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AREN4525 STUCTURAL CONCEPTS AND SYSTEMS FOR ARCHITECTS VICTOR E. SAOUMA SPRING 1997
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Draft LECTURE NOTES
AREN4525 STUCTURAL CONCEPTS AND SYSTEMS FOR ARCHITECTS VICTOR E. SAOUMA SPRING 1997
Dept. of Civil Environmental and Architectural Engineering University of Colorado, Boulder, CO 803090428 April 30, 1997
Draft 0{2
In order to invent a structure and to give it exact proportions, one must follow both the intuitive and the mathematical paths. Pier Luigi Nervi
Victor Saouma
Structural Concepts and Systems for Architects
Draft
Contents 1 INTRODUCTION
Science and Technology : : : : : : Structural Engineering : : : : : : : Structures and their Surroundings Architecture & Engineering : : : : Architectural Design Process : : : Architectural Design : : : : : : : : Structural Analysis : : : : : : : : : Structural Design : : : : : : : : : : Load Transfer Mechanisms : : : : Structure Types : : : : : : : : : : Structural Engineering Courses : : References : : : : : : : : : : : : : :
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2.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2.2 Vertical Loads : : : : : : : : : : : : : : : : : : : : : : : : : : : 2.2.1 Dead Load : : : : : : : : : : : : : : : : : : : : : : : : 2.2.2 Live Loads : : : : : : : : : : : : : : : : : : : : : : : : E 21 Live Load Reduction : : : : : : : : : : : : : : : : : : : 2.2.3 Snow : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2.3 Lateral Loads : : : : : : : : : : : : : : : : : : : : : : : : : : : 2.3.1 Wind : : : : : : : : : : : : : : : : : : : : : : : : : : : E 22 Wind Load : : : : : : : : : : : : : : : : : : : : : : : : 2.3.2 Earthquakes : : : : : : : : : : : : : : : : : : : : : : : : E 23 Earthquake Load on a Frame : : : : : : : : : : : : : : E 24 Earthquake Load on a Tall Building, (Schueller 1996) 2.4 Other Loads : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2.4.1 Hydrostatic and Earth : : : : : : : : : : : : : : : : : : E 25 Hydrostatic Load : : : : : : : : : : : : : : : : : : : : : 2.4.2 Thermal : : : : : : : : : : : : : : : : : : : : : : : : : : E 26 Thermal Expansion/Stress (Schueller 1996) : : : : : : 2.5 Other Important Considerations : : : : : : : : : : : : : : : : 2.5.1 Load Combinations : : : : : : : : : : : : : : : : : : : : 2.5.2 Load Placement : : : : : : : : : : : : : : : : : : : : : 2.5.3 Load Transfer : : : : : : : : : : : : : : : : : : : : : : : 2.5.4 Structural Response : : : : : : : : : : : : : : : : : : : 2.5.5 Tributary Areas : : : : : : : : : : : : : : : : : : : : :
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CONTENTS
3 STRUCTURAL MATERIALS 3.1 Steel : : : : : : : : : : : 3.1.1 Structural Steel : 3.1.2 Reinforcing Steel 3.2 Aluminum : : : : : : : : 3.3 Concrete : : : : : : : : : 3.4 Masonry : : : : : : : : : 3.5 Timber : : : : : : : : : 3.6 Steel Section Properties 3.7 Joists : : : : : : : : : :
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5.1 Reactions : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5.1.1 Equilibrium : : : : : : : : : : : : : : : : : : : : : : : : 5.1.2 Equations of Conditions : : : : : : : : : : : : : : : : : 5.1.3 Static Determinacy : : : : : : : : : : : : : : : : : : : : 5.1.4 Geometric Instability : : : : : : : : : : : : : : : : : : : 5.1.5 Examples : : : : : : : : : : : : : : : : : : : : : : : : : E 57 Simply Supported Beam : : : : : : : : : : : : : : : : : E 58 Three Span Beam : : : : : : : : : : : : : : : : : : : : E 59 Three Hinged Gable Frame : : : : : : : : : : : : : : : 5.2 Trusses : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5.2.1 Assumptions : : : : : : : : : : : : : : : : : : : : : : : 5.2.2 Basic Relations : : : : : : : : : : : : : : : : : : : : : : 5.2.3 Determinacy and Stability : : : : : : : : : : : : : : : : 5.2.4 Method of Joints : : : : : : : : : : : : : : : : : : : : : E 510 Truss, Method of Joints : : : : : : : : : : : : : : : : : 5.3 Shear & Moment Diagrams : : : : : : : : : : : : : : : : : : : 5.3.1 Theory : : : : : : : : : : : : : : : : : : : : : : : : : : 5.3.1.1 Design Sign Conventions : : : : : : : : : : : 5.3.1.2 Load, Shear, Moment Relations : : : : : : : 5.3.1.3 Moment Envelope : : : : : : : : : : : : : : : 5.3.1.4 Examples : : : : : : : : : : : : : : : : : : : : E 511 Simple Shear and Moment Diagram : : : : : : : : : : E 512 Frame Shear and Moment Diagram : : : : : : : : : : : E 513 Frame Shear and Moment Diagram; Hydrostatic Load E 514 Shear Moment Diagrams for Frame : : : : : : : : : : : E 515 Shear Moment Diagrams for Inclined Frame : : : : : : 5.3.2 Formulaes : : : : : : : : : : : : : : : : : : : : : : : : : 5.4 Flexure : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5.4.1 Basic Kinematic Assumption; Curvature : : : : : : : : 5.4.2 StressStrain Relations : : : : : : : : : : : : : : : : : : 5.4.3 Internal Equilibrium; Section Properties : : : : : : : : 5.4.3.1 Fx = 0; Neutral Axis : : : : : : : : : : : :
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4 Case Study I: EIFFEL TOWER 4.1 4.2 4.3 4.4 4.5
Materials, & Geometry : Loads : : : : : : : : : : Reactions : : : : : : : : Internal Forces : : : : : Internal Stresses : : : :
5 REVIEW of STATICS
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5.4.3.2 M = 0; Moment of Inertia : : : : : : : : : : : 5.4.4 Beam Formula : : : : : : : : : : : : : : : : : : : : : : : : E 516 Design Example : : : : : : : : : : : : : : : : : : : : : : : 5.4.5 Approximate Analysis : : : : : : : : : : : : : : : : : : : : E 517 Approximate Analysis of a Statically Indeterminate beam
6 Case Study II: GEORGE WASHINGTON BRIDGE 6.1 Theory : : : : : : : : 6.2 The Case Study : : : 6.2.1 Geometry : : 6.2.2 Loads : : : : 6.2.3 Cable Forces 6.2.4 Reactions : :
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7 A BRIEF HISTORY OF STRUCTURAL ARCHITECTURE Before the Greeks : : : : : : : : : : : : : : : : Greeks : : : : : : : : : : : : : : : : : : : : : : Romans : : : : : : : : : : : : : : : : : : : : : The Medieval Period (4771492) : : : : : : : : The Renaissance : : : : : : : : : : : : : : : : 7.5.1 Leonardo da Vinci 14521519 : : : : : 7.5.2 Brunelleschi 13771446 : : : : : : : : : 7.5.3 Alberti 14041472 : : : : : : : : : : : 7.5.4 Palladio 15081580 : : : : : : : : : : : 7.5.5 Stevin : : : : : : : : : : : : : : : : : : 7.5.6 Galileo 15641642 : : : : : : : : : : : : 7.6 Pre Modern Period, Seventeenth Century : : 7.6.1 Hooke, 16351703 : : : : : : : : : : : : 7.6.2 Newton, 16421727 : : : : : : : : : : : 7.6.3 Bernoulli Family 16541782 : : : : : : 7.6.4 Euler 17071783 : : : : : : : : : : : : 7.7 The preModern Period; Coulomb and Navier 7.8 The Modern Period (1857Present) : : : : : : 7.8.1 Structures/Mechanics : : : : : : : : : 7.8.2 Eiel Tower : : : : : : : : : : : : : : : 7.8.3 Sullivan 18561924 : : : : : : : : : : : 7.8.4 Roebling, 18061869 : : : : : : : : : : 7.8.5 Maillart : : : : : : : : : : : : : : : : : 7.8.6 Nervi, 18911979 : : : : : : : : : : : : 7.8.7 Khan : : : : : : : : : : : : : : : : : : 7.8.8 et al. : : : : : : : : : : : : : : : : : : :
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8 Case Study III: MAGAZINI GENERALI 8.1 8.2 8.3 8.4 8.5
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9 DESIGN PHILOSOPHIES and GUIDELINES 9.1 Safety Provisions : : : : : : : : : : : 9.2 Working Stress Method : : : : : : : 9.3 Ultimate Strength Method : : : : : : 9.3.1 y Probabilistic Preliminaries : 9.3.2 Discussion : : : : : : : : : : : 9.4 Example : : : : : : : : : : : : : : : : E 918 LRFD vs ASD : : : : : : : : 9.5 Design Guidelines : : : : : : : : : : :
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10.1 Nominal Strength : : : : : : : : : : : : : : : : : : 10.2 Failure Modes and Classi cation of Steel Beams : 10.3 Compact Sections : : : : : : : : : : : : : : : : : : 10.3.1 Bending Capacity of Beams : : : : : : : : 10.3.2 Design of Compact Sections : : : : : : : : 10.4 Partially Compact Section : : : : : : : : : : : : : 10.5 Slender Section : : : : : : : : : : : : : : : : : : : 10.6 Examples : : : : : : : : : : : : : : : : : : : : : : E 1019Z for Rectangular Section : : : : : : : : : E 1020Beam Design : : : : : : : : : : : : : : : :
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11.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : 11.1.1 Notation : : : : : : : : : : : : : : : : : : : : : 11.1.2 Modes of Failure : : : : : : : : : : : : : : : : 11.1.3 Analysis vs Design : : : : : : : : : : : : : : : 11.1.4 Basic Relations and Assumptions : : : : : : : 11.1.5 ACI Code : : : : : : : : : : : : : : : : : : : : 11.2 Cracked Section, Ultimate Strength Design Method : 11.2.1 Equivalent Stress Block : : : : : : : : : : : : 11.2.2 Balanced Steel Ratio : : : : : : : : : : : : : : 11.2.3 Analysis : : : : : : : : : : : : : : : : : : : : : 11.2.4 Design : : : : : : : : : : : : : : : : : : : : : : E 1121Ultimate Strength Capacity : : : : : : : : : : E 1122Beam Design I : : : : : : : : : : : : : : : : : E 1123Beam Design II : : : : : : : : : : : : : : : : : 11.3 Continuous Beams : : : : : : : : : : : : : : : : : : : 11.4 ACI Code : : : : : : : : : : : : : : : : : : : : : : : :
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10 BRACED ROLLED STEEL BEAMS
11 REINFORCED CONCRETE BEAMS
12 PRESTRESSED CONCRETE
12.1 Introduction : : : : : : : : : : : : : : 12.1.1 Materials : : : : : : : : : : : 12.1.2 Prestressing Forces : : : : : : 12.1.3 Assumptions : : : : : : : : : 12.1.4 Tendon Con guration : : : : 12.1.5 Equivalent Load : : : : : : : 12.1.6 Load Deformation : : : : : : 12.2 Flexural Stresses : : : : : : : : : : : E 1224Prestressed Concrete I Beam 12.3 Case Study: Walnut Lane Bridge : :
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13.1 Theory : : : : : : : : : : : : : : : : : : : : : : 13.1.1 Uniform Horizontal Load : : : : : : : E 1325Design of a Three Hinged Arch : : : : 13.2 Case Study: Salginatobel Bridge (Maillart) : 13.2.1 Geometry : : : : : : : : : : : : : : : : 13.2.2 Loads : : : : : : : : : : : : : : : : : : 13.2.3 Reactions : : : : : : : : : : : : : : : : 13.2.4 Internal Forces : : : : : : : : : : : : : 13.2.5 Internal Stresses : : : : : : : : : : : : 13.3 Structural Behavior of DeckStiened Arches
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14.1 Introduction : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.1.1 Beam Column Connections : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.1.2 Behavior of Simple Frames : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.1.3 Eccentricity of Applied Loads : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.2 Buildings Structures : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.2.1 Wall Subsystems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.2.1.1 Example: Concrete Shear Wall : : : : : : : : : : : : : : : : : : : : : 14.2.1.2 Example: Trussed Shear Wall : : : : : : : : : : : : : : : : : : : : : 14.2.2 Shaft Systems : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.2.2.1 Example: Tube Subsystem : : : : : : : : : : : : : : : : : : : : : : : 14.2.3 Rigid Frames : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.3 Approximate Analysis of Buildings : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.3.1 Vertical Loads : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.3.2 Horizontal Loads : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.3.2.1 Portal Method : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : E 1426Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads 14.4 Lateral De ections : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.4.1 Short Wall : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.4.2 Tall Wall : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.4.3 Walls and Lintel : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.4.4 Frames : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.4.5 Trussed Frame : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14.4.6 Example of Transverse De ection : : : : : : : : : : : : : : : : : : : : : : : : : 14.4.7 Eect of Bracing Trusses : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
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13 ThreeHinges ARCHES
14 BUILDING STRUCTURES
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Structural Concepts and Systems for Architects
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List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9
Types of Forces in Structural Elements (1D) : Basic Aspects of Cable Systems : : : : : : : : Basic Aspects of Arches : : : : : : : : : : : : Types of Trusses : : : : : : : : : : : : : : : : Variations in Post and Beams Con gurations Dierent Beam Types : : : : : : : : : : : : : Basic Forms of Frames : : : : : : : : : : : : : Examples of Air Supported Structures : : : : Basic Forms of Shells : : : : : : : : : : : : : :
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Approximation of a Series of Closely Spaced Loads Snow Map of the United States, ubc : : : : : : : : Loads on Projected Dimensions : : : : : : : : : : : Wind Map of the United States, (UBC 1995) : : : Eect of Wind Load on Structures(Schueller 1996) Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{10 Vibrations of a Building : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{12 Seismic Zones of the United States, (UBC 1995) : : : : : : : : : : : : : : : : : : : : : : : 2{13 Earth and Hydrostatic Loads on Structures : : : : : : : : : : : : : : : : : : : : : : : : : : 2{18 Load Placement to Maximize Moments : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{21 Load Transfer in R/C Buildings : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{22 Two Way Actions : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{23 Load Life of a Structure, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : : : : : 2{24 Concept of Tributary Areas for Structual Member Loading : : : : : : : : : : : : : : : : : 2{25
3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8
Stress Strain Curves of Concrete and Steel : : : : : : : : : : : : : : : : : : : : : Standard Rolled Sections : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : Residual Stresses in Rolled Sections : : : : : : : : : : : : : : : : : : : : : : : : Residual Stresses in Welded Sections : : : : : : : : : : : : : : : : : : : : : : : : In uence of Residual Stress on Average StressStrain Curve of a Rolled Section Concrete microcracking : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : W and C sections : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : prefabricated Steel Joists : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
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4.1 4.2 4.3 4.4 4.5
Eiel Tower (Billington and Mark 1983) : : : : : : : : : : : : : : : Eiel Tower Idealization, (Billington and Mark 1983) : : : : : : : : Eiel Tower, Dead Load Idealization; (Billington and Mark 1983) : Eiel Tower, Wind Load Idealization; (Billington and Mark 1983) : Eiel Tower, Wind Loads, (Billington and Mark 1983) : : : : : : :
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Eiel Tower, Reactions; (Billington and Mark 1983) : : : : : : : : Eiel Tower, Internal Gravity Forces; (Billington and Mark 1983) : Eiel Tower, Horizontal Reactions; (Billington and Mark 1983) : : Eiel Tower, Internal Wind Forces; (Billington and Mark 1983) : :
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5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10 5.11 5.12 5.13 5.14 5.15 5.16
Types of Supports : : : : : : : : : : : : : : : : : : : : : : : : : : : Inclined Roller Support : : : : : : : : : : : : : : : : : : : : : : : : Examples of Static Determinate and Indeterminate Structures : : : Geometric Instability Caused by Concurrent Reactions : : : : : : : Bridge Truss : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : A Statically Indeterminate Truss : : : : : : : : : : : : : : : : : : : X and Y Components of Truss Forces : : : : : : : : : : : : : : : : Sign Convention for Truss Element Forces : : : : : : : : : : : : : : Shear and Moment Sign Conventions for Design : : : : : : : : : : : Sign Conventions for 3D Frame Elements : : : : : : : : : : : : : : Free Body Diagram of an In nitesimal Beam Segment : : : : : : : Shear and Moment Forces at Dierent Sections of a Loaded Beam Slope Relations Between Load Intensity and Shear, or Between Shear and Moment Deformation of a Beam un Pure Bending : : : : : : : : : : : : : : : : : : : : : : : Elastic Curve from the Moment Diagram : : : : : : : : : : : : : : : : : : : : : : : Approximate Analysis of Beams : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9
Cable Structure Subjected to p(x) : : : : : : : : : : : : : : : : : : Longitudinal and Plan Elevation of the George Washington Bridge Truck Load : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : Dead and Live Loads : : : : : : : : : : : : : : : : : : : : : : : : : : Location of Cable Reactions : : : : : : : : : : : : : : : : : : : : : : Vertical Reactions in Columns Due to Central Span Load : : : : : Cable Reactions in Side Span : : : : : : : : : : : : : : : : : : : : : Cable Stresses : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : Deck Idealization, Shear and Moment Diagrams : : : : : : : : : : :
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Hamurrabi's Code : : : : : : : : : : : : : : : : : : : : : : : : Archimed : : : : : : : : : : : : : : : : : : : : : : : : : : : : : Pantheon : : : : : : : : : : : : : : : : : : : : : : : : : : : : : From Vitruvius Ten Books on Architecture, (Vitruvius 1960) Hagia Sophia : : : : : : : : : : : : : : : : : : : : : : : : : : : Florence's Cathedral Dome : : : : : : : : : : : : : : : : : : : Palladio's Villa Rotunda : : : : : : : : : : : : : : : : : : : : : Stevin : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : Galileo : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : Discourses Concerning Two New Sciences, Cover Page : : : : \Galileo's Beam" : : : : : : : : : : : : : : : : : : : : : : : : : Experimental Set Up Used by Hooke : : : : : : : : : : : : : : Isaac Newton : : : : : : : : : : : : : : : : : : : : : : : : : : : Philosophiae Naturalis Principia Mathematica, Cover Page : Leonhard Euler : : : : : : : : : : : : : : : : : : : : : : : : : : Coulomb : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : Nervi's Palazetto Dello Sport : : : : : : : : : : : : : : : : : :
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8.1 Magazzini Generali; Overall Dimensions, (Billington and Mark 1983) : : : : : : : : : : : : 8{2 8.2 Magazzini Generali; Support System, (Billington and Mark 1983) : : : : : : : : : : : : : : 8{2
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Magazzini Generali; Loads (Billington and Mark 1983) : : : : : : : : : : : : : : : : : : : : 8{3 Magazzini Generali; Beam Reactions, (Billington and Mark 1983) : : : : : : : : : : : : : : 8{3 Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983) : : : : : : : 8{4 Magazzini Generali; Internal Moment, (Billington and Mark 1983) : : : : : : : : : : : : : 8{4 Magazzini Generali; Similarities Between The Frame Shape and its Moment Diagram, (Billington and Mark 1983) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 8{5 8.8 Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark 1983) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 8{5 8.9 Magazzini Generali; Eect of Lateral Supports, (Billington and Mark 1983) : : : : : : : : 8{6 8.3 8.4 8.5 8.6 8.7
9.1 Load Life of a Structure : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9{2 9.2 Frequency Distributions of Load Q and Resistance R : : : : : : : : : : : : : : : : : : : : : 9{4 9.3 De nition of Reliability Index : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9{4 10.1 10.2 10.3 10.4 10.5 10.6 10.7
Lateral Bracing for Steel Beams : : : : : : : : : : : : : : : : : : : Failure of Steel beam; Plastic Hinges : : : : : : : : : : : : : : : : Failure of Steel beam; Local Buckling : : : : : : : : : : : : : : : Failure of Steel beam; Lateral Torsional Buckling : : : : : : : : : Stress distribution at dierent stages of loading : : : : : : : : : : Stressstrain diagram for most structural steels : : : : : : : : : : Nominal Moments for Compact and Partially Compact Sections :
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12.1 12.2 12.3 12.4 12.5 12.6
13.1 Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 13{2 13.2 Statics of a ThreeHinged Arch, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : 13{2 13.3 Two Hinged Arch, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : : : : : : : : : 13{3 13.4 Arch Rib Stiened with Girder or Truss, (Lin and Stotesbury 1981) : : : : : : : : : : : : 13{3 13.5 Salginatobel Bridge; Dimensions, (Billington and Mark 1983) : : : : : : : : : : : : : : : : 13{5 13.6 Salginatobel Bridge; Idealization, (Billington and Mark 1983) : : : : : : : : : : : : : : : : 13{6 13.7 Salginatobel Bridge; Hinges, (Billington and Mark 1983) : : : : : : : : : : : : : : : : : : : 13{6 13.8 Salginatobel Bridge; Sections, (Billington and Mark 1983) : : : : : : : : : : : : : : : : : : 13{7 13.9 Salginatobel Bridge; Dead Load, (Billington and Mark 1983) : : : : : : : : : : : : : : : : 13{8 13.10Salginatobel Bridge; Truck Load, (Billington and Mark 1983) : : : : : : : : : : : : : : : : 13{9 13.11Salginatobel Bridge; Total Vertical Load, (Billington and Mark 1983) : : : : : : : : : : : : 13{10
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13.12Salginatobel Bridge; Reactions, (Billington and Mark 1983) : : : : : : : : : : : 13.13Salganitobel Bridge; Shear Diagrams, (Billington and Mark 1983) : : : : : : : : 13.14Salginatobel Bridge; Live Load Moment Diagram, (Billington and Mark 1983) : 13.15Structural Behavior of Stiened Arches, (Billington 1979) : : : : : : : : : : : :
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14.2 Deformation of Flexible and Rigid Frames Subjected to Vertical and Horizontal Loads, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14{2 14.3 Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal Frames Subjected to Vertical and Horizontal Loads : : : : : : : : : : : : : : : : : : : : : : : : : : 14{3 14.4 Axial and Flexural Stresses : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14{4 14.5 Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : 14{6 14.6 Trussed Shear Wall : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14{8 14.7 Design Example of a Tubular Structure, (Lin and Stotesbury 1981) : : : : : : : : : : : : : 14{9 14.8 A Basic Portal Frame, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : : : : : : : 14{10 14.9 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments : : : : : : 14{12 14.10Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces : : : : 14{12 14.11Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments : : : : : 14{13 14.12Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear : : : : : : : : 14{14 14.13***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment : : : : : 14{15 14.14Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force : : : : 14{15 14.15Example; Approximate Analysis of a Building : : : : : : : : : : : : : : : : : : : : : : : : : 14{16 14.16Approximate Analysis of a Building; Moments Due to Vertical Loads : : : : : : : : : : : : 14{17 14.17Approximate Analysis of a Building; Shears Due to Vertical Loads : : : : : : : : : : : : : 14{18 14.18Approximate Analysis for Vertical Loads; SpreadSheet Format : : : : : : : : : : : : : : : 14{20 14.19Approximate Analysis for Vertical Loads; Equations in SpreadSheet : : : : : : : : : : : : 14{21 14.20Approximate Analysis of a Building; Moments Due to Lateral Loads : : : : : : : : : : : : 14{23 14.21Portal Method; SpreadSheet Format : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14{24 14.22Portal Method; Equations in SpreadSheet : : : : : : : : : : : : : : : : : : : : : : : : : : : 14{25 14.23Shear Deformation in a Short Building, (Lin and Stotesbury 1981) : : : : : : : : : : : : : 14{28 14.24Flexural Deformation in a Tall Building, (Lin and Stotesbury 1981) : : : : : : : : : : : : 14{28 14.25De ection in a Building Structure Composed of Two Slender Walls and Lintels, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 14{29 14.26Portal Method to Estimate Lateral Deformation in Frames, (Lin and Stotesbury 1981) : : 14{30 14.27Shear and Flexural De ection of a Rigid Frame Subsystem, (Lin and Stotesbury 1981) : : 14{31 14.28SideSway De ection from Unsymmetrical Vertical Load, (Lin and Stotesbury 1981) : : : 14{31 14.29Axial Elongation and Shortening of a Truss Frame, (Lin and Stotesbury 1981) : : : : : : 14{31 14.30Transverse De ection, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : : : : : : : 14{32 14.31Frame Rigidly Connected to Shaft, (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : 14{34 14.32Eect of Exterior Column Bracing in Buildings, (Lin and Stotesbury 1981) : : : : : : : : 14{35
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List of Tables 1.1 Structural Engineering Coverage for Architects and Engineers : : : : : : : : : : : : : : : : 1{12 1.2 tab:secae : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 1{12
2.10 2.11 2.12 2.13
Unit Weight of Materials : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{2 Weights of Building Materials : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{3 Average Gross Dead Load in Buildings : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{3 Minimum Uniformly Distributed Live Loads, (UBC 1995) : : : : : : : : : : : : : : : : : : 2{4 Wind Velocity Variation above Ground : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{7 Ce Coecients for Wind Load, (UBC 1995) : : : : : : : : : : : : : : : : : : : : : : : : : : 2{8 Wind Pressure Coecients Cq , (UBC 1995) : : : : : : : : : : : : : : : : : : : : : : : : : : 2{8 Importance Factors for Wind and Earthquake Load, (UBC 1995) : : : : : : : : : : : : : : 2{9 Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{10 Z Factors for Dierent Seismic Zones, ubc : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{12 S Site Coecients for Earthquake Loading, (UBC 1995) : : : : : : : : : : : : : : : : : : : 2{13 Partial List of RW for Various Structure Systems, (UBC 1995) : : : : : : : : : : : : : : : 2{15 Coecients of Thermal Expansion : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 2{19
3.1 3.2 3.3 3.4
Properties of Major Structural Steels : Properties of Reinforcing Bars : : : : : Joist Series Characteristics : : : : : : Joist Properties : : : : : : : : : : : : :
2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9
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5.1 Equations of Equilibrium : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5{3 5.2 Static Determinacy and Stability of Trusses : : : : : : : : : : : : : : : : : : : : : : : : : : 5{10 5.3 Section Properties : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 5{41 9.1 9.2 9.3 9.4
Allowable Stresses for Steel and Concrete : : : : : : : : : : : : : : : : : : : : : : : : : : : 9{3 Selected values for Steel and Concrete Structures : : : : : : : : : : : : : : : : : : : : : : 9{5 Strength Reduction Factors, : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9{6 Approximate Structural SpanDepth Ratios for Horizontal Subsystems and Components (Lin and Stotesbury 1981) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : 9{8
14.1 Columns Combined Approximate Vertical and Horizontal Loads : : : : : : : : : : : : : : 14{26 14.2 Girders Combined Approximate Vertical and Horizontal Loads : : : : : : : : : : : : : : : 14{27
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INTRODUCTION 1.1 Science and Technology \There is a fundamental dierence between science and and technology. Engineering or technology is the making of things that did not previously exist, whereas science is the discovering of things that have long existed. Technological results are forms that exist only because people want to make them, whereas scienti c results are informations of what exists independently of human intentions. Technology deals with the arti cial, science with the natural." (Billington 1985)
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1.2 Structural Engineering Structural engineers are responsible for the detailed analysis and design of: Architectural structures: Buildings, houses, factories. They must work in close cooperation with an architect who will ultimately be responsible for the design. Civil Infrastructures: Bridges, dams, pipelines, oshore structures. They work with transportation, hydraulic, nuclear and other engineers. For those structures they play the leading role. Aerospace, Mechanical, Naval structures: aeroplanes, spacecrafts, cars, ships, submarines to ensure the structural safety of those important structures. 2
1.3 Structures and their Surroundings 3
Structural design is aected by various environmental constraints: 1. Major movements: For example, elevator shafts are usually shear walls good at resisting lateral load (wind, earthquake). 2. Sound and structure interact: A dome roof will concentrate the sound A dish roof will diuse the sound 3. Natural light: A at roof in a building may not provide adequate light.
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A Folded plate will provide adequate lighting (analysis more complex). A bearing and shear wall building may not have enough openings for daylight. A Frame design will allow more light in (analysis more complex). 4. Conduits for cables (electric, telephone, computer), HVAC ducts, may dictate type of oor system. 5. Net clearance between columns (unobstructed surface) will dictate type of framing.
1.4 Architecture & Engineering Architecture must be the product of a creative collaboration of architects and engineers. 5 Architect stress the overall, rather than elemental approach to design. In the design process, they conceptualize a spaceform scheme as a total system. They are generalists. 6 The engineer, partly due to his/her education think in reverse, starting with details and without sucient regards for the overall picture. (S)he is a pragmatist who \knows everything about nothing". 7 Thus there is a conceptual gap between architects and engineers at all levels of design. 8 Engineer's education is more specialized and in depth than the architect's. However, engineer must be kept aware of overall architectural objective. 9 In the last resort, it is the architect who is the leader of the construction team, and the engineers are his/her servant. 10 A possible compromise might be an Architectural Engineer.
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1.5 Architectural Design Process Architectural design is hierarchical: Schematic: conceptual overall spaceform feasibility of basic schematic options. Collaboration is mostly between the owner and the architect. Preliminary: Establish basic physical properties of major subsystems and key components to prove design feasibility. Some collaboration with engineers is necessary. Final design: nal indepth design re nements of all subsystems and components and preparation of working documents (\blueprints"). Engineers play a leading role.
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1.6 Architectural Design Architectural design must respect various constraints: Functionality: In uence of the adopted structure on the purposes for which the structure was erected. Aesthetics: The architect often imposes his aesthetic concerns on the engineer. This in turn can place severe limitations on the structural system. Economy: It should be kept in mind that the two largest components of a structure are labors and materials. Design cost is comparatively negligible. 12
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Buildings may have dierent functions: Residential: housing, which includes lowrise (up tp 23 oors), midrise (up to 68 oors) and high rise buildings. Commercial: Oces, retail stores, shopping centers, hotels, restaurants. Industrial: warehouses, manufacturing. Institutional: Schools, hospitals, prisons, chruch, government buildings. Special: Towers, stadium, parking, airport, etc.
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1.7 Structural Analysis Given an existing structure subjected to a certain load determine internal forces (axial, shear, exural, torsional; or stresses), de ections, and verify that no unstable failure can occur. 15 Thus the basic structural requirements are: Strength: stresses should not exceed critical values: < f Stiness: de ections should be controlled: < max Stability: buckling or cracking should also be prevented
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1.8 Structural Design Given a set of forces, dimension the structural element. Steel/wood Structures Select appropriate section. Reinforced Concrete: Determine dimensions of the element and internal reinforcement (number and sizes of reinforcing bars).
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17 For new structures, iterative process between analysis and design. A preliminary design is made using rules of thumbs (best known to Engineers with design experience) and analyzed. Following design, we check for Serviceability: de ections, crack widths under the applied load. Compare with acceptable values speci ed in the design code. Failure (limit state): and compare the failure load with the applied load times the appropriate factors of safety. If the design is found not to be acceptable, then it must be modi ed and reanalyzed. 18 For existing structures rehabilitation, or veri cation of an old infrastructure, analysis is the most important component. 19 In summary, analysis is always required.
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Figure 1.1: Types of Forces in Structural Elements (1D)
1.9 Load Transfer Mechanisms From Strength of Materials, loads can be transferred through various mechanisms, Fig. 1.1 Axial: cables, truss elements, arches, membrane, shells Flexural: Beams, frames, grids, plates Torsional: Grids, 3D frames Shear: Frames, grids, shear walls.
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1.10 Structure Types 21
Structures can be classi ed as follows:
Tension & Compression Structures: only, no shear, exure, or torsion. Those are the most ecient types of structures. Cable (tension only): The high strength of steel cables, combined with the eciency of simple
tension, makes cables ideal structural elements to span large distances such as bridges, and dish roofs, Fig. 1.2. A cable structure develops its load carrying capacity by adjusting its shape so as to provide maximum resistance (form follows function). Care should be exercised in minimizing large de ections and vibrations. Arches (mostly compression) is a \reversed cable structure". In an arch, exure/shear is minimized and most of the load is transfered through axial forces only. Arches are used for large span roofs and bridges, Fig. 1.3
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Figure 1.2: Basic Aspects of Cable Systems
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Figure 1.3: Basic Aspects of Arches
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Trusses have pin connected elements which can transmit axial forces only (tension and compression). Elements are connected by either slotted, screwed, or gusset plate connectors. However, due to construction details, there may be secondary stresses caused by relatively rigid connections. Trusses are used for joists, roofs, bridges, electric tower, Fig. 1.4
Figure 1.4: Types of Trusses
Post and Beams: Essentially a support column on which a \beam" rests, Fig. 1.5, and 1.6. Beams: Shear, exure and sometimes axial forces. Recall that = McI is applicable only for shallow beams, i.e. span/depth at least equal to ve.
Whereas r/c beams are mostly rectangular or T shaped, steel beams are usually I shaped (if the top anges are not properly stiened, they may buckle, thus we must have stieners). Frames: Load is coplanar with the structure. Axial, shear, exure (with respect to one axis in 2D structures and with respect to two axis in 3D structures), torsion (only in 3D). The frame is composed of at least one horizontal member (beam) rigidly connected to vertical ones1. The vertical The precursor of the frame structures were the Post and Lintel where the post is vertical member on which the lintel 1
is simply posed.
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Figure 1.5: Variations in Post and Beams Con gurations
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VIERENDEEL TRUSS
OVERLAPPING SINGLESTRUT CABLESUPPORTED BEAM
TREESUPPORTED TRUSS
BRACED BEAM
CABLESTAYED BEAM
SUSPENDED CABLE SUPPORTED BEAM
BOWSTRING TRUSS
CABLESUPPORTED STRUTED ARCH OR CABLE BEAM/TRUSS
CABLESUPPORTED MULTISTRUT BEAM OR TRUSS
GABLED TRUSS
CABLESUPPORTED ARCHED FRAME
CABLESUPPORTED PORTAL FRAME
Figure 1.6: Dierent Beam Types
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members can have dierent boundary conditions (which are usually governed by soil conditions). Frames are extensively used for houses and buildings, Fig. 1.7.
Figure 1.7: Basic Forms of Frames
Grids and Plates: Load is orthogonal to the plane of the structure. Flexure, shear, torsion.
In a grid, beams are at right angles resulting in a twoway dispersal of loads. Because of the rigid connections between the beams, additional stiness is introduced by the torsional resistance of members. Grids can also be skewed to achieve greater eciency if the aspect ratio is not close to one. Plates are at, rigid, two dimensional structures which transmit vertical load to their supports. Used mostly for oor slabs. Folded plates is a combination of transverse and longitudinal beam action. Used for long span roofs. Note that the plate may be folded circularly rather than longitudinally. Folded plates are used mostly as long span roofs. However, they can also be used as vertical walls to support both vertical and horizontal loads. Membranes: 3D structures composed of a exible 2D surface resisting tension only. They are usually cablesupported and are used for tents and long span roofs Fig. 1.8.
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Figure 1.8: Examples of Air Supported Structures
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Shells: 3D structures composed of a curved 2D surface, they are usually shaped to transmit compressive axial stresses only, Fig. 1.9.
Figure 1.9: Basic Forms of Shells Shells are classi ed in terms of their curvature.
1.11 Structural Engineering Courses Structural engineering education can be approached from either one of two points of views, depending on the audience, ??.
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Architects Global Structure Approximate, \rules of thumbs" preliminary Structures Most Design Approximate Approach Emphasis Analysis
Engineers Elemental Component Exact, detailled Final Trusses, Frames Per code
Table 1.1: Structural Engineering Coverage for Architects and Engineers Table 1.2: tab:secae
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Architects: Start from overall design, and move toward detailed analysis. Emphasis on good understanding of overall structural behavior. Develop a good understanding of load transfer mechanism for most types of structures, cables, arches, beams, frames, shells, plates. Approximate analysis for most of them. Engineers: Emphasis is on the individual structural elements and not always on the total system. Focus on beams, frames (mostly 2D) and trusses. Very seldom are arches covered. Plates and shells are not even mentioned.
1.12 References Following are some useful references for structural engineering, those marked by y were consulted, and \borrowed from" in preparing the Lecture Notes or are particularly recommended.
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Structures for Architect
1. Ambrose, J., Building Structures, second Ed. Wiley, 1993. 2. Billington, D.P. Rober Maillart's Bridges; The Art of Engineering, Princeton University Pres, 1979. 3. yBillington, D.P., The Tower and the Bridge; The new art of structural engineering, Princeton University Pres,, 1983. 4. yBillington, D.P., Structures and the Urban Environment, Lectures Notes CE 262, Department of Civil Engineering, Princeton University, 1978 5. French, S., Determinate Structures; Statics, Strength, Analysis, Design, Delmar, 1996. 6. Gordon, J.E., Structures, or Why Things Do'nt Fall Down, Da Capo paperback, New York, 1978. 7. Gordon, J.E., The Science of Structures and Materials, Scienti c American Library, 1988. 8. Hawkes, N., Structures, the way things are built, MacMillan, 1990. 9. Levy, M. and Salvadori, M., Why Buildings Fall Down, W.W.Norton, 1992. 10. yLin, T.Y. and Stotesbury, S.D., Structural Concepts and Systems for Architects and Engineers, John Wiley, 1981. 11. yMainstone, R., Developments in Structural Form, Allen Lane Publishers, 1975. 12. Petroski, H., To Enginer is Human, Vintage Books, 1992. 13. ySalvadori, M. and Heller, R., Structure in Architecture; The Building of Buildings, Prentice Hall, Third Edition, 1986. 14. Salvadori, M. and Levy, M., Structural Design in Architecture, Prentice hall, Second Edition, 1981. 15. Salvadori, M., Why Buildings Stand Up; The Strength of Architecture, Norton Paperack, 1990. 16. ySandaker, B.N. and Eggen, A.P., The Structural Basis of Architecture, Whitney Library of Design, 1992. 17. ySchueller, W., The design of Building Structures, Prentice Hall, 1996.
Structures for Engineers 1. y Arbadi, F. Structural Analysis and Behavior, McGrawHill, Inc., 1991.
2. Biggs, J.M., Introduction to Structural Engineering; Analysis and Design, Prentice Hall, 1986.
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3. Hsieh, Y.Y., Elementary Theory of Structures, Third Edition, Prentice Hall, 1988. 4. Ghali, A., and Neville, A.M., Structural Analysis, Third Edition, Chapman and Hall, 1989 5. White, R. Gergely, P. and Sexmith, R., Structural Engineering; Combined Edition, John Wiley, 1976. 6. y Nilson, A., and Winter, G. Design of Concrete Structures, Eleventh Edition, McGraw Hill, 1991. 7. Galambos, T., Lin, F.J., and Johnston, B.G., Basic Steel Design with LRFD, Prentice Hall, 1996. 8. y Salmon C. and Johnson, J. Steel Structures, Third Edition, Harper Collins Publisher, 1990. 9. y Gaylord, E.H., Gaylord, C.N. and Stallmeyer, J.E., Design of Steel Structures, Third Edition, McGraw Hill, 1992. 10. Vitruvius, The Ten Books on Architecture, Dover Publications, 1960. 11. Palladio, A., The Four Books of Architecture, Dover Publication.
Codes
1. ACI31889, Building Code Requirements for Reinforced Concrete, American Concrete Institute 2. Load & Resistance Factor Design, Manual of Steel Construction, American Institute of Steel Construction. 3. Uniform Building Code, International Conference of Building Ocials, 5360 South Workman Road; Whittier, CA 90601 4. Minimum Design Loads in Buildings and Other Structures, ANSI A58.1, American National Standards Institute, Inc., New York, 1972.
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LOADS 2.1 Introduction The main purpose of a structure is to transfer load from one point to another: bridge deck to pier; slab to beam; beam to girder; girder to column; column to foundation; foundation to soil. 2 There can also be secondary loads such as thermal (in restrained structures), dierential settlement of foundations, PDelta eects (additional moment caused by the product of the vertical force and the lateral displacement caused by lateral load in a high rise building). 3 Loads are generally subdivided into two categories Vertical Loads or gravity load 1. dead load (DL) 2. live load (LL) also included are snow loads. Lateral Loads which act horizontally on the structure 1. Wind load (WL) 2. Earthquake load (EL) this also includes hydrostatic and earth loads.
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4 This distinction is helpful not only to compute a structure's load, but also to assign dierent factor of safety to each one. 5 For a detailed coverage of loads, refer to the Universal Building Code (UBC), (UBC 1995).
2.2 Vertical Loads For closely spaced identical loads (such as joist loads), it is customary to treat them as a uniformly distributed load rather than as discrete loads, Fig. 2.1
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LOADS P1
P2
P3
P4
P5
P6
P7
REPETITIVE JOIST LOADS ACTUAL DISCRETE LOADS ON SUPPORT BEAM
w LB/FT = TOTAL LOAD / SPAN
SUPPORT BEAM SPAN
ASSUMED EQUIVALENT UNIFORM LOAD TYPICAL SYSTEM OF JOISTS
Figure 2.1: Approximation of a Series of Closely Spaced Loads
2.2.1 Dead Load Dead loads (DL) consist of the weight of the structure itself, and other permanent xtures (such as walls, slabs, machinery). 8 For analysis purposes, dead loads can easily be determined from the structure's dimensions and density, Table 2.1
7
Material
lb=ft3 Aluminum 173 Brick 120 Concrete 145 Steel 490 Wood (pine) 40
kN=m3 27.2 18.9 33.8 77.0 6.3
Table 2.1: Unit Weight of Materials For steel structures, the weight per unit length of rolled sections is given in the AISC Manual of Steel Construction. 10 For design purposes, dead loads must be estimated and veri ed at the end of the design cycle. This makes the design process iterative. 11 Weights for building materials is given in Table 2.2 12 For preliminary design purposes the average dead loads of Table 2.3 can be used:
9
2.2.2 Live Loads Contrarily to dead loads which are xed and vertical, live loads (LL) are movable or moving and may be horizontal. 14 Occupancy load may be due to people, furniture, equipment. The loads are essentially variable point loads which can be placed anywhere.
13
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Structural Concepts and Systems for Architects
Draft 2.2 Vertical Loads
2{3 Material
Ceilings Channel suspended system Acoustical ber tile Floors Steel deck Concreteplain 1 in. Linoleum 1/4 in. Hardwood Roofs Copper or tin 5 ply felt and gravel Shingles asphalt Clay tiles Sheathing wood Insulation 1 in. poured in place Partitions Clay tile 3 in. Clay tile 10 in. Gypsum Block 5 in. Wood studs 2x4 (1216 in. o.c.) Plaster 1 in. cement Plaster 1 in. gypsum Walls Bricks 4 in. Bricks 12 in. Hollow concrete block (heavy aggregate) 4 in. 8 in. 12 in. Hollow concrete block (light aggregate) 4 in. 8 in. 12 in.
lb=ft2 1 1 210 12 1 4 15 6 3 914 3 2 17 40 14 2 10 5 40 120 30 55 80 21 38 55
Table 2.2: Weights of Building Materials
Material
lb=ft2 Timber 4050 Steel 5080 Reinforced concrete 100150 Table 2.3: Average Gross Dead Load in Buildings
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Draft 2{4
LOADS
In analysis load placement should be such that their eect (shear/moment) are maximized. 16 A statistical approach is used to determine a uniformly distributed static load which is equivalent to the weight of the maximum concentration of occupants. These loads are de ned in codes such as the Uniform Building Code or the ANSI Code, Table 2.4. 15
Use or Occupancy
lb=ft2 Assembly areas 50 Cornices, marquees, residential balconies 60 Corridors, stairs 100 Garage 50 Oce buildings 50 Residential 40 Storage 125250 Table 2.4: Minimum Uniformly Distributed Live Loads, (UBC 1995) For small areas (30 to 50 sq ft) the eect of concentrated load should be considered separately. 18 Since there is a small probability that the whole oor in a building be fully loaded, the UBC code speci es that the occupancy load for members supporting an area A larger than 150 ft2 (i.e. a column with a total tributary area, including oors above it, larger than 150 ft2 ) may be reduced by R where
17
R = r(A ; 150) 23:1 1 + DL LL
(2.1)
where r = :08 for oors, A is the supported area ( 2 ) DL and LL are the dead and live loads per unit area supported by the member. R can not exceed 40% for horizontal members and 60% for vertical ones. ft
Example 21: Live Load Reduction In a 10 story oce building with a column spacing of 16 ft in both directions, the total dead load is 60 psf, snow load 20 psf and live load 80 psf. what is the total live load and total load for which a column must be designed on the ground oor
Solution: 1. 2. 3. 4.
p
The tributary area is 16 16 = 256ft2 > 150 The reduction R for the roof is is R = :08(16 16 ; 150) = 8:48% ; 60 = 40:4% which is less than 60% p Maximum allowable reduction Rmax = 23:1 1 + 80 The reduced cumulative load for the column of each oor is Floor A A ; 150 R0 R% LL (100 ; R) LL=100
Victor Saouma
Roof 10 256 512 106 362 8.48 28.96 8.48 28.96 20 80 18.3 56.83
9 768 618 49.44 40.4 80 47.68
8 7 1024 1280 874 1130 69.92 90.40 40.4 40.4 80 80 47.68 47.68
6 1536 1386 110.88 40.4 80 47.68
5 1792 1642 131.36 40.4 80 47.68
4 2048 1898 151.84 40.4 80 47.68
3 2304 2154 172.32 40.4 80 47.68
2 2560 2410 192.8 40.4 80 47.68
Structural Concepts and Systems for Architects
Draft 2.3 Lateral Loads
2{5
The resulting design live load for the bottom column has been reduced from LLBefore = (20) (256) }2 + (9)(80) {z(256) }2 = 189,440 (2.2)  {z Roof 9 oors to LLReduced = (18 :3) {z(256) }2 + (9)(47:68){z (256) }2 = 114,540 (2.3)  Roof 9 oors 5. The total dead load is DL = (10)(60) (256) 2 (1;000) = 153:6 Kips, thus the total reduction ;268 100= 22% . in load is from 153:6+189:4 = 343 to 153:6+114:5 = 268:1 a reduction of 343343 psf
ft
psf
ft
psf
psf
ft
psf
ft
lbs
ft
lbs
k
k
lbs
k
2.2.3 Snow 19 Roof snow load vary greatly depending on geographic location and elevation. They range from 20 to 45 psf, Fig. 2.2.
Figure 2.2: Snow Map of the United States, ubc Snow loads are always given on the projected length or area on a slope, Fig. 2.3. 21 The steeper the roof, the lower the snow retention. For snow loads greater than 20 psf and roof pitches more than 20 the snow load p may be reduced by
20
2.3 Lateral Loads
p R = ( ; 20) 40 ; 0:5
(psf)
(2.4)
2.3.1 Wind
Wind load depend on: velocity of the wind, shape of the building, height, geographical location, texture of the building surface and stiness of the structure.
22
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Structural Concepts and Systems for Architects
Draft 2{6
LOADS LIVE LOAD DEAD LOAD
LE N G TH
RISE
WIND LOAD
RUN
Figure 2.3: Loads on Projected Dimensions Wind loads are particularly signi cant on tall buildings1. 24 When a steady streamline air ow of velocity V is completely stopped by a rigid body, the stagnation pressure (or velocity pressure) qs was derived by Bernouilli (17001782) qs = 21 V 2 (2.5) where the air mass density is the air weight divided by the accleration of gravity g = 32:2 ft/sec2 . At sea level and a temperature of 15oC (59oF), the ai weighs 0.0765 lb/ft3 this would yield a pressure of 23
3 (5280)ft/mile 2
qs = 21 (0:0765)lb/ft2 (32:2)ft/sec
or
(3600)sec/hr V
qs = 0:00256V 2
(2.6) (2.7)
where V is the maximum wind velocity (in miles per hour) and qs is in psf. V can be obtained from wind maps (in the United States 70 V 110), Fig. 2.4. 25 During storms, wind velocities may reach values up to or greater than 150 miles per hour, which corresponds to a dynamic pressure qs of about 60 psf (as high as the average vertical occupancy load in buildings). The primary design consideration for very high rise buildings is the excessive drift caused by lateral load (wind and 1
possibly earthquakes).
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2{7
Figure 2.4: Wind Map of the United States, (UBC 1995) 26
Wind pressure increases with height, Table 2.5. Height Zone (in feet) <30 30 to 49 50 to 99 100 to 499 500 to 1199 >1,200
20 15 20 25 30 35 40
WindVelocity Map Area 25 30 35 40 45 50 20 25 25 30 35 40 25 30 35 40 45 50 30 40 45 50 55 60 40 45 55 60 70 75 45 55 60 70 80 90 50 60 70 80 90 100
Table 2.5: Wind Velocity Variation above Ground Wind load will cause suction on the leeward sides, Fig. 2.6 28 This magnitude must be modi ed to account for the shape and surroundings of the building. Thus, the design base pressure (at 33.3 ft from the ground) p (psf) is given by 27
p = Ce Cq Iqs
(2.8)
The pressure is assumed to be normal to all walls and roofs and Ce Velocity Pressure Coecient accounts for height, exposure and gust factor. It accounts for the fact that wind velocity increases with height and that dynamic character of the air ow (i.e the wind pressure is not steady), Table 2.6. l Cq Pressure Coecient is a shape factor which is given in Table 2.7 for gabled frames. I Importance Factor as given by Table 2.8. where
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Figure 2.5: Eect of Wind Load on Structures(Schueller 1996)
Ce
Exposure D Open, at terrain facing large bodies of water C Flat open terrain, extending onehalf mile or open from the site in any full quadrant 0.621.80 B Terrain with buildings, forest, or surface irregularities 20 ft or more in height 1.392.34 1.062.19
Table 2.6: Ce Coecients for Wind Load, (UBC 1995) Windward Side Leeward Side Gabled Frames (V:H) Roof Slope <9:12 ;0:7 ;0:7 9:12 to 12:12 0:4 ;0:7 >12:12 0:7 ;0:7 Walls 0:8 ;0:5 Buildings (height < 200 ft) Vertical Projections height < 40 ft 1:3 ;1:3 height > 40 ft 1:4 ;1:4 Horizontal Projections ;0:7 ;0:7 Table 2.7: Wind Pressure Coecients Cq , (UBC 1995)
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2{9 I II III IV
Occupancy Category
Essential facilities Hazardous facilities Special occupancy structures Standard occupancy structures
Imprtance Factor I Earthquake Wind 1.25 1.25 1.00 1.00
1.15 1.15 1.00 1.00
Table 2.8: Importance Factors for Wind and Earthquake Load, (UBC 1995)
I Essential Facilities: Hospitals; Fire and police stations; Tanks; Emergency vehicle shelters,
standby powergenerating equipment; Structures and equipment in government. communication centers. II Hazardous Facilities: Structures housing, supporting or containing sucient quantities of toxic or explosive substances to be dangerous to the safety of the general public if released. III Special occupancy structure: Covered structures whose primary occupancy is public assembly, capacity > 300 persons. Buildings for schools through secondary or daycare centers, capacity > 250 persons. Buildings for colleges or adult education schools, capacity > 500 persons. Medical facilities with 50 or more resident incapacitated patients, but not included above Jails and detention facilities All structures with occupancy >5,000 persons. Structures and equipment in power generating stations and other public utilitiy facilities not included above, and required for continued operation. IV Standard occupancy structure: All structures having occupancies or functions not listed above.
29
For the preliminary design of ordinary buildings Ce = 1:0 and Cq = 1:3 may be assumed, yielding
p = (1:3):020256V 2 = :00333V 2
(2.9)
which corresponds to a pressure of 21 psf for a wind speed of 80 mph, Fig. 2.6, Table 2.9.
Example 22: Wind Load Determine the wind forces on the building shown on below which is built in St Louis and is surrouded by trees.
Solution:
1. From Fig. 2.4 the maximum wind velocity is St. Louis is 70 mph, since the building is protected we can take Ce = 0:7, I = 1:. The base wind pressure is qs = 0:00256 (70)2 = 12:54 psf.
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LOADS
Height Above Grade (ft) 015 20 25 30 40 60 80 100 120 160 200 300 400
Exposure B C Basic Wind Speed (mph) 70 10 11 12 12 14 17 18 20 21 23 25 29 32
80 13 14 15 16 18 22 24 26 28 30 33 37 41
70 17 18 19 20 21 25 27 28 29 31 33 36 38
80 23 24 25 26 28 33 35 37 38 41 43 47 50
Table 2.9: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures
400 Exposure B, 70 mph Exposure B, 80 mph Exposure C, 70 mph Exposure C, 80 mph
350
Height Above Grade (ft)
300
250
200
150
100
50
0
0
5
10
15 20 25 30 35 40 Approximate Design Wind Pressure (psf)
45
50
Figure 2.6: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures
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Structural Concepts and Systems for Architects
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2{11
2. The slope of the roof is 8:15=6.4:12 which gives Cq = ;0:7 for both the windward and leeward sides. The vertical walls have Cq = 0:8 for the winward side and Cq = ;0:5 for the leeward one. 3. Thus the applied pressure on the roof is p = 0:7 (;0:7) 12:54 = 6.14 psf that is the roof is subjected to uplift. 4. The winward wall, the pressure is 0:7 0:8 12:54 = 7.02 psf , and for the leeward wall 0:7 (;0:5) 12:54 = 4.39 psf (suction) , 5. The direction of the wind can change and hence each structural component must be designed to resist all possible load combinations. 6. For large structures which may be subjected to large wind loads, testing in a wind tunnel of the structure itself and its surroundings is often accomplished.
2.3.2 Earthquakes Buildings should be able to resist Minor earthquakes without damage Moderate earthquakes without structural damage but possibly with some nonstructural damages Major earthquakes without collapse but possibly with some structural damage as well as nonstructural damage This is achieved through an appropriate dynamic analysis. 31 For preliminary designs or for small structures an equivalent horizontal static load can be determined. 32 Actual loads depend on the following 1. Intensity of the ground acceleration (including soil/rock properties). 2. Dynamic properties of the building, such as its mode shapes and periods of vibration and its damping characteristics. 3. Mass of the building. 30
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LOADS
33 A critical factor in the dynamic response of a structure is the fundamental period of the structure's vibration (or rst mode of vibration). This is the time required for one full cycle of motion, Fig. 2.7. If the earthquake excitation has a frequency close to the one of the building, then resonance may occur. This should be avoided.
Figure 2.7: Vibrations of a Building Earthquake load manifests itself as a horizontal force due to the (primarily) horizontal inertia force (F = ma). 35 The horizontal force at each level is calculated as a portion of the base shear force V 34
V = ZIC RW W
(2.10)
where:
Z : Zone Factor: to be determined from Fig. 2.8 and Table 2.10. Seismic Zone 0 1 2A 2B 3 4 Z 0 0.075 0.15 0.2 0.3 0.4 Table 2.10: Z Factors for Dierent Seismic Zones, ubc
I : Importance Factor: which was given by Table 2.8. C : Design Response Spectrum given by
Victor Saouma
S C = 1T:25 2=3 2:75
(2.11)
Structural Concepts and Systems for Architects
Draft 2.3 Lateral Loads
2{13
Figure 2.8: Seismic Zones of the United States, (UBC 1995)
T is the fundamental period of vibration of the building in seconds. This can be determined from
either the free vibration analysis of the building, or estimated from the following empirical formula
T = Ct (hn )3=4
(2.12)
where: hn is the building height above base in ft. and
Ct 0.035 steel moment resisting frames Ct 0.030 reinforced concrete moment resisting frames and eccentrically braced frames Ct 0.020 all other buildings S : Site Coecient given by Table 2.11 Note that most of the damages in the 1990? earthquake Type Description S Factor S1 A soil pro le with either rocklike material or sti/dense soil less 1.0 than 200 ft. S2 Dense or sti soil exceeding 200 ft 1.2 S3 70 ft or more soil containing more than 20 ft of soft to medium sti 1.5 clay but not more than 40 ft. of soft clay. S4 Soil containing more than 40 ft of soft clay 2.0 Table 2.11: S Site Coecients for Earthquake Loading, (UBC 1995) in San Francisco occurred in the marina where many houses were built on soft soil. and
C RW 0:075
Victor Saouma
(2.13)
Structural Concepts and Systems for Architects
Draft 2{14
LOADS
RW is given by Table 2.12. W Load total structure load. The horizontal force V is distributed over the height of the building in two parts. The rst (applied only if T 0:7 sec.) is a concentrated force F1 equal to
36
Ft = 0:07TV 0:25V
(2.14)
is applied at the top of the building due to whiplash. The balance of the force V ; Ft is distributed as a triangular load diminishing to zero at the base. 37 Assuming a oor weight constant for every oor level, then the force acting on each one is given by (2.15) Fx = h +(Vh ;+Ft)h +x h = (V;n Fht )hx 1 2 n i=1 i where hi and hx are the height in ft above the base to level i, or x respectively. Note that it is assumed that all oors have also same width.
Example 23: Earthquake Load on a Frame Determine the approximate earthquake forces for the ductile hospital frame structure shown below. The DL for each oor is 200 lb/ft and the LL is 400 lb/ft. The structure is built on soft soil. Use DL plus 50%LL as the weight of each oor. The building is in zone 3.
Solution: 1. The fundamental period of vibration is
T = Ct (hn )3=4 = (0:030)(24)3=4 = 0:32 2. The C coecient is
sec.
S (1:25)(2:0) C = 1T:25 2=3 = (0:32)2=3 = 5:344 > 2:75
(2.16) (2.17)
use C = 2:75. 3. The other coecients are: Z =0.3; I =1.25; RW =12
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Structural Concepts and Systems for Architects
Draft 2.3 Lateral Loads
2{15
Structural System
Bearing wall system
Lightframed walls with shear panels Plywood walls for structures three stories or less All other lightframed walls Shear walls Concrete Masonry
Building frame system using trussing or shear walls)
Steel eccentrically braced ductile frame Lightframed walls with shear panels Plywood walls for structures three stories or less All other lightframed walls Shear walls Concrete Masonry Concentrically braced frames Steel Concrete (only for zones I and 2) Heavy timber
Momentresisting frame system
RW H (ft) 8 6
65 65
8 8
240 160
10
240
9 7
65 65
8 8
240 160
8 8 8
160 65
Special momentresisting frames (SMRF) Steel 12 Concrete 12 Concrete intermediate momentresisting frames (IMRF)only for zones 1 and 2 8 Ordinary momentresisting frames (OMRF) Steel 6 Concrete (only for zone 1) 5 Dual systems (selected cases are for ductile rigid frames only) Shear walls Concrete with SMRF 12 Masonry with SMRF 8 Steel eccentrically braced ductile frame 612 Concentrically braced frame 12 Steel with steel SMRF 10 Steel with steel OMRF 6 Concrete with concrete SMRF (only for zones 1 and 2) 9
N.L. N.L. 160 N.L. 160 160N.L. N. L. N.L. 160 
Table 2.12: Partial List of RW for Various Structure Systems, (UBC 1995)
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LOADS
4. Check
C = 2:75 = 0:23 > 0:075p RW 12
(2.18)
5. The total vertical load is
W = 2 ((200 + 0:5(400)) (20) = 16000
lbs
(2.19)
6. The total seismic base shear is
(0:3)(1:25)(2:75) = 0:086W V = ZIC = RW 12 = (0:086)(16000) = 1375 lbs
(2.20a) (2.20b)
7. Since T < 0:7 sec. there is no whiplash. 8. The load on each oor is thus given by
F2 = (1375)(24) 12 + 24 = 916.7 lbs
F1 = (1375)(12) 12 + 24 = 458.3 lbs
(2.21a) (2.21b)
Example 24: Earthquake Load on a Tall Building, (Schueller 1996) Determine the approximate critical lateral loading for a 25 storey, ductile, rigid space frame concrete structure in the short direction. The rigid frames are spaced 25 ft apart in the cross section and 20 ft in the longitudinal direction. The plan dimension of the building is 175x100 ft, and the structure is 25(12)=300 ft high. This oce building is located in an urban environment with a wind velocity of 70 mph and in seismic zone 4. For this investigation, an average building total dead load of 192 psf is used. Soil conditions are unknown.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2.3 Lateral Loads
2{17 470 k
25(12)=300’
2638 k
2(300)/3=200’
300/2=150’
1523 k
7(25)=175’
84000 k
3108 k
5(20)=100’
Solution: 1. The total building weight is
W = (0:1926) (100 175) ksf
2 25 storeys = 84; 000 k
ft
2. the fundamental period of vibration for a rigid frame is
T = Ct (hn )3=4 = 0:030(300)3=4 = 2:16 3. The C coecient is
sec.
> 0:7
sec.
p
p S (1:25)(1:5) C = 1T:25 2=3 = (2:16)2=3 = 1:12 2:75
(2.25)
W
6. The total seismic base shear along the critical short direction is (0:4)(1)(1:12) W = 0:037W V = ZIC RW W = (12) = (0:037)(84000) = 3108 kip sec.
, the whiplash eect must be considered
Ft = 0:07TV = (0:07)(2:16)(3108) = 470 le 0:25V = (0:25)(3108) = 777 k
Victor Saouma
(2.23) (2.24)
4. The other coecients are Z =0.4; I =1, RW =12 5. We check p C 1:12 R = 12 = 0:093 0:075
7. Since T > 0:7
(2.22)
k
(2.26a) (2.26b)
(2.27a) (2.27b)
Structural Concepts and Systems for Architects
Draft 2{18
LOADS
Hence the total triangular load is
V ; Ft = 3108 ; 470 = 2638
(2.28)
k
8. let us check if wind load governs. From Table xx we conservatively assume a uniform wind pressure of 29 psf resulting in a total lateral force of PW = (0:029) (175 300) 2 = 1523 < 3108 (2.29) The magnitude of the total seismic load is clearly larger than the total wind force. psf
ft
k
k
2.4 Other Loads
2.4.1 Hydrostatic and Earth 38
Structures below ground must resist lateral earth pressure.
q = K h
(2.30)
;sin is the pressure coecient, h is the height. where is the soil density, K = 11+sin 39 40
For sand and gravel = 120 lb= ft3 , and 30. If the structure is partially submerged, it must also resist hydrostatic pressure of water, Fig. 2.9.
Figure 2.9: Earth and Hydrostatic Loads on Structures
q = W h where W = 62:4
lbs
(2.31)
= 3. ft
Example 25: Hydrostatic Load Victor Saouma
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Draft 2.4 Other Loads
2{19
The basement of a building is 12 ft below grade. Ground water is located 9 ft below grade, what thickness concrete slab is required to exactly balance the hydrostatic uplift?
Solution:
The hydrostatic pressure must be countered by the pressure caused by the weight of concrete. Since p =
h we equate the two pressures and solve for h the height of the concrete slab (62 :4) = 3{z (12 ; 9) } =  water 3 h ) h = (62:4) = 3 (3) (12) 15.0 inch = 14 : 976 ' (150) = (150) = 3  {z } concrete lbs ft
lbs ft
lbs ft
lbs ft
ft
in/ft
ft
in
2.4.2 Thermal If a member is uniformly heated (or cooled) without restraint, then it will expand (or contract). This expansion is given by
41
(2.32)
l = lT where is the coecient of thermal expansion, Table 2.13
(/F ) Steel 6:5 10;6 Concrete 5:5 10;6 Table 2.13: Coecients of Thermal Expansion 42 43
If the member is restrained against expansion, then a compressive stress = ET is developed. To avoid excessive stresses due to thermal loading expansion joints are used in bridges and buildings.
Example 26: Thermal Expansion/Stress (Schueller 1996) A lowrise building is enclosed along one side by a 100 ftlong clay masonary ( = 3:6 10;6 in./in./oF, E = 2; 400; 000 psi) bearing wall. The structure was built at a temperature of 60oF and is located in the northern part of the United States where the temperature range is between 20o and +120oF.
Solution:
1. Assuming that the wall can move freely with no restraint from crosswalls and foundation, the wall expansion and contraction (summer and winter) are given by LSummer = TL = (3:6 10;6) = =oF (120 ; 60)o F (100) (12) = 0.26 (2.33a) LWinter = TL = (3:6 10;6) = =oF (;20 ; 60)oF (100) (12) = 0.35(2.33b)
Victor Saouma
in
in
in
in
ft
ft
in/ft
in/ft
in
in
Structural Concepts and Systems for Architects
Draft 2{20
LOADS
2. We now assume (conservatively) that the free movement cannot occur (L = 0) hence the resulting stress would be equal to = E" = E LL = E LTL = ET
Summer = ET = (2; 400; 000) 2 (3:6 10;6 ) = =oF (120 ; 60)oF = 518 lbs
in
lbs
in
in
in
Winter = ET = (2; 400; 000) 2 (3:6 10;6 ) = =oF (;20 ; 60)o F = 691 lbs
in
in
2
Tension(2.34a)
lbs
in
in
2
Compression (2.34b) (2.34c)
Note that the tensile stresses being beyond the masonary capacity, cracking will occur.
2.5 Other Important Considerations 2.5.1 Load Combinations
Live loads speci ed by codes represent the maximum possible loads. 45 The likelihood of all these loads occurring simultaneously is remote. Hence, building codes allow certain reduction when certain loads are combined together. 46 Furthermore, structures should be designed to resist a combination of loads. 47 Denoting D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature; H= soil: 48 For the load and resistance factor design (LRFD) method of concrete structures, the American Concrete Institute (ACI) Building design code (318) (318 n.d.) requires that the following load combinations be considered: 1. 1.4D+1.7L 2. 0.75(1.4D+1.7L+1.7W) 3. 0.9D+1.3W 4. 1.4D +1.7L+1.7H 5. 0.75(1.4D+1.4T+1.7L) 6. 1.4(D+T) whereas for steel structures, the American Institute of Steel Construction (AISC) code, (of Steel COnstruction 1986) requires that the following combinations be veri ed 1. 1.4D 2. 1.2D+1.6L+0.5(Lr or S) 3. 1.2D+0.5L (or 0.8W)+1.6(Lr or S) 4. 1.2D+0.5L+0.5(Lr or S)+1.3W 5. 1.2D+0.5L(or 0.2 S)+1.5E 44
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2{21
6. 0.9D+1.3W(or 1.5 E) Analysis can be separately performed for each of the basic loads (L, D, W, etc) and then using the principle of superposition the loads can be linearly combined (unless the elastic limit has been reached). 50 Loads are often characterized as Usual, Unusual and Extreme. 49
2.5.2 Load Placement Only the dead load is static. The live load on the other hand may or may not be applied on a given component of a structure. Hence, the load placement arrangement resulting in the highest internal forces (moment +ve or ve, shear) at dierent locations must be considered, Fig. 2.10.
51
Figure 2.10: Load Placement to Maximize Moments
2.5.3 Load Transfer Whereas we will be focusing on the design of a reinforced concrete or steel section, we must keep in mind the following: 1. The section is part of a beam or girder. 2. The beam or girder is really part of a three dimensional structure in which load is transmitted from any point in the structure to the foundation through any one of various structural forms.
52
Load transfer in a structure is accomplished through a \hierarchy" of simple exural elements which are then connected to the columns, Fig. 2.11 or by two way slabs as illustrated in Fig. 2.12.
53
2.5.4 Structural Response Under the action of the various forces and loadings described above, the structure must be able to respond with proper behavior, Fig. 9.1.
54
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LOADS
Figure 2.11: Load Transfer in R/C Buildings
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Structural Concepts and Systems for Architects
Draft 2.5 Other Important Considerations
2{23
Figure 2.12: Two Way Actions
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2{24
LOADS
Figure 2.13: Load Life of a Structure, (Lin and Stotesbury 1981)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2.5 Other Important Considerations
2{25
2.5.5 Tributary Areas 55 For preliminary analyses, the tributary area of a structural component will determine the total applied load.
1111111111 0000000000 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 1111111111 0000000000 0000000000 1111111111 0000000000 1111111111
00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 11111 00000
1111111111 0000000000 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111
1111111111 00000 00000 0000011111 11111 00000 0000011111 11111 00000 00000 11111 00000 0000011111 11111 00000 11111 0000011111 11111 00000 0000011111 11111 00000 0000011111 11111 00000
0000000000 1111111111 1111111111 0000000000 0000000000 1111111111 0000000000 1111111111
0000000000 1111111111 1111111111 0000000000 0000000000 1111111111 0000000000 1111111111
001111 11 000011 1111 00 11 11 00 0000 00 00 11 0000 1111 00 11 00 11 0000 1111 00 11 00 11 0000 1111 00 11 00 11 0000 1111 00 11 00 11 0000 1111 00 11 00 11 0000 1111 00 001111 11 000011 1111 00 11 11 11 00 0000 00
Figure 2.14: Concept of Tributary Areas for Structual Member Loading
Victor Saouma
Structural Concepts and Systems for Architects
Draft 2{26
LOADS
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 3
STRUCTURAL MATERIALS Proper understanding of structural materials is essential to both structural analysis and to structural design. 2 Characteristics of the most commonly used structural materials will be highlighted. 1
3.1 Steel
3.1.1 Structural Steel Steel is an alloy of iron and carbon. Its properties can be greatly varied by altering the carbon content (always less than 0.5%) or by adding other elements such as silicon, nickle, manganese and copper. 4 Practically all grades of steel have a Young Modulus equal to 29,000 ksi, density of 490 lb/cu ft, and a coecient of thermal expansion equal to 0:65 10;5 /deg F. 5 The yield stress of steel can vary from 40 ksi to 250 ksi. Most commonly used structural steel are A36 (yld = 36 ksi) and A572 (yld = 50 ksi), Fig. 3.1 6 Structural steel can be rolled into a wide variety of shapes and sizes. Usually the most desirable members are those which have a large section moduli (S ) in proportion to their area (A), Fig. 3.2. 7 Steel can be bolted, riveted or welded. 8 Sections are designated by the shape of their cross section, their depth and their weight. For example W 27 114 is a W section, 27 in. deep weighing 114 lb/ft. 9 Common sections are: S sections were the rst ones rolled in America and have a slope on their inside ange surfaces of 1 to 6. W or wide ange sections have a much smaller inner slope which facilitates connections and rivetting. W sections constitute about 50% of the tonnage of rolled structural steel. C are channel sections MC Miscellaneous channel which can not be classi ed as a C shape by dimensions. HP is a bearing pile section.
3
Draft 3{2
STRUCTURAL MATERIALS
Figure 3.1: Stress Strain Curves of Concrete and Steel
Figure 3.2: Standard Rolled Sections
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.1 Steel
3{3
M is a miscellaneous section. L are angle sections which may have equal or unequal sides. WT is a T section cut from a W section in two. 10
The section modulus Sx of a W section can be roughly approximated by the following formula
Sx wd=10 or Ix Sx d2 wd2 =20
(3.1)
and the plastic modulus can be approximated by
Zx wd=9 11
(3.2)
Properties of structural steel are tabulated in Table 3.1. ASTM Desig. A36
A500 A501 A529 A606 A611 A 709
Shapes Available Shapes and bars
Use
Riveted, bolted, welded; Buildings and bridges Cold formed welded and General structural purseamless sections; pose Riveted, welded or bolted; Hot formed welded and seam Bolted and welded less sections; Plates and bars in and less Building frames and thick; trusses; Bolted and welded Hot and cold rolled sheets; Atmospheric corrosion resistant Cold rolled sheet in cut Cold formed sections lengths Structural shapes, plates and Bridges bars 1 2
y (kksi)
u (kksi)
36 up through 8 in. (32 above 8.) Grade A: 33; Grade B: 42; Grade C: 46 36 42 4550 Grade C 33; Grade D 40; Grade E 80 Grade 36: 36 (to 4 in.); Grade 50: 50; Grade 100: 100 (to 2.5in.) and 90 (over 2.5 to 4 in.)
Table 3.1: Properties of Major Structural Steels Rolled sections, Fig. 3.3 and welded ones, Fig3.4 have residual stresses. Those originate during the rolling or fabrication of a member. The member is hot just after rolling or welding, it cools unevenly because of varying exposure. The area that cool rst become stier, resist contraction, and develop compressive stresses. The remaining regions continue to cool and contract in the plastic condition and develop tensile stresses. 13 Due to those residual stresses, the stressstrain curve of a rolled section exhibits a nonlinear segment prior to the theoretical yielding, Fig. 3.5. This would have important implications on the exural and axial strength of beams and columns.
12
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{4
STRUCTURAL MATERIALS
Maximum compressive stress, say 12 ksi average
Compression () ()
Tension (+) (+)
Figure 3.3: Residual Stresses in Rolled Sections
say 20 ksi say 12 ksi
_
+
+ +
+
say 40 ksi
20 ksi

say 35 ksi tension
+

Welded H say 20 ksi compression
Welded box
Figure 3.4: Residual Stresses in Welded Sections
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.1 Steel
.3
Fy Average stress P/A
3{5
Ideal coupon containing no residual stress
.2 Fp
Maximum residual compressive stress
.1
1
Members with residual stress
2 Average copressive strain
3
Shaded portion indicates area which has achieved a stress Fy
Figure 3.5: In uence of Residual Stress on Average StressStrain Curve of a Rolled Section
3.1.2 Reinforcing Steel Steel is also used as reinforcing bars in concrete, Table 3.2. Those bars have a deformation on their surface to increase the bond with concrete, and usually have a yield stress of 60 ksi1 .
14
Bar Designation No. 2 No. 3 No. 4 No. 5 No. 6 No. 7 No. 8 No. 9 No. 10 No. 11 No. 14 No. 18
Diameter Area Perimeter Weight (in.) ( 2) in lb/ft 2/8=0.250 0.05 0.79 0.167 3/8=0.375 0.11 1.18 0.376 4/8=0.500 0.20 1.57 0.668 5/8=0.625 0.31 1.96 1.043 6/8=0.750 0.44 2.36 1.5202 7/8=0.875 0.60 2.75 2.044 8/8=1.000 0.79 3.14 2.670 9/8=1.128 1.00 3.54 3.400 10/8=1.270 1.27 3.99 4.303 11/8=1.410 1.56 4.43 5.313 14/8=1.693 2.25 5.32 7.650 18/8=2.257 4.00 7.09 13.60 in
Table 3.2: Properties of Reinforcing Bars Steel loses its strength rapidly above 700 deg. F (and thus must be properly protected from re), and becomes brittle at ;30 deg. F 16 Steel is also used as wire strands and ropes for suspended roofs, cablestayed bridges, fabric roofs and other structural applications. A strand is a helical arrangement of wires around a central wire. A rope consists of multiple strands helically wound around a central plastic core, and a modulus of elasticity of 20,000 ksi, and an ultimate strength of 220 ksi. 17 Prestressing Steel cables have an ultimate strength up to 270 ksi.
15
1
Stirrups which are used as vertical reinforcement to resist shear usually have a yield stress of only 40 ksi.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{6
STRUCTURAL MATERIALS
3.2 Aluminum Aluminum is used whenever light weight combined with strength is an important factor. Those properties, along with its resistance to corrosion have made it the material of choice for airplane structures, light roof framing. 19 Aluminum members can be connected by riveting, bolting and to a lesser extent by welding. 20 Aluminum has a modulus of elasticity equal to 10,000 ksi (about three times lower than steel), a coecient of thermal expansion of 2:4 10;5 and a density of 173 = 3 . 21 The ultimate strength of pure aluminum is low (13,000 psi) but with the addition of alloys it can go up. 22 When aluminum is in contact with other metals in the presence of an electrolyte, galvanic corrosion may cause damage. Thus, steel and aluminum in a structure must be carefully separated by means of painting or a nonconductive material.
18
lbs
ft
3.3 Concrete Concrete is a mixture of Portland cement2 , water, and aggregates (usually sand and crushed stone). An ideal mixture is one in which: 1. A minimum amount of cementwater paste is used to ll the interstices between the particles of aggregates. 2. A minimum amount of water is provided to complete the chemical reaction with cement. In such a mixture, about 3/4 of the volume is constituted by the aggregates, and the remaining 1/4 being the cement paste. 24 Smaller particles up to 1/4 in. in size are called ne aggregates, and the larger ones being coarse aggregates. 25 Contrarily to steel to modulus of elasticity of concrete depends on the strength and is given by 23
p
(3.3)
p
(3.4)
E = 57; 000 fc0 or
E = 33 1:5 fc0 where both fc0 and E are in psi and is in = 3 . lbs
26
ft
Typical concrete (compressive) strengths range from 3,000 to 6,000 psi; However high strength
concrete can go up to 14,000 psi. 27 28 29 30
All concrete fail at an ultimate strain of 0.003, Fig. 3.1. Prepeak nonlinearity is caused by microcracking Fig. 3.6. The tensile strength of concrete ft0 is about 10% of the compressive strength. Density of normal weight concrete is 145 = 3 and 100 = 3 for lightweight concrete. lbs
ft
lbs
ft
Portland cement is a mixture of calcareous and argillaceous materials which are calcined in a kiln and then pulverized. When mixed with water, cement hardens through a process called hydration. 2
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.4 Masonry
3{7
εu linear .5f’c nonlinear f’c
Figure 3.6: Concrete microcracking Coecient of thermal expansion is 0:65 10;5 /deg F for normal weight concrete. 32 When concrete is poured (or rather placed), the free water not needed for the hydration process evaporates over a period of time and the concrete will shrink. This shrinkage is about 0.05% after one year (strain). Thus if the concrete is restrained, then cracking will occur3. 33 Concrete will also deform with time due to the applied load, this is called creep. This should be taken into consideration when computing the de ections (which can be up to three times the instantaneous elastic de ection). 31
3.4 Masonry Masonry consists of either natural materials, such as stones, or of manufactured products such as bricks and concrete blocks4, stacked and bonded together with mortar. 35 As for concrete, all modern structural masonry blocks are essentially compression members with low tensile resistance. 36 The mortar used is a mixture of sand, masonry cement, and either Portland cement or hydrated lime.
34
3.5 Timber Timber is one of the earliest construction materials, and one of the few natural materials with good tensile properties. 38 The properties of timber vary greatly, and the strength is time dependent. 39 Timber is a good shock absorber (many wood structures in Japan have resisted repeated earthquakes). 40 The most commonly used species of timber in construction are Douglas r, southern pine, hemlock and larch. 41 Members can be laminated together under good quality control, and exural strengths as high as 2,500 psi can be achieved. 37
For this reason a minimum amount of reinforcement is always necessary in concrete, and a 2% reinforcement, can reduce the shrinkage by 75%. Mud bricks were used by the Babylonians, stones by the Egyptians, and ice blocks by the Eskimos... 3
4
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{8
STRUCTURAL MATERIALS
3.6 Steel Section Properties 42
Dimensions and properties of rolled sections are tabulated in the following pages, Fig. 3.7.
Figure 3.7: W and C sections ==============
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.6 Steel Section Properties Designation W 36x848 W 36x798 W 36x720 W 36x650 W 36x588 W 36x527 W 36x485 W 36x439 W 36x393 W 36x359 W 36x328 W 36x300 W 36x280 W 36x260 W 36x245 W 36x230 W 36x256 W 36x232 W 36x210 W 36x194 W 36x182 W 36x170 W 36x160 W 36x150 W 36x135 W 33x619 W 33x567 W 33x515 W 33x468 W 33x424 W 33x387 W 33x354 W 33x318 W 33x291 W 33x263 W 33x241 W 33x221 W 33x201 W 33x169 W 33x152 W 33x141 W 33x130 W 33x118 W 30x581 W 30x526 W 30x477 W 30x433 W 30x391 W 30x357 W 30x326 W 30x292 W 30x261 W 30x235 W 30x211 W 30x191 W 30x173 W 30x148 W 30x132 W 30x124 W 30x116 W 30x108 W 30x 99 W 30x 90
Victor Saouma
A in 249.0 234.0 211.0 190.0 172.0 154.0 142.0 128.0 115.0 105.0 96.4 88.3 82.4 76.5 72.1 67.6 75.4 68.1 61.8 57.0 53.6 50.0 47.0 44.2 39.7 181.0 166.0 151.0 137.0 124.0 113.0 104.0 93.5 85.6 77.4 70.9 65.0 59.1 49.5 44.7 41.6 38.3 34.7 170.0 154.0 140.0 127.0 114.0 104.0 95.7 85.7 76.7 69.0 62.0 56.1 50.8 43.5 38.9 36.5 34.2 31.7 29.1 26.4 2
d in 42.45 41.97 41.19 40.47 39.84 39.21 38.74 38.26 37.80 37.40 37.09 36.74 36.52 36.26 36.08 35.90 37.43 37.12 36.69 36.49 36.33 36.17 36.01 35.85 35.55 38.47 37.91 37.36 36.81 36.34 35.95 35.55 35.16 34.84 34.53 34.18 33.93 33.68 33.82 33.49 33.30 33.09 32.86 35.39 34.76 34.21 33.66 33.19 32.80 32.40 32.01 31.61 31.30 30.94 30.68 30.44 30.67 30.31 30.17 30.01 29.83 29.65 29.53
3{9 bf
f
hc tw
2.0 2.1 2.3 2.5 2.7 3.0 3.2 3.5 3.8 4.2 4.5 5.0 5.3 5.7 6.1 6.5 3.5 3.9 4.5 4.8 5.1 5.5 5.9 6.4 7.6 2.4 2.6 2.8 3.0 3.3 3.6 3.8 4.2 4.6 5.0 5.7 6.2 6.8 4.7 5.5 6.0 6.7 7.8 2.3 2.5 2.7 2.9 3.2 3.5 3.7 4.1 4.6 5.0 5.7 6.3 7.0 4.4 5.3 5.7 6.2 6.9 7.8 8.5
12.5 13.2 14.5 16.0 17.6 19.6 21.0 23.1 25.8 28.1 30.9 33.3 35.6 37.5 39.4 41.4 33.8 37.3 39.1 42.4 44.8 47.8 50.0 52.0 54.1 15.2 16.6 18.2 19.7 21.7 23.8 25.8 28.8 31.2 34.5 36.1 38.7 41.9 44.7 47.2 49.6 51.7 54.5 13.7 15.1 16.6 18.0 19.9 21.8 23.7 26.5 29.0 32.5 34.9 38.0 41.2 41.5 43.9 46.2 47.8 49.6 51.9 57.5
2t
Ix in 67400 62600 55300 48900 43500 38300 34700 31000 27500 24800 22500 20300 18900 17300 16100 15000 16800 15000 13200 12100 11300 10500 9750 9040 7800 41800 37700 33700 30100 26900 24300 21900 19500 17700 15800 14200 12800 11500 9290 8160 7450 6710 5900 33000 29300 26100 23200 20700 18600 16800 14900 13100 11700 10300 9170 8200 6680 5770 5360 4930 4470 3990 3620 4
Sx in 3170 2980 2690 2420 2180 1950 1790 1620 1450 1320 1210 1110 1030 953 895 837 895 809 719 664 623 580 542 504 439 2170 1990 1810 1630 1480 1350 1230 1110 1010 917 829 757 684 549 487 448 406 359 1870 1680 1530 1380 1250 1140 1030 928 827 746 663 598 539 436 380 355 329 299 269 245 3
Iy in 4550 4200 3680 3230 2850 2490 2250 1990 1750 1570 1420 1300 1200 1090 1010 940 528 468 411 375 347 320 295 270 225 2870 2580 2290 2030 1800 1620 1460 1290 1160 1030 932 840 749 310 273 246 218 187 2530 2230 1970 1750 1550 1390 1240 1100 959 855 757 673 598 227 196 181 164 146 128 115 4
Sy in 501 467 414 367 328 289 263 235 208 188 171 156 144 132 123 114 86 77 68 62 58 53 49 45 38 340 308 276 247 221 200 181 161 146 131 118 106 95 54 47 43 38 33 312 278 249 222 198 179 162 144 127 114 100 90 80 43 37 34 31 28 24 22 3
Zx in 3830.0 3570.0 3190.0 2840.0 2550.0 2270.0 2070.0 1860.0 1660.0 1510.0 1380.0 1260.0 1170.0 1080.0 1010.0 943.0 1040.0 936.0 833.0 767.0 718.0 668.0 624.0 581.0 509.0 2560.0 2330.0 2110.0 1890.0 1700.0 1550.0 1420.0 1270.0 1150.0 1040.0 939.0 855.0 772.0 629.0 559.0 514.0 467.0 415.0 2210.0 1990.0 1790.0 1610.0 1430.0 1300.0 1190.0 1060.0 941.0 845.0 749.0 673.0 605.0 500.0 437.0 408.0 378.0 346.0 312.0 283.0 3
Zy in 799.0 743.0 656.0 580.0 517.0 454.0 412.0 367.0 325.0 292.0 265.0 241.0 223.0 204.0 190.0 176.0 137.0 122.0 107.0 97.7 90.7 83.8 77.3 70.9 59.7 537.0 485.0 433.0 387.0 345.0 312.0 282.0 250.0 226.0 202.0 182.0 164.0 147.0 84.4 73.9 66.9 59.5 51.3 492.0 438.0 390.0 348.0 310.0 279.0 252.0 223.0 196.0 175.0 154.0 138.0 123.0 68.0 58.4 54.0 49.2 43.9 38.6 34.7 3
Structural Concepts and Systems for Architects
Draft 3{10 Designation W 27x539 W 27x494 W 27x448 W 27x407 W 27x368 W 27x336 W 27x307 W 27x281 W 27x258 W 27x235 W 27x217 W 27x194 W 27x178 W 27x161 W 27x146 W 27x129 W 27x114 W 27x102 W 27x 94 W 27x 84 W 24x492 W 24x450 W 24x408 W 24x370 W 24x335 W 24x306 W 24x279 W 24x250 W 24x229 W 24x207 W 24x192 W 24x176 W 24x162 W 24x146 W 24x131 W 24x117 W 24x104 W 24x103 W 24x 94 W 24x 84 W 24x 76 W 24x 68 W 24x 62 W 24x 55 W 21x402 W 21x364 W 21x333 W 21x300 W 21x275 W 21x248 W 21x223 W 21x201 W 21x182 W 21x166 W 21x147 W 21x132 W 21x122 W 21x111 W 21x101 W 21x 93 W 21x 83 W 21x 73
Victor Saouma
STRUCTURAL MATERIALS A in 158.0 145.0 131.0 119.0 108.0 98.7 90.2 82.6 75.7 69.1 63.8 57.0 52.3 47.4 42.9 37.8 33.5 30.0 27.7 24.8 144.0 132.0 119.0 108.0 98.4 89.8 82.0 73.5 67.2 60.7 56.3 51.7 47.7 43.0 38.5 34.4 30.6 30.3 27.7 24.7 22.4 20.1 18.2 16.2 118.0 107.0 97.9 88.2 80.8 72.8 65.4 59.2 53.6 48.8 43.2 38.8 35.9 32.7 29.8 27.3 24.3 21.5 2
d in 32.52 31.97 31.42 30.87 30.39 30.00 29.61 29.29 28.98 28.66 28.43 28.11 27.81 27.59 27.38 27.63 27.29 27.09 26.92 26.71 29.65 29.09 28.54 27.99 27.52 27.13 26.73 26.34 26.02 25.71 25.47 25.24 25.00 24.74 24.48 24.26 24.06 24.53 24.31 24.10 23.92 23.73 23.74 23.57 26.02 25.47 25.00 24.53 24.13 23.74 23.35 23.03 22.72 22.48 22.06 21.83 21.68 21.51 21.36 21.62 21.43 21.24
bf
f
hc tw
2.2 2.3 2.5 2.7 3.0 3.2 3.5 3.7 4.0 4.4 4.7 5.2 5.9 6.5 7.2 4.5 5.4 6.0 6.7 7.8 2.0 2.1 2.3 2.5 2.7 2.9 3.2 3.5 3.8 4.1 4.4 4.8 5.3 5.9 6.7 7.5 8.5 4.6 5.2 5.9 6.6 7.7 6.0 6.9 2.1 2.3 2.5 2.7 2.9 3.2 3.5 3.9 4.2 4.6 5.4 6.0 6.5 7.1 7.7 4.5 5.0 5.6
12.3 13.4 14.7 15.9 17.6 19.2 20.9 22.9 24.7 26.6 29.2 32.3 33.4 36.7 40.0 39.7 42.5 47.0 49.4 52.7 10.9 11.9 13.1 14.2 15.6 17.1 18.6 20.7 22.5 24.8 26.6 28.7 30.6 33.2 35.6 39.2 43.1 39.2 41.9 45.9 49.0 52.0 50.1 54.6 10.8 11.8 12.8 14.2 15.4 17.1 18.8 20.6 22.6 24.9 26.1 28.9 31.3 34.1 37.5 32.3 36.4 41.2
2t
Ix in 25500 22900 20400 18100 16100 14500 13100 11900 10800 9660 8870 7820 6990 6280 5630 4760 4090 3620 3270 2850 19100 17100 15100 13400 11900 10700 9600 8490 7650 6820 6260 5680 5170 4580 4020 3540 3100 3000 2700 2370 2100 1830 1550 1350 12200 10800 9610 8480 7620 6760 5950 5310 4730 4280 3630 3220 2960 2670 2420 2070 1830 1600 4
Sx in 1570 1440 1300 1170 1060 970 884 811 742 674 624 556 502 455 411 345 299 267 243 213 1290 1170 1060 957 864 789 718 644 588 531 491 450 414 371 329 291 258 245 222 196 176 154 131 114 937 846 769 692 632 569 510 461 417 380 329 295 273 249 227 192 171 151 3
Iy in 2110 1890 1670 1480 1310 1170 1050 953 859 768 704 618 555 497 443 184 159 139 124 106 1670 1490 1320 1160 1030 919 823 724 651 578 530 479 443 391 340 297 259 119 109 94 82 70 34 29 1270 1120 994 873 785 694 609 542 483 435 376 333 305 274 248 93 81 71 4
Sy in 277 250 224 200 179 161 146 133 120 108 100 88 79 71 64 37 32 28 25 21 237 214 191 170 152 137 124 110 99 89 82 74 68 60 53 46 41 26 24 21 18 16 10 8 189 168 151 134 122 109 96 86 77 70 60 54 49 44 40 22 20 17 3
Zx in 1880.0 1710.0 1530.0 1380.0 1240.0 1130.0 1020.0 933.0 850.0 769.0 708.0 628.0 567.0 512.0 461.0 395.0 343.0 305.0 278.0 244.0 1550.0 1410.0 1250.0 1120.0 1020.0 922.0 835.0 744.0 676.0 606.0 559.0 511.0 468.0 418.0 370.0 327.0 289.0 280.0 254.0 224.0 200.0 177.0 153.0 134.0 1130.0 1010.0 915.0 816.0 741.0 663.0 589.0 530.0 476.0 432.0 373.0 333.0 307.0 279.0 253.0 221.0 196.0 172.0 3
Zy in 437.0 394.0 351.0 313.0 279.0 252.0 227.0 206.0 187.0 168.0 154.0 136.0 122.0 109.0 97.5 57.6 49.3 43.4 38.8 33.2 375.0 337.0 300.0 267.0 238.0 214.0 193.0 171.0 154.0 137.0 126.0 115.0 105.0 93.2 81.5 71.4 62.4 41.5 37.5 32.6 28.6 24.5 15.7 13.3 296.0 263.0 237.0 210.0 189.0 169.0 149.0 133.0 119.0 108.0 92.6 82.3 75.6 68.2 61.7 34.7 30.5 26.6 3
Structural Concepts and Systems for Architects
Draft 3.6 Steel Section Properties Designation W 21x 68 W 21x 62 W 21x 57 W 21x 50 W 21x 44 W 18x311 W 18x283 W 18x258 W 18x234 W 18x211 W 18x192 W 18x175 W 18x158 W 18x143 W 18x130 W 18x119 W 18x106 W 18x 97 W 18x 86 W 18x 76 W 18x 71 W 18x 65 W 18x 60 W 18x 55 W 18x 50 W 18x 46 W 18x 40 W 18x 35 W 16x100 W 16x 89 W 16x 77 W 16x 67 W 16x 57 W 16x 50 W 16x 45 W 16x 40 W 16x 36 W 16x 31 W 16x 26 W 14x730 W 14x665 W 14x605 W 14x550 W 14x500 W 14x455 W 14x426 W 14x398 W 14x370 W 14x342 W 14x311 W 14x283 W 14x257 W 14x233 W 14x211 W 14x193 W 14x176 W 14x159 W 14x145 W 14x132 W 14x120 W 14x109 W 14x 99 W 14x 90 W 14x 82
Victor Saouma
A in 20.0 18.3 16.7 14.7 13.0 91.5 83.2 75.9 68.8 62.1 56.4 51.3 46.3 42.1 38.2 35.1 31.1 28.5 25.3 22.3 20.8 19.1 17.6 16.2 14.7 13.5 11.8 10.3 29.4 26.2 22.6 19.7 16.8 14.7 13.3 11.8 10.6 9.1 7.7 215.0 196.0 178.0 162.0 147.0 134.0 125.0 117.0 109.0 101.0 91.4 83.3 75.6 68.5 62.0 56.8 51.8 46.7 42.7 38.8 35.3 32.0 29.1 26.5 24.1 2
3{11
hc d btff Ix Sx Iy Sy Zx Zy tw in in in in in in in 21.13 6.0 43.6 1480 140 65 16 160.0 24.4 20.99 6.7 46.9 1330 127 58 14 144.0 21.7 21.06 5.0 46.3 1170 111 31 9 129.0 14.8 20.83 6.1 49.4 984 94 25 8 110.0 12.2 20.66 7.2 53.6 843 82 21 6 95.4 10.2 22.32 2.2 10.6 6960 624 795 132 753.0 207.0 21.85 2.4 11.5 6160 564 704 118 676.0 185.0 21.46 2.6 12.5 5510 514 628 107 611.0 166.0 21.06 2.8 13.8 4900 466 558 96 549.0 149.0 20.67 3.0 15.1 4330 419 493 85 490.0 132.0 20.35 3.3 16.7 3870 380 440 77 442.0 119.0 20.04 3.6 18.0 3450 344 391 69 398.0 106.0 19.72 3.9 19.8 3060 310 347 61 356.0 94.8 19.49 4.2 21.9 2750 282 311 56 322.0 85.4 19.25 4.6 23.9 2460 256 278 50 291.0 76.7 18.97 5.3 24.5 2190 231 253 45 261.0 69.1 18.73 6.0 27.2 1910 204 220 39 230.0 60.5 18.59 6.4 30.0 1750 188 201 36 211.0 55.3 18.39 7.2 33.4 1530 166 175 32 186.0 48.4 18.21 8.1 37.8 1330 146 152 28 163.0 42.2 18.47 4.7 32.4 1170 127 60 16 145.0 24.7 18.35 5.1 35.7 1070 117 55 14 133.0 22.5 18.24 5.4 38.7 984 108 50 13 123.0 20.6 18.11 6.0 41.2 890 98 45 12 112.0 18.5 17.99 6.6 45.2 800 89 40 11 101.0 16.6 18.06 5.0 44.6 712 79 22 7 90.7 11.7 17.90 5.7 51.0 612 68 19 6 78.4 9.9 17.70 7.1 53.5 510 58 15 5 66.5 8.1 16.97 5.3 24.3 1490 175 186 36 198.0 54.9 16.75 5.9 27.0 1300 155 163 31 175.0 48.1 16.52 6.8 31.2 1110 134 138 27 150.0 41.1 16.33 7.7 35.9 954 117 119 23 130.0 35.5 16.43 5.0 33.0 758 92 43 12 105.0 18.9 16.26 5.6 37.4 659 81 37 10 92.0 16.3 16.13 6.2 41.2 586 73 33 9 82.3 14.5 16.01 6.9 46.6 518 65 29 8 72.9 12.7 15.86 8.1 48.1 448 56 24 7 64.0 10.8 15.88 6.3 51.6 375 47 12 4 54.0 7.0 15.69 8.0 56.8 301 38 10 3 44.2 5.5 22.42 1.8 3.7 14300 1280 4720 527 1660.0 816.0 21.64 2.0 4.0 12400 1150 4170 472 1480.0 730.0 20.92 2.1 4.4 10800 1040 3680 423 1320.0 652.0 20.24 2.3 4.8 9430 931 3250 378 1180.0 583.0 19.60 2.4 5.2 8210 838 2880 339 1050.0 522.0 19.02 2.6 5.7 7190 756 2560 304 936.0 468.0 18.67 2.8 6.1 6600 707 2360 283 869.0 434.0 18.29 2.9 6.4 6000 656 2170 262 801.0 402.0 17.92 3.1 6.9 5440 607 1990 241 736.0 370.0 17.54 3.3 7.4 4900 559 1810 221 672.0 338.0 17.12 3.6 8.1 4330 506 1610 199 603.0 304.0 16.74 3.9 8.8 3840 459 1440 179 542.0 274.0 16.38 4.2 9.7 3400 415 1290 161 487.0 246.0 16.04 4.6 10.7 3010 375 1150 145 436.0 221.0 15.72 5.1 11.6 2660 338 1030 130 390.0 198.0 15.48 5.5 12.8 2400 310 931 119 355.0 180.0 15.22 6.0 13.7 2140 281 838 107 320.0 163.0 14.98 6.5 15.3 1900 254 748 96 287.0 146.0 14.78 7.1 16.8 1710 232 677 87 260.0 133.0 14.66 7.1 17.7 1530 209 548 74 234.0 113.0 14.48 7.8 19.3 1380 190 495 68 212.0 102.0 14.32 8.5 21.7 1240 173 447 61 192.0 92.7 14.16 9.3 23.5 1110 157 402 55 173.0 83.6 14.02 10.2 25.9 999 143 362 50 157.0 75.6 14.31 5.9 22.4 882 123 148 29 139.0 44.8 2
4
3
4
3
3
3
Structural Concepts and Systems for Architects
Draft 3{12 Designation W 14x 74 W 14x 68 W 14x 61 W 14x 53 W 14x 48 W 14x 43 W 14x 38 W 14x 34 W 14x 30 W 14x 26 W 14x 22 W 12x336 W 12x305 W 12x279 W 12x252 W 12x230 W 12x210 W 12x190 W 12x170 W 12x152 W 12x136 W 12x120 W 12x106 W 12x 96 W 12x 87 W 12x 79 W 12x 72 W 12x 65 W 12x 58 W 12x 53 W 12x 50 W 12x 45 W 12x 40 W 12x 35 W 12x 30 W 12x 26 W 12x 22 W 12x 19 W 12x 16 W 12x 14 W 10x112 W 10x100 W 10x 88 W 10x 77 W 10x 68 W 10x 60 W 10x 54 W 10x 49 W 10x 45 W 10x 39 W 10x 33 W 10x 30 W 10x 26 W 10x 22 W 10x 19 W 10x 17 W 10x 15 W 10x 12
Victor Saouma
STRUCTURAL MATERIALS A in 21.8 20.0 17.9 15.6 14.1 12.6 11.2 10.0 8.9 7.7 6.5 98.8 89.6 81.9 74.1 67.7 61.8 55.8 50.0 44.7 39.9 35.3 31.2 28.2 25.6 23.2 21.1 19.1 17.0 15.6 14.7 13.2 11.8 10.3 8.8 7.7 6.5 5.6 4.7 4.2 32.9 29.4 25.9 22.6 20.0 17.6 15.8 14.4 13.3 11.5 9.7 8.8 7.6 6.5 5.6 5.0 4.4 3.5 2
d in 14.17 14.04 13.89 13.92 13.79 13.66 14.10 13.98 13.84 13.91 13.74 16.82 16.32 15.85 15.41 15.05 14.71 14.38 14.03 13.71 13.41 13.12 12.89 12.71 12.53 12.38 12.25 12.12 12.19 12.06 12.19 12.06 11.94 12.50 12.34 12.22 12.31 12.16 11.99 11.91 11.36 11.10 10.84 10.60 10.40 10.22 10.09 9.98 10.10 9.92 9.73 10.47 10.33 10.17 10.24 10.11 9.99 9.87
bf
f
hc tw
6.4 7.0 7.7 6.1 6.7 7.5 6.6 7.4 8.7 6.0 7.5 2.3 2.4 2.7 2.9 3.1 3.4 3.7 4.0 4.5 5.0 5.6 6.2 6.8 7.5 8.2 9.0 9.9 7.8 8.7 6.3 7.0 7.8 6.3 7.4 8.5 4.7 5.7 7.5 8.8 4.2 4.6 5.2 5.9 6.6 7.4 8.2 8.9 6.5 7.5 9.1 5.7 6.6 8.0 5.1 6.1 7.4 9.4
25.3 27.5 30.4 30.8 33.5 37.4 39.6 43.1 45.4 48.1 53.3 5.5 6.0 6.3 7.0 7.6 8.2 9.2 10.1 11.2 12.3 13.7 15.9 17.7 18.9 20.7 22.6 24.9 27.0 28.1 26.2 29.0 32.9 36.2 41.8 47.2 41.8 46.2 49.4 54.3 10.4 11.6 13.0 14.8 16.7 18.7 21.2 23.1 22.5 25.0 27.1 29.5 34.0 36.9 35.4 36.9 38.5 46.6
2t
Ix in 796 723 640 541 485 428 385 340 291 245 199 4060 3550 3110 2720 2420 2140 1890 1650 1430 1240 1070 933 833 740 662 597 533 475 425 394 350 310 285 238 204 156 130 103 89 716 623 534 455 394 341 303 272 248 209 170 170 144 118 96 82 69 54 4
Sx Iy Sy Zx Zy in in in in in 112 134 27 126.0 40.6 103 121 24 115.0 36.9 92 107 22 102.0 32.8 78 58 14 87.1 22.0 70 51 13 78.4 19.6 63 45 11 69.6 17.3 55 27 8 61.5 12.1 49 23 7 54.6 10.6 42 20 6 47.3 9.0 35 9 4 40.2 5.5 29 7 3 33.2 4.4 483 1190 177 603.0 274.0 435 1050 159 537.0 244.0 393 937 143 481.0 220.0 353 828 127 428.0 196.0 321 742 115 386.0 177.0 292 664 104 348.0 159.0 263 589 93 311.0 143.0 235 517 82 275.0 126.0 209 454 73 243.0 111.0 186 398 64 214.0 98.0 163 345 56 186.0 85.4 145 301 49 164.0 75.1 131 270 44 147.0 67.5 118 241 40 132.0 60.4 107 216 36 119.0 54.3 97 195 32 108.0 49.2 88 174 29 96.8 44.1 78 107 21 86.4 32.5 71 96 19 77.9 29.1 65 56 14 72.4 21.4 58 50 12 64.7 19.0 52 44 11 57.5 16.8 46 24 7 51.2 11.5 39 20 6 43.1 9.6 33 17 5 37.2 8.2 25 5 2 29.3 3.7 21 4 2 24.7 3.0 17 3 1 20.1 2.3 15 2 1 17.4 1.9 126 236 45 147.0 69.2 112 207 40 130.0 61.0 98 179 35 113.0 53.1 86 154 30 97.6 45.9 76 134 26 85.3 40.1 67 116 23 74.6 35.0 60 103 21 66.6 31.3 55 93 19 60.4 28.3 49 53 13 54.9 20.3 42 45 11 46.8 17.2 35 37 9 38.8 14.0 32 17 6 36.6 8.8 28 14 5 31.3 7.5 23 11 4 26.0 6.1 19 4 2 21.6 3.3 16 4 2 18.7 2.8 14 3 1 16.0 2.3 11 2 1 12.6 1.7 3
4
3
3
3
Structural Concepts and Systems for Architects
Draft 3.6 Steel Section Properties Designation W 8x 67 W 8x 58 W 8x 48 W 8x 40 W 8x 35 W 8x 31 W 8x 28 W 8x 24 W 8x 21 W 8x 18 W 8x 15 W 8x 13 W 8x 10 W 6x 25 W 6x 20 W 6x 15 W 6x 16 W 6x 12 W 6x 9 W 5x 19 W 5x 16 W 4x 13 M 14x 18 M 12x 12 M 12x 11 M 12x 10 M 10x 9 M 10x 8 M 10x 8 M 8x 6 M 6x 4 M 5x 19 S 24x121 S 24x106 S 24x100 S 24x 90 S 24x 80 S 20x 96 S 20x 86 S 20x 75 S 20x 66 S 18x 70 S 18x 55 S 15x 50 S 15x 43 S 12x 50 S 12x 41 S 12x 35 S 12x 32 S 10x 35 S 10x 25 S 8x 23 S 8x 18 S 7x 20 S 7x 15 S 6x 17 S 6x 12 S 5x 15 S 5x 10 S 4x 10 S 4x 8 S 3x 8 S 3x 6
Victor Saouma
A in 19.7 17.1 14.1 11.7 10.3 9.1 8.2 7.1 6.2 5.3 4.4 3.8 3.0 7.3 5.9 4.4 4.7 3.5 2.7 5.5 4.7 3.8 5.1 3.5 3.2 2.9 2.7 2.3 2.2 1.9 1.3 5.6 35.6 31.2 29.3 26.5 23.5 28.2 25.3 22.0 19.4 20.6 16.1 14.7 12.6 14.7 12.0 10.3 9.4 10.3 7.5 6.8 5.4 5.9 4.5 5.1 3.7 4.3 2.9 2.8 2.3 2.2 1.7 2
3{13
hc d btff Ix Sx Iy Sy tw in in in in in 9.00 4.4 11.1 272 60 89 21 8.75 5.1 12.4 228 52 75 18 8.50 5.9 15.8 184 43 61 15 8.25 7.2 17.6 146 36 49 12 8.12 8.1 20.4 127 31 43 11 8.00 9.2 22.2 110 28 37 9 8.06 7.0 22.2 98 24 22 7 7.93 8.1 25.8 83 21 18 6 8.28 6.6 27.5 75 18 10 4 8.14 8.0 29.9 62 15 8 3 8.11 6.4 28.1 48 12 3 2 7.99 7.8 29.9 40 10 3 1 7.89 9.6 40.5 31 8 2 1 6.38 6.7 15.5 53 17 17 6 6.20 8.2 19.1 41 13 13 4 5.99 11.5 21.6 29 10 9 3 6.28 5.0 19.1 32 10 4 2 6.03 7.1 21.6 22 7 3 2 5.90 9.2 29.2 16 6 2 1 5.15 5.8 14.0 26 10 9 4 5.01 6.9 15.8 21 9 8 3 4.16 5.9 10.6 11 5 4 2 14.00 7.4 60.3 148 21 3 1 12.00 6.8 62.5 72 12 1 1 11.97 7.3 63.6 65 11 1 1 11.97 9.1 68.0 62 10 1 1 10.00 6.5 58.4 39 8 1 0 9.95 7.4 59.3 34 7 1 0 9.99 7.8 65.0 33 7 0 0 8.00 6.0 53.8 18 5 0 0 6.00 5.4 47.0 7 2 0 0 5.00 6.0 11.2 24 10 8 3 24.50 3.7 26.4 3160 258 83 21 24.50 3.6 34.1 2940 240 77 20 24.00 4.2 28.3 2390 199 48 13 24.00 4.1 33.7 2250 187 45 13 24.00 4.0 42.1 2100 175 42 12 20.30 3.9 21.6 1670 165 50 14 20.30 3.8 26.2 1580 155 47 13 20.00 4.0 27.1 1280 128 30 9 20.00 3.9 34.1 1190 119 28 9 18.00 4.5 21.8 926 103 24 8 18.00 4.3 33.6 804 89 21 7 15.00 4.5 23.2 486 65 16 6 15.00 4.4 31.0 447 60 14 5 12.00 4.2 13.9 305 51 16 6 12.00 4.0 20.7 272 45 14 5 12.00 4.7 23.4 229 38 10 4 12.00 4.6 28.6 218 36 9 4 10.00 5.0 13.8 147 29 8 3 10.00 4.7 26.4 124 25 7 3 8.00 4.9 14.5 65 16 4 2 8.00 4.7 23.7 58 14 4 2 7.00 4.9 12.3 42 12 3 2 7.00 4.7 21.9 37 10 3 1 6.00 5.0 9.9 26 9 2 1 6.00 4.6 19.9 22 7 2 1 5.00 5.0 7.5 15 6 2 1 5.00 4.6 17.4 12 5 1 1 4.00 4.8 8.7 7 3 1 1 4.00 4.5 14.7 6 3 1 1 3.00 4.8 5.6 3 2 1 0 3.00 4.5 11.4 3 2 0 0 2
4
3
4
3
Zx in 70.2 59.8 49.0 39.8 34.7 30.4 27.2 23.2 20.4 17.0 13.6 11.4 8.9 18.9 14.9 10.8 11.7 8.3 6.2 11.6 9.6 6.3 24.9 14.3 13.2 12.2 9.2 8.2 7.7 5.4 2.8 11.0 306.0 279.0 240.0 222.0 204.0 198.0 183.0 153.0 140.0 125.0 105.0 77.1 69.3 61.2 53.1 44.8 42.0 35.4 28.4 19.3 16.5 14.5 12.1 10.6 8.5 7.4 5.7 4.0 3.5 2.4 2.0 3
Zy in 32.7 27.9 22.9 18.5 16.1 14.1 10.1 8.6 5.7 4.7 2.7 2.2 1.7 8.6 6.7 4.8 3.4 2.3 1.7 5.5 4.6 2.9 2.2 1.1 1.0 1.0 0.8 0.7 0.6 0.5 0.3 5.0 36.2 33.2 23.9 22.3 20.7 24.9 23.0 16.7 15.3 14.4 12.1 10.0 9.0 10.3 8.9 6.8 6.4 6.2 5.0 3.7 3.2 3.0 2.4 2.4 1.9 1.9 1.4 1.1 1.0 0.8 0.7 3
Structural Concepts and Systems for Architects
Draft 3{14
STRUCTURAL MATERIALS
A in C 15.x 50 14.7 C 15.x 40 11.8 C 15.x 34 10.0 C 12.x 30 8.8 C 12.x 25 7.3 C 12.x 21 6.1 C 10.x 30 8.8 C 10.x 25 7.3 C 10.x 20 5.9 C 10.x 15 4.5 C 9.x 20 5.9 C 9.x 15 4.4 C 9.x 13 3.9 C 8.x 19 5.5 C 8.x 14 4.0 C 8.x 12 3.4 C 7.x 15 4.3 C 7.x 12 3.6 C 7.x 10 2.9 C 6.x 13 3.8 C 6.x 11 3.1 C 6.x 8 2.4 C 5.x 9 2.6 C 5.x 7 2.0 C 4.x 7 2.1 C 4.x 5 1.6 C 3.x 6 1.8 C 3.x 5 1.5 C 3.x 4 1.2 Designation
Designation
d in 15. 15. 15. 12. 12. 12. 10. 10. 10. 10. 9. 9. 9. 8. 8. 8. 7. 7. 7. 6. 6. 6. 5. 5. 4. 4. 3. 3. 3. A in 11.00 8.44 6.43 5.75 7.69 6.48 5.25 3.98 11.00 9.73 8.44 7.11 6.43 5.75 5.06 4.36 3.65 7.98 6.94 5.86 5.31 4.75 4.18 3.61 3.03 4.50 3.42 2.87
2
L 8.0x4.0x1.000 L 8.0x4.0x0.750 L 8.0x4.0x0.563 L 8.0x4.0x0.500 L 7.0x4.0x0.750 L 7.0x4.0x0.625 L 7.0x4.0x0.500 L 7.0x4.0x0.375 L 6.0x6.0x1.000 L 6.0x6.0x0.875 L 6.0x6.0x0.750 L 6.0x6.0x0.625 L 6.0x6.0x0.563 L 6.0x6.0x0.500 L 6.0x6.0x0.438 L 6.0x6.0x0.375 L 6.0x6.0x0.313 L 6.0x4.0x0.875 L 6.0x4.0x0.750 L 6.0x4.0x0.625 L 6.0x4.0x0.563 L 6.0x4.0x0.500 L 6.0x4.0x0.438 L 6.0x4.0x0.375 L 6.0x4.0x0.313 L 6.0x3.5x0.500 L 6.0x3.5x0.375 L 6.0x3.5x0.313
Victor Saouma
2
bf
f
2t
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 wgt k=ft 37.40 28.70 21.90 19.60 26.20 22.10 17.90 13.60 37.40 33.10 28.70 24.20 21.90 19.60 17.20 14.90 12.40 27.20 23.60 20.00 18.10 16.20 14.30 12.30 10.30 15.30 11.70 9.80
hc tw
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Ix in 69.6 54.9 42.8 38.5 37.8 32.4 26.7 20.6 35.5 31.9 28.2 24.2 22.1 19.9 17.7 15.4 13.0 27.7 24.5 21.1 19.3 17.4 15.5 13.5 11.4 16.6 12.9 10.9 4
Ix in 404.0 349.0 315.0 162.0 144.0 129.0 103.0 91.2 78.9 67.4 60.9 51.0 47.9 44.0 36.1 32.6 27.2 24.2 21.3 17.4 15.2 13.1 8.9 7.5 4.6 3.8 2.1 1.9 1.7 Sx in 14.1 10.9 8.4 7.5 8.4 7.1 5.8 4.4 8.6 7.6 6.7 5.7 5.1 4.6 4.1 3.5 3.0 7.2 6.3 5.3 4.8 4.3 3.8 3.3 2.8 4.2 3.2 2.7 4
3
Sx in 53.8 46.5 42.0 27.0 24.1 21.5 20.7 18.2 15.8 13.5 13.5 11.3 10.6 11.0 9.0 8.1 7.8 6.9 6.1 5.8 5.1 4.4 3.6 3.0 2.3 1.9 1.4 1.2 1.1 Iy in 11.60 9.36 7.43 6.74 9.05 7.84 6.53 5.10 35.50 31.90 28.20 24.20 22.10 19.90 17.70 15.40 13.00 9.75 8.68 7.52 6.91 6.27 5.60 4.90 4.18 4.25 3.34 2.85 3
4
Iy in 11. 9.23 8.13 5.14 4.47 3.88 3.94 3.36 2.81 2.28 2.42 1.93 1.76 1.98 1.53 1.32 1.38 1.17 0.97 1.05 0.87 0.69 0.63 0.48 0.43 0.32 0.31 0.25 0.20 Sy in 3.94 3.07 2.38 2.15 3.03 2.58 2.12 1.63 8.57 7.63 6.66 5.66 5.14 4.61 4.08 3.53 2.97 3.39 2.97 2.54 2.31 2.08 1.85 1.60 1.35 1.59 1.23 1.04 4
3
Sy Zx Zy in in in 3.78 8.20 8.17 3.37 57.20 6.87 3.11 50.40 6.23 2.06 33.60 4.33 1.88 29.20 3.84 1.73 25.40 3.49 1.65 26.60 3.78 1.48 23. 3.19 1.32 19.30 2.71 1.16 15.80 2.35 1.17 16.80 2.47 1.01 13.50 2.05 0.96 12.50 1.95 1.01 13.80 2.17 0.85 10.90 1.73 0.78 9.55 1.58 0.78 9.68 1.64 0.70 8.40 1.43 0.63 7.12 1.26 0.64 7.26 1.36 0.56 6.15 1.15 0.49 5.13 0.99 0.45 4.36 0.92 0.38 3.51 0.76 0.34 2.81 0.70 0.28 2.26 0.57 0.27 1.72 0.54 0.23 1.50 0.47 0.20 1.30 0.40 Zx Zy in in 24.30 7.72 18.90 5.81 14.50 4.38 13.00 3.90 14.80 5.65 12.60 4.74 10.30 3.83 7.87 2.90 15.50 15.50 13.80 13.80 12.00 12.00 10.20 10.20 9.26 9.26 8.31 8.31 7.34 7.34 6.35 6.35 5.35 5.35 12.70 6.31 11.20 5.47 9.51 4.62 8.66 4.19 7.78 3.75 6.88 3.30 5.97 2.85 5.03 2.40 7.50 2.91 5.76 2.20 4.85 1.85 3
3
3
3
3
Structural Concepts and Systems for Architects
Draft 3.6 Steel Section Properties Designation L 5.0x5.0x0.875 L 5.0x5.0x0.750 L 5.0x5.0x0.625 L 5.0x5.0x0.500 L 5.0x5.0x0.438 L 5.0x5.0x0.375 L 5.0x5.0x0.313 L 5.0x3.5x0.750 L 5.0x3.5x0.625 L 5.0x3.5x0.500 L 5.0x3.5x0.438 L 5.0x3.5x0.375 L 5.0x3.5x0.313 L 5.0x3.5x0.250 L 5.0x3.0x0.625 L 5.0x3.0x0.500 L 5.0x3.0x0.438 L 5.0x3.0x0.375 L 5.0x3.0x0.313 L 5.0x3.0x0.250 L 4.0x4.0x0.750 L 4.0x4.0x0.625 L 4.0x4.0x0.500 L 4.0x4.0x0.438 L 4.0x4.0x0.375 L 4.0x4.0x0.313 L 4.0x4.0x0.250 L 4.0x3.5x0.500 L 4.0x3.5x0.438 L 4.0x3.5x0.375 L 4.0x3.5x0.313 L 4.0x3.5x0.250 L 4.0x3.0x0.500 L 4.0x3.0x0.438
Victor Saouma
A in 7.98 6.94 5.86 4.75 4.18 3.61 3.03 5.81 4.92 4.00 3.53 3.05 2.56 2.06 4.61 3.75 3.31 2.86 2.40 1.94 5.44 4.61 3.75 3.31 2.86 2.40 1.94 3.50 3.09 2.67 2.25 1.81 3.25 2.87 2
wgt k=ft 27.20 23.60 20.00 16.20 14.30 12.30 10.30 19.80 16.80 13.60 12.00 10.40 8.70 7.00 15.70 12.80 11.30 9.80 8.20 6.60 18.50 15.70 12.80 11.30 9.80 8.20 6.60 11.90 10.60 9.10 7.70 6.20 11.10 9.80
3{15 Ix Sx in in 17.8 5.2 15.7 4.5 13.6 3.9 11.3 3.2 10.0 2.8 8.7 2.4 7.4 2.0 13.9 4.3 12.0 3.7 10.0 3.0 8.9 2.6 7.8 2.3 6.6 1.9 5.4 1.6 11.4 3.5 9.4 2.9 8.4 2.6 7.4 2.2 6.3 1.9 5.1 1.5 7.7 2.8 6.7 2.4 5.6 2.0 5.0 1.8 4.4 1.5 3.7 1.3 3.0 1.0 5.3 1.9 4.8 1.7 4.2 1.5 3.6 1.3 2.9 1.0 5.1 1.9 4.5 1.7 4
3
Iy in 17.80 15.70 13.60 11.30 10.00 8.74 7.42 5.55 4.83 4.05 3.63 3.18 2.72 2.23 3.06 2.58 2.32 2.04 1.75 1.44 7.67 6.66 5.56 4.97 4.36 3.71 3.04 3.79 3.40 2.95 2.55 2.09 2.42 2.18 4
Sy in 5.17 4.53 3.86 3.16 2.79 2.42 2.04 2.22 1.90 1.56 1.39 1.21 1.02 0.83 1.39 1.15 1.02 0.89 0.75 0.61 2.81 2.40 1.97 1.75 1.52 1.29 1.05 1.52 1.35 1.16 0.99 0.81 1.12 0.99 3
Zx in 9.33 8.16 6.95 5.68 5.03 4.36 3.68 7.65 6.55 5.38 4.77 4.14 3.49 2.83 6.27 5.16 4.57 3.97 3.36 2.72 5.07 4.33 3.56 3.16 2.74 2.32 1.88 3.50 3.11 2.71 2.29 1.86 3.41 3.03 3
Zy in 9.33 8.16 6.95 5.68 5.03 4.36 3.68 4.10 3.47 2.83 2.49 2.16 1.82 1.47 2.61 2.11 1.86 1.60 1.35 1.09 5.07 4.33 3.56 3.16 2.74 2.32 1.88 2.73 2.42 2.11 1.78 1.44 2.03 1.79 3
Structural Concepts and Systems for Architects
Draft 3{16 Designation L 4.0x3.0x0.375 L 4.0x3.0x0.313 L 4.0x3.0x0.250 L 3.5x3.5x0.500 L 3.5x3.5x0.438 L 3.5x3.5x0.375 L 3.5x3.5x0.313 L 3.5x3.5x0.250 L 3.5x3.0x0.500 L 3.5x3.0x0.438 L 3.5x3.0x0.375 L 3.5x3.0x0.313 L 3.5x3.0x0.250 L 3.5x2.5x0.500 L 3.5x2.5x0.438 L 3.5x2.5x0.375 L 3.5x2.5x0.313 L 3.5x2.5x0.250 L 3.0x3.0x0.500 L 3.0x3.0x0.438 L 3.0x3.0x0.375 L 3.0x3.0x0.313 L 3.0x3.0x0.250 L 3.0x3.0x0.188 L 3.0x2.5x0.500 L 3.0x2.5x0.438 L 3.0x2.5x0.375 L 3.0x2.5x0.313 L 3.0x2.5x0.250 L 3.0x2.5x0.188 L 3.0x2.0x0.500 L 3.0x2.0x0.438 L 3.0x2.0x0.375 L 3.0x2.0x0.313 L 3.0x2.0x0.250 L 3.0x2.0x0.188 L 2.5x2.5x0.500 L 2.5x2.5x0.375 L 2.5x2.5x0.313 L 2.5x2.5x0.250 L 2.5x2.5x0.188 L 2.5x2.0x0.375 L 2.5x2.0x0.313 L 2.5x2.0x0.250 L 2.5x2.0x0.188 L 2.0x2.0x0.375 L 2.0x2.0x0.313 L 2.0x2.0x0.250 L 2.0x2.0x0.188 L 2.0x2.0x0.125 L 1.8x1.8x0.250 L 1.8x1.8x0.188 L 1.5x1.5x0.250 L 1.5x1.5x0.188 L 1.3x1.3x0.250 L 1.3x1.3x0.188 L 1.1x1.1x0.125 L 1.0x1.0x0.125
Victor Saouma
STRUCTURAL MATERIALS A wgt Ix Sx in k=ft in in 2.48 8.50 4.0 1.5 2.09 7.20 3.4 1.2 1.69 5.80 2.8 1.0 3.25 11.10 3.6 1.5 2.87 9.80 3.3 1.3 2.48 8.50 2.9 1.1 2.09 7.20 2.5 1.0 1.69 5.80 2.0 0.8 3.00 10.20 3.5 1.5 2.65 9.10 3.1 1.3 2.30 7.90 2.7 1.1 1.93 6.60 2.3 1.0 1.56 5.40 1.9 0.8 2.75 9.40 3.2 1.4 2.43 8.30 2.9 1.3 2.11 7.20 2.6 1.1 1.78 6.10 2.2 0.9 1.44 4.90 1.8 0.8 2.75 9.40 2.2 1.1 2.43 8.30 2.0 1.0 2.11 7.20 1.8 0.8 1.78 6.10 1.5 0.7 1.44 4.90 1.2 0.6 1.09 3.71 1.0 0.4 2.50 8.50 2.1 1.0 2.21 7.60 1.9 0.9 1.92 6.60 1.7 0.8 1.62 5.60 1.4 0.7 1.31 4.50 1.2 0.6 1.00 3.39 0.9 0.4 2.25 7.70 1.9 1.0 2.00 6.80 1.7 0.9 1.73 5.90 1.5 0.8 1.46 5.00 1.3 0.7 1.19 4.10 1.1 0.5 0.90 3.07 0.8 0.4 2.25 7.70 1.2 0.7 1.73 5.90 1.0 0.6 1.46 5.00 0.8 0.5 1.19 4.10 0.7 0.4 0.90 3.07 0.5 0.3 1.55 5.30 0.9 0.5 1.31 4.50 0.8 0.5 1.06 3.62 0.7 0.4 0.81 2.75 0.5 0.3 1.36 4.70 0.5 0.4 1.15 3.92 0.4 0.3 0.94 3.19 0.3 0.2 0.71 2.44 0.3 0.2 0.48 1.65 0.2 0.1 0.81 2.77 0.2 0.2 0.62 2.12 0.2 0.1 0.69 2.34 0.1 0.1 0.53 1.80 0.1 0.1 0.56 1.92 0.1 0.1 0.43 1.48 0.1 0.1 0.27 0.90 0.0 0.0 0.23 0.80 0.0 0.0 2
4
3
Iy in 1.92 1.65 1.36 3.64 3.26 2.87 2.45 2.01 2.33 2.09 1.85 1.58 1.30 1.36 1.23 1.09 0.94 0.78 2.22 1.99 1.76 1.51 1.24 0.96 1.30 1.18 1.04 0.90 0.74 0.58 0.67 0.61 0.54 0.47 0.39 0.31 1.23 0.98 0.85 0.70 0.55 0.51 0.45 0.37 0.29 0.48 0.42 0.35 0.27 0.19 0.23 0.18 0.14 0.11 0.08 0.06 0.03 0.02 4
Sy in 0.87 0.73 0.60 1.49 1.32 1.15 0.98 0.79 1.10 0.98 0.85 0.72 0.59 0.76 0.68 0.59 0.50 0.41 1.07 0.95 0.83 0.71 0.58 0.44 0.74 0.66 0.58 0.49 0.40 0.31 0.47 0.42 0.37 0.32 0.26 0.20 0.72 0.57 0.48 0.39 0.30 0.36 0.31 0.25 0.20 0.35 0.30 0.25 0.19 0.13 0.23 0.14 0.13 0.10 0.09 0.07 0.04 0.03 3
Zx in 2.64 2.23 1.82 2.68 2.38 2.08 1.76 1.43 2.63 2.34 2.04 1.73 1.41 2.53 2.26 1.97 1.67 1.36 1.93 1.72 1.50 1.27 1.04 0.79 1.88 1.68 1.47 1.25 1.02 0.78 1.78 1.59 1.40 1.19 0.97 0.75 1.31 1.02 0.87 0.71 0.55 0.99 0.84 0.69 0.53 0.63 0.54 0.44 0.34 0.23 0.34 0.26 0.24 0.19 0.16 0.13 0.07 0.06 3
Zy in 1.56 1.31 1.06 2.68 2.38 2.08 1.76 1.43 1.98 1.76 1.53 1.30 1.05 1.40 1.24 1.07 0.91 0.74 1.93 1.72 1.50 1.27 1.04 0.79 1.35 1.20 1.05 0.89 0.72 0.55 0.89 0.79 0.68 0.58 0.47 0.36 1.31 1.02 0.87 0.71 0.55 0.66 0.56 0.46 0.35 0.63 0.54 0.44 0.34 0.23 0.34 0.26 0.24 0.19 0.16 0.13 0.07 0.06 3
Structural Concepts and Systems for Architects
Draft 3.7 Joists
3{17
3.7 Joists 43 Steel joists, Fig. 3.8 look like shallow trusses (warren type) and are designed as simply supported uniformly loaded beams assuming that they are laterally supported on the top (to prevent lateral torsional buckling). The lateral support is often pro ded by the concrete slab it suppors. 44 The standard openweb joist designation consists of the depth, the series designation and the chord type. Three series are available for oor/roof construction, Table 3.3
Series Depth (in) K 830 LH 1848 DLH 5272
Span (ft) 860 2596 89120
Table 3.3: Joist Series Characteristics [Design Length = Span – 0.33 FT.] 4"
4"
4"
Span Figure 3.8: prefabricated Steel Joists Typical joist spacing ranges from 2 to 4 ft, and provides an ecient use of the corrugated steel deck which itself supports the concrete slab.
45
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{18 46
STRUCTURAL MATERIALS
For preliminary estimates of the joist depth, a depth to span ratio of 24 can be assumed, therefore
d L=2
(3.5)
where d is in inches, and L in ft. 47 Table 3.4 list the load carrying capacity of open web, Kseries steel joists based on a amximum allowable stress of 30 ksi. For each span, the rst line indicates the total safe uniformly distributed loadcarrying capacity in pounds per linear foot. Note that the dead load (including the one of the joist) must be substracted in order to determine the safe live load. The second line indicates the live load (pounds/linear foot) which will produce an approximate delection of L=360.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3.7 Joists Joint 8K1 10K1 12K1 Desig. Depth 8 10 12 (in.) W 5.1 5 5 (lbs/ft) Span (ft.) 8 550 550 9 550 550 10 550 550 480 550 11 532 550 377 542 12 444 550 550 288 455 550 13 377 479 550 225 363 510 14 324 412 500 179 289 425 15 281 358 434 145 234 344 16 246 313 380 119 192 282 17 277 336 159 234 18 246 299 134 197 19 221 268 113 167 20 199 241 97 142 21 218 123 22 199 106 23 181 93 24 166 81 25 26 27 28 29 30 31 32
3{19 12K3 12K5 14K1 14K3 14K4 14K6 16K2 16K3 16K4 16K5 16K6 16K7 16K9 12
12
14
14
14
14
16
16
16
16
16
5.7 7.1
5.2
6
6.7 7.7
5.5
6.3
7
7.5
8.1 8.6 10.0
550 550 511 475 448 390 395 324 352 272 315 230 284 197 257 170 234 147 214 128 196 113 180 100 166 88 154 79 143 70
550 550 550 507 550 467 495 404 441 339 395 287 356 246 322 212 293 184 268 160 245 141 226 124 209 110 193 98 180 88
550 550 550 507 550 467 550 443 530 397 475 336 428 287 388 248 353 215 322 188 295 165 272 145 251 129 233 115 216 103
550 550 512 488 456 409 408 347 368 297 333 255 303 222 277 194 254 170 234 150 216 133 200 119 186 106 173 95 161 86 151 78 142 71
550 550 550 526 508 456 455 386 410 330 371 285 337 247 308 216 283 189 260 167 240 148 223 132 207 118 193 106 180 96 168 87 158 79
550 550 550 526 550 490 547 452 493 386 447 333 406 289 371 252 340 221 313 195 289 173 268 155 249 138 232 124 216 112 203 101 190 92
550 550 550 526 550 490 550 455 550 426 503 373 458 323 418 282 384 248 353 219 326 194 302 173 281 155 261 139 244 126 228 114 214 103
550 550 550 526 550 490 550 455 550 426 548 405 498 351 455 307 418 269 384 238 355 211 329 188 306 168 285 151 266 137 249 124 233 112
550 550 550 510 550 463 543 428 476 351 420 291 374 245 335 207 302 177 273 153 249 132 227 116 208 101
550 550 550 510 550 463 550 434 550 396 550 366 507 317 454 269 409 230 370 198 337 172 308 150 282 132
550 550 550 507 550 467 550 443 550 408 550 383 525 347 475 299 432 259 395 226 362 199 334 175 308 56 285 139 265 124
16
550 550 550 526 550 490 550 455 550 426 550 406 550 385 507 339 465 298 428 263 395 233 366 208 340 186 317 167 296 151 277 137 259 124
16
550 550 550 526 550 490 550 455 550 426 550 406 550 385 550 363 550 346 514 311 474 276 439 246 408 220 380 198 355 178 332 161 311 147
Table 3.4: Joist Properties
Victor Saouma
Structural Concepts and Systems for Architects
Draft 3{20
STRUCTURAL MATERIALS
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 4
Case Study I: EIFFEL TOWER Adapted from (Billington and Mark 1983)
4.1 Materials, & Geometry The tower was built out of wrought iron, less expensive than steel,and Eiel had more expereince with this material, Fig. 4.1 1
Figure 4.1: Eiel Tower (Billington and Mark 1983)
Draft 4{2
Case Study I: EIFFEL TOWER The structure is essentially a tower, subjected to gravity and wind load. It is a relatively \light" structure, so dead load is small compared to the wind load. M gravity 3 To avoid overturning M wind had to be much higher than 1. This can be achieved either by: 1. Increase self weight (as in Washington's monument) 2. Increase the width of the base 3. Design support to resist tension. 4. Posttension the support.
2
The tower is 984 feet high, and 328 feet wide at the base. The large base was essential to provide adequate stability in the presence of wind load. 5 We can assume that the shape of the tower is parabolic. If we take the x axis to be along the vertical axis of symmetry and y the half width, then we know that at x = 984 the (half) width y = 0 and at x = 0 the half width is 328=2 = 164, thus the equation of the half width is given by
4
y = 164 984984; x  {z av(x)2
2
(4.1)
}
We recall from calculus that for y = v(x)
Thus for our problem
dy dv dy dx = dv dx d 2 dx ax = 2ax
(4.2a) (4.2b)
dy 984 ; x 1 dx = 2(164){z 984 }  ;{z 984 } dy dv
;x = 984 2; 952
(4.3a)
dv dx
Also
dy = tan ) = tan;1 dy dx dx where is the angle measured from the x axis to the tangent to the curve. 3
2
(4.3b) (4.4)
We now can tabulate the width and slope at various elevations
Victor Saouma
Structural Concepts and Systems for Architects
Draft 4.2 Loads
4{3
Width dy Location Height Width/2 Estimated Actual dx Support 0 164 328 .333 18.4o First platform 186 108 216 240 .270 15.1o second platform 380 62 123 110 .205 11.6o Intermediate platform 644 20 40 .115 6.6o Top platform 906 1 2 .0264 1.5o Top 984 0 0 0.000 0o 4 The tower is supported by four inclined supports, each with a cross section of 800 of the tower is shown in Fig. 4.2.
2.
in
ACTUAL CONTINUOUS CONNECTION
An idealization
IDEALIZED CONTINUOUS CONNECTION
ACTUAL POINTS OF CONNECTION
Figure 4.2: Eiel Tower Idealization, (Billington and Mark 1983)
4.2 Loads 5 6
The total weight of the tower is 18; 800 . The dead load is not uniformly distributed, and is approximated as follows, Fig. 4.3: k
Location Height Dead Weight Ground second platform 380 ft 15; 500 Second platformintermediate platform 264 ft 2; 200 intermediate platform  top 340 ft 1; 100 Total 984 ft 18; 800
k
k k k
From the actual width of the lower two platforms we can estimate the live loads (the intermediate and top platforms would have negligible LL in comparison):
7
kip 1st platform: (50) (240)2 2 (1;000) 2; 880 2 2 2nd platform: (50) (110) (1;000) 600 Total: 3; 480 Hence the total vertical load is Pvert = DL + LL = 18; 800 + 3; 480 = 22; 280 . psf
psf
ft
ft
lbs
k
lbs
k k
k
k
Victor Saouma
Structural Concepts and Systems for Architects
Draft 4{4
Case Study I: EIFFEL TOWER
Figure 4.3: Eiel Tower, Dead Load Idealization; (Billington and Mark 1983) The wind pressure is known to also have a parabolic distribution (maximum at the top), the cross sectional area over which the wind is acting is also parabolic (maximum at the base). Hence we will simplify our analysis by considering an equivalent wind force obtained from a constant wind pressure (force/length) and constant cross section Fig. 4.4: The pressure is assumed to be 2.6 k/ft, thus the lateral wind force is, Fig. 4.5 (4.5) Plat = (2:6) (984) = 2,560 acting at 984 2 = 492
8
k/ft
ft
k
ft
4.3 Reactions
Simplifying the three dimensional structure with 4 supports into a two dimensional one with two supports, the reactions can be easily determined for this statically determinate structure, Fig.4.6. Gravity Load
9
Pvert = 22; 280 ? grav Rvert = 22;2280 = 11,140
k
6
(4.6a) (4.6b)
Lateral Load Lateral Moment (we essentially have a cantilivered beam subjected to a uniform load). The
moment at a distance x from the support along the cantilevered beam subjected to a uniform pressure p is given by 2 (4.7) Mlat = p (L{z; x}) L ;2 x = p (L ;2 x)  {z } Force Moment arm
Victor Saouma
Structural Concepts and Systems for Architects
Draft 4.3 Reactions
4{5
Figure 4.4: Eiel Tower, Wind Load Idealization; (Billington and Mark 1983) TOTAL LOADS LOADS P
P=2560k Q
Q=22,280k
L/2
1111 0000 0000 H 1111 0000 1111 000 0 111 0000 1111 000 111 000 000 111 V111 0 000 111 000 111 000 111 000 111
REACTIONS M0
Figure 4.5: Eiel Tower, Wind Loads, (Billington and Mark 1983)
+
=
WINDWARD SIDE
VERTICAL FORCES
WIND FORCES
LEEWARD SIDE
TOTAL
Figure 4.6: Eiel Tower, Reactions; (Billington and Mark 1983)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 4{6
Case Study I: EIFFEL TOWER
Thus the lateral moment caused by the wind is parabolic. At the base (x = 0), the maximum moment is equal to 2 2 Mlat = p (L ;2 x) = (2:6) (984 2; 0) 2 = 1,260,000 ; (4.8) We observe that the shape of the moment diagram is also parabolic, just like the tower itself. This is not accidental, as nearly optimum structures have a shape which closely approximate their moment diagram (such as the varying depth of continuous long span bridges). To determine the resulting internal forces caused by the lateral (wind) moment, and since we have two supports (one under tension and the other under compression) we use 1; 260; 000 = 3,850 wind = Rvert = M (4.9) 6? d 328 Lateral Forces to be resisted by each of the two pairs. By symmetry, the lateral force will be equally divided among the two pairs of supports and will be equal to wind = 1,280 Rlat = (2; 560) (4.10) 2 k/ft
ft
k.ft
k.ft
k
ft
k
k
4.4 Internal Forces 10
First, a biref reminder cos = FFx (4.11a) sin = FFy (4.11b)
θ F
Fy
tan = FFy (4.11c) x
θ Fx
Internal forces are rst determined at the base. 12 Gravity load are rst considered, remember those are caused by the dead load and the live load, Fig. 4.7:
11
β=18.40 β=18.4 INCLINED INTERNAL FORCE: N CONSEQUENT HORIZONTAL COMPONENT: H
KNOWN VERTICAL COMPONENT: V
N
V
H FORCE POLYGON
Figure 4.7: Eiel Tower, Internal Gravity Forces; (Billington and Mark 1983)
Victor Saouma
V )N= V cos = N cos
(4.12a)
Structural Concepts and Systems for Architects
Draft 4.4 Internal Forces
4{7 ; 140 N = 11 cos 18:4o = 11,730 kip tan = H V ) H = V tan H = 11; 140 (tan 18:4o) = 3,700 kip
(4.12b)
k
(4.12c) (4.12d)
k
The horizontal forces which must be resisted by the foundations, Fig. 4.8.
H
H 3700 k
3700 k
Figure 4.8: Eiel Tower, Horizontal Reactions; (Billington and Mark 1983) Because the vertical load decreases with height, the axial force will also decrease with height. 14 At the second platform, the total vertical load is Q = 1; 100 + 2; 200 = 3; 300 and at that height the angle is 11:6o thus the axial force (per pair of columns) will be
13
k
3;300 k
2 Nvert = cos 11 :6o = 1; 685 o Hvert = 3; 300 2 (tan 11:6 ) = 339
(4.13a)
k
k
k
(4.13b)
Note that this is about seven times smaller than the axial force at the base, which for a given axial strength, would lead the designer to reduce (or taper) the crosssection. The horizontal force will be resisted by the axial forces in the second platform itself. 15 Wind Load: We now have determined at each pair of support the vertical and the horizontal forces caused by the wind load, the next step is to determine their axial components along the inclined leg, Fig. 4.9: wind wind Nc = ;Rvert cos ; Rlat sin o = ;(3; 850) (cos 18:4 ) ; (1; 280) (sin 18:40) k
= Nt = = =
Victor Saouma
4,050
k
Leeward wind ;R cos + Rlat sin o (3; 850) (cos 18:4 ) + (1; 280) (sin 18:40) 4,050 Winward k
wind vert
k
k
k
(4.14a) (4.14b) (4.14c) (4.14d) (4.14e) (4.14f)
Structural Concepts and Systems for Architects
Draft 4{8
Case Study I: EIFFEL TOWER
18.4
3,850 k
3,850 cos 18.4
1,280 sin 18.4
18.4
1,280 k
Figure 4.9: Eiel Tower, Internal Wind Forces; (Billington and Mark 1983)
4.5 Internal Stresses 16
The total forces caused by both lateral and gravity forces can now be determined: NLTotal = ;(11; 730) ;(4; 050) = 15,780 Leeward side 
{z
k
}
{z
k
k
}
gravity lateral Total ;{z050) } = 7,630 NW = ; (11;{z730) } +(4   gravity lateral k
k
k
Winward side
(4.15a) (4.15b)
We observe that even under wind load, the windward side is still under compression. 2. 17 In the idealization of the tower's geometry, the area of each pair of the simpli ed columns is 1; 600 and thus the maximum stresses will be determined from T ; 780 = 9.9 comp = NAL = ;1;15600 (4.16) 2 in
k
ksi
in
18
The strength of wrought iron is 45 ksi, hence the safety factor is stress 45 Safety Factor = ultimate actual stress = 9:9 = 4.5 ksi
ksi
Victor Saouma
(4.17)
Structural Concepts and Systems for Architects
Draft Chapter 5
REVIEW of STATICS To every action there is an equal and opposite reaction. Newton's third law of motion
5.1 Reactions In the analysis of structures (hand calculations), it is often easier (but not always necessary) to start by determining the reactions. 2 Once the reactions are determined, internal forces are determined next; nally, internal stresses and/or deformations (de ections and rotations) are determined last1 . 3 Reactions are necessary to determine foundation load. 4 Depending on the type of structures, there can be dierent types of support conditions, Fig. 5.1. Roller: provides a restraint in only one direction in a 2D structure, in 3D structures a roller may provide restraint in one or two directions. A roller will allow rotation. Hinge: allows rotation but no displacements. Fixed Support: will prevent rotation and displacements in all directions.
1
5.1.1 Equilibrium 5 6 7
Reactions are determined from the appropriate equations of static equilibrium. Summation of forces and moments, in a static system must be equal to zero2. In a 3D cartesian coordinate system there are a total of 6 independent equations of equilibrium: This is the sequence of operations in the exibility method which lends itself to hand calculation. In the stiness 1
method, we determine displacements rsts, then internal forces and reactions. This method is most suitable to computer implementation. In a dynamic system F = ma where m is the mass and a is the acceleration. 2
Draft 5{2
REVIEW of STATICS
Figure 5.1: Types of Supports Fx = Fy = Fz = 0 Mx = My = Mz = 0 8
(5.1)
In a 2D cartesian coordinate system there are a total of 3 independent equations of equilibrium: Fx = Fy = Mz = 0
(5.2)
For reaction calculations, the externally applied load may be reduced to an equivalent force3 . 10 Summation of the moments can be taken with respect to any arbitrary point. 11 Whereas forces are represented by a vector, moments are also vectorial quantities and are represented by a curved arrow or a double arrow vector. 12 Not all equations are applicable to all structures, Table 5.1 13 The three conventional equations of equilibrium in 2D: Fx ; Fy and Mz can be replaced by the independent moment equations MzA, MzB , MzC provided that A, B, and C are not colinear. 14 It is always preferable to check calculations by another equation of equilibrium. 15 Before you write an equation of equilibrium, 1. Arbitrarily decide which is the +ve direction 2. Assume a direction for the unknown quantities 3. The right hand side of the equation should be zero 9
3
However for internal forces (shear and moment) we must use the actual load distribution.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.1 Reactions
5{3
Structure Type
Beam, no axial forces 2D Truss, Frame, Beam Grid 3D Truss, Frame
Fx
Fy Fy
Fx
Fy
Alternate Set
Equations Fz Fz
Mz Mz
Mx My Mx My Mz
Beams, no axial Force MzA MzB 2 D Truss, Frame, Beam Fx MzA MzB MzA MzB MzC Table 5.1: Equations of Equilibrium If your reaction is negative, then it will be in a direction opposite from the one assumed. 16 Summation of external forces is equal and opposite to the internal ones. Thus the net force/moment is equal to zero. 17 The external forces give rise to the (nonzero) shear and moment diagram.
5.1.2 Equations of Conditions 18 If a structure has an internal hinge (which may connect two or more substructures), then this will provide an additional equation (M = 0 at the hinge) which can be exploited to determine the reactions. 19 Those equations are often exploited in trusses (where each connection is a hinge) to determine reactions. 20 In an inclined roller support with Sx and Sy horizontal and vertical projection, then the reaction R would have, Fig. 5.2.
Rx = Sy Ry Sx
(5.3)
Figure 5.2: Inclined Roller Support
5.1.3 Static Determinacy 21 In statically determinate structures, reactions depend only on the geometry, boundary conditions and loads.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{4
REVIEW of STATICS
If the reactions can not be determined simply from the equations of static equilibrium (and equations of conditions if present), then the reactions of the structure are said to be statically indeterminate. 23 The degree of static indeterminacy is equal to the dierence between the number of reactions and the number of equations of equilibrium (plus the number of equations of conditions if applicable), Fig. 5.3. 22
Figure 5.3: Examples of Static Determinate and Indeterminate Structures Failure of one support in a statically determinate system results in the collapse of the structures. Thus a statically indeterminate structure is safer than a statically determinate one. 25 For statically indeterminate structures, reactions depend also on the material properties (e.g. Young's and/or shear modulus) and element cross sections (e.g. length, area, moment of inertia).
24
5.1.4 Geometric Instability The stability of a structure is determined not only by the number of reactions but also by their arrangement. 27 Geometric instability will occur if: 1. All reactions are parallel and a nonparallel load is applied to the structure. 2. All reactions are concurrent, Fig. 5.4.
26
Figure 5.4: Geometric Instability Caused by Concurrent Reactions
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.1 Reactions
5{5
3. The number of reactions is smaller than the number of equations of equilibrium, that is a mechanism is present in the structure. 28 Mathematically, this can be shown if the determinant of the equations of equilibrium is equal to zero (or the equations are interdependent).
5.1.5 Examples 29 Examples of reaction calculation will be shown next. Each example has been carefully selected as it brings a dierent \twist" from the preceding one. Some of those same problems will be revisited later for the determination of the internal forces and/or de ections. Many of those problems are taken from Prof. Gerstle textbok Basic Structural Analysis.
Example 57: Simply Supported Beam Determine the reactions of the simply supported beam shown below.
Solution:
The beam has 3 reactions, we have 3 equations of static equilibrium, hence it is statically determinate. (+  ) Fx = 0; ) Rax ; 36 = 0 (+ 6) Fy = 0; ) Ray + Rdy ; 60 ; (4) (12) = 0 (+ ;) Mzc = 0; ) 12Ray ; 6Rdy ; (60)(6) = 0 k
k
or through matrix inversion (on your calculator) 2 4
1 0 0 0 1 1 0 12 ;6
38 < 5 :
Rax Ray Rdy
9 = ;
8 <
=:
36 108 360
9 = ;
8 <
):
k/ft
Rax Ray Rdy
9 = ;
ft
8 <
=:
36 56 52
k k k
9 = ;
Alternatively we could have used another set of equations:
Check:
(+ ;) Mza = 0; (60)(6) + (48)(12) ; (Rdy )(18) = 0 ) Rdy = 52 (+ ;) Mzd = 0; (Ray )(18) ; (60)(12) ; (48)(6) = 0 ) Ray = 56
Victor Saouma
k k
6 6
p
(+ 6) Fy = 0; ; 56 ; 52 ; 60 ; 48 = 0
Structural Concepts and Systems for Architects
Draft 5{6
REVIEW of STATICS
Example 58: Three Span Beam Determine the reactions of the following three spans beam
Solution:
We have 4 unknowns (Rax ; Ray ; Rcy and Rdy ), three equations of equilibrium and one equation of condition (Mb = 0), thus the structure is statically determinate. 1. Isolating ab: M ;b = 0; (9)(Ray ) ; (40)(5) = 0 ) Ray = 22.2 6 (+ ;) Ma = 0; (40)(4) ; (S )(9) = 0 ) S = 17.7 6 Fx = 0; ) Rax = 30 k
k
k
2. Isolating bd: (+ ;) Md = 0; ;(17:7)(18) ; (40)(15) ; (4)(8)(8) ; (30)(2) + Rcy (12) = 0 ) Rcy = 1;12236 = 103 6 ; (+ ) Mc = 0; ;(17:7)(6) ; (40)(3) + (4)(8)(4) + (30)(10) ; Rdy (12) = 0 ) Rdy = 20112:3 = 16.7 6 k
k
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.1 Reactions
5{7
3. Check
p
Fy = 0; 6; 22:2 ; 40 ; 40 + 103 ; 32 ; 30 + 16:7 = 0
Example 59: Three Hinged Gable Frame The threehinged gable frames spaced at 30 ft. on center. Determine the reactions components on the frame due to: 1) Roof dead load, of 20 of roof area; 2) Snow load, of 30 of horizontal projection; 3) Wind load of 15 of vertical projection. Determine the critical design values for the vertical and horizontal reactions. psf
psf
psf
Solution: 1. Due to symmetry, we will consider only the dead load on one side of the frame. 2. Due to symmetry, there is no vertical force transmitted by the hinge for snow and dead load. 3. Roof dead load per frame is p 1 302 + 152 1; 000 DL = (20) (30) = 20:2 ? psf
Victor Saouma
ft
ft
lbs/k
k
Structural Concepts and Systems for Architects
Draft 5{8
REVIEW of STATICS
4. Snow load per frame is
SL = (30)
psf
1 (30) (30) 1; 000 ft
ft
lbs/k
= 27:
k
?
5. Wind load per frame (ignoring the suction) is
WL = (15)
psf
1 (30) (35) 1; 000 ft
ft
lbs/k
= 15:75

k
6. There are 4 reactions, 3 equations of equilibrium and one equation of condition ) statically determinate. 7. The horizontal reaction H due to a vertical load V at midspan of the roof, is obtained by taking moment with respect to the hinge (+ ;) MC = 0; 15(V ) ; 30(V ) + 35(H ) = 0 ) H = 1535V = :429V
Substituting for roof dead and snow load we obtain A = VDL A HDL = A = VSL A HSL =
B VDL B HDL B VSL B HSL
= 20:2 6 = (:429)(20:2) = 8:66 = 27: 6 = (:429)(27:) = 11:58 k
k
k
k
8. The reactions due to wind load are B (+ ;) MA = 0; (15:75)( 20+15 2 ) ; VWL (60) = 0 ; B (+ ) MC = 0; HWL (35) ; (4:6)(30) = 0 A =0 (+  ) Fx = 0; 15:75 ; 3:95 ; HWL B A (+ 6) Fy = 0; VWL ; VWL = 0
B = 4:60 ) VWL B = 3:95 ) HWL A ) HWL = 11:80 A = ;4:60 ) VWL
k k
6 ?
k
k
9. Thus supports should be designed for
H = 8:66 + 11:58 + 3:95 V = 20:7 + 27:0 + 4:60 k
k
k
k
k
k
= 24.19 = 52.3
k
k
5.2 Trusses
5.2.1 Assumptions Cables and trusses are 2D or 3D structures composed of an assemblage of simple one dimensional components which transfer only axial forces along their axis. 31 Trusses are extensively used for bridges, long span roofs, electric tower, space structures. 32 For trusses, it is assumed that 1. Bars are pinconnected
30
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.2 Trusses
5{9
2. Joints are frictionless hinges4 . 3. Loads are applied at the joints only. A truss would typically be composed of triangular elements with the bars on the upper chord under compression and those along the lower chord under tension. Depending on the orientation of the diagonals, they can be under either tension or compression. 34 In a truss analysis or design, we seek to determine the internal force along each member, Fig. 5.5
33
Figure 5.5: Bridge Truss
5.2.2 Basic Relations
Sign Convention: Tension positive, compression negative. On a truss the axial forces are indicated as forces acting on the joints. StressForce: = PA StressStrain: = E" ForceDisplacement: " = LL Equilibrium: F = 0
5.2.3 Determinacy and Stability Trusses are statically determinate when all the bar forces can be determined from the equations of statics alone. Otherwise the truss is statically indeterminate. 36 A truss may be statically indeterminate with respect to the reactions or externally indeterminate and/or statically indeterminate with respect to the internal forces that is internally indeterminate. 37 a 2D truss is externally indeterminate if there are more than 3 reactions. 35
In practice the bars are riveted, bolted, or welded directly to each other or to gusset plates, thus the bars are not free to rotate and socalled secondary bending moments are developed at the bars. Another source of secondary moments is the dead weight of the element. 4
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{10
REVIEW of STATICS
Since each joint is pinconnected, we can apply M = 0 at each one of them. Furthermore, summation of forces applied on a joint must be equal to zero. 39 For a 2D truss we have 2 equations of equilibrium FX = 0 and FY = 0 which can be applied at each joint. For 3D trusses we would have three equations: FX = 0, FY = 0 and FZ = 0. 40 If we refer to j as the number of joints, R the number of reactions and m the number of members, then we would have a total of m + R unknowns and 2j (or 3j ) equations of statics (2D or 3D at each joint). If we do not have enough equations of statics then the problem is indeterminate, if we have too many equations then the truss is unstable, Table 5.2.
38
2D
3D
Static Indeterminacy
External R>3 R>6 Internal m + R > 2j m + R > 3j Unstable m + R < 2j m + R < 3j Table 5.2: Static Determinacy and Stability of Trusses Fig. 5.6 shows a truss with 4 reactions, thus it is externally indeterminate. This truss has 6 joints (j = 6), 4 reactions (R = 4) and 9 members (m = 9). Thus we have a total of m + R = 9 + 4 = 13 unknowns and 2 j = 2 6 = 12 equations of equilibrium, thus the truss is statically indeterminate.
41
Figure 5.6: A Statically Indeterminate Truss 42
There are two methods of analysis for statically determinate trusses 1. The Method of joints 2. The Method of sections
5.2.4 Method of Joints 43
The method of joints can be summarized as follows 1. Determine if the structure is statically determinate
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.2 Trusses
5{11
2. Compute all reactions 3. Sketch a free body diagram showing all joint loads (including reactions) 4. For each joint, and starting with the loaded ones, apply the appropriate equations of equilibrium (Fx and Fy in 2D; Fx , Fy and Fz in 3D).
5. Because truss elements can only carry axial forces, the resultant force (F~ = F~x + F~y ) must be along the member, Fig. 5.7.
F = Fx = Fy l lx ly
(5.4)
44 Always keep track of the x and y components of a member force (Fx , Fy ), as those might be needed later on when considering the force equilibrium at another joint to which the member is connected.
Figure 5.7: X and Y Components of Truss Forces This method should be used when all member forces should be determined. 46 In truss analysis, there is no sign convention. A member is assumed to be under tension (or compression). If after analysis, the force is found to be negative, then this would imply that the wrong assumption was made, and that the member should have been under compression (or tension). 47 On a free body diagram, the internal forces are represented by arrow acting on the joints and not as end forces on the element itself. That is for tension, the arrow is pointing away from the joint, and for compression toward the joint, Fig. 5.8.
45
Figure 5.8: Sign Convention for Truss Element Forces
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{12
REVIEW of STATICS
Example 510: Truss, Method of Joints Using the method of joints, analyze the following truss
Solution: 1. R = 3, m = 13, 2j = 16, and m + R = 2j 2. We compute the reactions
p
(+ ;) ME = 0; ) (20 + 12)(3)(24) + (40 + 8)(2)(24) + (40)(24) ; RAy (4)(24) = 0 ) RAy = 58 6 (+ ?) Fy = 0; ) 20 + 12 + 40 + 8 + 40 ; 58 ; REy = 0 ) REy = 62 6 k
k
3. Consider each joint separately: Node A: Clearly AH is under compression, and AB under tension.
(+ 6) Fy = 0; ) FAHy ; 58 = 0 FAH = lly (FAHy ) p ly = 32 l = 322 + 242 = 40 Compression ) FAH = 40 32 (58) = 72:5 (+  ) Fx = 0; ) ;FAHx + FAB = 0 FAB = llxy (FAHy ) = 24 32 (58) = 43:5 Tension
Victor Saouma
Structural Concepts and Systems for Architects
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5{13
Node B:
(+  ) Fx = 0; ) FBC = 43:5 Tension (+ 6) Fy = 0; ) FBH = 20 Tension
Node H:
(+  ) Fx = 0; ) FAHx ; FHCx ; FHGx = 0 p 24 43:5 ; p2424 (I) 2 +322 (FHC ) ; 242 +102 (FHG ) = 0 (+ 6) Fy = 0; ) FAHy + FHCy ; 12 ; FHGy ; 20 = 0 p 10 58 + p2432 2 +322 (FHC ) ; 12 ; 242 +102 (FHG ) ; 20 = 0 (II) Solving for I and II we obtain
FHC = ;7:5 Tension FHG = 52 Compression
Node E:
Fy = 0; ) FEFy = 62 ) FEF = p2432 = 77:5 2 2 (62) 24 (F+32 Fx = 0; ) FED = FEFx ) FED = 32 (62) = 46:5 EFy ) = 24 32
k k
The results of this analysis are summarized below
Victor Saouma
Structural Concepts and Systems for Architects
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REVIEW of STATICS
4. We could check our calculations by verifying equilibrium of forces at a node not previously used, such as D
5.3 Shear & Moment Diagrams 5.3.1 Theory
5.3.1.1 Design Sign Conventions Before we derive the ShearMoment relations, let us arbitrarily de ne a sign convention. 5 49 The sign convention adopted here, is the one commonly used for design purposes . With reference to Fig. 5.9 48
Figure 5.9: Shear and Moment Sign Conventions for Design 5
Later on, in more advanced analysis courses we will use a dierent one.
Victor Saouma
Structural Concepts and Systems for Architects
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5{15
2D: Load Positive along the beam's local y axis (assuming a right hand side convention), that is
positive upward. Axial: tension positive. Flexure A positive moment is one which causes tension in the lower bers, and compression in the upper ones. For frame members, a positive moment is one which causes tension along the inner side. Shear A positive shear force is one which is \up" on a negative face, or \down" on a positive one. Alternatively, a pair of positive shear forces will cause clockwise rotation. Torsion Counterclockwise Draftpositive 3D: Use double arrow vectors (and NOT curved arrows). Forces and moments (including torsions) are x 6M de ned with respect to a right hand side coordinate system, Fig. 5.10. * 6 y
* Tx * Mz
y
6M y 6 6 >

 z
> Figure 5.10: for 3D Frame Elements M Sign Conventions Tx
z
5.3.1.2 Load, Shear, Moment Relations 50 Let us (re)derive the basic relations between load, shear and moment. Considering an in nitesimal length dx of a beam subjected to a positive load6 w(x), Fig. 5.11. The in nitesimal section must also
Figure 5.11: Free Body Diagram of an In nitesimal Beam Segment be in equilibrium. 51 There are no axial forces, thus we only have two equations of equilibrium to satisfy Fy = 0 and Mz = 0. 52 Since dx is in nitesimally small, the small variation in load along it can be neglected, therefore we assume w(x) to be constant along dx. 53 To denote that a small change in shear and moment occurs over the length dx of the element, we add the dierential quantities dVx and dMx to Vx and Mx on the right face. 6
In this derivation, as in all other ones we should assume all quantities to be positive.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{16 54
or
REVIEW of STATICS
Next considering the rst equation of equilibrium (+ 6) Fy = 0 ) Vx + wx dx ; (Vx + dVx ) = 0
dV = w(x) dx
(5.5)
The slope of the shear curve at any point along the axis of a member is given by the load curve at that point. 55
Similarly
(+ ;) Mo = 0 ) Mx + Vx dx ; wx dx dx 2 ; (Mx + dMx ) = 0 Neglecting the dx2 term, this simpli es to
dM = V (x) dx
(5.6)
The slope of the moment curve at any point along the axis of a member is given by the shear at that point. 56
Alternative forms of the preceding equations can be obtained by integration
V =
Z
w(x)dx
V21 = Vx2 ; Vx1 =
Z
x2 x1
(5.7)
w(x)dx (5.8)
The change in shear between 1 and 2, V21 , is equal to the area under the load between x1 and x2 . and
M =
Z
V (x)dx
M21 = M2 ; M1 =
Z
x2 x1
(5.9)
V (x)dx (5.10)
The change in moment between 1 and 2, M21, is equal to the area under the shear curve between x1 and x2 . Note that we still need to have V1 and M1 in order to obtain V2 and M2 respectively. 58 Fig. 5.12 and 5.13 further illustrates the variation in internal shear and moment under uniform and concentrated forces/moment. 57
5.3.1.3 Moment Envelope For design, we often must consider dierent load combinations. 60 For each load combination, we should draw the shear, moment diagrams. and then we should use the Moment envelope for design purposes. 59
Victor Saouma
Structural Concepts and Systems for Architects
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5{17
Figure 5.12: Shear and Moment Forces at Dierent Sections of a Loaded Beam
Positive Constant
Negative Constant
Positive Increasing Positive Decreasing Negative Increasing Negative Decreasing
Positive Constant
Negative Constant
Positive Increasing Positive Decreasing Negative Increasing Negative Decreasing
Load
Shear
Shear
Moment
Figure 5.13: Slope Relations Between Load Intensity and Shear, or Between Shear and Moment
Victor Saouma
Structural Concepts and Systems for Architects
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REVIEW of STATICS
5.3.1.4 Examples Example 511: Simple Shear and Moment Diagram Draw the shear and moment diagram for the beam shown below
Solution:
The free body diagram is drawn below
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{19
Reactions are determined from the equilibrium equations (+ ) Fx = 0; ) ;RAx + 6 = 0 ) RAx = 6 (+ ;) MA = 0; ) (11)(4) + (8)(10) + (4)(2)(14 + 2) ; RFy (18) = 0 ) RFy = 14 (+ 6) Fy = 0; ) RAy ; 11 ; 8 ; (4)(2) + 14 = 0 ) RAy = 13 k
k
k
Shear are determined next. 1. 2. 3. 4.
At A the shear is equal to the reaction and is positive. At B the shear drops (negative load) by 11 to 2 . At C it drops again by 8 to ;6 . It stays constant up to D and then it decreases (constant negative slope since the load is uniform and negative) by 2 per linear foot up to ;14 . As a check, ;14 is also the reaction previously determined at F . is determined last: The moment at A is zero (hinge support). The change in moment between A and B is equal to the area under the corresponding shear diagram, or MB;A = (13)(4) = 52. etc... k
k
k
k
k
5.
Moment 1. 2. 3.
k
k
Example 512: Frame Shear and Moment Diagram Draw the shear and moment diagram of the following frame
Solution: Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{20
REVIEW of STATICS
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{21
Reactions are determined rst
(+ ) Fx = 0; ) RAx ; 45 (3)(15) = 0  {z }
) (+ ;) MA = 0; ) ) (+ 6) Fy = 0; ) )
load
RAx = 36
k
;
(3)(30)( 302 ) + 35 (3)(15) 30 + 29 ; 45 (3)(15) 122 ; 39RDy = 0 RFy = 52:96 RAy ; (3)(30) ; 35 (3)(15) + 52:96 = 0 RAy = 64:06 k
k
Shear: 1. For A ; B , the shear is constant, equal to the horizontal reaction at A and negative according to our previously de ned sign convention, VA = ;36 2. For member B ; C at B , the shear must be equal to the vertical force which was transmitted along A ; B , and which is equal to the vertical reaction at A, VB = 64:06. 3. Since B ; C is subjected to a uniform negative load, the shear along B ; C will have a slope equal to ;3 and in terms of x (measured from B to C ) is equal to VB;C (x) = 64:06 ; 3x k
4. The shear along C ; D is obtained by decomposing the vertical reaction at D into axial and shear components. Thus at D the shear is equal to 53 52:96 = 31:78 and is negative. Based on our sign convention for the load, the slope of the shear must be equal to ;3 along C ; D. Thus the shear at point C is such that Vc ; 53 9(3) = ;31:78 or Vc = 13:22. The equation for the shear is given by (for x going from C to D) V = 13:22 ; 3x k
5. We check our calculations by verifying equilibrium of node C p (+ ) Fx = 0 ) 53 (42:37) + 45 (13:22) = 25:42 + 10:58 = 36 p (+ 6) Fy = 0 ) 54 (42:37) ; 35 (13:22) = 33:90 ; 7:93 = 25:97
Moment:
1. Along A ; B , the moment is zero at A (since we have a hinge), and its slope is equal to the shear, thus at B the moment is equal to (;36)(12) = ;432 2. Along B ; C , the moment is equal to k.ft
Z
x
Z
x
VB;C (x)dx = ;432 + (64:06 ; 3x)dx 0 0 2 = ;432 + 64:06x ; 3 x2 which is a parabola. Substituting for x = 30, we obtain at node C : MC = ;432+64:06(30) ; 2 30 3 2 = 139:8 3. If we need to determine the maximum moment along B ; C , we know that dMdxB;C = 0 at the point where VB;C = 0, that is VB;C (x) = 64:06 ; 3x = 0 ) x = 643:06 = 25:0 . In other MB;C = MB +
k.ft
words, maximum moment occurs where the shear is zero. (25:0)2 Thus MBmax ;C = ;432 + 64:06(25:0) ; 3 2 = ;432 + 1; 601:5 ; 937:5 = 232
Victor Saouma
ft
k.ft
Structural Concepts and Systems for Architects
Draft 5{22
REVIEW of STATICS
4. Finally along C ; D, the moment varies quadratically (since we had a linear shear), the moment rst increases (positive shear), and then decreases (negative shear). The moment along C ; D is given by R R MC ;D = MC + 0x VC ;D (x)dx2 = 139:8 + 0x (13:22 ; 3x)dx = 139:8 + 13:22x ; 3 x2 which is a parabola. 2 Substituting for xp= 15, we obtain at node C MC = 139:8 + 13:22(15) ; 3 152 = 139:8 + 198:3 ; 337:5 = 0
Example 513: Frame Shear and Moment Diagram; Hydrostatic Load The frame shown below is the structural support of a ume. Assuming that the frames are spaced 2 ft apart along the length of the ume, 1. Determine all internal member end actions 2. Draw the shear and moment diagrams 3. Locate and compute maximum internal bending moments 4. If this is a reinforced concrete frame, show the location of the reinforcement.
Solution: The hydrostatic pressure causes lateral forces on the vertical members which can be treated as cantilevers xed at the lower end. The pressure is linear and is given by p = h. Since each frame supports a 2 ft wide slice of the
ume, the equation for w (pounds/foot) is w = (2)(62:4)(h) = 124:8h At the base w = (124:8)(6) = 749 = :749 Note that this is both the lateral pressure on the end walls as well as the uniform load on the horizontal members. lbs/ft
lbs/ft
Victor Saouma
k/ft
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{23
End Actions
1. Base force at B is FBx = (:749) 26 = 2:246 2. Base moment at B is MB = (2:246) 36 = 4:493 3. End force at B for member B ; E are equal and opposite. 4. Reaction at C is RCy = (:749) 162 = 5:99 Shear forces 1. Base at B the shear force was determined earlier and was equal to 2:246 . Based on the orientation of the x ; y axis, this is a negative shear. 2. The vertical shear at B is zero (neglecting the weight of A ; B ) 3. The shear to the left of C is V = 0 + (;:749)(3) = ;2:246 . 4. The shear to the right of C is V = ;2:246 + 5:99 = 3:744 Moment diagrams k
k.ft
k
k
k
k
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{24
REVIEW of STATICS
1. At the base: B M = 4:493 as determined above. 2. At the support C , Mc = ;4:493 + (;:749)(3)( 23 ) = ;7:864 3. The maximum moment is equal to Mmax = ;7:864 + (:749)(5)( 25 ) = 1:50 Design: Reinforcement should be placed along the bers which are under tension, that is on the side of the negative moment7 . The gure below schematically illustrates the location of the exural8 reinforcement. k.ft
k.ft
k.ft
Example 514: Shear Moment Diagrams for Frame That is why in most European countries, the sign convention for design moments is the opposite of the one commonly used in the U.S.A.; Reinforcement should be placed where the moment is \postive". Shear reinforcement is made of a series of vertical stirrups. 7
8
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{25
8’
12’
10’
30k
5k/ft
2k/ft
B
A
E
Vba 10k C
Vbc
bd M ba H
5’
VA
20k
Vbd
G
52.5k
M bc M bd 30k
15’ 0
0 0
D
650’k 450’k
HD
4k/ft
200’k 82.5k
VD
CHECK
30k
10k
5k/ft B
A
M bc
M ba
Vba
17.5k
2k/ft
B
Hbc
Hba
C Vbc
(10)+(2)(10) 30k
17.5k 17.55*x=0
22.5k
3.5’
22.5+(30)
10k
Vbc M bc 200’k
17.5(5)(8)
(10)(10)+(2)(10)(10)/2
Vba
52.5k 30.6’k
(17.5)(3.5)/2
20’k
(17.5)(3.5)/2+(22.5)(83.5)/2
(52.5)(12)+(20)
650’k
M ba
Vbd M bd
50k
Victor Saouma
20k
450’k (50)(15)[(4)(5)/2][(2)(15)/3)]
(50)(4)(15)/2
450’k
Hbd
20k
4k/ft
50k 82.5k
Structural Concepts and Systems for Architects
Draft 5{26
REVIEW of STATICS
Example 515: Shear Moment Diagrams for Inclined Frame
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{27
26k
26k
10’
13’
10’
13’ 13’
20k C
13 5
5
3
12
15’
4
D
B 2k/ft
20’ Ha
A
E
36’
20’
Ve
48.8k
2k/ft
60k
Fx F
800’k
z y
Fy
20k
60(2)(20)
20k
(20)(20)+(6020)(20)/2
19.2k 800k’
(60)(20)(2)(20)(20)/2
Va
0k’
x F/Fy=z/x F/Fx=z/y Fx/Fy=y/x
60k 2
11.1k
26.6k
800k’
8k
28.8k
20k
8k 12k
20k
778k’
7
29.3k
48.9k
’
777k
’k
1130
9 BC
.1
16 k
11 CD
(39.1)(12.5)
488’k 12 CD
14
13
k
1k
2k’
112
777k’ 48
8’
k
800’k
+60k
+20k
0’k 113
23 .1 k k k 8 6 . 5 . 6 0 2 39 .
(20)(12)/(13)=18.46 (19.2)(5)/(13)=7.38 (19.2)(12)/(13)=17.72 (26)(12)/(13)=24 (26.6)(13)/(12)=28.8 (26.6)(5)/(12)=11.1 (28.8)(4)/(5)=23.1 (28.8)(3)/(5)=17.28 (20)(4)/(5)=16 (20)(3)/(5)=12 (39.1)(5)/(4)=48.9 (39.1)(3)/(4)=29.3
k’
1,130(.58)(13) k’ 800+(25.4)(13) 1122
3.1
8 BC
777

25.42
.1k 2 39
1,122(26.6)(13)
k k 26.6 0.58 26 0.626
488+(23.1)(12.5)
+25
+25.4
10
23
8.7 17.7+ .4k
800’k
0k’
0k 39.1k
19.2k
(20)(15)/13=7.7
17.2
18.46k 7.38k
17.72k
6
16k
0k
24k
24k
10k
5
28.8k
778k’ 10k
26k
48.8k
4 1k
26k
3
23.
1
201010
19.2k
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{28
REVIEW of STATICS
5.3.2 Formulaes Adapted from (of Steel COnstruction 1986)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{29
1) Simple Beam; uniform Load L
x
w L R
R L / 2
R Vx
= V = w L2 ; x 2 at center Mmax = wL 8 Mx = wx 2 (L ;4 x) 5 wL max = 384 EI 3 2 3 x = 24wx EI (L ; 2Lx + x )
L / 2
V
Shear
V
M max.
Moment
2) Simple Beam; Unsymmetric Triangular Load R1 = V1 Max R2 = V2 Vx at x = :577L Mmax Mx at x = :5193L max x
= W3 = 2W 3 2 W = 3 ; Wx 2 L = :1283WL 2 2 = Wx 3L2 (L ; x ) 3
= :01304 WL EI 3
4 2 2 4 = 180Wx EIL2 (3x ; 10L x + 7L )
3) Simple Beam; Symmetric Triangular Load
R=V for x < L2 Vx at center Mmax for x < L2 Mx for x < L2 x max
Victor Saouma
W 2 W 2 2 2WL L2 (L ; 4x ) 6 2 1 2 x = Wx 2 ; 3 L2 2 22 = 480Wx 2 (5L ; 4x ) EIL 3 WL = 60 EI
= = =
Structural Concepts and Systems for Architects
Draft 5{30
REVIEW of STATICS
4) Simple Beam; Uniform Load Partially Distributed
Max when a < c Max when a > c when a < x < a + b when x < a when a < x < a + b when a + b < x at x = a + Rw1
R1 = V1 R2 = V2 Vx Mx Mx Mx Mmax
= = = = = = =
wb 2L (2c + b) wb 2L (2a + b) R1 ; w(x ; a)
R1 x R1 x ; w2 (x ; a)2 R2 (L ; x) R1 a + 2Rw1
5) Simple Beam; Concentrated Load at Center
max R1 = V1 R=V at x = L2 Mmax when x < L2 Mx whenx < L2 x at x = L2 max
wa (2L ; a) 2L 2P PL 4 Px 2Px 2 2 48EI3 (3L ; 4x ) PL = 48 EI = = = = =
6) Simple Beam; Concentrated Load at Any Point max when a < b max when a > b at x = a when x < a
at x =
Victor Saouma
q
R1 = V1 R2 = V2 Mmax Mx
at x = a a when x < a x a(a+2b) & a > b max 3
Pb L Pa L Pab L Pbx L2 2 Pa b = 3EIL 2 b2 ; x2 ) = 6Pbx EIL (L ; p Pab(a + 2b) 3a(a + 2b) = 27EIL = = = =
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{31
7) Simple Beam; Two Equally Concentrated Symmetric Loads
R=V = P Mmax = Pa max = 24Pa (3L2 ; 4a2 ) EI (3La ; 3a2 ; x2 ) when x < a x = 6Px EI 2 2 when a < x < L ; a x = 6Pa EI (3Lx ; 3x ; a )
8) Simple Beam; Two Equally Concentrated Unsymmetric Loads
max when a < b max when b < a when a < x < L ; b max when b < a max when a < b when x < a when a < x < L ; b
R1 = V1 R2 = V2 Vx M1 M2 Mx Mx
= = = = = = =
P ( L ; a + b) PL (L ; b + a) PL (b ; a) L R1 a R2 b R1 x R1 x ; P (x ; a)
9) Cantilevered Beam, Uniform Load R1 = V1 R2 = V2 Vx Mmax at x = 38 L M1 Mx x
at x = :4215L max
Victor Saouma
= = = = =
3 wL 85 8 wL
R1 ; wx wL2 98 wL2 128
2
= R1 x ; wx2 3 + 3 = 48wx EI4 (L ; 3Lx 2x ) wL = 185 EI
Structural Concepts and Systems for Architects
Draft 5{32
REVIEW of STATICS
10) Propped Cantilever, Concentrated Load at Center
R1 = V1 R2 = V2 at x = L Mmax when x < L2 Mx when L2 < x Mx at x = :4472L max
= = = =
5P 16P 11 316 PL 16 5Px 16
x = P L2 ; 11 16 3 = :009317 PL EI
11) Propped Cantilever; Concentrated Load R1 = V1 R2 = V2 at x = a M1 at x = L M2 at x = a a when a < :414L at x = L 3LL22+;aa22 max when :414L < a at x = L
q
a
2L+a
max
Pb2 (a + 2L) 2Pa L3 2 2 2L3 (3L ; a ) R1 a Pab (a + L) 2L2 2 3 b = 12Pa EIL32(3L +2 a3 ) (L ; a ) = 3Pa EI2(3rL2 ; a2 )2 a = Pab 6EI 2L + a2 = = = =
12) Beam Fixed at Both Ends, Uniform Load R = V = wL 2 L Vx = w 2 ;x 2 at x = 0 and x = L Mmax = wL 122 wL L at x = 2 M = 24 wL4 at x = L2 max = 384 EI 2 wx x = 24EI (L ; x)2
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{33
13) Beam Fixed at Both Ends; Concentrated Load
R = V = P2 at x = L2 Mmax = PL 8 when x < L2 Mx = P8 (4x ; L) PL3 at x = L2 max = 192 EI 2 Px L when x < 2 x = 48EI (3L ; 4x)
14) Cantilever Beam; Triangular Unsymmetric Load
R=V Vx at x = L Mmax Mx x at x = 0 max
= 83 W 2 = W Lx 2 = WL 3
2 = Wx 3LW2 = 60EIL2 (x5 ; 5L2x + 4L5 ) WL3 = 15 EI
15) Cantilever Beam; Uniform Load
R=V Vx Mx at x = L Mmax x
at x = 0 max
Victor Saouma
= wL = wx 2 = wx2 2 = wL 2 = 24wEI (x4 ; 4L3 x + 3L4 ) 4 = wL 8EI
Structural Concepts and Systems for Architects
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REVIEW of STATICS
16) Cantilever Beam; Point Load
R=V at x = L Mmax when a < x Mx at x = 0 max at x = a a when x < a x when a < x x
P Pb P (x ; a) Pb2 (3L ; b) 6EI3 = 3Pb EI2 = 6Pb EI (3L ;2 3x ; b) ; x) (3b ; L + x) = P (L6EI = = = =
17) Cantilever Beam; Point Load at Free End
R=V = P at x = L Mmax = PL Mx = Px 3 at x = 0 max = PL 3PEI (2L3 ; 3L2x + x3 ) x = 6EI
18) Cantilever Beam; Concentrated Force and Moment at Free End
R=V = P Mx = P L2 ; x at x = 0 and x = L Mmax = PL 2 PL3 at x = 0 max = 12 EI 2 P ( L ; x) ((L + 2x) x = 12EI
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.3 Shear & Moment Diagrams
5{35
19) Beam Overhanging One Support; Uniform Load Between Supports R = V = wL 2 Vx = w L2 ; x Mx = wx 2 (L4 ; x) 5wL at x = L2 max = 384 EI 3 x = 24wx (L ; 2Lx2 + x3 ) EI 3 x1 x when L < x < L + a x1 = wL 24EI
20) Beam Overhanging one Support; Concentrated Force
Max when a < b R1 = V1 = Pb L Pa Max when b < a R2 = V2 = L at x < aL Mx = Pbx L p q Pab (a + 2b) 3a(a + 2b) a ( a +2 b ) when a > b = at x = max 3 2 b2 27EIL Pa at x = a a = 3EIL 2 2 2 when x < a x = 6Pbx EIL (L ; b ; x ) L ; x) (2Lx ; x2 ; a2 ) when a < x x = Pa6(EIL 1 at L < x < L + a x1 = Pabx 6EIL (L + a)
21) Continuous Beam, Two Equal Spans; Concentrated Force
R1 = V1 R2 = V2 + V3 R3 = V3 V2 Mmax at x = L M1
Victor Saouma
= = = = = =
Pb 4L2 ; a(L + a) 4Pa L3 2 + b(L + a) 2 L 3 2L ; Pab 3 (L + a) Pa4L4L2 + b(L + a) 4L3 Pab 4L2 ; a(L + a) 4L3 Pab 4L2 (L + a)
Structural Concepts and Systems for Architects
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REVIEW of STATICS
22) Simple Beam, Uniform Load, End Moments M2 R1 = V1 = wL + M1 ; 2 L M1 ; M2 R2 = V2 = wL 2 ; L M2 Vx = w L2 ; x + M1 ; L ;M2 at x = L2 + M1wL wx M ; M 1 2 Mx = 2 (L ; x) + x ; M1 L s
b x
L2 ; M1 + M2 + M1 ; M2 2 4 wl w wx 3 M 4 M 1 3 = 24EI x ; 2L + wL ; wL2 x2 + 12wM1 x + L3 ; 8Mw1 L ; 4Mw2L
=
23) Simple Beam; Concentrated Force, End Moments R1 = V1 R2 = V2 at x = L2 M3 when x < L2 Mx when L2 < x Mx when x < L2 x
Victor Saouma
P + M1 ; M2 LM P2 ; M1 ; 2 2PL ML+ M ; 1 2 2 4 M2 x ; M = P2 + M1 ; 1 L M2 x ; M = P2 (L ; x) + M1 ; 1 L 2 Px = 48EI 3L ; 4x2 8( L ; x ) ; PL (M1 (2L ; x) + M2(L + x)) = = =
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{37
24) Beam Overhanging one Support; Uniform Load
at 0 < x < L at L < x< L + a at x = L2 1 ; La22 at x = L
at 0 < x < L at L < x < L + a at 0 < x < L x at L < x < L + a x1 R1 = V1 R2 = V2 + V3 at 0 < x < L at L < x< L + a 2 at x = L2 1 ; La 2 at x = L
V2 V3 Vx Vx1 M1 M2 Mx Mx1
at 0 < x < L at L < x < L + a at 0 < x < L x at L < x < L + a x1
5.4 Flexure
= = = = = = = = = = = = = =
R1 = V1 R2 = V2 + V3 V2 V3 Vx Vx1 M1 M2 Mx Mx1
w (L2 ; a2 ) 2wL 2 2L (L + a) wa w (L2 + a2 ) 2L R1 ; wx w(a ; x1 ) w 2 2 8L22 (L + a) (L ; a) = wa2 2 2 = wx 2wL (L ; a2 ; xL) = (a ; x ) = = = = = = =
2
wx (L4 ; 2L2x2 + Lx3 ; 2a2 L2 + 2a2 x2 ) 24 wxEIL 1 (4a2 L; L3 + 6a2 x ; 4ax2 + x3 ) 1 1 1 24 EI w (L2 ; a2 ) 2wL 2 2 L ( L + a) wa w (L2 + a2 ) 2L R1 ; wx w(a ; x1 ) w 2 2 8L22 (L + a) (L ; a) wa wx2 (L2 ; a2 ; xL) w2L(a ; x )2 2 wx 1 (L4 ; 2L2x2 + Lx3 ; 2a2 L2 + 2a2 x2 ) 24 EIL wx1 (4a2 L; L3 + 6a2x ; 4ax2 + x3 ) 1 1 1 24EI
1
5.4.1 Basic Kinematic Assumption; Curvature Fig.5.14 shows portion of an originally straight beam which has been bent to the radius by end couples M . support conditions, Fig. 5.1. It is assumed that plane crosssections normal to the length of the unbent beam remain plane after the beam is bent. 62 Except for the neutral surface all other longitudinal bers either lengthen or shorten, thereby creating a longitudinal strain "x. Considering a segment EF of length dx at a distance y from the neutral axis, its original length is EF = dx = d (5.11)
61
Victor Saouma
Structural Concepts and Systems for Architects
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REVIEW of STATICS O +ve Curvature, +ve bending
dθ ve Curvature, ve Bending
ρ M
M
Neutral Axis F’
E’
Y dA
E
F
X
Z
dx
Figure 5.14: Deformation of a Beam un Pure Bending and 63
d = dx
(5.12)
To evaluate this strain, we consider the deformed length E 0 F 0
E 0 F 0 = ( ; y)d = d ; yd = dx ; y dx
(5.13)
The strain is now determined from:
or after simpli cation
0 F 0 ; EF dx ; y dx E ; dx "x = EF = dx
(5.14)
"x = ; y
(5.15)
where y is measured from the axis of rotation (neutral axis). Thus strains are proportional to the distance from the neutral axis. 64 (Greek letter rho) is the radius of curvature. In some textbook, the curvature (Greek letter kappa) is also used where = 1 (5.16) thus,
"x = ;y
Victor Saouma
(5.17)
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{39
5.4.2 StressStrain Relations 65 So far we considered the kinematic of the beam, yet later on we will need to consider equilibrium in terms of the stresses. Hence we need to relate strain to stress. 66 For linear elastic material Hooke's law states
x = E"x
(5.18)
where E is Young's Modulus. 67 Combining Eq. with equation 5.17 we obtain
x = ;Ey
(5.19)
5.4.3 Internal Equilibrium; Section Properties Just as external forces acting on a structure must be in equilibrium, the internal forces must also satisfy the equilibrium equations. 69 The internal forces are determined by slicing the beam. The internal forces on the \cut" section must be in equilibrium with the external forces.
68
5.4.3.1 Fx = 0; Neutral Axis The rst equation we consider is the summation of axial forces. 71 Since there are no external axial forces (unlike a column or a beamcolumn), the internal axial forces must be in equilibrium. Z Fx = 0 ) x dA = 0 (5.20) 70
A
where x was given by Eq. 5.19, substituting we obtain Z
Z
A
x dA = ;
A
EydA = 0
(5.21a)
But since the curvature and the modulus of elasticity E are constants, we conclude that Z
A
ydA = 0
(5.22)
or the rst moment of the cross section with respect to the z axis is zero. Hence we conclude that the
neutral axis passes through the centroid of the cross section. 5.4.3.2 M = 0; Moment of Inertia
The second equation of internal equilibrium which must be satis ed is the summation of moments. However contrarily to the summation of axial forces, we now have an external moment to account for,
72
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{40
REVIEW of STATICS
the one from the moment diagram at that particular location where the beam was sliced, hence Z
Mz = 0; ;+ve; {z} M = ; x ydA Ext.  A{z } Int. where dA is an dierential area a distance y from the neutral axis. 73 Substituting Eq. 5.19 9 Z Z M = ; x ydA = M = E y2 dA A ; A x = ;Ey 74
(5.24)
We now pause and de ne the section moment of inertia with respect to the z axis as
I def = and section modulus as
75
(5.23)
Z
A
y2 dA
(5.25)
S def = Ic
(5.26)
Section properties for selected sections are shown in Table 5.3.
5.4.4 Beam Formula We now have the ingredients in place to derive one of the most important equations in structures, the beam formula. This formula will be extensively used for design of structural components. 77 We merely substitute Eq. 5.25 into 5.24,
76
M = E I =
Z
Z
a
A
y2 dA
y2 dA
9 > = > ;
M 1 EI = =
(5.27)
which shows that the curvature of the longitudinal axis of a beam is proportional to the bending moment M and inversely proportional to EI which we call exural rigidity. 78 Finally, inserting Eq. 5.19 above, we obtain x = ;Ey = ; My (5.28) x M I = EI Hence, for a positive y (above neutral axis), and a positive moment, we will have compressive stresses above the neutral axis. 79 Alternatively, the maximum ber stresses can be obtained by combining the preceding equation with Equation 5.26
x = ; M S
(5.29)
Example 516: Design Example Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{41
Y
Y
A x y Ix Iy
x
h
X y
= = = = =
x
bh
b 2h 2 bh3 12 hb3
h h’
A x y Ix Iy
X y
12
b’ b
b
= = = = =
bh ; b0 h0 b
2h 2
bh3 ;b0 h03 12 hb3 ;h0 b03 12
c
Y
Y
a
h
A = h(a2+b) y = h3((2aa++bb)) 3 a2 +4ab+b2 Ix = h (36( a+b)
X y
x
h
X y b
b
A x y Ix Iy
= = = = =
bh b2+c h3 3 bh3 36 2 bh (b ; bc + c2 ) 36
Y
Y
r X
A = r42 = d4 42 Ix = Iy = r4 = d64
t
r
X
A = 2rt = dt3 Ix = Iy = r3 t = d8 t
Y
b X b a
a
A = ab3 Ix = ab3 3 Iy = ba4 Table 5.3: Section Properties
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{42
REVIEW of STATICS
A 20 ft long, uniformly loaded, beam is simply supported at one end, and rigidly connected at the other. The beam is composed of a steel tube with thickness t = 0:25 . Select the radius such that max 18 , and max L=360. in
ksi
1 k/ft r
0.25’
20’
Solution: wL , and I = r3 t. 1. Steel has E = 29; 000 , and from above Mmax = wL8 , max = 185 EI 2. The maximum moment will be 2 (1) (20)2 2 = 50 Mmax = wL = 8 8 2
ksi
4
k/ft
(5.30)
ft
k.ft
3. We next seek a relation between maximum de ection and radius wL4 max = 185 3EI
I = r t
= = =
wL4
185Er3 t 4 4 3 3 3 (1) k/ft(20) ft (12) in = ft (185)(29;000) ksi(3:14)r3 (0:25) in 65:365
(5.31)
r
4. Similarly for the stress
= MS S = Ir I = r3 t
9 = ;
= rM2 t (50)
= (3:14)r2(12) (0:25) = 764 r2 k.ft
(5.32)
in/ft in
5. We now set those two values equal to their respective maximum
L = (20) (12) max = 360 360 ft
max = (18)
ksi
in/ft
= 0:67 = 65r:365 ) r = in
= 764 r2 ) r =
r
764 = 6.51 18
in
r 3
65:65 = 4:61 0:67
in
(5.33a) (5.33b)
5.4.5 Approximate Analysis M = = 1 ), we recall that that the moment is directly proportional From Fig. 5.14, and Eq. 5.27 ( EI to the curvature . 81 Thus, 80
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{43
1. A positive and negative moment would correspond to positive and negative curvature respectively (adopting the sign convention shown in Fig. 5.14). 2. A zero moment correspnds to an in ection point in the de ected shape. 82
Hence, for
Statically determinate structure, we can determine the de ected shape from the moment diagram, Fig. 5.15.
Figure 5.15: Elastic Curve from the Moment Diagram
Statically indeterminate structure, we can: 1. 2. 3. 4.
Plot the de ected shape. Identify in ection points, approximate their location. Locate those in ection points on the structure, which will then become statically determinate. Perform an approximate analysis.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{44
REVIEW of STATICS
Figure 5.16: Approximate Analysis of Beams
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{45
Example 517: Approximate Analysis of a Statically Indeterminate beam Perform an approximate analysis of the following beam, and compare your results with the exact solution.
20k 16’ 12’ 28’
28’
Solution: 20k 16’ 12’ 28’
22’
6’
28’
28’
20k Approximate Location of IP A
B 22’
C 6’
D 28’
1. We have 3 unknowns RA , RC , and RD , all in the vertical directions, and only two applicable equations of equilibrium (since we do not have any force in the x direction), thus the problem is statically indeterminare. 2. We sketch the anticipated de ected shape, and guess the location of the in ection point. 3. At that location, we place a hinge, and we now have an additional equation of condition at that location (M = 0).
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{46
REVIEW of STATICS
4. If we consider AB, and take the moments with respect to point B: (+ ;) MB = 0; (22)(RA ) ; (20)(22 ; 16) = 0 ) RA = 5.45
k
6
(5.34a)
5. If we now consider the entire beam:
(+ ;) MD = 0; (RA )(28 + 28) ; (20)(28 + 12) + (RC )(28) = 0 (5:45)(56) ; (20)(40) + (RC )(28) = 0 ) RC = 17.67 6 (+ 6) Fy = 0; RA ; 20 + Rc + RD = 0 5:45 ; 20 + 17:67 + RD = 0 ) RD = ;3:12 3.12 ? k
k
6. Check
(+ ;) MA = 0; (20)(16) ; (RC )(28) + (RD )(28 + 28) = 320 ; (17:67)(28) + (3:12)(56) = p 320 ; 494:76 + 174:72 = 0
(5.36a)
7. The moments are determined next
Mmax = RA a = (5:45)(16) = 87.2 M1 = RD L = (3:12)(28) = 87.36
(5.37a) (5.37b)
8. We now compare with the exact solution from Section 5.3.2, solution 21 where:L = 28, a = 16, b = 12, and P = 20 2 R1 = RA = 4Pb 4 L ; a ( L + a ) 3 L
2 ; (16)(28 + 16) = 6.64 = (20)(12) 4(28) 3 4(28) Pa R2 = RB = 2L3 2L2 + b(L + a) 2 + 12(28 + 16) = 15.28 2(28) = (20)(16) 3 2(28) R3 = RD = ; Pab 4L3 (L + a) = ; (20)(16)(12) 4(28)3 (28 + 16) = ;1.92 Mmax = R1 a = (6:64)(16) = 106.2 M1 = R3 L = (1:92)(28) = 53.8
(5.38a) (5.38b) (5.38c) (5.38d) (5.38e) (5.38f) (5.38g)
9. If we tabulate the results we have
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5.4 Flexure
5{47
Value Approximate Exact % Error RA 5.45 6.64 18 RC 17.67 15.28 16 RD 3.12 1.92 63 M1 87.36 53.8 62 Mmax 87.2 106.2 18 10. Whereas the correlation between the approximate and exact results is quite poor, one should not underestimate the simplicity of this method keeping in mind (an exact analysis of this structure would have been computationally much more involved). Furthermore, often one only needs a rough order of magnitude of the moments.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 5{48
REVIEW of STATICS
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 6
Case Study II: GEORGE WASHINGTON BRIDGE 6.1 Theory Whereas the forces in a cable can be determined from statics alone, its con guration must be derived from its deformation. Let us consider a cable with distributed load p(x) per unit horizontal projection of the cable length (thus neglecting the weight of the cable). An in nitesimal portion of that cable can be assumed to be a straight line, Fig. 6.1 and in the absence of any horizontal load we have
1
T V
H
V θ
w
x w(x)
H
v(x)
dv
ds
v
dx L
dx
θ V+dV
H T+dT
Figure 6.1: Cable Structure Subjected to p(x)
H =constant. Summation of the vertical forces yields (+ ?) Fy = 0 ) ;V + wdx + (V + dV ) = 0 (6.1a) dV + wdx = 0 (6.1b) where V is the vertical component of the cable tension at x (Note that if the cable was subjected to its own weight then we would have wds instead of wdx). Because the cable must be tangent to T , we have V tan = H (6.2)
Substituting into Eq. 6.1b yields
d (H tan ) = w d(H tan ) + wdx = 0 ) ; dx
(6.3)
Draft 6{2 2
Case Study II: GEORGE WASHINGTON BRIDGE
But H is constant (no horizontal load is applied), thus, this last equation can be rewritten as
d (tan ) = w ;H dx
(6.4)
dv which when substituted in Eq. 6.4 yields Written in terms of the vertical displacement v, tan = dx the governing equation for cables ;Hv00 = w (6.5)
3
4 For a cable subjected to a uniform load w , we can determine its shape by double integration of Eq. 6.5
;Hv0 = wx + C1 2 ;Hv = wx2 + C1 x + C2
(6.6a) (6.6b)
and the constants of integrations C1 and C2 can be obtained from the boundary conditions: v = 0 at x = 0 and at x = L ) C2 = 0 and C1 = ; wL 2 . Thus v = 2wH x(L ; x) (6.7) This equation gives the shape v(x) in terms of the horizontal force H , L ) we can solve for the horizontal force 5 Since the maximum sag h occurs at midspan (x = 2 2
H = wL 8h
(6.8)
we note the analogy with the maximum moment in a simply supported uniformly loaded beam M = Hh = wL8 2 . Furthermore, this relation clearly shows that the horizontal force is inversely proportional to the sag h, as h & H %. Finally, we can rewrite this equation as
r def = Lh wL = 8r H 6
(6.9a) (6.9b)
Eliminating H from Eq. 6.7 and 6.8 we obtain
2 v = 4h ; Lx 2 + Lx
(6.10)
Thus the cable assumes a parabolic shape (as the moment diagram of the applied load). 7 Whereas the horizontal force H is constant throughout the cable, the tension T is not. The maximum tension occurs at the support where the vertical component is equal to V = wL 2 and the horizontal one to H , thus s s 2 2 p wL 2 = H 1 + wL=2 Tmax = V 2 + H 2 = + H (6.11) 2 H
Victor Saouma
Structural Concepts and Systems for Architects
Draft 6.2 The Case Study
6{3
Combining this with Eq. 6.8 we obtain1 . p
Tmax = H 1 + 16r2 H (1 + 8r2 )
(6.12)
Had we assumed a uniform load w per length of cable (rather than horizontal projection), the equation would have been one of a catenary2.
8
w L v=H w cosh H 2 ; x
(6.13)
+h
The cable between transmission towers is a good example of a catenary.
6.2 The Case Study Adapted from (Billington and Mark 1983)
The George Washington bridge, is a suspension bridge spanning the Hudson river from New York City to New Jersey. It was completed in 1931 with a central span of 3,500 ft (at the time the world's longest span). The bridge was designed by O.H. Amman, who had emigrated from Switzerland. In 1962 the deck was stiened with the addition of a lower deck.
9
6.2.1 Geometry A longitudinal and plan elevation of the bridge is shown in For simplicity we will assume in our analysis that the two approaching spans are equal to 650 ft. 11 There are two cables of three feet diameter on each side of the bridge. The centers of each pair are 9 ft apart, and the pairs themselves are 106 ft apart. We will assume a span width of 100 ft. 12 The cables are idealized as supported by rollers at the top of the towers, hence the horizontal components of the forces in each side of the cable must be equal (their vertical components will add up). 13 The cables support the road deck which is hungby suspenders attached at the cables. The cables are made of 26,474 steel wires, each 0.196 inch in diameter. They are continuous over the tower supports and are rmly anchored in both banks by huge blocks of concrete, the anchors. I 14 Because the cables are much longer than they are thick (large L ), they can be idealized a perfectly
exible members with no shear/bending resistance but with high axial strength. 15 The towers are 578 ft tall and rest on concrete caissons in the river. Because of our assumption regarding the roller support for the cables, the towers will be subjected only to axial forces. 10
6.2.2 Loads The dead load is composed of the weight of the deck and the cables and is estimated at 390 and 400 psf respectively for the central and side spans respectively. Assuming an average width of 100 ft, this 2 3 Recalling that (a + b)n = an + nan; b + n n; an; b + or (1 + b)n = 1 + nb + n n; b + n n; n; b + ; p 1 Thus for b << 1, 1 + b = (1 + b) 2 1 + b
16
1
(
1
1)
2 2
2!
(
1)
2!
(
1)(
2)
3!
2
2
Derivation of this equation is beyond the scope of this course.
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Case Study II: GEORGE WASHINGTON BRIDGE
?? 377 ft
610 ft
327 ft
3,500 ft
650 ft
4,760 ft ELEVATION
N.J.
HUDSON RIVER
N.Y.
PLAN
Figure 6.2: Longitudinal and Plan Elevation of the George Washington Bridge
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would be equivalent to
DL = (390) (100) (1; 000) k
psf
ft
lbs
= 39
(6.14)
k/ft
for the main span and 40 for the side ones. 17 For highway bridges, design loads are given by the AASHTO (Association of American State Highway Transportation Ocials). The HS20 truck is often used for the design of bridges on main highways, Fig. 6.3. Either the design truck with speci ed axle loads and spacing must be used or the equivalent uniform load and concentrated load. This loading must be placed such that maximum stresses are produced. k/ft
Figure 6.3: Truck Load 18
With two decks, we estimate that there is a total of 12 lanes or
LL = (12)Lanes(:64) = =Lane = 7:68 k
ft
k/ft
8
k/ft
(6.15)
We do not consider earthquake, or wind loads in this analysis. 19 Final DL and LL are, Fig. 6.4: TL = 39 + 8 = 47 k/ft
6.2.3 Cable Forces 20
The thrust H (which is the horizontal component of the cable force) is determined from Eq. 6.8 2 H = wL8hcs
(3;500) = (47) (8)(327) = 220; 000 k/ft
2
ft
2
ft
k
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wD,S = 40 k/ft
Case Study II: GEORGE WASHINGTON BRIDGE
wD = 39 k/ft
wD,S = 40 k/ft
DEAD LOADS
wL = 8 k/ft
Figure 6.4: Dead and Live Loads
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From Eq. 6.12 the maximum tension is r = Tmax = = =
h = 327 = 0:0934 Lcs p 3;500
H 1 + 16pr2 (2; 200) 1 + (16)(0:0934)2 (2; 200) (1:0675) = 235,000 k
k
k
6.2.4 Reactions 21
Cable reactions are shown in Fig. 6.5. POINTS WITH REACTIONS TO CABLES
Figure 6.5: Location of Cable Reactions 22
The vertical force in the columns due to the central span (cs) is simply the support reaction, 6.6 wTOT = 39 + 8 = 47 k/ft B
A
REACTIONS AT TOP OF TOWER
POINT OF NO MOMENT
L = 3,500 FT
Figure 6.6: Vertical Reactions in Columns Due to Central Span Load
Vcs = 21 wLcs = 12 (47)
(3; 500) = 82; 250
k/ft
ft
k
Note that we can check this by determining the vector sum of H and V which should be equal to Tmax : p p Vcs + H = (82; 250) + (220; 000) = 235; 000 kp 2
2
2
2
(6.16) (6.17)
Along the side spans (ss), the total load is TL = 40 + 8 = 48 . We determine the vertical reaction by taking the summation of moments with respect to the anchor: (6.18a) MD = 0; ;+; hssH + (wss Lss) L2ss ; Vss Lss = 0
23
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= (377) (220; 000) + (48) k
24
k
(650) (650) 2 ; 650Vss = 0 Vss = 143; 200 ft
k/ft
ft
Hence the total axial force applied on the column is V = Vcs + Vss = (82; 250) + (143; 200) = 225,450 k
25
k
k
(6.18b) (6.18c)
(6.19)
k
The vertical reaction at the anchor is given by summation of the forces in the y direction, Fig. 6.7: (6.20a) (+ 6) Fy = 0; (wss Lss) + Vss + Ranchor = 0 ;(48) (650) + (143; 200) + Ranchor = 0 (6.20b) Ranchor = 112,000 ? (6.20c) (6.20d) k/ft
ft
k
k
225,450 k 220,000 k
112,000 k
Figure 6.7: Cable Reactions in Side Span 26
The axial force in the side cable is determined the vector sum of the horizontal and vertical reactions. q p ss 2 Tanchor = Ranchor + H 2 = (112; 000)2 + (220; 000)2 = 247; 000 (6.21a) q p ss Ttower = Vss2 + H 2 = (143; 200)2 + (220; 000)2 = 262,500 (6.21b) k
k
27
The cable stresses are determined last, Fig. 6.8: 2 (3:14)(0:196)2 Awire = D = 0:03017 2 4 = 4 Atotal = (4)cables(26; 474)wires/cable(0:03017) (220; 000) = 68:75 Central Span = H = A (3; 200) 2 T ss (262; 500) 2 = 82 ss = tower Side Span Tower tower = A (3; 200) 2 T ss (247; 000) 2 = 77:2 ss = Side Span Anchor tower = anchor A (3; 200) 2 in
k
ksi
in
in
ksi
in
in
ksi
in
Victor Saouma
(6.22a) 2 =wire = 3; 200 (6.22b) 2
in
in
(6.22c) (6.22d) (6.22e)
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73.4 ksi
81.9 ksi
68.75 ksi
77.2 ksi
Figure 6.8: Cable Stresses If the cables were to be anchored to a concrete block, the volume of the block should be at least equal (1;000) = = 747; 000 3 or a cube of approximately 91 ft to V = (112;000) 150 = 3
28
k
lbs
lbs ft
k
ft
The deck, for all practical purposes can be treated as a continuous beam supported by elastic springs with stiness K = AL=E (where L is the length of the supporting cable). This is often idealized as a beam on elastic foundations, and the resulting shear and moment diagrams for this idealization are shown in Fig. 6.9.
29
K=AL/E
Shear
Moment
Figure 6.9: Deck Idealization, Shear and Moment Diagrams
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Draft Chapter 7
A BRIEF HISTORY OF STRUCTURAL ARCHITECTURE If I have been able to see a little farther than some others, it was because I stood on the shoulders of giants. Sir Isaac Newton
1 More than any other engineering discipline, Architecture/Mechanics/Structures is the proud outcome of a of a long and distinguished history. Our profession, second oldest, would be better appreciated if we were to develop a sense of our evolution.
7.1 Before the Greeks Throughout antiquity, structural engineering existing as an art rather than a science. No record exists of any rational consideration, either as to the strength of structural members or as to the behavior of structural materials. The builders were guided by rules of thumbs and experience, which were passed from generation to generation, guarded by secrets of the guild, and seldom supplemented by new knowledge. Despite this, structures erected before Galileo are by modern standards quite phenomenal (pyramids, Via Appia, aqueducs, Colisseums, Gothic cathedrals to name a few). 3 The rst structural engineer in history seems to have been Imhotep, one of only two commoners to be dei ed. He was the builder of the step pyramid of Sakkara about 3,000 B.C., and yielded great in uence over ancient Egypt. 4 Hamurrabi's code in Babylonia (1750 BC) included among its 282 laws penalties for those \architects" whose houses collapsed, Fig. 7.1. 2
7.2 Greeks The greek philosopher Pythagoras (born around 582 B.C.) founded his famous school, which was primarily a secret religious society, at Crotona in southern Italy. At his school he allowed neither textbooks nor recording of notes in lectures, on pain of death. He taught until the age of 95, and is
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228. If a builder build a house for some one and complete it, he shall give him a fee of two shekels in money for each sar of surface. 229 If a builder build a house for some one, and does not construct it properly, and the house which he built fall in and kill its owner, then that builder shall be put to death. 230. If it kill the son of the owner the son of that builder shall be put to death. 231. If it kill a slave of the owner, then he shall pay slave for slave to the owner of the house. 232. If it ruin goods, he shall make compensation for all that has been ruined, and inasmuch as he did not construct properly this house which he built and it fell, he shall reerect the house from his own means. 233. If a builder build a house for some one, even though he has not yet completed it; if then the walls seem toppling, the builder must make the walls solid from his own means.
Figure 7.1: Hamurrabi's Code reported to have coined the term mathematics which means literally the \science of learning" (and also the word philosopher meaning \one who loves wisdom"). 6 Aristotle (384322 B.C.) was Dean of the Lyceum, a college just outside the city gates of Athens, and was a man of universal ability. He is credited with having written in more than 25 dierent elds of knowledge. One of the most in uential men of early civilization. 7 A pupil of Aristotle was Alexander the Great (356323 B.C.) who founded the city of Alexandria in 323. Upon his death, one of his generals Ptolemy I became Pharaoh and established a library. The library of Alexandria was founded with the private library of Aristotle as a nucleus, and later became the largest of the ancient world, containing about 700,000 scrolls. Many of these scrolls were subsequently brought to the attention of the western world through translations by the arabs. 8 Alexandria was also the seat of the rst university (with a reported enrollment of 14,000 students), and its rst professor of geometry was Euclid (315250 B.C.). 9 The greatest of the Greeks was Archimedes (287212) who was one of the greatest physicist of the ancient world and one of its greatest mathematician, Fig. 7.2. He is considered by many as the founder of mechanics because of his treatise \On Equilibrium". He introduced the concept of center of gravity. He refused to write about \practical stu" such as machines, catapults, spiral pumps, and others. It was one such invention (the lens) which kept the Roman armies at bay outside Syracuse for three years. When the city fell, he was supposed to have had his life spared. But the circumstances of his subsequent death are obscure. By some accounts he was killed by an ignorant soldier who disobeyed orders, and by other he was slain because he was too busy solving a mathematical problem to appear in front of the Roman consul and conqueror of Syracuse.
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Figure 7.2: Archimed
7.3 Romans Science made much less progress under the Romans than under the Greeks. The Romans apparently were more practical, and were not as interested in abstract thinking though they were excellent ghters and builders. 11 As the roman empire expanded, the Romans built great roads (some of them still in use) such as the Via Appia, Cassia, Aurelia; Also they built great bridges (such as the third of a mile bridge over the Rhine built by Caesars), and stadium (Colliseum). 12 One of the most notable Roman construction was the Pantheon, Fig. 7.3. It is the bestpreserved
10
Figure 7.3: Pantheon major edi ce of ancient Rome and one of the most signi cant buildings in architectural history. In shape it is an immense cylinder concealing eight piers, topped with a dome and fronted by a rectangular colonnaded porch. The great vaulted dome is 43 m (142 ft) in diameter, and the entire structure is lighted through one aperture, called an oculus, in the center of the dome. The Pantheon was erected by the Roman emperor Hadrian between AD 118 and 128. 13 Marcus Vitruvius Pollio (70?25 BC) was a Roman architect and engineer. He was an artillery engineer in the service of the rst Roman emperor, Augustus. His Ten Books on Architecture (Vitruvius
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1960) is the oldest surviving work on the subject and consists of dissertations on a wide variety of subjects relating to architecture, engineering, sanitation, practical hydraulics, acoustic vases, and the like. Much of the material appears to have been taken from earlier extinct treatises by Greek architects. Vitruvius's writings have been studied ever since the Renaissance as a thesaurus of the art of classical Roman architecture, Fig. 7.4.
Figure 7.4: From Vitruvius Ten Books on Architecture, (Vitruvius 1960)
7.4 The Medieval Period (4771492) This period, also called the Dark Ages, was marked by a general decline of civilization throughout Europe following the decline and fall of the western Roman Empire. 15 The eastern Roman Empire on the other hand was to continue, and the center of Greek life had by then been transferred to Constantinople. This city exerted great in uence throughout Asia Minor. 16 Hagia Sophia, also Church of the Holy Wisdom, Fig. 7.5, was the most famous Byzantine structure in Constantinople (now Istanbul). Built (53237) by Emperor Justinian I, its huge size and daring technical innovations make it one of the world's key monuments. The size of its dome though, 112 ft, was nevertheless smaller than the one of the Pantheon in Rome. 17 During that period, the Arabs carried the torch of knowledge, gave birth to algebra, translated some of the great books of the Library of Alexandria. 18 Architecture, was the most important and original art form during the Gothic period, (Anon. xx). The principal structural characteristics of Gothic architecture arose out of medieval masons' eorts to solve the problems associated with supporting heavy masonry ceiling vaults over wide spans. The problem was that the heavy stonework of the traditional arched barrel vault and the groin vault exerted a tremendous downward and outward pressure that tended to push the walls upon which the vault rested outward,
14
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Figure 7.5: Hagia Sophia thus collapsing them. A building's vertical supporting walls thus had to be made extremely thick and heavy in order to contain the barrel vault's outward thrust. Medieval masons solved this dicult problem about 1120 with a number of brilliant innovations. First and foremost they developed a ribbed vault, in which arching and intersecting stone ribs support a vaulted ceiling surface that is composed of mere thin stone panels. This greatly reduced the weight (and thus the outward thrust) of the ceiling vault, and since the vault's weight was now carried at discrete points (the ribs) rather than along a continuous wall edge, separate widely spaced vertical piers to support the ribs could replace the continuous thick walls. The round arches of the barrel vault were replaced by pointed (Gothic) arches which distributed thrust in more directions downward from the topmost point of the arch. Since the combination of ribs and piers relieved the intervening vertical wall spaces of their supportive function, these walls could be built thinner and could even be opened up with large windows or other glazing. A crucial point was that the outward thrust of the ribbed ceiling vaults was carried across the outside walls of the nave, rst to an attached outer buttress and then to a freestanding pier by means of a half arch known as a ying buttress. The ying buttress leaned against the upper exterior of the nave (thus counteracting the vault's outward thrust), crossed over the low side aisles of the nave, and terminated in the freestanding buttress pier, which ultimately absorbed the ceiling vault's thrust. These elements enabled Gothic masons to build much larger and taller buildings than their Romanesque predecessors and to give their structures more complicated ground plans. The skillful use of
ying buttresses made it possible to build extremely tall, thinwalled buildings whose interior structural system of columnar piers and ribs reinforced an impression of soaring verticality. 19 ViletLeDuc classical book, (le Duc 1977) provided an in depth study of Gothic architecture.
7.5 The Renaissance 20
During the Renaissance there was a major revival of interest in science and art.
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7.5.1 Leonardo da Vinci 14521519 21 Leonardo da Vinci was the most outstanding personality of that period (and of human civilization for that matter). He was not only a great artist (Mona Lisa), but also a great scientist and engineer. 22 He did not write books, but much information was found in his notebooks, one the most famous (Codex Leicester) was recently purchased by Bill Gates whose company Corbis made a CDROM from it. 23 He was greatly interested in mechanics, (Timoshenko 1982), and in one of his notes he states \Mechanics is the Paradise of mathematical science because here we come to the fruits of mathematics."
He was the rst to explore concepts of mechanics, since Archimedes, using a scienti c approach. He applied the principle of virtual displacements to analyze various systems of pulleys and levers. He appears to have developped a correct idea of the thrust produced by an arch. 25 In one of his manuscripts there is a sketch of two members on which a vertical load Q is acting and the question is asked: What forces are needed at a and b to have equilibrium? (From the dotted line parallelogram, in the sketch, it can be concluded that he had the right answer). 26 Leonardo also studied the strength of structural materials experimentally. He tried to determine the tensile strength of an iron wire of dierent length (size eect). 27 He also studied the load carrying capacity of a simply supported uniformly loaded beam and concluded that \the strength of the beam supported at both ends varies inversely as the length and directly as the width" (is this correct? how about the depth of the beam?). 28 For a cantilevered beam he stated \If a beam 2 braccia long supports 100 libre, a beam 1 braccia long will support 200" Finally, Leonardo brie y studied the strength of columns and found that \it varies inversely as its length, but directly as some ratio of its cross section." 24
Leonardo's was the rst indeed to attempt to apply statics in nding the forces acting in members of structures, friction and the rst to perform experiments to determine the strength of structural materials. 30 Interestingly, here is Leonardo's de nition of force, (Penvenuto 1991) \I say that force is a spiritual virtue, an invisible power, which, through accidental exterior violence, is caused by motion and placed and infused into bodies which are [thus] removed and deviated from their natural use, giving to such virtue an active life of marvelous power". 29
Unfortunatly, these important ndings, were buried in his notes, and engineers in the fteenth and sixteenth centuries continued, as in the Roman era, to x dimensions of structural elements by relying on experience and judgment.
31
7.5.2 Brunelleschi 13771446 Brunelleschi was a Florentine architect and one of the initiators of the Italian Renaissance. His revival of classical forms and his championing of an architecture based on mathematics, proportion, and perspective make him a key artistic gure in the transition from the Middle Ages to the modern era.
32
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33 He was born in Florence in 1377 and received his early training as an artisan in silver and gold. In 1401 he entered, and lost, the famous design competition for the bronze doors of the Florence Baptistery. He then turned to architecture and in 1418 received the commission to execute the dome of the un nished Gothic Cathedral of Florence, also called the Duomo. The dome, Fig. 7.6 a great innovation both
Figure 7.6: Florence's Cathedral Dome artistically and technically, consists of two octagonal vaults, one inside the other. Its shape was dictated by its structural needs one of the rst examples of architectural functionalism. Brunelleschi made a design feature of the necessary eight ribs of the vault, carrying them over to the exterior of the dome, where they provide the framework for the dome's decorative elements, which also include architectural reliefs, circular windows, and a beautifully proportioned cupola. This was the rst time that a dome created the same strong eect on the exterior as it did on the interior. 34 Completely dierent from the emotional, elaborate Gothic mode that still prevailed in his time, Brunelleschi's style emphasized mathematical rigor in its use of straight lines, at planes, and cubic spaces. This set the tone for many of the later buildings of the Florentine Renaissance. 35 His in uence on his contemporaries and immediate followers was very strong and has been felt even in the 20th century, when modern architects came to revere him as the rst great exponent of rational architecture.
7.5.3 Alberti 14041472 Alberti was an Italian architect and writer, who was the rst important art theorist of the Renaissance and among the rst to design buildings in a pure classical style based on a study of ancient Roman architecture.
36
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He was born in Genoa, the son of a Florentine noble. He received the best education available in the 15th century. He was pro cient in Greek, mathematics, and the natural sciences. As a poet, a philosopher, and one of the rst organists of his day, Alberti greatly in uenced his contemporaries. 38 Alberti's architectural training began with the study of antique monuments during his rst stay in Rome. Subsequently he joined the papal court in Florence, where he became intensely involved with the cultural life of the city. Probably at this time he became familiar with the mathematical laws of linear perspective, which Brunelleschi had studied. 39 Alberti took took an active part in the literary life of Florence and championed the literary use of Italian rather than the use of Latin. 40 In the late 1440s, Alberti began to work as an architect. Although his buildings rank among the best architecture of the Renaissance, he was a theoretical rather than a practical architect. He furnished the plans of his buildings but never supervised their construction. His De Re Aedi catoria (1485) was the rst printed work on architecture of the Renaissance. He also wrote books on sculpture, the family, government, and literature.
37
7.5.4 Palladio 15081580 Andrea Palladio was one of the most in uential architects in European history. He was born in Padua, and trained as a stonemason, he later joined the poet Trissino who took him to Rome, where Palladio studied and measured Roman architectural ruins; he also absorbed the treatises of Vitruvius. One outcome of these studies was Palladio's Antiquities of Rome (1554) (Palladio 19xx), the principal guidebook to Roman ruins for the next two centuries. 42 In and near Vicenza he designed many residences and public buildings. The best known of these are the Barbarano, Chieregati, Tiene, Porto, and Valmarana palaces and the Villa Capri, or Villa Rotunda. Although the historical antecedents of Palladio's style are the classically Romanin uenced High Renaissance works of architects such as the Italian Donato Bramante, Palladio's own use of classical motifs came through his direct, extensive study of Roman architecture. He freely recombined elements of Roman buildings as suggested by his own building sites and by contemporary needs, Fig. 7.7 At the same time he shared the Renaissance concern for harmonious proportion, and his facades have a noteworthy simplicity almost austerity and repose. 43 Palladio was the author of an important scienti c treatise on architecture, I Quattro Libri dell'Architettura, (Palladio 19xx) which was widely translated and in uenced many later architects. Its precise rules and formulas were widely utilized, especially in England, and were basic to the Palladian style, adopted by Inigo Jones, Christopher Wren, and other English architects, which preceded and in uenced the neoclassical architecture of the Georgian Style.
41
7.5.5 Stevin Stevin, Fig. 7.8, was a Dutch mathematician and engineer who founded the science of hydrostatics by showing that the pressure exerted by a liquid upon a given surface depends on the height of the liquid and the area of the surface. 45 Stevin was a bookkeeper in Antwerp, then a clerk in the tax oce at Brugge. After this he moved to Leiden where he rst attended the Latin school, then he entered the University of Leiden in 1583 (at the age of 35). While quartermaster in the Dutch army, Stevin invented a way of ooding the lowlands in the path of an invading army by opening selected sluices in dikes. He was an outstanding engineer who 44
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Figure 7.7: Palladio's Villa Rotunda
Figure 7.8: Stevin
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built windmills, locks and ports. He advised the Prince Maurice of Nassau on building forti cations for the war against Spain. 46 The author of 11 books, Stevin made signi cant contributions to trigonometry, geography, forti cation, and navigation. Inspired by Archimedes, Stevin wrote important works on mechanics. In his book De Beghinselen der Weeghconst in 1586 appears the theorem of the triangle of forces giving impetus to statics. In 1586 (3 years before Galileo) he reported that dierent weights fell a given distance in the same time.
7.5.6 Galileo 15641642 Galilei Galileo was born in Pisa in 1564. He received his early education in Latin, Greek and logic near Florence, Fig. 7.9. Just as his father had played an important role in the musical revolution 47
Figure 7.9: Galileo from medieval polyphony to harmonic modulation, Galileo came to see Aristotelian physical theology as limiting scienti c inquiry. 48 In 1581 he entered the University of Pisa to study medicine, but he soon turned to philosophy and mathematics, leaving the university without a degree in 1585. For a time he tutored privately and wrote on hydrostatics and natural motions, but he did not publish. 49 In 1589 he became professor of mathematics at Pisa, where he is reported to have shown his students the error of Aristotle's belief that speed of fall is proportional to weight, by dropping two objects of dierent weight simultaneously from the Leaning Tower, thus modern dynamics was born. His contract was not renewed in 1592, probably because he contradicted Aristotelian professors. The same year, he was appointed to the chair of mathematics at the University of Padua, where he remained until 1610. 50 In Padua he achieved great fame, and lecture halls capable of containing 2,000 students from all over Europe were used. In 1592 he wrote Della Scienza Meccanica in which various problems of statics were treated using the principle of virtual displacement. He subsequently became interested in astronomy and built one of the rst telescope through which he saw Jupiter and became an ardent proponent of the Copernican theory (which stated that the planets circle the sun as opposed to the Aristotelian and Ptolemaic assumptions that it was the sun which was circling Earth). This theory being condemned by the church, he received a semiocial warning to avoid theology and limit himself to physical reasoning. When he published his books dealing with the two ways of regarding the universe (which clearly favored the Copernican theory) he was called to Rome by the Inquisition, condemned and had to read his recantation (At the end of his process he murmured the famous e pur se muove).
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When he was almost seventy years old, his life shattered by the Inquisition, he retired to his villa near Florence and wrote his nal book, Discourses Concerning Two New Sciences, (Galilei 1974), Fig. 7.10. His rst science was the study of the forces that hold objects together the dialogue taking place
51
Figure 7.10: Discourses Concerning Two New Sciences, Cover Page in a shipyard, triggered by observations of craftsmen building the Venetian eet. His second science concerned local motions  laws governing the trajectory of projectiles. A portion of the book dealing with the mechanical properties of structural materials and with the strength of beams. Strength of Materials as a discipline was born. 52 He observed that if we make structures geometrically similar, then with increase of the dimensions they become weaker and weaker, One cannot reason from the small to the large, because many mechanical devices succeed on a small scale that cannot exist in great size. 53 It is interesting to note that when Galileo studied the strength of a cantilevered (wooden) beam with an applied load at the end, Fig. 7.11, he failed to properly understand the exact internal stress/strain
Figure 7.11: \Galileo's Beam" distribution. He determined that the stress is constant throughout the cross section (whereas as we know it varies linearly). 54 Galileo's lifelong struggle to free scienti c inquiry from restriction by philosophical and theological interference stands beyond science. Since the full publication of Galileo's trial documents in the 1870s,
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entire responsibility for Galileo's condemnation has customarily been placed on the Roman Catholic church. This conceals the role of the philosophy professors who rst persuaded theologians to link Galileo's science with heresy. An investigation into the astronomer's condemnation, calling for its reversal, was opened in 1979 by Pope John Paul II. In October 1992 a papal commission acknowledged the Vatican's error.
7.6 Pre Modern Period, Seventeenth Century 7.6.1 Hooke, 16351703
Hooke was best known for his study of elasticity but also original contributions to many other elds of science. 56 Hooke was born on the Isle of Wight and educated at the University of Oxford. He served as assistant to the English physicist Robert Boyle and assisted him in the construction of the air pump. In 1662 Hooke was appointed curator of experiments of the Royal Society and served in this position until his death. He was elected a fellow of the Royal Society in 1663 and was appointed Gresham Professor of Geometry at Oxford in 1665. After the Great Fire of London in 1666, he was appointed surveyor of London, and he designed many buildings. 57 Hooke anticipated some of the most important discoveries and inventions of his time but failed to carry many of them through to completion. He formulated the theory of planetary motion as a problem in mechanics, and grasped, but did not develop mathematically, the fundamental theory on which Newton formulated the law of gravitation. 58 His most important contribution was published in 1678 in the paper De Potentia Restitutiva. It contained results of his experiments with elastic bodies, and was the rst paper in which the elastic properties of material was discused, Fig. 7.12.
55
Figure 7.12: Experimental Set Up Used by Hooke \Take a wire string of 20, or 30, or 40 ft long, and fasten the upper part thereof to a nail, and to the other end fasten a Scale to receive the weights: Then with a pair of compasses take the distance of the bottom of the scale from the ground or oor underneath, and set down the
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said distance,then put inweights into the said scale and measure the several stretchings of the said string, and set them down. Then compare the several strtchings of the said string, and you will nd that they will always bear the same proportions one to the other that the weights do that made them". This became Hooke's Law = E". 59 Because he was concerned about patent rights to his invention, he did not publish his law when rst discovered it in 1660. Instead he published it in the form of an anagram \ceiinosssttuu" in 1676 and the solution was given in 1678. Ut tensio sic vis (at the time the two symbos u and v were employed interchangeably to denote either the vowel u or the consonant v), i.e. extension varies directly with force.
7.6.2 Newton, 16421727 60
Born on christmas day in the year of Galileo's death, Newton, Fig. 7.13 was Professor of Mathematics
Figure 7.13: Isaac Newton at Cambridge university. 61 In 1684 Newton's solitude was interrupted by a visit from Edmund Halley, the British astronomer and mathematician, who discussed with Newton the problem of orbital motion. Newton had also pursued the science of mechanics as an undergraduate, and at that time he had already entertained basic notions about universal gravitation. As a result of Halley's visit, Newton returned to these studies. 62 During the following two and a half years, Newton established the modern science of dynamics by formulating his three laws of motion. Newton applied these laws to Kepler's laws of orbital motion formulated by the German astronomer Johannes Kepler and derived the law of universal gravitation. Newton is probably best known for discovering universal gravitation, which explains that all bodies in space and on earth are aected by the force called gravity. He published this theory in his book Philosophiae Naturalis Principia Mathematica or simply Principia, in 1687, Fig. 7.14. This book marked a turning point in the history of science. 63 The Principia's appearance also involved Newton in an unpleasant episode with the English philosopher and physicist Robert Hooke. In 1687 Hooke claimed that Newton had stolen from him a central
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Figure 7.14: Philosophiae Naturalis Principia Mathematica, Cover Page idea of the book: that bodies attract each other with a force that varies inversely as the square of their distance. However, most historians do not accept Hooke's charge of plagiarism. 64 Newton also engaged in a violent dispute with Leibniz over priority in the invention of calculus. Newton used his position as president of the Royal Society to have a committee of that body investigate the question, and he secretly wrote the committee's report, which charged Leibniz with deliberate plagiarism. Newton also compiled the book of evidence that the society published. The eects of the quarrel lingered nearly until his death in 1727. 65 In addition to science, Newton also showed an interest in alchemy, mysticism, and theology. Many pages of his notes and writings particularly from the later years of his career are devoted to these topics. However, historians have found little connection between these interests and Newton's scienti c work.
7.6.3 Bernoulli Family 16541782 The Bernouilli family originally lived in Antwerp, but because of religious persecution, they left Holland and settled in Basel. Near the end of the seventeenth century this family produced outstanding mathematicians for more than a hundred years. Jacob and John were brothers. John was the father of Daniel, and Euler his pupil. 67 Whereas Galileo (and Mariotte) investigated the strength of beams (Strength), Jacob Bernoulli (16541705) made calculation of their de ection (Stiness) and did not contribute to our knowledge of physical properties. Jacob Bernouilli is also credited in being the rst to to have assumed that a bf plane section of a beam remains plane during bending, but assumed rotation to be with respect to the lower ber (as Galileo did) and this resulted in an erroneous solution (where is the exact location of the axis of rotation?). He also showed that the curvature at any point along a beam is proportional to the curvature of the de ection curve. 68 Daniel Bernoulli (17001782) rst postulated that a force can be decomposed into its equivalent (\Potentiis quibuscunque possunt substitui earundem aequivalentes". Another hypothesis de ned the sum of two \conspiring" forces applied to the same point. According to Bernoulli, this \necessary truth" follows from the metaphysical principle that the whole equalts the sum of its parts, (Penvenuto 1991). 66
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7.6.4 Euler 17071783 Leonhard Euler was born in Basel and early on caught the attention of John Bernoulli whose teaching was attracting young mathematicians from all over Europe, Fig. 7.15. He obtained his Master
69
Figure 7.15: Leonhard Euler at age 16, and before the age of 20 won a competition from the French Academy of Sciences. At age 20 he moved to the Russian Academy of Sciences in St Petersburg along with the two sons of John Bernoulli (Nicholas and Daniel). He was appointed professor of physics in 1730 and professor of mathematics in 1733. In 1741 he became professor of mathematics at the Berlin Academy of Sciences at the urging of the Prussian king Frederick the Great. Euler returned to St. Petersburg in 1766, remaining there until his death. 70 Although hampered from his late 20s by partial loss of vision and in later life by almost total blindness, Euler produced a number of important mathematical works and hundreds of mathematical and scienti c memoirs. In his Introduction to the Analysis of In nities, Euler gave the rst full analytical treatment of algebra, the theory of equations, trigonometry, and analytical geometry. In this work he treated the series expansion of functions and formulated the rule that only convergent in nite series can properly be evaluated. He also discussed threedimensional surfaces and proved that the conic sections are represented by the general equation of the second degree in two dimensions. Other works dealt with calculus, including the calculus of variations, number theory, imaginary numbers, and determinate and indeterminate algebra. Euler, although principally a mathematician, made contributions to astronomy, mechanics, optics, and acoustics. 71 In Russia he wrote a famous book in mechanics in which instead of applying the geometrical methods used by Newton, he introduced analytical methods. 72 As a mathematician, Euler was interested principally in the geometrical forms of elastic curves. He approached problems from the point of view of variational calculus and in the introduction of his book Methodus inveniendi lineas curva ... he stated Since the fabric of the universe is most perfect, and is the work of a most wise Creator, nothing whatsover takes place in the universe in which some relation o maximum an minimum does not appear. Therefore there is absolutely no doubt that every eect in the universe can be explained as satisfactorily from nal causes, by the aid of the method of maxima and minima, as it can from the eective causes themselves... 73
Euler obtained a near exact expression for the de ection of a cantilever subjected to a point load, and
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for the buckling load of a column.
7.7 The preModern Period; Coulomb and Navier 74
Coulomb (17361806) was a French military engineer, Fig. 7.16, as was the rst to publish the correct
Figure 7.16: Coulomb analysis of the ber stresses in exed beam with rectangular cross section (Sur une Application des Regles de maximis et minimis a quelques problemes de statique relatifs a l'architecture in 1773). He used Hooke's law, placed the neutral axis in its exact position, developed the equilibrium of forces on the cross section with external forces, and then correcly determined the stresses. He also worked on friction (\Coulomb friction") and on earth pressure. 75 Coulomb did also research on magnetism, friction, and electricity. In 1777 he invented the torsion balance for measuring the force of magnetic and electrical attraction. With this invention, Coulomb was able to formulate the principle, now known as Coulomb's law, governing the interaction between electric charges. In 1779 Coulomb published the treatise Theorie des machines simples (Theory of Simple Machines), an analysis of friction in machinery. After the war Coulomb came out of retirement and assisted the new government in devising a metric system of weights and measures. The unit of quantity used to measure electrical charges, the coulomb, was named for him. 76 Navier (17851836) Navier was educated at the Ecole Polytechnique and became a professor there in 1831. Whereas the famous memoir of Coulomb (1773) contained the correct solution to numerous important problems in mechanics of materials, it took engineers more than forty years to understand them correctly and to use them in practical application 77 In 1826 he published his Le cons (lecture notes) which is considered the rst great textbook in mechanics for engineering. In it he developed the rst general theory of elastic solids as well as the rst systematic treatment of the theory of structures. 78 It should be noted that no clear division existed between the theory of elasticity and the theory of structures until about the middle of the nineteenth century (Coulomb and Navier would today be considered professional structural engineers). 79 Three other structural engineers who pioneered the development of the theory of elasticity from that point on were Lame, Clapeyron and de SaintVenant. Lam'e published the rst book on elasticity in 1852, and credited Clapeyron for the theorem of equality between external and internal work. de SaintVenant was perhaps the greatest elasticians who according to Southwell \... combined with high mathematical ability an essentially practical outlook which gave direction to all his work". In 18556 he published his classical work on torsion, exure, and shear stresses.
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7.8 The Modern Period (1857Present) 7.8.1 Structures/Mechanics
From 1857 the evolution of a comprehensive theory of structures proceeded at astonishing rate now that the basic and requisite principles had been determined. 81 Great contributors in that period include: Maxwell ( rst analysis of indeterminate structures), Culmann (graphics statics), Mohr (Mohr's circle, indeterminate analysis), Castigliano (1st and 2nd theorems), Cross (moment distribution), Southwell (relaxation method).
80
7.8.2 Eiel Tower The Eiel Tower was designed and built by the French civil engineer Alexandre Gustave Eiel for the Paris World's Fair of 1889. The tower, without its modern broadcasting antennae, is 300 m (984 ft) high. The lower section consists of four immense arched legs set on masonry piers. The legs curve inward until they unite in a single tapered tower. Platforms, each with an observation deck, are at three levels; on the rst is also a restaurant. 83 The tower, constructed of about 6300 metric tons (about 7000 tons) of iron, has stairs and elevators. A meteorological station, a radio communications station, and a television transmission antenna, as well as a suite of rooms that were used by Eiel, are located near the top of the tower.
82
7.8.3 Sullivan 18561924 Sullivan wan an American architect, whose brilliant early designs for steelframe skyscraper construction led to the emergence of the skyscraper as the distinctive American building type. Through his own work, especially his commercial structures, and as the founder of what is now known as the Chicago school of architects, he exerted an enormous in uence on 20thcentury American architecture. His most famous pupil was the architect Frank Lloyd Wright, who acknowledged Sullivan as his master. 85 After studying architecture at the Massachusetts Institute of Technology, he spent a year in Paris at the E cole des BeauxArts and in the oce of a French architect. Settling in Chicago in 1875, he was employed as a draftsman, then in 1881 formed a partnership with Dankmar Adler. Together they produced more than 100 buildings. 86 Adler secured the clients and handled the engineering and acoustical problems, while Sullivan concerned himself with the architectural designs. One of their earliest and most distinguished joint enterprises was the tenstory Auditorium Building (188689) in Chicago. This famous showplace incorporated a hotel, an oce building, and a theater renowned for its superb acoustics. The Wainwright Building, also ten stories high, with a metal frame, was completed in 1891 in Saint Louis, Missouri. In 1895 the SullivanAdler partnership was dissolved, leading to a decline in Sullivan's practice. The Carson Pirie Scott (originally Schlesinger and Meyer) Department Store, Chicago, regarded by many as Sullivan's masterpiece, was completed in 1904. 87 His famous axiom, Form follows function became the touchstone for many in his profession. Sullivan, however, did not apply it literally. He meant that an architect should consider the purpose of the building as a starting point, not as a rigidly limiting stricture. 88 He also had tremendous respect for the natural world which played an enormous role in forging his theories about architecture (he spent all of his rst summers on his grandparents' farm in Massachusetts where he developed this love and respect for nature) expressed in his Autobiography of an Idea), 1924). 84
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7.8.4 Roebling, 18061869 John Augustus Roebling was an American civil engineer, who was one of the pioneers in the construction of suspension bridges. He was born in Germany, educated at the Royal Polytechnic School of Berlin and immigrated to the States in 1831. 90 In his rst job he was employed by the Pennsylvania Railroad Corp. to survey its route across the Allegheny Mountains between Harrisburg and Pittsburgh. He then demonstrated the practicability of steel cables in bridge construction and in 1841 established at Saxonburg the rst factory to manufacture steelwire rope in the U.S. 91 Roebling utilized steel cables in the construction of numerous suspension bridges and is generally considered one of the pioneers in the eld of suspensionbridge construction. He built railroad suspension bridges over the Ohio and Niagara rivers and completed plans for the Brooklyn Bridge shortly before his death. Roebling was the author of Long and Short Span Railway Bridges (1869).
89
7.8.5 Maillart From (Billington 1973)
Robert Maillart was born on February 6, 1872, in Bern, Switzerland, where his father, a Belgian citizen, was a banker. He studied civil engineering at the Federal Institute of Technology in Zurich and graduated in 1894. Ironically, one of his lowest grades was in bridge design, even though he is regarded today as one of the half dozen greatest bridge designers of the twentieth century. 93 For eight years following his graduation, he worked with dierent civil engineering organizations. In 1902, he founded his own rm for design and construction; thereafter, his business grew rapidly and expanded as far as Russia and Spain. In the summer of 1914, he took his wife and three children to Russia. Since the World War prevented their return to Switzerland, Maillart stayed and worked in Russia until 1919, when his business was liquidated by the Revolution. Forced to ee, he returned to Switzerland penniless and lonely, his wife having died in Russia. 94 Because of these misfortunes Maillart felt unable to take up the construction business again and henceforth concentrated on design alone. He opened an oce in Geneva in 1919 and branches in Bern and Zurich in 1924. 95 During the twenties he began to develop and modify his ideas of bridge design; and from 1930, when the Salginatobel and Landquart Bridges were completed, until his death in 1940, he produced over thirty bridge designs of extraordinary originality. Unfortunately, no Swiss municipality would accept his designs for prominent urban locations. 96 In 1936, he was elected an honorary member of the Royal institute of British Architects (R.I.B.A.) although he had never ocially acted as architect on any project. The 1941 rst edition of Space, Time and Architecture by art historian Siegfried Giedion introduced Maillart to a wide public in the U.S.A. Finally, Max Bill's 1949 book, Robert Maillart, with its photographs and commentary on nearly all Maillart's bridges powerfully presented him as an artist of rst rank.
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7.8.6 Nervi, 18911979 97 Pier Luigi Nervi was an Italian architect and engineer, whose technical innovations, particularly in the use of reinforced concrete, made possible aesthetically pleasing solutions to dicult structural problems.
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He attended the Civil Engineering School in Bologna and established his own rm in 1920. His rst major commission (a stadium in Florence, 1932) features cantilevered beams and a daringly exposed concrete structure. For airplane hangars he used reinforced concrete to cover enormous spans with a light but strong latticework. 99 Nervi considered himself primarily an engineer and technician, not an architect, and he strove primarily for strength through form. He maintained that the strong aesthetic appeal of his buildings was simply a byproduct of their structural correctness. His introduction of a versatile new type of reinforced concrete layers of ne steel mesh sprayed with cement mortar made possible one of his masterpieces, the Turin Exposition Hall (1949), in which the approximately 76m (250ft) corrugated lattice roof (only about 5 cm thick) creates an immense interior space as dramatic as a cathedral. 100 The best known and most in uential is probably his Palazetto dello Sport (Small Sport Palace, 1960, Rome), Fig. 7.17. Encircled by Yshaped supports and topped by a shallow scalloped concrete dome, 98
Figure 7.17: Nervi's Palazetto Dello Sport this building has become a paradigm of the 20thcentury sports arena.
7.8.7 Khan
Fazlur Khan was born in 1929 in Dacca India, (Anon. xx). After obtaining a B.A. in engineering from
the University of Dacca in 1950, Khan worked as assistant engineer for the India Highway Department and taught at the University of Dacca. Qualifying for a scholarship in 1952, he enrolled at the University of Illinois, ChampaignUrbana, where he received master's degrees in both applied mechanics and structural engineering and a Ph.D. in structural engineering. He returned brie y to Pakistan and won an important position as executive engineer of the Karachi Development Authority. Frustrated by administrative demands that kept him from design work, however, he returned to the United States and
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joined the prestigious architectural rm of Skidmore, Owings & Merrill in Chicago in 1955, eventually becoming a partner (1966). Among his many designs for skyscrapers are Chicago's John Hancock Center (1970) and the Sears Tower (1973), which are among the world's tallest buildings, and One Shell Plaza in Houston, Texas. The Sears Tower was his rst skyscraper to employ his "bundled tube" structural system, which consists of a group of narrow steel cylinders that are clustered together to form a thicker column. The system was innovative because it minimized the amount of steel needed for high towers, eliminated the need for internal wind bracing (since the perimeter columns carried the wind loadings), and permitted freer organization of the interior space. His later projects included the strikingly dierent Haj Terminal of the King Abdul Aziz International Airport, Jiddah, Saudi Arabia (197681), and King Abdul Aziz University, also in Jiddah (197778).
7.8.8
et al.
To name just a few of the most in uential Architects/Engineers: Menn, Isler, Candella, Torroja, Johnson, Pei, Calatrava, ...
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NEW ARCHITECTURE by
Felix Candela
Professor  University of Illinois Unauthorized reprint from (Billington 1973) The title of my lecture is New Architecture; but I cannot avoid the feeling that I have not too much to do with this subject. I don't think I can speak of my work as of any new architecture or even as architecture at all, and perhaps the same could be said of Maillart's work. But, of course, it all depends on what you consider architecture and there is not anymore a general consent about the meaning of this word, But, anyway, upon discussing this with David Billington, he told me it could be interesting to know what had been the in uence of Maillart on my development, and all of a sudden I realized that he may have been one of the strongest in uences at the critical moment in my career in which I was trying to become a builder of shells. But let me speak rst about my background, because it may be important to know what impels people to do things and the circumstances and diculties that they had to deal with in order to achieve their purposes. I was trained as an architect with, you may say, quite a backward kind of curriculum, in Madrid during the thirties. We had only one course in strength of materials, but it was a very good course, dealing mainly with theory of elasticity and following the classical and rigoristic French tradition. As you can imagine, most students considered the matter completely useless for their professional practice and, as it required some knowledge of mathematics, it was very dicult for most of them to pass the examination. This gave me opportunity to do some private tutoring to my classmates, which was a very instructive manner to make some money to pay for my studies. As a result, I became more familiar with the theoretical bases of the current methods of calculation of indeterminate structures. I discovered later that this modest background made me more knowledgeable on the matter than most practicing engineers whose training and interest tend to be directed towards mastering the accepted methods of analysis rather than questioning the basic hypotheses. Anyway, since I never had a high opinion of myself as an artist, I was more interested in the technical part of the curriculum and began to read extensively about structures. Among my lectures I found several French and German papers dealing with shells which were beginning to be in vogue at that time in Europe. Examples of such structures built in Germany and France could be found in magazines and Torroja was building the famous roof of the "Fronton Recoletos" in Madrid, with an unusual shape and a record span. Shells appeared to be an intriguing challenge for me, and I dreamed about the possibility of building some in the future. But my lack of experience and my youthful faith in the impressive wisdom displayed in learned magazines led me to believe that the key to shell design was in complete mathematical calculation, and I tried, rather unsucessfully, to understand and follow them and to make some sense of their results. I was not the only one misled and discouraged by the mathematical barrier so cunningly deployed by German engineers, a clever move which secured them practically an exclusive on the construction of barrel vaults for more than twenty years, hindering the normal employment of such structures during the same time. However, such was my enthusiasm with this mathematical approach that I managed to get a fellowship to go to Germany, hoping to learn something more from the German professors. But the outbreak of the Spanish Civil War saved me from such an ordeal. I could not leave the country and ended in Mexico after three years of military service, with no more baggage than my bare hands and no further addition to my academic background. After several years of general practice in Mexico, as draftsman, designer and contractor, I recalled my old fancy with shells and began to collect again papers on the subject. Whatever I learned from then on was to be the hard way, working alone, with no direct help from any university or engineering oce. But I am indebted to many people who did help me through their
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writings and Maillart was one of the foremost. I discovered him in Giedion's Space, Time and Architecture; and then I got Max Bill's book with its invaluable collection of Maillart's essays. I devoured his articles about "Reinforced Concrete Design and Calculation" (he was very careful to dierentiate the meaning of such words and to avoid the more than semantic confusion prevalent nowadays in Englishspeaking countries)., "The Engineer and the Authorities" which expresses his position in front of the establishment and "Mass and Quality in Reinforced Concrete Structures." Very short papers, indeed, but well provided with opinions, something I could rarely nd in other engineering articles. I learned later that to express personal opinions is considered bad taste among technical writers. Any discussion should be restricted to insigni cant details, but never touch fundamental dogmas, in a fashion curiously similar to what could be expected of the councils of the Church or the meetings of any Politbureau. But my attitude with respect to calculations of reinforced concrete structures was becoming unorthodox, being tired perhaps of performing long and tedious routines whose results were not always meaningful. Therefore, I found Maillart's thoughts delightfully sympathetic and encouraging. If a rebel was able to produce such beautiful and sound structures there could not be anything wrong with becoming also a rebel, which was besides, my only way to break the mystery surrounding shell analysis. Thus, I started to follow the bibliographic tread and met, through their writings, with Freudenthal, Johansen, Van der Broek, Kist, Saliger, Kacinczy and so many others who showed me there was more than a single and infallible manner to approach structural analysis. The discovery of rupture methods, with their emphasis on simple statics and their bearing on the actual properties of construction materials and their behavior in the plastic range, allowed me to trust in simpli ed procedures to understand and analyze the distribution of stresses in shell structures. It also helped me to get out of my naive belief in the indisputable truth of the printed word and to start reading with a new critical outlook. No longer did I need to believe whatever was in print, no matter how highsounding the name of the author. I could make my own judgements about what methods of stress analysis were better suited for my practice. Since I was working practically alone, I could not aord nor had time for complex calculations and did welcome Maillart's advice that simpler calculations are more reliable than complex ones, especially for somebody who builds his own structures. This was exactly my case and, since most structures I was building were of modest scale, I could control what was happening, check the results and con rm the accuracy of my judgement or correct my mistakes. In a way, I was working with full scale models. I understand that this was also true of Maillart who in many cases was the actual builder of his designs. Following the general trend to mess up issues, there has been a lot of speculation about the engineer as an artist and in some instances, like in the case of Nervi, about the engineer as an architect (as if the title of architect could confer, per se, artistic ability to its holder); but few people realize that the only way to be an artist in this dicult specialty of building is to be your own contractor. in countries like this, where the building industry has been thoroughly and irreversibly fragmented and the responsibility diluted among so many trades, it may be shocking to think of a contractor as an artist; but it is indeed the only way to have in your hands the whole set of tools or instruments to perform the forgotten art of building, to produce "works of art" which, by the way, was the common expression to designate a bridge in the old French engineering vocabulary. Implicit in the above statement is the fact that you have to be, besides your own structural designer and calculator and perhaps your own architect, also your own contractor, a very dicult proposition in some countries where such mergings of today's disparate professions may even be considered unlawful. This means, of course, that the price for being the master of your trade is to accept the whole responsibility for the good performance of the structure, and not too many people today would readily endorse such an awesome commitment. I am not advocating a return to the past; history is an irreversible process. I am simply stating that the Maillart phenomenom could not happen under today's situation of the industry. I like to think, however, that Maillart did not judge himself an artist. As Picasso said of himself "he was not looking for beauty; he found it". His main concerns must have been eciency and economy of means, since to be able to build one of his bridges he had to win a bidding competition and prove that
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he could do it cheaper than anybody else. But an ecient and economical structure has not necessarily to be ugly. Beauty has no price tag and there is never one single solution to an engineering problem. Therefore, it is always possible to modify the whole or the parts until the ugliness disappears. This aversion to ugliness is quite the opposite of the task of the professional artist who has to produce beauty as an obligation or of today's stararchitect who has to be original at any cost in each new project. Maillart's works did not need to be beautiful. This word did not even exist in the practical world of the serious citizens who had to judge his competitive bids. He achieved a beauty without need or purpose; just for the pure joy of it. The kind of joy that you can feel also in the works of Haydn or Vivaldi. They were simply enjoying what they were doing, and so was obviously Maillart. He did also possess that rare quality, source of artistic creation and of all invention, of being able to challenge the conventional wisdom and come up with the obvious solution, one, nevertheless, which nobody could think of before. I can imagine the ts of rage and jealousy of some of his contemporary colleagues at the sight of one of his bridges (Landquart or Schwandbach), in which the curved route is supported in a straight arched slab. The problem with this unusual combination  which, of course, looks perfectly logical after the fact  is that it was very dicult, if not impossible, to analyze with the methods available at that time. But Maillart would not take any unnecessary risk and rst he tried the soundness of his approximated calculations in a small example (the Halbkern Bridge) with a span of only fteen meters. This was his testing model which gave him rm ground from which to extrapolate at the next opportunity. I would like to insist at this moment cn something that everybody knows but which is easily forgotten; that all calculations, no matter how sophisticated and complex, can not be more than rough approximations of the natural phenomenon they try to represent by means of a mathematical model. The complexity, or even elegance, of such a model bears no relation at all with the degree of approximation. There is not such a thing as an exact method of structural analysis and, notwithstanding the popular belief in the letter of the codes, the accuracy of any calculation is still a question of personal judgement. This fortunate circumstance allows engineering to reach sometimes the highest category of art, to the despair of dull and in exible technicians. If I nd something lacking in this commendable conference in memory of one of the greatest engineers of all times it is that the side of Maillart's personality as a rebel, with his tireless and successful struggle against the establishment of his times, has not been suciently stressed.
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Case Study III: MAGAZINI GENERALI Adapted from (Billington and Mark 1983)
8.1 Geometry 1 This sotrage house, built by Maillart in Chiasso in 1924, provides a good example of the mariage between aesthetic and engineering. 2 The most strking feature of the Magazini Generali is not the structure itself, but rather the shape of its internal supporting frames, Fig. 8.1. 3 The frame can be idealized as a simply supported beam hung from two cantilever column supports. Whereas the beam itself is a simple structural idealization, the overhang is designed in such a way as to minimize the net moment to be transmitted to the supports (foundations), Fig. 8.2.
8.2 Loads The load applied on the frame is from the weights of the roof slab, and the frame itself. Given the space between adjacent frames is 14.7 ft, and that the roof load is 98 , and that the total frame weight is 13.6 kips, the total uniform load becomes, Fig. 8.3:
4
psf
qroof = (98) (14:7) = 1:4 (13:6) = 0:2 qframe = (63 :6) qtotal = 1:4 + 0:2 = 1.6 psf
ft
k
k/ft
ft
k/ft
k/ft
(8.1a) (8.1b) (8.1c)
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Case Study III: MAGAZINI GENERALI HINGE IDEALIZATION OF THIN SECTIONS
ACTUAL FRAME
ABSTRACTION OF MID SECTIONAS A SIMPLE BEAM
9.2 ft
63.6 ft
Figure 8.1: Magazzini Generali; Overall Dimensions, (Billington and Mark 1983)
B P
MB =B*d1
MP =P*d 2 d2
d1 B MR=MB M P
Figure 8.2: Magazzini Generali; Support System, (Billington and Mark 1983)
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Structural Concepts and Systems for Architects
Draft 8.3 Reactions
8{3 q ROOF = 1.4 k/ft + q FRAME = 0.2 k/ft q ROOF = 1.4 k/ft + q FRAME = 0.2 k/ft q TOTAL = 1.6 k/ft
Figure 8.3: Magazzini Generali; Loads (Billington and Mark 1983)
8.3 Reactions 5
Reactions for the beam are determined rst taking advantage of symmetry, Fig. 8.4:
W = (1:6) (63:6) = 102 R = W2 = 102 2 = 51 k/ft
ft
(8.2a) (8.2b)
k
k
We note that these reactions are provided by the internal shear forces. q TOTAL = 1.6 k/ft
63.6 ft
51 k
51 k
Figure 8.4: Magazzini Generali; Beam Reactions, (Billington and Mark 1983)
8.4 Forces The internal forces are pimarily the shear and moments. Those can be easily determined for a simply supported uniformly loaded beam. The shear varies linearly from 51 kip to 51 kip with zero at the center, and the moment diagram is parabolic with the maximum moment at the center, Fig. 8.5, equal to: 2 2 Mmax = qL8 = (1:6) 8(63:6) = 808 (8.3)
6
k/ft
ft
k.ft
The externally induced moment at midspan must be resisted by an equal and opposite internal moment. This can be achieved through a combination of compressive force on the upper bers, and tensile ones on the lower. Thus the net axial force is zero, however there is a net internal couple, Fig. 8.6.
7
Victor Saouma
Structural Concepts and Systems for Architects
Draft 8{4
Case Study III: MAGAZINI GENERALI
SHEAR FORCE
51 K 25 K L
0
x
L/2
25 K
MOMENT
51 K
Mmax
L 0
L/2
L/4
x
3L/4
Figure 8.5: Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983)
q TOTAL
A
C d
M
VA T
Figure 8.6: Magazzini Generali; Internal Moment, (Billington and Mark 1983)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 8.4 Forces
8{5 Mext = Cd ) C = Mdext
T = C = (808) (9:2)
k.ft ft
(8.4a)
= 88
(8.4b)
k
8 Because the frame shape (and thus d(x)) is approximately parabolic, and the moment is also parabolic, then the axial forces are constants along the entire frame, Fig. 8.7.
d
M MOMENT DIAGRAM
FRAME
CABLE :
FRAME :
CURVE OF DIAGRAM
SHAPE OF DIAGRAM
Figure 8.7: Magazzini Generali; Similarities Between The Frame Shape and its Moment Diagram, (Billington and Mark 1983) The axial force at the end of the beam is not balanced, and the 88 kip compression must be transmitted to the lower chord, Fig. 8.8. Fig. 8.9 This is analogous to the forces transmiited to the support by a
9
88 k
Tension 88 k
88 k
88 k
Compression Horizontal Component Tied Arch
Cable Force Axial Force Vertical Reaction
Figure 8.8: Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark 1983) tied arch. 10 It should be mentioned that when a rigorous computer analysis was performed, it was determined that the supports are contributing a compression force of about 8 kips which needs to be superimposed over the central values, Fig. 8.9.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 8{6
Case Study III: MAGAZINI GENERALI
FRAME ACTS AS A UNIT, UNLIKE THE ABSTRACTION
88 k  8 k = 96 k 16 k 88 k  8 k = 80 k
16 k
Figure 8.9: Magazzini Generali; Eect of Lateral Supports, (Billington and Mark 1983)
8.5 Internal Stresses The net compressive stress, for a top chord with a cross sectional area of 75 2 is equal to (88) = 1.17 = PA = (75) (8.5) 2 this is much lower than the allowable compressive stress of concrete which is about 1,350 ksi. It should be noted that if the frame was cast along with the roof (monolithic construction), than this stress would be even lower. 12 Since concrete has practically no tensile strength, the tensile force in the lower chord must be resisted by steel. The lower chord has 4 bars with 0.69 2 and 6 other bars with 0.58 2 , thus we have a total of As = 4(0:69) + 6(0:58) = 6:24 2 (8.6) Thus the steel stresses will be (8.7) = PA = (6(88) :24) 2 = 14.1 which is lower than the allowable steel stress. 11
in
k
ksi
in
in
in
in
k
ksi
in
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 9
DESIGN PHILOSOPHIES and GUIDELINES 9.1 Safety Provisions 1 Structures and structural members must always be designed to carry some reserve load above what is expected under normal use. This is to account for Variability in Resistance: The actual strengths (resistance) of structural elements will dier from those assumed by the designer due to: 1. Variability in the strength of the material (greater variability in concrete strength than in steel strength). 2. Dierences between the actual dimensions and those speci ed (mostly in placement of steel rebars in R/C). 3. Eect of simplifying assumptions made in the derivation of certain formulas. Variability in Loadings: All loadings are variable. There is a greater variation in the live loads than in the dead loads. Some types of loadings are very dicult to quantify (wind, earthquakes). Consequences of Failure: The consequence of a structural component failure must be carefully assessed. The collapse of a beam is likely to cause a localized failure. Alternatively the failure of a column is likely to trigger the failure of the whole structure. Alternatively, the failure of certain components can be preceded by warnings (such as excessive deformation), whereas other are sudden and catastrophic. Finally, if no redistribution of load is possible (as would be the case in a statically determinate structure), a higher safety factor must be adopted. 2 The purpose of safety provisions is to limit the probability of failure and yet permit economical structures. 3 The following items must be considered in determining safety provisions: 1. Seriousness of a failure, either to humans or goods. 2. Reliability of workmanship and inspection. 3. Expectation of overload and to what magnitude.
Draft 9{2
DESIGN PHILOSOPHIES and GUIDELINES
4. Importance of the member in the structure. 5. Chance of warning prior to failure. 4
Two major design philosophies have emerged 1. Working Stress Method 2. Ultimate Strength Method
9.2 Working Stress Method This is the simplest of the two methods, and the one which has been historically used by structural engineers. 6 Structural elements are designed for their service loads, and are dimensioned such that the stresses do not exceed some predesignated allowable strength, Fig. 9.1.
5
Figure 9.1: Load Life of a Structure In R/C this method was the one adopted by the ACI (American Concrete institute) code up to 1971, Working Stress Design Method (WSD). 8 The AISC (American Institute of Steel Construction) code refers to it as the Allowable Stress Design (ASD) and was used until 1986. 7
Victor Saouma
Structural Concepts and Systems for Architects
Draft 9.3 Ultimate Strength Method 9
9{3
In this method: 1. All loads are assumed to have the same average variability. 2. The entire variation of the loads and the strengths is placed on the strength side of the equation.
yld < all = F:S:
(9.1)
where F:S: is the factor of safety. 10 Major limitations of this method 1. An elastic analysis can not easily account for creep and shrinkage of concrete. 2. For concrete structures, stresses are not linearly proportional to strain beyond 0:45fc0 . 3. Safety factors are not rigorously determined from a probabilistic approach, but are the result of experience and judgment. 11
Allowable strengths are given in Table 9.1. Steel, AISC/ASD Tension, Gross Area Ft = 0:6Fy Tension, Eective Net Area Ft = 0:5Fu Bending Fb = 0:66Fy Shear Fv = 0:40Fy Concrete, ACI/WSD Tension 0 Compression 0:45fc0
Eective net area will be de ned in section ??.
Table 9.1: Allowable Stresses for Steel and Concrete
9.3 Ultimate Strength Method 9.3.1 12 13 14 15
y
Probabilistic Preliminaries
In this approach, it is assumed that the load Q and the resistance R are random variables. Typical frequency distributions of such random variables are shown in Fig. 9.2. The safety margin is de ned as Y = R ; Q. Failure would occur if Y < 0 Q and R can be combined and the result expressed logarithmically, Fig. 9.3.
R X = ln Q
(9.2)
Failure would occur for negative values of X 16 The probability of failure Pf is equal to the ratio of the shaded area to the total area under the curve in Fig. 9.3.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 9{4
DESIGN PHILOSOPHIES and GUIDELINES
Figure 9.2: Frequency Distributions of Load Q and Resistance R
Figure 9.3: De nition of Reliability Index
Victor Saouma
Structural Concepts and Systems for Architects
Draft 9.3 Ultimate Strength Method
9{5
If X is assumed to follow a Normal Distribution than it has a mean value X = ln QR and a m standard deviation . X 18 We de ne the safety index (or reliability index) as = 1 19 For standard distributions and for = 3:5, it can be shown that the probability of failure is Pf = 9;091 or 1:1 10;4. That is 1 in every 10,000 structural members designed with = 3:5 will fail because of either excessive load or understrength sometime in its lifetime. 20 Target values for are shown in Table 9.2. 17
Type of Load/Member
AISC
DL + LL; Members 3.0 DL + LL; Connections 4.5 DL + LL + WL; Members 3.5 DL + LL +EL; Members 1.75
ACI
Ductile Failure Sudden Failures
33.5 3.54
Table 9.2: Selected values for Steel and Concrete Structures Because the strengths and the loads vary independently, it is desirable to have one factor to account for variability in resistance, and another one for the variability in loads. 22 These factors are referred to as resistance factor and Load Factor respectively. The resistance factor is de ned as = RRm exp(;0:55 VR ) (9.3) n where RM RN and VR are the mean resistance, the nominal resistance (to be de ned later), and the coecient of variation of the resistance.
21
9.3.2 Discussion ACI refers to this method as the Strength Design Method, (previously referred to as the Ultimate Strength Method). 24 AISC refers to it as Load and Resistance Factor Design (LRFD). 25 Terms such as failure load should be avoided; it is preferable to refer to a structure's Limit State load. 1 26 The general form is (LRFDA4.1)
23
Rn i Qi
(9.4)
where is a strength reduction factor, less than 1, and must account for the type of structural element, Table 9.3. 1
Throughout the notes we will refer by this symbol the relevant design speci cation in the AISC code.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 9{6
DESIGN PHILOSOPHIES and GUIDELINES Type of Member
ACI
Axial Tension Flexure Axial Compression, spiral reinforcement Axial Compression, other Shear and Torsion Bearing on concrete
0.9 0.9 0.75 0.70 0.85 0.70
Tension, yielding Tension, fracture Compression Beams Fasteners, Tension Fasteners, Shear
0.9 0.75 0.85 0.9 0.75 0.65
AISC
Table 9.3: Strength Reduction Factors,
Rn is the nominal resistance (or strength). Rn is the design strength. i is the load factor corresponding to Qi and is greater than 1. i Qi is the required strength based on the factored load:
i is the type of load 27
The various factored load combinations which must be considered are
AISC
ACI
1. 2. 3. 4. 5. 6.
1.4D 1.2D+1.6L+0.5(Lr or S) 1.2D+0.5L (or 0.8W)+1.6(Lr or S) 1.2D+0.5L+0.5(Lr or S)+1.3W 1.2D+0.5L(or 0.2 S)+1.5E 0.9D+1.3W(or 1.5 E)
1. 2. 3. 4. 5. 6. 7.
1.4D+1.7L 0.75(1.4D+1.7L+1.7W) 0.9D+1.3W 1.05D+1.275W 0.9D+1.7H 1.4D +1.7L+1.7H 0.75(1.4D+1.4T+1.7L)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 9.4 Example
9{7
8. 1.4(D+T) where D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature; H= soil. We must select the one with the largest limit state load. 28 Thus, in this method, we must perform numerous analysis, one for each load, of a given structure. For trusses, this is best achieved if we use the matrix method, invert the statics matrix [B ], and multiply [B ];1 by each one of the load cases, (Refer to Section ??). For the WSD method, we need not perform more than one analysis in general. 29 Serviceability Limit States must be assessed under service loads (not factored). The most important ones being 1. De ections 2. Crack width (for R/C) 3. Stability
9.4 Example Example 918: LRFD vs ASD To illustrate the dierences between the two design approaches, let us consider the design of an axial member, subjected to a dead load of 100 and live load of 80 . Use A36 steel. k
k
ASD: We consider the total load P = 100 + 80 = 180 . From Table 9.1, the allowable stress is 0:6yld = 0:6 36 = 21:6 . Thus the required cross sectional area is k
ksi
180 = 8:33 A = 21 :6
2
in
USD we consider the largest of the two load combinations i Qi : 1:4D = 1:4(100) = 140 1:2D + 1:6L = 1:2(100) + 1:6(80) = 248
k k
From Table 9.3 = 0:9, and Rn = (0:9)Ayld . Hence, applying Eq. 9.4 the cross sectional area should be 2 A = i Qi = (0:248 9)(36) = 7:65 in
yld
Note that whereas in this particular case the USD design required a smaller area, this may not be the case for dierent ratios of dead to live loads.
9.5 Design Guidelines To assist in the preliminary design/dimensioning of structures, Table 9.4 provides average, maximum and typical spans for various types of structures.
30
Victor Saouma
Structural Concepts and Systems for Architects
Draft 9{8
DESIGN PHILOSOPHIES and GUIDELINES Average Max Typical Span Ft.
TIMBER
Plywood Planks Joists Beams Girders Gable bents Trusses I Beams Joists Plate and I girders Trusses Gable bents Arches span to rise Arches span to thickness Simple suspension (span to rise) Cable stayed
36 28 22 16 12 26 4 18 18 14 12 30 8 40 10 6
40 32 26 20 16 30 8 24 25 20 18 40 16 50 15 10
REINFORCED CONCRETE
35 26 1025 1530 2035 3050 30100 1560 1560 40100 4080 50120 80200 150300 150300
Solid slabs 28 32 1025 Slabs with drops and capitals 30 36 2035 Twoway slab on beams 30 36 2035 Wae slabs 20 24 3040 Joists 22 26 2545 Beams 16 20 1540 Girders 12 16 2060 Gable bents 24 30 4080 Arches span to rise 8 12 60150 Arches span to thickness 30 40 Cylindrical thin shell roof (Min. thickness may govern) Longitudinal span to Structural depth 12 15 5070 Transverse span to thickness 50 60 1230
PRESTRESSED CONCRETE
Solid slabs 40 44 2035 Slabs with drops 44 48 3545 Twoway slab on beams 44 48 3545 Wae slabs 28 32 3570 Cored slabs 36 40 3060 Joists 32 36 4060 Beams 24 28 3080 Girders 20 24 40120 Cylindrical thin shell roof (Min. thickness may govern) Longitudinal span to structural depth 15 20 60120 Transverse span to thickness 60 70 1535 Table 9.4: Approximate Structural SpanDepth Ratios for Horizontal Subsystems and Components (Lin and Stotesbury 1981)
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 10
BRACED ROLLED STEEL BEAMS 1 This chapter deals with the behavior and design of laterally supported steel beams according to the LRFD provisions. 2 A laterally stable beam is one which is braced laterally in the direction perpendicular to the plane of the web. Thus overall buckling of the compression ange as a column cannot occur prior to its full participation to develop the moment strength of the section. 3 If a beam is not laterally supported, Fig. 10.1, we will have a failure mode governed by lateral torsional
A) COMPOSITE BEAM
B) OTHER FRAMING
C) CROSS BRACING
Figure 10.1: Lateral Bracing for Steel Beams buckling. 4 By the end of this lecture you should be able to select the most ecient section (light weight with adequate strength) for a given bending moment and also be able to determine the exural strength of a given beam.
Draft 10{2
BRACED ROLLED STEEL BEAMS
10.1 Nominal Strength 5
The strength requirement for beams in load and resistance factor design is stated as
b Mn Mu
(10.1)
where:
b strength reduction factor; for exure 0:90 Mn nominal moment strength Mu factored service load moment.
The equations given in this chapter are valid for exural members with the following kinds of cross section and loading: 1. Doubly symmetric (such as W sections) and loaded in plane of symmetry 2. Singly symmetric (channels and angles) loaded in plane of symmetry or through the shear center parallel to the web1 .
6
10.2 Failure Modes and Classi cation of Steel Beams A beam is classi ed as laterally supported depending on Lb which is the distance between lateral supports (or unbraced length) and Lp .
7
Lb < Lp (10.2) 300 Lp = p r (10.3) Fy ; y r (10.4) ry = IAy where ry is the radius of gyration with respect to the (minor) y axis (as opposed to the major x axis). ksi
The strength of exural members is limited by: Plastic Hinge: at a particular cross section, Fig. 10.2. local buckling: of a crosssectional element (e.g. the web or the ange), Fig. 10.3. LateralTorsional buckling: of the entire member, Fig. 10.4. 8
Accordingly, the LRFD manual classi es steel sections as Compact sections: No local buckling can occur. Strength is based on the plastic moment. Partially compact sections: Where local buckling may occur Slender sections: where lateral torsional buckling may occur. We will cover only the rst two cases. 10 Shear should be checked, however with exception of short beams (and no selfrespecting architect will ever conceive such a thing:), exure generally controls. 9
1
More about shear centers in Mechanics of Materials II.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 10.2 Failure Modes and Classi cation of Steel Beams
10{3
w
σy

M=(wL2 )/8
+
σy
wu 
11 00 00 11 00 11
σy
+
Mp σy M p=(wL2 )/8
Figure 10.2: Failure of Steel beam; Plastic Hinges
tw
bf
hf hc
COMPACT
FLANGE BUCKLING
WEB BUCKLING
Figure 10.3: Failure of Steel beam; Local Buckling
Victor Saouma
Structural Concepts and Systems for Architects
Draft 10{4
BRACED ROLLED STEEL BEAMS LATERAL DEFLECTION AND TORSION OF THE COMPRESSION FLANGE
A
B LATERAL DEFLECTION AND TORSION OF THE COMPRESSION FLANGE
COMPRESSION FLANGE
B
A B A
Figure 10.4: Failure of Steel beam; Lateral Torsional Buckling
10.3 Compact Sections For compact sections, the mode of failure is the formation of a plastic hinge that is the section is fully plasti ed. Hence we shall rst examine the bending behavior of beams under limit load. Then we will relate this plastic moment to the design of compact sections.
11
10.3.1 Bending Capacity of Beams The stress distribution on a typical wide ange shape subjected to increasing bending moment is shown in Fig.10.5. In the service range (that is before we multiplied the load by the appropriate factors in the LRFD method) the section is elastic. This elastic condition prevails as long as the stress at the extreme ber has not reached the yield stress Fy . Once the strain " reaches its yield value "y , increasing strain induces no increase in stress beyond Fy . 12
Figure 10.5: Stress distribution at dierent stages of loading
Victor Saouma
Structural Concepts and Systems for Architects
Draft 10.3 Compact Sections
10{5
Figure 10.6: Stressstrain diagram for most structural steels When the yield stress is reached at the extreme ber, the nominal moment strength Mn, is referred to as the yield moment My and is computed as
13
(10.5)
Mn = My = SxFy
(assuming that bending is occurring with respect to the x ; x axis). 14 When across the entire section, the strain is equal or larger than the yield strain (" "y = Fy =Es ) then the section is fully plasti ed, and the nominal moment strength Mn is therefore referred to as the plastic moment Mp and is determined from
Mp = Fy
Z
where def
A
(10.6)
ydA = Fy Z
Z
(10.7)
Z = ydA
is the Plastic Section Modulus. 15 The plastic section modulus Z should not be confused with the elastic section modulus S de ned, Eq. 5.25 as
S = d=I 2 Z I def = y2 dA A
16
(10.8a) (10.8b)
The section modulus Sx of a W section can be roughly approximated by the following formula
Sx wd=10 or Ix Sx d2 wd2 =20
(10.9)
and the plastic modulus can be approximated by
Zx wd=9
Victor Saouma
(10.10)
Structural Concepts and Systems for Architects
Draft 10{6
BRACED ROLLED STEEL BEAMS
10.3.2 Design of Compact Sections 17
A section is compact if the following conditions are met: 1. Flanges are continuously connected to the web 2. Width to thickness ratios, known as the slenderness ratios, of the ange and the web must not exceed the limiting ratios p de ned as follows: Flange 2btff p p = p65Fy Web thwc p p = p640Fy
(10.11)
Note that 2btff and thwc are tabulated in Sect. 3.6. 18
The nominal strength Mn for laterally stable compact sections according to LRFD is (10.12)
Mn = Mp where:
19
Mp plastic moment strength = ZFy Z plastic section modulus Fy speci ed minimum yield strength Note that section properties, including Z values are tabulated in Section 3.6.
10.4 Partially Compact Section If the width to thickness ratios of the compression elements exceed the p values mentioned in Eq. 10.11 but do not exceed the following r , the section is partially compact and we can have local buckling.
20
where:
Flange: p < 2btff r p = p65Fy r = pF141 y ;Fr Web: p < thwc r p = p640Fy r = p970Fy
(10.13)
Fy bf tf hc
21
speci ed minimum yield stress in kksi width of the ange thickness of the ange unsupported height of the web which is twice the distance from the neutral axis to the inside face of the compression ange less the llet or corner radius. tw thickness of the web. Fr residual stress = 10:0 ksi for rolled sections and 16:5 ksi for welded sections. The nominal strength of partially compact sections according to LRFD is, Fig. 10.7
Mn = Mp ; (Mp ; Mr )( ;;p ) Mp r
Victor Saouma
p
(10.14)
Structural Concepts and Systems for Architects
Draft 10.5 Slender Section
10{7
Draft Mn
6 Mp
Compact
Partially Compact

Slender


Mr
p
r
Flanges
2tf
bf
p65F
pF141;F
Web
hc tw
p640F
p970F
y
y
y
r
y
Figure 10.7: Nominal Moments for Compact and Partially Compact Sections where:
Mr Residual Moment equal to (Fy ; Fr )S bf =2tf for Ishaped member anges and hc=tw for beam webs.
All other quantities are as de ned earlier. Note that we use the associated with the one being violated (or the lower of the two if both are).
22
10.5 Slender Section If the width to thickness ratio exceeds r values of ange and web, the element is referred to as slender compression element. Since the slender sections involve a dierent treatment, it will not be dealt here.
23
10.6 Examples Example 1019: Z for Rectangular Section Determine the plastic section modulus for a rectangular section, width b and depth d.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 10{8
BRACED ROLLED STEEL BEAMS
Solution: 1. The internal plastic moment is equal to 2 M = Fy b d2 d2 = Fy b d4
(10.15)
 {z }
Force
2. The yield stress, Fy , plastic moment Mp and plastic section modulus Z are related by:
3. Substituting, we get:
Z = MZp
(10.16)
p = Fy bd2 = bd2 Z=M 4 Fy 4Fy
(10.17)
Note that this is to be contrasted with the elastic section modulus S = bd62 .
Example 1020: Beam Design Select the lightest W or M section to carry a uniformly distributed dead load of 0:2 kip/ft superimposed (i.e., in addition to the beam weight) and 0:8 kip/ft live load. The simply supported span is 20 ft. The compression ange of the beam is fully supported against lateral movement. Select the sections for the following steels: A36; A572 Grade 50; and A572 Grade 65.
Solution:
Case 1: A36 Steel 1. Determine the factored load.
wD = 0:2 wL = 0:8 wu = 1:2wD + 1:6wL = 1:2(0:2) + 1:6(0:8) = 1:52 k/ft k/ft
Victor Saouma
k/ft
Structural Concepts and Systems for Architects
Draft 10.6 Examples
10{9
2. Compute the factored load moment Mu . For a simply supported beam carrying uniformly distributed load, Mu = wu L2 =8 = (1:52)(20)2=8 = 76 Assuming compact section, since a vast majority of rolled sections satisfy p for both the
ange and the web. The design strength b Mn is k.ft
b Mn = b Mp = b Zx Fy The design requirement is
b Mn = Mu
or, combing those two equations we have:
b Zx Fy = Mu 3. Required Zx is
Zx = MFu = 076(12) :90(36) = 28:1 b y
3
in
From the notes on Structural Materials, we select a W12X22 section which has a Zx = 29:3 (22)(12) Note that Zx is approximated by wd 9 = 9 = 29:3. 4. Check compact section limits p for the anges from the table
3
in
= 2btff = 4:7 p = p65Fy = p6536 = 10:8 > p and for the web:
= thwc = 41:8 p = p640Fy = p64036 = 107p
5. Check the Strength by correcting the factored moment Mu to include the self weight. Self weight of the beam W12X22 is 22 lb./ft. or 0.022 kip/ft
wD wu Mu Mn b Mn
= = = = =
0:2 + 0:022 = 0:222 1:2(0:222) + 1:6(0:8) = 1:55 (1:55)(20)2=8 = 77:3 3 Mp = Zx Fy = (29:3)(12) (36) = 87:9 p 0:90(87:9) = 79:1 > Mu k/ft
k/ft
k.ft
in
ksi
in/ft
k.ft
k.ft
Therefore use W12X22 section. 6. We nally check for the maximum distance between supports. r
r
Iy = 5 = 0:88 A 6:5 300 Lp = p ry Fy 300 = p 0:88 = 43 ry =
36
Victor Saouma
ft
in
(10.18a) (10.18b) (10.18c)
Structural Concepts and Systems for Architects
Draft 10{10
BRACED ROLLED STEEL BEAMS
Case 2: A572 Grade 65 Steel: 1. same as in case 1 2. same as in case 1 3. Required Zx = Mb Fuy = 076(12) :90(65) = 15:6 (14)(12) approximated by wd 9 = 9 = 18:7. 4. Check compact section limits p :
p p
= = = =
3
in
Select W12X14: Zx = 17:4
3
in
Note that Zx is
hc = 54:3 t640 pwFy = p64065 = 79:4p bf 2tf = 8:82 p65Fy = p6565 = 8:1 < Not Good
In this case the controlling limit state is local buckling of the ange. Since p < < r , as above, the section is classi ed as noncompact. 5. Check the strength: Since the section is noncompact, the strength is obtained by interpolation between Mp and Mr . For the anges:
r Mn Mp Mr Mn b Mn
= = = = = =
pF141 = p65141;10 = 19:0 y ;10 Mp ; (Mp ; Mr )( r;;pp ) Mp
ZxFy = (17:4)(12) (65) = 94:2 3 Sx (Fy ; Fr ) = (14:9)(12)(65;10) = 68:3 94:2 ; (94:2 ; 68:3) 198::80;;88::11 = 92:5 p 0:90(92:5) = 83:25 > Mu 3
in
ksi
k.ft
in/ft
in
ksi
in/ft
:
k.ft
k.ft
k.ft
Therefore provide W12X14 section.
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 11
REINFORCED CONCRETE BEAMS 11.1 Introduction Recalling that concrete has a tensile strength (ft0 ) about one tenth its compressive strength (fc0 ), concrete by itself is a very poor material for exural members. 2 To provide tensile resistance to concrete beams, a reinforcement must be added. Steel is almost universally used as reinforcement (longitudinal or as bers), but in poorer countries other indigenous materials have been used (such as bamboos). 3 The following lectures will focus exclusively on the exural design and analysis of reinforced concrete rectangular sections. Other concerns, such as shear, torsion, cracking, and de ections are left for subsequent ones. 4 Design of reinforced concrete structures is governed in most cases by the Building Code Requirements for Reinforced Concrete, of the American Concrete Institute (ACI318). Some of the most relevant provisions of this code are enclosed in this set of notes. 5 We will focus on determining the amount of exural (that is longitudinal) reinforcement required at a given section. For that section, the moment which should be considered for design is the one obtained from the moment envelope at that particular point.
1
11.1.1 Notation 6
In R/C design, it is customary to use the following notation
Draft 11{2 As b c d fc0 fr0 fs0 ft0 fy h
REINFORCED CONCRETE BEAMS
Area of steel Width Distance from top of compressive bers to neutral axis Distance from the top of the compressive bers to the centroid of the reinforcement Concrete compressive strength Concrete modulus of rupture Steel stress Concrete tensile strength Steel yield stress (equivalent to Fy in AISC) Height Steel ratio, Abds
11.1.2 Modes of Failure A reinforced concrete beam is a composite structure where concrete provides the compression and steel the tension. 8 Failure is initiated by, Fig. 11.5:
7
Steel Yielding
Concrete Crushing
Figure 11.1: Failure Modes for R/C Beams 1. Yielding of the steel when the steel stress reaches the yield stress (fs = fy ). This occurs if we do not have enough reinforcement that is the section is underreinforced. This will result in excessive rotation and deformation prior to failure. 2. Crushing of the concrete, when the concrete strain reaches its ultimate value ("c = "u = 0:003), ACI 318: 10.2.3. This occurs if there is too much reinforcement that is the section is overreinforced. This is a sudden mode of failure. 9 Ideally in an optimal (i.e. most ecient use of materials) design, a section should be dimensioned such that crushing of concrete should occur simultaneously with steel yielding. This would then be a balanced design. 10 However since concrete crushing is a sudden mode of failure with no prior warning, whereas steel yielding is often accompanied by excessive deformation (thus providing ample warning of an imminent failure), design codes require the section to be moderately underreinforced.
11.1.3 Analysis vs Design 11
In R/C we always consider one of the following problems:
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11.1 Introduction
11{3
Analysis: Given a certain design, determine what is the maximum moment which can be applied. Design: Given an external moment to be resisted, determine cross sectional dimensions (b and h) as well as reinforcement (As ). Note that in many cases the external dimensions of the beam (b and h) are xed by the architect.
12
We often consider the maximum moment along a member, and design accordingly.
11.1.4 Basic Relations and Assumptions In developing a design/analysis method for reinforced concrete, the following basic relations will be used, Fig. ??: 13
Compatibility
Equilibrium
C d
εy
T
T=C M_ext=Cd Figure 11.2: Internal Equilibrium in a R/C Beam 1. Equilibrium: of forces and moment at the cross section. 1) Fx = 0 or Tension in the reinforcement = Compression in concrete; and 2) M = 0 or external moment (that is the one obtained from the moment envelope) equal and opposite to the internal one (tension in steel and compression of the concrete). 2. Material Stress Strain: We recall that all normal strength concrete have a failure strain u = :003 in compression irrespective of fc0 . 14
Basic assumptions used:
Compatibility of Displacements: Perfect bond between steel and concrete (no slip). Note that those two materials do also have very close coecients of thermal expansion under normal temperature. Plane section remain plane ) strain is proportional to distance from neutral axis.
11.1.5 ACI Code 15 The ACI code is based on limit strength, or Mn Mu thus a similar design philosophy is used as the one adopted by the LRFD method of the AISC code, ACI318: 8.1.1; 9.3.1; 9.3.2 16 The required strength is based on (ACI318: 9.2)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11{4
REINFORCED CONCRETE BEAMS U = 1:4D + 1:7L (11.1) = 0:75(1:4D + 1:7L + 1:7W ) (11.2)
11.2 Cracked Section, Ultimate Strength Design Method
11.2.1 Equivalent Stress Block
17 In determining the limit state moment of a cross section, we consider Fig. 11.3. Whereas the strain distribution is linear (ACI318 10.2.2), the stress distribution is nonlinear because the stressstrain curve of concrete is itself nonlinear beyond 0:5fc0 . 18 Thus we have two alternatives to investigate the moment carrying capacity of the section, ACI318:
10.2.6
1. Use the nonlinear stress distribution. 2. Use a simpler equivalent stress distribution. The ACI code follows the second approach. Thus we seek an equivalent stress distribution such that: 1. The resultant force is equal 2. The location of the resultant is the same. We note that this is similar to the approach followed in determining reactions in a beam subjected to a distributed load when the load is replaced by a single force placed at the centroid.
19
Figure 11.3: Cracked Section, Limit State
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11.2 Cracked Section, Ultimate Strength Design Method 20
11{5
It was shown that the depth of the equivalent stress block is a function of fc0 :
1 = :85 if fc0 4; 000 1 0 = :85 ; (:05)(fc ; 4; 000) 1;000 if 4; 000 < fc0 < 8; 000
(11.3)
Figure 11.4: Whitney Stress Block
11.2.2 Balanced Steel Ratio Next we seek to determine the balanced steel ratio b such that failure occurs by simultaneous yielding of the steel fs = fy and crushing of the concrete "c = 0:003, ACI318: 10.3.2 We will separately consider the two failure possibilities: Tension Failure: we stipulate that the steel stress is equal to fy :
21
= Abds As fy = :85fc0 ab = :85fc0 b 1 c
) c = 0:85ffy0 d c 1
(11.4)
Compression Failure: where the concrete strain is equal to the ultimate strain; From the strain diagram
"c = 0:003
c :003 d = :003+"s
22
)c=
:003
d fs Es + :003
(11.5)
Balanced Design is obtained by equating Eq. 11.4 to Eq. 11.5 and by replacing by b and fs by fy : fy
= fs:003 d Es +:003 fs = fy = b
0:85fc0 1 d
When we replace Es by 29; 000
ksi
9 > = > ;
:003 b fy :85fc0 1 d = Efys + :003 d
we obtain 0 ; 000 b = :85 1 ffc 87;87000 +f y
y
(11.6)
This b corresponds to the only combination of b, d and As which will result in simultaneous yielding of the steel and crushing of the concrete, that is an optimal design.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11{6
REINFORCED CONCRETE BEAMS
23 Because we need to have ample warning against failure, hence we prefer to have an underreinforced section. Thus, the ACI code stipulates:
(11.7)
< :75b
In practice, depending on the relative cost of steel/concrete and of labour it is common to select lower values of . If < 0:5b (thus we will have a deeper section) then we need not check for de ection. 25 A minimum amount of reinforcement must always be used to prevent temperature and shrinkage cracks: (11.8) 200 24
min
26
fy
The ACI code adopts the limit state design method
MD = Mn > Mu
(11.9)
= b = 0:9
11.2.3 Analysis
Given As , b, d, fc0 , and fy determine the design moment: 1. act = Abds 0
87 2. b = (:85) 1 ffyc 87+ fy
3. If act < b (that is failure is triggered by yielding of the steel, fs = fy )
a = :A85sffc0yb ;From Equilibrium MD = As fy d ; a2
)
MD = As fy d ; 0:59 Afs0fby c  {z Mn
}
Combining this last equation with = Abds yields
MD = fy
bd2
1 ; :59 ffy0 c
(11.10)
4. y If act > b is not allowed by the code as this would be an overreinforced section which would fail with no prior warning. However, if such a section exists, and we need to determine its moment carrying capacity, then we have two unknowns: (a) Steel strain "s (which was equal to "y in the previous case) (b) Location of the neutral axis c. We have two equations to solve this problem Equilibrium: of forces s (11.11) c = :85Afs0fb c 1
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11.2 Cracked Section, Ultimate Strength Design Method
11{7
Strain compatibility: since we know that at failure the maximum compressive strain "c is equal to 0.003. Thus from similar triangles we have
c = :003 d :003 + "s
(11.12)
Those two equations can be solved by either one of two methods: (a) Substitute into one single equation (b) By iteration Once c and fs = E"s are determined then
MD = As fs d ; 21 c
(11.13)
11.2.4 Design We distinguish between two cases. The rst one has dimensions as well as steel area unknown, the second has the dimensions known (usually speci ed by the architect or by other constraints), and we seek As .
27
b, d and As unknowns and MD known:
1. We start by assuming , at most = :75b, and if de ection is of a concern (or steel too expensive), then we can select = 0:5b with b determined from Eq. 11.6
0 ; 000 = 0:75 :85 1 ffc 87;87000 +f y
2. From Eq. 11.10
MD = fy 1 ; :59 ffy0 bd2 
or
y
{z
R
c
}
f y R = fy 1 ; :59 f 0 c
(11.14)
which does not depend on unknown quantities1 . Then solve for bd2 : D bd2 = M R
(11.15)
3. solve for b and d (this will require either an assumption on one of the two, or on their ratio). 4. As = bd 1
Note analogy with Eq. 10.6 Mp = Fy Z for stell beams.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11{8
REINFORCED CONCRETE BEAMS
y b, d and Md known, As unknown: In this case there is no assurance that we can have a design
with b . If the section is too small, then it will require too much steel resulting in an overreinforced section. We will again have an iterative approach 1. Since we do not know if the steel will be yielding or not, use fs . 2. Assume an initial value for a (a good start is a = d5 ) 3. Assume initially that fs = fy 4. Check equilibrium of moments (M = 0)
M D ; fs d ; a2 5. Check equilibrium of forces in the x direction (Fx = 0) a = :A85sff0sb c 6. Check assumption of fs from the strain diagram "s :003 d;c d ; c = c ) fs = Es c :003 < fy where c = a1 . As =
(11.16) (11.17) (11.18)
7. Iterate until convergence is reached.
Example 1121: Ultimate Strength Capacity Determine the ultimate Strength of a beam with the following properties: b = 10 , d = 23 , 2 , f 0 = 4; 000 and fy = 60 . c
As = 2:35
Solution:
in
in
psi
2:35 = :0102 = Abds = (10)(23) 0 87 = :02885 > act p = :85 1 ffyc 87+87fy = (:85)(:85) 604 87+60 :35)(60) = 4:147 = :A85sffc0yb = (:(285)(4)(10) = (2:35)(60)(23 ; 4:147 2 ) = 2; 950 = Mn = (:9)(2; 950) = 2; 660 Note that from the strain diagram c = 0:a85 = 40::414 85 = 4:87 Alternative solution 2 1 ; :59act fy0 Mn = act fy bd f c = As fy d 1 ; :59act ffyc0 = 245 = (2:35)(60)(23) 1 ; (:59) 604 (:01021) = 2; 950 MD = Mn = (:9)(2; 950) = 2; 660
act b a Mn MD
in
ksi
in
k.in
k.in
in
k.in
k.ft
k.in
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11.2 Cracked Section, Ultimate Strength Design Method
11{9
Example 1122: Beam Design I Design a 15 ft beam to support a dead load of 1.27 k/ft, a live load of 2.44 k/ft using a 3,000 psi concrete and 40 ksi steel. Neglect beam weight
Solution:
wu = 1:4(1:27) + 1:7(22 :44) = 5:92 2 MD = (5:92) 0 8 (15) (12) = 2; 000 b = :85 1 ffyc 87+87fy 87 = 0:040 = (:85)(:85) 403 87+40 = :75b = :75(0:04) = 0:030 R = fy 1 ; :59 fy0 fc = (0:03)(40) 1: ; (0:59)(02:03) 403 = 0:917 2;000 d bd2 = M = 2; 423 3 R = (0:9)(0:917) k/ft
k/ft
ft
in/ft
k.in
k.in in
ksi
in
k
q
Assume b = 10 , this will give d = 2;10423 = 15:57 . We thus adopt b = 10 Finally, As = bd = (0:030)(10)(16) = 4:80 2 we select 3 bars No. 11 in
in
and d = 16
in
in
.
in
Example 1123: Beam Design II Select the reinforcement for a cross section with b = 11:5 ; d = 20 using fc0 = 3 ; and fy = 40
Md = 1; 600
Solution:
in
k.in
ksi
in
to support a design moment
ksi
1. Assume a = d5 = 205 = 4 and fs = fy 2. Equilibrium of moments: in
D As = f M (d ; a ) == y
2
1; 600 (:9)(40) (20 ; 42 ) k.in
ksi
in
= 2:47
2
in
3. Check equilibrium of forces:
:47) 2 (40) a = :A85sff0yb = (:(2 85)(3) (11:5) c in
ksi
ksi in
= 3:38
in
4. We originally assumed a = 4, at the end of this rst iteration a = 3:38, let us iterate again with a = 3:30
Victor Saouma
Structural Concepts and Systems for Architects
Draft 11{10
REINFORCED CONCRETE BEAMS
5. Equilibrium of moments:
D As = f M (d ; a ) == y
2
1; 600 (:9)(40) (20 ; 32:3 ) k.in
ksi
in
= 2:42
6. Check equilibrium of forces:
:42) 2 (40) a = :A85sff0yb = (:(2 85)(3) (11:5) c in
ksi
ksi in
= 3:3
in
in
2
p
7. we have converged on a. :42 = :011 8. Actual is act = (112:5)(20) 9. b is equal to 0 3 87 = :037 = ( : 85)( : 85) b = :85 1 ffc 87 87 +f 40 87 + 40 y
y
p
10. max = :75 = (0:75)(0:037) = :0278 > 0:011 thus fs = fy and we use As = 2:42
in
2
11.3 Continuous Beams Whereas coverage of continuous reinforced concrete beams is beyond the scope of this course, Fig. ?? illustrates a typical reinforcement in such a beam.
28
11.4 ACI Code Attached is an unauthorized copy of some of the most relevant ACI31889 design code provisions. 8.1.1  In design of reinforced concrete structures, members shall be proportioned for adequate strength in accordance with provisions of this code, using load factors and strength reduction factors speci ed in Chapter 9. 8.3.1  All members of frames or continuous construction shall be designed for the maximum eects of factored loads as determined by the theory of elastic analysis, except as modi ed according to Section 8.4. Simplifying assumptions of Section 8.6 through 8.9 may be used. p 8.5.1  Modulus of elasticity Ec for concrete may be taken as Wc1:533 fc0 ( psi) forpvalues of Wc between 90 and 155 lb per cu ft. For normal weight concrete, Ec may be taken as 57; 000 fc0 . 8.5.2  Modulus of elasticity Es for nonprestressed reinforcement may be taken as 29,000 psi. 9.1.1  Structures and structural members shall be designed to have design strengths at all sections at least equal to the required strengths calculated for the factored loads and forces in such combinations as are stipulated in this code. 9.2  Required Strength 9.2.1  Required strength U to resist dead load D and live load L shall be at least equal to U = 1:4D + 1:7L 9.2.2  If resistance to structural eects of a speci ed wind load W are included in design, the following combinations of D, L, and W shall be investigated to determine the greatest required strength U U = 0:75(1:4D + 1:7L + 1:7W )
Victor Saouma
Structural Concepts and Systems for Architects
Victor Saouma Stirrups
Straight top bar
(d) Straight and bent bar reinforcement
Exterior span
Bent bar at noncontinuous end
Straight bottom bar
No.3 stirrup support if necessary
(c) Straight bar reinforcement
Exterior span
Straight bottom bar
Stirrups
(b) Moment diagram under typical loading
(a) Deflected shape
Interior span
Interior column
Interior span
Interior column
Points of deflection
Straight bar
Bent bars
Cracks
Cracks
Interior span
Interior span
Reinforcement
Bottom bars
Bottom bars
Bent bar
Section through beam
Stirrups
Top bars
Section through beam
Stirrups
Top bars
Draft 11.4 ACI Code 11{11
Figure 11.5: Reinforcement in Continuous R/C Beams
Structural Concepts and Systems for Architects
Draft 11{12
REINFORCED CONCRETE BEAMS
where load combinations shall include both full value and zero value of L to determine the more severe condition, and U = 0:9D + 1:3W but for any combination of D, L, and W, required strength U shall not be less than Eq. (91). 9.3.1  Design strength provided by a member, its connections to other members, and its cross sections, in terms of exure, axial load, shear, and torsion, shall be taken as the nominal strength calculated in accordance with requirements and assumptions of this code, multiplied by a strength reduction factor . 9.3.2  Strength reduction factor shall be as follows: 9.3.2.1  Flexure, without axial load 0.90 9.4  Design strength for reinforcement Designs shall not be based on a yield strength of reinforcement fy in excess of 80,000 psi, except for prestressing tendons. 10.2.2  Strain in reinforcement and concrete shall be assumed directly proportional to the distance from the neutral axis, except, for deep exural members with overall depth to clear span ratios greater than 2/5 for continuous spans and 4/5 for simple spans, a nonlinear distribution of strain shall be considered. See Section 10.7. 10.2.3  Maximum usable strain at extreme concrete compression ber shall be assumed equal to 0.003. 10.2.4  Stress in reinforcement below speci ed yield strength fy for grade of reinforcement used shall be taken as Es times steel strain. For strains greater than that corresponding to fy , stress in reinforcement shall be considered independent of strain and equal to fy . 10.2.5  Tensile strength of concrete shall be neglected in exural calculations of reinforced concrete, except when meeting requirements of Section 18.4. 10.2.6  Relationship between concrete compressive stress distribution and concrete strain may be assumed to be rectangular, trapezoidal, parabolic, or any other shape that results in prediction of strength in substantial agreement with results of comprehensive tests. 10.2.7  Requirements of Section 10.2.5 may be considered satis ed by an equivalent rectangular concrete stress distribution de ned by the following: 10.2.7.1  Concrete stress of 0:85fc0 shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross section and a straight line located parallel to the neutral axis at a distance (a = 1 c) from the ber of maximum compressive strain. 10.2.7.2  Distance c from ber of maximum strain to the neutral axis shall be measured in a direction perpendicular to that axis. 10.2.7.3  Factor 1 shall be taken as 0.85 for concrete strengths fc0 up to and including 4,000 psi. For strengths above 4,000 psi, 1 shall be reduced continuously at a rate of 0.05 for each 1000 psi of strength in excess of 4,000 psi, but 1 shall not be taken less than 0.65. 10.3.2  Balanced strain conditions exist at a cross section when tension reinforcement reaches the strain corresponding to its speci ed yield strength fy just as concrete in compression reaches its assumed ultimate strain of 0.003. 10.3.3  For exural members, and for members subject to combined exure and compressive axial load when the design axial load strength (Pn ) is less than the smaller of (0:10fc0 Ag ) or (Pb ), the ratio of reinforcement p provided shall not exceed 0.75 of the ratio b that would produce balanced strain conditions for the section under exure without axial load. For members with compression reinforcement, the portion of b equalized by compression reinforcement need not be reduced by the 0.75 factor. 10.3.4  Compression reinforcement in conjunction with additional tension reinforcement may be used to increase the strength of exural members. 10.5.1  At any section of a exural member, except as provided in Sections 10.5.2 and 10.5.3, where positive reinforcement is required by analysis, the ratio provided shall not be less than that given by = 200 min
Victor Saouma
fy
Structural Concepts and Systems for Architects
Draft Chapter 12
PRESTRESSED CONCRETE 12.1 Introduction Beams with longer spans are architecturally more appealing than those with short ones. However, for a reinforced concrete beam to span long distances, it would have to have to be relatively deep (and at some point the self weight may become too large relative to the live load), or higher grade steel and concrete must be used. 2 However, if we were to use a steel with fy much higher than 60 ksi in reinforced concrete (R/C), then to take full advantage of this higher yield stress while maintaining full bond between concrete and steel, will result in unacceptably wide crack widths. Large crack widths will in turn result in corrosion of the rebars and poor protection against re. 3 One way to control the concrete cracking and reduce the tensile stresses in a beam is to prestress the beam by applying an initial state of stress which is opposite to the one which will be induced by the load. 4 For a simply supported beam, we would then seek to apply an initial tensile stress at the top and compressive stress at the bottom. In prestressed concrete (P/C) this can be achieved through prestressing of a tendon placed below the elastic neutral axis. 5 Main advantages of P/C: Economy, de ection & crack control, durability, fatigue strength, longer spans. 6 There two type of Prestressed Concrete beams: Pretensioning: Steel is rst stressed, concrete is then poured around the stressed bars. When enough concrete strength has been reached the steel restraints are released, Fig. 12.1. Postensioning: Concrete is rst poured, then when enough strength has been reached a steel cable is passed thru a hollow core inside and stressed, Fig. 12.2.
1
12.1.1 Materials P/C beams usually have higher compressive strength than R/C. Prestressed beams can have fc0 as high as 8,000 psi. 8 The importance of high yield stress for the steel is illustrated by the following simple example. If we consider the following:
7
Draft 12{2
PRESTRESSED CONCRETE
Vertical bulkhead
Harping holdup point
Harping holddown point Jacks
Anchorage
Prestressing bed slab
Continuous tendon
Precast Concrete element Tendon anchorage
Jacks
Support force
Casting bed
Jacks
Casting bed
Holddown force
Tendon
Figure 12.1: Pretensioned Prestressed Concrete Beam, (Nilson 1978)
Anchorage
Anchorage
Intermediate diaphragms
Jack
Beam
Jack
Tendon in conduct
Anchorage Jack
Slab
Wrapped tendon
Figure 12.2: Posttensioned Prestressed Concrete Beam, (Nilson 1978)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12.1 Introduction
12{3
1. An unstressed steel cable of length ls 2. A concrete beam of length lc 3. Prestress the beam with the cable, resulting in a stressed length of concrete and steel equal to ls0 = lc0 . 4. Due to shrinkage and creep, there will be a change in length lc = ("sh + "cr )lc
(12.1)
we want to make sure that this amout of deformation is substantially smaller than the stretch of the steel (for prestressing to be eective). ;3 = 5. Assuming ordinary steel: fs = 30 , Es = 29; 000 , "s = 2930 ;000 = 1:03 10 ksi
in
ksi
in
6. The total steel elongation is "s ls = 1:03 10;3ls 7. The creep and shrinkage strains are about "cr + "sh ' :9 10;3 8. The residual stres which is left in the steel after creep and shrinkage took place is thus (1:03 ; :90) 10;3(29 103) = 4
(12.2)
ksi
Thus the total loss is 3030;4 = 87% which is unacceptably too high. 9. Alternatively if initial stress was 150 after losses we would be left with 124 10. Note that the actual loss is (:90 10;3)(29 103 ) = 26 in each case ksi
ksi
or a 17% loss.
ksi
Having shown that losses would be too high for low strength steel, we will use Strands usually composed of 7 wires. Grade 250 or 270 ksi, Fig. 12.3.
9
0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 000000 111111 0000000 1111111 0000000 1111111 000000 111111 000000 111111 000000 111111 0000000 1111111 0000000 1111111 000000 111111 000000 111111 000000 111111 0000000 1111111 0000000 1111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 0000000 1111111 000000 111111 000000 111111 000000 111111 000000 111111 0000000 1111111 000000 111111 000000 111111 000000 111111 0000000 1111111 000000 111111 0000000 1111111 000000 111111 0000000 1111111 000000 111111 0000000 1111111 000000 111111 0000000 1111111 000000 111111 0000000 1111111 000000 111111 0000000 1111111 000000 111111 0000000 1111111 111111 000000 0000000 1111111
Figure 12.3: 7 Wire Prestressing Tendon
Tendon have diameters ranging from 1/2 to 1 3/8 of an inch. Grade 145 or 160 ksi. Wires come in bundles of 8 to 52. Note that yield stress is not well de ned for steel used in prestressed concrete, usually we take 1% strain as eective yield. 10 Steel relaxation is the reduction in stress at constant strain (as opposed to creep which is reduction of strain at constant stress) occrs. Relaxation occurs inde nitely and produces signi cant prestress loss. If we denote by fp the nal stress after t hours, fpi the initial stress, and fpy the yield stress, then
fp = 1 ; log t fpi ; :55 fpi 10 fpy
Victor Saouma
(12.3)
Structural Concepts and Systems for Architects
Draft 12{4
PRESTRESSED CONCRETE
12.1.2 Prestressing Forces 11
Prestress force \varies" with time, so we must recognize 3 stages: 1. Pj Jacking force. But then due to (a) friction and anchorage slip in posttension (b) elastic shortening in pretension is reduced to: 2. Pi Initial prestress force; But then due to time dependent losses caused by (a) relaxation of steel (b) shrinkage of concrete (c) creep of concrete is reduced to: 3. Pe Eective force
12.1.3 Assumptions 12
The following assumptions are made; 1. Materials are both in the elastic range 2. section is uncracked 3. sign convention: +ve tension, ;ve compression 4. Subscript 1 refers to the top and 2 to the bottom 5. I; S1 = cI1 ; S2 = cI2 6. e + ve if downward from concrete neutral axis
12.1.4 Tendon Con guration Through proper arrangement of the tendon (eccentricity at both support and midspan) various internal
exural stress distribution can be obtained, Fig. 12.4.
13
12.1.5 Equivalent Load An equivalent load for prestressing can be usually determined from the tendon con guration and the prestressing force, Fig. 12.5. 14
12.1.6 Load Deformation 15
The loaddeformation curve for a prestressed concrete beam is illustrated in Fig. 12.6.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12.2 Flexural Stresses
12{5 W 000 111 111 000 000 111 000 111 000 111
h
f’y
Q P
P
h/2
fc 00 11 11 00 00 11 00 11 00 11
+
fc
fc 000 111 111 000 000 111 000 111 000 111
=
2f c 00 11 11 00 00 11 00 11 00 11
fc =f t
2Q
P
P
2h/3
2Q P
P
h/2 h/3
2f c 0 11 00 0000000 1111111 00 11 0000000 00 +1111111 11 0000000 = 1111111 00 11 0000000 1111111 00 1111111 11 0000000 2f c 2f =2f t c
2f c 11 00 00 11 00 11 00 11 00 11 0
0 2f c 11 00 0000000 1111111 00 11 0000000 00 +1111111 11 0000000 = 1111111 00 11 0000000 1111111 00 1111111 11 0000000 2f c 2f t =2f c fc Midspan 000 111 000 + 111 = 0 000 111 000 111 000 111 Ends fc
2f c 11 00 00 11 00 11 00 11 00 11 0 fc 00 11 00 11 00 11 00 11 00 11 fc
0 00 11 00 11 00 11 00 11 00 11 2f c
Q P
P
h/2 h/3
fc 00 11 00 11 00 11 00 11 00 11 fc
+
f c 111 000 000 111 000 111 000 111 000 111
=
ft =f c
fc
Midspan +
0
fc 11 00 00 11 00 11 00 11 00 11
=
Ends
fc 11 00 00 11 00 11 00 11 00 11 fc
Figure 12.4: Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978)
12.2 Flexural Stresses We now identify the following 4 stages: Initial Stage when the beam is being prestressed (recalling that r2 = AIc 1. the prestressing force, Pi only 16
f1 = ; APi + PiIec1 = ; APi 1 ; ecr21 (12.4) c c P P ec P i i 2 i f2 = ; A ; I = ; A 1 + ecr22 (12.5) c c
2. Pi and the self weight of the beam M0 (which has to be acconted for the moment the beam cambers due to prestressing) 0 (12.6) f1 = ; APi 1 ; ecr21 ; M S1 c M P ec i 2 f2 = ; A 1 + r2 + S 0 (12.7)
c
2
Service Load when the prestressing force was reduced from Pi to Pe beacause of the losses, and the actual service (not factored) load is apllied 3. Pe and M0
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12{6
PRESTRESSED CONCRETE Member
Equivalent load on concrete from tendon Moment from prestressing
(a) P
P sin θ
P
θ
P sin θ
P cos θ
P cos θ 2 P sin θ
(b) θ
P
P sin θ
P
P sin θ
P cos θ
P cos θ
(c) Pe P
Pe P
P
e
P
(d) P
P
θ M
P sin θ
P sin θ
M
e P cos θ
P cos θ
(e) P P
P
P sin θ
P sin θ
P cos θ
None
P cos θ 2 P sin θ
(f) P P
None
(g) P
P
Figure 12.5: Determination of Equivalent Loads Load
Ultimate Steel yielding Service load limit including tolerable overload
Ru
ptu
re
Overload
Tn
Service load range
First cracking load fcr
Decompression
or higher cgs (f=0)
Balanced Full dead load
∆o ∆D
∆ pe ∆ pi
∆L
Deformation ∆ (deflection of camber) ∆ pi= Initial prestress camber ∆ pe= Effective prestress camber ∆ O= Selfweight deflection ∆ D= Dead load deflection ∆ L= Live load deflection
Figure 12.6: LoadDe ection Curve and Corresponding Internal Flexural Stresses for a Typical Prestressed Concrete Beam, (Nilson 1978)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12.2 Flexural Stresses
12{7 0 f1 = ; APe 1 ; ecr21 ; M S1 (12.8) c 0 f2 = ; APe 1 + ecr22 + M S (12.9)
c
2
4. Pe and M0 + MDL + MLL f1 = ; APe 1 ; ecr21 ; M0 + MSDL + MLL (12.10) c 1 M +M P ec + MLL (12.11) e 2 0 DL f2 = ; A 1 + r 2 + S
c
2
The internal stress distribution at each one of those four stages is illustrated by Fig. 12.7. Pi Ac
c1 e
Pi e c 1 Ic
11 00 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11
c2
11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111
Pi (1Ac
e c1 ) r2

111111111 000000000 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111
Pi (1+ Ac
Stage 2
Pe (1Ac
Stage 4
e c2 ) r2
e c1 Mo )r2 S1
000000 111111 111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111
Pe (1+ Ac
Mo S1
111 000 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111
+
+
Pi (1Ac
Md + Ml S1
Md + Ml S2
e c2 ) r2
e c1 Mo )r2 S1
111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111
Pi (1+ Ac
00000000000 11111111111 11111111111 00000000000 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111
e c2 Mo )+ r2 S2
Pi (1+ Ac
Mo S2

e c1 ) r2
111111111 000000000 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111
Pi e c 2 Ic
Pi Ac
Stage 1
Pi (1Ac
Pe (1Ac
e c2 Mo )+ r2 S2
e c1 Mt )r2 S1
111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111
Pe (1+ Ac
e c2 Mt )+ r2 S2
Figure 12.7: Flexural Stress Distribution for a Beam with Variable Eccentricity; Maximum Moment Section and Support Section, (Nilson 1978) 17
Those (service) exural stresses must be below those speci ed by the ACI code (where the subscripts
c, t, i and s refer to compression, tension, initial and service respectively):
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12{8
PRESTRESSED CONCRETE
0 permitted concrete compression stress at initial stage :60fp ci permitted concrete tensile stress at initial stage < 3 fci0 0 permitted concrete compressive stress at service stage :45 p pfc permitted concrete p tensile stress at initial stage 6 fc0 or 12 fc0 Note that fts can reach 12 fc0 only if appropriate de ection analysis is done, because section would be cracked. 18 Based on the above, we identify two types of prestressing: Full prestressing (pioneered by Freysinet), no tensile stresses, no crack, but there are some problems with excessive camber when unloaded. Partial prestressing (pioneered by Leonhardt, Abeles, Thurliman), cracks are allowed to occur (just as in R/C), and they are easier to control in P/C than in R/C.
fci fti fcs fts
The ACi code imposes the following limits on the steel stresses in terms of fpu which is the ultimate strength of the cable: Pj < :80fpuAs and Pi < :70fpuAs . No limits are speci ed for Pe .
19
Example 1224: Prestressed Concrete I Beam Adapted from (Nilson 1978)
The following I Beam has fc0 = 4; 000 , L = 40 ft, DL+LL =0.55 k/ft, concrete density = 150 lb/ft3 and multiple 7 wire strands with constant eccentricity e = 5:19 . Pi = 169 , and the total losses due to creep, shinkage, relaxation are 15%. psi
in
k
12" 4" 5"
2"
7"
6"
4"
24" 6"
7"
2" 5"
r2
4"
The section properties for this beam are Ic = 12; 000 4 , Ac = 176 2 , S1 = S2 = 1; 000 3 , = AI = 68:2 2 . Determine exural stresses at midspan and at support at initial and nal conditions. Solution: in
in
in
in
1. Prestressing force, Pi only
f1 = ; APi 1 ; ecr21 c 169 ; 000 1 ; (5:19)(12) = ;83 = ; 176 68:2 P ec f2 = ; Ai 1 + r22 c 169 ; 000 (5 : 19)(12) = ; 176 1 + 68:21 = 1; 837
(12.12a)
psi
Victor Saouma
(12.12b) (12.12c)
psi
(12.12d)
Structural Concepts and Systems for Architects
Draft 12.2 Flexural Stresses
12{9
2. Pi and the self weight of the beam M0 (which has to be acconted for the moment the beam cambers due to prestressing) (176) 2 (:150) = 3 = :183 (12.13a) w0 = (144) 2= 2 2 M0 = (:183)(40) = 36:6 (12.13b) 8 The exural stresses will thus be equal to: ; 000) = 439 f1w;20 = SM0 = (36:6)(12 (12.14) 1 ; 000 1;2 in
k
in
ft
k/ft
ft
k.ft
psi
0 f1 = ; APi 1 ; ecr21 ; M S c 1
(12.15a)
= ;83 ; 439 = ;522 p fti = 3 fc0 = +190p 0 f2 = ; APi 1 + ecr22 + M S
(12.15b) (12.15c) (12.15d)
psi
c
2
= ;1; 837 + 439 = ;1; 398 fci = :6fc0 = ;2; 400p
psi
(12.15e) (12.15f)
3. Pe and M0. If we have 15% losses, then the eective force Pe is equal to (1 ; 0:15)169 = 144 0 f1 = ; APe 1 ; ecr21 ; M S c 1 ; 000 1 ; (5:19)(12) ; 439 = ; 144176 68:2
= ;71 ; 439 = ;510
psi
0 f2 = ; APe 1 + ecr22 + M S c 2 144 ; 000 (5 : 19)(12) = ; 176 1 + 68:2 + 439 = ;1; 561 + 439 = ;1; 122
psi
k
(12.16a) (12.16b) (12.16c) (12.16d) (12.16e) (12.16f)
note that ;71 and ;1; 561 are respectively equal to (0:85)(;83) and (0:85)(;1; 837) respectively. 4. Pe and M0 + MDL + MLL 2 MDL + MLL = (0:55)(40) = 110 (12.17) 8 and corresponding stresses ; 000) = 1; 320 f1;2 = (110)(12 (12.18) 1; 000 Thus, (12.19a) f = ; Pe 1 ; ec1 ; M0 + MDL + MLL k.ft
psi
1
Victor Saouma
Ac
r2
S1
Structural Concepts and Systems for Architects
Draft 12{10
PRESTRESSED CONCRETE = ;510 ; 1; 320 = ;1; 830 fcs = :45fc0 = ;2; 700p f2 = ; APe 1 + ecr22 + M0 + MSDL + MLL c 2 = ;1; 122 + 1; 320 = +198 p fts = 6 fc0 = +380p
(12.19b) (12.19c) (12.19d)
psi
(12.19e) (12.19f)
psi
4
2 1122
+198
1 1398
3
1837
83 510 522
1830
5. The stress distribution at each one of the four stages is shown below.
12.3 Case Study: Walnut Lane Bridge Adapted from (Billington and Mark 1983)
The historical Walnut Lane Bridge ( rst major prestressed concrete bridge in the USA) is made of three spans, two side ones with lengths of 74 ft and a middle one of length 160 feet. Thirteen prestressed cocnrete beams are placed side by side to make up a total width of 44 fet of roadway and two 9.25 feet of sidewalk. In between the beams, and cast with them, are transverse stieners which connect the beams laterally, Fig. 12.8 20
12.3.1 CrossSection Properties 21
The beam cross section is shown in Fig. 12.9 and is simpli ed
Ac = 2(8" :9)(52) + (7)(61:2) = 1; 354 2 2 # 3 3 (52)(8 : 9) 79 8 : 9 I = 2 + (7)(61:2) + (52)(8:9) ; in
12 = = 1; 277 103 = h2 = 79 2 = 39:5
4
in
c1 = c2
Victor Saouma
in
2
2
12
(12.20a) (12.20b) (12.20c) (12.20d)
Structural Concepts and Systems for Architects
Draft 12.3 Case Study: Walnut Lane Bridge
12{11
80 ft CENTER LINE
ELEVATION OF BEAM HALF
9.25’
44 ’
ROAD
9.25’
SIDEWALK
BEAM CROSS SECTIONS
TRANSVERSE DIAPHRAGMS
CROSS  SECTION OF BRIDGE
52" 10" 3" 7"
TRANSVERSE DIAPHRAGM 10"
7"
3’3"
6’7" SLOTS FOR CABLES
6 1/2" 3 1/2" 7" 30"
CROSS  SECTION OF BEAM
Figure 12.8: Walnut Lane Bridge, Plan View
Victor Saouma
Structural Concepts and Systems for Architects
Draft 12{12
PRESTRESSED CONCRETE 52"
8.9"
22.5"
7"
22.5" 6’7" = 79"
61.2"
8.9"
SIMPLIFIED CROSS  SECTION OF BEAM
Figure 12.9: Walnut Lane Bridge, Cross Section
103 = 32; 329 S1 = S2 = Ic = 1; 277 39:5 1 ; 277 103 = 943: 2 I r2 = A = 1; 354
3
(12.20e)
in
(12.20f)
in
12.3.2 Prestressing Each beam is prestressed by two middle parabolic cables, and two outer horizontal ones along the
anges. All four have approximately the same eccentricity at midspan of 2.65 ft. or 31.8 inch. 23 Each prestressing cable is made up 64 wires each with a diameter of 0.27 inches. Thus the total area of prestressing steel is given by: 2 2 Awire = (d=2)2 = 3:14( 0:276 (12.21a) 2 ) = 0:0598 Acable = 64(0:0598) 2 = 3:83 2 (12.21b) 2 2 Atotal = 4(3:83) = 15:32 (12.21c)
22
in
in
in
in
in
in
Whereas the ultimate tensile strength of the steel used is 247 ksi, the cables have been stressed only to 131 ksi, thus the initial prestressing force Pi is equal to Pi = (131) (15:32) 2 = 2; 000 (12.22)
24
ksi
25
in
k
The losses are reported ot be 13%, thus the eective force is Pe = (1 ; 0:13)(2; 000) = 1; 740 k
Victor Saouma
k
(12.23)
Structural Concepts and Systems for Architects
Draft 12.3 Case Study: Walnut Lane Bridge
12{13
12.3.3 Loads The self weight of the beam is q0 = 1:72 . 27 The concrete (density=.15 = 3 ) road has a thickness of 0.45 feet. Thus for a 44 foot width, the total load over one single beam is 1 (44) (0:45) (0:15) = 3 = 0:23 (12.24) qr;tot = 13
26
k/ft
k
ft
ft
ft
k
ft
k/ft
Similarly for the sidewalks which are 9.25 feet wide and 0.6 feet thick: 1 (2)(9:25) (0:60) (0:15) = 3 = 0:13 qs;tot = 13 (12.25) We note that the weight can be evenly spread over the 13 beams beacause of the lateral diaphragms. 29 The total dead load is qDL = 0:23 + 0:13 = 0:36 (12.26) 28
ft
ft
k
ft
k/ft
k/ft
The live load is created by the trac, and is estimated to be 94 psf, thus over a width of 62.5 feet this gives a uniform live load of 1 (0:094) =ft2 (62:5) = 0:45 wLL = 13 (12.27)
30
k
31
ft
k/ft
Finally, the combined dead and live load per beam is
wDL+LL = 0:36 + 0:45 = 0:81
(12.28)
k/ft
12.3.4 Flexural Stresses 1. Prestressing force, Pi only
f1 = ; APi 1 ; ecr21 c
(12.29a)
106) 1 ; (31:8)(39:5) = 490: = ; (21; 354 943: ec P 2 i = ; A 1 + r2 c (2 106) 1 + (31:8)(39:5) = ;3; 445: = ; 1; 354 943:
(12.29b)
psi
f2
(12.29c) psi
(12.29d)
2. Pi and the self weight of the beam M0 (which has to be acconted for the moment the beam cambers due to prestressing) 2 M0 = (1:72)(160) = 5; 504 (12.30) 8 k.ft
The exural stresses will thus be equal to:
; 000) = 2; 043 f1w;20 = SM0 = (5; 50:4)(12 943: 1;2
Victor Saouma
psi
(12.31)
Structural Concepts and Systems for Architects
Draft 12{14
PRESTRESSED CONCRETE 0 f1 = ; APi 1 ; ecr21 ; M S c 1 = 490 ; 2; 043 = ;1; 553 p fti = 3 fc0 = +190p 0 f2 = APi 1 + ecr22 + M S
c
(12.32a) (12.32b) (12.32c) (12.32d)
psi
2
= ;3; 445 + 2; 043 = ;1; 402: fci = :6fc0 = ;2; 400p
(12.32e) (12.32f)
psi
3. Pe and M0 . If we have 13% losses, then the eective force Pe is equal to (1 ; 0:13)(2 106 ) = 1:74 106 lbs
0 f1 = ; APe 1 ; ecr21 ; M S c 1 6 (31 :5) ; 2; 043: = ;1; 616 1 : 74 10 = ; 1; 354 1 ; :8)(39 943: M ec P f2 = Ae 1 + r22 + S 0 c 2 6 1 : 74 10 (31 : 8)(39 : 5) = ; 1; 354 1 + 943: + 2; 043: = ;954:
(12.33a) psi
psi
(12.33b) (12.33c) (12.33d)
4. Pe and M0 + MDL + MLL 2
and corresponding stresses
MDL + MLL = (0:81)(160) = 2; 592 8 ; 000) = 962: f1;2 = (2; 592)(12 32; 329
k.ft
psi
(12.34) (12.35)
Thus, f1 = ; APe 1 ; ecr21 ; M0 + MSDL + MLL c 1 = ;1; 616 ; 962: = ;2; 578: fcs = :45fc0 = ;2; 700p f2 = APe 1 + ecr22 + M0 + MSDL + MLL psi
c
= ;954 + 962: = +8: p fts = 6 fc0 = +380p
Victor Saouma
2
psi
(12.36a) (12.36b) (12.36c) (12.36d) (12.36e) (12.36f)
Structural Concepts and Systems for Architects
Draft Chapter 13
ThreeHinges ARCHES 13.1 Theory
13.1.1 Uniform Horizontal Load In order to optimize deadload eciency, long span structures should have their shapes approximate the coresponding moment diagram, hence an arch, suspended cable, or tendon con guration in a prestressed concrete beam all are nearly parabolic, Fig. 13.1. 2 Long span structures can be built using at construction such as girders or trusses. However, for spans in excess of 100 ft, it is often more economical to build a curved structure such as an arch, suspended cable or thin shells. 3 Since the dawn of history, mankind has tried to span distances using arch construction. Essentially this was because an arch required materials to resist compression only (such as stone, masonary, bricks), and labour was not an issue. 4 The basic issues of static in arch design are illustrated in Fig. 13.2 where the vertical load is per unit horizontal projection (such as an external load but not a selfweight). Due to symmetry, the vertical reaction is simply V = wL 2 , and there is no shear across the midspan of the arch (nor a moment). Taking moment about the crown, L;L =0 (13.1) M = Hh ; wL 2 2 4
1
Solving for H
2
H = wL 8h
(13.2)
We recall that a similar equation was derived for arches., and H is analogous to the C ; T forces in a beam, and h is the overall height of the arch, Since h is much larger than d, H will be much smaller than C ; T in a beam. 5 Since equilibrium requires H to remain constant across thee arch, a parabolic curve would theoretically result in no moment on the arch section. 6 Threehinged arches are statically determinate structures which shape can acomodate support settlements and thermal expansion without secondary internal stresses. They are also easy to analyse through statics.
Draft 13{2
ThreeHinges ARCHES
2
M = w L /8
L w=W/L
IDEALISTIC ARCH SHAPE GIVEN BY MOMENT DIAGRAM
C RISE = h C BEAM +T W/2
MARM small C CT large BEAM T
C +T SAG = h
W/2
T
IDEALISTIC SUSPENSION SHAPE GIVEN BY MOMENT DIAGRAM
NOTE THAT THE "IDEAL" SHAPE FOR AN ARCH OR SUSPENSION SYSTEM IS EQUIVILENT TO THE DESIGN LOAD MOMENT DIAGRAM
Figure 13.1: Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Lin and Stotesbury 1981)
w
wL/2
w
H h
h
H
H
H = wL2 /8h
L
L/2 R
R V = wL/2
V
R = V 2+ H
2
V = wL/2 2
MCROWN = VL/2  wL /8  H h = 0 M BASE
2
= wL /8  H h = 0
Figure 13.2: Statics of a ThreeHinged Arch, (Lin and Stotesbury 1981)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13.1 Theory
13{3
An arch carries the vertical load across the span through a combination of axial forces and exural ones. A well dimensioned arch will have a small to negligible moment, and relatively high normal compressive stresses. 8 An arch is far more ecient than a beam, and possibly more economical and aesthetic than a truss in carrying loads over long spans. 9 If the arch has only two hinges, Fig. 13.3, or if it has no hinges, then bending moments may exist either at the crown or at the supports or at both places.
7
w
h’
w
M
h
h’ M base
H’
h
2
H’=wl /8h’< 2 wl /8h
APPARENT LINE OF PRESSURE WITH ARCH BENDING INCLUDING BASE
APPARENT LINE OF PRESSURE WITH ARCH BENDING EXCEPT AT THE BASE
H’
V
V
M crown
M base
h H
L V
V
Figure 13.3: Two Hinged Arch, (Lin and Stotesbury 1981) Since H varies inversely to the rise h, it is obvious that one should use as high a rise as possible. For a combination of aesthetic and practical considerations, a span/rise ratio ranging from 5 to 8 or perhaps as much as 12, is frequently used. However, as the ratio goes higher, we may have buckling problems, and the section would then have a higher section depth, and the arch advantage diminishes. 11 In a parabolic arch subjected to a uniform horizontal load there is no moment. However, in practice an arch is not subjected to uniform horizontal load. First, the depth (and thus the weight) of an arch is not usually constant, then due to the inclination of the arch the actual self weight is not constant. Finally, live loads may act on portion of the arch, thus the line of action will not necessarily follow the arch centroid. This last eect can be neglected if the live load is small in comparison with the dead load. p 2 2 V + H ), whereas at the crown 12 Since the greatest total force in the arch is at the support, (R = we simply have H , the crown will require a smaller section than the support.
10
w
h’
w
M
h
h’ M base
H’
h
2
H’=wl /8h’< 2 wl /8h
APPARENT LINE OF PRESSURE WITH ARCH BENDING INCLUDING BASE
APPARENT LINE OF PRESSURE WITH ARCH BENDING EXCEPT AT THE BASE
V
H’
M crown
M base
H
L V
h
V
Figure 13.4: Arch Rib Stiened with Girder or Truss, (Lin and Stotesbury 1981)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13{4
ThreeHinges ARCHES
Example 1325: Design of a Three Hinged Arch adapted from (Lin and Stotesbury 1981) A long arch 100 ft high and spanning 510 ft is to be designed for a garage and hotel building, using air rights over roads and highways.
100’ Garage and hotel Building
510’ It is necessary to determine preliminary dimensions for the size of the arch section. The arches are spaced 60 ft on centers and carry fourstory loading totaling 27 k/ft along each arch.
Solution:
1. To the initial DL and LL of 27 k/ft we add the arch own weight estimated to be 25% of the load, thus the total load is w = (1 + :25)27 = 33:7 33 (13.3) 2. We next determine the various forces: 2 (33)(510)2 H = wL (13.4a) 8h = 8(100) = 10; 700 (33)(510) = 8; 400 V = wL (13.4b) 2 = 2 k/ft
k
k
R =
3.
p
p
H 2 + V 2 = (10; 700)2 + (8; 400)2 = 13; 600
(13.4c) If we use concrete lled steel pipe for arch section, and selecting a pipe diameter of 6 ft with a thickness of 1/2 inch, then the steel cross sectional area is As = 2rt = Dt = (3:14)(6) (12) = (0:5) = 113 2 (13.5) The concrete area is 2 2 (13.6) Ac = D4 = (3:14) (6)4 2 (144) 2 =ftsq = 4; 070 2 Assuming that the steel has an allowable stress of 20 ksi and the concrete 2.5 ksi (noting that the strength of con ned concrete can be as high as three times the one of fc0 ), then the load carrySteel Ac (113)(20)ksi = 2,260 k ing capacity of each component is Concrete As (4; 070)(2:5)ksi = 10,180 k Total 12,440 kip which is o.k. for the crown section (H =10,700 k) but not quite for the abutments at R=13,600 k. This process of trial and error can be repeated until a satisfactory preliminary design is achieved. Furthermore, a new estimate for the arch self weight should be undertaken. ft
4.
ft
5.
6.
Victor Saouma
in
in
ft
in
k
in
in
Structural Concepts and Systems for Architects
Draft 13.2 Case Study: Salginatobel Bridge (Maillart)
13{5
13.2 Case Study: Salginatobel Bridge (Maillart) Adapted from (Billington and Mark 1983)
13.2.1 Geometry The Salginatobel bridge, perhaps the most famous and in uential structure of Maillart is located in high up in the Swiss Alps close to Shuders. 14 It is a three hinged pedestrian bridge which crosses a deep valley with a most beautiful shape which blends perfectly with its surrounding, Fig. 13.5
13
20 ft
20 ft
20 ft
87.5 ft
87.5 ft
20 ft
20 ft
42.6 ft
20 ft
295 ft
Figure 13.5: Salginatobel Bridge; Dimensions, (Billington and Mark 1983) The load supporting structure is the arch itself, whereas the bridge deck and the piers are transfering the vertical load into the arch. 16 The arch cross section is not constant, and can be idealized as in Fig. 13.6 17 The basic shape of the supporting structure is a three hinged arch as shown in Fig. 13.7 18 The arch is parabolic (which as we saw an the optimal shape which minimizes exure), and the cross section at the quarter point has an area of
15
At = 2[(0:62)(12:46) + (0:59)(12:17)] = 29:8 19
2
ft
2
in
(13.7)
Each ange has an area of
AF = (0:62)(12:46) = 7:73 2 = 1; 113 and the eective depth of the section is d = 12:79 ft, Fig. 13.8.
2
(13.8)
2 = 1; 735 in2
(13.9)
ft
20
= 4; 291
in
At the crown/hinge the section is rectangular with
Acr = (1:05)(11:48) = 12:05
Victor Saouma
ft
Structural Concepts and Systems for Architects
Draft 13{6
ThreeHinges ARCHES
42.6 ft
ACTUAL ARCH WITH CENTROID (DOTTED LINE)
IDEALIZATION (ONE DEMENSIONAL) 295 ft
Figure 13.6: Salginatobel Bridge; Idealization, (Billington and Mark 1983)
CONCRETE CORK PADS
HINGE
ACTUAL SPRINGING HINGE
IDEALIZATION
CORK PAD CONCRETE
HINGE HARD WOOD
ACTUAL CROWN HINGE
IDEALIZATION
Figure 13.7: Salginatobel Bridge; Hinges, (Billington and Mark 1983)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13.2 Case Study: Salginatobel Bridge (Maillart)
42.6 ft
13{7
ACTUAL ARCH
295 ft
SECTIONS
12.46 ft
0.59 ft
12.17 ft d=12.79 ft 13.41 ft
0.62 ft
0.62 ft
Figure 13.8: Salginatobel Bridge; Sections, (Billington and Mark 1983)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13{8
ThreeHinges ARCHES
13.2.2 Loads The dead load WD is assumed to be linearly distributed (even though it is greater where the arch is deeper, and the vertical members longer) and is equal to 1,680 kips, Fig. 13.9.
21
; 680) = 5:7 wD = WLD = (1(295)
(13.10)
k
k/ft
ft
wD = 5.7 k/ft WD = 1680 k
L = 295 ft
Figure 13.9: Salginatobel Bridge; Dead Load, (Billington and Mark 1983) For the sake of simplicity we will neglect the snow load (which is actually negligible compared to the dead load). 23 The live load is caused by trac, and we consider the case in which two trucks, each weighing 55 kips, are placed at the quarterpoint, Fig. 13.10. This placement of the load actually corresponds to one of the most critical loading arrangement. The total vertical load is shown in Fig. 13.11 22
13.2.3 Reactions 24
Reactions are easily determined from equilibrium, Fig. 13.15 VD = 1; 680 2 110 VL = 2 (+ ;) Mc (840)(147:5) ; (840)(73:75) ; HD (42:6)
= 840
= = = ) HD = (55)(147:5) ; (55)(73:75) ; HL (42:6) = ) HL = p 2 RD = (840) + (1; 455)2 = p RL = (55)2 + (95)2 =
Victor Saouma
(13.11a)
k
55 0 0 1; 455 0 95 1; 680 110 k
k
k
k
k
(13.11b) (13.11c) (13.11d) (13.11e) (13.11f) (13.11g) (13.11h) (13.11i)
Structural Concepts and Systems for Architects
Draft 13.2 Case Study: Salginatobel Bridge (Maillart)
13{9
PLAN
P = 55 k
ROADWAY
295 ft
P = 55 k
ARCH ABUTMENT
42.5 ft
Figure 13.10: Salginatobel Bridge; Truck Load, (Billington and Mark 1983)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13{10
ThreeHinges ARCHES
A
P = 55 k
B VB,D = 840 k
VA,D = 840 k
P = 55 k
B
A
DEAD LOAD
VA,L = 55 k
LIVE LOAD
42.6 ft
Q D = 1680 k
VB,L = 55 k
295 ft
Figure 13.11: Salginatobel Bridge; Total Vertical Load, (Billington and Mark 1983)
P
C
H
d=42.6 ft
A l/4=73.75 ft
V l/2=147.5 ft Figure 13.12: Salginatobel Bridge; Reactions, (Billington and Mark 1983)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13.2 Case Study: Salginatobel Bridge (Maillart)
13{11
13.2.4 Internal Forces 25
The shear diagrams for the dead, live and combined load is shown in Fig. 13.13. 840 k
+
0
3L/4 L/4
295 ft
55 k
SHEAR FORCE
SHEAR FORCE
420 k
x
L/2 420 k 840 k
295 ft
SHEAR FORCE
+ 895 k
L/2
0
L 55 k
295 ft
= + 475 k + 420 k L
0
x
 420 k  475 k  895 k 295 ft
Figure 13.13: Salganitobel Bridge; Shear Diagrams, (Billington and Mark 1983) At the quarter point the axial force can be expressed as: N = H cos + V sin (13.12) where tan = 2Ld (13.13) and = 16:1o at this location. The horizontal force for the dead and live loads was determined previously as 1; 455 and 95 kips respectively, and the vertical forces are obtained from the shear diagram, thus NDqr = (;1; 455) cos16:1o + (;420) sin 16:1o = ;1; 514 (13.14a) qr (13.14b) NL = (;95) cos 16:1o + (;55) sin 16:1o = ;106 and at the crown where there is no vertical force (and = 0)
26
k
k
NDcr = (;1; 455) cos0o + (;420) sin 0o = ;1; 455 NLcr = (;95) cos 0o + (;55) sin 0o = ;95
k
k
27 28
The uniform dead load will not produce a moment on the parabolic arch. The (point) live load will create a moment which can be decomposed into two parts, 1. Vertical load will cause a trapezoidal moment diagram, and the max moment is MLV = P2 L4 = 112o 295 4 = 4; 050 k.ft
(13.15a) (13.15b)
(13.16)
2. The second is caused by the horizontal reaction, and the resulting moment is MLH = Hd(x), since d varies parabolically, and H is constant, that second moment is parabolic with a peak value equal to MLV = Hdmax = (95)(32:6) = ;4; 050 (13.17) k.ft
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13{12
ThreeHinges ARCHES
at the quarter point
MLV = Hd1=4 = (95) 3(324 :6) = ;3; 040
(13.18)
k.ft
The overall bending moment diagram from the live loads is determined by simply adding those two components, Fig. 13.14.
29
P
L/4
+
4,050 kft
BENDING MOMENT
BENDING
L/4
3,040 kft
MOMENT
P
0
L/4
L/2
x 3L/4
L
295 ft
+ PL/4 = 4,050 kft
4,050 ft.k3,040 ft.k = 1,010 ft.k
MOMENT
BENDING
=
295 ft
Figure 13.14: Salginatobel Bridge; Live Load Moment Diagram, (Billington and Mark 1983) We observe that the actual shape of the arch follows this bending moment diagram for one of the most critical live load case. 31 The maximum moment at midspan is
30
MLmax = 4; 050 ; 3; 040 = 1; 010
(13.19)
k.ft
which would produce internal forces in the upper and lower anges equal to: max ; 010) = 79 Fint = MLd = (1(12 :8) k.ft
k
ft
(13.20)
13.2.5 Internal Stresses The axial stresses at the springlines were determined to be ;1; 680 and ;110 kips for the dead and live loads respectively. 2 , thus the axial stresses are 33 At the support the area of concrete is Ac = 2; 240 D = ;(1; 680) 1; 000 = ;750 sp (13.21a) (2; 240) 2 L = ;(110) 1; 000 = ;49 (13.21b) sp (2; 240) 2 s spTotal = ;750 ; 49 = ;799 (13.21c)
32
in
k
psi
in k
psi
in
psi
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13.3 Structural Behavior of DeckStiened Arches
13{13
At the crown, we repeat the same calculations, where the axial force is equal to the horizontal component of the reactions (1; 455) 1; 000 = ;839 crD = ; (13.22a) (1; 735) 2 (95) 1; 000 = ;55 crL = (1;; 735) (13.22b) 2 s crTotal = ;839 ; 55 = ;894 (13.22c)
34
k
psi
in k
psi
in
psi
The stresses at the quarter point are determined next. Note that we must include the eect of both axial and exural stresses (1; 514) 1; 000 + ;(106) 1; 000 ; (79) qrtop = ; (13.23a) 2 2 2 (4 ; 291) (4 ; 291) (1 ; 113)  {z }  {z }  {z } 35
k

DeadLoad
k
k
in
in
LiveLoad
{z
AxialStresses
in
}
Flexural
= ;353 ; 25 ; 71 ;449 (1; 514) 1; 000 + ;(106) 1; 000 (79) qrbot = ; (4 ; 291){z 2 } (4; 291){z 2 } (1; 113)  {z
(13.23b)
psi
k

DeadLoad
k
k
in
in
{z
LiveLoad
AxialStresses
= ;353 ; 25 + 71 ;307
}
2
in
Flexural
(13.23c)
}
psi
(13.23d)
13.3 Structural Behavior of DeckStiened Arches From (Billington 1979)
INCOMPLETE
The issue of unsymmetrical live load on a stiened or unstiened arch was also addressed by Maillart. As discussed in (Billington 1979) and illustrated by Fig. 13.15
36
Victor Saouma
Structural Concepts and Systems for Architects
Draft 13{14
ThreeHinges ARCHES
wL
∼∆/10
∆ wL
wL a wL Unstiffened Arch
wL
a
wL
0
wL a 2
wL a 2
Stiffened Arch
Figure 13.15: Structural Behavior of Stiened Arches, (Billington 1979)
Victor Saouma
Structural Concepts and Systems for Architects
Draft Chapter 14
BUILDING STRUCTURES 14.1 Introduction
14.1.1 Beam Column Connections 1
The connection between the beam and the column can be, Fig. 14.1: θb
θb
θb
θc
θc
θb
=
θc
θb
Flexible
=
θc
Rigid
θc
θ ) M=K( θ s s b c θ b = θc SemiFlexible
Figure 14.1: Flexible, Rigid, and SemiFlexible Joints
Flexible that is a hinge which can transfer forces only. In this case we really have cantiliver action only. In a exible connection the column and beam end moments are both equal to zero, Mcol = Mbeam = 0. The end rotation are not equal, col = 6 beam. Rigid: The connection is such that beam = col and moment can be transmitted through the connection. In a rigid connection, the end moments and rotations are equal (unless there is an externally applied moment at the node), Mcol = Mbeam = 6 0, col = beam. SemiRigid: The end moments are equal and not equal to zero, but the rotation are dierent. beam =6 col, Mcol = Mbeam = 6 0. Furthermore, the dierence in rotation is resisted by the spring Mspring = Kspring (col ; beam).
14.1.2 Behavior of Simple Frames For vertical load across the beam rigid connection will reduce the maximum moment in the beam (at the expense of a negative moment at the ends which will in turn be transferred to the columns).
2
Draft 14{2
BUILDING STRUCTURES
The advantages of a rigid connection are greater when the frame is subjected to a lateral load. Under those conditions, the connection will stien the structure and reduce the amount of lateral de ection, Fig. 14.2. 3
∆
H
θ
∆
H
θ
PI
V
PI
V
V
PI
V
PI
PI
PI
PI
PI
Figure 14.2: Deformation of Flexible and Rigid Frames Subjected to Vertical and Horizontal Loads, (Lin and Stotesbury 1981) 4 Fig. 14.3 illustrates the deformation, shear, moment and axial forces in frames with dierent boundary conditions under both vertical and horizontal loads.
14.1.3 Eccentricity of Applied Loads A concentric axial force P and moment M , applied on a support sytem (foundation, columns, prestressing) can be replaced by a static equivalent one in which the moment M is eliminated and the force P applied with an eccentricity 5
e= M P
(14.1)
The induced stresses can be decomposed into uniform (;P=L) (assuming a unit width) and linearly varying one ( = M=S ) and the end stresses are
6
min = ; PL ; max = ; PL +
(14.2a) (14.2b)
We note that the linearly varying stress distribution must satisfy two equilibrium requirements: F = 0, thus the neutral axis (where the stress is equal to zero) passes through the centroid of the section, and M = 0, i.e. Mint = Mext . max equals zero, then = P 7 If we seek the eccentricity ecr for which L
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.1 Introduction
14{3 2
Frame Type L
w
W=wL, M=wL/8, M’=Ph
Moment
Shear
Deformation w/2
Axial
M w/2 w/2
w/2
a h
b
P p
POST AND BEAM STRUCTURE M’
w/2
M w/2 w/2
w/2
c M’/L p
SIMPLE BENT FRAME
M’/L
M’
M’/L
d
w/2
M w/2 w/2
w/2
e M’/L
M’/L
M’
M’/L
f p THREEHINGE PORTAL
w/2
w/2
M
M M/h
M/h
M/h M’/2
M/L
h
M/L
p/2
THREEHINGE PORTAL
w/2
0.4M
0.4M
0.36M/h
w/2
M’/2
M/L
j
0.4M/h
0.64M
0.36M/h
w/2
M’/2
w/2
0.45M
0.45M
w/2 M’/4
0.5M’/L
0.68M/h
0.55M
0.68M/h
0.68M/h
w/2
M’/L
M’/L
p/2
TWOHINGE FRAME
w/2
M’/4 p/2
p/2
p/2
RIGID FRAME M’/4
M’/4
M’/2L
l
w/2
p/2 p/2
k
M/L
p/2 p/2
i
w/2
M’/2
M’/2L
g
w/2
Figure 14.3: Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal Frames Subjected to Vertical and Horizontal Loads
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{4
BUILDING STRUCTURES
The net tensile force due to the eccentric load is
8
T = 12 L2
(14.3)
If we want this net tensile force to be equal and opposite to the compressive force, then T = PL T = L4 P = A applied at 32 L2 from the centroid Thus the net internal moment is Mint = 2T 23 L2 = 2 P4 23 L2 = PL 6
(14.4) (14.5)
9 To satisfy the equilibrium equation, this internal moment must be equal and opposite to the external moment Mext = Pecr hence PL = Pe (14.6) 6  {zcr}
{z}
Mint
or
Mext
ecr = L6
(14.7)
in other words to avoid tensile stresses on either side, the resultant force P must be placed within the midle third kernel, Fig. 14.4 L/2
L/2
L/3
L/3
L/3
L/3
L/6
e
L/2
P
L/2
L/3
P L/3
L/3
L/3
L/3
L/3
L/3
L/3
P
L/6 L/3
L/3
L/6
e
P
P
P/A +
+
+
=
=
=
M/S
Figure 14.4: Axial and Flexural Stresses 10
This equation is fundamental in preventing tensile forces in
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.2 Buildings Structures
14{5
1. Prestressed concrete beams: If the prestressing cable is within the kernel (i.e middle third), then there will not be any tensile stresses caused by prestressing alone. 2. Foundations: If the eccentricity is within the middle kernel, then we have compressive stresses only under the foundation and no undesirable uplift. 3. Buildings: If the eccentricity of the vertical load is within the middle third, all columns will be loaded under compression only.
14.2 Buildings Structures There are three primary types of building systems: Wall Subsytem: in which very rigid walls made up of solid masonry, paneled or braced timber, or steel trusses constitute a rigid subsystem. This is only adequate for small rise buildings. Vertical Shafts: made up of four solid or trussed walls forming a tubular space structure. The tubular structure may be interior (housing elevators, staircases) and/or exterior. Most ecient for very high rise buildings. Rigid Frame: which consists of linear vertical components (columns) rigidly connected to sti horizontal ones (beams and girders). This is not a very ecient structural form to resist lateral (wind/earthquake) loads. 11
14.2.1 Wall Subsystems Whereas exterior wall provide enclosure and interior ones separation, both of them can also have a structural role in trnsfering vertical and horizontal loads. 13 Walls are constructed out of masonry, timber concrete or steel. 14 If the wall is braced by oors, then it can provide an excellent resitance to horizontal load in the plane of the wall (but not orthogonal to it). 15 When shearwalls subsytems are used, it is best if the center of orthogonal shear resistance is close to the centroid of lateral loads as applied. If this is not the case, then there will be torsional design problems.
12
14.2.1.1 Example: Concrete Shear Wall From (Lin and Stotesbury 1981)
We consider a reinforced concrete wall 20 ft wide, 1 ft thick, and 120 ft high with a vertical load of 400 k acting on it at the base. As a result of wind, we assume a uniform horizontal force of 0.8 kip/ft of vertical height acting on the wall. It is required to compute the exural stresses and the shearing stresses in the wall to resist the wind load, Fig. 14.5. 1. Maximum shear force and bending moment at the base 16
Vmax = wL = (0:8) (120) = 96 2 (0:8) (120)2 2 = 5; 760 Mmax = wL = 2 2 k.ft
k.ft
Victor Saouma
ft
(14.8a)
k
ft
k.ft
(14.8b)
Structural Concepts and Systems for Architects
Draft 14{6
BUILDING STRUCTURES w=0.8 k/ft
20’ 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 W11111 00000 00000 40011111 k 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111
H=96 k; M =5760 k’
60’
120’
1’
V f HORIZONTAL +f
2/(3d) +F
F
M
M
VERTICAL + FDL
11111 00000 11111 00000
+140 + 740 PSI + 600
7.7’ IN TENSION
Figure 14.5: Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981) 2. The resulting eccentricity is
3. The critical eccentricity is
(5; 760) = eActual = M P (400)
k.ft k
= 14:4
(14.9)
ft
ecr = L6 = (20)6 = 3:3 < eActual N.G.
(14.10)
ft
ft
thus there will be tension at the base. 4. The moment of inertia of the wall is
3 (1) (20) I = bh 12 = 12 ft
ft
3
= 667
5. The maximum exural stresses will be (5; 760) (10) = (86:5) max = Mc I = (667) 4 k.ft
4
(14.11)
ft
ft
ksf
ft
= (600)
psi
(14.12)
6. The average shearing stress is
(96) = VA = (1)(20)
k ft
2
= 4:8
ksf
= 33:3
psi
(14.13)
A concrete with nominal shear reinforcement can carry at least 100 psi in shear, those computed shear streses are permissible.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.2 Buildings Structures
14{7
7. At the base of the wall, the axial stresses will be ;(400) = (20) = (1)(20) 2 k
ksf
ft
= ;140
(14.14a)
psi
8. The maximum stresses will thus be:
1 = ;140 + 600 = 460 2 = ;140 ; 600 = ;740
psi psi
(Tension) (Compression)
(14.15a) (14.15b)
9. The compressive stress of 740 psi can easily be sustained by concrete, as to the tensile stress of 460 psi, it would have to be resisted by some steel reinforcement. 10. Given that those stresses are service stresses and not factored ones, we adopt the WSD approach, and use an allowable stress of 20 ksi, which in turn will be increased by 4=3 for seismic and wind load, all = 34 (20) = 26:7 (14.16) ksi
11. The stress distribution is linear, compression at one end, and tension at the other. The length of the tension area is given by (similar triangles) x = 20 ) x = 460 (20) = 7:7 (14.17) 460 460 + 740 460 + 740 ft
12. The total tensile force inside this triangular stress block is = 250 T = 12 (460) (7:7 12) (12)  {z } ksi
in
in
k
(14.18)
width
13. The total amount of steel reinforcement needed is (250) = 9:4 2 (14.19) As = (26 :7) This amount of reinforcement should be provided at both ends of the wall since the wind or eartquake can act in any direction. In addition, the foundations should be designed to resist tensile uplift forces (possibly using piles). k
in
ksi
14.2.1.2 Example: Trussed Shear Wall From (Lin and Stotesbury 1981) 17 We consider the same problem previously analysed, but use a trussed shear wall instead of a concrete one, Fig. 14.6. 1. Using the maximum moment of 5; 760 kipft (Eq. 14.8b), we can compute the compression and tension in the columns for a lever arm of 20 ft. 760) = 288 F = (5;(20) (14.20) k.ft
k
ft
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{8
BUILDING STRUCTURES
120’
20’
W 400 k
60’
H=96 k
24’
1 1.2
~1.6
V +FM
F
M
Figure 14.6: Trussed Shear Wall 2. If we now add the eect of the 400 kip vertical load, the forces would be C = ; (400) 2 ; 288 = ;488 T = ; (400) 2 + 288 = 88
(14.21a)
k
k
(14.21b)
k
k
3. The force in the diagonal which must resist a base shear of 96 kip is (similar triangles) p
p
F = (20)2 + (24)2 ) F = (20)2 + (24)2 (96) = 154 96 20 20
k
(14.22)
4. The design could be modi ed to have no tensile forces in the columns by increasing the width of the base (currently at 20 ft).
14.2.2 Shaft Systems 18 Vertical shearresisting shafts in buildings act as a tubular section and generally have a rectangular cross section. If there is only one shaft, it is generally located in the center and houses the elevators. If there are many shafts, then they should be symmetrically arranged. 19 If the shaft is relatively short and wide, with an aspect ratio under 1 or 2, then the dominant strcutral action is that of a sti shear resisting tube. If the aspect ratio is between 3 and 5, then the shear forces may not be the controlling criterion, and exure dominates.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.2 Buildings Structures
14{9
14.2.2.1 Example: Tube Subsystem From (Lin and Stotesbury 1981) 20
With reference to Fig. 14.7, the reinforced concrete shaft is 20 ft square, 120 ft high, and with 1 ft ~ 20 ’
20 ’
w = 0.8 k/ft
H = 96 k 60 ’
20 ’
~ 20 ’
111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111
120 ’
N.A.
Figure 14.7: Design Example of a Tubular Structure, (Lin and Stotesbury 1981) thick walls. It is subjected to a lateral force of 0.8 k/ft. 1. Comparing this structure with the one analysed in Sect. 14.2.1.1 the total vertical load acting on the base is now increased to V = 4(400) = 1; 600 (14.23) 2. As previously, the maximum moment and shear are 5; 760 and 96 respectively. 3. The moment of inertia for a tubular section is 3 (20)(20)3 (18)(18)3 4 I = bd (14.24) 12 = 12 ; 12 = 4; 600 4. The maximum exural stresses: (5; 760) (20=2) = 12:5 = 87 = fl = MC (14.25) I (4; 600) 4 k
k.ft
k
ft
k.ft
ft
ksf
psi
ft
5. The average shear stress is
(96) = VA = 2(20)(1)
k ft
2
= 2:4
ksf
6. The vertical load of 1,600 k produces an axial stress of ;(1; 600) = ;20 ax = PA = (4(20)(1) 2
= 17
k
ft
Victor Saouma
ksf
(14.26)
psi
= ;140
psi
(14.27)
Structural Concepts and Systems for Architects
Draft 14{10
BUILDING STRUCTURES
7. The total stresses are thus
= ax + fl 1 = ;140 + 87 = ;53 2 = ;140 ; 87 = ;227
(14.28a) (14.28b) (14.28c)
psi psi
thus we do not have any tensile stresses, and those stresses are much better than those obtained from a single shear wall.
14.2.3 Rigid Frames Rigid frames can carry both vertical and horizontal loads, however their analysis is more complex than for tubes. 22 The rigorous and exact analysis of a rigid frame can only be accomplished through a computer analysis. However, for preliminary design it is often sucient to perform approximate analyses. 23 There are two approximate methods for the analysis of rigid frames subjected to lateral loads: 1) Portal and 2) Cantilever method. 24 The portal frame method is based on the following major assumptions, Fig. 14.8: 21
L P
P L/2
P h/2
PI
h
H 1=P/2
H 2=P/2
h/2 V1 =P/(2L)
V2 =P/(2L)
Figure 14.8: A Basic Portal Frame, (Lin and Stotesbury 1981) 1. Each bay of a bent acts as a separate \portal" frame consisting of two adjacent columns and the connecting girder. 2. The point of in ection (zero moment) for all columns is at midheight 3. The point of in ection for all girders is at midspan. 4. For a multibay frame, the shears on the interior columns are equal and the shear in each exterior column is half the shear of an interior column. This method will be discussed in more details in the following section.
14.3 Approximate Analysis of Buildings 25
Despite the widespread availability of computers, approximate methods of analysis are justi ed by 1. Inherent assumption made regarding the validity of a linear elastic analysis vis a vis of an ultimate failure design.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.3 Approximate Analysis of Buildings
14{11
2. Ability of structures to redistribute internal forces. 3. Uncertainties in load and material properties 26 27 28 29
Vertical loads are treated separately from the horizontal ones. We use the design sign convention for moments (+ve tension below), and for shear (ccw +ve). Assume girders to be numbered from left to right. In all free body diagrams assume positivee forces/moments, and take algeebraic sums.
14.3.1 Vertical Loads The girders at each oor are assumed to be continuous beams, and columns are assumed to resist the resulting unbalanced moments from the girders. 31 Basic assumptions 1. Girders at each oor act as continous beams supporting a uniform load. 2. In ection points are assumed to be at (a) One tenth the span from both ends of each girder. (b) Midheight of the columns 3. Axial forces and deformation in the girder are negligibly small. 4. Unbalanced end moments from the girders at each joint is distributed to the columns above and below the oor. 30
Based on the rst assumption, all beams are statically determinate and have a span, Ls equal to 0.8 the original length of the girder, L. (Note that for a rigidly connected member, the in ection point is at 0.211 L, and at the support for a simply supported beam; hence, depending on the nature of the connection one could consider those values as upper and lower bounds for the approximate location of the hinge). 33 End forces are given by Maximum positive moment at the center of each beam is, Fig. 14.9
32
M + = 18 wL2s = w 81 (0:8)2 L2 = 0:08wL2
(14.29)
Maximum negative moment at each end of the girder is given by, Fig. 14.9 M left = M rgt = ; w2 (0:1L)2 ; w2 (0:8L)(0:1L) = ;0:045wL2
(14.30)
Girder Shear are obtained from the free body diagram, Fig. 14.10 V lft = wL 2
Victor Saouma
V rgt = ; wL 2
(14.31)
Structural Concepts and Systems for Architects
Draft 14{12
BUILDING STRUCTURES
w
Mrgt
lft
M
Vrgt Vlft 0.1L
0.1L
0.8L L
Figure 14.9: Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments
Pabove
Vrgti1
Vlfti
Pbelow
Figure 14.10: Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.3 Approximate Analysis of Buildings
14{13
Column axial force is obtained by summing all the girder shears to the axial force transmitted by the column above it. Fig. 14.10
lft P dwn = P up + Virgt ;1 ; Vi
(14.32)
Column Moment are obtained by considering the free body diagram of columns Fig. 14.11 h/2
h/2 Mcolabove
Mi1lft
Mi1rgt Vi1rgt
Vi1lft
Li1
Mirgt
Milft
Virgt
Vilft
Mcolbelow
Li h/2
h/2
Figure 14.11: Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments bot ; M rgt + M lft M top = Mabove i i;1
M bot = ;top
(14.33)
Column Shear Points of in ection are at midheight, with possible exception when the columns on the rst oor are hinged at the base, Fig. 14.11
top V = Mh 2
(14.34)
Girder axial forces are assumed to be negligible eventhough the unbalanced column shears above and below a oor will be resisted by girders at the oor.
14.3.2 Horizontal Loads We must dierentiate between low and high rise buildings. Low rise buidlings, where the height is at least samller than the hrizontal dimension, the de ected shape is characterized by shear deformations. High rise buildings, where the height is several times greater than its least horizontal dimension, the de ected shape is dominated by overall exural deformation. 34
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{14
BUILDING STRUCTURES
14.3.2.1 Portal Method Low rise buildings under lateral loads, have predominantly shear deformations. Thus, the approximate analysis of this type of structure is based on 1. Distribution of horizontal shear forces. 2. Location of in ection points.
35
36
The portal method is based on the following assumptions 1. In ection points are located at (a) Midheight of all columns above the second oor. (b) Midheight of oor columns if rigid support, or at the base if hinged. (c) At the center of each girder. 2. Total horizontal shear at the midheight of all columns at any oor level will be distributed among these columns so that each of the two exterior columns carry half as much horizontal shear as each interior columns of the frame.
Forces are obtained from Column Shear is obtained by passing a horizontal section through the midheight of the columns at each oor and summing the lateral forces above it, then Fig. 14.12 37
H/2
H
H
H/2
Figure 14.12: Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear P
lateral V ext = 2No.F of bays
V int = 2V ext
(14.35)
Column Moments at the end of each column is equal to the shear at the column times half the height of the corresponding column, Fig. 14.12
M top = V h2
Victor Saouma
M bot = ;M top
(14.36)
Structural Concepts and Systems for Architects
Draft 14.3 Approximate Analysis of Buildings
14{15 h/2
h/2 Mcol
Mi1lft
above
Mi1rgt Vi1rgt
Vi1lft
Li1/2
Li1/2
Mirgt
Milft
Virgt
Vilft
Mcolbelow
Li/2
Li/2
h/2
h/2
Figure 14.13: ***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment
Girder Moments is obtained from the columns connected to the girder, Fig. 14.13 above ; M below + M rgt Milft = Mcol col i;1
Mirgt = ;Milft
(14.37)
Girder Shears Since there is an in ection point at the center of the girder, the girder shear is obtained by considering the sum of moments about that point, Fig. 14.13
V lft = ; 2LM
V rgt = V lft
(14.38)
Column Axial Forces are obtained by summing girder shears and the axial force from the column above, Fig. ?? Pabove
Vrgti1
Vlfti
Pbelow
Figure 14.14: Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force
P = P above + P rgt + P lft
(14.39)
Example 1426: Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{16
BUILDING STRUCTURES 0.25K/ft
15
30
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Figure 14.15: Example; Approximate Analysis of a Building Draw the shear, and moment diagram for the following frame. Solution:
Vertical Loads 1. Top Girder Moments
= ;0:045w12L212 = ;(0:045)(0:25)(20)2 = 0:08w12L212 = (0:08)(0:25)(20)2 = M12lft = ;0:045w13L213 = ;(0:045)(0:25)(30)2 = 0:08w13L213 = (0:08)(0:25)(30)2 = M13lft = ;0:045w14L214 = ;(0:045)(0:25)(24)2 = 0:08w14L214 = (0:08)(0:25)(24)2 = M14lft 2. Bottom Girder Moments M9lft = ;0:045w9L29 = ;(0:045)(0:5)(20)2 M9cnt = 0:08w9 L29 = (0:08)(0:5)(20)2 M9rgt = M9lft M10lft = ;0:045w10L210 = ;(0:045)(0:5)(30)2 M10cnt = 0:08w10 L210 = (0:08)(0:5)(30)2 M10rgt = M11lft M11lft = ;0:045w12L212 = ;(0:045)(0:5)(24)2 M11cnt = 0:08w12 L212 = (0:08)(0:5)(24)2 M11rgt = M12lft 3. Top Column Moments M5top = +M12lft M5bot = ;M5top M6top = ;M12rgt + M13lft = ;(;4:5) + (;10:1) M6bot = ;M6top M7top = ;M13rgt + M14lft = ;(;10:1) + (;6:5) M7bot = ;M7top M8top = ;M14rgt = ;(;6:5) M8bot = ;M8top
M12lft M12cnt M12rgt M13lft M13cnt M13rgt M14lft M14cnt M14rgt
Victor Saouma
=; = =; =; = =; =; = =;
4:5 8:0 4:5 10:1 18:0 10:1 6:5 11:5 6:5
=; = =; =; = =; =; = =;
9:0 16:0 9:0 20:3 36:0 20:3 13:0 23:0 13:0
=; = =; = =; = = =;
4:5 4:5 5:6 5:6 3:6 3:6 6:5 6:5
k.ft k.ft k.ft k.ft k.ft k.ft
k.ft k.ft
k.ft
k.ft k.ft
k.ft k.ft k.ft k.ft k.ft k.ft k.ft
k.ft k.ft k.ft k.ft k.ft k.ft k.ft k.ft
Structural Concepts and Systems for Architects
Draft 14.3 Approximate Analysis of Buildings
14{17
4. Bottom Column Moments M1top = +M5bot + M9lft = 4:5 ; 9:0 M1bot = ;M1top M2top = +M6bot ; M9rgt + M10lft = 5:6 ; (;9:0) + (;20:3) M2bot = ;M2top M3top = +M7bot ; M10rgt + M11lft = ;3:6 ; (;20:3) + (;13:0) M3bot = ;M3top M4top = +M8bot ; M11rgt = ;6:5 ; (;13:0) M4bot = ;M4top
=; = =; = = =; = =;
4:5 4:5 5:6 5:6 3:6 3:6 6:5 6:5
k.ft k.ft k.ft k.ft k.ft k.ft k.ft k.ft
0.25K/ft
12
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+8.0’k
+18.0’k
4.5’k +16.0’k
9.0’k
9.0’k
4.5’k
+4.5’k
+5.6’k 4.5’k
+4.5’k
+5.6’k
5.6’
k
16’
24’ +11.5’k 6.5’k
k
+23.0’k
k
k k 13.0’ 20.2’k 20.2’
5.6’k
14’
4
10.1’k 10.1’k 6.5’ +32.0’
8
11 3
2
k
14
7
0.50K/ft
+3.6’k
13.0’k
+6.5’k
3.6’k 6.5’k k +3.6’ +6.5’ k
3.6’k
6.5’k
Figure 14.16: Approximate Analysis of a Building; Moments Due to Vertical Loads 5. Top Girder Shear
Victor Saouma
V12lft V12rgt V13lft V13rgt V14lft V14rgt
= = = = = =
w12 L12 = (0:25)(20) = 2 2 lft
;V12
w13 L13 = (0:25)(30) 2 2 lft
;V13
w14 L14 = (0:25)(24) 2 2 lft
;V14
=; = =; = =;
2:5 2:5 3:75 3:75 3:0 3:0
k k k k
k k
Structural Concepts and Systems for Architects
Draft 14{18
BUILDING STRUCTURES
6. Bottom Girder Shear
V9lft V9rgt V10lft V10rgt V11lft V11rgt 7. Column Shears
= = = = = =
V5 V6 V7 V8 V1 V2 V3 V4
+2.5K
+3.75
= = = = = = = =
w9 L9 = (0:5)(20) 2 2 lft
;V9
w10 L10 = (0:5)(30) 2 2 lft
;V10
w11 L11 = (0:5)(24) 2 2 lft
;V11
7 2
2 2
2
4 2
K
= =
2
k k k
0:46 0:81
k
k
K
+3.0
3.75K K
3.0K
+6.0
5.0K
0.56K
k k
2
M3top = 3:6 16 H3 2 2 M4top = 6:5 16 H
+7.5
0.64K
k
M8top = 6:5 = 0:93 k 14 H8 2 2 M1top = ;4:5 = ; 0:56 k 16 H1 2 2 M2top = ;5:6 = ; 0:70 k 16 H
2.5K +5.0
5:00 5:00 7:50 7:50 6:00 6:00
M5top = ;4:5 = ; 0:64 k 14 H5 2 2top M6 = ;5:6 = ; 0:80 k 14 H6 2 2 M7top = 3:6 = 0:52 k 14 H
K
K
= =; = =; = =;
6.0K
7.5K
0.80K
+0.51K
0.70K
+0.45K
+0.93K
+0.81K
Figure 14.17: Approximate Analysis of a Building; Shears Due to Vertical Loads
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.3 Approximate Analysis of Buildings
14{19
8. Top Column Axial Forces
P5 P6 P7 P8
= = = =
V12lft ;V12rgt + V13lft = ;(;2:50) + 3:75 ;V13rgt + V14lft = ;(;3:75) + 3:00 rgt ;V14
2:50 6:25 6:75 3:00
= = = =
9. Bottom Column Axial Forces P1 = P5 + V9lft = 2:50 + 5:0 P2 = P6 ; V10rgt + V9lft = 6:25 ; (;5:00) + 7:50 P3 = P7 ; V11rgt + V10lft = 6:75 ; (;7:50) + 6:0 P4 = P8 ; V11rgt = 3:00 ; (;6:00)
= = = =
k k k k
7:5 18:75 20:25 9:00 k
k k
k
Horizontal Loads, Portal Method 1. Column Shears
V5 V6 V7 V8 V1 V2 V3 V4
= = = = = = = =
15 (2)(3) 2(V5 ) = (2)(2:5) 2(V5 ) = (2)(2:5) V5 15+30 (2)(3) 2(V1 ) = (2)(7:5) 2(V1 ) = (2)(2:5) V1
= = = = = = = =
2:5 5 5 2:5 7:5 15 15 7:5
k
k k k k k k k
2. Top Column Moments
M5top M5bot M6top M6bot M7top M7bot M8top M8bot
= = = = = = = =
V1 H5 = (2:5)(14) 2 top 2
;M5
V6 H6 = (5)(14) 2 2
;upM6top
V7 H7 = (5)(14) 2 2
;upM7top
V8 H8 = (2:5)(14) 2 top 2
;M8
= =; = =; = =; = =;
17:5 17:5 35:0 35:0 35:0 35:0 17:5 17:5
k.ft k.ft k.ft k.ft k.ft k.ft k.ft k.ft
3. Bottom Column Moments
M1top M1bot M2top M2bot M3top M3bot M4top M4bot
Victor Saouma
= = = = = = = =
V1dwn H1 = (7:5)(16) = 2 top 2
;dwn M1
V2 H2 = (15)(16) 2 2
;dwn M2top V3 H3 2 ;dwn M3top V4 H4 2 ;M4top
= (15)(16) 2 =
(7:5)(16) 2
=; = =; = =; = =;
60 60 120 120 120 120 60 60
k.ft k.ft k.ft k.ft k.ft k.ft
k.ft k.ft
Structural Concepts and Systems for Architects
Draft 14{20
BUILDING STRUCTURES
Approximate Analysis Vertical Loads
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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Height 14 16
Span Load Load
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L1 20 0.25 0.5
L2 L3 30 24 0.25 0.25 0.5 0.5 MOMENTS Bay 1 Bay 2 Bay 3 Col Beam Column Beam Column Beam Col Lft Cnt Rgt Lft Cnr Rgt Lft Cnt Rgt AAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAAAAAA AAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAA 10.1 18.0 10.1AAAA AAAAAAAA AAAA AAAAAAAA AAAAA 4.5 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A 6.5 8.0 4.5 AAAA 11.5 6.5 AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA 4.5 AAAA 5.6 3.6 6.5 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA 4.5 AAAA 5.6 3.6 6.5 AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAA A AAAAAAAA AAAAAAAA AAAA AAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAA 20.3 36.0 20.3AAAA AAAAAAAAAAAA AAAA AAAAAAAA AAAAA 9.0 16.0 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A 13.0 23.0 13.0 AAAA 9.0 AAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAA A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA 4.5 AAAA 5.6 3.6 6.5 AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAA AA AAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAA AAA 4.5 AAAA 5.6 3.6 6.5 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA SHEAR Bay 1 Bay 2 Bay 3 Col Beam Column Beam Column Beam Col Lft Rgt Lft Rgt Lft Rgt AAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAA AAAA A AAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAA A AAAAAAAA AAAAAAAAAAAA 2.50 AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA 3.00 AAAAAAAA AAAAAAAA AAAAA 2.50AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA 3.75 AAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAA 3.75AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A 3.00 AAAA AAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAA AAAA A A AAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAA AAAAAAA AAAAAAAA AA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA 0.64 AAAA 0.80 0.52 0.93 AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAA A AAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAA A AAAAAAAA AAAAAAAAAAAA 5.00 AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA 6.00 AAAAAAAA AAAAAAAA AAAAA 5.00AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA 7.50 AAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAA 7.50AAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAA 6.00 AAAA AAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAA AAAA A A AAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAA AAAAAAA AAAAAAAA AA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 0.56 AAAA 0.70 0.46 0.81 AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AXIAL FORCE Bay 1 Bay 2 Bay 3 Col Beam Column Beam Column Beam Col AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA 0.00 0.00 0.00 A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA 2.50 6.25 6.75 3.00 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A 0.00 0.00 0.00 A AAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA 7.50 AAAA 18.75 20.25 9.00 AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
Figure 14.18: Approximate Analysis for Vertical Loads; SpreadSheet Format
Victor Saouma
Structural Concepts and Systems for Architects
Victor Saouma L1 20 0.25 0.5
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Q
Victor E. Saouma
=0.045*D5*D3^2 =0.08*D5*D3*D3 =+D13
=F13+I13+G12 =G14
=0.045*I5*I3^2 =0.08*I5*I3*I3 =+I13
=K13+N13+L12 =L14
=0.045*N5*N3^2 =0.08*N5*N3*N3 =+N13
=P13+Q12 =Q14
=+C28+D22
Bay 2 Beam 0
=+I3*I5/2
=I22
Column
=2*L14/A5
Bay 3 Beam 0
=+N3*N5/2
=N22
Col
=2*Q14/A5
0
=+G28F22+I22
=F20+I20
0
=+L28K22+N22
=K20+N20
0
=+Q28P22
=P20
AAAA AAA AAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAA AAAAAAAA AAAA AAAA AAAA AAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAA AAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAA AAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA A AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
=+D20
Column
AXIAL FORCE Bay 1 Col
Beam 0
=2*G14/A5
=D22
=2*C14/A5
=+D3*D5/2
AAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
Bay 2 Bay 3 Beam Column Beam Column Beam Col Lft Rgt Lft Rgt Lft Rgt AAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAA AAAA AAAA A A A AAAAAAAAAAAAAAAAAAAAAAA=I20 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA =+N3*N4/2 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA =+D3*D4/2 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA=D20 AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA =N20 AAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA=+I3*I4/2 AAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAA AAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAA =2*G11/A4 AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA =2*C11/A4 A =2*L11/A4 AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA=2*Q11/A4 AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA A
SHEAR Bay 1 Col
=+D13+C12 =C14
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAA AAAA AAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
Bay 2 Bay 3 Beam Column Beam Column Beam Col Lft Cnt Rgt AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAALft Cnr Rgt AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA Lft Cnt Rgt AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA =0.045*D4*D3^2 =0.08*D4*D3*D3 =+D10 =0.045*I4*I3^2 =0.08*I4*I3*I3 =+I10 =0.045*N4*N3^2 =0.08*N4*N3*N3 =N10 AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA A =+D10 =F10+I10 =K10+N10 =P10 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAA AAAA AAAA AAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAA=Q11 AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAA =L11 AAAA AAAA =C11 =G11 AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAA
MOMENTS Bay 1 Col
C
29 30
Span Load Load
B
24 25 26 27 28
22 23
16 17 18 19 20 21
13 14 15
A 1 2 3 Height 4 14 5 16 6 7 8 9 10 11 12
Approximate Analysis Vertical Loads
Draft 14.3 Approximate Analysis of Buildings 14{21
Figure 14.19: Approximate Analysis for Vertical Loads; Equations in SpreadSheet
Structural Concepts and Systems for Architects
Draft 14{22
BUILDING STRUCTURES
4. Top Girder Moments
M12lft M12rgt M13lft M13rgt M14lft M14rgt
= = = = = =
M5top ;M12lft M12rgt + M6top = ;17:5 + 35 ;M13lft M13rgt + M7top = ;17:5 + 35 ;M14lft
= =; = =; = =;
17:5 17:5 17:5 17:5 17:5 17:5
5. Bottom Girder Moments M9lft = M1top ; M5bot = 60 ; (;17:5) M9rgt = ;M9lft M10lft = M9rgt + M2top ; M6bot = ;77:5 + 120 ; (;35) M10rgt = ;M10lft M11lft = M10rgt + M3top ; M7bot = ;77:5 + 120 ; (;35) M11rgt = ;M11lft 6. Top Girder Shear
7. Bottom Girder Shear
V12lft V12rgt V13lft V13rgt V14lft V14rgt
= = = = = =
:5) = ;1:75 ; 2LM1212 = ; (2)(17 20 lft +V12 = ;1:75 lft 2 M (2)(17 : 5) 13 ; L13 = ; 30 = ;1:17 +V13lft = ;1:17 lft 2 M (2)(17 : 5) ; L1414 = ; 24 = ;1:46 +V14lft = ;1:46
k.ft k.ft k.ft k.ft k.ft k.ft
= =; = =; = =;
77:5 77:5 77:5 77:5 77:5 77:5
k.ft k.ft k.ft k.ft k.ft
lft
k k k k k k
:5) = ;7:75 = ; 2ML912 = ; (2)(77 20 = +V9lft = ;7:75 lft :5) = ;5:17 = ; 2LM1010 = ; (2)(77 30 = +V10lft = ;5:17 lft 2 M (2)(77 : 5) 11 = ; L11 = ; 24 = ;6:46 = +V11lft = ;6:46 8. Top Column Axial Forces (+ve tension, ve compression) P5 = ;V12lft = ;(;1:75) P6 = +V12rgt ; V13lft = ;1:75 ; (;1:17) = ;0:58 P7 = +V13rgt ; V14lft = ;1:17 ; (;1:46) = 0:29 P8 = V14rgt = ;1:46 9. Bottom Column Axial Forces (+ve tension, ve compression) P1 = P5 + V9lft = 1:75 ; (;7:75) = 9:5 P2 = P6 + V10rgt + V9lft = ;0:58 ; 7:75 ; (;5:17) = ;3:16 P3 = P7 + V11rgt + V10lft = 0:29 ; 5:17 ; (;6:46) = 1:58 P4 = P8 + V11rgt = ;1:46 ; 6:46 = ;7:66
V9lft V9rgt V10lft V10rgt V11lft V11rgt
k.ft
lft
k k k k k k
k
k
k
k
k
k
k
k
Design Parameters On the basis of the two approximate analyses, vertical and lateral load, we now seek the design parameters for the frame, Table 14.2.
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.3 Approximate Analysis of Buildings
15K
12
5
30K
14{23
13
6
9
10
1
20’
35’K +60’K
120’K
17.5’K
60’K
+17.5’K
17.5K +77.5’K
+17.5’K
120’K +17.5’K
16’
24’
35’K +120’K
60’K
+17.5’K
4
+35’K
+35’K
14’
11
30’
17.5’K +120’K
+60’K
8
3
2
+17.5’K
14
7
+77.5’K
17.5K
17.5K
77.5’K
77.5’K
+77.5’K
77.5’K
Figure 14.20: Approximate Analysis of a Building; Moments Due to Lateral Loads
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{24
BUILDING STRUCTURES
Portal Method
PORTAL.XLS
A 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
B
C
D
E
F
G
H
I
J
K
Victor E. Saouma
L
M
N
O
P
Q
R
S
PORTAL METHOD # of Bays
3
# of Storeys 2 Force Shear H Lat. Tot Ext Int H1
14 15 15 2.5
5
H2
16 30 45 7.5 15
L1 20
L2 L3 30 24 MOMENTS Bay 1 Bay 2 Bay 3 Col Beam Column Beam Column Beam Col Lft Rgt AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA Lft Rgt AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA Lft Rgt AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAA A 17.5 17.5 AAAA A AAAAAAAAAAAA AAAAAAAA AAA 17.5 17.5 AAAAAAAA AAAA 17.5 17.5AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA 17.5 AAAA 35.0 35.0 17.5 AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAA 17.5 AAAA 35.0 35.0 17.5 AAAAAAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAA 77.5 77.5 AAAAAAAA AAAA 77.5 77.5AAAAAAAA AAAAAAAA AAAAA 77.5 77.5 AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA A A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA 60.0 AAAA 120.0 120.0 60.0 AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA 60.0 AAAA 120.0 120.0 60.0 AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA SHEAR Bay 1 Bay 2 Bay 3 Col Beam Column Beam Column Beam Col Lft Rgt Lft Rgt Lft Rgt AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAA 1.46 1.46AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA 1.17 1.17 A 1.75 1.75 AAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA 2.50 AAAA 5.00 5.00 2.50 AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA 2.50 AAAA 5.00 5.00 2.50 AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAA A A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAA AAAAAAAA AAA AAAAAAAAAAAA AAAAAAAAAAAA A 7.75 7.75 AAAA A AAAAAAAAAAAA AAAAAAAA AAA 5.17 5.17 AAAAAAAA AAAA 6.46 6.46AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAA AAAA 7.50 AAAA 15.00 15.00 7.50 AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAA AAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA 7.50 AAAA 15.00 15.00 7.50 AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AXIAL FORCE Bay 1 Bay 2 Bay 3 Col Beam Column Beam Column Beam Col AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAA AAA AAAA AAAA AAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAA AAAAAAAA AAAA AAAAAAAA 0.00 0.00 0.00 A A AAAAAAAAAAAA AAAAAAAAAAAA AAA AAAAAAAAAAAA AAAA AAAAAAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAA AAAA AAAAAAAA A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA 1.75 0.58 0.29 1.46 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAA AAAAAAAA AAAA AAAAAAAA AAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA 0.00 0.00 0.00 AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA 9.50 AAAA 3.17 1.58 7.92 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA
Figure 14.21: Portal Method; SpreadSheet Format
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.3 Approximate Analysis of Buildings
14{25
Portal Method
PORTAL.XLS
A 1
PORTAL METHOD
A A A A A A
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A A A A A A
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A A A A A A
Victor E. Saouma
K
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AAAAAAAAA A AAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA A3 AL1 A AAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAA AA A A AAAAA 2 AAAA # of Bays L2 L3 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAA A A A A A AA A A A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAA AAAAAAAA AAAA AAAAAAAAA A AAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAA A AAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAA AA A A AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A A A A A A AA A A A A 3 A A A A A A20 A A AA30 A A A 24 A
A A A A A A A A A AA A A A A A A A A A A A A A A A AA A A A A A A A A A A A MOMENTS A A A A AA A A A A A A A A A A A A A A A AA A A A A A A AAAAAAAAAAA A A A A A A AA A A A A A A A A2 A A A AA A A A A Bay 1 Bay 2 AA Bay 3 AA A A A A A A A A A A A AAAAAAAAAAA A A AA A A A A A A A A A A A A A A A A A A Force A Shear Col Beam Column Beam Column Beam Col 6 A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A A A A AInt A Rgt ARgt A Rgt H Lat. AAA Tot Ext Lft Lft Lft 7 A A A A A AAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A AAAAAAAA AAAA=+H9 AAAAAAAA AAA=+J8+K9 AAAAAAAA AAA =+N8+O9 AAAAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAA A A =I8 AAAA =M8 AAAA =Q8 AAAA 8 AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAA A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA A A AAAAAAAA AAAAAAAA AAAAAAAA AAAA=+F9*B9/2 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA =+K9 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA=+H9 AAAAAAAA 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AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA A A A A A A AAAAAAAA AAAA=+F12 AAAAAAAA AA =+F12 AAAAAAAA AAA =+E12 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A A A A A 22 =+E12 AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAA=+K22 AAAAAAAAAAAAAAAAAAAAAAAAAA =+O22 AAAAAAAAAAAAAAAAAAAAAAAAAAA=+S22 A A A A A A 23 =+H22 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AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A A A A A AAAAAAAA AAAA0 AAAAAAAA AAA0 AAAAAAAA AAA 0 AAAAAAAA AA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A A A A A 27 AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA A A A A A AAAAAAAA AAAAAAAA AAAAAAAA AAAA=+J18M18 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA =+N18Q18 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA=+R18 AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAA AAAAAAAAAAAAAAAA 28 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA =I18 AAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAA AAAAA A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAA AAAAAAAAAAAAAAAA AAAA0 AAAAAAAA AAA0 AAAAAAAA AAA 0 AAAAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAA A A A A 29 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA A A A A A A 30 =+H28I21 A A A A A A AAAAAAAAAAAAAAAAAAAA=+K28+J21M21 AAAAAAAAAAAAAAAAAAAAAAAAAA =+O28+N21Q21 AAAAAAAAAAAAAAAAAAAAAAAAAAA=+S28+R21 A A A A A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA
4
5 # of Storeys
A A A A A A A A
A A A A A A A A
Figure 14.22: Portal Method; Equations in SpreadSheet
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{26
BUILDING STRUCTURES
Mem. 1 2 3 4 5 6 7 8
Vert.
Hor.
Moment 4.50 60.00 Axial 7.50 9.50 Shear 0.56 7.50 Moment 5.60 120.00 Axial 18.75 15.83 Shear 0.70 15.00 Moment 3.60 120.00 Axial 20.25 14.25 Shear 0.45 15.00 Moment 6.50 60.00 Axial 9.00 7.92 Shear 0.81 7.50 Moment 4.50 17.50 Axial 2.50 1.75 Shear 0.64 2.50 Moment 5.60 35.00 Axial 6.25 2.92 Shear 0.80 5.00 Moment 3.60 35.00 Axial 6.75 2.63 Shear 0.51 5.00 Moment 6.50 17.50 Axial 3.00 1.46 Shear 0.93 2.50
Design Values 64.50 17.00 8.06 125.60 34.58 15.70 123.60 34.50 15.45 66.50 16.92 8.31 22.00 4.25 3.14 40.60 9.17 5.80 38.60 9.38 5.51 24.00 4.46 3.43
Table 14.1: Columns Combined Approximate Vertical and Horizontal Loads
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.4 Lateral De ections
14{27 Mem. 9 10 11 12 13 14
ve Moment +ve Moment Shear ve Moment +ve Moment Shear ve Moment +ve Moment Shear ve Moment +ve Moment Shear ve Moment +ve Moment Shear ve Moment +ve Moment Shear
Vert.
Hor. Design
9.00 16.00 5.00 20.20 36.00 7.50 13.0 23.00 6.00 4.50 8.00 2.50 10.10 18.00 3.75 6.50 11.50 3.00
77.50 0.00 7.75 77.50 0.00 5.17 77.50 0.00 6.46 17.50 0.00 1.75 17.50 0.00 1.17 17.50 0.00 1.46
Values 86.50 16.00 12.75 97.70 36.00 12.67 90.50 23.00 12.46 22.00 8.00 4.25 27.60 18.00 4.92 24.00 11.50 4.46
Table 14.2: Girders Combined Approximate Vertical and Horizontal Loads
14.4 Lateral De ections Even at schematic or preliminary stages of design, it is important to estimate the lateral de ections of tall buildings for the following reasons 1. Lateral de ections are often limited by code requirements, for example < h=500 where h is the height of the story or of the building. This is important because occupants should not experience uncomfortable horizontal movements. 2. A building that de ects severly under lateral forces may have damage problems associated with vibration (as with vertical defelctions of beams). 3. Through the evaluation of de ection, one may also get some idea of the relative horizontal load carried by the various vertical subsystems in a building (i.e. how much is carried by the shaft compared to the frames). Since all systems are connected, they must move together and through their stiness (deformation per unit load) we can determine the contribution of each subsystem. 38
14.4.1 Short Wall In short structures (as with short beams), shear de ections, Fig. 14.23 dominates. For a concentrated load (14.40) 1:2V h
39
GA
where for concrete and steel G 25 E .
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{28
BUILDING STRUCTURES
000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111 000000000 111111111
V
∆
h
WALL ELEVATION
Figure 14.23: Shear Deformation in a Short Building, (Lin and Stotesbury 1981)
14.4.2 Tall Wall 40
Alternatively, in a tall building exural deformations, Fig. 14.24 are predominant. dominates.
∆
111111 000000 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111 000000 111111
w
h
WALL (OR TUBE) ELEVATION
Figure 14.24: Flexural Deformation in a Tall Building, (Lin and Stotesbury 1981) 4
8wh EI
(14.41)
and the moment of inertia I = bh12 for rectangular sections. 3
14.4.3 Walls and Lintel When two slender walls are connected by (heavy) lintels, the entire subsystem can be made to act as one cantilever supported by the foundations and de ections will be small. However if we have light lintels, their deformation is larger than those of the walls, Fig. 14.25.
41
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.4 Lateral De ections
14{29
LINTELS
w
0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111
L
111 000 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111
∆
α
LINTEL BENDING
h
α
α
RELATIONSHIP BETWEEN
a
WALL AND LINTEL
2 WALLS CONNECTED BY LINTELS
DEFORMATION
Figure 14.25: De ection in a Building Structure Composed of Two Slender Walls and Lintels, (Lin and Stotesbury 1981) 42
In this case de ections can be estimated from: 2 Mmax = wh2 T = C = Ma V=Lintel # of TLintels V L2 wall = lintel 12 EI
(14.42a) (14.42b) (14.42c) (14.42d) (14.42e)
and
h
(14.43)
14.4.4 Frames De ection of a rigid frame is essentially caused by shear between stories which produces vertical shears in the girders. From the portal method we can estimate those deformations, Fig. 14.26. 44 The deformation for the rst story at the exterior joint can be approximated from
43
3
VcolE h col = 12 EIcolE 2
(14.44a)
L h 2VcolE Lh gdr = V12gdr EIgdr = 12EIgdr
2
Eh totE = colE + gdr = Vcol 12E
Victor Saouma
(14.44b) 2
h 2L IcolE + Ig dr
(14.44c)
Structural Concepts and Systems for Architects
Draft 14{30
BUILDING STRUCTURES DUE TO GI
∆
h STORIES
DUE TO CO α θ
α
h
L
OVERALL FRAME ELEVATION
DEFORMATION OF ONE BE
Vcol
Vcol h/2
V gdr
L/2
h/2
L/2
L/2 Vgd
h/2
Vgdr
h/2 Vcol
V
MOMENT EQUILIBRIUM
INTERIOR JOINT
EXTERIOR JOINT
Vgdr (L)=Vcol (h)
Vgdr (L/2)=Vcol (h)
Figure 14.26: Portal Method to Estimate Lateral Deformation in Frames, (Lin and Stotesbury 1981) 45
For the interior joint:
3
VcolI h col = 12 EIcolI
(14.45a)
L2 h = 2VcolI Lh2 (14.45b) gdr = V12gdr EIgdr 12EIgdr 2 h V h L col I totI = colI + gdr = 12E I + I dr (14.45c) colI g and the total displacement will be (14.46) tot = n 2tot where n is the number of stories, and tot is for either the interior or exterior joints. 46 The two major sources of lateral de ection are the bending of column in resisting horizontal shear and girders in resisting vertical shear, Fig. 14.27. 47 A vertical unsymmetric load will cause lateral de ection in frames, Fig. 14.28.
14.4.5 Trussed Frame The cantilever de ection due to column shortening and lengthening (produced by overturning moment) is usually of secondary importance until the building is some 40 stories or higher,Fig. ??. 49 The total de ection at C is given by
48
= PPL AE
Victor Saouma
(14.47)
Structural Concepts and Systems for Architects
Draft 14.4 Lateral De ections
14{31 ∆M
∆S + ∆M
SHEAR EFFECT (RACKING)
SHORTENING
ELONGATION
∆S
OVERALL EFFECT
MOMENT EFFECT (OVERALL BENDING)
(RACKING + BENDING)
Figure 14.27: Shear and Flexural De ection of a Rigid Frame Subsystem, (Lin and Stotesbury 1981)
SIDE SWAY
P
Figure 14.28: SideSway De ection from Unsymmetrical Vertical Load, (Lin and Stotesbury 1981)
∆c
H1
C
C
1
∆
δT
δc
H2
ΣH
H3 H4
h
a P1
a T
C
Figure 14.29: Axial Elongation and Shortening of a Truss Frame, (Lin and Stotesbury 1981)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{32
BUILDING STRUCTURES
where: P is the force in any member due to loading on the whole system, L is the length of the member, A and E the corresponding cross sectional area and modulus of elasticity, P the force in the same member due to a unit (1) force applied in the direction of the de ection sought, and at the point in question. 50 Alternatively, we can neglect the web deformation and consider only the axial deformations in the colums: t + c h (14.48)
a Th t + c = 2 AE
(14.49)
14.4.6 Example of Transverse De ection Typical plan, elevation and oor section of a building are shown in Fig. 14.30. The lateral resiting 111 000 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111
13@12’=156’
CORE SHAFT
1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111
W=4.8 K/FT
CORE
MOMENT DIAGRAM
TOTAL M 58500 KFT 12’ TYP.
20’
LOAD
156’
51
20’ TRANSVERSE ELEVATION OF CORE
60’ TRANSVERSE ELEVATION OF BUILDING
1111 0000 0000 1111 0000 1111 0000 1111 1111 0000 1111 0000
20" TYP. 40’
COLUMN SECTION
20"
A
60’
A
20"
20’
20’
1111111111111111 0000000000000000 0000000000000000 1111111111111111 0000000000000000 1111111111111111 FLOOR PLAN
2.5’
5"
12" GIRDER SECTION
Figure 14.30: Transverse De ection, (Lin and Stotesbury 1981) elements are the center concrete shaft (20ft40ft in section and made up of four 12in walls) and the reinforcedprestressed concrete frames (made up of 12in30in T beams, Igdr = 3:64 4 , and 20 inch square reinforced concrete columns). 52 We consider a wind load of 4.8 k/ft in the transverse direction and make the following assumptions: 1. Colums are of uniform sectional properties and height for all stories. 2. Shaft walls are of uniform thickness for all stories. We neglect wall openings. 3. The wind load is uniform over the height of the building. ft
53
The solution proceeds as follows:
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.4 Lateral De ections
14{33
1. Determine the exural deformation of the top of the shaft (we may neglect shear deformations since the shaft is slender): 4
= 8wh EI 3 ; b2 d3 b 2 I = 1 d1 12 3 ; (39)(19)3 = (41)(21) 12 = 9; 400 4 E = 3 106 = 432; 000 (4:8) (156)4 4 = 0:087 = 8(432; 000) (9; 400) 4 = 0:087 = 1 p h 156 1; 800
(14.50a) (14.50b) (14.50c) (14.50d)
ft
psi
ksf
k.ft
(14.50e)
ft
ksf
ft
ft
(14.50f)
The h ratio is much less than 1/500 as permitted in most building codes, and s within the usual index for concrete buildings, which ranges between 1/1,000 and 1/2,500. If the wall thickness is reduced, and if door openings are considered, the de ection will be correspondingly smaller. The de ection due to moment increases rapidly at the top, the value of 1/1,800 indicates only the average drift index for the entire building, whereas the story drift index may be higher, especially for the top oor. 2. We next consider the de ection of the top of the frame. Assuming that each frame takes 1/9 of the total wind load and shear, and neglecting column shortening, then:
2 h 2L Eh = Vcol 12E IcolE + Ig dr 3 (20=12)(20=12)3 Icol = bh = 0:64 4 12 = 12 Igdr = 3:64 4 (156) = 41:7 =col ground = (4:8) Vcol I (2)(9) 2 (12) (41 : 7) (12) = 12(432; 000) + 2(60) (0:64) 4 (3:46) = 0:00116(18:8 + 34:7) = 0:062
(14.51a) (14.51b) (14.51c)
ft
ft
k.ft
(14.51d)
ft
k
ft
k
ksf
ft
ft
ft
4
ft
(14.51e) (14.51f)
3. Since the story drift varies with the shear in the story, which decreases linearly to the top, the average drift will be 0:062=2 = 0:31 per story and the de ection at top of the building is approximately (14.52) = (13)(0:031) = 0:40 which indicates a drift ratio of (0:4) = 1 (14.53a) Drift Ratio for Building = (156) 400 :062) = 1 Drift Ratio for Ground Floor = (0(12) (14.53b) 194 ft
ft
ft
ft
ft
ft
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{34
BUILDING STRUCTURES
4. Comparing the frame de ection of 0:40 with the shaft de ection of 0:087 , it is seen that the frame is about ve times more exible than the shaft. Furthermore, the frame would not be sti enough to carry all the lateral load by itlself. Proportioning the lateral load to the relative stinesses, the frame would carry about 1/6 of the load, and the remaining 5/6 will be carried by the shaft. Increasing the column size will stien the frame, but in order to be really eective, the girder stiness will also need to be increased, since thegirders contribute about 2/3 of the de ection. Then the frames can be made o carry a larger proportion of the load. Note that the de ected shapes of the shaft and the frames are quite dierent, so that the above simple comparison of top de ections is not an accurate assessment. Finally, we have not studied the eect of the shaft stiened by the exterior columns, which are rigidly connected to the shaft walls and will avt with the shaft as a unit, Fig. 14.31. This would ft
ft
60’ 20’
20’
20’


156’
1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 CORE 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111
w
COLUMNS PARTICIPATE
+ +
Figure 14.31: Frame Rigidly Connected to Shaft, (Lin and Stotesbury 1981) be quite eective as the horizontal oor diaphragms will hold and force them to de ect together. 5. In summary, this appears to be quite an ecient layout, further analysis would re ne and optimize it.
14.4.7 Eect of Bracing Trusses Through strategically located havy trusses at the top and possibly at the middle of a building we can brace the exterior columns against the core shaft. This will result in a framelike action in the shaft,
54
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14.4 Lateral De ections
14{35
equalize temperature shortening of vertical components, and reduce lateral de ections, Fig. 14.32. TUBE HAT FULL CANTILEVER DEFLECTION WIND HAT  TRUSS TRUSS
T TENSION
C
COMPRESSION
T
CORE
BRACING REDUCE OVERALL DEFLEC OF BUILDING
WITH BRACING EFFECT
HEIGHT
C
MID  HEIGHT BRACE
WITH CANTILEVER CORE BENDING
TIEDOWN
TOTAL RESISTANCE ARM IS INCREASED BY COL. ACTION
DEFLECTION
RESISTANCE ARM OF CORE SHAFT ONLY
Figure 14.32: Eect of Exterior Column Bracing in Buildings, (Lin and Stotesbury 1981)
Victor Saouma
Structural Concepts and Systems for Architects
Draft 14{36
BUILDING STRUCTURES
Victor Saouma
Structural Concepts and Systems for Architects
Draft
Bibliography 318, A. C.: n.d., Building Code Requirements for Reinforced Concrete, (ACI 31883), American Concrete Institute. Anon.: xx, Envyclopaedia Brittanica, University of Chicago. Billington, D.: 1973, in D. Billington, R. Mark and J. Abel (eds), The Maillart Papers; Second National Conference on Civil Engineering: History, Heritage and the Humanities, Department of Civil Engineering, Princeton University. Billington, D.: 1979, Robert Maillart's Bridges; The art of Engineering, Princeton University Press. Billington, D.: 1985, The Tower and the Bridge, xx. Billington, D. and Mark, R.: 1983, Structural studies, Technical report, Department of Civil Engineering, Princeton University. Galilei, G.: 1974, Two New Sciences, Including Centers of Gravity and Forces of Percussion, University of Wisconsin Press, Madison, Wisc. S. Drake translation. le Duc, V.: 1977, Entretiens sur L'Architecture, Pierre Mardaga, Bruxelles, Belgique. Lin, T. and Stotesbury, S.: 1981, Structural Concepts and Systems for Architects and Engineers, John Wiley. Nilson, A.: 1978, Design of Prestressed Concrete, John Wiley and Sons. of Steel COnstruction, A. I.: 1986, Manual of Steel Construction; Load and Resistance Factor Design, American Institute of Steel Construction. Palladio, A.: 19xx, The Four Books of Architecture, Dover Publications. Penvenuto, E.: 1991, An Introduction to the History of Structural Mechanics, SpringerVerlag. Schueller, W.: 1996, The design of Building Structures, Prentice Hall. Timoshenko, S.: 1982, History of Strength of Materials, Dover Publications. UBC: 1995, Uniform building code, Technical report, International COnference of Building Ocials. Vitruvius: 1960, The Ten Books on Architecture, Dover.