actuator Technologies 439

6.4 Actuator technologies Introduction The actuator component of the control system generates and applies the control forces at specific locations on the structure according to instructions from the controller. Over the past several decades, a number of force generation devices have been developed for a broad range of motion control applications. These devices can be described in terms of performance parameters such as response time, peak force, and operating requirements such as peak power and total energy demand. The ideal device is one that can deliver a large force in a short period of time for a small energy input. Civil structures generally require large control forces, on the order of a meganewton and, for seismic excitation, response times on the order of milliseconds. The requirement on peak force coupled with the constraint on energy demand is very difficult to achieve with a fully active force actuator system. There are force actuator systems that are capable of delivering large force, but they also have a high energy demand. Included in this group are hydraulic, electromechanical, and electromagnetic devices. All these types are based on very mature technology. There is considerable on-going research and development of new force actuators that have a low energy demand. One approach is based on modifying the physical makeup of the device in such a way that only a small amount of energy is required to produce a significant increase in force. Varying the orifice area of a viscous type damper is an example of this strategy. A typical viscous damper can deliver a force on the order of a meganewton. By adjusting the damping parameter, the damper can be transformed into a large scale force actuator. The second approach employs adaptive materials as the force generating mechanism. These materials respond to a low energy input by changing their properties and their state in a nonconventional manner which results in a force. Although these technologies are promising, the current devices can produce only low forces, on the order of a kilonewton, and therefore their applicability for civil structures is limited at this time. There are 2 issues that need to be addressed: i) how the force generation mechanism works, and ii) how the forces are applied to the structure. The first

440 Chapter 6: Introduction to Active Structural Motion Control issue is related to the physical make up and underlying physics of the device. The second question is concerned with how the device is attached to the structure so as to produce the “desired” control force. In what follows, the attachment issue is discussed first, then the state of the art for linear actuator technologies is reviewed, and lastly some adaptive material based actuators are described. Force application schemes The schematic drawing contained in Fig 6.9 shows the typical makeup of hydraulic, electro-mechanical, and electro-magnetic linear actuators. There are 2 basic elements, a piston and a mechanism that translates the piston linearly either by applying a force to one end or by moving the end with a gear mechanism. The interaction of the piston with an adjacent body produces a pair of contact forces F at the contact point and a corresponding reaction force at the actuator support. If the body moves under the action of F, the mechanism usually compensates for this motion such that the force remains constant until instructed by the controller to change the force magnitude.

u

Mechanism

Piston

F

Structure

Actuator Support F Fig. 6.9: Linear actuator

Consider the structural frame shown in Fig 6.10a. Suppose the objective is to apply a horizontal force at point A and there is no adjacent structure which could support the actuator. One option (Fig 6.10b) is to fasten a tendon to point A, pass it over a pulley attached to the base, and then connect it to a linear actuator which can generate a tensile force in the tendon. In this scheme, the actuator reaction force is transmitted directly to the base. A second option would be to place the actuator directly on the structure. The actuator reaction force is now

6.4 Actuator Technologies 441 transmitted to the structure; however, the other end of the piston needs to be restrained in order to generate a control force. If the restraining body is rigidly connected to the structure as shown in Fig 6.10c, the force system is selfequilibrating and the structure “feels” no lateral force. Member AC is in tension. In order to have a “non zero” lateral force acting on the structure, the restraining body must be allowed to move laterally. This objective can be achieved by attaching an auxiliary mass, ma, to the piston and supporting the mass on rollers. The mass moves with respect to the structure with an absolute acceleration equal to F/ma. One specifies the peak force and magnitude of the auxiliary mass, and designs the actuator so that it can provide the required force at that level of acceleration. Since the force is generated by driving the mass, this scheme is referred to as an active mass driver.

A

A

F Tendon Pulley F (b) Active tendon

(a) Active force

a

C F

A

(c) Self-equilibrating forces

ma

F

(d) Active mass driver

Fig. 6.10: Control force schemes The extension of these schemes to a multi-story structure is shown in Fig. 6.11. A linear actuator placed on a diagonal produces a set of self-equilibrating forces which impose a shearing action on the particular story to which it is attached. The other stories experience no deformation since the story shear due to this actuator is zero. It follows that one needs to incorporate active braces

442 Chapter 6: Introduction to Active Structural Motion Control throughout the structure in order to achieve global displacement control. Forces generated with active mass drivers are not self-equilibrating and consequently have more influence on the global displacement response.

F3 F3 T3

F3

F2

F2 T2

T1

(a) Active brace

F2

F1

F1 F1 (b) Active mass driver

Fig. 6.11: Control force schemes for a multi-story structure The previous examples relate to shear beam structures which require forces that act in the transverse direction. For bending beam problems, control force systems that produce bending moments are required. This action can be obtained with linear actuators placed on the upper and lower surfaces, as illustrated in Fig 6.12. The region between A and B is subjected to a constant moment equal to F d. Another scheme is shown in Fig 6.13. The actuator is attached to the beam with rods that provide the resistance to the piston motion, resulting in the selfequilibrating system that produces a triangular moment distribution over the region A-B-C. By combing a number of these actuator-rod configurations, one can generate a piecewise linear bending moment distribution.

6.4 Actuator Technologies 443

F d

F

a) A Fd

B Fd

b)

B

A (-) Fd

c)

Fig. 6.12: Constant moment field L

F/2

F/2 F

F a)

b) C

A

B

(-)

c)

FL/4

Fig. 6.13: Triangular moment field

444 Chapter 6: Introduction to Active Structural Motion Control Linear actuators generate control force systems composed of concentrated forces. For discrete structures such as frames, this type of force distribution is appropriate. However, for continuous structures such as beams and plates, a continuous force distribution is more desirable. One strategy that has been examined is based on using a adaptive material in the form of a thin plate. Fig 6.14 illustrates this approach for a continuous beam. Plates are attached by epoxy to the upper and lower surfaces. Applying a voltage to the plate generates a longitudinal strain. Since the plate is attached to the surface, motion of the plate is restrained and an interfacial shear stress τ ( t ) is generated. This stress produces a distributed control moment m c ( x, t ) equal to m c ( x, t ) = τb f d

(6.39)

where b f is the width of the plate. Spatial and temporal variation of the control force system is achieved by varying the voltage applied to the plate.

film τ beam

mo dx d

+ τ dx

Fig. 6.14: Moment generated by strain actuators Large scale linear actuators Referring back to Fig 6.9, a linear actuator can be considered to consist of a piston and a mechanism which applies a force to the piston and also controls the motion of the piston. This actuator type is the most widely available and extensively used, particularly for applications requiring a large force and short response time. The descriptors hydraulic, electro-mechanical, and electromagnetic refer to the nature of the force generation mechanisms. These devices generally have a high energy demand.

6.4 Actuator Technologies 445 Hydraulic systems generate the force by applying a pressure on the face of a piston head contained within a cylinder. Fig 6.15 illustrates this concept. Fluid is forced in or out of the cylinder through the orifice at C1 to compensate for the piston displacement and maintain a certain pressure. These systems have the highest force capacity of the linear actuator group, on the order of meganewtons (Dorey et al., 1996). Precise control movement and force can be achieved with a suitable control system. Protection against overload is provided by a pressure relief value. The disadvantages of this type of system are: the requirements for fluid storage systems, complex valves and pumps to regulate the flow and pressure, seals, and continuous maintenance. Durability of the seals and the potential for fluid spills are critical issues.

Fig. 6.15: Schematic cross section view - a hydraulic cylinder (Dorey et al, 1996) Electromechanical linear actuators generate the force by moving the piston with a gear mechanism that is driven by an electric motor. The motion, and therefore the force, is controlled by adjusting the power input to the motor. These devices are compact in size, environmentally safe, and economical. Figure 6.16 illustrates the various components of a linear electric actuator system (Raco, www.raco.de). The largest electric actuator that can be ordered “off the shelf” is rated for 600kN force.

446 Chapter 6: Introduction to Active Structural Motion Control

Positioning Control

Amplifier unit for motor

Servomotor

Electric Actuator

Fig. 6.16: Components of linear electric actuator systems (Raco, http//:www.raco.de) Hydraulic and electromechanical actuators are composed of many parts that are in contact with each other, and therefore have a high risk of breakdown. Since electromagnetic actuators are driven by magnetic forces, which do not require mechanical contact, they are theoretically more reliable. Small scale electromagnetic actuators with a force capacity ranging from tens of newtons to several kilonewtons are commercially available. In addition to their compact nature and low voltage and amperage requirements, their response time is low, on the order of milliseconds. These features are ideal for active force generation, and electromagnetic actuators are a popular choice for small scale structures. Large scale electromagnetic actuator technology is still in the research and development phase, and there is no commercial product with a force capacity in the meganewton range available at this time. However, since the force generation mechanism is so simple, it is reasonable to expect that economical designs will eventually emerge, and be viable candidates for controlling large scale structures. Figure 6.17 shows a schematic cross-sectional view of a design concept for a large scale electromagnetic actuator developed by Chaniotakis et al (99) at the Plasma Science and Fusion Center, Massachusetts Institute of Technology.

6.4 Actuator Technologies 447

driving coil

Fc

piston

field coil

Fem

R=Fem Fig. 6.17: Schematic view - MIT Electromagnetic actuator design The unit consists of a cylindrical shell housing and a piston. Two sets of axisymmetric electromagnets are used. The field coil is embedded in the cylinder, and generates a stationary magnetic field. A driving coil is attached to the piston which translates with respect to the housing. The force mechanism is based on the interaction between the magnetic field generated by the stationary field magnet and the current in the driving coil. For this particular design, the electromagnetic force is a linear function of the coil currents and is independent of the position of the piston. F em = cI f I d

(6.40)

where I f , I d are the currents in the field and driving coils, and c is a design parameter. The equation of motion for the piston relates the electromagnetic force, the inertia force for the piston mass, and the external contact force, F c . m p u˙˙ = F em – F c

(6.41)

When the device is used as an active mass driver, an auxiliary mass, m a , is

448 Chapter 6: Introduction to Active Structural Motion Control attached to the end of the piston, and F c is set equal to m a u˙˙ . The reaction at the base of the housing is equal to the electromagnetic force which acts on the field coil and is transmitted to the housing. There are several advantages to this concept: i) the response time is on the order of milliseconds, ii) there is minimal mechanical contact, and iii) the technology for controlling the current is mature. The disadvantages are: i) the current and voltage requirements for a force on the order of meganewton cannot be satisfied with conventional electrical power supply technology, and ii) there is minimal experience related to the design, fabrication, and operating performance of large scale electromagnetic actuators. Large scale adaptive configuration based actuators This category includes mechanical devices such as dampers, friction elements, and stiffness elements which have the ability to generate a large force by changing their physical makeup. Their distinguishing features are their low ratio of energy demand to force output, and their large scale force capacity. These features are very desirable for applications to large scale civil structures. Adaptive devices which have a low energy demand and also function as energy dissipation mechanisms are referred to as semi-active actuators since they behave like passive devices in the sense that they increase the stability of a structure and require no external energy. The key point is stability. A semi-active actuator will never destabilize a system whereas an active actuator may destabilize a system even though it has a low energy demand. The use of a variable orifice damper as a force actuator was suggested by Feng et al. (1990), and developed further by Shinozuka et al (1992). Kurata et al. (1994) implemented variable dampers in a large scale 3 story frame structure and Sack and Patten (1994) installed a hydraulic actuator with a controllable orifice on a bridge on interstate highway I-35 in Oklahoma. Figure 6.18 illustrates the concept of a variable orifice damper. The flow control valve is operated by a controller which in turn responds to the demand for a particular magnitude of force specified by the feedback control algorithm; the controller adjusts the valve opening according to the force demand. Since the valve motion is perpendicular to the flow, the force required to adjust the valve position is small, and therefore the energy demand is low. Kurata’s experiment required only 30 Watts to operate a valve. Letting v denote the velocity of the piston, the damping force, F, can be

6.4 Actuator Technologies 449 expressed as F= c v

(6.42)

where c is a function of v and the geometry of the valve. In the case of a passive damper, c is a prescribed function of v, and F is determined uniquely by specifying v. In the case of a variable damper, c is a function of the valve position as well as v. When used as an actuator, F is specified, v is observed, c is calculated with eqn (6.42), and the valve position is determined using the values of c and v. The limitation of this device is the dependency of the sense of F on the sense of v; F is always opposite in sense to v. Therefore, if the force required by the control algorithm at a particular time has the same sense as the observed velocity, the force demand cannot be met, and the actuator needs to be inactivated until a later time when the phase is reversed.

Fig. 6.18: Variable damping mechanism (Kajima Corporation, http://www.kajima.com) Small scale adaptive material based actuators Low force capacity electromechanical and electromagnetic linear actuators are standard “off-the-shelf” products offered by a number of suppliers. Our interest here is not with these devices but rather with a new generation of small

450 Chapter 6: Introduction to Active Structural Motion Control scale force actuators that utilize the unique properties of adaptive materials to generate the force. Research and development in this area was initiated by the aerospace industry as a potential solution for shape control of satellite arms and airplane control surfaces. As the technology evolved, other applications related to motion control of small scale structures such as robot arms and biomedical devices have occurred. Although the technology continues to evolve, and reliability is still a major concern, these devices are being seriously considered as candidates for force control where the required force level is on the order of a kilonewton. Brief descriptions of the various adaptive material based actuators are presented below. • Piezoelectric actuators Piezoelectric materials belong to the electrostrictive material category. When subjected to a voltage, they undergo a molecular transformation which results in the material extending (or contracting) in a manner similar to the Poisson effect for applied stress (Ikeda, 1997). Figure 6.19 illustrates this behavioral mode; a voltage V z in the Z direction produces extensional strains ε x and ε y in the X and Y directions. The opposite behavioral mode occurs when the material is stressed in the X-Y plane; a voltage V z' in the Z direction is generated by σ x and σ y . This behavior was first reported by Pierre and Jacques Curie in 1880. Pierre later won a Nobel price in Physics with his wife Marie for their work on radioactivity. Historically, piezoelectric materials have been used as strain sensors. Their use as actuators is more recent and stimulated primarily by the Aerospace Industry. (Crawley, 1987)

6.4 Actuator Technologies 451

VZ

Z

V' Z

εy

σy Y

X

εx

σx

Fig. 6.19: Piezoelectric electrical-mechanical interaction Piezoelectric actuators are fabricated with piezoceramic block type elements or piezopolymer films. Lead zirconate titanate (PZT) is the dominant piezoceramic composite used for sensors and actuators in the frequency range up to 10 6 Hz. Polyvinyldene fluoride (PVDF) is the most common piezoelectric film. Since it has a relatively low strength, PVDF is used mainly as a sensor, particularly for the high frequency range up to 10 9 Hz. The underlying principle is the same for both materials. The piezoelectric object is attached to a surface which restrains its motion. When a voltage is applied, the object tends to expand immediately, and contact forces between the object and the restraining medium are produced. Two actuator configurations have been developed. The first model is a conventional linear actuator, such as shown in Fig 6.20. Piezoceramic wafers are stacked vertically, bonded, enclosed in a protective housing, and fitted with electrical connectors. These devices can deliver large forces, on the order of 20 kN, with a response time of several milliseconds (Kinetic Ceramics Inc., www.kineticceramics.com).

452 Chapter 6: Introduction to Active Structural Motion Control

Fig. 6.20: Cylindrical piezoceramic linear actuator (Kinetic Ceramics Inc) The second configuration has the form of a thin plate as illustrated in Fig 6.21. Piezoceramic wafers are distributed over the area in a regular pattern. They may also be stacked through the thickness. This type of device is bonded to a surface, and applies a pair of self equilibrating tangential forces to the surface. The peak force depends on the applied voltage and degree of restraint. A force level of 500 N at 200 volts, and millisecond response, is the typical upper limit for off-the-shelf plate type piezoceramic actuators (Active Control Experts, www.acx.com).

impressed voltage

bond

wafer

bond

Fig. 6.21: Plate type piezoelectric actuator • Shape memory alloys Shape memory alloys are metal alloys which, if deformed inelastically at

6.4 Actuator Technologies 453 room temperature, return to their original shape when heated above a certain temperature. Figure 6.22 illustrates this behavior. The initial straight form is deformed inelastically at room temperature to the curved form. When the temperature is elevated, the curved form shifts back to the straight form, and remains in that form when the temperature is lowered to room temperature. If no further deformation is introduced, the straight form remains invariant during any subsequent thermal cycling.

a) Initial shape at room temperature

b) Deformed shape at room temperature

c) Shape at elevated temperature

d) Shape after cooling to room temperature Fig. 6.22: One Way Shape Memory Behavior The ability to return to its initial shape when heated is due to a phase transformation from martensite at room temperature to austenite at elevated temperature (Wayman et al., 1972). Inelastic deformation introduced during the martensite phase is eliminated when the state passes over to the austenite phase. The phase transitions are illustrated in Fig 6.23: As and Af define the temperatures for the start and finish of the transition from martensite to austenite for the case when the material is being heated; the corresponding temperatures for the cooling case are Ms and Mf. When T is greater than Af, the phase is austenite but it is possible to convert it back to martensite by applying stress. The quantity Md is the temperature beyond which austenite cannot be converted to martensite by

454 Chapter 6: Introduction to Active Structural Motion Control stress, i.e., the phase remains austenite for arbitrary applied stress.

As

Volume fraction martensite

Mf 1

0.5

0 Ms

Af

Md

Temperature, T Fig. 6.23: Martensitic transformation on cooling and heating The stress-strain behavior is strongly dependent on temperature. Figure 6.24 shows the limiting stress-strain curves for Nitinol, a nickel-titanium alloy (Jackson et al, 1972). For T<Mf, the lowest temperature for the fully martensitic phase, the material behaves like a typical ductile metal. Yielding occurs at about 150 MPa, and inelastic deformation is introduced. The behavior for T>Md is elastic up to 650 MPa, and the modulus, E1, is about 4 times larger than the initial value, E0. Between these limiting temperatures, there is a graduation in behavior between fully ductile to fully elastic.

6.4 Actuator Technologies 455

σ (MPa)

700 c’ 600

T > Md

500 T < Mf c’’

400 σ∗

b’

b

300 E1 200

100 E0

E0 a 0

2

4

c

6

8

10

ε (%)

Fig. 6.24: Effect of temperature on stress strain behavior of Nitinol (Jackson et al., 1972) Referring to Fig 6.24 the one way effect can be explained by tracking the response as a stress, σ * , is applied and then removed at T<Mf. This action produces the path a-b-c. Increasing the temperature at this time to T>Md shifts c

456 Chapter 6: Introduction to Active Structural Motion Control back to a since the behavior for T>Md is elastic. Further cycling of the temperature with no stress applied will not cause the point to shift from a. A different type of response is obtained when the stress is held constant and the temperature is cycled a number of times between Mf and Md. In this case, the path is b-b’-b. The effect of thermal cycling is to introduce 2 way shape memory. Repeating the scenario described above, starting at b and removing the stress shifts the b position to c. Then, increasing the temperature causes the position to shift to a. However, when the temperature is now lowered to Mf, the position shifts back to point c instead of remaining at a. Further thermal cycling at no stress results in the deformation switching between a and c. At high temperature, the material remembers the initial state; at low temperature it remembers the deformed state. Figure 6.25 is a modified version of Fig 6.22 which shows this behavioral mode. Applying the stress and subsequent thermal cycling trains the material to remember 2 shapes (Fremond et al, 1996).

a) Deformed shape at low temperature

b) Initial shape at elevated temperature Fig. 6.25: Two-way shape memory behavior The two-way shape memory behavior provides the basis for force actuation. If a trained shape memory alloy is restrained at low temperature so that it cannot deform, a force is generated when the alloy is heated since it wants to return to its initial undeformed shape. Referring back to the stress-strain plots in Fig 6.24, suppose the position at low temperature is point c. When heat is applied, the behavior is governed by the curve for T>Md. The material reacts as if it were subjected to the positive strain a-c, and the position jumps to c’. This behavior is “ideal”; the actual position is lower, such as point c’’, but the induced stress is still a significant value, on the order of 300 to 400 MPa.

6.4 Actuator Technologies 457 Nitinol alloys in the form of small diameter wires ( ≈ 0.4mm ) are used to assemble a force actuator. Heating is applied by passing an electric current through the wire. This process limits the response time to seconds vs. milliseconds for piezoelectric materials. Another limitation is the material cost; a typical price is on the order of $300/kg. Most of the applications of shape memory actuators are for products requiring low capacity, and where cost and response time are not critical issues. • Controllable fluids Controllable fluids are characterized by their ability to change from a fluid to a semi-solid in milliseconds when subjected to an electric or a magnetic field. This effect was identified by Winslow in 1949. Two materials belonging to this category are electrorheological (ER) and magnetorheological (MR) fluids. In their initial state, they behave as viscous fluids. Application of the field (electric or magnetic) introduces an additional plastic solid type behavior mode, and the response is now a combination of plastic and viscous actions. Figure 6.26 illustrates this transformation for the case where the material is idealized as a Bingham solid, which is defined as an ideal plastic solid in parallel with a linear viscous fluid. The stress-strain relation for a Bingham solid subjected to shear deformation has the form: τ = τ y sgn ( γ˙ ) + ηγ˙

(6.43)

where τ y denotes the yield stress and η is the viscosity. Experimental results show that the yield stress increases significantly with the field strength, f, up to a limiting value τ max . However, the viscosity is essentially constant (Carlson, J.D. et al, 1995). Therefore, it is reasonable to take η equal to a constant, η o , and interpret the material behavior as a combination of variable coulomb hysteretic damping and constant linear viscous damping. The equivalent linear material viscosity, η eq , is dependent on the amplitude and frequency of the shear deformation. For the case of periodic excitation, η eq is given by 4τ y (6.44) η eq = η o + ----------πΩγˆ where γˆ is the shear strain amplitude, and Ω is the excitation frequency.

458 Chapter 6: Introduction to Active Structural Motion Control

Saturation limit τ

f 2 > f1 η

f1

τy,max τy

f=0

γ˙

f= field strength measure

Fig. 6.26: Effect of electric and magnetic fields on the stress-strain relationship Devices containing controllable fluids can be used as either variable dampers or as semi-active force actuators. Figure 6.27 shows a schematic view of a prototype developed by the Lord Corporation (Lord Corp., www.lordcorp.com). The main cylinder houses the piston, the MR fluid, and the magnetic circuit. A small electromagnet is embedded in the piston head and supplied with current that generates the magnetic field across the annular orifice. Typical small scale versions have a force capacity of about 3 kN, millisecond response, and draw about 10 watts of power. A large scale version with capacity of 200 kN and millisecond response has also been produced (Spencer, B.F. et al, 1998).

6.4 Actuator Technologies 459

Fig. 6.27: Rheonetics SD-1000-2 MR damper (Pang, L. et al, 1998) Experimental results for quasi-static loading applied to the small scale version are plotted in Figure 6.28. These results show that the Bingham solid idealization is a reasonable approximation for the actual behavior, and also that the plastic yield force is limited by saturation of the fluid. For this device, saturation occurs in the vicinity of 1.5 amps. When used as a force actuator, the amperage is adjusted such that, for the observed value of velocity, the desired

460 Chapter 6: Introduction to Active Structural Motion Control force magnitude is generated. A controllable fluid type device is more effective than a variable orifice damper since the yield force is the primary component, and this force is independent of velocity. The low external power requirement, rapid response, and the potential for large capacity are attractive features. With further development, MR based actuators should be competitive solutions for large scale civil structures.

Fig. 6.28: Quasi-steady force velocity response for Rheonetics SD-1000-2 MR damper (Pang, L. et al., 1998)

6.5 Examples of existing large scale active structural control systems Kajima Corporation has pioneered the research, design, experimentation, and implementation of active control of large scale building structures (Sakamoto et al. 1994). Their work has been concerned with the following active motion control schemes: active mass driver (AMD), active variable stiffness (AVS), and hybrid mass damper (HMD). These schemes have been implemented in the set of buildings listed in Table 6.1. Other Japanese organizations such as Shimizu Corp., Takenaka Corp., and Mitsubishi have also carried out substantial research and

6.5 Examples of Existing Large Scale Active Structural Control Systems 461 implementation in the field of active structural control. In the USA, various government organizations such as NSF, ARO, AFOSR, ONR are funding research on active control of structures. A description of some of the implementations is given below. A detailed assessment of the performance of these structures during the earthquakes and strong winds that they have already encountered is found in Sakamoto et al. (1994). Table 6.1: Implementations of active and hybrid control systems in buildings Kyobashi Seiwa

KaTRI No.21

Completion date

Aug. 1989

Nov. 1990

July 1993

April 1994

Jan. 1995

Number of floors

11+1(BG)

3+2(BG)

14+2(BG)

52+5(BG)

29+3(BG)

Building height (m)

33.10

16.30

68.00

232.60

144.45

Total floor area (m2)

423.37

465.00

4,928.30

264,140.91

30,369.66

37.32

150.00

324.15

4,523.54

1,072.36

Control system

AMD

AVS

HMD

HMD

HMD

Type of disturbance

ME,SW

LE

ME,SW

ME,SW

ME,SW

2)

Typical floor area (m

Ando Shinjuku Nishikicho Park

Dowa Kasai Phoenix

Name of Building

(by Kajima Corporation) Note:

BG: Below ground level. ME: Moderate earthquake. LE: Large earthquake. SW: Strong wind.

AMD in Kyobashi Seiwa Building (Kajima 91-63E, 1991a) The Kyobashi Seiwa Building (see Fig 6.29) is a very slender building with a width of only 4m, a length of 12m, and a height of 33m (11 floors). It is constructed of rigidly connected steel frames consisting of box columns and H-shaped beams. The total structural weight is about 400 metric tons (1 metric ton equals 104 newtons). An active mass driver (AMD) system is installed on the top floor. The objective of the AMD is to reduce the maximum lateral response associated with 2 frequent earthquakes (i.e. peak ground acceleration of 10cm ⁄ s ) and strong winds (i.e. maximum speed of 20m ⁄ s ) to about one third of the uncontrolled value.

462 Chapter 6: Introduction to Active Structural Motion Control

a. AMD System installation

b.Schematic of AMD Fig. 6.29: AMD for Kyobashi Seiwa Building

6.5 Examples of Existing Large Scale Active Structural Control Systems 463 The AMD system is illustrated in Fig 6.29(b). The basic components are: • Sensors installed at several locations including: ground level, midheight, and the roof level to detect seismic motions and tremors at the ground level and in the building. • The control computer which analyzes each signal and issues a drive order. The control algorithm of the system is of the closed-loop control type where the active control force is determined through linear velocity feedback of the structure’s response. • Actuators which execute the control order and drive the masses. The hydraulic pressure source for the actuator consists of two pumps and an accumulator. One pump is small in comparison to the other; its function is to provide a minimum level of pressure continuously. The larger pump is activated when the earthquake occurs. • Two added masses driven by the actuators. Lateral vibration in the width direction is controlled with one mass (weighing about 4 metric tons, about 1% of the building weight) located at the center of the building. Torsional vibration is reduced with a second mass, weighing about 1 metric ton, located at one end of the structure. These masses are suspended by steel cables to reduce the frictional effects. The lag in the response is about 0.01s . AVS Control at Kajima Technical Research Institute (Kajima 91-65E, 1991b) An active variable stiffness (AVS) device is a non-resonant control type system which is designed to reduce the energy input to buildings from external excitation by continuously altering the building’s stiffness, based on the nature of the excitation, to avoid a resonant state. The major advantage of an AVS is that it can be activated and driven with a small amount of power, partially solving the energy requirement problem of active control. An AVS system was implemented in a steel structure, three stories high, weighing about 400 metric tons, located at the Kajima Technical Research Institute (see Fig 6.30). The lateral bracing system consists of steel braces placed in the transverse direction (gable side), and variable stiffness devices (VSD) installed between the brace tops and the lateral beams. These devices alter the building’s stiffness by shifting from the locked mode (engage: brace is effective) to the

464 Chapter 6: Introduction to Active Structural Motion Control unlocked mode (release: brace is ineffective). The variable stiffness range of the building was designed such that the natural resonant frequency is about 2.5Hz for the locked condition and about 1Hz for the unlocked condition. Auxiliary reinforcing braces are installed in the longitudinal direction to increase the building’s stiffness so that the control for the transverse direction is executed with minimum torsion.

a. View of building

6.5 Examples of Existing Large Scale Active Structural Control Systems 465

b.Schematic of locking device

c. Close up view of locking device Fig. 6.30: Kajima Technical Research Institute AVS System The AVS system consists of: • A measurement and control device consisting of an accelerometer placed on the first floor of the building which feeds the earthquake input into a motion analyzer. The analyzer consists of several special band-pass filters which approximate the response transfer characteristics to each stiffness type. Three stiffness types (eight stiffness types are possible) with mutually different resonant frequencies were selected for the building: all braces locked, all braces

466 Chapter 6: Introduction to Active Structural Motion Control unlocked, and only the braces in the bottom floor locked. Based on the filtered output, the control computer selects the instantaneous stiffness that yields the minimum building response. The control interval required to judge stiffness selection is 0.004s. To maximize the reliability and redundancy of the system, the control computer system is made up of a host computer which determines the operation of the entire system, and three controllers. As the host computer requires extremely fast data acquisition and processing, the real-time UNIX operating system was used. Personal computers utilizing MS-DOS were selected for the controllers which are placed between the host computer and each VSD. • The VSD switches over the connection condition between the brace and the beam in accordance with the signals from the control computer. A VSD consists of a two-ended-type enclosed hydraulic cylinder with a regulator valve inserted in the tube connecting the two cylinder chambers. The open/close function is controlled by oil movements, thus locking or unlocking the connection between the beam and the braces. Twenty watts of electric power per device is required for the valve function. The time needed to alter the lock/unlock condition is about 0.03s. • The emergency power source counteracts power blackout and enables the entire control system to continue to operate for at least 30 minutes even after sudden termination of the regular power supply. Moreover, in case of sudden power termination, the mechanism will cause the devices to automatically adopt a locked condition, thus increasing the building’s strength capacity. Under normal conditions, all the braces are kept in the locked condition. DUOX Active-Passive TMD in Ando Nishikicho Building (Kajima 93-82E, 1993) A combination of an AMD and a TMD, the DUOX system is installed in the Ando Nishikicho Building show in Fig 6.31. This structure consists of four main steel columns located at the corners, 14 stories above ground, and two basements. The building is located in an area of Tokyo that has mainly small, low-rise buildings, and consequently is susceptible to strong winds. The DUOX system was installed near the center of the top floor (i.e. at the building’s center of gravity) to control vibration in both horizontal directions.

6.5 Examples of Existing Large Scale Active Structural Control Systems 467

a. View of building

468 Chapter 6: Introduction to Active Structural Motion Control

b. Schematic of control system

c. Conceptual diagram of DUOX

6.5 Examples of Existing Large Scale Active Structural Control Systems 469

d. Bi-directional control Fig. 6.31: Nishikicho building motion control system

The DUOX system consists of: • Sensors installed in the basement and on the top floor of the building to monitor the ground acceleration and building motion. Sensors are also installed on the AMD and TMD to measure the motion of these devices. • A digital control computer which receives the feedback signals from the sensors, analyzes them, and determines the optimal control forces that will achieve the required control effect and also maintain the stroke of the AMD within the allowable range. • A passive TMD weighing 18 metric tons, about 0.8% of the weight of the above ground portion of the building (about 2,600 metric tons). The TMD is supported by laminated rubber bearings which provide the required stiffness. Oil dampers provide additional viscous damping in the system.

470 Chapter 6: Introduction to Active Structural Motion Control • Two AMD units, driven by AC servo motors and ball screws and weighing about 10% of the weight of the TMD (about 0.08% of the building’s weight). The units are mounted one on top of the other in a criss-cross manner to provide control in the two horizontal directions. The mass of the TMD moves out of phase with the building so that the building motion is always being retarded. The active mass is driven in the direction opposite to that of the TMD so as to magnify the motion of the TMD. When the building response falls within the allowable zone, the AMD operates to dampen out the motion of the TMD. ABS - 600 Ton Full-Scale Test Structure (Soong et al. 1991, Reinhorn et al. 1993) This 600 metric ton test structure, located in Tokyo, Japan, consists of a symmetric two-bay six-story building, 10m x 10m x 21.7m, constructed of rigidly connected steel frames composed of box columns and W-shaped beams with reinforced concrete slabs at each of the floors. The structure was intentionally designed to be a relatively flexible structure in order to simulate a typical high-rise building. The fundamental periods are 1s and 1.5s. The spectral properties of the structure obtained both analytically (with a shear beam model) and experimentally, were within 3% for all modes. A peak ground acceleration of 0.1g was taken as the design loading. The control forces were applied with an Active Bracing System (ABS). The ABS system consists of: • Velocity sensors, with an output range of ± 100cm ⁄ s , installed at the base, the first, the third, and the sixth floor, and an accelerometer, with 2 an output range of ± 1000cm ⁄ s , placed at the base of the structure. This data is fed into the controller and used as the state variables for feedback. • A PC-Limited 386 microcomputer facilitated with Analog-to-Digital (A/D) and Digital-to-Analog (D/A) boards is used for on-line computation. The control logic is implemented on an Intel 80386 25MHz processor equipped with an Intel 80387 25-MHz coprocessor. • Two algorithms based on classical linear optimal control theory were used. One control model works with the complete state vector and requires about 0.014s to process the data at each time point. The other

6.5 Examples of Existing Large Scale Active Structural Control Systems 471 control law is specialized for a reduced state vector corresponding to a 3DOF approximation which considers the translations of the first, third, and sixth floors as variables. The computation time for this algorithm is estimated at around 0.005s. • A biaxial active bracing system (ABS) consisting of a pair of hydraulic actuators (length 73.5cm, piston diameter 152.4mm, rod diameter 63.5mm) aligned in each of the principal directions and located at the first floor level. The actuators, monitored by two hydraulic servovalve controllers with auxiliary hydraulic accumulators, are capable of generating 685kN of control force in each direction with a time delay between the command signal and the actual actuator response of about 0.012s. The primary design parameters for the ABS are the required control force, the maximum displacement of the actuator ( ± 5cm ), and the actuator velocity (6.6cm/s). These parameters influence the actuator capacity, cylinder stroke, and flow rate requirement of the hydraulic servovalve. The servo-controlled system was designed to pump the hydraulic oil at a constant speed during control action, and the difference between the required and supplied oil flow is adjusted through the hydraulic accumulators. Each actuator is equipped with a linear variable displacement transducer (LVDT) having a range of ± 12mm , which is used to adjust the length of the brace via the servovalve loop. • Fail-safe features are incorporated in the hardware and software to protect the system in the event of a failure of one or more components of the system.

472

Introduction to Structural Motion Control

Problems 473

Problems Problem 6.1

b 0 = 20 kN/m

x L = 10 m

Consider the cantilever beam shown above. Assume transverse shear deformation is negligible. The desired state is constant curvature and u ( L ) ≤ L ⁄ 400 . a) Determine the bending rigidity distribution based on the equilibrium model, M(x) D B = ------------χ∗

(1)

b) Determine a “constant” value of D B using the least square procedure described below. The displacement function for the triangular loading is given by: b0 L 4 1 1 u 0 = ------------ x 2 1 – --- x + ------ x 3 = D B– 1 g ( x ) 6D B 2 20

(2)

where x = x ⁄ L . One forms the “displacement” error e ( x ) = u 0 ( x ) – u∗ ( x )

(3)

and the error functional, L

J =

∫0 e2dx

(4)

474 Chapter 6: Introduction to Active Structural Motion Control Let D B– 1 = f B . The value of f B is found by minimizing J with respect to f B . ∂J --------- = 0 ∂fB

(5)

Establish the expression for f B . c) Assume D B is specified. The displacement u 0 is now fixed. Suppose a vertical force, F , is now applied at x = L , and the magnitude of F can be adjusted. Determine a value for F by minimizing J with respect to F . The appropriate form of the error is e = u 0 + u c – u∗ , where F FL 3 x3 u c = ----------- x 2 – ----- = -------h ( x ) DB 2D B 3

(6)

Evaluate F for a representative range of D B . d) Let e i represent the displacement error at point x i . e i = u 0 ( x i ) + u c ( x i ) – u∗ ( x i )

(7)

Suppose one uses a “discrete” error functional consisting of N terms. 1 J = --2

N

∑ ei2

(8)

i=1

Derive the general expression for F corresponding to requiring J to be stationary. Evaluate F taking D B constant = 1/2 the value obtained in part b, N = 3 , and x i = 1/3, 2/3, and 1. Note: it is convenient to introduce matrix notation and express the error as: e =

e1 e2 = e + f F e3

(9)

1 J = --- e T e 2 e) Refer to part d. Consider 2 control forces, F 1 at x = L ⁄ 2 and F 2 at

Problems 475 x = L . Evaluate F 1 , F 2 for the conditions specified in part d. Compare the displacement solutions for part d and part e. f) Suppose one introduces weights for the error at the discrete points, x i , and also introduces a cost associated with the control force F applied at x = L. The expanded functional has the form, 1 J = --2

N

∑

1 q i e i2 + --- rF 2 2

(10)

i=1

where q i is the weight for e i and r is the cost weight for F . Increasing r places more weight on F vs. e . Similarly, increasing q i places more weight on e i . Derive the expression for F corresponding to requiring J to be stationary. Rework part d using this formulation, taking q 1 = q 2 = q 3 = 1 and r = r∗ ⁄ D B2 . Investigate the variation of F with r∗ .

Problem 6.2

b0

u

x L Consider a cantilever beam subjected to a uniform loading. Take D B = EI as constant and neglect transverse shear deformation. The deflected shape is given by: b0 L 4 u = ------------ [ x 2 ( 6 – 4x + x 2 ) ] 24EI

476 Chapter 6: Introduction to Active Structural Motion Control where x = x ⁄ L . Suppose the ‘‘desired’’ deflected shape is u = aLx 2 where a is a specified value. a)

M

x L Consider a concentrated moment, M , applied at x = L to ‘‘correct’’ the difference between the actual and desired deflected shape. How would you determine M ? b) m0

x L Suppose a uniform distributed moment, m 0 , is used as the control force system. How would you determine m 0 ?

Problems 477 Problem 6.3

v P

M

L

L

The structure show above consists of two single spans interconnected at the interior support with a force actuator system that can instantaneously apply a moment, M(t). The structure also has the capability to observe the transverse displacement at a limited number of points along both spans. Consider a force, P, which moves with velocity v along the span. Assume the velocity is sufficiently small such that the actual response of the structure can be approximated by the corresponding static response. Describe how you would establish the value of the actuator moment, M , such that the maximum transverse displacement at any time is less than a specified value, u*.

478

Introduction to Structural Motion Control

479

Chapter 7

Quasi-static Control Algorithms 7.1 Introduction to control algorithms Referring back to Fig 6.1, an active structural control system has 3 main components: i) a data acquisition system that collects observations on the excitation and response, ii) a controller that identifies the state of the structure and decides on a course of action and iii) a set of actuators that apply the actions specified by the controller. The decision process utilizes both information about how the structure responds to different inputs and optimization techniques to arrive at an “optimal” course of action. When this decision process is based on a specific procedure involving a set of prespecified operations, the process is said to be algorithmic, and the procedure is called a “control algorithm”. A non-adaptive control algorithm is time invariant, i.e., the procedure is not changed over the time period during which the structure is being controlled. Adaptive control algorithms have the ability to modify their decision making process over the time period, and can deal more effectively with unanticipated loadings. They also can upgrade their capabilities by incorporating a learning mechanism. This text is concerned primarily with time invariant control algorithms which are well established in the control literature. Adaptive control is an on-going research area which holds considerable promise but is not well defined at this time. A brief discussion is included here to provide an introduction to the topic. The topic addressed in this chapter is quasi-static control, i.e., where the

480 Chapter 7: Quasi-static Control Algorithms structural response to applied loading can be approximated as static response. Since time dependent effects are neglected, stiffness is the only quantity available for passive control. Active control combines stiffness with a set of pseudo-static control forces. The quasi-static case is useful for introducing fundamental concepts such as observability, controllability, and optimal control. Both continuous and discrete physical systems are treated. The next chapter considers time- invariant dynamic feedback control of multi-degree-of freedom structural systems. A combination of stiffness, damping, and time dependent forces is used for motion control of dynamic systems. The state-space formulations of the governing equations for SDOF and MDOF systems are used to discuss stability, controllability, and observability aspects of dynamically controlled systems. Continuous and discrete forms of the linear quadratic regulator (LQR) control algorithm are derived, and examples illustrating their application to a set of shear beam type buildings are presented. The effect of time delay in the stability of LQR control, and several other linear control algorithms are also discussed.

7.2 Active prestressing of a simply supported beam Passive prestressing The concept of introducing an initial stress in a structure to offset the stress produced by the design loading is known as prestressing. This strategy has been used for over 60 years to improve the performance of concrete structures, particularly beams. The approach is actually a form of quasi-static control, where the variables being controlled are the stresses. Figure 7.1 illustrates prestressing of a single span beam with a single cable. When the cable shape is parabolic, the tension introduced in the cable creates an “upward” uniform loading, wo, that is related to the tension by wo L 2 (7.1) ------------- = T 8d The initial moment distribution is parabolic, and the moment is negative according to the conventional notation.

7.2 Active Prestressing of a Simple Supported Beam 481

d cable L

d = distance between top and bottom location of the cable

wo

(-)

woL2/8

M+

P

PL/4

(+) M+ Fig. 7.1: Passive prestressing scheme

Suppose the design loading is a concentrated force that can act at any point on the span. The maximum positive moment due to the force occurs when the force acts at mid-span, and the resultant positive moment at mid-span is given by 2 L PL w o L (7.2) M --- = ------- – ------------ 2 8 4 The initial mid-span moment is negative and equal to wo L 2 L M --- = – ------------ 2 8

(7.3)

If the prestress level is selected such that wo L 2 1 PL ------------- = --- ------- = M * 2 4 8 which requires M* T = -------- ≡ T * d

(7.4)

(7.5)

482 Chapter 7: Quasi-static Control Algorithms then the maximum positive and negative moments are equal. The cross section can now be proportioned for M * , which is 1/2 the design moment corresponding to the case of no prestress. This reduction is the optimal value; taking T > T * will increase the initial moment beyond M * and result in the cross-section being controlled by the initial prestress. The limitation of this approach is the need to apply the total prestress loading prior to the application of the actual loading. Since the tension is not adjusted while the loading is being applied, the scheme can be viewed as a form of passive control. The best result that can be obtained with prestressing for this example is equal design moment values for the unloaded and loaded states. Active prestressing Suppose the cable tension can be adjusted at any time. The equivalent uniform upward loading due to the cable action can now be considered to be an active loading. Deforming w a as the equivalent active loading and noting eqn (7.1), the loading is related to the “active” tension force, T(t ), by 8d w a = ------ T ( t ) L 2

(7.6)

The time history of T(t ) can be established using simple static equilibrium relations. Given the spatial distribution and time history of the loading, T(t ) is determined such that the maximum moment at any time is less than the design moment M * for the cross-section. When the applied loading is uniformly distributed, the moment distributions are similar in form. The expression for the net positive moment has the form 1 M ( x, t ) = ( w ( t ) – w a ( t ) ) --- ( Lx – x 2 ) 2

(7.7)

Enforcing the constraint on the maximum moment, which occurs at mid-span, L2 ( w ( t ) – w a ( t ) ) ------ ≤ M * 8

(7.8)

results in the following control algorithm, 8 1. w a ( t ) = 0 for w ( t ) ≤ ------ M* ≡ w s L2 2. w a ( t ) = w ( t ) – w s for w ( t ) > w s

(7.9)

7.2 Active Prestressing of a Simple Supported Beam 483 No action needs to be taken until w reaches w s , since the maximum moment is less than M * . Above this load level, the active loading counteracts the difference between w ( t ) and w s . With active prestressing, the constraint imposed on the initial prestressing is eliminated. Theoretically, the total applied load can be carried by the active system for this example. This result is due to the fact that the moment distributions for the actual and active loadings have the same form. When these distributions are different, the effectiveness of active prestressing depends on the difference between the distributions. The following discussion addresses this point. Consider the case where the loading is a concentrated force that can act at any point on the span, and the prestressing action is provided by a single cable. The moment diagrams for the individual loadings are shown in Fig 7.2. When these distributions are combined, there is a local positive maximum at point B, the point of application of the load, and possibly also at another point, say C. Whether the second local negative maximum occurs depends on the level of prestressing. As w a is increased, the positive moment at B decreases, and the negative moment at C increases. For a given position of the loading, the control problem involves establishing whether w a can be selected such that the magnitudes of both local moment maxima are less than the prescribed target design value, M * , indicated in Fig 7.2. With passive prestressing, the optimal prestressing scheme produced a 50% reduction in the required design moment, i.e., it resulted in M * =0.5(PL/4). Whether an additional reduction can be achieved with active prestressing remains to be determined.

484 Chapter 7: Quasi-static Control Algorithms

P a

L –a

x ab P ----L

(+)

M+

M+

(-)

L2 – w a -----8

parabola + M*

M net

low prestress

A

B

–M*

C

D

M+

high prestress xc

Fig. 7.2: Active prestressing scheme for a concentrated load The net moment is given by Region A-B Lx – x 2 Px ( L – a ) M ( x, t ) = ------------------------ – w a ------------------ 2 L

(7.10)

7.2 Active Prestressing of a Simple Supported Beam 485 Region B-C-D Lx – x 2 Pa ( L – x ) M ( x, t ) = ------------------------ – w a ------------------ 2 L

(7.11)

Specializing eqn (7.10) for x = a leads to P 1 M + ≡ M ( a, t ) = a ( L – a ) --- – --- w a L 2

(7.12)

The location of the second maxima is established by differentiating eqn (7.11) with respect to x and setting the resulting expression equal to 0. This operation yields L Pa x c = --- + ---------2 Lw a The value for M at x = x c has the following form: Pa w a x c M - ≡ M ( x c, t ) = ( L – x c ) ------ – ------------ for x c < L L 2

(7.13)

(7.14)

When x c > L , the maximum negative moment occurs outside the span, and M – is taken as 0. The control algorithm is established by requiring M+ ≤ M *

(7.15)

M- ≤ M * for all a . Starting at a = 0 , no action is required as a increases until the moment due to the force P is equal to M * . The limiting value is denoted as a s , and determined with P ---a s ( L – a s ) = M * L

(7.16)

When a is greater than a s , the maximum positive moment, M + , is set equal to M* , P 1 a ( L – a ) --- – --- w a = M * L 2 Solving eqn (7.17) for w a leads to P M* w a = 2 --- – ------------------- L a ( L – a )

(7.17)

(7.18)

The last step involves checking whether for this value of w a , the maximum negative moment, M - , exceeds – M * .

486 Chapter 7: Quasi-static Control Algorithms It is convenient to work with dimensionless variables for x and M. x x = --L

M M = -------M*

(7.19)

where PL M * = f ------- 4

(7.20)

The factor, f, can be interpreted as the “reduction” due to prestressing. No prestress corresponds to f=1; passive prestress for this loading and prestressing scheme corresponds to f=0.5. Using this notation, a s is given by 1 a s = --- [ 1 – ( 1 – f ) 1 ⁄ 2 ] 2 The dimensionless form of eqn (7.18) is written as wa L 2 8d 8 2 ------------- ≡ -------- T = --- – ------------------- = g ( a, f ) * * f a ( 1 – a) M M Lastly, the dimensionless peak negative moment is expressed as M4a gx c -------- = M - = ( 1 – x c ) ------ – -------f 2 M*

(7.21)

(7.22)

(7.23)

where 1 4a x c = --- + -----2 fg

(7.24)

The peak negative moment is a function of the position coordinate, a , and the design moment reduction factor, f. Since M - must be less than 1 for all values of a between 0 and 0.5, the magnitude of f is constrained to be above a limiting value, fmin. Figure 7.3 shows plots of M - vs. a for a range of values f. For this case, the limiting value of f is equal to 0.26. Therefore, with active prestressing, the design moment can be reduced to 50% of the corresponding value for passive prestressing. The influence line for the cable tension required for the “optimal” active prestressing algorithm is plotted in Fig 7.4. Also plotted is the required tension corresponding to f=0.5, the optimal passive value. As expected, lowering the cross-sectional design moment results in an increase in the required cable tension. In order to arrive at an optimal design, the costs associated with the material (cross-section) and prestressing need to be considered.

7.2 Active Prestressing of a Simple Supported Beam 487

0

f=0.5

−0.2

f=0.4

−0.4

f=0.3

−0.6

M–

−0.8

−1

−1.2

f=0.2

−1.4

−1.6

−1.8

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.25

0.3

0.35

0.4

0.45

0.5

a⁄L

0

−0.2

−0.4

−0.6

−0.8

0.27 0.26 0.25

M– −1

−1.2

−1.4

−1.6

−1.8

0

0.05

0.1

0.15

0.2

a⁄L

Fig. 7.3: Influence lines for the peak negative moment

488 Chapter 7: Quasi-static Control Algorithms

6

5

f=0.26 4

8d ------------- T PL ⁄ 4

3

f=0.5 2

1

0

0

0.05

0.1

0.15

0.2

a⁄L

0.25

0.3

0.35

0.4

0.45

0.5

Fig. 7.4: Influence lines for the optimal cable tension Active prestressing with concentrated forces In this section, the use of concentrated forces to generate prestress moment fields is examined. The design loading is assumed to be a single concentrated force that can act anywhere along the span.

Example 7.1: A single force actuator Consider the structure shown in Fig (1). The active prestressing is provided by a single force, F, acting at mid-span. This loading produces 2 local moment maxima, M1 and M2. The moment at mid-span may be negative for certain combinations of a and F, and therefore it is necessary to check both M1 and M2 when selecting a value for the control force. Adopting the strategy discussed earlier, the control algorithm is based on the following requirements M1 ≤ M * M2 ≤ M *

(1)

7.2 Active Prestressing of a Simple Supported Beam 489 for all a where Pa ( L – a ) Fa M 1 = ----------------------- – -----L 2

(2)

Pa FL M 2 = ------ – ------2 4

(3)

and M * is the design moment for the cross-section.

a

P

L –a L/2

F a(L – a) P ------------------L

(+) M+

(-)

M+

FL – ------4 M1 M2

Figure 1

M+

490 Chapter 7: Quasi-static Control Algorithms Shifting to dimensionless variables, a a = --L PL M * = f ------4 Mi M i = -------M* F F = --P

(4)

transforms the equations to 4 aF M 1 = --- a ( 1 – a ) – -----f 2

(5)

1 M 2 = --- [ 2a – F ] f

(6)

No action is required for a ≤ a s where a s is defined by eqn (7.21). 4 a s ( 1 – a s ) --- ≡ 1 f

(7)

When a > a s , the force is selected such that M 1 ≡ 1 . f F = 2 1 – a – ------ 4a

(8)

The corresponding expression for M 2 is 4a – 2 1 M 2 = --------------- + -----f 2a

(9)

Figures 2 and 3 show the variation of M 2 and F with a and f. The limiting value of f for active prestressing is 0.345; when f>0.345, the negative moment at midspan is greater than the design moment, M * . For passive prestressing, the optimal solution is f=0.5. Shifting from passive to active control results in an additional 30 percent reduction in the allowable design moment.

7.2 Active Prestressing of a Simple Supported Beam 491

1

0.5

0.5

0

0.45 0.4

−0.5

0.35 −1 0.3 −1.5

M2

0.25

−2

−2.5

0.2

−3

−3.5

−4

0

0.05

0.1

0.15

0.2

0.25

a Figure 2

0.3

0.35

0.4

0.45

0.5

1.4

1.2 0.2 0.25

1

0.3 0.35

0.8

0.4

F

0.45

0.6

0.5 0.4

0.2

0

0

0.05

0.1

0.15

0.2

0.25

a Figure 3

0.3

0.35

0.4

0.45

0.5

492 Chapter 7: Quasi-static Control Algorithms

Example 7.2: Two force actuators The previous example showed that the effectiveness of a prestressing scheme depends on the difference between the moment distributions for the applied loading and the prestressing forces. In the case of a single control force applied at mid-span, the limiting condition occurs when the applied load is near the end support, where the difference in the moment distributions is a maximum. Increasing the number of control forces provides the capability to modify the “shape” of the “prestress” moment distribution to conform better with the applied moment field, and therefore increase the amount of prestressing that can be applied. This example illustrates the use of self-equilibrating control force systems which provide the maximum flexibility for adjusting the moment field. Consider the self-equilibrating force system shown in Fig. 1. This force system produces a bilinear moment field which is local, i.e., confined to the loaded region. Therefore, perturbing the control force magnitude, F, has no effect outside of this region.

F/2

F l

F/2 l

M+ Fl – ----2 Figure 1 Applying a set of these self-equilibrating systems results in a piecewize linear moment distribution. Figure 2 illustrates the case of 2 force systems located immediately adjacent to each other. The corresponding moment field is defined in terms of 2 force parameters, F1 and F2.

7.2 Active Prestressing of a Simple Supported Beam 493 Region A-B M1 M = – -------- ( x – x A ) l

(1)

Region B-C M1 M2 M = – -------- ( l + x B – x ) – -------- ( x – x B ) l l

(2)

Region C-D M2 M = – -------- ( l + x c – x ) l

(3)

F1 l M 1 = -------2

(4)

where F2 l M 2 = -------2

F1

F1 -----2 A

l

F2 -----2 B

F1 -----2 l

F2 C

F2 -----2 l

D M+

M1 x

(-)

M2

Figure 2 Example 7.1 treated the case of a single actuator deployed on a simply supported beam subjected to a single concentrated force that can act at any point on the beam. Suppose the control force system now consists of 2 local moment fields centered at the third points of the span. Figure 3 shows the 2 loading scenarios for this example. The moments at B,C, and D corresponding to the different loading scenarios are:

494 Chapter 7: Quasi-static Control Algorithms Region A-B (Fig 3a): M p = PL ( a ( 1 – a ) ) – 3aM 1

(5)

2 M B = --- PLa – M 1 3

(6)

1 M C = --- PLa – M 2 3

(7)

Region B-C (Fig 3b): M p = PL ( a ( 1 – a ) ) – M 1 ( 2 – 3a ) – M 2 ( 3a – 1 )

(8)

1 M B = --- PL ( 1 – a ) – M 1 3

(9)

1 M C = --- PLa – M 2 3

(10)

a = aL

P (+) B (-)

A

C M2

M1 MP

D

MB

MC M+

L/3

L/3

Figure 3a

L/3

7.2 Active Prestressing of a Simple Supported Beam 495

P

a = aL

(+) B (-)

A M1 MB

C

D

M2 MP

MC M+

L/3

L/3

L/3

Figure 3b Let M * represent the design moment. Expressing M * as a fraction of the maximum moment for the case where there is no prestressing, PL M * = f ------4

(11)

and working with dimensionless moments, M( ) M ( ) = -----------M*

(12)

transforms eqns (5) thru (10) to the following: Region A-B 4 M p = --- ( a ( 1 – a ) ) – 3aM 1 f

(13)

8 M B = ------a – M 1 3f

(14)

4 M C = ------a – M 2 3f

(15)

496 Chapter 7: Quasi-static Control Algorithms Region B-C 4 M p = --- a ( 1 – a ) – M 1 ( 2 – 3a ) – M 2 ( 3a – 1 ) f

(16)

4 M B = ------ ( 1 – a ) – M 1 3f

(17)

4 M C = ------a – M 2 3f

(18)

The control objective for this example is to limit the peak value of each of the dimensionless moment variables to be less than unity. MP ≤ 1

MB ≤ 1

MC ≤ 1

1 0 ≤ a ≤ --2

(19)

Since there are 3 constraints and only 2 control parameters, the problem is overconstrained. The strategy followed here is based on determining M 1 and M 2 using 2 of the constraints in eqn (19), and then adjusting f such that the third constraint is also satisfied. Figure 4 shows the variation in the moment measures with a , the coordinate defining the position of the load, corresponding to the following choice of constraints: Region A-B M P = +1 M C = +1

(20)

MB < 1 Region B-C M P = +1 M C = +1 – 4 ( 3a – 1 )

(21)

MB < 1 The minimum value of f is equal to 0.228. This value is controlled by the constraint on M B in the region A-B. For the case of a single actuator applied at

7.2 Active Prestressing of a Simple Supported Beam 497 mid-span, the optimum value for f was found to be 0.345. Applying 2 force actuators leads to an additional reduction in the “permissible” design moment. 4

3.5

3

f = 0.228 2.5

M1

2

1.5

1

0.5

0

0

0.05

0.1

0.15

0.2

a 0.25 Figure 4a

0.3

0.35

0.4

0.45

0.5

498 Chapter 7: Quasi-static Control Algorithms

4

3.5

3

2.5

M2

2

1.5

f = 0.228 1

0.5

0

0

0.05

0.1

0.15

0.2

0.25

a Figure 4b

0.3

0.35

0.4

0.45

0.5

0.3

0.35

0.4

0.45

0.5

1

0.5

0

MB −0.5

f = 0.228 −1

−1.5

0

0.05

0.1

0.15

0.2

a 0.25 Figure 4c

7.2 Active Prestressing of a Simple Supported Beam 499

1

0.8

0.6

f = 0.228

0.4

0.2

MC

0

−0.2

−0.4

−0.6

−0.8

−1

0

0.05

0.1

0.15

0.2

0.25

a Figure 4d

0.3

0.35

0.4

0.45

0.5

A general active prestressing methodology The discussion to this point has been concerned with a specific loading and a specific prestressing scheme. In what follows, a general methodology for dealing with the combination of an arbitrary design loading and prestressing scheme applied to a simply supported beam is described, and the control algorithm corresponding to a particular choice of error measure is formulated. This methodology is also applicable for displacement control which is discussed in the following section. Let M d ( x ) denote the moment due to the design loading, M c ( x ) the moment generated by the prestress system, and M(x) the net moment. By definition, M ( x ) = Md ( x ) + Mc ( x )

(7.25)

The design objective is to limit the magnitude of M ( x ) to be less than M * , the design moment for the cross-section.

500 Chapter 7: Quasi-static Control Algorithms Md + Mc ≤ M *

0≤x≤L

(7.26)

Equation (7.26) imposes a constraint on the magnitude of M c . – Md – M * ≤ Mc ≤ M * – Md

0≤x≤L

(7.27)

These limits establish the lower and upper bounds for M c ( x ) . Given M d ( x ) , one generates the limiting boundaries and then decides on a “target” distribution for Mc ( x ) . 3 Md 1 -------- 0 M* -1 -2

(a)

upper bound Mc 1 -------- 0 M* -1 -2

a

b

c

e

d

(b)

lower bound

Permissible region -4 1 M c* -------- 0 M * -1 -2

a

b

c

d

e

(c)

Optimum solution

Fig. 7.5: Limiting prestress moment fields Figure (7.5) illustrates the process of establishing the desired distribution for the control moment. The curves shown in Fig (7.5b) correspond to M c = – M d ± M * ; allowable values of M c are defined by the shaded area. The distribution corresponding to selecting the minimum allowable value of M c ( x ) at each x is plotted in Fig (7.5c). Assuming magnitude is the dominant measure,

7.2 Active Prestressing of a Simple Supported Beam 501 this distribution represents the optimal “target” prestress moment field. Let M c* ( x ) denote the desired “target” prestress moment field. Suppose the actual prestress moment field is a linear combination of r individual fields, r

Mc ( x ) =

∑

m jψ j(x)

(7.28)

j=1

where m j are moment amplitude parameters and ψ j ( x ) are dimensionless functions. The error associated with a specific choice of moment parameters is represented by the difference, e ( x ) , e ( x ) = M c* ( x ) – M c ( x )

(7.29)

Ideally, one wants e ( x ) = 0 for 0 ≤ x ≤ L . However this goal cannot be achieved when M c* is an arbitrary function, and it is necessary to work with an approximate error condition established using collocation, the least square method, or some other weighted residual scheme. The least square method is based on taking the integral of ( e ( x ) ) 2 as a measure of the accuracy of the approximation represented by eqn (7.28). This integral is denoted as J. 1 J = --2

L

∫

e 2 dx

(7.30)

0

In general, J is a function of the r moment parameters. Equations for these parameters are generated by requiring J to be stationary. ∂J = 0 ∂ mi

i = 1, 2, …, r

(7.31)

Expanding eqn (7.31) results in the following linear matrix equation, (7.32)

am = b where a, b, m are r’th order matrices, and the elements of a and b are: a ij =

∫

L

Ψ i ( x )Ψ j ( x ) dx

0

bi =

i, j = 1, 2…, r

L

∫

Ψ i ( x )M c* ( x ) dx 0

(7.33)

(7.34)

Given M c* ( x ) ,one determines b and then solves for m . This solution produces the least value for J, for a particular set of Ψ ‘s. A sense of convergence can be

502 Chapter 7: Quasi-static Control Algorithms obtained by expanding the set of basis functions, and comparing the corresponding values of J. It should be noted that the exact condition, e(x)=0 for 0<x<1, is generally not satisfied by eqn (7.32). This formulation works with continuous functions, and requires the evaluation of a set of integrals. It is more convenient to work with vectors rather than functions, since the computation reduces to matrix operations. Suppose the moment is monitored at n points within the interval 0 ≤ x ≤ L . The desired prestress moment vector is of order n × 1. M c* = { M c*, 1 , M c*, 2 , …, M c*, n }

(7.35)

Evaluating eqn (7.28) at these observation points leads to M c = Ψm

(7.36)

where Ψ is of order n × r . The error vector is taken to be the difference between M c* and M c , e = M c* – M c = M c* – Ψm

(7.37)

When r = n and the individual prestress moment fields are linearly independent, Ψ is non-singular and it is possible to determine an m that exactly satisfies e = 0 . m = Ψ – 1 M c*

(7.38)

When r ≠ n , a least square procedure can be used to establish an approximate solution for e = 0 . The error measure is taken as the norm of e . 1 J = --- e T e = J ( m ) 2

(7.39)

Requiring J to be stationary with respect to m leads to an equation having the same form as eqn (7.32), with a and b now given by a = ΨT Ψ

(7.40)

b = Ψ T M c*

(7.41)

The following example illustrates the application of the discrete formulation.

7.2 Active Prestressing of a Simple Supported Beam 503

Example 7.3: Multiple actuators Consider the design moment field shown in Fig (1). The longitudinal axis is discretized with 10 equal segments, resulting in 11 (n=11) observation points. Applying the criteria defined by eqn (7.27), and taking M * = 1 leads to the bounding curves for M c* plotted in Fig (2). The problem now consists of generating a prestress moment distribution which lies between these bounds. Suppose 4 self-equilibrating force system (r=4) that produce bilinear moment fields are applied at equally spaced interior points. The corresponding Ψ functions are shown in Fig (3a) and the typical field is plotted in Fig (3b). Since the prestress moment field is defined in terms of the moments at only 4 fixed points (3,5,7,9), and there are 9 interior points, the solution for the case of an arbitrary target distribution will be approximate. Various solutions for a particular target distribution are plotted in Fig (4). Curve (1) corresponds to taking m 1 = m 2 = – 1, m 3 = 0 , and m 4 = 1. Curve (2) is the least square solution for the 4 actuator system defined in Fig (3a). Curve (3) is based on the 7 actuator system defined in Fig (3c). The 3 additional actuators applied at points 2,4, and 6 eliminate the error up to point 6. Incorporating 2 more actuators at points 8 and 10 would completely eliminate the error associated with the 4 actuator system. It produces a moment field that is fully contained within the allowable zone and has the lowest cost as measured by the sum of the actuator moment magnitudes, Σm i . Md

3 2 1 1

2

3

4

5

6

7

8

9

-1 -2 Figure 1: Moment due to design loading

10

11

504 Chapter 7: Quasi-static Control Algorithms

M c* 3

upper bound

M* = 1

2 1 1

2

3

4

5

6

7

8

9

10

11

-1 -2

lower bound

-3 -4 Figure 2: Upper and lower bounds on the prestress moment field

7.2 Active Prestressing of a Simple Supported Beam 505 1

1

2

3

4

5

6

7

8

10

9

11

Ψ1

-1 1

1

2

3

4

5

6

7

8

9

10

11

1

2

3

4

5

6

7

8

9

10

11

2

3

4

5

6

7

8

9

10

11

Ψ2

-1 1 Ψ3

-1 1

1 -1

(a) r=4

1

1

5

3

m5 m1 2

3

m3

m2

m1

m6

4

m4

7 (b) r=4

m3 m2 m 7 5 7 6 (c) r=7

11

9

m4 8

9

10

11

Figure 3: Prestress moment fields for 4 and 7 actuators

Ψ4

506 Chapter 7: Quasi-static Control Algorithms

1

0.5

0

3

−0.5

2 −1

Target distribution

1

4 actuators (1)

−1.5

4 actuators (2) o...o... 7 actuators (3)

−2

−2.5

1

2

3

4

5

6

7

8

9

10

11

Figure 4: Prestress moment fields

7.3 Quasi-static displacement control of beams The previous section discussed how active prestressing can be applied to limit the magnitude of the bending moment in a beam subjected to a quasi-static loading. In this section, a procedure for controlling the quasi-static displacement profile of a beam is described. Both prestress and displacement control are based on the same least square error minimization algorithm which is generalized later in the next section. Consider a beam subjected to a loading that produces the displacement field u o ( x ) .Suppose the desired displacement profile is u * ( x ) , and the difference

7.3 Quasi-static Displacement Control of Beams 507 between u o ( x ) and u * ( x ) is to be reduced by applying a set of control force systems. Let u c ( x ) denote the displacement due to the control forces. The resultant error in displacement is denoted by e ( x ) , and is defined as e ( x ) = uo ( x ) + uc – u * ( x )

(7.42)

Assuming u c ( x ) is a function of r control force parameters, u c ( x ) = u c ( x, F 1, F 2, …, F r )

(7.43)

the issue now is how to establish approximate values for the force parameters. The least square procedure works with the square of the displacement error summed up over the domain, in this case the length of the beam. Both continuous and discrete forms of the least square error function are considered here. Continuous least square formulation In the continuous case, J is taken as 1 J = --2

L

∫

e 2 dx

(7.44)

0

and u c ( x ) is expressed as a linear combination of prescribed functions and unknown scalar force parameters, r

uc ( x ) =

∑

Fi φi ( x )

(7.45)

i=1

Substituting for u c ( x ) and e transforms J to a quadratic form in F, 1 J = --2

r

∫

L

( uo –

u * ) 2 dx

+

0

∑∫

L

i=1

r

1 + --2

r

∑∑ ∫ j = 1i = 1

( u o – u * )φ i dx F i +

0

(7.46)

L 0

Introducing the following notation,

φ i φ j dx F i F j

508 Chapter 7: Quasi-static Control Algorithms F = { Fi } b = { bi } a = [ a ij ] L

∫

a ij =

φ i φ j dx

(7.47)

0

L

bi =

∫

( – u o + u * )φ i dx 0

L

c =

∫

( u o – u * ) 2 dx 0

reduces eqn (7.46) to a matrix expression, 1 1 J = --- c – b T F + --- F T aF 2 2

(7.48)

The first and second differentials of J are r

dJ =

∑ i=1

∂J dF = dF T ( – b + aF ) ∂ Fi i

r

d2 J

=

r

∑ ∑ j=1

∂ ∂F j

i=1

∂J dF dF j = dF T adF ∂ Fi i

(7.49)

(7.50)

Requiring J to be stationary is enforced by setting the coefficient of dF T equal to 0 in eqn (7.49), and leads to a set of r equations for the r force parameters. dJ = 0 for arbitrary dF

⇒

aF = b

(7.51)

The stationary point corresponds to an absolute minimum value when a is a positive definite matrix. This property exists when the displacement functions, Ψ i ( x ) , are linearly independent. Discrete least square formulation The discrete least square formulation is based on selecting a displacement

7.3 Quasi-static Displacement Control of Beams 509 error vector, e , and taking J as 1 J = --- e T e 2

(7.52)

One forms e by specifying n points along the longitudinal axis and evaluating the difference between the various displacement components at these points. Using matrix notation, e is expressed as e = U o – U * + ΦF

(7.53)

where U o and U * are displacement vectors of order n × 1, and Φ is of order n × r . Substituting for e in eqn (7.52) and expressing J in the same form as for the continuous case 1 1 J = --- e T e = c – b T F + --- F T aF 2 2

(7.54)

leads to the definition equations for a , b and c . b = φ T ( – Uo + U * )

(7.55)

a = ΦT Φ

(7.56)

c = ( Uo – U * ) T ( Uo – U * )

(7.57)

The stationary requirement for J is still aF = b. Since a is now generated by pre-multiplying Φ by its transpose, the properties of a depend on the relative size of n and r. If n > r, a will be nonsingular provided that the columns of Φ are linearly independent. This condition is equivalent to requiring the rank of Φ to be equal to the number of its columns. If the rank of Φ is less than r, say r - s, then s control forces do not contribute to the error and need to be deleted. If n < r, a will be singular and a special solution procedure needs to be applied. Strang(1993) contains a detailed discussion of the pseudo-inverse procedure for solving aF = b when a is singular. This procedure is executed by MATLAB with the statement, F = pinv ( a )*b. The following example illustrates different scenarios for n vs. r.

510 Chapter 7: Quasi-static Control Algorithms

Example 7.4: Cantilever beam - Least square algorithm Consider the cantilever beam shown in Fig (1). Introducing the dimensionless variable x x = --L

(1)

and assuming the bending rigidity, D B , is constant, the expression for the initial and desired displacement fields are written as: wL 4 u o = ---------- g ( x ) DB

(2)

1 g ( x ) = ------ [ x 2 ( 6 – 4x + x 2 ) ] 24

(3)

L u * = ---x 2 α

(4)

where α is a dimensionless parameter.

u

F1

F2

L/2

w x L

Figure 1 Suppose a single control force F 1 is applied at x = L. The corresponding displacement distribution is

7.3 Quasi-static Displacement Control of Beams 511 L3 x3 u c = φ 1 F 1 = ----------- x 2 – ----- F 1 3 2D B

(5)

Substituting for the various functions in eqn (7.47) results in a 11 F 1 = b 1

(6)

L7 a 11 = -------- ( 0.0262 ) D B2

(7)

L5 wL 8 b 1 = – ---------- ( 0.0102 ) + ------------ ( 0.0722 ) αD B D B2

(8)

where

Solving for F 1 leads to: DB F 1 = ---------- ( 2.7576 ) – wL ( 0.3911 ) L2α

(9)

The error function associated with the solution is: wL 4 L e ( x ) = ---------- f ( x ) + --- g ( x ) DB α wL 4 = ---------- { 0.0545x 2 – 0.1015x 3 + 0.417x 4 } + DB L + --- { 0.3789x 2 – 0.4596x 3 } α Table 1 contains the errors at locations x = 0.5 and x = 1.0 .

(10)

512 Chapter 7: Quasi-static Control Algorithms

x = 0.5

x = 0.5

x = 1.0

x = 1.0

Case

f

g

f

g

1. Continuous least square F 1

0.0035

0.0373

-0.0053

-0.0808

2. Discrete least square F 1 and 2 obs. points

0.0048

0.0570

-0.0015

-0.0178

3. Discrete least square F 2 and 2 obs. points

-0.0049

0.1292

0.0020

-0.0521

4. Discrete least square F 1, F 2 and 2 obs. points

0

0

0

0

5. Discrete least square F 1, F 2 and 1 obs. point

0.0043

0.0703

0

0

Table 1 Comparison of displacement errors To illustrate the discrete formulation, observation points at x = 0.5 and x = 1.0 are selected. For this case, n = 2 and r = 1. Using eqns (1), (3), and (5), the terms defining e are: 17 --------

Uo –

U*

1 L --4- = ---------- – --D B 1 α --1 8 wL 4 384

(11)

5 -----

L 3 48

Φ = ------- DB 1 --3 Substituting in eqns (7.55) and (7.56) results in

(12)

7.3 Quasi-static Displacement Control of Beams 513 L6 a = a 11 = -------- ( 0.1220 ) D B2

(13)

wL 7 L4 b = b 1 = ---------- ( – 0.0463 ) + ------------ ( 0.3594 ) αD B D B2

(14)

DB F 1 = ---------- ( 2.9467 ) – wL ( 0.3795 ) αL 2

(15)

and

This solution approaches the continuous solution, eqn (9), as the number of observation points is increased. For example, the solution corresponding to 4 observation points located at x = 0.25, 0.50, 0.75, and 1.0 is DB F 1 = ---------- ( 2.8685 ) – wL ( 0.3842 ) αL 2

(16)

The location of the point of application of the control force also has an influence on the solution. Considering a single force, F 2 , applied at x = 0.5 , and working with 2 observation points at x = 0.5 and 1.0 generates the following least square solution, DB F 2 = ---------- ( 9.0997 ) – wL ( 1.1807 ) αL 2

(17)

Comparison of the errors listed in Table 1 indicates that applying the force at x = 1.0 is more effective for this structure/loading combination. Applying both F 1 and F 2 , and taking 2 observation points results in Φ being of order 2x2. In this case, one can set e = 0 and solve ΦF = U * – U o

(18)

for F . The least square formulation produces the same solution for F when Φ is square. Evaluating eqn (18) results in

514 Chapter 7: Quasi-static Control Algorithms F1 – 0.1964 D B 5.1429 = wL + ---------2- F – 0.5714 2 αL – 6.8571

(19)

Suppose control forces are applied at x = 0.5 and 1.0, and only one observation point is selected. In this case, n = 1 and r = 2. Considering the point to be at x = 1.0 , the corresponding error term defined by eqn (7.53) has the following form: L3 wL 4 L e ( x = 1 ) = ----------- – --- + ----------3D B 8D B α

5 L 3 F1 ------ ------- 48 D B F 2

(20)

e = U o – U c + ΦF Expanding a and b , 1 L3 2 a = ----------5 1 3D ----B 16

1 5 L3 ------ = ---------- 3D 5 16 B -----16

5 -----16 5 5 ------ ------ 16 16

1 L3 wL 4 L b = ----------- 5 – ----------- + --- 3D ------ 8D B B α 16

(21)

(22)

shows that a is singular; the second row is 5/16 times the first row. The same relationship holds for the elements of b . Therefore, in this case, there is only one independent equation in the set, aF = b, generated by requiring e T e to be stationary. Another equation has to be established. The pseudo inverse solution procedure starts by setting e equal to 0. Noting eqn (20), e(x = 1) = 0

⇒

DB 3 5 F 1 + ------ F 2 = 3 ---------- – --- wL 16 αL 2 8

(23)

An additional equation relating F 1 and F 2 is generated by requiring the mean square sum of the force magnitudes to be a minimum, considering eqn (23) to be a constraint between F 1 and F 2 . The steps are:

7.3 Quasi-static Displacement Control of Beams 515 a) Form the objective function 1 I = --- ( F 12 + F 22 ) 2

(24)

b) Form the first differential dI = F 1 dF 1 + F 2 dF 2 = 0

(25)

c) Establish the constraint between the differentials Noting eqn (23), 5 dF 1 + ------ dF 2 = 0 16

(26)

d) Establish the additional equation Substituting for dF 2 in terms of dF 1 in eqn (25), F + – 16 ------ F dF = 0 1 5 2 1

(27)

and requiring eqn (27) to be satisfied for arbitrary dF 1 yields 16 F 1 = ------ F 2 5

(28)

e) Solve for the primary variable Using eqn (28), the solution follows form eqn (23). DB 3D B 3 1 F 1 = ----------------------------- ----------- – --- wL = ---------- ( 2.7331 ) – wL ( 0.3416 ) 5 2 αL 2 8 αL 2 1 + ---- 16

(29)

5 F 2 = ------ F 1 16 Table 1 lists the error at x = 0.5 . The error at x = 1.0 was taken to be zero at the start of the solution process. One can obtain the solution defined by eqn (29) with the MATLAB statement, F = pinv ( a )*b.

516 Chapter 7: Quasi-static Control Algorithms The previous example illustrated the use of concentrated forces to adjust the deflected shape of a cantilever beam subjected to a uniform transverse loading. The approach involved selecting a set of points at which the displacements are monitored, and a set of control forces. Considering the displacement field to be continuous corresponds to taking an infinite number of observation points. In order to apply the procedure for an arbitrary loading, a finite number of points needs to be specified. The control algorithm is established using the discrete least square formulation. A square nonsingular set of equations is obtained when the number of observation points, n, is taken to be greater or equal to the number of control forces, r. If n is less than r, the problem is underspecified in the sense that additional equations have to be established. In this case, a pseudo inverse procedure can be applied. Extended least square formulation The standard least square formulation works with a weighted displacement error, e ( x ) , which represents the difference between the actual and target displacement fields. No priority is assigned to a specific location, say e ( x i ) , and also, no limit is placed on the magnitude of the control forces. This approach can be generalized to include priorities by introducing weighting functions for e ( x ) and F i , and employing an extended form of the error functional, J. Considering first the continuous case, the error at x is weighted with a function, q(x), and the error measure is taken as the integral of qe 2 , 1 J ⇒ J' = --2

∫

L

q ( x )e 2 dx

(7.58)

0

Setting q(x) = constant assigns equal priority in the interval 0 ≤ x ≤ L . Taking q(x) as a set of n Dirac delta functions centered at x = x i , i=1, 2, ..., n converts eqn (7.58) to the discrete form and allows for different weighting factors, 1 J' = --2

n

∑

q i e i2 ( x i )

(7.59)

i=1

The magnitudes of the control forces influence J’ indirectly through e ( x ) . Selecting the force magnitudes such that J’ is stationary corresponds to forcing the weighted displacement error to be a minimum. There is a “cost” associated with a

7.3 Quasi-static Displacement Control of Beams 517 particular choice of the force magnitudes, and this cost is not reflected by the stationary solution for J’. Consequently, as the displacement error is driven toward zero, the magnitude and therefore the “cost” of the forces may become excessive. This situation can be avoided by combining J’ with a force cost function, and requiring the sum to be stationary. A convenient choice for the force cost function is the weighted quadratic form, 1 Force cos t fuction = --2

r

∑

r j F 2j

(7.60)

j=1

where r j is a weight associated with F j . In general, r j > 0 . Combining this term with J’ leads to the extended functional, J 1 . 1 J 1 = --2

∫

L o

1 qe 2 dx + --2

r

∑

r j F 2j

(7.61)

j=1

The discrete form of J 1 is taken as 1 1 J 1 = --- e T Qe + --- F T RF 2 2

(7.62)

where e is the displacement error vector, and Q, R are diagonal matrices with nonnegative elements, Q = [ q i δ ij ]

i, j = 1, 2, …, n

(7.63)

R = [ r k δ kl ]

k, l = 1, 2, …, r

(7.64)

Substituting for e using eqn (7.53), J 1 expands to 1 1 J 1 = --- c 1 – b 1T F + --- F T a 1 F 2 2

(7.65)

where b1 = Φ T Q ( – Uo + U * )

(7.66)

a 1 = Φ T QΦ + R

(7.67)

c1 = ( Uo –U * ) T Q ( Uo – U * )

(7.68)

518 Chapter 7: Quasi-static Control Algorithms Requiring J 1 to be stationary with respect to F leads to a set of r equations for the r force variables, (7.69)

a1 F = b1

Although the order is the same as for the conventional formulation that considered only the displacement error, a 1 differs from a in 2 ways. Firstly, the quadratic term, Φ T Φ , is replaced with a weighted form, Φ T QQ . Secondly, positive quantities are added to the diagonal elements. The latter operation ensures that a 1 is non-singular for all values of n vs. r. Holding Q fixed and increasing the magnitude of R places more emphasis on lowering the cost of the control forces. Consequently, the displacement error will increase. Decreasing R with respect to Q has the opposite effect on the displacement error. Ideally, one wants minimum displacement error and minimum force cost, but that goal is impossible. Generating solutions corresponding to a range of Q and R values provides information concerning the sensitivity of the 2 weighting measures which is useful for selecting design values. The following example illustrates this computation.

Example 7.5: Example 7.4 revisited Example 7.4 considered observation points at x = 0.5 and 1.0, and a single control force applied at x = 1.0 . The example is reworked here using the extended least square algorithm. Starting with the basic data contained in eqns (11) and (12) of example 7.4, 17 --------

1 – --- L + --- 4 U o – U * = ---------- DB 1 α -- –1 8 wL 4 384

5 - L 3 ----Φ = ----------- 16 3D B 1 and applying eqns (7.66), (7.67), and (7.69) results in

(1)

7.3 Quasi-static Displacement Control of Beams 519

L4 85 5 wL 7 1 b 1 = ---------- ------ – --------- q 1 – q 2 + --------------- ------ q 1 + q 2 3D α 64 D B2 24 768 B

(2)

L6 5 2 a 1 = ----------- ------ q 1 + q 2 + r 1 9D B2 16

(3)

F 1 = a 1– 1 b 1

(4)

Suppose the q’s are equal to a common value, q . In this case, the coefficients simplify to L 4 23 wL 7 853 b 1 = ---------- – --------------- q + ------------ ------ q D B α 64 D B2 18432

(5)

L 6 281 a 1 = -------- ------------ q + r 1 D B2 2304

(6)

The first term on the right hand side of eqn (6) is a measure of the priority for the displacement error; the second term reflects the importance of the force cost. Since q is present in both b 1 and a 1 the force depends only on the ratio of r 1 to q. To establish the appropriate magnitude for r 1 ⁄ q , r 1 and q can be scaled such that the 2 terms on the right hand side of eqn (6) have the same order of magnitude. One possible approach is to take r 1 = 1 and scale q. r1 = 1 D B2 2304 ------- ------------ q * q = 6 L 281

(7)

where q * is a measure of the relative importance of the force cost vs. the displacement error. Using eqn (7) the expression for F 1 takes the form DB wl ( – 0.3794 ) + ---------- ( 2.9466 ) q * L2α F 1 = -------------------------------------------------------------------------------1 + q*

(8)

Equal weighting for displacement and force as assigned when q * = 1. Increasing

520 Chapter 7: Quasi-static Control Algorithms q * beyond 1 places more emphasis on the displacement error, and the solution approaches the solution corresponding to no constraint on the control force magnitude. Taking q * ≈ 0 ignores the displacement error, and the procedure will minimize the functional corresponding to only the force cost. Consequently, the solution will be F = 0.

The scaling operation on r and q described in example 7.5 can be applied to the general formulation represented by eqns (7.66) thru (7.69) in the following manner. Firstly, R is taken such that all its elements are between 0 and 1. This matrix is denoted as R * . R = [ r *j δ ij ] ≡ R *

(7.70)

where r *j are individual force cost “relative” weighting factors that are now of order 1. Secondly, Q is scaled with a factor, q , and expressed as Q = qQ *

(7.71)

The magnitude of q can be chosen such that both of the terms comprising a 1 have essentially the same magnitude. Using these definitions, eqn (7.69) is transformed to ( q ( Φ T Q * Φ ) + R * )F = qΦ T Q * ( – U o + U * )

(7.72)

Taking q to be equal to the inverse of the maximum element in the matrix product, Φ T Φ , is a convenient choice since now the elements of Q * can be considered to be of order 1. Sensitivity can be assessed by varying the elements of Q * and R * . Obviously Q * cannot be a null matrix, whereas R * = 0 is a limiting case. Setting R * = I and perturbing the elements of Q * is a reasonable strategy for assessing how the solution varies as the priorities with respect to the displacement error and force cost are shifted. Increasing Q * places more emphasis on the displacement error.

7.4 Quasi-static Control of MDOF Systems 521

7.4 Quasi-static control of MDOF systems Introduction The governing equations for the linear static behavior of an n’th order system controlled with r forces are written as: (7.73)

KU = P + E f F

where U contains the n displacement variables, K is the system stiffness matrix, P is the prescribed load vector, F contains the r control forces, and E f defines the distribution of the r control forces in the load vector. For n DOF and r control forces, E f is of order n x r. Figure 7.6 illustrates the form of E f for a particular set of control forces applied to a discrete shear beam model. u4 u3 u2 u1

F1 F2 F2

0 Ef = 0 1 0

–1 1 0 0

Fig. 7.6: Control force distribution matrix Selecting a control force system involves 2 decisions: firstly, the nature and location of the control forces which establishes E f and secondly, the control algorithm which determines the magnitude of the control force parameters corresponding to the choice of E f . The decision on E f depends in turn on the objective of adding control which usually involves limiting the response of certain variables. Therefore, the first step is to identify the performance measures. Given the performance measures, a set of response quantities which can be “observed” and then used to establish the magnitudes of the performance measures are identified. Also, once the performance measures are defined, the equilibrium equations are applied to establish the relationship between the “performance”

522 Chapter 7: Quasi-static Control Algorithms and the control force measures. Lastly, a mathematical procedure such as the least square method is employed to generate a control algorithm which is used to establish a quantitative estimate of the control forces. Selection of measures By definition performance measures are those quantities whose magnitudes are to be limited by the application of control forces. In the previous sections dealing with beam type structures, bending moments and displacements were selected. For a low rise building modelled as a shear beam, limiting the transverse shear deformation at certain story locations is the objective for motion based design. Assuming there are np performance measures, the individual measures are denoted by v i ( i = 1, 2, …, n p ) and included in the vector v .

u3 h

3 u2

h

2

h

u1

γ 1 v = γ 2

1 θ

Fig. 7.7: Deformations as performance measures Figure 7.7 illustrates the selection process for a single truss modelled as a 3 DOF shear beam. The performance measures are taken to be the transverse shear strains in the first and second stories. They are related to the displacements by u1 u2 – u1 (7.74) v 2 = -----------------v 1 = -----h h The general matrix form of eqn (7.74) is written as: v = BU

(7.75)

7.4 Quasi-static Control of MDOF Systems 523 where B can be interpreted as a transformation matrix which converts displacements measures into performance measures. Observation measures are those quantities which are monitored and used to establish the magnitude of the performance measures. They may be either deformations or displacements. Let w i denote the i’th observation measure, n ob the total number of observations, and w the observation vector. The key decisions here are selecting the number, the nature (i.e., displacement, deformation,...), and the location of the observations such that the performance measures are uniquely defined. Suppose w * is a particular choice of observation variables. Assuming the behavior is linear, v associated with this choice of w can be expressed as v

w*

= B1 w *

(7.76)

One evaluates the i’th row of B 1 using the definition equation for v i . If all the elements in row i are equal to 0, v i cannot be determined by this choice of w. Similarly, if all entries in column j are 0, w j contributes nothing to v and must be deleted from w * . Observability is the term used to characterize the relationship between v and w. A monitoring scheme is said to be observable when the rank of B 1 is equal to n p , the number of performance measures. In general, a necessary condition is n ob ≥ n p ; the sufficient condition is for B 1 to have at least n p linearly independent columns. Various scenarios are illustrated by the following example.

Example 7.6: Illustrative examples of observability Consider the physical model defined by Fig 7.7. There are 2 performance measures, the shear strains in the first 2 stories. Various cases are considered below. Observability for this example requires the rank of B to be 2. Case 1.

w * = { u 2, u 3 }

Applying eqns (7.74) leads to 1 B 1 = --- 0 0 h 1 0

(1)

524 Chapter 7: Quasi-static Control Algorithms The second column contains all zero’s since u 3 has no effect on either v 1 or v 2 . Case 2. w * = { u 1 } 1 B 1 = --- 1 h –1

(2)

The rank is 1. The problem here is the incomplete determination of γ 2 , which requires u 2 as well as u 1 , Case 3. w * = { u 1, u 2 } 1 B 1 = --- 1 0 h –1 1

(3)

Here, r(B1)=2. Observability is satisfied. Case 4. w * = { ε 1, ε 2 } where ε 1 and ε 2 are the extensional strains in the diagonal elements located in the first and second stories. Noting the definition equations for the extensions, h e 1 = u 1 cos θ = -----------ε 1 sin θ h e 2 = ( u 2 – u 1 ) cos θ = -----------ε 2 sin θ

(4)

and substituting for u 1 and u 2 in eqn (7.74) results in 1 B 1 = ------------------------ 1 0 sin θ cos θ 0 1

(5)

In this case, r(B1)=2 for arbitrary finite θ > 0 .

The last group of variables to be defined is the set of control forces represented by the vector F. Once the location of the forces is chosen, E f is known, and the equilibrium equations can be solved to determine the displacements generated by applying F. Using eqn (7.73), U due to F is given by

7.4 Quasi-static Control of MDOF Systems 525 U

F

= K –1 E f F

(7.77)

Then, substituting for U in eqn (7.75) leads to a relation between v and F. v F = BU F = ( BK – 1 E f )F = C f F

(7.78)

The coefficient matrix, Cf, is of order n p × r , and defines the effect of the control forces on the performance measures. If the elements in the i’th row of Cf are all zero, the choice of control forces has no influence on the i’th performance measure, i.e., v i cannot be controlled by F. Similarly, if all the elements in the j’th column of Cf are zero, F j contributes nothing and should be deleted from F. In the usual control scenario, the number of force parameters is less than the number of performance parameters. Requiring r(Cf)=r for n p > r ensures that the columns of Cf are linearly independent and eliminates the situation where Cf has a zero column. Controllability is defined for an individual performance measure, and requires the corresponding row of Cf to have at least one non-zero element. The following example illustrates some choices of control force systems for example 7.6

Example 7.7: Illustrative examples of controllability Consider the shear beam model defined in Fig (7.7) and discussed in example 7.6. There are 2 performance measures, and therefore one would not normally take more than 2 control forces. The forms of Cf corresponding to various force systems are listed below. Case 3 has an excess of control forces. Case 4 has a deficiency with respect to controllability; v 1 cannot be controlled with this scheme.

526 Chapter 7: Quasi-static Control Algorithms Case 1. F1 0 Ef = 0 1

Cf =

1 ---------D T1 1 ---------D T2

Case 2. F1

F2

0 1 Ef = 0 0 1 0

Cf =

1 1 ---------- ---------D T1 D T1 1 ---------D T2

0

Case 3. F1 F3 F2

010 Ef = 0 0 1 100

1 1 1 ---------- ---------- ---------D T1 D T1 D T1 Cf = 1 1 ---------- 0 ---------D T2 D T2

7.4 Quasi-static Control of MDOF Systems 527 Case 4. F1 F1

F2

F2

0 –1 E f = –1 1 1 0

0

0 Cf = 1 0 ---------D T2

Least square control algorithms Suppose the system is subjected to its design loading, and the values of the performance parameters resulting from this loading are observed with a monitoring scheme. Let v o denote the “observed” response, and v * the desired response. The difference between these vectors is the target response for the force control system. Taking v c as the response due to the control forces, the net performance error vector is e = vo + vc – v *

(7.79)

The goal is to select F such that e is less than the allowable error. Considering the behavior to be linear, and noting eqn (7.78), e is related to F by e = vo + C f F – v *

(7.80)

Equation (7.80) represents n p equations in r unknowns; Cf is of order n p × r . When n p = r , there is a unique F which satisfies e = 0. For all other combinations of n p and r, the solution for F is not unique and also may not satisfy e = 0. The least square solution for the case where n p > r is generated by forming the error norm, 1 J = --- e T e 2

(7.81)

and requiring J to be stationary with respect to F. This operation leads to a set of r equations, which are written as:

528 Chapter 7: Quasi-static Control Algorithms

aF = b a = C Tf C f

(7.82)

b = C Tf ( v * – v o ) The coefficient matrix will be non-singular when Cf satisfies the controllability requirement, r(Cf)=r. Since the square error norm is used, the solution of eqn (7.82) does not result in e = 0.

Example 7.8: Example 7.7 revisited The least square control algorithm is applied to the model considered in example 7.7. The following values are specified for the performance measures, 0.007 vo = 0.004

0.005 v* = 0.005

(1)

and the transverse shear rigidity is assumed to be constant. Case 1. Using eqn (1) of example 7.7, eqn (7.82) reduces to 2 1 -------- F 1 = -------- [ ( v 1* – v 1, 0 ) + ( v 2* – v 2, 0 ) ] 2 D DT T

(2)

The solution for F 1 and the corresponding error vector are F 1 ≡ 0.0005D T 0.0015 e = – 0.0015

(3)

Case 2 Since the number of control forces is equal to the number of performance measures, solving eqn (7.82) is equivalent to solving eqn (7.80) with e set equal to

7.4 Quasi-static Control of MDOF Systems 529 0. This result follows by writing eqn (7.82) as C Tf ( C f F – v * + v o ) = 0

(4)

and noting that the force vector which satisfies C f F = v * – vo

(5)

also satisfies (4). The solution is F 1 = 0.001D T F 2 = – 0.003D T

(6)

e = 0 Case 3 In this case, there are 3 force parameters and only 2 performance variables. The problem is undetermined. Setting e = 0 in eqn (7.80) provides 2 equations relating F 1 , F 2 , and F 3

C f F = v * – vo

⇒

1 -------- ( F 1 + F 2 + F 3 ) = – 0.002 DT 1 -------- ( F 1 + F 3 ) = 0.001 DT

(7)

Taking F 3 as the “excess” force parameter, and solving for F 1 and F 2 leads to F 1 = – F 3 + 0.001D T F 2 = – 0.003D T

(8)

The additional equation is obtained by requiring the norm of F to be stationary with respect to F 3 . For this example, the norm is taken as the Euclidean norm, I. 1 1 I = --- F T F = --- ( F 12 + F 22 + F 32 ) = I ( F 3 ) 2 2 Then

(9)

530 Chapter 7: Quasi-static Control Algorithms ∂I = 0 ∂ F3

⇒

F1

∂F 1

∂ F3

+ F3 = 0

⇒

F1 ( –1 ) + F3 = 0

(10)

Finally, the control force magnitudes are F 1 = ( 0.0005 )D T F 2 = ( – 0.003 )D T

(11)

F 3 = ( 0.0005 )D T Other possible solutions that satisfy e = 0 are 1 0 F = 10 – 3 D T – 3 and F = 10 – 3 D T – 3 0 1

(12)

The pseudo inverse solution given by (11) has the lowest norm, i.e., the smallest value of I.

The generalized version of the least square algorithm is based on the extended form of the error norm, 1 1 J 1 = --- qe T Q * e + --- F T R * F = J 1 ( F ) 2 2

(7.83)

where Q * and R * are diagonal weighting matrices with elements of order 1, and q is a scaling factor defined as q = 1 ⁄ ( maximum element in C Tf C f )

(7.84)

Requiring J 1 to be stationary with respect to F leads to r equations for F, a1 F = b1 a 1 = q C Tf Q * C f + R *

(7.85)

b 1 = q C Tf Q * ( v * – v o ) Different priorities are assigned by perturbing the individual elements of Q * and

7.4 Quasi-static Control of MDOF Systems 531 R * . If all the elements of R * are finite, a 1 will be non-singular for r > n p as well as for r ≤ n p , and it is not necessary to generate additional equations for F. However, since this property is due to including R * , the equations may be illconditioned if r > n p and R * is small. The preferred strategy is to take R * = I and perturb the elements of Q * . The following example illustrates different scenarios.

Example 7.9: Example 7.7 revisited with an extended least square algorithm Certain cases considered in example (7.8) are reworked here using the extended algorithm defined by eqn (7.85) and the same performance specifications as for example (7.8). Case 1 2 C Tf C f = -------D T2

(1)

D T2 1 q = --------------------------------------------------------------------------- = -------2 maximum element in C Tf C f

(2)

q 1* + q 2* a 1 = ------------------ + r 1* 2

(3)

DT b 1 = [ q 1* ( v 1* – v 1, 0 ) + q 2* ( v 2* – v 2, 0 ) ] -------2 DT = [ ( – 0.002 )q 1* + 0.001q 2* ] -------2 D T ( – 0.002 )q 1* + 0.001q 2* F 1 = -------- ----------------------------------------------------2 1 * --- ( q 1 + q 2* ) + r 1* 2

(4)

(5)

Taking equal weights for the displacement errors and the force cost ( q 1* = q 2* = r 1* = 1 ) reduces F 1 by 50%.

532 Chapter 7: Quasi-static Control Algorithms Case 3 212 1 C Tf C f = -------- 1 1 1 D T2 212

(6)

D T2 q = -------2

(7)

1 a 1 = --2

q 1* + q 2* + 2r 1*

q 1*

q 1* + q 2*

q 1*

q 1* + 2r 2*

q 1*

q 1* + q 2*

q 1*

q 1* + q 2* + 2r 3*

– 2q * + q * 1 2 10 – 3 D T b 1 = ------------------- – 2q 1* 2 – 2q 1* + q 2*

(8)

(9)

Note that the elements of a 1 are of the same order. When r 1* = r 3* = 0 , the third row of a 1 is equal to the first row, and a 1 is singular. The third row of b 1 is also equal to the first row, and a 1 F = b 1 represents only 2 independent equations in this case. The third equation relating the 3 control forces can be generated by requiring the norm, ( 1 ⁄ 2 )F T F , to be stationary. As mentioned earlier, the MATLAB function, F = pinv ( a 1 )*b 1 , generates this solution. The standard MATLAB statement, F = a 1 \b 1, will also generate a solution for this case, but the force norm will be greater than the value corresponding to the pseudo inverse procedure. Results for F and e based on taking q 1* = q 2* = q * , r 1* = r 2* = r 3* = 1, and letting q * range from 1 to ∞ are listed in Table 1. The last row contains the error for the case where no control forces are applied. Taking q * = ∞ places all the emphasis on the error reduction, and results in the maximum control force magnitude. Decreasing q * shifts some of the emphasis to reducing the control force.

7.4 Quasi-static Control of MDOF Systems 533

q*

10 3 F 1 ⁄ D T

10 3 F 2 ⁄ D T

10 3 F 3 ⁄ D T

10 3 e 1

10 3 e 2

∞

0.5

-3.0

0.5

0

0

∞ ( a1 \ b1 )

0

-3.0

1.0

0

0

16

0.3314

-2.3670

0.3314

0.2954

-0.3373

8

0.2264

-1.9623

0.2264

0.4906

-0.5472

4

0.1053

-1.4737

0.1053

0.7363

-0.7895

2

0

-1.0

0

1.0

-1.0

1

-0.0625

-0.625

-0.0625

1.25

-1.125

No control

0

0

0

2.0

-1.0

( pinv ( a 1 )*b 1 )

Table 1

534

Introduction to Structural Motion Control

535 Chapter 7: Quasi-static Control Algorithms

Problems Problem 7.1 A simply supported beam is to be subjected to a concentrated force which can be applied to any point along the longitudinal axis of the beam. The maximum moment that the beam can resist is 250kNm. Since the maximum moment due to the force is 500kNm, it is necessary to apply a force system which controls the magnitude of the maximum moment. Consider the force control scheme shown below. The control forces produce a piecewize - linear moment distribution defined by the nodal moments m 1 , m 2 , and m 3 . Take m 1 , m 2 , and m 3 as the “force” parameters a) Assuming the force is translated along the beam, determine the values of m 1 , m 2 , and m 3 corresponding to the 7 load points located at an interval of L/8 along the beam. b) Suppose the maximum value of m 2 that the force actuator can deliver is 200kNm. Assuming there are no restrictions on m 1 and m 3 , can the control force system prevent the moment in the beam from exceeding 250kNm? How would you modify the control scheme if this requirement cannot be satisfied?

x

P

L m2

m1

L = 20meters m3

L⁄4

L⁄4

L⁄4

L⁄4

Problems 536 Problem 7.2

b 0 = 20 kN/m

x L = 10 m

Consider the cantilever beam shown above. Assume transverse shear deformation is negligible. The desired state is constant curvature and u ( L ) ≤ L ⁄ 400 . a) Determine the bending rigidity distribution based on the equilibrium model, M(x) D B = ------------χ∗

(1)

b) Determine a “constant” value of D B using the least square procedure described below. The displacement function for the triangular loading is given by: 4

b0 L 1 1 u 0 = ------------ x 2 1 – --- x + ------ x 3 = D B– 1 g ( x ) 6D B 2 20

(2)

where x = x ⁄ L . One forms the “displacement” error e ( x ) = u 0 ( x ) – u∗ ( x )

(3)

and the error functional, L

J =

∫0 e2dx

(4)

Let D B– 1 = f B . The value of f B is found by minimizing J with respect to f B .

537 Chapter 7: Quasi-static Control Algorithms ∂J --------- = 0 ∂fB

(5)

Establish the expression for f B . c) Assume D B is specified. The displacement u 0 is now fixed. Suppose a vertical force, F , is now applied at x = L and the magnitude of F can be adjusted. Determine a value for F by minimizing J with respect to F . The appropriate form of the error is e = u 0 + u c – u∗ where FL 3 F x3 u c = ----------- x 2 – ----- = -------h ( x ) 2D B DB 3

(6)

Evaluate F for a representative range of D B . d) Let e i represent the displacement error at point x i . e i = u 0 ( x i ) + u c ( x i ) – u∗ ( x i )

(7)

Suppose one uses a “discrete” error functional consisting of N terms. 1 J = --2

N

∑ ei2

(8)

i=1

Derive the general expression for F corresponding to requiring J to be stationary. Evaluate F taking D B constant = 1/2 the value obtained in part b, N = 3 , and x i = 1/3, 2/3, and 1. Note: it is convenient to introduce matrix notation and express the error as: e =

e1 e2 = e + f F e3

(9)

1 J = --- e T e 2 e) Refer to part d. Consider 2 control forces, F 1 at x = L ⁄ 2 and F 2 at x = L . Evaluate F 1 , F 2 for the conditions specified in part d. Compare the displacement solutions for part d and part e.

Problems 538 Problem 7.3

u

b0

x L Consider a cantilever beam subjected to a uniform loading. Take D B = EI as constant and neglect transverse shear deformation. The deflected shape is given by: b0 L 4 u = ------------ [ x 2 ( 6 – 4x + x 2 ) ] 24EI where x = x ⁄ L . Suppose the ‘‘desired’’ deflected shape is u = aLx 2 where a is a specified value, and a concentrated moment, M , is applied at x = L to ‘‘correct’’ the difference between the actual and desired deflected shape. a) M

x L Determine an expression for M using the continuous least square error functional. b) Determine M using the discrete least square error functional. Take 10 equally spaced points in the interval 0 < x ≤ L . Compare the error norms for parts (a) and (b).

539 Chapter 7: Quasi-static Control Algorithms c) M

αL L Suppose M is applied at x = αL . Repeat part (b), taking α = 0.5 and 0.75. Compare the error norms corresponding to α = 0.5, 0.75 , and 1.0.

Problem 7.4 a) Consider the bending beam - outrigger system shown below. Assume the outriggers are infinitely stiff, the beam bending rigidity is constant, and the cables are initially tensioned to a level of T o . Suppose the cable tensions can be adjusted to counteract the effect of a lateral load. Consider a force, P, applied at x = H . Suppose the desired deflected shape is u = x 2 ⁄ ( 300H ) . Determine an estimate for ∆T . P, u C b

b B

T o + ∆T

H DB

A

T o – ∆T

2H -------3 x

Problems 540 b) Extend the analysis procedure of part (a) to deal with the case of 2 outriggers shown below. Assume ∆T 1 and ∆T 2 can be independently adjusted. P, u

C T o + ∆T 1

DB b

T o – ∆T 1 b H

B T o + ∆T 2

DB

T o – ∆T 2

H ---2 x

A

c) Refer to part (a). Suppose the outrigger is located a distance αH from the base. Determine an estimate for ∆T in terms of α . Consider α to vary from 0.25 to 0.75.

Problem 7.5

v P

Ma L

L

541 Chapter 7: Quasi-static Control Algorithms The structure shown above consists of two single spans interconnected at the interior support with a force actuator system that can instantaneously apply a moment, M a ( t ) . The structure also has the capability to observe the transverse displacement at a limited number of points along both spans. Consider a force, P , which moves with velocity v along the span. Assume the velocity is sufficiently small such that the actual response of the structure can be approximated by the corresponding static response. a) Describe how you would establish the time history of the actuator moment, M a , such that the maximum transverse displacement at any time in both spans is less than a specified value, u∗ . Take PL 3 u∗ = -----------96EI Note: PL 3 ⁄ ( 48EI ) is the maximum value of u due to P moving across the span with M a = 0 . b) Discuss how you would implement this control scheme.

Problem 7.6 Solve problem 7.2, part (d), using the extended error function. Take r 1* = 1 and q 1* = q 2* = q 3* = q * . Investigate the variation of F and the displacement error norm with q * .

Problem 7.7 a) Solve problem 7.4, part (b), considering a “force” cost as well as a displacement error norm. Generate solutions for a range of the weighting parameters. Assume equal relative weights for the displacements ( q 1 = q 2 = q* ) and the control forces ( r 1 = r 2 = r * ) . b) Problem 7.4 considered a single force applied at x = 1. Describe how you

Problems 542 would extend the solution scheme to handle an arbitrary lateral loading applied to member ABC.

Problem 7.8

12kN D T3 = 2400kN 8kN D T2 = 4000kN 4kN D T1 = 4800kN

Consider the 3 DOF shear beam shown above. Suppose the performance measures are the transverse shear strains and the desired value of strain is 1/300 for each segment. The 3 control force schemes shown below are proposed. Establish values for the control forces associated with each scheme using the extended error norm defined by eqn (7.83) and the following uniform weighting factors: r i* = 1

q *j = q *

q * = 1, 2, 5, 100

Also investigate how the norm of the strain error vector varies with q * for each scheme. Which scheme would you recommend? What is the basis for your decision?

543 Chapter 7: Quasi-static Control Algorithms

F3 F3

F1

F1 F2

F2

F2

F1

F1 (1)

(2)

(3)

Problem 7.9 Consider the discrete shear beam shown below. Assume the transverse shear rigidity is constant for all the elements. a) Suppose the goal is to have uniform transverse shear strain over the height. Formulate the control algorithms for the 2 control force schemes shown. Which scheme would you employ? b) Suppose the goal is to have a prescribed nodal displacement vector U * . Let U o denote the initial displacement vector. Formulate the control algorithms for the 2 schemes. Does scheme 1 still correspond to fully decentralized control? Elaborate on your answer.

Problems 544

F5 (5) h (4) h (3) h (2) h

F4

F1

F5

F4

F3

F2

F2

F3

F2

F1

(1) h (1)

(2)

545

Chapter 8

Dynamic Control Algorithms - Time invariant systems 8.1 Introduction This chapter extends the active control strategy introduced in the previous chapter to deal with time dependent loading. The discussion is restricted to time invariant systems, i.e., the case where the system properties and force feedback algorithm are constant over the duration of the time response. The material presented here is organized as follows. Firstly, the state-space formulation for a SDOF system is developed, and used to generate the free vibration response. This solution provides the basis for establishing a criterion for dynamic stability of a SDOF system. Secondly, linear negative feedback is introduced, and the topic of stability is revisited. The primary focus is on assessing the effect of time delay in applying the control force on the stability. Thirdly, the SDOF state-space formulation is specialized to deal with discrete time control, where the feedback forces are computed at discrete time points and held constant over time intervals. Stability for discrete time feedback with time delay is examined in detail, and a numerical procedure for determining the time increment corresponding to a stability transition is presented and illustrated with examples. Fourthly, the choice of the optimal magnitudes of the feedback

546 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems parameters is considered. Optimality is related to the magnitude of a quadratic performance index (LQR) which is taken as a time integral involving weighted response and control force terms. This approach is referred to as the “linear quadratic regulator problem” and leads to a time invariant linear relationship between the control forces and state variables. Fifthly, the state-space formulation is extended to MDOF systems. The modal properties for an arbitrary damping scheme are derived, and used to generate the governing equations expressed in terms of the modal coordinates. The last section deals with optimal feedback based on the LQR performance index generalized for MDOF systems. Examples which illustrate the sensitivity of the response and cost parameters to variations in the location and nature of the control forces, and weighting coefficients are presented.

8.2 State-space formulation - time invariant SDOF systems Governing equations

k

F m p

c ug

u + ug

Fig. 8.1: SDOF system. The dynamic response of the SDOF linear system shown in Fig. 8.1 is governed by the second order equation, mu˙˙ + cu˙ + ku = – ma g + p + F

(8.1)

where p is the applied external loading, F is the active force, and m, k, c are system parameters. Integrating eqn (8.1) in time, and enforcing the initial conditions on u and u˙ at t = 0 , one obtains the velocity and displacement as functions of time. These quantities characterize the state of the system in the sense that once u and u˙ are specified, the acceleration and internal forces can be determined by back substitution.

8.2 State-space Formulation-Time Invariant SDOF System 547 Rather than working with a second order equation, it is more convenient to transform eqn (8.1) to a set of first order equations involving the state variables u and u˙ . The new form is

du ------ = u˙ dt du˙ 1 1 c k ------ = – ---- u˙ + – ---- u + ( – 1 )a g + ---- p + ---- F m m m m dt

(8.2)

This form is called the state-space representation. The motivation for the statespace representation is mainly the reduced complexity in generating both analytical and numerical solutions. Matrix notation is convenient for expressing the state-space equations in a compact form. Defining X as the state vector, X = u = X (t) u˙

(8.3)

the matrix equilibrium equation is written as dX ------- = X˙ = AX + B f F + B g a g + B p p dt

(8.4)

where the various coefficient matrices are defined below. A =

0 1 k c – ---- – ---m m

Bf = Bp =

Bg =

0 1 ---m

0 –1

The initial conditions at t=0 are denoted by X o .

(8.5)

(8.6)

(8.7)

548 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems u(0) ≡ Xo u˙ ( 0 )

X(0) =

(8.8)

With this representation, the problem is reduced to solving a first order equation involving X . Free vibration uncontrolled response The free vibration uncontrolled response is governed by a reduced form of eqn (8.4) X˙ = AX

(8.9)

When A is constant, the general solution has the form X = Ve

λt

(8.10)

where V is an unspecified vector of order 2 and λ is a scalar. Substituting for X results in ( A – λI )V = 0

(8.11)

where I is the identity matrix. According to eqn (8.11), the eigenvalues of A define the frequency and damping characteristics of the free vibration response. Expanding A – λI = 0 , –λ 1 k c – ---- – ---- – λ m m

= 0

(8.12)

leads to the characteristic equation 2 c k λ + ----λ + ---- = 0 m m

(8.13)

and two eigenvalues 1 c k c 2 λ 1, 2 = --- – ---- ± i 4 ---- – ---- = λ R ± iλ I m m 2 m 2

(8.14)

Noting that k ⁄ m = ω and c ⁄ m = 2ξω , eqn (8.14) is identical to eqn (6.27) which was obtained from the second order equation. Given λ , eqn (8.11) can be solved for the eigenvectors which define the state-space modes. Since λ is complex, the eigenvectors occur as complex conjugates.

8.2 State-space Formulation-Time Invariant SDOF System 549

0 1 ˜ ±i = V R ± iV I = V 1, V 1 λI λR

V 1, 2 =

(8.15)

The total free vibration response is obtained by combining the 2 complex solutions such that the resulting expression is real. Starting with X = A1 e

λ1 t

V 1 + A2 e

λ˜ 1 t

V˜ 1

(8.16)

and taking 1 A 1 = --- ( A R + iA I ) 2 ˜ A = A

(8.17)

1

2

where A R and A I are real scalars, results in X (t) = e

λR t

{ ( A R V R – A I V I ) cos λ I t + ( – A R V I – A I V R ) sin λ I t }

(8.18)

The constants A R and A I are determined by enforcing the initial conditions on X at t=0. ˜ V˜ = A V – A V X ( 0 ) = A1 V 1 + A 1 1 R R I I ⇒ uo

=

u˙ o

AR AR λR – AI λI

(8.19)

⇒ AR = uo 1 A I = – ----- ( u˙ o + λ R u o ) λI Lastly, the solution for u(t) is given by the first scalar equation in eqn (8.18). u(t) = e

λR t

( A R cos λ I t – A I sin λ I t )

(8.20)

550 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems General solution - time invariant systems The general solution for an arbitrary loading can be expressed as a Duhamel integral involving a specialized form of the free vibration response. Considering first a first order scalar equation, (8.21)

y˙ = ay + g

where a is constant, and g is a function of t, the complete solution has the form y(t) = e

a ( t – to )

t

yo +

∫

e a ( t – τ ) g ( τ ) dτ

(8.22)

to

A similar form can be generated for the first order matrix equation, X˙ ( t ) = AX + G

(8.23)

The free vibration solution defined by eqn (8.18), can be expressed as X ( t ) = e At X o

(8.24)

where e At is defined by the following series: 1 1 e At = I + At + --- AAt 2 + … + ----- A n t n + … 2 n!

(8.25)

This matrix exponential function has the “same” property as the corresponding scalar function. d At ( e ) = Ae At dt Using eqn (8.24), the Duhamel integral matrix form of the total solution for eqn (8.4) is X (t) = e

A ( t – to )

t

Xo +

∫

e A ( t – τ ) G ( τ ) dτ

(8.26)

to

where G ( t ) = B f F + B g a g + Bp p The corresponding scalar form of the solution for u(t) is

(8.27)

8.2 State-space Formulation-Time Invariant SDOF System 551

u(t) = e

λR t

1 u cos λ I t + ----- ( u˙ o + λ R u o ) sin λ I t + o λ I

t

+

∫

0

(8.28)

1 λ (t – τ) p(τ) F(τ) ----- e R sin λ I ( t – τ ) – a g ( τ ) + ---------- + ----------- dτ λI m m

Equation (8.26) applies for an arbitrary linear time invariant system. It is convenient for establishing a discrete formulation of the governing equations. This topic is addressed in the next section.

Example 8.1: Equivalence of equations (8.18) and (8.24) Consider eqn (8.16). The total free vibration response is given by ˜ e λ˜ t V˜ X ( t ) = A 1 e λt V 1 + A 1 1 ˜ V˜ X (0) = A V + A 1

1

1

(1)

1

Noting eqn (8.11), the λ and V terms are related by AV 1 = λV 1 AV˜ = λ˜ V˜ 1

(2)

1

Expanding the product, e λt V 1 , and using eqn (2), leads to t2 e λt V 1 = V 1 + ( λV 1 ) t + λ ( λV 1 ) ---- + … 2 t2 = I + At + AA ---- + … V 1 2

(3)

= e At V 1 It follows that eqn (1) can be written as ˜ V˜ ) = e At X X ( t ) = e At ( A 1 V 1 + A 1 1 o

(4)

552 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Stability criterion Another advantage of the state-space representation is the ability to relate the stability of the physical system to the eigenvalues of A . A system is said to be stable when the motion resulting from some initial disturbance is bounded. Assuming the system state is X o at time t = 0 , stability requires X ( t ) – Xo ≤ ε

for all t

(8.29)

where ε defines the bound on the perturbation from X o . Equation (8.18) defines the general homogeneous solution for a SDOF time invariant system. The terms contained inside the brackets depend on the initial conditions and are bounded since the time dependency is harmonic. Therefore, it follows that the exponential term must be bounded. This requirement is satisfied when the exponent is negative, λR ≤ 0

(8.30)

In words, the real part of the eigenvalues of A must be equal to or less than zero. When λ R = 0 , the response is pure harmonic oscillation. A negative λ R produces a damped harmonic response. Plotting λ in the complex plane provides a geometric interpretation of the stability. For the SDOF case, there are two eigenvalues, λ = λ R ± iλ I c λ R = – -------- = – ξω 2m λI =

(8.31)

2 c 2 k ---- – -------- = ω 1 – ξ m 2m

Figure 8.2 shows the corresponding points in the complex plane. These points are referred to as poles. Undamped motion has poles on the imaginary axis. Holding stiffness constant and increasing c causes the poles to move along the circle of radius ω toward the critical damping point, ξ = 1 . With further increase in damping, the curves bifurcate with one branch heading in the negative (real axis)

8.2 State-space Formulation-Time Invariant SDOF System 553 direction, and the other toward the origin. Increasing the stiffness with c held constant moves the poles in the imaginary direction. With this terminology, the stability criterion requires all the poles corresponding to the eigenvalues of A to be on or to the left of the imaginary axis, as shown in Fig. 8.3. The uncontrolled SDOF system is, according to this definition, always stable since ξ ≤ 0 . k increases λ I ξ = 0 ω c increases ξ = 1

θ

λR

ω ξ = 0 Fig. 8.2: Poles for SDOF system.

λI

stable

unstable

λR

Fig. 8.3: Stability condition for SDOF system. Linear negative feedback The response of a SDOF time invariant system with negative linear feedback is governed by eqn (8.4) with F taken as a linear function of the state variables

554 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems F = –Kf X = – kd kv u u˙

(8.32)

Substituting for F , the governing equation is transformed to X˙ = A c X + B g a g + B p p

(8.33)

where 0 Ac = A – Bf Kf =

1

c kv k kd – ---- – ----- – ---- – ----m m m m

(8.34)

The general form of the free vibration solution of eqn (8.33) is X = Ve

λt

(8.35)

where λ and V are the eigenvalues and eigenvectors of A c , the modified coefficient matrix. They are related by [ A – B f K f – λI ]V = O

(8.36)

The eigenvalues of A c are written as λ = λR ±i λI c + kν λ R = – -------------- = – ξ eq ω eq 2m 2

λ I = ω eq 1 – ξ eq kd 1 ⁄ 2 ω eq = ω 1 + ----------ω2m ξ eq = ξ + ξ a

(8.37)

kν ω ξ a = ----------------- – ξ 1 – -------- 2mω eq ω eq Since k v is positive for negative feedback (note that the minus sign is incorporated in the definition equation, eqn (8.32)), the system is stable for arbitrary k v . Displacement feedback moves the poles in the imaginary direction, and consequently has no effect on the stability. It follows that increasing the negative

8.2 State-space Formulation-Time Invariant SDOF System 555 velocity feedback is a more appropriate mechanism for improving the stability of a SDOF system. Noting eqns (8.23) and (8.26), the total solution for time invariant linear feedback can be expressed in a form similar to eqn (8.26). X (t) = e

Ac ( t – to )

t

Xo +

∫

e

Ac ( t – τ )

to

( B g a g ( τ ) + B p p ( τ ) ) dτ

(8.38)

The identity established in example 8.1 is also applicable here. e

Ac ( t – to )

X o ≡ Ae

λ ( t – to )

˜

˜ e λ ( t – t o ) V˜ V+A

(8.39)

where A is a complex scalar, and ( λ, V ) are the solution of eqn (8.36). Requiring A ( t – to ) λ R ≤ 0 ensures that e c is bounded. The above discussion assumes there is no delay between observing the state and generating the force. In general, there is some delay and the force at time t is computed using data observed at an earlier time, t – t d . The force for linear negative feedback is taken as F ( t ) = –Kf X ( t – td )

(8.40)

where t d represents the delay time. Delay introduces additional terms in λ R and under certain conditions, can cause λ R to become positive and consequently, the system becomes unstable. Therefore, although ideal linear negative feedback is unconditionally stable, one needs to examine the potential destabilizing effect of delay for the actual control system. One procedure for investigating the effect of time delay on the stability of a controlled time invariant SDOF system is described in the following section. Additional approaches are discussed in later sections. Effect of time delay on feedback control Time delay in feedback control systems in the sum of the times required to execute the following actions: • acquire the data from sensors placed at different locations in the structure; • process the sensor data and calculate the control force;

556 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems • transmit the control force signal to the actuator; • ramp up the actuator to the desired force level. The resultant time delay affects the synchronization between the control force and the response of the system and may, under certain conditions, cause a significant degradation in the performance of the control system which could result to instability. In what follows, an analytical procedure for assessing the stability of a SDOF system with time delay is presented. This approach follows the method proposed by Agrawal et al. (1993). Results based on numerical simulations are included here to illustrate the effect of time delay on the response. Assuming the feedback control algorithm consists of a linear combination of displacement and velocity terms, and uses data associated with the time, t – t d , the feedback force at time t is written as kd kv 1 ----F ( t ) = – -----u ( t – t d ) – -----u˙ ( t – t d ) = – g d u ( t – t d ) – g v u˙ ( t – t d ) (8.41) m m m Substituting for F in eqn (8.1), the governing equation for this case is 2 p(t) u˙˙( t ) + 2ξωu˙ ( t ) + ω u ( t ) + g d u ( t – t d ) + g v u˙ ( t – t d ) = ---------m

(8.42)

The general form of the homogeneous solution of eqn (8.42) is u = Ae

λt

(8.43)

where A is an arbitrary constant and λ satisfies 2

λ + 2ξω + g v e

– λt d

2

λ + ω + gd e

– λt d

= 0

(8.44)

Letting λ = λ R ± iλ I represent the roots, the stability requirement is λ R ≤ 0 . Expressing λ R and λ I in a convenient form is complicated by the presence of the exponential terms. A first order approximation can be obtained by introducing the following expansion e

– λt d

1 1 = 1 – λt d + --- ( λt d ) 2 – --- ( λt d ) 3 + … 6 2

(8.45)

and retaining only the first 2 terms. The result is expressed in the same form as eqn (8.37) with ξ eq and ω eq replaced with modified terms.

8.2 State-space Formulation-Time Invariant SDOF System 557 λ R = – ξ' eq ω' eq (8.46)

2

λ I = ω' eq 1 – ξ' eq

The modified equivalent frequency and damping are related to the time delay by 2

1 + gd ⁄ ω ω' eq = ω -------------------------1 – gv td

1 ξ + ------- ( g v – g d t d ) 2ω ξ' eq = ---------------------------------------------------gd [ 1 – g v t d ] 1 + -----2 ω

(8.47)

(8.48)

Equation (8.46) is convenient for identifying behavioral trends. For no initial damping and no velocity feedback ( g v = 0 ), the approximation for λ R reduces to gd td k d td λ R = ------------ = -----------2ω 2ωm

(8.49)

Since λ R is positive, it follows that according to this first order approximation, displacement feedback with time delay produces unstable behavior for an initially undamped system ( ξ = 0 ) . The real part of λ corresponding to pure velocity feedback ( g d = 0 ) is estimated as kv gv ξ + -----------ξ + ------2ω 2ωm λ R = – ω -------------------- = – ω ---------------------1 – gv td kv td 1 – ---------m

(8.50)

This result suggests that the response is stable when ( k v t d ⁄ m ) < 1 . An improved approximation can be obtained by substituting the Pade

558 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems expansion e

– λt d

1 – λt d ⁄ 2 3 ≈ ------------------------- + O [ ( λt d ) ] 1 + λt d ⁄ 2

(8.51)

in eqn (8.44). The resulting expression is now a 3rd degree polynomial in λ td 3 td 2 g 2 2 ----- λ + 1 + t d ξω – -----v- λ + 2ξω + g v + ----- ( ω – g d ) λ + ω + g d = 0 2 2 2 (8.52) Introducing the notation for the equivalent frequency and damping defined earlier for the instantaneous feedback case (see eqn (8.37)), 2

2

ω eq = ω + g d 1 ξ eq ω eq = ξω + --- g ν 2 ξ eq = ξ + ξ a

(8.53)

transforms eqn (8.52) to td 3 2 ----- λ + [ 1 + ( 2ξω – ξ eq ω eq )t d ]λ + 2

(8.54) 2

2 ω eq 2 2ξ eq ω eq + t d ω – --------- λ + ω eq = 0 2 For pure displacement feedback, ξ eq ω eq = ξω . For pure velocity feedback, ω eq = ω and ξ a = k ν ⁄ ( 2ωm ) . Equation (8.54) is solved numerically for a specific SDOF system with a period of 5s , and the two limiting cases of pure displacement and pure velocity feedback with no initial damping. Figure 8.4 shows the movement of the poles for pure displacement feedback as a function of ω eq and t d . When t d = 0 , the poles move on the imaginary axis with increasing ω eq . As t d increases, the path shifts to the right, and when t d is about 1s , the direction is essentially along the positive real axis.

8.2 State-space Formulation-Time Invariant SDOF System 559

2.5 2

Imaginary axis, λ I

1.5

T = 5s ξ = 0% kv = 0

1 0.5 0

ω eq ω eq ω eq

td = 0 = 2.0ω = 1.5ω = 1.0ω

t d = 0.25

t d = 0.5 t d = 1.0

k+k 2 = -------------dω eq m

−0.5 −1

ω eq = 1.0ω ω eq = 1.5ω ω eq = 2.0ω td = 0

−1.5 −2 −2.5 −0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

Real axis, λ R

t d = 1.0 t d = 0.25 0.2

0.3

t d = 0.5 0.4

0.5

Fig. 8.4: Variation of pole profile with t d under pure displacement feedback. The effect of time delay for pure velocity feedback is illustrated by Fig. 8.5. For no time delay, increasing k ν moves the poles further back in the negative real half plane until the state of critical damping is reached. As t d increases, the paths tend to bend toward the positive real half plane, and eventually intersect the imaginary axis. For a given value of k ν , there may be a limiting time delay beyond which the system is unstable. These observations are based on an approximation and apply for a particular system, i.e. specific values of ω and ξ . An exact analysis of the instability problem is presented in the next section.

560 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

2.5 2

t d = 0.5

T = 5s ξ = 0%

t d = 1.0

Imaginary axis, λ I

1.5

t d = 0.25

1

0.5

td = 0

ξ a = 0.4 0.3

0.2

0.1 kν ξ a = -----------2ωm kd = 0

0 −0.5

td = 0

−1

t d = 0.25

ξ a = 0.4 0.3

0.2

0.1

−1.5

t d = 1.0

−2 −2.5 −1

t d = 0.5 −0.8

−0.6

−0.4

−0.2

0

Real axis, λ R

0.2

0.4

Fig. 8.5: Variation of pole profile with t d under pure velocity feedback.

Stability analysis for time delay Figure 8.5 shows that, for a given system, there is a particular value of t d which corresponds to a transition in the behavior of λ R . For t d greater than this value, λ R increases and eventually becomes positive. When λ R = 0 , there is a transition from stable to unstable behavior, and the corresponding value of t d defines the stability limit for the system, i.e., for the particular combination of k, c, k ν , and k d . Equation (8.44) is the definition equation for λ . Noting that the critical time delay corresponds to λ R = 0 , by expressing λ as λ = iΩ

(8.55)

where Ω is a real scalar, one can specialize eqn (8.44) for this case and determine the critical value of t d . Substituting for λ leads to 2

2

– Ω + i2ξωΩ + ω + g d e

– it d Ω

+ ig v Ωe

– it d Ω

= 0

(8.56)

8.2 State-space Formulation-Time Invariant SDOF System 561 Replacing the exponential term by e

– it d Ω

= cos ( t d Ω ) – i sin ( t d Ω )

(8.57)

yields 2

2

– Ω + ω + g d cos ( t d Ω ) + g v Ω sin ( t d Ω )

(8.58)

+ i [ 2ξωΩ – g d sin ( t d Ω ) + g v Ω cos ( t d Ω ) ] = 0 Equation (8.58) is satisfied when the real and imaginary terms vanish. Then, g d sin ( t d Ω ) – g v Ω cos ( t d Ω ) = 2ξωΩ 2

g d cos ( t d Ω ) + g v Ω sin ( t d Ω ) = Ω – ω

(8.59) 2

(8.60)

It remains to solve these equations for t d in terms of g d , g v , ω , and ξ . Squaring both sides and adding the equations eliminates the trigonometric terms and results in a quartic equation for Ω . 4

2 2

2

2

2

4

2

Ω + ( 4ξ ω – 2ω – g v )Ω + ω – g d = 0

(8.61)

The roots of eqn (8.61) are given by 2 2

2

2

2 2

2

2 2

( 4ξ ω – 2ω – g v ) ( 4ξ ω – 2ω – g v ) 4 2 Ω 1, 2 = − + – ------------------------------------------------- − + ---------------------------------------------------- – ( ω – g d ) 2 4 (8.62) Since the poles correspond to ± iΩ , only the positive value of Ω needs to be considered, resulting in two values of Ω . The next step is to determine t d Ω . Noting the trigonometric identities, td Ω 2 tan ---------2 sin ( t d Ω ) = -----------------------------------2 td Ω 1 + tan ---------2

(8.63)

562 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems 2 td Ω 1 – tan ---------2 cos ( t d Ω ) = -----------------------------------2 td Ω 1 + tan ---------2

(8.64)

eqn (8.59) can be expressed as td Ω 2 td Ω ( g v Ω – 2ξωΩ )tan ---------- + 2g d tan ---------- – ( g v Ω + 2ξωΩ ) = 0 2 2

(8.65)

The two roots of eqn (8.65) are td Ω – gd − + g d + g v Ω – 4ξ ω Ω tan ---------- = -----------------------------------------------------------------------------2 g v Ω – 2ξωΩ 2

2

2

2 2

2

(8.66)

Finally, the maximum time delay t d can be determined from max 2 2 2 2 2 2 − – 1 – g d + g d + g v Ω – 4ξ ω Ω 2 = ----tan -----------------------------------------------------------------------------td Ω g v Ω – 2ξωΩ max

(8.67)

The minimum positive value of t d is the maximum allowable time delay. max For the limiting case of pure velocity feedback control of an undamped system (i.e. g d = 0 and ξ = 0 ), eqn (8.59) reduces to g v cos ( t d Ω ) = 0

(8.68)

and it follows that td

π = ------2Ω max

(8.69)

The expression for Ω can be obtained from either eqn (8.60) or eqn (8.62), 2 Ω ---- = ξ a + 1 + ξ a ω

(8.70)

gν kν ξ a = ------- = -----------2ω 2ωm

(8.71)

where

Finally, the maximum delay can be expressed in terms of the fundamental period

8.2 State-space Formulation-Time Invariant SDOF System 563 of the uncontrolled system. td 1 max ----------------- = ---------------------------------------T 2 4 ξa + 1 + ξa

(8.72)

Figures 8.6 through 8.8 illustrate the effect of varying the displacement feedback, velocity feedback, system damping, and system period on the maximum allowable time delay. Figure 8.6 shows plots of the maximum allowable time delay for an undamped 5s SDOF system as a function of the active damping ratio ξ a (i.e. velocity feedback) for three values of displacement feedback k d . The central curve corresponds to k d = 0 . The lower curve corresponds to k d > 0 , i.e. leading to an increase in the frequency of the controlled system, and shows that t d = 0 for no active damping. Furthermore, Fig. 8.6 illustrates the effect of underestimating/overestimating the stiffness of the system on the maximum allowable time delay. If the model used to establish the system properties underestimates the stiffness, the actual limit on the time delay will be less than the predicted limit and stability may be a problem. Figure 8.7 shows the effect of the damping in the system on the maximum allowable time delay. In general, damping increases the allowable time delay. The effect of the fundamental period of the system on the maximum allowable time delay is illustrated by Fig. 8.8. As expected from eqn (8.20), the maximum allowable time delay increases with period and decreases with active damping. Finally, Fig. 8.9 illustrates the degradation in performance, with increasing time delay, of the SDOF system subjected to seismic excitation. Instability occurs beyond a time delay of 1.1s which, according to Fig. 8.7, corresponds to the maximum allowable time delay for this level of damping. The instability is due to the unbounded growth of the homogeneous solution, and will occur for any arbitrary external excitation.

564 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

2.5

T = 5s ξ = 0% 2

k 2 = ω 2 + ----dω eq m

1.5

ω

td

max

ω eq = 0.8ω

1

ω eq = 1.2ω 0.5

0 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

kv ξ a = ----------2ωm Fig. 8.6: Maximum allowable time delay t d as a function of k d and k v . max 2

ξ = 5% 10%15% 20%25%

T = 5s kd = 0

1.8 1.6 1.4

td

max

1.2 1

ξ = 0%

0.8 0.6 0.4 0.2 0 0

0.1

0.2

0.3 k v 0.4 ξ a = ----------2ωm Fig. 8.7: Maximum allowable time delay t d

max

0.5

0.6

as a function of k v and ξ .

8.2 State-space Formulation-Time Invariant SDOF System 565

1.5

ξ a = 0%

ξ = 0 kd = 0

10% 20% 30% 40% 50% 60% 70%

td

max

1

0.5

0 0

1

2

3

4

5

6

T - s Fig. 8.8: Maximum allowable time delay t d as a function of k v and T . max 10 9 8

T = 5s ξ = 2% kd = 0 ξ a = 20%

u ( td ) ⁄ u ( td = 0 )

7 6 5

El Centro Taft

4 3 2 1 0 0

0.5

td - s

1

1.5

Fig. 8.9: Degradation in performance of a SDOF system with time delay.

566 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

8.3 Discrete time formulation - SDOF system Governing equation The continuous state-space linear formulation considers X and F to be continuous functions of time that satisfy the following ordinary linear differential equation and initial conditions, X˙ = AX + B f F + B g a g + B p p X ( t = 0 ) = X o*

(8.73)

In the case of a SDOF system, the coefficients are second order matrices involving the system properties,

A =

0 1 k c – ---- – ---m m

B f = Bp =

Bg =

0 1 ---m

(8.74)

0 –1

The formulation presented in the previous section was based on the assumption that m, k, and c are constant. This restriction allows one to obtain eqn (8.26), the exact analytical solution expressed in terms of a convolution integral. A numerical integration procedure is required to evaluate the convolution integral when the loading is a complex function such as a ground acceleration time history. If the system parameters and/or the feedback parameters are time dependent, an exact analytic solution of the equilibrium equation can not be established, and one must resort to generating an approximate solution with a numerical procedure such as a finite difference method which works with values of the variables at discrete points in time. A discrete time approach is also necessary for real time feedback control, since the control force is computed using “observed” values for the response at discrete time points.

8.3 Discrete Time Formulation- SDOF System 567 One generates a discrete time formulation by subdividing the time domain into intervals, say t o – t 1 , t 1 – t 2 , . . . , t n – t f , and taking as unknowns the values of X ( t ) and F ( t ) at the discrete time points. The notation, X(t j) ≡ X j

(8.75)

F(t j) ≡ F j

is convenient for representing these discrete variables. Equation (8.73) is approximated at each time point by an algebraic equation relating the values of X and F at that point and neighboring points, and is used to estimate the value of X at a later time. In what follows, the procedure is illustrated using a simple approximation for eqn (8.73). When the system parameters are constant, the solution is given by eqn (8.26) which is listed below for convenience. t

X(t) = e

A ( t – to )

X ( to ) +

∫

t

e

A(t – τ)

to

B f F ( τ ) dτ +

∫

e

A(t – τ)

( B p p + B a g ( τ ) ) dτ (a) g

to

One can use this result to obtain an approximate solution between two time points, say t j and t j + 1 = t j + ∆t , by introducing assumptions for the variation of the force terms during the interval ∆t . The simplest model is based on using the values at the initial time t j . Taking the time limits as t = t j + 1 and t o = t j , assuming the force terms are equal to their value for t = t j , F(τ) = F(t j) = F j p(τ) = p(t j) = p j a g ( τ ) = a g ( t j ) = a g, j

(8.76)

At

and noting the expansion for e transforms the convolution integrals contained in eqn (a) to the following algebraic form, Xj + 1 = e

A∆t

X j + A – 1 ( e A∆t – I ) [ B g a g, j + B f F j + B p p j ]

(8.77)

568 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Equation (8.77) provides an estimate for X j + 1 based on data associated with the time point, t j . The first term on the right hand side is the exact free vibration response at t j + 1 , considering X j to be the initial state at t j . The remaining terms represent the contribution of the “constant” loading terms over the time interval, ∆t . Starting at t = 0 which corresponds to j = 0, one specifies X o and computes X 1 . This process is repeated until the desired time is reached. The computation will be bounded when eqn (8.75) is numerically stable. Stability is discussed in more detail later in the section. This approach can also be applied to an adaptive system. The mass is considered to be constant, and therefore B f and B g are constant. Stiffness and damping are assumed to be constant over a time interval, and to vary from one interval to another. Discrete values of stiffness and damping are defined for the time interval, t j ≤ t < t j + 1, as follows: k(t) = k(t j) ≡ k j c(t) = c(t j) ≡ c j

(8.78)

A(t) = A(t j) ≡ A j Since A is considered constant over an interval, eqn (a) is still applicable for the interval. The resulting form of the discrete equilibrium equation is obtained by replacing A with A j in eqn (8.77). Xj + 1 = e

A j ∆t

X j + A –j 1 ( e

A j ∆t

– I ) [ B g a g, j + B f F j + B p p j ]

(8.79)

Starting at j = 0, one forms A o and determines X 1 . Then, A is updated to A 1 and used to compute X 2 . This process is continued for successive time points. Linear negative feedback control The discrete formulation represented by eqn (8.79) assumes properties and forces are constant over a time interval and are updated at the starting point of the interval. A feedback law consistent with this assumption is tj ≤ t < tj + 1 F j = –K f , j X j where K f , j is the feedback gain matrix at t = t j .

(8.80)

8.3 Discrete Time Formulation- SDOF System 569 K f , j = k d ( t j)

kv ( t j )

(8.81)

Substituting for F transforms eqn (8.79) to Xj + 1 = e

A j ∆t

+ A j –1 ( e

– A –j 1 ( e

A j ∆t

A j ∆t

– I )B f K f X j

(8.82)

– I ) [ B g a g, j + B p p j ]

One can also derive this expression by specializing eqn (8.38). The time history response is generated by starting with X o , and computing X 1 , X 2 , ... etc. Key issues here are the specification of: the time interval; the magnitude and temporal distribution of the system stiffness and damping; and the feedback parameters, k d and k v . Stability analysis for time invariant linear feedback control The numerical stability of the discrete feedback control algorithm is determined by examining the nature of the homogeneous solution. The analysis presented here assumes the coefficient matrices are constant, i.e., the system is time invariant. Specializing eqn (8.79) for this case, and considering no external loading other than feedback leads to the governing equation X j + 1 = e A∆t X j + A – 1 ( e A∆t – I )B f F j

(8.83)

For no time delay, F j is taken as a linear function of X j . The effect of time delay is to shift the value of X used to compute F j back to X ( t j – ν∆t ) ≡ X j – ν . For j < ν , the control force is taken as 0. Fj = 0

j<ν

F j = –K f X j – ν

j≥ν

(8.84)

Substituting for F j , the discrete equilibrium equation allowing for time delay takes the following form X j + 1 = e A∆t X j X j + 1 = e A∆t X j – A – 1 ( e A∆t – I )B f K f X j – ν

0≤j<ν ν≤j

(8.85)

570 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Considering first the case of no time delay, the free vibration response is determined with X j + 1 = ( e A∆t – A – 1 ( e A∆t – I )B f K f )X j

j = 0, 1, 2, …

(8.86)

This equation applies when the control force is updated at the discrete time points, and assumed constant during the time intervals. The corresponding form for the case where the force is assumed to be continuously updated can be obtained by specializing eqn (8.38). Xj + 1 = e

( A – B f K f )∆t

(8.87)

Xj

The solution generated with eqn (8.87) is the “exact” damped solution. According to eqn (8.39), this solution is “bounded” for negative feedback applied to an initially stable system. The error introduced by approximating the control force is represented by the difference between the coefficient matrices. It remains to determine whether this “error” causes eqn (8.86) to became numerically unstable for particular combinations of K f and ∆t , and therefore to generate homogeneous solutions which increase rather than decrease with time. In what follows, a strategy for determining the numerical stability of first order matrix difference equations such as eqns (8.86) and (8.87) is presented. The approach is introduced using a simple scalar equation and is then generalized to handle an n’th order equation. Consider the first order differential equation, (8.88)

x˙ = ax + bf

where a and b are constants and f = f(t). The exact solution is determined using eqn (8.22). x(t) = e

a ( t – to )

x ( to ) +

∫

t

e a ( t – τ ) bf ( τ ) dτ

(8.89)

to

This solution is bounded when a<0. Specializing the limits for t leads to the difference form of (8.89)

8.3 Discrete Time Formulation- SDOF System 571

xj + 1 =

e a∆t

xj +

∫

tj + 1

e

a(t j + 1 – τ)

bf ( τ ) dτ

(8.90)

tj

The homogeneous solution is “exact”, and therefore is bounded when the system is stable, i.e., when a<0. A similar solution procedure can be followed for the case where the forcing term, f, is a continuous function of x. Assuming linear negative feedback, f ( t ) = – kx ( t )

k>0

(8.91)

transforms eqn (8.88) to x˙ = ( a – bk )x

(8.92)

The exact solution has the form x j + 1 = e ( a – bk )∆t x j

(8.93)

This solution is bounded when bk>a. Normally one starts with an initially stable system, (a<0), and adds negative feedback so that “boundness” is always ensured. Equation (8.93) is based on continuous feedback. The solution corresponding to the following discrete feedback f ( τ ) = – kx j tj ≤ τ ≤ tj + 1

(8.94)

is given by x j + 1 = cx j

(8.95)

bk c = e a∆t + ----- ( 1 – e a∆t ) a

(8.96)

where

Taking j=0, 1, 2, ... leads to x j = ( c ) j xo

(8.97)

572 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems For x j to be bounded, the absolute magnitude of c must be less than 1. c <1 ⇒

(8.98)

bk e a∆t + ----- ( 1 – e a∆t ) < 1 a Equation (8.98) represents a constraint on k and ∆t . When the system is initially stable, a is negative, a = –a

(8.99)

Specializing eqn (8.98) for this case, and noting that the negative value controls, results in kb ----- ( 1 – e – a ∆t ) – e – a ∆t < 1 a

(8.100)

which is written as: kb 1 + e – a ∆t k max b ----- < ------------------------ ≡ --------------a 1 – e – a ∆t a

(8.101)

Equation (8.101) defines the maximum value of k for a particular value of ∆t . The variation of k max b ⁄ a with a ∆t is shown in Fig 8.10. For small a ∆t , the curve tends to ∞ as ∆t → 0 ; for large a ∆t , the curve is bounded by k max b ⁄ a = 1 . If kb ⁄ a is less than 1, there is no restriction on the magnitude of ∆t . When, kb ⁄ a is greater than 1, the value of ∆t has to satisfy eqn (8.101). Using Fig (8.10), one determines the limiting value of a ∆t corresponding to the specified value of kb ⁄ a . The range of allowable values for a ∆t is between zero and this “limiting” value. For example, suppose kb ⁄ a is equal to 1.6. According to Fig (8.10), the maximum allowable value of a ∆t is 1.4, and the range is 0 < a ∆t < 1.4 .

8.3 Discrete Time Formulation- SDOF System 573

10

9

8

7

6

k max b ---------------5 a 4

3

eqn ( 8.101 ) 2

1

0

0

0.5

1

1.5

2

2.5

a ∆t

3

3.5

4

4.5

5

Fig. 8.10: Stability limits for x j + 1 = cx j According to the analysis presented above, the scalar equation, x j + 1 = cx j , is numerically stable when c < 1 . The matrix equation, X j + 1 = cX j , has a similar constraint involving the matrix, c. Observing that X j can be expressed as j

X j = c Xo

(8.102)

it follows that the elements of c j must be bounded as j → ∞ . A measure for c j is established by noting that the expression, j

Xj = ρ Ψ

(8.103)

where ρ and Ψ satisfy the characteristic equation for c, ( c – ρI )Ψ = 0

(8.104)

is a solution of X j + 1 = cX j . Suppose c is of order 2. There are 2 eigensolutions which may be real or complex. When the solution is complex, the individual solutions are complex conjugates, and the total real solution is a linear combination of these j solutions,

574 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems ˜ ρ˜ j Ψ ˜ X j = Aρ j Ψ + A

(8.105)

˜ are complex constants. These constants are evaluated using the where A , A initial conditions for X o . ˜Ψ ˜ X o = AΨ + A

(8.106)

Substituting for A and Ψ , A = A R + iA I Ψ = Ψ R + iΨ I

(8.107)

results in 2 scalar equations for A R and A I . 2 ( AR ΨR – AI ΨI ) = X o

(8.108)

The complex eigenvalue can be expressed as ρ = ρ R + iρ I = ρe iθ

(8.109)

where ρ is the magnitude of ρ and θ is the polar angle. 2 + ρ 2 ]1 ⁄ 2 ρ = [ ρR I ρI tan θ = -----ρR

(8.110)

Using the polar form for ρ , the j’th power is given by ρ j = ( ρ ) j e i ( jθ )

(8.111)

Introducing the expressions for A , Ψ , and ρ , the solution for X j expands to X j = ( ρ ) j [ 2 ( A R Ψ R – A I Ψ I ) cos ( jθ ) – 2 ( A R Ψ I + A I Ψ R ) sin ( jθ ) ]

(8.112)

j = 1, 2, … Since the terms within the brackets are bounded, it follows that X j is finite as j → ∞ when the modulus of ρ is less than unity, ρ<1

(8.113)

8.3 Discrete Time Formulation- SDOF System 575 Extending this analysis to the case where c is of order 2n, there may be n pairs of complex conjugates solutions having the same form as eqn (8.105), and the total solution is given by n

Xj =

∑

j ˜ ρ˜ lΨ ˜ ( Al ρl Ψl + A l l l)

l=1 n

=

∑

(8.114) ( ρ l ) j [ 2 ( A R, l Ψ R, l – A I, l Ψ I, l ) cos ( jθ l )

l=1

– 2 ( A R, l Ψ I, l + A I, l Ψ R, l ) sin ( jθ l ) ] Defining ρ max as the maximum modulus for the set of eigenvalues of c, the stability requirement is ρ max < 1

(8.115)

This requirement also applies when some of the eigenvalues are real. For this case, the maximum absolute value must be less than unity. In general, c depends on the nature of the finite difference approximation, the system parameters, and the magnitude of the time step, ∆t . For the case of time invariant negative linear discrete feedback with no time delay, the form of c follows from eqn (8.82) specialized for an external loading: c = e A∆t – A – 1 ( e A∆t – I )B f K f

(8.116)

Given A , B f , and K f , one selects a value for ∆t , determines the corresponding eigenvalues of c, and then adjusts the value of ∆t if ρ max is greater than 1. The complexity of the equation for c necessitates the use of a numerical eigenvalue routine such as the eig( ) function in MATLAB. The following example illustrates computational details and presents the stability limit for a SDOF system.

Example 8.2: Stability analysis - SDOF system with no time delay The various matrices for the SDOF case with negative velocity feedback

576 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems are:

A =

0

1

–ω 2

– 2ξω

– ωω 0 BfKf =

0

2ξ – -----ω 1

1 – -----ω2 0

ω ---ω – 2ξω

0

A∆t =

A –1 =

0 0 kν = 0 ----m

(1)

0 2ξ a ω

where ω is a dimensionless parameter. ∆t ω = ω∆t = 2π -----T

(2)

When ξ = 0 , the expansion of e A∆t can be expressed as 1 e A∆t = cos ωI + ---- sin ω A ω

(3)

Using eqn (3), one can obtain analytical expressions for the elements of c. The resulting form of c is

c =

cos ω – ω sin ω

1 ---- ( sin ω + 2ξ a ( cos ω – 1 ) ) ω cos ω – 2ξ a sin ω

(4)

Evaluating the eigenvalues of c leads to ρ 1, 2 = b 1 ± ( b 12 – b 2 ) 1 ⁄ 2 b 1 = cos ω – ξ a sin ω

(5)

b 2 = 1 – 2ξ a sin ω Equation (5) shows that the eigenvalues depend on the dimensionless parameter,

8.3 Discrete Time Formulation- SDOF System 577 ω = 2π ( ∆t ⁄ T ) , and the active damping ratio, ξ a . Figure 1 contains plots of the absolute values of the eigenvalues vs. ∆t ⁄ T for a representative range of ξ a . The curves intersect at ∆t ⁄ T = 0.5 , which corresponds to a transition in behavior. When ∆t ⁄ T ≥ 0.5 , the eigenvalues are complex and ρ > 1 . For the range 0 < ∆t ⁄ T < 0.5 , the eigenvalues start out as complex quantities, and then shift over to real quantities at a particular value of ∆t ⁄ T which depends on ξ a . The curve bifurcates at this value. When ξ a = 0 , the bifurcation occurs at ∆t ⁄ T = 0.5 . The intersection of the upper branch with ρ = 1 defines the limiting value of ∆t ⁄ T for the corresponding value of ξ a . This value corresponds to ρ = – 1 and is related to ξ a by cos ω – 2ξ a sin ω = – 1 . Although ρ = 1 for ξ = 0 and arbitrary ∆t ⁄ T , the modulus is greater than 1 for ξ a > 0 and ∆t ⁄ T > 0.5 . Therefore, ∆t ⁄ T = 0.5 represents the “limit” for an undamped system. 2

1.8

ξ = 0 kd = 0

1.6

ξ a = 0.4 0.3 0.2 0.1 0.0

1.4

1.2

ρ

ξ a = 0.0

1

0.1 0.2 0.3 0.4

0.8

0.6

0.4

0.2

0

0

0.1

0.2

0.3

0.4

0.5

∆t ⁄ T

0.6

0.7

0.8

0.9

1

Figure 1 The effect of including damping in the system is to offset the destabilizing effect of feedback. Figure 2 illustrates this trend. Increasing ξ lowers ρ and

578 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems decreases the range of ω for which ρ is greater than 1. When ξ a = 0 , c reduces to e A∆t . The eigenvalues of e A∆t are related to the eigenvalues of A by ρ = e λ∆t

(7)

Noting that λ = – ξω ± i ( 1 – ξ 2 )ω , eqn (7) expands to 2 1⁄2 ρ = e – ξω [ e ± i ( 1 – ξ ) ω ]

(8)

and it follows that ρ = e – ξω

(9)

Equation (9) explains the trend for ρ to decrease with increasing ω . The second term in eqn (8) is the source of the transition from real to complex eigenvalues. This shift occurs when ( 1 – ξ 2 ) 1 ⁄ 2 ω = π . 2

1.8

ξ = 0.05 kd = 0

1.6

1.4

ξ a = 0.4 0.3 0.2 0.1 0.0

1.2

ρ

ξ a = 0.0

1

0.1 0.2 0.3 0.4

0.8

0.6

0.4

0.2

0

0

0.1

0.2

0.3

0.4

0.5

∆t ⁄ T

a) ξ = 0.05

0.6

0.7

0.8

0.9

1

8.3 Discrete Time Formulation- SDOF System 579

2

1.8

ξ = 0.1 kd = 0

1.6

1.4

1.2

ρ

ξ a = 0.0

1

ξ a = 0.4 0.3 0.2 0.1 0.0

0.1 0.2 0.3

0.8

0.6

0.4

0.4

0.2

0

0

0.1

0.2

0.3

0.4

0.5

∆t ⁄ T

0.6

0.7

0.8

0.9

1

b) ξ = 0.1 Figure 2 The limits on ∆t ⁄ T depend on both ξ and ξ a . As Fig 2 shows, ξ has a significant effect on ρ , particularly in the region, ∆t ⁄ T > 0.5. When ξ = 0.1 , this region is stable even for ξ a = 0.4 , which represents significant active damping. A conservative strategy would be to require ∆t ⁄ T to be less than the value for which the upper branch first intersects ρ = 1 . These values are listed in Table 1. Table 1: Stability limit for ∆t ⁄ T ξ

ξa = 0

ξ a = 0.1

ξ a = 0.2

ξ a = 0.3

ξ a = 0.4

0

0.50

0.44

0.38

0.33

0.29

0.05

∞

0.48

0.38

0.33

0.29

0.10

∞

∞

0.39

0.34

0.29

Figure 3 illustrates the onset of instability for a SDOF system having

580 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems X ( 0 ) = { 1, 0 } . The discrete time history response is obtained by evaluating eqn (8.86) for j ranging from 0 to 100. Increasing ∆t from 0.33T to 0.34T causes the free vibration response of the initially displaced system to shift from stable to unstable behavior. 1

0.8

ξ = 0.1

Displacement, u

0.6

ξ a = 0.3

∆t ------ = 0.33 T kd = 0

0.4

0.2

0

−0.2

−0.4

0

10

20

30

40

50

60

Discrete time point a)

70

80

90

100

8.3 Discrete Time Formulation- SDOF System 581

5

4

ξ = 0.1 ξ a = 0.3 ∆t ------ = 0.34 T

3

Displacement, u

2

1

0

−1

−2

−3

−4

−5

0

10

20

30

40

50

60

70

80

90

100

Discrete time point b) Figure 3

The stability analysis allowing for time delay starts with eqn (8.86) which is written here in a simpler form. X j + 1 = c1 X J + c3 X j – ν

(8.117)

c 1 = e A∆t

(8.118)

c 3 = – A – 1 ( e A∆t – I )B f K f

(8.119)

where

The previous analysis corresponds to ν = 0 and c 1 + c 3 = c . Given ν and the damping parameters, ξ and ξ a , one needs to establish the limiting value of the

582 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems ratio of the time increment to the fundamental period, ∆t ⁄ T . One option is to employ numerical simulation and solve eqn (8.117) for a range of values of ∆t ⁄ T . Figure 8.11 illustrates the stability transition for ν = 1 and a specific set of parameters. The corresponding results for ν = 0 are shown in Figure 3 of example 8.2. In this case, the maximum time step is reduced by about 50%. The advantage of numerical simulation is that it allows arbitrary values of ξ , ξ a , and ν to be considered. The disadvantage is that the search can be inefficient if the starting point is not reasonably close to the solution. One can start with the solution for ν = 0 and decrease ∆t ⁄ T . Another option is to utilize a simpler formulation based on ξ = 0 to generate an estimate of the starting value. The latter option is described below. 1

ν = 1

0.8

ξ = 0.1 ξ a = 0.3 ∆t ------ = 0.15 T

Displacement, u

0.6

0.4

0.2

0

−0.2

−0.4

−0.6

0

50

100

150

200

a)

250

300

350

400

Discrete Time Points

8.3 Discrete Time Formulation- SDOF System 583

1500

ν = 1 ξ = 0.1 ξ a = 0.3 ∆t ------ = 0.16 T

Displacement, u

1000

500

0

−500

−1000

−1500

0

50

100

150

200

250

b)

300

350

400

Discrete Time Points

Fig. 8.11: Illustration of instability due to time delay The solution of eqn (8.117) can be expressed as X J = ρ jΨ

(8.120)

where ρ is a complex scalar, and Ψ is an unknown vector. In order for the solution to be stable, the absolute magnitude of ρ must be less that 1. Substituting for X j , eqn (8.117) takes the following form ( c 3 + ρ ν c 1 – ρ ν + 1 I )Ψ = 0

(8.121)

When ξ = 0 , the coefficient matrix reduces to (see example 8.2) ρ ν cos ω – ρ ν + 1 – ρ ν ω sin ω

sin ω cos ω – 1 ρ ν ------------ + 2ξ a ---------------------ω ω ρ ν cos ω – 2ξ

a

sin ω – ρ ν + 1

(8.122)

584 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems For Ψ to have a non-trivial value, the determinant of the coefficient matrix must vanish. Enforcing this requirement leads to an equation of degree ν + 2 in ρ . p ( ρ ) = ρ ν + 2 + aρ ν + 1 + ρ ν + bρ – b = 0

(8.123)

a = – 2 cos ω b = 2ξ a sin ω

(8.124)

where

Given ν and ξ a , one evaluates the roots of p ( ρ ) =0 for a representative range of ω = 2π ( ∆t ⁄ T ) . The following example illustrates this computation.

Example 8.3: Stability analysis -SDOF system with time delay Figure 1 contains plots of ρ vs. ∆t ⁄ T for a range of ξ a and ν = 1, 2 . As expected, increasing the time delay lowers the magnitude of the time interval at which instability occurs. The effect of time delay becomes more significant as the active damping is increased. Table 1 contains the limiting values of ∆t ⁄ T for ξ = 0 and ξ a ranging up to 0.4. Both active damping and time delay result in a decrease in the allowable time step. Including passive damping has the opposite effect on the time increment, i.e., it increases the allowable time increment. For example, the critical time step ratio for ν = 1 , ξ a = 0.3 is 0.15 for ξ = 0.1 vs. 0.124 for ξ = 0.0 .

8.3 Discrete Time Formulation- SDOF System 585

1.4

ξ a = 0.5

1.3

1.2

0.4

ξ = 0 ν = 1

0.3 0.2

1.1

0.1 ρ

1

0.0

0.9

0.8

0.7

0.6 0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

a) ν = 1

∆t ⁄ T

0.2

1.4

ξ a = 0.5 0.4 0.3

1.3

1.2

ξ = 0 ν = 2

0.2 0.1

1.1

ρ

1

0.9

0.8

0.7

0.6 0.04

0.06

0.08

0.1

0.12

b) ν = 2 Figure 1

0.14

0.16

0.18

0.2

∆t ⁄ T

586 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Table 1: Limiting values of ∆t ⁄ T for ξ = 0 Stability limit for ∆t ⁄ T ξa

ν = 0

ν = 1

ν = 2

0

0.50

0.500

0.5000

0.158

0.0955

0.151

0.0908

0.144

0.0860

0.137

0.0822

0.131

0.0780

0.124

0.0745

0.118

0.07085

0.113

0.0677

0.05 0.10

0.44

0.15 0.20

0.38

0.25 0.30

0.33

0.35 0.40

0.29

The actual time increment is governed by the responses of the monitoring and actuator systems. Given data on these response times, one can evaluate the expected ratio, ∆t ⁄ T , for a particular system and determine, using Table 1 as a starting point for numerical simulation, whether stability will be a problem.

8.4 Optimal linear feedback - time invariant SDOF systems Quadratic performance index The previous sections were concerned with presenting the state-space formulation and numerical procedures for evaluating the time history response of SDOF systems having linear time invariant feedback. Negative velocity feedback was shown to be equivalent to damping, and to result in stable behavior when it is applied in a continuous manner with no time delay. The effect of time delay can be destabilizing under certain conditions that depend on the magnitude of the feedback force and the delay time. Discrete time control works with response data at discrete time points and considers the feedback force to be constant over a time interval. Instability can also occur if the time interval exceeds a critical value that depends on the level of “feedback” induced damping and the time delay. Given

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 587 the feedback parameters, k d and k v , and the delay time, one can determine the limiting time interval using the procedure described in the previous section. The remaining issue that needs to be addressed here is the selection of the magnitude of the feedback parameters. A strategy similar to the approach followed for quasistatic control is introduced in this section. The key step is the formulation of a performance measure which provides a rational basis for comparing different choices for the feedback parameters and establishing the “optimal” solution. The quasi-static formulation presented in Chapter 7 was concerned with controlling a displacement profile, and worked with a quadratic performance measure formed by integrating the square of the displacement error over the structural spatial domain. This measure was extended to reflect the cost of the control forces by incorporating an additional quadratic force term, and introducing weights for the displacement and force terms. By adjusting the weights, one can assess the effect of assigning more emphasis to either the displacement error or the force magnitude. Dynamic control is concerned with controlling the response of the system over a specified time period. Starting at some initial state, X(0), the objective is to minimize the deviation between the actual response, X(t), and the desired response, X * ( t ) , by applying a control force F(t) over the time interval. Considering the integral of the square of the difference between the actual and desired responses to be the measure of the “closeness” of the solution, and including a cost associated with the control force leads to the following quadratic performance index for a SDOF system, 1 J = --2

∫

t*

{ q d ( u – u * ) 2 + q v ( u˙ – u˙ * ) 2 + rF 2 } dt

(8.125)

0

where t * is the time period over which the performance is being measured, and q d, q v, r are the weighting functions for the displacement, velocity and force terms. Specializing F for linear time invariant feedback transforms J to: 1 J = --2

∫

t*

{ q d ( u – u * ) 2 + q v ( u˙ – u˙ * ) 2 + r ( k d u + k v u˙ ) 2 } dt

(8.126)

0

where k d and k v are constants. When the system and weighting parameters are specified, J reduces to a

588 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems function of the feedback parameters, the external excitation, the desired response, and the time period. The optimal feedback for a specific combination of excitation, desired response, and time interval, is defined as the set of values for k d and k v which correspond to a stationary value of J specialized for these conditions. They are determined with the following equations, ∂ * J = 0 ∂ kd

∂ * J = 0 ∂ kv

(8.127)

where J * is J evaluated for specific u, u * , and t * . In general the response depends on k d and k v , as well as the system parameters and external excitation. One option is to evaluate J * numerically and use simulation to locate the “optimal” values for k d and k v . This approach is not very efficient since it requires ranging over both k d and k v . Another option is to apply variational calculus methodology and transform the stationary requirement to a differential equation. In the following section, an example illustrating the first option is presented. This example deals with the special case where u * is taken as 0, the response is due to an initial velocity at t=0, and the time interval extends out to infinity. These conditions are typical for the regulator problem which is concerned with applying control forces to take a system from an arbitrary non-zero initial state X o , to a final state X f at time t f . The regulator problem is encountered when a system is perturbed from its steady-state equilibrium position and there is a need to eliminate the unwanted perturbation as quickly as possible. Since the required force increases as the time interval decreases, it is necessary to include the force cost as well as the response magnitude in the performance index in order to obtain a realistic solution. Subsequent sections describe the variational calculus formulation of the stationary requirement for J as a differential equation, and illustrate its application. This derivation is extended to deal with multi-degree of freedom systems later in the Chapter. An example - linear quadratic regulator control algorithm Suppose a single degree of freedom system is subjected to an impulse at t=0. Considering only velocity feedback, the response is given by u˙ o λ t (8.128) u = -----e R sin λ I t λI

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 589 where u˙ o is the initial velocity, and the lambda terms depend on the system properties and the velocity feedback parameter. λ R = – ξ eq ω 2 ]1 ⁄ 2 λ I = ω [ 1 – ξ eq

(8.129)

kv ξ eq = ξ + ------------ = ξ + ξ a 2ωm Interpreting u as a disturbance from the steady-state position, u * = 0 , the objective is to reduce u to a small quantity, i.e., to regulate the response, in a specified time interval. The linear quadratic regulator control algorithm specialized for only velocity feedback is based on the following form of J, 1 J = --2

∫

∞

{ q d u 2 + q v u˙ 2 + r ( k v u˙ ) 2 } dt

(8.130)

0

and determines the optimal value of k v by enforcing the stationary requirement with respect to k v , ∂J = 0 ∂ kv

(8.131)

Substituting for u in eqn (8.130), considering the weighting functions to be constant, and expressing k v in terms of ξ a leads to qd 1 8ω ------- J = ------------------- -----(8.132) + q v + 4m 2 ω 2 rξ a2 u˙ 2 ( ξ + ξ ) ω2 o

a

The form of the terms within the square bracket suggest the following scale factors for the weights: qd = ω 2 qd qv = qv r r = -----------------4m 2 ω 2

(8.133)

With these definitions, all the superscripted weights have the units of (1/ velocity)2 and the performance index simplifies to ξ a2 qd + qv 8ω ------- J = ---------------- + r -------------(8.134) u˙ 2 ξ+ξ ξ+ξ o

a

a

Finally, enforcing the stationary requirement on J with respect to ξ a results in the

590 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems following equation for the optimal active damping ratio, ξa

qd + qv 1 / 2 = ξ 2 + ----------------–ξ r optimal

(8.135)

Figure 8.12 shows the variation of ξ a with the weighting parameters. Increasing r places more emphasis on minimizing the control force, and ξ a decreases. Increasing the displacement and velocity weights places more emphasis on decreasing the response, and the result is an increase in ξ a . 1

0.9

0.8

ξ = 0

0.7

ξ = 0.1 0.6

ξa

ξ = 0.2

0.5

opt 0.4

0.3

0.2

0.1

0

0

0.1

0.2

0.3

0.4

0.5

0.6

qd + qv qd + ω 2 qv ----------------- = -----------------------r 4ω 4 m 2 r

0.7

0.8

0.9

1

Fig. 8.12: Variation of the optimal velocity feedback with the weighting parameters The previous analysis considered the weighting functions to be constant over the time period during which the performance index is being evaluated. This restriction fixes the relative priorities on the various terms that contribute to the index. If the weights are allowed to vary with time, the priorities can be adjusted

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 591 as to place more emphasis on a certain term at a particular point in time. For example, one may want to suppress the response more aggressively later in the period. This shift in emphasis can be achieved by taking the weights as t q d = ω 2 ------ q d T* t q v = ------ q v T* r r = -----------------4m 2 ω 2

(8.136)

where T * is a time constant that defines the transition for the priority shift. It is convenient to express T * in terms of the fundamental period for the system and a scaling factor, α . 2πα T * = αT = ---------ω The performance index for this case has the following form, ξ a2 qd + qv 8ω 1 ------ J = ---------- 2q d + ---------------------- + r ------------- u˙ 2 4πα ( ξ + ξa ) 2 ξ + ξa o

(8.137)

(8.138)

Requiring J to be stationary with respect to ξ a results in a cubic polynomial for the optimal value of ξ a . 1 qd + qv ξ a ( ξ a + ξ ) ( ξ a + 2ξ ) – ---------- ----------------- = 0 2πα r

(8.139)

Figure 8.13 contains plots corresponding to ξ = 0 and various values of α . Also shown is the plot for the uniform (time) weighting case defined by eqn (8.135). Assuming q d , q v , and r are fixed, the effect of increasing α is to reduce the optimal value of ξ a . This trend is due to the decrease in magnitude of the weighted response during the initial phase of the response where the actual response magnitude is a maximum. Lowering this term decreases the demand on the control force, and consequently the required value of ξ a decreases.

592 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.18

1 α = --4

0.16

0.14

1 α = --2

eqn ( 8.135 )

0.12

ξa

α = 2

α = 1

ξ = 0

0.1

opt 0.08

0.06

0.04

0.02

0

0

0.005

0.01

0.015

0.02

0.025

qd + qv ----------------r a)

0.03

0.035

0.04

0.35

1 α = --4

0.3

1 α = --2

0.25

ξa

eqn ( 8.135 )

α = 1

0.2

α = 2

opt

ξ = 0

0.15

0.1

0.05

0

0

0.02

0.04

0.06

0.08

0.1

qd + qv ----------------r b)

0.12

0.14

0.16

0.18

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 593

0.5

1 α = --4

0.45

1 α = --2 α = 1

0.4

0.35

α = 2

ξ = 0

0.3

ξa

0.25

opt 0.2

eqn ( 8.135 ) 0.15

0.1

0.05

0

0

0.2

0.4

0.6

qd + qv ----------------r

0.8

1

1.2

1.4

c) Fig. 8.13: Variation of the optimal velocity feedback with the time scale factor, α The continuous-time algebraic Riccati equation The previous example illustrated the application of the linear quadratic regulator algorithm for a particular initial state and velocity feedback. In what follows, the performance index is expressed in matrix form, and the analysis is extended to be applicable for both displacement and velocity feedback, and arbitrary initial conditions. The scalar form of the performance index for the linear quadratic regular problem is obtained by setting u * = u˙ * = 0 in eqn (8.126) and taking t * = ∞ . 1 J = --2

∞

∫

0

[ q d u 2 + q v u˙ 2 + r ( k d u + k v u˙ ) 2 ] dt

(8.140)

594 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Introducing the state vector notation, J is expressed as 1 J = --2

∞

∫

X T ( Q + K Tf RK f )X T dt

(8.141)

0

where Q and R are diagonal weighting matrices, and K f is the linear state feedback matrix.

Q =

qd qv

(8.142)

R = r[1] K f = kd kv

The governing differential equation for free vibration response follows from eqn (8.33). X˙ = ( A – B f K f )X = A c X

(8.143)

Solutions of eqn (8.143) satisfy X ( t ) → 0 as t → ∞

(8.144)

when the feedback parameters ( k v, k d ) are selected such that the eigenvalues of A c have a negative real part. When the weighting matrices are taken to be constant, an explicit expression for J can be obtained by expressing the integrand as X T ( Q + K Tf RK f )X = –

d ( X T HX ) dt

(8.145)

where H is a symmetric positive-definite second order matrix. Noting eqn (8.143), the right hand side expands to d ( X T HX ) = X˙ T HX + X T H X˙ = X T ( A cT H + H A c )X dt

(8.146)

and it follows that A cT H + H A c = – ( Q + K Tf RK f ) With this change in variables, the integral reduces to

(8.147)

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 595 1 J = --- X T ( 0 )HX ( 0 ) 2

(8.148)

Requiring J to be stationary with respect to K f and noting that X ( 0 ) is independent of K f leads to δH = 0 . The differential of H is generated by operating on eqn (8.147). A cT δH + δH A c + δA cT H + HδA c = – δK Tf RK f – K Tf RδK f

(8.149)

Substituting for δA c , δA c = – B f δK f

(8.150)

and setting δH = 0 , the stationary condition takes the form δK Tf ( B Tf H – RK f ) + ( H B f – K Tf R )δK f = 0

(8.151)

For eqn (8.151) to be satisfied for arbitrary δK f , the coefficients must vanish. Then the optimal feedback is related to H by R – 1 B Tf H = K f optimal

(8.152)

The last step involves substituting for K f in eqn (8.147). This operation leads to the following definition equation for H, which is called the continuous time algebraic Riccati equation A T H + HA – H B f R – 1 B Tf H = – Q

(8.153)

Given the system parameters ( A, B f ) , one specifies the weights (Q and R), and solves for the elements of H. Note that by definition, H is symmetric and positive definite. With H known, the feedback matrix is determined with eqn (8.152). Substituting for A and B f , A =

0

1

–ω 2

– 2ξω

0 Bf = 1 ---m and expressing H as

(8.154)

596 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

H =

H 11 H 12 H 12 H 22

(8.155)

in eqn (8.153) yields the following 3 equations for the elements of H 1 2 --------- H + 2ω 2 H 12 – q d = 0 rm 2 12 1 --------- H H – H 11 + 2ξωH 12 + ω 2 H 22 = 0 rm 2 12 22

(8.156)

1 2 --------- H + 4ξωH 22 – 2H 12 – q v = 0 rm 2 22 The elements of H are also constrained by the requirement that H must be positive definite. H 11 > 0

H 22 > 0

2 >0 H 11 H 22 – H 12

(8.157)

Given H, the elements of K f are determined with eqn (8.152). H 12 k d = ---------rm H 22 k v = ---------rm

(8.158)

Solving for H 12 and H 22 , and enforcing the positive definite requirement leads to 1/2 qd H 12 = rω 2 m 2 – 1 + 1 + ---------------- rω 4 m 2

qv H 12 1 / 2 2 2 H 22 = 2rωm – ξ + ξ + -------------------- + -------------------- 4rω 2 m 2 2rω 2 m 2

(8.159)

The equivalent frequency and damping for the SDOF system with feedback corresponding to eqn (8.158) are determined with eqn (8.37) and (8.158).

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 597 1/4 kd 1 / 2 qd ω eq = ω 1 + ----------⇒ ω 1 + ----------------ω2m rω 4 m 2

(8.160) 1 / 2 1 / 2 kv qd qv ω 2 ω 1 ξ eq = ξ -------- + ----------------- ⇒ -------- ξ + -------------------- + --- – 1 + 1 + ---------------- ω eq ω eq 2mω eq 4rω 2 m 2 2 rω 4 m 2

Weighting the displacement introduces additional stiffness and increases the frequency. Weighting the velocity generates additional damping, but has no effect on the frequency. It follows that pure velocity feedback corresponds to q d = 0, q v ≠ 0 . The active damping ratio for this case is given by: H 22 q v 1 / 2 2 ξ a = ----------------- = – ξ + ξ + -------------------- 2rm 2 ω 4rω 2 m 2

(8.161)

This result coincides with eqn (8.135) which was generated by integrating the analytic solution for the impulse generated response. These solutions must coincide since the approach followed earlier is a simplified version of this procedure. The matrix formulation presented above is also applicable to multi-degree of freedom systems. MATLAB has a function called CARE (Continuous -time Algebraic Riccati Equation) which generates H, K f , and the eigenvalues of A – B f K f for specified values of A, B f , Q and R. One option for scaling Q and R is to use the factors defined by eqn (8.133). The corresponding forms are

Q =

qd qv

=

ω 2 qd 0 0

qv

(8.162)

r R = [ r ] = -----------------4m 2 ω 2 where q d , q v , and r range from 0 to ≈ 1. MATLAB’s default option is to assume R ≡ I . In this case, taking Q as:

598 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Q

= R≡I

4ω 4 m 2 q d

0

0

4m 2 ω 2 q v

(8.163)

maintains the same range ( 0 → 1 ) for q d and q v . The form of the continuous algebraic Riccati solution corresponding to r = 1 and q d , q v defined by eqn (8.163) is kd ----- = – 1 + [ 1 + 4q d ] 1 / 2 k kv kd 1 / 2 ------------ = ξ a = – ξ + ξ 2 + q v + -----2ωm 2k

(8.164)

A similar scaling strategy can be employed for a multi-degree-of-freedom system. In this case the magnitude of the weighting factor for each of the displacement and velocity variables can be independently assigned. The discrete time algebraic Riccati equation The discrete time formulation of the algebraic Riccati equation is based on a performance index that involves a summation of weighted response terms evaluated at discrete times. The discrete form of the performance index corresponding to the same assumptions introduced for the continuous time formulation is ∞

1 J = --2

∑

T XT j ( Q + K f RK f )X j

(8.165)

j=0

where X j is determined with eqn (8.86) which is written here as X j + 1 = c1 X j + c2 F j c 1 = e A∆t

c 2 = A – 1 ( e A∆t – I )B f

(8.166)

Taking F j = – K f X j , eqn (8.166) reduces to X j + 1 = ( c 1 – c 2 K f )X j = cX j

(8.167)

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 599 Finally, eqn (8.165) is expressed as ∞

1 T 1 T j j T ( c ) ( Q + K f RK f )c X o = --- X oT H X o J = --- X o 2 2 j = 0

∑

(8.168)

where H is a symmetrical positive definite matrix. Stability requires that c j approach 0 as j → ∞ . Noting this constraint, the expression for H reduces to (see problem 8.11) H – c T Hc = Q + K Tf RK f

(8.169)

Imposing the stationary requirement on J leads to 1 δJ = --- X oT δH X o = 0 for arbitrary δK f 2 ⇒

(8.170)

δH = 0 Operating on eqns (8.169) and (8.166), δH – c T δHc – δc T Hc – c T Hδc = δK Tf RK f + K Tf RδK f

(8.171)

δc = – c 2 δK f

(8.172)

and setting δH = 0 results in the following expression for the optimal feedback matrix Kf

optimal

= ( R + c 2T H c 2 ) – 1 c 2T H c 1

(8.173)

The final form of the discrete algebraic Riccati equation is H – c 1T H c 1 + ( c 1T H c 2 ) ( R + c 2T H c 2 ) – 1 ( c 1T H c 2 ) T = Q

(8.174)

Given m, k, c, and ∆t, c 1 and c 2 can be evaluated. One specifies the weighting matrices R and Q and then determines H. MATLAB has a function called DARE which computes H, K f , and the eigenvalues of c specialized for opt K f set equal to the optimal value. The scaling strategies discussed earlier for the continuous time case are also applicable for the discrete time formulation. DARE assumes the default value of R to be I and therefore eqn (8.163) can be used to determine the corresponding value of Q. The following example describes the

600 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems solution procedure in considerable detail.

Example 8.4: Solution of the discrete time algebraic Riccati equation for a SDOF system The undamped case is considered first. Noting the results presented in Example 8.2, the various coefficient matrices are:

c1 =

cos ω – ω sin ω

sin ω -----------ω cos ω

(1)

1 c 2 = ----------- 1 – cos ω mω 2 ω sin ω

(2)

1 f c 1T H c 2 = ----------- 1 mω 2 f 2

(3)

1 c 2T H c 2 = -------------- f 3 2 m ω4

(4)

∆t ω = ω∆t = 2π -----T

(5)

where

f 1 = H 11 ( cos ω ) ( 1 – cos ω ) + H 12 ( ω sin ω ) ( 2 cos ω – 1 ) – H 22 ( ω sin ω ) 2 sin ω ( 1 – cos ω ) f 2 = H 11 ---------------------------------------- + H 12 [ 2 ( sin ω ) 2 – 1 + cos ω ] ω + H 22 ( ω cos ω sin ω)

(6)

(7)

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 601 f 3 = H 11 [ 1 – cos ω ] 2 + H 12 [ 2ω sin ω ( 1 – cos ω ) ] + H 22 [ ω sin ω ] 2

(8)

The optimal feedback matrix is determined using eqn (8.173). K f = kd

1 k v = ----------------------------------------- f 1 f3 mω 2 r + -------------- m 2 ω 4

f2

(9)

Finally, the elements of H are obtained by expanding eqn (8.174). This step leads to the following 3 scalar equations: ( sin ω ) 2 H 11 + [ 2ω sin ω cos ω ]H 12 – [ ω 2 sin2 ω ]H 22 f 12 = q d – -----------------------------f 3 + ω4m2r sin ω cos ω – -------------------------- H 11 + [ 2 sin2 ω ]H 12 + [ ω sin ω cos ω ]H 22 ω –f1f2 = -----------------------------f 3 + ω4m2r

(10)

2 sin ω cos ω sin2 ω – --------------- H 11 – ------------------------------ H 12 + [ – cos2 ω ]H 22 ω ω2 f 22 = q v – -----------------------------f + ω4m2r 3

The above equations are reduced to a dimensionless form by scaling the various terms according to the following laws

602 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems f 1 = ω4m2r f 1

q d = ω 4 m 2 rq d

f 2 = ω3m2r f 2

q v = ω 2 m 2 rq v

f 3 = ω4m2r f 3

(11)

H 11 = ω 4 m 2 rH 11 H 12 = ω 3 m 2 rH 12 H 22 = ω 2 m 2 rH 22

Using eqn (11), the expression for the optimal feedback matrix takes the form ω2m K f = --------------- f 1 1 + f3

1 ---- f 2 ω

(12)

Equations (10) reduce to 2

) sin2 ω +

f1 2 sin ω cos ω H 12 = 4q d – --------------1 + f3

) sin2 ω +

f2 2 sin ω cos ω H 12 = – 4 q v + --------------1 + f3

( H 11 – H 22

2

( H 11 – H 22

(13)

f1f2 ( H 11 – H 22 ) sin ( ω cos ω – 2 sin2 ωH 12 ) = --------------1 + f3 Lastly, the f terms are determined with f 1 = H 11 cos ω ( 1 – cos ω ) + H 12 sin ω ( 2 cos ω – 1 ) – H 22 sin2 ω f 2 = H 11 sin ω ( 1 – cos ω ) + H 12 [ 2 sin2 ω – 1 + cos ω ] + H 22 sin ω cos ω

(14)

f 3 = H 11 ( 1 – cos ω ) 2 + H 12 ( 2 sin ω ( 1 – cos ω ) ) + H 22 sin2 ω One specifies the relative weights ( q d, q v ) , and the relative time step ratio, ω = 2π ( ∆t ⁄ T ) , solves eqn (13) for the elements of H , computes the optimal feedback with eqn (12), and lastly checks for stability by evaluating the

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 603 eigenvalues of c 1 – c 2 K f . As ∆t → 0, the solution for the discrete formulation approaches the solution for the continuous case. This can be shown by introducing second order approximations for cos ω , sin ω and letting ω approach 0. Assuming ω 2 << 1 , the trigonometric terms are replaced with cos ω ≈ 1 sin ω ≈ ω

(15)

The corresponding equations are f 1 ≈ ωH 12 + O ( ω 2 ) f 2 ≈ ωH 22 + O ( ω 2 )

(16)

f 3 ≈ ω 2 H 22 + O ( ω 3 ) and 2

ω 2 ( H 11 – H 22 ) + 2ωH 12 ≈ 4q d – ω 2 H 12 2

ω 2 ( H 11 – H 22 ) + 2ωH 12 ≈ – 4q v + ω 2 H 22

(17)

ω ( H 11 – H 22 ) – 2ω 2 H 12 ≈ ω 2 H 11 H 22 Based on eqn (17), the limiting solution is 2ωH 12 ≈ 4q d 2

2 ω 2 H 22 ≈ 4q v + 4q d – ω 2 H 12

(18)

≈ 4q v + 4q d ( 1 – q d ) Noting eqn (16), the limiting forms for f 1, f 2 are f 1 ≈ 2q d f 2 ≈ 2 { qv + qd ( 1 – qd ) } 1 / 2 Finally, the feedback terms are

(19)

604 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

K f = 2ω 2 m q d

2ωm { q v + q d ( 1 – q ) } 1 / 2 d

(20)

Setting q d = 0 produces pure velocity feedback with v

= 2ωm ( q v ) 1 / 2

(21)

The corresponding damping ratio is 1/2 qv kv ξ a = ------------ = ( q v ) 1 / 2 = -------------------- 4ω 2 m 2 r 2ωm

(22)

This result coincides with eqn (8.161). In the discrete case, setting q d = 0 does not produce pure velocity feedback for finite ω . However, when ω is small, f 1 is small and k d is usually neglected. To determine the behavior of the feedback parameters, DARE is used to solve the discrete algebraic Riccati equation for a set of damped SDOF systems, taking a range of values for q v and ∆t ⁄ T , with q d set to 0, q v scaled according to eqn (11), and r set to 1. The results for k d ⁄ k are plotted in Fig 1. Figure 2 shows the variation of k v ⁄ ( 2ωm ) .

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 605

0

ξ = 0.02 qd = 0

−0.05

q v = 4m 2 ω 2 q v

−0.1

kd ⁄ k

−0.15

q v = 0.01 −0.2

q v = 0.03 −0.25

q v = 0.05 q v = 0.07

−0.3

−0.35

0

0.1

0.2

0

0.3

∆t ⁄ T a)

0.4

0.5

0.7

q v = 4m 2 ω 2 q v

ξ = 0.05 qd = 0

−0.05

0.6

−0.1

q v = 0.01

kd ⁄ k

−0.15

q v = 0.03 −0.2

q v = 0.05 q v = 0.07

−0.25

−0.3

−0.35

0

0.1

0.2

0.3

∆t ⁄ T b)

0.4

0.5

0.6

0.7

606 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0

q v = 4m 2 ω 2 q v

ξ = 0.10 qd = 0

−0.05

q v = 0.01 q v = 0.03

kd ⁄ k

−0.1

q v = 0.05 −0.15

q v = 0.07

−0.2

−0.25

0

0.1

0.2

0.3

∆t ⁄ T c)

0.4

0.5

0.6

Figure 1: Displacement feedback predicted by the discrete algebraic Riccati equation for a SDOF system

0.7

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 607

k v ⁄ ( 2mω )

0.3

0.25

q v = 0.07

0.2

q v = 0.05

0.15

q v = 0.03

0.1

q v = 0.01

ξ = 0.02 qd = 0

q v = 4m 2 ω 2 q v

0.05

0

−0.05

−0.1

−0.15

−0.2

0

0.1

0.2

0.25

k v ⁄ ( 2mω )

q v = 0.05

0.15

q v = 0.03

0.1

q v = 0.01

∆t ⁄ T a)

0.4

0.5

0.6

0.7

q v = 4m 2 ω 2 q v

ξ = 0.05 qd = 0

q v = 0.07 0.2

0.3

0.05

0

−0.05

−0.1

−0.15

0

0.1

0.2

0.3

∆t ⁄ T b)

0.4

0.5

0.6

0.7

608 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.25

0.2

q v = 0.07 q v = 0.05

0.15

k v ⁄ ( 2mω )

q v = 4m 2 ω 2 q v

ξ = 0.10 qd = 0

q v = 0.03

0.1

q v = 0.01

0.05

0

−0.05

−0.1

0

0.1

0.2

0.3

∆t ⁄ T c)

0.4

0.5

0.6

0.7

Figure 2: Velocity feedback predicted by the discrete algebraic Riccati equation In general, the feedback parameters decrease with increasing damping in the “uncontrolled” system. As expected, increasing q v with r held constant places more emphasis on lowering the magnitude of the response and consequently the magnitudes of the feedback terms increase. As ∆t ⁄ T approaches 0, the solution tends toward the “continuous” solution given by eqn (8.160). As ∆t ⁄ T increases, k v decreases, and eventually becomes negative in the region of ∆t ⁄ T ≈ 0.35 . The feedback stiffness term is also negative. All these solutions are stable, i.e., the maximum modulus of the eigenvalues of c 1 – c 2 K f is less than unity. The pattern of behavior near ∆t ⁄ T = 0.5 was also observed in the stability analysis for the discrete state-space formulation presented in Section 8.3 It should be noted that the typical value for ∆t ⁄ T is small, on the order of 0.02. For this range, it is reasonable to take ( k d ⁄ k ) ≈ 0 . For seismic excitation, the minimum value of ∆t is controlled by the ground motion data. The typical time interval for acceleration

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 609 time history records is 0.02 sec.

Finite interval discrete time algebraic Riccati equation The algebraic Riccati equations presented in the previous sections are based on performance indices which involve integration over an infinite time interval. By considering the time interval to be finite, one can generate a spectrum of Riccati type equations corresponding to specific values of the time interval. The minimum value is equal to the time step used to generate the discrete time solution; the maximum value is equal to infinity. Eqn (8.174) corresponds to the latter choice. An equation corresponding to the first choice is derived below. The discrete quadratic cost function associated with the time interval between t j and t j + 1 is a convenient choice. 1 J j, j + 1 = --- [ X T QX + FT j R jF j] 2 j+1 j j+1

(8.175)

It allows for updating the system properties and weighting parameters at each discrete time and consequently is applicable for adaptive as well as time invariant systems. The equilibrium equation follows from eqn (8.166). X j + 1 = c 1, j X j + c 2, j F j

(8.176)

Requiring J j, j + 1 to be stationary with respect to F j , T δJ j, j + 1 = δX T j + 1 Q j X j + 1 + δF j RF j = 0

(8.177)

and noting that δX j + 1 = c 2, j δF j

(8.178)

results in R j F j + c 2T, j Q j X j + 1 = 0

(8.179)

Finally, substituting for X j + 1 yields the following linear feedback law for F j F j = – ( R j + c 2T, j Q j c 2, j ) – 1 ( c 2T, j Q j c 1, j )X j = – K f , j X j

(8.180)

This expression is similar to eqn (8.173) which is based on an infinite time

610 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems interval. Replacing H with Q reduces eqn (8.173) to eqn (8.180). Working with the latter equations is more convenient since it avoids solving the nonlinear Riccati equation. The following example illustrates the expanded form of K f for the undamped case.

Example 8.5: Example 8.4 revisited The various matrices derived in ex 8.4 are applicable for this case when H is replaced by Q. The resulting forms are (the subscript j is dropped here to simplify the notation): f1 1 K f = ----------------------- ----------f 3 mω 2 r + -------------m2ω4

f2 ----------mω 2

(1)

q d = ω 4 m 2 rq d

(2)

q v = ω 2 m 2 rq v

(3)

f1 ----------- = rmω 2 ( q d cos ω ( 1 – cos ω ) – q v sin2 ω ) mω 2

(4)

f2 ----------- = rmω ( q d sin ω ( 1 – cos ω ) + q v sin ω cos ω) mω 2

(5)

f3 r + -------------- = r ( 1 + q d ( 1 – cos ω ) 2 + q v sin2 ω ) m2ω4

(6)

Expressing K f as K f = kd kv

(7)

and using the above equations leads to the dimensionless form of the feedback parameters

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 611 q d cos ω ( 1 – cos ω ) – q v sin2 ω kd ----------- = -------------------------------------------------------------------------mω 2 1 + q d ( 1 – cos ω ) 2 + q v sin2 ω

(8)

q d sin ω ( 1 – cos ω ) + q v sin ω cos ω kv ------------ = ------------------------------------------------------------------------------------2mω 2 ( 1 + q d ( 1 – cos ω ) 2 + q v sin2 ω )

(9)

These equations show that k d depends mainly on q d , and k v on q v. The limiting forms for q d = 0 are: – sin2 ω q v kd ----------- = -----------------------------mω 2 1 + sin2 ω q v

(10)

sin ω cos ω q v kv ------------ = -------------------------------------2mω 2 ( 1 + sin2 ω q v)

(11)

Suppose ∆t ⁄ T = 0.02 . The corresponding values of the feedback parameters are: – 0.015775 q v kd ----------- = ------------------------------------1 + 0.015775 q v mω 2

(12)

0.1243q v kv ------------ = ---------------------------------2mω 2 + 0.03155q v

(13)

Taking q v = 1 results in an active damping ratio of ≈ 0.06 . The change in stiffness is on the order of 1.5%, and can be neglected.

Continuous time Riccati differential equation The algebraic Riccati equation is based on eqn (8.141), which involves an integral extending from t = 0 to t = ∞ . A more general formulation considers the time interval to be finite, and allows for a non-zero final state. Suppose the initial state at t = t o is X o . Let t f denote the final time, and X *f the desired final state. The generalized form of the performance index is written as

612 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

∫

tf

1 1 J = --( X T QX + F T RF ) dt + --- ( X f – X *f ) T S ( X f – X *f ) 2 t 2 o

(8.181)

where S is a symmetric positive semi-definite matrix which provides a measure of the error associated with the boundary conditions at t f . Given some initial state the optimal control force F(t) is determined by requiring J to be stationary. Although the solution procedure for t f ≠ ∞ is more complicated that for t f = ∞ , the optimal feedback law has the same form, F ( t ) = – ( R – 1 B f H )X

(8.182)

However, H must now satisfy a differential equation rather than an algebraic equation, H˙ + A T H + H A T – H B f R – 1 B Tf H = – Q

(8.183)

and boundary conditions at t = t f . H (t f ) = S

(8.184)

These equations are derived in the following section. In addition to eqn (8.183), the state vector and control force are constrained by X˙ = AX + B f F X ( to ) = X o

to ≤ t ≤ t f

(8.185)

Since the boundary condition on H is at t = t f , one has to solve for H backward in time, i.e., from t f toward t o . Given H, one solves for X and F by moving forward in time from t o . The nature of the solution near t f is similar to an exponential decay to a steady state solution. Since solving eqn (8.183) is not convenient, H(t) is approximated by the steady state solution obtained with the algebraic Riccati equation. This simplification is possible only when A, Q, and R are constant. Variational formulation of the continuous time Riccati equation For completeness, the derivation of the Riccati equation is presented here. The proof is based on the requirement that A , B f , Q , and R are bounded and piecewise-continuous in the interval t o ≤ t ≤ t f . Assuming an optimal control, F ,

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 613 exists, then an optimal trajectory, X , which satisfies ˙ X = AX + B f F

(8.186)

X ( to ) = X o also exists. The corresponding optimal cost functional is given by t

f T T T 1 J ( F ) = --- [ X f – X ∗ ] S [ X f – X ∗ ] + [ X QX + F RF ]dt 2 to

∫

(8.187)

The proof proceeds as follows. Let F = F + δF = F + εF˜

(8.188)

where ε is an arbitrary positive or negative scalar and F˜ is an arbitrary valued time function representing a perturbation of the control force from the optimal value. Substituting for F , eqn (8.185) expands to X˙ = AX + B f F + εB f F˜ X ( to ) = X o

(8.189)

The solution for X can be expressed in terms of the optimal trajectory and a perturbation due to the variation in the control force, ˜ X = X + δX = X + εX

(8.190)

˜ is related to the control perturbation F˜ by where the state perturbation X ˜˙ = AX ˜ + B F˜ X f ˜ (t ) = O X o

(8.191)

The constraint equation can be incorporated in the cost function by introducing a Lagrangian multiplier, β ( t ) , and treating β as a variable as well as X and F (Strang, 1993). The modified cost function is

614 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems T 1 J ( F ) = --- ( [ X – X f ] S [ X – X f ] ) 2 tf

+

∫

1--- X T QX + 1--- F T RF + β T [ AX + B F – X˙ ] dt f 2 2

(8.192)

to

Noting that tf

tf

∫

T T T β X˙ dt = β ( t f )X ( t f ) – β ( t o )X ( t o ) –

to

∫

T β˙ Xdt

(8.193)

to

one can write eqn (8.192) as T T T 1 J ( F ) = --- [ X ( t f ) – X ∗ ] S [ X ( t f ) – X ∗ ] – β ( t f )X ( t f ) + β ( t o )X ( t o ) 2 tf

+

∫

T T T 1 T 1 T --- X QX + β AX + β˙ X + --- F RF + β B f F dt 2 2

(8.194)

to

The cost for optimal control is T T T 1 J ( F ) = --- [ X ( t f ) – X ∗ ] S [ X ( t f ) – X ∗ ] – β ( t f )X ( t f ) + β ( t o )X ( t o ) 2 tf

+

∫

T T T 1 T 1 T --- X QX + β AX + β˙ X + --- F RF + β B f F dt 2 2

(8.195)

to

Since by definition, F and X are optimal, the cost difference must be positive, J(F ) – J(F ) ≥ 0

(8.196)

for all F , t o ≤ t ≤ t f . Substituting for F and X in terms of the perturbations, one obtains an expansion for the cost difference in terms of ε .

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 615 2

T T ε ˜T ˜ ( t ) + ---˜ (t ) - X ( t f )SX J ( F ) – J ( F ) = ε ( [ X ( t f ) – X ∗ ] S – β ( t f ) )X f f 2 tf

∫

T ˜ T T T T + ε ( [ X Q + β A + β˙ ]X + [ F R + β B f ]F˜ ) dt to

(8.197)

tf

∫

T ε2 ˜ T QX ˜ ]dt + ----- [ F˜ RF˜ + X 2 to

A necessary condition for eqn (8.196) to be satisfied is that the first order terms in ˜ , and F˜ . This requirement leads to a first order differential ε vanish for all ε , X equation for β T T T X Q + β A + β˙ = O T

(8.198)

T

[ X ( t f ) – X∗ ] S – β ( t f ) = O and the control law T

T

F R + β Bf = O ⇒

(8.199) –1

F = – R B Tf β The sufficient condition for eqn (8.196) is related to the second order terms. tf 2 ε ˜T ˜ ( t ) + [ F˜ T RF˜ + X ˜ T QX ˜ ]dt J ( F ) – J ( F ) = ----- X ( t f )SX f 2 to

∫

(8.200)

Since S and Q are positive semi-definite and R is positive definite, this term is always positive, and it follows that X and F represent the solution that corresponds to the minimum value of J(F). Equations (8.186) and (8.198) can be combined into a set of coupled linear vector differential equations involving X and β .

616 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

˙ X = β˙

–1 T Bf X T β

–B f R

A –Q

–A

(8.201)

The solution can be expressed as X ( t2 ) β ( t2 )

= Φ ( t 2, t 1 )

X ( t1 ) β ( t1 )

(8.202)

where t 1 and t 2 are arbitrary times, and Φ is a 2n x 2n state transition matrix of the form Φ 11 ( t 2, t 1 ) Φ 12 ( t 2, t 1 ) Φ ( t 2, t 1 ) = (8.203) Φ 21 ( t 2, t 1 ) Φ 22 ( t 2, t 1 ) Taking t 1 = t and t 2 = t f one obtains X ( t f ) = Φ 11 ( t f , t )X ( t ) + Φ 12 ( t f , t )β ( t ) β ( t f ) = Φ 21 ( t f , t )X ( t ) + Φ 22 ( t f , t )β ( t )

(8.204)

Using the boundary condition, β( t f ) = S[ X ( t f ) – X∗ ]

(8.205)

one can solve for β ( t ) in terms of X ( t ) [ SΦ 12 ( t f , t ) – Φ 22 ( t f , t ) ]β ( t ) = [ Φ 21 ( t f , t ) – SΦ 11 ( t f , t ) ]X ( t ) + SX * (8.206) The above relation is expressed as β ( t ) = H ( t )X ( t ) + G ( t )

(8.207)

Finally, substituting for β ( t ) in eqn (8.199) results in a linear state feedback control law –1 T

F ( t ) = –R B f [ H X + G ]

(8.208)

The equation for H is derived by operating on the governing equations. Differentiating eqn (8.207), one obtains

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 617 β˙ = H X˙ + H˙ X + G˙

(8.209)

Substituting for X˙ , β˙ expands to –1 T –1 T β˙ = [ H˙ + HA – H B f R B f H ]X + G˙ – H B f R B f G

(8.210)

Finally, introducing β and β˙ in eqn (8.198), –1 T

T

[ H˙ + HA + A H – H B f R B f H + Q ]X T

+ [A –

–1 T H B f R B f ]G

(8.211)

+ G˙ = O

and setting the coefficient of X to zero results in equations for H and G T –1 T H˙ + HA + A H – H B f R B f H + Q = O

(8.212)

T –1 T G˙ + [ A – H B f R B f ]G = O

(8.213)

The boundary conditions follow from the condition on β at t f and have the form H (t f ) = S

(8.214)

G ( t f ) = – SX ∗

For the regulator problem, X ∗ is taken as zero. Then, G = O and β = HX . As a final point, one can show that the minimum remaining cost at time t is equal to 1 T J ( F ) = --- [ X ( t )HX ( t ) ] 2

(8.215)

Example 8.6: Application to scalar case To illustrate the solution procedure, the scalar form of the LQR formulation is considered here. The equations for t o < t < t f are : X˙ – AX – B f F = O

(1)

618 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems β˙ + Aβ + QX = O

(2)

1 F = – ---- B f β R

(3)

At t = t o X ( to ) = X o

(4)

At t = t f β ( t f ) = S [ X ( t f ) – X∗ ]

(5)

Substituting for F transforms eqn (1) to 2

Bf X˙ – AX + ------ β = 0 R

(6)

For stability, A < 0 . The homogeneous solution has the form X = Ce β = De

λt

(7)

λt

(8)

where λ satisfies 2

QB f λ = A + ----------R 2

2

(9)

and the constants are related by λ+A C = – -------------- D Q

(10)

λ 1, 2 = ± λ

(11)

Then

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 619

2

λ =

QB f A + ----------R 2

(12)

A + λi C i = – --------------- D i = ν i D i Q

(13)

and the full solution is β = D1 e

λt

+ D2 e

X = ν1 D e 1

λt

– λt

(14)

+ ν2 D e

– λt

(15)

2

Imposing the end conditions results in two equations for D 1 and D 2 . ν1 D1 + ν2 D2 = X o

(16)

[ 1 – Sν 1 ]D 1 + [ 1 – Sν 2 ]D 2 e

– 2λt f

= –e

– λt f

SX∗

(17)

The case of finite t f is handled by expressing D 1 as D1 = D1 e

– λt f

(18)

This substitution transforms the solution to β = D1 e

–λ ( tf – t )

X = ν1 D1 e

+ D2 e

–λ ( tf – t )

– λt

(19)

+ ν2 D e 2

– λt

(20)

The first term applies near t f and the second term near t o . Enforcing the boundary conditions leads to – λt f 1 D 2 = ----- X o – ν 1 D 1 e ν2

(21)

620 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Xo – λt f – SX∗ – ------- [ 1 – Sν 2 ]e ν2 D 1 = -------------------------------------------------------------------------------ν1 – 2λt f [ 1 – ν 1 S ] – ----- [ 1 – ν 2 S ]e ν2

(22)

When λt f > 3 , the solutions uncouple as follows: 3 For 0 < t ≤ --λ β = D2 e X = ν2 β

– λt

Bf Bf F = – ----- β = – ---------- X R ν2 R

(23)

Xo D 2 ≈ ------ν2 2

QB f A–λ A A ν 2 = – ------------- = – ---- + ------- 1 + ----------2 Q Q Q A R 3 3 For --- < t < t f – --λ λ β = 0 X = 0 F = 0 3 For t f – --- < t ≤ t f λ

(24)

8.4 Optimal Linear Feedback-Time Invariant SDOF Systems 621 –λ ( tf – t )

β = D1 e X = ν1 β

Bf Bf F = – ----- β = – ---------- X R ν1 R

(25)

– SX∗ D 1 = -----------------1 – ν1 S 2

QB f A+λ A A ν 1 = – -------------- = – ---- – ------- 1 + ----------2 Q Q Q A R Stability of the controlled system requires A < 0 . Then, for the stable case ν2 > 0

(26)

ν1 < 0

The feedback is negative near t = 0 and positive near t = t f . Letting X∗ → 0 eliminates the solution in the region of t f . The algebraic Riccati equation approach for this set of equations starts with the scalar form of eqn (8.212) 2

–H

2

Bf ------ + 2AH + Q = 0 R

(27)

The roots are R H 1, 2 = ------ A ± A 2 Bf

2

QB f 1 + ----------2 RA

(28)

Enforcing the requirement that the uncontrolled system be stable, A<0 transforms eqn (28) to

(29)

622 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

2

QB f RA H 1, 2 = ----------- – 1 ± 1 + ----------2 2 Bf RA

(30)

Finally, the requirement that H > 0 eliminates the negative root, and one obtains RA H opt = ----------2 Bf

2

QB f 1 + ----------- – 1 2 RA

(31)

Referring back to eqn (23), the solution near t = 0 for the case where t f is large has the form – λt

X = Xoe X o – λt β = ------- e ν2

(32) 2

QB f A A ν 2 = – ---- + ------- 1 + ----------2 Q Q RA Taking A < 0 , the expression for ν 2 becomes 2

QB f A ν 2 = ------- 1 + 1 + ----------2 Q RA

(33)

Forming the ratio of β to X , β 1 ---- = ----X ν2

(34)

and substituting for ν 2 leads to RA 1 Q ----- = ----------------------------------------------- = ----------2 ν2 2 Bf QB f A 1 + 1 + ----------2 RA

2

QB f 1 + ----------- – 1 2 RA

(35)

8.5 State-space Formulation for MDOF Systems 623 Comparing eqn (35) with eqn (31) shows that H opt = 1 ⁄ ν 2 . The negative root of eqn (30) is equal to 1 ⁄ ν 1 , and corresponds to positive feedback in the region of t f .

8.5 State-space formulation for MDOF systems Notation and governing equations The material presented in the previous sections can be readily extended to the case of a multi-degree-of-freedom system. One has only to generalize the definition equations for the various matrices involved in the state-space representation. In what follows, the essential steps for an nth order linear system are discussed. Using the notation defined in Chapter 2, the equations for an nth order linear system subjected to seismic excitation and a set of r control forces applied at various locations on the system are written as ˙˙ + CU˙ + KU = – MEa + E F + P MU g f

(8.216)

where E f is an n x r matrix that defines the location of the control forces with respect to the degrees of freedom. Figure 8.14 illustrates the notation for the case of a lumped mass model of a shear beam having four degrees of freedom and two control forces. The initial conditions involve constraints on the displacements and velocities at time t = 0 . U ( 0 ) = Uo U˙ ( 0 ) = U˙ o

(8.217)

Following the approach of section 8.2, the state-space form for eqn (8.216) is taken as X˙ = AX + B f F + B g a g + B p P

(8.218)

624 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

u4

m4

F1 u3

m3

u1 u2

m2

U

F2 u1

=

u2 u3 u4

1 1 E = 1 1

0 0 Ef = 0 1

0 1 0 0

m1

Fig. 8.14: 4DOF system with two control forces where X is now a vector of order 2n, X =

U U˙

(8.219)

and the coefficient matrices are given by A =

O –1

I –1

(2n x 2n)

(8.220)

(2n x r)

(8.221)

(2n x 1)

(8.222)

(2n x 1)

(8.223)

–M K –M C

Bf =

Bg =

Bp =

O –1

M Ef

O –E

O M –1

8.5 State-space Formulation for MDOF Systems 625 Free vibration response - time invariant uncontrolled system Specializing eqn (8.218) for no external forcing leads to a set of 2n homogeneous first order differential equations. X˙ = AX

(8.224)

Considering A to be constant, the general solution has the form X = e λt V

(8.225)

where λ and V satisfy the following set of 2n homogeneous algebraic equations ( A – λI )V = 0

(8.226)

Substituting for A, eqn (8.226) expands to – λI

I

– M –1 K

– M – 1 C – λI

V = 0

The solution of eqn (8.227) has the form Φ V = λΦ

(8.227)

(8.228)

where λ and Φ are constrained by n conditions, λ 2 Φ + λM – 1 CΦ + M – 1 KΦ = 0

(8.229)

For the SDOF case, the complete solution of eqn (8.229) consists of 2 solutions involving a pair of complex roots for λ λ = λ1 λ = λ˜ 1

Φ = 1 Φ = 1

(8.230)

Example 3.13 dealt with a 2 DOF system. In this case, there are 4 solutions involving 2 pairs of complex roots:

626 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems λ = λ 1, λ˜ 1 and λ = λ , λ˜ 2

˜ Φ = Φ 1, Φ 1 (8.231) ˜ Φ = Φ 2, Φ 2

2

Then, for an n’th order system, it follows that there are 2n complex solutions involving n pairs of complex roots. λ = λ 1, λ˜ 1, λ 2, λ˜ 2, …, λ n, λ˜ n ˜ , Φ ,Φ ˜ , …, Φ , Φ ˜ Φ = Φ 1, Φ 2 n 2 n 1 ˜ ˜ Φ1 Φ1 Φn Φn V = , , …, , ˜ ˜ λ 1 Φ 1 λ˜ 1 Φ λ n Φ n λ˜ n Φ 1 n

(8.232)

Each term in eqn (8.232) is complex λ j = λ R, j + iλ I, j Φ j = Φ R, j + iΦ I, j

(8.233)

V j = V R, j + iV I, j The complete solution is obtained by combining the 2n solutions using ˜ . complex constants, A j and A j n

X =

∑

( A je

λ jt

˜ e λ˜ j t V˜ ) Vj + A j j

(8.234)

j=1

Substituting for A j and e

λ jt

,

A j = A R, j + iA I, j e

λ jt

= e

λ R, j t

( cos λ I t + i sin λ I t )

transforms eqn (8.234) to

(8.235)

8.5 State-space Formulation for MDOF Systems 627 n

∑

X (t) =

e

λ R, j t

[ ( A R, j V R, j – A I, j V I, j ) cos λ I, j t

j=1

(8.236)

– ( A R, j V I, j – A I, j V R, j ) sin λ I, j t ] One determines A R, j and A I, j using the initial conditions on X at time t=0. This computation is discussed in the next section. Stability requires the homogeneous solutions to be bounded. Noting eqn (8.236), a system having n DOF is stable when λ R, j < 0

j = 1, 2, …, n

(8.237)

The free vibration solution given by eqn (8.234) can also be expressed as X ( t ) = e At X ( 0 )

(8.238)

where A is defined by eqn (8.220) and e At is generated with the following series: 1 1 e At = I + At + --- AAt 2 + … + ----- A n t n + … 2 n!

(8.239)

Equation (8.238) is a generalized form of the SDOF solution. One can establish this result by specializing eqn (8.234) for t = 0, n

X (0) =

∑

˜ V˜ ) ( A jV j + A j j

(8.240)

j=1

and noting that e At V j = e e At V˜

j

= e

λ jt λ˜ j t

Vj V˜ j

Example 8.7: Free vibration solution for proportional damping Consider eqn (8.229) premultiplied by M.

(8.241)

628 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems λ 2 MΦ + λCΦ + KΦ = 0

(1)

Suppose C is a linear function of K and M C = αK + βM

(2)

Substituting for C results in 2+βλ λ ------------------- MΦ + KΦ = 0 1 + αλ

(3)

When the coefficient of MΦ is expressed as λ 2 + βλ ------------------- = – ω 2 1 + αλ

(4)

eqn (3) takes the form of the classical eigenvalue equation. KΦ = ω 2 MΦ

(5)

There are n eigensolutions to eqn (5) ω 1 2, ω 2 2, …, ω n 2 Φ 1, Φ 2, …, Φ n

(6)

All the eigenvectors are real vectors. Also, all the eigenvalues are positive real quantities. ωj 2 > 0

(7)

Given ω j 2 , one solves eqn (4) for λ j . The pair of complex roots are written as λ j, λ˜ j = – ξ j ω j ± i ω j { 1 – ξ 2j } 1 / 2

(8)

where the damping ratio is related to α and β by β + α ωj 2 ξ j = ------------------------2 ωj

(9)

8.5 State-space Formulation for MDOF Systems 629 In what follows, ω j is considered to be positive, and ω j is set equal to ω j . The eigenvectors, Φ j , are real and satisfy the following orthogonality relationships ΦT j M Φ k = δ jk m k

(10)

2 ΦT j KΦ k = δ jk ω k m k

(11)

Since Φ is real, the state eigenvector, V, has the following form Φj Φj 0 = Vj = + i λ jΦ j λ R, j Φ j λ I, j Φ j

(12)

Undamped free vibration response is the special case where α = β = 0 . The corresponding solution is λ j = ± iω j

(13)

Φ j 0 V j = + i 0 ω jΦ j

(14)

Example 8.8: General uncoupled damping The previous example dealt with the case where the damping matrix is a linear combination of the mass and stiffness matrices. This approach can be extended to deal with a more general form of C. Suppose C satisfies the following orthogonality conditions, ΦT j CΦ k = 2δ jk m k ω k ξ k

j = 1, 2, …, r

(1)

where ξ k are now considered to be independent parameters. The free vibration solution is the same as presented in example 8.7.

630 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems λ j = – ξ j ω j ± iω j [ 1 – ξ 2j ] 1 / 2 Φj Vj = λ jΦ j

(2)

KΦ j = ω 2j M Φ j The ability to independently specify the modal damping ratio for the first r modes by adjusting C is very convenient since it allows one to deal more effectively with the problem of suppressing the response of a particular subset of modes contained within this group. The form of C that satisfies eqn (1) is established by noting the following identities: n 2n –1 ΦT j M ( M K ) Φ k = ω j m j δ jk

(3)

n 2n + 2 m δ –1 ΦT j K ( M K ) Φk = ω j j jk

A linear combination of these matrices is a candidate solution. s

C =

∑

n

( an [ M ( M –1 K ) n ] + bn [ K ( M –1 K ) ] )

(4)

n=0

Taking s = 0 corresponds to Rayleigh damping. Taking s ≤ r allows one to specify r modal damping ratios. The coefficients for this case are determined with s

∑

– 1 + b ω 2n + 1 ) = 2ξ ( a n ω 2n j n j j

j = 1, 2, …, r

(5)

n=0

If both a and b are used, s = r/2. The basic problem with this approach is the form of C generated with eqn (3). The coupling between the off-diagonal elements extends beyond adjacent nodes when n ≥ 1. This pattern requires a deployment of dampers that involves connecting dampers to nodes that are not adjacent, e.g., a damper between floor 1 and floor 3. Realistically, the required pattern cannot be achieved.

8.5 State-space Formulation for MDOF Systems 631

Orthogonality properties of the state eigenvectors The eigenvector V j satisfies eqn (8.226). AV j = λ j V j

(8.242)

Since A is not symmetrical, V j is not orthogonal to V k . VT j V k ≠ δ jk

(8.243)

A set of vectors which are orthogonal to V can be established by considering the eigenvalue problem for the transpose of A: A T W j = λ *j W j

j = 1, 2, …, n

(8.244)

where λ * , W represent the eigenvalue and corresponding eigenvector for A T . Eqn (8.244) requires the determinant of the coefficient matrix to vanish. AT – λ*I = 0

(8.245)

Noting that the determinant of the transpose of a matrix is equal to the determinant of the original matrix, b ≡ bT

(8.246)

it follows that λ *j ≡ λ j

(8.247)

and eqn (8.244) is equivalent to AT W j = λ jW j

(8.248)

Starting with AV j = λ j V j AT W k = λk W k

k≠j

(8.249)

and premultiplying the first equation with W k and the second with V j , one obtains

632 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems W kT AV j = λ j W kT V j

(8.250)

T T VT j A W k = λk V j W k

(8.251)

Eqn (8.251) can be written as W kT AV j = λ k W kT V j

(8.252)

Then subtracting eqn (8.250) from (8.252) results in 0 = ( λ k – λ j )W kT V j

(8.253)

According to eqn (8.253), W k and V j are orthogonal W kT V j = 0

j≠k

for

(8.254)

The result for j = k is written as WT j Vj = f j

(8.255)

One can show that ˜ TV = 0 W k j W kT V˜ j = 0

for all j, k

(8.256)

and ˜ T V˜ = ˜f W j j j

(8.257)

Example 8.9: Initial conditions - free vibration response Consider the general free vibration solution defined by eqn (8.234). n

X (t) =

∑

( A je

λ jt

˜ e λ˜ j t V˜ ) Vj + A j j

(1)

j=1

The integration constants are determined using the initial conditions on X at t = 0.

8.5 State-space Formulation for MDOF Systems 633 n

X (0) =

∑

˜ V˜ ) ( A jV j + A j j

(2)

j=1

Noting the orthogonality relations, the expressions for the constants are: 1 A j = ----- W T X (0) fj j

(3)

1 ˜ T ˜ = ----W j X (0) A j ˜f j

Determination of W and f j One establishes W by solving the following eigenvalue problem A T W = λW

(8.258)

Substituting for A T , the coefficient matrix expands to – λI

– K M –1

I

– CM – 1 – λI

W = 0

(8.259)

Expressing W in partitioned form, W 1 W = W 2

(8.260)

and expanding eqn (8.259) leads to – λW 1 – K M – 1 W 2 = 0 W 1 – CM – 1 W 2 – λW 2 = 0

(8.261)

Solving the first equation for W 2 , W 2 = – λM K – 1 W 1

(8.262)

634 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems and substituting into the second equation results in W 1 = – λCK – 1 W 1 – λ 2 M K – 1 W 1

(8.263)

This equation can be transformed to a form similar to the eigen equation for Φ by expressing W 1 as (8.264)

W 1 = KW 1 The result is λ 2 M W 1 + λCW 1 + KW 1 = 0

(8.265)

Comparing this form with eqn (8.229) shows that W1 ≡ Φ

(8.266)

It follows that the forms of V j and W j are related by KΦ j Wj = –λ j M Φ j

Φj Vj = λ jΦ j

(8.267)

Using these expressions, the definition equation for f expands to T 2 T f j = WT j V j = Φ j KΦ j – λ j Φ j M Φ j T = – 2λ 2j Φ T j M Φ j – λ j Φ j CΦ j

(8.268)

When C is proportional to either K or M, Φ is real and f j is given by f j = m j ( ω 2j – λ 2j )

(8.269)

The value for no damping is f j = 2m j ω 2j . General Solution - time invariant system Following the approach employed for the SDOF case, the general solution for an arbitrary loading applied to a time invariant MDOF system can be expressed as a Duhamel integral involving the free vibration response. Noting eqns (8.238) and (8.239), the complete solution has the form

8.5 State-space Formulation for MDOF Systems 635

X (t) = e

A ( t – to )

Xo +

∫

t

to

e A ( t – τ ) ( B f F + B p P + B g a g ) dτ

(8.270)

where the coefficient matrices ( A, B f , B p, B g ) are defined by eqns (8.220) to (8.223). This equation is similar to the SDOF solution given by eqn (8.26); one has only to introduce the appropriate forms of A, B f , B p , and B g . The discrete time equilibrium equation for the MDOF case follows from eqn (8.77). X j + 1 = e A∆t X j + A – 1 ( e A∆t – I ) [ B g a g, j + B f F j + B p P j ]

(8.271)

Working with the full set of equations for an n’th order system requires dealing with matrices of order 2n. As n increases, the computational costs for e A∆t and A – 1 become excessive, and one needs to consider an alternative approach. In conventional dynamic analysis, this reduction in computational effort is achieved by expressing the state vector as a linear combination of prescribed modal vectors multiplied by unknown coordinate variables which are functions of time. Depending upon the nature of the loading, one can obtain a reasonably accurate solution by suitably selecting the participating modes so as to minimize the required number of modal coordinate functions, and consequently reduce the computational cost. Interpreting the total response as a superposition of modal responses is also useful from the perspective of active control. A key issue of active control is the optimal location of “active” forces so as to minimize the response of a particular mode. In what follows, the formulation of the governing equations in terms of modal coordinates is presented. This formulation is used later to establish the form of the LQR algorithm in terms of modal coordinates. Modal state space formulation - uncoupled damping The displacement vector is expressed as a linear combination of a subset of the eigenvectors for the undamped system scaled with functions of time. s

U(t) =

∑

q j ( t )Φ j

(8.272)

j=1

Assuming the system has n DOF, this expression will produce the exact solution when s is taken to be n. Substituting for U ( t ) in the force equilibrium equation,

636 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems premultiplying the result with Φ k , and noting the orthogonality conditions for Φ k , one obtains s equations having the following form, p

m k q˙˙k +

∑

( Φ kT CΦ j )q˙ j + m k ω k2 q k

j=1

= – Φ kT MEa g + Φ kT P + Φ kT E f F

k = 1, 2, …, s

(8.273)

These equations uncouple when C is orthogonal to Φ j . Assuming this condition is satisfied, and taking Φ kT CΦ j = 2δ jk m k ω k ξ k

(8.274)

results in q˙˙k + 2ξ k ω k q˙k + ω k2 q k 1 1 1 = – ------- Φ kT MEa g + ------- Φ kT P + ------- Φ kT E f F mk mk mk

k = 1, 2, …, s

(8.275)

The initial conditions for q k are 1 q k ( 0 ) = ------- Φ kT MU ( 0 ) mk 1 q˙k ( 0 ) = ------- Φ kT M U˙ ( 0 ) mk

(8.276)

When F = 0, these equations can be solved separately. The effect of feedback is to introduce coupling between the modal equations. In order to deal with feedback, the equations are transformed to the standard state-space form by defining q as the modal coordinate vector, q1 q2 q = … qs

s×1

(8.277)

8.5 State-space Formulation for MDOF Systems 637 and X m as the modal coordinate state vector, q Xm = q˙

2s × 1

(8.278)

Note that now X m contains only 2s modal coordinate terms. The equations are expressed in the same form as eqn (8.218), X˙ m = A m X m + B fm F + B pm P + B gm a g

(8.279)

where the “modal” forms of the coefficient matrices along with their sizes are: m = [ m j δ ij ]

(s × s)

(8.280)

Λ = [ ω 2j δ ij ]

(s × s)

(8.281)

Λ 1 = [ 2ξ j ω j δ ij ]

(s × s)

(8.282)

Φ = Φ1 Φ2 … Φs

(n × s)

(8.283)

( 2s × 2s )

(8.284)

( 2s × r )

(8.285)

( 2s × r )

(8.286)

Am =

B fm =

B pm =

B gm =

0 I –Λ –Λ1 0 m –1 Φ T E f

0 m –1 Φ T

0 – m – 1 Φ T ME

( 2s × 1 )

Lastly, the initial conditions follow from eqn (8.276)

(8.287)

638 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

q(0) m – 1 Φ T MU ( 0 ) X m(0) = = q˙ ( 0 ) m – 1 Φ T M U˙ ( 0 )

( 2s × 1 )

(8.288)

Feedback is introduced by expressing F as a linear function of the modal state vector (8.289)

F = – K fm X m = – k d q – k v q˙

The individual feedback matrices are of order r x s. For pure velocity feedback, k d is set equal to 0. Substituting for F in eqn (8.279) leads to the state equation specialized for continuous negative linear feedback. X˙ m = ( A m – B fm K fm )X m + B pm P + B gm a g

(8.290)

where

A m – B fm K fm =

0

I

– Λ – m –1 Φ T E f kd

– Λ1 – m –1 Φ T E f kv

(8.291)

The discrete form is similar to eqn (8.271). X m, j + 1 = e

A m ∆t

–1 ( e X m, j + A m

A m ∆t

– I ) [ B fm F j + B pm P j + B gm a g ]

(8.292)

F j = – K fm X m, j For the continuous case, the free vibration solution with feedback is stable when A m is stable and k v is positive. This condition is always satisfied. The discrete solution is stable provided that the modulus of the largest eigenvalue of c m is less than unity. ρ max ( c m ) < 1 cm = e

A m ∆t

–1 ( e – Am

(8.293) A m ∆t

– I )B fm K fm

(8.294)

Given the selection of the modes and the feedback matrix, K fm , one can establish the limit on ∆t by computing the eigenvalues of c m for a range of values of ∆t .

8.5 State-space Formulation for MDOF Systems 639 This computation is illustrated in example 8.14 Modal state space formulation - arbitrary damping When C is an arbitrary symmetric positive definite matrix, the expansion in terms of the eigenvectors for the undamped system and real modal coordinates does not lead to uncoupled modal equations. In this case, one has to work with an expansion involving complex modal coordinates and complex state eigenvectors. The state space vector is approximated as a linear combination of s modal vectors and coordinates, U(t) 1 X (t) = = --2 U˙ ( t )

s

∑

( q j V j + q˜ j V˜ j )

(8.295)

j=1

where q and V are complex quantities defined by q j ( t ) = q R, j ( t ) + iq I, j ( t )

Vj =

Φj

(8.296)

(8.297)

λ jΦ j

The expanded real forms for U ( t ) and U˙ ( t ) are s

U(t) =

∑

( q R, j Φ R, j – q I, j Φ I, j )

(8.298)

j=1 s

U˙ ( t ) =

∑

[ q R, j ( λ R, j Φ R, j – λ I, j Φ I, j ) – q I, j ( λ R, j Φ I, j + λ I, j Φ R, j ) ] (8.299)

j=1

When damping is uncoupled, Φ I, j = 0 . The fully undamped case has λ R, j = 0 , λ I, j = ω j , and Φ I, j = 0. The initial conditions for the modal coordinates are determined by specializing X ( t ) for t = 0.

640 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems s

1 X ( 0 ) = --2

∑

( q j ( 0 )V j + q˜ j ( 0 )V˜ j )

(8.300)

j=1

Premultiplying by W k , and noting the orthogonality relation between W k and V j , one obtains 2 q k ( 0 ) = -----W kT X ( 0 ) fk

(8.301)

Substituting for W k , the expression for q k ( 0 ) expands to 2λ k 2 q k ( 0 ) = -----Φ kT K U ( 0 ) – --------- Φ kT M U˙ ( 0 ) f fk k

k = 1, 2, …, s

(8.302)

Introducing the expression for X ( t ) in the state equilibrium equation, and premultiplying for W k results in a set of s complex scalar equations, 2 2 2 q˙k = λ k q k + -----W kT B f F + -----W kT B g a g + -----W kT B p P f f f k

k

k

(8.303)

k = 1, 2, …, s The terms involving F, a g , and P can be interpreted as complex modal forces associated with the k’th mode. Expanding the matrix products and substituting the notation, 2λ k 2 b f , k = -----W kT B f = – --------- Φ kT E f fk fk 2λ k 2 b g, k = -----W kT B g = --------- Φ kT ME fk fk

(8.304)

2λ k 2 b p, k = -----W kT B p = – --------- Φ kT fk fk reduces the governing modal equations to a simpler form, q˙k = λ k q k + b f , k F + b p, k P + b g, k a g

k = 1, 2, …, s

(8.305)

8.5 State-space Formulation for MDOF Systems 641 Since q k is complex, there are a total of 2s equations.The corresponding set of “real” equations for the k’th mode are: q˙R, k = λ R, k q R, k – λ I, k q I, k + b f R, k F + b pR, k P + b a gR, k g q˙I, k = λ I, k q R, k + λ R, k q I, k + b fI, k F + b pI, k P + b a gI, k g

(8.306)

k=1,2,...,s One establishes the initial conditions for q R and q I by expanding eqn (8.302). The resulting expressions are: 2 T + f T q R, k ( 0 ) = ---------------------------- [ ( f R, k Φ R ,k I, k Φ I, k )KU ( 0 ) 2 2 f R, k + f I, k

(8.307)

T – λ ΦT ) + f T T ˙ – { f R, k ( λ R, k Φ R ,k I, k I, k I, k ( λ R, k Φ I, k + λ I, k Φ R, k ) }MU ( 0 ) ]

f I, k q I, k ( 0 ) = – -----------q R, k + 2Φ IT, k KU ( 0 ) f R, k 2 T )M U ˙ (0) – ----------- ( λ R, k Φ IT, k + λ I, k Φ R ,k f R, k

(8.308)

Various specialized forms of these equations are presented in the examples listed below.

Example 8.10: Modal formulation - undamped case It is of interest to specialize the general formulation presented in this section for the case of no damping and compare this result with the solution obtained with the conventional formulation derived in the previous section. Setting C = 0 reduces the various terms to:

642 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Φ I, k = 0 λ R, k = 0 λ I, k = ω k f k = 2m k ω k2 i b f , k = – -------------- Φ kT E f mk ωk

(1)

i b p, k = – -------------- Φ kT mk ωk i b g, k = -------------- Φ kT ME mk ωk Substituting in eqns (8.307) and (8.308), leads to q˙R, k = – ω k q I, k

(2)

1 q˙I, k = ω k q R, k – -------------- Φ kT ( E f F + P – MEa g ) m ω

(3)

k k

Equation (2) shows that 1 q I, k = – ------ q˙R, k ωk

(4)

Then, eqn (3) can be written as 1 q˙˙R, k + ω k2 q R, k = ------- Φ kT ( E f F + P – MEa g ) mk

(5)

This result is identical to eqn (8.225).

Example 8.11: Modal formulation - uncoupled damping When the damping matrix is proportional to K and M, the eigenvector, Φ , is real. However, the eigenvalue is complex. The various terms specialized for this case are as follows:

8.5 State-space Formulation for MDOF Systems 643 Φ I, k = 0 λ R, k = – ξ k ω k λ I, k = ω k [ 1 – ξ k2 ] 1 / 2 f k = 2m k λ I, k ( λ I, k – iλ R, k ) = – ( 2m k λ I λ ) i λk i ----- = -------------------fk 2m k λ I, k i b f , k = – ----------------- Φ kT E f m k λ I, k

(1)

i b g, k = – ----------------- Φ kT ME m k λ I, k i b p, k = – ----------------- Φ kT m k λ I, k Since the forcing terms are pure imaginary, the equations simplify to q˙R, k = λ R, k q R, k – λ I, k q I, k 1 q˙I, k = λ I, k q R, k + λ R, k q I, k – ----------------- Φ kT ( E f F + P – MEa g ) m λ

(2)

k I, k

The initial conditions are 1 q R, k ( 0 ) = ------- Φ kT MU ( 0 ) mk 1 1 q I, k ( 0 ) = ---------- λ R, k q R, k ( 0 ) – ------- Φ T M U˙ ( 0 ) λ I, k mk

(3)

This example shows that q I, k involves a combination of q R, k and q˙R, k . The actual displacements are given by

644 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems s

U(t) =

∑

q R, j Φ j

j=1

U˙ ( t ) =

∑

(4)

s

s

( λ R, j q R, j – λ I, j q I, j )Φ j =

j=1

∑

q˙R, j Φ j

j=1

Example 8.12: Modal parameters - 4DOF system This example presents modal data for the 4 DOF system shown in Fig. 1. The element stiffness and damping values are listed in Table 1. The element stiffness factors are selected so that the displacement profile for the first mode is essentially linear. Two damping distributions are considered. Case 1 represents a low damping level for the first mode. The damping coefficients for case 2 are adjusted so that the sum, Σc j , is 3 times the corresponding sum for the first case. Since the internodal displacement for the first mode is essentially constant, this adjustment increases the modal damping ratio for the first mode by a factor of 3. More emphasis is placed on element 1 to exaggerate the non-proportional distribution and therefore increase the magnitude of the imaginary part of the modal vector.

1000kg

m4

u4

k 4, c 4 1000kg

m3

u3

k 3, c 3 1000kg

m2

u2

k 2, c 2 1000kg

m1 k 1, c 1

Figure 1

u1

8.5 State-space Formulation for MDOF Systems 645 Table 1 Element number

Element stiffness, k

Element damping, c (kN.s/m)

(kN/m)

case 1

case 2

1

1700

4

20

2

1400

3

7

3

1000

2

2

4

700

1

1

Σ = 10

Σ = 30

Scaled versions of the modal displacement profiles for the 2 damping distributions are plotted in Figures 2 and 3. The real parts are essentially identical. Since case 1 represents a low level of damping, the amplitude of the corresponding imaginary part is negligible in comparison to the real part which is of order 1. Case 2 results show a significant increase in the amplitude for the imaginary part. This shift is due to the large increment in c assigned to element 1. The modal periods and damping ratios are shown in Figures 4 and 5. There is essentially no change in the periods. As expected, all the modal damping ratios are higher for case 2. Lastly, to illustrate the effect of damping, time history responses were generated for the El-Centro ground excitation. The maximum values of the inter-nodal displacements for the elements are plotted in Figure 6. Increasing the damping reduces the absolute maximum value from 0.0258 to 0.0217, and also decreases the average value.

646 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Node number

4

3

Mode 2

Mode 3 2

Mode 1 Mode 4

1 −1

−0.8

−0.6

−0.4

−0.2

0

Amplitude case 1

0.2

0.4

0.6

0.8

1

Node number

4

3

Mode 2

Mode 3 2

Mode 1 Mode 4

1 −1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

Amplitude case 2 Figure 2: Modal displacement profile - real part

0.8

1

8.5 State-space Formulation for MDOF Systems 647

Node number

4

3

Mode 3 Mode 2

Mode 1 2

Mode 4

1 −0.02

−0.015

−0.01

−0.005

0

0.005

Amplitude case 1

0.01

0.015

0.02

0.025

Node number

4

Mode 2

3

Mode 1

Mode 3

2

Mode 4

1 −0.4

−0.3

−0.2

−0.1

0

Amplitude case 2

0.1

0.2

Figure 3: Modal displacement profile - imaginary part

0.3

648 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.5

0.45

Modal period, ( sec )

0.4

0.35

0.3

0.25

Damping case 1 0.2

Damping case 2 0.15

0.1

0.05

1

2

3

4

Mode number Figure 4: Modal period - No feedback

0.2

0.18

Modal damping ratio

0.16

0.14

0.12

0.1

Damping case 2

0.08

0.06

0.04

Damping case 1

0.02

0

1

2

Mode number

3

Figure 5: Modal damping ratio - No feedback

4

8.5 State-space Formulation for MDOF Systems 649

4

Element number

El Centro ground motion 3

Damping case 2

2

1 0.017

0.018

0.019

0.02

0.021

Damping case 1

0.022

0.023

0.024

0.025

0.026

Maximum element internodal displacement, (m) Figure 6: Element internodal displacement profile - No feedback

When F = 0, the problem reduces to solving a set of 2 equations for each mode. Including F couples the modal equations, and now one has to solve 2s simultaneous equations. In this case, it is convenient to shift over to a state-space type formulation. The state vector is defined in a similar way as for the previous formulations. Firstly, the real and imaginary modal coordinates are grouped separately, q R1 q R2 qR = … q Rs

q I1 q I2 qI = … q Is

Secondly, the “modal” state vector is taken as

(8.309)

650 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems qR Xm = qI

(8.310)

Lastly, the state equilibrium equation is expressed in the same form as for uncoupled damping X˙ m = A m X m + B fm F + B pm P + B gm a

(8.311)

where the corresponding form of the coefficient matrices are λ R = [ λ R, k δ jk ]

λ I = [ λ I, k δ jk ]

b fR, 1 b fR =

b pR =

b gR =

b fR, 2 …

b fI =

b fI, 2

b fR, s b pR, 1

b pI, 1

…

b pI =

b pI, 2 …

b pR, s

b pI, s

b gR, 1

b gI, 1

b gR, 2 … b gR, s

Am =

b fI, 1 … b fI, s

b pR, 2

λR

–λI

λI

λR

(8.312)

b gI =

b gI, 2 …

(8.313)

(8.314)

(8.315)

b gI, s

(8.316)

8.5 State-space Formulation for MDOF Systems 651

B fm =

b fR b fI

B pm =

b pR b pI

B gm =

b gR b gI

(8.317)

The initial conditions for X m are obtained with eqns (8.307) and (8.308). Negative linear feedback is taken as q F = – K fm X m = – k d k v R qI

(8.318)

Example 8.10 showed that q I is a linear function of q˙ R for no damping. Including damping results in q I being a function of both q˙ R and q R . For a lightly damped system, it is reasonable to assume q I depends only on q˙ R , and therefore to approximate pure velocity feedback by setting k d = 0 . The feedback term for this case reduces to:

B fm F =

0 0

– b fR k v q R –b k qI

(8.319)

fI v

Example 8.13: Modal response for example 8.12 The state space modal formulation is applied to the 4 DOF system with Case 1 damping defined in example 8.12. Results for the real and imaginary parts of the modal coordinates for the first 2 modes are plotted below. Comparison of the plots in Figure (1) shows that the response is dominated by the first mode. The ratio of the amplitudes for the first and second modes is approximately 30, which indicates that the second mode contributes about 3% to the total response for this combination of structure and excitation.

652 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.1

Modal coordinate, q R1

0.08

El Centro Damping Case 1

0.06

0.04

0.02

0

−0.02

−0.04

−0.06

−0.08

0

5

10

15

20

25

Time (sec) a) Mode 1

30

35

40

45

50

−3

4

x 10

Modal coordinate, q R2

3

El Centro Damping Case 1

2

1

0

−1

−2

−3

−4

0

5

10

15

20

25

Time (sec) b) Mode 2

30

35

40

45

Figure 1: Time history response of the real part of the modal coordinate

50

8.5 State-space Formulation for MDOF Systems 653

0.08

0.06

El Centro Damping Case 1

Modal coordinate, q I1

0.04

0.02

0

−0.02

−0.04

−0.06

−0.08

−0.1

0

5

10

15

20

25

30

35

40

45

50

Time (sec) a) Mode 1 −3

3

x 10

Modal coordinate, q I2

2

El Centro Damping Case 1

1

0

−1

−2

−3

0

5

10

15

20

25

Time (sec)

30

35

40

45

50

b) Mode 2 Figure 2: Time history response of the imaginary part of the modal coordinate

654 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Example 8.14: Modal response with feedback for example 8.12 This example illustrates the effect of linear negative velocity feedback on the modal damping ratios. The modal state space formulation defined by eqns (8.309) through (8.319) is applied to the 4 DOF system with Case 1 damping described in example 8.12. Two control force systems are considered. The first system is a single force applied at node 4; the second system consists of selfequilibrating sets of nodal forces applied on each element. Figure (1) shows the spatial distribution of the control forces. m4

u4

k 4, c 4 m3

F1 F2

u3

k 3, c 3 m2

F2 F3

u2

k 2, c 2 m1

F1

F1

F3 F4

u1

k 1, c 1 Scheme 1

Scheme 2

Figure 1 Given the feedback matrix, K fm , one determines F using eqn (8.318), F = – K fm X m

(1)

and then solves for X m with eqn (8.311). X˙ m = ( A m – B fm K fm )X m + B pm P + B gm a g

(2)

The frequency and damping parameters for the system with feedback are related to the eigenvalues of A m – B fm K fm , and depend on the control force scheme as

8.5 State-space Formulation for MDOF Systems 655 well as K fm . Particular forms of K fm obtained with the algebraic Riccati equation and specialized for pure velocity feedback (see eqn (8.319)) are used here to generate results for the 2 force schemes. This formulation is discussed in detail in the next section. Our objective here is to illustrate the variation in behavior associated with feedback systems. Figure (2) shows the modal damping ratios for the 4 DOF model with Case 1 damping, force scheme 1, and a particular choice for K fm . Feedback results in a significant increase (a factor of 10) in the damping ratio for the first mode. Since this response is dominated by the first mode, there also is a reduction in the displacement response. The relevant displacement quantities are plotted in Figs (2b) and (2c). The last plot shows the variation of the control force magnitude with time. Results for the control force scheme 2 are plotted in Figure (3). This combination of control forces and weighting produces essentially equal nodal damping ratios whereas scheme 1 places the priority on the first mode. Since the damping ratio is less, the displacement response is greater. There also is a reduction in the peak magnitudes of the 4 control forces. Modal damping ratio with feedback

0.24

0.22

Case 1 damping Model defined in example 8.12

0.2

0.18

0.16

0.14

0.12

0.1

0.08

0.06

0.04

1

2

3

Mode number a) Modal damping ratio with continuous feedback - no saturation limit

4

656 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

4

Element number

El Centro excitation

3

2

1 0.0092

0.0094

0.0096

0.0098

0.01

0.0102

0.0104

0.0106

0.0108

Maximum element internodal displacement b) Element internodal displacement profile

0.011

0.0112

0.04

Nodal displacement (m)

0.03

El Centro excitation 0.02

0.01

0

−0.01

−0.02

−0.03

−0.04

−0.05

0

5

10

15

Time (sec) c) Displacement of node 4

20

25

30

8.5 State-space Formulation for MDOF Systems 657

3000

El Centro excitation

Control force magnitude

2000

1000

0

−1000

−2000

−3000

−4000

Max Force = 4860 N −5000

0

500

Time step number

1000

1500

d) Control force time history Figure 2: Modal response for control force scheme 1

Modal damping ratio with feedback

0.135

Case 1 damping Model defined in example 8.12

0.13

0.125

0.12

0.115

0.11

1

2

3

Mode number a) Modal damping ratio with continuous feedback - no saturation limit

4

658 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Element number

4

3

2

1 0.0122

0.0124

0.0126

0.0128

0.013

0.0132

0.0134

0.0136

0.0138

0.014

0.0142

Maximum element internodal displacement b) Element internodal displacement profile 0.06

El Centro excitation

Nodal displacement (m)

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

Time (sec) c) Displacement of node 4

20

25

30

8.5 State-space Formulation for MDOF Systems 659

Maximum control force magnitude (Newtons)

3800

El Centro excitation 3700

3600

3500

3400

3300

3200

3100

3000

2900

1

2

3

4

Control force number d) Control force time history Figure 3: Modal response for control force scheme 2

Stability analysis - discrete modal formulation The discrete form of the state equilibrium equation expressed in terms of modal coordinates is used here to evaluate the effect of the time interval and the delay time on the stability. Allowing for linear feedback, the governing equation for free vibration response is X m, j + 1 = e

A m ∆t

–1 ( e X m, j + A m

A m ∆t

– I )B fm F j

(8.320)

Time delay is introduced by taking F j as Fj = 0 F j = – K fm X m, j – ν

0≤j<ν j≥ν

(8.321)

660 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems With this specification for F j , the response equations take the following form X m, j + 1 = c m1 X m, j X m, j + 1 = c m1 X m, j + c m3 X m, j – ν

0≤j<ν j≥ν

(8.322)

where c m1 = e c m3 =

A m ∆t

– 1 ( e A m ∆t – Am

(8.323) – I )B fm K fm

The stability analysis for the SDOF system showed that time delay lowered the limiting value of the time interval. Therefore, one can obtain an upper bound estimate for ∆t by considering no time delay. For this case, eqn (8.322) simplifies to X m, j + 1 = ( c m1 + c m3 )X m, j = c m X m, j

(8.324)

Stability requires the modulus of the largest eigenvalue of c m to be less than 1. ρ max ( c m ) < 1

(8.325)

The complexity of c m necessitates the use of a numerical computational procedure. Example 8.2 illustrates the computational details for a SDOF system with negative velocity feedback. In general, one ranges over ∆t to determine the value corresponding to ρ max ( c m ) = 1. When there is time delay, the eigenvalue problem is of order ν + 1 , ( c m3 + ρ ν c m1 – ρ ν + 1 I )Ψ = 0

(8.326)

For the SDOF case, it was possible to expand the determinant of the coefficient matrix and obtain a polynomial expansion for ρ . However, a numerical procedure is still required to evaluate the roots. For MDOF systems, this hybrid approach is not feasible and one must resort to a fully numerical approach. Given ∆t and ν , eqn (8.322) can be used to generate the corresponding free vibration response. By decreasing ∆t in increments from the critical value for no time delay, one can converge on the limiting value for the specified time delay. This solution strategy is efficient when the number of modal DOF is small, on the order of 5.

8.5 State-space Formulation for MDOF Systems 661 The computational cost for the exponential matrix increases nonlinearly with the number of modes and eventually becomes excessive. In this case, one can resort to an approximate formulation which assumes the modal equations are fully uncoupled. This assumption reduces the stability analysis for a problem involving s modes to s individual SDOF stability analyses. The approximate stability formulation proceeds as follows. Firstly the feedback force is considered to involve only q I . s

F =

∑

k v, j q I, j

(8.327)

j=1

Secondly, b fR, k is taken as a null matrix in eqn (8.306). These conditions correspond to pure velocity feedback for a system with uncoupled damping. Lastly, the feedback terms in the k’th equation are assumed to involve only q I, k . With these assumptions, eqn (8.306) takes the form q˙R, k = λ R, k q R, k – λ I, k q I, k q˙I, k = λ I, k q R, k + λ R, k q I, k – b k q I, k

(8.328)

where b k = b fI, k k v, k

(8.329)

The remaining steps are the same as for the SDOF formulation presented in section 8.3. Introducing matrix notation, eqn (8.328) is expressed as q˙ k = A k q k – B k q k

(8.330)

q R, k qk = q I, k

(8.331)

where

662 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Ak =

Bk =

λ R, k – λ I, k λ I, k λ R, k

(8.332)

0

(8.333)

bk

The discrete form allowing for time delay is written as q k, j + 1 = c k1 q k, j + c k3 q k, j – ν

(8.334)

where the c matrices are now 2 x 2. c k1 = e

A k ∆t

c k3 = – A k– 1 ( e

(8.335) A k ∆t

– I )B k

(8.336)

One determines the limiting value of ∆t for each k ranging from 1 to s, the total number of modes selected to represent the solution. The lowest value controls the change of the time interval.

Example 8.15: Stability analysis for example 8.14 This example examines the effect of time delay on the response of the 4 DOF system considered in Example 8.14. Modal properties corresponding to the 2 control force schemes are listed below in Table 1. The time increment is considered to be equal to 0.02 secs. An estimate of the limiting value of ν for each force scheme can be obtained using the data contained in Table 1 of Example 8.3. These results are for ξ = 0 , and provide a lower bound estimate, ν o . The limiting value of ν for ξ > 0 is higher than ν o . For convenience, the table is listed below as Table 2.

8.5 State-space Formulation for MDOF Systems 663 Table 1: Force scheme 1

Force scheme 2

ξa

ξa

0.013

0.212

0.100

0.102

0.029

0.045

0.096

0.129

0.155

0.046

0.005

0.078

0.095

0.211

0.071

0.000

0.060

Mode

T

∆t -----T

ξ

1

0.498

0.040

2

0.197

3 4

Table 2: Limiting values of ∆t ⁄ T for ξ = 0 (Example 8.3, Table 1) Stability limit for ∆t ⁄ T ξa

ν = 0

ν = 1

ν = 2

0

0.50

0.500

0.5000

0.158

0.0955

0.151

0.0908

0.144

0.0860

0.137

0.0822

0.131

0.0780

0.124

0.0745

0.118

0.07085

0.113

0.0677

0.05 0.10

0.44

0.15 0.20

0.38

0.25 0.30

0.33

0.35 0.40

0.29

For force scheme 1, the first and second modes have the larger values of ξ a and therefore control the delay. Assuming ξ = 0 , Table 2 indicates that mode 2 is close to instability for ν = 2 (∆t ⁄ T = 0.102 vs. 0.096). The value for mode 1 is closer to 3. Since ξ a is ≈ 0 for modes 3 and 4, they are not critical even though the corresponding time ratios are higher. Force scheme 2 produces some damping for modes 3 and 4, and since these time ratios are the highest, they control the stability limit. Allowing for some increase due to initial damping, the critical value for ν is between 1 and 2 for modes 3, 4, and >2 for modes 1, 2. The least critical mode is mode 1 since it has the lowest time ratio. Figure 1 contains results generated for control force scheme 1, with ν

664 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems taken equal to 2. They are in close agreement with the solution for ν = 0 presented in Example 8.14. Increasing ν to 3 produces the responses plotted in Figure 2. Mode 2 has gone critical, and dominates the response whereas mode 1 remains at essentially the same peak level as for ν = 2 . These results are based on limiting the peak magnitude of the control force to 15,000N, which is 3 times the peak value for ν = 2 . Limiting F simulates saturation of the force actuators. Comparison of the power input plots shows that shifting ν from 2 to 3 transforms the energy dissipation process, resulting in energy being supplied to the system and a periodic forced vibration response following the seismic induced response. Reducing the peak magnitude down to 5,000N, the value required for ν = 0 (and also ν = 2 ), leads to the responses plotted in Figure 3. The magnitudes have been reduced, but the system is still responding as if it were subjected to a periodic forced vibration as well as a seismic excitation. Decreasing F max further will not eliminate this additional motion, as illustrated by the plot for F = 2,500 and 1,000N. Results for force scheme 2 are plotted in Figs 5 and 6. Mode 4 controls the allowable delay. For ν = 1 , there is a negligible effect on the response. For ν = 2 , mode 1 is essentially unchanged, while mode 2 shows some additional motion due to delay. Two out of 4 of the control forces reach the prescribed peak value of 15,000N. The response corresponding to F max = 4,000N shown in Fig 7 exhibits similar features.

8.5 State-space Formulation for MDOF Systems 665

0.04

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

Nodal displacement (m)

0.03

0.02

0.01

0

−0.01

−0.02

−0.03

−0.04

0

5

10

15

20

25

30

35

40

45

50

Time (sec) a) Displacement of node 4 −3

Modal coordinate response

4

x 10

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

3

2

1

0

−1

−2

−3

0

5

10

15

20

25

30

35

40

45

Time (sec) b) Time history response of the real part of the coordinate for mode 2

50

666 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

−4

2.5

x 10

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

Modal coordinate response

2

1.5

1

0.5

0

−0.5

−1

−1.5

−2

0

5

10

15

20

25

Time (sec)

30

35

40

45

50

c) Time history response of the real part of the coordinate for mode 4 5000

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

Control force magnitude

4000

3000

2000

1000

0

−1000

−2000

−3000

−4000

−5000

0

500

1000

1500

2000

Time step number d) Time history response of the magnitude of the control force

2500

8.5 State-space Formulation for MDOF Systems 667

Power supplied by the control force

500

0

−500

Force scheme 1 Damping case 1 El Centro ν = 2 Fmax = 15000N

−1000

−1500

−2000

−2500

−3000

−3500

0

500

1000

1500

Time step number e) Time history response of the input power Figure 1

2000

2500

668 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.15

Nodal displacement (m)

0.1

0.05

0

−0.05

−0.1

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 15000N

−0.15

−0.2

0

5

10

15

20

25

30

35

40

45

50

35

40

45

50

Time (sec) a) Displacement of node 4 0.1

Modal coordinate response

0.08

0.06

0.04

0.02

0

−0.02

−0.04

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 15000N

−0.06

−0.08

−0.1

0

5

10

15

20

25

30

Time (sec) b) Time history response of the real part of the coordinate for mode 2

8.5 State-space Formulation for MDOF Systems 669

−4

2.5

x 10

Modal coordinate response

2

1.5

1

0.5

0

−0.5

−1

−1.5

−2

−2.5

0

5

10

15

20

25

30

35

40

45

Time (sec) c) Time history response of the real part of the coordinate for mode 4

50

4

Control force magnitude

1.5

x 10

1

0.5

0

−0.5

−1

−1.5

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 15000N 0

500

1000

1500

Time step number

d) Control force magnitude

2000

2500

670 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

4

5

Power supplied by the control force

4

3

x 10

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 15000N

2

1

0

−1

−2

−3

0

500

1000

1500

Time step number

e) Input power Figure 2

2000

2500

8.5 State-space Formulation for MDOF Systems 671

0.06

Nodal displacement (m)

Force scheme 1 Damping case 1 El Centro 0.04 ν = 3 Fmax = 5000N 0.02

0

−0.02

−0.04

−0.06

0.03

Modal coordinate response

0.02

0

5

10

15

20

25

30

35

40

45

50

15

20

25

30

35

40

45

50

Time (sec) a) Displacement of node 4

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 5000N

0.01

0

−0.01

−0.02

−0.03

0

5

10

Time (sec) b) Time history response of the real part of the coordinate for mode 2

672 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

−4

2.5

x 10

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 5000N

Modal coordinate response

2

1.5

1

0.5

0

−0.5

−1

−1.5

−2

0

5

10

15

20

25

30

35

40

45

50

Time (sec) c) Time history response of the real part of the coordinate for mode 4

Power supplied by the control force

6000

4000

2000

0

−2000

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 5000N

−4000

−6000

0

500

1000

1500

Time step number d) Time history response of the input power Figure 3

2000

2500

8.5 State-space Formulation for MDOF Systems 673

0.08

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 1500N

Nodal displacement (m)

0.06

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

30

Time (sec) a) Displacement of node 4

0.08

40

45

50

Force scheme 1 Damping case 1 El Centro ν = 3 Fmax = 1000N

0.06

Nodal displacement (m)

35

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

30

Time (sec) b) Displacement of node 4 Figure 4

35

40

45

50

674 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.05

Force scheme 2 Damping case 1 El Centro ν = 1 Fmax = 15000N

0.04

Nodal displacement (m)

0.03

0.02

0.01

0

−0.01

−0.02

−0.03

−0.04

−0.05

0

5

10

15

20

25

30

Time (sec) a) Displacement of node 4

35

40

45

50

−4

2.5

x 10

Force scheme 2 Damping case 1 El Centro ν = 1 Fmax = 15000N

Modal coordinate response

2

1.5

1

0.5

0

−0.5

−1

−1.5

−2

0

5

10

15

20

25

30

35

40

45

Time (sec) b) Time history response of the real part of the coordinate for mode 4

50

8.5 State-space Formulation for MDOF Systems 675

Maximum control force magnitude

3800

Force scheme 2 Damping case 1 El Centro ν = 1 Fmax = 15000N

3700

3600

3500

3400

3300

3200

3100

3000

2900

1

2

3

Control force number c) Peak values of the control forces Figure 5

4

676 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.06

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

Nodal displacement (m)

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

30

35

40

35

40

45

50

Time (sec) a) Displacement of node 4 −3

Modal coordinate response

3

x 10

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

2

1

0

−1

−2

−3

0

5

10

15

20

25

30

45

Time (sec) b) Time history response of the real part of the coordinate for mode 4

50

8.5 State-space Formulation for MDOF Systems 677

−3

4

x 10

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

Modal coordinate response

3

2

1

0

−1

−2

−3

0

5

10

15

20

25

Time (sec)

30

35

40

45

50

c) Time history response of the real part of the coordinate for mode 2 0.06

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

Modal coordinate response

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

30

35

40

45

Time (sec) d) Time history response of the real part of the coordinate for mode 1

50

678 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

4

Maximum control force magnitude

1.5

x 10

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 15000N

1.4

1.3

1.2

1.1

1

0.9

0.8

0.7

0.6

0.5

1

2

3

Control force number e) Peak values of the control forces Figure 6

4

8.5 State-space Formulation for MDOF Systems 679

0.06

Force scheme 2 Damping case 1 El Centro ν = 2 Fmax = 4000N

Nodal displacement (m)

0.04

0.02

0

−0.02

−0.04

−0.06

0

5

10

15

20

25

Time (sec)

30

35

40

45

50

Figure 7: Displacement of node 4

Controllability of a particular modal response The governing equation for the response of the k’th mode is given by eqn (8.305). 2λ k q˙k = λ k q k + b f , k F = λ k q k – --------- Φ kT E f F f

(8.337)

k

By definition, the j’th column of E f defines the nodal force pattern corresponding to F j . Premultiplication by Φ kT generates the equivalent modal force for mode k. If Φ kT is orthogonal to the j’th column of E f , F j will have no effect on the response of the k’th mode. Therefore, for mode k to be controllable, the nodal location of the active control forces must satisfy the following constraint Φ kT E f ≠ 0

(8.338)

680 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems When damping is uncoupled, Φ k real, and this constraint is obvious. One cannot apply a control force at a null point of a mode.

Example 8.16: Controllability analysis for a 20 DOF Model Controllability is illustrated using a 20 DOF shear beam model having constant mass and stiffness. Figure 1 shows the first 4 modal displacement profiles which are real in this case since there is no damping. The corresponding internodal displacement profiles are plotted in Fig 2. The latter curves are scaled versions of the element shear deformation profiles. 20

18

Node number

16

14

12

10

8

6

4

No feedback m = 1000kg k = 1700000N ⁄ m

2

1 0 −1

−0.8

−0.6

−0.4

−0.2

0

2 0.2

3

4 0.4

Amplitude Figure 1: Modal displacement profiles

0.6

0.8

1

8.5 State-space Formulation for MDOF Systems 681

20

18

16

Element number

14

12

10

8

6

4

2

0 −0.8

1 −0.6

−0.4

−0.2

Amplitude

0

2 0.2

3

4 0.4

0.6

Figure 2: Scaled modal element shear deformation profiles Suppose the j’th control force system consists of a single force applied at one of the 20 nodes, say node l. If node l is a null point for a particular mode, the j’th force has no effect on the response of that mode. Considering Fig 1, the first mode has no null points, and it follows that this mode can be controlled by applying the force at any of the 20 nodes. The second mode, has a null point between nodes 13 and 14. Therefore, these nodes are ineffective. The optimum locations for controlling mode 2 are nodes 7 and 20 which correspond to the maximum amplitudes. Mode 3 has null points adjacent to nodes 8 and 16. Nodes 4,12, and 20 are the optimal force locations. Lastly, mode 4 has null points adjacent to nodes 6, 12, and 18. The corresponding optimal locations are nodes 3, 9, 15, and 20. These results are summarized below in Table 1.

682 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Table 1 Nodal location - Single Control Force Mode Optimal

Not allowed

1

20

1

2

7, 20

13, 14

3

4, 12, 20

8, 16

4

3, 9, 15, 20

6, 12, 18

Another option is for the j’th control force system to consist of a set of selfequilibrating force applied at adjacent nodes, say l and l+1. The product of the k’th modal vector, Φ k , and the j’th column of E f is proportional to the shear deformation for the element located between nodes l and l+1. Therefore, if this element corresponds to a null point for the modal element shear deformation profile, control force system j has no effect on the response of mode k. It follows that null points are to be avoided, and the elements corresponding to the peak values are the optimal locations for the pair of self-equilibrating control forces. Considering element j to be between nodes j and j-1, and the forces are applied at the ends of an element, the constraints and optimal locations for this case follow from Fig 2, and are listed in Table 2. Table 2

Mode

Element location - Self-equilibrating Force Optimal

Not allowed

1

1

20

2

1, 14

7

3

1, 9, 17

5, 13

4

1, 6, 12, 18

3, 4, 9, 10, 15

Observability of a particular modal response Feedback is implemented by observing the response, computing the

8.5 State-space Formulation for MDOF Systems 683 control forces with an algorithm and then applying these forces with actuators. When the general modal formulation is used, F is taken as a linear function of the modal coordinates, (8.339)

F = – kd qR – kv qI

Assuming s modes are considered, each q is of order s x 1. If both terms are retained, 2s observations are required to uniquely define the modal response. Let y 1, y 2, …, y N denote the observations. These observations may be displacements, velocities, strains, strain rates, or some other measured that are linearly related to the state variables. The k’th observation is expressed as U yk = Dk = Dk X U˙

(8.340)

where D k depends on the nature and location of the observation represented by y k . One can interpret D k as defining the nodal variables that y k is monitoring. Substituting for the state vector in terms of the full set of n coordinates leads to the following exact expression for y k n

1 y k = --2

∑

n

D k ( q j V j + q˜ j V˜ j ) =

j=1

∑

D k ( V R, j q R, j – V I, j q I, j )

(8.341)

j=1

If D k is orthogonal to V R, j , q R, j does not contribute to y k and it follows that y k does not observe q R, j . A similar statement applies for q I, j . Therefore, for the j’th mode to be observable, at least one of the set of k D’s cannot be orthogonal to the modal vectors for those modes which are included in the modal approximation for the solution. Assuming s modes are retained and N observations are made at a particular instant in time, eqn (8.341) is approximated by y ≈ DV R q R – DV I q I

(8.342)

where y is of order N x 1; q R, q I are of order s x 1; the j’th row of D contains D j ; and the k’th column of V contains V k . Observability for the modal coordinates selected to approximate the solution requires each column of the products, DV R

684 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems and DV I , to have at least one non-zero term. This constraint can be expressed in terms of the rank of the products. rank ( DV R ) = s rank ( DV I ) = s

(8.343)

Since DV is of order N x s, it follows that N ≥ s . If N = 2s and the rank constraint is satisfied, the solution for q R and q I is unique. Otherwise, a least square type solution has to be generated. When damping is uncoupled, the linear velocity feedback law expressed in terms of modal coordinates simplifies to F = – k v q˙

(8.344)

where q˙ and U˙ are related by U˙ ≈ Φq˙

(8.345)

Suppose the observations involve modal velocities. Expressing y as ˙ y = D' U

(8.346)

and substituting for U˙ results in y = ( D'Φ )q˙

(8.347)

Assuming s modes are retained, observability requires D'Φ j ≠ 0 for j = 1, 2, …, s . This constraint is equivalent to the condition that the observation points are not located at null points for the j’th modal vector. Requiring that rank of D'Φ to be equal to s satisfies the observability condition. Since it also requires N ≥ s , the solution is either unique (N = s), or over-determined.

Example 8.17: Observability analysis for a 20 DOF model Suppose 4 nodes are retained, and the observed variables are nodal velocities. At least 4 observations are necessary when damping is uncoupled. For arbitrary damping, 8 observations are required. If damping is uncoupled, the

8.6 Optimal Linear Feedback - MDOF Time Invariant System 685 constraint on the location of the observation points is that they do not coincide with null of the modal displacement profiles. The most desirable locations are those which coincide with the maximum modal amplitudes. This choice eliminates the possibility of an ill-conditioned set of equations for the modal velocities defined by eqn (8.347). The unacceptable and optimal locations for the velocity observations are listed below. Table 1 Modal response quantity

Nodal location - Single Control Force Optimal

Not allowed

q˙1

20

1

q˙2

7, 20

13, 14

q˙3

4, 12, 20

8, 16

q˙4

3, 9, 15, 20

6, 12, 18

One needs to select at least 4 observation nodes. Taking 7, 9, 12, and 20, and applying eqn (8.347) leads to the following set of equations for the first 4 modal velocities, u˙ 7

0.5114 u˙ 9 = 0.6367 0.7959 u˙ 12 1.0000 u˙ 20

1.0000 0.8792 0.3741 – 0.9941

0.4440 – 0.3019 – 0.9941 0.9824

q˙ – 0.5757 1 – 0.9941 q˙2 0.1528 q˙3 – 0.9650 q˙ 4

(1)

u˙ 7 0.3973 – 0.3375 u˙ 9 0.1761 u˙ 12 – 0.0975 u˙ 20

(2)

Inverting eqn (1) results in q˙1

0.6124 q˙2 = 0.7397 0.8881 q˙3 0.7767 q˙4

– 0.6095 – 0.0967 – 0.7186 – 1.2636

0.8513 0.0262 – 0.2170 0.6343

686 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

8.6 Optimal linear feedback - MDOF time invariant systems The formulation of the linear quadratic regulator problem developed for a time-invariant SDOF system in Section 8.4 is extended here to MDOF systems. Since matrix notation was used for the SDOF case, the only difference between the SDOF and MDOF formulations is the form of the weighting matrices, Q and R, contained in the quadratic performance index. In what follows, the LQR control algorithm corresponding to the modal state-space formulation is derived for both the continuous and discrete time feedback scenarios. The role of Q in this case is to assign relative weights to the different modal responses. Weighting the different control forces is achieved by adjusting the elements of R. Examples illustrating the sensitivity of the modal damping due to feedback as a function of Q and R are presented. Lastly, various issues involved in “selecting” a control force scheme are also discussed. Continuous time modal formulation The governing equation for this formulation is eqn (8.311). For convenience, the relevant equations are listed below. X˙ m = A m X m + B fm F + B pm P + B gm a g

(a)

where X m contains the real and imaginary parts of the modal coordinates, qR Xm = qI

(b)

and the coefficient matrices depend on the modal properties of the MDOF system and the nodal distribution of the control forces defined by E f . Negative linear feedback is taken as q F = – K fm X m = – k d k v R qI

(c)

An expression for K fm is established by requiring the following performance index to be stationary, 1 J = --2

∫

∞

T QX + F T QF ) dt (Xm m 0

(8.348)

8.6 Optimal Linear Feedback - MDOF Time Invariant System 687 where Q and R are taken as diagonal weighting matrices,

Q =

Qd 0 (8.349)

0 Qv

R = [ r i δ ij ] When Q and R are constant, the solution is K fm = R – 1 B Tfm H = k d

kv

(8.350)

where H is determined by T H + H A – H B R –1 B T H = Q Am m fm fm

(8.351)

Just as for the SDOF case, k d ≡ 0 when Q d is taken as a null matrix. This choice avoids the potential instability associated with displacement feedback. Given A m , B fm , R, and Q, one can determine K fm and the eigenvalues of A m – B fm K fm with the MATLAB function CARE. These eigenvalues define the modal damping ratios and frequencies for the system corresponding to the particular choice of control force locations, ( E f ) , and weighting matrices. Given E f , one can generate a range of modal damping ratio distributions by varying the individual weighting factors contained in Q and R. To avoid a potential instability due to displacement feedback, Q d is taken as a null matrix. The form of Q v is a generalized version of eqn (8.163) which applies for the SDOF case. Q v = [ 4m i2 ω i2 w i δ ij ]

(8.352)

where m i is the modal mass, ω i is the modal frequency, and w i is the relative weighting for the i’th mode. With this scaling law, w i and r j are if O(1). Increasing w n places more emphasis on reducing the response of mode n. Increasing r m places more emphasis on reducing the “cost” for the m’th control force. Both of these perturbations result in changes in the magnitudes of the modal damping ratios. In addition to these parameters, the modal damping distribution is also influenced by the nature of the control force scheme. A pair of self-equilibrating control forces is equivalent to “material” damping which tends to produce

688 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems damping that increases with the modal frequency. A single control force applied at a node is similar to mass proportional damping, and results in damping that decreases with the modal frequency. The force schemes considered in Example 8.14 exhibited these behavioral trends. Continuous time LQR control is useful for establishing an initial estimate of the control force system properties required to achieve certain performance objectives such as limiting the peak deformations and accelerations in a structure subject to constraints on the external power requirement and the peak value of the control forces. This “estimate” is then evaluated using the discrete time formulation specialized for the particular time step selected, and the design is modified if necessary. The effect of time delay is also considered at this stage. Discrete time modal formulation The discrete time free vibration formulation allowing for feedback is based on eqn (8.320) which is expressed here as X m, j + 1 = c m1 X m, j + c m2 F j

(8.353)

where c m1 = e c m2 =

A m ∆t

– 1 ( e A m ∆t – Am

(8.354) I )B fm

For the discrete time case, the quadratic performance index is taken as ∞

1 J = --2

∑

T QX T (Xm ,j m, j + F j RF j )

(8.355)

j=0

with Q, R defined by eqn (8.349). Expressing F j as F j = – K fm X m, j

(8.356)

and requiring J to be stationary with respect to K fm leads to the discrete algebraic Riccati equation, –1 T T T Hc T T H – c m1 m1 + ( c m1 H c m2 ) ( R + c m2 H c m2 ) ( c m1 H c m2 ) = Q

(8.357)

8.6 Optimal Linear Feedback - MDOF Time Invariant System 689 and the following expression for the optimal feedback, K fm

optimal

T H c ) –1 ( c T H c ) T = ( R + c m2 m2 m1 m2

(8.358)

This solution can be generated with the MATLAB function DARE. Substituting for F j , the free vibration response with optimal feedback is governed by X m, j + 1 = c m1 – c m2 K fm X j = c m X j opt

(8.359)

The frequency and damping properties of the discrete model with feedback are related to the eigenvalues of c m . Suppose s modes are considered. There are s pairs of complex conjugates, ρ = ρ 1, ρ˜ 1, ρ 2, ρ˜ 2, …, ρ s, ρ˜ s

(8.360)

The j’th pair is expressed in polar form. ρ j ρ˜ j = ρ j e

± iθ j

(8.361)

With this notation, the modal periods and corresponding damping ratios are given by δj ξ j = ------------------------------[ δ 2j + θ 2j ] 1 / 2 2π∆t T j = ------------θj

(8.362)

δ j = – ln ( ρ j ) These equations are useful for comparing discrete vs. continuous feedback. Although the actual control system is based on discrete time feedback, it is convenient to work with the continuous time formulation during the preliminary design phase which is concerned with selecting the location and nature of the control forces, and estimating the relative weighting factors in order to satisfy the specified performance requirements. Shifting from continuous to discrete time feedback with time delay changes the response characteristics, such as the modal damping, and can be potentially destabilizing. For this formulation, the free vibration response becomes unstable when ρ > 1, which corresponds to a

690 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems negative value for ξ . Examples illustrating various behavioral aspects of MDOF systems controlled with the LQR algorithm are presented in the following section. Application studies - LQR control The 4 DOF model defined in Fig 8.15 is utilized to illustrate the design of LQR based control force systems. Versions of this model were used in the preceding examples to demonstrate instability. The focus here is on modal damping introduced by feedback. Both global and self-equilibrating force schemes (Fig 8.15 b, c) are considered. m4

u4

k 4, c 4 m3

F1

F1

F1 F2

u3

k 3, c 3 m2

F2 F3

u2

F2

k 2, c 2 m1

F3 F4

u1

k 1, c 1 a)

b)

c)

Node/Element

m (kg)

k (kN/m)

c (kNs/m)

1

1000

1700

4

2

1000

1400

3

3

1000

1000

2

4

1000

700

1

Fig. 8.15: Definition of model and control force schemes Case 1: Global forcing The scheme shown in (b) involves specifying 2 force weighting parameters ( r 1 and r 2 ) in R and 4 modal velocity weighting parameters ( w 1 → w 4 ) in Q v . Suppose the design objective is a uniform distribution of the peak internodal displacements for a specified dynamic excitation, such as an earthquake. Starting

8.6 Optimal Linear Feedback - MDOF Time Invariant System 691 with uniform weighting, one can adjust the r’s and w’s separately until the desired uniform displacement profile is obtained. The strategy followed here is based on first perturbing the force weights to obtain a reasonable level of damping, then adjusting the modal weights to obtain essentially uniform modal damping, and finally scaling the r’s to shift the average value of the peak internodal displacement to the desired value. Results for the first step are shown in Figure 8.16 a). Starting with w and r set to unity, the r’s are reduced to 0.5 and 0.25. The modal damping for the first mode (period ≈ 0.5 sec) is essentially doubled, while the change in the corresponding damping ratios for the third and fourth modes is small. There is close agreement between the continuous and discrete feedback results for all but the second mode.

Modal damping ratio with feedback

0.09

0.085

0.08

0.075

0.07

0.065

0.06

w = 1 r1 = r2 = 1

0.055

0.05

0.045 0.05

Continuous feedback Discrete feedback 0.1

0.15

0.2

0.25

0.3

Modal period (sec) a)

0.35

0.4

0.45

0.5

692 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.13

Modal damping ratio with feedback

0.12

0.11

0.1

0.09

0.08

0.07

w = 1 r 1 = r 2 = 0.5

0.06

0.05

0.04 0.05

Continuous feedback Discrete feedback 0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

Modal period (sec) b) Modal damping ratio with feedback

0.18

0.16

0.14

0.12

0.1

0.08

w = 1 r 1 = r 2 = 0.25

0.06

Continuous feedback Discrete feedback 0.04 0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

Modal period (sec) c) Fig. 8.16: Sensitivity of the modal damping ratio to the control force weighting factor, r

8.6 Optimal Linear Feedback - MDOF Time Invariant System 693 Taking r = 0.25 as a first trial value for r, the modal weightings are adjusted to increase the damping ratios for the higher modes. Fig 8.17 shows results for a particular w distribution which places the primary emphasis on modes 3 and 4. The modal damping is nearly uniform, but note that there is a significant difference between the continuous and discrete damping ratios. This weighting scheme produces the peak internodal displacement profile plotted in Fig 8.17 b. 0.21

Modal damping ratio with feedback

0.2

0.19

0.18

0.17

w1 = 1 0.16

w2 = 4

0.15

w 3 = 70

0.14

w 4 = 50 r 1 = r 2 = 0.25

0.13

0.12 0.05

Continuous feedback Discrete feedback 0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Modal period (sec) a) Modal damping ratio with continuous and discrete feedback - no saturation limit

694 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

4

Element number

3

El Centro excitation w1 = 1 w2 = 4 2

w 3 = 70 w 4 = 50 r 1 = r 2 = 0.25 Maximum displacement Average displacement

1 0.01

0.0105

0.011

0.0115

0.012

0.0125

0.013

Maximum element internodal displacement (m)

0.0135

b) Element internodal profile Fig. 8.17: Modal damping and peak response for initial weighting scheme Holding the w values constant and decreasing r generates the results shown in Fig 8.18. The average damping in increased, and the average peak displacement is decreased. One continues this process until the “design” displacement value is obtained. In addition to motion based design requirements, there are also constraints on the peak values of the control forces and the peak power required. For this design the peak quantities are: F 1, max = 4.234kN F 2, max = 1.783kN Peak Power = 3.772kNm ⁄ s Average internodal displacement = 0.011m Optimal design involves a consideration of all of these requirements, and assigning priorities for the multiple objectives.

8.6 Optimal Linear Feedback - MDOF Time Invariant System 695

Modal damping ratio with feedback

0.28

0.26

0.24

0.22

w1 = 1 w2 = 4

0.2

w 3 = 70 0.18

w 4 = 50 r 1 = r 2 = 0.125

0.16

Continuous feedback Discrete feedback 0.14 0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

Modal period (sec) a) Modal damping ratio with continuous and discrete feedback - no saturation limit

Element number

4

3

El Centro excitation w1 = 1 w2 = 4 2

w 3 = 70 w 4 = 50 r 1 = r 2 = 0.125 Maximum displacement Average displacement

1 0.0085

0.009

0.0095

0.01

0.0105

0.011

0.0115

Maximum element internodal displacement (m) b) Element internodal displacement profile Fig. 8.18:

0.012

696 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Case 2: Self-equilibrating control force The force scheme defined by Fig 8.15c involves 4 control forces. Starting with uniform weightings, the scale factors are adjusted until the modal damping ratio distribution is essentially the same as obtained with the final weighting scheme of Case 1. Figure 8.19 contains the final results for Case 2. Since the modal damping distributions are nearly identical, internodal displacement profiles are also in close agreement. The peak values of the control forces and power for this control force scheme are F 1, max = 3.818kN F 2, max = 5.074kN F 3, max = 4.849kN F 4, max = 4.449kN Peak Power = 3.599kNm ⁄ s Average internodal displacement = 0.0108m This scheme requires a larger force, 5.074kN vs. 4.234kN. The peak power required depends on the equivalent damping, and since this quantity is essentially the same, it follows that the power requirements will also be close.

Modal damping ratio with feedback

8.6 Optimal Linear Feedback - MDOF Time Invariant System 697

0.34

w1 = 8

0.32

w2 = 5

0.3

w 3 = 12

0.28

w 4 = 25

0.26

r 1 = r 2 = r 3 = r 4 = 0.35

0.24

0.22

0.2

0.18

0.16

Continuous feedback Discrete feedback

0.14 0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

Modal period (sec) a) Modal damping ratio with continuous and discrete feedback - no saturation limit

Element number

4

3

El Centro excitation w1 = 8 w2 = 5 2

w 3 = 12 w 4 = 25 r 1 = r 2 = r 3 = r 4 = 0.35 Maximum displacement Average displacement

1 0.0085

0.009

0.0095

0.01

0.0105

0.011

0.0115

0.012

Maximum element internodal displacement (m) b) Element internodal displacement profile Fig. 8.19: Modal damping and peak response - Self-equilibrating control force system

698 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Example 8.18: Control force design studies for a 20 DOF shear beam A 20 DOF shear beam with constant mass (1000kg) and constant element damping (10,000Nsec/m) is considered. An estimate for the element stiffness distribution is selected such that the element shear deformation profile for the first mode is essentially uniform, and the average element relative displacement response due to the El-Centro seismic excitation is approximately 0.0125m. Figure 1 shows the modal properties and response for this choice of stiffness and no feedback control. There is close agreement between the actual and desired deformation for nodes 1 through 13. Beyond this point, the difference increases rapidly and exhibits an exponential type growth pattern, similar to the internodal displacement profiles for modes 3 and 4. This result indictates that the contribution of the higher modes is dominating the response in the upper region.

20

18

Element number

16

14

12

10

8

6

4

2

0

0

0.5

1

1.5

2

2.5

3

3.5 6

x 10

Element stiffness-newtons per meter a) Element shear stiffness distribution Figure 1 Properties for no iteration and no feedback

8.6 Optimal Linear Feedback - MDOF Time Invariant System 699

20

18

16

Node number

14

12

10

8

6

4

2

0 −1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Amplitude b) Modal displacement profile-real part 20

18

16

Element number

14

12

10

8

6

4

2

0 −0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

Amplitude c) Element internodal modal displacement profile-real part Figure 1 Properties for no iteration and no feedback

1.2

1.4

700 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.14

Modal damping ratio

0.12

0.1

0.08

0.06

0.04

0.02

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period d) Modal damping ratio without feedback 20 maximum average design value

18

16

Element number

14

12

10

8

6

4

2

0 0.012

0.014

0.016

0.018

0.02

0.022

0.024

Maximum element internodal displacement e) Element internodal displacement profile Figure 1 Properties for no iteration and no feedback

0.026

0.028

8.6 Optimal Linear Feedback - MDOF Time Invariant System 701

Various options are possible. One can alter the deformation profiles for modes 3 and 4 by modifying the stiffness in the upper region so that the gradients are decreased. Another option is to work with the initial stiffness and incorporate addtional damping with feedback control . These approaches generate the maximum values for stiffness (option 1) and damping (option 2). Combining these approaches results in intermediate values for these parameters. Figure 2 contains the results generated by iterating on the element stiffness according to the following algorithm maximum element displacement k new = k old ------------------------------------------------------------------------------------------- desired element displacement

(1)

The computation proceeds as follows. Using the initial stiffness, the time history reponse due to El-Centro is generated, and the peak value of inter-nodal displacement is determined for each element.Equation (1) is applied to update each element stiffness, and the complete analysis is then repeated. Covergence is quite rapid for this example. After 2 cycles, the correction process has essentially reached the final state. Figure 2c shows the modified element displacement profiles after 2 iterations. The peak value for element 20, the most critical location, has been reduced by approximately 33%. This correcton results in a significant improvement in the element reponse profiles plotted in Fig 2d. The “bulge” in the upper region has been eliminated.

702 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

20 initial iterated 18

16

Element number

14

12

10

8

6

4

2

0

0

0.5

1

1.5

2

2.5

3

3.5

4 6

x 10

Iterated element stiffness-newtons per meter (a) Two iterations 20 initial iterated 18

16

Element number

14

12

10

8

6

4

2

0

0

0.5

1

1.5

2

2.5

3

3.5

4 6

x 10

Iterated element stiffness-newtons per meter (b) Four iterations Figure 2 Iterated stiffness with no feedback

8.6 Optimal Linear Feedback - MDOF Time Invariant System 703

20

18

16

Element number

14

12

10

8

6

4

2

0 −0.4

−0.2

0

0.2

0.4

0.6

0.8

Amplitude-internodal modal displacement profile (real part) (c) Two iterations Figure 2 Iterated stiffness with no feedback

1

704 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

20 maximum average design value

18

16

12

10

8

6

4

2

0 0.0115

0.012

0.0125

0.013

Maximum element internodal displacement (d) Two iterations 20 maximum average design value

18

16

Element number

Element number

14

14

12

10

8

6

4

2

0 0.0118

0.012

0.0122

0.0124

0.0126

0.0128

Maximum element internodal displacement ( e) Four iterations Figure 2 Iterated stiffness with no feedback

0.013

8.6 Optimal Linear Feedback - MDOF Time Invariant System 705 Following the second approach, 4 self-equilibrating pairs of nodal forces are applied on elements 17,18,19 ,and 20. Figure 3 defines the notation for these forces. Results based on the initial stiffness and the following set of weighting coeffcients are plotted in Fig 4. w 1 = 0 w 2 =1 w 3 = 3 w 4 = 5 r 1 = r 2 = r 3 = r 4 = 0.125 These coefficients were selected to focus the control mainly on modes 3 and 4. Even though the modal damping for these modes is increased significantly, these still is a substantial difference between the actual and desired response in the upper zone.

20

F1

19

F1

F2

F2

F3

18 17 16

1

Fig 3

F3 F4

F4

706 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.5 continuous discrete−Riccati no feedback

0.45

Modal damping ratio

0.4

0.35

0.3

0.25

0.2

0.15

0.1

0.05

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period a) Modal damping ration with and without continuous or discrete feedback-no Saturation limit 20 maximum average design value

18

16

Element number

14

12

10

8

6

4

2

0 0.011

0.012

0.013

0.014

0.015

0.016

0.017

Maximum element internodal displacement b) Element internodal displacement profile Figure 4 Initial stiffness and feedback

0.018

0.019

8.6 Optimal Linear Feedback - MDOF Time Invariant System 707

Maximum control force magnitude

3800

3700

3600

3500

3400

3300

3200

3100

3000

1

2

3

4

Control force number c) Maximum value of the control force Figure 4 Initial stiffness and feedback Results based on iterating once on the stiffness, and then applying feedback to the “modifed ” system are plotted in Fig. 5. The weighting coefficients for this case are: w 1 = 0 w 2 =1 w 3 = 3 r 1 = r 2 = r 3 = r 4 = 1.0

w4 = 5

(3)

Increasing the r values reduces the equvialent damping, as shown in Fig 5a. However since the stiffness was corrected, the net effect is a significantly improved element displacement profile and lower magnitudes for the control forces.

708 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

0.14 continuous discrete−Riccati no feedback

Modal damping ration

0.12

0.1

0.08

0.06

0.04

0.02

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period a) Modal damping ration with and without continuous or discrete feeback-no saturation limit 20 maximum average design value

18

Element number

16

14

12

10

8

6

4

2

0 0.0114

0.0116

0.0118

0.012

0.0122

0.0124

0.0126

0.0128

Maximum element internodal displacement Element internodal displacement profile Figure 5 Single iteration on stiffness and then feedback

0.013

8.6 Optimal Linear Feedback - MDOF Time Invariant System 709

1500

Maximum control force magnitude

1490

1480

1470

1460

1450

1440

1430

1420

1

2

3

4

Control force number c) Maximum value of the control forces 20 initial iterated 18

16

Elememt number

14

12

10

8

6

4

2

0

0

0.5

1

1.5

2

2.5

3

3.5

4 6

x 10

Iterated element stiffness-newtons per meter d) Iterated element shear stiffness distribution Figure 5 Single iteration on stiffness and then feedback

710 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Applying a single control force at node 20 produces the results shown in Fig 6. The corresponding weights and peak value of the control force are w 1 = 0 w 2 =1 w 3 = 3 r1 = 3 F 1 max = 1719N

w4 = 5

(4)

0.1 continuous discrete−Riccati no feedback

0.09

0.08

Element number

0.07

0.06

0.05

0.04

0.03

0.02

0.01

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period a) Modal damping ratio with and without continuous or discrete feedback-no saturation limit Figure 6 Single iteration on stiffness and a single control force

8.6 Optimal Linear Feedback - MDOF Time Invariant System 711

20 maximum average design value

18

16

Element number

14

12

10

8

6

4

2

0 0.0114

0.0116

0.0118

0.012

0.0122

0.0124

0.0126

Maximum element internodal displacement b) Element internodal displacement profile Figure 6 Single iteration on stiffness and a single control force This example illustrates that there is no unique solution. One can vary the stiffness, damping, and active feedback control scheme to adjust the response. In order to determine the “optional”solution, cost measures need to be assigned to each of the paremeters. Simulation studies, as illustrated here, provide the data on sensitivities which allows for a more informed decision as to the final design.

Example 8.19: Alternate choice of response measures The previous examples are based on the specification of either nodal velocities or the first derivative of the modal coordinates as the response measures included in the performance index for the LQR control algorithm. Other response measures, such as element shear deformation rate, can also be selected. One needs only to specify the relationship between the alternate measures and the state vector, and specify weights for the alternate measures. Defining Y as the (1)

712 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems vector containing the alternate measures, the relationship is expressed as either: Y = DX (2) or Y = DX m (3) depending on whether the nodal or modal formulation is used. Taking the index T as Y Q 0 Y and substituting for Y leads to T

Q = D Q0 D The remaining steps are the same as for the standard formulation. This formulation is applied to the system considered in example 8.18. The reponse measures are taken as the shear deformation rates for elements 18,19, and 20. Uniform weighting ( w 1 = w 2 = w 3 = 1 ) is used for the element deformations. The location of the control force is the same as for example 8.18. Figure 1 shows some of the results for this case. These plots correspond to Fig 6 of example 8.18. The peak value of the control force is 1200 N for Fig1b vs. 1719 N for Fig 6. 0.1 continuous discrete−Riccati no feedback

0.09

Modal damping ratio

0.08

0.07

0.06

0.05

0.04

0.03

0.02

0.01

0 0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

Modal period a) Modal damping ratio with and without continuous or discrete feedback-

8.6 Optimal Linear Feedback - MDOF Time Invariant System 713 no saturation limit Figure 1 Solution based on using deformation rates for elements 18,19,20 and a single control force at node 20 . 20 maximum average design value

18

16

Element number

14

12

10

8

6

4

2

0 0.01

0.0105

0.011

0.0115

0.012

0.0125

0.013

0.0135

Maximum element internodal displacement b) Element internodal displacement profile Fig 1 Solution based on using deformation rates for elements 18,19,20 and a single control force at node 20.

714

Introduction to Structural Motion Control

Problems 715

Problems Problem 8.1 Refer to eqn (8.28). Consider a SDOF system which is initially at rest ( u 0 = u˙ 0 = 0 ). a) Determine u ( t ) for the case of a sinusoidal ground acceleration, a g = aˆ g sin Ωt . b) Discuss how you would develop a numerical procedure for evaluating at various times such as t1 , t2 , ... etc. u Problem 8.2 Show that the error in the Pade approximation, (eqn 8.51), is of order ( ( λ td ) . 3

e

– λt d

1 1 – --- ( λt d ) 2 = -------------------------- + Error 1 1 + --- ( λt d ) 2

Problem 8.3 Consider a single degree-of-freedom system with continuous pure velocity feedback. Suppose the natural period is 1 second. Use eqn(8.54) to determine the maximum allowable time delay corresponding to the following values for ξ and ξa : ξ = 0.05 ξ a = 0.05, 0.10, 0.20, 0.30 Problem 8.4 Verify eqn(8.77). Problem 8.5 Consider eqn(8.73). Integrating this equation between

t j and t j + 1 leads

716 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems to tj + 1

X j + 1 –X j =

∫

( AX + B f F + B g a g + B p p) dt

tj

Suppose the integrand is assumed to vary linearly over the time interval, and the coefficient matrices A, B f, … ,are constant. a) Derive the expression for X j + 1 corresponding to these conditions. Compare this result with eqn(8.77). Comment on the nature of the error. b) Specialize a) for negative linear feedback, and compare with eqn(8.82) . c) Specialize b) for no time delay and free vibration response. Compare with eqn(8.86). Define the stability requirement for this approximation. Problem 8.6 A SDOF system has the following properties: m = 1000 kg k = 70, 000 N ⁄ m c = 320 N ⋅ s ⁄ m Using the data contained in example 8.2, estimate the limiting time step for each of the following values of the feedback parameters: kd = 0 k v = 0, 800, 1600, 3200, 6400 N⋅ s ⁄ m Problem 8.7 Apply numeral simulation to eqn(8.117) and establish an estimate for the stability limit of the following systems: m = 1000 kg k = 60, 000N ⁄ m c = 750 N⋅ s ⁄ m kd = 0 k v = 3000 N ⋅ s ⁄ m υ = 0, 1, 2

Problems 717

Example 8.3 contains data for c=0. The STABILITY option of MOTIONLAB solves eqn (8.117) for given values of the system properties and the time ratio, ∆t ⁄ T . Problem 8.8 Refer to Example 8.3. Extend Table 1 to include ξ = 0.05 . Consider υ = 0, 1, 2 and ξ a = 0, 0.1, 0.2, 0.3 . Problem 8.9 Verify eqn(8.158) using the CARE function of MATLAB. Take: m = 1000 kg c = 1

k = 60, 000N ⁄ m

c = 750N ˙ s ⁄ m

q v = 4ω 2 m 2 q˙˙v

qd = 0

and consider q v to have the following values: 0,0.5,1.0. Problem 8.10 Refer to eqn (8.160).Will the LQR algorithm ever produce an unstable system? Problem 8.11 Consider eqn(8.168). Let D = Q + KTf RK f n

Sn =

∑C

j, T

DC

j

j=0

Noting the identity, T

C Sn C – Sn = – D + C

n + 1, T

DC

and the limit condition, j

C →0

as

j→∞

n+1

718 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems derive eqn(8.169) Problem 8.12 Refer to Figs 1 and 2 of example 8.4. Suppose the time ratio ∆t ⁄ T is determined by the external loading, and is equal to 0.1. Suggest a value for q v such that ξ a is close to 0.2 when ξ = 0.05 . Problem 8.13 Consider the following system m = 1000 kg k = 60, 000 N ⁄ m c=1000 N.s ⁄ m Suppose ∆t = 0.02s . Select the parameters for discrete time feedback control such that the effective damping ratio is equal to 0.2, Use Figs 1 and 2 of example 8.4 to obtain an initial estimate, and the discrete time Riccati equation option of MOTIONLAB to refine the estimate. Note that the solution tends toward the continuous time feedback case as ∆t ⁄ T approaches 0. Problem 8.14 Refer to example 8.5. Compare the expressions for k d and k v corresponding to q d = 0 with the continuous time Riccati solution defined by eqn(8.161). Use the discrete time Riccati solution for ξ = 0.02 listed in example 8.4 to compare the values of q v required to produce ( k v ⁄ 2ωm ) = 0.2 for 2 time increments, ∆t ⁄ T = 0.02 and 0.1. Problem 8.15 Rework Problem 8.13 using the finite interval discrete time algebraic Riccati equation. Note that the weighting factors for the finite interval formulation are different than the corresponding weighing factors for the discrete time algebraic Riccati equation. Problem 8.16 Verify eqn(8.236).

Problems 719 Problem 8.17 Refer to eqn(5) of example 8.7. Show that the eigenvetors are real when K and M are symmetric and the eigenvalues are positive real quantities when K and M are positive definite. Problem 8.18 T

Show that W j V k = 0 for j ≠ k by substituting for W and V using eqn (8.267) and noting the definition equation for φ j , λ j . 2

λ j Mφ j + λ j Cφ j + Kφ j = 0 Problem 8.19 Refer to eqn(2) of example 8.11. Reduce these equations to a single equation in terms of q R and compare with the equation obtained with the conventional modal formulation(e.g, see eqn(8.275). Is there any advantage to working with the state-space formulation vs the conventional modal formulation? Problem 8.20 Consider a 20 DOF system having the following constant properties: mass = 1000 kg element stiffness = 1800 kN ⁄ m element damping = 25 kN ⋅ s ⁄ m a) Determine the modal properties (period, damping, modal displacement profile, modal intermodal displacement profile) for the first 4 modes. b) Determine the response for node 20 due to the Kobe ground excitation using the complete state-space formulation. Take ∆t = 0.02 sec. c) Repeat part b) using the modal state-space formulation and only the first 4 modes. Compare the time history and Fourier components for node 20 with the corresponding results obtained in b). Also compare the peak values of the modal coordinates.

720 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems Problem 8.21 Consider a SDOF system having the following properties: m = 10, 000 kg k = 400, 000 N ⁄ m c=2500 N ⋅ s ⁄ m a) Using the LQR control algorithm, establish values for the weighting parameters such that the effective damping for continuous velocity feedback has the following values: ξ eq = 0.05, 0.1, 0.2 . b) Take ∆t = 0.02 seconds. Evaluate the discrete time damping ratios for a). c) Using the finite time interval control algorithm, establish values for the weighting parameters such that the discrete time damping ratios are the same as found in b). d) Using the model properties corresponding to ξ eq = 0.1 established in a), determine the maximum values of the displacement, control force, and power associated with the El-Centro accelerogram. Use ∆t = 0.02 . e) Repeat d) for the Kobe accelerogram. f) Repeat d) for the Mexico City 1 accelerogram. Problem 8.22 Consider a 4 DOF system having the following properties: Node/Element

m (kg)

k (kN/m)

c (kNs/m)

1

1000

1700

4

2

1000

1400

3

3

1000

1000

2

4

1000

700

1

Suppose a single control force is applied at the top node. Using the LQR algorithm, select the weighting parameters which result in a value of the damping ratio for discrete feedback equal to 0.2 for the first mode. Take ∆t = 0.02 seconds, and apply the following strategies:

Problems 721 a) Use the conventional state-space formulation and weight the nodal velocities uniformly. b) Use the modal state-space formulation and weight the first derivative of the modal coordinates uniformly. c) Use the conventional state-space formulation and weight the element deformation rates uniformly. d) Repeat c) using the modal state-space formulation. Problem 8.23 Consider a 5 DOF shear beam with the following constant mass and stiffness properties: m = 10, 000 kg k = 350, 000 N ⁄ m

a) Assuming uniform element viscous damping, determine the magnitude of element damping such that the first mode damping ratio is 0.02. b) Apply a single control force at mode 5. Assuming all 5 modes are retained, and they are weighted equally, determine the weighting parameters such that the equivalent damping for continuous feedback is 0.15 for the first mode. c) Determine the corresponding damping ratio for discrete time feedback. Take ∆t = 0.02 seconds. d) Investigate the effect of delay on the free vibration reponse of the modal coordinates due to an initial displacement. Use ∆t = 0.02 seconds and the parameters established in part b. Problem 8.24 Consider the following 5 DOF systems: Node/Element

m (kg)

k (kN/m)

c (kNs/m)

1

1000

2000

5

2

1000

1700

5

722 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

Node/Element

m (kg)

k (kN/m)

c (kNs/m)

3

1000

1400

5

4

1000

1000

5

5

1000

700

5

a) Suppose control forces are applied at all 5 nodes. Determine the modal coordinate weighting parameters such that the equivalent damping ratio corresponding to continuous feedback is equal to 0.15 for the first mode. Assume uniform weighting. b) Suppose self-equilibrating sets of control forces are applied on all 5 elements and the weighting is applied to the element deformation time rates. Determine the weights such that the first mode damping ratio is 0.15. Assume uniform weighting. c) Apply the Northridge earthquake to each system and compare 1. the internode displacement profiles 2. the peak power 3. the peak value of the control forces Problem 8.25 Consider a 10 DOF shear beam with constant mass, element stiffness, and element damping. Take m=10000 kg . a) Determine the stiffness and damping constants such that the properties for the first mode are Period = 1 sec ond Damping ratio = 0.02

b) Select an active control force scheme which provides a damping ratio of 0.2 for the first mode. c) Apply the Kobe ground acceleration to the system defined in b). Examine the responses of the first 3 modes. Generate both the time histories and the Fourier components.

Problems 723 d) If the design objective is to have uniform peak element shear deformation throughout the system, what design modifications would you suggest? Illustrate your strategy for the case where the target value of the relative inter-nodal displacement is 0.0125m. Problem 8.26 Consider a 10 DOF shear beam with constant nodal mass equal to 10000kg. a) Select a parabolic distribution of element stiffness and a constant element viscous damping so that the period for the first mode is 1 second, and the modal damping ratio is 0.02. b) Carry out 2 cycles of iteration on the element stiffness using the ElCentro ground excitation and 0.0125m as the desired value of internodal displacement. c) Incorporate active control in the system obtained in part b. Select the weighting parameters such that the modal damping ratios ( for continuous feedback ) for the first 3 modes are approximately equal to 0.15. Consider a global forcing scheme and weight the modal coordinate velocities. d) Repeat c) using self-equilibrating control force schemes and weight the modal coordinate velocities. e) Repeat d) using the internodal element displacements performance measures. (See Example 8.19).

as the

Problem 8.27 Consider the bending beam-outrigger system shown below. Assume the outriggers are inifintely stiff, the beam bending rigidity is constant, and the cables are initially tensioned to a level of T 0 . Suppose the cable tensions can be continuously adjusted to counteract the effect of lateral load.

724 Chapter 8: Dynamic Control Algorithm - Time Invariant Systems

C

m1=a( ρ m H) m1

b

b B

T o + ∆T

m1 3H/2

D B, ρ m T o – ∆T u(x ,t)

A

H x

a) Take the lateral displacement and rotation at points B, C as the degreesof-freedom and establish the corresponding matrix form of the equilibrium equations. Work with "lumped" masses, rotatory inertias, and loads. b) Develop the state-space formulation for a) c) Describe how you would implement linear velocity feedback control. d) Suppose the bending rigidity is specified, and the critical dynamic loading is a uniform periodic excitation. Discuss how you would "calibrate" the feedback parameters for the case where the design objective is to limit the maximum acceleration. Illustrate your strategy.

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