Lectures in Structural Engineering Part I

Draft DRAFT LECTURE NOTES CVEN 3525/3535 STUCTURAL ENGINEERING I c VICTOR E. SAOUMA Spring 1999 Dept. of Civil E...

Author: Victor E Saouma

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DRAFT

LECTURE NOTES CVEN 3525/3535 STUCTURAL ENGINEERING I

c VICTOR

E. SAOUMA

Spring 1999

Dept. of Civil Environmental and Architectural Engineering University of Colorado, Boulder, CO 80309-0428

June 5, 2001

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Blank Page

Victor Saouma

Structural Engineering

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0–3 PREFACE

Whereas there are numerous excellent textbooks covering Structural Analysis, or Structural Design, I felt that there was a need for a single reference which • Provides a succinct, yet rigorous, coverage of Structural Engineering. • Combines, as much as possible, Analysis with Design. • Presents numerous, carefully selected, example problems. in a properly type set document. As such, and given the reluctance of undergraduate students to go through extensive verbage in order to capture a key concept, I have opted for an unusual format, one in which each key idea is clearly distinguishable. In addition, such a format will hopefully foster group learning among students who can easily reference misunderstood points. Finally, whereas all problems have been taken from a variety of references, I have been very careful in not only properly selecting them, but also in enhancing their solution through appropriate ﬁgures and LATEX typesetting macros.

Victor Saouma

Structural Engineering

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Structural Engineering can be characterized as the art of molding materials we don’t really understand into shapes we cannot really analyze so as to withstand forces we cannot really assess in such a way that the public does not really suspect.

-Really Unknown Source

Victor Saouma

Structural Engineering

Draft Contents I

STRUCTURAL ENGINEERING I; CVEN 3525

1 INTRODUCTION 1.1 Structural Engineering . . . . . . . 1.2 Structures and their Surroundings 1.3 Architecture & Engineering . . . . 1.4 Architectural Design Process . . . 1.5 Architectural Design . . . . . . . . 1.6 Structural Analysis . . . . . . . . . 1.7 Structural Design . . . . . . . . . . 1.8 Load Transfer Elements . . . . . . 1.9 Structure Types . . . . . . . . . . 1.10 Structural Engineering Courses . . 1.11 References . . . . . . . . . . . . . .

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1–1 1–1 1–1 1–1 1–2 1–2 1–2 1–2 1–3 1–3 1–10 1–12

2 LOADS 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Vertical Loads . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Dead Load . . . . . . . . . . . . . . . . . . . . 2.2.2 Live Loads . . . . . . . . . . . . . . . . . . . . E 2-1 Live Load Reduction . . . . . . . . . . . . . . . 2.2.3 Snow . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Lateral Loads . . . . . . . . . . . . . . . . . . . . . . . 2.3.1 Wind . . . . . . . . . . . . . . . . . . . . . . . E 2-2 Wind Load . . . . . . . . . . . . . . . . . . . . 2.3.2 Earthquakes . . . . . . . . . . . . . . . . . . . . E 2-3 Earthquake Load on a Frame . . . . . . . . . . E 2-4 Earthquake Load on a Tall Building, (Schueller 2.4 Other Loads . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Hydrostatic and Earth . . . . . . . . . . . . . . E 2-5 Hydrostatic Load . . . . . . . . . . . . . . . . . 2.4.2 Thermal . . . . . . . . . . . . . . . . . . . . . . E 2-6 Thermal Expansion/Stress (Schueller 1996) . . 2.4.3 Bridge Loads . . . . . . . . . . . . . . . . . . . 2.4.4 Impact Load . . . . . . . . . . . . . . . . . . . 2.5 Other Important Considerations . . . . . . . . . . . . 2.5.1 Load Combinations . . . . . . . . . . . . . . . . 2.5.2 Load Placement . . . . . . . . . . . . . . . . . 2.5.3 Structural Response . . . . . . . . . . . . . . . 2.5.4 Tributary Areas . . . . . . . . . . . . . . . . .

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2–1 2–1 2–1 2–1 2–2 2–4 2–4 2–6 2–6 2–9 2–11 2–14 2–15 2–16 2–16 2–16 2–17 2–17 2–18 2–18 2–18 2–18 2–19 2–19 2–22

3 STRUCTURAL MATERIALS 3.1 Steel . . . . . . . . . . . . . . 3.1.1 Structural Steel . . . . 3.1.2 Reinforcing Steel . . . 3.2 Aluminum . . . . . . . . . . .

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3–1 3–1 3–1 3–2 3–5

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4 EQUILIBRIUM & REACTIONS 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . 4.2 Equilibrium . . . . . . . . . . . . . . . . . . . . . 4.3 Equations of Conditions . . . . . . . . . . . . . . 4.4 Static Determinacy . . . . . . . . . . . . . . . . . E 4-1 Statically Indeterminate Cable Structure 4.5 Geometric Instability . . . . . . . . . . . . . . . . 4.6 Examples . . . . . . . . . . . . . . . . . . . . . . E 4-2 Simply Supported Beam . . . . . . . . . . E 4-3 Parabolic Load . . . . . . . . . . . . . . . E 4-4 Three Span Beam . . . . . . . . . . . . . E 4-5 Three Hinged Gable Frame . . . . . . . . E 4-6 Inclined Supports . . . . . . . . . . . . . . 4.7 Arches . . . . . . . . . . . . . . . . . . . . . . . .

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4–1 . 4–1 . 4–2 . 4–3 . 4–3 . 4–3 . 4–5 . 4–5 . 4–5 . 4–6 . 4–7 . 4–8 . 4–10 . 4–11

5 TRUSSES 5.1 Introduction . . . . . . . . . . . . 5.1.1 Assumptions . . . . . . . 5.1.2 Basic Relations . . . . . . 5.2 Trusses . . . . . . . . . . . . . . 5.2.1 Determinacy and Stability 5.2.2 Method of Joints . . . . . E 5-1 Truss, Method of Joints . 5.2.2.1 Matrix Method . E 5-2 Truss I, Matrix Method . E 5-3 Truss II, Matrix Method . 5.2.3 Method of Sections . . . . E 5-4 Truss, Method of Sections 5.3 Case Study: Stadium . . . . . . .

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Concrete . . . . . . . . . . . . . . . . Masonry . . . . . . . . . . . . . . . . Timber . . . . . . . . . . . . . . . . Steel Section Properties . . . . . . . 3.6.1 ASCII File with Steel Section Joists . . . . . . . . . . . . . . . . .

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3–5 3–7 3–7 3–7 3–17 3–19

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5–1 5–1 5–1 5–1 5–3 5–3 5–3 5–5 5–7 5–10 5–11 5–13 5–13 5–14

6 CABLES 6.1 Funicular Polygons . . . . . . . . . . . . E 6-1 Funicular Cable Structure . . . . 6.2 Uniform Load . . . . . . . . . . . . . . . 6.2.1 qdx; Parabola . . . . . . . . . . . 6.2.2 † qds; Catenary . . . . . . . . . . 6.2.2.1 Historical Note . . . . . E 6-2 Design of Suspension Bridge . . . 6.3 Case Study: George Washington Bridge 6.3.1 Geometry . . . . . . . . . . . . . 6.3.2 Loads . . . . . . . . . . . . . . . 6.3.3 Cable Forces . . . . . . . . . . . 6.3.4 Reactions . . . . . . . . . . . . .

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6–1 6–1 6–1 6–3 6–3 6–5 6–6 6–7 6–9 6–9 6–10 6–10 6–11

7 DESIGN PHILOSOPHIES of ACI and AISC CODES 7.1 Safety Provisions . . . . . . . . . . . . . . . . . . . . . . 7.2 Working Stress Method . . . . . . . . . . . . . . . . . . 7.3 Ultimate Strength Method . . . . . . . . . . . . . . . . . 7.3.1 The Normal Distribution . . . . . . . . . . . . . 7.3.2 Reliability Index . . . . . . . . . . . . . . . . . . 7.3.3 Discussion . . . . . . . . . . . . . . . . . . . . . .

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7–1 7–1 7–2 7–2 7–2 7–4 7–5

Victor Saouma

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Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–7 E 7-1 LRFD vs ASD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–7

8 DESIGN I 8.1 Case Study: Eiﬀel Tower . . . . . . . . . . . . . . . . . . 8.1.1 Materials, & Geometry . . . . . . . . . . . . . . 8.1.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . 8.1.3 Reactions . . . . . . . . . . . . . . . . . . . . . . 8.1.4 Internal Forces . . . . . . . . . . . . . . . . . . . 8.1.5 Internal Stresses . . . . . . . . . . . . . . . . . . 8.2 Magazini Generali by Maillart . . . . . . . . . . . . . . . 8.2.1 Geometry . . . . . . . . . . . . . . . . . . . . . . 8.2.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . 8.2.3 Reactions . . . . . . . . . . . . . . . . . . . . . . 8.2.4 Forces . . . . . . . . . . . . . . . . . . . . . . . . 8.2.5 Internal Stresses . . . . . . . . . . . . . . . . . . 8.3 Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.1 Buildings Structures . . . . . . . . . . . . . . . . 8.3.1.1 Wall Subsystems . . . . . . . . . . . . . 8.3.1.1.1 Example: Concrete Shear Wall 8.3.1.1.2 Example: Trussed Shear Wall 8.3.1.2 Shaft Systems . . . . . . . . . . . . . . 8.3.1.2.1 Example: Tube Subsystem . . 8.3.1.3 Rigid Frames . . . . . . . . . . . . . . . 8.4 Design of a Three Hinged Arch . . . . . . . . . . . . . . 8.5 Salginatobel Bridge (Maillart) . . . . . . . . . . . . . . . 8.5.1 Geometry . . . . . . . . . . . . . . . . . . . . . . 8.5.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . 8.5.3 Reactions . . . . . . . . . . . . . . . . . . . . . . 8.5.4 Internal Forces . . . . . . . . . . . . . . . . . . . 8.5.5 Internal Stresses . . . . . . . . . . . . . . . . . . 8.5.6 Structural Behavior of Deck-Stiﬀened Arches . .

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8–1 . 8–1 . 8–1 . 8–2 . 8–5 . 8–6 . 8–7 . 8–8 . 8–8 . 8–8 . 8–9 . 8–10 . 8–11 . 8–12 . 8–12 . 8–12 . 8–13 . 8–15 . 8–15 . 8–16 . 8–17 . 8–17 . 8–18 . 8–18 . 8–20 . 8–21 . 8–21 . 8–24 . 8–25

9 STEEL TENSION MEMBERS 9.1 Introduction . . . . . . . . . . . . . . . . . . . 9.2 Geometric Considerations . . . . . . . . . . . 9.2.1 Areas . . . . . . . . . . . . . . . . . . 9.2.1.1 Net Area, An . . . . . . . . . E 9-1 Net Area . . . . . . . . . . . . . . . . E 9-2 Net Area, Angle . . . . . . . . . . . . 9.2.1.2 Eﬀective Net Area, Ae . . . . E 9-3 Reduction Factor . . . . . . . . . . . . 9.2.2 Stiﬀness . . . . . . . . . . . . . . . . . 9.3 LRFD Design of Tension Members . . . . . . 9.3.1 Tension Failure . . . . . . . . . . . . . 9.3.2 Block Shear Failure . . . . . . . . . . 9.3.3 Summary . . . . . . . . . . . . . . . . E 9-4 Load Capacity of Angle . . . . . . . . E 9-5 Shear Rupture . . . . . . . . . . . . . E 9-6 Shear Rupture . . . . . . . . . . . . . E 9-7 Complete Analysis/Design of a Truss . 9.4 †† Computer Aided Analysis . . . . . . . . .

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10 INTERNAL FORCES IN STRUCTURES 10.1 Design Sign Conventions . . . . . . . . . . . . . . . . . . . . . 10.2 Load, Shear, Moment Relations . . . . . . . . . . . . . . . . . 10.3 Moment Envelope . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4.1 Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . E 10-1 Simple Shear and Moment Diagram . . . . . . . . . . E 10-2 Sketches of Shear and Moment Diagrams . . . . . . . 10.4.2 Frames . . . . . . . . . . . . . . . . . . . . . . . . . . E 10-3 Frame Shear and Moment Diagram . . . . . . . . . . . E 10-4 Frame Shear and Moment Diagram; Hydrostatic Load E 10-5 Shear Moment Diagrams for Frame . . . . . . . . . . . E 10-6 Shear Moment Diagrams for Inclined Frame . . . . . . 10.4.3 3D Frame . . . . . . . . . . . . . . . . . . . . . . . . . E 10-7 3D Frame . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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10–1 . 10–1 . 10–3 . 10–4 . 10–4 . 10–4 . 10–4 . 10–6 . 10–7 . 10–8 . 10–10 . 10–13 . 10–14 . 10–15 . 10–15 . 10–18

11 ARCHES and CURVED STRUCTURES 11.1 Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1.1 Statically Determinate . . . . . . . . . . . . . . . . . . . . . . . . . . E 11-1 Three Hinged Arch, Point Loads. (Gerstle 1974) . . . . . . . . . . . E 11-2 Semi-Circular Arch, (Gerstle 1974) . . . . . . . . . . . . . . . . . . . 11.1.2 Statically Indeterminate . . . . . . . . . . . . . . . . . . . . . . . . . E 11-3 Statically Indeterminate Arch, (Kinney 1957) . . . . . . . . . . . . . 11.2 Curved Space Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 11-4 Semi-Circular Box Girder, (Gerstle 1974) . . . . . . . . . . . . . . . 11.2.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2.1.2 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . E 11-5 Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974)

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11–1 . 11–1 . 11–4 . 11–4 . 11–4 . 11–6 . 11–6 . 11–9 . 11–9 . 11–10 . 11–11 . 11–11 . 11–12

12 DEFLECTION of STRUCTRES; Geometric Methods 12.1 Flexural Deformation . . . . . . . . . . . . . . . . . . . 12.1.1 Curvature Equation . . . . . . . . . . . . . . . . 12.1.2 Diﬀerential Equation of the Elastic Curve . . . . 12.1.3 Moment Temperature Curvature Relation . . . . 12.2 Flexural Deformations . . . . . . . . . . . . . . . . . . . 12.2.1 Direct Integration Method . . . . . . . . . . . . . E 12-1 Double Integration . . . . . . . . . . . . . . . . . 12.2.2 Curvature Area Method (Moment Area) . . . . . 12.2.2.1 First Moment Area Theorem . . . . . . 12.2.2.2 Second Moment Area Theorem . . . . . E 12-2 Moment Area, Cantilevered Beam . . . . . . . . E 12-3 Moment Area, Simply Supported Beam . . . . . 12.2.2.3 Maximum Deﬂection . . . . . . . . . . E 12-4 Maximum Deﬂection . . . . . . . . . . . . . . . . E 12-5 Frame Deﬂection . . . . . . . . . . . . . . . . . . E 12-6 Frame Subjected to Temperature Loading . . . . 12.2.3 Elastic Weight/Conjugate Beams . . . . . . . . . E 12-7 Conjugate Beam . . . . . . . . . . . . . . . . . . 12.3 Axial Deformations . . . . . . . . . . . . . . . . . . . . . 12.4 Torsional Deformations . . . . . . . . . . . . . . . . . .

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12–1 . 12–1 . 12–1 . 12–3 . 12–3 . 12–4 . 12–4 . 12–4 . 12–5 . 12–5 . 12–5 . 12–8 . 12–8 . 12–10 . 12–10 . 12–11 . 12–12 . 12–14 . 12–15 . 12–17 . 12–17

Victor Saouma

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13 ENERGY METHODS; Part I 13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Real Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.1 External Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2.2 Internal Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 13-1 Deﬂection of a Cantilever Beam, (Chajes 1983) . . . . . . . . . . . . . . . . . . . 13.3 Virtual Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ∗ 13.3.1 External Virtual Work δW . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ∗ 13.3.2 Internal Virtual Work δU . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 13-2 Beam Deﬂection (Chajes 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 13-3 Deﬂection of a Frame (Chajes 1983) . . . . . . . . . . . . . . . . . . . . . . . . . E 13-4 Rotation of a Frame (Chajes 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . E 13-5 Truss Deﬂection (Chajes 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 13-6 Torsional and Flexural Deformation, (Chajes 1983) . . . . . . . . . . . . . . . . . E 13-7 Flexural and Shear Deformations in a Beam (White, Gergely and Sexmith 1976) E 13-8 Thermal Eﬀects in a Beam (White et al. 1976) . . . . . . . . . . . . . . . . . . . E 13-9 Deﬂection of a Truss (White et al. 1976) . . . . . . . . . . . . . . . . . . . . . . . E 13-10Thermal Defelction of a Truss; I (White et al. 1976) . . . . . . . . . . . . . . . . E 13-11Thermal Deﬂections in a Truss; II (White et al. 1976) . . . . . . . . . . . . . . . E 13-12Truss with initial camber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 13-13Prestressed Concrete Beam with Continously Variable I (White et al. 1976) . . . 13.4 *Maxwell Betti’s Reciprocal Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.5 Summary of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13–1 . 13–1 . 13–1 . 13–1 . 13–2 . 13–3 . 13–4 . 13–6 . 13–6 . 13–9 . 13–9 . 13–11 . 13–12 . 13–13 . 13–14 . 13–15 . 13–16 . 13–17 . 13–18 . 13–19 . 13–20 . 13–21 . 13–24 . 13–24

14 BRACED ROLLED STEEL BEAMS 14.1 Review from Strength of Materials . . . . . . . . . . . . 14.1.1 Flexure . . . . . . . . . . . . . . . . . . . . . . . 14.1.2 Shear . . . . . . . . . . . . . . . . . . . . . . . . 14.2 Nominal Strength . . . . . . . . . . . . . . . . . . . . . . 14.3 Flexural Design . . . . . . . . . . . . . . . . . . . . . . . 14.3.1 Failure Modes and Classiﬁcation of Steel Beams 14.3.1.1 Compact Sections . . . . . . . . . . . . 14.3.1.2 Partially Compact Section . . . . . . . 14.3.1.3 Slender Section . . . . . . . . . . . . . . 14.3.2 Examples . . . . . . . . . . . . . . . . . . . . . . E 14-1 Shape Factors, Rectangular Section . . . . . . . E 14-2 Shape Factors, T Section . . . . . . . . . . . . . E 14-3 Beam Design . . . . . . . . . . . . . . . . . . . . 14.4 Shear Design . . . . . . . . . . . . . . . . . . . . . . . . 14.5 Deﬂections . . . . . . . . . . . . . . . . . . . . . . . . . 14.6 Complete Design Example . . . . . . . . . . . . . . . . .

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14–1 . 14–1 . 14–1 . 14–3 . 14–4 . 14–5 . 14–5 . 14–5 . 14–6 . 14–7 . 14–7 . 14–7 . 14–8 . 14–9 . 14–11 . 14–11 . 14–12

15 COLUMN STABILITY 15.1 Introduction; Discrete Rigid Bars . . . . . 15.1.1 Single Bar System . . . . . . . . . 15.1.2 Two Bars System . . . . . . . . . . 15.1.3 ‡Analogy with Free Vibration . . . 15.2 Continuous Linear Elastic Systems . . . . 15.2.1 Lower Order Diﬀerential Equation 15.2.2 Higher Order Diﬀerential Equation 15.2.2.1 Derivation . . . . . . . . 15.2.2.2 Hinged-Hinged Column . 15.2.2.3 Fixed-Fixed Column . . . 15.2.2.4 Fixed-Hinged Column . . 15.2.3 Eﬀective Length Factors K . . . . 15.3 Inelastic Columns . . . . . . . . . . . . . .

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15–1 . 15–1 . 15–1 . 15–2 . 15–4 . 15–5 . 15–6 . 15–6 . 15–6 . 15–8 . 15–8 . 15–9 . 15–10 . 15–12

Victor Saouma

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16 STEEL COMPRESSION MEMBERS 16.1 AISC Equations . . . . . . . . . . . . . . . . . 16.2 LRFD Equations . . . . . . . . . . . . . . . . . 16.3 Examples . . . . . . . . . . . . . . . . . . . . . 16.3.1 Evaluation . . . . . . . . . . . . . . . . E 16-1 Unbraced Column Evaluation, (?) . . . E 16-2 Braced Column Evaluation, (?) . . . . . E 16-3 Braced Column Evaluation, (?) . . . . . 16.3.2 Design . . . . . . . . . . . . . . . . . . . E 16-4 Column Design, (?) . . . . . . . . . . . E 16-5 Design of Braced Column, (?) . . . . . . E 16-6 Design of a Column, (?) . . . . . . . . . E 16-7 Column Design Using AISC Charts, (?)

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16–1 . 16–1 . 16–2 . 16–6 . 16–6 . 16–6 . 16–6 . 16–7 . 16–8 . 16–8 . 16–9 . 16–9 . 16–10

17 STEEL CONNECTIONS 17.1 Bolted Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.1.1 Types of Bolts . . . . . . . . . . . . . . . . . . . . . . . . . 17.1.2 Types of Bolted Connections . . . . . . . . . . . . . . . . . 17.1.3 Nominal Strength of Individual Bolts . . . . . . . . . . . . . 17.1.4 Concentric Loads (Tension Connections) . . . . . . . . . . . 17.1.4.1 AISC Requirements . . . . . . . . . . . . . . . . . 17.1.4.2 Examples . . . . . . . . . . . . . . . . . . . . . . . E 17-1 Analysis of Bolted Connections (Salmon and Johnson 1990) E 17-2 Design of Bolted Connections (Salmon and Johnson 1990) . 17.1.5 Eccentric Connection (Shear Connections) . . . . . . . . . . 17.2 Welded Connections . . . . . . . . . . . . . . . . . . . . . . . . . .

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17–1 . 17–1 . 17–1 . 17–1 . 17–4 . 17–5 . 17–5 . 17–6 . 17–6 . 17–8 . 17–9 . 17–9

18 UNBRACED ROLLED STEEL BEAMS 18.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 18.2 Background . . . . . . . . . . . . . . . . . . . . . . 18.3 AISC Equations . . . . . . . . . . . . . . . . . . . 18.3.1 Dividing values . . . . . . . . . . . . . . . . 18.3.2 Governing Moments . . . . . . . . . . . . . 18.4 Examples . . . . . . . . . . . . . . . . . . . . . . . 18.4.1 Veriﬁcation . . . . . . . . . . . . . . . . . . E 18-1 Adequacy of an unbraced beam, (?) . . . . E 18-2 Adequacy of an unbraced beam, II (?) . . . 18.4.2 Design . . . . . . . . . . . . . . . . . . . . . E 18-3 Design of Laterally Unsupported Beam, (?) 18.5 Summary of AISC Governing Equations . . . . . .

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19 Beam Columns, (Unedited) 19.1 Potential Modes of Failures . . . . . . . . 19.2 AISC Speciﬁcations . . . . . . . . . . . . 19.3 Examples . . . . . . . . . . . . . . . . . . 19.3.1 Veriﬁcation . . . . . . . . . . . . . E 19-1 Veriﬁcation, (?) . . . . . . . . . . . E 19-2 8.2, (?) . . . . . . . . . . . . . . . 19.3.2 Design . . . . . . . . . . . . . . . . E 19-3 Design of Steel Beam-Column, (?)

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19–1 . 19–1 . 19–1 . 19–1 . 19–1 . 19–1 . 19–4 . 19–6 . 19–7

II

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STRUCTURAL ENGINEERING II; CVEN 3235

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20 ARCHES and CURVED STRUCTURES 20.1 Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.1.1 Statically Determinate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 20-1 Three Hinged Arch, Point Loads. (Gerstle 1974) . . . . . . . . . . . . . . . . . . Victor Saouma

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E 20-2 Semi-Circular Arch, (Gerstle 1974) . . . . . . . . . . . . . . . . . . . 20.1.2 Statically Indeterminate . . . . . . . . . . . . . . . . . . . . . . . . . E 20-3 Statically Indeterminate Arch, (Kinney 1957) . . . . . . . . . . . . . 20.2 Curved Space Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 20-4 Semi-Circular Box Girder, (Gerstle 1974) . . . . . . . . . . . . . . . 20.2.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2.1.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2.1.2 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . E 20-5 Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974)

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21 STATIC INDETERMINANCY; FLEXIBILITY METHOD 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 The Force/Flexibility Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Short-Cut for Displacement Evaluation . . . . . . . . . . . . . . . . . . . . . . . 21.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 21-1 Steel Building Frame Analysis, (White et al. 1976) . . . . . . . . . . . . E 21-2 Analysis of Irregular Building Frame, (White et al. 1976) . . . . . . . . E 21-3 Redundant Truss Analysis, (White et al. 1976) . . . . . . . . . . . . . . E 21-4 Truss with Two Redundants, (White et al. 1976) . . . . . . . . . . . . . E 21-5 Analysis of Nonprismatic Members, (White et al. 1976) . . . . . . . . . E 21-6 Fixed End Moments for Nonprismatic Beams, (White et al. 1976) . . . E 21-7 Rectangular Frame; External Load, (White et al. 1976) . . . . . . . . . E 21-8 Frame with Temperature Eﬀects and Support Displacements, (White et al. 1976) . . . . . . . . . . . . . E 21-9 Braced Bent with Loads and Temperature Change, (White et al. 1976) 22 APPROXIMATE FRAME ANALYSIS 22.1 Vertical Loads . . . . . . . . . . . . . . . . . . . . . 22.2 Horizontal Loads . . . . . . . . . . . . . . . . . . . 22.2.1 Portal Method . . . . . . . . . . . . . . . . E 22-1 Approximate Analysis of a Frame subjected

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23 KINEMATIC INDETERMINANCY; STIFFNESS METHOD 23.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.1.1 Stiﬀness vs Flexibility . . . . . . . . . . . . . . . . . . . . . . . . . . 23.1.2 Sign Convention . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.2 Degrees of Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.2.1 Methods of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.3 Kinematic Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.3.1 Force-Displacement Relations . . . . . . . . . . . . . . . . . . . . . . 23.3.2 Fixed End Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.3.2.1 Uniformly Distributed Loads . . . . . . . . . . . . . . . . . 23.3.2.2 Concentrated Loads . . . . . . . . . . . . . . . . . . . . . . 23.4 Slope Deﬂection; Direct Solution . . . . . . . . . . . . . . . . . . . . . . . . 23.4.1 Slope Deﬂection Equations . . . . . . . . . . . . . . . . . . . . . . . 23.4.2 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.4.3 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.4.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 23-1 Propped Cantilever Beam, (Arbabi 1991) . . . . . . . . . . . . . . . E 23-2 Two-Span Beam, Slope Deﬂection, (Arbabi 1991) . . . . . . . . . . . E 23-3 Two-Span Beam, Slope Deﬂection, Initial Deﬂection, (Arbabi 1991) E 23-4 dagger Frames, Slope Deﬂection, (Arbabi 1991) . . . . . . . . . . . . 23.5 Moment Distribution; Indirect Solution . . . . . . . . . . . . . . . . . . . . 23.5.1 Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23.5.1.1 Sign Convention . . . . . . . . . . . . . . . . . . . . . . . . 23.5.1.2 Fixed-End Moments . . . . . . . . . . . . . . . . . . . . . . 23.5.1.3 Stiﬀness Factor . . . . . . . . . . . . . . . . . . . . . . . . .

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23.5.1.4 Distribution Factor (DF) . . . . . . . . . . . 23.5.1.5 Carry-Over Factor . . . . . . . . . . . . . . . 23.5.2 Procedure . . . . . . . . . . . . . . . . . . . . . . . . . 23.5.3 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . 23.5.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . E 23-5 Continuous Beam, (Kinney 1957) . . . . . . . . . . . . E 23-6 Continuous Beam, Simpliﬁed Method, (Kinney 1957) . E 23-7 Continuous Beam, Initial Settlement, (Kinney 1957) . E 23-8 Frame, (Kinney 1957) . . . . . . . . . . . . . . . . . . E 23-9 Frame with Side Load, (Kinney 1957) . . . . . . . . . E 23-10Moment Distribution on a Spread-Sheet . . . . . . . .

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23–16 23–16 23–17 23–17 23–18 23–18 23–20 23–22 23–23 23–27 23–29

24 REINFORCED CONCRETE BEAMS; Part I 24.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 24.1.1 Notation . . . . . . . . . . . . . . . . . . . . . 24.1.2 Modes of Failure . . . . . . . . . . . . . . . . 24.1.3 Analysis vs Design . . . . . . . . . . . . . . . 24.1.4 Basic Relations and Assumptions . . . . . . . 24.1.5 ACI Code . . . . . . . . . . . . . . . . . . . . 24.2 Cracked Section, Ultimate Strength Design Method . 24.2.1 Equivalent Stress Block . . . . . . . . . . . . 24.2.2 Balanced Steel Ratio . . . . . . . . . . . . . . 24.2.3 Analysis . . . . . . . . . . . . . . . . . . . . . 24.2.4 Design . . . . . . . . . . . . . . . . . . . . . . E 24-1 Ultimate Strength Capacity . . . . . . . . . . E 24-2 Beam Design I . . . . . . . . . . . . . . . . . E 24-3 Beam Design II . . . . . . . . . . . . . . . . . 24.3 ACI Code . . . . . . . . . . . . . . . . . . . . . . . .

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25 REINFORCED CONCRETE BEAMS; Part II 25.1 T Beams, (ACI 8.10) . . . . . . . . . . . . . . . . 25.1.1 Review . . . . . . . . . . . . . . . . . . . . 25.1.2 Design, (balanced) . . . . . . . . . . . . . . E 25-1 T Beam; Moment Capacity I . . . . . . . . E 25-2 T Beam; Moment Capacity II . . . . . . . . E 25-3 T Beam; Design . . . . . . . . . . . . . . . 25.2 Doubly Reinforced Rectangular Beams . . . . . . . 25.2.1 Tests for fs and fs . . . . . . . . . . . . . . 25.2.2 Moment Equations . . . . . . . . . . . . . . E 25-4 Doubly Reinforced Concrete beam; Review E 25-5 Doubly Reinforced Concrete beam; Design .

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26 DIRECT STIFFNESS METHOD 26.1 Introduction . . . . . . . . . . . . . . 26.1.1 Structural Idealization . . . . 26.1.2 Structural Discretization . . . 26.1.3 Coordinate Systems . . . . . 26.1.4 Sign Convention . . . . . . . 26.1.5 Degrees of Freedom . . . . . 26.2 Stiﬀness Matrices . . . . . . . . . . . 26.2.1 Truss Element . . . . . . . . 26.2.2 Beam Element . . . . . . . . 26.2.3 2D Frame Element . . . . . . 26.2.4 Remarks on Element Stiﬀness 26.3 Direct Stiﬀness Method . . . . . . . 26.3.1 Orthogonal Structures . . . . E 26-1 Beam . . . . . . . . . . . . .

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26.3.2 Local and Global Element Stiﬀness Matrices ([k(e) ] [K(e) ]) 26.3.2.1 2D Frame . . . . . . . . . . . . . . . . . . . . . . . 26.3.3 Global Stiﬀness Matrix . . . . . . . . . . . . . . . . . . . . 26.3.3.1 Structural Stiﬀness Matrix . . . . . . . . . . . . . 26.3.3.2 Augmented Stiﬀness Matrix . . . . . . . . . . . . 26.3.4 Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . 26.3.5 Boundary Conditions, [ID] Matrix . . . . . . . . . . . . . . 26.3.6 LM Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.3.7 Assembly of Global Stiﬀness Matrix . . . . . . . . . . . . . E 26-2 Assembly of the Global Stiﬀness Matrix . . . . . . . . . . . 26.3.8 Algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . E 26-3 Direct Stiﬀness Analysis of a Truss . . . . . . . . . . . . . . E 26-4 Analysis of a Frame with MATLAB . . . . . . . . . . . . . E 26-5 Analysis of a simple Beam with Initial Displacements . . . 26.4 Computer Program Organization . . . . . . . . . . . . . . . . . . . 26.5 Computer Implementation with MATLAB . . . . . . . . . . . . . . 26.5.1 Program Input . . . . . . . . . . . . . . . . . . . . . . . . . 26.5.1.1 Input Variable Descriptions . . . . . . . . . . . . . 26.5.1.2 Sample Input Data File . . . . . . . . . . . . . . . 26.5.1.3 Program Implementation . . . . . . . . . . . . . . 26.5.2 Program Listing . . . . . . . . . . . . . . . . . . . . . . . . 26.5.2.1 Main Program . . . . . . . . . . . . . . . . . . . . 26.5.2.2 Assembly of ID Matrix . . . . . . . . . . . . . . . 26.5.2.3 Element Nodal Coordinates . . . . . . . . . . . . . 26.5.2.4 Element Lengths . . . . . . . . . . . . . . . . . . . 26.5.2.5 Element Stiﬀness Matrices . . . . . . . . . . . . . 26.5.2.6 Transformation Matrices . . . . . . . . . . . . . . 26.5.2.7 Assembly of the Augmented Stiﬀness Matrix . . . 26.5.2.8 Print General Information . . . . . . . . . . . . . 26.5.2.9 Print Load . . . . . . . . . . . . . . . . . . . . . . 26.5.2.10 Load Vector . . . . . . . . . . . . . . . . . . . . . 26.5.2.11 Nodal Displacements . . . . . . . . . . . . . . . . 26.5.2.12 Reactions . . . . . . . . . . . . . . . . . . . . . . . 26.5.2.13 Internal Forces . . . . . . . . . . . . . . . . . . . . 26.5.2.14 Plotting . . . . . . . . . . . . . . . . . . . . . . . . 26.5.2.15 Sample Output File . . . . . . . . . . . . . . . . .

27 COLUMNS 27.1 Introduction . . . . . . . . . . . . . . . . . 27.1.1 Types of Columns . . . . . . . . . 27.1.2 Possible Arrangement of Bars . . . 27.2 Short Columns . . . . . . . . . . . . . . . 27.2.1 Concentric Loading . . . . . . . . . 27.2.2 Eccentric Columns . . . . . . . . . 27.2.2.1 Balanced Condition . . . 27.2.2.2 Tension Failure . . . . . . 27.2.2.3 Compression Failure . . . 27.2.3 ACI Provisions . . . . . . . . . . . 27.2.4 Interaction Diagrams . . . . . . . . 27.2.5 Design Charts . . . . . . . . . . . E 27-1 R/C Column, c known . . . . . . . E 27-2 R/C Column, e known . . . . . . . E 27-3 R/C Column, Using Design Charts

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28 DESIGN II 28.1 Frames . . . . . . . . . . . . . . . . . . . . . . . . 28.1.1 Beam Column Connections . . . . . . . . 28.1.2 Behavior of Simple Frames . . . . . . . . 28.1.3 Eccentricity of Applied Loads . . . . . . . 28.1.4 Design of a Statically Indeterminate Arch 28.1.5 Temperature Changes in an Arch . . . . .

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29 INFLUENCE LINES (unedited) 30 ELEMENTS of STRUCTURAL RELIABILITY 30.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.2 Elements of Statistics . . . . . . . . . . . . . . . . . . . . . . . . 30.3 Distributions of Random Variables . . . . . . . . . . . . . . . . . 30.3.1 Uniform Distribution . . . . . . . . . . . . . . . . . . . . . 30.3.2 Normal Distribution . . . . . . . . . . . . . . . . . . . . . 30.3.3 Lognormal Distribution . . . . . . . . . . . . . . . . . . . 30.3.4 Beta Distribution . . . . . . . . . . . . . . . . . . . . . . . 30.3.5 BiNormal distribution . . . . . . . . . . . . . . . . . . . . 30.4 Reliability Index . . . . . . . . . . . . . . . . . . . . . . . . . . . 30.4.1 Performance Function Identiﬁcation . . . . . . . . . . . . 30.4.2 Deﬁnitions . . . . . . . . . . . . . . . . . . . . . . . . . . 30.4.3 Mean and Standard Deviation of a Performance Function 30.4.3.1 Direct Integration . . . . . . . . . . . . . . . . . 30.4.3.2 Monte Carlo Simulation . . . . . . . . . . . . . . 30.4.3.3 Point Estimate Method . . . . . . . . . . . . . . 30.4.3.4 Taylor’s Series-Finite Diﬀerence Estimation . . . 30.4.4 Overall System Reliability . . . . . . . . . . . . . . . . . . 30.4.5 Target Reliability Factors . . . . . . . . . . . . . . . . . . 30.5 Reliability Analysis . . . . . . . . . . . . . . . . . . . . . . . . . .

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30–1 . 30–1 . 30–1 . 30–3 . 30–3 . 30–3 . 30–4 . 30–4 . 30–4 . 30–4 . 30–4 . 30–5 . 30–6 . 30–6 . 30–6 . 30–9 . 30–9 . 30–10 . 30–10 . 30–12

Structural Engineering

Draft List of Figures 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

Types of Forces in Structural Elements (1D) . Basic Aspects of Cable Systems . . . . . . . . Basic Aspects of Arches . . . . . . . . . . . . Types of Trusses . . . . . . . . . . . . . . . . Variations in Post and Beams Conﬁgurations Diﬀerent Beam Types . . . . . . . . . . . . . Basic Forms of Frames . . . . . . . . . . . . . Examples of Air Supported Structures . . . . Basic Forms of Shells . . . . . . . . . . . . . . Sequence of Structural Engineering Courses .

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1–3 1–4 1–5 1–6 1–7 1–8 1–9 1–10 1–11 1–11

2.1 2.2 2.3 2.4 2.5 2.6 2.7

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2–2 2–4 2–5 2–5 2–6 2–7

2.8 2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17 2.18

Approximation of a Series of Closely Spaced Loads . . . . . . . . . . . . . . . . . . . . . Snow Map of the United States, ubc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Loads on Projected Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vertical and Normal Loads Acting on Inclined Surfaces . . . . . . . . . . . . . . . . . . Wind Map of the United States, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . . Eﬀect of Wind Load on Structures(Schueller 1996) . . . . . . . . . . . . . . . . . . . . . Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Vibrations of a Building . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Seismic Zones of the United States, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . Earth and Hydrostatic Loads on Structures . . . . . . . . . . . . . . . . . . . . . . . . . Truck Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Load Placement to Maximize Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . Load Life of a Structure, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . Concept of Tributary Areas for Structural Member Loading . . . . . . . . . . . . . . . . One or Two Way actions in Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Load Transfer in R/C Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Two Way Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example of Load Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2–10 2–11 2–12 2–17 2–18 2–20 2–20 2–21 2–21 2–23 2–24 2–24

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9

Stress Strain Curves of Concrete and Steel . . . . . . Standard Rolled Sections . . . . . . . . . . . . . . . Residual Stresses in Rolled Sections . . . . . . . . . Residual Stresses in Welded Sections . . . . . . . . . Inﬂuence of Residual Stress on Average Stress-Strain Concrete Stress-Strain curve . . . . . . . . . . . . . . Concrete microcracking . . . . . . . . . . . . . . . . W and C sections . . . . . . . . . . . . . . . . . . . . prefabricated Steel Joists . . . . . . . . . . . . . . .

4.1 4.2 4.3 4.4

Types of Supports . . . . . . . . . . . . . . . . . . . . . . . . . Inclined Roller Support . . . . . . . . . . . . . . . . . . . . . . Examples of Static Determinate and Indeterminate Structures . Geometric Instability Caused by Concurrent Reactions . . . . .

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3–1 3–2 3–3 3–3 3–4 3–6 3–6 3–8 3–20

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4–1 4–3 4–4 4–5

0–2

Draft

LIST OF FIGURES

5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10

Types of Trusses . . . . . . . . . . . . . . Bridge Truss . . . . . . . . . . . . . . . . A Statically Indeterminate Truss . . . . . X and Y Components of Truss Forces . . Sign Convention for Truss Element Forces Direction Cosines . . . . . . . . . . . . . . Forces Acting on Truss Joint . . . . . . . Complex Statically Determinate Truss . . Florence Stadium, Pier Luigi Nervi (?) . . Florence Stadioum, Pier Luigi Nervi (?) .

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5–2 5–2 5–4 5–4 5–5 5–8 5–8 5–9 5–14 5–15

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11

Cable Structure Subjected to q(x) . . . . . . . . . . . . . . . . . . Catenary versus Parabola Cable Structures . . . . . . . . . . . . . Leipniz’s Figure of a catenary, 1690 . . . . . . . . . . . . . . . . . . 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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6–3 6–6 6–7 6–9 6–10 6–11 6–11 6–12 6–13 6–13 6–14

7.1 7.2 7.3 7.4 7.5

Load Life of a Structure . . . . . . . . . . . . . . . . . . . . . . . . . . Normalized Gauss Distribution, and Cumulative Distribution Function Frequency Distributions of Load Q and Resistance R . . . . . . . . . . Deﬁnition of Reliability Index . . . . . . . . . . . . . . . . . . . . . . . Probability of Failure in terms of β . . . . . . . . . . . . . . . . . . . .

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7–2 7–3 7–4 7–5 7–5

8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16

Eiﬀel Tower (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . Eiﬀel Tower Idealization, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . Eiﬀel Tower, Dead Load Idealization; (Billington and Mark 1983) . . . . . . . . . . . . . Eiﬀel Tower, Wind Load Idealization; (Billington and Mark 1983) . . . . . . . . . . . . . Eiﬀel Tower, Wind Loads, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . Eiﬀel Tower, Reactions; (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . Eiﬀel Tower, Internal Gravity Forces; (Billington and Mark 1983) . . . . . . . . . . . . . Eiﬀel Tower, Horizontal Reactions; (Billington and Mark 1983) . . . . . . . . . . . . . . Eiﬀel Tower, Internal Wind Forces; (Billington and Mark 1983) . . . . . . . . . . . . . . Magazzini Generali; Overall Dimensions, (Billington and Mark 1983) . . . . . . . . . . . Magazzini Generali; Support System, (Billington and Mark 1983) . . . . . . . . . . . . . Magazzini Generali; Loads (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . Magazzini Generali; Beam Reactions, (Billington and Mark 1983) . . . . . . . . . . . . . Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983) . . . . . . Magazzini Generali; Internal Moment, (Billington and Mark 1983) . . . . . . . . . . . . Magazzini Generali; Similarities Between The Frame Shape and its Moment Diagram, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Magazzini Generali; Eﬀect of Lateral Supports, (Billington and Mark 1983) . . . . . . . Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . Trussed Shear Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Design Example of a Tubular Structure, (Lin and Stotesbury 1981) . . . . . . . . . . . . A Basic Portal Frame, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . Salginatobel Bridge; Dimensions, (Billington and Mark 1983) . . . . . . . . . . . . . . . Salginatobel Bridge; Idealization, (Billington and Mark 1983) . . . . . . . . . . . . . . . Salginatobel Bridge; Hinges, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . Salginatobel Bridge; Sections, (Billington and Mark 1983) . . . . . . . . . . . . . . . . .

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8–1 8–3 8–3 8–4 8–4 8–5 8–6 8–6 8–7 8–8 8–9 8–9 8–9 8–10 8–10

8.17 8.18 8.19 8.20 8.21 8.22 8.23 8.24 8.25 8.26

Victor Saouma

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. 8–11 . 8–11 . 8–12 . 8–13 . 8–15 . 8–16 . 8–17 . 8–19 . 8–19 . 8–19 . 8–20

Structural Engineering

LIST OF FIGURES

0–3

Draft 8.27 8.28 8.29 8.30 8.31 8.32 8.33

Salginatobel Bridge; Dead Load, (Billington and Mark 1983) . . . . . . . . . . Salginatobel Bridge; Truck Load, (Billington and Mark 1983) . . . . . . . . . . Salginatobel Bridge; Total Vertical Load, (Billington and Mark 1983) . . . . . . Salginatobel Bridge; Reactions, (Billington and Mark 1983) . . . . . . . . . . . Salganitobel Bridge; Shear Diagrams, (Billington and Mark 1983) . . . . . . . . Salginatobel Bridge; Live Load Moment Diagram, (Billington and Mark 1983) . Structural Behavior of Stiﬀened Arches, (Billington 1979) . . . . . . . . . . . .

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. 8–20 . 8–22 . 8–23 . 8–23 . 8–23 . 8–24 . 8–25

9.1 9.2 9.3 9.4 9.5 9.6

Stress Concentration Around Circular Hole Hole Sizes . . . . . . . . . . . . . . . . . . . Eﬀect of Staggered Holes on Net Area . . . Gage Distances for an Angle . . . . . . . . . Net and Gross Areas . . . . . . . . . . . . . Tearing Failure Limit State . . . . . . . . .

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10.1 10.2 10.3 10.4 10.5 10.6

Shear and Moment Sign Conventions for Design . . . . . . . . . . . . . . . . . . . . Sign Conventions for 3D Frame Elements . . . . . . . . . . . . . . . . . . . . . . . Free Body Diagram of an Inﬁnitesimal Beam Segment . . . . . . . . . . . . . . . . Shear and Moment Forces at Diﬀerent Sections of a Loaded Beam . . . . . . . . . Slope Relations Between Load Intensity and Shear, or Between Shear and Moment Inclined Loads on Inclined Members . . . . . . . . . . . . . . . . . . . . . . . . . .

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. 10–2 . 10–2 . 10–3 . 10–4 . 10–5 . 10–8

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9–2 9–3 9–3 9–4 9–9 9–9

11.1 Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . 11.3 Two Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Arch Rib Stiﬀened with Girder or Truss, (Lin and Stotesbury 1981) . . . . . . . . . . . 11.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Semi-Circular three hinged arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.7 Semi-Circular three hinged arch; Free body diagram . . . . . . . . . . . . . . . . . . . . 11.8 Statically Indeterminate Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.9 Statically Indeterminate Arch; ‘Horizontal Reaction Removed . . . . . . . . . . . . . . . 11.10Semi-Circular Box Girder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.11Geometry of Curved Structure in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.12Free Body Diagram of a Curved Structure in Space . . . . . . . . . . . . . . . . . . . . . 11.13Helicoidal Cantilevered Girder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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11–2 11–2 11–3 11–3 11–4 11–5 11–6 11–7 11–8 11–9 11–11 11–12 11–13

12.1 12.2 12.3 12.4 12.5 12.6 12.7

Curvature of a ﬂexural element . . . . . . . . . . . . . Moment Area Theorems . . . . . . . . . . . . . . . . . Sign Convention for the Moment Area Method . . . . Areas and Centroid of Polynomial Curves . . . . . . . Maximum Deﬂection Using the Moment Area Method Conjugate Beams . . . . . . . . . . . . . . . . . . . . . Torsion Rotation Relations . . . . . . . . . . . . . . .

13.1 Load Deﬂection Curves . . . 13.2 Strain Energy Deﬁnition . . . 13.3 Deﬂection of Cantilever Beam 13.4 Real and Virtual Forces . . . 13.5 Torsion Rotation Relations . 13.6 . . . . . . . . . . . . . . . . 13.7 . . . . . . . . . . . . . . . . 13.8 . . . . . . . . . . . . . . . . 13.9 . . . . . . . . . . . . . . . . 13.10 . . . . . . . . . . . . . . . . 13.11 . . . . . . . . . . . . . . . . 13.12 . . . . . . . . . . . . . . . . 13.13 . . . . . . . . . . . . . . . . Victor Saouma

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12–2 12–6 12–7 12–7 12–10 12–15 12–17

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13–2 13–3 13–4 13–5 13–7 13–10 13–10 13–11 13–12 13–13 13–14 13–15 13–17

Structural Engineering

0–4

LIST OF FIGURES

Draft

13.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–18 13.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–19 13.16*(correct 42.7 to 47.2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–22 14.1 14.2 14.3 14.4 14.5

Bending of a Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . Stress distribution at diﬀerent stages of loading . . . . . . . . . . . Stress-strain diagram for most structural steels . . . . . . . . . . . Flexural and Shear Stress Distribution in a Rectangular Beam . . Local (ﬂange) Buckling; Flexural and Torsional Buckling Modes (Lulea University) . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.6 W Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14.7 Nominal Moments for Compact and Partially Compact Sections . . 14.8 AISC Requirements for Shear Design . . . . . . . . . . . . . . . . .

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14–2 14–2 14–3 14–4

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14–5 14–6 14–7 14–12

15.1 Stability of a Rigid Bar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–1 15.2 Stability of a Rigid Bar with Initial Imperfection . . . . . . . . . . . . . . . . . . . . . . . 15–2 15.3 Stability of a Two Rigid Bars System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–3 15.4 Two DOF Dynamic System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–4 15.5 Euler Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–5 15.6 Simply Supported Beam Column; Diﬀerential Segment; Eﬀect of Axial Force P . . . . . . 15–7 15.7 Solution of the Tanscendental Equation for the Buckling Load of a Fixed-Hinged Column 15–10 15.8 Column Eﬀective Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–11 15.9 Frame Eﬀective Lengths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–11 15.10Column Eﬀective Length in a Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–12 15.11Standard Alignment Chart (AISC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–13 15.12Inelastic Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–13 15.13Euler Buckling, and SSRC Column Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . 15–14 16.1 SSRC Column Curve and AISC Critical Stresses . . . . . . . . . . . . . . . . . . . . . . . 16–2 16.2 Fcr versus KL/r According to LRFD, for Various Fy . . . . . . . . . . . . . . . . . . . . . 16–5 17.1 17.2 17.3 17.4 17.5 17.6 17.7

Examples of Bolted Connections . . . . . . . . . . . . . . . . Stress Transfer by Shear and Bearing in a Bolted Connection Stress Transfer in a Friction Type Bolted Connection . . . . . Modes of Failure of Bolted Connections . . . . . . . . . . . . Number of Shearing Planes m in Bolted Connections . . . . . Bearing Strength Related to End Distance . . . . . . . . . . . Basic Types of Weld . . . . . . . . . . . . . . . . . . . . . . .

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17–2 17–2 17–3 17–3 17–4 17–4 17–9

18.1 I Shaped Beam in Slightly Buckled Position . . . . . . . . . . . . . . . . . . . . . . . . . . 18–1 18.2 Nominal Strength Mn of “Compact” Sections as Aﬀected by Lateral-Torsional Buckling . 18–3 18.3 Design of Laterally Unsupported Steel Beam . . . . . . . . . . . . . . . . . . . . . . . . . 18–7 19.1 Rigid Frame Subjected to Lateral Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . 19–4 20.1 Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.2 Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . 20.3 Two Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . 20.4 Arch Rib Stiﬀened with Girder or Truss, (Lin and Stotesbury 1981) . . . . . . . . . . . 20.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.6 Semi-Circular three hinged arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.7 Semi-Circular three hinged arch; Free body diagram . . . . . . . . . . . . . . . . . . . . 20.8 Statically Indeterminate Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.9 Statically Indeterminate Arch; ‘Horizontal Reaction Removed . . . . . . . . . . . . . . . 20.10Semi-Circular Box Girder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.11Geometry of Curved Structure in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . 20.12Free Body Diagram of a Curved Structure in Space . . . . . . . . . . . . . . . . . . . . . Victor Saouma

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20–2 20–2 20–3 20–3 20–4 20–5 20–6 20–7 20–8 20–9 20–11 20–12

Structural Engineering

LIST OF FIGURES

0–5

Draft

20.13Helicoidal Cantilevered Girder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20–13 21.1 Statically Indeterminate 3 Cable Structure . . . . 21.2 Propped Cantilever Beam . . . . . . . . . . . . . 21.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.7 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.10 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.11 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.13 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.14 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.15 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.16Deﬁnition of Flexibility Terms for a Rigid Frame 21.17 . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.18 . . . . . . . . . . . . . . . . . . . . . . . . . . .

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. 21–2 . 21–3 . 21–6 . 21–6 . 21–7 . 21–8 . 21–9 . 21–10 . 21–12 . 21–13 . 21–14 . 21–16 . 21–16 . 21–18 . 21–19 . 21–19 . 21–22 . 21–23

22.1 Uniformly Loaded Beam and Frame with Free or Fixed Beam Restraint . . . . . . . . . . 22–2 22.2 Uniformly Loaded Frame, Approximate Location of Inﬂection Points . . . . . . . . . . . . 22–3 22.3 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments . . . . . . 22–4 22.4 Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces . . . . 22–5 22.5 Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments . . . . . 22–6 22.6 Horizontal Force Acting on a Frame, Approximate Location of Inﬂection Points . . . . . . 22–7 22.7 Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear . . . . . . . . 22–8 22.8 ***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment . . . . . 22–9 22.9 Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force . . . . 22–10 22.10Example; Approximate Analysis of a Building . . . . . . . . . . . . . . . . . . . . . . . . . 22–10 22.11Free Body Diagram for the Approximate Analysis of a Frame Subjected to Vertical Loads 22–11 22.12Approximate Analysis of a Building; Moments Due to Vertical Loads . . . . . . . . . . . . 22–13 22.13Approximate Analysis of a Building; Shears Due to Vertical Loads . . . . . . . . . . . . . 22–15 22.14Approximate Analysis for Vertical Loads; Spread-Sheet Format . . . . . . . . . . . . . . . 22–16 22.15Approximate Analysis for Vertical Loads; Equations in Spread-Sheet . . . . . . . . . . . . 22–17 22.16Free Body Diagram for the Approximate Analysis of a Frame Subjected to Lateral Loads 22–18 22.17Approximate Analysis of a Building; Moments Due to Lateral Loads . . . . . . . . . . . . 22–20 22.18Portal Method; Spread-Sheet Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22–22 22.19Portal Method; Equations in Spread-Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . 22–23 23.1 23.2 23.3 23.4 23.5 23.6 23.7 23.8 23.9

Sign Convention, Design and Analysis . . . . . . . . . . Independent Displacements . . . . . . . . . . . . . . . . Total Degrees of Freedom for various Type of Elements Flexural Problem Formulation . . . . . . . . . . . . . . . Illustrative Example for the Slope Deﬂection Method . . Slope Deﬂection; Propped Cantilever Beam . . . . . . . Two-Span Beam, Slope Deﬂection . . . . . . . . . . . . Two Span Beam, Slope Deﬂection, Moment Diagram . . Frame Analysis by the Slope Deﬂection Method . . . . .

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23–2 23–2 23–4 23–5 23–9 23–10 23–11 23–12 23–13

24.1 Cracked Section, Limit State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–4 24.2 Whitney Stress Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24–4 25.1 25.2 25.3 25.4

T Beams . . . . . . . . . . . . . . . . . . . . T Beam as Rectangular Section . . . . . . . T Beam Strain and Stress Diagram . . . . . Decomposition of Steel Reinforcement for T

Victor Saouma

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25–1 25–2 25–2 25–3

Structural Engineering

0–6

LIST OF FIGURES

Draft 25.5 25.6 25.7 25.8

Doubly Reinforced Beams; Strain and Stress Diagrams . . . . Diﬀerent Possibilities for Doubly Reinforced Concrete Beams Strain Diagram, Doubly Reinforced Beam; is As Yielding? . . Summary of Conditions for top and Bottom Steel Yielding . .

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25–7 25–7 25–8 25–9

26.1 Global Coordinate System . . . . . . . . . . . . . . . . . 26.2 Local Coordinate Systems . . . . . . . . . . . . . . . . . 26.3 Sign Convention, Design and Analysis . . . . . . . . . . 26.4 Total Degrees of Freedom for various Type of Elements 26.5 Dependent Displacements . . . . . . . . . . . . . . . . . 26.6 Examples of Active Global Degrees of Freedom . . . . . 26.7 Problem with 2 Global d.o.f. θ1 and θ2 . . . . . . . . . . 26.8 2D Frame Element Rotation . . . . . . . . . . . . . . . . 26.9 *Frame Example (correct K32 and K33 ) . . . . . . . . . 26.10Example for [ID] Matrix Determination . . . . . . . . . 26.11Simple Frame Anlysed with the MATLAB Code . . . . 26.12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26.13Simple Frame Anlysed with the MATLAB Code . . . . 26.14ID Values for Simple Beam . . . . . . . . . . . . . . . . 26.15Structure Plotted with CASAP . . . . . . . . . . . . . .

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26–3 26–4 26–4 26–4 26–5 26–6 26–9 26–13 26–14 26–16 26–17 26–19 26–23 26–26 26–49

27.1 27.2 27.3 27.4 27.5 27.6 27.7

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27–1 27–2 27–2 27–3 27–4 27–5 27–8

Types of columns . . . . . . . . . . . . . . . . Tied vs Spiral Reinforcement . . . . . . . . . Possible Bar arrangements . . . . . . . . . . . Sources of Bending . . . . . . . . . . . . . . . Load moment interaction diagram . . . . . . Strain and Stress Diagram of a R/C Column Column Interaction Diagram . . . . . . . . .

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28.1 Flexible, Rigid, and Semi-Flexible Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.2 Deformation of Flexible and Rigid Frames Subjected to Vertical and Horizontal Loads, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.3 Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal Frames Subjected to Vertical and Horizontal Loads . . . . . . . . . . . . . . . . . . . . . . . . . 28.4 Axial and Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.5 Design of a Statically Indeterminate Arch . . . . . . . . . . . . . . . . . . . . . . . . . . 28.6 Normal and Shear Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. 28–1 . 28–2 . . . .

28–3 28–4 28–5 28–7

30.1 Normalized Gauss Distribution, and Cumulative Distribution Function . . . . . . . . . . . 30–3 30.2 Deﬁnition of Reliability Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–5 30.3 Probability of Failure in terms of β . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30–11

Victor Saouma

Structural Engineering

Draft List of Tables 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12 2.13

Unit Weight of Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Weights of Building Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Average Gross Dead Load in Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . Minimum Uniformly Distributed Live Loads, (UBC 1995) . . . . . . . . . . . . . . . . . Wind Velocity Variation above Ground . . . . . . . . . . . . . . . . . . . . . . . . . . . . Ce Coeﬃcients for Wind Load, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . . . Wind Pressure Coeﬃcients Cq , (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . . . Importance Factors for Wind and Earthquake Load, (UBC 1995) . . . . . . . . . . . . . Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Z Factors for Diﬀerent Seismic Zones, ubc . . . . . . . . . . . . . . . . . . . . . . . . . . S Site Coeﬃcients for Earthquake Loading, (UBC 1995) . . . . . . . . . . . . . . . . . . Partial List of RW for Various Structure Systems, (UBC 1995) . . . . . . . . . . . . . . Coeﬃcients of Thermal Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2–2 2–3 2–3 2–3 2–7 2–8 2–8 2–8

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2–9 2–12 2–13 2–13 2–17

3.1 3.2 3.3

Properties of Major Structural Steels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–3 Properties of Reinforcing Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–4 Joist Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–21

4.1

Equations of Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–2

5.1

Static Determinacy and Stability of Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . 5–3

7.1 7.2 7.3

Allowable Stresses for Steel and Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–3 Selected β values for Steel and Concrete Structures . . . . . . . . . . . . . . . . . . . . . . 7–6 Strength Reduction Factors, Φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–6

9.1 9.2

Usual Gage Lengths g, g1 , g2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–5 Eﬀective Net Area Ae for Bolted and Welded Connections . . . . . . . . . . . . . . . . . . 9–7

12.1 Conjugate Beam Boundary Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–15 13.1 Possible Combinations of Real and Hypothetical Formulations . . . . . . . . . . . . . . . . 13–5 13.2 k Factors for Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–9 13.3 Summary of Expressions for the Internal Strain Energy and External Work . . . . . . . . 13–25 15.1 Essential and Natural Boundary Conditions for Columns . . . . . . . . . . . . . . . . . . . 15–8 16.1 Design Stress φFcr for Fy =36 ksi in Terms of rKL . . . . . . . . . . . . . . . . . . . . . . 16–3 min KL 16.2 Design Stress φFcr for Fy =50 ksi in Terms of rmin . . . . . . . . . . . . . . . . . . . . . . 16–4 16.3 Stiﬀness Reduction Factors (AISC) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16–5 17.1 Nominal Areas of Standard Bolts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17–1 18.1 Unbraced Beam Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–9 18.2 Summary of AISC Governing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18–11

0–2

LIST OF TABLES

Draft

19.1 Values of m for use in Beam-Column Design Equation, (from a paper in AISC Engineering Journal by Uang, Wattar, and Leet (1990)) . . . . . . . . . . . . . . . . . . . . . . . . . . 19–7

L

g1 (x)g2 (x)dx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–5

21.1 Table of 0

21.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21–15 21.3 Displacement Computations for a Rectangular Frame . . . . . . . . . . . . . . . . . . . . 21–20 22.1 Columns Combined Approximate Vertical and Horizontal Loads . . . . . . . . . . . . . . 22–21 22.2 Girders Combined Approximate Vertical and Horizontal Loads . . . . . . . . . . . . . . . 22–24 23.1 Stiﬀness vs Flexibility Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23–1 23.2 Degrees of Freedom of Diﬀerent Structure Types Systems . . . . . . . . . . . . . . . . . . 23–3 26.1 26.2 26.3 26.4

Example of Nodal Deﬁnition . Example of Element Deﬁnition Example of Group Number . . Degrees of Freedom of Diﬀerent

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28.1 28.2 28.3 28.4 28.5 28.6 28.7 28.8 28.9

Geometry of the Arch . . . . . . . Calculation of Horizontal Force . . Moment at the Centers of the Ribs Values of Normal and Shear Forces Design of the Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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26–2 26–2 26–2 26–7

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28–5 28–6 28–7 28–8 28–9 28–10 28–10 28–11 28–11

30.1 Tabulated Results of a Simple Monte-Carlo Simulation. . . . . . . . . . . . . . . . . . . . 30–9 30.2 Selected β values for Steel and Concrete Structures . . . . . . . . . . . . . . . . . . . . . . 30–12

Victor Saouma

Structural Engineering

Draft

Part I

STRUCTURAL ENGINEERING I; CVEN 3525

Draft

Draft Chapter 1

INTRODUCTION 1.1 1

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, oﬀshore 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.

1.2 2

Structures and their Surroundings

Structural design is aﬀected 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 diﬀuse the sound 3. Natural light: • A ﬂat roof in a building may not provide adequate light. • 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 ﬂoor system. 5. Net clearance between columns (unobstructed surface) will dictate type of framing.

1.3 3

Architecture & Engineering

Architecture must be the product of a creative collaboration of architects and engineers.

4 Architect stress the overall, rather than elemental approach to design. In the design process, they conceptualize a space-form scheme as a total system. They are generalists.

1–2

Draft

INTRODUCTION

5 The engineer, partly due to his/her education think in reverse, starting with details and without suﬃcient regards for the overall picture. (S)he is a pragmatist who “knows everything about nothing”.

6

Thus there is a conceptual gap between architects and engineers at all levels of design.

7 Engineer’s education is more specialized and in depth than the architect’s. However, engineer must be kept aware of overall architectural objective. 8 In the last resort, it is the architect who is the leader of the construction team, and the engineers are his/her servant.

9

A possible compromise might be an Architectural Engineer.

1.4 10

Architectural Design Process

Architectural design is hierarchical:

Schematic: conceptual overall space-form 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: ﬁnal in-depth design reﬁnements of all subsystems and components and preparation of working documents (“blue-prints”). Engineers play a leading role.

1.5 11

Architectural Design

Architectural design must respect various constraints:

Functionality: Inﬂuence 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.

1.6

Structural Analysis

12 Given an existing structure subjected to a certain load determine internal forces (axial, shear, ﬂexural, torsional; or stresses), deﬂections, and verify that no unstable failure can occur.

13

Thus the basic structural requirements are:

Strength: stresses should not exceed critical values: σ < σf Stiﬀness: deﬂections should be controlled: ∆ < ∆max Stability: buckling or cracking should also be prevented

1.7 14

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). Victor Saouma

Structural Engineering

1.8 Load Transfer Elements

1–3

Draft

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

15

Serviceability: deﬂections, crack widths under the applied load. Compare with acceptable values speciﬁed in the design code. Failure: 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 modiﬁed and reanalyzed. For existing structures rehabilitation, or veriﬁcation of an old infrastructure, analysis is the most important component.

16

17

In summary, analysis is always required.

1.8 18

Load Transfer Elements

From Strength of Materials, Fig. 1.1

Figure 1.1: Types of Forces in Structural Elements (1D) Axial: cables, truss elements, arches, membrane, shells Flexural: Beams, frames, grids, plates Torsional: Grids, 3D frames Shear: Frames, grids, shear walls.

1.9 19

Structure Types

Structures can be classiﬁed as follows:

Victor Saouma

Structural Engineering

1–4

INTRODUCTION

Draft

Tension & Compression Structures: only, no shear, ﬂexure, or torsion Cable (tension only): The high strength of steel cables, combined with the eﬃciency of simple tension, makes cables ideal structural elements to span large distances such as bridges, and dish roofs, Fig. 1.2

Figure 1.2: Basic Aspects of Cable Systems Arches (mostly compression) is a “reversed cable structure”. In an arch, we seek to minimize ﬂexure and transfer the load through axial forces only. Arches are used for large span roofs and bridges, Fig. 1.3 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 Post and Beams: Essentially a support column on which a “beam” rests, Fig. 1.5, and 1.6. Victor Saouma

Structural Engineering

1.9 Structure Types

1–5

Draft

Figure 1.3: Basic Aspects of Arches

Victor Saouma

Structural Engineering

1–6

INTRODUCTION

Draft

Figure 1.4: Types of Trusses

Victor Saouma

Structural Engineering

1.9 Structure Types

1–7

Draft

Figure 1.5: Variations in Post and Beams Conﬁgurations

Victor Saouma

Structural Engineering

1–8

INTRODUCTION

Draft

VIERENDEEL TRUSS

OVERLAPPING SINGLE-STRUT CABLE-SUPPORTED BEAM

TREE-SUPPORTED TRUSS

BRACED BEAM

CABLE-STAYED BEAM

SUSPENDED CABLE SUPPORTED BEAM

BOWSTRING TRUSS

CABLE-SUPPORTED STRUTED ARCH OR CABLE BEAM/TRUSS

CABLE-SUPPORTED MULTI-STRUT BEAM OR TRUSS

GABLED TRUSS

CABLE-SUPPORTED ARCHED FRAME

CABLE-SUPPORTED PORTAL FRAME

Figure 1.6: Diﬀerent Beam Types

Victor Saouma

Structural Engineering

1.9 Structure Types

1–9

Draft

Beams: Shear, ﬂexure and sometimes axial forces. Recall that σ = beams, i.e. span/depth at least equal to ﬁve.

Mc I

is applicable only for shallow

Whereas r/c beams are mostly rectangular or T shaped, steel beams are usually I shaped (if the top ﬂanges are not properly stiﬀened, they may buckle, thus we must have stiﬀeners). Frames: Load is co-planar with the structure. Axial, shear, ﬂexure (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 members can have diﬀerent 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 two-way dispersal of loads. Because of the rigid connections between the beams, additional stiﬀness is introduced by the torsional resistance of members. Grids can also be skewed to achieve greater eﬃciency if the aspect ratio is not close to one. Plates are ﬂat, rigid, two dimensional structures which transmit vertical load to their supports. Used mostly for ﬂoor 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. 1 The precursor of the frame structures were the Post and Lintel where the post is vertical member on which the lintel is simply posed.

Victor Saouma

Structural Engineering

1–10

INTRODUCTION

Draft

Membranes: 3D structures composed of a ﬂexible 2D surface resisting tension only. They are usually cable-supported and are used for tents and long span roofs Fig. 1.8.

Figure 1.8: Examples of Air Supported Structures Shells: 3D structures composed of a curved 2D surface, they are usually shaped to transmit compressive axial stresses only, Fig. 1.9. Shells are classiﬁed in terms of their curvature.

1.10 20

Structural Engineering Courses

Structural engineering education can be approached from either one of two points of views:

Architectural: Start from overall design, and move toward detailed analysis. Education: Elemental rather than global approach. Emphasis is on the individual structural elements and not always on the total system. CVEN3525 will seek a balance between those two approaches. This is only the third of a long series of courses which can be taken in Structural Engineering, Fig. 1.10

21

Victor Saouma

Structural Engineering

1.10 Structural Engineering Courses

1–11

Draft

Figure 1.9: Basic Forms of Shells

Figure 1.10: Sequence of Structural Engineering Courses

Victor Saouma

Structural Engineering

1–12

Draft 1.11

INTRODUCTION

References

Following are some useful references for structural engineering, those marked by † were consulted, and “borrowed from” in preparing the Lecture Notes:

22

Structural Art 1. Billington, D.P., The Tower and the Bridge; The new art of structural engineering, Princeton University Pres,, 1983. Structural Engineering 1. Biggs, J.M., Introduction to Structural Engineering; Analysis and Design, Prentice Hall, 1986. 2. Gordon, J.E., Structures, or Why Things Do’nt Fall Down, Da Capo paperback, New York, 1978 3. Mainstone, R., Developments in Structural Form, Allen Lane Publishers, 1975. Structural Engineering, Architectural Analysis and Design 1. Ambrose, J., Building Structures, second Ed. Wiley, 1993. 2. Salvadori, M. and Heller, R., Structure in Architecture; The Building of Buildings, Prentice Hall, Third Edition, 1986. 3. Salvadori, M. and Levy, M., Structural Design in Architecture, Prentice hall, Second Edition, 1981. 4. Salvadori, M., Why Buildings Stand Up; The Strength of Architecture, Norton Paperack, 1990. 5. Lin, T.Y. and Stotesbury, S.D., Structural Concepts and Systems for Architects and Engineers, John Wiley, 1981. 6. † White, R. Gergely, P. and Sexmith, R., Structural Engineering; Combined Edition, John Wiley, 1976. 7. Sandaker, B.N. and Eggen, A.P., The Structural Basis of Architecture, Whitney Library of Design, 1992. Structural Analysis 1. † Arbadi, F. Structural Analysis and Behavior, McGraw-Hill, Inc., 1991. 2. Hsieh, Y.Y., Elementary Theory of Structures, Third Edition, Prentice Hall, 1988. 3. Ghali, A., and Neville, A.M., Structural Analysis, Third Edition, Chapman and Hall, 1989 Structural Design 1. † Nilson, A., and Winter, G. Design of Concrete Structures, Eleventh Edition, McGraw Hill, 1991. 2. † Salmon C. and Johnson, J. Steel Structures, Third Edition, Harper Collins Publisher, 1990. 3. † Gaylord, E.H., Gaylord, C.N. and Stallmeyer, J.E., Design of Steel Structures, Third Edition, McGraw Hill, 1992. Codes 1. ACI-318-89, 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 Oﬃcials, 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.

Victor Saouma

Structural Engineering

Draft Chapter 2

LOADS 2.1

Introduction

1 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), diﬀerential settlement of foundations, P-Delta eﬀects (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. 4 This distinction is helpful not only to compute a structure’s load, but also to assign diﬀerent 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

6 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

2.2.1

Dead Load

7 Dead loads (DL) consist of the weight of the structure itself, and other permanent ﬁxtures (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

2–2

LOADS

Draft 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 Material Aluminum Brick Concrete Steel Wood (pine)

lb/ft3 173 120 145 490 40

kN/m3 27.2 18.9 33.8 77.0 6.3

Table 2.1: Unit Weight of Materials 9 For steel structures, the weight per unit length of rolled sections is given in the AISC Manual of Steel Construction.

For design purposes, dead loads must be estimated and veriﬁed at the end of the design cycle. This makes the design process iterative.

10

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:

2.2.2

Live Loads

Contrarily to dead loads which are ﬁxed and vertical, live loads (LL) are movable or moving and may be horizontal.

13

Occupancy load may be due to people, furniture, equipment. The loads are essentially variable point loads which can be placed anywhere.

14

15

In analysis load placement should be such that their eﬀect (shear/moment) are maximized.

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 deﬁned in codes such as the Uniform Building Code or the ANSI Code, Table 2.4.

16

17

For small areas (30 to 50 sq ft) the eﬀect of concentrated load should be considered separately.

Since there is a small probability that the whole ﬂoor in a building be fully loaded, the UBC code speciﬁes that the occupancy load for members supporting an area A larger than 150 ft2 (i.e. a column with a total tributary area, including ﬂoors above it, larger than 150 ft2 ) may be reduced by R where DL R = r(A − 150) ≤ 23.1 1 + (2.1) LL

18

Victor Saouma

Structural Engineering

2.2 Vertical Loads

Draft

2–3 lb/ft2

Material

Ceilings Channel suspended system Acoustical ﬁber tile Floors Steel deck Concrete-plain 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 (12-16 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.

1 1 2-10 12 1 4 1-5 6 3 9-14 3 2 17 40 14 2 10 5 40 120 30 55 80 21 38 55

Table 2.2: Weights of Building Materials Material Timber Steel Reinforced concrete

lb/ft2 40-50 50-80 100-150

Table 2.3: Average Gross Dead Load in Buildings Use or Occupancy Assembly areas Cornices, marquees, residential balconies Corridors, stairs Garage Oﬃce buildings Residential Storage

lb/ft2 50 60 100 50 50 40 125-250

Table 2.4: Minimum Uniformly Distributed Live Loads, (UBC 1995) Victor Saouma

Structural Engineering

2–4

LOADS

Draft

where r = .08 for ﬂoors, A is the supported area ( ft2 ) 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.

Example 2-1: Live Load Reduction In a 10 story oﬃce 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 ﬂoor Solution: 1. The tributary area is 16 × 16 = 256f t2 > 150

√

2. The reduction R is R = .08(16 × 16 − 150) = 8.48% √ 3. Maximum allowable reduction Rmax = 23.1 1 + 60 80 = 40.4% which is less than 60% 4. The reduced cumulative load for the column of each ﬂoor is Floor Cumulative R (%) Cumulative LL Cumulative R× LL

Roof 8.48 20 18.3

10 16.96 80 66.4

9 25.44 80 59.6

8 33.92 80 52.9

7 42.4 80 46.08

6 51.32 80 38.9

5 59.8 80 32.2

4 60 80 32

3 60 80 32

2 60 80 32

Total 740 410

The resulting design live load for the bottom column has been reduced from 740 Kips to 410 Kips . 5. The total dead load is DL = (10)(60) = 600 Kips, thus the total reduction in load is

740−410 740+600

×

100= 25% .

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

20

Snow loads are always given on the projected length or area on a slope, Fig. 2.3.

Victor Saouma

Structural Engineering

2.2 Vertical Loads

2–5

Draft

LIVE LOAD DEAD LOAD

LE N G TH

RISE

WIND LOAD

RUN

Figure 2.3: Loads on Projected Dimensions The steeper the roof, the lower the snow retention. For snow loads greater than 20 psf and roof pitches α more than 20◦ the snow load p may be reduced by p R = (α − 20) − 0.5 (psf) (2.2) 40

21

22

Other examples of loads acting on inclined surfaces are shown in Fig. 2.4.

L1

L

2

w1 w3 w1 Snow Load

2

w

w2 Wind Load

θ

w3 Roof Dead Load L L = 1 2 cos θ

Figure 2.4: Vertical and Normal Loads Acting on Inclined Surfaces

Victor Saouma

Structural Engineering

2–6

LOADS

Draft 2.3

2.3.1

Lateral Loads Wind

Wind load depend on: velocity of the wind, shape of the building, height, geographical location, texture of the building surface and stiﬀness of the structure.

23

24

Wind loads are particularly signiﬁcant on tall buildings1 .

When a steady streamline airﬂow of velocity V is completely stopped by a rigid body, the stagnation pressure (or velocity pressure) qs was derived by Bernouilli (1700-1782)

25

qs =

1 2 ρV 2

(2.3)

where the air mass density ρ is the air weight divided by the acceleration of gravity g = 32.2 ft/sec2 . At sea level and a temperature of 15o C (59o F), the air weighs 0.0765 lb/ft3 this would yield a pressure of 1 (0.0765)lb/ft3 qs = 2 (32.2)ft/sec2

(5280)ft/mile V (3600)sec/hr

2 (2.4)

or qs = 0.00256V 2

(2.5)

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. 2.5.

Figure 2.5: Wind Map of the United States, (UBC 1995) 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).

26

27

28

Wind pressure increases with height, Table 2.5. Wind load will cause suction on the leeward sides, Fig. 2.7 1 The

primary design consideration for very high rise buildings is the excessive drift caused by lateral load (wind and possibly earthquakes).

Victor Saouma

Structural Engineering

2.3 Lateral Loads

2–7

Draft

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

Wind-Velocity 25 30 35 20 25 25 25 30 35 30 40 45 40 45 55 45 55 60 50 60 70

Map 40 30 40 50 60 70 80

Area 45 35 45 55 70 80 90

50 40 50 60 75 90 100

Table 2.5: Wind Velocity Variation above Ground

Figure 2.6: Eﬀect of Wind Load on Structures(Schueller 1996)

Victor Saouma

Structural Engineering

2–8

LOADS

Draft

This magnitude must be modiﬁed to account for the shape and surroundings of the building. Thus, the design pressure p (psf) is given by

29

(2.6)

p = Ce Cq Iqs The pressure is assumed to be normal to all walls and roofs and

Ce Velocity Pressure Coeﬃcient accounts for height, exposure and gust factor. It accounts for the fact that wind velocity increases with height and that dynamic character of the airﬂow (i.e the wind pressure is not steady), Table 2.6. Ce 1.39-2.34 1.06-2.19

Exposure D C

0.62-1.80

B

Open, ﬂat terrain facing large bodies of water Flat open terrain, extending one-half mile or open from the site in any full quadrant Terrain with buildings, forest, or surface irregularities 20 ft or more in height

Table 2.6: Ce Coeﬃcients for Wind Load, (UBC 1995) Cq Pressure Coeﬃcient is a shape factor which is given in Table 2.7 for gabled frames. Windward Side Gabled Frames (V:H) Roof Slope <9:12 −0.7 9:12 to 12:12 0.4 >12:12 0.7 Walls 0.8 Buildings (height < 200 ft) Vertical Projections height < 40 ft 1.3 height > 40 ft 1.4 Horizontal Projections −0.7

Leeward Side −0.7 −0.7 −0.7 −0.5 −1.3 −1.4 −0.7

Table 2.7: Wind Pressure Coeﬃcients Cq , (UBC 1995) I Importance Factor as given by Table 2.8. where

I II III IV

Occupancy Category Essential facilities Hazardous facilities Special occupancy structures Standard occupancy structures

Importance Factor I Earthquake Wind 1.25 1.15 1.25 1.15 1.00 1.00 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 power-generating equipment; Structures and equipment in government. communication centers. II Hazardous Facilities: Structures housing, supporting or containing suﬃcient quantities of toxic or explosive substances to be dangerous to the safety of the general public if released. Victor Saouma

Structural Engineering

2.3 Lateral Loads

2–9

Draft

III Special occupancy structure: Covered structures whose primary occupancy is public assembly, capacity > 300 persons. Buildings for schools through secondary or day-care 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 utility facilities not included above, and required for continued operation. IV Standard occupancy structure: All structures having occupancies or functions not listed above.

30

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

(2.7)

which corresponds to a pressure of 21 psf for a wind speed of 80 mph, Fig. 2.7, Table 2.9.

Height Above Grade (ft) 0-15 20 25 30 40 60 80 100 120 160 200 300 400

Exposure B C Basic Wind Speed (mph) 70 80 70 80 10 13 17 23 11 14 18 24 12 15 19 25 12 16 20 26 14 18 21 28 17 22 25 33 18 24 27 35 20 26 28 37 21 28 29 38 23 30 31 41 25 33 33 43 29 37 36 47 32 41 38 50

Table 2.9: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures

Example 2-2: Wind Load Determine the wind forces on the building shown on below which is built in St Louis and is surrounded by trees. Solution: 1. From Fig. 2.5 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.

Victor Saouma

Structural Engineering

2–10

LOADS

Draft 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 2.7: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures

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 windward 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.

Victor Saouma

Structural Engineering

2.3 Lateral Loads

2–11

Draft 2.3.2 31

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. For preliminary designs or for small structures an equivalent horizontal static load can be determined.

32

33

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.

A critical factor in the dynamic response of a structure is the fundamental period of the structure’s vibration (or ﬁrst mode of vibration). This is the time required for one full cycle of motion, Fig. 2.8. If the earthquake excitation has a frequency close to the one of the building, then resonance may occur. This should be avoided.

34

Figure 2.8: Vibrations of a Building Earthquake load manifests itself as a horizontal force due to the (primarily) horizontal inertia force (F = ma).

35

36

The horizontal force at each level is calculated as a portion of the base shear force V

V = Victor Saouma

ZIC W RW

(2.8) Structural Engineering

2–12

LOADS

Draft where:

Z: Zone Factor: to be determined from Fig. 2.9 and Table 2.10.

Figure 2.9: Seismic Zones of the United States, (UBC 1995)

Seismic Zone Z

0 0

1 0.075

2A 0.15

2B 0.2

3 0.3

4 0.4

Table 2.10: Z Factors for Diﬀerent Seismic Zones, ubc I: Importance Factor: which was given by Table 2.8. C: Design Response Spectrum given by

C=

1.25S ≤ 2.75 T 2/3

(2.9)

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

(2.10)

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 S: Site Ct 0.020 all other buildings Coeﬃcient given by Table 2.11 Note that most of the damages in the 1990? earthquake in San Francisco occurred in the marina where many houses were built on soft soil. and C ≥ 0.075 RW Victor Saouma

(2.11)

Structural Engineering

2.3 Lateral Loads

2–13

Draft Type S1 S2 S3 S4

Description A soil proﬁle with either rock-like material or stiﬀ/dense soil less than 200 ft. Dense or stiﬀ soil exceeding 200 ft 70 ft or more soil containing more than 20 ft of soft to medium stiﬀ clay but not more than 40 ft. of soft clay. Soil containing more than 40 ft of soft clay

S Factor 1.0 1.2 1.5 2.0

Table 2.11: S Site Coeﬃcients for Earthquake Loading, (UBC 1995)

Structural System

RW Bearing wall system Light-framed walls with shear panels Plywood walls for structures three stories or less 8 All other light-framed walls 6 Shear walls Concrete 8 Masonry 8 Building frame system using trussing or shear walls) Steel eccentrically braced ductiel frame 10 Light-framed walls with shear panels Plywood walls for structures three stories or less 9 All other light-framed walls 7 Shear walls Concrete 8 Masonry 8 Concentrically braced frames Steel 8 Concrete (only for zones I and 2) 8 Heavy timber 8 Moment-resisting frame system Special moment-resisting frames (SMRF) Steel 12 Concrete 12 Concrete intermediate moment-resisting frames (IMRF)only for zones 1 and 2 8 Ordinary moment-resisting 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 6-12 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

H (ft)

65 65 240 160 240 65 65 240 160 160 65

N.L. N.L. 160 -

N.L. 160 160-N.L. N. L. N.L. 160 -

Table 2.12: Partial List of RW for Various Structure Systems, (UBC 1995)

Victor Saouma

Structural Engineering

2–14

LOADS

Draft

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 ﬁrst (applied only if T ≥ 0.7 sec.) is a concentrated force F1 equal to

37

Ft = 0.07T V ≤ 0.25V

(2.12)

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. 38

Assuming a ﬂoor weight constant for every ﬂoor level, then the force acting on each one is given by Fx =

(V − Ft )hx (V − Ft )hx = h1 + h2 + · · · + hn Σni=1 hi

(2.13)

where hi and hn are the height in ft above the base to level i, n or x respectively

Example 2-3: Earthquake Load on a Frame Determine the approximate earthquake forces for the ductile hospital frame structure shown below. The DL for each ﬂoor 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 ﬂoor. 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 sec. 2. The C coeﬃcient is C=

1.25S (1.25)(2.0) = = 5.344 > 2.75 T 2/3 (0.32)2/3

(2.14)

(2.15)

use C = 2.75. 3. The other coeﬃcients are: Z =0.3; I=1.25; RW =12 4. Check

√ C 2.75 = 0.23 > 0.075 = RW 12

(2.16)

W = 2 ((200 + 0.5(400)) (20) = 16000 lbs

(2.17)

5. The total vertical load is

Victor Saouma

Structural Engineering

2.3 Lateral Loads

2–15

Draft

6. The total seismic base shear is V

=

ZIC (0.3)(1.25)(2.75) = 0.086W = RW 12

(2.18-a)

= (0.086)(16000) = 1375 lbs

(2.18-b)

7. Since T < 0.7 sec. there is no whiplash. 8. The load on each ﬂoor is thus given by F2

=

F1

=

(1375)(24) = 916.7 lbs 12 + 24 (1375)(12) = 458.3 lbs 12 + 24

(2.19-a) (2.19-b)

Example 2-4: 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 oﬃce 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. 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) psf(100 × 175) ft2 × 25 storeys = 84, 000 k Victor Saouma

(2.20)

Structural Engineering

2–16

LOADS

Draft

2. the fundamental period of vibration for a rigid frame is √ T = Ct (hn )3/4 = 0.030(300)3/4 = 2.16 sec. > 0.7 sec. 3. The C coeﬃcient is

√ 1.25S (1.25)(1.5) = = 1.12 ≤ 2.75 T 2/3 (2.16)2/3

C=

(2.21)

(2.22)

4. The other coeﬃcients are Z=0.4; I=1, RW =12 5. We check

√ C 1.12 = = 0.093 ≥ 0.075 RW 12

(2.23)

6. The total seismic base shear along the critical short direction is V

=

ZIC (0.4)(1)(1.12) W = 0.037W W = RW (12)

(2.24-a)

=

(0.037)(84000) = 3108 kip

(2.24-b)

7. Since T > 0.7 sec., the whiplash eﬀect must be considered Ft

= 0.07T V = (0.07)(2.16)(3108) = 470 k le 0.25V = (0.25)(3108) = 777 k

✛

(2.25-a) (2.25-b)

Hence the total triangular load is V − Ft = 3108 − 470 = 2638 k

(2.26)

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) psf(175 × 300) ft2 = 1523 k < 3108 k

(2.27)

The magnitude of the total seismic load is clearly larger than the total wind force.

2.4 2.4.1 39

Other Loads Hydrostatic and Earth

Structures below ground must resist lateral earth pressure. (2.28)

q = Kγh where γ is the soil density, K =

1−sin Φ 1+sin Φ

is the pressure coeﬃcient, h is the height.

40

For sand and gravel γ = 120 lb/ ft3 , and Φ ≈ 30◦ .

41

If the structure is partially submerged, it must also resist hydrostatic pressure of water, Fig. 2.10.

q = γW h

(2.29)

where γW = 62.4 lbs/ft3 .

Example 2-5: Hydrostatic Load Victor Saouma

Structural Engineering

2.4 Other Loads

2–17

Draft

Figure 2.10: Earth and Hydrostatic Loads on Structures 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 = γh we equate the two pressures and solve for h the height of the concrete slab (62.4) lbs/ft3 × (12 − 9) ft =

water 3 (150) lbs/ft3 × h ⇒ h = (62.4) lbs/ft3 (3) ft(12) in/ft = 14.976 in 15.0 inch (150) lbs / ft

concrete

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

42

∆l = αl∆T

(2.30)

where α is the coeﬃcient of thermal expansion, Table 2.13

Steel Concrete

α (/◦F ) 6.5 × 10−6 5.5 × 10−6

Table 2.13: Coeﬃcients of Thermal Expansion

43

If the member is restrained against expansion, then a compressive stress σ = Eα∆T is developed.

44

To avoid excessive stresses due to thermal loading expansion joints are used in bridges and buildings.

Example 2-6: Thermal Expansion/Stress (Schueller 1996) A low-rise building is enclosed along one side by a 100 ft-long clay masonry (α = 3.6×10−6 in./in./o F, E = 2, 400, 000 psi) bearing wall. The structure was built at a temperature of 60o F and is located in the northern part of the United States where the temperature range is between -20o and +120oF. Solution:

Victor Saouma

Structural Engineering

2–18

LOADS

Draft

1. Assuming that the wall can move freely with no restraint from cross-walls and foundation, the wall expansion and contraction (summer and winter) are given by ∆LSummer ∆LWinter

=

α∆T L = (3.6 × 10−6 ) in/ in/o F (120 − 60)o F (100) ft(12) in/ft = 0.26 (2.31-a) in

=

α∆T L = (3.6 × 10−6 ) in/ in/o F (−20 − 60)o F (100) ft(12) in/ft = -0.35 (2.31-b) in

2. We now assume (conservatively) that the free movement cannot occur (∆L = 0) hence the resulting α∆T L = Eα∆T stress would be equal to σ = Eε = E ∆L L =E L σ Summer

=

Eα∆T = (2, 400, 000)

σ Winter

=

Eα∆T = (2, 400, 000)

lbs

(3.6 in2 lbs in2

× 10−6 ) in/ in/o F (120 − 60)o F = 518

lbs

Tension (2.32-a)

in2 lbs

(3.6 × 10−6 ) in/ in/o F (−20 − 60)o F = -691

in2

Compression (2.32-b) (2.32-c)

Note that the tensile stresses being beyond the masonry capacity, cracking will occur.

2.4.3

Bridge Loads

For highway bridges, design loads are given by the AASHTO (Association of American State Highway Transportation Oﬃcials). The HS-20 truck is used for the design of bridges on main highways, Fig. 6.5. Either the design truck with speciﬁed 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.

Figure 2.11: Truck Load

2.4.4

2.5 2.5.1 45

Impact Load

Other Important Considerations Load Combinations

Live loads speciﬁed by codes represent the maximum possible loads.

Victor Saouma

Structural Engineering

2.5 Other Important Considerations

2–19

Draft

The likelihood of all these loads occurring simultaneously is remote. Hence, building codes allow certain reduction when certain loads are combined together.

46

47

Furthermore, structures should be designed to resist a combination of loads.

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:

49

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 veriﬁed 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 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

51

Loads are often characterized as Usual, Unusual and Extreme.

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 diﬀerent locations must be considered, Fig. 2.12.

52

2.5.3

Structural Response

Under the action of the various forces and loadings described above, the structure must be able to respond with proper behavior, Fig. 7.1.

53

Victor Saouma

Structural Engineering

2–20

LOADS

Draft

Figure 2.12: Load Placement to Maximize Moments

Figure 2.13: Load Life of a Structure, (Lin and Stotesbury 1981)

Victor Saouma

Structural Engineering

2.5 Other Important Considerations

2–21

Draft

0000000000 1111111111 1111111111 0000000000 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111 1111111111 0000000000 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111

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

1111111111 0000000000 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111

1111111111 0000000000 0000000000 1111111111 0000000000 1111111111 0000000000 1111111111

11 00 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 00 11 0000 00 11 00 11 0000 1111 00 11 00 11 0000 1111 00 11 11 001111 000011 00

Figure 2.14: Concept of Tributary Areas for Structural Member Loading

a

b a/b > 2 one way slab

b a/b <2 Two way slab

Figure 2.15: One or Two Way actions in Slabs

Victor Saouma

Structural Engineering

2–22

Draft 2.5.4

LOADS

Tributary Areas

For preliminary analysis, the tributary area of a structural component will determine the total applied load.

54

For a slab simply supported over four linear supports, we may have a one way or tow way action if the ratio a/b is greater or smaller than two respectively, Fig. 2.15.

55

Whereas we will be focusing on the design of a reinforced concrete or steel section, we must keep in mind the following:

56

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. Load transfer in a structure is accomplished through a “hierarchy” of simple ﬂexural elements which are then connected to the columns, Fig. 2.16 or by two way slabs as illustrated in Fig. 2.17.

57

58

An example of load transfer mechanism is shown in Fig. 2.18.

Victor Saouma

Structural Engineering

2.5 Other Important Considerations

Draft

2–23

Figure 2.16: Load Transfer in R/C Buildings

Victor Saouma

Structural Engineering

2–24

LOADS

Draft

Figure 2.17: Two Way Actions

B

A

C

D b=12’; a=4’; b/a=12/4=3>2 One way slab 2

(40) lb/ft (4) ft= 160 lb/ft=0.16 k/ft b B

G (0.16)(12)/2=0.96 k 0.96 k

LL=40 lbs/ft 2

12 ft 0.96 k

0.96 k

H G

F

a

a

0.48 k

E a

0.48 k B

A 4 ft

C

D

4 ft

4 ft

0.48+0.96=1.44 k

1.44 k 2

(2)(40)lb/ft (3)ft=0.24 k/ft

b=10’; a=6’; b/a<2 Two way slab B

A

3’

4’

3’

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 00000000 11111111 00000000 11111111 0000 1111 00000000 11111111 00000000 11111111 0000 1111 00000000 11111111 00000000 11111111 0000 1111 00000000 H 11111111 00000000 11111111 0000 G 1111 3’

G

B

(0.24)(2+3/2)=0.84k 0.84 k

3’

3’

4’

0.84k

0.84k

0.42k

D

A 1.8k

2

(40)lb/ft (3)ft=0.12k/ft 0.42k

B 3’

3’

C 3’

3’

3’

3’

1.8k

(0.42)+(0.84)+(0.12)[6/2+3/2]=1.8k

3’

Figure 2.18: Example of Load Transfer

Victor Saouma

Structural Engineering

Draft Chapter 3

STRUCTURAL MATERIALS 1 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.

3.1 3.1.1

Steel Structural Steel

3 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, nickel, manganese and copper.

Practically all grades of steel have a Young Modulus equal to 29,000 ksi, density of 490 lb/cu ft, and a coeﬃcient of thermal expansion equal to 0.65 × 10−5 /deg F.

4

5 The yield stress of steel can vary from 40 ksi to 250 ksi. Most commonly used structural steel are A36 (σyld = 36 ksi) and A572 (σyld = 50 ksi), Fig. 3.6

Figure 3.1: Stress Strain Curves of Concrete and Steel 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.

3–2

STRUCTURAL MATERIALS

Draft

Figure 3.2: Standard Rolled Sections

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 ﬁrst ones rolled in America and have a slope on their inside ﬂange surfaces of 1 to 6. W or wide ﬂange 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 classiﬁed as a C shape by dimensions. HP is a bearing pile section. 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. Properties of structural steel are tabulated in Table 3.1. Rolled sections, Fig. 3.3 and welded ones, Fig 3.4 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 ﬁrst become stiﬀer, resist contraction, and develop compressive stresses. The remaining regions continue to cool and contract in the plastic condition and develop tensile stresses.

10

Due to those residual stresses, the stress-strain curve of a rolled section exhibits a non-linear segment prior to the theoretical yielding, Fig. 3.5. This would have important implications on the ﬂexural and axial strength of beams and columns.

11

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 .

12

1 Stirrups

which are used as vertical reinforcement to resist shear usually have a yield stress of only 40 ksi.

Victor Saouma

Structural Engineering

3.1 Steel

3–3

Draft ASTM Desig. A36

Shapes Available

Shapes and bars

A500

Cold formed welded seamless sections;

A501

Hot formed welded and seamless sections; Plates and bars 12 in and less thick;

A529

and

A606

Hot and cold rolled sheets;

A611

Cold rolled sheet in cut lengths Structural shapes, plates and bars

A 709

Use

σy (ksi)

Riveted, bolted, welded; Buildings and bridges General structural purpose Riveted, welded or bolted; Bolted and welded

36 up through 8 in. (32 above 8.) Grade A: 33; Grade B: 42; Grade C: 46

Building frames and trusses; Bolted and welded Atmospheric corrosion resistant Cold formed sections Bridges

σu (ksi)

36 42

45-50 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

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 Engineering

3–4

STRUCTURAL MATERIALS

Draft Ideal coupon containing no residual stress

.3

Average stress P/A

Fy

.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: Inﬂuence of Residual Stress on Average Stress-Strain Curve of a Rolled Section

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 (in.) 2/8=0.250 3/8=0.375 4/8=0.500 5/8=0.625 6/8=0.750 7/8=0.875 8/8=1.000 9/8=1.128 10/8=1.270 11/8=1.410 14/8=1.693 18/8=2.257

Area ( in2 ) 0.05 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56 2.25 4.00

Perimeter in 0.79 1.18 1.57 1.96 2.36 2.75 3.14 3.54 3.99 4.43 5.32 7.09

Weight lb/ft 0.167 0.376 0.668 1.043 1.5202 2.044 2.670 3.400 4.303 5.313 7.650 13.60

Table 3.2: Properties of Reinforcing Bars

Victor Saouma

Structural Engineering

3.2 Aluminum

3–5

Draft

Steel loses its strength rapidly above 700 deg. F (and thus must be properly protected from ﬁre), and becomes brittle at −30 deg. F2 .

13

Steel is also used as wire strands and ropes for suspended roofs, cable-stayed 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. 14

15

Prestressing Steel cables have an ultimate strength up to 270 ksi.

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.

16

17

Aluminum members can be connected by riveting, bolting and to a lesser extent by welding.

Aluminum has a modulus of elasticity equal to 10,000 ksi (about three times lower than steel), a coeﬃcient of thermal expansion of 2.4 × 10−5 and a density of 173 lbs/ft3 .

18

The ultimate strength of pure aluminum is low (13,000 psi) but with the addition of alloys it can go up.

19

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.

20

3.3

Concrete

Concrete is a mixture of Portland cement3 , water, and aggregates (usually sand and crushed stone). An ideal mixture is one in which:

21

1. A minimum amount of cement-water paste is used to ﬁll 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. Smaller particles up to 1/4 in. in size are called ﬁne aggregates, and the larger ones being coarse aggregates.

22

23

Contrarily to steel to modulus of elasticity of concrete depends on the strength and is given by E = 57, 000 fc (3.1)

or E = 33γ 1.5 where both

fc

fc

(3.2)

3

and E are in psi and γ is in lbs/ft .

Typical concrete (compressive) strengths range from 3,000 to 6,000 psi; However high strength concrete can go up to 14,000 psi.

24

25

26

All concrete fail at an ultimate strain of 0.003, Fig. 3.6. Pre-peak nonlinearity is caused by micro-cracking Fig. 3.7. 2 This

brittleness is even more accentuated in high strength steel, and has been the cause of some bridge collapses during winter nights. (CVEN 6831 Fracture Mechanics) 3 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.

Victor Saouma

Structural Engineering

3–6

STRUCTURAL MATERIALS

Draft

Figure 3.6: Concrete Stress-Strain curve

f’c non-linear

.5f’c linear

εu

Figure 3.7: Concrete microcracking

Victor Saouma

Structural Engineering

3.4 Masonry

3–7

Draft 27

The tensile strength of concrete ft is about 10% of the compressive strength.

28

Density of normal weight concrete is 145 lbs/ft3 and 100 lbs/ft3 for lightweight concrete.

29

Coeﬃcient of thermal expansion is 0.65 × 10−5 /deg F for normal weight concrete.

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 occur4 . 30

Concrete will also deform with time due to the applied load, this is called creep. This should be taken into consideration when computing the deﬂections (which can be up to three times the instantaneous elastic deﬂection).

31

3.4

Masonry

Masonry consists of either natural materials, such as stones, or of manufactured products such as bricks and concrete blocks5 , stacked and bonded together with mortar.

32

As for concrete, all modern structural masonry blocks are essentially compression members with low tensile resistance.

33

34

The mortar used is a mixture of sand, masonry cement, and either Portland cement or hydrated lime.

3.5

Timber

Timber is one of the earliest construction materials, and one of the few natural materials with good tensile properties.

35

36

The properties of timber vary greatly, and the strength is time dependent.

37

Timber is a good shock absorber (many wood structures in Japan have resisted repeated earthquakes).

The most commonly used species of timber in construction are Douglas ﬁr, southern pine, hemlock and larch.

38

Members can be laminated together under good quality control, and ﬂexural strengths as high as 2,500 psi can be achieved.

39

3.6 40

Steel Section Properties

Dimensions and properties of rolled sections are tabulated in the following pages, Fig. 3.8.

4 For this reason a minimum amount of reinforcement is always necessary in concrete, and a 2% reinforcement, can reduce the shrinkage by 75%. 5 Mud bricks were used by the Babylonians, stones by the Egyptians, and ice blocks by the Eskimos...

Victor Saouma

Structural Engineering

3–8

STRUCTURAL MATERIALS

Draft

Figure 3.8: W and C sections

Victor Saouma

Structural Engineering

3.6 Steel Section Properties

Draft Designation

W 44x285 W 44x248 W 44x224 W 44x198 W 40x328 W 40x298 W 40x268 W 40x244 W 40x221 W 40x192 W 40x655 W 40x593 W 40x531 W 40x480 W 40x436 W 40x397 W 40x362 W 40x324 W 40x297 W 40x277 W 40x249 W 40x215 W 40x199 W 40x183 W 40x167 W 40x149 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

3–9

A

d

tw

bf

tf

in2

in 44.02 43.62 43.31 42.91 40.00 39.69 39.37 39.06 38.67 38.20 43.62 42.99 42.34 41.81 41.34 40.95 40.55 40.16 39.84 39.69 39.38 38.98 38.67 38.98 38.59 38.20 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

in 1.025 0.865 0.785 0.710 0.910 0.830 0.750 0.710 0.710 0.710 1.970 1.790 1.610 1.460 1.340 1.220 1.120 1.000 0.930 0.830 0.750 0.650 0.650 0.650 0.650 0.630 2.520 2.380 2.165 1.970 1.790 1.610 1.500 1.360 1.220 1.120 1.020 0.945 0.885 0.840 0.800 0.760 0.960 0.870 0.830 0.765 0.725 0.680 0.650 0.625 0.600 1.970 1.810 1.650 1.520 1.380 1.260 1.160 1.040 0.960 0.870 0.830 0.775 0.715 0.670 0.635 0.605 0.580 0.550 1.970 1.790 1.630 1.500 1.360 1.240 1.140 1.020 0.930 0.830 0.775 0.710 0.655 0.650 0.615 0.585 0.565 0.545 0.520 0.470

in 11.810 11.810 11.810 11.810 17.910 17.830 17.750 17.710 17.710 17.710 16.870 16.690 16.510 16.360 16.240 16.120 16.020 15.905 15.825 15.830 15.750 15.750 15.750 11.810 11.810 11.810 18.130 17.990 17.775 17.575 17.400 17.220 17.105 16.965 16.830 16.730 16.630 16.655 16.595 16.550 16.510 16.470 12.215 12.120 12.180 12.115 12.075 12.030 12.000 11.975 11.950 16.910 16.750 16.590 16.455 16.315 16.200 16.100 15.985 15.905 15.805 15.860 15.805 15.745 11.500 11.565 11.535 11.510 11.480 16.200 16.020 15.865 15.725 15.590 15.470 15.370 15.255 15.155 15.055 15.105 15.040 14.985 10.480 10.545 10.515 10.495 10.475 10.450 10.400

in 1.770 1.575 1.415 1.220 1.730 1.575 1.415 1.260 1.065 0.830 3.540 3.230 2.910 2.640 2.400 2.200 2.010 1.810 1.650 1.575 1.420 1.220 1.065 1.220 1.025 0.830 4.530 4.290 3.900 3.540 3.230 2.910 2.680 2.440 2.200 2.010 1.850 1.680 1.570 1.440 1.350 1.260 1.730 1.570 1.360 1.260 1.180 1.100 1.020 0.940 0.790 3.540 3.270 2.990 2.720 2.480 2.280 2.090 1.890 1.730 1.570 1.400 1.275 1.150 1.220 1.055 0.960 0.855 0.740 3.540 3.230 2.950 2.680 2.440 2.240 2.050 1.850 1.650 1.500 1.315 1.185 1.065 1.180 1.000 0.930 0.850 0.760 0.670 0.610

83.8 72.8 65.8 58.0 96.4 87.6 78.8 71.7 64.8 56.5 192.0 174.0 156.0 140.0 128.0 116.0 106.0 95.3 87.4 81.3 73.3 63.3 58.4 53.7 49.1 43.8 249.0 234.0 211.0 190.0 172.0 154.0 142.0 128.0 115.0 105.0 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

Victor Saouma

bf 2tf 3.3 3.7 4.2 4.8 5.2 5.7 6.3 7.0 8.3 10.7 2.4 2.6 2.8 3.1 3.4 3.7 4.0 4.4 4.8 5.0 5.5 6.5 7.4 4.8 5.8 7.1 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

hc tw 38.0 44.9 49.4 54.9 37.6 41.2 45.6 48.1 48.1 48.1 17.4 19.1 21.2 23.4 25.5 28.0 30.5 34.2 36.8 41.2 45.6 52.6 52.6 52.6 52.6 54.3 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

Ix

Sx

rx

Iy

Sy

ry

Zx

Zy

J

in4 24600 21400 19200 16700 26800 24200 21500 19200 16600 13500 56500 50400 44300 39500 35400 32000 28900 25600 23200 21900 19500 16700 14900 13300 11600 9780 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

in3 1120 983 889 776 1340 1220 1090 983 858 708 2590 2340 2090 1890 1710 1560 1420 1280 1170 1100 992 858 769 682 599 512 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

in 17.1 17.2 17.1 16.9 16.7 16.6 16.5 16.4 16.0 15.5 17.2 17.0 16.9 16.8 16.6 16.6 16.5 16.4 16.3 16.4 16.3 16.2 16.0 15.7 15.3 14.9 16.4 16.4 16.2 16.0 15.9 15.8 15.6 15.6 15.5 15.4 15.3 15.2 15.1 15.0 15.0 14.9 14.9 14.8 14.6 14.6 14.5 14.5 14.4 14.3 14.0 15.2 15.1 14.9 14.8 14.7 14.7 14.5 14.4 14.4 14.3 14.1 14.1 14.0 13.7 13.5 13.4 13.2 13.0 13.9 13.8 13.7 13.5 13.5 13.4 13.2 13.2 13.1 13.0 12.9 12.8 12.7 12.4 12.2 12.1 12.0 11.9 11.7 11.7

in4 490 435 391 336 1660 1490 1320 1170 988 770 2860 2520 2200 1940 1720 1540 1380 1220 1090 1040 926 796 695 336 283 229 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

in3 83 74 66 57 185 167 149 132 112 87 339 302 266 237 212 191 173 153 138 132 118 101 88 57 48 39 501 467 414 367 328 289 263 235 208 188 171 156 144 132 123 114 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

in 2.42 2.44 2.44 2.41 4.15 4.12 4.09 4.04 3.90 3.69 3.86 3.81 3.75 3.72 3.67 3.65 3.61 3.57 3.54 3.58 3.56 3.54 3.45 2.50 2.40 2.29 4.27 4.24 4.18 4.12 4.07 4.02 3.98 3.95 3.90 3.87 3.84 3.83 3.81 3.78 3.75 3.73 2.65 2.62 2.58 2.56 2.55 2.53 2.50 2.47 2.38 3.98 3.94 3.89 3.85 3.81 3.79 3.74 3.71 3.69 3.66 3.63 3.59 3.56 2.50 2.47 2.43 2.39 2.32 3.86 3.80 3.75 3.71 3.68 3.65 3.61 3.58 3.54 3.52 3.49 3.46 3.43 2.28 2.25 2.23 2.19 2.15 2.10 2.09

in3 1310.0 1150.0 1030.0 902.0 1510.0 1370.0 1220.0 1100.0 967.0 807.0 3060.0 2750.0 2450.0 2180.0 1980.0 1790.0 1630.0 1460.0 1330.0 1250.0 1120.0 963.0 868.0 781.0 692.0 597.0 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

in3 135.0 118.0 105.0 90.0 286.0 257.0 229.0 203.0 172.0 135.0 541.0 481.0 422.0 374.0 334.0 300.0 270.0 239.0 215.0 204.0 182.0 156.0 137.0 89.6 76.0 62.2 799.0 743.0 656.0 580.0 517.0 454.0 412.0 367.0 325.0 292.0 265.0 241.0 223.0 204.0 190.0 176.0 137.0 122.0 107.0 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

in4 60.0 40.7 30.0 20.1 74.2 56.3 41.1 30.4 21.2 13.7 596.0 451.0 329.0 245.0 186.0 142.0 109.0 79.4 61.2 51.1 37.7 24.4 18.1 19.6 14.0 9.6 1269.0 1073.0 804.0 600.0 453.0 330.0 260.0 195.0 143.0 109.0 84.5 64.2 52.6 41.5 34.6 28.6 53.3 39.8 28.0 22.2 18.4 15.1 12.4 10.1 7.0 567.0 444.0 338.0 256.0 193.0 149.0 115.0 84.4 65.0 48.5 35.8 27.5 20.5 17.7 12.4 9.7 7.4 5.3 537.0 405.0 307.0 231.0 174.0 134.0 103.0 74.9 53.8 40.0 27.9 20.6 15.3 14.6 9.7 8.0 6.4 5.0 3.8 2.9

Structural Engineering

3–10

STRUCTURAL MATERIALS

Draft 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 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

A

d

tw

bf

tf

in2

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 21.13 20.99 21.06 20.83 20.66 22.32 21.85 21.46 21.06 20.67 20.35 20.04 19.72 19.49 19.25 18.97 18.73 18.59 18.39 18.21 18.47 18.35 18.24 18.11 17.99 18.06 17.90 17.70 16.97 16.75 16.52 16.33 16.43 16.26 16.13 16.01 15.86 15.88 15.69 22.42 21.64 20.92 20.24 19.60 19.02 18.67 18.29 17.92 17.54 17.12 16.74 16.38 16.04 15.72 15.48

in 1.970 1.810 1.650 1.520 1.380 1.260 1.160 1.060 0.980 0.910 0.830 0.750 0.725 0.660 0.605 0.610 0.570 0.515 0.490 0.460 1.970 1.810 1.650 1.520 1.380 1.260 1.160 1.040 0.960 0.870 0.810 0.750 0.705 0.650 0.605 0.550 0.500 0.550 0.515 0.470 0.440 0.415 0.430 0.395 1.730 1.590 1.460 1.320 1.220 1.100 1.000 0.910 0.830 0.750 0.720 0.650 0.600 0.550 0.500 0.580 0.515 0.455 0.430 0.400 0.405 0.380 0.350 1.520 1.400 1.280 1.160 1.060 0.960 0.890 0.810 0.730 0.670 0.655 0.590 0.535 0.480 0.425 0.495 0.450 0.415 0.390 0.355 0.360 0.315 0.300 0.585 0.525 0.455 0.395 0.430 0.380 0.345 0.305 0.295 0.275 0.250 3.070 2.830 2.595 2.380 2.190 2.015 1.875 1.770 1.655 1.540 1.410 1.290 1.175 1.070 0.980 0.890

in 15.255 15.095 14.940 14.800 14.665 14.545 14.445 14.350 14.270 14.190 14.115 14.035 14.085 14.020 13.965 10.010 10.070 10.015 9.990 9.960 14.115 13.955 13.800 13.660 13.520 13.405 13.305 13.185 13.110 13.010 12.950 12.890 12.955 12.900 12.855 12.800 12.750 9.000 9.065 9.020 8.990 8.965 7.040 7.005 13.405 13.265 13.130 12.990 12.890 12.775 12.675 12.575 12.500 12.420 12.510 12.440 12.390 12.340 12.290 8.420 8.355 8.295 8.270 8.240 6.555 6.530 6.500 12.005 11.890 11.770 11.650 11.555 11.455 11.375 11.300 11.220 11.160 11.265 11.200 11.145 11.090 11.035 7.635 7.590 7.555 7.530 7.495 6.060 6.015 6.000 10.425 10.365 10.295 10.235 7.120 7.070 7.035 6.995 6.985 5.525 5.500 17.890 17.650 17.415 17.200 17.010 16.835 16.695 16.590 16.475 16.360 16.230 16.110 15.995 15.890 15.800 15.710

in 3.540 3.270 2.990 2.720 2.480 2.280 2.090 1.930 1.770 1.610 1.500 1.340 1.190 1.080 0.975 1.100 0.930 0.830 0.745 0.640 3.540 3.270 2.990 2.720 2.480 2.280 2.090 1.890 1.730 1.570 1.460 1.340 1.220 1.090 0.960 0.850 0.750 0.980 0.875 0.770 0.680 0.585 0.590 0.505 3.130 2.850 2.620 2.380 2.190 1.990 1.790 1.630 1.480 1.360 1.150 1.035 0.960 0.875 0.800 0.930 0.835 0.740 0.685 0.615 0.650 0.535 0.450 2.740 2.500 2.300 2.110 1.910 1.750 1.590 1.440 1.320 1.200 1.060 0.940 0.870 0.770 0.680 0.810 0.750 0.695 0.630 0.570 0.605 0.525 0.425 0.985 0.875 0.760 0.665 0.715 0.630 0.565 0.505 0.430 0.440 0.345 4.910 4.520 4.160 3.820 3.500 3.210 3.035 2.845 2.660 2.470 2.260 2.070 1.890 1.720 1.560 1.440

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 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

Victor Saouma

bf 2tf 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 6.0 6.7 5.0 6.1 7.2 2.2 2.4 2.6 2.8 3.0 3.3 3.6 3.9 4.2 4.6 5.3 6.0 6.4 7.2 8.1 4.7 5.1 5.4 6.0 6.6 5.0 5.7 7.1 5.3 5.9 6.8 7.7 5.0 5.6 6.2 6.9 8.1 6.3 8.0 1.8 2.0 2.1 2.3 2.4 2.6 2.8 2.9 3.1 3.3 3.6 3.9 4.2 4.6 5.1 5.5

hc tw 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 43.6 46.9 46.3 49.4 53.6 10.6 11.5 12.5 13.8 15.1 16.7 18.0 19.8 21.9 23.9 24.5 27.2 30.0 33.4 37.8 32.4 35.7 38.7 41.2 45.2 44.6 51.0 53.5 24.3 27.0 31.2 35.9 33.0 37.4 41.2 46.6 48.1 51.6 56.8 3.7 4.0 4.4 4.8 5.2 5.7 6.1 6.4 6.9 7.4 8.1 8.8 9.7 10.7 11.6 12.8

Ix

Sx

rx

Iy

Sy

ry

Zx

Zy

J

in4 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 1480 1330 1170 984 843 6960 6160 5510 4900 4330 3870 3450 3060 2750 2460 2190 1910 1750 1530 1330 1170 1070 984 890 800 712 612 510 1490 1300 1110 954 758 659 586 518 448 375 301 14300 12400 10800 9430 8210 7190 6600 6000 5440 4900 4330 3840 3400 3010 2660 2400

in3 1570 1440 1300 1170 1060 970 884 811 742 674 624 556 502 455 411 345 299 267 243 213 1290 1170 1060 957 864 789 718 644 588 531 491 450 414 371 329 291 258 245 222 196 176 154 131 114 937 846 769 692 632 569 510 461 417 380 329 295 273 249 227 192 171 151 140 127 111 94 82 624 564 514 466 419 380 344 310 282 256 231 204 188 166 146 127 117 108 98 89 79 68 58 175 155 134 117 92 81 73 65 56 47 38 1280 1150 1040 931 838 756 707 656 607 559 506 459 415 375 338 310

in 12.7 12.6 12.5 12.3 12.2 12.1 12.0 12.0 11.9 11.8 11.8 11.7 11.6 11.5 11.4 11.2 11.0 11.0 10.9 10.7 11.5 11.4 11.3 11.1 11.0 10.9 10.8 10.7 10.7 10.6 10.5 10.5 10.4 10.3 10.2 10.1 10.1 10.0 9.9 9.8 9.7 9.6 9.2 9.1 10.2 10.0 9.9 9.8 9.7 9.6 9.5 9.5 9.4 9.4 9.2 9.1 9.1 9.1 9.0 8.7 8.7 8.6 8.6 8.5 8.4 8.2 8.1 8.7 8.6 8.5 8.4 8.4 8.3 8.2 8.1 8.1 8.0 7.9 7.8 7.8 7.8 7.7 7.5 7.5 7.5 7.4 7.4 7.2 7.2 7.0 7.1 7.1 7.0 7.0 6.7 6.7 6.7 6.6 6.5 6.4 6.3 8.2 8.0 7.8 7.6 7.5 7.3 7.3 7.2 7.1 7.0 6.9 6.8 6.7 6.6 6.6 6.5

in4 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 65 58 31 25 21 795 704 628 558 493 440 391 347 311 278 253 220 201 175 152 60 55 50 45 40 22 19 15 186 163 138 119 43 37 33 29 24 12 10 4720 4170 3680 3250 2880 2560 2360 2170 1990 1810 1610 1440 1290 1150 1030 931

in3 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 16 14 9 8 6 132 118 107 96 85 77 69 61 56 50 45 39 36 32 28 16 14 13 12 11 7 6 5 36 31 27 23 12 10 9 8 7 4 3 527 472 423 378 339 304 283 262 241 221 199 179 161 145 130 119

in 3.66 3.61 3.57 3.52 3.48 3.45 3.42 3.40 3.37 3.33 3.32 3.29 3.26 3.24 3.21 2.21 2.18 2.15 2.12 2.07 3.41 3.36 3.33 3.28 3.23 3.20 3.17 3.14 3.11 3.08 3.07 3.04 3.05 3.01 2.97 2.94 2.91 1.99 1.98 1.95 1.92 1.87 1.38 1.34 3.27 3.23 3.19 3.15 3.12 3.09 3.05 3.02 3.00 2.98 2.95 2.93 2.92 2.90 2.89 1.84 1.83 1.81 1.80 1.77 1.35 1.30 1.26 2.95 2.91 2.88 2.85 2.82 2.79 2.76 2.74 2.72 2.70 2.69 2.66 2.65 2.63 2.61 1.70 1.69 1.69 1.67 1.65 1.29 1.27 1.22 2.51 2.49 2.47 2.46 1.60 1.59 1.57 1.57 1.52 1.17 1.12 4.69 4.62 4.55 4.49 4.43 4.38 4.34 4.31 4.27 4.24 4.20 4.17 4.13 4.10 4.07 4.05

in3 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 1010.0 915.0 816.0 741.0 663.0 589.0 530.0 476.0 432.0 373.0 333.0 307.0 279.0 253.0 221.0 196.0 172.0 160.0 144.0 129.0 110.0 95.4 753.0 676.0 611.0 549.0 490.0 442.0 398.0 356.0 322.0 291.0 261.0 230.0 211.0 186.0 163.0 145.0 133.0 123.0 112.0 101.0 90.7 78.4 66.5 198.0 175.0 150.0 130.0 105.0 92.0 82.3 72.9 64.0 54.0 44.2 1660.0 1480.0 1320.0 1180.0 1050.0 936.0 869.0 801.0 736.0 672.0 603.0 542.0 487.0 436.0 390.0 355.0

in3 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 24.4 21.7 14.8 12.2 10.2 207.0 185.0 166.0 149.0 132.0 119.0 106.0 94.8 85.4 76.7 69.1 60.5 55.3 48.4 42.2 24.7 22.5 20.6 18.5 16.6 11.7 9.9 8.1 54.9 48.1 41.1 35.5 18.9 16.3 14.5 12.7 10.8 7.0 5.5 816.0 730.0 652.0 583.0 522.0 468.0 434.0 402.0 370.0 338.0 304.0 274.0 246.0 221.0 198.0 180.0

in4 499.0 391.0 297.0 225.0 169.0 131.0 101.0 78.8 61.0 46.3 37.0 26.5 19.5 14.7 10.9 11.2 7.3 5.3 4.0 2.8 456.0 357.0 271.0 205.0 154.0 119.0 91.7 67.3 51.8 38.6 31.0 24.1 18.5 13.4 9.5 6.7 4.7 7.1 5.3 3.7 2.7 1.9 1.7 1.2 298.0 225.0 174.0 130.0 101.0 75.2 54.9 41.3 31.1 23.9 15.4 11.3 9.0 6.8 5.2 6.0 4.3 3.0 2.5 1.8 1.8 1.1 0.8 177.0 135.0 104.0 79.7 59.3 45.2 34.2 25.4 19.4 14.7 10.6 7.5 5.9 4.1 2.8 3.5 2.7 2.2 1.7 1.2 1.2 0.8 0.5 7.7 5.4 3.6 2.4 2.2 1.5 1.1 0.8 0.5 0.5 0.3 1450.0 1120.0 870.0 670.0 514.0 395.0 331.0 273.0 222.0 178.0 136.0 104.0 79.1 59.5 44.6 34.8

Structural Engineering

3.6 Steel Section Properties

Draft 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

3–11

A

d

tw

bf

tf

in2

in 9.00 8.75 8.50 8.25 8.12 8.00 8.06 7.93 8.28 8.14 8.11 7.99 7.89 6.38 6.20 5.99 6.28 6.03 5.90 5.15 5.01 4.16 14.00 12.00 11.97 11.97 10.00 9.95 9.99 8.00 6.00 5.00 24.50 24.50 24.00 24.00 24.00 20.30 20.30 20.00 20.00 18.00 18.00 15.00 15.00 12.00 12.00 12.00 12.00 10.00 10.00 8.00 8.00 7.00 7.00 6.00 6.00 5.00 5.00 4.00 4.00 3.00 3.00

in 0.570 0.510 0.400 0.360 0.310 0.285 0.285 0.245 0.250 0.230 0.245 0.230 0.170 0.320 0.260 0.230 0.260 0.230 0.170 0.270 0.240 0.280 0.215 0.177 0.160 0.149 0.157 0.141 0.130 0.135 0.114 0.316 0.800 0.620 0.745 0.625 0.500 0.800 0.660 0.635 0.505 0.711 0.461 0.550 0.411 0.687 0.462 0.428 0.350 0.594 0.311 0.441 0.271 0.450 0.252 0.465 0.232 0.494 0.214 0.326 0.193 0.349 0.170

in 8.280 8.220 8.110 8.070 8.020 7.995 6.535 6.495 5.270 5.250 4.015 4.000 3.940 6.080 6.020 5.990 4.030 4.000 3.940 5.030 5.000 4.060 4.000 3.065 3.065 3.250 2.690 2.690 2.690 2.281 1.844 5.003 8.050 7.870 7.245 7.125 7.000 7.200 7.060 6.385 6.255 6.251 6.001 5.640 5.501 5.477 5.252 5.078 5.000 4.944 4.661 4.171 4.001 3.860 3.662 3.565 3.332 3.284 3.004 2.796 2.663 2.509 2.330

in 0.935 0.810 0.685 0.560 0.495 0.435 0.465 0.400 0.400 0.330 0.315 0.255 0.205 0.455 0.365 0.260 0.405 0.280 0.215 0.430 0.360 0.345 0.270 0.225 0.210 0.180 0.206 0.182 0.173 0.189 0.171 0.416 1.090 1.090 0.870 0.870 0.870 0.920 0.920 0.795 0.795 0.691 0.691 0.622 0.622 0.659 0.659 0.544 0.544 0.491 0.491 0.426 0.426 0.392 0.392 0.359 0.359 0.326 0.326 0.293 0.293 0.260 0.260

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

Victor Saouma

bf 2tf 4.4 5.1 5.9 7.2 8.1 9.2 7.0 8.1 6.6 8.0 6.4 7.8 9.6 6.7 8.2 11.5 5.0 7.1 9.2 5.8 6.9 5.9 7.4 6.8 7.3 9.1 6.5 7.4 7.8 6.0 5.4 6.0 3.7 3.6 4.2 4.1 4.0 3.9 3.8 4.0 3.9 4.5 4.3 4.5 4.4 4.2 4.0 4.7 4.6 5.0 4.7 4.9 4.7 4.9 4.7 5.0 4.6 5.0 4.6 4.8 4.5 4.8 4.5

hc tw 11.1 12.4 15.8 17.6 20.4 22.2 22.2 25.8 27.5 29.9 28.1 29.9 40.5 15.5 19.1 21.6 19.1 21.6 29.2 14.0 15.8 10.6 60.3 62.5 63.6 68.0 58.4 59.3 65.0 53.8 47.0 11.2 26.4 34.1 28.3 33.7 42.1 21.6 26.2 27.1 34.1 21.8 33.6 23.2 31.0 13.9 20.7 23.4 28.6 13.8 26.4 14.5 23.7 12.3 21.9 9.9 19.9 7.5 17.4 8.7 14.7 5.6 11.4

Ix

Sx

rx

Iy

Sy

ry

Zx

Zy

J

in4 272 228 184 146 127 110 98 83 75 62 48 40 31 53 41 29 32 22 16 26 21 11 148 72 65 62 39 34 33 18 7 24 3160 2940 2390 2250 2100 1670 1580 1280 1190 926 804 486 447 305 272 229 218 147 124 65 58 42 37 26 22 15 12 7 6 3 3

in3 60 52 43 36 31 28 24 21 18 15 12 10 8 17 13 10 10 7 6 10 9 5 21 12 11 10 8 7 7 5 2 10 258 240 199 187 175 165 155 128 119 103 89 65 60 51 45 38 36 29 25 16 14 12 10 9 7 6 5 3 3 2 2

in 3.7 3.7 3.6 3.5 3.5 3.5 3.5 3.4 3.5 3.4 3.3 3.2 3.2 2.7 2.7 2.6 2.6 2.5 2.5 2.2 2.1 1.7 5.4 4.6 4.6 4.6 3.8 3.8 3.8 3.1 2.4 2.1 9.4 9.7 9.0 9.2 9.5 7.7 7.9 7.6 7.8 6.7 7.1 5.8 5.9 4.6 4.8 4.7 4.8 3.8 4.1 3.1 3.3 2.7 2.9 2.3 2.5 1.9 2.0 1.6 1.6 1.1 1.2

in4 89 75 61 49 43 37 22 18 10 8 3 3 2 17 13 9 4 3 2 9 8 4 3 1 1 1 1 1 0 0 0 8 83 77 48 45 42 50 47 30 28 24 21 16 14 16 14 10 9 8 7 4 4 3 3 2 2 2 1 1 1 1 0

in3 21 18 15 12 11 9 7 6 4 3 2 1 1 6 4 3 2 2 1 4 3 2 1 1 1 1 0 0 0 0 0 3 21 20 13 13 12 14 13 9 9 8 7 6 5 6 5 4 4 3 3 2 2 2 1 1 1 1 1 1 1 0 0

in 2.12 2.10 2.08 2.04 2.03 2.02 1.62 1.61 1.26 1.23 0.88 0.84 0.84 1.52 1.50 1.46 0.97 0.92 0.90 1.28 1.27 1.00 0.72 0.53 0.54 0.58 0.48 0.43 0.47 0.42 0.36 1.19 1.53 1.57 1.27 1.30 1.34 1.33 1.36 1.16 1.19 1.08 1.14 1.03 1.07 1.03 1.06 0.98 1.00 0.90 0.95 0.80 0.83 0.73 0.77 0.68 0.70 0.62 0.64 0.57 0.58 0.52 0.52

in3 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

in3 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

in4 5.1 3.3 2.0 1.1 0.8 0.5 0.5 0.3 0.3 0.2 0.1 0.1 0.0 0.5 0.2 0.1 0.2 0.1 0.0 0.3 0.2 0.2 0.1 0.1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.3 12.8 10.1 7.6 6.0 4.9 8.4 6.6 4.6 3.6 4.2 2.4 2.1 1.5 2.8 1.8 1.1 0.9 1.3 0.6 0.6 0.3 0.4 0.2 0.4 0.2 0.3 0.1 0.1 0.1 0.1 0.0

Structural Engineering

3–12

Draft Designation

C 15.x 50 C 15.x 40 C 15.x 34 C 12.x 30 C 12.x 25 C 12.x 21 C 10.x 30 C 10.x 25 C 10.x 20 C 10.x 15 C 9.x 20 C 9.x 15 C 9.x 13 C 8.x 19 C 8.x 14 C 8.x 12 C 7.x 15 C 7.x 12 C 7.x 10 C 6.x 13 C 6.x 11 C 6.x 8 C 5.x 9 C 5.x 7 C 4.x 7 C 4.x 5 C 3.x 6 C 3.x 5 C 3.x 4 MC18.x 58 MC18.x 52 MC18.x 46 MC18.x 43 MC13.x 50 MC13.x 40 MC13.x 35 MC13.x 32 MC12.x 50 MC12.x 45 MC12.x 40 MC12.x 35 MC12.x 31 MC12.x 11 MC10.x 41 MC10.x 34 MC10.x 29 MC10.x 25 MC10.x 22 MC10.x 8 MC10.x 7 MC 9.x 25 MC 9.x 24 MC 8.x 23 MC 8.x 21 MC 8.x 20 MC 8.x 19 MC 8.x 9 MC 7.x 23 MC 7.x 19 MC 6.x 18 MC 6.x 15 MC 6.x 16 MC 6.x 15 MC 6.x 12

STRUCTURAL MATERIALS A

d

tw

bf

tf

in2

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. 18. 18. 18. 18. 13. 13. 13. 13. 12. 12. 12. 12. 12. 12. 10. 10. 10. 10. 10. 10. 10. 9. 9. 8. 8. 8. 8. 8. 7. 7. 6. 6. 6. 6. 6.

in 0.716 0.520 0.400 0.510 0.387 0.282 0.673 0.526 0.379 0.240 0.448 0.285 0.233 0.487 0.303 0.220 0.419 0.314 0.210 0.437 0.314 0.200 0.325 0.190 0.321 0.184 0.356 0.258 0.170 0.700 0.600 0.500 0.450 0.787 0.560 0.447 0.375 0.835 0.712 0.590 0.467 0.370 0.190 0.796 0.575 0.425 0.380 0.290 0.170 0.152 0.450 0.400 0.427 0.375 0.400 0.353 0.179 0.503 0.352 0.379 0.340 0.375 0.316 0.310

in 3.716 3.520 3.400 3.170 3.047 2.942 3.033 2.886 2.739 2.600 2.648 2.485 2.433 2.527 2.343 2.260 2.299 2.194 2.090 2.157 2.034 1.920 1.885 1.750 1.721 1.584 1.596 1.498 1.410 4.200 4.100 4.000 3.950 4.412 4.185 4.072 4.000 4.135 4.012 3.890 3.767 3.670 1.500 4.321 4.100 3.950 3.405 3.315 1.500 1.127 3.500 3.450 3.502 3.450 3.025 2.978 1.874 3.603 3.452 3.504 3.500 3.000 2.941 2.497

in 0.650 0.650 0.650 0.501 0.501 0.501 0.436 0.436 0.436 0.436 0.413 0.413 0.413 0.390 0.390 0.390 0.366 0.366 0.366 0.343 0.343 0.343 0.320 0.320 0.296 0.296 0.273 0.273 0.273 0.625 0.625 0.625 0.625 0.610 0.610 0.610 0.610 0.700 0.700 0.700 0.700 0.700 0.309 0.575 0.575 0.575 0.575 0.575 0.280 0.202 0.550 0.550 0.525 0.525 0.500 0.500 0.311 0.500 0.500 0.475 0.385 0.475 0.475 0.375

14.7 11.8 10.0 8.8 7.3 6.1 8.8 7.3 5.9 4.5 5.9 4.4 3.9 5.5 4.0 3.4 4.3 3.6 2.9 3.8 3.1 2.4 2.6 2.0 2.1 1.6 1.8 1.5 1.2 17.1 15.3 13.5 12.6 14.7 11.8 10.3 9.4 14.7 13.2 11.8 10.3 9.1 3.1 12.1 9.9 8.4 7.3 6.4 2.5 1.9 7.5 7.0 6.7 6.3 5.9 5.5 2.5 6.7 5.6 5.3 4.5 4.8 4.4 3.5

Victor Saouma

bf 2tf 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 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 0 0 0 0 0 0

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 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 0 0 0 0 0 0

Ix

Sx

rx

Iy

Sy

ry

Zx

Zy

J

in4 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 1.7 676.0 627.0 578.0 554.0 314.0 273.0 252.0 239.0 269.0 252.0 234.0 216.0 203.0 55.4 158.0 139.0 127.0 110.0 103.0 32.0 22.1 88.0 85.0 63.8 61.6 54.5 52.5 23.3 47.5 43.2 29.7 25.4 26.0 25.0 18.7

in3 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 75.1 69.7 64.3 61.6 48.4 42.0 38.8 36.8 44.9 42.0 39.0 36.1 33.8 9.2 31.5 27.8 25.3 22.0 20.5 6.4 4.4 19.6 18.9 16.0 15.4 13.6 13.1 5.8 13.6 12.3 9.9 8.5 8.7 8.3 6.2

in 5.24 5.44 5.62 4.29 4.43 4.61 3.42 3.52 3.66 3.87 3.22 3.40 3.48 2.82 2.99 3.11 2.51 2.60 2.72 2.13 2.22 2.34 1.83 1.95 1.47 1.56 1.08 1.12 1.17 6.29 6.41 6.56 6.64 4.62 4.82 4.95 5.06 4.28 4.36 4.46 4.59 4.71 4.22 3.61 3.75 3.89 3.87 3.99 3.61 3.40 3.43 3.48 3.09 3.13 3.05 3.09 3.05 2.67 2.77 2.37 2.38 2.33 2.37 2.30

in4 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 17.80 16.40 15.10 14.40 16.50 13.70 12.30 11.40 17.40 15.80 14.30 12.70 11.30 0.38 15.80 13.20 11.40 7.35 6.50 0.33 0.11 7.65 7.22 7.07 6.64 4.47 4.20 0.63 7.29 6.11 5.93 4.97 3.82 3.51 1.87

in3 3.78 3.37 3.11 2.06 1.88 1.73 1.65 1.48 1.32 1.16 1.17 1.01 0.96 1.01 0.85 0.78 0.78 0.70 0.63 0.64 0.56 0.49 0.45 0.38 0.34 0.28 0.27 0.23 0.20 5.32 5.07 4.82 4.69 4.79 4.26 3.99 3.81 5.65 5.33 5. 4.67 4.39 0.31 4.88 4.38 4.02 3. 2.80 0.27 0.12 3.02 2.93 2.84 2.74 2.05 1.97 0.43 2.85 2.57 2.48 2.03 1.84 1.75 1.04

in 0.87 0.89 0.90 0.76 0.78 0.80 0.67 0.68 0.69 0.71 0.64 0.66 0.67 0.60 0.62 0.63 0.56 0.57 0.58 0.52 0.53 0.54 0.49 0.49 0.45 0.45 0.42 0.41 0.40 1.02 1.04 1.06 1.07 1.06 1.08 1.10 1.11 1.09 1.09 1.10 1.11 1.12 0.35 1.14 1.16 1.17 1. 1. 0.37 0.24 1.01 1.01 1.03 1.03 0.87 0.87 0.50 1.05 1.04 1.06 1.05 0.89 0.89 0.73

in3 68.20 57.20 50.40 33.60 29.20 25.40 26.60 23. 19.30 15.80 16.80 13.50 12.50 13.80 10.90 9.55 9.68 8.40 7.12 7.26 6.15 5.13 4.36 3.51 2.81 2.26 1.72 1.50 1.30 94.60 86.50 78.40 74.40 60.50 50.90 46.20 43.10 56.10 51.70 47.30 42.80 39.30 11.60 38.90 33.40 29.60 25.80 23.60 7.86 5.73 23.20 22.20 18.80 18. 16.20 15.40 6.91 16.20 14.30 11.50 9.82 10.20 9.69 7.38

in3 8.17 6.87 6.23 4.33 3.84 3.49 3.78 3.19 2.71 2.35 2.47 2.05 1.95 2.17 1.73 1.58 1.64 1.43 1.26 1.36 1.15 0.99 0.92 0.76 0.70 0.57 0.54 0.47 0.40 9.94 9.13 8.42 8.10 10.10 8.57 7.95 7.60 10.20 9.35 8.59 7.91 7.44 0.64 8.71 7.51 6.83 5.21 4.86 0.55 0.25 5.23 5.05 4.88 4.71 3.57 3.44 0.88 4.86 4.34 4.14 3.28 3.18 3. 1.79

in4 2.67 1.46 1.02 0.87 0.54 0.37 1.23 0.69 0.37 0.21 0.43 0.21 0.17 0.44 0.19 0.13 0.27 0.16 0.10 0.24 0.13 0.08 0.11 0.06 0.08 0.04 0.07 0.04 0.03 2.81 2.03 1.45 1.23 2.98 1.57 1.14 0.94 3.24 2.35 1.70 1.25 1.01 0.06 2.27 1.21 0.79 0.64 0.51 0.04 0.02 0.69 0.60 0.57 0.50 0.44 0.38 0.06 0.63 0.41 0.38 0.22 0.34 0.29 0.15

Structural Engineering

3.6 Steel Section Properties

Draft Designation

WT18.x180. WT18.x164. WT18.x150. WT18.x140. WT18.x130. WT18.x123. WT18.x115. WT18.x128. WT18.x116. WT18.x105. WT18.x 97. WT18.x 91. WT18.x 85. WT18.x 80. WT18.x 75. WT18.x 68. WT17.x177. WT17.x159. WT17.x146. WT17.x132. WT17.x121. WT17.x111. WT17.x101. WT17.x 85. WT17.x 76. WT17.x 71. WT17.x 65. WT17.x 59. WT15.x118. WT15.x106. WT15.x 96. WT15.x 87. WT15.x 74. WT15.x 66. WT15.x 62. WT15.x 58. WT15.x 54. WT15.x 50. WT14.x109. WT14.x 97. WT14.x 89. WT14.x 81. WT14.x 73. WT14.x 65. WT14.x 57. WT14.x 51. WT14.x 47. WT14.x 42. WT12.x 88. WT12.x 81. WT12.x 73. WT12.x 66. WT12.x 59. WT12.x 52. WT12.x 52. WT12.x 47. WT12.x 42. WT12.x 38. WT12.x 34. WT12.x 31. WT12.x 28. WT11.x 83. WT11.x 74. WT11.x 66. WT11.x 61. WT11.x 56. WT11.x 51. WT11.x 47. WT11.x 42. WT11.x 37. WT11.x 34. WT11.x 31. WT11.x 29. WT11.x 25. WT11.x 22. WT 9.x 72. WT 9.x 65. WT 9.x 60. WT 9.x 53. WT 9.x 49. WT 9.x 43. WT 9.x 38. WT 9.x 36. WT 9.x 33. WT 9.x 30. WT 9.x 28. WT 9.x 25. WT 9.x 23. WT 9.x 20. WT 9.x 18.

3–13

A

d

tw

bf

tf

in2

in 18.70 18.54 18.37 18.26 18.13 18.04 17.95 18.71 18.56 18.34 18.25 18.17 18.08 18.00 17.92 17.77 17.77 17.58 17.42 17.26 17.09 16.97 16.84 16.91 16.75 16.65 16.55 16.43 15.65 15.47 15.34 15.22 15.34 15.15 15.09 15.01 14.91 14.82 14.22 14.06 13.90 13.80 13.69 13.81 13.65 13.55 13.46 13.35 12.63 12.50 12.37 12.24 12.13 12.03 12.26 12.15 12.05 11.96 11.86 11.87 11.78 11.24 11.03 10.91 10.84 10.76 10.68 10.81 10.72 10.62 10.56 10.49 10.53 10.41 10.33 9.74 9.63 9.48 9.36 9.30 9.19 9.10 9.23 9.18 9.12 9.06 8.99 9.03 8.95 8.85

in 1.120 1.020 0.945 0.885 0.840 0.800 0.760 0.960 0.870 0.830 0.765 0.725 0.680 0.650 0.625 0.600 1.160 1.040 0.960 0.870 0.830 0.775 0.715 0.670 0.635 0.605 0.580 0.550 0.830 0.775 0.710 0.655 0.650 0.615 0.585 0.565 0.545 0.520 0.830 0.750 0.725 0.660 0.605 0.610 0.570 0.515 0.490 0.460 0.750 0.705 0.650 0.605 0.550 0.500 0.550 0.515 0.470 0.440 0.415 0.430 0.395 0.750 0.720 0.650 0.600 0.550 0.500 0.580 0.515 0.455 0.430 0.400 0.405 0.380 0.350 0.730 0.670 0.655 0.590 0.535 0.480 0.425 0.495 0.450 0.415 0.390 0.355 0.360 0.315 0.300

in 16.730 16.630 16.655 16.595 16.550 16.510 16.470 12.215 12.120 12.180 12.115 12.075 12.030 12.000 11.975 11.950 16.100 15.985 15.905 15.805 15.860 15.805 15.745 11.500 11.565 11.535 11.510 11.480 15.055 15.105 15.040 14.985 10.480 10.545 10.515 10.495 10.475 10.450 14.115 14.035 14.085 14.020 13.965 10.010 10.070 10.015 9.990 9.960 12.890 12.955 12.900 12.855 12.800 12.750 9.000 9.065 9.020 8.990 8.965 7.040 7.005 12.420 12.510 12.440 12.390 12.340 12.290 8.420 8.355 8.295 8.270 8.240 6.555 6.530 6.500 11.220 11.160 11.265 11.200 11.145 11.090 11.035 7.635 7.590 7.555 7.530 7.495 6.060 6.015 6.000

in 2.010 1.850 1.680 1.570 1.440 1.350 1.260 1.730 1.570 1.360 1.260 1.180 1.100 1.020 0.940 0.790 2.090 1.890 1.730 1.570 1.400 1.275 1.150 1.220 1.055 0.960 0.855 0.740 1.500 1.315 1.185 1.065 1.180 1.000 0.930 0.850 0.760 0.670 1.500 1.340 1.190 1.080 0.975 1.100 0.930 0.830 0.745 0.640 1.340 1.220 1.090 0.960 0.850 0.750 0.980 0.875 0.770 0.680 0.585 0.590 0.505 1.360 1.150 1.035 0.960 0.875 0.800 0.930 0.835 0.740 0.685 0.615 0.650 0.535 0.450 1.320 1.200 1.060 0.940 0.870 0.770 0.680 0.810 0.750 0.695 0.630 0.570 0.605 0.525 0.425

52.7 48.2 44.1 41.2 38.2 36.0 33.8 37.7 34.1 30.9 28.5 26.8 25.0 23.5 22.1 19.9 52.1 46.7 42.8 38.7 35.4 32.5 29.5 24.8 22.4 20.8 19.2 17.3 34.5 31.0 28.1 25.4 21.7 19.4 18.2 17.1 15.9 14.5 31.9 28.5 26.1 23.7 21.5 18.9 16.8 15.0 13.8 12.4 25.8 23.9 21.5 19.3 17.2 15.3 15.1 13.8 12.4 11.2 10.0 9.1 8.1 24.4 21.6 19.4 17.9 16.3 14.9 13.7 12.2 10.7 10.0 9.1 8.4 7.4 6.5 21.0 19.1 17.5 15.6 14.3 12.7 11.2 10.4 9.6 8.8 8.1 7.3 6.8 5.9 5.2

Victor Saouma

bf 2tf 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 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 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 0 0 0

hc tw 14.1 15.4 16.7 17.8 18.7 19.7 20.7 16.9 18.7 19.6 21.2 22.4 23.9 25.0 26.0 27.1 12.9 14.4 15.6 17.2 18.1 19.3 21.0 22.4 23.6 24.8 25.8 27.3 16.2 17.4 19.0 20.6 20.8 22.0 23.1 23.9 24.8 26.0 14.6 16.2 16.7 18.4 20.0 19.9 21.3 23.5 24.7 26.3 14.4 15.3 16.6 17.8 19.6 21.6 19.6 20.9 22.9 24.5 26.0 25.1 27.3 12.4 13.0 14.4 15.6 17.1 18.8 16.2 18.2 20.6 21.8 23.5 23.2 24.7 26.8 11.0 11.9 12.3 13.6 15.0 16.7 18.9 16.2 17.8 19.3 20.6 22.6 22.3 25.5 26.8

Ix

Sx

rx

Iy

Sy

ry

Zx

Zy

J

in4 1500.0 1350.0 1230.0 1140.0 1060.0 995.0 934.0 1200.0 1080.0 985.0 901.0 845.0 786.0 740.0 698.0 637.0 1320.0 1160.0 1050.0 943.0 871.0 799.0 725.0 649.0 592.0 552.0 513.0 469.0 674.0 610.0 549.0 497.0 466.0 421.0 396.0 373.0 349.0 322.0 502.0 444.0 414.0 372.0 336.0 323.0 289.0 258.0 239.0 216.0 319.0 293.0 264.0 238.0 212.0 189.0 204.0 186.0 166.0 151.0 137.0 131.0 117.0 226.0 204.0 181.0 166.0 150.0 135.0 144.0 127.0 110.0 103.0 93.8 90.4 80.3 71.1 142.0 127.0 119.0 104.0 93.8 82.4 71.8 78.2 70.7 64.7 59.5 53.5 52.1 44.8 40.1

in3 104.0 94.1 86.1 80.0 75.1 71.0 67.0 87.4 78.5 73.1 67.0 63.1 58.9 55.8 53.1 49.7 96.8 85.8 78.3 70.2 65.8 60.8 55.5 51.1 47.4 44.7 42.1 39.2 55.1 50.5 45.7 41.7 40.6 37.4 35.3 33.7 32.0 30.0 45.2 40.3 38.2 34.4 31.2 31.0 28.3 25.3 23.8 21.9 32.2 29.9 27.2 24.8 22.3 20.0 22.0 20.3 18.3 16.9 15.6 15.6 14.1 25.5 23.7 21.1 19.3 17.5 15.8 17.9 15.7 13.8 12.9 11.9 11.8 10.7 9.7 18.5 16.7 15.9 14.1 12.7 11.2 9.8 11.2 10.1 9.3 8.6 7.8 7.8 6.7 6.2

in 5.33 5.29 5.27 5.25 5.26 5.26 5.25 5.66 5.63 5.65 5.62 5.62 5.61 5.61 5.62 5.66 5.03 4.99 4.97 4.94 4.96 4.96 4.95 5.12 5.14 5.15 5.18 5.20 4.42 4.43 4.42 4.42 4.63 4.66 4.66 4.67 4.69 4.71 3.97 3.95 3.98 3.96 3.95 4.13 4.15 4.14 4.16 4.18 3.51 3.50 3.50 3.52 3.51 3.51 3.67 3.67 3.67 3.68 3.70 3.79 3.80 3.04 3.08 3.06 3.04 3.03 3.01 3.25 3.22 3.21 3.20 3.21 3.29 3.30 3.31 2.60 2.58 2.60 2.59 2.56 2.55 2.54 2.74 2.72 2.71 2.71 2.70 2.77 2.76 2.79

in4 786.00 711.00 648.00 599.00 545.00 507.00 470.00 264.00 234.00 206.00 187.00 174.00 160.00 147.00 135.00 113.00 729.00 645.00 581.00 517.00 466.00 420.00 375.00 155.00 136.00 123.00 109.00 93.60 427.00 378.00 336.00 299.00 113.00 98.00 90.40 82.10 73.00 63.90 352.00 309.00 278.00 248.00 222.00 92.20 79.40 69.60 62.00 52.80 240.00 221.00 195.00 170.00 149.00 130.00 59.70 54.50 47.20 41.30 35.20 17.20 14.50 217.00 188.00 166.00 152.00 137.00 124.00 46.40 40.70 35.30 32.40 28.70 15.30 12.50 10.30 156.00 139.00 126.00 110.00 100.00 87.60 76.20 30.10 27.40 25.00 22.50 20.00 11.30 9.55 7.67

in3 94.00 85.50 77.80 72.20 65.90 61.40 57.10 43.20 38.60 33.80 30.90 28.80 26.60 24.60 22.50 18.90 90.60 80.70 73.10 65.50 58.80 53.20 47.60 27.00 23.60 21.30 18.90 16.30 56.80 50.10 44.70 39.90 21.70 18.60 17.20 15.70 13.90 12.20 49.90 44.10 39.40 35.40 31.70 18.40 15.80 13.90 12.40 10.60 37.20 34.20 30.30 26.50 23.20 20.30 13.30 12.00 10.50 9.18 7.85 4.90 4.15 35.00 30.00 26.70 24.60 22.20 20.20 11.00 9.75 8.51 7.83 6.97 4.67 3.82 3.18 27.70 24.90 22.50 19.70 18.00 15.80 13.80 7.89 7.22 6.63 5.97 5.35 3.72 3.17 2.56

in 3.86 3.84 3.83 3.81 3.78 3.75 3.73 2.65 2.62 2.58 2.56 2.55 2.53 2.50 2.47 2.38 3.74 3.71 3.69 3.66 3.63 3.59 3.56 2.50 2.47 2.43 2.39 2.32 3.52 3.49 3.46 3.43 2.28 2.25 2.23 2.19 2.15 2.10 3.32 3.29 3.26 3.24 3.21 2.21 2.18 2.15 2.12 2.07 3.04 3.05 3.01 2.97 2.94 2.91 1.99 1.98 1.95 1.92 1.87 1.38 1.34 2.98 2.95 2.93 2.92 2.90 2.89 1.84 1.83 1.81 1.80 1.77 1.35 1.30 1.26 2.72 2.70 2.69 2.66 2.65 2.63 2.61 1.70 1.69 1.69 1.67 1.65 1.29 1.27 1.22

in3 187.00 168.00 153.00 142.00 133.00 125.00 118.00 156.00 140.00 131.00 120.00 113.00 105.00 100.00 95.50 94.30 174.00 154.00 140.00 125.00 116.00 107.00 97.70 90.80 84.50 79.80 75.60 74.80 98.20 89.50 80.80 73.40 72.20 66.80 63.10 60.40 57.70 57.40 81.10 71.80 67.60 60.80 55.00 55.10 50.40 45.00 42.40 39.20 57.80 53.30 48.20 43.90 39.20 35.10 39.20 36.10 32.50 30.10 27.90 28.40 25.60 46.30 42.40 37.60 34.30 31.00 27.90 31.80 28.00 24.40 22.90 21.10 21.20 20.80 17.60 34.00 30.50 28.70 25.20 22.60 19.90 17.30 20.00 18.00 16.50 15.30 13.80 13.90 12.00 12.00

in3 146.00 132.00 120.00 112.00 102.00 94.90 88.10 68.60 61.00 53.50 48.90 45.40 41.90 38.60 35.50 29.80 141.00 125.00 113.00 101.00 90.90 82.10 73.40 42.20 37.00 33.50 29.70 25.70 87.50 77.20 68.90 61.40 34.00 29.20 27.00 24.60 22.00 19.30 77.00 67.90 60.80 54.50 48.80 28.80 24.70 21.70 19.40 16.60 57.30 52.60 46.60 40.70 35.70 31.20 20.70 18.80 16.30 14.30 12.30 7.87 6.67 53.90 46.30 41.10 37.80 34.10 30.90 17.40 15.30 13.30 12.20 10.90 7.42 6.09 5.09 42.70 38.30 34.60 30.20 27.60 24.20 21.10 12.30 11.20 10.30 9.27 8.29 5.85 4.97 4.03

in4 54.30 42.10 32.00 26.20 20.70 17.30 14.30 26.60 19.80 13.90 11.10 9.19 7.51 6.17 5.04 3.48 57.20 42.10 32.40 24.20 17.90 13.70 10.20 8.80 6.16 4.84 3.67 2.64 19.90 13.90 10.30 7.61 7.30 4.85 3.98 3.21 2.49 1.88 18.50 13.20 9.74 7.31 5.44 5.60 3.65 2.64 2.01 1.40 12.00 9.22 6.70 4.74 3.35 2.35 3.50 2.62 1.84 1.34 0.93 0.85 0.59 11.90 7.69 5.62 4.47 3.40 2.60 3.01 2.16 1.51 1.22 0.51 0.88 0.57 0.38 9.70 7.30 5.30 3.73 2.92 2.04 1.41 1.74 1.36 1.08 0.83 0.61 0.61 0.40 0.25

Structural Engineering

3–14

STRUCTURAL MATERIALS

Draft Designation

WT 8.x 50. WT 8.x 45. WT 8.x 39. WT 8.x 34. WT 8.x 29. WT 8.x 25. WT 8.x 23. WT 8.x 20. WT 8.x 18. WT 8.x 16. WT 8.x 13. WT 7.x365. WT 7.x333. WT 7.x303. WT 7.x275. WT 7.x250. WT 7.x228. WT 7.x213. WT 7.x199. WT 7.x185. WT 7.x171. WT 7.x156. WT 7.x142. WT 7.x129. WT 7.x117. WT 7.x106. WT 7.x 97. WT 7.x 88. WT 7.x 80. WT 7.x 73. WT 7.x 66. WT 7.x 60. WT 7.x 55. WT 7.x 50. WT 7.x 45. WT 7.x 41. WT 7.x 37. WT 7.x 34. WT 7.x 31. WT 7.x 27. WT 7.x 24. WT 7.x 22. WT 7.x 19. WT 7.x 17. WT 7.x 15. WT 7.x 13. WT 7.x 11. WT 6.x168. WT 6.x153. WT 6.x140. WT 6.x126. WT 6.x115. WT 6.x105. WT 6.x 95. WT 6.x 85. WT 6.x 76. WT 6.x 68. WT 6.x 60. WT 6.x 53. WT 6.x 48. WT 6.x 44. WT 6.x 40. WT 6.x 36. WT 6.x 33. WT 6.x 29. WT 6.x 27. WT 6.x 25. WT 6.x 23. WT 6.x 20. WT 6.x 18. WT 6.x 15. WT 6.x 13. WT 6.x 11. WT 6.x 10. WT 6.x 8. WT 6.x 7. WT 5.x 56. WT 5.x 50. WT 5.x 44. WT 5.x 39. WT 5.x 34. WT 5.x 30. WT 5.x 27. WT 5.x 25. WT 5.x 23. WT 5.x 20. WT 5.x 17. WT 5.x 15. WT 5.x 13. WT 5.x 11. WT 5.x 10. WT 5.x 9. WT 5.x 8. WT 5.x 6.

A

d

tw

bf

tf

in2

in 8.48 8.38 8.26 8.16 8.22 8.13 8.06 8.01 7.93 7.94 7.84 11.21 10.82 10.46 10.12 9.80 9.51 9.34 9.15 8.96 8.77 8.56 8.37 8.19 8.02 7.86 7.74 7.61 7.49 7.39 7.33 7.24 7.16 7.08 7.01 7.16 7.09 7.02 6.95 6.96 6.89 6.83 7.05 6.99 6.92 6.95 6.87 8.41 8.16 7.93 7.70 7.53 7.36 7.19 7.01 6.86 6.70 6.56 6.45 6.36 6.26 6.19 6.13 6.06 6.09 6.03 6.09 6.03 5.97 6.25 6.17 6.11 6.16 6.08 5.99 5.95 5.68 5.55 5.42 5.30 5.20 5.11 5.05 4.99 5.05 4.96 4.86 5.24 5.16 5.09 5.12 5.05 4.99 4.93

in 0.585 0.525 0.455 0.395 0.430 0.380 0.345 0.305 0.295 0.275 0.250 3.070 2.830 2.595 2.380 2.190 2.015 1.875 1.770 1.655 1.540 1.410 1.290 1.175 1.070 0.980 0.890 0.830 0.745 0.680 0.645 0.590 0.525 0.485 0.440 0.510 0.450 0.415 0.375 0.370 0.340 0.305 0.310 0.285 0.270 0.255 0.230 1.775 1.625 1.530 1.395 1.285 1.180 1.060 0.960 0.870 0.790 0.710 0.610 0.550 0.515 0.470 0.430 0.390 0.360 0.345 0.370 0.335 0.295 0.300 0.260 0.230 0.260 0.235 0.220 0.200 0.755 0.680 0.605 0.530 0.470 0.420 0.370 0.340 0.350 0.315 0.290 0.300 0.260 0.240 0.250 0.240 0.230 0.190

in 10.425 10.365 10.295 10.235 7.120 7.070 7.035 6.995 6.985 5.525 5.500 17.890 17.650 17.415 17.200 17.010 16.835 16.695 16.590 16.475 16.360 16.230 16.110 15.995 15.890 15.800 15.710 15.650 15.565 15.500 14.725 14.670 14.605 14.565 14.520 10.130 10.070 10.035 9.995 8.060 8.030 7.995 6.770 6.745 6.730 5.025 5.000 13.385 13.235 13.140 13.005 12.895 12.790 12.670 12.570 12.480 12.400 12.320 12.220 12.160 12.125 12.080 12.040 12.000 10.010 9.995 8.080 8.045 8.005 6.560 6.520 6.490 4.030 4.005 3.990 3.970 10.415 10.340 10.265 10.190 10.130 10.080 10.030 10.000 8.020 7.985 7.960 5.810 5.770 5.750 4.020 4.010 4.000 3.960

in 0.985 0.875 0.760 0.665 0.715 0.630 0.565 0.505 0.430 0.440 0.345 4.910 4.520 4.160 3.820 3.500 3.210 3.035 2.845 2.660 2.470 2.260 2.070 1.890 1.720 1.560 1.440 1.310 1.190 1.090 1.030 0.940 0.860 0.780 0.710 0.855 0.785 0.720 0.645 0.660 0.595 0.530 0.515 0.455 0.385 0.420 0.335 2.955 2.705 2.470 2.250 2.070 1.900 1.735 1.560 1.400 1.250 1.105 0.990 0.900 0.810 0.735 0.670 0.605 0.640 0.575 0.640 0.575 0.515 0.520 0.440 0.380 0.425 0.350 0.265 0.225 1.250 1.120 0.990 0.870 0.770 0.680 0.615 0.560 0.620 0.530 0.435 0.510 0.440 0.360 0.395 0.330 0.270 0.210

14.7 13.1 11.3 9.8 8.4 7.4 6.6 5.9 5.3 4.6 3.8 107.0 97.8 88.9 80.9 73.5 66.9 62.6 58.5 54.4 50.3 45.7 41.6 37.8 34.2 31.0 28.4 25.9 23.4 21.3 19.4 17.7 16.0 14.6 13.2 12.0 10.9 10.0 9.0 7.8 7.1 6.3 5.6 5.0 4.4 3.8 3.3 49.4 44.8 41.0 37.0 33.9 30.9 27.9 25.0 22.4 20.0 17.6 15.6 14.1 12.8 11.6 10.6 9.5 8.5 7.8 7.3 6.6 5.9 5.2 4.4 3.8 3.2 2.8 2.4 2.1 16.5 14.7 12.9 11.3 10.0 8.8 7.9 7.2 6.6 5.7 4.8 4.4 3.8 3.2 2.8 2.5 2.2 1.8

Victor Saouma

bf 2tf 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 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 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 0 0 0 0 0 0 0

hc tw 12.1 13.5 15.6 18.0 16.5 18.7 20.6 23.3 24.1 25.8 28.4 1.9 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.7 4.0 4.4 4.9 5.3 5.8 6.4 6.9 7.7 8.4 8.8 9.7 10.9 11.8 13.0 11.2 12.7 13.7 15.2 15.4 16.8 18.7 19.8 21.5 22.7 24.1 26.7 2.7 3.0 3.2 3.5 3.8 4.1 4.6 5.1 5.6 6.1 6.8 8.0 8.8 9.4 10.3 11.3 12.4 13.5 14.1 13.1 14.5 16.5 18.1 20.9 23.6 20.9 23.1 24.7 27.2 5.2 5.8 6.5 7.4 8.4 9.4 10.6 11.6 11.2 12.5 13.6 14.8 17.0 18.4 17.7 18.4 19.2 23.3

Ix

Sx

rx

Iy

Sy

ry

Zx

Zy

J

in4 76.8 67.2 56.9 48.6 48.7 42.3 37.8 33.1 30.6 27.4 23.5 739.0 622.0 524.0 442.0 375.0 321.0 287.0 257.0 229.0 203.0 176.0 153.0 133.0 116.0 102.0 89.8 80.5 70.2 62.5 57.8 51.7 45.3 40.9 36.4 41.2 36.0 32.6 28.9 27.6 24.9 21.9 23.3 20.9 19.0 17.3 14.8 190.0 162.0 141.0 121.0 106.0 92.1 79.0 67.8 58.5 50.6 43.4 36.3 32.0 28.9 25.8 23.2 20.6 19.1 17.7 18.7 16.6 14.4 16.0 13.5 11.7 11.7 10.1 8.7 7.7 28.6 24.5 20.8 17.4 14.9 12.9 11.1 10.0 10.2 8.8 7.7 9.3 7.9 6.9 6.7 6.1 5.4 4.3

in3 11.4 10.1 8.6 7.4 7.8 6.8 6.1 5.3 5.1 4.6 4.1 95.4 82.1 70.6 60.9 52.7 45.9 41.4 37.6 33.9 30.4 26.7 23.5 20.7 18.2 16.2 14.4 13.0 11.4 10.2 9.6 8.6 7.6 6.9 6.2 7.1 6.3 5.7 5.1 4.9 4.5 4.0 4.2 3.8 3.5 3.3 2.9 31.2 27.0 24.1 20.9 18.5 16.4 14.2 12.3 10.8 9.5 8.2 6.9 6.1 5.6 5.0 4.5 4.1 3.8 3.5 3.8 3.4 3.0 3.2 2.8 2.4 2.6 2.3 2.0 1.8 6.4 5.6 4.8 4.0 3.5 3.0 2.6 2.4 2.5 2.2 1.9 2.2 1.9 1.7 1.7 1.6 1.5 1.2

in 2.28 2.27 2.24 2.22 2.41 2.40 2.39 2.37 2.41 2.45 2.47 2.62 2.52 2.43 2.34 2.26 2.19 2.14 2.10 2.05 2.01 1.96 1.92 1.88 1.84 1.81 1.78 1.76 1.73 1.71 1.73 1.71 1.68 1.67 1.66 1.85 1.82 1.81 1.80 1.88 1.87 1.86 2.04 2.04 2.07 2.12 2.14 1.96 1.90 1.86 1.81 1.77 1.73 1.68 1.65 1.62 1.59 1.57 1.53 1.51 1.50 1.49 1.48 1.47 1.50 1.51 1.60 1.58 1.57 1.76 1.75 1.75 1.90 1.90 1.92 1.92 1.32 1.29 1.27 1.24 1.22 1.21 1.19 1.18 1.24 1.24 1.26 1.45 1.44 1.46 1.54 1.56 1.57 1.57

in4 93.10 81.30 69.20 59.50 21.60 18.60 16.40 14.40 12.20 6.20 4.80 2360.00 2080.00 1840.00 1630.00 1440.00 1280.00 1180.00 1090.00 994.00 903.00 807.00 722.00 645.00 576.00 513.00 466.00 419.00 374.00 338.00 274.00 247.00 223.00 201.00 181.00 74.20 66.90 60.70 53.70 28.80 25.70 22.60 13.30 11.70 9.79 4.45 3.50 593.00 525.00 469.00 414.00 371.00 332.00 295.00 259.00 227.00 199.00 172.00 151.00 135.00 120.00 108.00 97.50 87.20 53.50 47.90 28.20 25.00 22.00 12.20 10.20 8.66 2.33 1.88 1.41 1.18 118.00 103.00 89.30 76.80 66.80 58.10 51.70 46.70 26.70 22.50 18.30 8.35 7.05 5.71 2.15 1.78 1.45 1.09

in3 17.90 15.70 13.40 11.60 6.06 5.26 4.67 4.12 3.50 2.24 1.74 264.00 236.00 211.00 189.00 169.00 152.00 141.00 131.00 121.00 110.00 99.40 89.70 80.70 72.50 65.00 59.30 53.50 48.10 43.70 37.20 33.70 30.60 27.60 25.00 14.60 13.30 12.10 10.70 7.16 6.40 5.65 3.94 3.45 2.91 1.77 1.40 88.60 79.30 71.30 63.60 57.50 51.90 46.50 41.20 36.40 32.10 28.00 24.70 22.20 19.90 17.90 16.20 14.50 10.70 9.58 6.97 6.21 5.51 3.73 3.12 2.67 1.16 0.94 0.71 0.59 22.60 20.00 17.40 15.10 13.20 11.50 10.30 9.34 6.65 5.64 4.60 2.87 2.44 1.99 1.07 0.89 0.72 0.55

in 2.51 2.49 2.47 2.46 1.60 1.59 1.57 1.57 1.52 1.17 1.12 4.69 4.62 4.55 4.49 4.43 4.38 4.34 4.31 4.27 4.24 4.20 4.17 4.13 4.10 4.07 4.05 4.02 4.00 3.98 3.76 3.74 3.73 3.71 3.70 2.48 2.48 2.46 2.45 1.92 1.91 1.89 1.55 1.53 1.49 1.08 1.04 3.47 3.42 3.38 3.34 3.31 3.28 3.25 3.22 3.19 3.16 3.13 3.11 3.09 3.07 3.05 3.04 3.02 2.51 2.48 1.96 1.94 1.93 1.54 1.52 1.51 0.85 0.82 0.77 0.75 2.68 2.65 2.63 2.60 2.59 2.57 2.56 2.54 2.01 1.98 1.94 1.37 1.36 1.33 0.87 0.84 0.81 0.79

in3 20.70 18.10 15.30 13.00 13.80 12.00 10.80 9.43 8.93 8.27 8.12 211.00 182.00 157.00 136.00 117.00 102.00 91.70 82.90 74.40 66.20 57.70 50.40 43.90 38.20 33.40 29.40 26.30 22.80 20.20 18.60 16.50 14.40 12.90 11.50 13.20 11.50 10.40 9.16 8.87 8.00 7.05 7.45 6.74 6.25 5.89 5.20 68.40 59.10 51.90 44.80 39.40 34.50 29.80 25.60 22.00 19.00 16.20 13.60 11.90 10.70 9.49 8.48 7.50 6.97 6.46 6.90 6.12 5.30 5.71 4.83 4.20 4.63 4.11 3.72 3.32 13.40 11.40 9.65 8.06 6.85 5.87 5.05 4.52 4.65 3.99 3.48 4.01 3.39 3.02 3.10 2.90 3.03 2.50

in3 27.40 24.00 20.50 17.70 9.43 8.16 7.23 6.37 5.42 3.52 2.74 408.00 365.00 326.00 292.00 261.00 234.00 217.00 201.00 185.00 169.00 152.00 137.00 123.00 110.00 99.00 90.20 81.40 73.00 66.30 56.60 51.20 46.40 41.80 37.80 22.40 20.30 18.50 16.40 11.00 9.82 8.66 6.07 5.32 4.49 2.77 2.19 137.00 122.00 110.00 97.90 88.40 79.70 71.30 63.00 55.60 49.00 42.70 37.50 33.70 30.20 27.20 24.60 22.00 16.30 14.60 10.70 9.50 8.41 5.73 4.78 4.08 1.83 1.49 1.13 0.95 34.60 30.50 26.50 22.90 20.00 17.50 15.70 14.20 10.10 8.59 7.01 4.42 3.75 3.05 1.68 1.40 1.15 0.87

in4 3.85 2.72 1.78 1.19 1.10 0.76 0.65 0.40 0.27 0.23 0.13 714.00 555.00 430.00 331.00 255.00 196.00 164.00 135.00 110.00 88.30 67.50 51.80 39.30 29.60 22.20 17.30 13.20 9.84 7.56 6.13 4.67 3.55 2.68 2.03 2.53 1.94 1.51 1.10 0.97 0.73 0.52 0.40 0.28 0.19 0.18 0.10 120.00 92.00 70.90 53.50 41.60 32.20 24.40 17.70 12.80 9.22 6.43 4.55 3.42 2.54 1.92 1.46 1.09 1.05 0.79 0.89 0.66 0.48 0.37 0.23 0.15 0.15 0.09 0.05 0.04 7.50 5.41 3.75 2.55 1.78 1.23 0.91 0.69 0.75 0.49 0.29 0.31 0.20 0.12 0.12 0.08 0.05 0.03

Structural Engineering

3.6 Steel Section Properties

Draft Designation

3–15 bf 2tf

A

d

tw

bf

tf

in2 9.8 8.6 7.1 5.9 5.1 4.6 4.1 3.5 3.1 2.6 2.2 1.9 1.5 3.7 2.9 2.2 2.4 1.8 1.3 2.8 2.3 1.9

in 4.50 4.38 4.25 4.13 4.06 4.00 4.03 3.96 4.14 4.07 4.05 3.99 3.94 3.19 3.10 2.99 3.14 3.02 2.95 2.58 2.51 2.08

in 0.570 0.510 0.400 0.360 0.310 0.285 0.285 0.245 0.250 0.230 0.245 0.230 0.170 0.320 0.260 0.230 0.260 0.230 0.170 0.270 0.240 0.280

in 8.280 8.220 8.110 8.070 8.020 7.995 6.535 6.495 5.270 5.250 4.015 4.000 3.940 6.080 6.020 5.990 4.030 4.000 3.940 5.030 5.000 4.060

in 0.935 0.810 0.685 0.560 0.495 0.435 0.465 0.400 0.400 0.330 0.315 0.255 0.205 0.455 0.365 0.260 0.405 0.280 0.215 0.430 0.360 0.345

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

9. 6. 5. 3. 2. 9.

2.5 1.7 1.3 1.0 0.6 2.8

7.00 6.00 5.00 4.00 3.00 2.50

0.215 0.177 0.157 0.135 0.114 0.316

4.000 3.065 2.690 2.281 1.844 5.003

0.270 0.225 0.206 0.189 0.171 0.416

ST12.x 61. ST12.x 53. ST12.x 50. ST12.x 45. ST12.x 40. ST10.x 48. ST10.x 43. ST10.x 38. ST10.x 33. ST 9.x 35. ST 9.x 27. ST 8.x 25. ST 8.x 21. ST 6.x 25. ST 6.x 20. ST 6.x 18. ST 6.x 16. ST 5.x 18. ST 5.x 13. ST 4.x 12. ST 4.x 9. ST 4.x 10. ST 4.x 8. ST 3.x 9. ST 3.x 6. ST 3.x 7. ST 3.x 5. ST 2.x 5. ST 2.x 4. ST 2.x 4. ST 2.x 3.

17.8 15.6 14.7 13.2 11.7 14.1 12.7 11.0 9.7 10.3 8.0 7.3 6.3 7.3 6.0 5.2 4.7 5.2 3.7 3.4 2.7 2.9 2.3 2.5 1.8 2.2 1.5 1.4 1.1 1.1 0.8

12.25 12.25 12.00 12.00 12.00 10.15 10.15 10.00 10.00 9.00 9.00 7.50 7.50 6.00 6.00 6.00 6.00 5.00 5.00 4.00 4.00 3.50 3.50 3.00 3.00 2.50 2.50 2.00 2.00 1.50 1.50

0.800 0.620 0.745 0.625 0.500 0.800 0.660 0.635 0.505 0.711 0.461 0.550 0.411 0.687 0.462 0.428 0.350 0.594 0.311 0.441 0.271 0.450 0.252 0.465 0.232 0.494 0.214 0.326 0.193 0.349 0.170

8.050 7.870 7.245 7.125 7.000 7.200 7.060 6.385 6.225 6.251 6.001 5.640 5.501 5.477 5.252 5.078 5.000 4.944 4.661 4.171 4.001 3.860 3.662 3.565 3.332 3.284 3.004 2.796 2.663 2.509 2.330

HP14.x117. HP14.x102. HP14.x 89. HP14.x 73. HP13.x100. HP13.x 87. HP13.x 73. HP13.x 60. HP12.x 84. HP12.x 74. HP12.x 63. HP12.x 53. HP10.x 57. HP10.x 42. HP 8.x 36.

34.4 30.0 26.1 21.4 29.4 25.5 21.6 17.5 24.6 21.8 18.4 15.5 16.8 12.4 10.6

14.21 14.01 13.83 13.61 13.15 12.95 12.75 12.54 12.28 12.13 11.94 11.78 9.99 9.70 8.02

0.805 0.705 0.615 0.505 0.765 0.665 0.565 0.460 0.685 0.605 0.515 0.435 0.565 0.415 0.445

14.885 14.785 14.695 14.585 13.205 13.105 13.005 12.900 12.295 12.215 12.125 12.045 10.225 10.075 8.155

WT 4.x 34. WT 4.x 29. WT 4.x 24. WT 4.x 20. WT 4.x 18. WT 4.x 16. WT 4.x 14. WT 4.x 12. WT 4.x 11. WT 4.x 9. WT 4.x 8. WT 4.x 7. WT 4.x 5. WT 3.x 13. WT 3.x 10. WT 3.x 8. WT 3.x 8. WT 3.x 6. WT 3.x 5. WT 3.x 10. WT 3.x 8. WT 2.x 7. MT MT MT MT MT MT

7.x 6.x 5.x 4.x 3.x 3.x

Victor Saouma

hc tw

Ix

Sx

rx

Iy

Sy

ry

Zx

Zy

J

5.6 6.2 7.9 8.8 10.2 11.1 11.1 12.9 13.8 15.0 14.0 15.0 20.2 7.8 9.6 10.8 9.6 10.8 14.6 7.0 7.9 5.3

in4 10.9 9.1 6.8 5.7 4.8 4.3 4.2 3.5 3.9 3.4 3.3 2.9 2.2 2.3 1.8 1.4 1.7 1.3 0.9 1.0 0.9 0.5

in3 3.0 2.6 2.0 1.7 1.4 1.3 1.3 1.1 1.2 1.0 1.1 1.0 0.7 0.9 0.7 0.6 0.7 0.6 0.4 0.5 0.4 0.3

in 1.05 1.03 0.99 0.99 0.97 0.97 1.01 1.00 1.12 1.14 1.22 1.23 1.20 0.79 0.77 0.80 0.84 0.86 0.84 0.61 0.60 0.52

in4 44.30 37.50 30.50 24.50 21.30 18.50 10.80 9.14 4.89 3.98 1.70 1.37 1.05 8.53 6.64 4.66 2.21 1.50 1.10 4.56 3.75 1.93

in3 10.70 9.13 7.52 6.08 5.31 4.64 3.31 2.81 1.85 1.52 0.85 0.68 0.53 2.81 2.21 1.56 1.10 0.75 0.56 1.82 1.50 0.95

in 2.12 2.10 2.08 2.04 2.03 2.02 1.62 1.61 1.26 1.23 0.88 0.84 0.84 1.52 1.50 1.45 0.97 0.92 0.90 1.28 1.27 1.00

in3 6.29 5.25 3.94 3.25 2.71 2.39 2.38 1.98 2.11 1.86 1.91 1.74 1.27 1.68 1.29 1.03 1.25 1.01 0.72 0.97 0.80 0.62

in3 16.30 13.90 11.40 9.25 8.06 7.04 5.05 4.29 2.84 2.33 1.33 1.08 0.83 4.28 3.36 2.37 1.70 1.16 0.86 2.76 2.29 1.46

in4 2.52 1.66 0.98 0.56 0.38 0.27 0.27 0.17 0.14 0.09 0.07 0.04 0.02 0.23 0.12 0.05 0.11 0.05 0.02 0.15 0.09 0.08

0 0 0 0 0 0

30.1 31.3 29.2 26.9 23.5 5.6

13.1 6.6 3.5 1.6 0.6 1.0

2.7 1.6 1.0 0.6 0.3 0.5

2.27 1.95 1.62 1.28 0.94 0.62

1.32 0.49 0.31 0.17 0.08 3.93

0.66 0.32 0.23 0.15 0.09 1.57

0.72 0.53 0.48 0.42 0.36 1.19

4.88 2.89 1.81 1.01 0.52 1.03

1.16 0.58 0.41 0.26 0.16 2.66

0.06 0.03 0.02 0.01 0.01 0.17

1.090 1.090 0.870 0.870 0.870 0.920 0.920 0.795 0.795 0.691 0.691 0.622 0.622 0.659 0.659 0.545 0.544 0.491 0.491 0.425 0.425 0.392 0.392 0.359 0.359 0.326 0.326 0.293 0.293 0.260 0.260

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 0 0

13.2 17.0 14.1 16.8 21.1 10.8 13.1 13.6 17.0 10.9 16.8 11.6 15.5 7.0 10.3 11.7 14.3 6.9 13.2 7.3 11.8 6.1 10.9 5.0 10.0 3.8 8.7 4.3 7.3 2.8 5.7

259.0 216.0 215.0 190.0 162.0 143.0 125.0 109.0 93.1 84.7 62.4 40.6 33.0 25.2 18.9 17.2 14.9 12.5 7.8 5.0 3.5 3.4 2.2 2.1 1.3 1.3 0.7 0.5 0.3 0.2 0.1

30.1 24.1 26.3 22.6 18.7 20.3 17.2 15.8 12.9 14.0 9.6 7.7 6.0 6.1 4.3 4.0 3.3 3.6 2.1 1.8 1.1 1.4 0.8 1.0 0.6 0.7 0.4 0.3 0.2 0.2 0.1

3.82 3.72 3.83 3.79 3.72 3.18 3.14 3.15 3.10 2.87 2.79 2.35 2.29 1.85 1.78 1.83 1.78 1.56 1.45 1.22 1.14 1.07 0.99 0.92 0.83 0.76 0.68 0.58 0.53 0.43 0.38

41.70 38.50 23.80 22.50 21.10 25.10 23.40 14.90 13.80 12.10 10.40 7.85 7.19 7.85 6.78 4.94 4.68 4.18 3.39 2.15 1.86 1.59 1.32 1.15 0.91 0.83 0.61 0.45 0.38 0.29 0.23

10.40 9.80 6.58 6.31 6.04 6.97 6.63 4.66 4.43 3.86 3.47 2.78 2.61 2.87 2.58 1.95 1.87 1.69 1.46 1.03 0.93 0.82 0.72 0.65 0.55 0.51 0.41 0.32 0.29 0.23 0.19

1.53 1.57 1.27 1.30 1.34 1.33 1.36 1.16 1.19 1.08 1.14 1.03 1.07 1.03 1.06 0.98 1.00 0.90 0.95 0.80 0.83 0.73 0.77 0.68 0.70 0.62 0.64 0.57 0.58 0.52 0.52

54.50 43.30 47.50 41.10 33.60 36.90 31.10 28.60 23.40 25.10 17.30 14.00 10.80 11.00 7.71 7.12 5.94 6.58 3.70 3.19 2.07 2.47 1.48 1.85 1.01 1.34 0.65 0.59 0.38 0.35 0.20

18.10 16.60 12.00 11.20 10.40 12.50 11.60 8.37 7.70 7.21 6.07 5.01 4.54 5.19 4.45 3.41 3.22 3.11 2.49 1.84 1.59 1.48 1.23 1.18 0.93 0.94 0.69 0.57 0.48 0.41 0.33

6.38 5.04 3.76 3.01 2.43 4.15 3.30 2.28 1.78 2.05 1.18 1.05 0.77 1.39 0.87 0.54 0.45 0.63 0.30 0.27 0.17 0.22 0.12 0.18 0.08 0.16 0.06 0.06 0.04 0.04 0.02

0.805 0.705 0.615 0.505 0.765 0.665 0.565 0.460 0.685 0.610 0.515 0.435 0.565 0.420 0.445

9.2 10.5 11.9 14.4 8.6 9.9 11.5 14.0 9.0 10.0 11.8 13.8 9.0 12.0 9.2

14.2 16.2 18.5 22.6 13.6 15.7 18.4 22.7 14.2 16.0 18.9 22.3 13.9 18.9 14.2

1220.0 1050.0 904.0 729.0 886.0 755.0 630.0 503.0 650.0 569.0 472.0 393.0 294.0 210.0 119.0

172.0 150.0 131.0 107.0 135.0 117.0 98.8 80.3 106.0 93.8 79.1 66.8 58.8 43.4 29.8

5.96 5.92 5.88 5.84 5.49 5.45 5.40 5.36 5.14 5.11 5.06 5.03 4.18 4.13 3.36

443.00 380.00 326.00 261.00 294.00 250.00 207.00 165.00 213.00 186.00 153.00 127.00 101.00 71.70 40.30

59.50 51.40 44.30 35.80 44.50 38.10 31.90 25.50 34.60 30.40 25.30 21.10 19.70 14.20 9.88

3.59 3.56 3.53 3.49 3.16 3.13 3.10 3.07 2.94 2.92 2.88 2.86 2.45 2.41 1.95

194.00 169.00 146.00 118.00 153.00 131.00 110.00 89.00 120.00 105.00 88.30 74.00 66.50 48.30 33.60

91.40 78.80 67.70 54.60 68.60 58.50 48.80 39.00 53.20 46.60 38.70 32.20 30.30 21.80 15.20

8.02 5.40 3.60 2.01 6.25 4.12 2.54 1.39 4.24 2.98 1.83 1.12 1.97 0.81 0.77

Structural Engineering

3–16

Draft

Designation

L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L L

8.0x8.0x1.125 8.0x8.0x1.000 8.0x8.0x0.875 8.0x8.0x0.750 8.0x8.0x0.625 8.0x8.0x0.563 8.0x8.0x0.500 8.0x6.0x1.000 8.0x6.0x0.875 8.0x6.0x0.750 8.0x6.0x0.625 8.0x6.0x0.563 8.0x6.0x0.500 8.0x6.0x0.438 8.0x4.0x1.000 8.0x4.0x0.750 8.0x4.0x0.563 8.0x4.0x0.500 7.0x4.0x0.750 7.0x4.0x0.625 7.0x4.0x0.500 7.0x4.0x0.375 6.0x6.0x1.000 6.0x6.0x0.875 6.0x6.0x0.750 6.0x6.0x0.625 6.0x6.0x0.563 6.0x6.0x0.500 6.0x6.0x0.438 6.0x6.0x0.375 6.0x6.0x0.313 6.0x4.0x0.875 6.0x4.0x0.750 6.0x4.0x0.625 6.0x4.0x0.563 6.0x4.0x0.500 6.0x4.0x0.438 6.0x4.0x0.375 6.0x4.0x0.313 6.0x3.5x0.500 6.0x3.5x0.375 6.0x3.5x0.313 5.0x5.0x0.875 5.0x5.0x0.750 5.0x5.0x0.625 5.0x5.0x0.500 5.0x5.0x0.438 5.0x5.0x0.375 5.0x5.0x0.313 5.0x3.5x0.750 5.0x3.5x0.625 5.0x3.5x0.500 5.0x3.5x0.438 5.0x3.5x0.375 5.0x3.5x0.313 5.0x3.5x0.250 5.0x3.0x0.625 5.0x3.0x0.500 5.0x3.0x0.438 5.0x3.0x0.375 5.0x3.0x0.313 5.0x3.0x0.250 4.0x4.0x0.750 4.0x4.0x0.625 4.0x4.0x0.500 4.0x4.0x0.438 4.0x4.0x0.375 4.0x4.0x0.313 4.0x4.0x0.250 4.0x3.5x0.500 4.0x3.5x0.438 4.0x3.5x0.375 4.0x3.5x0.313 4.0x3.5x0.250 4.0x3.0x0.500 4.0x3.0x0.438 4.0x3.0x0.375 4.0x3.0x0.313 4.0x3.0x0.250 3.5x3.5x0.500 3.5x3.5x0.438 3.5x3.5x0.375 3.5x3.5x0.313 3.5x3.5x0.250 3.5x3.0x0.500 3.5x3.0x0.438 3.5x3.0x0.375 3.5x3.0x0.313 3.5x3.0x0.250 3.5x2.5x0.500 3.5x2.5x0.438 3.5x2.5x0.375 3.5x2.5x0.313 3.5x2.5x0.250 3.0x3.0x0.500 3.0x3.0x0.438 3.0x3.0x0.375 3.0x3.0x0.313 3.0x3.0x0.250 3.0x3.0x0.188 3.0x2.5x0.500 3.0x2.5x0.438 3.0x2.5x0.375 3.0x2.5x0.313 3.0x2.5x0.250 3.0x2.5x0.188 3.0x2.0x0.500 3.0x2.0x0.438 3.0x2.0x0.375 3.0x2.0x0.313 3.0x2.0x0.250 3.0x2.0x0.188 2.5x2.5x0.500 2.5x2.5x0.375 2.5x2.5x0.313 2.5x2.5x0.250 2.5x2.5x0.188 2.5x2.0x0.375 2.5x2.0x0.313

Victor Saouma

STRUCTURAL MATERIALS A in2 16.70 15.00 13.20 11.40 9.61 8.68 7.75 13.00 11.50 9.94 8.36 7.56 6.75 5.93 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 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 2.48 2.09 1.69 3.25 2.87 2.48 2.09 1.69 3.00 2.65 2.30 1.93 1.56 2.75 2.43 2.11 1.78 1.44 2.75 2.43 2.11 1.78 1.44 1.09 2.50 2.21 1.92 1.62 1.31 1.00 2.25 2.00 1.73 1.46 1.19 0.90 2.25 1.73 1.46 1.19 0.90 1.55 1.31

wgt k/f t 56.90 51.00 45.00 38.90 32.70 29.60 26.40 44.20 39.10 33.80 28.50 25.70 23.00 20.20 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 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 8.50 7.20 5.80 11.10 9.80 8.50 7.20 5.80 10.20 9.10 7.90 6.60 5.40 9.40 8.30 7.20 6.10 4.90 9.40 8.30 7.20 6.10 4.90 3.71 8.50 7.60 6.60 5.60 4.50 3.39 7.70 6.80 5.90 5.00 4.10 3.07 7.70 5.90 5.00 4.10 3.07 5.30 4.50

Ix in4 98.0 89.0 79.6 69.7 59.4 54.1 48.6 80.8 72.3 63.4 54.1 49.3 44.3 39.2 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 17.8 15.7 13.6 11.3 10.0 8.7 7.4 13.9 12.0 10.0 8.9 7.8 6.6 5.4 11.4 9.4 8.4 7.4 6.3 5.1 7.7 6.7 5.6 5.0 4.4 3.7 3.0 5.3 4.8 4.2 3.6 2.9 5.1 4.5 4.0 3.4 2.8 3.6 3.3 2.9 2.5 2.0 3.5 3.1 2.7 2.3 1.9 3.2 2.9 2.6 2.2 1.8 2.2 2.0 1.8 1.5 1.2 1.0 2.1 1.9 1.7 1.4 1.2 0.9 1.9 1.7 1.5 1.3 1.1 0.8 1.2 1.0 0.8 0.7 0.5 0.9 0.8

Sx in3 17.5 15.8 14.0 12.2 10.3 9.3 8.4 15.1 13.4 11.7 9.9 8.9 8.0 7.1 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 5.2 4.5 3.9 3.2 2.8 2.4 2.0 4.3 3.7 3.0 2.6 2.3 1.9 1.6 3.5 2.9 2.6 2.2 1.9 1.5 2.8 2.4 2.0 1.8 1.5 1.3 1.0 1.9 1.7 1.5 1.3 1.0 1.9 1.7 1.5 1.2 1.0 1.5 1.3 1.1 1.0 0.8 1.5 1.3 1.1 1.0 0.8 1.4 1.3 1.1 0.9 0.8 1.1 1.0 0.8 0.7 0.6 0.4 1.0 0.9 0.8 0.7 0.6 0.4 1.0 0.9 0.8 0.7 0.5 0.4 0.7 0.6 0.5 0.4 0.3 0.5 0.5

rx in 2.42 2.44 2.45 2.47 2.49 2.50 2.50 2.49 2.51 2.53 2.54 2.55 2.56 2.57 2.52 2.55 2.58 2.59 2.22 2.24 2.25 2.27 1.80 1.81 1.83 1.84 1.85 1.86 1.87 1.88 1.89 1.86 1.88 1.90 1.90 1.91 1.92 1.93 1.94 1.92 1.94 1.95 1.49 1.51 1.52 1.54 1.55 1.56 1.57 1.55 1.56 1.58 1.59 1.60 1.61 1.62 1.57 1.59 1.60 1.61 1.61 1.62 1.19 1.20 1.22 1.23 1.23 1.24 1.25 1.23 1.24 1.25 1.26 1.27 1.25 1.25 1.26 1.27 1.28 1.06 1.07 1.07 1.08 1.09 1.07 1.08 1.09 1.10 1.11 1.09 1.09 1.10 1.11 1.12 0.90 0.90 0.91 0.92 0.93 0.94 0.91 0.92 0.93 0.94 0.94 0.95 0.92 0.93 0.94 0.95 0.96 0.97 0.74 0.75 0.76 0.77 0.78 0.77 0.78

Iy in4 98.00 89.00 79.60 69.70 59.40 54.10 48.60 38.80 34.90 30.70 26.30 24.00 21.70 19.30 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 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 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

Sy in3 17.50 15.80 14.00 12.20 10.30 9.34 8.36 8.92 7.94 6.92 5.88 5.34 4.79 4.23 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 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 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

ry in 2.42 2.44 2.45 2.47 2.49 2.50 2.50 1.73 1.74 1.76 1.77 1.78 1.79 1.80 1.03 1.05 1.07 1.08 1.09 1.10 1.11 1.13 1.80 1.81 1.83 1.84 1.85 1.86 1.87 1.88 1.89 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.17 0.97 0.99 1.00 1.49 1.51 1.52 1.54 1.55 1.56 1.57 0.98 0.99 1.01 1.01 1.02 1.03 1.04 0.81 0.83 0.84 0.85 0.85 0.86 1.19 1.20 1.22 1.23 1.23 1.24 1.25 1.04 1.05 1.06 1.07 1.07 0.86 0.87 0.88 0.89 0.90 1.06 1.07 1.07 1.08 1.09 0.88 0.89 0.90 0.90 0.91 0.70 0.71 0.72 0.73 0.74 0.90 0.90 0.91 0.92 0.93 0.94 0.72 0.73 0.74 0.74 0.75 0.76 0.55 0.55 0.56 0.57 0.57 0.58 0.74 0.75 0.76 0.77 0.78 0.58 0.58

Zx in3 31.60 28.50 25.30 22.00 18.60 16.80 15.10 27.30 24.20 21.10 17.90 16.20 14.50 12.80 24.30 18.90 14.50 13.00 14.80 12.60 10.30 7.87 15.50 13.80 12.00 10.20 9.26 8.31 7.34 6.35 5.35 12.70 11.20 9.51 8.66 7.78 6.88 5.97 5.03 7.50 5.76 4.85 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 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

Zy in3 31.60 28.50 25.30 22.00 18.60 16.80 15.10 16.20 14.40 12.50 10.50 9.52 8.51 7.50 7.72 5.81 4.38 3.90 5.65 4.74 3.83 2.90 15.50 13.80 12.00 10.20 9.26 8.31 7.34 6.35 5.35 6.31 5.47 4.62 4.19 3.75 3.30 2.85 2.40 2.91 2.20 1.85 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 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

J in4 7.13 5.08 3.46 2.21 1.30 0.96 0.68 4.35 2.96 1.90 1.12 0.82 0.58 0.40 3.68 1.61 0.70 0.50 1.47 0.87 0.46 0.20 3.68 2.51 1.61 0.95 0.70 0.50 0.34 0.22 0.13 2.07 1.33 0.79 0.58 0.42 0.28 0.18 0.11 0.40 0.17 0.10 2.07 1.33 0.79 0.42 0.28 0.18 0.11 1.11 0.66 0.35 0.24 0.15 0.09 0.05 0.61 0.32 0.22 0.14 0.08 0.04 1.02 0.61 0.32 0.22 0.14 0.08 0.04 0.30 0.21 0.13 0.08 0.04 0.28 0.19 0.12 0.07 0.04 0.28 0.19 0.12 0.07 0.04 0.26 0.18 0.11 0.07 0.04 0.23 0.16 0.10 0.06 0.03 0.23 0.16 0.10 0.06 0.03 0.01 0.21 0.15 0.09 0.06 0.03 0.01 0.19 0.13 0.09 0.05 0.03 0.01 0.19 0.08 0.05 0.03 0.01 0.07 0.04

Structural Engineering

3.6 Steel Section Properties

Draft 3.6.1

3–17

ASCII File with Steel Section Properties

Available is an ASCII ﬁle (PC/DOS or Sun/UNIX (/usr/src/public/aisc) containing the following section properties (All dimensions are inches and kips). A Area BF Width of ﬂange BTF ”bf/2tf” the ratio of the ﬂange width to twice the ﬂange thickness CW Cw, warping constant D Depth, actual DAF ”d/Af” the ratio of the depth to the compression ﬂange area DI Inner diameter of pipe DN Nominal depth DSG Shape designation i.e. ”W” or ”C” DTW ”d/tw” the ratio of the depth to the web thickness EO ”eo” distance from outside face of web to shear center FYP ”Fy’” the theoretical maximum yield stress based on the width-thickness ratio of 1/2 the unstiﬀened compression ﬂange, beyond which a particular shape is not compact. (Speciﬁed as 0 if greater than 65 ksi) FYPPPA ”Fy”’” (ASD) the theoretical maximum yield stress based on the depth-thickness ratio of the web below which a particular shape may be considered compact for any condition of combined bending and axial stresses. (Speciﬁed as 0 if greater than 65 ksi) FYPPPL ”Fy”’” (LRFD) see above H ”H” ﬂexural constant LRFD Formula (A-E3-9) H38 ”H” ﬂexural constant, double angles 3/8” back to back H34 ”H” ﬂexural constant, double angles 3/4” back to back HL Horizontal leg or side dimension HTW ”hc/tw”, ratio of assumed web depth for stability to thickess of web JSHP JUMBO SHAPE (”J”) see note below NT Tensile group number per ASTM A6 NX1 ”X1” beam buckling factor LRFD Formula (F1-8) OD Outer diameter of pipe QF ”Qf” statical moment at point in ﬂange QW ”Qw” statical moment at mid-depth

Victor Saouma

Structural Engineering

3–18

Draft RA RI RO RO38 RO34 RT

RX RY RY38 RY34 RZ SX SY SW T TF TW VL WGT WNO XB XI XJ XK X2 YB YI ZX ZY

STRUCTURAL MATERIALS

Minimum ﬁllet radius, detailing value (in.); ﬂange toe value for M, S, C, MC shapes Minimum ﬁllet radius, design ”ro” shear center coordinate ”ro” double angles 3/8” back to back ”ro” double angles 3/4” back to back The radius of gyration of a section comprising the compression ﬂange plus 1/3 of the compression web area, taken about an axis in the plane of the web ”rx” radius of gyration about axis XX ”ry” radius of gyration about axis YY ”ry” double angles 3/8” back to back ”ry” double angles 3/4” back to back ”rz” radius of gyration about principal axis ”Sx” elastic section modulus about axis XX ”Sy” elastic section modulus about axis YY ”Sw” warping statical moment Thickness of leg or wall Thickness of ﬂange Thickness of web Vertical leg or side dimension Nominal weight ”Wno” normalized warping function ”x” centroid of section in X direction from outer web ”Ix” moment of inertia about axis XX ”J” torsional constant ”k” distance from outer face of ﬂange to web toe ﬁllet of rolled shape ”X2” beam buckling factor LRFD Formula (F1-9) ”y” centroid of section in Y direction from outer top ﬂange face ”Iy” moment of inertia about axis YY ”Zx” plastic modulus about axis XX ”Zy” plastic modulus about axis YY

program form c c c c c c c

test program to open the AISC1.PRN file, read a single record and print the data from that record to the screen. aisc1.bin contains contains 658 records (sections) with 38 variables in each record. character*2 dsg, jshp real dn, wgt, a, d, tw, bf, tf, xk, 1 btf, fyp, htw, dtw, fypppl, fypppa, x1, x2, rt, daf, 2 ri, ra, nt, xi, sx, rx, yi, sy, ry, zx, zy, xj, cw, 3 wno, sw, qf, qw, ro

c c c

c c c 100

c c c

--------------------------------open aisc file --------------------------------open(unit=1,file=’aisc1.bin’, 1 status=’old’,access=’direct’,recl=152) ----------------------------------prompt user to select record (line) ----------------------------------write(*,*) ’ Enter line number to read: (0 to quit) >’ read(*,*) line if(line .eq. 0) stop if(line.gt.658)then write(*,*)’ Invalid line number, must be less than 658’ goto 100 endif -----------------------------------read a single record from aisc file ------------------------------------

Victor Saouma

Structural Engineering

3.7 Joists

3–19

Draft

read(1,rec=line) dsg, dn, wgt, jshp, a, d, tw, bf, tf, xk, btf, fyp, htw, dtw, fypppl, fypppa, x1, x2, rt, ri, ra, nt, xi, sx, rx, yi, sy, ry, zx, zy, xj, wno, sw, qf, qw, ro c ------------------------c write data to screen c ------------------------write(*,999,err=1000) dsg, dn, wgt, jshp, a, d, tw, bf, tf, 1 btf, fyp, htw, dtw, fypppl, fypppa, x1, x2, rt, 2 ri, ra, nt, xi, sx, rx, yi, sy, ry, zx, zy, xj, 3 wno, sw, qf, qw, ro c -----c goto beginning c -----goto 100 c--------------------------------------------------------c error message if read attempt is past last record (658) c--------------------------------------------------------1000 write(*,*) ’ line number is past last record (658) ’ goto 100 999 format(a2,2x,f3.0,2x,f4.0,a2,6(2x,1pe10.3)) 1 2 3

daf, cw,

xk, daf, cw,

end

3.7 41

Joists

Table 3.3 list the Joist properties.

Victor Saouma

Structural Engineering

3–20

STRUCTURAL MATERIALS

4"

4"

Span

[Design Length = Span – 0.33 FT.]

4"

Draft

Figure 3.9: prefabricated Steel Joists

Victor Saouma

Structural Engineering

3.7 Joists

3–21

Draft Joint 8K1 Desig. Depth 8 (in.) ≈ W 5.1 (lbs/ft) Span (ft.) 8 550 550 9 550 550 10 550 480 11 532 377 12 444 288 13 377 225 14 324 179 15 281 145 16 246 119 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

10K1 12K1 12K3 12K5 14K1 14K3 14K4 14K6 16K2 16K3 16K4 16K5 16K6 16K7 16K9 10

12

12

12

14

14

14

14

16

16

16

16

16

16

16

5

5

5.7

7.1

5.2

6

6.7

7.7

5.5

6.3

7

7.5

8.1

8.6

10.0

550 550 550 510 500 425 434 344 380 282 336 234 299 197 268 167 241 142 218 123 199 106 181 93 166 81

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

550 550 511 475 448 390 395 324 352 272 315 230 284 197 257 170 234 147 214 128 196 113 180 100 166 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

550 550 550 507 550 467 550 443 550 408 550 383 525 347 475 299 432 259 395 226 362 199 334 175 308 56 285 139 265 124

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

550 550 550 526 550 490 550 455 550 426 550 406 550 385 507 339 465 298 428 263 395 233 366 208 340 186 317 167 296 151 277 137 259 124

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

550 550 550 542 550 455 479 363 412 289 358 234 313 192 277 159 246 134 221 113 199 97

Table 3.3: Joist Properties

Victor Saouma

Structural Engineering

3–22

Draft

STRUCTURAL MATERIALS

Victor Saouma

Structural Engineering

Draft Chapter 4

EQUILIBRIUM & REACTIONS To every action there is an equal and opposite reaction. Newton’s third law of motion

4.1

Introduction

1 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; ﬁnally, deformations (deﬂections and rotations) are determined last1 .

3

Reactions are necessary to determine foundation load.

4

Depending on the type of structures, there can be diﬀerent types of support conditions, Fig. 4.1.

Figure 4.1: Types of Supports 1 This

is the sequence of operations in the ﬂexibility method which lends itself to hand calculation. In the stiﬀness method, we determine displacements ﬁrsts, then internal forces and reactions. This method is most suitable to computer implementation.

4–2

EQUILIBRIUM & REACTIONS

Draft

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.

4.2

Equilibrium

5

Reactions are determined from the appropriate equations of static equilibrium.

6

Summation of forces and moments, in a static system must be equal to zero2 .

7

8

In a 3D cartesian coordinate system there are a total of 6 independent equations of equilibrium: ΣFx = ΣFy = ΣFz = 0 (4.1) ΣMx = ΣMy = ΣMz = 0 In a 2D cartesian coordinate system there are a total of 3 independent equations of equilibrium:

ΣFx

9

10

=

ΣFy

= ΣMz

=

(4.2)

0

For reaction calculations, the externally applied load may be reduced to an equivalent force3 . Summation of the moments can be taken with respect to any arbitrary point.

Whereas forces are represented by a vector, moments are also vectorial quantities and are represented by a curved arrow or a double arrow vector.

11

12

Not all equations are applicable to all structures, Table 4.1 Structure Type Beam, no axial forces 2D Truss, Frame, Beam Grid 3D Truss, Frame Beams, no axial Force 2 D Truss, Frame, Beam

Equations ΣFx

ΣFy ΣFy

ΣFz ΣFx ΣFy ΣFz Alternate Set ΣMzA ΣMzB ΣFx ΣMzA ΣMzB A ΣMz ΣMzB ΣMzC

ΣMz ΣMz ΣMx ΣMx

ΣMy ΣMy

ΣMz

Table 4.1: Equations of Equilibrium The three conventional equations of equilibrium in 2D: ΣFx , ΣFy and ΣMz can be replaced by the independent moment equations ΣMzA , ΣMzB , ΣMzC provided that A, B, and C are not colinear.

13

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 2 In

a dynamic system ΣF = ma where m is the mass and a is the acceleration. for internal forces (shear and moment) we must use the actual load distribution.

3 However

Victor Saouma

Structural Engineering

4.3 Equations of Conditions

4–3

Draft

If your reaction is negative, then it will be in a direction opposite from the one assumed. Summation of all external forces (including reactions) is not necessarily zero (except at hinges and at points outside the structure).

16

Summation of external forces is equal and opposite to the internal ones. Thus the net force/moment is equal to zero.

17

18

The external forces give rise to the (non-zero) shear and moment diagram.

4.3

Equations of Conditions

If a structure has an internal hinge (which may connect two or more substructures), then this will provide an additional equation (ΣM = 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. 4.2.

21

Rx Sy = Ry Sx

(4.3)

Figure 4.2: Inclined Roller Support

4.4

Static Determinacy

In statically determinate structures, reactions depend only on the geometry, boundary conditions and loads.

22

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

24 the degree of static indeterminacy is equal to the diﬀerence between the number of reactions and the number of equations of equilibrium, Fig. 4.3.

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 structures4 , 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).

26

4 Which

will be studied in CVEN3535.

Victor Saouma

Structural Engineering

4–4

EQUILIBRIUM & REACTIONS

Draft

Figure 4.3: Examples of Static Determinate and Indeterminate Structures Example 4-1: Statically Indeterminate Cable Structure A rigid plate is supported by two aluminum cables and a steel one. Determine the force in each cable5 .

If the rigid plate supports a load P, determine the stress in each of the three cables. Solution: 1. We have three unknowns and only two independent equations of equilibrium. Hence the problem is statically indeterminate to the ﬁrst degree. right left = PAl ΣMz = 0; ⇒ PAl ΣFy = 0; ⇒ 2PAl + PSt = P

Thus we eﬀectively have two unknowns and one equation. 2. We need to have a third equation to solve for the three unknowns. This will be derived from the compatibility of the displacements in all three cables, i.e. all three displacements must be equal:  σ = PA  PL ⇒ ∆L = ε = ∆L L  AE σ ε =indeterminate 5 This example problem will be the only statically E problem analyzed in CVEN3525. Victor Saouma

Structural Engineering

4.5 Geometric Instability

Draft

4–5 PAl L PSt L PAl (EA)Al = ⇒ = EAl AAl ESt ASt PSt (EA)St ∆Al

∆St

or −(EA)St PAl + (EA)Al PSt = 0 3. Solution of this system of two equations with two unknowns yield: 2 1 PAl P = −(EA)St (EA)Al PSt 0 −1 PAl 2 1 P ⇒ = PSt −(EA) St (EA)Al 0 1 P (EA)Al −1 = 0 2(EA)Al + (EA)St (EA)St 2

Determinant

4.5

Geometric Instability

The stability of a structure is determined not only by the number of reactions but also by their arrangement.

27

28

Geometric instability will occur if: 1. All reactions are parallel and a non-parallel load is applied to the structure. 2. All reactions are concurrent, Fig. ??.

Figure 4.4: Geometric Instability Caused by Concurrent Reactions 3. The number of reactions is smaller than the number of equations of equilibrium, that is a mechanism is present in the structure. Mathematically, this can be shown if the determinant of the equations of equilibrium is equal to zero (or the equations are inter-dependent).

29

4.6

Examples

Examples of reaction calculation will be shown next. Each example has been carefully selected as it brings a diﬀerent “twist” from the preceding one. Some of those same problems will be revisited later for the determination of the internal forces and/or deﬂections. Many of those problems are taken from Prof. Gerstle textbok Basic Structural Analysis.

30

Example 4-2: Simply Supported Beam Victor Saouma

Structural Engineering

4–6

EQUILIBRIUM & REACTIONS

Draft

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. (+ ✲ ) ΣFx = 0; ⇒ (+ ✻) ΣFy = 0; ⇒ (+ ✛) ✁ ΣMzc = 0; ⇒

Rax − 36 k = 0 Ray + Rdy − 60 k − (4) k/ft(12) ft = 0 12Ray − 6Rdy − (60)(6) = 0

        1 0 0  Rax   36   Rax   36 k   0 1 1  108 Ray Ray 56 k = ⇒ =         0 12 −6 360 Rdy Rdy 52 k 

or

Alternatively we could have used another set of equations: (+ ✛) ✁ ΣMza = 0; (60)(6) + (48)(12) − (Rdy )(18) = 0 ⇒ Rdy = 52 k ✻ (+ ✛) ✁ ΣMzd = 0; (Ray )(18) − (60)(12) − (48)(6) = 0 ⇒ Ray = 56 k ✻ Check: ) ΣFy = 0; ; 56 − 52 − 60 − 48 = 0 (+ ✻

√

Example 4-3: Parabolic Load Determine the reactions of a simply supported beam of length L subjected to a parabolic load x 2 w = w0 L

Victor Saouma

Structural Engineering

4.6 Examples

4–7

Draft

Solution: Since there are no axial forces, there are two unknowns and two equations of equilibrium. Considering an inﬁnitesimal element of length dx, weight dW , and moment dM : x=L x 2 w0 dx ×x −(Rb )(L) = 0 (+ ✛) ✁ ΣMa = 0; x=0 L w

dW

dM

M L4 = 14 w0 L ⇒ Rb = L1 w0 4L 2 x=L 2 x 1 w0 dx = 0 (+ ✻ ) ΣFy = 0; Ra + w0 L − L x=0 4 Rb w0 L3 L2 3

⇒ Ra =

− 14 w0 L

=

1 12 w0 L

Example 4-4: 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 (ΣMb = 0), thus the structure is statically determinate. Victor Saouma

Structural Engineering

4–8

EQUILIBRIUM & REACTIONS

Draft

1. Isolating ab:

ΣM ✛✁b = 0; (9)(Ray ) − (40)(5) = 0 (+ ✛) ✁ ΣMa = 0; (40)(4) − (S)(9) = 0 ΣFx = 0;

⇒ Ray = 22.2 k ✻ ⇒ S = 17.7 k ✻ ⇒ Rax = 30 k ✛

2. Isolating bd: (+ ✛) ✁ ΣMd = 0; −(17.7)(18) − (40)(15) − (4)(8)(8) − (30)(2) + Rcy (12) = 0 ⇒ Rcy = 1,236 12 = 103 k ✻ (+ ✛) ✁ ΣMc = 0; −(17.7)(6) − (40)(3) + (4)(8)(4) + (30)(10) − Rdy (12) = 0 ⇒ Rdy = 201.3 12 = 16.7 k ✻ 3. Check ΣFy = 0; ✻; 22.2 − 40 − 40 + 103 − 32 − 30 + 16.7 = 0

√

Example 4-5: Three Hinged Gable Frame The three-hinged gable frames spaced at 30 ft. on center. Determine the reactions components on the frame due to: 1) Roof dead load, of 20 psf of roof area; 2) Snow load, of 30 psf of horizontal projection; 3) Wind load of 15 psf of vertical projection. Determine the critical design values for the vertical and horizontal reactions.

Victor Saouma

Structural Engineering

4.6 Examples

4–9

Draft

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 DL = (20) psf(30) ft

302 + 152 ft

1 lbs/k = 20.2 k ❄ 1, 000

4. Snow load per frame is SL = (30) psf(30) ft(30) ft

1 lbs/k = 27. k ❄ 1, 000

5. Wind load per frame (ignoring the suction) is W L = (15) psf(30) ft(35) ft

1 lbs/k = 15.75 k✲ 1, 000

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 (+ ✛) ✁ ΣMC = 0; 15(V ) − 30(V ) + 35(H) = 0

⇒ H=

15V 35

= .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

= = = =

(.429)(20.2) = (.429)(27.)

=

20.2 k ✻ 8.66 k✲ 27. k ✻ 11.58 k✲

8. The reactions due to wind load are (+ ✛) ✁ ΣMA (+ ✛) ✁ ΣMC (+ ✲ ) ΣFx (+ ✻) ΣFy Victor Saouma

= 0; = 0; = 0; = 0;

B (15.75)( 20+15 2 ) − VW L (60) = 0 B HW L (35) − (4.6)(30) = 0 A 15.75 − 3.95 − HW L = 0 B A VW L − VW L = 0

B ⇒ VW L = 4.60 k ✻ B ✛ ⇒ HW L = 3.95 k A ⇒ HW L = 11.80 k ✛ A ❄ ⇒ VW L = −4.60 k

Structural Engineering

4–10

EQUILIBRIUM & REACTIONS

Draft

9. Thus supports should be designed for H = 8.66 k + 11.58 k + 3.95 k V = 20.7 k + 27.0 k + 4.60 k

= 24.19 k = 52.3 k

Example 4-6: Inclined Supports Determine the reactions of the following two spans beam resting on inclined supports.

Solution: A priori we would identify 5 reactions, however we do have 2 equations of conditions (one at each inclined support), thus with three equations of equilibrium, we have a statically determinate system. (+ ✛) ✁ ΣMb = 0; (Ray )(20) − (40)(12) − (30)(6) + (44.72)(6) − (Rcy )(12) = 0 ⇒ 20Ray = 12Rcy + 391.68 (+ ✲ ) ΣFx = 0; 34 Ray − 22.36 − 43 Rcy = 0 ⇒ Rcy = 0.5625Ray − 16.77 Solving for those two equations: 20 −12 Ray 391.68 Ray 14.37 k ✻ = ⇒ = −8.69 k ❄ 0.5625 −1 Rcy Rcy 16.77 The horizontal components of the reactions at a and c are Rax Rcx

= =

3 4 14.37 4 3 8.69

= =

10.78 k✲ −11.59 k✲

Finally we solve for Rby (+ ✛) ✁ ΣMa = 0; (40)(8) + (30)(14) − (Rby )(20) + (44.72)(26) + (8.69)(32) = 0 ⇒ Rby = 109.04 k ✻ We check our results

√ ) ΣFy = 0; 14.37 − 40 − 30 + 109.04 − 44.72 − 8.69 = 0√ (+ ✻ (+ ✲ ) ΣFx = 0; 10.78 − 22.36 + 11.59 =0

Victor Saouma

Structural Engineering

4.7 Arches

Draft 4.7 31

4–11

Arches

See Section ??.

Victor Saouma

Structural Engineering

4–12

Draft

EQUILIBRIUM & REACTIONS

Victor Saouma

Structural Engineering

Draft Chapter 5

TRUSSES 5.1

Introduction

5.1.1

Assumptions

1 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.

2

Cables can carry only tensile forces, trusses can carry tensile and compressive forces.

3

Cables tend to be ﬂexible, and hence, they tend to oscillate and therefore must be stiﬀened.

4

Trusses are extensively used for bridges, long span roofs, electric tower, space structures.

5

For trusses, it is assumed that 1. Bars are pin-connected 2. Joints are frictionless hinges1 . 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.

6

7

Fig. 5.1 illustrates some of the most common types of trusses.

8 It can be easily determined that in a Pratt truss, the diagonal members are under tension, while in a Howe truss, they are in compression. Thus, the Pratt design is an excellent choice for steel whose members are slender and long diagonal member being in tension are not prone to buckling. The vertical members are less likely to buckle because they are shorter. On the other hand the Howe truss is often preferred for for heavy timber trusses.

9

In a truss analysis or design, we seek to determine the internal force along each member, Fig. 5.2

5.1.2

Basic Relations

Sign Convention: Tension positive, compression negative. On a truss the axial forces are indicated as forces acting on the joints. Stress-Force: σ =

P A

Stress-Strain: σ = Eε 1 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 so-called secondary bending moments are developed at the bars. Another source of secondary moments is the dead weight of the element.

5–2

TRUSSES

Draft

Figure 5.1: Types of Trusses

Figure 5.2: Bridge Truss

Victor Saouma

Structural Engineering

5.2 Trusses

5–3

Draft

Force-Displacement: ε =

∆L L

Equilibrium: ΣF = 0

5.2 5.2.1

Trusses 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.

10

A truss may be statically/externally determinate or indeterminate with respect to the reactions (more than 3 or 6 reactions in 2D or 3D problems respectively).

11

12

A truss may be internally determinate or indeterminate, Table 5.1.

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 5.1.

13

2D 3D Static Indeterminacy External R>3 R>6 Internal m + R > 2j m + R > 3j Unstable m + R < 2j m + R < 3j Table 5.1: Static Determinacy and Stability of Trusses If m < 2j − 3 (in 2D) the truss is not internally stable, and it will not remain a rigid body when it is detached from its supports. However, when attached to the supports, the truss will be rigid.

14

Since each joint is pin-connected, we can apply ΣM = 0 at each one of them. Furthermore, summation of forces applied on a joint must be equal to zero.

15

For 2D trusses the external equations of equilibrium which can be used to determine the reactions are ΣFX = 0, ΣFY = 0 and ΣMZ = 0. For 3D trusses the available equations are ΣFX = 0, ΣFY = 0, ΣFZ = 0 and ΣMX = 0, ΣMY = 0, ΣMZ = 0.

16

For a 2D truss we have 2 equations of equilibrium ΣFX = 0 and ΣFY = 0 which can be applied at each joint. For 3D trusses we would have three equations: ΣFX = 0, ΣFY = 0 and ΣFZ = 0.

17

Fig. 5.3 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.

18

19

There are two methods of analysis for statically determinate trusses 1. The Method of joints 2. The Method of sections

5.2.2 20

Method of Joints

The method of joints can be summarized as follows 1. Determine if the structure is statically determinate 2. Compute all reactions

Victor Saouma

Structural Engineering

5–4

TRUSSES

Draft

Figure 5.3: A Statically Indeterminate Truss 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 (ΣFx and ΣFy in 2D; ΣFx , ΣFy and ΣFz in 3D). 5. Because truss elements can only carry axial forces, the resultant force (F7 = F7x + F7y ) must be along the member, Fig. 5.4. Fx F Fy = = L Lx Ly

(5.1)

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.

21

Figure 5.4: X and Y Components of Truss Forces 22

This method should be used when all member forces must be determined.

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).

23

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. 5.5.

24

Victor Saouma

Structural Engineering

5.2 Trusses

5–5

Draft

Figure 5.5: Sign Convention for Truss Element Forces

Example 5-1: 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 ✛ (+ ✁) ΣME = 0; ⇒ (20 + 12)(3)(24) + (40 + 8)(2)(24) + (40)(24) − RAy (4)(24) = 0 ⇒ RAy = 58 k ✻ ) ΣFy = 0; ⇒ 20 + 12 + 40 + 8 + 40 − 58 − REy = 0 (+ ❄ ⇒ REy = 62 k ✻ Victor Saouma

Structural Engineering

5–6

TRUSSES

Draft

3. Consider each joint separately: Node A: Clearly AH is under compression, and AB under tension.

(+ ✻ ) ΣFy = 0; ⇒ −FAHy + 58 = 0 FAH = lly (FAHy ) ly = 32 ⇒ FAH = 40 32 (58) = 72.5 k Compression (+ ✲ ) ΣFx = 0; ⇒ −FAHx + FAB = 0 24 x FAB = L Ly (FAHy ) = 32 (58) = 43.5 k Tension

l=

√

322 + 242 = 40

Node B:

(+ ✲ ) ΣFx = 0; ⇒ FBC ) ΣFy = 0; ⇒ FBH (+ ✻

= =

43.5 k Tension 20 k Tension

Node H:

(+ ✲ ) ΣFx = 0; ⇒ FAHx − FHCx − FHGx = 0 43.5 − √2424 (FHC ) − √2424 (FHG ) = 0 2 +322 2 +102 (+ ✻ ) ΣFy = 0; ⇒ FAHy + FHCy − 12 − FHGy − 20 = 0 58 + √2432 (FHC ) − 12 − √2410 (FHG ) − 20 = 0 2 +322 2 +102 This can be most conveniently written as 0.6 0.921 FHC −7.5 = −0.8 0.385 FHG 52

(5.2)

Solving we obtain FHC FHG Victor Saouma

= =

−7.5 k Tension 52 k Compression Structural Engineering

5.2 Trusses

5–7

Draft Node E:

ΣFy = 0; ⇒ FEFy = 62 ΣFx = 0; ⇒ FED = FEFx

⇒ FEF = ⇒ FED =

√

242 +322 (62) 32 24 24 (F ) EFy = 32 (62) 32

= 77.5 k = 46.5 k

The results of this analysis are summarized below

4. We could check our calculations by verifying equilibrium of forces at a node not previously used, such as D

5.2.2.1 25

26

27

Matrix Method

This is essentially the method of joints cast in matrix form2 . We seek to determine the Statics Matrix [B] such that P F BF F BF R = BRF BRR R 0

(5.3)

This method can be summarized as follows 1. Select a coordinate system 2. Number the joints and the elements separately 2 Writing

a computer program for this method, would be an acceptable project.

Victor Saouma

Structural Engineering

5–8

TRUSSES

Draft 3. Assume

(a) All member forces to be positive (i.e. tension) (b) All reactions to be positive 4. Compute the direction cosines at each node j and for each element e, Fig. 5.6

Y α - ve β + ve

α + ve β + ve X

α - ve β - ve

α + ve β - ve

Figure 5.6: Direction Cosines αej βje

= =

Lx L Ly L

F

e

5. Write the two equations of equilibrium at each joint j in terms of the unknowns (member forces and reactions), Fig. 5.7

e e e Fy = L y F = β F L

e e e Fx = Lx F = α F L

Figure 5.7: Forces Acting on Truss Joint elements e αj Fe + Rxj + Pxj = 0 (+ ✲ ) ΣFx = 0; ⇒ Σ#of e=1 elements e βj Fe + Ryj + Pyj = 0 (+ ✻) ΣFy = 0; ⇒ Σ#of e=1

6. Invert the matrix to compute {F } and {R} The advantage of this method, is that once the [B] matrix has been inverted, we can readily reanalyze the same structure for diﬀerent load cases. With the new design codes in which dead loads and live loads are separately factored (Chapter 7), this method can save substantial reanalysis eﬀort. Furthermore, when deﬂections are determined by the virtual force method (Chapter 13), two analysis with two diﬀerent loads are required.

28

This method may be the only one appropriate to analyze statically determinate trusses which solution’s defy the previous two methods, Fig. 5.8.

29

30

Element e connecting joint i to j will have αei = −αej , and βie = −βje

The matrix [B] will have 2j rows and m + r columns. It can only be inverted if ti is symmetric (i.e 2j = m + r, statically determinate).

31

32

An algorithm to implement this method in simple computer programs:

Victor Saouma

Structural Engineering

5.2 Trusses

5–9

Draft

Figure 5.8: Complex Statically Determinate Truss 1. Prepare input data: (a) Nodal information: Node 1 2 3

x(1) 0. 10. 5.

x(2) 0. 0. 5.

P (1) 0. 0. 0.

P (2) 0. 0. -10.

R(1) 1 0 0

R(2) 1 1 0

where x(1), x(2), P (1), and P (2) are the x and y coordinates; the x and y component of applied nodal load. R(1), R(2) correspond to the x and y boundary conditions, they will be set to 1 if there is a corresponding reaction, and 0 otherwise. (b) Element Connectivity Element 1 2 3

Node(1) 1 2 3

Node(2) 2 3 1

2. Determine the size of the matrix (2 times the number of joints) and initialize a square matrix of this size to zero. 3. Assemble the ﬁrst submnatrix of the Statics matrix (a) Loop over each element (e), determine Lix , Liy (as measured from the ﬁrst node), L, αei = βie =

Liy e e e e L , αj = −αi , βj = −βi , where αei in row 2i − 1 column e, βie in

(b) Store 2j column e.

Lix L

i and j are the end nodes of element e. row 2i column e, αej in row 2j − 1 column e, βje in row

4. Loop over each node: (a) Store Rxi (reaction boundary condition for node i along x) in row 2i − 1 column e + 2i − 1. Similarly, store Ryi (reaction boundary condition for node i along y) in row 2i column e + 2i. (b) Store Pxi (Load at node i along x axis) in vector P row 2i − 1, similarly, store Pyi in row 2i. 5. Invert the matrix B, multiply it with the load vector P and solve for the unknown member forces and reactions. % % Initialize the statics matrix % b(1:2*npoin,1:2*npoin)=0.; % Victor Saouma

Structural Engineering

5–10

Draft

TRUSSES

% Determine direction cosines and insert them in b % for ielem=1:nelem nod1=lnods(ielem,1); nod2=lnods(ielem,2); lx=coord(nod2,1)-coord(nod1,1); ly=coord(nod2,2)-coord(nod1,2); l_elem(ielem)= alpha_i=... beta_i=... b(...,...)=...; b(...,...)=...; alpha_j=... beta_j=... b(...,...)=... b(...,...)=... end % % Boundary conditions % nbc=0; for inode=1:npoin for ibc=1:2 if id(inode,ibc)==1 nbc=nbc+1; b(...,...)=1.; end end end

Example 5-2: Truss I, Matrix Method Determine all member forces for the following truss

Solution: 1. We ﬁrst determine the directions cosines

Victor Saouma

Structural Engineering

5.2 Trusses

Draft

5–11 Member 1

Nodes 1-2

α

β

α11 = 1 2

2-3

3

3-1

α

β Node 2 α12 = −1 β21 = 0 Node 3 α23 = .707 β32 = −.707 Node 1 α31 = .707 β13 = .707

Node 1 β11 = 0

Node 2 α22 = −.7071 β22 = .707 Node 3 α33 = −.707 β33 = −.707

2. Next we write the equations of equilibrium

Node 1 Node 2 Node 3

ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy

=0 =0 =0 =0 =0 =0

F1 α11  β11  1  α2  1  β2   0 0 

F2 0 0 α22 β22 α23 β32

F3 α31 β13 0 0 α33 β33

R1x 1 0 0 0 0 0

R1y 0 1 0 0 0 0

R2y  0     0     0   1     0     0

F1 F2 F3 R1x R1y R2y

               

F1 F2 F3 R1x R1y R2y

{F }

              

        +

      

0 0 0 0 0 −10

{P }

              

=0

Or Node 1 ΣFx ΣFy Node 2 ΣFx ΣFy Node 3 ΣFx ΣFy

=0 =0 =0 =0 =0 =0

       

1 0 .707 0 0 .707 −1 −.707 0 0 .707 0 0 .707 −.707 0 −.707 −.707

1 0 0 0 0 0

0 1 0 0 0 0

0 0 0 1 0 0

[B]

Inverting [B] we obtain from F = [B]−1 P  F1     F  2   F3 R1x     R    1y R2y

{F }

       

              

+

      

0 0 0 0 0 −10

{P }

              

=0

  5         −7.07       −7.07 = 0             5          5        

Example 5-3: Truss II, Matrix Method Set up the statics matrix for the truss of Example 5-1 using the matrix method Solution: First we number the joints and the elements as shown below.

Victor Saouma

Structural Engineering

5–12

TRUSSES

Draft

Then we determine the direction cosines 24 242 +322 √ 32 242 +322 √ 10 242 +102 √ 24 242 +102 √

# 1 # 2 # 3 # 4 # 5 # 6 # 7 # 8

ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy

           

1 −1

= = = =

3 5 4 5 10 26 24 26

= 0.6 = 0.8 = 0.38 = 0.92

.6 .8

1 1

1 1 −1

−.6 .8

1 −1

.6 .8

1

1 1 −1 −.6 −.8

−1

−.6 .8 .6 −.8

−1 −.6 −.8

−1

.6 −.8

1

−.92 −.38 .92 .38

−.92 .38 .92 −.38

[B]

  F1   FF23  F4   F5   F6      FF78  9   FF10  F11    FF12  13   R1    R1xy R5 y ?

                      

+

                      

0 0 0 −20 0 −40 0 −40 0 0 0 0 0 −8 0 −12

                      

√

Using Mathematica we would have b={ { 1, 0., 0., 0.,0.6, 0., 0., 0., {0., 0., 0., 0.,0.8, 0., 0., 0., {-1, 1, 0., 0., 0., 0., 0., 0., {0.,0., 0., 0., 0., 1, 0., 0., {0.,-1, 1, 0., 0., 0.,-0.6, 0., {0., 0., 0., 0., 0., 0., 0.8, 1., {0., 0., -1, 1, 0., 0., 0., 0., {0., 0., 0., 0., 0., 0., 0., 0., {0., 0., 0., -1, 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., 0., 0., 0., 0., 0., 0., 0., {0., 0., 0., 0., 0., 0., 0., -1, {0., 0., 0., 0., -0.6, 0., 0.6, 0., {0., 0., 0., 0.,-0.8, -1,-0.8, 0., } p={0, 0, 0, -20, 0, -40 , 0, -40, m=Inverse[b].p

0., 0., 0., 0., 0., 1, 0., 0.}, 0., 0., 0., 0., 0., 0., 1, 0.}, 0., 0., 0., 0., 0., 0., 0., 0.}, 0., 0., 0., 0., 0., 0., 0., 0.}, 0.6, 0., 0., 0., 0., 0., 0., 0.}, 0.8, 0., 0., 0., 0., 0., 0., 0.}, 0., 0., 0., 0., 0., 0., 0., 0.}, 0., 1., 0., 0., 0., 0., 0., 0.}, 0., 0.,-0.6, 0., 0., 0., 0., 0.}, 0., 0., 0.8, 0., 0., 0., 0., 1.}, -0.6, 0., 0.6, 0.,-0.92, 0., 0., 0.}, -0.8,-1.,-0.8, 0., 0.38, 0., 0., 0.}, 0., 0., 0.,-0.92, 0.92, 0., 0., 0.}, 0., 0., 0.,-0.38,-0.38, 0., 0., 0.}, 0., 0., 0., 0.92, 0., 0., 0., 0.}, 0., 0., 0., 0.38, 0., 0., 0., 0.} 0, 0,

0, 0, 0, -8,

0, -12}

And the result will be

Victor Saouma

Structural Engineering

=0

5.2 Trusses

5–13

Draft

Out[3]= {-43.5, -43.5, -46.5, -46.5, 72.5, -20., -7.66598, -31.7344, >

-2.66598, -40., 77.5, 52.2822, 52.2822, -1.22125 10

-14 , -58., -62.}

which correspond to the unknown element internal forces and external reactions.

5.2.3

Method of Sections

When only forces in selected members (away from loaded joints) is to be determined, this method should be used.

33

34

This method can be summarized as follows 1. “Cut” the truss into two substructures by an imaginary line (not necessarily straight) such that it will at least intersect the member for which force is to be determined. 2. Consider either one of the two substructures as the free body 3. Each substructure must remain in equilibrium. Apply the equations of equilibrium (a) Summation of moments about a particular point (usually the intersection of 2 cut members) would permit the determination of other member forces (b) Summation of forces is usually used to determine forces in inclined members

Example 5-4: Truss, Method of Sections Determine FBC and FHG in the previous example. Solution: Cutting through members HG, HC and BC, we ﬁrst take the summation of forces with respect to H:

Victor Saouma

Structural Engineering

5–14

TRUSSES

Draft (+ ✛) ✁ ΣMH = 0

⇒ RAy (24) − FBC (32) = 0 FBC = 24 32 (58) = 43.5 k Tension (+ ✛) ✁ ΣMC = 0; ⇒ (58)(24)(2) − (20 + 12)(24) − FHGx (32) − FHGy (24) = 0 2770 − 768 − (32)(FHG ) √2424 − (24)(FHG ) √2410 =0 2 +102 2 +102 2, 000 − (29.5)FHG − (9.2)FHG = 0 ⇒ FHG = 52 k Compression

5.3

Case Study: Stadium B

A

3m

D C 1m E

2m G F

3m

0.5 m 7.5 m

Figure 5.9: Florence Stadium, Pier Luigi Nervi (?)

Victor Saouma

Structural Engineering

5.3 Case Study: Stadium

5–15

Draft

800 kN 1

2

1

j=6

2

3

3m

m=9 r=3

3 6

1m

4 4

5

8 2m

7 9

5 111 000 000 111 000 111

3m

6 0000 1111 1111 0000 0000 1111

0.5 m 7.5 m

Figure 5.10: Florence Stadioum, Pier Luigi Nervi (?)

Victor Saouma

Structural Engineering

5–16

Draft

TRUSSES

Victor Saouma

Structural Engineering

Draft Chapter 6

CABLES 6.1

Funicular Polygons

1 A cable is a slender ﬂexible member with zero or negligible ﬂexural stiﬀness, thus it can only transmit tensile forces1 . 2 The tensile force at any point acts in the direction of the tangent to the cable (as any other component will cause bending).

3

Its strength stems from its ability to undergo extensive changes in slope at the point of load application.

Cables resist vertical forces by undergoing sag (h) and thus developing tensile forces. The horizontal component of this force (H) is called thrust.

4

5

The distance between the cable supports is called the chord.

6

The sag to span ratio is denoted by r=

h l

(6.1)

7 When a set of concentrated loads is applied to a cable of negligible weight, then the cable deﬂects into a series of linear segments and the resulting shape is called the funicular polygon.

If a cable supports vertical forces only, then the horizontal component H of the cable tension T remains constant.

8

Example 6-1: Funicular Cable Structure Determine the reactions and the tensions for the cable structure shown below.

1 Due

to the zero ﬂexural rigidity it will buckle under axial compressive forces.

6–2

CABLES

Draft

Solution: We have 4 external reactions, however the horizontal ones are equal and we can use any one of a number of equations of conditions in addition to the three equations of equilibrium. First, we solve for Ay , Dy and H = Ax = Dx . For this problem we could use the following 3 equations of static equilibrium ΣFx = ΣFy = ΣM = 0, however since we do not have any force in the x direction, the second equation is of no avail. Instead we will consider the following set ΣFy = ΣMA = ΣMD = 0 1. First we solve for Dy (+ ✛) ✁ ΣMA = 0; ⇒ 12(30) + 6(70) − Dy (100) = 0 ⇒ Dy = 7.8 k

(6.2)

2. Then we solve for Ay (+ ✻) ΣFy = 0; ⇒ Ay − 12 − 6 + 7.8 = 0 ⇒ Ay = 10.2 k

(6.3)

3. Solve for the horizontal force (+ ✛) ✁ ΣMB = 0; ⇒ Ay (30) − H(6) = 0 ⇒ H = 51 k

(6.4)

4. Now we can solve for the sag at point C (+ ✛) ✁ ΣMC = 0 ⇒ −DY (30) + H(hc ) = 0 ⇒ hc =

30(7.8) 30Dy = = 4.6 ft H 51

5. We now solve for the cable internal forces or tractions in this case 6 = 0.200 ⇒ θA = 11.31 deg TAB ; tan θA = 30 H 51 = = 51.98 k = cos θA 0.981 6 − 4.6 TBC ; = 0.035 ⇒ θB = 2 deg tan θB = 40 H 51 = = 51.03 k = cos θB 0.999 4.6 TCD ; = 0.153 ⇒ θC = 8.7 deg tan θC = 30 H 51 = = 51.62 k = cos θC 0.988

Victor Saouma

(6.5)

(6.6-a) (6.6-b) (6.6-c) (6.6-d) (6.6-e) (6.6-f)

Structural Engineering

6.2 Uniform Load

6–3

Draft 6.2

6.2.1

Uniform Load qdx; Parabola

9 Whereas the forces in a cable can be determined from statics alone, its conﬁguration must be derived from its deformation. Let us consider a cable with distributed load p(x) per unit horizontal projection of the cable length2 . An inﬁnitesimal 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 H =constant. Summation of the vertical

T V

H

V θ

q(x)

x q(x)

H

y(x)

dy

ds

y

dx

ds L

θ dx

V+dV

y’

H T+dT

x h x’

y L/2

Figure 6.1: Cable Structure Subjected to q(x) forces yields ) ΣFy = 0 ⇒ −V + qdx + (V + dV ) = (+ ❄

0

(6.7-a)

dV + qdx =

0

(6.7-b)

where V is the vertical component of the cable tension at x3 . Because the cable must be tangent to T , we have V tan θ = (6.8) H 10

Substituting into Eq. 6.7-b yields d(H tan θ) + qdx = 0

(6.9)

or

d (H tan θ) = q dx Since H is constant (no horizontal load is applied), this last equation can be rewritten as −

−H

d (tan θ) = q dx

(6.11)

dy dx

which when substituted in Eq. 6.11 yields

Written in terms of the vertical displacement y, tan θ = the governing equation for cables

11

2 Thus 3 Note

(6.10)

d2 y q =− dx2 H

(6.12)

neglecting the weight of the cable that if the cable was subjected to its own weight then we would have qds instead of pdx.

Victor Saouma

Structural Engineering

6–4

CABLES

Draft

For a cable subjected to a uniform load p. we can determine its shape by double integration of Eq. 6.12

12

−Hy

=

−Hy

=

qx + C1 qx2 + C1 x + C2 2

(6.13-a) (6.13-b)

To solve for C1 and C2 this last equation must satisfy the boundary conditions: y = 0 at x = 0 and at x = L ⇒ C2 = 0 and C1 = − qL 2 . Thus q (6.14) Hy = x(L − x) 2 13 This equation gives the shape y(x) in terms of the horizontal force H, it can be rewritten in terms of the maximum sag h which occurs at midspan, hence at x = L2 we would have4

qL2 8

Hh =

(6.15)

This relation clearly shows that the horizontal force is inversely proportional to the sag h, as h H . Furthermore, this equation can be rewritten as

14

8h qL = H L

(6.16)

and combining this equation with Eq. 6.1 we obtain

15

(6.17)

4hx (L − x) L2

(6.18)

Combining Eq. 6.14 and 6.15 we obtain y=

16

qL = 8r H

If we shift the origin to midspan, and reverse y, then

y=

4h 2 x L2

(6.19)

Thus the cable assumes a parabolic shape (as the moment diagram of the applied load). The maximum tension occurs at the support where the vertical component is equal to V = qL 2 and the horizontal one to H, thus 2 2 qL/2 qL Tmax = V 2 + H 2 = + H2 = H 1 + (6.20) 2 H

17

Combining this with Eq. 6.17 we obtain5 .

Tmax = H 4 Note

M=

1 + 16r2 ≈ H(1 + 8r2 )

(6.21)

the analogy between this equation and the maximum moment in a simply supported uniformly loaded beam

qL2 . 8

5 Recalling

that (a + b)n = an + nan−1 b + √ 1 2 Thus for b << 1, 1 + b = (1 + b) 2 ≈ 1 + 2b

Victor Saouma

n(n−1) n−2 2 a b 2!

+ · or (1 + b)n = 1 + nb +

n(n−1)b2 2!

+

n(n−1)(n−2)b3 3!

+ · · ·;

Structural Engineering

6.2 Uniform Load

6–5

Draft 6.2.2

† qds; Catenary

Let us consider now the case where the cable is subjected to its own weight (plus ice and wind if any). We would have to replace qdx by qds in Eq. 6.7-b

18

dV + qds = 0

(6.22)

The diﬀerential equation for this new case will be derived exactly as before, but we substitute qdx by qds, thus Eq. 6.12 becomes d2 y q ds =− 2 dx H dx 19

But ds2 = dx2 + dy 2 , hence:

q d2 y =− 2 dx H

1+

(6.23)

dy dx

2 (6.24)

solution of this diﬀerential equation is considerably more complicated than for a parabola. 20

We let dy/dx = p, then

dp q =− 1 + p2 dx H

rearranging

dp =− 1 + p2

(6.25)

q dx H

(6.26)

From Mathematica (or handbooks), the left hand side is equal to dp = loge (p + 1 + p2 ) 1 + p2 Substituting, we obtain loge (p + 1 + p2 ) p + 1 + p2 1 + p2 1+p

2

p

y

21

(6.27)

qx + C1 = − H

(6.28-a)

A

= e

A

= −p + e

(6.28-b) A

(6.28-c)

= p − 2pe + e e2A − 1 eA − e−A = sinh A = = 2eA 2 qx dy = sinh − + C1 = H dx qx qx H = sinh − + C1 dx = − cosh − + C1 + C2 H q H 2

A

2A

(6.28-d) (6.28-e) (6.28-f) (6.28-g)

To determine the two constants, we set dy dx dy dx ⇒0 ⇒y

Victor Saouma

L at x = 2 qx q H = − sinh − + C1 H q H q L q L = sinh − + C1 ⇒ C1 = H2 H2 q L H − x + C2 = − cosh q H 2 = 0

(6.29-a) (6.29-b) (6.29-c) (6.29-d) Structural Engineering

6–6

CABLES

Draft

At midspan, the sag is equal to h, thus H q L L cosh − + C2 q H 2 2 H = h+ q

= −

h C2

22

(6.30-a) (6.30-b)

If we move the origin at the lowest point along the cable at x = x − L/2 and y = h − y, we obtain q q y = cosh x −1 H H

23

(6.31)

This equation is to be contrasted with 6.19, we can rewrite those two equations as: q 1 q 2 y= x Parabola H 2 H q q y = cosh x − 1 Catenary H H

(6.32-a) (6.32-b)

The hyperbolic cosine of the catenary can be expanded into a Taylor power series as qy 1 qx 2 1 qx 4 1 qx 6 = + + + ... H 2 H 24 H 720 H

(6.33)

The ﬁrst term of this development is identical as the formula for the parabola, and the other terms constitute the diﬀerence between the two. The diﬀerence becomes signiﬁcant only for large qx/H, that is for large sags in comparison with the span, Fig. 6.3. 3 Parabola Catenary 2.5

2

1.5

1

0.5

0 −2

−1.5

−1

−0.5

0 qx/H

0.5

1

1.5

2

Figure 6.2: Catenary versus Parabola Cable Structures

6.2.2.1

Historical Note

It should be mentioned that solution of this problem constitued one of the major mathematical/Mechanics challenges of the early 18th century. Around 1684, diﬀerential and integral calculus took their ﬁrst eﬀective forms, and those powerful new techniques allowed scientists to tackle complex problems for the ﬁrst time, (Penvenuto 1991). One of these problems was the solution to the catenary problem as presented by Jakob Bernouilli. Immediately thereafter, Leibniz presented a solution based on inﬁnitesimal calculus, another one was presented by Huygens. Finally, the brother of the challenger, Johann Bernoulli did also present a solution.

24

Victor Saouma

Structural Engineering

6.2 Uniform Load

6–7

Draft

Huygens solution was complex and relied on geometrical arguments. The one of Leibniz was ellegant and correct (y/a = (bx/a + b−x/a )/2 (we recognize Eq. 6.31 albeit written in slightly diﬀerent form, Fig. 6.3. Finally, Bernoulli presented two correct solution, and in his solution he did for the ﬁrst time express 25

Figure 6.3: Leipniz’s Figure of a catenary, 1690 equations of equilibrium in diﬀerential form. Example 6-2: Design of Suspension Bridge Design the following 4 lanes suspension bridge by selecting the cable diameters assuming an allowable cable strength σall of 190 ksi. The bases of the tower are hinged in order to avoid large bending moments.

The total dead load is estimated at 200 psf. Assume a sag to span ratio of Solution:

1 5

1 = 1. The dead load carried by each cable will be one half the total dead load or p1 = 12 (200) psf(50) ft 1,000 5.0 k/ft

Victor Saouma

Structural Engineering

6–8

CABLES

Draft

2. Using the HS 20 truck, the uniform additional load per cable is p2 = (2)lanes/cable(.64)k/ft/lane = 1.28 k/ft/cable. Thus, the total design load is p1 + p2 = 5 + 1.28 = 6.28 k/ft 3. The thrust H is determined from Eq. 6.15 H

pl2 8h (6.28) k/ft(300)2 ft2 (8) 300 5 ft

= = =

4. From Eq. 6.21 the maximum tension is Tmax

(6.34-a) (6.34-b)

1, 177 k

(6.34-c)

1 + 16r2

(6.35-a)

=

H

=

2 1 (1, 177) k 1 + (16) 5

(6.35-b)

1, 507 k

=

(6.35-c)

5. Note that if we used the approximate formula in Eq. 6.21 we would have obtained Tmax

H(1 + 8r2 ) 2 1 1, 177 1 + 8 5

= = =

(6.36-a) (6.36-b)

1, 554 k

(6.36-c)

or 3% diﬀerence! 6. The required cross sectional area of the cable along the main span should be equal to Tmax 1, 507 k = = 7.93 in2 σall 190 ksi

A= which corresponds to a diameter d=

!

4A = π

!

(4)(7.93) = 3.18 in π

7. We seek next to determine the cable force in AB. Since the pylon can not take any horizontal force, we should have the horizontal component of Tmax (H) equal and opposite to the horizontal √ component of TAB or

TAB H

TAB

=

(100)2 +(120)2 100

thus (100)2 + (120)2 = (1, 177)(1.562) = 1, 838 k =H 100

the cable area should be A= which corresponds to a diameter

1, 838 k = 9.68 in2 190 ksi

! d=

(6.37)

(4)(9.68) = 3.51 in π

8. To determine the vertical load acting on the pylon, we must add the vertical components of Tmax and of TAB (VBC and VAB respectively). We can determine VBC from H and Tmax , thus 120 (1, 177) + (1, 507)2 − (1, 177)2 = 1, 412 + 941 = 2, 353 k (6.38) P = 100 Using A36 steel with an allowable stress of 21 ksi, the cross sectional area of the tower should be 2 A = 2,353 21 = 112 in . Note that buckling of such a high tower might govern the ﬁnal dimensions. Victor Saouma

Structural Engineering

6.3 Case Study: George Washington Bridge

6–9

Draft

9. If the cables were to be anchored to a concrete block, the volume of the block should be at least equal to (1, 412) k(1, 000) V = = 9, 413 ft3 150 lbs/ft3 or a cube of approximately 21 ft

6.3

Case Study: George Washington Bridge

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 stiﬀened with the addition of a lower deck.

26

6.3.1 27

Geometry

A longitudinal and plan elevation of the bridge is shown in Fig. 6.4. For simplicity we will assume in

?? 377 ft

610 ft

327 ft

3,500 ft

650 ft

4,760 ft ELEVATION

N.J.

HUDSON RIVER

N.Y.

PLAN

Figure 6.4: Longitudinal and Plan Elevation of the George Washington Bridge our analysis that the two approaching spans are equal to 650 ft. 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.

28

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).

29

The cables support the road deck which is hung by 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 ﬁrmly anchored in both banks by huge blocks of concrete, the anchors.

30

Because the cables are much longer than they are thick (small LI ), they can be idealized as perfectly ﬂexible members with no shear/bending resistance but with high axial strength.

31

Victor Saouma

Structural Engineering

6–10

CABLES

Draft

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.

32

6.3.2

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 would be equivalent to

33

DL = (390) psf(100) ft

k

(1, 000) lbs

= 39 k/ft

(6.39)

for the main span and 40 k/ft for the side ones. For highway bridges, design loads are given by the AASHTO (Association of American State Highway Transportation Oﬃcials). The HS-20 truck is often used for the design of bridges on main highways, Fig. 6.5. Either the design truck with speciﬁed 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. 34

Figure 6.5: Truck Load

35

With two decks, we estimate that there is a total of 12 lanes or LL = (12)Lanes(.64) k/ ft/Lane = 7.68 k/ft ≈ 8 k/ft

(6.40)

We do not consider earthquake, or wind loads in this analysis. 36

Final DL and LL are, Fig. 6.6: T L = 39 + 8 = 47 k/ft

6.3.3 37

Cable Forces

The thrust H (which is the horizontal component of the cable force) is determined from Eq. 6.15 H

= = =

Victor Saouma

wL2cs 8h (47) k/ft(3, 500)2 ft2 (8)(327) ft 220, 000 k

(6.41-a) (6.41-b) (6.41-c) Structural Engineering

6.3 Case Study: George Washington Bridge

6–11

Draft

wD,S = 40 k/ft

wD = 39 k/ft

wD,S = 40 k/ft

DEAD LOADS

wL = 8 k/ft

Figure 6.6: Dead and Live Loads From Eq. 6.21 the maximum tension is r Tmax

327 h = = 0.0934 Lcs 3, 500 = H 1 + 16r2 = (220, 000) k 1 + (16)(0.0934)2

=

= (220, 000) k(1.0675) = 235,000 k

6.3.4 38

(6.42-a) (6.42-b) (6.42-c) (6.42-d)

Reactions

Cable reactions are shown in Fig. 6.7. POINTS WITH REACTIONS TO CABLES

Figure 6.7: Location of Cable Reactions 39

The vertical force in the columns due to the central span (cs) is simply the support reaction, 6.8 Vcs =

Victor Saouma

1 1 wLcs = (47) k/ft(3, 500) ft = 82, 250 k 2 2

(6.43) Structural Engineering

6–12

CABLES

Draft

wTOT = 39 + 8 = 47 k/ft B

A

REACTIONS AT TOP OF TOWER

POINT OF NO MOMENT

L = 3,500 FT

Figure 6.8: Vertical Reactions in Columns Due to Central Span Load Note that we can check this by determining the vector sum of H and V which should be equal to Tmax : √ 2 + H2 = Vcs (82, 250)2 + (220, 000)2 = 235, 000 k

(6.44)

Along the side spans (ss), the total load is T L = 40 + 8 = 48 k/ft. We determine the vertical reaction by taking the summation of moments with respect to the anchor:

40

Lss − Vss Lss 2 (650) ft = (377) k(220, 000) k + (48) k/ft(650) ft − 650Vss 2 Vss ΣMD = 0; ✛+; ✁ hss H + (wss Lss )

=

0

(6.45-a)

=

0

(6.45-b)

=

143, 200 k

(6.45-c)

Note: that we have used equilibrium to determine the vertical component of the cable force. It would 377 have been wrong to determine Vss from Vss = 220, 000 650 as we did in the previous example, because the cable is now loaded. We would have to determine the shape of the cable and the tangent at the support. Beginiing with Eq. 6.13-b:

41

−Hy

=

−y

=

1 2 wx + C1 x + C2 2 w x2 C1 C2 + x+ H 2 H H

(6.46-a) (6.46-b) (6.46-c)

At x = 0, y = 0, thus C2 = 0; and at x = 650, y = −377; with H = 220, 000 k and w = 48 =

−

(48) k C1 x2 − x ft(220, 000) k (220, 000) k

(6.47-a)

=

−1.091 × 10−4 x2 − 4.545 × 10−6 C1 x

(6.47-b)

y|x=650 C1

= =

−377 = −1.091 × 10−4 (650)2 − 4.545 × 10−6 C1 (650) 112, 000

(6.47-c) (6.47-d)

y dy dx

=

−1.091 × 10−4 x2 − 0.501x

(6.47-e)

=

−2.182 × 10−4 x − 0.501

(6.47-f)

=

−0.1418 − 0.501 = −0.6428 =

=

(220, 000)(0.6428) = 141, 423 k

y

" dy "" dx "x=650 V

V H

(6.47-g) (6.47-h)

which is only 1% diﬀerent. 42

Hence the total axial force applied on the column is V = Vcs + Vss = (82, 250) k + (143, 200) k = 225,450 k

Victor Saouma

(6.48)

Structural Engineering

6.3 Case Study: George Washington Bridge

6–13

Draft 43

The vertical reaction at the anchor is given by summation of the forces in the y direction, Fig. 6.9: (+ ✻) ΣFy = 0; (wss Lss ) + Vss + Ranchor −(48) k/ft(650) ft + (143, 200) k + Ranchor Ranchor

= 0

(6.49-a)

= 0

(6.49-b)

=

112,000 k ❄

(6.49-c) (6.49-d)

225,450 k 220,000 k

112,000 k Figure 6.9: Cable Reactions in Side Span

44

45

The axial force in the side cable is determined the vector sum of the horizontal and vertical reactions. # ss Tanchor = + H 2 = (112, 000)2 + (220, 000)2 = 247, 000 k R2 (6.50-a) anchor # ss 2 + H2 = = Vss (143, 200)2 + (220, 000)2 = 262,500 k (6.50-b) Ttower

The cable stresses are determined last, Fig. 6.10: Awire

=

Atotal

=

Central Span σ

=

ss Side Span Tower σtower

=

ss Side Span Anchor σtower

=

(3.14)(0.196)2 πD2 = = 0.03017 in2 (6.51-a) 4 4 2 (4)cables(26, 474)wires/cable(0.03017) in2 /wire = 3, 200 in(6.51-b) H (220, 000) k = = 68.75 ksi (6.51-c) A (3, 200) in2 ss Ttower (262, 500) in2 = 82 ksi (6.51-d) = A (3, 200) in2 ss Tanchor (247, 000) in2 = = 77.2 ksi (6.51-e) A (3, 200) in2

73.4 ksi

81.9 ksi

68.75 ksi

77.2 ksi

Figure 6.10: Cable Stresses Victor Saouma

Structural Engineering

6–14

CABLES

Draft

If the cables were to be anchored to a concrete block, the volume of the block should be at least equal to V = (112,000) k(1,000)3 lbs/ k = 747, 000 ft3 or a cube of approximately 91 ft

46

150

lbs/ft

The deck, for all practical purposes can be treated as a continuous beam supported by elastic springs with stiﬀness 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. 6.11.

47

K=AL/E

Shear

Moment

Figure 6.11: Deck Idealization, Shear and Moment Diagrams

Victor Saouma

Structural Engineering

Draft Chapter 7

DESIGN PHILOSOPHIES of ACI and AISC CODES 7.1

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 diﬀer 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. Diﬀerences between the actual dimensions and those speciﬁed (mostly in placement of steel rebars in R/C). 3. Eﬀect 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 diﬃcult 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. The purpose of safety provisions is to limit the probability of failure and yet permit economical structures.

2

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. 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

7–2

DESIGN PHILOSOPHIES of ACI and AISC CODES

Draft 7.2

Working Stress Method

5 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. 7.1.

Figure 7.1: Load Life of a Structure 7 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.

9

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. σ < σall =

σyld F.S.

(7.1)

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 0.45fc . 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 7.1.

7.3 7.3.1

Ultimate Strength Method The Normal Distribution

The normal distribution has been found to be an excellent approximation to a large class of distributions, and has some very desirable mathematical properties:

12

Victor Saouma

Structural Engineering

7.3 Ultimate Strength Method

Draft ∗

7–3

Steel, AISC/ASD Tension, Gross Area Ft = 0.6Fy Tension, Eﬀective Net Area∗ Ft = 0.5Fu Bending Fb = 0.66Fy Shear Fv = 0.40Fy Concrete, ACI/WSD Tension 0 Compression 0.45fc

Eﬀective net area will be deﬁned in section 9.2.1.2. Table 7.1: Allowable Stresses for Steel and Concrete 1. f (x) is symmetric with respect to the mean µ. 2. f (x) is a “bell curve” with inﬂection points at x = µ ± σ. 3. f (x) is a valid probability distribution function as: ∞ f (x) = 1

(7.2)

−∞

4. The probability that xmin < x < xmax is given by:

xmax

P (xmin < x < xmax ) =

f (x)dx

(7.3)

xmin

13

The general normal (or−2.5 Gauss) is given Fig. 1.0 30.1: 1.5 −3.0 −2.0distribution −1.5 −1.0 −0.5 0.0by,0.5

2.0

2.5

3.0

0.40 0.35 0.30

PDF

0.25

2σ

0.20 0.15 0.10 0.05 0.5

1.0

1.5

2.0

2.5

3.0

0.5

1.0

1.5

2.0

2.5

3.0

CDF

0.00 −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0.0 µ 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0.0 x

Figure 7.2: Normalized Gauss Distribution, and Cumulative Distribution Function

1 x−µ 2 1 e− 2 [ σ ] φ(x) = √ 2πσ

Victor Saouma

(7.4)

Structural Engineering

7–4

DESIGN PHILOSOPHIES of ACI and AISC CODES

Draft 14

A normal distribution N (µ, σ 2 ) can be normalized by deﬁning

y=

x−µ σ

(7.5)

and y would have a distribution N (0, 1): y2 1 φ(y) = √ e− 2 2π

7.3.2

(7.6)

Reliability Index

15

In this approach, it is assumed that the load Q and the resistance R are random variables.

16

Typical frequency distributions of such random variables are shown in Fig. 7.3.

Figure 7.3: Frequency Distributions of Load Q and Resistance R

17

The safety margin is deﬁned as Y = R − Q. Failure would occur if Y < 0

18

Q and R can be combined and the result expressed logarithmically, Fig. 7.4. X = ln

R Q

(7.7)

Failure would occur for negative values of X The probability of failure Pf is equal to the ratio of the shaded area to the total area under the curve in Fig. 7.4. R 20 If X is assumed to follow a Normal Distribution than it has a mean value X = ln Q and a m standard deviation σ. 19

21

We deﬁne the safety index (or reliability index) as β =

X σ

1 For standard distributions and for β = 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 β = 3.5 will fail because of either excessive load or understrength sometime in its lifetime. 22

Reliability indices are a relative measure of the current condition and provide a qualitative estimate of the structural performance.

23

Structures with relatively high reliable indices will be expected to perform well. If the value is too low, then the structure may be classiﬁed as a hazard.

24

Victor Saouma

Structural Engineering

7.3 Ultimate Strength Method

7–5

Draft

Figure 7.4: Deﬁnition of Reliability Index

10

10

10

10

4

3

2

1

Good

Above Average

Below Average

10

5

Poor

1/(Probability of Failure)

10

6

Unsatisfactory

10

Probability of Failure in terms of β

7

Hazardous

10

0

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

β

Figure 7.5: Probability of Failure in terms of β

25

Target values for β are shown in Table 30.2, and in Fig. 30.3

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.

26

These factors are referred to as resistance factor Φ and Load Factor α respectively. The resistance factor is deﬁned as Rm exp(−0.55βVR ) (7.8) Φ= Rn

27

where RM RN and VR are the mean resistance, the nominal resistance (to be deﬁned later), and the coeﬃcient of variation of the resistance.

7.3.3

Discussion

28 ACI refers to this method as the Strength Design Method, (previously referred to as the Ultimate Strength Method).

Victor Saouma

Structural Engineering

7–6

DESIGN PHILOSOPHIES of ACI and AISC CODES

Draft

Type of Load/Member AISC DL + LL; Members DL + LL; Connections DL + LL + WL; Members DL + LL +EL; Members ACI Ductile Failure Sudden Failures

β 3.0 4.5 3.5 1.75 3-3.5 3.5-4

Table 7.2: Selected β values for Steel and Concrete Structures

29

AISC refers to it as Load and Resistance Factor Design (LRFD).

Terms such as failure load should be avoided; it is preferable to refer to a structure’s Limit State load.

30

31

The general form is (LRFD-A4.1)1 ΦRn ≥ Σαi Qi

(7.9)

where Φ is a strength reduction factor, less than 1, and must account for the type of structural element, Table 7.3. Type of Member

Φ ACI

Axial Tension Flexure Axial Compression, spiral reinforcement Axial Compression, other Shear and Torsion Bearing on concrete AISC Tension, yielding Tension, fracture Compression Beams Fasteners, Tension Fasteners, Shear

0.9 0.9 0.75 0.70 0.85 0.70 0.9 0.75 0.85 0.9 0.75 0.65

Table 7.3: Strength Reduction Factors, Φ Rn is the nominal resistance (or strength). ΦRn is the design strength. αi is the load factor corresponding to Qi and is greater than 1. Σαi Qi is the required strength based on the factored load: i is the type of load 32

The various factored load combinations which must be considered are 1 Throughout

the notes we will refer by this symbol the relevant design speciﬁcation in the AISC code.

Victor Saouma

Structural Engineering

7.4 Example

7–7

Draft AISC

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 6. 0.9D+1.3W(or 1.5 E) ACI 1. 1.4D+1.7L 2. 0.75(1.4D+1.7L+1.7W) 3. 0.9D+1.3W 4. 1.05D+1.275W 5. 0.9D+1.7H 6. 1.4D +1.7L+1.7H 7. 0.75(1.4D+1.4T+1.7L) 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. 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 5.2.2.1). For the WSD method, we need not perform more than one analysis in general. 33

Serviceability Limit States must be assessed under service loads (not factored). The most important ones being

34

1. Deﬂections 2. Crack width (for R/C) 3. Stability 35

Most code requirements on strength are written as:

Φx Rn > Qu

(7.10)

where Rn is the nominal strength of the member, Qu is the ultimate (factored) load eﬀect (force, moment, shear, torque). Design load is also at times deﬁned as

Rd =

7.4

Qu Φx

(7.11)

Example

Example 7-1: LRFD vs ASD Victor Saouma

Structural Engineering

7–8

Draft

DESIGN PHILOSOPHIES of ACI and AISC CODES

To illustrate the diﬀerences between the two design approaches, let us consider the design of an axial member, subjected to a dead load of 100 k and live load of 80 k. Use A36 steel. ASD: We consider the total load P = 100 + 80 = 180 k. From Table 7.1, the allowable stress is 0.6σyld = 0.6 ∗ 36 = 21.6 ksi. Thus the required cross sectional area is A=

180 = 8.33 in2 21.6

USD we consider the largest of the two load combinations = Σαi Qi : 1.4D 1.2D + 1.6L =

1.4(100) = 140 k 1.2(100) + 1.6(80) = 248 k

✛

From Table 7.3 Φ = 0.9, and ΦRn = (0.9)Aσyld . Hence, applying Eq. 7.9 the cross sectional area should be Σαi Qi 248 = 7.65 in2 A= = Φσyld (0.9)(36) Note that whereas in this particular case the USD design required a smaller area, this may not be the case for diﬀerent ratios of dead to live loads.

Victor Saouma

Structural Engineering

Draft Chapter 8

DESIGN I 8.1

Case Study: Eiﬀel Tower

Adapted from (Billington and Mark 1983)

8.1.1

Materials, & Geometry

1 The tower was built out of wrought iron, less expensive than steel,and Eiﬀel had more experience with this material, Fig. 8.1

Figure 8.1: Eiﬀel Tower (Billington and Mark 1983) 2 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.

3

To avoid overturning

M gravity 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

8–2

DESIGN I

Draft

3. Design support to resist tension. 4. Post-tension the support.

4 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

y = 164

2 984 − x 984

(8.1)

av(x)2

We recall from calculus that for y = v(x)

6

dy dx d ax2 dx

=

dy dv dv dx

(8.2-a)

= 2ax

(8.2-b)

Thus for our problem dy dx

984 − x 1 = 2(164) − 984 984

dy dv

=

(8.3-a)

dv dx

984 − x 2, 952

(8.3-b)

Also

dy dy = tan β ⇒ β = tan−1 dx dx where β is the angle measured from the x axis to the tangent to the curve. 7

(8.4)

We now can tabulate the width and slope at various elevations

Location Support First platform second platform Intermediate platform Top platform Top

Height 0 186 380 644 906 984

Width/2 164 108 62 20 1 0

Width Estimated Actual 328 216 240 123 110 40 2 0

dy dx

.333 .270 .205 .115 .0264 0.000

β 18.4o 15.1o 11.6o 6.6o 1.5o 0o

The tower is supported by four inclined supports, each with a cross section of 800 in2 . An idealization of the tower is shown in Fig. 8.2.

8

8.1.2 9

10

Loads

The total weight of the tower is 18, 800 k. The dead load is not uniformly distributed, and is approximated as follows, Fig. 8.3:

Location Ground- second platform Second platform-intermediate platform intermediate platform - top Victor Saouma Total

Height 380 ft 264 ft 340 ft 984 ft

Dead Weight 15, 500 k 2, 200 k 1, 100 k 18,Structural 800 k Engineering

8.1 Case Study: Eiﬀel Tower

8–3

Draft

ACTUAL CONTINUOUS CONNECTION

IDEALIZED CONTINUOUS CONNECTION

ACTUAL POINTS OF CONNECTION

Figure 8.2: Eiﬀel Tower Idealization, (Billington and Mark 1983)

Figure 8.3: Eiﬀel Tower, Dead Load Idealization; (Billington and Mark 1983)

Victor Saouma

Structural Engineering

8–4

DESIGN I

Draft

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):

11

1st platform:

kip (50) psf(240)2 ft2 (1,000) lbs

2nd platform: Total:

2

(50) psf(110) ft

2

2, 880 k

k

(1,000)

600 k 3, 480 k

lbs

Hence the total vertical load is Pvert = DL + LL = 18, 800 + 3, 480 = 22, 280 k. 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. 8.4: The pressure is assumed to be 2.6 k/ft, thus the

12

Figure 8.4: Eiﬀel Tower, Wind Load Idealization; (Billington and Mark 1983) lateral wind force is, Fig. 8.5

LOADS

TOTAL LOADS

P

Q

P=2560k

00000 11111

L/2

H0 00000 11111 Q=22,280k 00000 11111 00 11 00 11 00 11 V0 11 00

REACTIONS

11 00 00 11 00 11

M0

Figure 8.5: Eiﬀel Tower, Wind Loads, (Billington and Mark 1983)

Plat = (2.6) k/ft(984) ft = 2,560 k

Victor Saouma

acting at

984 = 492 ft 2

(8.5)

Structural Engineering

8.1 Case Study: Eiﬀel Tower

8–5

Draft 8.1.3

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. 8.6.

13

+

=

WINDWARD SIDE

VERTICAL FORCES

WIND FORCES

LEEWARD SIDE

TOTAL

Figure 8.6: Eiﬀel Tower, Reactions; (Billington and Mark 1983) Gravity Load Pvert grav Rvert

= 22, 280 ❄ 22, 280 = = 11,140 k ✻ 2

(8.6-a) (8.6-b)

Lateral Load Lateral Moment (we essentially have a cantilevered 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 L−x (L − x)2 =p Mlat = p(L − x) 2 2

Force Moment arm

(8.7)

Thus the lateral moment caused by the wind is parabolic. At the base (x = 0), the maximum moment is equal to Mlat = p

(984 − 0)2 2 (L − x)2 = (2.6) k/ft ft = 1,260,000 k.ft ✛✁ 2 2

(8.8)

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 wind =± Rvert

1, 260, 000 k.ft M = = ± 3,850 k ✻❄ d 328 ft

(8.9)

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 = Rlat

Victor Saouma

(2, 560) k = 1,280 k 2

✛

(8.10)

Structural Engineering

8–6

DESIGN I

Draft 8.1.4 14

Internal Forces

First, a brief reminder θ F

Fy θ

cos θ

=

sin θ

=

tan θ

=

Fx (8.11-a) F Fy (8.11-b) F Fy (8.11-c) Fx

Fx 15

Internal forces are ﬁrst determined at the base.

Gravity load are ﬁrst considered, remember those are caused by the dead load and the live load, Fig. 8.7:

16

0

β=18.4

β=18.4 INCLINED INTERNAL FORCE: N CONSEQUENT HORIZONTAL COMPONENT: H

N

KNOWN VERTICAL COMPONENT: V

V

H FORCE POLYGON

Figure 8.7: Eiﬀel Tower, Internal Gravity Forces; (Billington and Mark 1983)

cos β

=

N

=

tan β

=

H

=

V V ⇒N = N cos β 11, 140 k = 11,730 kip cos 18.4o H ⇒ H = V tan β V 11, 140 k(tan 18.4o ) = 3,700 kip

(8.12-a) (8.12-b) (8.12-c) (8.12-d)

The horizontal forces which must be resisted by the foundations, Fig. 8.8.

H

H

3700 k

3700 k

Figure 8.8: Eiﬀel Tower, Horizontal Reactions; (Billington and Mark 1983)

17

Because the vertical load decreases with height, the axial force will also decrease with height.

18

At the second platform, the total vertical load is Q = 1, 100 + 2, 200 = 3, 300 k and at that height the

Victor Saouma

Structural Engineering

8.1 Case Study: Eiﬀel Tower

8–7

Draft

angle is 11.6o thus the axial force (per pair of columns) will be Nvert

3,300 2

=

k

= 1, 685 k (8.13-a) cos 11.6o 3, 300 k (tan 11.6o) = 339 k Hvert = (8.13-b) 2 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 cross-section. The horizontal force will be resisted by the axial forces in the second platform itself. 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. 8.9:

19

18.4

3,850 k

3,850 cos 18.4

18.4

1,280 sin 18.4

1,280 k

Figure 8.9: Eiﬀel Tower, Internal Wind Forces; (Billington and Mark 1983) Nc

Nt

8.1.5 20

wind wind = −Rvert cos β − Rlat sin β o = −(3, 850) k(cos 18.4 ) − (1, 280) k(sin 18.40 )

(8.14-a) (8.14-b)

=

(8.14-c)

-4,050 k Leeward

= −R cos β + R sin β o = (3, 850) k(cos 18.4 ) + (1, 280) k(sin 18.40 )

(8.14-d) (8.14-e)

=

(8.14-f)

wind vert

wind lat

4,050 k Winward

Internal Stresses

The total forces caused by both lateral and gravity forces can now be determined: NLTotal

= −(11, 730) k −(4, 050) k = -15,780 k Leeward side

gravity lateral

(8.15-a)

Total NW

= −(11, 730) k +(4, 050) k = -7,630 k Winward side

gravity lateral

(8.15-b)

We observe that even under wind load, the windward side is still under compression. In the idealization of the tower’s geometry, the area of each pair of the simpliﬁed columns is 1, 600 in2 . and thus the maximum stresses will be determined from

21

σcomp = 22

NLT −15, 780 k = -9.9 ksi = A 1, 600 in2

(8.16)

The strength of wrought iron is 45 ksi, hence the safety factor is Safety Factor =

Victor Saouma

45 ksi ultimate stress = = 4.5 actual stress 9.9 ksi

(8.17) Structural Engineering

8–8

DESIGN I

Draft 8.2

Magazini Generali by Maillart

Adapted from (Billington and Mark 1983)

8.2.1

Geometry

This storage house, built by Maillart in Chiasso in 1924, provides a good example of the marriage between aesthetic and engineering.

23

The most striking feature of the Magazini Generali is not the structure itself, but rather the shape of its internal supporting frames, Fig. 8.10.

24

HINGE IDEALIZATION OF THIN SECTIONS

ACTUAL FRAME

ABSTRACTION OF MID SECTIONAS A SIMPLE BEAM

9.2 ft

63.6 ft

Figure 8.10: Magazzini Generali; Overall Dimensions, (Billington and Mark 1983) 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. 8.11.

25

8.2.2

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 psf, and that the total frame weight is 13.6 kips, the total uniform load becomes, Fig. 8.12:

26

qroof qf rame qtotal

Victor Saouma

= (98) psf(14.7) ft = 1.4 k/ft (13.6) k = = 0.2 k/ft (63.6) ft = 1.4 + 0.2 = 1.6 k/ft

(8.18-a) (8.18-b) (8.18-c)

Structural Engineering

8.2 Magazini Generali by Maillart

8–9

Draft

B P

MB =B*d1

MP =P*d 2 d2

d1 B MR=MB -M P

Figure 8.11: Magazzini Generali; Support System, (Billington and Mark 1983) 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 8.12: Magazzini Generali; Loads (Billington and Mark 1983)

8.2.3 27

Reactions

Reactions for the beam are determined ﬁrst taking advantage of symmetry, Fig. 8.13: W

=

R

=

(1.6) k/ft(63.6) ft = 102 k 102 W = = 51 k 2 2

(8.19-a) (8.19-b)

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 8.13: Magazzini Generali; Beam Reactions, (Billington and Mark 1983)

Victor Saouma

Structural Engineering

8–10

DESIGN I

Draft 8.2.4

Forces

The internal forces are primarily 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. 8.14, equal to: (1.6) k/ft(63.6) ft2 qL2 = = 808 k.ft (8.20) Mmax = 8 8

28

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 8.14: Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983) 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 ﬁbers, and tensile ones on the lower. Thus the net axial force is zero, however there is a net internal couple, Fig. 8.15.

29

q TOTAL

A

C d

M

VA T

Figure 8.15: Magazzini Generali; Internal Moment, (Billington and Mark 1983)

Mext T =C Victor Saouma

= Cd ⇒ C = =

Mext d

(808) k.ft = ± 88 k (9.2) ft

(8.21-a) (8.21-b) Structural Engineering

8.2 Magazini Generali by Maillart

8–11

Draft

30 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. 8.16.

d

M

FRAME

MOMENT DIAGRAM

CABLE :

FRAME :

CURVE OF DIAGRAM

SHAPE OF DIAGRAM

Figure 8.16: 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. 8.17. Fig. 8.18 This is analogous to the forces transmitted to the support by a

31

88 k

Tension 88 k

88 k

88 k

Compression Horizontal Component Tied Arch

Cable Force Axial Force Vertical Reaction

Figure 8.17: Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark 1983) tied arch. 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. 8.18.

32

8.2.5 33

Internal Stresses

The net compressive stress, for a top chord with a cross sectional area of 75 in2 is equal to σ=

(88) k P = = 1.17 ksi A (75) in2

(8.22)

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.

Victor Saouma

Structural Engineering

8–12

DESIGN I

Draft 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 8.18: Magazzini Generali; Eﬀect of Lateral Supports, (Billington and Mark 1983) 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 in2 and 6 other bars with 0.58 in2 , thus we have a total of (8.23) As = 4(0.69) + 6(0.58) = 6.24 in2

34

Thus the steel stresses will be σ=

P (88) k = = 14.1 ksi A (6.24) in2

(8.24)

which is lower than the allowable steel stress.

8.3 8.3.1 35

Buildings Buildings Structures

There are three primary types of building systems:

Wall Subsystem: 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 eﬃcient for very high rise buildings. Rigid Frame: which consists of linear vertical components (columns) rigidly connected to stiﬀ horizontal ones (beams and girders). This is not a very eﬃcient structural form to resist lateral (wind/earthquake) loads. 8.3.1.1

Wall Subsystems

Whereas exterior wall provide enclosure and interior ones separation, both of them can also have a structural role in transferring vertical and horizontal loads.

36

Victor Saouma

Structural Engineering

8.3 Buildings

8–13

Draft 37

Walls are constructed out of masonry, timber concrete or steel.

If the wall is braced by ﬂoors, then it can provide an excellent resistance to horizontal load in the plane of the wall (but not orthogonal to it).

38

When shear-walls subsystem 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.

39

8.3.1.1.1

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 ﬂexural stresses and the shearing stresses in the wall to resist the wind load, Fig. 8.19.

40

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

M

-F

M

VERTICAL + FDL

11111 00000 00000 11111

+140 + 740 PSI + 600

7.7’ IN TENSION

Figure 8.19: Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981) 1. Maximum shear force and bending moment at the base Vmax

=

Mmax

=

wL = (0.8) k.ft(120) ft = 96 k (0.8) k.ft(120)2 ft2 wL2 = = 5, 760 k.ft 2 2

(8.25-a) (8.25-b)

2. The resulting eccentricity is eActual =

(5, 760) k.ft M = = 14.4 ft P (400) k

(8.26)

3. The critical eccentricity is ecr =

(20) ft L = = 3.3 ft < eActual N.G. 6 6

(8.27)

thus there will be tension at the base. Victor Saouma

Structural Engineering

8–14

DESIGN I

Draft

4. The moment of inertia of the wall is (1) ft(20) ft3 bh3 = = 667 ft4 12 12

I=

(8.28)

5. The maximum ﬂexural stresses will be σmax = ±

(5, 760) k.ft(10) ft Mc = = ±(86.5) ksf = ±(600) psi I (667) ft4

(8.29)

6. The average shearing stress is τ=

(96) k V = = 4.8 ksf = 33.3 psi A (1)(20) ft2

(8.30)

A concrete with nominal shear reinforcement can carry at least 100 psi in shear, those computed shear stresses are permissible. 7. At the base of the wall, the axial stresses will be σ=

−(400) k = (20) ksf =≈ −140 psi (1)(20) ft2

(8.31-a)

8. The maximum stresses will thus be: σ1

=

−140 + 600 = 460 psi

σ2

=

−140 − 600 = −740 psi

(Tension)

(8.32-a)

(Compression)

(8.32-b)

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, 4 σall = (20) = 26.7 ksi (8.33) 3 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) 20 460 x = ⇒x= (20) = 7.7 ft 460 460 + 740 460 + 740

(8.34)

12. The total tensile force inside this triangular stress block is T =

1 (460) ksi(7.7 × 12) in (12) in = 250 k 2

(8.35)

width

13. The total amount of steel reinforcement needed is As =

(250) k = 9.4 in2 (26.7) ksi

(8.36)

This amount of reinforcement should be provided at both ends of the wall since the wind or earthquake can act in any direction. In addition, the foundations should be designed to resist tensile uplift forces (possibly using piles).

Victor Saouma

Structural Engineering

8.3 Buildings

8–15

Draft 120’

20’

W 400 k

60’

H=96 k

24’

1 1.2

~1.6

V +FM

-F

M

Figure 8.20: Trussed Shear Wall 8.3.1.1.2

Example: Trussed Shear Wall

From (Lin and Stotesbury 1981)

We consider the same problem previously analyzed, but use a trussed shear wall instead of a concrete one, Fig. 8.20.

41

1. Using the maximum moment of 5, 760 kip-ft (Eq. 8.25-b), we can compute the compression and tension in the columns for a lever arm of 20 ft. F =±

(5, 760) k.ft = ±288 k (20) ft

(8.37)

2. If we now add the eﬀect of the 400 kip vertical load, the forces would be C

=

T

=

(400) k − 288 = −488 k 2 (400) k + 288 = 88 k − 2 −

(8.38-a) (8.38-b)

3. The force in the diagonal which must resist a base shear of 96 kip is (similar triangles) (20)2 + (24)2 (20)2 + (24)2 F = ⇒F = (96) = 154 k 96 20 20

(8.39)

4. The design could be modiﬁed to have no tensile forces in the columns by increasing the width of the base (currently at 20 ft). 8.3.1.2

Shaft Systems

Vertical shear-resisting 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.

42

If the shaft is relatively short and wide, with an aspect ratio under 1 or 2, then the dominant structural action is that of a stiﬀ shear resisting tube. If the aspect ratio is between 3 and 5, then the shear forces may not be the controlling criterion, and ﬂexure dominates.

43

Victor Saouma

Structural Engineering

8–16

DESIGN I

Draft 8.3.1.2.1 44

Example: Tube Subsystem

From (Lin and Stotesbury 1981)

With reference to Fig. 8.21, the reinforced concrete shaft is 20 ft square, 120 ft high, and with 1 ft ~ 20 ’

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

w = 0.8 k/ft

H = 96 k 60 ’

~ 20 ’

120 ’

N.A.

Figure 8.21: 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 analyzed in Sect. 8.3.1.1.1 the total vertical load acting on the base is now increased to V = 4(400) = 1, 600 k (8.40) 2. As previously, the maximum moment and shear are 5, 760 k.ft and 96 k respectively. 3. The moment of inertia for a tubular section is I =Σ

(20)(20)3 (18)(18)3 bd3 = − = 4, 600 ft4 12 12 12

4. The maximum ﬂexural stresses: (5, 760) k.ft(20/2) ft MC =± σf l = ± = ±12.5 ksf = ±87 psi I (4, 600) ft4

(8.41)

(8.42)

5. The average shear stress is τ=

(96) k V = 2.4 ksf = 17 psi = A 2(20)(1) ft2

(8.43)

6. The vertical load of 1,600 k produces an axial stress of σax =

−(1, 600) k P = = −20 ksf = −140 psi A (4(20)(1) ft2

(8.44)

7. The total stresses are thus σ

=

σax + σf l

(8.45-a)

σ1

=

−140 + 87 = −53 psi

(8.45-b)

σ2

=

−140 − 87 = −227 psi

(8.45-c)

thus we do not have any tensile stresses, and those stresses are much better than those obtained from a single shear wall. Victor Saouma

Structural Engineering

8.4 Design of a Three Hinged Arch

8–17

Draft 8.3.1.3

Rigid Frames

Rigid frames can carry both vertical and horizontal loads, however their analysis is more complex than for tubes.

45

The rigorous and exact analysis of a rigid frame can only be accomplished through a computer analysis. However, for preliminary design it is often suﬃcient to perform approximate analysis.

46

There are two approximate methods for the analysis of rigid frames subjected to lateral loads: 1) Portal and 2) Cantilever method.

47

48

The portal frame method is based on the following major assumptions, Fig. 8.22: L P

P

P

L/2

h/2

PI

h

H 1=P/2

H 2=P/2

h/2 V1 =P/(2L)

V2 =P/(2L)

Figure 8.22: 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 inﬂection (zero moment) for all columns is at midheight 3. The point of inﬂection 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.

8.4

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 four-story loading totaling 27 k/ft along each arch. 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 (8.46) w = (1 + .25)27 = 33.7 ≈ 33 k/ft Victor Saouma

Structural Engineering

8–18

DESIGN I

Draft

2. We next determine the various forces: H V R

(33)(510)2 wL2 = = 10, 700 k 8h 8(100) (33)(510) wL = = 8, 400 k = 2 2 H 2 + V 2 = (10, 700)2 + (8, 400)2 = 13, 600 k =

=

(8.47-a) (8.47-b) (8.47-c)

3. If we use concrete-ﬁlled 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 = 2πrt = πDt = (3.14)(6) ft(12) in/ ft(0.5) in = 113 in2

(8.48)

4. The concrete area is Ac = π

(6)2 2 D2 = (3.14) ft (144) in2 /f tsq = 4, 070 in2 4 4

(8.49)

5. Assuming that the steel has an allowable stress of 20 ksi and the concrete 2.5 ksi (noting that the strength of conﬁned concrete can be as high as three times the one of fc ), then the load carry(113)(20)ksi = 2,260 k Steel Ac 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. 6. 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.

8.5

Salginatobel Bridge (Maillart)

Adapted from (Billington and Mark 1983)

8.5.1

Geometry

The Salginatobel bridge, perhaps the most famous and inﬂuential structure of Maillart is located in high up in the Swiss Alps close to Shuders.

49

It is a three hinged pedestrian bridge which crosses a deep valley with a most beautiful shape which blends perfectly with its surrounding, Fig. 8.23

50

The load supporting structure is the arch itself, whereas the bridge deck and the piers are transferring the vertical load into the arch.

51

52

The arch cross section is not constant, and can be idealized as in Fig. 8.24

53

The basic shape of the supporting structure is a three hinged arch as shown in Fig. 8.25

The arch is parabolic (which as we saw an the optimal shape which minimizes ﬂexure), and the cross section at the quarter point has an area of

54

At = 2[(0.62)(12.46) + (0.59)(12.17)] = 29.8 ft2 = 4, 291 in2 55

(8.50)

Each ﬂange has an area of AF = (0.62)(12.46) = 7.73 ft2 = 1, 113 in2

(8.51)

and the eﬀective depth of the section is d = 12.79 ft, Fig. 8.26. 56

At the crown/hinge the section is rectangular with Acr = (1.05)(11.48) = 12.05 ft2 = 1, 735 in2

Victor Saouma

(8.52) Structural Engineering

8.5 Salginatobel Bridge (Maillart)

Draft

20 ft

20 ft

87.5 ft

87.5 ft

20 ft

20 ft

20 ft

42.6 ft

20 ft

8–19

295 ft

Figure 8.23: Salginatobel Bridge; Dimensions, (Billington and Mark 1983)

42.6 ft

ACTUAL ARCH WITH CENTROID (DOTTED LINE)

IDEALIZATION (ONE DEMENSIONAL) 295 ft

Figure 8.24: 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 8.25: Salginatobel Bridge; Hinges, (Billington and Mark 1983) Victor Saouma

Structural Engineering

8–20

DESIGN I

42.6 ft

Draft 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 8.26: Salginatobel Bridge; Sections, (Billington and Mark 1983)

8.5.2

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. 8.27.

57

wD =

(1, 680) k WD = = 5.7 k/ft L (295) ft

(8.53)

wD = 5.7 k/ft WD = 1680 k

L = 295 ft

Figure 8.27: 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).

58

The live load is caused by traﬃc, and we consider the case in which two trucks, each weighing 55 kips, are placed at the quarter-point, Fig. 8.28. This placement of the load actually corresponds to one

59

Victor Saouma

Structural Engineering

8.5 Salginatobel Bridge (Maillart)

8–21

Draft

of the most critical loading arrangement. The total vertical load is shown in Fig. 8.29

8.5.3 60

Reactions

Reactions are easily determined from equilibrium, Fig. 8.33 1, 680 2 110 VL = 2 (+ ✛) ✁ ΣMc (840)(147.5) − (840)(73.75) − HD (42.6)

= =

⇒ HD

=

VD =

=

840 k

(8.54-a)

=

55 k

(8.54-b)

(55)(147.5) − (55)(73.75) − HL (42.6) = RD

8.5.4

⇒ HL 2 = (840) + (1, 455)2 RL = (55)2 + (95)2

0 0

(8.54-c) (8.54-d)

1, 455 k 0

(8.54-e) (8.54-f)

=

95 k

(8.54-g)

=

1, 680 k

(8.54-h)

=

110 k

(8.54-i)

Internal Forces

61

The shear diagrams for the dead, live and combined load is shown in Fig. 8.31.

62

At the quarter point the axial force can be expressed as: N = H cos α + V sin α

(8.55)

where

2d (8.56) L and α = 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 tan α =

qr ND

=

(−1, 455) cos 16.1o + (−420) sin 16.1o = −1, 514 k

(8.57-a)

NLqr

=

(−95) cos 16.1o + (−55) sin 16.1o = −106 k

(8.57-b)

and at the crown where there is no vertical force (and α = 0) cr ND

=

(−1, 455) cos 0o + (−420) sin 0o = −1, 455 k

(8.58-a)

NLcr

=

(−95) cos 0o + (−55) sin 0o = −95 k

(8.58-b)

63

The uniform dead load will not produce a moment on the parabolic arch.

64

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 =

11o 295 PL = = 4, 050 k.ft 2 4 2 4

(8.59)

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 (8.60) MLV = Hdmax = (95)(32.6) = −4, 050 k.ft Victor Saouma

Structural Engineering

8–22

DESIGN I

Draft

PLAN

P = 55 k

ROADWAY

295 ft

P = 55 k

ARCH ABUTMENT

42.5 ft

Figure 8.28: Salginatobel Bridge; Truck Load, (Billington and Mark 1983)

Victor Saouma

Structural Engineering

8.5 Salginatobel Bridge (Maillart)

8–23

Draft

A

P = 55 k

B VB,D = 840 k

VA,D = 840 k

P = 55 k

B

A

DEAD LOAD

42.6 ft

Q D = 1680 k

VA,L = 55 k

LIVE LOAD

VB,L = 55 k

295 ft

Figure 8.29: 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 8.30: Salginatobel Bridge; Reactions, (Billington and Mark 1983)

840 k

+

0

3L/4 L/4

295 ft

L/2

55 k

SHEAR FORCE

SHEAR FORCE

420 k

x

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 8.31: Salganitobel Bridge; Shear Diagrams, (Billington and Mark 1983)

Victor Saouma

Structural Engineering

8–24

DESIGN I

Draft

at the quarter point MLV = Hd1/4 = (95)

3(32.6) = −3, 040 k.ft 4

(8.61)

The overall bending moment diagram from the live loads is determined by simply adding those two components, Fig. 8.32.

65

P

L/4

+

-4,050 k-ft

BENDING MOMENT

BENDING

L/4

-3,040 k-ft

MOMENT

P

0

L/4

L/2

x 3L/4

L

295 ft

+ PL/4 = 4,050 k-ft

4,050 ft.k-3,040 ft.k = 1,010 ft.k

MOMENT

BENDING

=

295 ft

Figure 8.32: 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.

66

67

The maximum moment at midspan is MLmax = 4, 050 − 3, 040 = 1, 010 k.ft

(8.62)

which would produce internal forces in the upper and lower ﬂanges equal to: Fint = ±

8.5.5

(1, 010) k.ft MLmax = = ±79 k d (12.8) ft

(8.63)

Internal Stresses

68 The axial stresses at the springlines were determined to be −1, 680 and −110 kips for the dead and live loads respectively.

69

At the support the area of concrete is Ac = 2, 240 in2 , thus the axial stresses are D σsp

=

L σsp

=

σs spT otal

−(1, 680) k 1, 000 = −750 psi (2, 240) in2 −(110) k 1, 000 = −49 psi (2, 240) in2

= −750 − 49 = −799 psi

(8.64-a) (8.64-b) (8.64-c)

At the crown, we repeat the same calculations, where the axial force is equal to the horizontal component of the reactions

70

D σcr

Victor Saouma

=

−(1, 455) k 1, 000 = −839 psi (1, 735) in2

(8.65-a) Structural Engineering

8.5 Salginatobel Bridge (Maillart)

Draft

8–25

L σcr

=

−(95) k 1, 000 = −55 psi (1, 735) in2

(8.65-b)

σs cr T otal

=

−839 − 55 = −894 psi

(8.65-c)

The stresses at the quarter point are determined next. Note that we must include the eﬀect of both axial and ﬂexural stresses

71

top σqr

=

−(1, 514) k −(106) k (79) k 1, 000 + 1, 000 − 2 2 (4, 291) in (4, 291) in (1, 113) in2

DeadLoad LiveLoad Flexural

(8.66-a)

AxialStresses

= −353 − 25 − 71 −449 psi bot σqr

=

(8.66-b)

−(1, 514) k −(106) k (79) k 1, 000 + 1, 000 2 2 (4, 291) in (4, 291) in (1, 113) in2

DeadLoad LiveLoad Flexural

(8.66-c)

AxialStresses

= −353 − 25 + 71 −307 psi

8.5.6

(8.66-d)

Structural Behavior of Deck-Stiﬀened Arches

From (Billington 1979) INCOMPLETE The issue of unsymmetrical live load on a stiﬀened or unstiﬀened arch was also addressed by Maillart. As discussed in (Billington 1979) and illustrated by Fig. 8.33

72

wL

∼∆/10

∆ wL

wL a wL Unstiffened Arch

wL

a

wL

0

wL a 2

wL a 2

Stiffened Arch

Figure 8.33: Structural Behavior of Stiﬀened Arches, (Billington 1979)

Victor Saouma

Structural Engineering

8–26

Draft

DESIGN I

Victor Saouma

Structural Engineering

Draft Chapter 9

STEEL TENSION MEMBERS 1 This will be your ﬁrst exposure to structural design. Contrarily to structural analysis where you have one and only one exact solution, there are typically many acceptable and correct alternative designs. 2 Another important diﬀerence, is that in analysis you have so far focused on the application of one equation (such as the ﬂexural stress in a beam) or of a single very well deﬁned sequential process (such as truss analysis). In design, (and very much so in the design of tension steel members), you will have to satisfy more than one equation. 3 In almost all cases, design is governed by a code. In this course we adopt the LRFD method of the AISC code. Hence, a design would have to simultaneously satisfy all the relevant code provisions. 4 In analysis you did not have to concern yourself with details on how the structure is put together, evidently in design you would have to take this into account. To complicate matters, you would have to use some form of common sense to make certain decisions which govern the design process. 5 Finally, design is loosely deﬁned as a process in which you either select new members, or you verify whether an existing design is satisfactory.

9.1

Introduction

6 Tension members are encountered mostly in truss structures (bridges, roof trusses, transmission towers, wind bracing of multistoried buildings).

7

We will separately cover the design of compression members due to the possibility of buckling.

8 Structural elements are connected together to form a structure. Connections fall into one of two categories With fasteners, bolts, rivets; or without fasteners, welds. Fasteners require drilling or punching a hole. 9 It can be shown that a circular hole in a uniformly stressed plate under σ, will have a stress concentration factor equal to 3, Fig. 9.1. That is around the hole, σmax = 3σ 1 .

9.2 10

Geometric Considerations

Before we consider the design of tension members, due consideration must be given to 1. The deﬁnition of the cross sectional area which will be used to dimension our design 2. Stiﬀness of the member 1 If

the circle becomes an ellipse, than the stress concentration factor becomes greater than three; At the limit if the ellipse collapses into a crack, then we would have theoretically an inﬁnite stress. This is why one should always avoid sharp corners in a structure.

9–2

STEEL TENSION MEMBERS

Draft

Figure 9.1: Stress Concentration Around Circular Hole

9.2.1

Areas

The presence of a hole, and the resulting stress concentration must be accounted for in design. This is accomplished by introducing the concept of eﬀective areas.

11

The eﬀective area Ae is deﬁned in terms of the net area An when the load is transmitted by bolts or rivets, and gross area Ag for welded members:

12

Ae Ae

= =

U An U Ag

Bolts, Rivets Welds

(9.1)

where U is a reduction factor, and Ag is the gross area of the member. 9.2.1.1

Net Area, An

13

The net area must be deﬁned whenever we have holes due to bolts or rivets.

14

Holes are usually punched 1/16 in. larger than the diameter of the bolt or rivet.

1 When holes are punched, a small zone (∆R ≈ 32 in) around it is damaged, thus the total width to 1 in. be deducted in determining the net area is ∆D = 16

15

Thus with respect to the fastener of diameter D we punched a hole of diameter D + 1 additional 16 was damaged by the process, thus for net area calculation we must subtract fastener diameter, Fig. 9.2.

16

1 16 , and an 1 8 from the

Accordingly, for a plate of width w and thickness t, and with a single hole which accomodates a bolt of diameter D, the net area would be

17

1 An = wt − D + t 8

(9.2)

Whenever there is a need for more than two rivets or bolts, the holes are not aligned but rather staggered. Thus there is more than one potential failure line which can be used to determine the net area.

18

19

For staggered holes, we deﬁne the net are in terms of, Fig. 9.3,

pitch s, longitudinal center to center spacing between two consecutive holes. gage g, transverse center to center spacing between two adjacent holes. Victor Saouma

Structural Engineering

9.2 Geometric Considerations

9–3

Draft

11111111111111111111111111 00000000000000000000000000 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 00000000000000000000000 11111111111111111111111 00000000000000000000000000 11111111111111111111111111 Hole size D 00000000000000000000000000 11111111111111111111111111 D+

1 16

D+

1 8

Bolt size

Effective hole size

Figure 9.2: Hole Sizes

Figure 9.3: Eﬀect of Staggered Holes on Net Area

Victor Saouma

Structural Engineering

9–4

STEEL TENSION MEMBERS

Draft

The net area is determined as above, however we subtract Length Correction = An

s2 4g

(9.3)

1 # of holes D+ = Ag − Σ1 8

2

# of correct. s t + Σ1 t 4g

(9.4)

In this example the two possible failure lines have the following net lengths: 1 A-B =Length of A-B −(width of hole + 16 in.) 1 A-C =Length of A-B −2(width of hole + 16 in.) +

s2 4g

The minimum net area will obviously be equal to the net length multiplied by the plate thickness. 20 When holes are staggered on two legs of an angle, the gage length g is obtained by using a length between the centers of the holes measured along the centerline of the angle thickness, i.e., the distance t t A-B in Fig. 9.4. Thus the gage distance is g = ga − + gb − or 2 2

Figure 9.4: Gage Distances for an Angle

g = ga + gb − t

(9.5)

It should be mentioned that every rolled angle has a standard value for the location of holes (i.e. gage distances ga and gb ) depending on the length of the leg, Table 9.1.

21

Example 9-1: Net Area Determine the minimum net area of the plate shown below assuming as shown.

Victor Saouma

15 16

in.-diameter holes are located

Structural Engineering

9.2 Geometric Considerations

9–5

Draft

Leg g g1 g2

8 1 42 3 3

7 4 1 22 3

6 1 32 1 24 1 22

5 3 2 3 14

4 1 22

1

32 2

3 3 14

1

22 3 18

2 1 18

3

14 1

1

3

1

12

18

14

1

7 8

7 8

3 4

5 8

Table 9.1: Usual Gage Lengths g, g1 , g2 Solution: We consider various paths: % $ 1 (0.25) = 2.50 in2 12 − 2 15 16 + 16 & ' (2.125)2 (2.125)2 1 A − B − D : 12 − 3 15 (0.25) = 2.43 in2 16 + 16 + 4(2.5) + 4(4) & ' (2.125)2 (1.875)2 1 A − B − C : 12 − 3 15 (0.25) = 2.4 in2 16 + 16 + 4(2.5) + 4(4) A−D :

(9.6-a) (9.6-b) (9.6-c)

Thus path A-B-C controls.

Example 9-2: Net Area, Angle Determine the net area An for the angle shown below if

15 16

in. diameter holes are used.

Solution: For the net area calculation, the angle may be visualized as being ﬂattened into a plate as shown below.

Victor Saouma

Structural Engineering

9–6

STEEL TENSION MEMBERS

Draft

and An = Ag − Dt +

s2 t 4g

(9.7)

Where D is the width to be deducted for the hole, We consider the following paths: A−C : A−B−C :

9.2.1.2

1 4.75 − 2 15 16 + 16 0.5 & 2 (3) 1 4.75 − 3 15 16 + 16 0.5 + 4(2.5) +

2

(3) 4(4.25)

'

= 3.75 in2 0.5 = 3.96 in

2

(9.8-a) (9.8-b)

Eﬀective Net Area, Ae

Often tension members are connected at the end by connection to some but not all of the cross sectional elements. For instance, an angle section may be connected through one leg only. Thus, away from the connection we used the gross or net area to resist the full tensile load, but right at the connection, only one leg is connected and the other is for all practical purposes unstressed.

22

To account for this, the AISC code stipulates that the net area must be reduced by U , Table 9.2 where Ae = U An or Ae = U Ag .

23

Example 9-3: Reduction Factor Determine the reduction factor U to be used in computing the eﬀective net area of a W14x82 section connected by plates at its two ﬂanges. There are three bolts along each connection line. Solution: Since only two (the ﬂanges) of the three elements (the third one being the web) are connected, U < 1. b We ﬁrst check for the proﬁle ratio2 df . √ bf 10.13 = = 0.71 > 0.67 d 14.31

(9.9)

Thus from Table 9.2, U = 0.9

9.2.2 24

Stiﬀness

Tension members which are too long may be 1. Too ﬂexible during erection of the structure (excessive and possibly permanent deformations when carried due to their own weight). 2 Section

properties are taken from Sect. 3.6.

Victor Saouma

Structural Engineering

9.2 Geometric Considerations

9–7

Draft

Type of members

Minimum Number of Fasteners/line Fastened Rolled Sections Members having all cross sectional elements con1 (or welds)

Special Requirements

Ae

An

nected to transmit tensile force W, M, or S shapes with ﬂange widths not less

3 (or welds)

b d

≥ 0.67

0.9An

than 2/3 the depth, and structural tees cut from these shapes, provided the connection is to the ﬂanges. Bolted or riveted connections shall have no fewer than three fasteners per line in the direction of stress W, M, or S not meeting above conditions, structural tees cut from these shapes and all other

3 (or welds)

0.85An

shapes, including built-up cross-sections. Bolted or riveted connections shall have no fewer than three fasteners per line in the direction of stress Structural tees cut from W, M, or S connected at

3 (or welds)

b d

≥ 0.67

0.9An

ﬂanges only All members with bolted or riveted connections

2

0.75An

having only two fasteners per line in the direction of stress

Welded Plates Welds Welds Welds

l > 2w 2w > l > 1.5w l/w.1

Ag 0.87Ag 0.75Ag

b Flange width; d section depth; l Weld length; w Plate width; Table 9.2: Eﬀective Net Area Ae for Bolted and Welded Connections

Victor Saouma

Structural Engineering

9–8

STEEL TENSION MEMBERS

Draft

2. Subjected to excessive lateral deﬂections and vibrations once assembled.

25

To limit this, the AISC code imposes a constraint on the element stiﬀness.

P Recall that stiﬀness is deﬁned as K = ∆ , and for ﬂexural member it will be shown later that M 4EI Kf = θ = L where I is the moment of inertia.

26

Since # for steel E, the elastic modulus is constant, and given the deﬁnition of the radius of gyration I L , the AISC code (LRFD-B7) speciﬁes that the slenderness ratio rmin should not exceed 300 r= A

27

L < 300 rmin

(9.10)

This limitation does not apply to rods in tension. In applying this requirement, the higher slenderness ratio based on the two principal axes must be used, i.e. rLx and rLy .

28

9.3 29

LRFD Design of Tension Members

Recall the general form of the structural safety requirement under LRFD (Eq. 7.9): ΦRn ≥ Σαi Qi

(9.11)

Φt T n ≥ T u

(9.12)

For tension members, this equation becomes

where

30

= = =

resistance factor relating to tensile strength nominal strength of a tension member factored load on a tension member

From Table 7.3, Φt is 0.75 and 0.9 for fracture and yielding failure respectively.

9.3.1 31

Φt Tn Tu

Tension Failure

The design strength Φt Tn is the smaller of that based on, Fig. 9.5 1. Yielding in the gross section: We can not allow yielding of the gross section, because this will result in unacceptable elongation of the entire member.

Φt Tn = Φt Fy Ag = 0.90Fy Ag

(9.13)

or 2. Fracture in the net section: Yielding is locally allowed, because Ae is applicable only on a small portion of the element. Local excessive elongation is allowed, however fracture must be prevented. This mode of failure usually occurs if there is insuﬃcient distance behind the pin.

Φt Tn = Φt Fu Ae = 0.75Fu Ae

Victor Saouma

(9.14)

Structural Engineering

9.3 LRFD Design of Tension Members

Draft

9–9

o o o o o o o o

o o o o o o o o

o

o

o o o

o

0.75FuA e

0.9FyA g

o o o

0.75FuA e

Figure 9.5: Net and Gross Areas

9.3.2

Block Shear Failure

32

For bolted connections, tearing failure may occur and control the strength of the tension member.

33

For instance, with reference to Fig. 9.6 the angle tension member attached to the gusset plate3 may

00000 11111 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

0000 1111 0000 1111 1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

0000 1111 1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

Figure 9.6: Tearing Failure Limit State have a tearing failure along a-b-c. In this case the shear strength along a-b plus the tensile strength along bc contribute to resist this particular failure mode. This mode of failure is uncommon in tension members; However this failure mode commonly controls the design of bolted end connections to the thin webs of beams.

34

LRFD-J4 indicates that the following two equations must be considered to account for either one of the two combinations

35

Shear yielding-tension fracture: fracture on the net area in tension followed by yielding on the gross 3A

gusset plate is a plate at the intersection of members to which they are connected.

Victor Saouma

Structural Engineering

9–10

STEEL TENSION MEMBERS

Draft

section in shear (9.15)

Φt Tn = 0.75(0.6Fy Avg + Fu Ant )

Shear fracture -tension yielding: Fracture on the net section in shear followed by yielding on the gross section in tension. (9.16)

Φt Tn = 0.75(0.6Fu Ans + Fy Atg ) where

Avg = Atg = Ans = Ant = and the largest

36

gross area subjected to shear yielding gross area subjected to tensile yielding net (gross - area of holes) area subjected to shear fracture net (gross - area of holes) area subjected to tensile fracture of the two will control.

Note that a conservative simpliﬁcation of the above two equations is given below (9.17)

Φt Tn = 0.75(0.6Fu )Ans

In this equation Ans is the area along the entire failure path (including both tension and shear failure), whereas in Eq. 9.15 and 9.16 Ans is the net area where actual shear failure occurs (not including tension failure). Note that in the preceding equation, we assumed the shear yield stress τy = 0.6Fy , and the shear tensile strength τu = 0.6Fu .

37

9.3.3 38

Summary

The design procedure for tension member can be summarized as follows 1. Determine the net area An from Section 9.2.1.1; if there are staggered holes, use Eq. 9.3. 2. Determine the eﬀective area Ae from Section 9.2.1.2, and U from Table 9.2. 3. The limit load will be the lowest of the following Tension Failure: by taking the lowest of the following (a) Yielding (σyld ) in the gross section, Eq. 9.13 (b) Fracture (σu ) in the eﬀective section, Eq. 9.14 Tearing: By checking either one of the following (a) Eq. 9.17 (b) Eq. 9.15 and 9.16, take the largest value. 4. Check for slenderness, Eq. 9.10

Example 9-4: Load Capacity of Angle Determine the service load capacity in tension for an L6x4x 21 of A572 grade 50 steel connected with in. diameter bolts as shown below. Assume a live load to dead load ratio of 3.0. Since there are many fasteners, there is no need to check for shear failure. 7 8

Victor Saouma

Structural Engineering

9.3 LRFD Design of Tension Members

9–11

Draft

Solution: 1. The net area is given by (a) Section 1-1:

An = Ag − 1 hole = 4.75 −

7 1 + 8 8

0.50 = 4.25 in2

(9.18)

(b) Section 1-2:

An

= =

s2 Ag − 2 holes + t 4g 7 1 (2)2 + (0.5) = 3.95 in2 4.75 − 2 0.50 + 8 8 4(2.5)

(9.19-a) (9.19-b)

2. Since not both legs are connected, From Table 9.2 U = 0.85 and Ae = U An = (0.85)(3.95) = 3.36 in2

(9.20)

3. Checking yielding failure of the gross area (Eq. 9.13) Φt T n

= Φt Fy Ag

(9.21-a)

= 0.90(50)(4.75) = 214 k

(9.21-b)

4. Checking for fracture in the net section, Eq. 9.14 Φt T n

=

Φt Fu Ae

(9.22-a)

=

0.75(65)(3.36) = 163.8 k

(9.22-b)

5. Thus the controlling Φt Tn is 163.8 Kips 6. Next we must determine the factored load, From Section 7.3.3 we consider ΦTn = 1.2D + 1.6L, since L = 3D, this reduces to ΦTn = 1.2D + 1.6(3D) = 6.0D or ΦTn

Victor Saouma

= 163.8 = 6.0D

(9.23-a)

D

=

(9.23-b)

L

= (3)(27.3) = 81.9 k

27.3 k

(9.23-c)

Structural Engineering

9–12

STEEL TENSION MEMBERS

Draft Example 9-5: Shear Rupture

Investigate the shear rupture failure mode on the angle L4x4x 41 attached with three bolts to a 38 in. gusset plate. The material is A36 steel.

7 8

in. diameter

Solution: 1. Checking yielding failure of the gross area (Eq. 9.13) Φt T n

=

Φt Fy Ag

(9.24-a)

= =

0.90(36)(1.94) 62.9 k

(9.24-b) (9.24-c)

2. Checking for fracture in the net section, Eq. (9.14) Φt T n

= = =

Φt Fu Ae = Φt Fu U An

(9.25-a)

7 1 0.75(58)(0.85)[1.94 − ( + )(0.25)] 8 8 62.5 k

(9.25-b) (9.25-c)

3. Thus the controlling Φt Tn is 62.5 Kips 4. The block shear failure along path a-b-c investigated according to LRFD-J4 using Eq. 9.17 gives Ans

Φt T n

= (length a-b-c less three holes)thickness 7 1 = 7.5 + 1.5 − (3)( + ) 0.25 8 8

(9.26-a) (9.26-b)

= 1.5 in2

(9.26-c)

= Φ(0.6Fu )Ans = 0.75(0.6)(58)(1.5)

(9.26-d) (9.26-e)

=

(9.26-f)

39.15 k

Thus, block shear gives a lower design strength than yielding or fracture. 5. Had we instead used the larger of Eq. 9.15 and 9.16 we would have obtained

Victor Saouma

Structural Engineering

9.3 LRFD Design of Tension Members

9–13

Draft

(a) Shear yielding-tension fracture: Φt T n

= Φt (0.6Fy Avg + Fu Ant ) (9.27-a) 1 7 1 = 0.75 (0.6)(36) [(2 × 3) + 1.5] (0.25) + (58) 1.5 − ( + ) (0.25) (9.27-b) 2 8 8 = 41.25 k (9.27-c)

(b) Shear fracture -tension yielding: Φt T n

= = =

Φt (0.6Fu Ans + Fy Atg ) (9.28-a) 7 1 (9.28-b) 0.75 (0.6)(58) 7.5 − ( + )(2.5) (0.25) + (36)(1.5)(0.25) 8 8 42.75 k

(9.28-c)

6. Hence, when the more correct equations are used, the strength is controlled by block shear; Φt Tn = 43.8 k which is less than the fracture strength of 62.5 k. 7. This general result is expected when minimum number of connectors are used along with thin elements such as 14 in. or less.

Example 9-6: Shear Rupture Determine the service load that can be carried by this C15x33.9 section (Ag = 9.96 in2 ), assuming that the load is 20% dead load, and 80% live load. A36 steel with 3/4 in. diameter bolts.

C15X33.9

3" 15"

t =0.4" w

3" 3"

1.5"

3"

3"

Solution: 1. Checking yielding failure of the gross area (Eq. 9.13) Φt T n

= Φt Fy Ag

(9.29-a)

= 0.90(36)(9.96) = 322 k

(9.29-b) (9.29-c)

2. Checking for fracture in the net section, Eq. (9.14) An Ae Φt T n

Victor Saouma

3 1 = = 9.96 − 4( + )(0.4) = 9.96 − 4(0.875)(0.4) = 8.56 in2 4 8 = U An = 0.85(8.56) = 7.276 in2

(9.30-b)

= Φt Fu Ae = Φt Fu U An

(9.30-c)

= 0.75(58)(0.85)(7.276)] = 269. k

(9.30-d)

(9.30-a)

Structural Engineering

9–14

STEEL TENSION MEMBERS

Draft

3. Thus the controlling Φt Tn is 269 Kips 4. The block shear failure according to LRFD-J4 using Eq. 9.17 gives Ans

=

[2(7.5) + 9 − 8(0.875)]0.4 = 6.80 in2

(9.31-a)

Φt T n

= =

Φ(0.6Fu )Ans 0.75(0.6)(58)(6.8)

(9.31-b) (9.31-c)

=

177.5 k

(9.31-d)

Thus, block shear gives a lower design strength than yielding or fracture. 5. Had we instead used the larger of Eq. 9.15 and 9.16 we would have obtained (a) Shear yielding-tension fracture: Avg

=

2(7.5)(0.4) = 6.00 in2

(9.32-a)

Ant

=

[9 − 3(0.875)](0.4) = 2.55 in2

(9.32-b)

Φt T n

= =

Φt (0.6Fy Avg + Fu Ant ) 0.75[(0.6)(36)(6.0) + (58)(2.55)] = 208 in2

(9.32-c) (9.32-d) (9.32-e)

(b) Shear fracture -tension yielding: Ans

=

2[7.5 − (2.5)(0.875)](0.4) = 4.25 in2

(9.33-a)

Atg Φt T n

= =

(9)(0.4) = 3.60 in2 Φt (0.6Fu Ans + Fy Atg )

(9.33-b) (9.33-c)

=

0.75[(0.6)(58)(4.25) + (36)(3.6)] = 208 k

✛

(9.33-d)

6. Thus the design tension T − d is 208 kips 7. Finally we determine the service loads DL = 0.2P LL = 0.8P

(9.34-a) (9.34-b)

1.4DL = (1.4)(0.2)P = 0.28P 1.2DL + 1.6LL = (1.2)(0.2P ) + (1.6)(0.8P ) = 1.52P Φt T n DL

✛

(9.34-c) (9.34-d)

= 208 k = 1.52P ⇒ P = 137 k

(9.34-e)

= (0.2)(137) = 27 k

(9.34-f)

LL = (0.8)(137) = 109 k

(9.34-g)

Note that P is the sum of the service loads (DL +LL).

Example 9-7: Complete Analysis/Design of a Truss Design the tension members of a welded roof truss for the industrial building made of A36 steel. Trusses are space 20 ft center to center. Roof loads: 1) Built-up rooﬁng: 5.5 psf; 2) Decking: 2.5 psf; 3) Purlins: 3.0 psf; 4) Truss (estimate) 3.0 psf. Snow load 40 psf. Trolley load on bottom chord: 6.4 k at L2 and L2 ; and 12.3 k at L5 . Victor Saouma

Structural Engineering

9.3 LRFD Design of Tension Members

9–15

Draft

Solution: Loading: First we must determine the joint load in order to analyze the truss. 1. The total roof load is 5.5 + 2.5 + 3.0 + 3.0 = 14.0 psf

(9.35)

Top chord loads: D

= (14.0) psf(20) ft = 280 lbs/ft

(9.36-a)

S

= (40.0) psf(20) ft = 800 lbs/ft

(9.36-b)

2. Next we must translate this vertical uniform load into joint loads on the truss U0 , U0 : Snow Load: Roof Load: U1 , U1

: Snow Load: Roof Load:

U2 , U2 , U3 , U3 , U4 , U4 , U4

: Snow Load: Roof Load:

1 1 2 (6.42)(800) 1,000 1 1 2 (6.42)(280) 1,000 1 2 (6.42 1 2 (6.42

1 + 6.00)(800) 1,000 1 + 6.00)(280) 1,000 1 (6.00)(800) 1,000 1 (6.00)(280) 1,000

= 2.57 k (9.37-a) = 0.9 k (9.37-b) = 4.97 k (9.37-c) = 1.74 k (9.37-d) = 4.80 k (9.37-e) = 1.68 k (9.37-f)

Analysis: With the load deﬁned, we are now ready to perform a structural analysis of the truss: 1. We have three types of loads (Dead, Snow, and Live), and according to the AISC code (Sect. 7.3.3 we must consider the following three load conditions: (a) 1.4 D (b) 1.2D+1.6L+0.5S (c) 1.2D+1.6S+0.5L 2. We have 22 joints, 3 reactions and 41 members; 22(2) = 41 + 3 determinate.

√

thus the truss is statically

3. In this case, the structure is symmetrical. That is there is no horizontal displacements at joints L5 and U5 , however those joints undergo a vertical displacement. Hence, we can analyze the left half of the truss, and assume that we do have roller supports along the y axis at joints Victor Saouma

Structural Engineering

9–16

STEEL TENSION MEMBERS

Draft

L5 and U5 . This substructure will have 3 reactions, 21 elements, and 12 joints. Thus, it is statically determinate. Note that at joint U5 we will apply only one half of the load due to symmetry.

4. Hence, we shall analyze the structure for each load (D, S, and L), and then determine the various load combinations. This is best achieved by the Matrix method described in Sect. 5.2.2.1. 5. We renumber the truss as shown below.

6. Let us determine the direction cosines for each member4 . Member

joint

∆x

∆y

L

α

β

Joint

∆x

∆y

α

β

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11

6.42 0 6.42 6.42 0. 6.0 6.0 6.0 0. 6.0 6.0 6.0 0. 6.0 6.0 6.0 0. 6.0 6.0 6.0 0.

0. 4.50 -4.50 0.54 5.04 0. -5.04 .501 5.54 0. -5.54 .501 6.04 0. -6.04 .501 6.54 0. -6.54 .501 7.04

6.42 4.50 7.84 6.44 5.04 6.0 7.84 6.02 5.54 6.0 8.17 6.02 6.04 6.0 8.51 6.02 6.54 6.0 8.88 6.02 7.04

1.0 0. .82 1.0 0. 1.0 .77 1.0 0. 1.0 .73 1.0 0. 1.0 .71 1.0 0. 1.0 .68 1.0 0.

0. 1.0 -.57 .08 1.0 0. -.64 .08 1.0 0. -.68 .08 1.0 0. -.71 .08 1.0 0. -.74 .08 1.0

3 2 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12

-6.42 0. -6.42 -6.42 0. -6.0 -6.0 -6.0 0. -6.0 -6.0 -6.0 0. -6.0 -6.0 -6.0 0. -6.0 -6.0 -6.0 0.

0. -4.50 4.50 -0.54 -5.04 0. 5.04 -.501 -5.54 0. 5.54 -.501 -6.04 0. 6.04 -.501 -6.54 0. 6.54 -.501 -7.04

-1.0 0. -.82 -1.0 0. -1.0 -.77 -1.0 0. -1.0 -.73 -1.0 0. -1.0 -.71 -1.0 0. -1.0 -.68 -1.0 0.

0. -1.0 .57 -.08 -1.0 0. .64 -.08 -1.0 0. .68 -.08 -1 0. .71 -.08 -1.0 0. .74 -.08 -1.0

7. Having determined the directions cosines,and assuming all members under tension, the statics [B] matrix can be assembled5 . [B] = [BL |BR ] 4 Note

that the upper chord have an elevation of y = 4.5 + .08356x with respect to L0 . that in assembling the matrix, each column corresponds to an unknown internal element force, or an external reaction. Each row corresponds to an equation of equilibrium. To facilitate the construction of the matrix, equations of equilibrium at each joint are lumped into a “row” containing two lines. The ﬁrst line corresponds to ΣFx , and the second u one to ΣFy . Thus, the ﬁrst line contains αu j , and the second βj where the subscript corresponds to the joint (or “row” number). The superscript corresponds to the column number or unknown quantity (member force or reaction). αu j and βju are taken from the table above. 5 Note

Victor Saouma

Structural Engineering

9.3 LRFD Design of Tension Members

9–17

Draft or

# 1 # 2 # 3 # 4 # 5 # 6

[BL ] =

# 7 # 8 # 9 # 10 # 11 # 12

# 1 # 2 # 3 # 4 # 5 # 6

[BR ] =

# 7 # 8 # 9 # 10 # 11 # 12

ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy

ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy ΣFx ΣFy

                   

1. 0.

0. 1. 0. −1.

−1. 0.

 .82 −.57 −.82 .57

1. .08

−1. −.08

0. 1. 0. −1.

1. 0. −.77 .64 .77 −.64

−1. 0.

1. .08 0. 1. 0. −1.

−1. −.08

1. 0. .73 −.68 −.73 .68

−1. 0.

−1. −.08

                   

1.

0. 1. 0. −1.

1. 0. .71 −.71 −.71 .71

−1. 0.

1. .08 0. 1. 0. −1.

−1. −.08

1. 0.

−1. 0.

.68 −.74 −.68 .74

1.0 .08

−1. −.08

8. We will have three vectors of joint loads  P     1x 0.   P1y   0.           0. P2x             P2y −0.9 k              0. P3x           0.    P 3y        0.      P4x         −1.74 k    P   4y      0.        P5x    0.       P5y       0.    P6x −1.68 k = α +α P6y 1 2 0.      0. P7x             0. P7y              −1.68 k P8x           0.   P8y         0.      P9x        0.      P9y         −1.68 k      P10x      0.      P10y      0.           P 0.   12x P12y

1. .08

−0.84

k

Roof

     k          k        

0. 0. 0. −2.57 0. 0. 0. −4.97 0. 0. 0. −4.8 0. 0. 0. −4.8 0. 0. 0. −4.8 0. 0. 0. −2.4

k

     k          k        k

0. 1. 0. −1.

+α3

                                      

1. 1.

0. 0. 0. 0. 0. 0. 0. 0. 0. −6.4 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. −6.15 0. 0.

                                       

                  

F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F12 F13 F14 F15 F16 F17 F18 F19 F20 F21 R1y R11x R12x

                                      

                                      

Crane (Live Load)

Snow

9. There are many tools available to solve this problem. We will use Mathematica. Below is a listing of our input ﬁle. (* define the statics matrix *) b={ { 1., 0. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 1. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 1. , 0., 0.}, { 0., 0. , .82, 1. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., -1.,-.57, .08, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, {-1., 0.,-.82, 0., 0. , 1. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., .57, 0., 1. , 0. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.},

Victor Saouma

Structural Engineering

9–18

Draft

STEEL TENSION MEMBERS

{ 0., 0., 0., -1., 0. , 0.,-.77, 1. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0.,-.08, -1., 0., .64, .08, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., -1., .77, 0., 0. , 1. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0.,-.64, 0., 1. , 0. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0.,-1. , 0. , 0.,.73 , 1., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0.,-.08, -1., 0.,-.68, .08, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., -1.,-.73, 0., 0., 1. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0. ,.68 , 0., 1., 0. , 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., -1., 0., 0., .71, 1., 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,-.08, -1., 0.,-.71,.08 , 0., 0., 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., -1.,-.71, 0., 0. , 1. , 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0. , .71, 0., 1. , 0. , 0., 0., 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., -1., 0. , 0.,.68 , 1.0, 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,-.08, -1., 0.,-.74, .08, 0., 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., -1.,-.68, 0., 0. , 0., 1. , 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0. , .74, 0., 1. , 0., 0., 0.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,-1. , 0. , 0., 0., 1.}, { 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,-.08, -1., 0., 0., 0.} } (* define the joint load vectors (roof, snow, and live) *) r=-{0., 0., 0., -0.9, 0., 0., 0., -1.74, 0., 0., 0., -1.68, 0., 0., 0., -1.68, 0., 0., 0., -1.68, 0., 0., 0., -0.84} s=-{0., 0., 0., -2.57, 0., 0., 0., -4.97, 0., 0., 0., -4.8, 0., 0., 0., -4.8 , 0., 0., 0., -4.8 , 0., 0., 0., -2.4} l=-{0., 0., 0., 0., 0., 0., 0., 0., 0., -6.4, 0., 0., 0., 0., 0., 0., 0., 0., 0., 0., 0.,-6.15, 0., 0.} (* Invert the statics matrix b *) bm1=Inverse[b] (* Solve for the roof, snow, and live load forces *) rf=N[bm1.r,2] sf=N[bm1.s,2] lf=N[bm1.l,2] Print[" Dead (roof) Load Forces/Reactions"] Print[rf] Print[" Live (Crane) Load Forces/Reactions"] Print[lf] Print[" Snow Load Forces/Reactions"] Print[sf] (* Consider various load combinations ber AISC code *) lc1=N[1.4 rf,2] lc2=N[1.2 rf + 1.6 lf +0.5 sf,2] lc3=N[1.2 rf + 0.5 lf + 1.6 sf,2] Print[" Load Combination 1: 1.4 D"] Print[lc1] Print[" Load Combination 2: 1.2 D + 1.6 L +0.5 S"] Print[lc2] Print[" Load Combination 3: 1.2 D +0.5 L +1.6 S"] Print[lc3] Print[" Get the Maximum load of each member"] maxvec=Map[Max,Transpose[{lc1,lc2,lc3}]] Print[" Get the Minimum load of each member"] minvec=Map[Min,Transpose[{lc1,lc2,lc3}]] (* Function definitions to prune results *) pos[x_]:=If[x>0,x,0] neg[x_]:=If[x<0,x,0] (* Get the vector with Maximum and Minimum forces *) tension=Map[pos,maxvec] compression=Map[neg,minvec]

Victor Saouma

Structural Engineering

9.3 LRFD Design of Tension Members

9–19

Draft

Print[" Vector of Tensile Forces"] Print[tension] Print[" Vector of Compressive Forces"] Print[compression] (* Define a function which will give the cross sectional area *) area[x_]:=If[x>0,x/(0.9*36),"-"] (* Det ermine the area for each element which is under tension *) xarea=N[Map[area,tempos],2] Print[" Cross sectional areas for elements under tension"] Print[xarea]

10. The following reactions are obtained Load Roof Snow Crane

R1y 8.5 24 13.

R11x 19. 53. 38

R12x -19. -53. -38

√ We observe that ΣFx = 0 , we check for ΣFy = 0 for the roof load (Never ever trust any computer analysis unless you check at least equilibrium, and the order of magnitude √ of the deﬂections). The external joint load is 0.9 + 1.74 + 1.68 + 1.68 + 1.68 + 0.84 = 8.52 . Design:

1. The above results can now be summarized in the following table Member 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

L0 − L1 U0 − L 0 U0 − L 1 U0 − U1 U1 − L 1 L1 − L2 U1 − L 2 U1 − U2 U2 − L 2 L2 − L3 U2 − L 3 U2 − U3 U3 − L 3 L3 − L4 U3 − L 4 U3 − U4 U4 − L 4 L4 − L5 U4 − L 5 U4 − U5 U5 − L 5

Roof 0. -8.5 12. -9.8 -6.8 9.8 7.3 -15. -4.6 15. 4. -18. -2.7 18. 1.4 -19. -0.97 19. -0.89 -19. 0.66

Load Snow 0. -24. 34. -28. -20. 28. 21. -44. -13. 44. 11. -52. -7.8 52. 3.9 -55. -2.8 55. -2.5 -53. 1.9

Live 0. -13. 20. -16. -11. 16. 16. -29. -3.9 29. 5.2 -32. -3.6 32. 4.6 -36. -3.3 36. 4.2 -38. 3.1

Load Cases 1 2 3 0. 0. 0. -12. -35. -55. 17. 52. 79. -14. -43. -65. -9.6 -30. -45. 14. 43. 65. 10. 38. 50. -22. -72. -100. -6.5 -14. -29. 22. 72. 100. 5.6 15. 26. -26. -83. -120. -3.8 -10. -18. 26. 83. 120. 1.9 9.8 10. -27. -90. -130. -1.4 -6.9 -7.3 27. 90. 130. -1.2 5.1 -3.1 -26. -93. -130. 0.92 6. 5.3

Design Load Compr. Tens. 0 0 -55. 0 0 79. -65. 0 -45. 0 0. 65. 0. 50. -100. 0 -29. 0 0 100. 0 26. -120. 0 -18. 0 0 120. 0 10. -130. 0 -7.3 0 0 130. -3.1 5.1 -130. 0 0 6.

2. We now focus on the design of members with tensile forces (Note that member U4 − L5 ) can be subjected to both tensile and compressive forces).

Victor Saouma

Structural Engineering

9–20

Draft

STEEL TENSION MEMBERS Member L0 − L1 L1 − L2 L2 − L3 L3 − L4 L4 − L5 U0 − L 1 U1 − L 2 U2 − L 3 U3 − L 4 U4 − L 5 U5 − L 5

Des. Load kip 65. 100. 120. 130. 79. 50. 26. 10. 5.1 6.

Reqd. Area in2 2.01 3.09 3.70 4.01 2.44 1.54 0.80 0.31 0.16 0.19

Section WT 7x15 WT 7x15 WT 7x15 WT 7x15 WT 7x15 1 2L3x2 2 x 41 2L2x2x 41 3 2L2x2x 16 3 2L2x2x 16 3 2L2x2x 16 3 2L2x2x 16

Area in2 4.42 4.42 4.42 4.42 4.42 2.63 1.88 1.43 1.43 1.43 1.43

L in. 77. 72. 72. 72 72 94. 94. 98. 102. 107 84

rmin in. 1.49 1.49 1.49 1.49 1.49 0.75 0.61 0.62 0.62 0.62 0.62

L rmin

52 48 48 48 48 125 154 158 165 172 135

Comments: 1. We designed only the members under tension (Note that U4 − L5 ) would also have to be designed for compression. 2. The same size section was used for the entire lower chord. A smaller section could have been used for L0 − L2 , however the cost saving in steel would probably be oﬀset by the additional labor cost. 3. Similarly, additional savings would result in using diﬀerent sections for the inclined members. 3 4. Again we could have used angles smaller than 2L2x2x 16 , for instance 2L2x2x 81 , however it is 3 preferable to avoid thicknesses smaller than 16 in. because of welding.

5. Angles with legs smaller than 2 in. could have been used, however those may not always be available, and they may result in a slenderness ratio exceeding 300. 6. The joint details of a truss of this kind are usually worked out by the fabricator. How to review them and approve them, the Engineer must know how to design them. 7. Design of steel connections will be covered in subsequent courses (note that however we have seen above how to review some aspects of this design).

9.4

†† Computer Aided Analysis

In more advanced Structural Analysis courses (CVEN4525), you will be exposed to the computer aided analysis of structures. Using such a program (which will be made available later to you), following is the input data ﬁle: *TITLE Design of Welded truss *CONTROL 12 21 1 0 6 2 *COORD 1 0.0 0.0 2 0.0 4.5 3 6.42 0.0 4 6.42 5.04 5 12.42 0.0 6 12.42 5.54 7 18.42 0.0 8 18.42 6.04 9 24.42 0.0 10 24.42 6.54 11 30.42 0.0 12 30.42 7.04 *BOUND 1 0 1 11 1 0 12 1 0 *ELEM 1 1 3 1 2 1 2 1

Victor Saouma

0

Structural Engineering

9.4 †† Computer Aided Analysis

9–21

Draft 3 2 3 1 4 2 4 1 5 3 4 1 6 3 5 1 7 4 5 1 8 4 6 1 9 5 6 1 10 5 7 1 11 6 7 1 12 6 8 1 13 7 8 1 14 7 9 1 15 8 9 1 16 8 10 1 17 9 10 1 18 9 11 1 19 10 11 1 20 10 12 1 21 11 12 1 *PROP 1 30000. 1. *LOAD 3 3 Roof Load 6 0 0 2 0. -0.9 4 0. -1.74 6 0. -1.68 8 0. -1.68 10 0. -1.68 12 0. -0.84 Snow Load 6 0 0 2 0. -2.57 4 0. -4.97 6 0. -4.8 8 0. -4.8 10 0. -4.8 12 0. -2.4 Crane load 2 0 0 5 0. -6.4 11 0. -6.15 *COMB LC 1 1 1 1.4 LC 2 3 1 1.2 2 0.5 3 1.6 LC 3 3 1 1.2 2 1.6 3 0.5

The output of such an analysis would be as follow. Compare those results with the one obtained by inverting the statics matrix. *********************************** I S D S INTEGRATED STRUCTURAL DESIGN SYSTEM *********************************** DEPT. OF CIVIL ENGINEERING, UNIV. OF COLORADO, BOULDER, CO Version 0.2 July 1991

NUMBER OF JOINTS -----12

NUMBER OF MEMBERS ------21

NUMBER OF PROPERTY GROUPS --------1

NUMBER OF ELASTIC SPRINGS --------0

NUMBER OF LOADING CASES --------6

ASSOCIATED D.O.F. NUMBERS AND COORDINATES ----------------------------------------JOINT DELTA-X DELTA-Y GLOBAL GLOBAL NUMBER D.O.F.# D.O.F.# X-COOR Y-COOR ------ ------- ------- ------ -----1 1 0 0.0 0.0 2 2 3 0.0 4.5 3 4 5 6.4 0.0 4 6 7 6.4 5.0 5 8 9 12.4 0.0 6 10 11 12.4 5.5 7 12 13 18.4 0.0 8 14 15 18.4 6.0 9 16 17 24.4 0.0

Victor Saouma

STRUCTURE TYPE CODE NUMBER --------2

LOAD CASE:

5

DISPLACEMENTS ------------DEFLECTIONS JOINT ----------------------NUMBER DELTA-X DELTA-Y ------ ----------- ----------1 -6.8194E-02 0.0000E+00 2 4.6579E-02 -6.3711E-03 3 -6.8194E-02 -1.9852E-01 4 5.2197E-02 -2.0454E-01 5 -5.7975E-02 -3.5366E-01 6 4.7734E-02 -3.5703E-01 7 -4.0982E-02 -4.6059E-01 8 3.6636E-02 -4.6314E-01 9 -2.1247E-02 -5.2493E-01 10 2.0455E-02 -5.2659E-01 11 0.0000E+00 -5.4697E-01 12 0.0000E+00 -5.4536E-01 REACTIONS --------JOINT FORCE IN X FORCE IN Y NUMBER DIRECTION DIRECTION ------ ---------- ---------1 0.000E+00 4.247E+01 11 1.090E+02 0.000E+00 12 -1.090E+02 0.000E+00

Structural Engineering

9–22

STEEL TENSION MEMBERS

Draft 10 11 12

18 0 0

19 20 21

24.4 30.4 30.4

6.5 0.0 7.0

LOAD CASE:

MEMBER INCIDENCES AND PROPERTY GROUP ID NUMBERS ----------------------------------------------MEMBER START END PROPERTY NUMBER JOINT JOINT GROUP # ------ ----- ----- -------1 1 3 1 2 1 2 1 3 2 3 1 4 2 4 1 5 3 4 1 6 3 5 1 7 4 5 1 8 4 6 1 9 5 6 1 10 5 7 1 11 6 7 1 12 6 8 1 13 7 8 1 14 7 9 1 15 8 9 1 16 8 10 1 17 9 10 1 18 9 11 1 19 10 11 1 20 10 12 1 21 11 12 1

REACTIONS --------JOINT FORCE IN X FORCE IN Y NUMBER DIRECTION DIRECTION ------ ---------- ---------1 0.000E+00 5.544E+01 11 1.251E+02 0.000E+00 12 -1.251E+02 0.000E+00

MEMBER LOADS -----------MEMBER LOAD JOINT AXIAL NUMBER CASE NUMBER LOAD ------ ---- ------ ---------

MEMBER PROPERTY GROUP DEFINITIONS --------------------------------PROPERTY YOUNG"S AREA OF GROUP # MODULUS X-SECTION -------- --------- --------1 3.0000E+04 1.0000E+00

1

1 2

NUMBER OF APPLIED LOAD CASES ---------3

NUMBER OF COMBINED LOAD CASES ---------3

3 4 5 6

load case: Roof Load

1 2

APPLIED JOINT LOADS ------------------JOINT FORCE IN X FORCE IN Y NUMBER DIRECTION DIRECTION ------ ---------- ---------2 0.000E+00 -9.000E-01 4 0.000E+00 -1.740E+00 6 0.000E+00 -1.680E+00 8 0.000E+00 -1.680E+00 10 0.000E+00 -1.680E+00 12 0.000E+00 -8.400E-01

1 2 3 4 5 6

3 load case: Snow Load

2

1 2 3

APPLIED JOINT LOADS ------------------JOINT FORCE IN X FORCE IN Y NUMBER DIRECTION DIRECTION ------ ---------- ---------2 0.000E+00 -2.570E+00 4 0.000E+00 -4.970E+00 6 0.000E+00 -4.800E+00 8 0.000E+00 -4.800E+00 10 0.000E+00 -4.800E+00 12 0.000E+00 -2.400E+00

4 5 6

4

1 2 3

load case: Crane load

3

4 5

APPLIED JOINT LOADS ------------------JOINT FORCE IN X FORCE IN Y NUMBER DIRECTION DIRECTION ------ ---------- ---------5 0.000E+00 -6.400E+00 11 0.000E+00 -6.150E+00

6

5

1 2 3

R E S U L T

OF

M E S H

O P T I M I Z A T I O N 4

BANDWIDTH PROFILE

INITIAL MESH 2 21

OPTIMIZED MESH 2 21

5 6

OLD NODES : NEW NODES :

1 1

2 2

3 3

4 4

5 5

6 6

7 7

8 8

9 9

10 10

11 11

12 12

6

1 2

SKYLINE SOLVER STATISTICS AVERAGE BANDWIDTH............................... NUMBER OF EQUATIONS............................. NUMBER OF NON-ZERO ELEMENTS IN STIFFNESS MATRIX. MAXIMUM CAPACITY OF EACH BLOCK..................

Victor Saouma

6

DISPLACEMENTS ------------DEFLECTIONS JOINT ----------------------NUMBER DELTA-X DELTA-Y ------ ----------- ----------1 -8.2746E-02 0.0000E+00 2 5.5835E-02 -8.3164E-03 3 -8.2746E-02 -2.4162E-01 4 6.2249E-02 -2.4915E-01 5 -6.9944E-02 -4.2656E-01 6 5.6902E-02 -4.3183E-01 7 -4.9587E-02 -5.5737E-01 8 4.3305E-02 -5.6085E-01 9 -2.5489E-02 -6.3314E-01 10 2.3701E-02 -6.3466E-01 11 0.0000E+00 -6.5504E-01 12 0.0000E+00 -6.5373E-01

3 4 21 101 101

4 5

1 3 1 3 1 3 1 3 1 3 1 3

0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00 0.000E+00

1 2 1 2 1 2 1 2 1 2 1 2

8.520E+00 -8.520E+00 2.434E+01 -2.434E+01 1.255E+01 -1.255E+01 1.193E+01 -1.193E+01 4.247E+01 -4.247E+01 5.544E+01 -5.544E+01

2 3 2 3 2 3 2 3 2 3 2 3

-1.185E+01 1.185E+01 -3.386E+01 3.386E+01 -1.952E+01 1.952E+01 -1.659E+01 1.659E+01 -6.239E+01 6.239E+01 -7.817E+01 7.817E+01

2 4 2 4 2 4 2 4 2 4 2 4

9.741E+00 -9.741E+00 2.783E+01 -2.783E+01 1.604E+01 -1.604E+01 1.364E+01 -1.364E+01 5.127E+01 -5.127E+01 6.424E+01 -6.424E+01

3 4 3 4 3 4 3 4 3 4 3 4

6.804E+00 -6.804E+00 1.944E+01 -1.944E+01 1.121E+01 -1.121E+01 9.525E+00 -9.525E+00 3.581E+01 -3.581E+01 4.487E+01 -4.487E+01

3 5 3 5 3 5 3 5 3 5

-9.706E+00 9.706E+00 -2.773E+01 2.773E+01 -1.599E+01 1.599E+01 -1.359E+01 1.359E+01 -5.109E+01 5.109E+01

Structural Engineering

9.4 †† Computer Aided Analysis

Draft

NUMBER OF BLOCK(S).............................. MAXIM. CAPACITY FOR CONTROL INFORMATION......... ESTIMATE OF UNUSED STORAGE FOR EACH BLOCK.......

1 21 11

LENGTH OF STIFFNESS ARRAY....................... TOTAL LENGTH OF ARRAY AVAILABLE................. TOTAL NUMBER OF I/O PERFORMED...................

101 49106 0

9–23

7

6

3 5

-6.401E+01 6.401E+01

1

4 5 4 5 4 5 4 5 4 5 4 5

-7.173E+00 7.173E+00 -2.049E+01 2.049E+01 -1.587E+01 1.587E+01 -1.004E+01 1.004E+01 -4.424E+01 4.424E+01 -4.933E+01 4.933E+01

4 6 4 6 4 6 4 6 4 6 4 6

1.525E+01 -1.525E+01 4.357E+01 -4.357E+01 2.823E+01 -2.823E+01 2.135E+01 -2.135E+01 8.526E+01 -8.526E+01 1.021E+02 -1.021E+02

5 6 5 6 5 6 5 6 5 6 5 6

4.613E+00 -4.613E+00 1.318E+01 -1.318E+01 3.805E+00 -3.805E+00 6.459E+00 -6.459E+00 1.822E+01 -1.822E+01 2.853E+01 -2.853E+01

5 7 5 7 5 7 5 7 5 7 5 7

-1.520E+01 1.520E+01 -4.342E+01 4.342E+01 -2.814E+01 2.814E+01 -2.128E+01 2.128E+01 -8.497E+01 8.497E+01 -1.018E+02 1.018E+02

6 7 6 7 6 7 6 7 6 7 6 7

-3.966E+00 3.966E+00 -1.133E+01 1.133E+01 -5.145E+00 5.145E+00 -5.553E+00 5.553E+00 -1.866E+01 1.866E+01 -2.546E+01 2.546E+01

6 8 6 8 6 8 6 8 6 8 6 8

1.818E+01 -1.818E+01 5.193E+01 -5.193E+01 3.203E+01 -3.203E+01 2.545E+01 -2.545E+01 9.902E+01 -9.902E+01 1.209E+02 -1.209E+02

7 8 7 8 7 8 7 8 7 8 7 8

2.691E+00 -2.691E+00 7.688E+00 -7.688E+00 3.490E+00 -3.490E+00 3.767E+00 -3.767E+00 1.266E+01 -1.266E+01 1.727E+01 -1.727E+01

7 9 7 9 7 9 7 9 7 9 7 9

-1.811E+01 1.811E+01 -5.175E+01 5.175E+01 -3.192E+01 3.192E+01 -2.536E+01 2.536E+01 -9.867E+01 9.867E+01 -1.205E+02 1.205E+02

8 9 8 9 8 9

-1.316E+00 1.316E+00 -3.759E+00 3.759E+00 -4.544E+00 4.544E+00

2 3 4

load case: LC 1

4 5 6

NUMBER OF LOAD CASES COMBINED ---------1

8

1 2

LOAD CASE COMBINATION ID NUMBER FACTOR --------- ----------1 1.400

3 4 5 6

load case: LC 2

5 9

NUMBER OF LOAD CASES COMBINED ---------3

1 2 3 4

LOAD CASE COMBINATION ID NUMBER FACTOR --------- ----------1 1.200 2 0.500 3 1.600

5 6

10

1 2

load case: LC 3

6

3 4

NUMBER OF LOAD CASES COMBINED ---------3

5 6

11 LOAD CASE COMBINATION ID NUMBER FACTOR --------- ----------1 1.200 2 1.600 3 0.500

1 2 3 4 5

LOAD CASE:

1

DISPLACEMENTS ------------DEFLECTIONS JOINT ----------------------NUMBER DELTA-X DELTA-Y ------ ----------- ----------1 -1.2412E-02 0.0000E+00 2 8.3395E-03 -1.2780E-03 3 -1.2412E-02 -3.6280E-02 4 9.2804E-03 -3.7423E-02 5 -1.0470E-02 -6.3848E-02 6 8.4821E-03 -6.4700E-02 7 -7.4305E-03 -8.3526E-02 8 6.4357E-03 -8.4067E-02 9 -3.8080E-03 -9.4769E-02 10 3.4968E-03 -9.4973E-02 11 0.0000E+00 -9.7803E-02 12 0.0000E+00 -9.7640E-02

6

12

1 2 3 4 5 6

13

1 2 3

REACTIONS --------JOINT FORCE IN X FORCE IN Y NUMBER DIRECTION DIRECTION ------ ---------- ---------1 0.000E+00 8.520E+00 11 1.840E+01 0.000E+00 12 -1.840E+01 0.000E+00

4 5 6

14

1 2

LOAD CASE:

2 3

DISPLACEMENTS ------------DEFLECTIONS JOINT ----------------------NUMBER DELTA-X DELTA-Y ------ ----------- ----------1 -3.5460E-02 0.0000E+00 2 2.3826E-02 -3.6510E-03 3 -3.5460E-02 -1.0365E-01 4 2.6515E-02 -1.0692E-01 5 -2.9914E-02 -1.8242E-01 6 2.4234E-02 -1.8485E-01 7 -2.1229E-02 -2.3864E-01 8 1.8387E-02 -2.4018E-01

Victor Saouma

4 5 6

15

1 2 3

Structural Engineering

9–24

STEEL TENSION MEMBERS

Draft 9 10 11 12

-1.0880E-02 9.9907E-03 0.0000E+00 0.0000E+00

-2.7076E-01 -2.7134E-01 -2.7943E-01 -2.7896E-01

4 5 6

REACTIONS --------JOINT FORCE IN X FORCE IN Y NUMBER DIRECTION DIRECTION ------ ---------- ---------1 0.000E+00 2.434E+01 11 5.258E+01 0.000E+00 12 -5.258E+01 0.000E+00

16

1 2 3 4 5

LOAD CASE:

3 6

DISPLACEMENTS ------------DEFLECTIONS JOINT ----------------------NUMBER DELTA-X DELTA-Y ------ ----------- ----------1 -2.2231E-02 0.0000E+00 2 1.5411E-02 -1.8825E-03 3 -2.2231E-02 -6.4475E-02 4 1.7377E-02 -6.6357E-02 5 -1.9034E-02 -1.1615E-01 6 1.5899E-02 -1.1685E-01 7 -1.3407E-02 -1.5065E-01 8 1.2325E-02 -1.5136E-01 9 -7.0236E-03 -1.7239E-01 10 7.0396E-03 -1.7310E-01 11 0.0000E+00 -1.8119E-01 12 0.0000E+00 -1.8045E-01

17

3 4 5 6

18

3 4 5 6

19 4

DISPLACEMENTS ------------DEFLECTIONS JOINT ----------------------NUMBER DELTA-X DELTA-Y ------ ----------- ----------1 -1.7376E-02 0.0000E+00 2 1.1675E-02 -1.7892E-03 3 -1.7376E-02 -5.0792E-02 4 1.2993E-02 -5.2392E-02 5 -1.4658E-02 -8.9387E-02 6 1.1875E-02 -9.0580E-02 7 -1.0403E-02 -1.1694E-01 8 9.0100E-03 -1.1769E-01 9 -5.3311E-03 -1.3268E-01 10 4.8956E-03 -1.3296E-01 11 0.0000E+00 -1.3692E-01 12 0.0000E+00 -1.3670E-01 REACTIONS --------JOINT FORCE IN X FORCE IN Y NUMBER DIRECTION DIRECTION ------ ---------- ---------1 0.000E+00 1.193E+01 11 2.576E+01 0.000E+00 12 -2.576E+01 0.000E+00

Victor Saouma

1 2

REACTIONS --------JOINT FORCE IN X FORCE IN Y NUMBER DIRECTION DIRECTION ------ ---------- ---------1 0.000E+00 1.255E+01 11 3.787E+01 0.000E+00 12 -3.787E+01 0.000E+00

LOAD CASE:

1 2

1 2 3 4 5 6

20

1 2 3 4 5 6

21

1 2

8 9 8 9 8 9

-1.842E+00 1.842E+00 -1.073E+01 1.073E+01 -9.865E+00 9.865E+00

8 10 8 10 8 10 8 10 8 10 8 10

1.911E+01 -1.911E+01 5.459E+01 -5.459E+01 3.524E+01 -3.524E+01 2.675E+01 -2.675E+01 1.066E+02 -1.066E+02 1.279E+02 -1.279E+02

9 10 9 10 9 10 9 10 9 10 9 10

9.333E-01 -9.333E-01 2.667E+00 -2.667E+00 3.224E+00 -3.224E+00 1.307E+00 -1.307E+00 7.611E+00 -7.611E+00 6.999E+00 -6.999E+00

9 11 9 11 9 11 9 11 9 11 9 11

-1.904E+01 1.904E+01 -5.440E+01 5.440E+01 -3.512E+01 3.512E+01 -2.666E+01 2.666E+01 -1.062E+02 1.062E+02 -1.274E+02 1.274E+02

10 11 10 11 10 11 10 11 10 11 10 11

9.413E-01 -9.413E-01 2.689E+00 -2.689E+00 -4.064E+00 4.064E+00 1.318E+00 -1.318E+00 -4.028E+00 4.028E+00 3.400E+00 -3.400E+00

10 12 10 12 10 12 10 12 10 12 10 12

1.847E+01 -1.847E+01 5.276E+01 -5.276E+01 3.800E+01 -3.800E+01 2.585E+01 -2.585E+01 1.093E+02 -1.093E+02 1.256E+02 -1.256E+02

11 12 11 12

-6.936E-01 6.936E-01 -1.982E+00 1.982E+00

Structural Engineering

Draft Chapter 10

INTERNAL FORCES IN STRUCTURES 1 This chapter will start as a review of shear and moment diagrams which you have studied in both Statics and Strength of Materials, and will proceed with the analysis of statically determinate frames, arches and grids. 2 By the end of this lecture, you should be able to draw the shear, moment and torsion (when applicable) diagrams for each member of a structure. 3 Those diagrams will subsequently be used for member design. For instance, for ﬂexural design, we will consider the section subjected to the highest moment, and make sure that the internal moment is equal and opposite to the external one. For the ASD method, the basic beam equation (derived in Strength of Materials) σ = MC I , (where M would be the design moment obtained from the moment diagram) would have to be satisﬁed. 4 Some of the examples ﬁrst analyzed in chapter 4 (Reactions), will be revisited here. Later on, we will determine the deﬂections of those same problems.

10.1

Design Sign Conventions

5

Before we (re)derive the Shear-Moment relations, let us arbitrarily deﬁne a sign convention.

6

The sign convention adopted here, is the one commonly used for design purposes1 .

7

With reference to Fig. 10.1

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 ﬁbers, and compression in the upper ones. Alternatively, moments are drawn on the compression side (useful to keep in mind for frames). 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 positive 3D: Use double arrow vectors (and NOT curved arrows). Forces and moments (including torsions) are deﬁned with respect to a right hand side coordinate system, Fig. ??. 1 Later

on, in more advanced analysis courses we will use a diﬀerent one.

10–2

INTERNAL FORCES IN STRUCTURES

Draft

+ Axial Force

+ve Load

+ +ve Shear

+ve Moment

Figure 10.1: Shear and Moment Sign Conventions for Design

My

Tx

X

Mz

Y

My Mz

Z

Tx Figure 10.2: Sign Conventions for 3D Frame Elements

Victor Saouma

Structural Engineering

10.2 Load, Shear, Moment Relations

10–3

Draft 10.2

Load, Shear, Moment Relations

8 Let us (re)derive the basic relations between load, shear and moment. Considering an inﬁnitesimal length dx of a beam subjected to a positive load2 w(x), Fig. 10.3. The inﬁnitesimal section must also

Figure 10.3: Free Body Diagram of an Inﬁnitesimal Beam Segment be in equilibrium. There are no axial forces, thus we only have two equations of equilibrium to satisfy ΣFy = 0 and ΣMz = 0.

9

Since dx is inﬁnitesimally small, the small variation in load along it can be neglected, therefore we assume w(x) to be constant along dx.

10

To denote that a small change in shear and moment occurs over the length dx of the element, we add the diﬀerential quantities dVx and dMx to Vx and Mx on the right face.

11

12

Next considering the ﬁrst equation of equilibrium (+ ✻ ) ΣFy = 0 ⇒ Vx + wx dx − (Vx + dVx ) = 0

or dV = w(x) dx

(10.1)

The slope of the shear curve at any point along the axis of a member is given by the load curve at that point. 13

Similarly (+ ✛) ✁ ΣMO = 0 ⇒ Mx + Vx dx − wx dx

dx − (Mx + dMx ) = 0 2

Neglecting the dx2 term, this simpliﬁes to dM = V (x) dx

(10.2)

The slope of the moment curve at any point along the axis of a member is given by the shear at that point. 14

Alternative forms of the preceding equations V = w(x)dx

(10.3)

∆V21

=

Vx2 − Vx1 =

x2

w(x)dx

(10.4)

x1 2 In

this derivation, as in all other ones we should assume all quantities to be positive.

Victor Saouma

Structural Engineering

10–4

INTERNAL FORCES IN STRUCTURES

Draft

The change in shear between 1 and 2, ∆V1−2 , is equal to the area under the load between x1 and x2 .

and

M

=

V (x)dx

(10.5)

∆M21

= M2 − M1 =

x2

V (x)dx

(10.6)

x1

The change in moment between 1 and 2, ∆M21 , is equal to the area under the shear curve between x1 and x2 . 15

Note that we still need to have V1 and M1 in order to obtain V2 and M2 .

Similar relations will be determined later (Chapter 12) between curvature 12.30).

16

M EI

and rotation θ (Eq.

Fig. 10.4 and 10.5 further illustrates the variation in internal shear and moment under uniform and concentrated forces/moment.

17

Figure 10.4: Shear and Moment Forces at Diﬀerent Sections of a Loaded Beam

10.3 18

Moment Envelope

For design, we often must consider diﬀerent load combinations.

For each load combination, we should draw the shear, moment diagrams. and then we should use the Moment envelope for design purposes.

19

10.4

Examples

10.4.1

Beams

Example 10-1: Simple Shear and Moment Diagram Victor Saouma

Structural Engineering

10.4 Examples

Draft

10–5 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 10.5: Slope Relations Between Load Intensity and Shear, or Between Shear and Moment Draw the shear and moment diagram for the beam shown below

Solution: The free body diagram is drawn below

Victor Saouma

Structural Engineering

10–6

Draft

INTERNAL FORCES IN STRUCTURES

Reactions are determined from the equilibrium equations (+ )✛ΣFx = 0; ⇒ −RAx + 6 = 0 ⇒ RAx = 6 k (+ ✛) ✁ ΣMA = 0; ⇒ (11)(4) + (8)(10) + (4)(2)(14 + 2) − RFy (18) = 0 ⇒ RFy = 14 k (+ ✻) ΣFy = 0; ⇒ RAy − 11 − 8 − (4)(2) + 14 = 0 ⇒ RAy = 13 k Shear are determined next. 1. At A the shear is equal to the reaction and is positive. 2. At B the shear drops (negative load) by 11 k to 2 k. 3. At C it drops again by 8 k to −6 k. 4. It stays constant up to D and then it decreases (constant negative slope since the load is uniform and negative) by 2 k per linear foot up to −14 k. 5. As a check, −14 k is also the reaction previously determined at F . Moment is determined last: 1. The moment at A is zero (hinge support). 2. The change in moment between A and B is equal to the area under the corresponding shear diagram, or ∆MB−A = (13)(4) = 52. 3. etc...

Example 10-2: Sketches of Shear and Moment Diagrams Victor Saouma

Structural Engineering

10.4 Examples

Draft

10–7

For each of the following examples, sketch the shear and moment diagrams.

10.4.2

Frames

Inclined loads on inclined members are often mishandled. With reference to Fig. 10.6 we would have the following relations

20

Victor Saouma

Structural Engineering

10–8

INTERNAL FORCES IN STRUCTURES

Draft L w1

ly

lx

LX w4

θ l

w2

W

l

LY

w

w3

l l

θ

l

y

l

θ y

lx

x

Figure 10.6: Inclined Loads on Inclined Members

w

=

W

=

lx l ly l

ly lx ly lx + w2 + w3 + w4 l l l l ly lx ly lx w1 L + w2 L + w3 LY + w4 LX l l l l

w1

(10.7) (10.8)

=

cos θ

(10.9)

=

sin θ

(10.10)

Example 10-3: Frame Shear and Moment Diagram Draw the shear and moment diagram of the following frame

Solution:

Victor Saouma

Structural Engineering

10.4 Examples

10–9

Draft

Reactions are determined ﬁrst 4 )✛ΣFx = 0; ⇒ RAx − (3)(15) = 0 5 load ⇒ RAx = 36 k 9 12 3 4 30 (+ ✛) − (3)(15) −39RDy = 0 ✁ ΣMA = 0; ⇒ (3)(30)( 2 ) + (3)(15) 30 + 5 2 5 2

(+

CDY

CDX

⇒ RDy = 52.96 k ) ΣFy = 0; ⇒ RAy − (3)(30) − 35 (3)(15) + 52.96 = 0 (+ ✻ ⇒ RAy = 64.06 k Victor Saouma

Structural Engineering

10–10

INTERNAL FORCES IN STRUCTURES

Draft Shear:

1. For A − B, the shear is constant, equal to the horizontal reaction at A and negative according to our previously deﬁned sign convention, VA = −36 k 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 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 35 52.96 = 31.78 k 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 5. We check our calculations by verifying equilibrium of node C (+ (+

)✛ΣFx = 0 ) ΣFy = 0 ✻

⇒ ⇒

3 5 (42.37) + 4 5 (42.37) −

4 5 (13.22) 3 5 (13.22)

√ = 25.42 + 10.58 = 36 √ = 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 k.ft 2. Along B − C, the moment is equal to x x MB−C = MB + VB−C (x)dx = −432 + (64.06 − 3x)dx 0

=

2

−432 + 64.06x − 3 x2

0

which is a parabola. Substituting for x = 30, we obtain at node C: MC = −432 + 64.06(30) − 2 3 302 = 139.8 k.ft B−C 3. If we need to determine the maximum moment along B − C, we know that dMdx = 0 at the 64.06 point where VB−C = 0, that is VB−C (x) = 64.06 − 3x = 0 ⇒ x = 3 = 25.0 ft. In other words, maximum moment occurs where the shear is zero. 2

max = −432 + 64.06(25.0) − 3 (25.0) = −432 + 1, 601.5 − 937.5 = 232 k.ft Thus MB−C 2

4. Finally along C − D, the moment varies quadratically (since we had a linear shear), the moment ﬁrst increases (positive shear), and then decreases (negative shear). The moment along C − D is given by (x (x MC−D = MC + 0 VC−D (x)dx = 139.8 + 0 (13.22 − 3x)dx 2 = 139.8 + 13.22x − 3 x2 which is a parabola. 2 Substituting for x√= 15, we obtain at node C MC = 139.8 + 13.22(15) − 3 152 = 139.8 + 198.3 − 337.5 = 0

Example 10-4: Frame Shear and Moment Diagram; Hydrostatic Load Victor Saouma

Structural Engineering

10.4 Examples

10–11

Draft

The frame shown below is the structural support of a ﬂume. Assuming that the frames are spaced 2 ft apart along the length of the ﬂume, 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 ﬁxed at the lower end. The pressure is linear and is given by p = γh. Since each frame supports a 2 ft wide slice of the ﬂume, the equation for w (pounds/foot) is w

= =

(2)(62.4)(h) 124.8h lbs/ft

At the base w = (124.8)(6) = 749 lbs/ft = .749 k/ft Note that this is both the lateral pressure on the end walls as well as the uniform load on the horizontal members.

Victor Saouma

Structural Engineering

10–12

Draft

INTERNAL FORCES IN STRUCTURES

End Actions 1. Base force at B is FBx = (.749) 26 = 2.246 k 2. Base moment at B is MB = (2.246) 36 = 4.493 k.ft 3. End force at B for member B − E are equal and opposite. 4. Reaction at C is RCy = (.749) 16 2 = 5.99 k Shear forces 1. Base at B the shear force was determined earlier and was equal to 2.246 k. 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 k. 4. The shear to the right of C is V = −2.246 + 5.99 = 3.744 k Moment diagrams 1. At the base: B M = 4.493 k.ft as determined above. 2. At the support C, Mc = −4.493 + (−.749)(3)( 23 ) = −7.864 k.ft 3. The maximum moment is equal to Mmax = −7.864 + (.749)(5)( 52 ) = 1.50 k.ft

Victor Saouma

Structural Engineering

10.4 Examples

Draft

10–13

Design: Reinforcement should be placed along the ﬁbers which are under tension, that is on the side of the negative moment3 . The ﬁgure below schematically illustrates the location of the ﬂexural4 reinforcement.

Example 10-5: Shear Moment Diagrams for Frame 3 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”. 4 Shear reinforcement is made of a series of vertical stirrups.

Victor Saouma

Structural Engineering

10–14

INTERNAL FORCES IN STRUCTURES

Draft

8’

12’

10’ 2k/ft

30k

5k/ft

B

A

Vba C

Hbd

M ba

E

Vbc

10k

5’ 20k

VA

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

2k/ft 10k

5k/ft B

A 17.5k

Hba M ba

Vba

B

Hbc M bc

C Vbc

(10)+(2)(10) 30k

17.5-5x=0

-22.5k

3.5’

-22.5+(-30)

10k

Vbc

17.5k

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)(8-3.5)/2

(-52.5)(12)+(-20)

-650’k

M ba

Vbd M bd

50k

20k

450’k (50)(15)-[(4)(5)/2][(2)(15)/3)]

(50)-(4)(15)/2

450’k

Hbd

20k

4k/ft

50k 82.5k

Example 10-6: Shear Moment Diagrams for Inclined Frame Victor Saouma

Structural Engineering

10.4 Examples

10–15

Draft

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

800’k

F

0k’

60k 2

7.69k

1k

20k

(20)(15)/13=7.7 0k’

0k 39.1k

CD 7

29.3k

48.9k

’

777k

’k 1130

800’k

9 B-C

13

1

-16 k

7k ’

1,130-(.58)(13) k’ 800+(25.4)(13) 1122

3.1

8 B-C

77

-2

.1k -2 39.

1,122-(26.6)(13)

k 26.6k -0.58 6 - -0.6-26 25.42

11 C-D

(39.1)(12.5)

488’k 12 C-D

14

k

-23 .1 k 6.6k k 8 5 . -0 -2 -39 .

0’k 113 1k

2k’

112

(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 777k’

48

8’

k

800’k

+60k

+20k

+25.4

10

-23

488+(23.1)(12.5)

+25

8k 12k

778k’

BC

19.2k 7.7 17.7+ .4k

6

16k

11.1k

28.8k

20k

18.46k 7.38k

5

17.2

23.

0k

26.6k

800k’

48.8k

4

28.8k

778k’ 10k

24k

24k

3

20-10-10

26k 26k 10k

17.72k

F/Fy=z/x F/Fx=z/y Fx/Fy=y/x

ED 1

y x

AB 19.2k

z

Fy

20k

60-(2)(20)

20k

(20)(20)+(60-20)(20)/2

19.2k

800k’

(60)(20)-(2)(20)(20)/2

Va

10.4.3

3D Frame

Example 10-7: 3D Frame

Victor Saouma

Structural Engineering

10–16

INTERNAL FORCES IN STRUCTURES

Draft

1. The frame has a total of 6 reactions (3 forces and 3 moments) at the support, and we have a total of 6 equations of equilibrium, thus it is statically determinate. 2. Each member has the following internal forces (deﬁned in terms of the local coordinate system of each member x − y − z ) Member Member

Axial Nx

C −D B−C A−B

√ √

Internal Forces Shear Moment Vy Vz My Mz √ √ √ √ √ √ √ √ √ √

Torsion T x √ √

3. The numerical calculations for the analysis of this three dimensional frame are quite simple, however the main complexity stems from the diﬃculty in visualizing the inter-relationships between internal forces of adjacent members. 4. In this particular problem, rather than starting by determining the reactions, it is easier to determine the internal forces at the end of each member starting with member C − D. Note that temporarily we adopt a sign convention which is compatible with the local coordinate systems. C-D ΣFy ΣFz ΣMy ΣMz

=0 =0 =0 =0

⇒ ⇒ ⇒ ⇒

VyC = (20)(2) VzC MyC MzC

ΣFx ΣFy ΣMy ΣMz ΣTx

=0 =0 =0 =0 =0

⇒ ⇒ ⇒ ⇒ ⇒

NxB VyB MyB MzB TxB

B-C

Victor Saouma

= +40kN = +60kN = −(60)(2) = −120kN.m = (20)(2) 22 = +40kN.m

= VzC = VyC = MyC C = V y (4) = (40)(4) = −MzC

= −60kN = +40kN = −120kN.m = +160kN.m = −40kN.m

Structural Engineering

10.4 Examples

10–17

Draft A-B

ΣFx ΣFy ΣMy ΣMz ΣTx

⇒ ⇒ ⇒ ⇒ ⇒

=0 =0 =0 =0 =0

NxA VyA MyA MzA TxA

= VyB = NxB = TxB = MzB + NxB (4) = 160 + (60)(4) = MyB

= +40kN = +60kN = +40kN.m = +400kN.m = −120kN.m

The interaction between axial forces N and shear V as well as between moments M and torsion T is clearly highlighted by this example. 120 kN-m

y’

60

C B

40 kN

160 kN m -m 40 kN N- kN k 40 60 120 kN-m 40 kN N 120 kN-m k 60 -m kN 40 160 kN m 160 -m kN 0 4 kN -m N 40 kN k 60 120 kN-m 120 kN-m

120 kN-m -m kN 40 kN 60

-m

N 0k

4

y’

x’

kN

60

kN

40 120 kN-m 40 kN

40 kN

x’

z’

60

m

kN

y’

40

20 kN/m

kN -m

60

kN

kN

120 kN-m

40 kN kN 60

-m

40

kN

160

x’

kN

-m

z’

y’

-m

N 0k

4

Victor Saouma

400

kN N 120 kN-m-m k 60 40 kN

Structural Engineering

10–18

INTERNAL FORCES IN STRUCTURES

Draft

y’ C

B

y’

x’

y’

x’

C

C

B

40 z’

40 160

120

60

z’

z’

D M

x’ V

40

y’

V

M

C

T

40 z’

120 D x’

x’

B

x’

60

B

40 A y’

A z’ V

10.5 21

y’

x’ B

160

120 A

400 z’ M

T

Arches

See section ??.

Victor Saouma

Structural Engineering

Draft Chapter 11

ARCHES and CURVED STRUCTURES 1

This chapter will concentrate on the analysis of arches.

2 The concepts used are identical to the ones previously seen, however the major (and only) diﬀerence is that equations will be written in polar coordinates. 3 Like cables, arches can be used to reduce the bending moment in long span structures. Essentially, an arch can be considered as an inverted cable, and is transmits the load primarily through axial compression, but can also resist ﬂexure through its ﬂexural rigidity.

4

A parabolic arch uniformly loaded will be loaded in compression only.

5 A semi-circular arch uniﬁrmly loaded will have some ﬂexural stresses in addition to the compressive ones.

11.1

Arches

6 In order to optimize dead-load eﬃciency, long span structures should have their shapes approximate the coresponding moment diagram, hence an arch, suspended cable, or tendon conﬁguration in a prestressed concrete beam all are nearly parabolic, Fig. 20.1. 7 Long span structures can be built using ﬂat 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. 8 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. 9 The basic issues of static in arch design are illustrated in Fig. 20.2 where the vertical load is per unit horizontal projection (such as an external load but not a self-weight). 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, wL L L − M = Hh − =0 (11.1) 2 2 4

Solving for H H=

wL2 8h

(11.2)

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

11–2

ARCHES and CURVED STRUCTURES

Draft 2

M = w L /8

L w=W/L

IDEALISTIC ARCH SHAPE GIVEN BY MOMENT DIAGRAM

C RISE = h -C BEAM +T W/2

M-ARM small C C-T 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 11.1: 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

= wL2 /8 - H h = 0

Figure 11.2: Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981)

Victor Saouma

Structural Engineering

11.1 Arches

11–3

Draft

than C − T in a beam. Since equilibrium requires H to remain constant across thee arch, a parabolic curve would theoretically result in no moment on the arch section.

10

Three-hinged 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.

11

An arch carries the vertical load across the span through a combination of axial forces and ﬂexural ones. A well dimensioned arch will have a small to negligible moment, and relatively high normal compressive stresses.

12

An arch is far more eﬃcient than a beam, and possibly more economical and aesthetic than a truss in carrying loads over long spans.

13

If the arch has only two hinges, Fig. 20.3, or if it has no hinges, then bending moments may exist either at the crown or at the supports or at both places.

14

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 11.3: 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.

15

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 eﬀect can be neglected if the live load is small in comparison with the dead load. √ 17 Since the greatest total force in the arch is at the support, (R = V 2 + H 2 ), whereas at the crown we simply have H, the crown will require a smaller section than the support. 16

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 11.4: Arch Rib Stiﬀened with Girder or Truss, (Lin and Stotesbury 1981)

Victor Saouma

Structural Engineering

11–4

ARCHES and CURVED STRUCTURES

Draft 11.1.1

Statically Determinate

Example 11-1: Three Hinged Arch, Point Loads. (Gerstle 1974) Determine the reactions of the three-hinged arch shown in Fig. 20.5

Figure 11.5: Solution: Four unknowns, three equations of equilibrium, one equation of condition ⇒ statically determinate. (+ ✛) ✁ ΣMzC

=

(+ ✲ ) ΣFx (+ ✻) ΣFy (+ ✛) ✁ ΣMzB

= = =

0; (RAy )(140) + (80)(3.75) − (30)(80) − (20)(40) + RAx (26.25) ⇒ 140RAy + 26.25RAx 0; 80 − RAx − RCx 0; RAy + RCy − 30 − 20 0; (Rax )(60) − (80)(30) − (30)(20) + (RAy )(80) ⇒ 80RAy + 60RAx

Solving those four equations simultaneously we have:     140 26.25 0 0  RAy   2, 900        0  80 R 1 0 1 Ax   =  1 R 0 1 0    Cy  50      3, 000 RCx 80 60 0 0

 RAy    RAx ⇒    RCy    RCx    

 15.1 k    29.8 k =   34.9 k     50.2 k    

= = = = = =

0 2.900 0 0 0 3, 000 (11.3)

      

We can check our results by considering the summation with respect to b from the right: √ ✛ (+ ✁) ΣMzB = 0; −(20)(20) − (50.2)(33.75) + (34.9)(60) = 0

(11.4)

(11.5)

Example 11-2: Semi-Circular Arch, (Gerstle 1974) Determine the reactions of the three hinged statically determined semi-circular arch under its own dead weight w (per unit arc length s, where ds = rdθ). 20.6 Solution: I Reactions The reactions can be determined by integrating the load over the entire structure Victor Saouma

Structural Engineering

11.1 Arches

11–5

Draft

dP=wRdθ

θ

B

B

r

R

R

θ

A

θ R cosθ

A

C

Figure 11.6: Semi-Circular three hinged arch 1. Vertical Reaction is determined ﬁrst: (+ ✛) ✁ ΣMA

θ=π

0; −(Cy )(2R) +

=

wRdθ R(1 + cos θ) = 0 dP moment arm θ=π wR wR θ=π [θ − sin θ] |θ=0 (1 + cos θ)dθ = 2 2 θ=0 wR [(π − sin π) − (0 − sin 0)] 2 π 2 wR

(11.6-a)

θ=0

⇒ Cy

= = =

2. Horizontal Reactions are determined next ✛ (+ ✁) ΣMB

=

0; −(Cx )(R) + (Cy )(R) −

θ= π 2 θ=0

⇒ Cx

= =

wR π wR − 2 2 π 2 − 1 wR

θ= π 2

cos θdθ = θ=0

(11.6-b)

wRdθ

θ R cos

dP

moment arm

=0

(11.7-a)

π π π θ= π wR − wR[sin θ] |θ=02 = wR − wR( − 0) 2 2 2 (11.7-b)

By symmetry the reactions at A are equal to those at C II Internal Forces can now be determined, Fig. 20.7. 1. Shear Forces: Considering the free body diagram of the arch, and summing the forces in the radial direction (ΣFR = 0): θ π π wRdα sin θ + V = 0 (11.8) −( − 1)wR cos θ + wR sin θ − α=0 2

2 Cx

$

Cy

⇒ V = wR ( π2 − 1) cos θ + (θ − π2 ) sin θ

%

(11.9)

2. Axial Forces: Similarly, if we consider the summation of forces in the axial direction (ΣFN = 0): θ ( π2 − 1)wR sin θ + π2 wR cos θ − wRdα cos θ + N = 0 (11.10) α=0

$ % ⇒ N = wR (θ − π2 ) cos θ − ( π2 − 1) sin θ

(11.11)

3. Moment: Now we can consider the third equation of equilibrium (ΣMz = 0): (+ ✛) ✁ ΣM

( π2 − 1)wR · R sin θ − π2 wR2 (1 − cos θ) + θ wRdα · R(cos α − cos θ) + M = 0 α=0

⇒ M = wR2 Victor Saouma

$π

2 (1

− sin θ) + (θ − π2 ) cos θ

(11.12) %

(11.13)

Structural Engineering

11–6

ARCHES and CURVED STRUCTURES

Draft

dP=wRdα θ

M

r

V

R sin θ

N

R cos θ

C x=(π /2-1)wR

R(1-cos θ) R cos α

y

α

C =π /2 wR

θ

R cos(θ −α)

dα

Figure 11.7: Semi-Circular three hinged arch; Free body diagram III Deﬂection are determined last 1. The real curvature φ is obtained by dividing the moment by EI ' wR2 & π π M = (1 − sin θ) + (θ − ) cos θ EI EI 2 2

φ=

(11.14)

1. The virtual force δP will be a unit vertical point in the direction of the desired deﬂection, causing a virtual internal moment δM =

R [1 − cos θ − sin θ] 2

0≤θ≤

π 2

(11.15)

p 2. Hence, application of the virtual work equation yields: 1 ·∆

= 2

dx

δP

11.1.2

' R π wR2 & π (1 − sin θ) + (θ − ) cos θ · · [1 − cos θ − sin θ] Rdθ 2 2 θ=0 EI

2

π 2

M EI

=

% wR4 $ 2 7π − 18π − 12 16EI

=

.0337 wR EI

=φ

4

δM

(11.16-a)

Statically Indeterminate

Example 11-3: Statically Indeterminate Arch, (Kinney 1957) Victor Saouma

Structural Engineering

11.1 Arches

11–7

Draft

Determine the value of the horizontal reaction component of the indicated two-hinged solid rib arch, Fig. 20.8 as caused by a concentrated vertical load of 10 k at the center line of the span. Consider shearing, axial, and ﬂexural strains. Assume that the rib is a W24x130 with a total area of 38.21 in2 , that it has a web area of 13.70 in2 , a moment of inertia equal to 4,000 in4 , E of 30,000 k/in2 , and a shearing modulus G of 13,000 k/in2 .

Figure 11.8: Statically Indeterminate Arch Solution:

1. Consider that end C is placed on rollers, as shown in Fig. ?? A unit ﬁctitious horizontal force is applied at C. The axial and shearing components of this ﬁctitious force and of the vertical reaction at C, acting on any section θ in the right half of the rib, are shown at the right end of the rib in Fig. 13-7. 2. The expression for the horizontal displacement of C is 1 ∆Ch = 2

B

δM C

M ds + 2 EI

B

δV C

V ds + 2 Aw G

B

δN C

N ds AE

(11.17)

δP

3. From Fig. 20.9, for the rib from C to B,

Victor Saouma

M

=

δM

=

V

=

δV

=

N

=

P (100 − R cos θ) 2 1(R sin θ − 125.36) P sin θ 2 cos θ P cos θ 2

(11.18-a) (11.18-b) (11.18-c) (11.18-d) (11.18-e) Structural Engineering

11–8

ARCHES and CURVED STRUCTURES

Draft

Figure 11.9: Statically Indeterminate Arch; ‘Horizontal Reaction Removed δN ds

= =

− sin θ Rdθ

(11.18-f) (11.18-g)

4. If the above values are substituted in Eq. 20.17 and integrated between the limits of 0.898 and π/2, the result will be ∆Ch = 22.55 + 0.023 − 0.003 = 22.57 (11.19) 5. The load P is now assumed to be removed from the rib, and a real horizontal force of 1 k is assumed to act toward the right at C in conjunction with the ﬁctitious horizontal force of 1 k acting to the right at the same point. The horizontal displacement of C will be given by B B B M V N δChCh = 2 ds + 2 ds + 2 ds (11.20-a) δM δV δN EI A G AE w C C C = 2.309 + 0.002 + 0.002 = 2.313 in (11.20-b) 6. The value of the horizontal reaction component will be HC =

∆Ch 22.57 = 9.75 k = δChCh 2.313

(11.21)

7. If only ﬂexural strains are considered, the result would be HC =

22.55 = 9.76 k 2.309

(11.22)

Comments 1. For the given rib and the single concentrated load at the center of the span it is obvious that the eﬀects of shearing and axial strains are insigniﬁcant and can be disregarded. 2. Erroneous conclusions as to the relative importance of shearing and axial strains in the usual solid rib may be drawn, however, from the values shown in Eq. 20.19. These indicate that the eﬀects of the shearing strains are much more signiﬁcant than those of the axial strains. This is actually the case for the single concentrated load chosen for the demonstration, but only because the rib does not approximate the funicular polygon for the single load. As a result, the shearing components on most sections of the rib are more important than would otherwise be the case. Victor Saouma

Structural Engineering

11.2 Curved Space Structures

11–9

Draft

3. The usual arch encountered in practice, however, is subjected to a series of loads, and the axis of the rib will approximate the funicular polygon for these loads. In other words, the line of pressure is nearly perpendicular to the right section at all points along the rib. Consequently, the shearing components are so small that the shearing strains are insigniﬁcant and are neglected. 4. Axial strains, resulting in rib shortening, become increasingly important as the rise-to-span ratio of the arch decreases. It is advisable to determine the eﬀects of rib shortening in the design of arches. The usual procedure is to ﬁrst design the rib by considering ﬂexural strains only, and then to check for the eﬀects of rib shortening.

11.2

Curved Space Structures

Example 11-4: Semi-Circular Box Girder, (Gerstle 1974) Determine the reactions of the semi-circular cantilevered box girder shown in Fig. 20.10 subjected x

x wRd α θ dα α

O R

B

V ΖΜ r

Τ θ

C A

A y

y

z

z

Figure 11.10: Semi-Circular Box Girder to its own weight w. Solution: I Reactions are again determined ﬁrst From geometry we have OA = R, OB = R cos θ, AB = OA − BO = R − R cos θ, and BP = R sin θ. The moment arms for the moments with respect to the x and y axis are BP and AB respectively. Applying three equations of equilibrium we obtain

−

⇒

FzA = wRπ

(11.23-a)

(wRdθ)(R sin θ) = 0

⇒

MxA = 2wR2

(11.23-b)

(wRdθ)R(1 − cos θ) = 0

⇒

MyA = −wR2 π

(11.23-c)

θ=0 θ=π

MxA −

θ=π

wRdθ = 0

FzA

θ=0 θ=π

MyA − θ=0

II Internal Forces are determined next 1. Shear Force:

(+ ✻) ΣFz = 0 ⇒ V −

Victor Saouma

θ

wRdα = 0 ⇒ V = wrθ

(11.24)

0

Structural Engineering

11–10

ARCHES and CURVED STRUCTURES

Draft

2. Bending Moment:

θ

ΣMR = 0 ⇒ M −

(wRdα)(R sin α) = 0 ⇒ M = wR2 (1 − cos θ)

(11.25)

(wRdα)R(1 − cos α) = 0 ⇒ T = −wR2 (θ − sin θ)

(11.26)

0

3. Torsion:

θ

ΣMT = 0 ⇒ + 0

III Deﬂection are determined last we assume a rectangular cross-section of width b and height d = 2b and a Poisson’s ratio ν = 0.3. 1. Noting that the member will be subjected to both ﬂexural and torsional deformations, we seek to determine the two stiﬀnesses. 3

3

b(2b) = 2. The ﬂexural stiﬀness EI is given by EI = E bd 12 = E 12

2Eb4 3

= .667Eb4 .

3

3. The torsional stiﬀness of solid rectangular sections J = kb d where b is the shorter side of the E E = 2(1+.3) = section, d the longer, and k a factor equal to .229 for db = 2. Hence G = 2(1+ν) 4 4 .385E, and GJ = (.385E)(.229b ) = .176Eb . 4. Considering both ﬂexural and torsional deformations, and replacing dx by rdθ: π π M T δP Rdθ ∆ = δM Rdθ δT + EI GJ z 0 0 ∗

δW Torsion Flexure

δU

(11.27)

∗

where the real moments were given above. 5. Assuming a unit virtual downward force δP = 1, we have δM δT

= R sin θ

(11.28-a)

= −R(1 − cos θ)

(11.28-b)

6. Substituting these expression into Eq. 20.27 wR2 π wR2 ∆ = (R sin θ) (1 − cos θ) Rdθ + 1 EI GJ 0 δP

4 π

= =

=

11.2.1

M

δM

π 0

(θ − sin θ) R(1 − cos θ) Rdθ

δT

T

1 (θ − θ cos θ − sin θ + sin θ cos θ) dθ (sin θ − sin θ cos θ) + .265

wR EI 0 wR4 ( 2. + 18.56 ) EI Flexure Torsion 4

20.56 wR EI

(11.29-a)

Theory

ßAdapted from (Gerstle 1974) Because space structures may have complicated geometry, we must resort to vector analysis1 to determine the internal forces.

18

19

In general we have six internal forces (forces and moments) acting at any section. 1 To

which you have already been exposed at an early stage, yet have very seldom used it so far in mechanics!

Victor Saouma

Structural Engineering

11.2 Curved Space Structures

11–11

Draft 11.2.1.1

Geometry

In general, the geometry of the structure is most conveniently described by a parameteric set of equations y = f2 (θ); z = f3 (θ) (11.30) x = f1 (θ);

20

as shown in Fig. 20.11. the global coordinate system is denoted by X − Y − Z, and its unit vectors are

Figure 11.11: Geometry of Curved Structure in Space denoted2 i, j, k. The section on which the internal forces are required is cut and the principal axes are identiﬁed as N − S − W which correspond to the normal force, and bending axes with respect to the Strong and Weak axes. The corresponding unit vectors are n, s, w.

21

22

The unit normal vector at any section is given by n=

dxi + dyj + dzk dxi + dyj + dzk = ds (dx2 + dy 2 + dz 2 )1/2

(11.31)

The principal bending axes must be deﬁned, that is if the strong bending axis is parallel to the XY plane, or horizontal (as is generally the case for gravity load), then this axis is normal to both the N and Z axes, and its unit vector is n×k (11.32) s= |n×k|

23

24

The weak bending axis is normal to both N and S, and thus its unit vector is determined from w = n×s

11.2.1.2 25

(11.33)

Equilibrium

For the equilibrium equations, we consider the free body diagram of Fig. 20.12 an applied load P 2 All

vectorial quantities are denoted by a bold faced character.

Victor Saouma

Structural Engineering

11–12

ARCHES and CURVED STRUCTURES

Draft

Figure 11.12: Free Body Diagram of a Curved Structure in Space is acting at point A. The resultant force vector F and resultant moment vector M acting on the cut section B are determined from equilibrium ΣF = 0;

P + F = 0;

F = −P

(11.34-a)

ΣM = 0; L×P + M = 0; M = −L×P B

(11.34-b)

where L is the lever arm vector from B to A. The axial and shear forces N, Vs and Vw are all three components of the force vector F along the N, S, and W axes and can be found by dot product with the appropriate unit vectors:

26

N Vs

= =

F·n F·s

(11.35-a) (11.35-b)

Vw

=

F·w

(11.35-c)

Similarly the torsional and bending moments T, Ms and Mw are also components of the moment vector M and are determined from

27

T

=

M·n

(11.36-a)

Ms Mw

= =

M·s M·w

(11.36-b) (11.36-c)

Hence, we do have a mean to determine the internal forces. In case of applied loads we summ, and for distributed load we integrate.

28

Example 11-5: Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974) Determine the internal forces N, Vs , and Vw and the internal moments T, Ms and Mw along the helicoidal cantilevered girder shown in FIg. 20.13 due to a vertical load P at its free end. Solution: 1. We ﬁrst determine the geometry in terms of the angle θ x = R cos θ; Victor Saouma

y = R sin θ;

z=

H θ π

(11.37) Structural Engineering

11.2 Curved Space Structures

Draft

11–13

Figure 11.13: Helicoidal Cantilevered Girder

Victor Saouma

Structural Engineering

11–14

ARCHES and CURVED STRUCTURES

Draft

2. To determine the unit vector n at any point we need the derivatives: dx = −R sin θdθ;

dy = R cos θdθ;

dz =

H dθ π

(11.38)

and then insert into Eq. 20.31 n = =

−R sin θi + R cos θj + H/πk %1/2 R2 sin2 θ + R2 cos2 θ + (H/π)2 1 $ %1/2 [sin θi + cos θj + (H/πR)k] 1 + (H/πR)2

$

(11.39-a) (11.39-b)

K

Since the denominator depends only on the geometry, it will be designated by K. 3. The strong bending axis lies in a horizontal plane, and its unit vector can thus be determined from Eq. 20.32: " " " i j k "" 1 "" H " (11.40-a) n×k = − sin θ cos θ πR " K "" 0 0 1 " 1 = (cos θi + sin θj) (11.40-b) K and the absolute magnitude of this vector |k×n| =

1 K,

and thus

s = cos θi + sin θj

(11.41)

4. The unit vector along the weak axis is determined from Eq. 20.33 " " " i j k "" 1 "" cos θ sin θ 0 "" w = s×n = K "" H " − sin θ cos θ πR 1 H H = sin θi − cos θj + k K πR πR

(11.42-a)

(11.42-b)

5. With the geometry deﬁnition completed, we now examine the equilibrium equations. Eq. 20.34-a and 20.34-b. ΣF = 0; F ΣMb = 0; M where

= −P = −L×P

(11.43-a) (11.43-b)

θ L = (R − R cos θ)i + (0 − R sin θ)j + 0 − H k π

and L×P =

" " i " R "" (1 − cos θ) " 0

j − sin θ 0

k − πθ H R P

" " " " " "

(11.44)

(11.45-a)

P R[− sin θi − (1 − cos θ)j]

(11.45-b)

M = P R[sin θi + (1 − cos θ)j]

(11.46)

= and

Victor Saouma

Structural Engineering

11.2 Curved Space Structures

11–15

Draft

6. Finally, the components of the force F = −P k and the moment M are obtained by appropriate dot products with the unit vectors

Victor Saouma

N

=

1 H P πR F·n = − K

(11.47-a)

Vs

=

F·s = 0

(11.47-b)

Vw

=

1 F·w = − K P

(11.47-c)

T

=

M·n = − PKR (1 − cos θ)

(11.47-d)

Ms

=

M·s = P R sin θ

(11.47-e)

Mw

=

M·w =

PH πK (1

− cos θ)

(11.47-f)

Structural Engineering

11–16

Draft

ARCHES and CURVED STRUCTURES

Victor Saouma

Structural Engineering

Draft Chapter 12

DEFLECTION of STRUCTRES; Geometric Methods 1 Deﬂections of structures must be determined in order to satisfy serviceability requirements i.e. limit deﬂections under service loads to acceptable values (such as ∆ L ≤ 360). 2 Later on, we will see that deﬂection calculations play an important role in the analysis of statically indeterminate structures. 3 We shall focus on ﬂexural deformation, however the end of this chapter will review axial and torsional deformations as well.

4

Most of this chapter will be a review of subjects covered in Strength of Materials.

5 This chapter will examine deﬂections of structures based on geometric considerations. Later on, we will present a more pwerful method based on energy considerations.

12.1

Flexural Deformation

12.1.1

Curvature Equation

6 Let us consider a segment (between point 1 and point 2), Fig. 12.1 of a beam subjected to ﬂexural loading.

The slope is denoted by θ, the change in slope per unit length is the curvature κ, the radius of curvature is ρ.

7

8

From Strength of Materials we have the following relations ds = ρdθ ⇒

9

10

(12.1)

We also note by extension that ∆s = ρ∆θ As a ﬁrst order approximation, and with ds ≈ dx and κ=

11

dθ 1 = ds ρ

dy dx

= θ Eq. 12.1 becomes

1 dθ d2 y = = 2 ρ dx dx

(12.2)

† Next, we shall (re)derive the exact expression for the curvature. From Fig. 12.1, we have tan θ =

dy dx

(12.3)

12–2

DEFLECTION of STRUCTRES; Geometric Methods

Draft

Figure 12.1: Curvature of a ﬂexural element Deﬁning t as t= and combining with Eq. 12.3 we obtain

12

Applying the chain rule to κ =

dθ ds

ds

=

13

=

(12.5)

we have dθ dt dt ds

2 + dy 2 !dx

= t

(12.4)

θ = tan−1 t

κ= ds can be rewritten as

dy dx

1+

dy dx

dy dx

2

    dx  ds =  

(12.6)

1 + t2 dx

(12.7)

Next combining Eq. 12.6 and 12.7 we obtain κ θ dθ dt

= = =

dθ √ dt dt 1+t2 dx −1

tan

1 1+t2

t

  

κ

=

dt dx

=

1 √ 1 dt 1+t2 1+t2 dx 2 d y dx2

    

d2 y dx2

κ= 2 32 dy 1 + dx

(12.8)

Thus the slope θ, curvature κ, radius of curvature ρ are related to the y displacement at a point x along a ﬂexural member by

14

d2 y dx2

κ= 2 32 dy 1 + dx

Victor Saouma

(12.9)

Structural Engineering

12.1 Flexural Deformation

12–3

Draft 15

If the displacements are very small, we will have

κ=

12.1.2

dy dx

<< 1, thus Eq. 12.9 reduces to

d2 y 1 = dx2 ρ

(12.10)

Diﬀerential Equation of the Elastic Curve

Again with reference to Figure 12.1 a positive dθ at a positive y (upper ﬁbers) will cause a shortening of the upper ﬁbers ∆u = −y∆θ (12.11)

16

17

This equation can be rewritten as lim

∆s→0

∆u ∆θ = −y lim ∆s→0 ∆s ∆s

(12.12)

du dθ = −y dx dx

(12.13)

and since ∆s ≈ ∆x

ε

Combining this with Eq. 12.10 ε 1 =κ=− ρ y

(12.14)

This is the fundamental relationship between curvature (κ), elastic curve (y), and linear strain (ε). 18

Note that so far we made no assumptions about material properties, i.e. it can be elastic or inelastic.

19

For the elastic case: εx σ

= =

σ E

− My I

ε=−

My EI

(12.15)

Combining this last equation with Eq. 12.14 yields M dθ d2 y 1 = = 2 = ρ dx dx EI

(12.16)

This fundamental equation relates moment to curvature. 20

Combining this equation with the moment-shear-force relations determined in the previous chapter 2 dV d M = w(x) dx (12.17-a) dM = V (x) dx2 dx

we obtain d4 y w(x) = 4 EI dx

12.1.3 21

(12.18)

Moment Temperature Curvature Relation

Assuming linear variation in temperature

Victor Saouma

∆T

=

α∆TT dx Top

(12.19-a)

∆B

=

α∆TB dx Bottom

(12.19-b)

Structural Engineering

12–4

DEFLECTION of STRUCTRES; Geometric Methods

Draft 22

Next, considering d2 y ε M =− = 2 dx EI y

In this case, we can take ε = α∆TT at y =

h 2

(12.20)

or ε = α(TT − TB ) at y = h thus

d2 y α(TT − TB ) dθ M = 2 = =− dx dx EI h

12.2

Flexural Deformations

12.2.1

Direct Integration Method

(12.21)

Equation 12.18 lends itself naturally to the method of double integration which was presented in Strength of Materials

23

Example 12-1: Double Integration Determine the deﬂection at B for the following cantilevered beam

Solution: At: 0 ≤ x ≤

2L 3

1. Moment Equation EI

5 wL d2 y x − wL2 = Mx = dx2 3 18

(12.22)

EI

dy wL 2 5 = x − wL2 x + C1 dx 6 18

(12.23)

2. Integrate once

However we have at x = 0,

dy dx

= 0, ⇒ C1 = 0

3. Integrate twice wL 3 5wL2 2 x − x + C2 18 36 Again we have at x = 0, y = 0, ⇒ C2 = 0 EIy =

Victor Saouma

(12.24)

Structural Engineering

12.2 Flexural Deformations

12–5

Draft At:

≤x≤L

2L 3

1. Moment equation EI

5 d2 y wL 2L x − 2L 2 3 x − wL )( ) = M = − w(x − x dx2 3 18 3 2

(12.25)

dy wL 2 2L 3 5 w = x − wL2 x − (x − ) + C3 dx 6 18 6 3

(12.26)

2. Integrate once EI

Applying the boundary condition at x = the left, ⇒ C3 = 0

2L 3 ,

we must have

dy dx

equal to the value coming from

3. Integrating twice 2L 4 wL 3 5 w x − wL2 x2 − (x − ) + C4 18 36 24 3 Again following the same argument as above, C4 = 0 EIy =

(12.27)

Substituting for x = L we obtain y=

163 wL4 1944 EI

(12.28)

12.2.2

Curvature Area Method (Moment Area)

12.2.2.1

First Moment Area Theorem

24

From equation 12.16 we have M dθ = dx EI

this can be rewritten as (note similarity with

dV dx

(12.29)

= w(x)).

θ21 = θ2 − θ1 =

x2

x2

dθ = x1

x1

M dx EI

(12.30)

or with reference to Figure 12.2 First Area Moment Theorem: The change in slope from point 1 to point 2 on a beam is equal to the area under the M/EI diagram between those two points. 12.2.2.2

Second Moment Area Theorem

Similarly, with reference to Fig. 12.2, we deﬁne by t21 the distance between point 2 and the tangent at point 1. For an inﬁnitesimal distance ds = ρdθ and for small displacements M dt = dθ(x2 − x) (x2 − x)dx (12.31) dt = dθ M = EI dx EI

25

To evaluate t21

x2

t21 =

dt = x1

x2

M (x2 − x)dx EI x1

(12.32)

or Second Moment Area Theorem: The tangent distance t21 between a point, 2, on the beam and the tangent of another point, 1, is equal to the moment of the M/EI diagram between points 1 and 2, with respect to point 2. Victor Saouma

Structural Engineering

12–6

Draft

DEFLECTION of STRUCTRES; Geometric Methods

Figure 12.2: Moment Area Theorems

Victor Saouma

Structural Engineering

12.2 Flexural Deformations

12–7

Draft

Figure 12.3: Sign Convention for the Moment Area Method

Figure 12.4: Areas and Centroid of Polynomial Curves

Victor Saouma

Structural Engineering

12–8

DEFLECTION of STRUCTRES; Geometric Methods

Draft 26

The sign convention is as shown in Fig. 12.3

27

Fig. 12.4 is a helpful tool to determine centroid and areas.

Example 12-2: Moment Area, Cantilevered Beam Determine the deﬂection of point A

Solution: EItA/C =

Thus, ∆A =

1 2

2 5L 9L −2wL2 5L 1 −wL2 29wL4 + =− (L) 5 2 3 2 3 2 4 24

area area

Moment wrt A Moment wrt A

(12.33)

29wL4 24EI

Example 12-3: Moment Area, Simply Supported Beam Determine ∆C and θC for the following example

Victor Saouma

Structural Engineering

12.2 Flexural Deformations

12–9

Draft

Solution: Deﬂection ∆C is determined from ∆C = c c = c c” − c”c, c”c = tC/B , and c c” =  

tA/B

Victor Saouma

tA/B 2

       5P a3 3P a 2a 3P a a 3a 1  3a  = + = a+  EI  4 2 3 4 2 3  2EI 

    A1 A2 +A3

Moment Moment

from geometry

(12.34)

Structural Engineering

12–10

DEFLECTION of STRUCTRES; Geometric Methods

Draft

Figure 12.5: Maximum Deﬂection Using the Moment Area Method This is positive, thus above tangent from B Pa 2a 2a 1 P a3 tC/B = = EI 2 2 3 3EI

(12.35)

Positive, thus above the tangent from B Finally, ∆C =

P a3 11 P a3 5P a3 − = 4EI 3EI 12 EI

(12.36)

Rotation θC is θBC θBC θB

12.2.2.3

= = =

 θB − θC ⇒ θC = θB − θBC  5P a3 Pa 2a P a2 A3 ⇒ θC = − =  2EI(4a) 2 2 8EI tA/B

(12.37)

L

Maximum Deﬂection

A joint along the beam will have the maximum (or minimum) relative deﬂection if So we can determine ∆max if x is known, Fig. 12.5. To determine x:

28

dy dx

= 0 or θ = 0

1. Compute tC/A 2. θA =

tC/A L

3. θB = 0 and θAB = θA − θB , thus θAB = θA . Hence, compute θAB in terms of x using First Theorem. 4. Equate items 2 and 3, then solve for x.

Example 12-4: Maximum Deﬂection

Victor Saouma

Structural Engineering

12.2 Flexural Deformations

12–11

Draft

Determine the deﬂection at D, and the maximum deﬂection at B

Solution: Deﬂection at D: 9 = tD/A − tC/A 5 L 1 −4P L = (L) 2 5 3 3 2P L = − 15EI 1 −4P L 17L 1 −4P L 4L 8L = (L) + 2 5 15 2 5 5 15 3 234 P L = − 375 EI

∆D EItC/A tC/A EItD/A tD/A

(12.38-a) (12.38-b) (12.38-c) (12.38-d) (12.38-e)

Substituting we obtain ∆D = −

48 P L3 125 EI

(12.39)

Maximum deﬂection at B:

tC/A

= −

θA

=

θAB

=

θA

=

∆max

=

⇒ ∆max

=

2P L3 15EI

(12.40-a)

tC/A 2P L2 =− L 15EI 1 1 4P x 2 P x2 (x) = − EI 2 5 5 EI 2 2 P x2 L 2 PL =− ⇒x= √ θAB ⇒ − 15 EI 5 EI 3 4P x x 2 L x at x = √ 5EI 2 3 3 3 4P L √ 45 3EI

(12.40-b) (12.40-c) (12.40-d) (12.40-e) (12.40-f)

Example 12-5: Frame Deﬂection Victor Saouma

Structural Engineering

12–12

Draft

DEFLECTION of STRUCTRES; Geometric Methods

Complete the following example problem

Solution: xxxx

Example 12-6: Frame Subjected to Temperature Loading Neglecting axial deformation, compute displacement at A for the following frame

Solution:

1. First let us sketch the deformed shape Victor Saouma

Structural Engineering

12.2 Flexural Deformations

12–13

Draft

2. BC ﬂexes ⇒ θB = θC = 0 3. Rigid hinges at B and C with no load on AB and CD 4. Deﬂection at A ∆A AA

= =

AA = AA + A A ∆B = ∆C = |θC |h2

(12.41-a) (12.41-b)

A A

=

|θB |h1

(12.41-c)

5. We need to compute θB & θC θB

=

tC/B L

θC

=

θCB + θB or θC =

(12.42-a) tB/C L

(12.42-b)

6. In order to apply the curvature are theorem, we need a curvature (or moment diagram). 1 TB − TT M = α( )= ρ h EI 7. tC/B = A

(12.43)

L L 1 A A or θB = − (-ve) ⇒ |θB | = A = 2 2 L 2 2

8. θCB θC

= =

A θCB + θB

θC = A −

A A = (+ve) 2 2

(12.44)

(12.45)

9. From above, ∆A

= = A =

|θC |h2 + |θB |h1 = A 2 (h2 + h1 ) α(TB −TT ) L h

A 2 h2

+

A 2 h1

 

L ∆A = α(TB − TT )  h

1 (h2 + h1 ) 2

(12.46)

10. Substitute = (6.5 × 10−6 )(200 − 60)

∆A

(20)(12) 1 ( )(10 + 25)(12) (16) 2

= 2.87 in

(12.47-a) (12.47-b)

11. Other numerical values: θB

= = =

1 α(TB − TT ) A = L 2 2 h 1 (200 − 60) (6.5 × 10−6 ) (20)(12) 2 (16) dy << 1 .00683rad. (.39 degrees) = dx

θC =

(12.48-a) (12.48-b) (12.48-c)

Note: sin(.00683) = .006829947 and tan(.00683) = .006830106 M EI

= =

⇒ρ Victor Saouma

=

1 (TB − TT ) =α ρ h (200 − 60) (6.5 × 10−6 ) = 5.6875 × 10−5 16 1 = 1.758 × 104 in = 1, 465 ft 5.6875 × 10−5

(12.49-a) (12.49-b) (12.49-c)

Structural Engineering

12–14

DEFLECTION of STRUCTRES; Geometric Methods

Draft

12. In order to get M , we need E & I. Note the diﬀerence with other statically determinate structures; the stiﬀer the beam, the higher the moment; the higher the moment, the higher the stress? NO!!

13. σ =

My I

=

EI y ρ I

=

Ey ρ

14. ρ is constant ⇒ BC is on arc of circle M is constant & 2 C x2 + dx + e

M EI

d2 y dx2

=

⇒

d2 y dx2

=

M EI

= C ⇒ y =

15. The slope is a parabola, (Why?) d2 y

3 M dy d2 y 1 2 = = dx 1 + ( )2 ) 2 ≈ 2 ρ EI ( dx dx

29

(12.50)

Let us get curvature from the parabola slope and compare it with ρ y dy dx d2 y dx2

=

cx2 + dx + e 2

(12.51-a)

=

cx + d

(12.51-b)

=

c

(12.51-c)

at x = 0, y = 0, thus e = 0 at x = 0, dy x = θB = −.00683 thus d = −.00683 dy at x = 20 ft, dx = θC = .00683 thus c(20) − .00683 = .00683 thus c = 6.83 × 10−4 thus y dy dx d2 y dx2 κ ρ

= 6.83 × 10−4

x2 − 10x 2

(12.52-a)

= 6.83 × 10−4 (x − 10)

(12.52-b)

= 6.83 × 10−4

(12.52-c)

= 6.83 × 10−4 1 = 1, 464f t as expected! = 6.83 × 10− 4

(12.52-d) (12.52-e)

If we were to use the exact curvature formula

1

d2 y 3 dx2 )2 dy 2 + ( dx )

=

(12.53-a)

3

[1 + (.00683)2] 2

⇒ρ

12.2.3

6.83 × 10−4

=

6.829522 × 10−4

(12.53-b)

=

1464.23f t (compared with 1465ft)

(12.53-c)

Elastic Weight/Conjugate Beams V12 =

x2

V and M wdx V = wdx + C1

θ12 =

V dx

t21 =

xx12

M12 = x1

Victor Saouma

M=

V dx + C2

θ and y 1 1 dx θ = dx + C1 ρ ρ x1

x2

x2

θdx

y=

θdx + C2

x1

Structural Engineering

12.2 Flexural Deformations

12–15

Draft

Figure 12.6: Conjugate Beams

29

M 1 =κ= ρ EI

Load q

≡

curvature

Shear V

≡

slope θ

(12.55)

This leads to the following Moment M

≡

deﬂection y

(12.56)

(12.54)

30 Since V & M can be conjugated from statics, by analogy θ & y can be thought of as the V & M of M elastic weight. a ﬁctitious beam (or conjugate beam) loaded by EI 31

What about Boundary Conditions? Table 12.1, and Fig. 12.6. Actual Beam Hinge θ = 0 Fixed End θ=0 Free End θ = 0 Interior Hinge θ = 0 Interior Support θ = 0

y y y y y

=0 =0 = 0 = 0 =0

V V V V V

= 0 =0 = 0 = 0 = 0

Conjugate Beam M = 0 “Hinge” M = 0 Free end M = 0 Fixed end M = 0 Interior support M = 0 Interior hinge

Table 12.1: Conjugate Beam Boundary Conditions Whereas the Moment area method has a well deﬁned basis, its direct application can be sometimes confusing.

32

Alternatively, the moment area method was derived from the moment area method, and is a far simpler method to remember and use in practice when simple “back of the envelope” calculations are required.

33

Note that we can only have distributed load, and that the load the load is positive for a positive moment, and negative for a negative moment. “Shear” and “Moment” diagrams should be drawn accordingly.

34

Units of the “distributed load” w∗ are FEIL (force time length divided by EI). Thus the “Shear” would L2 L3 and the “moment” would have units of (w∗ × L) × L or FEI . Recalling have units of w∗ × L or FEI that EI has units of F L−2 L4 = F L2 , we observe that indeed the “shear” corresponds to a rotation in radians and the “moment” to a displacement.

35

Victor Saouma

Structural Engineering

12–16

DEFLECTION of STRUCTRES; Geometric Methods

Draft

Example 12-7: Conjugate Beam Analyze the following beam.

Solution: 3 equations of equilibrium and 1 equation of condition = 4 = number of reactions. Deﬂection at D = Shear at D of the corresponding conjugate beam (Reaction at D) Take AC and ΣM with respect to C 4P L L L RA (L) − = 0 5EI 2 3 2P L2 ⇒ RA = 15EI (Slope in real beam at A) As computed before! Let us draw the Moment Diagram for the conjugate beam x 4 2 2 x P L x− x M = EI 15 5 2 3 2 2 2 P L x − x3 = EI 15 15 2P 2 L x − x3 = 15EI Point of Maximum Moment (∆max ) occurs when

dM dx

(12.57-b)

(12.58-a) (12.58-b) (12.58-c)

=0

2P L dM = (L2 − 3x2 ) = 0 ⇒ 3x2 = L2 ⇒ x = √ dx 15EI 3 Victor Saouma

(12.57-a)

(12.59)

Structural Engineering

12.3 Axial Deformations

12–17

Draft

Figure 12.7: Torsion Rotation Relations as previously determined x

=

⇒M

= =

L √ 3 2 2P L L L3 √ − √ 15EI 3 3 3 4P L3 √ 45 3EI

(12.60-a) (12.60-b) (12.60-c)

as before.

12.3

Axial Deformations

Statics: σ =

P A

Material: σ = Eε Kinematics: Σ =

∆ L

∆=

12.4

PL AE

(12.61)

Torsional Deformations

Since torsional eﬀects are seldom covered in basic structural analysis, and students may have forgotten the derivation of the basic equations from the Strength of Material course, we shall brieﬂy review the basic equations.

36

Assuming a linear elastic material, and a linear strain (and thus stress) distribution along the radius of a circular cross section subjected to torsional load, Fig. 12.7 we have: ρ τmax dA ρ (12.62) T = A c area arm stress

F orce

37

=

τmax c

torque

ρ2 dA

(12.63)

A

J

τmax Victor Saouma

=

Tc J

(12.64) Structural Engineering

12–18

DEFLECTION of STRUCTRES; Geometric Methods

Draft

Note the analogy of this last equation with σ = Mc Iy 38 ρ2 dA is the polar moment of inertia J for circular cross sections and is equal to: A

J

ρ2 (2πρdρ)

ρ dA =

=

0

A 4

=

c

2

πd4 πc = 2 32

(12.65)

Having developed a relation between torsion and shear stress, we now seek a relation between torsion and torsional rotation. Considering Fig. 12.7-b, we look at the arc length BD   dΦ γmax dΦ T γmax dx = dΦc ⇒ dx = c τmax dΦ = (12.66) = τmax dx Gc γmax = G  dx GJ τmax = TJC

39

40

Finally, we can rewrite this last equation as

T dx =

T =

Victor Saouma

GJ L Φ

GjdΦ and obtain: (12.67)

Structural Engineering

Draft Chapter 13

ENERGY METHODS; Part I 13.1

Introduction

Energy methods are powerful techniques for both formulation (of the stiﬀness matrix of an element1 ) and for the analysis (i.e. deﬂection) of structural problems.

1

2

We shall explore two techniques: 1. Real Work 2. Virtual Work (Virtual force)

13.2 3

Real Work

We start by revisiting the ﬁrst law of thermodynamics: The time-rate of change of the total energy (i.e., sum of the kinetic energy and the internal energy) is equal to the sum of the rate of work done by the external forces and the change of heat content per unit time. d dt (K

+ U ) = We + H

(13.1)

where K is the kinetic energy, U the internal strain energy, We the external work, and H the heat input to the system. 4 For an adiabatic system (no heat exchange) and if loads are applied in a quasi static manner (no kinetic energy), the above relation simpliﬁes to:

We = U

(13.2)

5 Simply stated, the ﬁrst law stipulates that the external work must be equal to the internal strain energy due to the external load.

13.2.1 6

External Work

The external work is given by, Fig. 13.1 We 1 More

about this in Matrix Structural Analysis.

∆f

0 θf

=

P d∆

=

(13.3) M dθ

0

13–2

ENERGY METHODS; Part I

Draft P

M 1

1 1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

P

K

dW

∆

K ∆

P

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 00 11 00 11 00 11

K

θ

dW

M

M θ

Figure 13.1: Load Deﬂection Curves for point loads and concentrated moments respectively. 7

For linear elastic systems, we have for point loads  P = K∆  ∆f We = K P d∆  We =

∆f

∆d∆ = 0

1 K∆2f 2

(13.4)

0

When this last equation is combined with Pf = K∆f we obtain

We =

1 Pf ∆f 2

(13.5)

We =

1 M f θf 2

(13.6)

where K is the stiﬀness of the structure. 8

Similarly for an applied moment we have

13.2.2

Internal Work

9 Considering an inﬁnitesimal element from an arbitrary structure subjected to uniaxial state of stress, the strain energy can be determined with reference to Fig. 13.2. The net force acting on the element while deformation is taking place is P = σx dydz. The element will undergo a displacement u = εx dx. Thus, for a linear elastic system, the strain energy density is dU = 12 σε. And the total strain energy will thus be 1 (13.7) U= ε Eε dVol 2 Vol

σ

10

When this relation is applied to various structural members it would yield:

Victor Saouma

Structural Engineering

13.2 Real Work

13–3

Draft

Figure 13.2: Strain Energy Deﬁnition

Axial Members: U

=

σ ε dV

= = =

Torsional Members:

εσ dVol Vol 2

P A P AE

          

Adx

L

U= 0

P2 dx 2AE

(13.8)

U

=

U

=

      Vol σ      1  γ Gγ d Vol  xy xy 2 

 Vol  1 2

ε Eε dVol

τxy

τxy γxy dVol J

              

= TJr τxy = G = rdθdrdx r 2π = r2 dθ dr o

L

U= 0

T2 dx 2GJ

(13.9)

0

Flexural Members: U

=

σx ε

= = dVol = Iz

=

  ε Eε    Vol σ     Mz y   I 1 2

z

Mz y EIz

dA dx y 2 dA

         

L

U= 0

M2 dx 2EIz

(13.10)

A

Example 13-1: Deﬂection of a Cantilever Beam, (Chajes 1983) Determine the deﬂection of the cantilever beam, Fig. 13.3 with span L under a point load P applied at its free end. Assume constant EI. Solution:

We

Victor Saouma

=

1 P ∆f 2

(13.11-a)

Structural Engineering

13–4

ENERGY METHODS; Part I

Draft

P

1 0 0 1 0 1 1111111111111111111111111 0000000000000000000000000 x

M=PL Figure 13.3: Deﬂection of Cantilever Beam

13.3

U

=

M

=

U

=

1 P ∆f 2

=

∆f

=

L

M2 dx 0 2EI −P x P2 L 2 P 2 L3 x dx = 2EI 0 6EI 2 3 P L 6EI P L3 3EI

(13.11-b) (13.11-c) (13.11-d) (13.11-e) (13.11-f)

Virtual Work

A severe limitation of the method of real work is that only deﬂection along the externally applied load can be determined.

11

12

A more powerful method is the virtual work method.

The principle of Virtual Force (VF) relates force systems which satisfy the requirements of equilibrium, and deformation systems which satisfy the requirement of compatibility

13

In any application the force system could either be the actual set of external loads dp or some virtual force system which happens to satisfy the condition of equilibrium δp. This set of external forces will induce internal actual forces dσ or internal virtual forces δσ compatible with the externally applied load.

14

15 Similarly the deformation could consist of either the actual joint deﬂections du and compatible internal deformations dε of the structure, or some virtual external and internal deformation δu and δε which satisfy the conditions of compatibility.

16

It is often simplest to assume that the virtual load is a unit load.

Thus we may have 4 possible combinations, Table 13.1: where: d corresponds to the actual, and δ (with an overbar) to the hypothetical values. This table calls for the following observations

17

1. The second approach is the same one on which the method of virtual or unit load is based. It is simpler to use than the third as a internal force distribution compatible with the assumed virtual Victor Saouma

Structural Engineering

13.3 Virtual Work

Draft

1 2 3 4

13–5 Force External Internal dp dσ δp δσ dp dσ δp δσ

Deformation External Internal du dε du dε δu δε δu δε

IVW

Formulation

∗

δU δU

Flexibility Stiﬀness

Table 13.1: Possible Combinations of Real and Hypothetical Formulations force can be easily obtained for statically determinate structures. This approach will yield exact solutions for statically determinate structures. 2. The third approach is favored for statically indeterminate problems or in conjunction with approximate solution. It requires a proper “guess” of a displacement shape and is the basis of the stiﬀness method. Let us consider an arbitrary structure and load it with both real and virtual loads in the following sequence, Fig. 13.4. For the sake of simplicity, let us assume (or consider) that this structure develops only axial stresses.

18

P1

1 0 0 1

11 00 00 11 00 11 111 000 000 111 000 111 000 111 000 111

δε

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

δ∆

1 0 0 1

11 00 00 11 00 11

δσ 111 000 000 111 000 111 000 111 000 111

∆1

σ

ε

111 000 000 111 000 111 000 111 000 111

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111

∆

δσ

Virtual System

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111

P

δ∆ σ

δε

Real Load

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

δP

δσ

∆1

11111 00000 00000 11111 00000 11111 00000 11111 00000 11111

ε

1111 0000 0000 1111 0000 1111 0000 1111 δ∆ 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 δε

P1

δσ

∆1 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111 00000 11111

ε

δ∆+∆

δP δP

1 0 0 1

11 00 00 11 00 11

δP P

σ

1111 0000 0000 1111 0000 1111 0000 ∆ 1111 0000 1111 0000 1111 0000 1111 0000 1111 0000 ε 1111

δP

1

δσ

1111 0000 0000 1111 0000 1111 0000 1111 ∆ 0000 1111 0000 1111 0000 1111 ε

Virtual + Real Loads

Inter.

Ext.

δσ ε = δP ∆ Virtual Real

Figure 13.4: Real and Virtual Forces 1. If we apply the virtual load, then 1 1 δP δ∆ = 2 2 Victor Saouma

δσδεdVol

(13.12)

dVol

Structural Engineering

13–6

ENERGY METHODS; Part I

Draft

2. Load with the real (applied) load, since the external work must be equal to the internal strain energy over the entire volume, then: 1 1 P1 ∆1 = σIdVol (13.13) 2 2 dVol 3. We now immagine that the virtual load was ﬁrst applied, and we then apply the real (actual) load on top of it, then the total work done is 1 1 1 1 δP δ∆ + P1 ∆1 + δP (13.14) ∆ = 2 Vol δσδεdVol + 2 Vol σεdVol + Vol δσεdVol 2 2

4. Since the strain energy and work done must be the same whether the loads are applied together or separately, we obtain, from substracting the sum of Eqs. 13.13 and 13.12 from 13.14 and generalizing, we obtain (δσε + δτ γ) dVol = δP ∆ (13.15) ∗

δW δU

∗

This last equation is the key to the method of virtual forces. The left hand side is the internal virtual ∗ strain energy δU 2 . Similarly the right hand side is the external virtual work.

19

13.3.1

External Virtual Work δW

∗ ∗

20

The general expression for the external virtual work δW is ∗

δW =

)n i=1

(∆i )δP i +

)n i=1

(13.16)

(θi )δM i

for distributed load, point load, and concentrated moment respectively. 21

Recall that all overbar quantities are virtual and the other ones are the real.

13.3.2 22

Internal Virtual Work δU

∗

The general expression for the internal virtual work is

(δσε + δτ γ) dVol = δP ∆ ∗

δW δU

(13.17)

∗

We will generalize it to diﬀerent types of structural members. We will ﬁrst write the equations independently of the material stress strain relation, and then we will rewrite those same equations for a linear elastic system.

23

General : Axial Members: δU

∗

L

=

δσεdVol 0

dVol =

Victor Saouma

Adx

  

δU = A ∗

L

δσεdx

(13.18)

0

Structural Engineering

13.3 Virtual Work

13–7

Draft

Figure 13.5: Torsion Rotation Relations Torsional Members: With reference to Fig. 13.5 Note that in torsion the “strain” is γ the rate of change of rotation of the cross section about the longitudinal axis x.  ∗ δU = δτ xy γxy dVol     Vol  L  ∗ δT = δτ xy rdA δU = ( δτ xy rdA) θdx (13.19)  A 0 A   dθ

 γxy = r dx   δT dVol = dAdx L dθ ∗ ⇒ δU = δT dx dx 0 Shear Members:

 δτ xy γxy dVol     Vol ∗ δU = δV = δτ xy dA    A  dVol = dAdx δU

∗

=

L 0

(

∗

A

δτ xy dA) γxy dx

δV L

⇒ δU =

(13.20)

δV γxy dx 0

Flexural Members: δU

∗

δM

=

δσ x εx dVol δσ x ydA ⇒

=

φ = φy =

ε y

ε

A

δM = y

A

          δσ x dA  

L

dVol =

dAdx 0

          

∗

L

δU =

δM φdx

(13.21)

0

A

Linear Elastic Systems Should we have a linear elastic material

σ = Eε

(13.22)

then the previous equations can be rewritten as: 2 We

use the * to distinguish it from the internal virtual strain energy obtained from the virtual displacement method

δU .

Victor Saouma

Structural Engineering

13–8

ENERGY METHODS; Part I

Draft

Axial Members: δU

∗

=

δσ ε dV

  εδσdVol     Vol

δP A P AE

= = =

    

Adx

∗

L

δU =

P dx AE

δP 0

(13.23)

∆

Note that for a truss where we have n members, the above expression becomes ∗

δU = Σn1 δP i Shear Members: δU

∗

τxy γxy dVol λ

Pi Li Ai Ei

(13.24)

= = = = def

=

 δτ xy γxy dVol     Vol   V  kA  τxy G

       

Adx k 2 dydz

∗ δU = λ

L

δV 0

V dx GA

(13.25)

γxy

A

24

Note that the exact expression for the shear stress is τ=

VQ Ib

(13.26)

where Q is the moment of the area from the external ﬁbers to y with respect to the neutral axis; For a rectangular section, this yields τ

= = =

VQ Ib V Ib 6V bh3

(13.27-a) h/2

by dy =

y

h2 −4 4

V 2I

h2 − y2 4

(13.27-b) (13.27-c)

and we observe that the shear stress is zero for y = h/2 and maximum at the neutral axis V where it is equal to 1.5 bh . 25

To determine the form factor λ of a rectangular section  2 τ = VIbQ A h 2   k = − y  2I 4 = k VA 2 h/2 k 2 dydz bh = b h  λ  − y2  Q = by dy = A A 2 4 y

    

λ = 1.2

(13.28)

26

Thus, the form factor λ may be taken as 1.2 for rectangular beams of ordinary proportions.

27

For I beams, k can be also approximated by 1.2, provided A is the area of the web.

Torsional Members: δU

∗

=

δτ xy Vol

τxy dVol G γxy

τxy = γxy = dVol = Victor Saouma

J

=

Tr τJ

xy

G

rdθdrdx r 2π r2 dθ dr o

0

                      

∗

L

δU = 0

T δT dx GJ δσ

(13.29)

ε

Structural Engineering

13.3 Virtual Work

13–9

Draft 28

Note the similarity with the corresponding equation for shear deformation.

29

The torsional stiﬀness of cylindrical sections is given by J =

30

The torsional stiﬀness of solid rectangular sections is given by

πd4 32 .

J = kb3 d

(13.30)

where b is the shorter side of the section, d the longer, and k a factor given by Table 13.2. d/b k

1.0 0.141

1.5 0.196

2.0 0.229

2.5 0.249

3.0 0.263

4.0 0.281

5.0 0.291

10 0.312

∞ 0.333

Table 13.2: k Factors for Torsion

31

Recall from strength of materials that

G=

Flexural Members: δU

E 2(1 + ν)

(13.31)

∗

  ε Eδε dV ol     Vol   δσ   Mz y  I

=

σx = ε = dVol = Iz

=

z

Mz y EIz

         

dA dx y 2 dA

∗

L

δU =

δM 0

M dx EIz

(13.32)

Φ

A

13.3.3

Examples

Example 13-2: Beam Deﬂection (Chajes 1983) Determine the deﬂection at point C in Fig. 13.6 E = 29, 000 ksi, I = 100 in4 . Solution: For the virtual force method, we need to have two expressions for the moment, one due to the real load, and the other to the (unit) virtual one, Fig. 13.7. Element AB BC

x=0 A C

M 15x − x2 −x2

δM −0.5x −x

Applying Eq. 13.32 we obtain ∆C δP = δW

∗

M δM dx EI 0 z

(1)∆C

L

δU

(−0.5x) 0

Victor Saouma

∗

20

=

(13.33-a)

(15x − x2 ) dx + EI

10

(−x) 0

−x2 dx EI

(13.33-b)

Structural Engineering

13–10

ENERGY METHODS; Part I

Draft

Figure 13.6:

15x -x 2

-0.5x

-x Figure 13.7:

.

Victor Saouma

Structural Engineering

13.3 Virtual Work

13–11

Draft

= ∆C

= =

2, 500 EI (2, 500) k2 − ft3 (1, 728) in3 / ft3 (29, 000) ksi(100) in4 1.49 in

(13.33-c) (13.33-d) (13.33-e)

Example 13-3: Deﬂection of a Frame (Chajes 1983) Determine both the vertical and horizontal deﬂection at A for the frame shown in Fig. 13.8. E = 200 × 106 kN/ m2 , I = 200 × 106 mm4 .

Figure 13.8: . Solution: To analyse this frame we must determine analytical expressions for the moments along each member for the real load and the two virtual ones. One virtual load is a unit horizontal load at A, and the other a unit vertical one at A also, Fig. 13.9. Element AB BC CD

x=0 A B C

M 0 50x 100

δM v x 2+x 4

δM h 0 0 −x

Note that moments are considered positive when they produce compression on the inside of the frame.

Victor Saouma

Structural Engineering

13–12

ENERGY METHODS; Part I

Draft

50 kN

1 kN

x

50x

x

+100 +

+4

1 kN

+

x +

-x 1 kN 100 kN-m

4 kN-m

50 kN

5 kN-m

1 kN

Figure 13.9: Substitution yields: ∆v δP = δW

∗

0

= = =

M δM dx EIz

δU

(13.34-a)

∗

2 5 (0) 50x 100 dx + dx + dx (x) (2 + x) (4) EI EI EI 0 0 0 2, 333 kN m3 EI (2, 333) kN m3 (103 )4 mm4 / m4 (200 × 106 ) kN/ m2 (200 × 106 ) mm4

(1)∆v

L

2

= 0.058 m = 5.8 cm

(1)∆h

= = =

δU

(13.34-c) (13.34-d) (13.34-e)

Similarly for the horizontal displacement: L M ∆h δP = δM dx EI z 0 ∗

δW

(13.35-a)

∗

2 5 (0) 50x 100 dx + dx + dx (0) (0) (−x) EI EI EI 0 0 0 −1, 250 kN m3 EI (−1, 250) kN m3 (103 )4 mm4 / m4 (200 × 106 ) kN/ m2 (200 × 106 ) mm4

(13.34-b)

2

= −0.031 m = −3.1 cm

(13.35-b) (13.35-c) (13.35-d) (13.35-e)

Example 13-4: Rotation of a Frame (Chajes 1983) Determine the rotation of joint C for frame shown in Fig. 13.10. E = 29, 000 ksi, I = 240 in4 . Solution: In this problem the virtual force is a unit moment applied at joint C, δM e . It will cause an internal moment δM i Victor Saouma

Structural Engineering

13.3 Virtual Work

13–13

Draft

Figure 13.10: Element AB BC CD

x=0 A B D

M 0 30x − 1.5x2 0

δM 0 −0.05x 0

Note that moments are considered positive when they produce compression on the outside of the frame. Substitution yields: θC δM e δW

=

∗

M δM dx EIz 0

=

(1)θC

= =

L

δU

(13.36-a)

∗

(30x − 1.5x2 ) dx k2 ft3 EI 0 (1, 000)(144) (29, 000)(240) 20

(−0.05x)

(13.36-b) (13.36-c)

−0.021 radians

(13.36-d)

Example 13-5: Truss Deﬂection (Chajes 1983) Determine the vertical deﬂection of joint 7 in the truss shown in Fig. 13.11. E = 30, 000 ksi. Solution: Two analyses are required. One with the real load, and the other using a unit vertical load at joint 7. Results for those analysis are summarized below. Note that advantage was taken of the symmetric load and structure. Member

A in

1&4 10 & 13 11 & 12 5&9 6&8 2& 3 7 Total Victor Saouma

2 2 2 1 1 2 1

2

L

P

δP

ft

k

k

25 20 20 15 25 20 15

-50 40 40 20 16.7 -53.3 0

-0.083 0.67 0.67 0 0.83 -1.33 0

δP P L A

n

nδP P L A

518.75 268.0 268.0 0 346.5 708.9 0

2 2 2 2 2 2 1

1,037.5 536.0 536.0 0 693.0 1,417.8 0 4,220.3 Structural Engineering

13–14

ENERGY METHODS; Part I

Draft

Figure 13.11: Thus from Eq. 13.24 we have:

∆δP = δW

∗

L

P dx δP AE 0

δU

= 1∆ = =

(13.37-a)

∗

PL AE (4, 220.3)(12) 30, 000

(13.37-b)

ΣδP

(13.37-c)

1.69 in

(13.37-d)

Example 13-6: Torsional and Flexural Deformation, (Chajes 1983) Determine the vertical deﬂection at A in the structure shown in Fig. 13.12. E = 30, 000 ksi, I = 144 in4 , G = 12, 000 ksi, J = 288 in4 Solution: 1. In this problem we have both ﬂexural and torsional deformation. Hence we should determine the internal moment and torsion distribution for both the real and the unit virtual load. 2. Then we will use the following relation δW

∗

δP ∆A =

δU

∗

M T dx + δT dx δM EI GJ

Torsion ﬂexure

3. The moment and torsion expressions are given by Victor Saouma

Structural Engineering

13.3 Virtual Work

13–15

Draft

Figure 13.12: Element AB BC 4. Substituting,

x=0 A B

M 10x 15x

δM x x

T 0 50

δT 0 5

M T dx + δT dx 5 GJ 5 5 EI 10x 15x 50 dx + dx + dx x x (5) GJ 0 EI 0 EI 0 (1,042)(1,728) (1,250)(1,728) (30,000)(144) + (12,000)(288) 0.417 + 0.625 1.04 in

δP ∆A

=

1∆A

=

δM

= = =

Example 13-7: Flexural and Shear Deformations in a Beam (White et al. 1976) Determine the delection of a cantilevered beam, of length L subjected to an end force P due to both ﬂexural and shear deformations. Assume G = 0.4E, and a square solid beam cross section. Solution: 1. The virtual work equation is δP ·∆ = 1

L

δM dφ + 0

=

L

δV γxy dx

(13.38-a)

0

M dx + λ δM EI 0 L

L

δV λ 0

V dx GA

(13.38-b)

2. The ﬁrst integral yields for M = P x, and δM = (1)(x)

Victor Saouma

L

M dx = δM EI 0 =

L P x2 dx EI 0 P L3 /3EI

(13.39-a) (13.39-b) Structural Engineering

13–16

ENERGY METHODS; Part I

Draft

3. The second integral represents the contribution of the shearing action to the total internal virtual work and hence to the total displacement. 4. Both the real shear V and virtual shear δV are constant along the length of the member, hence

L

δV λ 0

λ V dx = GA GA

L

1(P )dx = 0

λP L GA

(13.40)

5. Since λ= 1.2 for a square beam; hence 1·∆ = =

P L3 1.2P L + 3EI GA P L L2 3.6 + = 3E I 0.4A

(13.41-a) PL 3E

&

L2 I

+

9 A

' (13.41-b)

6. For a square beam of dimension h I= then

h4 12

and A = h2

(13.42)

P L 12L2 9 3P L 1.33L2 ∆= + 2 = +1 3E h4 h Eh2 h2

(13.43)

7. Choosing L = 20 ft and h = 1.5 ft (L/h = 13.3) 3P L [1.33 ∆= Eh2

20 1.5

2 + 1] =

3P L Eh2 (237

+ 1)

(13.44)

Thus the ﬂexural deformation is 237 times the shear displacement. This comparison reveals why we normally neglect shearing deformation in beams. As the beam gets shorter or deeper, or as L/h decreases, the ﬂexural deformation decreases relative to the shear displacement. At L/h = 5, the ﬂexural deformation has reduced to 1.33(5)2 = 33 times the shear displacement.

Example 13-8: Thermal Eﬀects in a Beam (White et al. 1976) The cantilever beam of example 13-7 is subjected to a thermal environment that produces a temperature change of 70◦ C at the top surface and 230◦ C at the bottom surface, Fig. 13.13. If the beam is a steel, wide ﬂange section, 2 m long and 200 mm deep, what is the angle of rotation, θ1 , at the end of beam as caused by the temperature eﬀect? The original uniform temperature of the beams was 40◦ C. Solution: 1. The external virtual force conforming to the desired real displacement θ1 is a moment δM = 1 at the tip of the cantilever, producing an external work term of moment times rotation. The internal virtual force system for this cantilever beam is a uniform moment δM int = 1. 2. The real internal deformation results from: (a) the average beam temperature of 150◦ C, which is 110◦ C above that of the original temperature, and (b) the temperature gradient of 160◦ C across the depth of the beam. 3. The ﬁrst part of the thermal eﬀect produces only a lengthening of the beam and does not enter into the work equation since the virtual loading produces no axial force corresponding to an axial change in length of the beam. 4. The second eﬀect (thermal gradient) produces rotation dφ, and an internal virtual work term of (L 0 δM dφ. Victor Saouma

Structural Engineering

13.3 Virtual Work

13–17

Draft

Figure 13.13: 5. We determine the value of dφ by considering the extreme ﬁber thermal strain as shown above. The angular rotation in the length dx is the extreme ﬁber thermal strain divided by half the beam depth, or dφ =

α∆T dx (11.7 × 10−6 )(80)dx 936 × 10−6 ε = = = = (9.36 × 10−6 )dx h/2 h/2 100 100

(13.45)

6. Using the virtual work equation 1 · θ1

L

0 2,000

=

δM dφ

=

1(9.36 × 10−6 )dx

(13.46-a) (13.46-b)

0

= =

+0.01872 0.01872 ✛✁ radians

(13.46-c) (13.46-d)

7. This example raises the following points: 1. The value of θ1 would be the same for any shape of 200 mm deep steel beam that has its neutral axis of bending at middepth. 2. Curvature is produced only by thermal gradient and is independent of absolute temperature values. 3. The calculation of rotations by the method of virtual forces is simple and straightforward; the applied virtual force is a moment acting at the point where rotation is to be calculated. 4. Internal angular deformation dφ has been calculated for an eﬀect other than load-induced stresses. The extension of the method of virtual forces to treat inelastic displacements is obvious – all we need to know is a method for determining the inelastic internal deformations.

Example 13-9: Deﬂection of a Truss (White et al. 1976)

Victor Saouma

Structural Engineering

13–18

ENERGY METHODS; Part I

Draft 60 k

120 k 7

4 3 1

4

5 5

12’

1

2

12’

12’

3

Figure 13.14: Determine the deﬂection at node 2 for the truss shown in Fig. 13.14. Solution:

Member 1 2 3 4 5 6 7

δP kips +0.25 +0.25 -0.56 +0.56 +0.56 -0.56 -0.50

P, kips +37.5 +52.5 -83.8 +16.8 -16.8 -117.3 -45.0

L, ft 12 12 13.42 13.42 13.42 13.42 12

A, in2 5.0 5.0 5.0 5.0 5.0 5.0 5.0

E, ksi 10 × 103 10 × 103 10 × 103 10 × 103 10 × 103 10 × 103 10 × 103

PL δP AE

+22.5 × 10−4 +31.5 × 10−4 +125.9 × 10−4 +25.3 × 10−4 −25.3 × 10−4 +176.6 × 10−4 +54.0 × 10−4 +410.5 × 10−4

The deﬂection is thus given by δP ∆ =

7 ) 1

∆ =

δP

PL AE

(13.47-a)

(410.5 × 10−4 )(12 in/ ft) = 0.493 in

(13.47-b)

Example 13-10: Thermal Defelction of a Truss; I (White et al. 1976)

Victor Saouma

Structural Engineering

13.3 Virtual Work

13–19

Draft

The truss in example 13-9 the preceding example is built such that the lower chords are shielded from the rays of the sun. On a hot summer day the lower chords are 30◦ F cooler than the rest of the truss members. What is the magnitude of the vertical displacement at joint 2 as a result of this temperature diﬀerence? Solution: 1. The virtual force system remains identical to that in the previous example because the desired displacement component is the same. 2. The real internal displacements are made up of the shortening of those members of the truss that are shielded from the sun. 3. Both bottom chord members 1 and 2 thus shorten by ∆l = α(∆T )(L) = (0.0000128) in/ in/oF (30)o (12) ft(12) in/ft = 0.0553 in

(13.48)

4. Then, 1·∆ =

)

=

δP (∆L) = .25(−0.0553) + .25(−0.0553) = −0.0276

−0.0276 in ✻

(13.49-a) (13.49-b)

5. The negative sign on the displacement indicates that it is in opposite sense to the assumed direction of the displacement; the assumed direction is always identical to the direction of the applied virtual force. 6. Note that the same result would be obtained if we had considered the internal displacements to be made up of the lengthening of all truss members above the bottom chord.

Example 13-11: Thermal Deﬂections in a Truss; II (White et al. 1976) A six-panel highway bridge truss, Fig. 13.15 is constructed with sidewalks outside the trusses so that

Figure 13.15: the bottom chords are shaded. What will be the vertical deﬂection component of the bottom chord at the center of the bridge when the temperature of the bottom chord is 40◦ F (∆T ) below that of the top chord, endposts, and webs? (coeﬃcient of steel thermal expansion is α=0.0000065 per degree F.) Solution: 1. The deﬂection is given by

= ∆δP δW

∗

L

P PL dx = ΣδP = ΣδP ∆L = ΣδP α∆T L δP AE AE 0

δU

Victor Saouma

(13.50)

∗

Structural Engineering

13–20

ENERGY METHODS; Part I

Draft

where ∆L is the temperature change in the length of each member, and δP are the member virtual internal forces. 2. Taking advantage of symmetry: Member 4 5 6 7 8 9 10 11

L (ft) 35 21 21 0 35 28 35 0

(0.0000065)(40)L α∆T L +0.00910 +0.00546 +0.00546 0 +0.00910 +0.00728 +0.00910 0

δP (k) + 0.625 + 0.75 +1.13 0 -0.625 +0.5 -0.625 0

δP ∆L +0.00568 +0.00409 +0.00616 0 -0.00568 +0.00364 -0.00568 0 +0.00821

3. Hence, the total deﬂection is ∆ = (2)(0.00821)(12) in/ft = +0.20 in ✻

(13.51)

4. A more eﬃcient solution would have consisted in considering members 1,2, and 3 only and apply a ∆T = −40, we would obtain the same displacement. 5. Note that the forces in members 1, 2, and 3 (-0.75, -0.375, and -0.375 respectively) were not included in the table because the corresponding ∆T = 0. 6. A simpler solution would have ∆T = −40 in members 1, 2, and 3 thus, Member 1 2 3

L (ft) 21 21 21

(0.0000065)(40)L α∆T L -0.00546 -0.00546 -0.00546

δP (k) -0.75 -0.375 -0.375

δP ∆L +0.004095 +0.0020475 +0.0020475 +0.00819

∆ = (2)(0.00819)(12) in/ft = +0.20 in ✻

(13.52)

Example 13-12: Truss with initial camber It is desired to provide 3 in. of camber at the center of the truss shown below

by fabricating the endposts and top chord members additionally long. How much should the length of each endpost and each panel of the top chord be increased? Solution: 1. Assume that each endpost and each section of top chord is increased 0.1 in. Victor Saouma

Structural Engineering

13.3 Virtual Work

13–21

Draft

Member 1 2 3

δP int +0.625 +0.750 +1.125

∆L +0.1 +0.1 +0.1

δP int ∆L +0.0625 +0.0750 +0.1125 +0.2500

Thus, (13.53)

(2)(0.250) = 0.50 in

2. Since the structure is linear and elastic, the required increase of length for each section will be 3.0 (13.54) (0.1) = 0.60 in 0.50 3. If we use the practical value of 0.625 in., the theoretical camber will be (6.25)(0.50) = 3.125 in 0.1

(13.55)

Example 13-13: Prestressed Concrete Beam with Continously Variable I (White et al. 1976)

A prestressed concrete beam is made of variable depth for proper location of the straight pretensioning tendon, Fig. 13.16. Determine the midspan displacement (point c) produced by dead weight of the girder. The concrete weights 23.6 kN/m3 and has E = 25, 000 MPa (N/mm2 ). The beam is 0.25 m wide. Solution: 1. We seek an expression for the real moment M , this is accomplished by ﬁrst determining the reactions, and then considering the free body diagram. 2. We have the intermediary resultant forces R1 R2

= (0.25)x(0.26)m3 (23.6)kN/m3 = 3.54x = f rac12(0.25)x(0.04x)m3 (23.6)kN/m3 = 0.118x2

(13.56-a) (13.56-b)

Hence, M (x)

= 47.2x − 3.54x

x

− 0.118x2

2 2 = 47.2x − 1.76x − 0.0393x3

x 3

(13.57-a) (13.57-b)

3. The moment of inertia of the rectangular beam varies continously and is given, for the left half of the beam, by 1 3 1 I(x) = bh = (0.25)(0.6 + 0.04x)3 (13.58) 12 12 1 48

4. Thus, the real angle changes produced by dead load bending are dφ =

M 47.2x − 1.76x2 − 0.0393x3 1 dx = dx EI (0.6 + 0.04x)3 E 48

(13.59)

5. The virtual force system corresponding to the desired displacement is shown above with δM = (1/2)x for the left half of the span. Since the beam is symmetrical, the virtual work equations can be evaluated for only one half of the beam and the ﬁnal answer is then obtained by multiplying the half-beam result by two. Victor Saouma

Structural Engineering

13–22

Draft

ENERGY METHODS; Part I

Figure 13.16: *(correct 42.7 to 47.2)

Victor Saouma

Structural Engineering

13.3 Virtual Work

13–23

Draft

6. The direct evaluation of the integral

δM dφ is diﬃcult because of the complexity of the expression for * dφ. Hence we shall use a numerical procedure, replacing the δM (M/EI)dx with the δM (M/EI)∆x, where each quantity in the summation is evaluated at the center of the interval ∆x and held constant over the interval length. As ∆x becomes very short, the solution approaches the exact answer. 7. An interval length of 1 meter, giving 10 elements in the half length of the beam, is chosen to establish an accurate result. 8. The internal virtual work quantity is then L/2 ) δM M M dx (∆x) (13.60-a) δM EI EI 0 ) δM M (∆x) (13.60-b) = h3 E 0.25 12 48 ) δM M (∆x) (13.60-c)

E h3 9. The summation for the 10 elements in the left half of the beam gives Segment 1 2 3 4 5 6 7 8 9 10

x 0.5 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5

h3 0.238 0.288 0.343 0.405 0.475 0.551 0.636 0.729 0.831 0.941

h 0.62 0.66 0.70 0.74 0.78 0.82 0.86 0.90 0.94 0.98

M 23.2 66.7 106.4 150 173 200 222 238 250 256

δM 0.25 0.75 1.25 1.75 2.25 2.75 3.25 3.75 4.25 4.75

δM M h3

24 174 388 648 820 998 1,134 1,224 1,279 1,292 7,981

10. The SI units should be checked for consistency. Letting the virtual force carry the units of kN, the virtual moment δM has the units of m·kN, and the units of the equation ∆c = are

1 1 kN

) δM M EI

∆x

1 ( m · kN)( m · kN) m = mm m= ( MN/ m2 ) m4 1, 000

(13.62)

48 δM M dx 2 (7, 981)(1) = 30.6mm EI 25, 000

(13.63)

kN

11. Then

L 0

(13.61)

and the deﬂection at midspan is ∆c = 30.6 mm

(13.64)

12. Acceptably accurate results may be obtained with considerably fewer elements (longer intervals ∆x). * Using four elements with centers at 2, 5, 8, and 10, the (δM M/h3 )∆x is ) = 3(174) + 3(820) + 3(1, 224) + 1(1, 292) = 7, 946 (13.65) which is only 0.4% lower than the 10 element solution. If we go to two elements, 3 and 8, we obtain a summation of 5(388) + 5(1, 224) = 8, 060, which is 1% high. A one element solution, with x = 5 m and h = 0.8 m, gives a summation of 9,136, which is 14.4% high and much less accurate than the 2 element solution. 13. Finally, it should be noted that the calculations involved in this example are essentially identical to those necessary in the moment area method.

Victor Saouma

Structural Engineering

13–24

ENERGY METHODS; Part I

Draft 13.4 24

*Maxwell Betti’s Reciprocal Theorem

If we consider a beam with two points, A, and B, we seek to determine 1. The deﬂection at A due to a unit load at B, or fAB 2. The deﬂection at B due to a unit load at A, or fBA

25

Applying the theorem of vitual work fAB

fBA

L

=

δM A

MB dx EI

(13.66-a)

δM B

MA dx EI

(13.66-b)

0 L

= 0

But since the both the real and the virtual internal moments are caused by a unit load, both moments are numerically equal δM A δM B

= =

MA MB

(13.67-a) (13.67-b)

thus we conclude that fBA = fBA

(13.68)

or The displacement at a point B on a structure due to a unit load acting at point A is equal to the displacement of point A when the unit load is acting at point B. 26 Similarly The rotation at a point B on a structure due to a unit couple moment acting at point A is equal to the rotation at point A when a unit couple moment is acting at point B. 27 And The rotation in radians at point B on a structure due to a unit load acting at point A is equal to the displacement of point A when a unit couple moment is acting at point B.

These theorems will be used later on in justifying the symmetry of the stiﬀness matrix, and in construction of inﬂuence lines using the M¨ uller-Breslau principle.

28

13.5

Summary of Equations

Victor Saouma

Structural Engineering

13.5 Summary of Equations

13–25

Draft

∗

U

L

Axial 0

P2 dx 2AE

Shear

...

L

Flexure 0

L

Torsion 0

P M

2

M dx 2EIz T2 dx 2GJ

W Σi 21 Pi ∆i Σi 21 Mi θi

L

w

w(x)v(x)dx 0

Virtual Force δU General Linear L L P δσεdx δP dx AE 0 0 δσ ε

L

δV γxy dx 0L δM φdx

...

0

L 0

L

L

δT θdx 0

0

M δM dx EIz δσ ε

T δT dx GJ δσ

∗

ε

Virtual Force δW Σi δP i ∆i Σ δM i θi L i δw(x)v(x)dx 0

Table 13.3: Summary of Expressions for the Internal Strain Energy and External Work

Victor Saouma

Structural Engineering

13–26

Draft

ENERGY METHODS; Part I

Victor Saouma

Structural Engineering

Draft Chapter 14

BRACED ROLLED STEEL BEAMS This chapter deals with the behavior and design of laterally supported steel beams according to the LRFD provisions.

1

2

If a beam is not laterally supported, we will have a failure mode governed by lateral torsional buckling.

3 By the end of this lecture you should be able to select the most eﬃcient section (light weight with adequate strength) for a given bending moment and also be able to determine the ﬂexural strength of a given beam.

14.1

Review from Strength of Materials

14.1.1

Flexure

Fig.14.1 shows portion of an originally straight beam which has been bent to the radius ρ by end couples M , thus the segment is subjected to pure bending. It is assumed that plane cross-sections normal to the length of the unbent beam remain plane after the beam is bent. Therefore, considering the cross-sections AB and CD a unit distance apart, the similar sectors Oab and bcd give

4

ε=

y ρ

(14.1)

where y is measured from the axis of rotation (neutral axis). Thus strains are proportional to the distance from the neutral axis. 5 The corresponding variation in stress over the cross-section is given by the stress-strain diagram of the material rotated 90o from the conventional orientation, provided the strain axis ε is scaled with the distance y (Fig.14.1). The bending moment M is given by M = yσdA (14.2)

where dA is an diﬀerential area a distance y (Fig.14.1) from the neutral axis. Thus the moment M can be determined if the stress-strain relation is known. 6

If stress is proportional to strain, σ = Eε, Equations 14.1 and 14.2 give EI M = Eρ y 2 dA = ρ σI = EIε y = y = σS

where S=

I y

(14.3)

14–2

BRACED ROLLED STEEL BEAMS

Draft

O

ρ

C c d y b

A M

a B

M

D

Figure 14.1: Bending of a Beam is the Elastic Section Modulus. Note that for doubly symmetric sections, we have two section modulus Sx and Sy . 7 The stress distribution on a typical wide-ﬂange shape subjected to increasing bending moment is shown in Fig.14.2. 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 ﬁber has not reached the yield stress Fy . Once the strain ε reaches its yield value εy , increasing strain induces no increase in stress beyond Fy .

Figure 14.2: Stress distribution at diﬀerent stages of loading When the yield stress is reached at the extreme ﬁber, the nominal moment strength Mn , is referred to as the yield moment My and is computed as

8

Mn = My = Sx Fy

(14.4)

(assuming that bending is occurring with respect to the x − x axis). When across the entire section, the strain is equal or larger than the yield strain (ε ≥ εy = Fy /Es ) then the section is fully plastiﬁed, and the nominal moment strength Mn is therefore referred to as the

9

Victor Saouma

Structural Engineering

14.1 Review from Strength of Materials

Draft

F

Elastic Region

14–3

Plastic Region

F

y

Stress

Strain

ε

Figure 14.3: Stress-strain diagram for most structural steels plastic moment Mp and is determined from Mp = Fy

(14.5)

ydA = Fy Z A

Z=

(14.6)

ydA

is the Plastic Section Modulus. 10 The ratio Mp /My is a property of the cross-sectional shape and is independent of the material properties, it is referred to as the Shape Factor

ξ=

Mp Z = My S

(14.7)

The shape factors of some regular cross-sectional shapes bending about their major axis are: 1. Rectangular section = 1.50 2. I section = 1.11 3. Circular section = 1.70

14.1.2

Shear

Examining the stress distribution in FIg. 14.4, if there is a variation of moment along the inﬁnitesimal segment dx (thus a shear force is present), we determine the total force acting over the shaded area on the left of the section dx, on the right, and then take the diﬀerence which must be equal to the shear force along the top surface of the shaded zone

11

σx dA =

Victor Saouma

F1

=

F2

=

τ bdx

=

My dA I My dA I A (M + dM )y dA I A F2 − F1

(14.8-a) (14.8-b) (14.8-c) (14.8-d) Structural Engineering

14–4

BRACED ROLLED STEEL BEAMS

Draft

b

τ h/2

M

M+dM

O

h h/2

y1 00000 11111 11111 00000 00000 11111 00000 11111 00000 11111 00000 11111 11111 00000

y 00000000 11111111 11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 11111111 00000000 11111111 00000000

σx

dx

z

dA

τ

max

y

Figure 14.4: Flexural and Shear Stress Distribution in a Rectangular Beam = τ

=

dM y dA I A dM 1 ydA dx Ib A

(14.8-e) (14.8-f) (14.8-g)

substituting V = dM/dx we now obtain τ

=

VQ Ib

Q =

(14.9)

ydA (14.10) A

this is the shear formula, and Q is the ﬁrst moment of the area. 12

For a rectangular beam, we will have a parabolic shear stress distribution.

For a W section, it can be easily shown that about 95% of the shear force is carried by the web, and that the average shear stress is within 10% of the actual maximum shear stress.

13

14.2

Nominal Strength

The strength requirement for beams in load and resistance factor design is stated as φb Mn ≥ Mu where: φb Mn Mu

(14.11)

strength reduction factor; for ﬂexure 0.90 nominal moment strength factored service load moment.

The equations given in this chapter are valid for ﬂexural members with the following kinds of cross section and loading:

14

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 . 1 More

about shear centers in Mechanics of Materials II.

Victor Saouma

Structural Engineering

14.3 Flexural Design

14–5

Draft 14.3

Flexural Design

14.3.1

Failure Modes and Classiﬁcation of Steel Beams

15

The strength of ﬂexural members is limited by:

Plastic Hinge: at a particular cross section. local buckling: of a cross-sectional element (e.g. the web or the ﬂange), Lateral-Torsional buckling: of the entire member. Fig. 14.5 illustrates some of the buckling mode possible in a rolled section.

Figure 14.5: Local (ﬂange) Buckling; Flexural and Torsional Buckling Modes in a Rolled Section, (Lulea University)

16

Accordingly, the LRFD manual classiﬁes 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 ﬁrst two cases. 14.3.1.1

Compact Sections

For compact sections, the mode of failure is the formation of a plastic hinge that is the section is fully plastiﬁed. Hence we shall ﬁrst examine the bending behavior of beams under limit load. Then we will relate this plastic moment to the design of compact sections.

17

18

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 ﬂange and the web must not exceed the limiting ratios λp deﬁned as follows: Flange Web Note that

bf 2tf

Victor Saouma

and

hc tw

bf 2tf hc tw

≤ λp ≤ λp

λp = √65

Fy

640 λp = √

(14.12)

Fy

are tabulated in Sect. 3.6.

Structural Engineering

14–6

BRACED ROLLED STEEL BEAMS

Draft 19

The nominal strength Mn for laterally stable compact sections according to LRFD is (14.13)

Mn = Mp where: Mp Z Fy 20

plastic moment strength = ZFy plastic section modulus speciﬁed minimum yield strength

Note that section properties, including Z values are tabulated in Section 3.6.

14.3.1.2

Partially Compact Section

If the width to thickness ratios of the compression elements exceed the λp values mentioned in Eq. 14.12 but do not exceed the following λr , the section is partially compact and we can have local buckling.

21

Flange: λp < Web:

λp <

bf 2tf hc tw

≤ λr

λp = √65

λr = √ 141

≤ λr

λp =

970 λr = √

Fy 640 √ Fy

Fy −Fr

(14.14)

Fy

where, Fig. 14.6: Fy speciﬁed minimum yield stress in ksi bf width of the ﬂange tf thickness of the ﬂange hc unsupported height of the web which is twice the distance from the neutral axis to the inside face of the compression ﬂange less the ﬁllet or corner radius. tw thickness of the web. Fr residual stress = 10.0 ksi for rolled sections and 16.5 ksi for welded sections.

Figure 14.6: W Section 22

The nominal strength of partially compact sections according to LRFD is, Fig. 14.7 Mn = Mp − (Mp − Mr )(

where: Mr λ

λ − λp ) ≤ Mp λr − λp

(14.15)

Residual Moment equal to (Fy − Fr )S bf /2tf for I-shaped member ﬂanges and hc /tw for beam webs.

Victor Saouma

Structural Engineering

14.3 Flexural Design

Draft

14–7 Mn Compact

Partially Compact

Slender

Mp

Mr

λp Flanges

λ r

bf

65

141

2tf

Fy

Fy - Fr

640 Fy

970 Fy

h

Web

λ

c

tw

Figure 14.7: Nominal Moments for Compact and Partially Compact Sections All other quantities are as deﬁned earlier. Note that we use the λ associated with the one being violated (or the lower of the two if both are).

23

14.3.1.3

Slender Section

If the width to thickness ratio exceeds λr values of ﬂange and web, the element is referred to as slender compression element. Since the slender sections involve a diﬀerent treatment, it will not be dealt here.

24

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 ﬂange as a column cannot occur prior to its full participation to develop the moment strength of the section.

25

26

The AISC code requires that Lb

≤

Lp

=

Lp 300 rmin Fy

(14.16) (14.17)

where Lb is the maximum unbraced length of the compression ﬂange, and rmin is the radius of gyration with respect to the weak axis. This is an emperical equation, thus units are in inches, ksi, and ft.

14.3.2

Examples

Example 14-1: Shape Factors, Rectangular Section Determine the shape factor for a rectangular beam of width b and depth d.

Victor Saouma

Structural Engineering

14–8

BRACED ROLLED STEEL BEAMS

Draft

Solution: From Eq. 14.7 the shape factor is given byξ = My f

= =

 

f ydA A y Fy d/2

Mp My

 = Fy 2y d

The moment at ﬁrst yield is

d/2

⇒ My = 2 0

2Fy 2 bd2 y bdy = Fy d 6

The plastic moment is given by:

Mp =

d/2

f ydA = 2 A

Fy bydy = Fy 0

bd2 4

Thus, the shape factor is 2

Fy bd4 Mp ξ= = 2 = 1.5 My Fy bd6

Example 14-2: Shape Factors, T Section Determine the shape factor of the inverted T-section shown in ﬁg. below.

Solution:

1. Determine the position of the Elastic Neutral Axis: Taking moment about the top, the depth of the neutral axis is given by y¯ =

Victor Saouma

(10)(1)(9.5) + (9)(1)(4.5) = 7.13 cm (10)(1) + (9)(1)

Structural Engineering

14.3 Flexural Design

14–9

Draft

2. Determine the moment of inertia about the elastic neutral axis: (1)(9)3 (10)(1)3 + (9)(1)(7.13 − 4.5)2 + + (10)(1) ((10 − 7.13) − 0.5)2 = 180 cm4 I= 12 12 3. Determine My : My = Fy

I 180.0 = 25.25Fy = Fy ymax 7.13

4. Determine the position of Plastic Neutral Axis: Plastic neutral axis divides the section in such a way that, the area above it is equal to area below it. In other words it is also called as the bf Equal Area Axis Total area = (10)(1) + (9)(1) = 19 cm2 Half area = 9.5 cm2 Thus the plastic neutral axis passes through the horizontal seat at 0.95 cm. from the bottom. 5. Determine Mp : Taking moments of the resultant forces C1 and C2 with respect to the location of the resultant force of the lower tensile forces we have: 0.95 0.05 0.95 ] + [(Fy )(10)(0.05)][ + ] = 45.475Fy Mp = [(Fy )(1)(9)][4.5 + 0.05 + 2 2 2 6. Shape Factor Mp 45.475Fy = = 1.8 My 25.25Fy

ξ=

Example 14-3: 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 ﬂange 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 wL wu

= = = =

0.2 k/ft 0.8 k/ft 1.2wD + 1.6wL 1.2(0.2) + 1.6(0.8) = 1.52 k/ft

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 k.ft Assuming compact section, since a vast majority of rolled sections satisfy λ ≤ λp for both the ﬂange and the web. The design strength φb Mn is φb Mn = φb Mp = φb Zx Fy The design requirement is φb Mn = Mu or, combing those two equations we have: φb Zx Fy = Mu Victor Saouma

Structural Engineering

14–10

BRACED ROLLED STEEL BEAMS

Draft

3. Required Zx is Mu 76(12) = = 28.1 in3 φb Fy 0.90(36)

Zx =

From the notes on Structural Materials, we select a W12X22 section which has a Zx = 29.3 in3 (22)(12) Note that Zx is approximated by wd = 29.3. 9 = 9 4. Check compact section limits λp for the ﬂanges from the table λ=

bf 2tf

= 4.7 = √65 =

λp and for the web: λ=

Fy

hc tw

λp

= =

√65 36

41.8 640 √ = Fy

√ = 10.8 > λ

640 √ 36

= 107

√

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

= = =

0.2 + 0.022 = 0.222 k/ft 1.2(0.222) + 1.6(0.8) = 1.55 k/ft (1.55)(20)2 /8 = 77.3 k.ft

Mn

=

Mp = Zx Fy =

φb Mn

=

0.90(87.9) = 79.1 k.ft > Mu

(29.3) in3 (36) (12) in/ft

ksi

√

= 87.9 k.ft

Therefore use W12X22 section. 6. We ﬁnally check for the maximum distance between supports. ! ! Iy 5 = = 0.88 in rmin = A 6.5 300 Lp = rmin Fy 300 = √ 0.88 = 43 ft 36

(14.18-a) (14.18-b) (14.18-c)

Case 2: A572 Grade 65 Steel: 1. same as in case 1 2. same as in case 1 Mu φb Fy by wd 9

3. Required Zx = approximated

= =

76(12) 0.90(65) = 15.6 (14)(12) = 18.7. 9

in3 Select W12X14: Zx = 17.4 in3 Note that Zx is

4. Check compact section limits λp : λ = λp = λ = λp =

hc tw = 54.3 640 640 √ = √ 65 Fy bf 2tf = 8.82 √65 = √65 65 Fy

= 79.4

√

= 8.1 < λ Not Good

In this case the controlling limit state is local buckling of the ﬂange. Since λp < λ < λr , as above, the section is classiﬁed as non-compact.

Victor Saouma

Structural Engineering

14.4 Shear Design

14–11

Draft

5. Check the strength: Since the section is non-compact, the strength is obtained by interpolation between Mp and Mr . For the ﬂanges: λr

=

√ 141

Mn

=

Mp − (Mp − Mr )( λr −λpp ) ≤ Mp

Mp

=

Zx Fy =

Mr

=

Mn φb Mn

= =

Fy −10

=

√ 141 65−10

= 19.0 λ−λ

3

(17.4) in (65) ksi = 94.2 k.ft (12) in/ft (14.9) in3 (65−10) ksi Sx (Fy − Fr ) = = 68.3 k.ft. (12) in/ft 8.8−8.1 = 92.5 k.ft 94.2 − (94.2 − 68.3) 19.0−8.1

0.90(92.5) = 83.25 k.ft > Mu

√

Therefore provide W12X14 section.

14.4

Shear Design

Except for very short span beams, shear seldom govern. Thus sections should ﬁrst be selected on the basis of ﬂexural reauirements, and shear should be checked next.

27

28

as for beams Vd = φv Vn ;

29

(14.19)

φv = 0.9

There are three failure modes possible:

Web yielding Inelastic web buckling Elastic web buckling 30

The AISC requirements for shear design are summarized below Web Yielding Inel. Web Buckling Elast. Web Buckling

h tw

≤

√418

Vn = 0.6Fyw Aw

Fyw

√418

Fyw

<

h tw

≤

√418

√523

Fyw √523 Fyw

Vn = 0.6Fyw Aw <

h tw

Vn =

132,000Aw

Fyw h tw

( thw )2 (14.20)

and are shown in Fig. ??.

14.5

Deﬂections

Deﬂections of beams, under live load, are often curtailed in order not to impair serviceability (plaster cracking, excessive vibration).

31

The AISC code states in Sect. L3.1 that deformations of structural members and systems due to service loads shall not impair the serviceability of the structure. However no speciﬁc limits are given, as those would depend on the type of structire.

32

Victor Saouma

Structural Engineering

14–12

BRACED ROLLED STEEL BEAMS

Draft

40.0

Fy=36 ksi Fy=54 ksi

35.0

Vn/Aw [ksi]

30.0

25.0

20.0

15.0

10.0 50.0

60.0

70.0

80.0

90.0

100.0

h/tw

Figure 14.8: AISC Requirements for Shear Design

33

In lieu of practical design experience, it is customary to limit the live-load central span deﬂection to ∆≤

L 360

(14.21)

It should be recalled that the midspan deﬂection of a simply supported, uniformly loaded beam is equal to 5 wL4 (14.22) ∆= 384 EIx

34

14.6

Complete Design Example

A simply supported beam with a span of 20 ft supports a dead load of 1.0 k/ft, and a live load of 1.5 k/ft in addition to its own weight (estimate dead load to 40 lbs/ft). Select the most economical W shape using A-36, determine the maximum distance for lateral bracing of the compression ﬂanges, and check for deﬂection which should not exceed L/360. WD

= 1.0 + 0.04 = 1.04 k/ft

(14.23-a)

WL Wu1

= 1.5 k/ft = 1.4wd = 1.4(1.04) = 1.456 k/ft

(14.23-b) (14.23-c)

Wu2 Mu Vu Zx bf 2tf h tw h tw φv Vn

= 1.2wd + 1.6wL = 1.2(1.04) + 1.6(1.5) = 3.648 k/ft ✛ 3.65(20)2(12) wu L2 = = 2, 189 k.in = 8 8 (3.65)(20) wu L = = 36.5 k = 2 2 Mu 2, 189 = 67.6 in3 ⇒ W16x40, Zx = 72.9 in3 = = φb Fy (0.9)(36) √ 65 = 6.9 < √ = 10.8 36 √ 640 = 46.6 < √ = 106.7 36 √ 418 = 46.6 < √ = 69.7 36 √ = 0.9(0.6Fy ) dtw = (0.9)(0.6)(36)(16.01)(0.305) = 94.9 k > 36.5

(14.23-d) (14.23-e) (14.23-f) (14.23-g) (14.23-h) (14.23-i) (14.23-j) (14.23-k)

Aw

Victor Saouma

Structural Engineering

14.6 Complete Design Example

Draft

5 wL L4 5 (1.5/12) k/ in(20)4 (12)4 in4 = = 0.36 in 384 EIx 384 (29, 000) ksi(518) in4 √ (20)(12) = = 0.67 in > 0.36 in 360 = 1.57 in 300rmin (300)(1.57) √ = = = 78.5 in Fy 36

∆ =

L 360 rmin Lp

Victor Saouma

14–13

(14.23-l) (14.23-m) (14.23-n) (14.23-o)

Structural Engineering

14–14

Draft

BRACED ROLLED STEEL BEAMS

Victor Saouma

Structural Engineering

Draft Chapter 15

COLUMN STABILITY 15.1

Introduction; Discrete Rigid Bars

15.1.1

Single Bar System

1 Let us begin by considering a rigid bar connected to the support by a spring and axially loaded at the other end, Fig. 15.1. Taking moments about point A:

P

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 L1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

B

k

A

11111 00000 00000 11111 00000 11111 00000 11111

0P 1 1 0 0 1 0 1 0 1 0 1 0 0 1 ∆ 1 0 1 00000 11111 B 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0θ 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 0 1 00000 11111 A 0 1 00000 11111 0 1 00000 11111

P Unstable Equilibrium k L

θ (stable) sin θ Neutral Equilibrium

Stable Equilibrium k L

11111 00000 00000 11111 00000 11111 00000 11111

1111 0000 0000 1111 0000 1111 0000 1111

111 000 000 111 000 111 000 111

Stable

Neutral

111 000 000 111 000 111

θ

Unstable

Figure 15.1: Stability of a Rigid Bar ΣMA = ∆ = P θL − kθ = k (P − ) = L

P ∆ − kθ = 0 Lθ for small rotation

(15.1-a) (15.1-b)

0

(15.1-c)

0

(15.1-d)

hence we have three possibilities: Stable Equilibrium: for P < k/L, θ = 0 Neutral Equilibrium: for P = k/L, and θ can take any value

15–2

COLUMN STABILITY

Draft

Unstable equilibrium: for P > k/L, θ = 0 2

Thus we introduce a critical load deﬁned by Pcr =

3

k L

(15.2)

For large rotation, we would have ΣMA = ∆ = Pcr

=

P ∆ − kθ = 0 L sin(θ)P θL − kθ = 0 k θ L sin(θ)

(15.3-a) (15.3-b) (15.3-c)

Because there is more than one possible path (θ) when P = Pcr , we call this point a biﬀurcation point.

4

5 If we now assume that there is an initial imperfection in the column, i.e. the column is initially “crooked”, Fig. 15.2, then

ΣMA

=

Pcr

=

P Lθ − k(θ − θ0 ) = 0 θ0 k (1 − ) L θ

(15.4-a) (15.4-b)

P P

∆

L

θ0

k

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 00000 11111 00000 11111 00000 11111 θ 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 A 00000 11111

B

Perfect System

k L

k (1L

θ0 θ

)

θ0

θ

00000 11111 11111 00000 00000 11111 00000 11111

Figure 15.2: Stability of a Rigid Bar with Initial Imperfection

15.1.2 6

Two Bars System

Next we consider the two rigid bar problem illustrated in Fig. 15.3.

Victor Saouma

ΣMB

=

⇒ −θ1 + θ2

=

ΣMA

=

⇒ 2θ1 − θ2

=

P Lθ2 − k(θ2 − θ1 ) = 0 PL θ2 k P Lθ1 + k(θ2 − θ1 ) − kθ1 = 0 PL θ1 k

(15.5-a) (15.5-b) (15.5-c) (15.5-d) Structural Engineering

15.1 Introduction; Discrete Rigid Bars

Draft

P P C

L B k

L

A k

15–3

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 0000000 1111111 C 1 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 L 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 θ 0000000 1111111 0000000 1111111 2 0000000 1111111 P 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 0 1 0000000 1111111 B 0000 1111 0000B 1111 0000 1111 0000 k(θ2 - θ)1 1111 0000k(θ2 1111 0000 1111 0000 1111 0000 1111 0000 1111 P 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 A 1111 kθ 0000 1111 0 1 1 0 1 0 1 0 1 0 1 0 1 P 0 1 0 1 0 1

0 P1 0 1 0 1

P

0000000 1111111 0000 C 1111 1111111 0000000 0000 1111 θ1 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 θ2 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 0000000 1111111 0000 1111 B 0000000 1111111 0000 1111 L 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 θ 0000 1111 1 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

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

θ)1

1 0 0 1 0 1 0 1 0 1 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 1.618 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000000 1111111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 1 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

0 P 1 1 0 0 1 0 1 0 1 0 1 0 1 0 1 000 111 000 111 000 111 000 111 000 111 000 111 000 111 .618 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 0000 1111 0000 1111 0000 1111 0000 1111 0000 1111 1 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

Figure 15.3: Stability of a Two Rigid Bars System

7

Those two equations can be cast in matrix form θ1 2 −1 θ1 =λ θ2 θ2 −1 1

where λ = P L/k, this is an eigenvalue formulation and can be rewritten as 2−λ −1 θ1 0 =λ θ2 −1 1−λ 0

(15.6)

(15.7)

This is a homogeneous system of equation, and it can have a non-zero solution only if its determinant is equal to zero " " " 2−λ −1 "" " = 0 (15.8-a) " −1 1−λ " 2 − λ − 2λ + λ2 − 1

= 0

(15.8-b)

λ − 3λ + 1

= 0

(15.8-c)

=

(15.8-d)

2

λ1,2

8

√ √ 3± 5 3± 9−4 = 2 2

Hence we now have two critical loads: Pcr1

=

Pcr2

=

√ k 3− 5 k = 0.382 2√ L L k 3+ 5 k = 2.618 2 L L

(15.9) (15.10)

9 We now seek to determine the deformed shape for each of the ﬁrst critical loads. It should be noted that since the column will be failing at the critical buckling load, we can not determine the absolute values of the deformations, but rather the shape of the buckling column. √ 3− 5 (15.11-a) λ1 = 2 + , √ 2 − 3−2 5 θ1 0 −1 √ = (15.11-b) θ2 0 −1 1 − 3− 5

2

Victor Saouma

Structural Engineering

15–4

COLUMN STABILITY

Draft

+

√ 1+ 5 2

−1

,

−1 1−

√ −1+ 5 2

1.618 −1 −1 0.618

θ1 θ2

=

θ1 θ2

=

0 0 0 0

(15.11-c) (15.11-d)

we now arbitrarily set θ1 = 1, then θ2 = 1/0.618 = 1.618, thus the ﬁrst eigenmode is

θ1 θ2

=

1 1.618

(15.12)

10

Note that we can determine the deformed shape upon buckling but not the geometry.

11

Finally, we examine the second mode shape loads +

√

2 − 3+2 −1 + √

1− 5 2

−1

,

−1 √ 1 − 3+2 5

5

,

−1 1−

√ −1− 5 2

−0.618 −1 −1 −1.618

θ1 θ2 θ1 θ2 θ1 θ2

=

λ2

= = =

√ 3+ 5 2 0 0 0 0 0 0

(15.13-a) (15.13-b) (15.13-c) (15.13-d)

we now arbitrarily set θ1 = 1, then θ2 = −1/1.618 = −0.618, thus the second eigenmode is

15.1.3

θ1 θ2

=

1 −0.618

(15.14)

‡Analogy with Free Vibration

The problem just considered bears great ressemblence with the vibration of a two degree of freedom mass spring system, Fig. 15.4.

12

u1 k =k

k =k

1

2

m=m 1

m2= 2m

k 3= k

u2 1

1

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

k u1

u

.. mu

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

u2

k(u -u ) 2

.. 2m u ku

u

1

2

2

2

Figure 15.4: Two DOF Dynamic System Each mass is subjected to an inertial force equals to the mass times the acceleration, and the spring force:

13

2m¨ u2 + ku2 + k(u2 − u1 ) = Victor Saouma

0

(15.15-a) Structural Engineering

15.2 Continuous Linear Elastic Systems

Draft

or in matrix form

m¨ u1 + ku1 + k9u2 − u1 0 =

14

15–5

m 0

0

(15.15-b)

¨1 u 0 2k −k u1 + ¨2 u 2m −k 2k u2

M

K

¨ U

The characteristic equation is |K − λM| where λ = ω 2 , " " 2h − λ −h " " −h 2h − 2λ

(15.16)

U

and ω is the natural frequency. " " "=0 "

(15.17)

where h = k/m. This reduces to the polynomial 3 λ2 − 3hλ + h2 = 0 2

(15.18)

0.796 k/m 1.538 k/m

(15.19-a)

√ Solving, λ = (3 ± − 3)h/2 or ω1

=

ω1

=

(15.19-b)

To ﬁnd the mode shapes φ1 and φ2 (relative magnitudes of the DOF) we substitute in the characteristic equation and set the ﬁrst element equal to 1: 1.000 1.000 and φ1 = (15.20) φ1 = 1.3660 −0.3660

15

15.2

Continuous Linear Elastic Systems

16

Column buckling theory originated with Leonhard Euler in 1744.

17

An initially straight member is concentrically loaded, and all ﬁbers remain elastic until buckling occur.

For buckling to occur, it must be assumed that the column is slightly bent as shown in Fig. 15.5. Note, in reality no column is either perfectly straight, and in all cases a minor imperfection is present.

18

P

x

x and y are principal axes

x

Slightly bent position L

y, v

Figure 15.5: Euler Column Two sets of solutions will be presented, in both cases the equation of equilibrium is written for the deformed element.

19

Victor Saouma

Structural Engineering

15–6

COLUMN STABILITY

Draft 15.2.1

Lower Order Diﬀerential Equation

20 At any location x along the column, the imperfection in the column compounded by the concentric load P , gives rise to a moment (15.21) Mz = P v

Note that the value of y is irrelevant. 21

Recalling that

d2 v Mz = dx2 EI upon substitution, we obtain the following diﬀerential equation

(15.22)

d2 v P v=0 − dx2 EI 22

Letting k 2 =

P EI ,

(15.23)

the solution to this second-order linear diﬀerential equation is v = −A sin kx − B cos kx

23

(15.24)

The two constants are determined by applying the boundary conditions 1. v = 0 at x = 0, thus B = 0 2. v = 0 at x = L, thus A sin kL = 0

(15.25)

This last equation can be satisﬁed if: 1) A = 0, that is there is no deﬂection; 2) kL = 0, that is no applied load; or 3) kL = nπ (15.26) 2 cr = nπ or Thus buckling will occur if PEI L

24

Pcr =

n2 π 2 EI L2

25 The fundamental buckling mode, i.e. a single curvature deﬂection, will occur for n = 1; Thus Euler critical load for a pinned column is

Pcr =

26

π 2 EI L2

(15.27)

π2 E 2

(15.28)

The corresponding critical stress is σcr =

L

rmin

where I = 27

2 . Armin

Note that buckling will take place with respect to the weakest of the two axis.

15.2.2

Higher Order Diﬀerential Equation

15.2.2.1

Derivation

In the preceding approach, the buckling loads were obtained for a column with speciﬁed boundary conditons. A second order diﬀerential equation, valid speciﬁcally for the member being analyzed was used.

28

Victor Saouma

Structural Engineering

15.2 Continuous Linear Elastic Systems

15–7

Draft

In the next approach, we derive a single fourth order equation which will be applicable to any column regardelss of the boundary conditions.

29

Considering a beam-column subjected to axial and shear forces as well as a moment, Fig. 15.6 (note analogy with cable structure, Fig. 6.1), taking the moment about i for the beam segment and assuming

30

w w(x)

1111 0000 0000 1111

1111 0000 0000 1111

y, u

w

M

P

P

x

P δv δx

dx

V+ δ V dx δx

i

P dx

P M+δ M dx δx

θi

i j

θj

P

Figure 15.6: Simply Supported Beam Column; Diﬀerential Segment; Eﬀect of Axial Force P the angle

dv dx

between the axis of the beam and the horizontal axis is small, leads to dv (dx)2 dV dM dx + w + V + dx − P M− M+ dx = 0 dx 2 dx dx

(15.29)

Note that the ﬁrst underbraced term is identical to the one used in earlier derivation of the beam’s diﬀerential equation. neglecting the terms in dx2 which are small, and then diﬀerentiating each term with respect to x, we obtain d2 v d2 M dV −P 2 =0 − (15.30) 2 dx dx dx

31

32

However, considering equilibrium in the y direction gives dV = −w dx

33

From beam theory, neglecting axial and shear deformations, we have M = −EI

34

(15.31)

d2 v dx2

(15.32)

Substituting Eq. 15.31 and 15.32 into 15.30, and assuming a beam of uniform cross section, we obtain EI #

substituting λ =

P EI ,

d4 v d2 v − P =w dx4 dx2

(15.33)

and w = 0 we ﬁnally obtain 2 d4 v 2d v + λ =0 dx4 dx2

(15.34)

Again, we note that by considering equilibrium in the deformed state we have introduced the second term to what would otherwise be the governing diﬀerential equation for ﬂexural members (beams). Finally, we note the analogy between this equation, and the governing diﬀerential equation for a cable structure, Eq. 6.12.

35

36

The general solution of this fourth order diﬀerential equation to any set of boundary conditions is v = C1 + C2 x + C3 sin λx + C4 cos λx

Victor Saouma

(15.35)

Structural Engineering

15–8

COLUMN STABILITY

Draft

Essential (Dirichlet) v dv dx

Natural (Neumann) d3 v (V ) 3 dx 2 d v (M ) dx2

Table 15.1: Essential and Natural Boundary Conditions for Columns The constants C s are obtained from the boundary condtions. For columns, those are shown in Table ?? The essential boundary conditions are associated with displacement, and slope, the natural ones with shear and moments (through their respective relationships with the displacement).

37

We note that at each node, we should have two boundary conditions, all combinations are possible except pairs from the same raw (i.e. we can not have known displacement and shear, or known slope and moment).

38

15.2.2.2

Hinged-Hinged Column

If we consider again the stability of a hinged-hinged column, the boundary conditions are displacement 2 (v) and moment ( ddxv2 EI) equal to zero at both ends1 , or

39

v v

= =

0, 0,

d2 v 2 dx d2 v dx2

= 0 = 0

at x = at x =

0 L

(15.36)

substitution of the two conditions at x = 0 leads to C1 = C4 = 0. From the remaining conditions, we obtain C3 sin λL + C2 L −C3 k 2 sin λL

= 0 = 0

(15.37-a) (15.37-b)

these relations are satisﬁed either if C2 = C3 = 0 or if sin λL = C2 = 0. The ﬁrst alternative leads to the trivial solution of equilibrium at all loads, and the second to λL = nπ for n = 1, 2, 3 · · ·. The critical load is Pcr =

n2 π 2 EI L2

(15.38)

which was derived earlier using the lower order diﬀerential equation. 40

The shape of the buckled column is

nπx (15.39) L and only the shape (but not the geometry) can be determined. In general, we will assume C3 = 1 and plot y. y = C3 sin

15.2.2.3

Fixed-Fixed Column

We now consider a column which is restrained against rotation at both ends, the boundary conditions are given by:

41

v(0)

= 0 = C1 + C4

1 It

(15.40-a)

v (0) = 0 = C2 + C3 λ v(L) = 0 = C1 + C2 L + C3 sin λL + C4 cos λL

(15.40-b) (15.40-c)

v (L) = 0 = C2 + C3 λ cos(λL) − C4 sin(λL)

(15.40-d)

will be shown in subsequent courses, that the former BC is an essential B.C., and the later a natural B.C.

Victor Saouma

Structural Engineering

15.2 Continuous Linear Elastic Systems

Draft

those equations can be set  1  1   1 0

15–9

in matrix form 0 1 L 1

  0 1 C1        C2 λ 0  C3  sin λL cos λL       C4 λ cos λL −λ sin λL " " " " 1 0 0 1 " " " " 1 1 λ 0 " " " 1 L sin λL cos λL "" " " 0 1 λ cos λL −λ sin λL "

  0       0 =   0     0

= 0

(15.41-b)

The determinant is obtain from

" " sin λL " " λ cos λL

(15.41-a)

" " " " " " " " " "

=

0 (15.42-a)

=

0 (15.42-b)

−2λ + λ2 L sin λL + 2λ cos λL

=

0 (15.42-c)

" " " " " 1 " 1 λ 0 " " " " L sin λL cos λL "" − "" 1 " " 1 λ cos λL −λ sin λL " " 0 " " " " " L cos λL "" cos λL "" "" 1 sin λL " − λ" + −λ sin λL " 1 −λ sin λL " " 0 λ cos λL

1 λ L sin λL 1 λ cos λL " " " " " − λ" 1 L " " 0 1

The ﬁrst solution, λ = 0 is a trivial one, and the next one λL sin λL + 2 cos λL − 2 λL sin λL

= 0 = 2(1 − cos λL)

(15.43-a) (15.43-b)

=

2nπ

(15.44-a)

=

2nπ L

(15.44-b)

The solution to this transcendental equation is !

λL

P EI thus the critical load and stresses are given by

42

Pcr

=

σcr

=

4n2 π 2 EI L2 2 2 4n π E (L/r)2

(15.45) (15.46)

The deﬂected shape (or eigenmodes) can be obtained by substituting the value of λ into the c s.

15.2.2.4

Fixed-Hinged Column

Next we consider a column with one end ﬁxed (at x = L), and one end hinged (at x = 0). The boundary conditions are d2 v v = 0, dx = 0 at x = 0 2 (15.47) dv = 0 at x = L v = 0, dx The ﬁrst two B.C. yield C1 = C4 = 0, and the other two

43

sin λL − λL cos λL = 0

(15.48)

But since cos λL can not possibly be equal to zero, the preceding equation can be reduced to tan λL = λL

(15.49)

which is a transcendental algebraic equation and can only be solved numerically. We are essentially looking at the intersection of y = x and y = tan x, Fig. 15.7 and the smallest positive root is λL = 4.4934, P since k 2 = EI , the smallest critical load is Pcr =

(4.4934)2 π2 EI = EI L2 (0.699L)2

(15.50)

Note that if we were to solve for x such that v,xx = 0 (i.e. an inﬂection point), then x = 0.699L. Victor Saouma

Structural Engineering

15–10

COLUMN STABILITY

Draft 10.0 8.0 6.0 4.0 2.0 0.0 −2.0 −4.0 −6.0 −8.0 −10.0 0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

9.0

10.0

Figure 15.7: Solution of the Tanscendental Equation for the Buckling Load of a Fixed-Hinged Column

15.2.3

Eﬀective Length Factors K

Recall that the Euler buckling load was derived for a pinned column. In many cases, a column will have diﬀerent boundary conditions. It can be shown that in all cases, the buckling load would be given by

44

Pcr =

π 2 EI

(15.51)

(KL)2

where K is called eﬀective length factor, and KL is the eﬀective length. and σcr =

π2 E 2

(15.52)

KL rmin

The ratio rKL is termed the slenderness ratio. Note that rmin should be the smallest radius of min gyration in the unbraced direction(s). The eﬀective length, can only be determined by numerical or approximate methods, and is the distance between two adjacent (real or virtual) inﬂection points, Fig. 15.8, 15.9

45

The most widely used charts for the eﬀective length determination are those produced by the Structural Stability Research Council. The alignment chart, for an individual column, Fig. 15.10 is shown in Fig. 15.11. It should be noted that this chart assumes that all members are still in the elastic range.

46

The use of the alignment chart involces computing G at each end of the column using the following formula * Ic (15.53) Ga = * LI c

47

g

Lg

where Ga is the stiﬀness at end a of the column, Ic , Ig are the moment of inertias of the columns and girders respectively. The summation must include only those members which are rigidly connected to that joint and lying in the plane for which buckling is being considered. Victor Saouma

Structural Engineering

15.2 Continuous Linear Elastic Systems

15–11

Draft

P

P P

P

KL = 0.7L KL = L

L

KL =

1 L L 2

KL < L

L

P P

P

P (a) End rotations unrestrained

(b) End rotations fully restrained

(c) One end restrained, other unrestrained

(d) Partially restrained at each end

Figure 15.8: Column Eﬀective Lengths

∆ P

P

P

P

KL 2 L

L 0.7L
KL>2L

∆

(a) Braced frame, hinged base P

P

L

0.5L
(b) Unbraced frame, hinged base P

P

L L
(c) Braced frame, fixed base

(d) Unbraced frame, fixed base

Figure 15.9: Frame Eﬀective Lengths

Victor Saouma

Structural Engineering

15–12

COLUMN STABILITY

Draft

Figure 15.10: Column Eﬀective Length in a Frame Hence, once Ga and Gb are determined, those values are connected by a straight line in the appropriate chart, and k is the point of intersection of that line with the midle axis.

48

49

Alternatively :-) GA + GB GA GB π 2 π/K 2 tan π/2K =1 + 1− + 4 K 2 tan π/K π/K

(15.54)

(Ref. McGuire P. 467).

15.3 50

Inelastic Columns

There are two limiting loads for a column 1. Yielding of the gross section Pcr = Fy Ag , which occirs in short stiﬀ columns 2. Elastic (Euler) buckling, Eq. 15.51 Pcr =

π 2 EI , (KL)2

in long slender columns.

Those two expression are asymptotic values for actual column buckling. Intermediary failure loads are caused by the presence of residual stresses which in turn give rise to inelastic buckling.

51

Inelastic buckling buckling occurs when the stresses (average) have not yet reached the yield stress, and is based on the tangent modulus Et which is lower than the initial modulus E.

52

53

Residual stresses (caused by uneven cooling) will initiate inelastic buckling, Fig. 15.12.

The inelastic buckling curve has to be asymptotic to both the Euler elastic buckling (for slender columns), and to the yield stress (for stiﬀ columns).

54

The Structural Stability Research Council (SSRC) has proposed a parabolic curve which provides a transition between elastic buckling and yielding, thus accounting for the presence of residual stresses

55

Victor Saouma

Structural Engineering

15.3 Inelastic Columns

15–13

Draft

Sidesway Inhibited ∞ 50. 10.

Ga

K

Sidesway Uninhibited Gb ∞ 50. 10.

1.0

5.

5. 0.9

3.

3.

2.

∞ 100. 50.

Ga

K

Gb ∞ 100. 50.

∞ 20. 10.

30. 20.

5.

10. 9. 8. 7. 6.

3.

30. 20.

4.

2. 0.8

1.

1.

0.8 0.7 0.6

0.8 0.7 0.6

0.7

0.5

0.5

0.4

0.4

0.3

0.3 0.6

0.2

10. 9. 8. 7. 6.

5.

5. 2.

4.

4.

3.

3.

2.

2. 1.5

0.2 1.

0.1

1.

0.1

0.

0.5

0.

0

1.

0

Figure 15.11: Standard Alignment Chart (AISC)

σ

σ

Euler Buckling (Linear Elastic) 2

σ=πE

2

( kLr ) σ

yd

yd

ET < E

Inelastic Buckling

Effect of Residual Stresses

σr

Gross Yielding

σ

Proportional Limit

Proportional Limit

E

ε

kL r

Figure 15.12: Inelastic Buckling

Victor Saouma

Structural Engineering

15–14

COLUMN STABILITY

Draft

and the resulting inelastic buckling, Fig. 15.13. + σcr = σy

σy 1− 2 4π E

KL rmin

2 , (15.55)

Fy=36 ksi; E=29,000 ksi 50 Euler SSRC

Stress σ [ksi]

40

30

20

10

0

0

20

40

60

80 100 120 140 Slenderness Ratio KL/rmin

160

180

200

Figure 15.13: Euler Buckling, and SSRC Column Curve

Victor Saouma

Structural Engineering

Draft Chapter 16

STEEL COMPRESSION MEMBERS 1 This chapter will cover the elementary analysis and design of steel column according to the LRFD provisions.

16.1

AISC Equations

By introducing a slenderness parameter λc (not to be confused with the slenderness ratio) deﬁned as

2

λ2c

λc

=

=

Fy Fy = π2 E Euler Fcr KL 2 KL rmin

!

(16.1-a)

rmin

Fy π2 E

(16.1-b)

Hence, this parameter accounts for both the slenderness ratio as well as steel material properties. This new parameter is a more suitable one than the slenderness ratio (which was a delimeter for elastic buckling) for the inelastic buckling. 3

Equation 15.55 becomes Fcr λ2 =1− c Fy 4

4

When λc >

for λc ≤

√ 2

(16.2)

√ 2, then Euler equation applies Fcr 1 = 2 Fy λc

for λc ≥

√ 2

(16.3)

Hence the ﬁrst equation is based on inelastic buckling (with gross yielding as a limiting case), and the second one on elastic buckling. The two curves are tangent to each other. 5

The above equations are valid for concentrically straight members.

6 The previous two equations, were modiﬁed to account for an initial out-of-straightness of about 1/1500. This will result in, Fig. 16.1:

Fcr

=

Fcr

=

2

0.658λc Fy for λc ≤ 1.5 0.877 Fy for λc > 1.5 λ2c

(16.4) (16.5)

16–2

STEEL COMPRESSION MEMBERS

Draft where

KL λc = πr

!

Fy E

(16.6)

Fy=36 ksi; E=29,000 ksi 50 SSRC 2 λ c 0.6584 ; λc<1.5 2 0.877λc Fy; λc>1.5

Stress σ [ksi]

40

Elastic Buckling

30

Inelastic Buckling

20

λ

λ

10

0

0

20

40

60

c

c

>1.5

<1.5

80 100 120 140 Slenderness Ratio KL/rmin

160

180

200

Figure 16.1: SSRC Column Curve and AISC Critical Stresses Fcr can readily be determined from Figure 16.2 or from Table 16.1 and 16.2.

7

8 The AISC code introduces a reduction factor Q into Eq. 16.7-a and 16.7-b when certan width/thickness limitations (LRFD Table B5.1) are not satisﬁed, and local buckling may occur prior to overall buckling.

Fcr

=

Fcr

=

2 0.658Qλc Fy for λc Q ≤ 1.5 0.877 F for λ Q > 1.5 y c λ2c

(16.7-a) (16.7-b)

since Q = 1 for all rolled H sections (standard W, S, and M shapes) it will be neglected in this course. 9 Finally, the AISC introduces a stiﬀness reduction factor (SRF) for columns when the nomographs are used. Such a reduction would account for the discrepancy between elastic buckling (assumed in the nomographs) and inelastic buckling which occur for relatively high Pu /A values, Table 16.3.

16.2 10

LRFD Equations

The strength requirement for columns in load and resistance factor design is stated as φc Pn ≥ Pu

where: φc strength reduction factor, 0.85 Pn nominal strength=Ag Fcr Fcr Eq. 16.7-a and 16.7-b Pu factored service load. Victor Saouma

(16.8)

Structural Engineering

16.2 LRFD Equations

16–3

Draft

Kl rmin

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 31 32 33 34 35 36 37 38 39 40

λc 0.01 0.02 0.03 0.04 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.24 0.25 0.26 0.27 0.28 0.29 0.30 0.31 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.40 0.41 0.43 0.44 0.45

φc Fcr 30.60 30.59 30.59 30.57 30.56 30.54 30.52 30.50 30.47 30.44 30.41 30.37 30.33 30.29 30.24 30.19 30.14 30.08 30.02 29.96 29.90 29.83 29.76 29.69 29.61 29.53 29.45 29.36 29.27 29.18 29.09 28.99 28.90 28.79 28.69 28.58 28.47 28.36 28.25 28.13

Kl rmin

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

λc 0.46 0.47 0.48 0.49 0.50 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.70 0.71 0.72 0.73 0.74 0.75 0.76 0.77 0.79 0.80 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.89 0.90

φc Fcr 28.01 27.89 27.76 27.63 27.51 27.37 27.24 27.10 26.97 26.83 26.68 26.54 26.39 26.25 26.10 25.94 25.79 25.63 25.48 25.32 25.16 24.99 24.83 24.66 24.50 24.33 24.16 23.99 23.82 23.64 23.47 23.29 23.11 22.94 22.76 22.58 22.40 22.21 22.03 21.85

Kl rmin

81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

λc 0.91 0.92 0.93 0.94 0.95 0.96 0.98 0.99 1.00 1.01 1.02 1.03 1.04 1.05 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 1.26 1.27 1.28 1.29 1.30 1.31 1.32 1.33 1.35

φc Fcr 21.66 21.48 21.29 21.11 20.92 20.73 20.54 20.35 20.17 19.98 19.79 19.60 19.41 19.22 19.03 18.84 18.65 18.46 18.27 18.08 17.89 17.70 17.51 17.32 17.13 16.94 16.75 16.56 16.37 16.18 16.00 15.81 15.62 15.44 15.25 15.07 14.88 14.70 14.52 14.34

Kl rmin

121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160

λc 1.36 1.37 1.38 1.39 1.40 1.41 1.42 1.44 1.45 1.46 1.47 1.48 1.49 1.50 1.51 1.53 1.54 1.55 1.56 1.57 1.58 1.59 1.60 1.61 1.63 1.64 1.65 1.66 1.67 1.68 1.69 1.70 1.72 1.73 1.74 1.75 1.76 1.77 1.78 1.79

Table 16.1: Design Stress φFcr for Fy =36 ksi in Terms of

Victor Saouma

φc Fcr 14.16 13.98 13.80 13.62 13.44 13.27 13.09 12.92 12.74 12.57 12.40 12.23 12.06 11.88 11.71 11.54 11.37 11.20 11.04 10.89 10.73 10.58 10.43 10.29 10.15 10.01 9.87 9.74 9.61 9.48 9.36 9.23 9.11 9.00 8.88 8.77 8.66 8.55 8.44 8.33

Kl rmin

161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200

λc 1.81 1.82 1.83 1.84 1.85 1.86 1.87 1.88 1.90 1.91 1.92 1.93 1.94 1.95 1.96 1.97 1.99 2.00 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.09 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.18 2.19 2.20 2.21 2.22 2.23 2.24

φc Fcr 8.23 8.13 8.03 7.93 7.84 7.74 7.65 7.56 7.47 7.38 7.30 7.21 7.13 7.05 6.97 6.89 6.81 6.73 6.66 6.59 6.51 6.44 6.37 6.30 6.23 6.17 6.10 6.04 5.97 5.91 5.85 5.79 5.73 5.67 5.61 5.55 5.50 5.44 5.39 5.33

KL rmin

Structural Engineering

16–4

STEEL COMPRESSION MEMBERS

Draft

Kl rmin

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 31 32 33 34 35 36 37 38 39 40

λc 0.01 0.03 0.04 0.05 0.07 0.08 0.09 0.11 0.12 0.13 0.15 0.16 0.17 0.19 0.20 0.21 0.22 0.24 0.25 0.26 0.28 0.29 0.30 0.32 0.33 0.34 0.36 0.37 0.38 0.40 0.41 0.42 0.44 0.45 0.46 0.48 0.49 0.50 0.52 0.53

φc Fcr 42.50 42.49 42.47 42.45 42.42 42.39 42.35 42.30 42.25 42.19 42.13 42.05 41.98 41.90 41.81 41.71 41.61 41.51 41.39 41.28 41.15 41.02 40.89 40.75 40.60 40.45 40.29 40.13 39.97 39.79 39.62 39.43 39.25 39.06 38.86 38.66 38.45 38.24 38.03 37.81

Kl rmin

41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80

λc 0.54 0.56 0.57 0.58 0.59 0.61 0.62 0.63 0.65 0.66 0.67 0.69 0.70 0.71 0.73 0.74 0.75 0.77 0.78 0.79 0.81 0.82 0.83 0.85 0.86 0.87 0.89 0.90 0.91 0.93 0.94 0.95 0.96 0.98 0.99 1.00 1.02 1.03 1.04 1.06

φc Fcr 37.58 37.36 37.13 36.89 36.65 36.41 36.16 35.91 35.66 35.40 35.14 34.88 34.61 34.34 34.07 33.79 33.51 33.23 32.95 32.66 32.38 32.09 31.79 31.50 31.21 30.91 30.61 30.31 30.01 29.70 29.40 29.09 28.79 28.48 28.17 27.86 27.55 27.24 26.93 26.62

Kl rmin

81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120

λc 1.07 1.10 1.11 1.12 1.14 1.15 1.16 1.18 1.19 1.20 1.22 1.23 1.24 1.26 1.27 1.28 1.30 1.31 1.32 1.33 1.35 1.36 1.37 1.39 1.40 1.41 1.43 1.44 1.45 1.47 1.48 1.49 1.51 1.52 1.53 1.55 1.56 1.57 1.59

φc Fcr 26.31 25.99 25.68 25.37 25.06 24.75 24.44 24.13 23.82 23.51 23.20 22.89 22.58 22.27 21.97 21.66 21.36 21.06 20.76 20.46 20.16 19.86 19.57 19.27 18.98 18.69 18.40 18.11 17.83 17.55 17.26 16.98 16.71 16.42 16.13 15.86 15.59 15.32 15.07 14.82

Kl rmin

121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160

λc 1.60 1.61 1.63 1.64 1.65 1.67 1.68 1.69 1.71 1.72 1.73 1.74 1.76 1.77 1.78 1.80 1.81 1.82 1.84 1.85 1.86 1.88 1.89 1.90 1.92 1.93 1.94 1.96 1.97 1.98 2.00 2.01 2.02 2.04 2.05 2.06 2.08 2.09 2.10 2.11

Table 16.2: Design Stress φFcr for Fy =50 ksi in Terms of

Victor Saouma

φc Fcr 14.57 14.33 14.10 13.88 13.66 13.44 13.23 13.02 12.82 12.62 12.43 12.25 12.06 11.88 11.71 11.54 11.37 11.20 11.04 10.89 10.73 10.58 10.43 10.29 10.15 10.01 9.87 9.74 9.61 9.48 9.36 9.23 9.11 9.00 8.88 8.77 8.66 8.55 8.44 8.33

Kl rmin

161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200

λc 2.13 2.14 2.15 2.17 2.18 2.19 2.21 2.22 2.23 2.25 2.26 2.27 2.29 2.30 2.31 2.33 2.34 2.35 2.37 2.38 2.39 2.41 2.42 2.43 2.45 2.46 2.47 2.48 2.50 2.51 2.52 2.54 2.55 2.56 2.58 2.59 2.60 2.62 2.63 2.64

φc Fcr 8.23 8.13 8.03 7.93 7.84 7.74 7.65 7.56 7.47 7.38 7.30 7.21 7.13 7.05 6.97 6.89 6.81 6.73 6.66 6.59 6.51 6.44 6.37 6.30 6.23 6.17 6.10 6.04 5.97 5.91 5.85 5.79 5.73 5.67 5.61 5.55 5.50 5.44 5.39 5.33

KL rmin

Structural Engineering

16.2 LRFD Equations

16–5

Draft

80 70 Fy=36 ksi Fy=50 ksi Fy=60 ksi Fy=70 ksi Fy=100 ksi

φ Fcr (ksi)

60 50 40 30 20 10 0

0

20

40

60

80 100 120 140 Slenderness ratio, KL/r

160

180

200

Figure 16.2: Fcr versus KL/r According to LRFD, for Various Fy

Pu /A

Fy 36 ksi

42 41 40 39 38 37 36 35 34 33 32 31 30 29 28 27

0.05 0.14 0.22 0.30

Pu /A 50 ksi 0.03 0.09 0.16 0.21 0.27 0.33 0.38 0.44 0.49 0.53 0.58 0.63 0.67 0.71 0.75 0.79

26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11

Fy 36 ksi 0.38 0.45 0.52 0.58 0.65 0.70 0.76 0.81 0.85 0.89 0.92 0.85 0.97 0.99 1.00 1.00

50 ksi 0.82 0.85 0.88 0.90 0.93 0.95 0.97 0.98 0.99 1.00 1.00 1.00 1.00 1.00 1.00 1.00

Table 16.3: Stiﬀness Reduction Factors (AISC) .

Victor Saouma

Structural Engineering

16–6

STEEL COMPRESSION MEMBERS

Draft 16.3

Examples

16.3.1

Evaluation

In this class of problems, we seek to assess the load carrying capacity of a given column. This, in contrast to design problems, is a straightforward application of the governing equations.

11

Example 16-1: Unbraced Column Evaluation, (?) Calculate the maximum factored load Pu that a 20 ft. long, unbraced A36 W 14x53 column can hold. (K = 1.0) Solution: 1. The W14x53 section has, (Sect. 3.6) A = 15.6 in2 , rx = 5.89 in, and ry = 1.92 in. 2. Since the column is not braced, the weak axis will control (1.0)(20) ft(12) in/ft KL = = 125 dimensionless ry (1.92) in 3. The slenderness parameter KL λc = πr

!

125 Fy = E π

!

36, 000 = 1.401 < 1.5 29 × 106

4. Since λc = 1.401 < 1.5, inelastic behavior controls and we use Eq. 16.7-a 2

2

Fcr = 0.658λc Fy = 0.6581.401 (36) ksi = 15.81 ksi 5. The design capacity of the section is thus Pu = φc Pn = φc Ag Fcr = (0.85)(15.6) in2 (15.81) ksi = 209.6 k

Example 16-2: Braced Column Evaluation, (?) Calculate the maximum load, Pu for a 20 ft. long, A36 W24x104 column with K = 0.8. Then recalculate if the column is braced at midheight in the weak axis (still assume K = 0.8). Solution: Unbraced Case: 1. The W24x104 section has, (Sect. 3.6) A = 30.6 in2 , rx = 10.1 in, and ry = 2.91 in. Since the column is not braced, the weak axis will control (0.8)(20) ft(12) in/ft KL = = 65.98 ry (2.91) in 2. The slenderness parameter KL λc = πr

!

65.98 Fy = E π

!

36, 000 = .7399 < 1.5 29 × 106

3. Since λc = 1.401 < 1.5 inelastic behavior controls and we use Eq. 16.7-a 2

2

Fcr = 0.658λc Fy = 0.658.7399 (36) ksi = 28.62 ksi 4. The design capacity of the section is thus Pu = φc Pn = φc Ag Fcr = (0.85)(30.6) in2 (28.62) ksi = 744.4 k Victor Saouma

Structural Engineering

16.3 Examples

16–7

Draft Braced Case:

5. We compute the slenderness ratios for both axis KL rx

(0.8)(20) ft(12) (10.1) in (0.8)(10) ft(12) (2.91) in

=

KL ry

=

in/ft in/ft

= 19.01 = 32.99 ✛

6. The slenderness parameter KL λc = πr

!

32.99 Fy = E π

!

36, 000 = .369 < 1.5 29 × 106

7. Since λc = 1.401 < 1.5 inelastic behavior controls and we use Eq. 16.7-a 2

2

Fcr = 0.658λc Fy = 0.658.369 (36) ksi = 34 ksi 8. The design capacity of the section is thus Pu = φc Pn = φc Ag Fcr = (0.85)(30.6) in2 (34) ksi = 884.4 k 9. We observe that bracing has increased the column capacity by

888.4−744.4 744.4

= 18.8%

Example 16-3: Braced Column Evaluation, (?) Calculate the maximum load, Pu for a 20 ft. long, A36 W12x79 column. The column is braced along its weak axis at ﬁve feet from the top and the bottom. The column is pinned at both ends. Solution: 1. The W12x79 section has, (Sect. 3.6) A = 23.2 in2 , rx = 5.34 in, and ry = 3.05 in. 2. The unbraced lengths along the weak axis are 5 ft. at the ends and 10 ft. in the middle, however the unbraced length along the strong axis is 20 ft. 3. We compute the slenderness ratios for both axis KL rx KL ry

(1.0)(20) ft(12) (5.34) in (1.0)(10) ft(12) (3.05) in

= =

in/ft in/ft

= 44.9

✛

= 39.3

4. The slenderness parameter KL λc = πr

!

44.9 Fy = E π

!

36, 000 = .503 < 1.5 29 × 106

5. Since λc = 1.401 < 1.5 inelastic behavior controls and we use Eq. 16.7-a 2

2

Fcr = 0.658λc Fy = 0.658.503 (36) ksi = 32.4 ksi 6. The design capacity of the section is thus Pu = φc Pn = φc Ag Fcr = (0.85)(23.2) in2 (32.4) ksi = 638.9 k

Victor Saouma

Structural Engineering

16–8

STEEL COMPRESSION MEMBERS

Draft 16.3.2

Design

Design problems are inherently more diﬃcult than veriﬁcation ones, and a trial and error procedure must be followed.

12

1. Assume a design a design stress, φc Fcr , typically around 25-30 ksi = 2. Using the assumed φc Fcr , determine Atrial g

Pu φc Fcr

3. Select a trial member on the basis of Atrial g 4. Determine Fcr of trial member, and compare φc Pn with Pu . 5. If (a) φc Pn ≈ Pu , selection is appropriate (b) φc Pn > Pu , select a smaller section (c) φc Pn < Pu , select a larger section 13

Note that often the depth of the member is speciﬁed. Example 16-4: Column Design, (?)

Find the most economical W12 section to hold a compressive load of 120 kips. Steel is A36, K = 1.0, L = 16 feet. Live load to dead load ratio is 1.0 and the column is unbraced. Solution: 1. Since the dead load and live loads are evenly split, from Sect. 7.3.3 Pu = 1.6(60) + 1.2(60) = 168 k 2. The eﬀective length KL=16 feet, we assume a design stress of φc Fcr = 25 ksi and = Atrial g

Pu 168 = 6.72 in2 = φc Fcr 25

3. Select W12x26 as a trial member (A = 7.75 in2 ) KL ry

=

λc

= =

Fcr φc Pn

(1.0)(16) ft(12) in/ft #(1.51) in Fy KL πr #E 36,000 127.1 π 29×106 2 0.658λc Fy 1.422

=

127.1

=

1.42 < 1.5

= = 0.658 (36) ksi = = φc Ag Fcr = (0.85)(7.65) in2 (15.4) ksi =

15.4 ksi 100.1 k

4. Since the design capacity of this section is less than the required one, we need a bigger section, try W12x40, (A = 11.8 in2 ). KL ry

=

λc

=

Fcr φc Pn

= = =

(1.0)(16) ft(12) #(1.93) in 36,000 99.5 π 29×106 1.1162

in/ft

=

99.5

=

1.116 < 1.5

0.658 (36) ksi = φc Ag Fcr (0.85)(11.8) in2 (21.37) ksi =

21.37 ksi 214.3 k

5. Since W 12x40 has a design capacity greater than the required one, it is a satisfactory design. Because there is a sizeable gap between the required strength (168 kips) and the design strength (214 kips), we could try a smaller section, W12x35. If we did, it will turn out that this section is not satisfactory. Hence, we select W12x40 . Victor Saouma

Structural Engineering

16.3 Examples

16–9

Draft Example 16-5: Design of Braced Column, (?)

Design an economical W section to hold a factored load Pu of 795 kips. The column is 20 ft. long and is assumed to have pin-pin end conditions. The column is braced at midheight in the weak direction and is made of A36 steel. Solution: 1. Since the column is braced at midheight in the weak direction, the eﬀective length is 10 ft. in the weak direction, and 20 ft in the strong one. 2. Assuming a design stress φc Fcr = 30 ksi and Atrial = g

Pu 795 = 26.5 in2 = φc Fcr 30

3. Select W14x90 as a trial member (A = 26.5 in2 ) KL ry KL rx

= =

(1.0)(10) ft(12) (3.7) in (1.0)(20) ft(12) (6.14) in

in/ft in/ft

=

32.4

=

39.1

✛

4. Using Table 16.1, for a slenderness ratio of 39.1, the design stress is 28.23 ksi, thus the design capacity is φc Pn = φc Fcr Ag = (28.23) ksi(26.5) in2 = 748.1 k 5. Since this is slightly less than the required capacity (795 kips), we need a bigger section and try W14x99 (1.0)(10) ft(12) in/ft KL = = 32.34 ry (3.71) in KL rx

=

(1.0)(20) ft(12) (6.17) in

in/ft

=

38.9

✛

6. Using Table 16.1, for a slenderness ratio of 38.9, the design stress is 28.27 ksi, thus the design capacity is φc Pn = φc Fcr Ag = (28.27) ksi(29.1) in2 = 822.6 k 7. Hence, select W14x99 .

Example 16-6: Design of a Column, (?) Select the lightest W section of A36 steel to serve as a pinned-end main member column 16 ft. long to carry an axial compression load of 95 kips dead load and 100 kips live load in a braced structure, as shown below. Design and indicate the ﬁrst three choices. P

Hinged

A 36 Steel 16’ 0"

Hinged

P

Victor Saouma

Structural Engineering

16–10

STEEL COMPRESSION MEMBERS

Draft

Section W8 X 48 W10 X 49 W12 X 50 W14 X 53

Area (sq.in.) 14.1 14.4 14.7 15.6

KL/ry 92 76 98 100

Fcr (kksi) 23.0 26.6 21.7 21.3

φc Pn (kips) 276 326 271 282

First choice Second choice Not good Third choice

Solution: 1. Compute factored load. Pu = 1.2D + 1.6L = 1.2(95) + 1.6(100) = 274 k 2. We simply summarize the calculations for this example

Example 16-7: Column Design Using AISC Charts, (?) Select the lightest W section of A36 steel to serve as a main member 30 ft. long to carry compression load of 50 kips dead load and 110 kips live load in a braced structure, as shown below. The member is assumed pinned at top and bottom and in addition has weak direction support at mid-height. Potential buckled shape if KyLy/r y governs

P

P Assume hinged top and bottom for x- and y-axis bending A 36 Steel

15’ 0"

Assume hinged at mid-height for y-axis bending only Potential buckled shape if KxLx/r x governs

15’ 0"

Bracing P (a)

P (b)

Solution: 1. Compute the factored loads: Pu = 1.2D + 1.6L = 1.2(50) + 1.6(110) = 236 k 2. Select from LRFD Column Load Tables. The eﬀective length factors K for buckling in either the strong or the weak direction equal unity; i.e., Kx = Ky = 1.0. Since the AISC Column Load tables are computed assuming (KL/r)y controls, enter these tables with the eﬀective length (KL)y . Thus enter with, Pu = 236 k; (KL)y = 15 ft. Starting with the shallow W8 sections and working towards the deeper sections, ﬁnd W8 X 40 φc Pn = 238 k rx /ry = 1.73 W10 X 45 φc Pn = 267 k rx /ry = 2.15 W12 X 45 φc Pn = 257 k rx /ry = 2.65 Victor Saouma

Structural Engineering

16.3 Examples

16–11

Draft

3. Since the actual support conditions are such that (KL)x = 2(KL)y , if rx /ry ≥ 2, weak axis controls and tabular loads give the correct answer. Thus, W10 X 45 and W12 X 45 are both acceptable. 4. Since rx /ry for W8 X 40 is less than 2, strong axis bending controls. The strength may be obtained from the tables by entering with the equivalent (KL)y that corresponds to (KL)x = 30f t. 5. For equal strength with respect to the x and y axes, (KL)x rx

=

(KL)y

=

(KL)y ry (KL)x 30 = 17.3 ft = rx /ry 1.73

(16.9-a) (16.9-b)

For (KL)y = 17.3 ft., the W8 X 40 has a design strength φc Pn of only 208 kips; therefore it is not acceptable. 6. Check of section. It is always advisable to make a ﬁnal check of the apparent acceptable solution:W10 X 45, (KL/r)y φc Fcr φc Pn = φc Fcr Ag

= 15(12)/(2.10) = 89.6

(16.10-a)

= 20.1 ksi √ = (20.1)(13.3) = 267 k > 236 k

(16.10-b) (16.10-c)

Therefore use W10 X 45

Victor Saouma

Structural Engineering

16–12

Draft

STEEL COMPRESSION MEMBERS

Victor Saouma

Structural Engineering

Draft Chapter 17

STEEL CONNECTIONS 17.1

Bolted Connections

1 Bolted connections, Fig. 17.1 are increasingly used instead of rivets (too labor intensive) and more often than welds (may cause secondary cracks if not properly performed).

17.1.1

Types of Bolts

2

Most common high strength bolts are the A325 and A490.

3

A325 is made from heat-treated medium carbon steel, min. Fyb ≈ 81 − 92 ksi, Fub = 120 ksi

A490 is a higher strength manufactured from an alloy of steel, min. Fyb ≈ 115 − 130 ksi, and Fub = 150 ksi

4

5

Most common diameters are 3/4”, 7/8” for building constructions; 7/8” and 1” for bridges, Table 17.1.

Bolt Diameter (in.) 5/8 3/4 7/8 1 1 1/8 1 1/4

Nominal Area (in2 ) 0.3068 0.4418 0.6013 0.7854 0.9940 1.2272

Table 17.1: Nominal Areas of Standard Bolts

17.1.2 6

Types of Bolted Connections

There are two types of bolted connections:

Bearing type which transmits the load by a combination of shear and bearing on the bolt, Fig. 17.2. Slip-critical transmits load by friction, Fig. 17.3. In addition of providing adequate at ultimate load, it must not slip during service loads. 7 Possible failure modes (or “limit states”) which may control the strength of a bolted connection are shown in Fig. 17.4.

17–2

STEEL CONNECTIONS

Draft Butt Splice

Lap Splice

Axial Shear

Hanger Connection Bracket Connection

Tension

Eccentric Shear 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 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 00 11 00 11

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 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 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11

111 000 000 111 000 111

1 0 0 1 0 1 0 1

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111

1111 0000 0000 1111 0000 1111

111 000 000 111 000 111

Standard Beam Connection

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 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 00 11 00 11

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111

1111 0000 0000 1111 0000 1111 0000 1111 0000 1111

Unstiffened Seat Connection

Stiffened Seat Connection

Structural Tee Connection

Shear and Tension

Figure 17.1: Examples of Bolted Connections

11 00 00 11 00 11 00 11 11 00

P

P

11 00 00 11 00 11 P

P

P

11 00 00 11 00 11 00 11 00 11

P

P

P

P

11 00 00 11 00 11

P

Figure 17.2: Stress Transfer by Shear and Bearing in a Bolted Connection

Victor Saouma

Structural Engineering

17.1 Bolted Connections

17–3

Draft

High-strength bolt P

P

T P µT T

µT

P= µT

T P T

Figure 17.3: Stress Transfer in a Friction Type Bolted Connection

* Checked Through Design of Plates

P

11 00 00 11 00 11 00 11 00 11 11 00

P

Shear Failure 11 00 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11

111 000 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111 000 111

11 00 11 00

Shear Failure of Plate *

Bearing Failure Tensile Failure

Bearing Failure of Plate

00 11 11 00 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11 00 11

Bending Failure

Tensile Failure of Plate *

Figure 17.4: Modes of Failure of Bolted Connections

Victor Saouma

Structural Engineering

17–4

STEEL CONNECTIONS

Draft 17.1.3

Nominal Strength of Individual Bolts

Tensile Strength The nominal strength Rn of one fastener in tension is Rn = Fub An

(17.1)

where Fub is the tensile strength of the bolt, and An the area through the threaded portion, also known as “tensile stress area”. The ratio of the tensile stress area to the gross area Ag ranges from 0.75 to 0.79. Shear Strength is given by Rn = mAb Fn

(17.2)

where m is the number of shear planes, Fig. 17.5, Ab the gross area of the bolt, and Fn is the nominal shear strength of bolts P

11 00 00 11 00 11 000 111 111 000 111 000

11 00 00 11 00 11 00 11 00 11 00 11 00 11 00 11

P

111 000 111 000

m=1

m=2

Figure 17.5: Number of Shearing Planes m in Bolted Connections Bearing Strength Rn is the force applied against the side of the hole to split or tear the plate, Fig. 17.6. The larger the end distance L, the less likely that of having a splitting failure. Tearing will occur along lines 1-1 and 2-2. In the most conservative case, α is taken as zero, thus thickness t

thickness t α

1

1 1

P

P d

Failure Line Resisting Shear Stresses

α

2

2

d/2 L

Figure 17.6: Bearing Strength Related to End Distance d p Rn = 2t L − τ 2 u

(17.3)

where: τup is the shear strength of the plate ≈ 0.7Fu Fu is the tensile strength of the plate material d is the nominal bolt diameter. Thus this equation transforms into Rn

= =

which can be approximated as

d 2t L − (0.70Fu ) 2 L 1 1.40Fu dt − d 2

L = LtFu Rn = Fu dt d

(17.4-a) (17.4-b)

(17.5)

which applies to the bolt closest to the edge for the design of single bolt connection, or two or more in line force, when the end distance is less than 1.5d. Victor Saouma

Structural Engineering

17.1 Bolted Connections

17–5

Draft 17.1.4

Concentric Loads (Tension Connections)

17.1.4.1

AISC Requirements

8

AISC RFD requirements for fasteners must satisfy φRn ≥ Σαi Qi

(17.6)

where φ is the strength resistance factor, Rn the nominal resistance (strength), αi the load magniﬁcation factor, and Qi the actual loads. 9

The most important relevant requirements are summarized below:

Shear Strength No threads in shear plane (LRFD-J3.3) φRn = φ(0.6Fub )mAb

(17.7)

where φ = 0.65, and m the number of shear planes, A − b the gross-sectional area across the unthreaded shank of the bolt. Threads in shear plane (LRFD-J3.3) φRn = φ(0.45Fub )mAb

(17.8)

where φ = 0.65, and m the number of shear planes, Ab the gross-sectional area across the unthreaded shank of the bolt. Tension Strength (LRFD-J3.3) φRn = φFub (0.75)Ab

(17.9)

where φ = 0.75. Bearing Strength (LRFD-J3.6). Note that for Bearing strength, Fu is the one of the plate and not the bolt. 1. Usual conditions (standard holes or short-slotted holes, end distance not less than 1.5d, bolt spacing center to center not less than 3d, and with two or more bolts in the line of force: φRn = φ(2.4dtFu )

(17.10)

where φ=0.75, d is the nominal diameter of the bolt (not at threads), t thickness of connected part. 2. Long-slotted holes perpendicular to the direction of load transmission, end distance not less than 1.5d, bolt center-to-center spacing not less than 3d and with two or more bolts in line of force (LRFD J3-1b) φRn = φ(2.0dtFu )

(17.11)

where φ=0.75, d is the nominal diameter of the bolt (not at threads), t thickness of connected part. Victor Saouma

Structural Engineering

17–6

STEEL CONNECTIONS

Draft

3. For the bolt closest to the edge and when Eqs 17.10 and 17.11 are not satisﬁed, then (LRFD J3-1c) (17.12)

φRn = φLtFu

where L is the end distance of force, from the center of a standard hole or oversized hole, or from the mid-width of a slotted hole, to an edge of a connected part. Minimum Spacing of bolts in the direction of transmitted forces must be at least 3 diameters when Eq. 17.10 and 17.11 are used. Under all other conditions Eq. 17.5 applies. Solving for L from the center of the fastener to the edge of the adjacent hole L≥

Rn Fu t

(17.13)

then adding the radius dh /2, and replacing Rn by P/φ, gives the minimum center to center spacing Spacing ≥

dh P + φFu t 2

(17.14)

where φ = 0.75 Minimum End Distance When Eq. 17.10 and 17.11 are used, then the minimum end distances must be at least 1 1/2 diameters. If higher bearing strengths are used, then Eq. 17.5 gives L≥

Rn Fu t

(17.15)

L≥

P φFu t

(17.16)

thus LRFDJ3.10 results in

where φ=0.75 In all cases Fub is the tensile strength of the bolt (120 ksi for A325; 150 ksi for A490). 17.1.4.2

Examples

Example 17-1: Analysis of Bolted Connections (Salmon and Johnson 1990) Compute the tensile service load capacity for the bearing-type connection shown below if a) the bolt threads are excluded from the shear plane and b) the bolt threads are included in the shear plane. 7/8 in diameter A325 bolts in standard holes, and A572 Grade 50 steel plates are used. LL=3 DL Plates 5/8 x 6 T

11 00 00 11 00 11 00 11 00 11 00 11

11 00 00 11 00 11 00 11 00 11 00 11

T

T

T

1 1/2"

1 1/2" 3"

Victor Saouma

7/8" hole A325 Bolts A572 Grade 50 Steel Plates

Structural Engineering

17.1 Bolted Connections

17–7

Draft Solution:

1. Determine the strength based on tension members: Ag

=

An

=

Ae φTn

= =

φTn

6(0.625) = 3.75 in2 7 1 + 6−2 (0.625) = 2.50 in2 8 8 U An = 1(2.50) = 2.50 in2 φFy Ag = (0.9)(50)(3.75) = 169 k

= φ Fu Ae = (0.75)(65)(2.5) = 122 k

(17.17-a) (17.17-b) (17.17-c) (17.17-d)

✛

(17.17-e)

2. Shear strength φRn

= φ(0.60Fub )mAb

(17.18-a)

= 0.65(0.60)(120)(1)(0.6013) = 28.1 k/bolt

(17.18-b)

3. Bearing strength for 1.5d end distance and 3d center to center spacing is φRn

=

φ(2.4Fu dt)

(17.19-a)

=

0.75(2.4)(65)(0.875)(0.625) = 64 k/bolt

(17.19-b)

note that for the bearing strength, Fu is the ultimate strength of the plate. 4. Since bearing strength is larger than shear strength, shear controls and the total design strength based on the fasteners is φTn = No. of fasteners Rn = 4(28.1) = 112 k

(17.20)

5. The single shear design strength (112 kip) is smaller than the plate design strength (122 kip) then φTn =112 kips 6. The end distance L from the center of the hole to the end of the plate should be, Eq. 17.16 L≥

√ P 28.1 = = 0.92 in φFu t 0.75(65)(0.625)

(17.21)

7. Shear block failure was not checked, it should have been 8. The design strength is related to the service load φTn

≥

Tu = 112 = D = L = Tmax

=

Tu

(17.22-a)

1.2D + 1.6L 1.2D + 1.6(3D) = 6.0D

(17.22-b) (17.22-c)

18.7 k 3D = 56.1 k

(17.22-d) (17.22-e)

D + L = 18.7 + 56.1 = 74.8 k

(17.22-f)

9. If threads are included in the shear plane, the design strength per bolt in single shear is φRn

Victor Saouma

= φ(0.45Fub )mAb = 0.65(0.45)(120)(1)(0.6013) = 21.1 k/bolt

(17.23-a) (17.23-b)

Structural Engineering

17–8

STEEL CONNECTIONS

Draft

10. Strength in single shear governs also φTn

=

84.4 = D = L = Tmax

=

No. of fasteners Rn = 4(21.1) = 84.4 k

(17.24-a)

1.2D + 1.6(3D) = 6.0D 14.1 k

(17.24-b) (17.24-c)

3D = 42.3 k

(17.24-d)

D + L = 14.1 + 42.4 = 56.4 k

(17.24-e)

Example 17-2: Design of Bolted Connections (Salmon and Johnson 1990) Determine the number of 3/4 in diameter A325 bolts in standard size holes required to carry 7 kips dead load and 43 kips live load on the plates shown below using A36 steel. Assume the portion of the double lap splice is a bearing-type connection with threads excluded from the shear plane and a double row of bolts. 1/4X10 Plates T

T/2 T/2

T

T 10"

Solution: 1. The center plate controls the tension strength (10)(0.25) = 2.50 in2 3 1 + 10 − 2 0.25 = 2.06 in2 4 8

Ag

=

An

=

Ae φTn

= =

U An = (1)(2.06) = 2.06 in2 φFy Ag = 0.9(36)(2.50) = 81 k

φTn Tu

= =

φFu Ae = 0.75(58)(2.06) = 90 k √ 1.2D + 1.6L = 1.2(7) + 1.6(43) = 77 k < φTn

(17.25-a) (17.25-b) (17.25-c) (17.25-d)

✛

(17.25-e) (17.25-f)

2. The shear strength of each A325 bolt is φRn = φ(0.60Fub )mAb = 0.65(0.60)(120)(2)(0.4418) = 41.4 k/bolt

(17.26)

3. The design strength in bearing on the 1/4 in plate is φRn = φ(2.4Fu dt) = 0.75(2.4)(58)(0.75)(0.25) = 19.6 k/bolt

✛

(17.27)

4. Thus the total number of bolts required is No. of bolts = Victor Saouma

77 = 3.9 19.6

(17.28) Structural Engineering

17.2 Welded Connections

17–9

Draft

5. The end distance is obtained from Eq. 17.16 P

=

L

≥

77 = 19.3 k 4 P 19.3 = = 1.77 in φFu t 0.75(58)(0.25)

(17.29-a) (17.29-b)

6. Select the bolt arrangement shown below a

2"

11111111 00000000 00000000 11111111 00000000 11111111 00000000 11111111 c 11111111 00000000

b

6"

2"

T f 11111111 00000000 e 00000000 11111111 00000000 11111111 00000000 11111111 11111111 00000000

d

2"

3"

7. We must check shear rupture since the plate is a thin one and heavily loaded Shear Yielding-Tension Fracture Tn

Shear Fracture-Tension Yielding Tn

=

0.60Fy Avg + Fu Ant

= =

0.60(36)(10)0.25 + 58(4 − 13/16)0.25 (17.30-b) 54 + 46 = 100 k (17.30-c)

= =

0.60Fu Ans + Fy Atg (17.30-d) 0.60(58)[10 − 3(13/16)]0.25 + 36(4)0.25(17.30-e) (17.30-f) 66 + 36 = 102 k ✛

=

(17.30-a)

(17.30-g) Note that we have used 3/4 + 1/16 for the hole diameter instead of 3/4 + 1/8 8. The design strength φTn gives φTn = 0.75(102) = 77 k

(17.31)

which equals the factored load Tu = 77 kips that must be supported. 9. Use 2 in. edge distance along the sides, 3 in. longitudinal spacing, and 2 in. end distance, with 4 A325 bolts.

17.1.5

Eccentric Connection (Shear Connections)

17.2

Welded Connections

10

Basic types of welds are shown in Fig. 17.7

t

L

L

b t

45

o

t 45

o

Figure 17.7: Basic Types of Weld

Victor Saouma

Structural Engineering

17–10

Draft

STEEL CONNECTIONS

Victor Saouma

Structural Engineering

Draft Chapter 18

UNBRACED ROLLED STEEL BEAMS 18.1

Introduction

1 In a previous chapter we have examined the behavior of laterally supported beams. Under those conditions, the potential modes of failures were either the formation of a plastic hinge (if the section is compact), or local buckling of the ﬂange or the web (partially compact section). 2 Rarely are the compression ﬂange of beams entirely free of all restraint, and in general there are two types of lateral supports:

1. Continuous lateral support by embedment of the compression ﬂange in a concrete slab. 2. Lateral support at intervals through cross beams, cross frames, ties, or struts. 3 Now that the beam is not laterally supported, we ought to consider a third potential mode of failure, lateral torsional buckling.

18.2

Background

4 Whereas it is beyond the scope of this course to derive the governing diﬀerential equation for ﬂexural torsional buckling (which is covered in either Mechanics of Materials II or in Steel Structures), we shall review some related topics in order to understand the AISC equations later on

5

There are two types of torsion:

Saint-Venant’s torsion: or pure torsion (torsion is constant throughout the length) where it is assumed that the cross-sectional plane prior to the application of torsion remains plane, and only rotation occurs. Warping torsion: out of plane eﬀects arise when the ﬂanges are laterally displaced during twisting. Compression ﬂange will bend in one direction laterally while its tension ﬂange will bend in another. In this case part of the torque is resisted by bending and the rest by Saint-Venant’s torsion. 6

With reference to Fig. 18.1, the governing torsional diﬀerential equation is

Insert Fig. 9.4.1 from Steel Structures by Salmon and Johnson Figure 18.1: I Shaped Beam in Slightly Buckled Position

ECw

d2 φ M02 d4 φ − Gj − φ=0 dz 4 dz 2 EIy

(18.1)

18–2

UNBRACED ROLLED STEEL BEAMS

Draft

where: Cw Torsional warping constant E Elastic modulus G Shear modulus J Saint-Venant Torsional constant M0 Applied moment in the yz plane Cw and J are section properties tabulated in the AISC manual. 7

Solution of this equation is Mcr

π = Cb L

πE L

2 Cw Iy + EIy GJ

(18.2)

where Cb is a factor to account for moment gradient.

18.3

AISC Equations

18.3.1

Dividing values

We identify three failure modes depending on Lb which is the distance between lateral supports (or unbraced length) Failure Mode Conditions Lb < Lp “very short” Plastic hinge Lp < Lb < Lr “short” Inelastic lateral torsional buckling “long” Elastic lateral torsional buckling Lr < Lb and

8

where: Lp Lr ry Fyw Fr

Lp

=

Lr

=

300 ry Fy , ksi ! # ry X1 2 1 + 1 + X2 (Fyf − Fr ) Fyf − Fr

(18.3) (18.4)

Limit dividing line for plastic hinge formation Dividing line between inelastic and elastic lateral torsional buckling Radius of gyration with respect to the “weak” axis, [in.] Yield stress of the ﬂange, [ksi] Compressive residual stress, usually taken as 10 ksi. for rolled sections, and 16.5 ksi for welded shapes.

and

X1

=

X2

=

!

EGJA 2 2 Cw Sx 4 Iy GJ π Sx

(18.5-a) (18.5-b)

are not really physical properties but a combination of cross-sectional ones which simpliﬁes the writing of Eq. 18.4. Those values are tabulated in the AISC manual. 9

The ﬂexural eﬃciency of the member increases when X1 decreases and/or X2 increases.

18.3.2

Governing Moments

1. Lb < Lp : “very short” Plastic hinge Mn = Mp = Zx Fy

Victor Saouma

(18.6)

Structural Engineering

18.3 AISC Equations

18–3

Draft

2. Lp < Lb < Lr : “short” inelastic lateral torsional buckling

Mn = Cb Mp − (Mp − Mr ) and

Cb = 1.75 + 1.05

M1 M2

Lb − Lp Lr − Lp

+ 0.3

M1 M2

≤ Mp

(18.7)

2 ≤ 2.3

(18.8)

1 where M1 is the smaller and M2 is the larger end moment in the unbraced segment M M2 is negative when the moments cause single curvature (i.e. one of them is clockwise, the other counterclockwise), hence the most severe loading case with constant M gives Cb = 1.75 − 1.05 + 0.3 = 1.0.

Mr is the moment strength available for service load when extreme ﬁber reach the yield stress Fy ; Mr = (Fy − Fr )Sx

(18.9)

3. Lr < Lb “long” elastic lateral torsional buckling, and the critical moment is the same as in Eq.18.2

Mcr

π = Cb Lb

πE Lb

2 Cw Iy + EIy GJ ≤ Cb Mr and Mp

(18.10)

or if expressed in terms of X1 and X2 Mcr

10

√ 2 Cb Sx X1 2 . .1 + X1 X2 ≤ C M = / 2 b r Lb ry 2 Lryb

(18.11)

Fig. 18.2 summarizes the governing equations.

Mn

Mp

Mr

✻Mp = Zx Fy Plastic Lateral Torsional Buckling Analysis ✛ ✲✛ ✲ 1 2 s s ❅ Inelastic & ' ❅ L −Lp ❅ Mn = Cb Mp − (Mp − Mr ) Lrb −Lp ❅ ❅ 3 ❅s Elastic ! Mcr =

Cb Sx X1

√

2

Lb ry

1+

✲ Lp =

Lr = !

300 √ ry Fy

ry X1 Fyf −Fr

1+

#

1 + X2 Fyf − Fr

X12 X2

L 2

2

b ry

Lb

2

Figure 18.2: Nominal Strength Mn of “Compact” Sections as Aﬀected by Lateral-Torsional Buckling

Victor Saouma

Structural Engineering

18–4

UNBRACED ROLLED STEEL BEAMS

Draft 18.4

Examples

18.4.1

Veriﬁcation

Example 18-1: Adequacy of an unbraced beam, (?) Verify the adequacy of a 15 ft W12X87 beam, subjected to a uniform load of 3k/ft and which is braced only at the ends. The load is 60% live and 40% dead and unfactored. Steel is A36 and Cb = 1.0. Solution: 1. The factored load and moments are wu

=

Mu

=

1.2(0.4)(3) + 1.6(0.6)(3) = 4.32 k/ft (4.32)(15)2 wu L2 = = 121.5 k.ft 8 8

(18.12-a) (18.12-b)

2. Next we check if the W12X87 is compact, From Eq. 14.12 bf = 7.5 < 2tf

√ 65 √ = 10.8 36

(18.13-a)

λp

hc = 18.9 < tw

√ 640 √ = 106. 36

(18.13-b)

λp

Note that those values are taken from the AISC manual. 3. The unbraced length is Lb =15 ft. 4. From AISC manual, ry

=

3.07 in

(18.14-a)

X1 X2

= =

3, 880 ksi 0.000586

(18.14-b) (18.14-c)

5. Determine Lp from Eq. 18.3 Lp

= =

300 ry Fy 300 √ (3.07) = 153.5 in = 12.8 ft 36

(18.15-a) (18.15-b)

6. Determine Lr from Eq. 18.4 Lr

= = =

Victor Saouma

!

# 2 1 + 1 + X2 (Fyf − Fr ) ! # (3.07) in(3, 880) ksi 2 1 + 1 + 0.000586 (36 − 10) (36 − 10) ksi 676.6 in = 56.4 ft ry X1 Fyf − Fr

(18.16-a) (18.16-b) (18.16-c)

Structural Engineering

18.4 Examples

18–5

Draft

7. Hence Lp < Lb < Lr , and failure is caused by inelastic lateral torsional buckling, and we must use Eq. 18.7 Lb − Lp (18.17-a) ≤ Mp Mn = Cb Mp − (Mp − Mr ) Lr − Lp 1 = 396 k.ft (18.17-b) Mp = Zx Fy = (36) ksi(132) in3 (12) in/ft 1 = 255.7 k.ft Mr = (Fy − Fr )Sx = (36 − 10)(118) (18.17-c) (12) √ 15 − 12.8 Mn = 1.0 396 − (396 − 255.7) (18.17-d) = 388.9 k.ft ≤ 396 k.ft 56.4 − 12.8 8. The moment capacity will then be φb Mn = (0.9)(388.9) k.ft = 350 k.ft > 121.5 k.ft

√

(18.18)

Example 18-2: Adequacy of an unbraced beam, II (?) Check the design of a 25 ft. long W10X33 beam which is subjected to a factored load (including self weight) of 1.2 kips/ft. The beam is unbraced and made from A242 steel (Fy =50 ksi). Assume Cb = 1.0. Solution: 1. The factored moment is

(1.2)(25)2 wu L2 = = 93.75 k.ft 8 8 2. Check if the W10X33 is compact, From Eq. 14.12 Mu =

bf = 9.1 < 2tf

√ 65 √ = 9.19 50

(18.19)

(18.20-a)

λp

hc = 27.1 < tw

√ 640 √ = 90.5 50

(18.20-b)

λp

3. The unbraced length is Lb =25 ft. 4. From AISC manual, ry X1

= =

1.94 in 2, 710 ksi

(18.21-a) (18.21-b)

X2

=

0.00251

(18.21-c)

5. Determine Lp from Eq. 18.3 Lp

= =

Victor Saouma

300 ry Fy 300 √ (1.94) = 82.3 in = 6.85 ft 50

(18.22-a) (18.22-b)

Structural Engineering

18–6

UNBRACED ROLLED STEEL BEAMS

Draft

6. Determine Lr from Eq. 18.4 Lr

!

# 1 + 1 + X2 (Fyf − Fr )2 ! # (1.94)(2, 710) 2 1 + 1 + 0.00251 (50 − 10) = (50 − 10) = 236.5 in = 19.7 ft

=

ry X1 Fyf − Fr

(18.23-a) (18.23-b) (18.23-c)

7. Since Lb > Lr , failure is caused by elastic lateral torsional buckling, and we must use Eq. 18.8

Mcr

=

=

√ 2 Cb Sx X1 2 . .1 + X1 X2 / 2 Lb ry 2 Lryb √ 2 (1.0)(35.0)(2, 710) 2 . .1 + (2, 710) (0.00251) 2 / (25)(12) 1.94 2 (25)(12) 1.94

= 1, 021 k.in = 85.1 k.ft

(18.24-a)

(18.24-b) (18.24-c)

8. The moment capacity will then be φb Mn = (0.9)(85.1) k.ft = 76.6 k.ft < 93.75 k.ft NG

(18.25)

Hence the section is not adequate.

18.4.2

Design

Seeking a design where failure is through a plastic hinge, the objective is to have Lb < Lp . This can be achieved by setting Lb = Lp , or from Eq. 18.3

11

rymin

Lb Fy = 300

(18.26)

and from Eq. 18.6 Zxreq =

Mn Fy

(18.27)

12 If Lb is so large that the previous design objectives can not be met, then the bam will fail by either inelastic or elastic lateral torsional buckling. Recalling that the limiting cases for the former is Lp < Lb < Lr , and that the corresponding limiting moments Mp and Mr are given by Eq. 18.6 and 18.9 respectively, then the upper and lower limits would be

Zxreq

<

Sxreq

>

Mn Fy Mn Fy − Fr

(18.28-a) (18.28-b)

Hence, if Lb is close to Lp , then we should select a section using Zxreq , and if Lb is close to Lr , then we select a section close to Sxreq . Alternatively, one can use the AISC design charts. In those charts, the design moment φb Mn is plotted in terms of Lb for a given Fy and Cb = 1.0. In those charts, the most economical section by

13

Victor Saouma

Structural Engineering

18.4 Examples

18–7

Draft

weight are diﬀerentiated from the others (solid versus dashed lines), and limiting Lp and Lr values are highlighted by closed and open circles. To use those charts, one would enter Lb and φb Mn , where those two values intersect, then any beam listed to the right or above will satisfy the necessary requirements (assuming that Cb = 1.0).

14

Note that since all of the strength equations are linear in Cb , we can use Lb /Cb as a substitute of Lb with Cb = 1 15

Example 18-3: Design of Laterally Unsupported Beam, (?) Select an economical W section for the beam shown in Fig. 18.3. Lateral support is provided at the vertical supports, concentrated load points, and at the end of the cantilever. The 26-kip load contains 6 kips dead load and the 11.5-kip load includes 4 kips dead load; the remainder is live load. Use A36 steel. Use W24 section.

DL: 6 LL: 20

24’ A

LS

LS

460 +

DL 4 LL: 7.5

28’ B

LS

21’ C

LS

LL on 52 ft span, not on cantilever

70

-

101

LL on cantilever, not on 52 f span

353 Factored bending moment envelope due to superimposed load Figure 18.3: Design of Laterally Unsupported Steel Beam Solution: 1. The moment envelope resulting from the factored load is ﬁrst obtained. It contains two loading scenario; a) DL+LL on the 52-ft span, and no LL on the cantilever; and b) DL+LL on the cantilever and no LL on the 52 ft span: Wu1

= 1.2(6) + 1.6(20) = 39.2 k

(18.29-a)

Wu2

= 1.2(4) + 1.6(7.5) = 16.8 k

(18.29-b)

2. For the ﬁrst loading case, the maximum factored moment at the cantilever support and under the point load Mu Mu Victor Saouma

= (1.2)(4)(21) = 101 k.ft 24 1 = 460 k.ft = (39.2)(24)(58) − (101) 2 52

(18.30-a) (18.30-b) Structural Engineering

18–8

UNBRACED ROLLED STEEL BEAMS

Draft

3. For the second load case, the maximum factored moment at the cantilever support is Mu = (16.8)(21) = 353 k.ft

(18.31)

4. Three laterally unbraced segments must be considered since each length is diﬀerent and is subjected to a diﬀerent maximum factored moment Mu . By inspection, segment A controls over segment C; both the moment and the unbraced length are larger on segment A, and the moment gradient is the same (i.e. Cb =1.75 for both). 5. Segment B has a larger Cb 2 M1 M1 Cb = 1.75 + 1.05 ≤ 2.3 (18.32-a) + 0.3 M2 M2 2 0 0 A Cb = 1.75 + 1.05 = 1.75 (18.32-b) + 0.3 460 460 2 +101 +101 CbB = 1.75 + 1.05 = 2.0 ≤ 2.3 (18.32-c) + 0.3 +460 +460 which is beneﬁcial, but also a larger Lb , hence we should design both segment A and B. 6. For segment A, Mu =460 ft-kips, and Lb = 24 ft. The required Mn is Mn =

Mu 460 = 511 k.ft = φb 0.9

(18.33)

7. Aiming for a plastic failure, we use Eq. 18.26 to determine rymin and Zxmin √ Lb Fy (24)(12) 36 = = 5.76 in rymin = 300 300 Mn (511) k.ft(12) in/ft = = 170 in3 Zxmin = Fy (36) ksi

(18.34-a) (18.34-b)

8. From the AISC manual we have the following Section W24x492

ry 3.41

Zx 1,550

Sx 1,290

Lp

Lr

W24x76 W24x68 W24x55

1.92 1.87 1.34

200. 177. 134

176. 154. 114.

8.0 7.8 5.6

23.4 22.4 16.5

Remarks Largest W24 section, ry NG and section too stiﬀ (very large Zx ) Zx OK but ry NG Zx OK but ry NG Smallest W24 section

Hence it is quite clear that failure will have to be through inelastic or elastic torsional buckling. 9. We try W24x76. 10. Is the section compact? bf = 6.66 < 2tf

√ 65 √ = 10.8 36

hc = 49 < tw

√ 640 √ = 106 36

(18.35-a)

λp

(18.35-b)

λp

Then we can proceed, 11. From Eq. 18.3 and 18.4:

Victor Saouma

Lp

=

Lr

=

300 1 300 ry = √ 1.92 = 8.0 ft 12 Fy 36 ! # ry X1 2 1 + 1 + X2 (Fyf − Fr ) Fyf − Fr

(18.36-a) (18.36-b) Structural Engineering

18.4 Examples

18–9

Draft

! 1 (1.92)(1, 760) = 12 36 − 10 = 23.4 ft

1+

# 2 1 + (0.0186) (36 − 10)

(18.36-c) (18.36-d)

12. Since Lr < Lb , failure is governed by elastic lateral torsional buckling, and we use Eq. 18.11 1 = 381. k.ft (12)

Mr

=

(Fy − Fr )Sx = (36 − 10)(176)

Cb Mr

=

Mp

=

Lb ry

=

Mcr

=

(1.75)(381) = 667 k.ft (200)(36) = 600 k.ft Zx Fy = 12 (24)(12) = 150 = 1.92 √ 2 Cb Sx X1 2 . .1 + X1 X2 ≤ C M / 2 b r Lb ry 2 Lryb √ (1, 760)2(0.0186) 1 (1.75)(176)(1, 760) 2 1+ (12) 150 2(150)2

=

⇒ Mn φb Mn

(18.37-a) (18.37-b) (18.37-c) (18.37-d) (18.37-e)

(18.37-f)

=

643.14 k.ft ≤ Cb Mr but > Mp

(18.37-g)

= =

Mp = 600 k.ft √ (0.9)(600) = 540 k.ft

(18.37-h) (18.37-i)

13. Hence the W24x76 is a satisfactory design. 14. In an attempt to automate the design procedure through a spread-sheet, and seek the most economical design (smallest weight), we consider table 18.1. and note that W24x68 is the most economical Section

Lb /ry

W24.x76. W24.x68. W24.x62. W24.x55. W21.x83. W21.x73. W21.x68. W21.x62. W18.x97. W18.x86. W18.x76. W16.x100. W16.x89. W16.x77. W14.x109. W14.x99. W14.x90. W12.x120. W12.x106. W12.x96. W10.x112. W10.x100. W8.x67. W8.x58. W6.x25. W6.x20.

150.00 154.01 208.70 214.93 157.38 159.12 160.00 162.71 108.68 109.51 110.34 114.74 115.66 116.60 77.21 77.63 77.84 92.01 92.60 93.20 107.46 108.68 135.85 137.14 189.47 192.00

Lp ft. 8.0 7.8 5.8 5.6 7.6 7.5 7.5 7.4 11.0 11.0 10.9 10.5 10.4 10.3 15.5 15.5 15.4 13.0 13.0 12.9 11.2 11.0 8.8 8.8 6.3 6.3

Lr ft. 23.4 22.4 17.2 16.6 24.9 23.5 22.8 21.7 38.1 35.5 33.3 42.1 38.6 34.9 62.7 58.2 54.1 75.5 67.2 61.4 86.4 77.4 64.0 56.0 31.3 25.6

Mp ft-kips 600. 531. 459. 402. 588. 516. 480. 432. 633. 558. 489. 594. 525. 450. 576. 519. 471. 558. 492. 441. 441. 390. 211. 179. 57. 45.

Mr ft-kips 381. 334. 284. 247. 371. 327. 303. 275. 407. 360. 316. 379. 336. 290. 375. 340. 542. 353. 314. 284. 273. 243. 131. 113. 36. 29.

Cb Mr ft-kips 667. 584. 497. 432. 648. 573. 531. 482. 713. 629. 554. 664. 588. 508. 656. 595. 762. 618. 550. 497. 478. 425. 229. 197. 63. 51.

Eq. 18.7 ft-kips 0. 0. 0. 0. 669. 0. 0. 0. 918. 792. 679. 879. 759. 632. 945. 846. 762. 914. 798. 709. 722. 632. 330. 276. 74. 53.

Mine ft-kips 0. 0. 0. 0. 588. 0. 0. 0. 633. 558. 489. 594. 525. 450. 576. 519. 471. 558. 492. 441. 441. 390. 211. 179. 57. 45.

Eq. 18.8 ft-kips 643. 523. 298. 239. 0. 555. 491. 414. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.

Mel. ft-kips 600. 523. 298. 239. 0. 516. 480. 414. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.

φb M n ft-kips 540. 471. 268. 215. 529. 464. 432. 372. 570. 502. 440. 535. 473. 405. 518. 467. 424. 502. 443. 397. 397. 351. 190. 161. 51. 40.

Table 18.1: Unbraced Beam Design design. Should we increase the depth, then the weight increases (as expected for a beam). Victor Saouma

Structural Engineering

18–10

UNBRACED ROLLED STEEL BEAMS

Draft

15. Next, we must consider segment B, it has a steeper moment gradient than segment A, i.e. Cb > 1.75 2 +101 +101 Cb = 1.75 + 1.05 = 2.0 ≤ 2.3 (18.38) + 0.3 +460 +460

however the laterally unbraced length of 28 ft is now longer. 16. Check the W24x68 ! # ry X1 2 Lr = 1 + 1 + X2 (Fyf − Fr ) Fyf − Fr ! # 1 (1.87)(1, 590) = 1 + 1 + (0.0290) (36 − 10)2 12 36 − 10 = 22.4 ft Mp = φb Zx Fy 1 = 531 k.ft = (177)(36) 12 Since Lr < Lb , the beam fails by linear elastic torsional buckling and

(18.39-b)

1 = 333.7 k.ft (12)

(18.40-a)

Mr

=

(Fy − Fr )Sx = (36 − 10)(154)

Cb Mr

=

Mcr

=

(2.0)(333.7) = 667 k.ft √ 2 Cb Sx X1 2 . .1 + X1 X2 ≤ C M / b r 2 Lb ry 2 Lryb

Lb ry

=

=

= =

(18.39-a)

(18.39-c) (18.39-d) (18.39-e)

(18.40-b) (18.40-c)

(28)(12) = 154. 1.87

(18.40-d)

√ (1, 590)2 (0.0290) 1 (2.0)(154)(1, 590) 2 1+ (12) 154 2(154)2 468 k.ft ≤ Cb Mr and > Mp

(18.40-e) (18.40-f)

17. We observe that Mcr = 468 k.ft for segment B, and 523 k.ft for segment A. Hence, segment B controls. 18. Checking for the ultimate moment φb Mn = (0.9)(468) = 421 < Mu (460)

(18.41-a)

19. Hence, we must increase the section to W24x76. From above Lr = 23.4 ft. since Lr < Lb , failure is governed by elastic lateral torsional buckling, and we use Eq. 18.11 =

(Fy − Fr )Sx = (36 − 10)(176)

Cb Mr

=

Mp

=

Lb ry

=

Mcr

=

(2.0)(381) = 762 k.ft (200)(36) = 600 k.ft Zx Fy = 12 (28)(12) = 175 = 1.92 √ 2 Cb Sx X1 2 . .1 + X1 X2 ≤ C M / 2 b r Lb ry 2 Lryb √ (1, 760)2 (0.0186) 1 (2.0)(176)(1, 760) 2 1+ (12) 175 2(175)2

= = φb Mn Victor Saouma

1 = 381. k.ft (12)

Mr

=

581 k.ft ≤ Cb Mr and > Mp √ (0.9)(581) = 523 k.ft

(18.42-a) (18.42-b) (18.42-c) (18.42-d) (18.42-e)

(18.42-f) (18.42-g) (18.42-h)

Structural Engineering

18.5 Summary of AISC Governing Equations

18–11

Draft

20. Note that Mc r is 581 for segment B, and was 600 for segment A. 21. Use W24x76

18.5

Summary of AISC Governing Equations

Table 18.2 summarizes the AISC design equations covered so far. Braced Beams ≤ λp Mn = ZFy hc ≤ λ p tw b λp < 2tff ≤ λr bf 2tf

Compact

λ−λ

Mn = Mp − (Mp − Mr )( λr −λpp ) ≤ Mp

Partially compact

Plastic Hinge Inelastic lateral torsional buckling Elastic lateral torsional buckling

λp < thwc ≤ λr Unbraced Beams Lb < Lp Mn = Mp&= Zx Fy ' L −L ≤ Mp Lp < Lb < Lr Mn = Cb Mp − (Mp − Mr ) Lrb −Lpp √ ! 2 X X Lr < Lb Mcr = Cb SxLXb 1 2 1 + 1L 22 ≤ Cb Mr 2

ry

Elastic Buckling Inelastic Buckling

b ry

Columns 2 λc ≤ 1.5 Fcr = 0.658λc Fy λc > 1.5 Fcr = 0.877 λ2 Fy c

Table 18.2: Summary of AISC Governing Equations

Victor Saouma

Structural Engineering

18–12

Draft

UNBRACED ROLLED STEEL BEAMS

Victor Saouma

Structural Engineering

Draft Chapter 19

Beam Columns, (Unedited) UNEDITED Examples taken from Structural Steel Design; LRFD, by T. Burns, Delmar Publishers, 1995

19.1

Potential Modes of Failures

19.2

AISC Speciﬁcations Pu 8 + φc Pn 9

19.3

Examples

19.3.1

Veriﬁcation

Mux Muy + φb Mnx φb Mny Pu Mux + 2φc Pn φb Mnx

Pu ≥ .20 (19.1) φc Pn Pu ≤ 1.0 if ≤ .20 (19.2) φc Pn

≤ 1.0 if

Example 19-1: Veriﬁcation, (?) A W 12 × 120 is used to support the loads and moments as shown below, and is not subjected to sidesway. Determine if the member is adequate and if the factored bending moment occurs about the weak axis. The column is assumed to be perfectly pinned (K = 1.0) in both the strong and weak directions and no bracing is supplied. Steel is A36 and assumed Cb = 1.0.

19–2

Beam Columns, (Unedited)

Draft

Solution: 1. Calculate the slenderness ratios in both the weak and strong directions (even though when unbraced in both directions, the weak axis will of course be critical). Klx rx Kly ry

= =

(1.0)(15) ft(12) in/ft = 32.67 (5.51) in (1.0)(15) ft.(12) in/ft) = 57.50 (3.13) in

✛

2. Using Table 16.1, we can ﬁnd the design axial stress (φc Fcr ) as 25.71 ksi. And, multiplying that with the area of the section, we can ﬁnd the design axial capacity (φc Pn ) as follows: φc Pn

= =

φc Fcr Ag (25.71) ksi(35.3) in2 = 907.6 k

3. Since the factored axial load is 400 kips, the ratio of

Pu φc Pn

is as follows:

√ (400) k = .44 > .20 (907.6) k therefore we must use Eq. as the beam-column analysis. Pu Mux 8 Muy + + ≤ 1.0 φc Pn 9 φb Mnx φb Mny 4. To start, calculate the design moment capacity (φb Mn ) of the W 12 × 120. The beam is compact and the unbraced length, lb , is 15 feet. This value of unbraced length is then compared to limits referred to as Lp and Lr to determine whether plastic behavior can be developed or whether buckling behavior controls the bending behavior. The values for Lp and Lr can be calculated using Eq. 6-1 and 6-2 respectively or they can be found in the beam section of the LRFD manual. In either case these values for a W 12 × 120 are: Lp Lr

Victor Saouma

= 13 ft = 75.5 ft

Structural Engineering

19.3 Examples

19–3

Draft

5. Since our unbraced length falls between these two values, the beam will be controlled by inelastic buckling, and the nominal moment capacity Mn can be calculated from Eq. 6-5. Using this equation we must ﬁrst calculate the plastic and elastic moment capacity values, Mp and Mr . Mp Mr

Fy Zx = (36) ksi(186) in3 (12)ftin = 558 k.ft (Fyw − 10) ksiSx (36 − 10) ksi(163) in3 = 353.2 k.ft

= =

6. Using Eq. 6-5 (assuming Cb is equal to 1.0) we ﬁnd: Mn

=

Mn

= =

l −L

Cb [Mp − (Mp − Mr )] Lbr −Lpp

(15−13) ft 1.0[558 − (558 − 453.2)] k.ft (75.5−15) ft 551.4 k.ft

7. Therefore the design moment capacity is as follows: φb Mn = 0.90(551.4) k.ft = 496.3 k.ft 8. Now consider the eﬀects of moment magniﬁcation on this section. Based on the alternative method and since the member is not subjected to sidesway (Mlt = 0) Mu B1

= =

Cm Cm Pu Pe

= = = =

1 .60 − .4 M M2 −50 .60 − .4 50 = 1.0 400 k π 2 EAg (Euler’s Buckling Load Equation) Kl 2

Pe

=

π 2 (29,000 ksi)(35.3 32.672

B1 Mnt Cm u 1− P Pe

r

in2 )

= 9, 456 k

9. Therefore, calculating the B1 magniﬁer we ﬁnd: B1 =

1 1−

(400) k (9,456) k

= 1.044

Calculating the ampliﬁed moment as follows: Mu Mu

= =

B1 Mnt 1.044(50) k.ft = 52.2 k.ft

Therefore the adequacy of the section is calculated from Eq. as follows: Pu 8 Mu x φc Pn + 9 φb Mn x (52.2) 94000 k + 89 (496.3) k.ft (907.6) k k.ft

≤ 1.0 =

.53 < 1.0

√

1 When dealing with unbraced frames, the procedure gets more involved since the calculation of sidesway eﬀects (sidesway moments and sidesway axial loads) is required to utilize the alternate method that we perform to estimate the secondary eﬀects. This generally means that a single load case on an unbraced frame that causes sidesway must be analyzed as if it were two separate load cases—one that would have a negligible sidesway eﬀect (typically the vertical gravity loads) and one that would cause sidesway (typically the lateral loads). This is illustrated in Figure 19.1 2 As the forces and moments for each of these “broken-down” load cases are calculated for a member, the same basic procedure is then followed as it was in the ﬁrst example.

Victor Saouma

Structural Engineering

19–4

Beam Columns, (Unedited)

Draft

Figure 19.1: Rigid Frame Subjected to Lateral Loading

3

The analysis of frames and frames subject to sidesway is not within the scope of this course.

You should be aware that as unbraced members are encountered, the B2 magniﬁer will become active as the frame exhibits sidesway behavior. The following example illustrates this situation.

4

Example 19-2: 8.2, (?) In the unbraced frame shown below, determine the adequacy of either W 14 × 90 columns under the factored loads given. The steel is A36 and assume that the columns have a Ky = 1.0 and Kx = 1.5 The lateral load is causing bending to occur about the x-axis.

Solution: 1. Typically, we would start this problem by separating the loads into those that cause sidesway eﬀects (the lateral load) and those that have little inﬂuence on such eﬀects (the vertical loads). As shown below, the eﬀects on the columns from the lateral load and the vertical load can be computed separately by various standard techniques of analysis such as moment distribution.

Victor Saouma

Structural Engineering

19.3 Examples

19–5

Draft

2. From these techniques, we ﬁnd that each column is under approximately 20 kips of factored axial load and a moment of 40 kip-feet in no sidesway case. And, they are under 20 kips of factored axial load (diﬀerent directions) and a moment of 200 kip-feet (assuming pinned supports) in the sidesway case. 3. To begin the evaluation, calculate the column section’s design axial capacity φc Pn . Calculating the slenderness ratios about both axis (although the weak axis will control in this case): Kl ry Kl rx

= =

1.0(240) in (3.70) in 1.5(240) in (6.14) in

= 64.8 = 58.6

4. Since the weak axis controls, ﬁnd the design axial stress φc Fcr that can be taken from Table 16.1. In this case the design axial stress corresponding to the critical slenderness ratio is 24.54 ksi. This would make the design axial capacity of this section as follows: φc Pn φc Pn Pu 5. Since φPc Pun = section.

40 k 650.3 k

= φc Fcr Ag = (24.54) ksi(26.5) in2 ) = 650.3 k = (20) k (no sidesway) + (20) k (sidesway) = 40 k

= .061 < .20 Eq. controls and should be used to check the adequacy of the

6. Calculate the design bending capacity of these sections about the strong axis. The unbraced length of this column is assumed to be 20 feet and the nominal moment capacity depends on what bending behavior controls in this case. Compare the unbraced length (20 ft.) to the Lp and Lr limits that were discussed in Chapter 6. 7. These values can be calculated from Equations 6-1 and 6-2 or taken from the LRFD manual. Since the Lp value for a W 14 × 90 is 15.4 feet and the Lr , value is 54.1 feet, the section’s capacity is controlled by Case 2 (compact beams exhibiting inelastic lateral torsional buckling). 8. The capacity of this section is found by calculating Eq. 6-5 (see Chapter 6) as follows and notice that Cb is equal to 1.67 using Mmax = 40, Ma = 10, Mb = 20, and Mc = 30. 9. Before calculating the nominal moment capacity from Eq. 6-5, ﬁrst calculate the plastic and elastic moment capacity Mp and Mr . Mp Mr Victor Saouma

= = =

Fy Zx = (36) ksi(157) in3 (12)ftin = 471 k.ft (Fyw − 10 ksi)Sx (36 − 10) ksi(143) in3 = 309.8 k.ft Structural Engineering

19–6

Beam Columns, (Unedited)

Draft

Calculating Eq. 6-5: Mn

= 1.67[(471) k.ft − [(471) k.ft(309.8) k.ft)((20 − 15.4) ft(54.1 − 15.4) ft)] = 791.5 k.ft

10. However because the nominal moment capacity cannot be greater than the plastic moment capacity, use 471 ft.-kips which is the plastic moment capacity. The design moment capacity is: φb Mn = 0.90(471) k.ft = 423.9 k.ft 11. After this we can calculate the magniﬁcation factors B1 and B2 , remembering that B1 is used for the nonsway eﬀects and B2 for the sway eﬀects. In this case, the axis of bending has been stated to be the x-axis and the eﬀective length factor K was given as 1.5. Therefore for B1 B1

=

Cm Cm

= =

Pe

=

Pu

= =

Cm u (1− P Pe

M1 .60 − .4 M 2 1 .6 since M M2 = 0 π 2 EAg

2 ( Kl r ) 2206 k 40 k

=

π 2 E(26.5) 58.62

in2

12. Therefore, B1 =

.6 1−

(40) k (2,206) k

= .61 use 1.0

13. Calculating B2 from Eq. 8-5: B2

=

1 u 1− ΣP ΣPe

ΣPu ΣPe B2

= = =

2(20) k = 40 k (for both columns in the story) (2, 206) k(2) columns = 4, 412 k 1 40 1 − 4,412 = 1.01

14. Calculating the nominal moment capacity about the x-axis, Mus , from Equation 8-1 as follows: Mu Mux

= B1 Mnt + B2 Mlt = 1.0(40) k.ft + 1.01(200) k.ft = 242 k.ft

15. Calculating the adequacy per Eq. : Pu 2φc Pn

+

Mux φb Mnx

(40) k = (2)(650.3) k+ = .60 < 1.0

(242) k.ft (423.9) k.ft

16. Therefore the section is good. Notice that the calculation of Pe for the B1 and B2 magniﬁers is always taken about the axis of bending, in this case the strong axis.

19.3.2

Design

5 The design of beam-columns is truly a trial and error procedure and depends a great deal on the designer’s experience. If the designer can select a good trial section at the very beginning, the work in obtaining a safe and economical member is greatly simpliﬁed although, in any case, the process will ultimately converge on a solution.

The most popular method of selecting a beam-column is the equivalent axial load method. This method replaces the factored axial load and moment that are applied to the column with a ﬁctitious concentric

6

Victor Saouma

Structural Engineering

19.3 Examples

19–7

Draft

axial load. This ﬁctitious concentric axial load will be larger than the factored axial load but will produce approximately the same maximum eﬀect.

7 The following equation is used to convert the factored bending moment into an estimated axial load, P . The actual factored axial load Pu , is added to P resulting in a ﬁctitious axial load called the equivalent factored axial load Puequiv . This idea is shown below:

Pu + P = Puequiv where Pu P Puequiv

actual factored axial load estimated axial load caused by moment Equivalent factored axial load

8 Although there are equations that would estimate the equivalent factored axial load based on the assumptions used by the interaction formulas separately (Eq. and ), it is much faster to use the following combined approximation developed from these formulas. This approximation is as follows:

(19.3)

Puequiv = Pu + Mux m + Muy mU

In this formula, Pu is again the actual factored load expressed in kips and Mux and Muy are the bending moments expressed in kip-feet. 9 If bending should occur about one axis only, the corresponding term regarding the other axis will then drop out. The value of m is taken from Table 19.1.

10

12

14

All Shapes

2.0

1.9

1.8

Preliminary Beam-Column Design Fy = 36 ksi, Fy = 50 ksi Values of m 36 ksi 16 18 20 22 and 10 12 14 over 1st Approximation 1.7 1.6 1.5 1.3 1.9 1.8 1.7

W4 W6 W8 W8 W10 W12 W14

3.1 3.2 2.8 2.5 2.1 1.7 1.5

2.3 2.7 2.5 2.3 2.0 1.7 1.5

1.7 2.1 2.1 2.2 1.9 1.6 1.4

1.4 1.7 1.8 2.0 1.8 1.5 1.4

Fy KL(f t)

Subsequent 1.1 1.0 1.4 1.2 1.5 1.3 1.8 1.6 1.7 1.6 1.5 1.4 1.3 1.3

Approximation 0.8 2.4 1.8 1.0 2.8 2.2 1.1 2.5 2.2 1.4 2.4 2.2 1.4 2.0 1.9 1.3 1.7 1.6 1.2 1.5 1.4

1.4 1.7 1.8 2.0 1.8 1.5 1.4

50 ksi 16 18

20

22 and over

1.6

1.4

1.3

1.2

1.1 1.4 1.5 1.7 1.7 1.5 1.3

1.0 1.1 1.3 1.5 1.5 1.4 1.3

0.9 1.0 1.2 1.3 1.4 1.3 1.2

0.8 0.9 1.1 1.2 1.3 1.2 1.2

Table 19.1: Values of m for use in Beam-Column Design Equation, (from a paper in AISC Engineering Journal by Uang, Wattar, and Leet (1990)) . The value of U , which is the bending moment conversion factor, is taken from the column load tables that are shown in Appendix C. (A full set of the column load tables can be found in the column section of the LRFD manual as well).

10

11 The procedure using this approximate formula is an iterative process that begins by assuming the value of m from Table 8-1 and a typical value of U (usually about 2.0). After a section is selected, the section is checked against the applicable AISC interactions formulas (Eq. and ). With successive trials the values of m and U can be reﬁned to a point where their values begin to stabilize.

Victor Saouma

Structural Engineering

19–8

Beam Columns, (Unedited)

Draft

Example 19-3: Design of Steel Beam-Column, (?)

Design the most economical W 12 section subjected to the loads and moments shown below. The value of K = 1.0 in both strong and weak directions and steel is A36. Assume this member to be part of a frame that is not subjected to sidesway. Solution:

Using the equivalent load formula and choosing the ﬁrst trial values of m = 1.9 as suggested In Table 8-1 and U = 2.0 (approximate), the equivalent load is as follows: = = =

Puequiv Puequiv

Pu + Mux m + Muy mU (300) k + (120) k.ft(1.9) + (80) k.ft(1.9)(2.0) 832 k

From Column Load tables found in Appendix C for Kl =12 ft. try a W 12 x 106 (with an axial design strength of 853 kips) Checking this section with the interaction formulas we ﬁnd the following: Pu Since φP = (300) k = .352 > .20, the Eq. will control. n (853) k Calculating the design moment capacity about each axis we ﬁnd the following: Lp = 13.0 ft Since this value is greater than 12 ft. which is this W 12 x 106’s unbraced length, the section is able to develop as plastic capacity. Therefore, φb Mnx

=

φb Mny

=

0.90(36) ksi(164) in3

1 = 442.8 k.ft in/ft 3 ft .90(36) ksi(75.1) in 12 in = 202.8 k.ft 12

Calculating the magniﬁer, B1 , similar to what was done previously, we would ﬁnd the following: Cmx Cmy Kl rx Kl ry

.6 − .4 −120 = 1.0 120 1.0( similarly) 26.3(λc = .295) 46.3(λc = .519)

= = = =

Calculating the Euler buckling load as follows:

Victor Saouma

Pe

=

Pex Pex

= =

Pey Pey B1x B1y

= = = =

Ag Fy λ2 c (31.2)

2

in (36) .2952

ksi

(12, 907) k (based on λc = .295) 2 (31.2) in (36) ksi .5192

(4, 162) k (based on λc = .519) 1 300 − 12,907 = 1.023 1 1 300 − = 1.078 1 4,162

Structural Engineering

19.3 Examples

19–9

Draft

This would make the magniﬁed factored as shown below: Mu Mux Muy

= = =

B1 Mnt (120) k.ft(1.023) = 122.76 k.ft (80) k.ft(1.078) = 86.24 k.ft

Applying AISC Eq. to the above information:

8 (300) k + (853) k 9

(86.24) k.ft (122.76) k.ft + (442.8) k.ft (202.8) k.ft

√ = .976 < 1.0

which works well and therefore is a very good choice!

Victor Saouma

Structural Engineering

19–10

Draft

Beam Columns, (Unedited)

Victor Saouma

Structural Engineering

Draft

Part II

STRUCTURAL ENGINEERING II; CVEN 3235

Draft

Draft Chapter 20

ARCHES and CURVED STRUCTURES 1

This chapter will concentrate on the analysis of arches.

2 The concepts used are identical to the ones previously seen, however the major (and only) diﬀerence is that equations will be written in polar coordinates. 3 Like cables, arches can be used to reduce the bending moment in long span structures. Essentially, an arch can be considered as an inverted cable, and is transmits the load primarily through axial compression, but can also resist ﬂexure through its ﬂexural rigidity.

4

A parabolic arch uniformly loaded will be loaded in compression only.

5 A semi-circular arch uniﬁrmly loaded will have some ﬂexural stresses in addition to the compressive ones.

20.1

Arches

6 In order to optimize dead-load eﬃciency, long span structures should have their shapes approximate the coresponding moment diagram, hence an arch, suspended cable, or tendon conﬁguration in a prestressed concrete beam all are nearly parabolic, Fig. 20.1. 7 Long span structures can be built using ﬂat 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. 8 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. 9 The basic issues of static in arch design are illustrated in Fig. 20.2 where the vertical load is per unit horizontal projection (such as an external load but not a self-weight). 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, wL L L − M = Hh − =0 (20.1) 2 2 4

Solving for H H=

wL2 8h

(20.2)

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

20–2

ARCHES and CURVED STRUCTURES

Draft 2

M = w L /8

L w=W/L

IDEALISTIC ARCH SHAPE GIVEN BY MOMENT DIAGRAM

C RISE = h -C BEAM +T W/2

M-ARM small C C-T 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 20.1: 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

= wL2 /8 - H h = 0

Figure 20.2: Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981)

Victor Saouma

Structural Engineering

20.1 Arches

20–3

Draft

than C − T in a beam. Since equilibrium requires H to remain constant across thee arch, a parabolic curve would theoretically result in no moment on the arch section.

10

Three-hinged 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.

11

An arch carries the vertical load across the span through a combination of axial forces and ﬂexural ones. A well dimensioned arch will have a small to negligible moment, and relatively high normal compressive stresses.

12

An arch is far more eﬃcient than a beam, and possibly more economical and aesthetic than a truss in carrying loads over long spans.

13

If the arch has only two hinges, Fig. 20.3, or if it has no hinges, then bending moments may exist either at the crown or at the supports or at both places.

14

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 20.3: 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.

15

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 eﬀect can be neglected if the live load is small in comparison with the dead load. √ 17 Since the greatest total force in the arch is at the support, (R = V 2 + H 2 ), whereas at the crown we simply have H, the crown will require a smaller section than the support. 16

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 20.4: Arch Rib Stiﬀened with Girder or Truss, (Lin and Stotesbury 1981)

Victor Saouma

Structural Engineering

20–4

ARCHES and CURVED STRUCTURES

Draft 20.1.1

Statically Determinate

Example 20-1: Three Hinged Arch, Point Loads. (Gerstle 1974) Determine the reactions of the three-hinged arch shown in Fig. 20.5

Figure 20.5: Solution: Four unknowns, three equations of equilibrium, one equation of condition ⇒ statically determinate. (+ ✛) ✁ ΣMzC

=

(+ ✲ ) ΣFx (+ ✻) ΣFy (+ ✛) ✁ ΣMzB

= = =

0; (RAy )(140) + (80)(3.75) − (30)(80) − (20)(40) + RAx (26.25) ⇒ 140RAy + 26.25RAx 0; 80 − RAx − RCx 0; RAy + RCy − 30 − 20 0; (Rax )(60) − (80)(30) − (30)(20) + (RAy )(80) ⇒ 80RAy + 60RAx

Solving those four equations simultaneously we have:     140 26.25 0 0  RAy   2, 900        0  80 R 1 0 1 Ax   =  1 R 0 1 0    Cy  50      3, 000 RCx 80 60 0 0

 RAy    RAx ⇒    RCy    RCx    

 15.1 k    29.8 k =   34.9 k     50.2 k    

= = = = = =

0 2.900 0 0 0 3, 000 (20.3)

      

We can check our results by considering the summation with respect to b from the right: √ ✛ (+ ✁) ΣMzB = 0; −(20)(20) − (50.2)(33.75) + (34.9)(60) = 0

(20.4)

(20.5)

Example 20-2: Semi-Circular Arch, (Gerstle 1974) Determine the reactions of the three hinged statically determined semi-circular arch under its own dead weight w (per unit arc length s, where ds = rdθ). 20.6 Solution: I Reactions The reactions can be determined by integrating the load over the entire structure Victor Saouma

Structural Engineering

20.1 Arches

20–5

Draft

dP=wRdθ

θ

B

B

r

R

R

θ

A

θ R cosθ

A

C

Figure 20.6: Semi-Circular three hinged arch 1. Vertical Reaction is determined ﬁrst: (+ ✛) ✁ ΣMA

θ=π

0; −(Cy )(2R) +

=

wRdθ R(1 + cos θ) = 0 dP moment arm θ=π wR wR θ=π [θ − sin θ] |θ=0 (1 + cos θ)dθ = 2 2 θ=0 wR [(π − sin π) − (0 − sin 0)] 2 π 2 wR

(20.6-a)

θ=0

⇒ Cy

= = =

2. Horizontal Reactions are determined next ✛ (+ ✁) ΣMB

=

0; −(Cx )(R) + (Cy )(R) −

θ= π 2 θ=0

⇒ Cx

= =

wR π wR − 2 2 π 2 − 1 wR

θ= π 2

cos θdθ = θ=0

(20.6-b)

wRdθ

θ R cos

dP

moment arm

=0

(20.7-a)

π π π θ= π wR − wR[sin θ] |θ=02 = wR − wR( − 0) 2 2 2 (20.7-b)

By symmetry the reactions at A are equal to those at C II Internal Forces can now be determined, Fig. 20.7. 1. Shear Forces: Considering the free body diagram of the arch, and summing the forces in the radial direction (ΣFR = 0): θ π π wRdα sin θ + V = 0 (20.8) −( − 1)wR cos θ + wR sin θ − α=0 2

2 Cx

$

Cy

⇒ V = wR ( π2 − 1) cos θ + (θ − π2 ) sin θ

%

(20.9)

2. Axial Forces: Similarly, if we consider the summation of forces in the axial direction (ΣFN = 0): θ ( π2 − 1)wR sin θ + π2 wR cos θ − wRdα cos θ + N = 0 (20.10) α=0

$ % ⇒ N = wR (θ − π2 ) cos θ − ( π2 − 1) sin θ

(20.11)

3. Moment: Now we can consider the third equation of equilibrium (ΣMz = 0): (+ ✛) ✁ ΣM

( π2 − 1)wR · R sin θ − π2 wR2 (1 − cos θ) + θ wRdα · R(cos α − cos θ) + M = 0 α=0

⇒ M = wR2 Victor Saouma

$π

2 (1

− sin θ) + (θ − π2 ) cos θ

(20.12) %

(20.13)

Structural Engineering

20–6

ARCHES and CURVED STRUCTURES

Draft

dP=wRdα θ

M

r

V

R sin θ

N

R cos θ

C x=(π /2-1)wR

R(1-cos θ) R cos α

y

α

C =π /2 wR

θ

R cos(θ −α)

dα

Figure 20.7: Semi-Circular three hinged arch; Free body diagram III Deﬂection are determined last 1. The real curvature φ is obtained by dividing the moment by EI ' wR2 & π π M = (1 − sin θ) + (θ − ) cos θ EI EI 2 2

φ=

(20.14)

1. The virtual force δP will be a unit vertical point in the direction of the desired deﬂection, causing a virtual internal moment δM =

R [1 − cos θ − sin θ] 2

0≤θ≤

π 2

(20.15)

p 2. Hence, application of the virtual work equation yields: 1 ·∆

= 2

dx

δP

20.1.2

' R π wR2 & π (1 − sin θ) + (θ − ) cos θ · · [1 − cos θ − sin θ] Rdθ 2 2 θ=0 EI

2

π 2

M EI

=

% wR4 $ 2 7π − 18π − 12 16EI

=

.0337 wR EI

=φ

4

δM

(20.16-a)

Statically Indeterminate

Example 20-3: Statically Indeterminate Arch, (Kinney 1957) Victor Saouma

Structural Engineering

20.1 Arches

20–7

Draft

Determine the value of the horizontal reaction component of the indicated two-hinged solid rib arch, Fig. 20.8 as caused by a concentrated vertical load of 10 k at the center line of the span. Consider shearing, axial, and ﬂexural strains. Assume that the rib is a W24x130 with a total area of 38.21 in2 , that it has a web area of 13.70 in2 , a moment of inertia equal to 4,000 in4 , E of 30,000 k/in2 , and a shearing modulus G of 13,000 k/in2 .

Figure 20.8: Statically Indeterminate Arch Solution:

1. Consider that end C is placed on rollers, as shown in Fig. ?? A unit ﬁctitious horizontal force is applied at C. The axial and shearing components of this ﬁctitious force and of the vertical reaction at C, acting on any section θ in the right half of the rib, are shown at the right end of the rib in Fig. 13-7. 2. The expression for the horizontal displacement of C is 1 ∆Ch = 2

B

δM C

M ds + 2 EI

B

δV C

V ds + 2 Aw G

B

δN C

N ds AE

(20.17)

δP

3. From Fig. 20.9, for the rib from C to B,

Victor Saouma

M

=

δM

=

V

=

δV

=

N

=

P (100 − R cos θ) 2 1(R sin θ − 125.36) P sin θ 2 cos θ P cos θ 2

(20.18-a) (20.18-b) (20.18-c) (20.18-d) (20.18-e) Structural Engineering

20–8

ARCHES and CURVED STRUCTURES

Draft

Figure 20.9: Statically Indeterminate Arch; ‘Horizontal Reaction Removed δN ds

= =

− sin θ Rdθ

(20.18-f) (20.18-g)

4. If the above values are substituted in Eq. 20.17 and integrated between the limits of 0.898 and π/2, the result will be ∆Ch = 22.55 + 0.023 − 0.003 = 22.57 (20.19) 5. The load P is now assumed to be removed from the rib, and a real horizontal force of 1 k is assumed to act toward the right at C in conjunction with the ﬁctitious horizontal force of 1 k acting to the right at the same point. The horizontal displacement of C will be given by B B B M V N δChCh = 2 ds + 2 ds + 2 ds (20.20-a) δM δV δN EI A G AE w C C C = 2.309 + 0.002 + 0.002 = 2.313 in (20.20-b) 6. The value of the horizontal reaction component will be HC =

∆Ch 22.57 = 9.75 k = δChCh 2.313

(20.21)

7. If only ﬂexural strains are considered, the result would be HC =

22.55 = 9.76 k 2.309

(20.22)

Comments 1. For the given rib and the single concentrated load at the center of the span it is obvious that the eﬀects of shearing and axial strains are insigniﬁcant and can be disregarded. 2. Erroneous conclusions as to the relative importance of shearing and axial strains in the usual solid rib may be drawn, however, from the values shown in Eq. 20.19. These indicate that the eﬀects of the shearing strains are much more signiﬁcant than those of the axial strains. This is actually the case for the single concentrated load chosen for the demonstration, but only because the rib does not approximate the funicular polygon for the single load. As a result, the shearing components on most sections of the rib are more important than would otherwise be the case. Victor Saouma

Structural Engineering

20.2 Curved Space Structures

20–9

Draft

3. The usual arch encountered in practice, however, is subjected to a series of loads, and the axis of the rib will approximate the funicular polygon for these loads. In other words, the line of pressure is nearly perpendicular to the right section at all points along the rib. Consequently, the shearing components are so small that the shearing strains are insigniﬁcant and are neglected. 4. Axial strains, resulting in rib shortening, become increasingly important as the rise-to-span ratio of the arch decreases. It is advisable to determine the eﬀects of rib shortening in the design of arches. The usual procedure is to ﬁrst design the rib by considering ﬂexural strains only, and then to check for the eﬀects of rib shortening.

20.2

Curved Space Structures

Example 20-4: Semi-Circular Box Girder, (Gerstle 1974) Determine the reactions of the semi-circular cantilevered box girder shown in Fig. 20.10 subjected x

x wRd α θ dα α

O R

B

V ΖΜ r

Τ θ

C A

A y

y

z

z

Figure 20.10: Semi-Circular Box Girder to its own weight w. Solution: I Reactions are again determined ﬁrst From geometry we have OA = R, OB = R cos θ, AB = OA − BO = R − R cos θ, and BP = R sin θ. The moment arms for the moments with respect to the x and y axis are BP and AB respectively. Applying three equations of equilibrium we obtain

−

⇒

FzA = wRπ

(20.23-a)

(wRdθ)(R sin θ) = 0

⇒

MxA = 2wR2

(20.23-b)

(wRdθ)R(1 − cos θ) = 0

⇒

MyA = −wR2 π

(20.23-c)

θ=0 θ=π

MxA −

θ=π

wRdθ = 0

FzA

θ=0 θ=π

MyA − θ=0

II Internal Forces are determined next 1. Shear Force:

(+ ✻) ΣFz = 0 ⇒ V −

Victor Saouma

θ

wRdα = 0 ⇒ V = wrθ

(20.24)

0

Structural Engineering

20–10

ARCHES and CURVED STRUCTURES

Draft

2. Bending Moment:

θ

ΣMR = 0 ⇒ M −

(wRdα)(R sin α) = 0 ⇒ M = wR2 (1 − cos θ)

(20.25)

(wRdα)R(1 − cos α) = 0 ⇒ T = −wR2 (θ − sin θ)

(20.26)

0

3. Torsion:

θ

ΣMT = 0 ⇒ + 0

III Deﬂection are determined last we assume a rectangular cross-section of width b and height d = 2b and a Poisson’s ratio ν = 0.3. 1. Noting that the member will be subjected to both ﬂexural and torsional deformations, we seek to determine the two stiﬀnesses. 3

3

b(2b) = 2. The ﬂexural stiﬀness EI is given by EI = E bd 12 = E 12

2Eb4 3

= .667Eb4 .

3

3. The torsional stiﬀness of solid rectangular sections J = kb d where b is the shorter side of the E E = 2(1+.3) = section, d the longer, and k a factor equal to .229 for db = 2. Hence G = 2(1+ν) 4 4 .385E, and GJ = (.385E)(.229b ) = .176Eb . 4. Considering both ﬂexural and torsional deformations, and replacing dx by rdθ: π π M T δP Rdθ ∆ = δM Rdθ δT + EI GJ z 0 0 ∗

δW Torsion Flexure

δU

(20.27)

∗

where the real moments were given above. 5. Assuming a unit virtual downward force δP = 1, we have δM δT

= R sin θ

(20.28-a)

= −R(1 − cos θ)

(20.28-b)

6. Substituting these expression into Eq. 20.27 wR2 π wR2 ∆ = (R sin θ) (1 − cos θ) Rdθ + 1 EI GJ 0 δP

4 π

= =

=

20.2.1

M

δM

π 0

(θ − sin θ) R(1 − cos θ) Rdθ

δT

T

1 (θ − θ cos θ − sin θ + sin θ cos θ) dθ (sin θ − sin θ cos θ) + .265

wR EI 0 wR4 ( 2. + 18.56 ) EI Flexure Torsion 4

20.56 wR EI

(20.29-a)

Theory

ßAdapted from (Gerstle 1974) Because space structures may have complicated geometry, we must resort to vector analysis1 to determine the internal forces.

18

19

In general we have six internal forces (forces and moments) acting at any section. 1 To

which you have already been exposed at an early stage, yet have very seldom used it so far in mechanics!

Victor Saouma

Structural Engineering

20.2 Curved Space Structures

20–11

Draft 20.2.1.1

Geometry

In general, the geometry of the structure is most conveniently described by a parameteric set of equations y = f2 (θ); z = f3 (θ) (20.30) x = f1 (θ);

20

as shown in Fig. 20.11. the global coordinate system is denoted by X − Y − Z, and its unit vectors are

Figure 20.11: Geometry of Curved Structure in Space denoted2 i, j, k. The section on which the internal forces are required is cut and the principal axes are identiﬁed as N − S − W which correspond to the normal force, and bending axes with respect to the Strong and Weak axes. The corresponding unit vectors are n, s, w.

21

22

The unit normal vector at any section is given by n=

dxi + dyj + dzk dxi + dyj + dzk = ds (dx2 + dy 2 + dz 2 )1/2

(20.31)

The principal bending axes must be deﬁned, that is if the strong bending axis is parallel to the XY plane, or horizontal (as is generally the case for gravity load), then this axis is normal to both the N and Z axes, and its unit vector is n×k (20.32) s= |n×k|

23

24

The weak bending axis is normal to both N and S, and thus its unit vector is determined from w = n×s

20.2.1.2 25

(20.33)

Equilibrium

For the equilibrium equations, we consider the free body diagram of Fig. 20.12 an applied load P 2 All

vectorial quantities are denoted by a bold faced character.

Victor Saouma

Structural Engineering

20–12

ARCHES and CURVED STRUCTURES

Draft

Figure 20.12: Free Body Diagram of a Curved Structure in Space is acting at point A. The resultant force vector F and resultant moment vector M acting on the cut section B are determined from equilibrium ΣF = 0;

P + F = 0;

F = −P

(20.34-a)

ΣM = 0; L×P + M = 0; M = −L×P B

(20.34-b)

where L is the lever arm vector from B to A. The axial and shear forces N, Vs and Vw are all three components of the force vector F along the N, S, and W axes and can be found by dot product with the appropriate unit vectors:

26

N Vs

= =

F·n F·s

(20.35-a) (20.35-b)

Vw

=

F·w

(20.35-c)

Similarly the torsional and bending moments T, Ms and Mw are also components of the moment vector M and are determined from

27

T

=

M·n

(20.36-a)

Ms Mw

= =

M·s M·w

(20.36-b) (20.36-c)

Hence, we do have a mean to determine the internal forces. In case of applied loads we summ, and for distributed load we integrate.

28

Example 20-5: Internal Forces in an Helicoidal Cantilevered Girder, (Gerstle 1974) Determine the internal forces N, Vs , and Vw and the internal moments T, Ms and Mw along the helicoidal cantilevered girder shown in FIg. 20.13 due to a vertical load P at its free end. Solution: 1. We ﬁrst determine the geometry in terms of the angle θ x = R cos θ; Victor Saouma

y = R sin θ;

z=

H θ π

(20.37) Structural Engineering

20.2 Curved Space Structures

Draft

20–13

Figure 20.13: Helicoidal Cantilevered Girder

Victor Saouma

Structural Engineering

20–14

ARCHES and CURVED STRUCTURES

Draft

2. To determine the unit vector n at any point we need the derivatives: dx = −R sin θdθ;

dy = R cos θdθ;

dz =

H dθ π

(20.38)

and then insert into Eq. 20.31 n = =

−R sin θi + R cos θj + H/πk %1/2 R2 sin2 θ + R2 cos2 θ + (H/π)2 1 $ %1/2 [sin θi + cos θj + (H/πR)k] 1 + (H/πR)2

$

(20.39-a) (20.39-b)

K

Since the denominator depends only on the geometry, it will be designated by K. 3. The strong bending axis lies in a horizontal plane, and its unit vector can thus be determined from Eq. 20.32: " " " i j k "" 1 "" H " (20.40-a) n×k = − sin θ cos θ πR " K "" 0 0 1 " 1 = (cos θi + sin θj) (20.40-b) K and the absolute magnitude of this vector |k×n| =

1 K,

and thus

s = cos θi + sin θj

(20.41)

4. The unit vector along the weak axis is determined from Eq. 20.33 " " " i j k "" 1 "" cos θ sin θ 0 "" w = s×n = K "" H " − sin θ cos θ πR 1 H H = sin θi − cos θj + k K πR πR

(20.42-a)

(20.42-b)

5. With the geometry deﬁnition completed, we now examine the equilibrium equations. Eq. 20.34-a and 20.34-b. ΣF = 0; F ΣMb = 0; M where

= −P = −L×P

(20.43-a) (20.43-b)

θ L = (R − R cos θ)i + (0 − R sin θ)j + 0 − H k π

and L×P =

" " i " R "" (1 − cos θ) " 0

j − sin θ 0

k − πθ H R P

" " " " " "

(20.44)

(20.45-a)

P R[− sin θi − (1 − cos θ)j]

(20.45-b)

M = P R[sin θi + (1 − cos θ)j]

(20.46)

= and

Victor Saouma

Structural Engineering

20.2 Curved Space Structures

20–15

Draft

6. Finally, the components of the force F = −P k and the moment M are obtained by appropriate dot products with the unit vectors

Victor Saouma

N

=

1 H P πR F·n = − K

(20.47-a)

Vs

=

F·s = 0

(20.47-b)

Vw

=

1 F·w = − K P

(20.47-c)

T

=

M·n = − PKR (1 − cos θ)

(20.47-d)

Ms

=

M·s = P R sin θ

(20.47-e)

Mw

=

M·w =

PH πK (1

− cos θ)

(20.47-f)

Structural Engineering

20–16

Draft

ARCHES and CURVED STRUCTURES

Victor Saouma

Structural Engineering

Draft Chapter 21

STATIC INDETERMINANCY; FLEXIBILITY METHOD All the examples in this chapter are taken verbatim from White, Gergely and Sexmith

21.1

Introduction

1 A statically indeterminate structure has more unknowns than equations of equilibrium (and equations of conditions if applicable).

2

The advantages of a statically indeterminate structures are: 1. Lower internal forces 2. Safety in redundancy, i.e. if a support or members fails, the structure can redistribute its internal forces to accomodate the changing B.C. without resulting in a sudden failure.

3

Only disadvantage is that it is more complicated to analyse.

4

Analysis mehtods of statically indeterminate structures must satisfy three requirements

Equilibrium Force-displacement (or stress-strain) relations (linear elastic in this course). Compatibility of displacements (i.e. no discontinuity) 5

This can be achieved through two classes of solution

Force or Flexibility method; Displacement or Stiﬀness method 6 The ﬂexibility method is ﬁrst illustrated by the following problem of a statically indeterminate cable structure in which a rigid plate is supported by two aluminum cables and a steel one. We seek to determine the force in each cable, Fig. 21.1

1. We have three unknowns and only two independent equations of equilibrium. Hence the problem is statically indeterminate to the ﬁrst degree. 2. Applying the equations of equilibrium right left ΣMz = 0; ⇒ PAl = PAl

ΣFy = 0; ⇒ 2PAl + PSt = P Thus we eﬀectively have two unknowns and one equation.

(21.1-a)

21–2

Draft

STATIC INDETERMINANCY; FLEXIBILITY METHOD

Figure 21.1: Statically Indeterminate 3 Cable Structure 3. We need to have a third equation to solve for the three unknowns. This will be derived from the compatibility of the displacements in all three cables, i.e. all three displacements must be equal: (21.2) DAl = DSt 4. Finally, those isplacements are obtained from the Force-Displacement relations:  σ = PA  PL ⇒ ∆L = ε = ∆L L  AE σ ε = E PAl L PSt L PAl (EA)Al = ⇒ = EAl AAl ESt ASt PSt (EA)St DAl

(21.3)

(21.4)

DSt

or −(EA)St PAl + (EA)Al PSt = 0

(21.5)

5. Solution of Eq. 21.1-a and 21.5 yields PAl P 2 1 = PSt 0 −(EA)St (EA)Al −1 PAl 2 1 P ⇒ = PSt −(EA) St (EA)Al 0 1 P (EA)Al −1 = 0 2(EA)Al + (EA)St (EA)St 2

Determinant 6. We observe that the solution of this sproblem, contrarily to statically determinate ones, depends on the elastic properties. 7

Another example is the propped cantiliver beam of length L, Fig. 21.2 1. First we remove the roller support, and are left with the cantilever as a primary structure. 2. We then determine the deﬂection at point B due to the applied load P using the virtual force method M 1.D = δM dx (21.6-a) EI L/2 L/2 PL −px = dx + + P x (−x)dx (21.6-b) 0 − EI 2 0 0

Victor Saouma

Structural Engineering

21.1 Introduction

21–3

Draft

P

C

B

L/2

A

L/2

x

x P

D

Primary Structure Under Actual Load

-PL

PL/2

f BB

Primary Structure Under Redundant Loading

-(1)L/2

1

QL/2

+ Bending Moment Diagram

PL/4

Figure 21.2: Propped Cantilever Beam

Victor Saouma

Structural Engineering

21–4

Draft

STATIC INDETERMINANCY; FLEXIBILITY METHOD

= = =

PL x + P x2 dx 2 0 "L/2 2 P x3 "" 1 P Lx + EI 4 3 "0 1 EI

L/2

(21.6-c) (21.6-d)

5 P L3 48 EI

(21.6-e)

3. We then apply a unit load at point B and solve for the displacement at B using the virtual force method M 1fBB = dx (21.7-a) δM EI L/2 x = (x) dx (21.7-b) EI 0 (1)L3 (21.7-c) = 24EI 4. Then we argue that the displacement at point B is zero, and hence the displacement fBB should be multiplied by RB such that RB fBB = D (21.8) to ensure compatibility of displacements, hence RB =

21.2

D = fBB

5 3 48 P L EI (1)L3 24EI

=

5 2P

(21.9)

The Force/Flexibility Method

8 Based on the previous two illustrative examples, we now seek to develop a general method for the linear elastic analysis of statically indeterminate structures.

1. Identify the degree of static indeterminancy (exterior and/or interior) n. 2. Select n redundant unknown forces and/or couples in the loaded structure along with n corresponding releases (angular or translation). 3. The n releases render the structure statically determinate, and it is called the primary structure. 4. Determine the n displacements in the primary structure (with the load applied) corresponding to the releases, Di . 5. Apply a unit force at each of the releases j on the primary structure, without the external load, and determine the displacements in all releases i, we shall refer to these displacements as the ﬂexibility coeﬃcients, fij , i.e. displacement at release i due to a unit force at j 6. Write the compatibility of displacement relation   f11 f12 · · · f1n   R1  f21 f22 · · · f2n   R2    ··· ··· ··· ···  ···   fn1 fn2 · · · fnn Rn

[f ]

R

   D1            D2 =    ···         Dn D

D10 D20 ··· Dn0

      

(21.10)

D0i

and 0 1 [R] = [f ]−1 D − D0i

(21.11)

Note that D0i is the vector of initial displacements, which is usually zero unless we have an initial displacement of the support (such as support settlement). Victor Saouma

Structural Engineering

21.3 Short-Cut for Displacement Evaluation

21–5

Draft

7. The reactions are then obtained by simply inverting the ﬂexibility matrix.

9

Note that from Maxwell-Betti’s reciprocal theorem, the ﬂexibility matrix [f ] is always symmetric.

21.3

Short-Cut for Displacement Evaluation

Since deﬂections due to ﬂexural eﬀects must be determined numerous times in the ﬂexibility method, Table 21.1 may simplify some of the evaluation of the internal strain energy. You are strongly discouraged to use this table while still at school!.

10

g2 (x) a

a ❍❍ ❍ L

a

✥✥✥ b

g1 (x)

L

L

L

Lac

Lac 2

Lc(a+b) 2

Lac 2

Lac 3

Lc(2a+b) 6

Lac 2

Lac 6

Lc(a+2b) 6

La(c+d) 2

La(2c+d) 6

La(2c+d)+Lb(c+2d) 6

La(c+4d+e) 6

La(c+2d) 6

La(c+2d)+Lb(2d+e) 6

c c ❍❍ ❍ L ✟✟ c ✟ L ✥✥✥ d c L c

d e L

L

Table 21.1: Table of

g1 (x)g2 (x)dx 0

21.4

Examples

Example 21-1: Steel Building Frame Analysis, (White et al. 1976) A small, mass-produced industrial building, Fig. 21.3, is to be framed in structural steel with a typical cross section as shown below. The engineer is considering three diﬀerent designs for the frame: (a) for poor or unknown soil conditions, the foundations for the frame may not be able to develop any dependable horizontal forces at its bases. In this case the idealized base conditions are a hinge at one of the bases and a roller at the other; (b) for excellent soil conditions with properly designed foundations, the bases of the frame legs will have no tendency to move horizontally, and the idealized base condition is that of hinges at both points A and D; and (c) a design intermediate to the above cases, with a steel tie member capable of carrying only tension running between points A and D in the ﬂoor of the building. The foundations would not be expected to provide any horizontal restraint for this latter case, and the hinge-roller details at points A and D would apply. Critical design loads for a frame of this type are usually the gravity loads (dead load + snow load) and the combination of dead load and wind load. We will restrict our attention to the ﬁrst combination, and will use a snow load of 30 psf and an estimated total dead load of 20 psf. With frames spaced at 15 ft on centers along the length of the building, the design load is 15 (30 + 20) = 750 lb/ft.

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Structural Engineering

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Figure 21.3: If the frame is made of steel beam sections 21 in. deep and weighing 62 lb/ft of length (W 21 × 62), and the tie member for design (c) is tentatively chosen as a 2-in.2 bar, determine the bending moment diagrams for the three designs and discuss the alternate solutions. Solution: Structure a This frame is statically determinate since it has three possible unknown external forces acting on it, and the bending moment is shown in Fig. 21.6-a. Structure b Hinging both legs of the frame results in another unknown force, making the structure statically indeterminate to the ﬁrst degree (one redundant). 1. A lateral release at point A is chosen. with the redundant shearing force R1 . The displacement D1Q in the primary structure, as a result of the real loading, is shown in Figure 21.4-a. D1Q is computed by virtual work. 2. The virtual force system in Figure 21.4-b produces a virtual bending moment δM , which is uniform

Figure 21.4: across the top member of the frame. The virtual moment acting through the real angle changes gives the internal work term 40 M dx (21.12) δM δM dφ = EI 0

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Structural Engineering

21.4 Examples

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Draft

3. Equating this to the external virtual work of 1 · D1Q , we have

40

1 · D1Q = 0

(12)(1/2)(.75)(Lx − x2 ) dx EI

(21.13)

or

48, 000 k ft3 → (21.14) EI 4. The equation of consistent displacement is D1Q +f11 R1 = 0. The ﬂexibility coeﬃcient f11 is computed by applying a unit horizontal force at the release and determining the displacement at the same point. 5. It is seen that the real loading and the virtual loading are identical for this calculation, and 12 2 20 2 x dx 12 dx + (21.15) 1 · f11 = 2 EI EI 0 0 D1Q =

or f11 = 6. Solving for R1

6, 912 k ft3 → EI

(21.16)

1 [48, 000 + 6, 912R1] = 0 EI

(21.17)

R1 = 6.93 k ←

(21.18)

or 7. The bending moment diagram is given below Structure c The frame with the horizontal tie between points A and D has three unknown external forces. However, the structure is statically indeterminate to the ﬁrst degree since the tie member provides one degree of internal redundancy. 8. The logical release to choose is a longitudinal release in the tie member, with its associated longitudinal displacement and axial force. 9. The primary structure Fig. 21.5 is the frame with the tie member released. The compatibility

Figure 21.5: equation is based on the fact that the displacement at the release must be zero; that is, the relative displacement of the two sections of the tie at the point of release must be zero, or D1Q + f11 R1 = 0 Victor Saouma

(21.19) Structural Engineering

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STATIC INDETERMINANCY; FLEXIBILITY METHOD

where D1Q = displacement at release 1 in the primary structure, produced by the loading, f11 = relative displacement at release 1 for a unit axial force in the tie member, R1 = force in the tie member in the original structure. 10. Virtual work is used to determine both displacement terms. 11. The value of D1Q is identical to the displacement D1Q computed for structure (b) because the tie member has no forces (and consequently no deformations) in the primary structure. Thus D1Q =

(48, 000)(1, 728) = 2.08 in✲ (30 · 103 )(1, 327)

(21.20)

12. The ﬂexibility coeﬃcient f11 is composed of two separate eﬀects: a ﬂexural displacement due to the ﬂexibility of the frame, and the axial displacement of the stressed tie member. The virtual and real loadings for this calculation are shown in Figure 21.5. The virtual work equation is 12 2 20 x1 dx1 (12)2 dx pL + (21.21-a) 1 · f11 = 2 + δP EI EI EA 0 0 6, 912 1(1)(40) = + (21.21-b) EI EA 40(12) (6, 912)(1, 728) + (21.21-c) = (30 · 103 )(1, 327) (30 · 103 )(2) = 0.300 + 0.008 = 0.308 (21.21-d) and f11 = 0.308in./k

(21.22)

13. The equation of consistent deformation is D1Q + f11 R1 = 0

(21.23)

2.08 − .308R1 = 0

(21.24)

R1 = 6.75 k (tension)

(21.25)

or or 14. The two displacement terms in the equation must carry opposite signs to account for their diﬀerence in direction. Comments The bending moment in the frame, Figure 21.6-c diﬀers only slightly from that of structure (b). In other words, the tie member has such high axial stiﬀness that it provides nearly as much restraint as the foundation of structure (b). Frames with tie members are used widely in industrial buildings. A lesson to be learned here is that it is easy to provide high stiﬀness through an axially loaded member.

Figure 21.6: The maximum moment in frames (b) and (c) is about 55% of the maximum moment in frame (a). This eﬀect of continuity and redundancy is typical – the positive bending moments in the members are lowered while the joint moments increase and a more economical design can be realized. Finally, we should notice that the vertical reactions at the bases of the columns do not change with the degree of horizontal restraint at the bases. A question to ponder is “Does this type of reaction behavior occur in all frames, or only in certain geometrical conﬁgurations?” Victor Saouma

Structural Engineering

21.4 Examples

21–9

Draft Example 21-2: Analysis of Irregular Building Frame, (White et al. 1976)

The structural steel frame for the Church of the Holy Spirit, Penﬁeld, New York is shown below. In this example we will discuss the idealization of the structure and then determine the forces and bending moments acting on the frame. Solution: 1. A sectional view of the building is given in Figure 21.8-a.

Figure 21.7: The two main horizontal members of the frame are supported at points A and D by masonry walls. 2. The connection used at these points is not intended to transmit axial forces from the frame to the wall; accordingly, the axial forces in the horizontal members are assumed to be zero and the joints at A and D are idealized as rollers that transmit vertical forces only. 3. The base joint E is designed to resist both horizontal and vertical loads, but not moment, and is assumed to be a hinge. 4. Finally, joints B and C are designed to provide continuity and will be taken as rigid; that is, the angles of intersection of the members at the joint do not change with applied loading. 5. The frame is simpliﬁed for analysis by removing the small 4-in. wide ﬂange members EF and F G and replacing their load eﬀect by applying the roof load which acts on EF directly to the segment AG. 6. The idealized frame is shown in Figure 21.8-b. 7. The dead load on the higher portion of the frame is wAB = 25 psf times the frame spacing of 13.33 ft, or wAB = 25(13.33) = 334 lb/ft along the frame. 8. The dead load on CD is less because the weight of the frame member is substantially smaller, and the dead load is about 19 psf, or wCD = 19(13.33) = 254 lb/ft of frame. 9. Snow load is 35 psf over both areas, or w = 35(13.33) = 467 lb/ft. 10. Total loads are then

Victor Saouma

Structural Engineering

21–10

Draft

STATIC INDETERMINANCY; FLEXIBILITY METHOD

Figure 21.8:

Victor Saouma

Structural Engineering

21.4 Examples

21–11

Draft

Member AB: Member CD:

w = 334 + 467 = 800 lb/ft = 0.8 k/ft w = 254 + 467 = 720 lb/ft = 0.72 k/ft

11. The frame has four unknown reaction components and therefore has one redundant. Although several diﬀerent releases are possible, we choose an angular (bending) release at point B. 12. The resulting primary structure is shown in Figure 21.8-c, where the redundant quantity R1 is the bending moment at point B. 13. The equation of compatibility is (21.26) θ1Q + θ11 R1 = 0 where θ1Q is the relative angular rotation corresponding to release 1 as produced by the actual loading, and θ11 is the ﬂexibility coeﬃcient for a unit moment acting at the release. 14. From virtual work we have M dx (21.27) 1 · θ1Q = δM dφ = δM EI

and

1 · θ11 =

δM dφ =

δM

M dx EI

(21.28)

where m is a real unit load and δM and M are deﬁned in Figure 21.8-d and e. 15. Then

θ1Q

= + =

1 (EI)AB

0

61.42

x (24.6x − 0.4x2 )dx 61.42

(21.29-a)

21.33 1 x (7.68x − 0.36x2 )dx (EI)CD 0 21.33 7750 295 + (EI)AB (EI)CD

(21.29-b) (21.29-c)

16. with IAB = 4, 470 in.4 and ICD = 290 in.4 θ1Q =

2.75 1.733 1.0107 + = E E E

(21.30)

17. Similarly θ11

=

x 2 dx 61.42 0 21.33 8 1 x 2 1 + dx + (1)2 dx (EI)CD 0 21.33 (EI)BC 0 1 (EI)AB

18. with IBC = 273 in4 θ11 =

61.42

20.5 7.11 8.0 0.0585 + + = (EI)AB (EI)CD (EI)BC E

(21.31-a) (21.31-b)

(21.32)

Note that the numerators of θ1Q and θ11 have the units k-ft2 /in4 . 19. Applying the compatibility equation, 2.75 0.0585 + R1 = 0 E E

(21.33)

and the bending moment at point B is R1 = −47.0 ft-k . The reactions and moments in the structure are given in Figure 21.8-f.

Example 21-3: Redundant Truss Analysis, (White et al. 1976) Victor Saouma

Structural Engineering

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Draft

STATIC INDETERMINANCY; FLEXIBILITY METHOD

Figure 21.9: Determine the bar forces in the steel truss shown below using the force method of analysis. The truss is part of a supporting tower for a tank, and the 20 kN horizontal load is produced by wind loading on the tank. Solution: 1. Applying the criteria for indeterminacy, 2 × 4 = 8 equations, 6 members + 3 reactions ⇒ one degree of indeterminacy. A longitudinal release in any of the six bars may be chosen. 2. Because the truss members carry only axial load, a longitudinal release is identical to actually cutting the member and removing its axial force capability from the truss. 3. In analyzing trusses with double diagonals it is both convenient and customary to select the release in one of the diagonal members because the resulting primary structure will be the conventional truss form to which we are accustomed. 4. Choosing the diagonal member BC for release, we cut it and remove its axial stiﬀness from the structure. The primary structure is shown in Figure 21.9-b. 5. The analysis problem reduces to applying an equation of compatibility to the changes in length of the release member. The relative displacement D1Q of the two cut ends of member BC, as produced by the real loading, is shown in Figure 21.9-c. 6. The displacement is always measured along the length of the redundant member, and since the redundant is unstressed at this stage of the analysis, the displacement D1Q is equal to the relative displacement of joint B with respect to joint C. 7. This displacement must be eliminated by the relative displacements of the cut ends of member BC when the redundant force is acting in the member. The latter displacement is written in terms of the axial ﬂexibility coeﬃcient f11 , and the desire equation of consistent deformation is D1Q + f11 R1 = 0

(21.34)

1 · D1Q = ΣδP (∆l) = ΣδP (P L/AE)

(21.35)

8. The quantity D1Q is given by

Victor Saouma

Structural Engineering

21.4 Examples

21–13

where δP and P are given in Figure 21.9-d and c, respectively. 9. Similarly, f11 = ΣδP (P L/AE)

(21.36)

Draft

10. Evaluating these summations in tabular form: Member AB BC CD AC AD BC

P 0 0 +20 +20 −28.28 0

δP −0.707 −0.707 −0.707 −0.707 +1 +1

L 3 3 3 3 4.242 4.242

δP P L 0 0 −42.42 −42.42 −119.96 0 -204.8

δP P L 1.5 1.5 1.5 1.5 4.242 4.242 14.484

11. Since A = constant for each member D1Q = ΣδP then

−204.8 PL 14.484 =− andf11 = AE AE AE

(21.37)

1 [−204.8 + 14.484R1] = 0 AE

(21.38)

12. The solution for the redundant force value is R1 = 14.14 kN . 13. The ﬁnal values for forces in each of the truss members are given by superimposing the forces due to the redundant and the forces due to the real loading. 14. The real loading forces are shown in Figure 21.9-c while the redundant force eﬀect is computed by multiplying the member forces in Figure 21.9-d by 2.83, the value of the redundant. 15. It is informative to compare the member forces from this solution to the approximate analysis obtained by assuming that the double diagonals each carry half the total shear in the panel. The comparison is given in Figure 21.10; it reveals that the approximate analysis is the same as the exact

Figure 21.10: analysis for this particular truss. The reason for this is that the stiﬀness provided by each of the diagonal members (against “shear” deformation of the rectangular panel) is the same, and therefore they each carry an equal portion of the total shear across the panel. 16. How would this structure behave if the diagonal members were very slender?

Example 21-4: Truss with Two Redundants, (White et al. 1976) Victor Saouma

Structural Engineering

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STATIC INDETERMINANCY; FLEXIBILITY METHOD

Another panel with a second redundant member is added to the truss of Example 21-3 and the new truss is supported at its outermost lower panel points, as shown in Figure 21.11. The truss, which is similar in form to trusses used on many railway bridges, is to be analyzed for bar forces under the given loading. Solution: 1. The twice redundant truss is converted to a determinate primary structure by releasing two members of the truss; we choose two diagonals (DB and BF ). 2. Releasing both diagonals in a single panel, such as members AE and DB, is inadmissible since it leads to an unstable truss form. 3. The member forces and required displacements for the real loading and for the two redundant forces

Figure 21.11: in members DB and BF are given in Figure 21.11. 4. Although the real loading ordinarily stresses all members of the entire truss, we see that the unit forces corresponding to the redundants stres only those members in the panel that contains the redundant; all other bar forces are zero. 5. Recognizing this fact enables us to solve the double diagonal truss problem more rapidly than a frame with multiple redundants. 6. The virtual work equations for computing the six required displacements (two due to load and four ﬂexibilities) are PL 1 · D1Q = ΣδP 1 (21.39-a) AE PL 1 · D2Q = ΣδP 2 (21.39-b) AE P 1L 1 · f11 = ΣδP 1 (21.39-c) AE P 1L 1 · f21 = ΣδP 2 (21.39-d) AE = f21 by the reciprocal theorem (21.39-e) f12 Victor Saouma

Structural Engineering

21.4 Examples

21–15

Draft

1 · f22

= ΣδP 2

P 2L AE

(21.39-f)

7. If we assume tension in a truss member as postive, use tensile unit loads when computing the ﬂexibility coeﬃcients corresponding to the redundants, and let all displacement terms carry their own signs, then in the solution for the redundants a positive value of force indicates tension while a negative value means the member is in compression. 8. The calculation of f22 involves only the six members in the left panel of the truss; f21 involves only member BE. 9. The simple procedures used for performing the displacement analyses, as summarized in tabular form in Table 21.2, leads one quickly to the compatibility equations which state that the cut ends of both Member AB BC CF EF DE AD AE BE CE BD BF

P -9.5 -9.5 -9.5 0 +4 -5.5 +7.78 .15 +13.43 0 0

P1 -0.707 0 0 0 -0.707 -0.707 +1 -0.707 0 +1 0

P2 0 -0.707 -0.707 -0.707 0 0 0 -0.707 +1 0 +1

L 120 120 120 120 120 120 170 120 170 170 170

D1Q δP 1 P L +806 0 0 0 -340 +466 +1,322 +1,272 0 0 0 +3,528

D2Q δP 2 P L 0 +806 +806 0 0 0 0 +1272 +2,280 0 0 +5,164

f11 δP 1 P 1 L 60 0 0 0 60 60 170 60 0 170 0 +580

f21 δP 2 P 1 L 0 0 0 0 0 0 0 60 0 0 0 +60

f22 δP 2 P 2 L 0 60 60 60 0 0 0 60 170 0 170 +580

Table 21.2: redundant members must match (there can be no gaps or overlaps of members in the actual structure). 10. The equations are

or

1 AE

D1Q + f11 R1 + f12 R2

= 0

(21.40-a)

D2Q + f21 R1 + f22 R2

= 0

(21.40-b)

580 60 60 580

R1 R2

=−

1 AE

3, 528 5, 164

(21.41)

and R1

=

5.20 k

(21.42-a)

R2

=

−8.38 k

(21.42-b)

11. The ﬁnal set of forces in the truss is obtained by adding up, for each member, the three separate eﬀects. In terms of the forces shown in Figure 21.11 and Table 21.2, the force in any member is given by F = P + R1 P 1 + R2 P 2 . The ﬁnal solution is given in Figure 21.11-e. The truss is part of a supporting tower for a tank, and the 20 kN horizontal load is produced by wind loading on the tank. 12. NOTE: A mixture of internal redundant forces and external redundant reactions is no more diﬃcult than the preceding example. Consider the two-panel truss of this example modiﬁed by the addition of another reaction component at joint E, Figure 21.12. The three releases for this truss can be chosen from a number of possible combinations: diagonals DB and BF and the reaction at E; the same two diagonals and the reaction at F ; the same diagonals and the top chord member BC, etc. The only requirement to be met is that the primary structure be stable and statically determinate. 13. For any set of releases there are four new displacement components: the displacement at the third release resulting from the actual load on the primary structure, and the three ﬂexibility coeﬃcients f31 , f32 , and f33 . Judicious choice of releases often results in a number of the ﬂexibility coeﬃcients being zero Victor Saouma

Structural Engineering

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Figure 21.12:

Example 21-5: Analysis of Nonprismatic Members, (White et al. 1976) The nonprismatic beam of Figure 21.13-a is loaded with an end moment MA at its hinged end A. Determine the moment induced at the ﬁxed end B by this loading.

Figure 21.13: Solution: 1. The beam has one redundant force; we select MB as the redundant R1 , and obtain the primary structure shown in Figure 21.13-c. It can be shown that the ﬂexibility co-eﬃcients for unit moments applied at each end are those shown in Fig. 21.13-d and e, with a sign convention of counterclockwise as positive. 2. The equation of consistent displacements at B is −

3 l MA l + R1 = 0 8EI 16 EI

(21.43)

and the value of MB is Victor Saouma

Structural Engineering

21.4 Examples

21–17

Draft

2 MA (21.44) 3 3. The resulting moment diagram is given in Figure 21.13-f. We note that the inﬂection point is 0.40l from the ﬁxed end. If the beam had a uniform value of I across its span, the inﬂection point would be L/3 from the ﬁxed end. Thus the inﬂection point shifts toward the section of reduced stiﬀness. 4. The end rotation θA is given by l 11 MA l 5 MA l 1 2 − MA = (21.45) θa = 16 EI 8 3 EL 48 EI MB = R1 =

5. The ratio of applied end moment to rotation. MA /θA , is called the rotational stiﬀness and is EI 48 EI MA = 4.364 = θA 11 l l

(21.46)

6. If we now reverse the boundary conditions, making A ﬁxed and B hinged, and repeat the analysis for an applied moment MB , the resulting moment diagram will be as given in Figure 21.13-h. The moment induced at end A is only 40% of the applied end moment MB . The inﬂection point is 0.286l from the ﬁxed end A. The corresponding end rotation θB in Figure 21.13-g is θB = 7. The rotational stiﬀness

MB θB

11 MB l 80 EI

(21.47)

is EI MB 80 EI = 7.272 = θB 11 l l

(21.48)

8. A careful comparison of the rotational stiﬀnesses, and of the moment diagrams in Figures 21.13-f and h, illustrate the fact that ﬂexural sections of increased stiﬀness attract more moment, and that inﬂection points always shift in the direction of decreased stiﬀness. 9. The approach illustrated here may be used to determine moments and end rotations in any type of nonprismatic member. The end rotations needed in the force analysis may be calculated by either virtual work or moment area (or by other methods). Complex variations in EI are handled by numerical integration of the virtual work equation or by approximating the resultant M/EI areas and their locations in the moment area method.

Example 21-6: Fixed End Moments for Nonprismatic Beams, (White et al. 1976) The beam of example 21-5, with both ends ﬁxed, is loaded with a uniform load w, Figure 21.14-a. Determine the ﬁxed end moments MA and MB . Solution: 1. The beam has two redundant forces and we select MA and MB . Releasing these redundants, R1 and R2 , the primary structure is as shown in Figure 21.14-c. 2. The equations of consistent deformations are D1Q + f11 R1 + f12 R2

= 0

(21.49-a)

D2Q + f21 R1 + f22 R2

= 0

(21.49-b)

where R1 is MA and R2 is MB . 3. The values of D1Q and D2Q , the end rotations produced by the real loading on the primary structure, can be computed by the virtual work method. 4. The ﬂexibility coeﬃcients are also separately derived (not yet in these notes) and are given in Figures 21.14-d and e of the previous example.

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Structural Engineering

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STATIC INDETERMINANCY; FLEXIBILITY METHOD

Figure 21.14: 5. We deﬁne counterclockwise end moments and rotations a positive and obtain 5 l wl3 −0.352 R1 − 81 16 = 3 R2 EI − 18 EI +0.0273 16

(21.50)

from which R1

=

MA = 0.0742wl2

(21.51-a)

R2

=

MB = −0.0961wl2

(21.51-b)

6. The stiﬀer end of the beam attracts 30% more than the ﬂexible end. 7. For a prismatic beam with constant I, the ﬁxed end moments are equal in magnitude (MA = −MB = wl2 /12) and intermediate in value between the two end moments determined above. 8. Fixed end moments are an essential part of indeterminate analysis based on the displacement (stiﬀness) method and will be used extensively in the Moment Distribution method.

Example 21-7: Rectangular Frame; External Load, (White et al. 1976) Solution: 1. The structure is statically indeterminate to the third degree, and the displacements (ﬂexibility terms) are shown in Fig. 21.15 2. In order to evaluate the 9 ﬂexility terms, Fig. 21.16 we refer to Table 21.3 3. Substituting h = 10 ft, L = 20 ft, and EIb = EIc = EI, the ﬂexibility matrix then becomes   2, 667 3, 000 −300 1  3, 000 6, 667 −400  [f ] = (21.52) EI −300 −400 40 and the vector of displacements for the primary structure is   −12, 833  1  −31, 333 {D} =  EI  1, 800

(21.53)

where the units are kips and feet. Victor Saouma

Structural Engineering

21.4 Examples

21–19

Draft

Figure 21.15:

Figure 21.16: Deﬁnition of Flexibility Terms for a Rigid Frame

Victor Saouma

Structural Engineering

Victor Saouma

D3Q

D2Q

D1Q

f33

f32

f31

f23

f22

❆❆

✥✥✥

✁✁

✥✥✥

✥✥✥

✄✄ −h

2

❆❆

❆❆

❛❛ ❛

❛❛ ❛

❛❛ ❛

✥✥✥

✁✁

✥✥✥

✁✁

+ h(2h+10L−20) 2

− Lh(2h+10L−20) 2

(2h+15L−30) 6

+h

−hL

2

− h2

−hL

+L2 h

+ h2L

2

(

2

(30−10L) 6

− L(20−5L) 2

+L

+ Lh(20−5L) 2

+L

2

− L2

−Lh

2

− L2

3

+ L3

2

+ L2 h

−Lh

2

+ L2 h

+Lh2

δM M dx BC

2

2

0

0

0

+h

0

− h2

0

0

0

− h2

0

3

+ h3

CD

2

(2h+15L−30) 6EIc

− h(2h+10L−20) 2EIc

−

+

+

+

2h + EI c

− Lh(2h+10L−20) 2EIc

−h

−

−

−

+

+

−

+

+

hL − EI c

2

h − EI c

hL − EI c

2

h +L EIc

2

h L + 2EI c

2

h − EI c

2

h L + 2EI c

3

M δM EI dx Total

2h + 3EI c

(

Table 21.3: Displacement Computations for a Rectangular Frame

✄✄

✄✄

2

✥✥✥

f21

− h2

❆❆

✁✁

f13

2

+ h2L

✥✥✥

❆❆

✁✁

f12

3

+ h3

❆❆

✁✁

❆❆

✁✁

AB

f11

M

δM

D

L(20−5L) 2EIb

L2 (30−10L) 6EIb

Lh(20−5L) 2EIb

L EIb

L2 2EIb

Lh EIb

L2 2EIb

L3 3EIb

L2 h 2EIb

Lh EIb

L2 h 2EIb

Lh2 EIb

Draft

21–20 STATIC INDETERMINANCY; FLEXIBILITY METHOD

Structural Engineering

21.4 Examples

21–21

Draft

4. The inverse of the ﬂexibility matrix is 

[f ]−1

 2.40 0.00 18.00 = 10−3 EI  0.00 0.375 3.750  18.000 3.750 197.5

(21.54)

5. Hence the reactions are determined from         2.40 0.00 18.00  −1.60   R1   −12, 833  1 +5.00 R2 −31, 333 = 10−3 EI  0.00 0.375 3.750  {R} = =      EI  −7.00 R3 1, 800 18.000 3.750 197.5

(21.55)

Example 21-8: Frame with Temperature Eﬀectsand Support Displacements, (White et al. 1976)

The single bay frame, of example 21-7, has a height h = 10 ft and span L = 20 ft and its two suports rigidly connected and is constructed of reinforced concrete. It supports a roof and wall partitions in such a manner that a linear temperature variation occurs across the depth of the frame members when inside and outside temperatures diﬀer. Assume the member depth is constant at 1 ft, and that the structure was built with ﬁxed bases A and D at a temperature of 85◦ F. The temperature is now 70◦ F inside and 20◦ F outside. We wish to determine the reactions at D under these conditions. Assume that the coeﬃcient of linear expansion of reinforced concrete is α = 0.0000055/◦F. Solution: 1. Our analysis proceeds as before, using Equation 21.11 with the [D] vector interpreted appropriately. The three releases shown in Fig. 21.16 will be used. 2. The ﬁrst stage in the analysis is the computation of the relative displacements D1∆ , D2∆ , D3∆ of the primary structure caused by temperature eﬀects. These displacements are caused by two eﬀects: axial shortening of the members because of the drop in average temperature (at middepth of the members), and curvature of the members because of the temperature gradient. 3. In the following discussion the contributions to displacements due to axial strain are denoted with a single prime () and those due to curvature by a double prime (). 4. Consider the axial strain ﬁrst. A unit length of frame member shortens as a result of the temperature decrease from 85◦ F to 45◦ F at the middepth of the member. The strain is therefore α∆T = (0.0000055)(40) = 0.00022

(21.56)

5. The eﬀect of axial strain on the relative displacements needs little analysis. The horizontal member shortens by an amount (0.00022)(20) = 0.0044 ft. The shortening of the vertical members results in no relative displacement in the vertical direction 2. No rotation occurs. 6. We therefore have D1∆ = −0.0044 ft, D2∆ = 0, and D3∆ = 0. 7. The eﬀect of curvature must also be considered. A frame element of length dx undergoes an angular strain as a result of the temperature gradient as indicated in Figure 21.17. The change in length at an extreme ﬁber is I = α∆T dx = 0.0000055(25)dx = 0.000138dx (21.57) 8. with the resulting real rotation of the cross section dφ = I/0.5 = 0.000138dx/0.5 = 0.000276dx radians

(21.58)

9. The relative displacements of the primary structure at D are found by the virtual force method. 10. A virtual force δQ is applied in the direction of the desired displacement and the resulting moment diagram δM determined. Victor Saouma

Structural Engineering

21–22

STATIC INDETERMINANCY; FLEXIBILITY METHOD

Draft

Figure 21.17: 11. The virtual work equation

δQD =

δM dφ

(21.59)

is used to obtain each of the desired displacements D. 12. The results, which you should verify, are D1∆

=

0.0828 ft

(21.60-a)

D2∆

=

0.1104 ft

(21.60-b)

D3∆

=

−0.01104 radians

(21.60-c)

13. Combining the eﬀects of axial and rotational strain, we have D1∆ D2∆ D3∆

= = =

D1∆ + D1∆ D2∆ + D2∆ D3∆ + D3∆

= = =

0.0784 ft 0.1104 ft −0.01104 radians

(21.61)

14. We now compute the redundants caused by temperature eﬀects: [R] = [f ]−1 (−[D])

(21.62)



   R1 −0.0784 $ %  R2  = 10−3 EI 18.0 3.75 197.5  −0.1104  = +0.0106 · 10−3 EI +0.355 R3 +0.01104

(21.63)

where the units are feet and kips. 15. You should construct the moment diagram for this structure using the values of the redundants found in the analysis. 16. Notice that the stiﬀness term EI does not cancel out in this case. Internal forces and reactions in a statically indeterminate structure subject to eﬀects other than loads (such as temperature) are dependent on the actual stiﬀnesses of the structure. 17. The eﬀects of axial strain caused by forces in the members have been neglected in this analysis. This is usual for low frames where bending strain dominates behavior. To illustrate the signiﬁcance of this assumption, consider member BC. We have found R1 = 10.6 · 10−6 EI k. The tension in BC has this same value, resulting in a strain for the member of 10.6 · 10−6 EI/EA. For a rectangular member, I/A = (bd3 /12)(bd) = d2 /12. In our case d = 1 ft, therefore the axial strain is 10.6 · 10−6 (0.0833) = 8.83 · 10−7 , which is several orders of magnitude smaller than the temperature strain computed for the same member. We may therefore rest assured that neglecting axial strain caused by forces does not aﬀect the values of the redundants in a signiﬁcant manner for this structure. 18. Now consider the eﬀects of foundation movement on the same structure. The indeterminate frame behavior depends on a structure that we did not design: the earth. The earth is an essential part of nearly all structures, and we must understand the eﬀects of foundation behavior on structural behavior. Victor Saouma

Structural Engineering

21.4 Examples

21–23

Draft

For the purposes of this example, assume that a foundation study has revealed the possibility of a clockwise rotation of the support at D of 0.001 radians and a downward movement of the support at D of 0.12 ft. We wish to evaluate the redundants R1 , R2 , and R3 caused by this foundation movement. 19. No analysis is needed to determine the values of D1∆ , D2∆ , and D3∆ for the solution of the redundants. These displacements are found directly from the support movements, with proper consideration of the originally chosen sign convention which deﬁned the positive direction of the relative displacements. From the given support displacements, we ﬁnd D1∆ = 0, D2∆ = +0.12 ft, and D3∆ = −0.001 radians. Can you evaluate these quantities for a case in which the support movements occurred at A instead of D? 20. The values of the redundants is given by [R] = [f ]−1 (−[D])

(21.64)

 R1 $ % −0.12 18.0 −6  R2  = 10−3 EI 18.0 3.75 197.5 = 10 EI −252.5 +0.001 R3 

(21.65)

with units in kips and feet. 21. A moment diagram may now be constructed, and other internal force quantities computed from the now known values of the redundants. The redundants have been valuated separately for eﬀects of temperature and foundation settlement. These eﬀects may be combined with those due to loading using the principle of superposition.

Example 21-9: Braced Bent with Loads and Temperature Change, (White et al. 1976)

The truss shown in Figure 21.18 reperesents an internal braced bent in an enclosed shed, with lateral loads of 20 kN at the panel points. A temperature drop of 30◦ C may occur on the outer members (members 1-2, 2-3, 3-4, 4-5, and 5-6). We wish to analyze the truss for the loading and for the temperature eﬀect. Solution: 1. The ﬁrst step in the analysis is the deﬁnition of the two redundants. The choice of forces in diagonals 2-4 and 1-5 as redundants facilitates the computations because some of the load eﬀects are easy to analyze. Figure ??-b shows the deﬁnition of R1 and R2 . 2. The computations are organized in tabular form in Table 21.4. The ﬁrst column gives the bar forces P in the primary structure caused by the actual loads. Forces are in kN. Column 2 gives the force in each bar caused by a unit load (1 kN) corresponding to release 1. These are denoted p1 and also represent the bar force q¯1 /δQ1 caused by a virtual force δQ1 applied at the same location. Column 3 lists the same quantity for a unit load and for a virtual force δQ2 applied at release 2. These three columns constitute a record of the truss analysis needed for this problem. 3. Column 4 gives the value of L/EA for each bar in terms of Lc /EAc of the vertical members. This is useful because the term L/EA cancels out in some of the calculations. 4. The method of virtual work is applied directly to compute the displacements D1Q and D2Q corresponding to the releases and caused by the actual loads. Apply a virtual force δQ1 at release 1. The internal virtual forces q¯1 are found in column 2. The internal virtual work q¯1 ∆l is found in column 5 as the product of columns 1, 2, and 4. The summation of column 5 is D1q = −122.42 Lc /EAc . Similarly, column 6 is the product of columns 1, 3, and 4, giving D2Q = −273.12 Lc /EAc . 5. The same method is used to compute the ﬂexibilities fij . In this case the real loading is a unit load corresponding ﬁrst to release 1 leading to f11 , and f21 , and then to release 2 leading to f12 and f22 . Column 7 shows the computation for f11 . It is the product of column 2 representing force due to the real unit load with column 2 representing force due to a virtual load δQ1 at the same location (release 1) multiplied by column 4 to include the Lc /EAc term. Column 8 derives from columns 2, 3, and 4 and leads to f21 . Columns 9 and 10 are the computations for the remaining ﬂexibilities.

Victor Saouma

Structural Engineering

Victor Saouma

multiply by 1-2 60.0 2-3 20.00 3-4 0 4-5 0 5-6 −20.00 6-1 40.00 2-5 20.00 1-5 0 2-6 −56.56 2-4 0 3-5 −28.28

P

0 −0.707 −0.707 −0.707 0 0 −0.707 0 0 1.00 1.00

p1 −0.707 0 0 0 −0.707 −0.707 −0.707 1.00 1.00 0 0

p2

L/EA Lc /EAc 1 1 2 1 1 2 2 2.828 2.828 2.828 2.838

D1Q q¯1 P L/EA Lc /EAc 0 −14.14 0 0 0 0 −28.28 0 0 0 −80.00 −122.42

D2Q q¯2 P L/EA Lc /EAc −42.42 0 0 0 14.14 −56.56 −28.28 0 −160.00 0 0 −273.12

f11 q¯1 p1 L/EA Lc /EAc 0 0.50 1.00 0.50 0 0 1.00 0 0 2.83 2.83 8.66

f21 q¯2 p2 L/EA Lc /EAc 0 0 0 0 0 0 1.00 0 0 0 0 1.00

f12 q¯p2 L/EA Lc /EAc 0 0 0 0 0 0 1.00 0 0 0 0 1.00

f22 q¯2 p2 L/EA Lc /EAc 0.50 0 0 0 0.50 1.00 1.00 2.83 2.83 0 0 8.66

∆/temp Lc −0.0003 −0.0003 −0.0003 −0.0003 −0.0003 0 0 0 0 0 0

D1∆ q¯1 ∆/ 10−4 Lc 0 2.12 2.12 2.12 0 0 0 0 0 0 0 6.36

D2∆ q¯2 ∆/ 10−4 Lc 2.12 0 0 0 2.12 0 0 0 0 0 0 4.24

Draft

21–24 STATIC INDETERMINANCY; FLEXIBILITY METHOD

Structural Engineering

21.4 Examples

21–25

Draft

Figure 21.18: 6. We have assumed that a temperature drop of 30◦ C occurs in the outer members. The corresponding length changes are found in column 11. Again using the virtual work method, column 12 tabulates the internal virtual work of virtual forces q¯1 through displacements ∆l where for each bar, ∆l = αl∆T . Column 12 is therefore the product of columns 2 and 11. The summation of the elements of column 12 is the displacement D1 corresponding to release 1. Column 13 repeats this process for D2 corresponding to release 2. 7. The tabulated information provides the necessary terms for a matrix solution of the problem. We have

f

=

Dq

=

D∆

=

therefore f −1 =

8.66 1.00 Lc /EAc 1.00 8.66 −122.42 Lc /EAc −273.12 6.36 (10)−4 Lc 4.24

0.117 −0.0134 −0.0134 0.117

(21.66-a) (21.66-b) (21.66-c)

EAc /Lc

(21.67)

8. The redundant forces due to the applied loading are R

= =

f −1 (−DQ ) $ % 10.66 122.42 −0.0134 0.117 EAc /Lc Lc /EAc = 30.32 273.12

(21.68-a) (21.68-b)

9. thus R1 = 10.66 kN, R2 = 30.32 kN. 10. The redundant forces due to the temperature drop are R = f −1 (−D∆ ) $ % −6.87 −6.36 10−5 EAc 10−4 Lc = = −0.0134 0.117 EAc /Lc −4.11 −4.24 11. Thus with E = 200 kN/mm2 , Ac = 500 mm2 , we have R1 = R2 = Victor Saouma

6.87(10−5)(200)(500) = 4.11(10

−5

)(200)(500) =

−6.87kN

(21.69-a)

−4.11kN

(21.69-b) Structural Engineering

21–26

Draft

STATIC INDETERMINANCY; FLEXIBILITY METHOD

12. Using the redundant forces from each of these analyses, the remainder of the bar forces are computed by simple equilibrium. The information in Table 21.4 contains the basis for such computations. The bar force in any bar is the force of column 1 added to that in column 2 multiplied by R1 plus that in column 3 multiplied by R2 . This follows from the fact that columns 2 and 3 are bar forces caused by a force of unity corresponding to each of the redundants. The results of the calculations are shown in Figure ??-c for the applied loading and ??-d for the temperature drop. The forces caused by the temperature drop are similar in magnitude to those caused by wind load in this example. Temperature diﬀerences, shrinkage, support settlement, or tolerance errors can cause important eﬀects in statically indeterminate structures. These stresses are self-limiting, however, in the sense that if they cause yielding or some ductile deformation failure does not necessarily follow, rather relief from th

Victor Saouma

Structural Engineering

Draft Chapter 22

APPROXIMATE FRAME ANALYSIS 1

Despite the widespread availability of computers, approximate methods of analysis are justiﬁed by 1. Inherent assumption made regarding the validity of a linear elastic analysis vis a vis of an ultimate failure design. 2. Ability of structures to redistribute internal forces. 3. Uncertainties in load and material properties

2

Vertical loads are treated separately from the horizontal ones.

3

We use the design sign convention for moments (+ve tension below), and for shear (ccw +ve).

4

Assume girders to be numbered from left to right.

5

In all free body diagrams assume positivee forces/moments, and take algeebraic sums.

6 The key to the approximate analysis method is our ability to sketch the deﬂected shape of a structure and identify inﬂection points. 7 We begin by considering a uniformly loaded beam and frame. In each case we consider an extreme end of the restraint: a) free or b) restrained. For the frame a relativly ﬂexible or stiﬀ column would be analogous to a free or ﬁxed restrain on the beam, Fig. 22.1.

22.1 8

Vertical Loads

With reference to Fig. 22.1, we now consider an intermediary case as shown in Fig. 22.2.

9 With the location of the inﬂection points identiﬁed, we may now determine all the reactions and internal forces from statics.

If we now consider a multi-bay/multi-storey frame, the girders at each ﬂoor are assumed to be continuous beams, and columns are assumed to resist the resulting unbalanced moments from the girders, we may make the following assumptions

10

1. Girders at each ﬂoor act as continous beams supporting a uniform load. 2. Inﬂection points are assumed to be at (a) One tenth the span from both ends of each girder. (b) Mid-height of the columns 3. Axial forces and deformation in the girder are negligibly small.

22–2

APPROXIMATE FRAME ANALYSIS

Draft

w

111 000 000 111

1 0 0 1 0 1 0 1 0 1 0 1

111 000 000 111

1 0 0 1 0 1 0 1 0 1 0 1 2

wL/24 2

+

wL/12

-

-

0.21 L

L

0.21 L

L

1111111111111111 0000000000000000 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 000 111 000 111 000 111 000 111

111111111111111111 000000000000000000 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 000000000000000000 111111111111111111 0000 1111 000 111 0000 1111 000 111

111 000 000 111 000 111

1111 0000 1111 0000 1111 0000

111 000 000 111 000 111

111 000 000 111 111 000

Figure 22.1: Uniformly Loaded Beam and Frame with Free or Fixed Beam Restraint

Victor Saouma

Structural Engineering

22.1 Vertical Loads

22–3

Draft

11111111111111111 00000000000000000 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 00000000000000000 11111111111111111 000 111 000 111 00000000000000000 11111111111111111 111 000 111 000 111 111 000 000

0.5H

0.5H

1111111111111111 0000000000000000 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 0000000000000000 1111111111111111 000 111 0000 1111 0000000000000000 1111111111111111 000 111 1111 0000 111 1111 000 0000

111 000 000 111 000 111

1111 0000 0000 1111 0000 1111

111 000 000 111 000 111

0.1 L

111 000 000 111 000 111

0.1 L

Figure 22.2: Uniformly Loaded Frame, Approximate Location of Inﬂection Points

Victor Saouma

Structural Engineering

22–4

APPROXIMATE FRAME ANALYSIS

Draft

4. Unbalanced end moments from the girders at each joint is distributed to the columns above and below the ﬂoor.

Based on the ﬁrst 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 inﬂection 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).

11

12

End forces are given by

Maximum positive moment at the center of each beam is, Fig. 22.3

w

Mrgt

lft

M

Vrgt Vlft 0.1L

0.1L

0.8L L

Figure 22.3: Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments

M+ =

1 1 wL2s = w (0.8)2 L2 = 0.08wL2 8 8

(22.1)

Maximum negative moment at each end of the girder is given by, Fig. 22.3 w w M lef t = M rgt = − (0.1L)2 − (0.8L)(0.1L) = −0.045wL2 2 2

(22.2)

Girder Shear are obtained from the free body diagram, Fig. 22.4 V lf t =

Victor Saouma

wL 2

V rgt = −

wL 2

(22.3)

Structural Engineering

22.2 Horizontal Loads

22–5

Draft

Pabove

Vrgti-1

Vlfti

Pbelow

Figure 22.4: Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces Column axial force is obtained by summing all the girder shears to the axial force transmitted by the column above it. Fig. 22.4 rgt P dwn = P up + Vi−1 − Vilf t

(22.4)

Column Moment are obtained by considering the free body diagram of columns Fig. 22.5 rgt bot M top = Mabove − Mi−1 + Milf t

M bot = −M top

(22.5)

Column Shear Points of inﬂection are at mid-height, with possible exception when the columns on the ﬁrst ﬂoor are hinged at the base, Fig. 22.5

V =

M top h 2

(22.6)

Girder axial forces are assumed to be negligible eventhough the unbalanced column shears above and below a ﬂoor will be resisted by girders at the ﬂoor.

22.2

Horizontal Loads

Again, we begin by considering a simple frame subjected to a horizontal force, Fig. 22.6. depending on the boundary conditions, we will have diﬀerent locations for the inﬂection points.

13

For the analysis of a multi-bays/multi-storeys frame, we must diﬀerentiate between low and high rise buildings.

14

Low rise buidlings, where the height is at least samller than the hrizontal dimension, the deﬂected shape is characterized by shear deformations. Victor Saouma

Structural Engineering

22–6

APPROXIMATE FRAME ANALYSIS

Draft

h/2

h/2 Mcolabove

lft i-1

M

Mi-1rgt Vi-1rgt

Vi-1lft

Li-1

Mirgt

Milft

Virgt

Vilft

Mcolbelow

Li h/2

h/2

Figure 22.5: Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments

Victor Saouma

Structural Engineering

22.2 Horizontal Loads

22–7

Draft

P

PH 2

PH 2

PH 4

PH 4

α I

I

H P 2

P 2 111 000 111 000

111 000 111 000

PH L

PH L L

P α I

I

H

P 2 000 111 000 111 000 111

PH 2L

000 111

P 2

000 111 111 PH 000 4 PH 2L

PH 4

111 000 000 111

111 000 000 111

PH 4

PH 4

L

Figure 22.6: Horizontal Force Acting on a Frame, Approximate Location of Inﬂection Points

Victor Saouma

Structural Engineering

22–8

APPROXIMATE FRAME ANALYSIS

Draft

High rise buildings, where the height is several times greater than its least horizontal dimension, the deﬂected shape is dominated by overall ﬂexural deformation.

22.2.1

Portal Method

Low rise buildings under lateral loads, have predominantly shear deformations. Thus, the approximate analysis of this type of structure is based on

15

1. Distribution of horizontal shear forces. 2. Location of inﬂection points. 16

The portal method is based on the following assumptions 1. Inﬂection points are located at (a) Mid-height of all columns above the second ﬂoor. (b) Mid-height of ﬂoor columns if rigid support, or at the base if hinged. (c) At the center of each girder. 2. Total horizontal shear at the mid-height of all columns at any ﬂoor 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.

17

Forces are obtained from

Column Shear is obtained by passing a horizontal section through the mid-height of the columns at each ﬂoor and summing the lateral forces above it, then Fig. 22.7

H/2

H

H

H/2

Figure 22.7: Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear * V ext =

Victor Saouma

F lateral

2No. of bays

V int = 2V ext

(22.7)

Structural Engineering

22.2 Horizontal Loads

22–9

Draft

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. 22.7 M top = V

h 2

M bot = −M top

(22.8)

Girder Moments is obtained from the columns connected to the girder, Fig. 22.8

h/2

h/2 Mcolabove

lft i-1

M

Mi-1rgt Vi-1rgt

Vi-1lft

Li-1/2

Li-1/2

Mirgt

Milft

Virgt

Vilft

Mcolbelow

Li/2

Li/2

h/2

h/2

Figure 22.8: ***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment

rgt above below Milf t = Mcol − Mcol + Mi−1

Mirgt = −Milf t

(22.9)

Girder Shears Since there is an inﬂection point at the center of the girder, the girder shear is obtained by considering the sum of moments about that point, Fig. 22.8 V lf t = −

2M L

V rgt = V lf t

(22.10)

Column Axial Forces are obtained by summing girder shears and the axial force from the column above, Fig. ?? P = P above + P rgt + P lf t

Victor Saouma

(22.11)

Structural Engineering

22–10

APPROXIMATE FRAME ANALYSIS

Draft Pabove

Vrgti-1

Vlfti

Pbelow

Figure 22.9: Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force In either case, you should always use a free body diagram in conjunction with this method, and never rely on a blind application of the formulae.

18

Example 22-1: Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads

Draw the shear, and moment diagram for the following frame. Solution:

0.25K/ft

15K

5

30K

12

6

9 1

13 0.50

K/ft

10

14

30’

8

14’

11 3

2

20’

7

4

16’

24’

Figure 22.10: Example; Approximate Analysis of a Building

Vertical Loads The analysis should be conducted in conjunction with the free body diagram shown in Fig. 22.11.

Victor Saouma

Structural Engineering

22.2 Horizontal Loads

22–11

6.5 0.51

0.8

6.0 13.0

6.5

0.45

0.7 4.5

6.0

7.5 3.6

0.7

0.56

0.93 3.6

13.0

20.2

5.6

3.6

20.2

7.5

5.0 9.0

9.0 4.5 0.56

0.93

0.51 5.6

5.0

4.5

6.5

3.6

0.8

0.64

3.0

3.0

3.75

5.6

0.64

6.5

10.1

4.5

10.1

3.75

2.5 4.5

4.5

2.5

Draft

0.81

0.45 5.6

3.6

0.81 6.5

Figure 22.11: Free Body Diagram for the Approximate Analysis of a Frame Subjected to Vertical Loads

Victor Saouma

Structural Engineering

22–12

APPROXIMATE FRAME ANALYSIS

Draft

1. Top Girder Moments lf t M12 cnt M12 rgt M12 lf t M13 cnt M13 rgt M13 lf t M14 cnt M14 rgt M14

= = = = = = = = =

−0.045w12L212 = −(0.045)(0.25)(20)2 0.08w12 L212 = (0.08)(0.25)(20)2 lf t M12 −0.045w13L213 = −(0.045)(0.25)(30)2 0.08w13 L213 = (0.08)(0.25)(30)2 lf t M13 −0.045w14L214 = −(0.045)(0.25)(24)2 0.08w14 L214 = (0.08)(0.25)(24)2 lf t M14

=− = =− =− = =− =− = =−

4.5 k.ft 8.0 k.ft 4.5 k.ft 10.1 k.ft 18.0 k.ft 10.1 k.ft 6.5 k.ft 11.5 k.ft 6.5 k.ft

−0.045w9 L29 = −(0.045)(0.5)(20)2 0.08w9 L29 = (0.08)(0.5)(20)2 M9lf t −0.045w10 L210 = −(0.045)(0.5)(30)2 0.08w10 L210 = (0.08)(0.5)(30)2 lf t M11 −0.045w12 L212 = −(0.045)(0.5)(24)2 0.08w12 L212 = (0.08)(0.5)(24)2 lf t M12

=− = =− =− = =− =− = =−

9.0 k.ft 16.0 k.ft 9.0 k.ft 20.3 k.ft 36.0 k.ft 20.3 k.ft 13.0 k.ft 23.0 k.ft 13.0 k.ft

2. Bottom Girder Moments M9lf t M9cnt M9rgt lf t M10 cnt M10 rgt M10 lf t M11 cnt M11 rgt M11

= = = = = = = = =

3. Top Column Moments M5top M5bot M6top M6bot M7top M7bot M8top M8bot

= = = = = = = =

lf t +M12 −M5top rgt lf t −M12 + M13 = −(−4.5) + (−10.1) top −M6 rgt lf t −M13 + M14 = −(−10.1) + (−6.5) −M7top rgt −M14 = −(−6.5) −M8top

=− = =− = =− = = =−

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

4. Bottom Column Moments M1top M1bot M2top M2bot M3top M3bot M4top M4bot

= = = = = = = =

+M5bot + M9lf t = 4.5 − 9.0 −M1top lf t +M6bot − M9rgt + M10 = 5.6 − (−9.0) + (−20.3) top −M2 rgt lf t +M7bot − M10 + M11 = −3.6 − (−20.3) + (−13.0) top −M3 rgt +M8bot − M11 = −6.5 − (−13.0) top −M4

=− = =− = = =− = =−

4.5 k.ft 4.5 k.ft 5.6 k.ft 5.6 k.ft 3.6 k.ft 3.6 k.ft 6.5 k.ft 6.5 k.ft

5. Top Girder Shear lf t V12 rgt V12 lf t V13 rgt V13 lf t V14 rgt V14

Victor Saouma

= = = = = =

w12 L12 2 lf t −V12 w13 L13 2 lf t −V13 w14 L14 2 lf t −V14

=

(0.25)(20) 2

=

(0.25)(30) 2

=

(0.25)(24) 2

= =− = =− = =−

2.5 k 2.5 k 3.75 k 3.75 k 3.0 k 3.0 k

Structural Engineering

22.2 Horizontal Loads

22–13

Draft

0.25K/ft

12

5

6

13 0.50

9

10

1

20’

30’

+8.0’k

+18.0’k

-4.5’k +16.0’k

-9.0’

k

-20.2’

+4.5’k

+5.6’k -4.5’k

+4.5’k

+5.6’k

4

-5.6’k

-5.6’

k

16’

24’ +11.5’k -6.5’k

k

+23.0’

+32.0’k k

14’

11

-10.1’k -10.1’k -6.5’

-9.0’k

-4.5’k

8

3

2

-4.5’k

14

7

K/ft

k

k -20.2’k -13.0’

+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 22.12: Approximate Analysis of a Building; Moments Due to Vertical Loads

Victor Saouma

Structural Engineering

22–14

APPROXIMATE FRAME ANALYSIS

Draft

6. Bottom Girder Shear V9lf t V9rgt lf t V10 rgt V10 lf t V11 rgt V11

= w92L9 = (0.5)(20) 2 = −V9lf t = w102L10 = (0.5)(30) 2 lf t = −V10 = w112L11 = (0.5)(24) 2 lf t = −V11

7. Column Shears V5

=

V6

=

V7

=

V8

=

V1

=

V2

=

V3

=

V4

=

M5top H5 2 M6top H6 2 M7top H7 2 M8top H8 2 M1top H1 2 M2top H2 2 M3top H3 2 M4top H4 2

= =− = =− = =−

5.00 k 5.00 k 7.50 k 7.50 k 6.00 k 6.00 k

−4.5

= − 0.64 k

−5.6

= − 0.80 k

=

3.6

=

0.52 k

=

6.5

=

0.93 k

= =

14 2 14 2

14 2 14 2

−4.5

= − 0.56 k

−5.6

= − 0.70 k

=

3.6

=

0.46 k

=

6.5

=

0.81 k

= =

16 2 16 2

16 2 16 2

8. Top Column Axial Forces P5 P6 P7 P8

= = = =

lf t V12 rgt lf t −V12 + V13 = −(−2.50) + 3.75 rgt lf t −V13 + V14 = −(−3.75) + 3.00 rgt −V14

= = = =

2.50 6.25 6.75 3.00

k k k k

9. Bottom Column Axial Forces P1 P2 P3 P4

= = = =

P5 + V9lf t = 2.50 + 5.0 rgt P6 − V10 + V9lf t = 6.25 − (−5.00) + 7.50 rgt lf t P7 − V11 + V10 = 6.75 − (−7.50) + 6.0 rgt P8 − V11 = 3.00 − (−6.00)

= = = =

7.5 k 18.75 k 20.25 k 9.00 k

Horizontal Loads, Portal Method 1. Column Shears V5 V6 V7 V8 V1 V2 V3 V4

Victor Saouma

= = = = = = = =

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 k 5k 5k 2.5 k 7.5 k 15 k 15 k 7.5 k

Structural Engineering

22.2 Horizontal Loads

22–15

Draft

+2.5K

+3.75K

+3.0K

-2.5K K

+7.5

+5.0K

-3.75K K

+6.0

-5.0K

-0.64K

-0.56K

-3.0K

-0.80K

-0.70K

-6.0K

-7.5K

+0.51K

+0.45K

+0.93K

+0.81K

Figure 22.13: Approximate Analysis of a Building; Shears Due to Vertical Loads

Victor Saouma

Structural Engineering

22–16

APPROXIMATE FRAME ANALYSIS

Draft 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

A

B

Height 14 16

Span Load Load

C

D

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F

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H

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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 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA 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 AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA 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 AAAAAAAA AAAAAAAA AAAAAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA -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 AA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAA 4.5 AAAA 5.6 -3.6 -6.5 AAAA AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAAAAAA AAAA AAAAAAAA AAAA A A -9.0 16.0 AAAAAAAAAAAA AAAAAAAA AAAA -20.3 36.0 -20.3AAAA AAAAAAAAAAAA AAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA -13.0 23.0 -13.0 AAAA -9.0 AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA 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 -4.5 AAAA -5.6 3.6 6.5 AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAA AA AAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA 4.5 AAAA 5.6 -3.6 -6.5 AAAAAAAA AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA 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 AAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAAAAAA AAAA -2.50 AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAAA 2.50AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAA 3.75 AAAA AAAAAAAAAAAA -3.00 AAAAAAAA AAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA -3.75AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAA A 3.00 AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 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 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA -0.64 AAAA -0.80 0.52 0.93 AA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAA A AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA A AAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAAAAAA AAAA -5.00 AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAAA 5.00AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAA 7.50 AAAA AAAAAAAAAAAA -6.00 AAAAAAAA AAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA -7.50AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAA A 6.00 AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 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 -0.56 AAAA -0.70 0.46 0.81 AA AAAAAAAA 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 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 AAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA A 0.00 0.00 0.00 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 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 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 AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA 2.50 AAAA 6.25 6.75 3.00 AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA A AAAAAAAA AAAAAAAA AAAAAAAA AAAA 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 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 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 22.14: Approximate Analysis for Vertical Loads; Spread-Sheet Format

Victor Saouma

Structural Engineering

Victor Saouma L1 20 0.25 0.5

D

E

F

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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

=+G28-F22+I22

=-F20+I20

0

=+L28-K22+N22

=-K20+N20

0

=+Q28-P22

=-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

22.2 Horizontal Loads 22–17

Figure 22.15: Approximate Analysis for Vertical Loads; Equations in Spread-Sheet

Structural Engineering

35

35

30 60

120

60

77.5

35

120

7.5

7.5

5

35

5

77.5

77.5

17.5

77.5

2.5

17.5 5

5

77.5

2.5

17.5

2.5

60 15

15

120 15

2.5

77.5

17.5

17.5

17.5

15

17.5

17.5

Draft

17.5

APPROXIMATE FRAME ANALYSIS

17.5

22–18

15

120

7.5

7.5

60

Figure 22.16: Free Body Diagram for the Approximate Analysis of a Frame Subjected to Lateral Loads

Victor Saouma

Structural Engineering

22.2 Horizontal Loads

22–19

Draft

2. Top Column Moments M5top M5bot M6top M6bot M7top M7bot M8top M8bot

= = = = = = = =

V1 H5 = (2.5)(14) 2 2 top −M5 V6 H6 = (5)(14) 2 2 top −M6 up V7 H7 = (5)(14) 2 2 −M7top V8up H8 = (2.5)(14) 2 2 top −M8

= =− = =− = =− = =−

17.5 k.ft 17.5 k.ft 35.0 k.ft 35.0 k.ft 35.0 k.ft 35.0 k.ft 17.5 k.ft 17.5 k.ft

3. Bottom Column Moments M1top M1bot

= =

M2top M2bot

= =

M3top M3bot

= =

M4top M4bot

= =

V1dwn H1 2 −M1top V2dwn H2 2 −M2top V3dwn H3 2 −M3top V4dwn H4 2 −M4top

=

(7.5)(16) 2

= 60 k.ft = − 60 k.ft

=

(15)(16) 2

= 120 k.ft = − 120 k.ft

=

(15)(16) 2

= 120 k.ft = − 120 k.ft

=

(7.5)(16) 2

= 60 k.ft = − 60 k.ft

4. Top Girder Moments lf t M12 rgt M12 lf t M13 rgt M13 lf t M14 rgt M14

= = = = = =

M5top lf t −M12 rgt M12 + M6top = −17.5 + 35 lf t −M13 rgt M13 + M7top = −17.5 + 35 lf t −M14

= =− = =− = =−

17.5 k.ft 17.5 k.ft 17.5 k.ft 17.5 k.ft 17.5 k.ft 17.5 k.ft

5. Bottom Girder Moments M9lf t M9rgt lf t M10 rgt M10 lf t M11 rgt M11

= = = = = =

M1top − M5bot = 60 − (−17.5) −M9lf t M9rgt + M2top − M6bot = −77.5 + 120 − (−35) lf t −M10 rgt M10 + M3top − M7bot = −77.5 + 120 − (−35) lf t −M11

= =− = =− = =−

77.5 k.ft 77.5 k.ft 77.5 k.ft 77.5 k.ft 77.5 k.ft 77.5 k.ft

6. Top Girder Shear 2M lf t

lf t V12 rgt V12

= =

12 − L12 = − (2)(17.5) 20 lf t +V12

= =

−1.75 k −1.75 k

lf t V13 rgt V13

= =

13 − L13 = − (2)(17.5) 30 lf t +V13

= =

−1.17 k −1.17 k

lf t V14 rgt V14

= =

14 − L14 = − (2)(17.5) 24 lf t +V14

= =

−1.46 k −1.46 k

V9lf t V9rgt

= =

− L12 = − (2)(77.5) 20 9 +V9lf t

= =

−7.75 k −7.75 k

lf t V10 rgt V10

= =

10 − L10 = − (2)(77.5) 30 lf t +V10

= =

−5.17 k −5.17 k

lf t V11 rgt V11

= =

11 − L11 = − (2)(77.5) 24 lf t +V11

= =

−6.46 k −6.46 k

2M lf t

2M lf t

7. Bottom Girder Shear

Victor Saouma

2M lf t

2M lf t

2M lf t

Structural Engineering

22–20

APPROXIMATE FRAME ANALYSIS

Draft

15K

12

5

30K

13

6

9

10

1

20’

+17.5’K

-35’K +60’K

-120’K +17.5’K

-120’K

+77.5’

+77.5’

K

+77.5’

-77.5’K

-17.5’K

-60’K

+17.5’K

-17.5K 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

-17.5K

-17.5K

-77.5’K

-77.5’K

K

Figure 22.17: Approximate Analysis of a Building; Moments Due to Lateral Loads

Victor Saouma

Structural Engineering

22.2 Horizontal Loads

22–21

Draft

8. Top Column Axial Forces (+ve tension, -ve compression) P5 P6 P7 P8

= = = =

lf t −V12 = rgt lf t +V12 − V13 = −1.75 − (−1.17) = rgt lf t +V13 − V14 = −1.17 − (−1.46) = rgt V14 = −1.46 k

−(−1.75) k −0.58 k 0.29 k

9. Bottom Column Axial Forces (+ve tension, -ve compression) P1 P2 P3 P4

= = = =

P5 + V9lf t = 1.75 − (−7.75) rgt P6 + V10 + V9lf t = −0.58 − 7.75 − (−5.17) rgt lf t P7 + V11 + V10 = 0.29 − 5.17 − (−6.46) rgt P8 + V11 = −1.46 − 6.46

= = = =

9.5 k −3.16 k 1.58 k −7.66 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 22.2. Mem.

1

2

3

4

5

6

7

8

Moment Axial Shear Moment Axial Shear Moment Axial Shear Moment Axial Shear Moment Axial Shear Moment Axial Shear Moment Axial Shear Moment Axial Shear

Vert.

Hor.

4.50 7.50 0.56 5.60 18.75 0.70 3.60 20.25 0.45 6.50 9.00 0.81 4.50 2.50 0.64 5.60 6.25 0.80 3.60 6.75 0.51 6.50 3.00 0.93

60.00 9.50 7.50 120.00 15.83 15.00 120.00 14.25 15.00 60.00 7.92 7.50 17.50 1.75 2.50 35.00 2.92 5.00 35.00 2.63 5.00 17.50 1.46 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 22.1: Columns Combined Approximate Vertical and Horizontal Loads

Victor Saouma

Structural Engineering

22–22

APPROXIMATE FRAME ANALYSIS

Draft 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 AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA 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 A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA 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 AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAA -17.5 AAAA -35.0 -35.0 -17.5 AAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA 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 AAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA 60.0 AAAA 120.0 120.0 60.0 AAAA AAAAAAAA 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 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAA -60.0 AAAA -120.0 -120.0 -60.0 AAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA 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 AAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAA AAAAAAAAA -1.75 -1.75 AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA -1.46 -1.46AAAAAAAA AAAAAAAAAAAA AAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA -1.17 -1.17 AAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA 2.50 AAAA 5.00 5.00 2.50 AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA 2.50 AAAA 5.00 5.00 2.50 AAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAA AAAAAAAA AAAA AAA AAAA AAAA AAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A A AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA -5.17 -5.17 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA -6.46 -6.46AAAAAAAA AAAAAAAAAAAA AAAAA AAAAAAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAA -7.75 -7.75 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAAAA AAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA 7.50 AAAA 15.00 15.00 7.50 AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAA AAAA AAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA 7.50 AAAA 15.00 15.00 7.50 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA 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 AAAA AAAA 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 AAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAA AAAA AAAA AAAA AAAA A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAA 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 22.18: Portal Method; Spread-Sheet Format

Victor Saouma

Structural Engineering

22.2 Horizontal Loads

22–23

Draft

Portal Method

PORTAL.XLS

A 1 PORTAL METHOD

A A A A A A

B

A A A A A A

C

A A A A A A

D

A A A A A A

E

A A A A A A

F

A A A A A

G

A A A A A

H

A A A A A A

I

A A A A A A

J

A A A A A A

Victor E. Saouma

K

A A A A A

L

AA AA AA AA AA AA

M

A A A A A A

N

A A A A A A

O

A A A A A

P

A A A A A A

Q

A A A A A

R

A A A A A A

S

AAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAA A3 AL1 A AAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAA AA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAA 2 AAAA # of Bays L2 L3 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAA A A A A A A A AA A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAA AAAAAA AAAAAAAAA AAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAA A AAAAAAAA A AAAAAAAAAAAAAAAAAAAAAAA AA AAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A 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 AA 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 MOMENTS A A A A AA 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 A A A A A A A A Bay 1 Bay 2 AA Bay 3 AA A A A2 A A A AA 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 Force AA Shear Col Beam Column Beam Column Beam Col 6 A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A A A A A A A A A H Lat. AA Tot Ext Lft Lft Lft 7 AInt A Rgt A Rgt A Rgt A AAAAAAAAAAA AAAAAAAAAAAAAAAA A A AAAAAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA AAAAAAAAAAAAAA AA AAAA A A AAAAAAAAAAAAAAAAAAAA=+H9 AAAAAAAAAAAAAAAAAAAAAAAAAAA =+N8+O9 AAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA=+J8+K9 A A =-I8 AAAA =-M8 AAAA =-Q8 AAAA 8 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAAAAAAAA AAAA A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA A A AAAAAAAAAAAAAAAA=+F9*B9/2 AAAAAAAAAAAAAAAAAAAAAAA=+H9 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAA A AAAAAAAAAAAAAAAAA =+E9*B9/2 H1AAAAAAAAAAAAAAAAAAAAAAAAAAA 14 AAAAA 15 AAAAAAAAAA =+C9 =+D9/(2*$F$2) 9 AAAA AAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA A=2*E9 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA =+K9 AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA A A AAAA AAAAAAAAAAAAAAAAAAAA=-K9 AAAAAAAAAAAAAAAAAAAAAAAAAAA=+H10 A A AAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAA =+K10 =-H9 10 AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAA A A AAAA AAAA AAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAAAA AA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA A A AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA AAAA AA AAAA A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA =+O12-O10+N11 =-Q11 AAAAAAAA AAAAAAAA AAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA=+K12-K10+J11 =-M11 AAAA A A 11 AAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA=+H12-H10 =-I11 AAAA AAAAAAAAAAAAAA A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA AAAA A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA A A AAAAAAAAAAAAAAAA=+F12*B12/2 AAAAAAAAAAAAAAAAAAAAAAA=+H12 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AA AAAA A A H2 16 30 =SUM($C$9:C12) =+D12/(2*$F$2) =2*E12 =+E12*B12/2 =+K12 12 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA A A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA A A AAAA AAAAAAAAAAAAAAAAAAAA=-K12 AAAAAAAAAAAAAAAAAAAAAAAAAAA=+H13 A A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA =+K13 =-H12 13 AAAA AAAA A A AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAA AAA AAAAAAAAAAA AAAA AAAA AAAAAAAA AAAAAAAA AAAAAA AAAA A A A A AA A A 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 A A A A A A AA A A A A A A A A A A A A 14 A SHEAR A A A A AA A A A A A A A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA 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 A A Bay 1 Bay 2 AA Bay 3 15 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A A A AA 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 A A A A Col Beam Column Beam Column Beam Col 16 A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A A A A A A A A A A A A A A A A A A A A A Lft Rgt Lft Rgt Lft Rgt 17 A A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAA=-2*I8/I$3 =+I18 AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA =-2*Q8/Q$3 AAAAAAAAAAAAAA AA A A A A A A =-2*M8/M$3 =+M18 =+Q18 18 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAAAA A A A A A A AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAA AAAAAA AA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAA=+F9 AAAAAAAAAAAAAAAAAAAAAAAAAAA =+E9 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA =+F9 A A A A A A =+E9 19 AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAA AAAA AAAA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAA=+K19 AAAAAAAAAAAAAAAAAAAAAAAAAAA=+S19 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA =+O19 A A A A A A =+H19 20 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA AAAAAAAAAAAAAA AA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAA=-2*I11/I$3 =+I21 AAAAAAAAAAAAAAAAAAAAAAAAAAA=-2*M11/M$3 =+M21AAAAAAAAAAAAAAAAAAAAAAAAAAA =-2*Q11/Q$3 AAAAAAAAAAAAAA A A A A A A =+Q21AAAA 21 AAAAAAAAAAAA AAAAAAAAAAAAAAAAAAA AAAAAAAAAA AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA AAAA A A A A A A AAAAAAAA AAAA AAAAAAAA AAA AAAA A A A A A A =+E12 22 AAAAAAAA AAAAAAAA AAAAAAAA=+F12 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA =+E12 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA =+F12 AAAAAAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAA=+K22 AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA=+S22 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA =+O22 A A A A A A =+H22 23 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AA AAAA A A A A A A AAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA A A A A AA A AAAAAA A A A A AAAAAAA A A 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 A A AA A A A A A A A A A A A A 24 A AXIAL FORCE A A A A AA A A A A A A A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A A A A A A A A A A AA 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 A A Bay 1 Bay 2 AA Bay 3 25 A A A AA A A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA A A A A A A A A A A A A Col Beam Column Beam Column Beam Col 26 A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA A A A A A A AAAAAAAA AAAA0 AAAAAAAA AAA 0 AAAAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAA AAAA A A A A A A 0 27 AAAAAAAA AAAAAAAA AAAAAAAA AAAA AAAA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA A A A A A A AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA AAAA A A A AAAAAAAAAAAAAAAAAAAA=+J18-M18 AAAAAAAAAAAAAAAAAAAAAAAAAAA=+R18 AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA A AAAAAAAAAAAAAAAA AAAAAAAAA AAAAAAAAAAAA =-I18 =+N18-Q18 28 AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAAAAAA AAAA AAAA AAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAA AAAAAA AAAAAAAAA AAAAAAAAAAAAAAAAAAAAA A AAAAAAAAAAAAAAAA AAAAAAAAA AAAAAAAAAAAA AAAAAAAA AAAA0 AAAAAAAA AAA 0 AAAAAAAA AA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAA AAAAAA AAAA A A A A A 0 29 AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAAA AAA AAAA AAAA A A A A A A AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAAAAAAA A A A A A A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA AAAAAAAAAAAAAAAAAAAA AAAA AAAAAAAAAAAAAAAAAAAAAAAAAAA AAA A A A A A A AAAA AAAAAAAAAAAAAAAAAAAA=+K28+J21-M21 AAAAAAAAAAAAAAAAAAAAAAAAAAA=+S28+R21 A A A A A A AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAAAA AAAAAA AA =+O28+N21-Q21 =+H28-I21 30 AAAAAAAAAAAA AAAAAAAAAAAAAAAAAAA A A A A A A AAAAAAAA AAAA AAAAAAAA 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

A A A A

Figure 22.19: Portal Method; Equations in Spread-Sheet

Victor Saouma

Structural Engineering

22–24

APPROXIMATE FRAME ANALYSIS

Draft

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.

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

Design 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 22.2: Girders Combined Approximate Vertical and Horizontal Loads

Victor Saouma

Structural Engineering

Draft Chapter 23

KINEMATIC INDETERMINANCY; STIFFNESS METHOD 23.1

Introduction

23.1.1

Stiﬀness vs Flexibility

1

There are two classes of structural analysis methods, Table 23.1:

Flexibility: where the primary unknown is a force, where equations of equilibrium are the starting point, static indeterminancy occurs if there are more unknowns than equations, and displacements of the entire structure (usually from virtual work) are used to write an equation of compatibility of displacements in order to solve for the redundant forces. Stiﬀness: method is the counterpart of the ﬂexibility one. Primary unknowns are displacements, and we start from expressions for the forces written in terms of the displacements (at the element level) and then apply the equations of equilibrium. The structure is considered to be kinematically indeterminate to the nth degree where n is the total number of independent displacements. From the displacements, we then compute the internal forces.

Primary Variable (d.o.f.) Indeterminancy Force-Displacement Governing Relations Methods of analysis

Flexibility Forces Static Displacement(Force)/Structure Compatibility of displacement “Consistent Deformation”

Stiﬀness Displacements Kinematic Force(Displacement)/Element Equilibrium Slope Deﬂection; Moment Distribution

Table 23.1: Stiﬀness vs Flexibility Methods 2 In the ﬂexibility method, we started by releasing as many redundant forces as possible in order to render the structure statically determinate, and this made it quite ﬂexible. We then applied an appropriate set of forces such that kinematic constraints were satisifed. 3 In the stiﬀness method, we follow a diﬀerent approach, we stiﬀen the structure by constraining all the displacements, hence making it kinematically determinate, and then we will release all the constraints in such a way to satisfy equilibrium.

23–2

Draft

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

In the slope deﬂection method, all constraints are released simultaneously, thus resulting in a linear system of n equations with n unknowns. In the Moment Distribution method, we release the constraints one at a time and essentially solve for the system of n equations iteratively. 4

23.1.2

Sign Convention

5 The sign convention in the stiﬀness method is diﬀerent than the one previously adopted in structural analysis/design, Fig. 26.3. 6 In the stiﬀness method the sign convention adopted is consistent with the prevailing coordinate system. Hence, we deﬁne a positive moment as one which is counter-clockwise at the end of the element, Fig. 26.3.

Figure 23.1: Sign Convention, Design and Analysis

23.2 7

Degrees of Freedom

A degree of freedom (d.o.f.) is an independent generalized nodal displacement of a node.

8 The displacements must be linearly independent and thus not related to each other. For example, a roller support on an inclined plane would have three displacements (rotation θ, and two translations u and v), however since the two displacements are kinematically constrained, we only have two independent displacements, Fig. 26.5.

Figure 23.2: Independent Displacements 9 We note that we have been referring to generalized displacements, because we want this term to include translations as well as rotations. Depending on the type of structure, there may be none, one or more than one such displacement.

10

The types of degrees of freedom for various types of structures are shown in Table 26.4

Victor Saouma

Structural Engineering

23.3 Kinematic Relations

Draft

23–3

Type

Node 1 1 Dimensional Fy1 , Mz2

{p}

Node 2

Fy3 , Mz4

Beam {δ}

v1 , θ2 2 Dimensional Fx1

{p}

v3 , θ4 Fx2

Truss {δ} {p}

u1 Fx1 , Fy2 , Mz3

u2 Fx4 , Fy5 , Mz6

Frame {δ}

u1 , v2 , θ3 3 Dimensional Fx1 ,

{p}

u4 , v5 , θ6 Fx2

Truss {δ} {p}

u1 , Fx1 , Fy2 , Fy3 , Tx4 My5 , Mz6

u2 Fx7 , Fy8 , Fy9 , Tx10 My11 , Mz12

{δ}

u1 , v2 , w3 , θ4 , θ5 θ6

u7 , v8 , w9 , θ10 , θ11 θ12

Frame

Table 23.2: Degrees of Freedom of Diﬀerent Structure Types Systems Fig. 26.4 also shows the geometric (upper left) and elastic material (upper right) properties associated with each type of element.

11

23.2.1 12

Methods of Analysis

There are three methods for the stiﬀness based analysis of a structure

Slope Deﬂection: (Mohr, 1892) Which results in a system of n linear equations with n unknowns, where n is the degree of kinematic indeterminancy (i.e. total number of independent displacements/rotation). Moment Distribution: (Cross, 1930) which is an iterative method to solve for the n displacements and corresponding internal forces in ﬂexural structures. Direct Stiﬀness method: (˜ 1960) which is a formal statement of the stiﬀness method and cast in matrix form is by far the most powerful method of structural analysis. The ﬁrst two methods lend themselves to hand calculation, and the third to a computer based analysis.

23.3

Kinematic Relations

23.3.1

Force-Displacement Relations

Whereas in the ﬂexibility method we sought to obtain a displacement in terms of the forces (through virtual work) for an entire structure, our starting point in the stiﬀness method is to develop a set of relationship for the force in terms of the displacements for a single element.      V1  − − − −  v1            M1 θ1 − − − −    = (23.1) V2  v2  − − − −           M2 θ2 − − − −

13

Victor Saouma

Structural Engineering

23–4

Draft

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Figure 23.3: Total Degrees of Freedom for various Type of Elements

Victor Saouma

Structural Engineering

23.3 Kinematic Relations

23–5

Draft

We start from the diﬀerential equation of a beam, Fig. 23.4 in which we have all positive known displacements, we have from strength of materials

14

θ1 M2

V2

V1 M1

θ2

v2

v1 2

1 L Figure 23.4: Flexural Problem Formulation

d2 v = M1 − V1 x + m(x) (23.2) dx2 where m(x) is the moment applied due to the applied load only. It is positive when counterclockwise. M = −EI

15

Integrating twice

where f (x) = 16

17

18

(

−EIv

=

−EIv

= (

m(x)dx, and g(x) =

(23.4)

= =

−EIθ1 −EIv1

(23.5)

Applying the boundary conditions at x = L and combining with the expressions for C1 and C2 v = θ2 −EIθ2 = M1 L − 12 V1 L2 + f (L) − EIθ1 (23.6) ⇒ v = v2 −EIv2 = 12 M1 L2 − 16 V1 L3 + g(L) − EIθ1 L − EIv1 Since equilibrium of forces and moments must be satisﬁed, we have: (L 0

V1 + q + V2 = 0

M1 − V1 L + m(L) + M2 = 0

(23.7)

p(x)dx, thus V1 =

20

(23.3)

f (x)dx.

Applying the boundary conditions at x = 0 v = θ1 C1 ⇒ v = v1 C2

where q =

19

1 M1 x − V1 x2 + f (x) + C1 2 1 1 2 M1 x − V1 x3 + g(x) + C1 x + C2 2 6

1 (M1 + M2 ) + m(L) L L

V2 = −(V1 + q)

Substituting V1 into the expressions for θ2 and v2 in Eq. 23.6 and rearranging 2EIz 2 z M1 − M2 = 2EI L θ1 + L θ2 + m(L) − L f (L) 6EIz 6EIz 6EIz 2M1 − M2 = L θ1 − L2 v1 − L2 v2 + m(L) − L62 g(L)

(23.8)

(23.9)

Solving those two equations, we obtain:

Victor Saouma

Structural Engineering

23–6

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Draft M1

2EIz 6EIz (2θ1 + θ2 ) − (v2 − v1 ) + M1F 2 L L

=

II

I

M2

(23.10)

2EIz 6EIz (θ1 + 2θ2 ) − (v2 − v1 ) + M2F 2 L L

=

(23.11)

II

I

where M1F

=

M2F

=

2 [Lf (L) − 3g(L)] L2 % 1 $ − 2 L2 m(L) − 4Lf (L) + 6g(L) L

(23.12-a) (23.12-b)

M1F and M2F are the ﬁxed end moments for θ1 = θ2 = 0 and v1 = v2 = 0, that is ﬁxed end moments. They can be obtained either from the analysis of a ﬁxed end beam, or more readily from the preceding two equations. In Eq. 23.10 and 23.11 we observe that the moments developed at the end of a member are caused by: I) end rotation and displacements; and II) ﬁxed end members.

21

22

Finally, we can substitute those expressions in Eq. 23.8 V1

6EIz 12EIz (θ1 + θ2 ) − (v2 − v1 ) + V1F 2 3 L L

=

II

I

V2

(23.13)

6EIz 12EIz = − 2 (θ1 + θ2 ) + (v2 − v1 ) + V2F 3 L L

(23.14)

II

I

where

23

V1F

=

V2F

=

6 [Lf (L) − 2g(L)] L3 6 − 3 [Lf (L) − 2g(L)] + q L

(23.15-a) (23.15-b)

The relationships just derived enable us now to determine the stiﬀness matrix of a beam element.  V1    M1 V2    M2



v1 z V1 12EI L3 6EIz  M 1 L2 12EIz V2  − L3 6EIz M 2 L2

      

=

23.3.2

θ1 6EIz L2 4EIz L z − 6EI L2 2EIz L

ke

v2 z − 12EI L3 6EIz − L2 12EIz L3 z − 6EI L2

θ2 6EIz L2 2EIz L z − 6EI L2 4EIz L



  v1       θ1   v2       θ2

(23.16)

Fixed End Actions

As mentioned above, the end actions developed in a member involve the end displacements, rotations, and the in-span loads. In-spans loads exhibit themselves in the form of ﬁxed-end forces.

24

The ﬁxed-end actions can be determined from the equations derived above, or by analyzing a ﬁxed-end beam under the applied loads.

25

26

Note that in both cases, the load has to be assumed positive, i.e. pointing up for a beam.

The equations derived for calculating the ﬁxed-end actions can be summarized as follows. We recall that, with the x axis directed to the right, positive loads and shear forces act upward and positive

27

Victor Saouma

Structural Engineering

23.3 Kinematic Relations

23–7

Draft

moments are counterclockwise. To calculate the ﬁxed end actions the only thing we need is an expression for the moment of the applied loads (without the end reactions) in the analysis sign convention. Thus with m(x)

=

f (x)

=

moment due to the applied loads at section x

(23.17-a)

m(x)dx

(23.17-b)

f (x)dx

(23.17-c)

p(x)dx = total load on the span

(23.17-d)

g(x)

=

q

=

and

23.3.2.1 28

M1F

=

M2F

=

V1F

=

V2F

=

2 [Lf (L) − 3g(L)] L2 % 1 $ − 2 L2 m(L) − 4Lf (L) + 6g(L) L 6 [Lf (L) − 2g(L)] L3 6 − 3 [Lf (L) − 2g(L)] − q L

(23.18) (23.19) (23.20) (23.21)

Uniformly Distributed Loads

For a uniformly distributed load w over the entire span, 1 1 1 m(x) = − wx2 ; f (x) = − wx3 ; g(x) = − wx4 ; q = wL 2 6 24

29

(23.22)

Substituting M1F M2F V1F V2F

23.3.2.2

2 2 1 1 3 4 wL wL L − − 3 − = − wL 12 L2 6 24 1 1 1 1 = − 2 L2 − wL2 − 4L − wL3 + 6 − wL4 = L 2 6 24 6 1 1 3 4 wL wL = L − − 2 − = − wL 2 L3 6 24 6 1 1 = − 3 L − wL3 − 2 − wL4 − wL = − wL 2 L 6 24 =

(23.23-a) wL2 12

(23.23-b) (23.23-c) (23.23-d)

Concentrated Loads

For a concentrated load we can use the unit step function to ﬁnd m(x). For a concentrated load P acting at a from the left-hand end with b = L − a,

30

m(x) = f (x) = g(x) =

−P (x − a)Ha − 21 P (x − a)2 Ha − 61 P (x − a)3 Ha

gives m(L) = −P b f (L) = − 12 P b2 g(L) = − 16 P b3

(23.24)

where we deﬁne Ha = 0 if x < a, and Ha = 1 if x ≥ a. 31

and q M1F

= P 2 2 1 2 1 3 = L − Pb − 3 − Pb == − PLb2 a 2 L 2 6

Victor Saouma

(23.25-a) (23.25-b) Structural Engineering

23–8

Draft M2F

KINEMATIC INDETERMINANCY; STIFFNESS METHOD 1 1 2 1 3 Pb 2 = − 2 L (−P b) − 4L − P b + 6 − P b = 2 L2 − 2Lb + b2 L 2 6 1 =

V1F V2F

32

P ba2 L2

(23.25-c) (23.25-d)

2 6 1 2 1 3 P b2 P b P b = (3L − 2b) = − PLb3 (3a + b) L − − 2 − = − L3 2 6 L3 2 6 1 2 1 3 = − L − Pb − 2 − Pb + P = − PLa3 (a + 3b) L3 2 6

(23.25-e) (23.25-f)

If the load is applied at midspan (a = B = L/2), then the previous equation reduces to M1F M2F V1F V2F

PL 8 PL = 8 P = − 2 P = − 2 = −

(23.26) (23.27) (23.28) (23.29)

23.4

Slope Deﬂection; Direct Solution

23.4.1

Slope Deﬂection Equations

33 In Eq. 23.10 and 23.11 if we let ∆ = v2 − v1 (relative displacement), ψ = ∆/L (rotation of the chord of the member), and K = I/L (stiﬀness factor) then the end equations are:

34

M1

=

2EK(2θ1 + θ2 − 3ψ) + M1F

(23.30)

M2

=

2EK(θ1 + 2θ2 − 3ψ) + M2F

(23.31)

Note that ψ will be positive if counterclockwise, negative otherwise.

From Eq. 23.30 and 23.31, we note that if a node has a displacement ∆, then both moments in the adjacent element will have the same sign. However, the moments in elements on each side of the node will have diﬀerent sign.

35

23.4.2

Procedure

To illustrate the general procedure, we consider the two span beam in Fig. 23.5 under the applied load, we will have three rotations θ1 , θ2 , and θ3 (i.e. three degrees of freedom) at the supports. Separating the spans from the supports, we can write the following equilibrium equations for each support

36

M12 M21 + M23

= 0 = 0

(23.32-a) (23.32-b)

M32

= 0

(23.32-c)

The three equilibrium equations in turn can be expressed in terms of the three unknown rotations, thus we can analyze this structure (note that in the slope deﬂection this will always be the case).

37

38

Using equations 23.30 and 23.31 we obtain

Victor Saouma

M12

F = 2EK12 (2θ1 + θ2 ) + M12

(23.33-a)

M21 M23

= 2EK12 (θ1 + 2θ2 ) + = 2EK23 (2θ2 + θ3 )

F M21

(23.33-b) (23.33-c)

M32

= 2EK23 (θ2 + 2θ3 )

(23.33-d) Structural Engineering

23.4 Slope Deﬂection; Direct Solution

Draft

23–9

P

1 111 000 000 111

2

3

1111 0000 1111 0000

111 000 111 000

a

L

L1 P

M 12

M21

2

M23

M 32

2

1

3

Figure 23.5: Illustrative Example for the Slope Deﬂection Method

39

Substituting into the equations of equilibrium, we obtain 

2  K12 0

   MF 1 0  θ1    − 2EK1212 F M21 2(K12 + K23 ) K23  θ2 =     − 2E 1 2 θ3 0

Stiﬀness Matrix

    

(23.34)

Where the ﬁxed end moment can be separately determined. Once the rotations are determined, we can then determine the moments from the slope deﬂection equation Eq. 23.30.

40

The computational requirements of this method are far less than the one involved in the ﬂexibility method (or method of consistent deformation).

41

23.4.3 42

Algorithm

Application of the slope deﬂection method requires the following steps: 1. Sketch the deﬂected shape. 2. Identify all the unknown support degrees of freedom (rotations and deﬂections). 3. Write the equilibrium equations at all the supports in terms of the end moments. 4. Express the end moments in terms of the support rotations, deﬂections and ﬁxed end moments. 5. Substitute the expressions obtained in the previous step in the equilibrium equations. 6. Solve the equilibrium equations to determine the unknown support rotation and/or deﬂections. 7. Use the slope deﬂection equations to determine the end moments. 8. Draw the moment diagram, careful about the diﬀerence in sign convention between the slope deﬂection moments and the moment diagram.

Victor Saouma

Structural Engineering

23–10

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Draft 23.4.4

Examples

Example 23-1: Propped Cantilever Beam, (Arbabi 1991) Find the end moments for the beam of Fig. 23.6 20 kN

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1

11111 00000 00000 11111

1

2

10 m

5m

3

Figure 23.6: Slope Deﬂection; Propped Cantilever Beam Solution: 1. The beam is kinematically indeterminate to the third degree (θ2 , ∆3 , θ3 ), however by replacing the the overhang by a ﬁxed end moment equal to 100 kN.m at support 2, we reduce the degree of kinematic indeterminancy to one (θ2 ). 2. The equilbrium relation is (23.35) M21 − 100 = 0 3. The members end moments in terms of the rotations are (Eq. 23.30 and 23.31) M12 M21

2 EIθ2 10 4 EIθ2 = 2EK12 (θ1 + 2θ2 ) = 10 = 2EK12 (2θ1 + θ2 ) =

(23.36-a) (23.36-b)

4. Substituting into the equilibrium equations θ2 = or M12 =

10 250 M21 = 4EI EI

(23.37)

10 2 2 (2)EI(50) EIθ2 = EI M21 = = 50 kN.m 10 10 4EI (10)EI

(23.38)

Example 23-2: Two-Span Beam, Slope Deﬂection, (Arbabi 1991) Draw the moment diagram for the two span beam shown in Fig. 23.8 Solution: 1. The unknowns are θ1 , θ2 , and θ3 2. The equilibrium relations are

Victor Saouma

M21 + M23

= 0

(23.39-a)

M32

= 0

(23.39-b)

Structural Engineering

23.4 Slope Deﬂection; Direct Solution

23–11

Draft

5 kips 2 kip/ft θ 2

1 0 0 1 0 1 0 1 0 1 0 1 0 1

11 00 11 00

1

2

20’

111 000 111 000

15’

θ 3

3

15’

0 40.97

79.52

Figure 23.7: Two-Span Beam, Slope Deﬂection 3. The ﬁxed end moments are given by Eq. 23.23-b and 23.26 (−2)(20)2 wL2 =− = 66.67 k.ft 12 12 (−5)(30) PL =− = 18.75 k.ft =− 8 8

F M12

=

F −M21 =−

(23.40-a)

F M23

=

F −M32

(23.40-b)

4. The members end moments in terms of the rotations are (Eq. 23.30 and 23.31) M12

=

M21

=

M23

= =

M32

= =

2EI EI F θ2 + 66.67 θ2 + M12 = L1 10 4EI EI F F θ2 − 66.67 2EK12 (2θ2 ) + M21 = θ2 + M21 = L1 5 2EI F F 2EK23 (2θ2 + θ3 ) + M23 = (2θ2 + θ3 ) + M23 L2 EI EI θ2 + θ3 + 18.75 7.5 15 2EI F F 2EK23 (θ2 + 2θ3 ) + M32 = (θ2 + 2θ3 ) + M32 L2 EI EI θ2 + θ3 − 18.75 15 7.5 F 2EK12 (θ2 ) + M12 =

(23.41-a) (23.41-b)

(23.41-c)

(23.41-d) (23.41-e)

5. Substituting into the equilibrium equations EI EI θ2 − 66.67 + θ2 + 5 7.5 EI θ2 + 15 or

Victor Saouma

EI θ3 + 18.75 = 0 15 EI θ3 − 18.75 = 0 7.5

5 1 718.8 θ2 EI = 1 2 θ3 281.25

Stiﬀness Matrix

(23.42-a) (23.42-b)

(23.43)

Structural Engineering

23–12

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Draft

which will give EIθ2 = 128.48 and EIθ3 = 76.38 6. Substituting for the moments M12

=

M21

=

M23

=

M32

=

12.85 + 66.67 = 79.52 k.ft 128.48 − 66.67 = −40.97 k.ft 5 √ 128.48 76.38 + + 18.75 = 40.97 k.ft 7.5 15 √ 128.48 76.38 + − 18.75 = 0 k.ft 15 7.5

(23.44-a) (23.44-b) (23.44-c) (23.44-d)

The ﬁnal moment diagram is also shown in Fig. 23.8. We note that the midspan moment has to be separately computed from the equations of equilibrium in order to complete the diagram.

43

Example 23-3: Two-Span Beam, Slope Deﬂection, Initial Deﬂection, (Arbabi 1991) Determine the end moments for the previous problem if the middle support settles by 6 inches, Fig. 23.8.

1 0 0 1 0 1 0 1 0 1 0 1 0 1

111 000 000 111

11 00 00 11

1

6"

3

2

20’

15’

15’

Figure 23.8: Two Span Beam, Slope Deﬂection, Moment Diagram Solution: 1. Since we are performing a linear elastic analysis, we can separately analyze the beam for support settlement, and then add then add the moments to those due to the applied loads. 2. The unknowns are θ1 , θ2 , and θ3 3. The equilibrium relations are M21 + M23 M32

= 0 = 0

(23.45-a) (23.45-b)

4. The members end moments in terms of the rotations are (Eq. 23.30 and 23.31) ∆ 3EI EI θ2 + M12 = 2EK12 θ2 − 3 = L12 10 400 ∆ 3EI EI θ2 + M21 = 2EK12 2θ2 − 3 = L12 5 400 ∆ EI EI EI M23 = 2EK23 2θ2 + θ3 − 3 θ2 + θ3 + = L23 7.5 15 300 ∆ EI EI EI θ2 + θ3 + M32 = 2EK23 θ2 + 2θ3 − 3 = L23 15 7.5 300 Victor Saouma

(23.46-a) (23.46-b) (23.46-c) (23.46-d)

Structural Engineering

23.4 Slope Deﬂection; Direct Solution

23–13

Draft

5. Substituting into the equilibrium equations EI 3EI EI EI EIθ2 + + θ3 + 5 400 15 300 EI EI 5EI θ2 + θ3 + 15 7.5 300 100 20 θ2 EI = EI 20 40 θ3

Stiﬀness Matrix

or

=

0

(23.47-a)

=

0

(23.47-b)

− 13 4 −1

(23.48)

−1+ 5.5

5.5 which will give θ2 = − 180 = −0.031 radians and θ3 = 40 9 = −0.0097 radians 6. Thus the additional moments due to the settlement are

M12

=

M21

=

M23

=

M32

=

3EI EI (−0.031) + = 0.0044EI 10 400 3EI EI (−0.031) + = 0.0013EI 5 400 EI EI EI (−0.031) + (0.0097) + = 0.0015EI 7.5 15 300 √ EI EI EI θ2 + (0.0097) + = 0. 15 7.5 300

(23.49-a) (23.49-b) (23.49-c) (23.49-d)

Example 23-4: dagger Frames, Slope Deﬂection, (Arbabi 1991) Determine the end moments for the frame shown in Fig. 23.9.

θ2 ∆2

θ3 ∆

3 kips/ft

2

2 3

5

5’ 10 kips 5’ θ4 1

4

20’

6’

Figure 23.9: Frame Analysis by the Slope Deﬂection Method Solution: 1. The eﬀect of the 35 cantilever can be included by replacing it with its end moment. M3 = −wL Victor Saouma

L = −(3)(6)(3) = −54 k.ft 2

(23.50) Structural Engineering

23–14

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Draft

2. The unknowns displacements and rotations are • ∆2 and θ2 at joint 2. • θ3 and θ4 at joints 3 and 4.

We observe that due to the lack of symmetry, there will be a lateral displacement in the frame, and neglecting axial deformations, ∆2 = ∆3 . 3. The equilibrium relations are M21 + M23 M32 + M34

= 0 = −54

(23.51-a) (23.51-b)

M43 = 0 V12 + V43 − 10 = 0

(23.51-c) (23.51-d)

Thus we have four unknown displacements and four equations. However, the last two equations are in terms of the shear forces, and we need to have them in term of the end moments, this can be achieved through the following equilibrium relations V12

=

V43

=

M12 + M21 + 50 L12 M34 + M43 L34

(23.52-a) (23.52-b)

Hence, all four equations are now in terms of the moments. 4. The ﬁxed end moments fro member 23 are F M21 F M23

−(10)(10) PL =− = 12.5 k.ft 8 8 −(3)(20)2 wL2 =− = 100 k.ft = − 12 12 = −

(23.53-a) (23.53-b)

5. The members end moments in terms of the rotations are (Eq. 23.30 and 23.31) ∆2 F M12 = 2EK12 θ2 − 3 = 0.2EI(θ2 − 0.3∆2 ) + 12.5 + M12 L12 ∆2 F = 0.2EI(2θ2 − 0.3∆2 ) − 12.5 M21 = 2EK12 2θ2 − 3 + M21 L21

(23.54-a) (23.54-b)

=

F 2EK23 (2θ2 + θ3 ) + M23 = 0.1EI(2θ2 + θ3 ) + 100.

(23.54-c)

M32

=

(23.54-d)

M34

=

M43

=

2EK32 (θ2 + 2θ3 ) + = 0.1EI(θ2 + 2θ3 ) − 100. 3∆2 2EK34 2θ3 + θ4 − = 0.2EI(2θ3 + θ4 − 0.3∆2 ) L34 3∆2 2EK43 θ3 + 2θ4 − = 0.2EI(θ3 + 2θ4 − 0.3∆2 ) L34

M23

F M32

(23.54-e) (23.54-f)

6. Substituting into the equilibrium equations and dividing by EI 6θ2 + θ3 − 0.6∆2 θ2 + 6θ3 + 2θ4 − 0.6∆2 θ3 + 2θ4 − 0.3∆2

875 EI 460 = EI = 0 = −

(23.55-a) (23.55-b) (23.55-c)

and the last equilibrium equation is obtained by substituting V12 and V43 and multiplying by 10/EI: θ2 + θ3 + θ4 − 0.4∆2 = − Victor Saouma

83.3 EI

(23.56) Structural Engineering

23.5 Moment Distribution; Indirect Solution

Draft or



6 1 0 −0.6  1 6 2 −0.6 EI   0 1 2 −0.3 1 1 1 −0.4

Stiﬀness Matrix

which will give

 θ2    θ3 θ4    ∆2

   θ2   θ3    θ4   ∆2

23–15   −875       1 460 = 0   EI        −83.3    

  −294.8      1  68.4 = −240.6   EI        −1, 375.7

(23.57)

   

(23.58)

7. Substitution into the slope deﬂection equations gives the end-moments  M12     M21    M23 M32     M34    M43

  36.0        −47.88        47.88 =      −115.80             61.78     0        

23.5

Moment Distribution; Indirect Solution

23.5.1

Background

(23.59)

The moment distribution is essentially a variation of the slope deﬂection method, however rather than solving a system of n linear equations directly, the solution is achieved iteratively through a successive series of operations.

43

The method starts by locking all the joints, and then unlock each joint in succession, the internal moments are then “distributed” and balanced until all the joints have rotated to their ﬁnal (or nearly ﬁnal) position.

44

45

In order to better understand the method, some key terms must ﬁrst be deﬁned.

23.5.1.1

Sign Convention

The sign convention is the same as the one adopted for the slope deﬂection method, counter-clockwise moment atelement’s end is positive.

46

23.5.1.2

Fixed-End Moments

Again ﬁxed end moments are the same set of forces deﬁned in the slope deﬂection method for a beam which is rigidly connected at both ends.

47

Consistent with the sign convention, the ﬁxed end moments are the moments caused by the applied load at the end of the beam (assuming it is rigidly connected).

48

23.5.1.3

Stiﬀness Factor

We deﬁne the stiﬀness factor as the moment required to rotate the end of a beam by a unit angle of one radian, while the other end is ﬁxed. From Eq. 23.10, we set θ2 = v1 = v2 = 0, and θ1 = 1, this will

49

Victor Saouma

Structural Engineering

23–16

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Draft yield

K=

50

4EI Far End Fixed L

(23.60)

We note that this is slightly diﬀerent than the deﬁnition given in the slope deﬂection method (I/L).

51 If the far end of the beam is hinged rather than ﬁxed, then we will have a reduced stiﬀness factor. From Eq. 23.10 and 23.11, with M2 = v1 = v2 = 0, we obtain

M2 ⇒ θ2

2EIz (θ1 + 2θ2 ) = 0 L θ1 = − 2 =

(23.61-a) (23.61-b)

Substituting into M1 M1 θ2

= =

2EIz L − θ21

Comparing this reduced stiﬀness factor

52

(2θ1 + θ2 )

3EI L

M1 =

Far End Pinned

(23.62)

with the stiﬀness of a beam, we deﬁne the reduced stiﬀness

Reduced Stiﬀness Factor =

23.5.1.4

3EI L θ1

3 Kred = Far End Pinned K 4

(23.63)

Distribution Factor (DF)

If a member is applied to a ﬁxed-connection joint where there is a total of n members, then from equilibrium: M = M1 + M2 + · · · + Mn (23.64)

53

However, from Eq. 23.30, and assuming the other end of the member to be ﬁxed, then M = K1 θ + K2 θ + · · · + Kn θ

(23.65)

or DFi =

Ki Mi = M ΣKi

(23.66)

54 Hence if a moment M is applied at a joint, then the portion of M carried by a member connected to this joint is proportional to the distribution factor. The stiﬀer the member, the greater the moment carried.

55

Similarly, DF = 0 for a ﬁxed end, and DF = 1 for a pin support.

23.5.1.5 56

Carry-Over Factor

Again from Eq. 23.30, and 23.31 M1 M2

= =

2EK(2θ1 + θ2 − 3ψ) + M1F 2EK(θ1 + 2θ2 − 3ψ) + M2F

(23.67-a) (23.67-b)

we observe that if one end of the beam is restrained (θ2 = ψ = 0), and there is no member load, then the previous equations reduce to 1 M1 = 2EK(2θ1 ) (23.68) ⇒ M2 = M1 M2 = 2EK(θ1 ) 2 Victor Saouma

Structural Engineering

23.5 Moment Distribution; Indirect Solution

23–17

Draft

Hence in this case the carry-over factor represents the fraction of M that is “carried over” from the rotating end to the ﬁxed one.

57

CO =

58

(23.69)

If the far end is pinned there is no carry over.

23.5.2 59

1 Far End Fixed 2

Procedure

The general procedure of the Moment Distribution method can be described as follows: 1. Constrain all the rotations and translations. 2. Apply the load, and determine the ﬁxed end moments (which may be caused by element loading, or support translation). F F 3. At any given joint i equilibrium is not satisﬁed Mlef t = Mright , and the net moment is Mi

4. We enforce equilibrium by applying at the node −Mi , in other words we balance the forces at the node. 5. How much of Mi goes to each of the elements connected to node i depends on the distribution factor. 6. But by applying a portion of −Mi to the end of a beam, while the other is still constrained, from Eq. 23.30, half of that moment must also be carried over to the other end. 7. We then lock node i, and move on to node j where these operations are repeated (a) Sum moments (b) Balance moments (c) Distribute moments (K, DF ) (d) Carry over moments (CO) (e) lock node 8. repeat the above operations until all nodes are balanced, then sum all moments. 9. The preceding operations can be easily carried out through a proper tabulation. If an end node is hinged, then we can use the reduced stiﬀness factor and we will not carry over moments to it.

60

Analysis of frame with unsymmetric loading, will result in lateral displacements, and a two step analysis must be performed (see below).

61

23.5.3

Algorithm

1. Calculate the stiﬀness (K = 4EI/L, however this can often be simpliﬁed to I/L) factor for all the members and the distribution factors at all the joints. 2. If a member AB is pinned at B, then K AB = 3EI/L, and K BA = 4EI/L. Thus, we must apply the reduced stiﬀness factor to K AB only and not to K BA . 3. The carry-over factor is

1 2

for members with constant cross-section.

4. Find the ﬁxed-end moments for all the members. Note that even if the end of a member is pinned, determine the ﬁxed end moments as if it was ﬁxed. 5. Start out by ﬁxing all the joints, and release them one at a time. Victor Saouma

Structural Engineering

23–18

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Draft

6. If a node is pinned, start by balancing this particular node. If no node is pinned, start from either end of the structure. 7. Distribute the unbalanced moment at the released joint 8. Carry over the moments to the far ends of the members (unless it is pinned). 9. Fix the joint, and release the next one.

10. Continue releasing joints until the distributed moments are insigniﬁcant. If the last moments carried over are small and cannot be distributed, it is better to discard them so that the joints remain in equilibrium. 11. Sum up the moments at each end of the members to obtain the ﬁnal moments.

23.5.4

Examples

Example 23-5: Continuous Beam, (Kinney 1957) Solve for moments at A and B by moment distribution, using (a) the ordinary method, and (b) the simpliﬁed method.

10k

A

1 0 0 1 0 1 0 1 0 1 0 1 0 1 0 1

12’

11 00

B

10’

12.9

25.8

3.22

3.22 6.29

10’

11 00

C

0

3.71

37.1 12.9

9.51 M

25.8 Solution:

Victor Saouma

Structural Engineering

23.5 Moment Distribution; Indirect Solution

23–19

Draft

1. For this example the ﬁxed-end moments are computed as follows: F MBC

=

F MCB

=

PL (10)(20) = = +25.0 k.ft 8 8 −25.0 k.ft

(23.70-a) (23.70-b)

2. Since the relative stiﬀness is given in each span, the distribution factors are

DFAB

=

KAB ΣK

DFBA

=

KBA ΣK

DFBC

=

KBC ΣK

DFCB

=

KCB ΣK

5 = 0, ∞+5 5 = = 0.625, 8 3 = = 0.375, 8 3 = = 1. 3

(23.71-a)

=

(23.71-b) (23.71-c) (23.71-d)

3. The balancing computations are shown below. Joint Member K DF FEM

Total

A AB 5 0

B

C Balance BC CB 3 3 0.375 1 +25.0 -25.0 ✛ ❄ ✑ +12.5 ✲+25.0 C ✛ ✰ -14.1 ❄ -7.0 -11.7 B -23.4✑ ✛ ❄C +3.5 +7.0 ✑ ✛ ✲ ✰ -1.3 ❄ -1.1 -0.6 B -2.2 ✑ ✛ ❄C +0.6 +0.3 ✑ ✛ ✰ -0.1 ❄ -0.1 B -0.2 ✑ -12.9 -25.8 +25.8 0

CO

BA 5 0.625

BC AB; CB BC AB; CB BC AB

4. The above solution is that referred to as the ordinary method, so named to designate the manner of handling the balancing at the simple support at C. It is known, of course, that the ﬁnal moment must be zero at this support because it is simple. 5. Consequently, the ﬁrst step is to balance the ﬁxed-end moment at C to zero. The carry-over is then made immediately to B. When B is balanced, however, a carry-over must be made back to C simply because the relative stiﬀness of BC is based on end C of this span being ﬁxed. It is apparent, however, that the moment carried back to C (in this case, -7.0) cannot exist at this joint. Accordingly, it is immediately balanced out, and a carry-over is again made to B, this carry-over being considerably smaller than the ﬁrst. Now B is again balanced, and the process continues until the numbers involved become too small to have any practical value. 6. Alternatively, we can use the simpliﬁed method. It was previously shown that if the support at C is simple and a moment is applied at B, then the resistance of the span BC to this moment is reduced to three-fourths of the value it would have had with C ﬁxed. Consequently, if the relative stiﬀness of span BC is reduced to three-fourths of the value given, it will not be necessary to carry over to C. Joint Member K DF FEM

Total

Victor Saouma

A AB 5 0

B

C Balance CB 3 3.00 4 ×3 = 2.25 0.31 1 +25.0 -25.0 ✛ +25.0 ❄ C ✑ +12.5 ✛ ❄ ✰ ✑ -12.9 -11.7 B -25.8 -12.9 -25.8 +25.8 0 BA 5 0.69

CO

BC

BC AB

Structural Engineering

23–20

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Draft

7. From the standpoint of work involved, the advantage of the simpliﬁed method is obvious. It should always be used when the external (terminal) end of a member rests on a simple support, but it does not apply when a structure is continuous at a simple support. Attention is called to the fact that when the opposite end of the member is simply supported, the reduction factor for stiﬀness is always 34 for a prismatic member but a variable quantity for nonprismatic members. 8. One valuable feature of the tabular arrangement is that of dropping down one line for each balancing operation and making the carry-over on the same line. This practice clearly indicates the order of balancing the joints, which in turn makes it possible to check back in the event of error. Moreover, the placing of the carry-over on the same line with the balancing moments deﬁnitely decreases the chance of omitting a carry-over. 9. The correctness of the answers may in a sense be checked by verifying that ΣM = 0 at each joint. However, even though the ﬁnal answers satisfy this equation at every joint, this in no way a check on the initial ﬁxed-end moments. These ﬁxed-end moments, therefore, should be checked with great care before beginning the balancing operation. Moreover, it occasionally happens that compensating errors are made in the balancing, and these errors will not be apparent when checking ΣM = 0 at each joint. 10. To draw the ﬁnal shear and moment diagram, we start by drawing the free body diagram of each beam segment with the computed moments, and then solve from statics for the reactions: 12, 9 + 25.8 − 12VA = 0 25.8 +

VA + VBL (10)(10) − 20VBR

=0 =0

6.29 + VC − 10 = 0 −VBL − VBR + RB = 0 Check: RA + RB + RC − 10 + MBC

⇒ VA = RA = 3.22 k ❄

(23.72-a)

⇒ ⇒

(23.72-b) (23.72-c)

VBL VBR

= 3.22 k ✻ = 6.29 k ✻ ⇒ VC = RC 3.71 k ✻ ⇒ RB = 9.51 k ✻ =

−3.22 + 9.51 + 3.71 − 10 = 0

= (3.71)(10) = 37.1 k.ft

√

(23.72-d) (23.72-e) (23.72-f) (23.72-g)

Example 23-6: Continuous Beam, Simpliﬁed Method, (Kinney 1957) Using the simpliﬁed method of moment distribution, ﬁnd the moments in the following continuous beam. The values of I as indicated by the various values of K, are diﬀerent for the various spans. Determine the values of reactions, draw the shear and bending moment diagrams, and sketch the deﬂected structure.

Solution: 1. Fixed-end moments: F = −(0.5)(10) = −5.0 k.ft MAO

(23.73)

For the 1 k load:

Victor Saouma

F MAB

=

F MBA

=

P ab2 (1)(5)(152 ) = = +2.8 k.ft 2 L 202 P a2 b (1)(52 )(15) = = −0.9 k.ft L2 202

(23.74-a) (23.74-b)

Structural Engineering

23.5 Moment Distribution; Indirect Solution

23–21

Draft

For the 4 k load: F MAB

PL 8

=

=

(4)(20) 8

= +10.0 k.ft

(23.75-a)

= −10.0 k.ft

F MBA

(23.75-b)

For the uniform load: F MCD

=

F MDC

=

(0.2)(152 ) wL2 = = +3.8 k.ft 12 12 −3.8 k.ft

(23.76-a) (23.76-b)

2. The balancing operation is shown below Joint Member K DF FEM

Total

A AO 0 0 -5.0

-5.0

B C D AB BA BC CB CD DC 3 20 60 60 40 40 4 (20) = 15 1 0.2 0.8 0.6 0.4 0 +12.8 -10.9 +3.8 -3.8 ✲ -7.8 -3.9 ◗ s +11.9 ✲+5.9 ◗ +2.9 ❄ ◗ ✛ ❄ s-3.9 ✲-1.9 ◗ ✑ -2.9 ✲ -5.8 ✰ +2.3 ❄ +1.1 +0.6 ✑ ◗ ✛ s-0.4 ✲-0.2 ◗ -0.7 ❄ ✑ -0.3 ❄ ✰ ✑ +0.1 +0.2 +5.0 -11.2 +11.2 +0.5 -0.5 -5.9

Balance

CO

A B C B C B

BA CB DC; BC CB DC; BC

3. The only new point in this example is the method of handling the overhanging end. It is obvious that the ﬁnal internal moment at A must be 5.0 k.ft and, accordingly, the ﬁrst step is to balance out 7.8 k.ft of the ﬁxed-end moment at AB, leaving the required 5.0 k.ft for the internal moment at AB. Since the relative stiﬀness of BA has been reduced to three-fourths of its original value, to permit considering the support at A as simple in the balancing, no carry-over from B to A is required. 4. The easiest way to determine the reactions is to consider each span as a free body. End shears are ﬁrst determined as caused by the loads alone on each span and, following this, the end shears caused by the end moments are computed. These two shears are added algebraically to obtain the net end shear for each span. An algebraic summation of the end shears at any support will give the total reaction.

Victor Saouma

Structural Engineering

23–22

Draft

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Example 23-7: Continuous Beam, Initial Settlement, (Kinney 1957) For the following beam ﬁnd the moments at A, B, and C by moment distribution. The support at C settles by 0.1 in. Use E = 30, 000 k/in2 .

Solution: 1. Fixed-end moments: Uniform load:

F MBA

(5)(202 ) wL2 = = +167 k.ft 12 12 = −167 k.ft

F MCD

=

F MAB

=

(23.77-a) (23.77-b)

Concentrated load:

F MDC

Victor Saouma

(10)(30) PL = = +37.5 k.ft 8 8 = −37.5 k.ft

(23.78-a) (23.78-b) Structural Engineering

23.5 Moment Distribution; Indirect Solution

23–23

Draft

Moments caused by deﬂection: F MBC

=

F MCB

=

F MCD

=

F MDC

=

6EI∆ 6(30, 000)(1, 200)(0.1) = = +1, 500 k.in = +125 k.ft L2 (120)2

(23.79-a)

+125 k.ft 6EI∆ (6)(30, 000)(7, 200)(0.1) = = 1, 000 k.in = −83 k.ft 2 L (360)2

(23.79-b)

−83 k.ft

(23.79-d)

(23.79-c)

2. Moment distribution Joint Member K DF FEM Load FEM ∆

Total

A AB 10 0 +167

B

D DC 3 20 4 (20) = 15 0.6 1 +38 -38 +125 +125 -83 -83 ✛ +60 +121 ✑ ✛ ✰ -84 ❄ -28 -56 ✑ ✑ ✰ +35 ❄✲ +17 +17 ✛ +35✑ ✛ ❄ ◗◗ -3 -7 s -10 ✑ ✛ ✰ +1 ❄ +1 +2 ✑ +185 -130 +130 +79 -79 0 BA 10 0.5 -167

C

BC 10 0.5

CB 10 0.4

Balance

CO

D C B C B

CD BC AB; CB BC

CD

The ﬁxed-end moments caused by a settlement of supports have the same sign at both ends of each span adjacent to the settling support. The above computations have been carried to the nearest k.ft, which for moments of the magnitudes involved, would be suﬃciently close for purposes of design.

Example 23-8: Frame, (Kinney 1957) Find all moments by moment distribution for the following frame Draw the bending moment diagram and the deﬂected structure.

Solution:

Victor Saouma

Structural Engineering

23–24

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Draft

1. The ﬁrst step is to perform the usual moment distribution. The reader should fully understand that this balancing operation adjusts the internal moments at the ends of the members by a series of corrections as the joints are considered to rotate, until ΣM = 0 at each joint. The reader should also realize that during this balancing operation no translation of any joint is permitted. 2. The ﬁxed-end moments are F MBC

=

F MCB

=

(18)(12)(62 ) = +24 k.ft 182 (18)(6)(122 ) = −48 k.ft 182

(23.80-a) (23.80-b)

3. Moment distribution Joint Member K DF FEM

Total

A AB 10 0

B

C D BC CB CD DC 20 20 15 15 0.667 0.571 0.429 0 +24.0 -48.0 ✛ ✲ +10.3 ✑ +13.7 ✲ +27.4 +20.6 ✛ ❄ ✰ ✑ -12.5◗ -6.3 -25.1 -12.6 ✛ ❄ ◗ s +5.4 ✲ +2.7 ✑ +3.6 ✲ +7.1 ✛ ❄ ✰ ✑ -1.2 -0.6 -2.4 -1.2 ✛ ❄ ◗◗ s +0.5 ✲ +0.02 +0.7 +0.3 ✑ ❄ ✰ ✑ -0.1 -0.2 -6.9 -13.9 +13.9 -26.5 +26.5 +13.2

Balance

CO

FEM C B C B C B

DC; BC AB; CB BC; DC AB; CB BC; DC

BA 10 0.333

4. The ﬁnal moments listed in the table are correct only if there is no translation of any joint. It is therefore necessary to determine whether or not, with the above moments existing, there is any tendency for side lurch of the top of the frame. 5. If the frame is divided into three free bodies, the result will be as shown below.

Inspection of this sketch indicates that if the moments of the ﬁrst balance exist in the frame, there is a net force of 1.53 − 0.80 = 0.73 k tending to sway the frame to the left. In order to prevent side-sway, and thus allow these moments to exist (temporarily, for the purpose of the analysis), it is necessary that an imaginary horizontal force be considered to act to the right at B or C. This force is designated as the artiﬁcial joint restraint (abbreviated as AJR) and is shown below.

6. This illustration now shows the complete load system which would have to act on the structure if the ﬁnal moments of the ﬁrst balance are to be correct. The AJR, however, cannot be permitted to remain, Victor Saouma

Structural Engineering

23.5 Moment Distribution; Indirect Solution

23–25

Draft

and thus its eﬀect must be cancelled. This may be accomplished by ﬁnding the moments in the frame resulting from a force equal but opposite to the AJR and applied at the top. 7. Although it is not possible to make a direct solution for the moments resulting from this force, they may be determined indirectly. Assume that some unknown force P acts on the frame, as shown below

and causes it to deﬂect laterally to the left, without joint rotation, through some distance ∆. Now, regardless of the value of P and the value of the resulting ∆, the ﬁxed-end moments induced in the ends of the columns must be proportional to the respective values of KM . 62

Recalling that the ﬁxed end moment is M F = 6EI L∆2 = 6EKm ∆, where Km = ∆ ⇒

F MAB F MDC

= =

F F MDC MAB = 6EKm 6EKm AB Km 10 = DC Km 15

I L2

=

K L

we can write (23.81-a) (23.81-b)

These ﬁxed-end moments could, for example, have the values of −10 and −15 k.ft or −20 and −30, or −30 and −45, or any other combination so long as the above ratio is maintained. The proper procedure is to choose values for these ﬁxed-end moments of approximately the same order of magnitude as the original ﬁxed-end moments due to the real loads. This will result in the same accuracy for the results of the balance for the side-sway correction that was realized in the ﬁrst balance for the real loads. Accordingly, it will be assumed that P , and the resulting ∆, are of such magnitudes as to result in the ﬁxed-end moments shown below 63

8. Obviously, ΣM = 0 is not satisﬁed for joints B and C in this deﬂected frame. Therefore these joints must rotate until equilibrium is reached. The eﬀect of this rotation is determined in the distribution below

Victor Saouma

Structural Engineering

23–26

Draft

Joint Member K DF FEM

Total

KINEMATIC INDETERMINANCY; STIFFNESS METHOD A AB 10 0 -30.0

B

D CD DC 15 15 0.429 0 -45.0 -45.0 ✑ ✰ +19.2 ❄✲+9.6 +12.9 ✛ +25.8✑ ✑ ✛ ✲ ✰ +11.4 ❄ +5.7 +2.8 +5.7✑ ✛ ❄ ◗◗ ✲ -1.2 -3.3 s -2.4 -1.6 ✑ ✛ ✲ ❄ ✰ ✑ +0.5 +0.5◗ +0.2 +1.1 ❄ ◗ s -0.2 ✲ -0.1 -0.3 -27.0 -23.8 +23.8 +28.4 -28.4 -36.7 BA 10 0.333 -30.0

C

BC 20 0.667

Balance

CO

CB 20 0.571

C B C B C

BC; AB; BC; AB;

DC CB DC CB

9. During the rotation of joints B and C, as represented by the above distribution, the value of ∆ has remained constant, with P varying in magnitude as required to maintain ∆. 10. It is now possible to determine the ﬁnal value of P simply by adding the shears in the columns. The shear in any member, without external loads applied along its length, is obtained by adding the end moments algebraically and dividing by the length of the member. The ﬁnal value of P is the force necessary to maintain the deﬂection of the frame after the joints have rotated. In other words, it is the force which will be consistent with the displacement and internal moments of the structure as determined by the second balancing operation. Hence this ﬁnal value of P will be called the consistent joint force (abbreviated as CJF). 11. The consistent joint force is given by CJF =

+27.0 + 23.8 28.4 + 36.7 + = 1.95 + 2.50 = 4.45 k 26 26

(23.82)

and inspection clearly indicates that the CJF must act to the left. 12. Obviously, then, the results of the last balance above are moments which will exist in the frame when a force of 4.45 k acts to the left at the top level. It is necessary, however, to determine the moments resulting from a force of 0.73 k acting to the left at the top level, and some as yet unknown factor “z” times 4.45 will be used to represent this force acting to the left.

13. The free body for the member BC is shown above. ΣH = 0 must be satisﬁed for this ﬁgure, and if forces to the left are considered as positive, the result is 4.45z − 0.73 = 0, and z = +0.164.

(23.83)

If this factor z = +0.164 is applied to the moments obtained from the second balance, the result will be the moments caused by a force of 0.73 k acting to the left at the top level. If these moments are then added to the moments obtained from the ﬁrst balance, the result will be the ﬁnal moments for the frame, the eﬀect of the AJR having been cancelled. This combination of moments is shown below. Joint Member M from 1st balance z × M from 2nd balance Final moments

A AB -6.9 -4.4 -11.3

B BA -13.9 -3.9 -17.8

BC +13.9 +3.9 +17.8

C CB -26.5 +4.7 -21.8

CD +26.5 -4.7 +21.8

D DC +13.2 -6.0 +7.2

14. If the ﬁnal moments are correct, the shears in the two columns of the frame should be equal and opposite to satisfy ΣH = 0 for the entire frame. This check is expressed as +11.3 + 17.8 −21.8 − 7.2 + = 0, 26 26

(23.84)

+1.12 − 1.11 = 0(nearly)

(23.85)

and

Victor Saouma

Structural Engineering

23.5 Moment Distribution; Indirect Solution

Draft

23–27

The signs of all moments taken from the previous table have been reversed to give the correct signs for the end moments external to the columns. It will be remembered that the moments considered in moment distribution are always internal for each member. However, the above check actually considers each column as a free body and so external moments must be used. 15. The moment under the 18 k load is obtained by treating BC as a free body:

M18 = (5.77)(12) − 17.8 = +51.5 k.ft

(23.86)

16. The direction of side-lurch may be determined from the obvious fact that the frame will always lurch in a direction opposite to the AJR. If required, the magnitude of this side lurch may be found. The procedure which follows will apply.

A force P of suﬃcient magnitude to result in the indicated column moments and the lurch ∆ was applied to the frame. During the second balance this value of ∆ was held constant as the joints B and C rotated, and the value of P was considered to vary as necessary. The ﬁnal value of P was found to be 4.45 k. Since ∆ was held constant, however, its magnitude may be determined from the equation M = 6EI∆/L2 , where M is the ﬁxed-end moment for either column, I is the moment of inertia of that column, and L is the length. This ∆ will be the lurch for 4.45 k acting at the top level. For any other force acting horizontally, ∆ would vary proportionally and thus the ﬁnal lurch of the frame would be the factor z multiplied by the ∆ determined above.

Example 23-9: Frame with Side Load, (Kinney 1957) Find by moment distribution the moments in the following frame

Victor Saouma

Structural Engineering

23–28

Draft

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Solution: The ﬁrst balance will give the results shown AB -7.2

BA -14.6

BC +14.6

CB -22.5

CD -22.5

DC 0

A check of the member BC as a free body for ΣH = 0 will indicate that an AJR is necessary as follows: AJR + 0.84 − 0.87 − 5.0 = 0

(23.87)

AJR = +5.03 in the direction assumed

(23.88)

from which The values of KM for the two columns are shown, with KM for column CD being K/2L because of the pin at the bottom. The horizontal displacement ∆ of the top of the frame is

assumed to cause the ﬁxed-end moments shown there. These moments are proportional to the values of KM and of approximately the same order of magnitude as the original ﬁxed-end moments due to the real loads. The results of balancing out these moments are AB -34.4

BA -28.4

CJF =

BC +28.4

CB +23.6

CD -23.6

+34.4 + 28.4 + 23.6 = 3.32 k 26

DC 0 (23.89)

and 5.03 − z(3.32) = 0,

(23.90)

z = +1.52.

(23.91)

from which The ﬁnal results are Victor Saouma

Structural Engineering

23.5 Moment Distribution; Indirect Solution

Draft

AB -7.2 -52.1 -59.3

M from 1st balance z × M 2nd balance Final moments

BA -14.6 -43.0 -57.6

23–29 BC +14.6 +43.0 +57.6

CB -22.5 +35.8 +13.3

CD +22.5 -35.8 -13.3

DC 0 0 0

If these ﬁnal moments are correct, the sum of the column shears will be 5.0 k: Sum of column shears: ΣV =

59.3 + 57.6 + 13.3 = 5.01 k 26

(23.92)

The 5 k horizontal load acting at C enters into the problem only in connection with the determination of the AJR. If this load had been applied to the column CD between the ends, it would have resulted in initial ﬁxed-end moments in CD and these would be computed in the usual way. In addition, such a load would have entered into the determination of the AJR, since the horizontal reaction of CD against the right end of BC would have been computed by treating CD as a free body.

Example 23-10: Moment Distribution on a Spread-Sheet Analyse the following frame

12’

8’ 20 k C

B

I = cst 15’

10’ D

A Solution:

Victor Saouma

Structural Engineering

23–30

KINEMATIC INDETERMINANCY; STIFFNESS METHOD

Draft

Sheet1

Artificially Restrained Structure A B C D AB BA BC CB CD DC Balance Lenght 15 15 20 20 10 10 EI 30 30 30 30 30 30 "arbitrary" K 2.0 2.0 1.5 1.5 3.0 3.0 DF 1.0 0.6 0.4 0.3 0.7 1.0 FEM 38.4 -57.6 -11.0 -21.9 -16.5 -8.2 B 11.0 21.9 43.9 21.9 C -3.1 -6.3 -4.7 -2.4 B 0.4 0.8 1.6 0.8 C -0.1 -0.2 -0.2 -0.1 B 0.0 0.0 0.1 0.0 C Total -14.2 -28.4 28.5 -45.5 45.5 22.8 . Horizontal displacement Delta

Lenght EI K DF FEM

A B AB BA 15 15 30 30 2.0 2.0 1.0 0.6 -8.0 -8.0 2.3 4.6 -0.8

Total

-6.5

1st z 2nd Final

AB -14.2 -9.6 -23.9

C BC CB CD 20 20 10 30 30 30 1.5 1.5 3.0 0.4 0.3 0.7 -18.0 3.4 1.7 2.7 5.4 10.9 -1.6 -1.2 -0.6 0.1 0.2 0.4 -0.1 0.0 -5.0 5.0 6.8 -6.8 Final Moments BA BC CB CD -28.4 28.5 -45.5 45.5 -7.5 7.5 10.0 -10.0 -35.9 35.9 -35.5 35.5

D DC Balance 10 30 "arbitrary" 3.0 1.0 -18.0 B 5.4 C B 0.2 C B -12.4 .

CO

V 6EI/L^2 M^F H_A -2.84 AB 0.80 -8 H_D 6.826355685 DC 1.80 -18 AJR 3.98 Delta -10 Delta to the left, will cause -ve M^F Frame sways to the right

AB, CB BC, DC AB, CB BC, DC AB, CB BC, DC

CO

AB, CB BC, DC AB, CB BC, DC AB, CB

H_A H_D CJF z

V -0.77 -1.9 -2.68 1.49

DC 22.8 -18.4 4.4

Page 1

Victor Saouma

Structural Engineering

Draft Chapter 24

REINFORCED CONCRETE BEAMS; Part I 24.1

Introduction

Recalling that concrete has a tensile strength (ft ) about one tenth its compressive strength (fc ), concrete by itself is a very poor material for ﬂexural members.

1

2 To provide tensile resistance to concrete beams, a reinforcement must be added. Steel is almost universally used as reinforcement (longitudinal or as ﬁbers), but in poorer countries other indigenous materials have been used (such as bamboos). 3 The following lectures will focus exclusively on the ﬂexural design and analysis of reinforced concrete rectangular sections. Other concerns, such as shear, torsion, cracking, and deﬂections 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 (ACI-318). 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 ﬂexural (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.

24.1.1 6

Notation

In R/C design, it is customary to use the following notation As b c d fc fr fs ft fy h ρ

Area of steel Width Distance from top of compressive ﬁbers to neutral axis Distance from the top of the compressive ﬁbers 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

24–2

Draft 24.1.2

REINFORCED CONCRETE BEAMS; Part I

Modes of Failure

7 A reinforced concrete beam is a composite structure where concrete provides the compression and steel the tension.

8

Failure is initiated by 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 under-reinforced. This will result in excessive rotation and deformation prior to failure. 2. Crushing of the concrete, when the concrete strain reaches its ultimate value (εc = εu = 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 eﬃcient 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.

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 under-reinforced.

10

24.1.3 11

Analysis vs Design

In R/C we always consider one of the following problems:

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 ﬁxed by the architect. 12

We often consider the maximum moment along a member, and design accordingly.

24.1.4

Basic Relations and Assumptions

In developing a design/analysis method for reinforced concrete, the following basic relations will be used:

13

1. Equilibrium: of forces and moment at the cross section. 1) ΣFx = 0 or Tension in the reinforcement = Compression in concrete; and 2) ΣM = 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 Iu = .003 in compression irrespective of fc . 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 coeﬃcients of thermal expansion under normal temperature. Plane section remain plane ⇒ strain is proportional to distance from neutral axis.

Victor Saouma

Structural Engineering

24.2 Cracked Section, Ultimate Strength Design Method

24–3

Draft 24.1.5

ACI Code

The ACI code is based on limit strength, or ΦMn ≥ Mu thus a similar design philosophy is used as the one adopted by the LRFD method of the AISC code, ACI-318: 8.1.1; 9.3.1; 9.3.2

15

16

The required strength is based on (ACI-318: 9.2) U

17

=

1.4D + 1.7L

(24.1)

=

0.75(1.4D + 1.7L + 1.7W ) (24.2)

We should consider the behaviors of a reinforced concrete section under increasing load: 1. Section uncracked 2. Section cracked, elastic 3. Section cracked, limit state

The second analysis gives rise to the Working Stress Design (WSD) method (to be covered in Structural Engineering II), and the third one to the Ultimate Strength Design (USD) method.

24.2

Cracked Section, Ultimate Strength Design Method

24.2.1

Equivalent Stress Block

In determining the limit state moment of a cross section, we consider Fig. 24.1. Whereas the strain distribution is linear (ACI-318 10.2.2), the stress distribution is non-linear because the stress-strain curve of concrete is itself non-linear beyond 0.5fc .

18

Thus we have two alternatives to investigate the moment carrying capacity of the section, ACI-318: 10.2.6

19

1. Use the non-linear 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:

20

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. 21

From the actual stress distribution we have the following relations C M

= T = As fs = As fy = T (d − βc)

We select an equivalent rectangular stress distribution, often referred to as the Whitney Stress Block such that the constant stress is equal to 0.85fc and is applied over a depth a = β1 c, ACI-318: 10.2.7

22

Comparing the actual and the equivalent stress distributions, the two requirements previously stated translate into the following mathematical relations

23

αfc bc βc Victor Saouma

= 0.85fc bβ1 c = 12 β1 c Structural Engineering

24–4

REINFORCED CONCRETE BEAMS; Part I

Draft

Figure 24.1: Cracked Section, Limit State Those two equations yield β1 α β1

= 2β = .85

We have two equations and three unknowns (α, β1 , and β). Thus we need to use test data to solve this problem1 . From experimental tests, the following relations are obtained

24

fc (ppsi) α β β1

< 4,000 .72 .425 .85

5,000 .68 .400 .80

6,000 .64 .375 .75

7,000 .60 .350 .70

8,000 .56 .325 .65

Thus we have a general equation for β1 (ACI-318 10.2.7.3): β1

= =

.85 1 .85 − (.05)(fc − 4, 000) 1,000

if fc ≤ 4, 000 if 4, 000 < fc < 8, 000

(24.3)

Figure 24.2: Whitney Stress Block 1 This approach is often used in Structural Engineering. Perform an analytical derivation, if the number of unknowns then exceeds the number of equations, use experimental data.

Victor Saouma

Structural Engineering

24.2 Cracked Section, Ultimate Strength Design Method

24–5

Draft 24.2.2

Balanced Steel Ratio

Next we seek to determine the balanced steel ratio ρb such that failure occurs by simultaneous yielding of the steel fs = fy and crushing of the concrete εc = 0.003, ACI-318: 10.3.2 We will separately consider the two failure possibilities:

25

Tension Failure: we stipulate that the steel stress is equal to fy : ρfy ρ = Abds ⇒c= d As fy = .85fc ab = .85fc bβ1 c 0.85fc β1

(24.4)

Compression Failure: where the concrete strain is equal to the ultimate strain; From the strain diagram .003 εc = 0.003 d (24.5) ⇒ c = fs c .003 = + .003 d .003+εs Es

26

Balanced Design is obtained by equating Eq. 24.4 to Eq. 24.5 and by replacing ρ by ρb and fs by fy :  ρfy .003 d  fs  ρf 0.85fc β1 d = +.003 .003 Es b y d d = fy fs = fy β  .85f c 1  Es + .003 ρ = ρb

When we replace Es by 29, 000 ksi we obtain ρb = .85β1

fc 87, 000 fy 87, 000 + fy

(24.6)

This ρb 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. Because we need to have ample warning against failure, hence we prefer to have an under-reinforced section. Thus, the ACI code stipulates:

27

ρ < .75ρb

(24.7)

In practice, depending on the relative cost of steel/concrete and of labour it is common to select lower values of ρ. If ρ < 0.5ρb (thus we will have a deeper section) then we need not check for deﬂection.

28

29 A minimum amount of reinforcement must always be used to prevent temperature and shrinkage cracks:

ρmin ≥

30

200 fy

(24.8)

The ACI code adopts the limit state design method

MD = ΦMn > Mu

24.2.3

Φ = Φb = 0.9

(24.9)

Analysis

Given As , b, d, fc , and fy determine the design moment: 1. ρact =

As bd f

2. ρb = (.85)β1 fyc Victor Saouma

87 87+fy

Structural Engineering

24–6

REINFORCED CONCRETE BEAMS; Part I

Draft

3. If ρact < ρb (that is failure is triggered by yielding of the steel, fs = fy ) 2 As fy As fy From Equilibrium a = .85f b c = Φ A f M d − 0.59 D s y fc b MD = ΦAs fy d − a2

Mn

Combining this last equation with ρ =

MD

As bd

yields

fy = Φρfy bd 1 − .59ρ fc 2

(24.10)

4. † If ρact > ρb is not allowed by the code as this would be an over-reinforced 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 εs (which was equal to εy in the previous case) (b) Location of the neutral axis c. We have two equations to solve this problem Equilibrium: of forces c=

As fs .85fc bβ1

(24.11)

Strain compatibility: since we know that at failure the maximum compressive strain εc is equal to 0.003. Thus from similar triangles we have c .003 = d .003 + εs

(24.12)

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 = Eεs are determined then β1 c MD = ΦAs fs d − 2

24.2.4

(24.13)

Design

We distinguish between two cases. The ﬁrst one has dimensions as well as steel area unknown, the second has the dimensions known (usually speciﬁed by the architect or by other constraints), and we seek As .

31

b, d and As unknowns and MD known: 1. We assume that ρ = .75ρb with ρb determined from Eq. 24.6 f 87, 000 ρ = 0.75 .85β1 c fy 87, 000 + fy Often times, a designer will take a smaller portion of ρb if the relative cost of steel is higher than the one of concrete, resulting in deeper sections. The ratio may range from 0.4 to 0.75.

Victor Saouma

Structural Engineering

24.2 Cracked Section, Ultimate Strength Design Method

Draft

2. From Eq. 24.10

24–7

fy MD = Φ ρfy 1 − .59ρ bd2 fc

R

or

fy R = ρfy 1 − .59ρ fc

(24.14)

which does not depend on unknown quantities2 . Then solve for bd2 : bd2 =

MD ΦR

(24.15)

Note that in most case R is around 1 ksi 3. solve for b and d (this will require either an assumption on one of the two, or on their ratio). 4. As = ρbd dag b, d and Md known, As unknown: In this case there is no assurance that we can have a design with ρb . If the section is too small, then it will require too much steel resulting in an over-reinforced 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 (ΣM = 0) As =

MD Φfs d − a2

(24.16)

5. Check equilibrium of forces in the x direction (ΣFx = 0) a=

As fs .85fc b

(24.17)

6. Check assumption of fs from the strain diagram .003 d−c εs = ⇒ fs = Es .003 < fy d−c c c where c =

(24.18)

a β1 .

7. Iterate until convergence is reached.

Example 24-1: Ultimate Strength Capacity Determine the ultimate Strength of a beam with the following properties: b = 10 in, d = 23 in, As = 2.35 in2 , fc = 4, 000 psi and fy = 60 ksi. Solution: As bd

2.35 (10)(23) = .0102 f 87 4 87 ρb = .85β1 fyc 87+f = (.85)(.85) 60 87+60 = y As fy (2.35)(60) = (.85)(4)(10) = 4.147 in a = .85f cb 2 Note analogy with Eq. 14.5 M = F Z for stell beams. y Mn = p (2.35)(60)(23 − 4.147 2 ) = 2, 950 k.in MD = ΦMn = (.9)(2, 950) = 2, 660 k.in

ρact

Victor Saouma

=

=

.02885 > ρact

√

Structural Engineering

24–8

REINFORCED CONCRETE BEAMS; Part I

Draft

Note that from the strain diagram c= Alternative solution Mn

= = =

4.414 a = = 4.87 in 0.85 0.85

f ρact fy bd2 1 − .59ρact fy c f As fy d 1 − .59ρact fy c % $ (2.35)(60)(23) 1 − (.59) 60 4 (.01021) = 2, 950 k.in = 245 k.ft

Example 24-2: 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 MD ρb ρ R bd2

= 1.4(1.27) + 1.7(2.44) = 5.92 k/ft (5.92)

k/ft(15)2 ft2

= (12) in/ft = 2, 000 k.in 8 fc 87 = .85β1 fy 87+fy 3 87 = (.85)(.85) 40 87+40 = 0.040 = .75ρb = .75(0.04) = 0.030 f = ρfy 1 − .59ρ fy % $ c = 0.917 ksi = (0.03)(40) 1. − (0.59)(0.03) 40 3 2 2,000 Md k.in in = ΦR = (0.9)(0.917) = 2, 423 in3 k

# Assume b = 10 in, this will give d = 2,423 10 = 15.57 in. We thus adopt b = 10 in and d = 16 in. Finally, As = ρbd = (0.030)(10)(16) = 4.80 in2 we select 3 bars No. 11

Example 24-3: Beam Design II Select the reinforcement for a cross section with b = 11.5 in; d = 20 in to support a design moment Md = 1, 600 k.in using fc = 3 ksi; and fy = 40 ksi Solution: 1. Assume a =

d 5

=

20 5

= 4 in and fs = fy

2. Equilibrium of moments: As =

1, 600 k.in MD == = 2.47 in2 Φfy (d − a2 ) (.9)(40) ksi(20 − 42 ) in

3. Check equilibrium of forces: a=

Victor Saouma

(2.47) in2 (40) ksi As fy = = 3.38 in .85fc b (.85)(3) ksi(11.5) in Structural Engineering

24.3 ACI Code

24–9

Draft

4. We originally assumed a = 4, at the end of this ﬁrst iteration a = 3.38, let us iterate again with a = 3.30

5. Equilibrium of moments: As =

1, 600 k.in MD a == Φfy (d − 2 ) (.9)(40) ksi(20 −

3.3 2 )

in

= 2.42 in2

6. Check equilibrium of forces: a=

√ (2.42) in2 (40) ksi As fy = = 3.3 in .85fc b (.85)(3) ksi(11.5) in

7. we have converged on a. 8. Actual ρ is ρact =

2.42 (11.5)(20)

= .011

9. ρb is equal to

87 fc 87 3 = .037 = (.85)(.85) fy 87 + fy 40 87 + 40 √ = .75ρ = (0.75)(0.037) = .0278 > 0.011 thus fs = fy . ρb = .85β1

10. ρmax

24.3

ACI Code

Attached is an unauthorized copy of some of the most relevant ACI-318-89 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 Φ speciﬁed in Chapter 9. 8.3.1 - All members of frames or continuous construction shall be designed for the maximum eﬀects of factored loads as determined by the theory of elastic analysis, except as modiﬁed according to Section 8.4. Simplifying assumptions of Section 8.6 through 8.9 may be used. 8.5.1 - Modulus of elasticity Ec for concrete may be taken as Wc1.5 33 fc ( psi) for values of Wc between 90 and 155 lb per cu ft. For normal weight concrete, Ec may be taken as 57, 000 fc . 8.5.2 - Modulus of elasticity Es for non-prestressed 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 eﬀects of a speciﬁed 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 ) 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. (9-1). 9.3.1 - Design strength provided by a member, its connections to other members, and its cross sections, in terms of ﬂexure, 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 Φ. Victor Saouma

Structural Engineering

24–10

Draft

REINFORCED CONCRETE BEAMS; Part I

9.3.2 - Strength reduction factor Φ 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 ﬂexural members with overall depth to clear span ratios greater than 2/5 for continuous spans and 4/5 for simple spans, a non-linear distribution of strain shall be considered. See Section 10.7. 10.2.3 - Maximum usable strain at extreme concrete compression ﬁber shall be assumed equal to 0.003. 10.2.4 - Stress in reinforcement below speciﬁed 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 ﬂexural 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 satisﬁed by an equivalent rectangular concrete stress distribution deﬁned by the following: 10.2.7.1 - Concrete stress of 0.85fc 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 = β1 c) from the ﬁber of maximum compressive strain. 10.2.7.2 - Distance c from ﬁber of maximum strain to the neutral axis shall be measured in a direction perpendicular to that axis. 10.2.7.3 - Factor β1 shall be taken as 0.85 for concrete strengths fc up to and including 4,000 psi. For strengths above 4,000 psi, β1 shall be reduced continuously at a rate of 0.05 for each 1000 psi of strength in excess of 4,000 psi, but β1 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 speciﬁed yield strength fy just as concrete in compression reaches its assumed ultimate strain of 0.003. 10.3.3 - For ﬂexural members, and for members subject to combined ﬂexure and compressive axial load when the design axial load strength (ΦPn ) is less than the smaller of (0.10fc Ag ) or (ΦPb ), the ratio of reinforcement p provided shall not exceed 0.75 of the ratio ρb that would produce balanced strain conditions for the section under ﬂexure without axial load. For members with compression reinforcement, the portion of ρb 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 ﬂexural members. 10.5.1 - At any section of a ﬂexural member, except as provided in Sections 10.5.2 and 10.5.3, where positive reinforcement is required by analysis, the ratio ρ provided shall not be less than that given by ρmin =

Victor Saouma

200 fy

Structural Engineering

Draft Chapter 25

REINFORCED CONCRETE BEAMS; Part II 25.1 1

T Beams, (ACI 8.10)

Equivalent width for uniform stress, Fig. 25.1 must satisfy the following requirements (ACI 8.10.2):

b hf

bw

be

Figure 25.1: T Beams

25–2

REINFORCED CONCRETE BEAMS; Part II

Draft 1.

1 2 (b

− bw ) ≤ 8hf

2. b < 4bw for isolated T beams only 3. hf > 4. b < 2

bw 2

L 4

Two possibilities: 1. Neutral axis within the ﬂanges (c < hf ) ⇒ rectangular section of width b, Fig. 25.2. 2. Neutral axis in the web (c > hf ) ⇒ T beam.

b

Figure 25.2: T Beam as Rectangular Section

b

εu

hf

.85f’c a=β1c

c

d

As fy

εs

Figure 25.3: T Beam Strain and Stress Diagram 3 For T beams, we have a large concrete area, start by assuming that failure will occur by steel yielding, Fig. 25.3.

4

The approach consists in decomposing As into 2 components 1. Asf → resists compression force in (b − bw )hf 2. (As − Asf ) → resists compression force in bw c

Victor Saouma

Structural Engineering

25.1 T Beams, (ACI 8.10)

25–3

Draft 25.1.1 5

Review

Given, b, d, hf , As , fc , fy , determine the moment capacity M , Fig. 25.4.

=

+

As

Asf

As-Asf

Figure 25.4: Decomposition of Steel Reinforcement for T Beams 6 The moment is obtained from Flanges:

Asf

=

Mn1

=

.85fc (b−bw )hf fy h Asf fy (d − 2f

ΣF = 0 ) ΣM = 0

Web: a = Mn2 =

(As −Asf )fy .85fc bw

ΣF = 0 (As − Asf )fy (d − a2 ) ΣM = 0

Total moment: Mn = Mn1 + Mn2

25.1.2 7

Design, (balanced)

Let us derive an expression for ρb and use it for design c d

As fy

u = εuε+ε StrainCompatibility y = .85fc β1 cbw + .85fc(b − bw )hf ΣF = 0

Asf fy

thus, Asf y ρw ρf

= = =

 .85fc β1 cbw + Asf fy  As bw d Asf bwd

ρwb

=

ρb + ρf

Hence, ρmax

=

.75(ρb + ρf ) (25.2)

ρb

εu ρw = .85 β1 +ρf  fy εu + εy fc

(25.1)

Example 25-1: T Beam; Moment Capacity I For the following beam: As = 8 # 11 ( 12.48 in2 ); fc =3,000 psi; fy = 50,000 psi. Determine Mn

Victor Saouma

Structural Engineering

25–4

REINFORCED CONCRETE BEAMS; Part II

Draft

30"

7"

36"

14"

Solution: 1. Check requirements for isolated T sections (a) be = 30 in should not exceed 4bw = 4(14) = 56 in √ (b) hf ≥ b2u ⇒ 7 ≥ 14 2

√

2. Assume Rectangular section a=

(12.48)(50) = 8.16 in > hf (0.85)(3)(30)

3. For a T section

.85fc hf (b−bw ) fy (.85)(3)(7)(30−14) 50 5.71 (14)(36) = .0113

Asf

=

ρf Asw ρw

= = 5.71 in2 = = 12.48 − 5.71 = 6.77 in2 12.48 = (14)(36) = .025

ρb

f

87 = .85β1 fyc 87+f y 3 87 = (.85)(.85) 50 87+50 = .0275

4. Maximum permissible ratio ρmax

= .75(ρb + ρf ) √ = .75(.0275 + .0113) = .029 > ρw

5. The design moment is then obtained from Mn1 a Mn2 Md

= (5.71)(50)(36 − 72 ) = 9, 280 k.in (6.77)(50) = (.85)(3)(14) = 9.48 in = (6.77)(50)(36 − 9.48 2 ) = 10, 580 k.in = (.9)(9, 280 + 10, 580) = 17, 890 k.in → 17, 900 k.in

Example 25-2: T Beam; Moment Capacity II

Victor Saouma

Structural Engineering

25.1 T Beams, (ACI 8.10)

25–5

Draft

Determine the moment capacity of the following section, assume ﬂange dimensions to satisfy ACI requirements; As = 6#10 = 7.59 in2 ; fc = 3 ksi; fy =60 ksi. 28"

6"

26"

10"

As = 6#10= 7.59in2 Solution: Assume rectangular beam ρ

=

ρb

=

a

=

Asf Asw ρw ρf ρmax Mn1 As − Asf a Mn2 Md

= = = = = = = = = =

7.59 (28)(26)

= .0104 3 87 (.85)(.85) 60 87+60 = .0214 > ρ ⇒ fs = fy (7.59)(60) (.85)(3)(28) = (.85)(3)(18)(6) 60

6.37 in > 6 in ⇒ T beam

= 4.59 in2 7.59 − 4.59 = 3.00 in2 7.59 (26)(10) = .0292 4.59 (26)(10) = .0177 .75(.0214 + .0177) = .0294 > .0292 ⇒ Ductile failure (4.59)(60)(26 − 3) = 6, 330 k.in 7.59 − 4.59 = 3. in2 (3)(60) (.85)(3)(10) = 7.07 in (3.00)(60)(26 − 7.07 2 ) = 4, 050 k.in (.9)(6, 330 + 4, 050) = 9, 350 k.in

Example 25-3: T Beam; Design given L = 24 ft; fy = 60 ksi; fc = 3 ksi; Md = 6, 400 k.in; Design a R/C T beam.

Victor Saouma

Structural Engineering

25–6

REINFORCED CONCRETE BEAMS; Part II

Draft

3"

20"

11" 47"

Solution: 1. Determine eﬀective ﬂange width: − bw ) ≤ 8hf 16hf + bw = (16)(3) + 11 = 59 in L 24 = 72 in 4 = 4 12 Center Line spacing = 47 in 1 2 (b

2. Assume a = 3 in As

=

a

=

      

6,400 Md = 0.9)(60)(20− = 3 φfy (d− a 2) 2) As fy (6.4)(60) (.85)fc b = (.85)(3)(47) = 3.20

b = 47 in

6.40 in2 in > hf

3. Thus a T beam analysis is required. Asf

=

.85fc (b−bw )hf fy

Md1 Md2

= =

φAsf fy (d − 2f ) = (.90)(4.58)(60)(20 − 32 ) = 4, 570 k.in Md − Md1 = 6, 400 − 4, 570 = 1, 830 k.in

=

(.85)(3)(47−11)(3) 60

= 4.58 in2

h

d 5

4. Now, this is similar to the design of a rectangular section. Assume a = As − Asf = 5. check

=

20 5

= 4. in

1, 830 = 1.88 in2 (.90)(60) 20 − 42

a

=

1.88)(60) (.85)(3)(11)

As ρw ρf

= = =

ρb ρmax

= =

4.58 + 1.88 = 6.46 in2 6.46 (11)(20) = .0294 4.58 (11)(20) = .0208 3 87 (.85)(.85) 60 87+60 = .0214 √ .75(.0214 + .0208) = .0316 > ρw

= 4.02 in ≈ 4.00

6. Note that 6.46 in2 (T beam) is close to As = 6.40 in2 if rectangular section was assumed.

25.2 8

Doubly Reinforced Rectangular Beams

Negative steel reinforcement is needed to 1. Increase internal moment resistance capacity (not very eﬃcient)

Victor Saouma

Structural Engineering

25.2 Doubly Reinforced Rectangular Beams

25–7

Draft

2. Support stirups 3. Reverse moments (moving load) 4. Provide ductility (earthquake) 5. Reduce creep (long term deﬂections)

9

Approach will again be based on a strain compatibility analysis & equilibrium equation, Fig. 25.5.

εu

b

d’ A’s εs

d

c

.85f’c

.85f’c

A’sf’s

ε’s

A’sf’y

=

d-d’

+ A’sfy

Asfs

As

a

(As-A’s)fs

Figure 25.5: Doubly Reinforced Beams; Strain and Stress Diagrams

10

If ρ ≤ ρmax = .75ρb we can disregard compression steel

11

As for T beams, we decompose the tension steel into two components 1. As to resist the force in the top steel (assuming both yield) 2. As − As to resist compression in the concrete.

and we deﬁne ρ =

25.2.1 12

As bd

(25.3)

Tests for fs and fs

Diﬀerent possibilities: Fig. 25.6

1 yes

no

As yield?

yes

I f s=fy f’s=fy

A’s yield?

no

II fs=fy f’s< fy

3

yes

A’s yield?

III fs< fy f’s=fy

no

IV fs< fy f’s< fy

{

2

not allowed by code Figure 25.6: Diﬀerent Possibilities for Doubly Reinforced Concrete Beams

Victor Saouma

Structural Engineering

25–8

REINFORCED CONCRETE BEAMS; Part II

Draft

Test 1 Is As yielding? Assuming εs = εy , and fs = fy , we have from the strain diagram εs fs c From equilibrium:

= εu − dd (εu + εy ) = Es εs u = εuε+ε d y

ρbdfy = ρ bdfs + .85fc β1 bc

Combining: ρb = ρ1 = ρ

fs εu f + .85 c β1 fy fy εu + εy

ρb

ρb

=

ρ1 = ρ

fs

fy

+ ρb

(25.4)

fs (25.5)Note that 0.75 premultiplies only one term as in the other fy failure is ipso facto by yielding. We also note the similarity with ρmax of T Beams (where 0.75 premultiplied both terms).

thus ρmax

=

0.75ρb + ρ

Test 2 fs = fy is fs = fy ? We set εs = εy , and from the strain diagram

εu

ε’s=εy

εs>εy Figure 25.7: Strain Diagram, Doubly Reinforced Beam; is As Yielding? c= from equilibrium

εu d εu − εy

ρbdfy = ρ bdfy + .85fc β1 cb

combining ρmin ≡≡ ρ2 = ρ + .85β1

fc d 87 fy d 87 − fy

(25.6)

which corresponds to the minimum amount of steel to ensure yielding of compression steel at failure. Thus, if ρ < ρmin then fs < fy . Test 3 fs < fy , is fs = fy ? From strain diagram: c εs Victor Saouma

= =

εu εu −εy d d−c c−d εy

Structural Engineering

25.2 Doubly Reinforced Rectangular Beams

25–9

Draft

From equilibrium

ρbdfs = ρ bdfy + .85fc β1 bc

combining ρ = ρ3 =

f c c − d ρ + .85β1 c d−c fy d

(25.7)

Summary of the tests are shown in Fig. 25.8

1

2

3

ρcy

ρ

ρ

f’s< fy

f’s = fy

II

I

fs = fy III

ρ

f’s< fy fs < fy IV

Figure 25.8: Summary of Conditions for top and Bottom Steel Yielding

25.2.2

Moment Equations

Case I fs = fy and fs = fy , (small bottom and top reinforcement ratios) As fy a

= As fy + .85fc ab (As −As )fy = .85f b c

a + As fy (d − d ) MnI = .85fc ab d − 2

(25.8)

Case II We have fs = fy and fs < fy (small bottom and large top reinforcement ratios, most common case) εs fs As fy

= εu c−d c = Es εs = As fs + .85fc bβ1 c

solve for c and fs by iteration. Using a = β1 c a + As fs (d − d ) MnII = .85fc ab d − 2

(25.9)

Case III fs < fy and fs = fy (large bottom and small top reinforcement ratios, rare) εs fs As fs a Victor Saouma

= = = =

εu d−c c Es εs As fy + .85fc ab β1 c Structural Engineering

25–10

REINFORCED CONCRETE BEAMS; Part II

Draft solve for a

a MnIII = .85fc ab d − + As fy (d − d ) 2

(25.10)

Case IV fs < fy and fs < fy (large bottom and top reinforcement ratios, rare) εs εs As fs a

= = = =

εu c−d c εu d−c c As fs + .85fc ab β1 c

solve for a a + As fs (d − d ) MnIV = .85fc ab d − 2

(25.11)

Note that in most beams of normal size and proportions, it will be found that fs < fy when fs = fy . We nevertheless use As in order to ensure ductility, stiﬀness and support for the stirrups.

13

Example 25-4: Doubly Reinforced Concrete beam; Review Given, fc = 4, 000 psi, fy = 60,000 psi, As = 3 (1.56) = 4.68 in2 , As = 4 (1.56) = 6.24 in2 , determine the moment carrying capacity of the following beam.

Solution: ρb ρ ρ

= = =

f

(.85)β1 fyc 6.24 (16)(27.3) 4.68 (16)(27.3)

87 87+fy

4 87 = (.85)(.85) 60 87+60 = .0285 = .0143 = .0107

We can not check yet if ρ < ρmax because fs is yet unknown ρmin

= =

f

u ρ + .85 fyc β1 dd εuε−ε y 4 3 .003 .0107 + (.85) 60 (.85) 27.3 .003− 60

= .0278 > ρ

29,000

Hence ρ < ρmin < ρb .0143 < .0278 < .0285 Victor Saouma

Structural Engineering

25.2 Doubly Reinforced Rectangular Beams

25–11

Draft

and thus fs = fy and fs < fy and we have case II

solve for c

fs

= = As fy = (6.24)(60) = 374.4 = fs =

Es εu c−d = (29, 000)(.003) c−3 c c c−3 87 c As fs + .85fc bβ1 c (4.68)fs + (.85)(4)(16)(.85)c 4.68fs + 46.24c −9.9c + 80.2

Note that if we were to plott those two equations, 50

25

2

3

4

5

6

-25

-50

-75

-100

We note that f s increases with c from the strain diagram, but fs decreases with c from equilibrium. Graphically the solution is around 4.9. Combining those two equations1 c2 + .7085c − 26.42 = 0 we obtain c = 4.80 in a = 0.85(4.8) = 4.078 in, and fs = (.003)(29, 000) 4.80−3 4.80 = 32.6 ksi Substituting into the moment equation Mn = .85fc ab d − a2 + As fs (d − d ) + (4.68)(32.62)(27.3 − 3) = (.85)(4)(4.078)(16) 27.3 − 4.078 2 = 9, 313 k.in Md = 0.9(9, 313) = 8, 382 k.in = 699 k.ft Check ρmax

f

= .75ρb + fys ρ = (.75)(.0285) +

32.6 60 (.0107)

= .027

√

ρ

Example 25-5: Doubly Reinforced Concrete beam; Design Given Md = 505 k.ft, fc = 4 ksi, fy = 60 ksi, b = 12 in, h = 24.5 in, d = 21 in, and d = 2.5 in, determine the reinforcement As and possibly As . Solution: Md ρb

= (505)(12) = 6, 060 k.in f 87 4 87 = .85β1 fyc 87+f = (.85)(.85) 60 87+60 = .0285 y ρmax = .75ρb = (.75)(.0285) = .0213 Amax = (.0213)(12)(21) = 5.37 in2 s 1 In this problem, unfortunately As fy (5.37)(60) iterative method if we were to start with a = d5 . a = .85f =an(.85)(4)(12) = 7.89diverges in cb = 4, 943 k.in < 6, 060 k.in Mmax = (0.9)As fy d − a2 = (.9)(5.37)(60) 21 − 7.89 2 Victor Saouma Structural Engineering

25–12

REINFORCED CONCRETE BEAMS; Part II

Draft

Thus compression steel is required. Assuming that fs = fy Md2 As

= =

⇒ As

=

As

=

6, 060 − 4, 943 = 1, 117 k.in 1,117 Md2 2 φfy (d−d ) = (0.9)(60)(21−2.5) = 1.12 in 1.12 in2 5.37 + 1.12 = 6.49 in2

check that fs = fy ρ ρ ρmin

= =

1.12 (12)(21) 6.49 (12)(21)

= .00444 = .0257 f

u = ρ + .85β1 fyc dd εuε−ε y √ 4 2.5 87 = .00444 + (.85)(.85) 60 21.0 87−60 = .0229 < ρ(.0257)

Note that if it turns out that fs < fy , then we will need to make an assumption on As (such as As = A2s , as we will have three equations (2 of equilibrium and one of strain compatibility) and four unknowns (As , As , fs and c).

Victor Saouma

Structural Engineering

Draft Chapter 26

DIRECT STIFFNESS METHOD 26.1

Introduction

26.1.1

Structural Idealization

1 Prior to analysis, a structure must be idealized for a suitable mathematical representation. Since it is practically impossible (and most often unnecessary) to model every single detail, assumptions must be made. Hence, structural idealization is as much an art as a science. Some of the questions confronting the analyst include:

1. Two dimensional versus three dimensional; Should we model a single bay of a building, or the entire structure? 2. Frame or truss, can we neglect ﬂexural stiﬀness? 3. Rigid or semi-rigid connections (most important in steel structures) 4. Rigid supports or elastic foundations (are the foundations over solid rock, or over clay which may consolidate over time) 5. Include or not secondary members (such as diagonal braces in a three dimensional analysis). 6. Include or not axial deformation (can we neglect the axial stiﬀness of a beam in a building?) 7. Cross sectional properties (what is the moment of inertia of a reinforced concrete beam?) 8. Neglect or not haunches (those are usually present in zones of high negative moments) 9. Linear or nonlinear analysis (linear analysis can not predict the peak or failure load, and will underestimate the deformations). 10. Small or large deformations (In the analysis of a high rise building subjected to wind load, the moments should be ampliﬁed by the product of the axial load times the lateral deformation, P − ∆ eﬀects). 11. Time dependent eﬀects (such as creep, which is extremely important in prestressed concrete, or cable stayed concrete bridges). 12. Partial collapse or local yielding (would the failure of a single element trigger the failure of the entire structure?). 13. Load static or dynamic (when should a dynamic analysis be performed?). 14. Wind load (the lateral drift of a high rise building subjected to wind load, is often the major limitation to higher structures). 15. Thermal load (can induce large displacements, specially when a thermal gradient is present.). 16. Secondary stresses (caused by welding. Present in most statically indeterminate structures).

26–2

DIRECT STIFFNESS METHOD

Draft 26.1.2

Structural Discretization

2 Once a structure has been idealized, it must be discretized to lend itself for a mathematical representation which will be analyzed by a computer program. This discretization should uniquely deﬁne each node, and member. 3 The node is characterized by its nodal id (node number), coordinates, boundary conditions, and load (this one is often deﬁned separately), Table 26.1. Note that in this case we have two nodal coordinates,

Node No. 1 2 3 4

Coor. X Y 0. 0. 5. 5. 20. 5. 25. 2.5

X 1 0 0 1

B. C. Y 1 0 0 1

Z 0 0 0 1

Table 26.1: Example of Nodal Deﬁnition and three degrees of freedom (to be deﬁned later) per node. Furthermore, a 0 and a 1 indicate unknown or known displacement. Known displacements can be zero (restrained) or non-zero (as caused by foundation settlement). 4

The element is characterized by the nodes which it connects, and its group number, Table 26.2. Element No. 1 2 3

From Node 1 3 3

To Node 2 2 4

Group Number 1 2 2

Table 26.2: Example of Element Deﬁnition 5 Group number will then deﬁne both element type, and elastic/geometric properties. The last one is a pointer to a separate array, Table 26.3. In this example element 1 has element code 1 (such as beam element), while element 2 has a code 2 (such as a truss element). Material group 1 would have diﬀerent elastic/geometric properties than material group 2.

Group No. 1 2 3

Element Type 1 2 1

Material Group 1 1 2

Table 26.3: Example of Group Number 6 From the analysis, we ﬁrst obtain the nodal displacements, and then the element internal forces. Those internal forces vary according to the element type. For a two dimensional frame, those are the axial and shear forces, and moment at each node. 7 Hence, the need to deﬁne two coordinate systems (one for the entire structure, and one for each element), and a sign convention become apparent.

26.1.3 8

Coordinate Systems

We should diﬀerentiate between 2 coordinate systems:

Victor Saouma

Structural Engineering

26.1 Introduction

26–3

Draft

Global: to describe the structure nodal coordinates. This system can be arbitrarily selected provided it is a Right Hand Side (RHS) one, and we will associate with it upper case axis labels, X, Y, Z, Fig. 26.1 or 1,2,3 (running indeces within a computer program).

Figure 26.1: Global Coordinate System Local: system is associated with each element and is used to describe the element internal forces. We will associate with it lower case axis labels, x, y, z (or 1,2,3), Fig. 26.2. The x-axis is assumed to be along the member, and the direction is chosen such that it points from the 1st node to the 2nd node, Fig. 26.2.

9

10

Two dimensional structures will be deﬁned in the X-Y plane.

26.1.4

Sign Convention

The sign convention in structural analysis is completely diﬀerent than the one previously adopted in structural analysis/design, Fig. 26.3 (where we focused mostly on ﬂexure and deﬁned a positive moment as one causing “tension below”. This would be awkward to program!).

11

In matrix structural analysis the sign convention adopted is consistent with the prevailing coordinate system. Hence, we deﬁne a positive moment as one which is counter-clockwise, Fig. 26.3

12

13

Fig. 26.4 illustrates the sign convention associated with each type of element.

Fig. 26.4 also shows the geometric (upper left) and elastic material (upper right) properties associated with each type of element.

14

26.1.5 15

Degrees of Freedom

A degree of freedom (d.o.f.) is an independent generalized nodal displacement of a node.

The displacements must be linearly independent and thus not related to each other. For example, a roller support on an inclined plane would have three displacements (rotation θ, and two translations u

16

Victor Saouma

Structural Engineering

26–4

Draft

DIRECT STIFFNESS METHOD

Figure 26.2: Local Coordinate Systems

Figure 26.3: Sign Convention, Design and Analysis

Figure 26.4: Total Degrees of Freedom for various Type of Elements Victor Saouma

Structural Engineering

26.2 Stiﬀness Matrices

26–5

Draft

and v), however since the two displacements are kinematically constrained, we only have two independent displacements, Fig. 26.5.

Figure 26.5: Dependent Displacements We note that we have been referring to generalized displacements, because we want this term to include translations as well as rotations. Depending on the type of structure, there may be none, one or more than one such displacement. It is unfortunate that in most introductory courses in structural analysis, too much emphasis has been placed on two dimensional structures, and not enough on either three dimensional ones, or two dimensional ones with torsion.

17

In most cases, there is the same number of d.o.f in local coordinates as in the global coordinate system. One notable exception is the truss element. In local coordinate we can only have one axial deformation, whereas in global coordinates there are two or three translations in 2D and 3D respectively for each node.

18

Hence, it is essential that we understand the degrees of freedom which can be associated with the various types of structures made up of one dimensional rod elements, Table 26.4.

19

20

This table shows the degree of freedoms and the corresponding generalized forces.

We should distinguish between local and global d.o.f.’s. The numbering scheme follows the following simple rules:

21

Local: d.o.f. for a given element: Start with the ﬁrst node, number the local d.o.f. in the same order as the subscripts of the relevant local coordinate system, and repeat for the second node. Global: d.o.f. for the entire structure: Starting with the 1st node, number all the unrestrained global d.o.f.’s, and then move to the next one until all global d.o.f have been numbered, Fig. 26.6.

26.2

Stiﬀness Matrices

26.2.1

Truss Element

22

From strength of materials, the force/displacement relation in axial members is σ Aσ P

= EI AE ∆ = L

Hence, for a unit displacement, the applied force should be equal to other end must be equal and opposite. 23

(26.1)

1

AE L .

From statics, the force at the

The truss element (whether in 2D or 3D) has only one degree of freedom associated with each node.

Victor Saouma

Structural Engineering

26–6

Draft

DIRECT STIFFNESS METHOD

Figure 26.6: Examples of Active Global Degrees of Freedom

Victor Saouma

Structural Engineering

26.2 Stiﬀness Matrices

Draft

26–7

Type

Node 1

Node 2 1 Dimensional Fy3 , Mz4

{p}

Fy1 , Mz2

{δ}

v1 , θ2

{p}

Fx1

{δ} {p}

u1 Fx1 , Fy2 , Mz3

u2 Fx4 , Fy5 , Mz6

{δ} {p}

u1 , v2 , θ3 Tx1 , Fy2 , Mz3

u4 , v5 , θ6 Tx4 , Fy5 , Mz6

{δ}

θ1 , v2 , θ3

{p}

Fx1 ,

{δ} {p}

u1 , Fx1 , Fy2 , Fy3 , Tx4 My5 , Mz6

u2 Fx7 , Fy8 , Fy9 , Tx10 My11 , Mz12

{δ}

u1 , v2 , w3 , θ4 , θ5 θ6

u7 , v8 , w9 , θ10 , θ11 θ12

Beam v3 , θ4 2 Dimensional Fx2

Truss

Frame

Grid θ4 , v5 , θ6 3 Dimensional Fx2

Truss

Frame

[k] (Local)

[K] (Global)

4×4

4×4

2×2

4×4

6×6

6×6

6×6

6×6

2×2

6×6

12 × 12

12 × 12

Table 26.4: Degrees of Freedom of Diﬀerent Structure Types Systems Hence, from Eq. 26.1, we have

[kt ] =

26.2.2

AE L

 u1 u2  p1 1 −1 1  p 2 −1

(26.2)

Beam Element

Using Equations 23.10, 23.11, 23.13 and 23.14 we can determine the forces associated with each unit displacement.

24

[kb ] =

25

 V1 Eq. M 1 Eq.  V2 Eq. M2 Eq.

v1 23.13(v1 23.10(v1 23.14(v1 23.11(v1

= 1) = 1) = 1) = 1)

Eq. Eq. Eq. Eq.

θ1 23.13(θ1 23.10(θ1 23.14(θ1 23.11(θ1

= 1) = 1) = 1) = 1)

Eq. Eq. Eq. Eq.

v2 23.13(v2 23.10(v2 23.14(v2 23.11(v2

= 1) = 1) = 1) = 1)

Eq. Eq. Eq. Eq.

θ2 23.13(θ2 23.10(θ2 23.14(θ2 23.11(θ2

 = 1) = 1)   = 1)  = 1)

(26.3)

The stiﬀness matrix of the beam element (neglecting shear and axial deformation) will thus be 

[kb ] =

Victor Saouma

v1 z V1 12EI L3 6EIz  M 1 L2 12EIz V2  − L3 6EIz M 2 L2

θ1 6EIz L2 4EIz L z − 6EI L2 2EIz L

v2 z − 12EI L3 6EIz − L2 12EIz L3 z − 6EI L2

θ2 6EIz L2 2EIz L z − 6EI L2 4EIz L

      

(26.4)

Structural Engineering

26–8

DIRECT STIFFNESS METHOD

Draft 26

We note that this is identical to Eq.23.16  V1    M1 V2    M2

      

=

v1  z V1 12EI L3 6EIz  M 1 L2 z V2 − 12EI L3 6EIz M2 L2

θ1 6EIz L2 4EIz L z − 6EI L2 2EIz L

v2 z − 12EI L3 6EIz − L2

12EIz L3 z − 6EI L2

θ2 6EIz L2 2EIz L z − 6EI L2 4EIz L

  v1        θ1    v2      θ2

(26.5)

k(e)

26.2.3

2D Frame Element

The stiﬀness matrix of the two dimensional frame element is composed of terms from the truss and beam elements where kb and kt refer to the beam and truss element stiﬀness matrices respectively.

27

[k2df r ] =

u  t1 P1 k11 V1  0 M 1 0  t P2  k21 V2 0 M2 0

v1 0 b k11 b k21 0 b k31 b k41

θ1 0 b k12 b k22 0 b k32 b k42

u2 t k12 0 0 t k22 0 0

v2 0 b k13 b k23 0 b k33 b k43

θ2  0 b  k14  b  k24  0   b  k34 b k44

(26.6)

Thus, we have:

[k2df r ] =

26.2.4

 u1 P1 EA L V1  0 M 0 1  EA P2 − L  V2 0 M 0 2

v1 0

θ1 0

12EIz L3 6EIz L2

6EIz L2 4EIz L

0

0

z − 12EI L3

z − 6EI L2

6EIz L2

2EIz L

u2 − EA L 0 0 EA L

0 0

v2 0 z − 12EI L3 6EIz − L2 0

12EIz L3 z − 6EI L2

θ2  0  6EIz  L2  2EIz   L 0    − 6EI 2 L  4EIz 

(26.7)

L

Remarks on Element Stiﬀness Matrices

Singularity: All the derived stiﬀness matrices are singular, that is there is at least one row and one column which is a linear combination of others. For example in the beam element, row 4 = −row 1; and L times row 2 is equal to the sum of row 3 and 6. This singularity (not present in the ﬂexibility matrix) is caused by the linear relations introduced by the equilibrium equations which are embedded in the formulation. Symmetry: All matrices are symmetric due to Maxwell-Betti’s reciprocal theorem, and the stiﬀness ﬂexibility relation. 28

More about the stiﬀness matrix properties later.

26.3

Direct Stiﬀness Method

26.3.1

Orthogonal Structures

As a “vehicle” for the introduction to the stiﬀness method let us consider the problem in Fig 26.7-a, and recognize that there are only two unknown displacements, or more precisely, two global d.o.f: θ1 and θ2 . 29

30

If we were to analyse this problem by the force (or ﬂexibility) method, then

Victor Saouma

Structural Engineering

26.3 Direct Stiﬀness Method

26–9

Draft

Figure 26.7: Problem with 2 Global d.o.f. θ1 and θ2

Victor Saouma

Structural Engineering

26–10

DIRECT STIFFNESS METHOD

Draft

1. We make the structure statically determinate by removing arbitrarily two reactions (as long as the structure remains stable), and the beam is now statically determinate. 2. Assuming that we remove the two roller supports, then we determine the corresponding deﬂections due to the actual laod (∆B and ∆C ). 3. Apply a unit load at point B, and then C, and compute the deﬂections fij at note i due to a unit force at node j. 4. Write the compatibility of displacement equation RB ∆1 0 fBB fBC − = fCB fCC RC ∆2 0

(26.8)

5. Invert the matrix, and solve for the reactions 31

We will analyze this simple problem by the stiﬀness method. 1. The ﬁrst step consists in making it kinematically determinate (as opposed to statically determinate in the ﬂexibility method). Kinematically determinate in this case simply means restraining all the d.o.f. and thus prevent joint rotation, Fig 26.7-b. 2. We then determine the ﬁxed end actions caused by the element load, and sum them for each d.o.f., Fig 26.7-c: ΣFEM1 and ΣFEM2 . 3. In the third step, we will apply a unit displacement (rotation in this case) at each degree of freedom at a time, and in each case we shall determine the reaction forces, K11 , K21 , and K12 , K22 respectively. Note that we use [K], rather than k since those are forces in the global coordinate system, Fig 26.7-d. Again note that we are focusing only on the reaction forces corresponding to a global degree of freedom. Hence, we are not attempting to determine the reaction at node A. 4. Finally, we write the equation of equilibrium at each node:

M1 M2

=

ΣFEM1 ΣFEM2

+

K11 K21

K12 K22

θ1 θ2

(26.9)

Note that the FEM being on the right hand side, they are detemined as the reactions to the applied load. Strictly speaking, it is a load which should appear on the left hand side of the equation, and are the nodal equivalent loads to the element load (more about this later).

32

As with the element stiﬀness matrix, each entry in the global stiﬀness matrix Kij , corresponds to the internal force along d.o.f. i due to a unit displacement (generalized) along d.o.f. j (both in global coordinate systems).

33

Example 26-1: Beam Considering the previous problem, Fig. 26.7-a, let P1 = 2P , M = P L, P2 = P , and P3 = P , Solve for the displacements. Solution: 1. Using the previously deﬁned sign convention: ΣFEM1

PL P1 L P2 L 2P L P L + =− = − + =− 8 8 8 8 8 BA

ΣFEM2

PL = − 8

(26.10)

BC

(26.11)

CB

Victor Saouma

Structural Engineering

26.3 Direct Stiﬀness Method

26–11

Draft

4EI AB BC 2. If it takes 4EI L (k44 ) to rotate AB (Eq. 26.4) and L (k22 ) to rotate BC, it will take a total force 8EI of L to simultaneously rotate AB and BC, (Note that a rigid joint is assumed). 3. Hence, K11 which is the sum of the rotational stiﬀnesses at global d.o.f. 1. will be equal to K11 = 8EI L ; 2EI BC similarly, K21 = L (k42 ) . 2EI BC BC 4. If we now rotate dof 2 by a unit angle, then we will have K22 = 4EI L (k22 ) and K12 = L (k42 ) . 5. The equilibrium relation can thus be written as: P L 8EI 2EI − 8 PL θ1 L L = + (26.12) 2EI 4EI θ2 0 −PL 8 L L M F EM K ∆

or

P L + P8L + P8L

=

8EI L 2EI L

2EI L 4EI L

θ1 θ2

(26.13)

We note that this matrix corresponds to the structure’s stiﬀness matrix, and not the augmented one. 6. The two by two matrix is next inverted 8EI 2EI −1 3 17 P L2 2 P L + P8L θ1 L L 112 EI 2 (26.14) = 2EI 4EI = 5 PL θ2 + P8L − 112 L L EI 7. Next we need to determine both the reactions and the internal forces. 8. Recall that for each element {p} = [k]{δ}, and in this case {p} = {P} and {δ} = {∆} for element AB. The element stiﬀness matrix has been previously derived, Eq. 26.4, and in this case the global and local d.o.f. are the same. 9. Hence, the equilibrium equation for element AB, at the element level, can be written as:    12EI   2P   6EI 6EI 0 − 12EI p1       3 2 3 2 L L L L        6EI     2P2L  4EI 2EI  0 p2 − 6EI 2 2 L L L L 8   =  12EI + (26.15) 12EI 2P  0 p3  − L3 − 6EI − 6EI          L2 L3 L2 2 2        6EI 2EI 4EI 17 P L p4 − 6EI − 2P8 L 2 L2 L L L 112 EI

{p} [k] FEM {δ } solving p1

p2

p3

10. Similarly, for element BC:    12EI 6EI p1   L3 L2      6EI 4EI p2 2 L L  =  12EI p3  − L3 − 6EI    L2   6EI 2EI p4 L2 L

p4 =

− 12EI L3 − 6EI L2 12EI L3 − 6EI L2

107 56 P

6EI L2 2EI L − 6EI L2 4EI L

31 56 P L

       

5 56 P

5 14 P L

0

   

17 P L2 112 EI

0 2

5 PL − 112 EI

  

    +

  

(26.16)

P 2 PL 8 P 2 − P8L

      

(26.17)

or p1

p2

p3

p4 =

7 8P

9 14 P L

− P7

0

(26.18)

11. This simple example calls for the following observations: 1. Node A has contributions from element AB only, while node B has contributions from both AB and BC. = pBC eventhough they both correspond to a shear force at node B, the 12. We observe that pAB 3 1 diﬀerence betweeen them is equal to the reaction at B. Similarly, pAB = pBC due to the externally 4 2 applied moment at node B. 2. From this analysis, we can draw the complete free body diagram, Fig. 26.7-e and then the shear and moment diagrams which is what the Engineer is most interested in for design purposes.

Victor Saouma

Structural Engineering

26–12

DIRECT STIFFNESS METHOD

Draft 26.3.2

Local and Global Element Stiﬀness Matrices ([k(e) ] [K(e) ])

In the previous section, in which we focused on orthogonal structures, the assembly of the structure’s stiﬀness matrix [K(e) ] in terms of the element stiﬀness matrices was relatively straight-forward.

34

The determination of the element stiﬀness matrix in global coordinates, from the element stiﬀness matrix in local coordinates requires the introduction of a transformation.

35

This section will examine the 2D transformation required to obtain an element stiﬀness matrix in global coordinate system prior to assembly (as discussed in the next section).

36

37

38

Recalling that {p} =

[k(e) ]{δ}

(26.19)

{P} =

(e)

(26.20)

[K

]{∆}

Let us deﬁne a transformation matrix [Γ(e) ] such that: {δ}

=

{p} =

[Γ(e) ]{∆} (e)

[Γ

(26.21)

]{P}

(26.22)

Note that we use the same matrix Γ(e) since both {δ} and {p} are vector quantities (or tensors of order one). 39

Substituting Eqn. 26.21 and Eqn. 26.22 into Eqn. 26.19 we obtain [Γ(e) ]{P} = [k(e) ][Γ(e) ]{∆}

(26.23)

{P} = [Γ(e) ]−1 [k(e) ][Γ(e) ]{∆}

(26.24)

premultiplying by [Γ(e) ]−1

40

But since the rotation matrix is orthogonal, we have [Γ(e) ]−1 = [Γ(e) ]T and {P} = [Γ(e) ]T [k(e) ][Γ(e) ]{∆}

(26.25)

[K(e) ]

[K(e) ] = [Γ(e) ]T [k(e) ][Γ(e) ]

(26.26)

which is the general relationship between element stiﬀness matrix in local and global coordinates. 26.3.2.1

2D Frame

The vector rotation matrix is deﬁned in terms of 9 direction cosines of 9 diﬀerent angles. However for the 2D case, Fig. 26.8, we will note that four angles are interrelated (lxX , lxY , lyX , lyY ) and can all be expressed in terms of a single one α, where α is the direction of the local x axis (along the member from the ﬁrst to the second node) with respect to the global X axis. The remaining 5 terms are related to another angle, β, which is between the Z axis and the x-y plane. This angle is zero because we select an orthogonal right handed coordinate system. Thus, the rotation matrix can be written as:       cos α sin α 0 cos α cos( π2 − α) 0 lxX lxY lxZ cos α 0  =  − sin α cos α 0  (26.27) [γ] =  lyX lyY lyZ  =  cos( π2 + α) 0 0 1 0 0 1 lzX lzY lzZ

41

and we observe that the angles are deﬁned from the second subscript to the ﬁrst, and that counterclockwise angles are positive.

Victor Saouma

Structural Engineering

26.3 Direct Stiﬀness Method

26–13

Draft

Figure 26.8: 2D Frame Element Rotation

42

The element rotation matrix [Γ(e) ] will then be given by    cos α sin α 0 p1  0         − sin α cos α 0 p 0  2        0 0 1 p3 0 =  p4  0 0 0 cos α          p − sin α 0 0 0      5  0 0 0 p6 0

(e) [Γ ]

26.3.3

0 0 0 sin α cos α 0

0 0 0 0 0 1

               

P1 P2 P3 P4 P5 P6

              

(26.28)

Global Stiﬀness Matrix

The physical interpretation of the global stiﬀness matrix K is analogous to the one of the element, i.e. If all degrees of freedom are restrained, then Kij corresponds to the force along global degree of freedom i due to a unit positive displacement (or rotation) along global degree of freedom j.

43

For instance, with reference to Fig. 26.9, we have three global degrees of freedom, ∆1 , ∆2 , and θ3 . and the global (restrained or structure’s) stiﬀness matrix is   K11 K12 K13 (26.29) K =  K21 K22 K23  K31 K32 K33

44

and the ﬁrst column corresponds to all the internal forces in the unrestrained d.o.f. when a unit displacement along global d.o.f. 1 is applied. 26.3.3.1

Structural Stiﬀness Matrix

The structural stiﬀness matrix is assembled only for those active dgrees of freedom which are active (i.e unrestrained). It is the one which will be inverted (or rather decomposed) to determine the nodal displacements.

45

26.3.3.2

Augmented Stiﬀness Matrix

The augmented stiﬀness matrix is expressed in terms of all the dof. However, it is partitioned into two groups with respective subscript ‘u’ where the displacements are known (zero otherwise), and t where the loads are known. √ Ktt Ktu Pt ∆t ? (26.30) √ = Ru ? Kut Kuu ∆u

46

Victor Saouma

Structural Engineering

26–14

DIRECT STIFFNESS METHOD

Draft

Figure 26.9: *Frame Example (correct K32 and K33 ) We note that Ktt corresponds to the structural stiﬀness matrix. 47

The ﬁrst equation enables the calculation of the unknown displacements. ∆t = K−1 tt (Pt − Ktu ∆u )

48

(26.31)

The second equation enables the calculation of the reactions

Ru = Kut ∆t + Kuu ∆u

(26.32)

For internal book-keeping purpose, since we are assembling the augmented stiﬀness matrix, we proceed in two stages:

49

1. First number all the global unrestrained degrees of freedom 2. Then number all the global restrained degrees of freedom (i.e. those with known displacements, zero or otherwise) and multiply by -11 .

26.3.4

Internal Forces

The element internal forces (axia and shear forces, and moment at each end of the member) are determined from

50

(e)

pint = k(e) δ (e)

(26.33)

1 An

alternative scheme is to separately number the restrained dof but assign a negative number. This will enable us later on to distinguish the restrained from unrestrained dof.

Victor Saouma

Structural Engineering

26.3 Direct Stiﬀness Method

Draft

26–15

(e)

at the element level where pint is the six by six array of internal forces, k(e) the element stiﬀness matrix in local coordinate systems, and δ (e) is the vector of nodal displacements in local coordinate system. Note that this last array is obrained by ﬁrst identifying the displacements in global coordinate system, and then premultiplying it by the transofrmation matrix to obtain the displacements in local coordinate system.

26.3.5

Boundary Conditions, [ID] Matrix

51 Because of the boundary condition restraints, the total structure number of active degrees of freedom (i.e unconstrained) will be less than the number of nodes times the number of degrees of freedom per node.

52

To obtain the global degree of freedom for a given node, we need to deﬁne an [ID] matrix such that: ID has dimensions l × k where l is the number of degree of freedom per node, and k is the number of nodes). ID matrix is initialized to zero. 1. At input stage read ID(idof,inod) of each degree of freedom for every node such that: 0 if unrestrained d.o.f. ID(idof, inod) = 1 if restrained d.o.f.

(26.34)

2. After all the node boundary conditions have been read, assign incrementally equation numbers (a) First to all the active dof (b) Then to the other (restrained) dof. (c) Multiply by -1 all the passive dof. Note that the total number of dof will be equal to the number of nodes times the number of dof/node NEQA. 3. The largest positive global degree of freedom number will be equal to NEQ (Number Of Equations), which is the size of the square matrix which will have to be decomposed. 53

For example, for the frame shown in Fig. 26.10: 1. The input data ﬁle may contain: Node No. 1 2 3 4

[ID]T 000 110 000 100

2. At this stage, the [ID] matrix is equal to: 

0 ID =  0 0

1 0 1 0 0 0

 1 0  0

3. After we determined the equation numbers, we would have:   1 −1 5 −3 ID =  2 −2 6 8  3 4 7 9

Victor Saouma

(26.35)

(26.36)

Structural Engineering

26–16

DIRECT STIFFNESS METHOD

Draft

Figure 26.10: Example for [ID] Matrix Determination

26.3.6

LM Vector

The LM vector of a given element gives the global degree of freedom of each one of the element degree of freedom’s. For the structure shown in Fig. 26.10, we would have:

54

LM LM LM

26.3.7

= −1 −2 4 5 6 7 element 1 (2 → 3) = 5 6 7 1 2 3 element 2 (3 → 1) = 1 2 3 −3 8 9 element 3 (1 → 4)

Assembly of Global Stiﬀness Matrix

As for the element stiﬀness matrix, the global stiﬀness matrix [K] is such that Kij is the force in degree of freedom i caused by a unit displacement at degree of freedom j.

55

Whereas this relationship was derived from basic analysis at the element level, at the structure level, this term can be obtained from the contribution of the element stiﬀness matrices [K(e) ] (written in global coordinate system).

56

For each Kij term, we shall add the contribution of all the elements which can connect degree of freedom i to degree of freedom j, assuming that those forces are readily available from the individual element stiﬀness matrices written in global coordinate system.

57

Kij is non-zero if and only if degree of freedom i and degree of freedom j are connected by an element or share a node.

58

There are usually more than one element connected to a dof. Hence, individual element stiﬀness matrices terms must be added up.

59

Because each term of all the element stiﬀness matrices must ﬁnd its position inside the global stiﬀness matrix [K], it is found computationally most eﬀective to initialize the global stiﬀness matrix [KS ](N EQA×N EQA ) to zero, and then loop through all the elements, and then through each entry of (e) the respective element stiﬀness matrix Kij . 60

(e)

The assignment of the element stiﬀness matrix term Kij (note that e, i, and j are all known since we are looping on e from 1 to the number of elements, and then looping on the rows and columns of the

61

Victor Saouma

Structural Engineering

26.3 Direct Stiﬀness Method

26–17

Draft

S element stiﬀness matrix i, j) into the global stiﬀness matrix Kkl is made through the LM vector (note that it is k and l which must be determined).

Since the global stiﬀness matrix is also symmetric, we would need to only assemble one side of it, usually the upper one.

62

63

Contrarily to the previous method, we will assemble the full augmented stiﬀness matrix. Example 26-2: Assembly of the Global Stiﬀness Matrix As an example, let us consider the frame shown in Fig. 26.11. 50kN

4 kN/m 1 0 0 1 0 1 0 1

8m 3m 111 000 000 111 000 111 000 111 000 111

7.416 m

8m

Figure 26.11: Simple Frame Anlysed with the MATLAB Code The ID matrix is initially set to:



1 [ID] =  1 1

 0 1 0 1  0 1

(26.37)

We then modify it to generate the global degrees of freedom of each node:   −4 1 −7 [ID] =  −5 2 −8  −6 3 −9

(26.38)

Finally the LM vectors for the two elements (assuming that Element 1 is deﬁned from node 1 to node 2, and element 2 from node 2 to node 3): 2 3 −4 −5 −6 1 (26.39) [LM ] = 1 2 3 −7 −8 −9 Let us simplfy the operation by designating the element stiﬀness matrices in global coordinates as follows:

K (1)

=

−4  −4 A11 −5  A21 −6  A31 1   A41 2  A51 3 A61

−5 A12 A22 A32 A42 A52 A62

−6 A13 A23 A33 A43 A53 A63

1 A14 A24 A34 A44 A54 A64

2 A15 A25 A35 A45 A55 A65

3  A16 A26   A36   A46   A56  A66

1 1 B11 2   B21 3   B31 −7  B41 −8 B51 −9 B61

2 B12 B22 B32 B42 B52 B62

3 B13 B23 B33 B43 B53 B63

−7 B14 B24 B34 B44 B54 B64

−8 B15 B25 B35 B45 B55 B65

−9  B16 B26   B36   B46   B56  B66



K (2)

Victor Saouma

=

(26.40-a)

(26.40-b)

Structural Engineering

26–18

Draft

We note that for each element we have shown the Now, we assemble the global stiﬀness matrix  A44 + B11 A45 + B12 A46 + B13  A54 + B21 A55 + B22 A56 + B23   A64 + B31 A65 + B32 A66 + B33   A14 A15 A16  K= A A26 A 24 25   A A A36 34 35   B B B43 41 42   B51 B52 B53 B61 B62 B63

DIRECT STIFFNESS METHOD corresponding LM vector. A41 A51 A61 A11 A21 A31 0 0 0

A42 A52 A62 A12 A22 A32 0 0 0

A43 A53 A63 A13 A23 A33 0 0 0

B14 B24 B34 0 0 0 B44 B54 B64

B15 B25 B35 0 0 0 B45 B55 B65

B16 B26 B36 0 0 0 B46 B56 B66

             

(26.41)

We note that some terms are equal to zero because we do not have a connection between the corresponding degrees of freedom (i.e. node 1 is not connected to node 3).

26.3.8 64

Algorithm

The direct stiﬀness method can be summarized as follows:

Preliminaries: First we shall 1. Identify type of structure (beam, truss, grid or frame) and determine the (a) Number of spatial coordinates (1D, 2D, or 3D) (b) Number of degree of freedom per node (local and global) (c) Number of cross-sectional and material properties 2. Determine the global unrestrained and restrained degree of freedom equation numbers for each node, Update the [ID] matrix (which included only 0’s and 1’s in the input data ﬁle). Analysis : 1. For each element, determine (a) Vector LM relating local to global degree of freedoms. (b) Element stiﬀness matrix [k(e) ] (c) Angle α between the local and global x axes. (d) Rotation matrix [Γ(e) ] (e) Element stiﬀness matrix in global coordinates [K(e) ] = [Γ(e) ]T [k(e) ][Γ(e) ] 2. Assemble the augmented stiﬀness matrix [K(S) ] of unconstrained and constrained degree of freedom’s. 3. Extract [Ktt ] from [K(S) ] and invert (or decompose into into [Ktt ] = [L][L]T where [L] is a lower triangle matrix. 4. Assemble load vector {P} in terms of nodal load and ﬁxed end actions. 5. Backsubstitute and obtain nodal displacements in global coordinate system. 6. SOlve for the reactions. 7. For each element, transform its nodal displacement from global to local coordinates {δ} = [Γ(e) ]{∆}, and determine the internal forces [p] = [k]{δ}. 65

Some of the prescribed steps are further discussed in the next sections. Example 26-3: Direct Stiﬀness Analysis of a Truss

Victor Saouma

Structural Engineering

26.3 Direct Stiﬀness Method

26–19

Draft

4

4

5

5 1

1

6 8

3

2

50k

7

2

12’

3

100k

16’

16’

Figure 26.12: Using the direct stiﬀness method, analyse the truss shown in Fig. 26.12. Solution: 1. Determine the structure ID matrix Node # 1 2 3 4 5 ID

=

Bound. X 0 0 1 0 0

0 0 1 0

1 0 1 0

Cond. Y 1 0 1 0 0 0 0

(26.42-a)

N ode 1 2 3 4 1 2 −2 4 −1 3 −3 5

=

5 6 7

(26.42-b)

2. The LM vector of each element is evaluated next LM 1 LM 2 LM 3 LM 4 LM 5 LM 6 LM 7 LM 8

= 1 = 1 = 2 = 4

−1 4 −1 2

5 3

4 5 6 7 = −2 −3 4 5 = 2 3 6 7 = 2 3 0 0 = −2 −3 6 7 3 5

(26.43-a) (26.43-b) (26.43-c) (26.43-d) (26.43-e) (26.43-f) (26.43-g) (26.43-h)

3. Determine the element stiﬀness matrix of each element in the global coordinate system noting that for a 2D truss element we have [K (e) ] = =

Victor Saouma

[Γ(e) ]T [k(e) ][Γ(e) ]  2 cs c 2 EA  cs s  2 L  −c −cs −cs −s2

 −c2 −cs −cs −s2   c2 cs  cs s2

(26.44-a) (26.44-b)

Structural Engineering

26–20

DIRECT STIFFNESS METHOD

Draft

x2 −x1 L ;

where c = cos α =

Element 1 L = 20 , c =

Y2 −Y1 L

s = sin α =

16−0 20

12−0 20

= 0.8, s =

= 0.6,

EA L

=

(30,000

ksi)(10 in2 ) 20

= 15, 000 k/ft.

1 −1 4 5  1 9600 7200 −9600 −7200 −1 5400 −7200 −5400   7200   4 −9600 −7200 9600 7200  5 −7200 −5400 7200 5400 

[K1 ] =

EA L

Element 2 L = 16 , c = 1 , s = 0 ,

Element 3 L = 12 , c = 0 , s = 1 ,

[K3 ] =

Element 4 L = 16 , c = 1 , s = 0 ,

[K4 ] =

Element 5 L = 20 , c =

−16−0 20

EA L

2  2 0 3 0 4 0 5 0 EA L

3 0 25, 000 0 −25, 000

3  0 0  0 0

(26.46)

4 0 0 0 0

5  0 −25, 000    0 25, 000

(26.47)

= 18, 750 k/ft.

4  4 18, 750 5 0  6 −18, 750 7 0

5 0 0 0 0

6 −18, 750 0 18, 750 0 EA L

7  0 0  0 0

(26.48)

= 15, 000 k/ft.

−2 −3 4 5   −2 9600 −7200 −9600 7200 −3 7200 −5400   −7200 5400   4 −9600 7200 9600 −7200  5 7200 −5400 −7200 5400

Element 6 L = 20 , c = 0.8 , s = 0.6 ,

[K6 ] =

2 −18, 750 0 18, 750 0

= 25, 000 k/ft.

= − 0.8 , s = 0.6 ,

[K5 ] =

EA L

[K7 ] =

EA L

(26.49)

= 15, 000 k/ft.

2 3 6 7   2 9600 7200 −9600 −7200 3 5400 −7200 −5400   7200   6 −9600 −7200 9600 7200  7 −7200 −5400 7200 5400

Element 7 L = 16 , c = 1 , s = 0 ,

Victor Saouma

= 18, 750 k/ft.

1 −1  1 18, 750 0 −1 0 0  2  −18, 750 0 3 0 0

[K2 ] =

(26.45)

(26.50)

= 18, 750 k/ft.

2  2 18, 750 3 0  0 −18, 750 0 0

3 0 0 0 0

0 −18, 750 0 18, 750 0

0  0 0  0 0

(26.51)

Structural Engineering

26.3 Direct Stiﬀness Method

Draft

26–21

EA L

Element 8 L = 12 , c = 0 , s = 1 ,

 −2 −3  6  7

[K8 ] =

= 25, 000 k/ft. −2 0 0 0 0

−3 0 25, 000 0 −25, 000

6 7  0 0 0 −25, 000    0 0 0 25, 000

(26.52)

4. Assemble the global stiﬀness matrix in k/ft Note that we are not assembling the augmented stiﬀness matrix, but rather its submatrix [Ktt ].  0   9600 + 18, 750  −18, 750 0 −9600 −7200 0 0    0 9600 + (2) 18, 750 7200 0 0 −9600 −7200   −100k  5400 + 25, 000 0 −25, 000 −7200 −5400   0 18, 750 + (2)9600 7200 − 7200 −18, 750 0 =   0 SYMMETRIC 25, 000 + 5400(2) 0 0      50k  18, 750 + 9600 7200 0

25, 000 + 5400 (26.53)

5. convert to k/in and simplify    2362.5 −1562.5 0 −800 −600 0 0 0         3925.0 600 0 0 −800 −600 0           2533.33 0 −2083.33 −600 −450    −100  3162.5 0 −1562.5 0 0 =      2983.33 0 0   SYMMETRIC  0        2362.5 600 50        2533.33 0 6. Invert stiﬀness matrix and solve for displacements    U1  −0.0223 in            U 0.00433 in    2         V3   −0.116 in   = U4 −0.0102 in       V    5   −0.0856 in       −0.00919 in  U   6        V7 −0.0174 in

                   

 U1    U2     V3   U4  V5     U6     V7 (26.54)

                  

7. Solve for member internal forces (in this case axial forces) in local coordinate systems   U1       u1 V1 c s −c −s = u2 U2  −c −s c s      V2

(26.55)

(26.56)

Element 1

p1 p2

1 = (15, 000 k/ft)( =

52.1 k −52.1 k

1 ft ) 12 in

  −0.0223       0.8 0.6 −0.8 −0.6 0 (26.57-a) −0.8 −0.6 0.8 0.6  −0.0102      −0.0856

Compression

(26.57-b)

Element 2

Victor Saouma

p1 p2

2 = 18, 750 k/ft(

1 ft ) 12 in

  −0.0233       1 0 −1 0 0 (26.58-a) −1 0 1 0   0.00433     −0.116 Structural Engineering

u1 u2 v3 u4 v5 u6 v7

    

26–22

DIRECT STIFFNESS METHOD

Draft

=

−43.2 k 43.2 k

Tension

(26.58-b)

Element 3

3

p1 p2

= 25, 000 k/ft( =

−63.3 k 63.3 k

1 ft ) 12 in

0 1 0 0 −1 0

  0.00433       −1 −0.116 1  −0.0102      −0.0856

Tension

(26.59-a)

(26.59-b)

Element 4

4

p1 p2

=

18, 750 k/ft(

=

−1.58 k 1.58 k

1 ft ) 12 in

1 0 −1 −1 0 1

  −0.0102       0 −0.0856 0  −0.00919      −0.0174

Tension

(26.60-a)

(26.60-b)

Element 5

5

p1 p2

1 ft −0.8 0.6 0.8 −0.6 −0.0102 = 15, 000 k/ft( ) (26.61-a) 0.8 −0.6 −0.8 0.6 −0.0856 12 in 54.0 k = Compression (26.61-b) −54.0 k

Element 6

p1 p2

6 = =

   −0.116  1 ft 0.8 0.6 −0.8 −0.6 −0.00919 (26.62-a) 15, 000 k/ft( ) −0.8 −0.6 0.8 0.6   12 in −0.0174 −60.43 k Tension (26.62-b) 60.43 k

Element 7

p1 p2

7

1 ft 1 0 ) −1 0 12 in 6.72 k Compression = −6.72 k = 18, 750 k/ft(

   −0.116  −1 0 0 1 0   0

(26.63-a)

(26.63-b)

Element 8

p1 p2

8 =

25, 000 k/ft(

= Victor Saouma

36.3 k −36.3 k

1 ft ) 12 in

0 1 0 0 −1 0

Compression

−1 1

   

0 0 −0.00919    −0.0174

      

(26.64-a)

(26.64-b) Structural Engineering

26.3 Direct Stiﬀness Method

26–23

Draft

8. Determine the structure’s MAXA vector  1 3  2 5   4  [K] =     

9 8 7 6

14 13 12 11 10

 19 18 17 16 15

25 24 23 22 21 20

        

  1        2           4  6 MAXA =    10          15       20

(26.65)

Thus, 25 terms would have to be stored.

Example 26-4: Analysis of a Frame with MATLAB The simple frame shown in Fig. 26.13 is to be analysed by the direct stiﬀness method. Assume: E = 200, 000 MPa, A = 6, 000 mm2 , and I = 200 × 106 mm4 . The complete MATLAB solution is shown below along with the results. 50kN

4 kN/m 00 11 11 00 00 11 00 11 00 11

8m 3m 111 000 000 111 000 111 000 111 000 111

7.416 m

8m

Figure 26.13: Simple Frame Anlysed with the MATLAB Code

% zero the matrices k=zeros(6,6,2); K=zeros(6,6,2); Gamma=zeros(6,6,2); % Structural properties units: mm^2, mm^4, and MPa(10^6 N/m) A=6000;II=200*10^6;EE=200000; % Convert units to meter and kN A=A/10^6;II=II/10^12;EE=EE*1000; % Element 1 i=[0,0];j=[7.416,3]; [k(:,:,1),K(:,:,1),Gamma(:,:,1)]=stiff(EE,II,A,i,j); % Element 2 i=j;j=[15.416,3]; [k(:,:,2),K(:,:,2),Gamma(:,:,2)]=stiff(EE,II,A,i,j); % Define ID matrix ID=[ -4 1 -7; -5 2 -8; -6 3 -9]; % Determine the LM matrix LM=[ -4 -5 -6 1 2 3; 1 2 3 -7 -8 -9]; % Assemble augmented stiffness matrix Kaug=zeros(9); Victor Saouma

Structural Engineering

26–24

Draft

DIRECT STIFFNESS METHOD

for elem=1:2 for r=1:6 lr=abs(LM(elem,r)); for c=1:6 lc=abs(LM(elem,c)); Kaug(lr,lc)=Kaug(lr,lc)+K(r,c,elem); end end end % Extract the structures Stiffness Matrix Ktt=Kaug(1:3,1:3); % Determine the fixed end actions in local coordinate system fea(1:6,1)=0; fea(1:6,2)=[0,8*4/2,4*8^2/12,0,8*4/2,-4*8^2/12]’; % Determine the fixed end actions in global coordinate system FEA(1:6,1)=Gamma(:,:,1)*fea(1:6,1); FEA(1:6,2)=Gamma(:,:,2)*fea(1:6,2); % FEA_Rest for all the restrained nodes FEA_Rest=[0,0,0,FEA(4:6,2)’]; % Assemble the load vector for the unrestrained node P(1)=50*3/8;P(2)=-50*7.416/8-fea(2,2);P(3)=-fea(3,2); % Solve for the Displacements in meters and radians Displacements=inv(Ktt)*P’ % Extract Kut Kut=Kaug(4:9,1:3); % Compute the Reactions and do not forget to add fixed end actions Reactions=Kut*Displacements+FEA_Rest’ % Solve for the internal forces and do not forget to include the fixed end actions dis_global(:,:,1)=[0,0,0,Displacements(1:3)’]; dis_global(:,:,2)=[Displacements(1:3)’,0,0,0]; for elem=1:2 dis_local=Gamma(:,:,elem)*dis_global(:,:,elem)’; int_forces=k(:,:,elem)*dis_local+fea(1:6,elem) end function [k,K,Gamma]=stiff(EE,II,A,i,j) % Determine the length L=sqrt((j(2)-i(2))^2+(j(1)-i(1))^2); % Compute the angle theta (carefull with vertical members!) if(j(1)-i(1))~=0 alpha=atan((j(2)-i(2))/(j(1)-i(1))); else alpha=-pi/2; end % form rotation matrix Gamma Gamma=[ cos(alpha) sin(alpha) 0 0 0 0; -sin(alpha) cos(alpha) 0 0 0 0; 0 0 1 0 0 0; 0 0 0 cos(alpha) sin(alpha) 0; 0 0 0 -sin(alpha) cos(alpha) 0; 0 0 0 0 0 1]; % form element stiffness matrix in local coordinate system EI=EE*II; EA=EE*A; k=[EA/L, 0, 0, -EA/L, 0, 0; 0, 12*EI/L^3, 6*EI/L^2, 0, -12*EI/L^3, 6*EI/L^2; Victor Saouma

Structural Engineering

26.3 Direct Stiﬀness Method

26–25

Draft

0, 6*EI/L^2, 4*EI/L, 0, -6*EI/L^2, 2*EI/L; -EA/L, 0, 0, EA/L, 0, 0; 0, -12*EI/L^3, -6*EI/L^2, 0, 12*EI/L^3, -6*EI/L^2; 0, 6*EI/L^2, 2*EI/L, 0, -6*EI/L^2, 4*EI/L]; % Element stiffness matrix in global coordinate system K=Gamma’*k*Gamma; This simple proigram will produce the following results: Displacements = 0.0010 -0.0050 -0.0005 Reactions = 130.4973 55.6766 13.3742 -149.2473 22.6734 -45.3557

int_forces =

int_forces =

141.8530 2.6758 13.3742 -141.8530 -2.6758 8.0315

149.2473 9.3266 -8.0315 -149.2473 22.6734 -45.3557

We note that the internal forces are consistent with the reactions (specially for the second node of element 2), and amongst themselves, i.e. the moment at node 2 is the same for both elements (8.0315).

Example 26-5: Analysis of a simple Beam with Initial Displacements The full stiﬀness matrix of a beam element is given by

[K] =

v1  V1 12EI/L3 2 M1  6EI/L 3 V2  −12EI/L M2 6EI/L2

θ1 6EI/L2 4EI/L −6EI/L2 2EI/L

v2 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

θ2  6EI/L2 2EI/L   −6EI/L2  4EI/L

(26.66)

This matrix is singular, it has a rank 2 and order 4 (as it embodies also 2 rigid body motions). 66

We shall consider 3 diﬀerent cases, Fig. 26.14

Cantilivered Beam/Point Load 1. The element stiﬀness matrix is

k=

Victor Saouma

−3  −3 12EI/L3 −4 6EI/L2 1 −12EI/L3 2 6EI/L2

−4 6EI/L2 4EI/L −6EI/L2 2EI/L

1 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

2  6EI/L2 2EI/L  −6EI/L2 4EI/L

Structural Engineering

26–26

DIRECT STIFFNESS METHOD

Draft

Figure 26.14: ID Values for Simple Beam 2. The structure stiﬀness matrix is assembled 1  1 12EI/L2 2  −6EI/L2 −3 −12EI/L3 −4 −6EI/L2

K=

2 −6EI/L2 4EI/L 6EI/L2 2EI/L

3. The global matrix can be rewritten as √    12EI/L2 −6EI/L2  −P√  2 −6EI/L 4EI/L 0 = −12EI/L3 6EI/L2  R3 ?  2 −6EI/L

R4 ?

2EI/L

−3 −12EI/L3 6EI/L2 12EI/L3 6EI/L2

−12EI/L3 6EI/L2 12EI/L3 6EI/L2

−4  −6EI/L2 2EI/L  6EI/L2 4EI/L

−6EI/L2 2EI/L 6EI/L2 4EI/L

   ∆1 ?  θ2 ?  √  ∆3√  θ4

4. Ktt is inverted (or actually decomposed) and stored in the same global matrix     

L3 /3EI

L2 /2EI

−12EI/L3

−6EI/L2

L2 /2EI

L/EI

6EI/L2

2EI/L

−12EI/L3 −6EI/L2

6EI/L2 2EI/L

12EI/L3 6EI/L2

6EI/L2 4EI/L

  

5. Next we compute the equivalent load, Pt = Pt − Ktu ∆u , and overwrite Pt by    L3 /3EI L2 /2EI −12EI/L3 −6EI/L2 −P      2 0 L/EI 6EI/L2 2EI/L Pt − Ktu ∆u = −  L /2EI    0   −12EI/L3 6EI/L2 12EI/L3 6EI/L2 0 −6EI/L2 2EI/L 6EI/L2 4EI/L   −P     =

0 0 0

 

Victor Saouma

0

 

6. Now we solve for the displacement ∆t = K−1 tt Pt ,    L3 /3EI L2 /2EI ∆1      2 θ2 L/EI =  L /2EI    0   −12EI/L3 6EI/L2 2 0 2EI/L  −6EI/L  3 /3EI −P L      

=

Pt   −P     0    0   

  

−P L2 /2EI 0 0

and overwrite Pt by ∆t   −12EI/L3 −6EI/L2 −P      0 6EI/L2 2EI/L   0    12EI/L3 6EI/L2 6EI/L2

4EI/L

0

  

Structural Engineering

26.3 Direct Stiﬀness Method

26–27

Draft

7. Finally, we solve for the reactions, Ru = Kut ∆tt + Kuu ∆u , and overwrite ∆u by Ru    −P L3 /3EI  3 L3 /3EI L2 /2EI −12EI/L3 −6EI/L2   −P L /3EI  2   −P L2 /2EI   L/EI 6EI/L2 2EI/L    L /2EI −P L2 /2EI  =  −12EI/L3 6EI/L2 12EI/L3 6EI/L2  R 3        0   R4 −6EI/L2 2EI/L 6EI/L2 4EI/L 0   −P L3 /3EI    −P L2 /2EI  =

 

        

 

P PL

Simply Supported Beam/End Moment 1. The element stiﬀness matrix is −3  −3 12EI/L3 1  6EI/L2 −4 −12EI/L3 2 6EI/L2

k=

1 6EI/L2 4EI/L −6EI/L2 2EI/L

−4 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

2  6EI/L2 2EI/L  −6EI/L2 4EI/L

2. The structure stiﬀness matrix is assembled 1  1 4EI/L 2  2EI/L −3 6EI/L2 −4 −6EI/L2

K=

2 2EI/L 4EI/L 6EI/L2 −6EI/L2

−3 6EI/L2 6EI/L2 12EI/L3 −12EI/L3

3. The global stiﬀness matrix can be rewritten as  √  0    4EI/L 2EI/L 6EI/L2   M√   2EI/L 4EI/L 6EI/L2 = 6EI/L2 6EI/L2 12EI/L3     R3 ?   −6EI/L2 −6EI/L2 −12EI/L3 R4 ?

4. Ktt is inverted



L3 /3EI

  −L/6EI  2

6EI/L −6EI/L2

5. We compute the equivalent    0 Pt − Ktu ∆u

=

=

      

M 0 0 0 M 0 0

−6EI/L2 −6EI/L2 −12EI/L3 12EI/L3

−L/6EI

6EI/L2

−6EI/L2

L/3EI

6EI/L2

−6EI/L2

6EI/L2 −6EI/L2

12EI/L3 −12EI/L3

−12EI/L3 12EI/L3

load, Pt = Pt − Ktu ∆u , and overwrite   3 L /3EI −L/6EI 6EI/L2    L/3EI 6EI/L2 −  −L/6EI    6EI/L2 6EI/L2 12EI/L3 −6EI/L2 −6EI/L2 −12EI/L3   

Victor Saouma

  

   θ1 ?  θ ? 2  √  ∆3 √  ∆4

    Pt by Pt −6EI/L2 −6EI/L2 −12EI/L3 12EI/L3

  0     M    0    0

 

6. Solve for the displacements, ∆t = K−1 tt Pt , and overwrite Pt    L3 /3EI −L/6EI 6EI/L2 θ   1    θ2 L/3EI 6EI/L2 =  −L/6EI    0   6EI/L2 6EI/L2 12EI/L3 2 2 0 −6EI/L −12EI/L3  −6EI/L    −M L/6EI    

=

−4  −6EI/L2 2 −6EI/L  −12EI/L3 12EI/L3

M L/3EI 0 0

by ∆t

  0      M −6EI/L2   0    −12EI/L3 −6EI/L2

12EI/L3

0

  

Structural Engineering

26–28

DIRECT STIFFNESS METHOD

Draft

7. Solve for the reactions, Rt = Kut ∆tt + Kuu ∆u , and overwrite   −M L/6EI  L3 /3EI −L/6EI 6EI/L2  2  M L/3EI   −L/6EI L/3EI 6EI/L   = 6EI/L2 12EI/L3 R1  6EI/L2     2 2 3 −6EI/L

R2

−6EI/L

−12EI/L

  −M L/6EI     M L/3EI  

=

M/L

  

−M/L

∆u by Ru −6EI/L2 −6EI/L2 −12EI/L3 12EI/L3

  −M L/6EI     M L/3EI   0    0

        

  

Cantilivered Beam/Initial Displacement and Concentrated Moment 1. The element stiﬀness matrix is −2  −2 12EI/L3 −3 6EI/L2 −4 −12EI/L3 1 6EI/L2

k=

−3 6EI/L2 4EI/L −6EI/L2 2EI/L

−4 −12EI/L3 −6EI/L2 12EI/L3 −6EI/L2

1  6EI/L2 2EI/L  −6EI/L2 4EI/L

−3 2EI/L 6EI/L2 4EI/L −6EI/L2

−4  −6EI/L2 −12EI/L3  −6EI/L2 12EI/L3

2. The structure stiﬀness matrix is assembled 1  1 4EI/L −2 6EI/L2 −3 2EI/L −4 −6EI/L2

K=

3. The global matrix can be  √    M  R2 ? =  R3 ?  R4 ?

−2 6EI/L2 12EI/L3 6EI/L2 −12EI/L3

rewritten as 4EI/L 6EI/L2 2EI/L −6EI/L2

6EI/L2 12EI/L3 6EI/L2 −12EI/L3

2EI/L 6EI/L2 4EI/L −6EI/L2

−6EI/L2 −12EI/L3 −6EI/L2 12EI/L3

   θ1 ?√  ∆2√   θ3 √  ∆4

4. Ktt is inverted (or actually decomposed) and stored in the same global matrix   L/4EI

 6EI/L2 

2EI/L −6EI/L2

6EI/L2

2EI/L

−6EI/L2

12EI/L3

6EI/L2

−12EI/L3 −6EI/L2 12EI/L3

6EI/L2 −12EI/L3

4EI/L −6EI/L2

 

5. Next we compute the equivalent load, Pt = Pt − Ktu ∆u , and overwrite Pt by Pt      θ1  L/4EI 6EI/L2 2EI/L −6EI/L2    0  M    6EI/L2  3 2 3 0 12EI/L 6EI/L −12EI/L  Pt − Ktu ∆u = − 0   2EI/L 6EI/L2 4EI/L −6EI/L2   00    ∆ −6EI/L2 −12EI/L3 −6EI/L2 12EI/L3 ∆0   0 2   M + 6EI∆ /L   =

 

0 0 ∆0

Victor Saouma

 

0 0 ∆0

  

 

6. Now we solve for the displacements, ∆t = K−1 tt Pt , and overwrite Pt by ∆t     0 2 L/4EI 6EI/L2 2EI/L −6EI/L2    θ1    M + 6EI∆ /L   2 3 2 3 0 12EI/L 6EI/L −12EI/L  0 =  6EI/L   2EI/L 6EI/L2 4EI/L −6EI/L2 0  00    0 ∆ −6EI/L2 −12EI/L3 −6EI/L2 12EI/L3 ∆   0   M L/4EI + 3∆ /2L  

=

   

    

 

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26.4 Computer Program Organization

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Draft

7. Finally, we solve for the reactions, Rt    M L/4EI + 3∆0 /2L        R2  =  R3        R4

=

26.4 66

= Kut ∆tt + Kuu ∆u , and overwrite ∆u by Ru L/4EI

6EI/L2

2EI/L

−6EI/L2

6EI/L2

12EI/L3

6EI/L2

−12EI/L3

2EI/L

6EI/L2

4EI/L

−6EI/L2

−6EI/L2

−12EI/L3

−6EI/L2

12EI/L3

  M L/4EI + 3∆0 /2L   

    

        

   

0 0

∆0 M L/4EI + 3∆0 /2L 3M/2L − 3EI∆0 /L3

M/2 − 3EI∆0 /L2     −3M/2L + 3EI∆0 /L3

    

        

Computer Program Organization

The main program should, 1. Read (a) title card (b) control card which should include: i. ii. iii. iv. v.

Number of nodes Number of elements Type of structure: beam, grid, truss, or frame; (2D or 3D) Number of diﬀerent element properties Number of load cases

2. Determine: (a) Number of spatial coordinates for the structure (b) Number of local and global degrees of freedom per node 3. For each node read: (a) Node number (b) Boundary conditions of each global degree of freedom [ID] (c) Spatial coordinates Note that all the above are usually written on the same “data card” 4. For each element, read: (a) Element number (b) First and second node (c) Element Property number 5. For each element property group read the associated elastic and cross sectional characteristics. Note these variables will depend on the structure type. 6. Determine the vector ∆u which stores the initial displacements. 7. Loop over all the elements and for each one: Victor Saouma

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(a) Retrieve its properties

(b) Determine the length (c) Call the appropriate subroutines which will determine: i. The stiﬀness matrix in local coordinate systems [k(e) ]. ii. The angle α and the transformation matrix [Γ(e) ]. 8. Assembly of the global stiﬀness matrix (a) Initialize the global stiﬀness matrix to zero (b) Loop through each element, e, and for each element: i. Retrieve its stiﬀness matrix (in local coordinates) [k(e) ] and transformation matrix [Γ(e) ]. ii. Compute the element stiﬀness matrix in global coordinates from [K(e) ] = [Γ(e) ]T [k(e) ][Γ(e) ]. iii. Deﬁne the {LM} array of the element iv. Loop through each row i and column j of the element stiﬀness matrix, and for those degree of freedom not equal to zero, add the contributions of the element to the structure’s stiﬀness matrix K S [LM (i), LM (j)] = K S [LM (i), LM (j)] + K (e) [i, j] 9. Extract the structure’s stiﬀness matrix [Ktt ] from the augmented stiﬀness matrix. 10. Invert the structure’s stiﬀness matrix (or decompose it). 11. For each load case: (a) Determine the nodal equivalent loads (ﬁxed end actions), if any. (b) Assemble the load vector (c) Load assembly (once for each load cae) once the stiﬀness matrix has been decomposed, than the main program should loop through each load case and, i. Initialize the load vector (of length NEQ) to zero. ii. Read number of loaded nodes. For each loaded node store the non-zero values inside the load vector (using the [ID] matrix for determining storage location). iii. Loop on all loaded elements: A. Read element number, and load value B. Compute the ﬁxed end actions and rotate them from local to global coordinates. C. Using the LM vector, add the ﬁxed end actions to the nodal load vector (unless the corresponding equation number is zero, ie. restrained degree of freedom). D. Store the ﬁxed end actions for future use. (d) Apply Eq. 26.31 to determine the nodal displacements ∆t = K−1 tt (Pt − Ktu ∆u ) (e) Apply Eq. 26.32 to determine the nodal reactions Rt = Kut ∆t + Kuu ∆u (f) Determine the internal forces (axial, shear and moment) i. For each element retrieve: A. nodal coordinates B. rotation matrix [Γ(e) ]. C. element stiﬀness matrix [k(e) ].

4 5 ii. Compute nodal displacements in local coordinate system from δ (e) = [Γ(e) ] {∆} 5 4 iii. Compute element internal forces from {p} = [k(e) ] δ (e) iv. If the element is loaded, add corresponding ﬁxed end actions v. print the interior forces

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Computer Implementation with MATLAB

You will be required, as part of your term project, to write a simple MATLAB (or whatever other language you choose) program for the analysis of two dimensional frames with nodal load and initial displacement, as well as element load.

67

To facilitate the task, your instructor has taken the liberty of taking a program written by Mr. Dean Frank (as part of his term project with this instructor in the Advanced Structural Analysis course, Fall 1995), modiﬁed it with the aid of Mr. Pawel Smolarki, and is making available most, but not all of it to you. Hence, you will be expected to ﬁrst familiarize yourself with the code made available to you, and then complete it by essentially ﬁlling up the missing parts.

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26.5.1

Program Input

From Dean Frank’s User’s Manual It is essential that the structure be idealized such that it can be discretized. This discretization should deﬁne each node and element uniquely. In order to decrease the required amount of computer storage and computation it is best to number the nodes in a manner that minimizes the numerical separation of the node numbers on each element. For instance, an element connecting nodes 1 and 4, could be better deﬁned by nodes 1 and 2, and so on. As it was noted previously, the user is required to have a decent understanding of structural analysis and structural mechanics. As such, it will be necessary for the user to generate or modify an input ﬁle input.m using the following directions. Open the ﬁle called input.m and set the existing variables in the ﬁle to the appropriate values. The input ﬁle has additional helpful directions given as comments for each variable. After setting the variables to the correct values, be sure to save the ﬁle. Please note that the program is case-sensitive.

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In order for the program to be run, the user must supply the required data by setting certain variables in the ﬁle called indat.m equal to the appropriate values. All the user has to do is open the text ﬁle called indat.txt, ﬁll in the required values and save the ﬁle as indat.m in a directory within MATAB’s path. There are helpful hints within this ﬁle. It is especially important that the user keep track of units for all of the variables in the input data ﬁle. All of the units MUST be consistent. It is suggested that one always use the same units for all problems. For example, always use kips and inches, or kilonewtons and millimeters.

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26.5.1.1

Input Variable Descriptions

A brief description of each of the variables to be used in the input ﬁle is given below: npoin This variable should be set equal to the number of nodes that comprise the structure. A node is deﬁned as any point where two or more elements are joined. nelem This variable should be set equal to the number of elements in the structure. Elements are the members which span between nodes. istrtp This variable should be set equal to the type of structure. There are six types of structures which this program will analyze: beams, 2-D trusses, 2-D frames, grids, 3-D trusses, and 3-D frames. Set this to 1 for beams, 2 for 2D-trusses, 3 for 2D- frames, 4 for grids, 5 for 3D-trusses, and 6 for 3D-frames. An error will occur if it is not set to a number between 1 and 6. Note only istrp=3 was kept. nload This variable should be set equal to the number of diﬀerent load cases to be analyzed. A load case is a speciﬁc manner in which the structure is loaded. ID (matrix) The ID matrix contains information concerning the boundary conditions for each node. The number of rows in the matrix correspond with the number of nodes in the structure and the number of columns corresponds with the number of degrees of freedom for each node for that type of structure type. The matrix is composed of ones and zeros. A one indicates that the degree of freedom is restrained and a zero means it is unrestrained. nodecoor (matrix) This matrix contains the coordinates (in the global coordinate system) of the nodes in the structure. The rows correspond with the node number and the columns correspond with the global coordinates x, y, and z, respectively. It is important to always include all three coordinates for each node even if the structure is only two- dimensional. In the case of a two-dimensional structure, the z-coordinate would be equal to zero. 71

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lnods (matrix) This matrix contains the nodal connectivity information. The rows correspond with the element number and the columns correspond with the node numbers which the element is connected from and to, respectively. E,A,Iy (arrays) These are the material and cross-sectional properties for the elements. They are arrays with the number of terms equal to the number of elements in the structure. The index number of each term corresponds with the element number. For example, the value of A(3) is the area of element 3, and so on. E is the modulus of elasticity, A is the cross-sectional area, Iy is the moment of inertia about the y axes Pnods This is an array of nodal loads in global degrees of freedom. Only put in the loads in the global degrees of freedom and if there is no load in a particular degree of freedom, then put a zero in its place. The index number corresponds with the global degree of freedom. Pelem This an array of element loads, or loads which are applied between nodes. Only one load between elements can be analyzed. If there are more than one element loads on the structure, the equivalent nodal load can be added to the nodal loads. The index number corresponds with the element number. If there is not a load on a particular member, put a zero in its place. These should be in local coordinates. a This is an array of distances from the left end of an element to the element load. The index number corresponds to the element number. If there is not a load on a particular member, put a zero in its place. This should be in local coordinates. w This is an array of distributed loads on the structure. The index number corresponds with the element number. If there is not a load on a particular member, put a zero in its place. This should be in local coordinates dispflag Set this variable to 1 if there are initial displacements and 0 if there are none. initial displ This is an array of initial displacements in all structural degrees of freedom. This means that you must enter in values for all structure degrees of freedom, not just those restrained. For example, if the structure is a 2D truss with 3 members and 3 node, there would be 6 structural degrees of freedom, etc. If there are no initial displacements, then set the values equal to zero. angle This is an array of angles which the x-axis has possibly been rotated. This angle is taken as positive if the element has been rotated towards the z-axis. The index number corresponds to the element number. drawflag Set this variable equal to 1 if you want the program to draw the structure and 0 if you do not. 26.5.1.2

Sample Input Data File

The contents of the input.m ﬁle which the user is to ﬁll out is given below: %********************************************************************************************** % Scriptfile name: indat.m (EXAMPLE 2D-FRAME INPUT DATA) % % Main Program: casap.m % % This is the main data input file for the computer aided % structural analysis program CASAP. The user must supply % the required numeric values for the variables found in % this file (see user’s manual for instructions). % % By Dean A. Frank % CVEN 5525 - Term Project % Fall 1995 % % Edited by Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %********************************************************************************************** % HELPFUL INSTRUCTION COMMENTS IN ALL CAPITALS % SET NPOIN EQUAL TO THE NUMBER OF NODES IN THE STRUCTURE npoin=3; % SET NELEM EQUAL TO THE NUMBER OF ELEMENTS IN THE STRUCTURE

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Draft nelem=2;

% SET NLOAD EQUAL TO THE NUMBER OF LOAD CASES nload=1; % INPUT THE ID MATRIX CONTAINING THE NODAL BOUNDARY CONDITIONS (ROW # = NODE #) ID=[1 1 1; 0 0 0; 1 1 1]; % INPUT THE NODE COORDINATE (X,Y) MATRIX, NODECOOR (ROW # = NODE #) nodecoor=[ 0 0; 7416 3000; 15416 3000 ]; % INPUT THE ELEMENT CONNECTIVITY MATRIX, LNODS (ROW # = ELEMENT #) lnods=[ 1 2; 2 3 ]; % INPUT THE MATERIAL PROPERTIES ASSOCIATED WITH THIS TYPE OF STRUCTURE % PUT INTO ARRAYS WHERE THE INDEX NUMBER IS EQUAL TO THE CORRESPONDING ELEMENT NUMBER. % COMMENT OUT VARIABLES THAT WILL NOT BE USED E=[200 200]; A=[6000 6000]; Iz=[200000000 200000000]; % % % % % % % % %

INPUT THE LOAD DATA. NODAL LOADS, PNODS SHOULD BE IN MATRIX FORM. THE COLUMNS CORRESPOND TO THE GLOBAL DEGREE OF FREEDOM IN WHICH THE LOAD IS ACTING AND THE THE ROW NUMBER CORRESPONDS WITH THE LOAD CASE NUMBER. PELEM IS THE ELEMENT LOAD, GIVEN IN A MATRIX, WITH COLUMNS CORRESPONDING TO THE ELEMENT NUMBER AND ROW THE LOAD CASE. ARRAY "A" IS THE DISTANCE FROM THE LEFT END OF THE ELEMENT TO THE LOAD, IN ARRAY FORM. THE DISTRIBUTED LOAD, W SHOULD BE IN MATRIX FORM ALSO WITH COLUMNS = ELEMENT NUMBER UPON WHICH W IS ACTING AND ROWS = LOAD CASE. ZEROS SHOULD BE USED IN THE MATRICES WHEN THERE IS NO LOAD PRESENT. NODAL LOADS SHOULD BE GIVEN IN GLOBAL COORDINATES, WHEREAS THE ELEMENT LOADS AND DISTRIBUTED LOADS SHOULD BE GIVEN IN LOCAL COORDINATES.

Pnods=[18.75 -46.35 0]; Pelem=[0 0]; a=[0 0]; w=[0 4/1000]; % IF YOU WANT THE PROGRAM TO DRAW THE STUCTURE SET DRAWFLAG=1, IF NOT SET IT EQUAL TO 0. % THIS IS USEFUL FOR CHECKING THE INPUT DATA. drawflag=1; % END OF INPUT DATA FILE

26.5.1.3

Program Implementation

In order to ”run” the program, open a new MATLAB Notebook. On the ﬁrst line, type the name of the main program CASAP and evaluate that line by typing ctrl-enter. At this point, the main program reads the input ﬁle you have just created and calls the appropriate subroutines to analyze your structure. In doing so, your input data is echoed into your MATLAB notebook and the program results are also displayed. As a note, the program can also be executed directly from the MATAB workspace window, without Microsoft Word.

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Structural Engineering

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Draft 26.5.2

Program Listing

26.5.2.1

Main Program

DIRECT STIFFNESS METHOD

%********************************************************************************************** %Main Program: casap.m % % This is the main program, Computer Aided Structural Analysis Program % CASAP. This program primarily contains logic for calling scriptfiles and does not % perform calculations. % % All variables are global, but are defined in the scriptfiles in which they are used. % % Associated scriptfiles: % % (for all stuctures) % indat.m (input data file) % idrasmbl.m % elmcoord.m % draw.m % % (3 - for 2D-frames) % length3.m % stiffl3.m % trans3.m % assembl3.m % loads3.m % disp3.m % react3.m % % By Dean A. Frank % CVEN 5525 % Advanced Structural Analysis - Term Project % Fall 1995 % % Edited by Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %********************************************************************************************** % COMMENT CARDS ARE IN ALL CAPITALS % SET NUMERIC FORMAT format short e % CLEAR MEMORY OF ALL VARIABLES clear % INITIALIZE OUTPUT FILE fid = fopen(’casap.out’, ’wt’); % SET ISTRTP EQUAL TO THE NUMBER CORRESPONDING TO THE TYPE OF STRUCTURE: % 3 = 2DFRAME istrtp=3; % READ INPUT DATA SUPPLIED BY THE USER indat % REASSAMBLE THE ID MATRIX AND CALCULATE THE LM VECTORS % CALL SCRIPTFILE IDRASMBL idrasmbl % ASSEMBLE THE ELEMENT COORDINATE MATRIX elmcoord % 2DFRAME CALCULATIONS

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26.5 Computer Implementation with MATLAB

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Draft

% CALCULATE THE LENGTH AND ORIENTATION ANGLE, ALPHA FOR EACH ELEMENT % CALL SCRIPTFILE LENGTH3.M length3 % CALCULATE THE 2DFRAME ELEMENT STIFFNESS MATRIX IN LOCAL COORDINATES % CALL SCRIPTFILE STIFFL3.M stiffl3 % CALCULATE THE 2DFRAME ELEMENT STIFFNESS MATRIX IN GLOBAL COORDINATES % CALL SCRIPTFILE TRANS3.M trans3 % ASSEMBLE THE GLOBAL STRUCTURAL STIFFNESS MATRIX % CALL SCRIPTFILE ASSEMBL3.M assembl3 %

PRINT STRUCTURAL INFO

print_general_info % LOOP TO PERFORM ANALYSIS FOR EACH LOAD CASE for iload=1:nload print_loads % DETERMINE THE LOAD VECTOR IN GLOBAL COORDINATES % CALL SCRIPTFILE LOADS3.M loads3 % CALCULATE THE DISPLACEMENTS % CALL SCRIPTFILE DISP3.M disp3 % CALCULATE THE REACTIONS AT THE RESTRAINED DEGREES OF FREEDOM % CALL SCRIPTFILE REACT3.M react3 % CALCULATE THE INTERNAL FORCES FOR EACH ELEMENT intern3 % END LOOP FOR EACH LOAD CASE end % DRAW THE STRUCTURE, IF USER HAS REQUESTED (DRAWFLAG=1) % CALL SCRIPTFILE DRAW.M draw st=fclose(’all’); % END OF MAIN PROGRAM (CASAP.M) disp(’Program completed! - See "casap.out" for complete output’);

26.5.2.2

Assembly of ID Matrix

%************************************************************************************************ %SCRIPTFILE NAME: IDRASMBL.M % %MAIN FILE : CASAP % %Description : This file re-assambles the ID matrix such that the restrained % degrees of freedom are given negative values and the unrestrained

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% degrees of freedom are given incremental values begining with one % and ending with the total number of unrestrained degrees of freedom. % % By Dean A. Frank % CVEN 5525 % Advanced Structural Analysis - Term Project % Fall 1995 % % Edited by Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %************************************************************************************************ % TAKE CARE OF SOME INITIAL BUSINESS: TRANSPOSE THE PNODS ARRAY Pnods=Pnods.’; % SET THE COUNTER TO ZERO count=1; negcount=-1; % REASSEMBLE THE ID MATRIX if istrtp==3 ndofpn=3; nterm=6; else error(’Incorrect structure type specified’) end % SET THE ORIGINAL ID MATRIX TO TEMP MATRIX orig_ID=ID; % REASSEMBLE THE ID MATRIX, SUBSTITUTING RESTRAINED DEGREES OF FREEDOM WITH NEGATIVES, % AND NUMBERING GLOBAL DEGREES OF FREEDOM for inode=1:npoin for icoord=1:ndofpn if ID(inode,icoord)==0 ID(inode,icoord)=count; count=count+1; elseif ID(inode,icoord)==1 ID(inode,icoord)=negcount; negcount=negcount-1; else error(’ID input matrix incorrect’) end end end % CREATE THE LM VECTORS FOR EACH ELEMENT for ielem=1:nelem LM(ielem,1:ndofpn)=ID(lnods(ielem,1),1:ndofpn); LM(ielem,(ndofpn+1):(2*ndofpn))=ID(lnods(ielem,2),1:ndofpn); end % END OF IDRASMBL.M SCRIPTFILE

26.5.2.3

Element Nodal Coordinates

%********************************************************************************************** %SCRIPTFILE NAME: ELEMCOORD.M % %MAIN FILE : CASAP % %Description : This file assembles a matrix, elemcoor which contains the coordinates % of the first and second nodes on each element, respectively. %

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Draft

% By Dean A. Frank % CVEN 5525 % Advanced Structural Analysis - Term Project % Fall 1995 % % Edited by Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %********************************************************************************************** % ASSEMBLE THE ELEMENT COORDINATE MATRIX, ELEMCOOR FROM NODECOOR AND LNODS for ielem=1:nelem elemcoor(ielem,1)=nodecoor(lnods(ielem,1),1); elemcoor(ielem,2)=nodecoor(lnods(ielem,1),2); %elemcoor(ielem,3)=nodecoor(lnods(ielem,1),3); elemcoor(ielem,3)=nodecoor(lnods(ielem,2),1); elemcoor(ielem,4)=nodecoor(lnods(ielem,2),2); %elemcoor(ielem,6)=nodecoor(lnods(ielem,2),3); end % END OF ELMCOORD.M SCRIPTFILE

26.5.2.4

Element Lengths

%********************************************************************************************** % Scriptfile name : length3.m (for 2d-frame structures) % % Main program : casap.m % % When this file is called, it computes the length of each element and the % angle alpha between the local and global x-axes. This file can be used % for 2-dimensional elements such as 2-D truss, 2-D frame, and grid elements. % This information will be useful for transformation between local and global % variables. % % Variable descriptions: (in the order in which they appear) % % nelem = number of elements in the structure % ielem = counter for loop % L(ielem) = length of element ielem % elemcoor(ielem,4) = xj-coordinate of element ielem % elemcoor(ielem,1) = xi-coordinate of element ielem % elemcoor(ielem,5) = yj-coordinate of element ielem % elemcoor(ielem,2) = yi-coordinate of element ielem % alpha(ielem) = angle between local and global x-axes % % By Dean A. Frank % CVEN 5525 - Term Project % Fall 1995 % % Edited by Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %********************************************************************************************** % COMPUTE THE LENGTH AND ANGLE BETWEEN LOCAL AND GLOBAL X-AXES FOR EACH ELEMENT for ielem=1:nelem L(ielem)= XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX alpha(ielem)= XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX % END OF LENGTH3.M SCRIPTFILE

26.5.2.5

Element Stiﬀness Matrices

%********************************************************************************************** % Scriptfile name: stiffl3.m (for 2d-frame structures) %

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DIRECT STIFFNESS METHOD

% Main program: casap.m % % When this file is called, it computes the element stiffenss matrix % of a 2-D frame element in local coordinates. The element stiffness % matrix is calculated for each element in the structure. % % The matrices are stored in a single matrix of dimensions 6x6*i and % can be recalled individually later in the program. % % Variable descriptions: (in the order in which the appear) % % ielem = counter for loop % nelem = number of element in the structure % k(ielem,6,6)= element stiffness matrix in local coordinates % E(ielem) = modulus of elasticity of element ielem % A(ielem) = cross-sectional area of element ielem % L(ielem) = lenght of element ielem % Iz(ielem) = moment of inertia with respect to the local z-axis of element ielem % % By Dean A. Frank % CVEN 5525 - Term Project % Fall 1995 % %********************************************************************************************** for ielem=1:nelem k(1:6,1:6,ielem)=... XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX end % END OF STIFFL3.M SCRIPTFILE

26.5.2.6

Transformation Matrices

%********************************************************************************************** % Scriptfile name : trans3.m (for 2d-frame structures) % % Main program : casap.m % % This file calculates the rotation matrix and the element stiffness % matrices for each element in a 2D frame. % % Variable descriptions: (in the order in which they appear) % % ielem = counter for the loop % nelem = number of elements in the structure % rotation = rotation matrix containing all elements info % Rot = rotational matrix for 2d-frame element % alpha(ielem) = angle between local and global x-axes % K = element stiffness matrix in global coordinates % k = element stiffness matrix in local coordinates % % By Dean A. Frank % CVEN 5525 - Term Project % Fall 1995 % %********************************************************************************************** % CALCULATE THE ELEMENT STIFFNESS MATRIX IN GLOBAL COORDINATES % FOR EACH ELEMENT IN THE STRUCTURE for ielem=1:nelem % SET UP THE ROTATION MATRIX, ROTATAION rotation(1:6,1:6,ielem)=... XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX ktemp=k(1:6,1:6,ielem); % CALCULATE THE ELEMENT STIFFNESS MATRIX IN GLOBAL COORDINATES Rot=rotation(1:6,1:6,ielem); K(1:6,1:6,ielem)=

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XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX end % END OF TRANS3.M SCRIPTFILE

26.5.2.7

Assembly of the Augmented Stiﬀness Matrix

%********************************************************************************************** % Scriptfile name : assembl3.m (for 2d-frame structures) % % Main program : casap.m % % This file assembles the global structural stiffness matrix from the % element stiffness matrices in global coordinates using the LM vectors. % In addition, this file assembles the augmented stiffness matrix. % % Variable Descritpions (in order of appearance): % % ielem = Row counter for element number % nelem = Number of elements in the structure % iterm = Counter for term number in LM matrix % LM(a,b) = LM matrix % jterm = Column counter for element number % temp1 = Temporary variable % temp2 = Temporary variable % temp3 = Temporary variable % temp4 = Temporary variable % number_gdofs = Number of global dofs % new_LM = LM matrix used in assembling the augmented stiffness matrix % aug_total_dofs = Total number of structure dofs % K_aug = Augmented structural stiffness matrix % Ktt = Structural Stiffness Matrix (Upper left part of Augmented structural stiffness matrix) % Ktu = Upper right part of Augmented structural stiffness matrix % Kut = Lower left part of Augmented structural stiffness matrix % Kuu = Lower rigth part of Augmented structural stiffness matrix % % % By Dean A. Frank % CVEN 5525 - Term Project % Fall 1995 % % Edited by Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %**********************************************************************************************

% RENUMBER DOF INCLUDE ALL DOF, FREE DOF FIRST, RESTRAINED NEXT new_LM=LM; number_gdofs=max(LM(:)); new_LM(find(LM<0))=number_gdofs-LM(find(LM<0)); aug_total_dofs=max(new_LM(:)); % ASSEMBLE THE AUGMENTED STRUCTURAL STIFFNESS MATRIX K_aug=zeros(aug_total_dofs); for ielem=1:nelem XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX Tough one! XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX end % SET UP SUBMATRICES FROM THE AUGMENTED STIFFNESS MATRIX Ktt= XXXXXXXXXXXXXXXXXXXXXX COMPLETE Ktu= XXXXXXXXXXXXXXXXXXXXXX COMPLETE Kut= XXXXXXXXXXXXXXXXXXXXXX COMPLETE Kuu= XXXXXXXXXXXXXXXXXXXXXX COMPLETE % END OF ASSEMBL3.M SCRIPTFILE

Victor Saouma

XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

Structural Engineering

26–40

Draft 26.5.2.8

DIRECT STIFFNESS METHOD

Print General Information

%********************************************************************************************** % Scriptfile name : print_general_info.m % % Main program : casap.m % % Prints the general structure info to the output file % % By Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %********************************************************************************************** fprintf(fid,’\n\nNumber of Nodes: %d\n’,npoin); fprintf(fid,’Number of Elements: %d\n’,nelem); fprintf(fid,’Number of Load Cases: %d\n’,nload); fprintf(fid,’Number of Restrained dofs: %d\n’,abs(min(LM(:)))); fprintf(fid,’Number of Free dofs: %d\n’,max(LM(:))); fprintf(fid,’\nNode Info:\n’); for inode=1:npoin fprintf(fid,’ Node %d (%d,%d)\n’,inode,nodecoor(inode,1),nodecoor(inode,2)); freedof=’ ’; if(ID(inode,1))>0 freedof=strcat(freedof,’ X ’); end if(ID(inode,2))>0 freedof=strcat(freedof,’ Y ’); end if(ID(inode,3))>0 freedof=strcat(freedof,’ Rot’); end if freedof==’ ’ freedof=’ none; node is fixed’; end fprintf(fid,’ Free dofs:%s\n’,freedof); end fprintf(fid,’\nElement Info:\n’); for ielem=1:nelem fprintf(fid,’ Element %d (%d->%d)’,ielem,lnods(ielem,1),lnods(ielem,2)); fprintf(fid,’ E=%d A=%d Iz=%d \n’,E(ielem),A(ielem),Iz(ielem)); end

26.5.2.9

Print Load

%********************************************************************************************** % Scriptfile name : print_loads.m % % Main program : casap.m % % Prints the current load case data to the output file % % By Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %********************************************************************************************** Load_case=iload if iload==1 fprintf(fid,’\n_________________________________________________________________________\n\n’); end fprintf(fid,’Load Case: %d\n\n’,iload); fprintf(fid,’ Nodal Loads:\n’); for k=1:max(LM(:)); %WORK BACKWARDS WITH LM MATRIX TO FIND NODE# AND DOF LM_spot=find(LM’==k); elem=fix(LM_spot(1)/(nterm+1))+1; dof=mod(LM_spot(1)-1,nterm)+1; node=lnods(elem,fix(dof/4)+1);

Victor Saouma

Structural Engineering

26.5 Computer Implementation with MATLAB

26–41

Draft

switch(dof) case {1,4}, dof=’Fx’; case {2,5}, dof=’Fy’; otherwise, dof=’ M’; end %PRINT THE DISPLACEMENTS if Pnods(k)~=0 fprintf(fid,’ Node: %2d %s = %14d\n’,node, dof, Pnods(k)); end

end fprintf(fid,’\n Elemental Loads:\n’); for k=1:nelem fprintf(fid,’ Element: %d Point load = %d at %d from left\n’,k,Pelem(k),a(k)); fprintf(fid,’ Distributed load = %d\n’,w(k)); end fprintf(fid,’\n’);

26.5.2.10

Load Vector

%********************************************************************************************** % Scriptfile name: loads3.m (for 2d-frame structures) % % Main program: casap.m % % When this file is called, it computes the fixed end actions for elements which % carry distributed loads for a 2-D frame. % % Variable descriptions: (in the order in which they appear) % % ielem = counter for loop % nelem = number of elements in the structure % b(ielem) = distance from the right end of the element to the point load % L(ielem) = length of the element % a(ielem) = distance from the left end of the element to the point load % Ffl = fixed end force (reaction) at the left end due to the point load % w(ielem) = distributed load on element ielem % L(ielem) = length of element ielem % Pelem(ielem) = element point load on element ielem % Mfl = fixed end moment (reaction) at the left end due to the point load % Ffr = fixed end force (reaction) at the right end due to the point load % Mfr = fixed end moment (reaction) at the right end due to the point load % feamatrix_local = matrix containing resulting fixed end actions in local coordinates % feamatrix_global = matrix containing resulting fixed end actions in global coordinates % fea_vector = vector of fea’s in global dofs, used to calc displacements % fea_vector_abs = vector of fea’s in every structure dof % dispflag = flag indicating initial displacements % Ffld = fea (vert force) on left end of element due to initial disp % Mfld = fea (moment) on left end of element due to initial disp % Ffrd = fea (vert force) on right end of element due to initial disp % Mfrd = fea (moment) on right end of element due to initial disp % fea_vector_disp = vector of fea’s due to initial disp, used to calc displacements % fea_vector_react = vector of fea’s due to initial disp, used to calc reactions % % By Dean A. Frank % CVEN 5525 - Term Project % Fall 1995 % %********************************************************************************************** % CALCULATE THE FIXED END ACTIONS AND INSERT INTO A MATRIX IN WHICH THE ROWS CORRESPOND % WITH THE ELEMENT NUMBER AND THE COLUMNS CORRESPOND WITH THE ELEMENT LOCAL DEGREES % OF FREEDOM for ielem=1:nelem b(ielem)=L(ielem)-a(ielem); Ffl=((w(ielem)*L(ielem))/2)+((Pelem(ielem)*(b(ielem))^2)/(L(ielem))^3)*(3*a(ielem)+b(ielem)); Mfl=((w(ielem)*(L(ielem))^2))/12+(Pelem(ielem)*a(ielem)*(b(ielem))^2)/(L(ielem))^2; Ffr=((w(ielem)*L(ielem))/2)+((Pelem(ielem)*(a(ielem))^2)/(L(ielem))^3)*(a(ielem)+3*b(ielem));

Victor Saouma

Structural Engineering

26–42

Draft

DIRECT STIFFNESS METHOD

Mfr=-((w(ielem)*(L(ielem))^2))/12+(Pelem(ielem)*a(ielem)*(b(ielem))^2)/(L(ielem))^2; feamatrix_local(ielem,1:6)=[0 Ffl Mfl 0 Ffr Mfr]; % ROTATE THE LOCAL FEA MATRIX TO GLOBAL feamatrix_global=... XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX end % CREATE A LOAD VECTOR USING THE LM MATRIX % INITIALIZE FEA VECTOR TO ALL ZEROS for idofpn=1:ndofpn fea_vector(idofpn,1)=0; end for ielem=1:nelem for idof=1:6 if ielem==1 if LM(ielem,idof)>0 fea_vector(LM(ielem,idof),1)=feamatrix_global(idof,ielem); end elseif ielem>1 if LM(ielem,idof)>0 fea_vector(LM(ielem,idof),1)=fea_vector(LM(ielem,1))+feamatrix_global(idof,ielem); end end end end for ielem=1:nelem for iterm=1:nterm if feamatrix_global(iterm,ielem)==0 else if new_LM(ielem,iterm)>number_gdofs fea_vector_react(iterm,1)=feamatrix_global(iterm,ielem); end end end end % END OF LOADS3.M SCRIPTFILE

26.5.2.11

Nodal Displacements

%********************************************************************************************** % Scriptfile name : disp3.m (for 2d-frame structures) % % Main program : casap.m % % When this file is called, it computes the displacements in the global % degrees of freedom. % % Variable descriptions: (in the order in which they appear) % % Ksinv = inverse of the structural stiffness matrix % Ktt = structural stiffness matrix % Delta = vector of displacements for the global degrees of freedom % Pnods = vector of nodal loads in the global degrees of freedom % fea_vector = vector of fixed end actions in the global degrees of freedom % % By Dean A. Frank % CVEN 5525 - Term Project % Fall 1995 % % Edited by Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only

Victor Saouma

Structural Engineering

26.5 Computer Implementation with MATLAB

26–43

Draft

% %********************************************************************************************** % CREATE A TEMPORARY VARIABLE EQUAL TO THE INVERSE OF THE STRUCTURAL STIFFNESS MATRIX Ksinv=inv(Ktt); % CALCULATE THE DISPLACEMENTS IN GLOBAL COORDINATES Delta= XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX %

PRINT DISPLACEMENTS WITH NODE INFO

fprintf(fid,’ Displacements:\n’); for k=1:size(Delta,1) %WORK BACKWARDS WITH LM MATRIX TO FIND NODE# AND DOF LM_spot=find(LM’==k); elem=fix(LM_spot(1)/(nterm+1))+1; dof=mod(LM_spot(1)-1,nterm)+1; node=lnods(elem,fix(dof/4)+1); switch(dof) case {1,4}, dof=’delta X’; case {2,5}, dof=’delta Y’; otherwise, dof=’rotate ’; end %PRINT THE DISPLACEMENTS fprintf(fid,’ (Node: %2d %s) %14d\n’,node, dof, Delta(k)); end fprintf(fid,’\n’); % END OF DISP3.M SCRIPTFILE

26.5.2.12

Reactions

%********************************************************************************************** % Scriptfile name : react3.m (for 2d-frame structures) % % Main program : casap.m % % When this file is called, it calculates the reactions at the restrained degrees of % freedom. % % Variable Descriptions: % % Reactions = Reactions at restrained degrees of freedom % Kut = Upper left part of aug stiffness matrix, normal structure stiff matrix % Delta = vector of displacements % fea_vector_react = vector of fea’s in restrained dofs % % % By Dean A. Frank % CVEN 5525 - Term Project % Fall 1995 % % Edited by Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %********************************************************************************************** % CALCULATE THE REACTIONS FROM THE AUGMENTED STIFFNESS MATRIX Reactions= XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX fprintf(fid,’ Reactions:\n’); for k=1:size(Reactions,1) %WORK BACKWARDS WITH LM MATRIX TO FIND NODE# AND DOF LM_spot=find(LM’==-k); elem=fix(LM_spot(1)/(nterm+1))+1;

Victor Saouma

Structural Engineering

26–44

Draft

DIRECT STIFFNESS METHOD

dof=mod(LM_spot(1)-1,nterm)+1; node=lnods(elem,fix(dof/4)+1); switch(dof) case {1,4}, dof=’Fx’; case {2,5}, dof=’Fy’; otherwise, dof=’M ’; end %PRINT THE REACTIONS fprintf(fid,’ (Node: %2d %s) %14d\n’,node, dof, Reactions(k));

end fprintf(fid,’\n’); % END OF REACT3.M SCRIPTFILE

26.5.2.13

Internal Forces

%********************************************************************************************** % Scriptfile name : intern3.m (for 2d-frame structures) % % Main program : casap.m % % When this file is called, it calculates the internal forces in all elements % freedom. % % By Pawel Smolarkiewicz, 3/16/99 % Simplified for 2D Frame Case only % %********************************************************************************************** Pglobe=zeros(6,nelem); Plocal=Pglobe; fprintf(fid,’ Internal Forces:’); %LOOP FOR EACH ELEMENT for ielem=1:nelem %FIND ALL 6 LOCAL DISPLACEMENTS elem_delta=zeros(6,1); for idof=1:6 gdof=LM(ielem,idof); if gdof<0 elem_delta(idof)= XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX else elem_delta(idof)= XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX end end %SOLVE FOR ELEMENT FORCES (GLOBAL) Pglobe(:,ielem)=K(:,:,ielem)*elem_delta+feamatrix_global(:,ielem); %ROTATE FORCES FROM GLOBAL TO LOCAL COORDINATES %ROTATE FORCES TO LOCAL COORDINATES Plocal(:,ielem)= XXXXXXXXXXXXXXXXXXXXXX COMPLETE XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX %PRINT RESULTS fprintf(fid,’\n Element: %2d\n’,ielem); for idof=1:6 if idof==1 fprintf(fid,’ At Node: %d\n’,lnods(ielem,1)); end if idof==4 fprintf(fid,’ At Node: %d\n’,lnods(ielem,2)); end switch(idof) case {1,4}, dof=’Fx’; case {2,5}, dof=’Fy’; otherwise, dof=’M ’; end fprintf(fid,’ (Global : %s ) %14d’,dof, Pglobe(idof,ielem));

Victor Saouma

Structural Engineering

26.5 Computer Implementation with MATLAB

Draft

fprintf(fid,’

26–45

(Local : %s ) %14d\n’,dof, Plocal(idof,ielem));

end end fprintf(fid,’\n_________________________________________________________________________\n\n’);

26.5.2.14

Plotting

%************************************************************************************************ % SCRIPTFILE NAME : DRAW.M % % MAIN FILE : CASAP % % Description : This file will draw a 2D or 3D structure (1D structures are generally % boring to draw). % % Input Variables : nodecoor - nodal coordinates % ID - connectivity matrix % drawflag - flag for performing drawing routine % % By Dean A. Frank % CVEN 5525 % Advanced Structural Analysis - Term Project % Fall 1995 % % (with thanks to Brian Rose for help with this file) % %************************************************************************************************ % PERFORM OPERATIONS IN THIS FILE IF DRAWFLAG = 1 if drawflag==1 if istrtp==1 drawtype=2; elseif istrtp==2 drawtype=2; elseif istrtp==3 drawtype=2; elseif istrtp==4 drawtype=2; elseif istrtp==5 drawtype=3; elseif istrtp==6 drawtype=3; else error(’Incorrect structure type in indat.m’) end

ID=orig_ID.’; % DRAW 2D STRUCTURE IF DRAWTYPE=2 if drawtype==2 % RETREIVE NODAL COORDINATES x=nodecoor(:,1); y=nodecoor(:,2); %IF 2D-TRUSS, MODIFY ID MATRIX if istrtp==2 for ipoin=1:npoin if ID(1:2,ipoin)==[0;0] ID(1:3,ipoin)=[0;0;0] elseif ID(1:2,ipoin)==[0;1] ID(1:3,ipoin)=[0;1;0] elseif ID(1:2,ipoin)==[1;1] ID(1:3,ipoin)=[1;1;0] end

Victor Saouma

Structural Engineering

26–46

Draft

DIRECT STIFFNESS METHOD

end end

%if size(ID,1)==2 % ID=[ID;zeros(1,size(ID,2))]; %end % IF GRID, SET ID=ZEROS if istrtp==4 ID=ID*0; end % SET UP FIGURE handle=figure; margin=max(max(x)-min(x),max(y)-min(y))/10; axis([min(x)-margin, max(x)+margin, min(y)-margin, max(y)+margin]) axis(’equal’) hold on % CALC NUMBER OF NODES, ETC. number_nodes=length(x); number_elements=size(lnods,1); number_fixities=size(ID,2); axislimits=axis; circlesize=max(axislimits(2)-axislimits(1),axislimits(4)-axislimits(3))/40; % DRAW SUPPORTS for i=1:number_fixities % DRAW HORIZ. ROLLER if ID(:,i)==[0 1 0]’ plot(sin(0:0.1:pi*2)*circlesize/2+x(i),cos(0:0.1:pi*2)*circlesize/2-circlesize/2+y(i),’r’) % DRAW PIN SUPPORT elseif ID(:,i)==[1 1 0]’ plot([x(i),x(i)-circlesize,x(i)+circlesize,x(i)],[y(i),y(i)-circlesize,y(i)-circlesize,y(i)],’r’) % DRAW HOERIZ. ROLLER SUPPORT elseif ID(:,i)==[0 1 1]’ plot([x(i)+circlesize*2,x(i)-circlesize*2],[y(i),y(i)],’r’); plot(sin(0:0.1:pi*2)*circlesize/2+x(i),cos(0:0.1:pi*2)*circlesize/2-circlesize/2+y(i),’r’) plot(sin(0:0.1:pi*2)*circlesize/2+x(i)+circlesize,cos(0:0.1:pi*2)*circlesize/2-circlesize/2+y(i),’r’) plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize,cos(0:0.1:pi*2)*circlesize/2-circlesize/2+y(i),’r’) % DRAW VERT. ROLLER SUPPORT elseif ID(:,i)==[1 0 0]’ plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize*.5,cos(0:0.1:pi*2)*circlesize/2,’r’) % DRAW ROLLER SUPPORT WITH NO ROTATION elseif ID(:,i)==[1 0 1]’ plot([x(i),x(i)],[y(i)+circlesize*2,y(i)-circlesize*2],’r’); plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize*.5,cos(0:0.1:pi*2)*circlesize/2,’r’) plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize*.5,cos(0:0.1:pi*2)*circlesize/2+circlesize,’r’) plot(sin(0:0.1:pi*2)*circlesize/2+x(i)-circlesize*.5,cos(0:0.1:pi*2)*circlesize/2-circlesize,’r’) end % DRAW HORIZ. PLATFORM if min(ID(:,i)==[0 1 0]’) | min(ID(:,i)==[1 1 0]’) | min(ID(:,i)==[0 1 1]’) plot([x(i)-circlesize*2,x(i)+circlesize*2],[y(i)-circlesize,y(i)-circlesize],’r’) plot([x(i)-circlesize*1.5,x(i)-circlesize*2],[y(i)-circlesize,y(i)-circlesize*1.5],’r’) plot([x(i)-circlesize*1,x(i)-circlesize*2],[y(i)-circlesize,y(i)-circlesize*2],’r’) plot([x(i)-circlesize*.5,x(i)-circlesize*1.5],[y(i)-circlesize,y(i)-circlesize*2],’r’)

Victor Saouma

Structural Engineering

26.5 Computer Implementation with MATLAB

26–47

Draft

plot([x(i)+circlesize*0,x(i)-circlesize*1],[y(i)-circlesize,y(i)-circlesize*2],’r’) plot([x(i)+circlesize*.5,x(i)-circlesize*(0.5)],[y(i)-circlesize,y(i)-circlesize*2],’r’) plot([x(i)+circlesize*1,x(i)+circlesize*0],[y(i)-circlesize,y(i)-circlesize*2],’r’) plot([x(i)+circlesize*1.5,x(i)+circlesize*.5],[y(i)-circlesize,y(i)-circlesize*2],’r’) plot([x(i)+circlesize*2,x(i)+circlesize*1],[y(i)-circlesize,y(i)-circlesize*2],’r’) plot([x(i)+circlesize*2,x(i)+circlesize*1.5],[y(i)-circlesize*1.5,y(i)-circlesize*2],’r’) % DRAW FIXED SUPPORT

elseif ID(:,i)==[1 1 1]’ plot([x(i)-circlesize*2,x(i)+circlesize*2],[y(i)-circlesize,y(i)-circlesize]+circlesize,’r’) plot([x(i)-circlesize*1.5,x(i)-circlesize*2],[y(i)-circlesize,y(i)-circlesize*1.5]+circlesize,’r’) plot([x(i)-circlesize*1,x(i)-circlesize*2],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’) plot([x(i)-circlesize*.5,x(i)-circlesize*1.5],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’) plot([x(i)+circlesize*0,x(i)-circlesize*1],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’) plot([x(i)+circlesize*.5,x(i)-circlesize*(0.5)],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’) plot([x(i)+circlesize*1,x(i)+circlesize*0],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’) plot([x(i)+circlesize*1.5,x(i)+circlesize*.5],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’) plot([x(i)+circlesize*2,x(i)+circlesize*1],[y(i)-circlesize,y(i)-circlesize*2]+circlesize,’r’) plot([x(i)+circlesize*2,x(i)+circlesize*1.5],[y(i)-circlesize*1.5,y(i)-circlesize*2]+circlesize,’r’) % DRAW VERT. PLATFORM elseif min(ID(:,i)==[1 0 0]’) | min(ID(:,i)==[1 0 1]’) xf=[x(i)-circlesize,x(i)-circlesize*2]; yf=[y(i),y(i)- circlesize]; plot(xf,yf,’r’) plot(xf,yf+circlesize*.5,’r’) plot(xf,yf+circlesize*1,’r’) plot(xf,yf+circlesize*1.5,’r’) plot(xf,yf+circlesize*2,’r’) plot([x(i)-circlesize*1.5,x(i)-circlesize*2],[y(i)+circlesize*2,y(i)+ circlesize*1.5],’r’) plot(xf,yf-circlesize*.5,’r’) plot(xf,yf-circlesize*1,’r’) plot([x(i)-circlesize,x(i)-circlesize*1.5],[y(i)-circlesize*1.5,y(i)- circlesize*2],’r’) plot([xf(1),xf(1)],[y(i)+circlesize*2,y(i)-circlesize*2],’r’); end end % DRAW ELEMENTS for i=1:number_elements plot([x(lnods(i,1)),x(lnods(i,2))],[y(lnods(i,1)),y(lnods(i,2))],’b’); if i==1 end end % DRAW JOINTS for i=1:number_nodes if ~max(ID(:,i)) plot(x(i),y(i),’mo’) end end % DRAW ELEMENT NUMBERS for i=1:number_elements set(handle,’DefaultTextColor’,’blue’) text( (x(lnods(i,1))+x(lnods(i,2)))/2+circlesize,(y(lnods(i,1))+y(lnods(i,2)))/2+circlesize,int2str(i)) end % DRAW JOINT NUMBERS for i=1:number_nodes set(handle,’DefaultTextColor’,’magenta’) text(x(i)+circlesize,y(i)+circlesize,int2str(i)) end if exist(’filename’) title(filename) end

Victor Saouma

Structural Engineering

26–48

Draft

DIRECT STIFFNESS METHOD

hold off set(handle,’DefaultTextColor’,’white’) % DRAW 3D STRUCTURE IF DRAWTYPE=3 elseif drawtype==3 % RETREIVE NODE COORIDINATES x=nodecoor(:,1); y=nodecoor(:,2); z=nodecoor(:,3); % SET UP FIGURE handle=figure; margin=max([max(x)-min(x),max(y)-min(y),max(z)-min(z)])/10; axis([min(x)-margin, max(x)+margin, min(y)-margin, max(y)+margin, min(z)-margin, max(z)+margin]) axis(’equal’) hold on % RETREIVE NUMBER OF NODES, ETC. number_nodes=length(x); number_elements=size(lnods,1); axislimits=axis; circlesize=max([axislimits(2)-axislimits(1),axislimits(4)-axislimits(3),axislimits(6)-axislimits(5)])/40; % DRAW ELEMENTS for i=1:number_elements plot3([x(lnods(i,1)),x(lnods(i,2))],[y(lnods(i,1)),y(lnods(i,2))],[z(lnods(i,1)),z(lnods(i,2))],’b’); end % DRAW JOINTS for i=1:number_nodes plot3(x(i),y(i),z(i),’mo’) end % DRAW ELEMENT NUMBERS

for i=1:number_elements set(handle,’DefaultTextColor’,’blue’) text( (x(lnods(i,1))+x(lnods(i,2)))/2+circlesize,(y(lnods(i,1))+y(lnods(i,2)))/2+circlesize,(z(lnods(i,1))+z(lnods(i,2) end % DRAW JOINT NUMBERS for i=1:number_nodes set(handle,’DefaultTextColor’,’magenta’) text(x(i)+circlesize,y(i)+circlesize,z(i)+circlesize,int2str(i)) end if exist(’filename’) title(filename) end xlabel(’x’) ylabel(’y’) zlabel(’z’) % DRAW GROUND X=x; Y=y; Z=z; X=axislimits(1)-margin:margin:axislimits(2)+margin; Y=axislimits(3)-margin:margin:axislimits(4)+margin; Z=zeros(length(X),length(Y)); mesh(X,Y,Z)

Victor Saouma

Structural Engineering

26.5 Computer Implementation with MATLAB

26–49

Draft size(X) size(Y) size(Z)

hold off set(handle,’DefaultTextColor’,’white’) hAZ=uicontrol(’style’,’slider’,’position’,[.7 .95 .3 .05],’units’,’normalized’,’min’,0,’max’,360,... ’callback’,’[az,el]=view; az=get(gco,’’val’’); view(az,el);’); hEL=uicontrol(’style’,’slider’,’position’,[.7 .89 .3 .05],’units’,’normalized’,’min’,0,’max’,180,... ’callback’,’[az,el]=view; el=get(gco,’’val’’); view(az,el);’);

hdef=uicontrol(’style’,’pushbutton’,’callback’,’view(-37.5, 30)’,’position’,[.88 .83 .12 .05],’units’,’normalized’,’Strin %set(handle,’units’,’normalized’) %text(.68,.95,’azimuth’) %text(.68,.89,’elevation’) end end % END OF DRAW.M SCRIPTFILE

26.5.2.15

Sample Output File

CASAP will display ﬁgure 26.15. 8000

6000

4000 2

2000

2

3

1

1 0

−2000

−4000

−6000

0

2000

4000

6000

8000

10000

12000

14000

16000

Figure 26.15: Structure Plotted with CASAP Number Number Number Number Number

of of of of of

Nodes: 3 Elements: 2 Load Cases: 1 Restrained dofs: 6 Free dofs: 3

Node Info: Node 1 (0,0) Free dofs: none; node is fixed Node 2 (7416,3000) Free dofs: X Y Rot Node 3 (15416,3000) Free dofs: none; node is fixed Element Info: Element 1 (1->2) Element 2 (2->3)

Victor Saouma

E=200 A=6000 Iz=200000000 E=200 A=6000 Iz=200000000

Structural Engineering

26–50

DIRECT STIFFNESS METHOD

Draft

_________________________________________________________________________ Load Case: 1 Nodal Loads: Node: 2 Fx = 1.875000e+001 Node: 2 Fy = -4.635000e+001 Elemental Loads: Element: 1 Point load = 0 at 0 from left Distributed load = 0 Element: 2 Point load = 0 at 0 from left Distributed load = 4.000000e-003 Displacements: (Node: 2 delta X) 9.949820e-001 (Node: 2 delta Y) -4.981310e+000 (Node: 2 rotate ) -5.342485e-004 Reactions: (Node: 1 (Node: 1 (Node: 1 (Node: 3 (Node: 3 (Node: 3

Fx) 1.304973e+002 Fy) 5.567659e+001 M ) 1.337416e+004 Fx) -1.492473e+002 Fy) 2.267341e+001 M ) -4.535573e+004

Internal Forces: Element: 1 At Node: 1 (Global : (Global : (Global : At Node: 2 (Global : (Global : (Global : Element: 2 At Node: 2 (Global (Global (Global At Node: 3 (Global (Global (Global

Fx ) Fy ) M )

1.304973e+002 5.567659e+001 1.337416e+004

(Local : Fx ) (Local : Fy ) (Local : M )

1.418530e+002 2.675775e+000 1.337416e+004

Fx ) -1.304973e+002 Fy ) -5.567659e+001 M ) 8.031549e+003

(Local : Fx ) -1.418530e+002 (Local : Fy ) -2.675775e+000 (Local : M ) 8.031549e+003

: Fx ) 1.492473e+002 : Fy ) 9.326590e+000 : M ) -8.031549e+003

(Local : Fx ) 1.492473e+002 (Local : Fy ) 9.326590e+000 (Local : M ) -8.031549e+003

: Fx ) -1.492473e+002 : Fy ) 2.267341e+001 : M ) -4.535573e+004

(Local : Fx ) -1.492473e+002 (Local : Fy ) 2.267341e+001 (Local : M ) -4.535573e+004

_________________________________________________________________________

Victor Saouma

Structural Engineering

Draft Chapter 27

COLUMNS 27.1

Introduction

1 Columns resist a combination of axial P and ﬂexural load M , (or M = P e for eccentrically applied load).

27.1.1

Types of Columns

Types of columns, Fig. 27.1 Tied column tie steel main longitudinal steel reinforcement

Spiral column

Composite column

Pipe column

Figure 27.1: Types of columns

2

Lateral reinforcement, Fig. 27.2 1. Restrains longitudinal steel from outward buckling 2. Restrains Poisson’s expansion of concrete 3. Acts as shear reinforcement for horizontal (wind & earthquake) load 4. Provide ductility

very important to resist earthquake load.

27.1.2 3

Possible Arrangement of Bars

Bar arrangements, Fig. 27.3

27–2

COLUMNS

Draft P Spiral

tied

X

δ

Figure 27.2: Tied vs Spiral Reinforcement

4 bars

6 bars

8 bars

Corner column

10 bars 12 bars

Wall column

16 bars 14 bars

Figure 27.3: Possible Bar arrangements

Victor Saouma

Structural Engineering

27.2 Short Columns

27–3

Draft 27.2

Short Columns

27.2.1

Concentric Loading

4

No moment applied,

Elastic Behaviour P

= fc Ac + fs As = fc (Ac + nAs )

Ultimate Strength Pd Pn

5

= φPn = .85fc Ac + fy As

note: 1. 0.85 is obtained also from test data 2. Matches with beam theory using rect. stress block 3. Provides an adequate factor of safety

27.2.2 5

Eccentric Columns

Sources of ﬂexure, Fig. 27.4

e

ML

P

MR

Figure 27.4: Sources of Bending 1. Unsymmetric moments M L = M R 2. uncertainty of loads (must assume a minimum eccentricity) 3. unsymmetrical reinforcement 6

Types of Failure, Fig. 27.5 1. large eccentricity of load ⇒ failure by yielding of steel 2. small eccentricity of load ⇒ failure by crushing of concrete 3. balanced condition

7

Assumptions As = As ; ρ =

Victor Saouma

As bd

=

As bd

= fs = fy Structural Engineering

27–4

COLUMNS

Draft

Pn P0

e=0

Compression failure range Radial lines show Mn constant e= Pn

e small

Load path for givin e

Tension failure range

eb

e=

8

e large

Mn

M0

Figure 27.5: Load moment interaction diagram 27.2.2.1

Balanced Condition

There is one speciﬁc eccentricity eb =

8

M P

such that failure will be triggered by simultaneous

1. steel yielding 2. concrete crushing From the strain diagram (and compatibility of concrete and steel strains), Fig. 27.6

9

εc εy c

= .003 fy = Es εu = d= εu + εy

(27.1-a) (27.1-b) .003 fy Es

+ .003

d

(27.1-c) (27.1-d)

Furthermore, εs c − d

=

⇒ εs

=

εc c c − d εc c

(27.2-a) (27.2-b)

thus the compression steel will be yielding (i.e. εs = εy ) for εc = .003 and d = 2 in if c > 6 in 10

Equilibrium: Pn fs As

= = =

 .85fc ab + As fy − As fs  a a fy  c b As

Pn = .85f cb = β1 c b = fy.003 d Es

+.003

    

Pn,b = .85β1 fc bd

.003 fy Es

+ .003

(27.3)

or Pnb = .85β1 fc bd

Victor Saouma

87, 000 fy + 87, 000

(27.4)

Structural Engineering

27.2 Short Columns

27–5

Draft

d’ d h/2 As

A’s b

εcs

εs A sf y

ε’s Pn

A sf s

c

c A sf s

0.85f’c A’sf s

A’sf y

a e e’

Figure 27.6: Strain and Stress Diagram of a R/C Column

Victor Saouma

Structural Engineering

27–6

COLUMNS

Draft

To obtain Mnb we take moment about centroid of tension steel As of internal forces, this must be equal and opposite to the externally applied moment, Fig. 27.6.

11

a Mnb = Pnb eb = .85fc ab(d − ) + As fy (d − d ) 2

M ext

12

(27.5)

Mint

Note: Internal moments due to As fy and As fy cancel each other for symmetric columns.

27.2.2.2

Tension Failure

Case I, e is known and e > eb In this case a and Pn are unknowns, and for failure to be triggered by fy in As we must have e > eb . Can still assume As fy = As fy ΣFy = 0 ⇒ Pn ΣM = 0 ⇒ Pn e

= .85fc ab ⇒ a =

Pn .85fc b

(27.6-a)

a = Pn (d − ) + As fy (d − d ) 2

(27.6-b)

Two approaches 1. Solve iteratively for those two equations (a) (b) (c) (d) (e)

Assume a (a < h2 ) From strain compatibility solve for fsc , center steel stress if applicable. ΣFy = 0 ⇒ solve for Pn ΣM = 0 with respect to tensile reinforcement,⇒ solve for Pn If no convergence among the two Pn , iterate by solving for a from ΣFy = 0

2. Combine them into a quadratic equation in Pn   2   d e e e Pn = .85fc bd −ρ − ( − 1) + + 2ρ µ 1 − 1− +  d d d d  where ρ = µ =

(27.7)

As As bd = bd fy .85fc

In this case, we only have two unknown, Pn and fs . a fs fs

= def

= =

C

=

Pn

=

Mn

=

β1 c

(27.8-a)

fy

(27.8-b)

c − d εc Es c 0.85fc ab

≤ fy

(27.8-c) (27.8-d)

C + As fs − As fy h h−a h + as f s − d + As fs d − C 2 2 2

(27.8-e) (27.8-f)

Note this approach is favoured when determining the interaction diagram.

Victor Saouma

Structural Engineering

27.2 Short Columns

27–7

Draft 27.2.2.3

Compression Failure

Case I e is known and e < eb ; Pn , a and fs are unknown Case II c is known and c < cb ; Pn is unknown Compression failure occurs if e < eb ⇒ εu = .003, assume fs = fy , and fs < fy From geometry c

= =

fs Es

d−c εu d ⇒ fs = Es εu c + εu

Es εu

d−

a β1

(27.9-a)

a β1

.85fc ab + As fy − As fs (27.9-b) a Pn e = .85fc ab(d − ) + As fy (d − d ) (27.9-c) 2 this would yield a cubic equation in Pn , which can be solved analytically or by iteration Pn

1. 2. 3. 4. 5.

=

Assume a (a h) Solve for ΣM = 0 with respect to tensile reinforcement & solve for Pn From strain compatibility solve for fs Check that ΣFy = 0 & solve for a If ai+1 = ai go to step 2

In this case a

= β1 c

fs

= εc Es

fs C Pn Mn

27.2.3

(27.10-a)

d−c c c − d = εc Es c = 0.85fc ab

≤ fy

(27.10-b)

≤ fy

(27.10-c) (27.10-d)

As fs

= C + As fs + h h−a h + As fs − d + As fs d − = C 2 2 2

(27.10-e) (27.10-f)

ACI Provisions

Case II: c is known and c > cb ; fs , fs , and Pn are unknown 1. ρmin ρmax ρs φ φ

= = = = =

1% 8% A f 0.45( Agc − 1) fyc 0.7 for tied columns 0.75 for spiral columns

ACI 10.9.1 ACI 10.5

(27.11)

2. where minimum ratio of spiral reinforcement ρs Ag gross area of section Ac area of core 3. A minimum of 4 bars for tied circular and rect 4. A minimum of 6 bars for spirals(10.9.2) 5. φ increases linearly to 0.9 as φPn decreases from 0.10fc Ag or φP0 , whichever is smaller, to zero (ACI 9.3.2). 6. Maximum strength is 0.8φP0 for tied columns (φ = 0.7) and 0.85φP0 for spirally reinforced columns (φ = 0.75). Victor Saouma

Structural Engineering

27–8

COLUMNS

Draft 27.2.4 13

Interaction Diagrams

Each column is characterized by its own interaction diagram, Fig. 27.7

P

Tied: Pn(max)=0.80P0 Spirally reinf: Pn(max)=0.85P0

P0 Pn(max)

Compression control region

1

A

e ure

Pn-Mn

φPn(max)

(Mn’Pn)

min

Pu-Mu (Mnb’Pnb)

e

ed

nc

la Ba

fail

Tension control region

eb

0.10f’cAg

φMn

0

Mn

M

Figure 27.7: Column Interaction Diagram

27.2.5

Design Charts

To assist in the design of R.C. columns, design charts have been generated by ACI in term of non As +As fy χe Pn Mn and µ = .85f dimensionalized parameters χ = bhf vs bh2 f = h for various ρt where ρt = bh

14

c

c

c

Example 27-1: R/C Column, c known A 12 by 20 in. column is reinforced with four No. 4 bars of area 1.0 in2 each, at each corner. = 3.5 ksi, fy = 50 ksi, d = 2.5 in. Determne: 1) Pb and Mb ; 2) The load and moment for c = 5 in; 3) load and moment for c = 18 in. Solution: fc

Balanced Conditions is derived by revisiting the fundamental equations, rather than mere substitution into previously derived equation. d

=

cb

=

h − d = 20 − 2.5 = 17.5 in .003 .003 d = 50 17.5 = 11.1 in fy 29,000 + .003 E + .003

(27.12-a) (27.12-b)

s

a fs

= def

=

fs

=

C

=

Pnb

=

Mnb

= =

Victor Saouma

β1 cb = (0.85)(11.1) = 9.44 in

(27.12-c)

fy = 50 ksi (27.12-d) c−d 11.1 − 2.5 εc = (29, 000)( (0.003) = 67.4 ksi > fy ⇒ fs = 50 ksi(27.12-e) Es c 11.1 (27.12-f) 0.85fc ab = (0.85)(3.5)(9.44)(12) = 337 k C + As fs − As fs = 337 + (2.0)(50) + (2.0)(−50) = 337 k a Pnb e = .85fc ab(d − ) + As fy (d − d ) 2 9.44 337 17.5 − + (2.0)(50)(17.5 − 2.5) = 3, 280 k.in = 273 k.ft 2

(27.12-g) (27.12-h) (27.12-i)

Structural Engineering

27.2 Short Columns

Draft

eb

=

27–9 3, 280 = 9.72 in 337

(27.12-j)

Tension failure, c = 5 in fs fs

def

= =

= a

=

fy = 50 ksi c − d εc Es c

(27.13-a) ≤ fy

(27.13-b)

5.0 − 2.5 = 43.5 ksi 5.0 β1 c = 0.85(5.0) = 4.25 in

(27.13-c)

(0.003)(29, 000)

(27.13-d)

0.85fc ab

C

= =

(0.85)(3.5)(4.25)(12) = 152 k

(27.13-e) (27.13-f)

Pn

=

C + As fs − As fy

(27.13-g)

=

152 + (2.0)(43.5) − (2.0)(50) = 139 k (27.13-h) h h−a h + as fs − d + As fs d − C about section centroid (27.13-i) 2 2 2 20 20 20 − 4.25 − 2.5 + (2.0)(50) 17.5 − + (2.0)(43.5) (27.13-j) (152) 2 2 2

Mn

= =

e

=

2, 598 k.in = 217 k.ft

(27.13-k)

=

2, 598 = 18.69 in 139

(27.13-l)

Compression failure, c = 18 in a

=

fs

= =

fs

= =

C Pn

Mn

=

As fs

(27.14-b) (27.14-c) (27.14-d) (27.14-e) (27.14-f)

C+

=

(27.14-h) 546 + (2.0)(50) + (−2.0)(2) = 650 k h h−a h + As fs − d + As fs d − C about section centroid (27.14-i) 2 2 2 20 20 20 − 15.3 − 2.5 + (2.0)(−2.0) 17.5 − + (2.0)(50) (27.14-j) (546) 2 2 2

=

+ As fs

(27.14-a)

=

=

e

β1 c = 0.85(18) = 15.3 in d−c ≤ fy εc Es c 17.5 − 18.0 = −2 ksi As is under compression (0.003)(29, 000) 18.0 c−d ≤ fy εc Es c 18.0 − 2.5 = 75 ksi > fy ⇒ fs = 50 ksi (0.003)(29, 000) 18.0 0.85fc ab = (0.85)(3.5)(15.3)(12) = 546 k

(27.14-g)

=

2, 000 k.in = 167 k.ft

(27.14-k)

=

2, 000 = 3.07 in 650

(27.14-l)

Example 27-2: R/C Column, e known Victor Saouma

Structural Engineering

27–10

COLUMNS

Draft

For the following column, determine eb , Pb , Mb fc = 3, 000 psi and fy = 40, 000 psi. The area of each bar is 1.56 in2 . 12"

20" 3"

3"

24"

ε

c y

.003

Cc

Balanced Condition: fy 40 = .001379 = Es 29, 000 εu (.003)(21) d= = 14.4 in εu + εy .003 + .001379 β1 cb = (.85)(14.4) = 12.2 in .85fc ab = (.85)(3)(12.2)(20) = 624 k .003 (14.4 − 12) = .0005 14.4 (29, 000)(.0005) = 15 ksi center bars

εy

=

cb

=

a Cc

= =

εsc

=

fsc

=

Cs

= (.0005)(29, 000)(2)(1.56) = 45.2 k

(27.15-g)

= 624 + 45.2 = 671 k

(27.15-h)

Pnb

(27.15-a) (27.15-b) (27.15-c) (27.15-d) (27.15-e) (27.15-f)

assuming that both edge bars are yielding and cancel each other. Alternatively Pnb = 624 + (2 × 1.56)[15 − (.85)(3)] − (4 × 1.56)(.85)(3) = 647 k Note the diﬀerence

671−647 647

(27.16)

× 100 = 3.7%

Taking moment about centroid of section Mnb

eb

= Pnb e

a + Asc fy (9) + As (fy − .85fc )(12 − d ) = .85fc ab 12 − 2 12.2 = (.85)(3)(12.2)(20) 12 − + 4(1.56)(9)(40) 2 +4(1.56)(40 − .85 × 3)(12 − 3)

(27.17-a) (27.17-b)

(27.17-c)

= 3, 671 + 2, 246 + 2, 103

(27.17-d)

=

8, 020 k.in; 668 k.ft

(27.17-e)

=

8, 020 = 12.4 in 647

(27.17-f)

e= .1 h e = (.1)(24) = 2.4 in < eb ⇒ failure by compression. Pn , a and fs are unknown

Victor Saouma

Structural Engineering

27.2 Short Columns

27–11

Draft

12

h-c-d’

εs

.003

23.5

1. Assume a = 20 in c=

a 20 = 23.5 in = β1 .85

(27.18)

2. For center steel (from geometry) εsc c − h2

=

.003 c

(27.19-a)

⇒ εsc

=

(27.19-b)

fsc

=

c − h2 .003 c Es εsc c− c

(27.19-d)

=

Es

=

29, 000

h 2

(27.19-c)

.003

23.5 − 12 .003 = 42.5 ksi > fy ⇒ fsc = fy 23.5

(27.19-e)

3. Take moment about centroid of tensile steel bar a h = 0.85fc ab(d − ) + As fy (h − 2d ) + Asc fy ( − d ) (27.20-a) 2 2 20 (27.20-b) Pn (9 + 2.4) = (.85)(3)(20)(20)(21 − ) + 4(1.56)(40)(24 − 6) + 2(1.56)(40)(9) 2 = 11, 220 + 4, 493 + 259.7 (27.20-c) Pn e

⇒ Pn

= 1, 476 k

(27.20-d)

4. Get εs in tension bar

=

.003 c .003 (24 − 3 − 23.5) 23.5 −.000319

=

Eεs = (29, 000)(0.000319) = 9.25 ksi

εs h − d − 23.5

=

⇒ εs

=

fs

(27.21-a) (27.21-b) (27.21-c) (27.21-d)

5. Take ΣF = 0 to check assumption of a Pn = 1, 476 k = 1, 476 =

e=h

⇒a

=

Pn

=

Mn

=

0.85fc ab + As fy + Asc fsc + As fs (.85)(3)(a)(20) + (6)(1.56)(40) + (4)(1.56)(9.25)

(27.22-a) (27.22-b)

51a + 374.4 + 57.7 √ 20.4 in

(27.22-c) (27.22-d)

1, 476 k

(27.22-e)

(1, 476)(2.4) = 3, 542 k.in = 295 k.ft

(27.22-f)

1. In this case e = 24 in > eb ⇒ failure by tension. Pn and a are unknown.

Victor Saouma

Structural Engineering

27–12

COLUMNS

Draft

2. Assume a = 7.9 in ⇒ c =

a β1

=

7.9 .85

= 9.3 in

3. Steel stress at centroid c .003 ⇒ εsc ⇒ fsc

12 − c εsc 12 − 9.3 .003 = .00087 = 9.3 = (29, 000)(0.00087) = 25.3 ksi

=

(27.23-a) (27.23-b) (27.23-c)

4. Iterate ΣF = 0 ⇒ Pn

= (.85)fc ab + Asc fsc

(27.24-a)

= (.85)(3)(7.9)(20) − 2(1.56)(25.3) (27.24-b) = 403 − 79 = 324 k (27.24-c) a ΣM = 0 ⇒ Pn (e + h/2 − d ) = .85fc ab(d − ) + As fy (d − d ) 2 d − d ) (27.24-d) −Asc fsc ( 2 7.9 Pn (24 + 9) = (.85)(3)(7.9)(20)(21 − ) + 4(1.56)(40)(21 − 3) 2 +2(1.56)(25.3)(9) (27.24-e) Pn (33) = 6, 870 + 4, 493 − 710 = 10, 653 k.in √ ⇒ Pn = 323 k

(27.24-f) (27.24-g)

5. Determine Mn Mn = Pn e = (323)(24) = 7, 752 k.in = 646 k.ft

(27.25)

% e is known solve for P_n and M_n % % input data b=20;h=24;A_s=4*1.56; Ap_s=A_s;A_sc=A_s/2;fp_c=3;f_y=40;E_s=29000;dp=3; d=h-dp; % solve for balanced conditions first epsi_y=f_y/E_s; c_b=0.003*d/(0.003+epsi_y); a=0.85*c_b; C_c=0.85*fp_c*a*b; epsi_sc=0.003*(c_b-h/2)/c_b; f_sc=E_s*epsi_sc; C_s=epsi_sc*E_s*A_s; P_nb=C_c+A_sc*f_sc M_nb=0.85*fp_c*a*b*(h/2-a/2)+Ap_s*f_y*(h/2-dp)+Ap_s*(f_y-0.85*fp_c)*(h/2-dp) e_b=M_nb/P_nb % % e=0.1h failure by compression; assume a=20 % e=0.1*h; error=1.; a=17; iter=0; while abs(error) > 0.1 iter=iter+1; c=a/0.85; epsi_sc=0.003*(c-h/2)/c ; f_sc=E_s*epsi_sc; if(f_sc>f_y) f_sc=f_y; end M_n=0.85*fp_c*a*b*(d-a/2)+Ap_s*f_y*(h-2*dp)+A_sc*f_y*(h/2-dp); P_n=M_n/(h/2-dp+e); epsi_s=0.003*(h-dp-c)/c; f_s=abs(E_s*epsi_s); a_new=(P_n-Ap_s*f_y-A_sc*f_sc-A_s*f_s)/(0.85*fp_c*b); error=(a_new-a)/(a_new);

Victor Saouma

Structural Engineering

27.2 Short Columns

Draft

27–13

a=a_new; end iter P_n M_n=P_n*e % % e=h; failure by tension % e=h; error=1.; a=10; iter=0; while abs(error) > 0.1 iter=iter+1; c=a/0.85 epsi_sc=0.003*(c-h/2)/c ; f_sc=E_s*epsi_sc; if(f_sc>f_y) f_sc=f_y; end P_n_1=0.85*fp_c*a*b+A_sc*f_sc P_n_2=(0.85*fp_c*a*b*(d-a/2)+Ap_s*f_y*(d-dp)+A_sc*f_sc*(d-dp)/2)/(e+h/2-dp); a_new=(P_n_2-A_sc*f_sc)/(0.85*fp_c*b); error=(a_new-a)/a_new; a=a_new; end iter P_n_1 M_n=P_n_1*e %================================ Balanced Conditions d=21 epsi_y=0.0014 c_b=14.3858 a=12.2280 C_c=623.6256 epsi_sc=4.9754e-04 f_sc= 14.4286 C_s= 90.0343 P_nb= 668.6427 M_nb= 8.0203e+03 e_b= 11.9948 Failure by Compression e=2.4000 error=1 a=17 iter=0 iter=1 c=20 epsi_sc=0.0012 05 f_sc=34.8000 M_n=1.6454e+04 P_n=1.4433e+03 epsi_s=1.5000e-04 f_s=4.3500 a_new=20.7445 error=0.1805 a= 20.7445 iter= 2 c=24.4053 epsi_sc=0.0015 f_sc=44.2224 f_sc=40 M_n=1.6860e+04 P_n=1.4789e+03 epsi_s=-4.1859e-04 f_s=12.1392 Failure by Tension e=24 error=1

Victor Saouma

Structural Engineering

27–14

COLUMNS

Draft a=10 iter=0 iter=1 c=11.7647 epsi_sc=-6.0000ef_sc=-1.7400 P_n_1=504.5712 P_n_2=381.9376 a_new=7.5954 error=-0.3166 a=7.5954 iter=2 c=8.9358 epsi_sc=-0.0010 f_sc=-29.8336 P_n_1=294.2857 P_n_2=312.6867 a_new=7.9562 error=0.0453 a=7.9562 iter=2 P_n_1=294.2857

Example 27-3: R/C Column, Using Design Charts Design the reinforcement for a column with h = 20 in, b = 12 in, d = 2.5 in, fc = 4, 000 psi, fy = 60, 000 psi, to support PDL = 56 k, PLL = 72 k, MDL = 88 k.ft, MLL = 75 k.ft, Solution: 1. Ultimate loads Pu Mu

= =

(1.4)(56) + (1.7)(72) = 201 k (1.4)(88) + (1.7)(75) = 251 k.ft

⇒ Pn ⇒ Mn

= =

201 0.7 251 0.7

= 287 k = 358 k.ft

(27.26)

2. Chart parameters e h

=

γ

=

κ = κ

e h

=

(358)(12) = 0.75 (287)(20) 20 − (2)(2.5) = 0.75 ⇒ interpolate between A3 and A4 20 Pn 287 = 0.3 = bhfc (12)(20)(4)

(27.27-b)

(0.3)(0.75) = 0.225

(27.27-d)

(27.27-a)

(27.27-c)

3. Interpolating between A.3 and A.4 ⇒ ρt µ = 0.4 4. Reinforcement ρt µ

= =

At bh fy .85fc

2 At =

(0.4)(b)(h)(.85)(fc ) 1 = 5.45 in2 (27.28-a) = (0.4)(12)(20)(.85)(4) fy (60) (27.28-b)

⇒ use 4 # 9 & 2 # 8, ⇒ At = 5.57 in2

Victor Saouma

Structural Engineering

Draft Chapter 28

DESIGN II 28.1

Frames

28.1.1

Beam Column Connections

1

The connection between the beam and the column can be, Fig. 28.1:

θb

θb

θb

θc

θc

θb

=

θc

θb

Flexible

=

θc

Rigid

θc

θ ) M=K( θ s s b- c θ b = θc Semi-Flexible

Figure 28.1: Flexible, Rigid, and Semi-Flexible Joints Flexible that is a hinge which can transfer forces only. In this case we really have cantiliver action only. In a ﬂexible connection the column and beam end moments are both equal to zero, Mcol = Mbeam = 0. The end rotation are not equal, θcol = θbeam . Rigid: The connection is such that θbeam = θcol 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 = 0, θcol = θbeam . Semi-Rigid: The end moments are equal and not equal to zero, but the rotation are diﬀerent. θbeam = θcol , Mcol = Mbeam = 0. Furthermore, the diﬀerence in rotation is resisted by the spring Mspring = Kspring (θcol − θbeam ).

28.1.2

Behavior of Simple Frames

2 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). 3 The advantages of a rigid connection are greater when the frame is subjected to a lateral load. Under those conditions, the connection will stiﬀen the structure and reduce the amount of lateral deﬂection, Fig. 28.2. 4 Fig. 28.3 illustrates the deformation, shear, moment and axial forces in frames with diﬀerent boundary conditions under both vertical and horizontal loads.

28–2

Draft

DESIGN II ∆

H

∆

H

θ

θ

PI

V

PI

V

V

PI

V

PI

PI

PI

PI

PI

Figure 28.2: Deformation of Flexible and Rigid Frames Subjected to Vertical and Horizontal Loads, (Lin and Stotesbury 1981)

28.1.3

Eccentricity of Applied Loads

5 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

e=

M P

(28.1)

The induced stresses can be decomposed into uniform (−P/L) (assuming a unit width) and linearly varying one (σ = M/S) and the end stresses are

6

σ min σ max

P −σ L P = − +σ L = −

(28.2-a) (28.2-b)

We note that the linearly varying stress distribution must satisfy two equilibrium requirements: ΣF = 0, thus the neutral axis (where the stress is equal to zero) passes through the centroid of the section, and ΣM = 0, i.e. Mint = Mext . 7

If we seek the eccentricity ecr for which σ max equals zero, then σ =

8

The net tensile force due to the eccentric load is T =

P L

1 L σ 2 2

(28.3)

If we want this net tensile force to be equal and opposite to the compressive force, then T = σL T = PL 4 P σ = A applied at 23 L2 from the centroid

(28.4)

Thus the net internal moment is Mint = 2T

Victor Saouma

P 2L PL 2L =2 = 32 432 6

(28.5)

Structural Engineering

28.1 Frames

28–3

Draft 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 THREE-HINGE PORTAL

w/2

-w/2

M

M M/h

-M/h

M/h M’/2

-M/L

h

M’/2

p/2

THREE-HINGE 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

TWO-HINGE FRAME

w/2

M’/4

p/2

RIGID FRAME M’/4

M’/4

M’/2L

p/2 p/2

-M’/2L

l

w/2

p/2 p/2

k

M/L

p/2 p/2

i

w/2

-M/L

g

w/2

Figure 28.3: Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal Frames Subjected to Vertical and Horizontal Loads

Victor Saouma

Structural Engineering

28–4

DESIGN II

Draft

9 To satisfy the equilibrium equation, this internal moment must be equal and opposite to the external moment Mext = P ecr hence PL = P ecr (28.6) 6

Mext

Mint

or ecr =

L 6

(28.7)

in other words to avoid tensile stresses on either side, the resultant force P must be placed within the midle third kernel, Fig. 28.4 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 28.4: Axial and Flexural Stresses

10

This equation is fundamental in preventing tensile forces in 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.

28.1.4

Design of a Statically Indeterminate Arch

Adapted from (Kinney 1957) Design a two-hinged, solid welded-steel arch rib for a hangar. The moment of inertia of the rib is to vary as necessary. The span, center to center of hinges, is to be 200 ft. Ribs are to be placed 35 ft center to center, with a rise of 35 ft. Roof deck, purlins and rib will be assumed to weight 25 lb/ft2 on roof surface, and snow will be assumed at 40 lb/ft2 of this surface. Twenty purlins will be equally spaced around the rib. 1. The center line of the rib will be taken as the segment of a circle. By computation the radius of this circle is found to be 160.357 ft, and the length of the arc AB to be 107.984 ft. 2. For the analysis the arc AB will be considered to be divided into ten segments, each with a length of 10.798 ft. Thus a concentrated load is applied to the rib by the purlins framing at the center of each segment. (The numbered segments are indicated in Fig. ??. 3. Since the total dead and snow load is 65 lb/ft2 of roof surface, the value of each concentrated force will be (28.8) P = 10.798 × 35 × 65 = 24.565 k Victor Saouma

Structural Engineering

28.1 Frames

28–5

Draft

Figure 28.5: Design of a Statically Indeterminate Arch 4. The computations necessary to evaluate the coodinates of the centers of the various segments, referred to the hinge at A, are shown in Table 28.1. Also shown are the values of ∆x, the horizontal projection of the distance between the centers of the several segments. 5. If experience is lacking and the designing engineer is therefore at a loss as to the initial assumptions regarding the sectional variation along the rib, it is recommended that the ﬁrst analysis be based on the assumption of a constant moment of inertia. Even the experienced engineer will often ﬁnd that this is the best procedure. 6. The value of H having thus been determined, moments and axial thrusts are computed throughout the rib. Cross-sectional areas at the various sections are then designed for these moments and axial thrusts. A new value for H, as well as revised values for moments and axial thrust throughout the rib, are determined for the new rib with the varying cross section. Stresses are checked in the rib, and the whole process is repeated if necessary. This procedure will be followed in the present case. 7. By the general method, assuming that end A is on rollers, ∆Ah + HA δAhAh = 0

(28.9)

8. Deﬂections will be computed by virtual work, and thus this expression becomes )

δM

) M M ∆S + HA δM ∆S = 0 EI EI

(28.10)

9. Since E and ∆S are constant and, for the ﬁrst analysis, the moment of inertia is constant, the ﬁnal expression for HA is * M δM HA = − * (28.11) δM M All the computations necessary for the ﬁrst evaluation of HA are shown in Table 28.2 yielding * 2 M δM 2 × 2, 158, 100 = −333.97k (28.12) =− HA = − 2 × 6461.9 2δM M 10. The moments at the centers of the various segments of the rib, using the approximate value just determined for HA , are computed in Table 28.3, where a moment is considered to be positive if it tends to cause compression on the upper surface of the rib. Victor Saouma

Structural Engineering

28–6

DESIGN II

Draft Segment

α

sin α

cos α

x = R sin α (ft)

A 1 2 3 4 5 6 7 8 9 10 Crown

38◦ 34 49 36◦ 39 05 32◦ 47 36 28◦ 56 07 25◦ 04 38 21◦ 13 09 17◦ 21 40 13◦ 30 11 09◦ 38 42 05◦ 47 13 01◦ 55 44

0.62361 0.59695 0.54161 0.48382 0.42384 0.36194 0.29840 0.23350 0.16754 0.10083 0.03367

0.78173 0.80229 0.84063 0.87516 0.90574 0.93220 0.95444 0.97236 0.98586 0.99490 0.99943

100.000 95.725 86.851 77.584 67.966 58.040 47.851 37.443 26.866 16.169 5.399

y = R cos α (ft)

x = 100 − x (ft)

y = y − 125.36 (ft)

∆x (ft)

125.36 128.65 134.80 140.34 145.24 149.48 153.05 155.92 158.09 159.54 160.27 160.36

0 4.275 13.149 22.416 32.034 41.960 52.149 62.557 73.134 83.831 94.601 100.000

0 3.29 9.44 14.98 19.88 24.13 27.69 30.57 32.73 34.18 34.91 35.00

4.275 8.874 9.267 9.618 9.926 10.189 10.408 10.577 10.697 10.770 5.399

Table 28.1: Geometry of the Arch

Segment A 1 2 3 4 5 6 7 8 9 10 Crown *

(2)

(3)

x (ft) 0.0 4.275 13.149 22.416 32.034 41.960 52.149 62.557 73.134 83.831 94.601 100.00

y = δM (ft · k) 0.0 3.29 9.44 14.98 19.88 24.13 27.69 30.57 32.73 34.18 34.91 35.00

(4) Shear to right (k) 245.650 221.085 196.520 171.955 147.390 122.825 98.260 73.695 49.130 24.565 0.000

(5)

(6)

∆x (ft)

M increment (ft · k)

(7) M simple beam (ft· k)

4.275 8.874 9.267 9.618 9.926 10.189 10.408 10.577 10.697 10.770 5.399

1,050 1,962 1,821 1,654 1,463 1,251 1,023 779 526 265 0

1,050 3,010 4,830 6,490 7,950 9,200 10,220 11,000 11,530 11,790 11,790

(8)

(9)

M δM

δM M

3,500 28,400 72,400 129,100 191,800 254,800 312,400 360,000 394,100 411,600

10.9 89.2 224.4 395.4 582.2 767.0 934.4 1,071.4 1,168.4 1,218.6

2,158,100

6,461.9

Table 28.2: Calculation of Horizontal Force

Victor Saouma

Structural Engineering

28.1 Frames

28–7

Draft Segment A 1 2 3 4 5 6 7 8 9 10 Crown

y (ft) 0 3.29 9.44 14.98 19.88 24.13 27.69 30.57 32.73 34.18 34.19 35.00

M simple beam (ft · k)

HA y (ft · k)

Total M at segment (ft · k)

1,050 3,010 4,830 6,490 7,950 9,200 10,220 11,000 11,530 11,790 11,790

–1,100 –3,150 –5,000 –6,640 –8,060 –9,250 –10,210 –10,930 –11,420 –11,660 –11,690

–50 –140 –170 –150 –110 –50 +10 +70 +110 +130 +100

Table 28.3: Moment at the Centers of the Ribs 11. The thrust N , normal to the cross section at the center of each segment, must be computed before the section requirements can be determined. As a matter of interest the various values of S, the total shear on the cross section, will also be computed for the center of each segment. 12. N and S are obtained by combining components of H and the vertical shear V at the center of each segment, a combination which is shown in Fig. 28.6 From this ﬁgure it is apparent that os α

Hc

α

in α

Hs

Vc

V α

os α

H

in α

Vs

α

Figure 28.6: Normal and Shear Forces S

= V cos α − H sin α

(28.13-a)

N

= V sin α + G cos α

(28.13-b)

13. Values of S and N are computed in Table . The values of S are included in support of the statement previously made to the eﬀect that shearing forces are small in the usual arch rib. 14. Inspection of the values for the moment M in Table 28.3 and for the thrust N in Table 28.3 will indicate that segment 3 is critical in the region of negative moment and segment 10 in the region of positive moment. The necessary sections for the rib, at the centers of these two segments, must be determined in accordance with the speciﬁcations of the American Institute of Steel Construction before Table 28.5 can be completed. 15. Flanges are 12 in. × 1 in., and the allowable stress for bending is FB =20 k/in2 . The web thickness is 34 in., and the allowable axial stress1 is FA =17,000-0.485 L/r. 1 In

this equation the value of L is 10.798, since ribs are laterally braced at purlin framing points.

Victor Saouma

Structural Engineering

28–8

Draft Segment A 1 2 3 4 5 6 7 8 9 10 Crown

DESIGN II V (k) 245.6 221.1 196.5 172.0 147.4 122.8 98.3 73.7 49.1 24.6 0.0

V cos α 192 177 165 150 133 114 94 72 48 244 0

− H sin α -208 -199 -181 -162 -142 -121 -100 -78 -56 -34 -11

= = = = = = = = = = = =

S (k) -16 -22 -16 -12 -9 -7 -6 -6 -8 -10 -11

V sin α 153 132 106 83 62 44 29 17 8 2 0

+H cos α + 261 +268 +281 +292 +303 +311 +319 +325 +329 +332 +334 +334

= = = = = = = = = = = = =

N (k) 414 400 387 375 365 355 348 342 337 334 334 334

Table 28.4: Values of Normal and Shear Forces 16. Thickness of ﬂange and web plates and width of web plates are matters of judgment. The total depth determined for the center of segment 3 is used from the center of this segment to the end of the rib. The depth determined for the center of segment 10 is arbitrarily used for the rib depth at the crown. The total depth of the rib from the center of segment 3 to the crown is made to vary linerarly. The adequacies of the sections thus determined for the centers of the several segments are checked in Table 28.5. 17. It is necessary to recompute the value of HA because the rib now has a varying moment of inertia. Equation 28.11 must be altered to include the I of each segment and is now written as * M δM /I (28.14) HA = − * δM M /I 18. The revised value for HA is easily determined as shown in Table 28.6. Note that the values in column (2) are found by dividing the values of M δM for the corresponding segments in column (8) of Table 28.2 by the total I for each segment as shown in Table 28.5. The values in column (3) of Table 28.6 are found in a similar manner from the values in column (9) of Table 28.2. The simple beam moments in column (5) of Table 28.6 are taken directly from column (7) of Table 28.2. 19. Thus the revised HA is * 2 M δM /I = −f rac2 × 1, 010.852 × 3.0192 = −334.81k HA = − * 2 δM M /I

(28.15)

20. The revised values for the axial thrust N at the centers of the various segments are computed in Table 28.7. 21. The sections previously designed at the centers of the segments are checked for adequacy in Table 28.8. From this table it appears that all sections of the rib are satisfactory. This cannot be deﬁnitely concluded, however, until the secondary stresses caused by the deﬂection of the rib are investigated.

28.1.5

Temperature Changes in an Arch

Adapted from (Kinney 1957) Determine the eﬀects of rib shortening and temperature changes in the arch rib of Fig. 28.5. Consider a temperature drop of 100◦ F. 1. In order to determine the eﬀects of rib shortening it is ﬁrst necessary to ﬁnd the horizontal displacement of end A which will result if the elastic axial strains, which occur in the actual loaded arch, are introduced into the rib while end A is on rollers. It is then necessary to ﬁnd the horizontal Victor Saouma

Structural Engineering

Victor Saouma

Segment A 1 2 3 4 5 6 7 8 9 10 Crown

(3)

Area (in2 ) 39.75 39.75 39.75 39.75 39.05 38.35 37.65 36.95 36.24 35.55 34.85 34.50

(2)

Total depth (in.) 23.00 23.00 23.00 23.00 22.07 21.13 20.20 19.27 18.32 17.40 16.47 16.00 I (in4 ) 3483 3483 3483 3483 3172 2865 2585 2329 2071 1838 1631 1527

(4)

50 140 170 150 110 50 10 70 110 130 100

M (ft · k)

(5)

2.0 5.9 6.8 6.3 4.9 2.3 0.5 3.7 6.3 7.9 6.3

(k/in2 )

Allowable Fa (k/in2 ) 15.88 15.88 15.88 15.88 15.90 15.92 15.94 15.96 15.97 15.99 16.01 16.03

Mc I

N A

(k/in2 ) 10.4 10.1 9.7 9.4 9.3 9.2 9.2 9.3 9.3 9.4 9.6 9.7

(9)

(8) fb =

(7) fa =

Table 28.5: Design of the Arch

Axial thrust N (k) 414 400 387 375 365 355 348 342 337 334 334 334

(6)

0.66 0.63 0.61 0.60 0.59 0.58 0.58 0.58 0.58 0.59 0.60 0.60

fa Fa

(10)

+

0.10 0.28 0.34 0.32 0.24 0.11 0.02 0.19 0.31 0.40 0.32

fb Fb

(11)

=

Decimal of capacity 0.66 0.73 0.89 0.94 0.91 0.82 0.70 0.61 0.77 0.90 1.00 0.92

(12)

Draft

28.1 Frames 28–9

Structural Engineering

28–10

Draft

DESIGN II (1)

(2)

(3)

(4)

Segment

Mn I

m2 I

1 2 3 4 5 6 7 8 9 10 Crown *

1.00 8.16 20.79 40.67 66.95 98.57 134.13 173.82 214.41 252.35

0.0031 0.0256 0.0644 0.1247 0.2032 0.2967 0.4012 0.5174 0.6357 0.7472

1010.85

3.0192

Revised HA y (ft · k) 1,100 3,160 5,020 6,660 8,080 9,270 10,240 10,960 11,450 11,690 11,720

(5) M simple beam (ft · k) 1,050 3,010 4,830 6,490 7,950 9,200 10,220 11,000 11,530 11,790 11,790

(6) Revised moment (ft · k) -50 -150 -190 -170 -130 -70 -20 +40 +80 +100 +70

Table 28.6: Segment

V sin α

A 1 2 3 4 5 6 7 8 9 10 Crown

153 132 106 83 62 44 29 17 8 2 0

+

H cos α

=

262 269 282 293 303 312 320 326 330 333 335 335

Revised N (k) 415 401 388 376 365 356 349 343 338 335 335 335

Table 28.7: N A

Segment A 1 2 3 4 5 6 7 8 9 10 Crown

(k/in2 ) 10.4 10.1 9.8 9.5 9.4 9.3 9,3 9.3 9.3 9.4 9.6 9.7

Mc I

(k/in2 ) 2.0 6.0 7.6 7.1 5.7 3.3 1.0 2.1 4.5 6.1 4.4

N/A FA

0.66 0.64 0.62 0.60 0.59 0.58 0.58 0.58 0.58 0.59 0.60 0.61

+

Mc /I FB

0.11 0.30 0.38 0.36 0.29 0.17 0.05 0.11 0.23 0.31 0.23

=

Decimal of capacity 0.66 0.75 0.92 0.98 0.95 0.87 0.75 0.63 0.69 0.82 0.91 0.84

Table 28.8: Victor Saouma

Structural Engineering

28.1 Frames

28–11

Draft

N (k) 401 388 376 365 356 349 343 338 335 335

Segment 1 2 3 4 5 6 7 8 9 10 *

δN = cos α (k) 0.8023 0.8406 0.8752 0.9057 0.9322 0.9544 0.9724 0.9859 0.9949 0.9994

A (in2 ) 39.75 39.75 39.75 39.05 38.35 37.65 36.95 36.24 35.55 34.85

N δN A

8.07 8.21 8.30 8.49 8.68 8.86 9.04 9.22 9.41 9.60 87.88

δN A

2

0.0162 0.0178 0.0193 0.0210 0.0227 0.0242 0.0256 0.0268 0.0286 0.0278 0.2300

Table 28.9: displacement of end A, caused by a horizontal force of 1 k acting at A, considering both ﬂexural and axial strains in the rib. The horizontal displacement of A, resulting from the rib shortening caused by the real loads, is given by ) N δN ∆L (28.16) AE where N is the axial thrust taken from Table 28.7, δN is the axial thrust caused by a 1 k ﬁctitious horizontal force acting at A and equal to cos α in Table 28.1, ∆L is the length of each segment, and A is the right sectional area. 2. The horizontal displacement of A, resulting from the rib shortening caused by a 1 k horizontal force acting at A, is given by ) δN N ∆L (28.17) AE 3. The necessary computations for evaluating these two deﬂections of A are shown in Table 28.9. 4. The horizontal displacement of A, resulting from the rib shortening caused by the real load, is )

δN

2 × 87.88 × 10.798 × 12 N ∆L = = 0.760 in AE 30, 000

(28.18)

5. That resulting from the rib shortening caused by a 1 k horizontal load at A is )

δN

2 × 0.2300 × 10.798 × 12 N ∆L = = 0.002 in AE 30, 000

(28.19)

6. The relative value of the horizontal displacement of A, resulting from the ﬂexural strain caused by a 1 k horizontal force at A, has previously been determined in column (3) of Table ??. The absolute value for this displacement is 2 × 3.019 × 10.798 × 1728 × 1728 = 3.755 in . 30, 000

(28.20)

7. The total value for the displacement of A due to both ﬂexural and axial strains resulting from a 1 k horizontal force at A is therefore 3.755 + 0.002 = 3.757 in

(28.21)

8. The reduction in HA due to rib shortening is given by 0.760 = 0.20 k 3.757 Victor Saouma

(28.22) Structural Engineering

28–12

DESIGN II

Draft

9. An additional reduction in HA will result from any drop in the temperature of the rib below that temperature at which the arch is erected. This occurs, of course, because some additional shortening of the rib will result. The decrease in the length of the horizontal projection of the rib resulting from a temperature drop of 100◦F will be δL = 0.0000065 × 100 × 200 × 12 = 1.56 in

10. The resulting decrease in HA will be

(28.23)

1.56 = 0.42 k 3.757

(28.24)

11. The total decrease in HA due to both rib shortening and temperature drop is 0.20 + 0.42 = 0.62 k

(28.25)

This will have the eﬀect of increasing the moment near the center of the span. 12. Inspection of column (6) in Table 28.8 indicates that segment 10 should be checked for possible overstress as the result of this increase. The increase in moment at segment 10 will be given by ∆M = ∆HA (yfor segment10) = 0.62 × 34.91 = 21.6 ft · k

(28.26)

13. The moment of 100 ft·k, as previously determined for segment 10, is accurate only in the ﬁrst two places, and hence the revised total moment for segment 10 will be 100 + 20 = 120 ft · k

(28.27)

14. The thrust of 335 k, previously determined for segment 10, is accurate to three places. The change in HA due to rib shortening and temperature change must therefore be rounded to 1 k in order to determine the corrected thrust in segment 10. This corrected thrust will be 335 − 1 = 334 k

(28.28)

15. The extreme ﬁber stress is f=

Mc 120 × 8.23 × 12 2 = = 7.25 k/in I 1631

(28.29)

16. The section at segment 10 is checked, as indicated in Table 28.8, by 334/34.85 7.25 N/A Mc /I + = 0.60 + 0.36 = 0.96 + = FA FB 16.01 20

(28.30)

and obviously the section at segment 10 is adequate. 17. No consideration has been given to the eﬀects of a temperature rise, since such a rise can occur only during the summer months when no snow loading is possible. Consequently, this phase of the problem has no practical signiﬁcance. 18. The method of analysis just demonstrated will apply equally well to a reinforced concrete or a timber two-hinged arch. As previously stated, however, very few reinforced concrete arches are built with two hinges. 19. Of special note here is the fact that, in addition to axial strain and temperature drop, the shrinkage of the concrete in a reinforced concrete rib will also cause rib shortening. The shrinkage coeﬃcient should be determined for the mix to be used and applied in the analysis in the same general way as demonstrated above for the temperature coeﬃcient.

Victor Saouma

Structural Engineering

Draft Chapter 29

INFLUENCE LINES (unedited) UNEDITED An inﬂuence line is a diagram whose ordinates are the values of some function of the structure (reaction, shear, moment, etc.) as a unit load moves across the structure. The shape of the inﬂuence line for a function (moment, shear, reaction, etc.) can be obtained by removing the resistance of the structure to that function, at the section where the inﬂuence line is desired, and applying an internal force associated with that function at the section so as to produce a unit deformation at the section. The deformed shape that the structure will take represents the shape of the inﬂuence line.

29–2

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INFLUENCE LINES (unedited)

Victor Saouma

Structural Engineering

Draft Chapter 30

ELEMENTS of STRUCTURAL RELIABILITY 30.1

Introduction

C Traditionally, evaluations of structural adequacy have been expressed by safety factors SF = D , where C is the capacity (i.e. strength) and D is the demand (i.e. load). Whereas this evaluation is quite simple to understand, it suﬀers from many limitations: it 1) treats all loads equally; 2) does not diﬀerentiate between capacity and demands respective uncertainties; 3) is restricted to service loads; and last but not least 4) does not allow comparison of relative reliabilities among diﬀerent structures for diﬀerent performance modes. Another major deﬁciency is that all parameters are assigned a single value in an analysis which is then deterministic.

1

2 Another approach, a probabilistic one, extends the factor of safety concept to explicitly incorporate uncertainties in the parameters. The uncertainties are quantiﬁed through statistical analysis of existing data or judgmentally assigned. 3 This chapter will thus develop a procedure which will enable the Engineer to perform a reliability based analysis of a structure, which will ultimately yield a reliability index. This is turn is a “universal” indicator on the adequacy of a structure, and can be used as a metric to 1) assess the health of a structure, and 2) compare diﬀerent structures targeted for possible remediation.

30.2

Elements of Statistics

4 Elementary statistics formulaes will be reviewed, as they are needed to properly understand structural reliability.

When a set of N values xi is clustered around a particular one, then it may be useful to characterize the set by a few numbers that are related to its moments (the sums of integer powers of the values):

5

Mean: estimates the value around which the data clusters. µ=

N 1 ) xi N i=1

(30.1)

it is the arithmetic average of all the data points. Expected Value: If data are not available, an expected value is assigned based on experience and judgment, and E(x) = µx . Both the mean and the expected values are termed ﬁrst moment, and they correspond to the centroid of a probability density distribution. (∞ = −∞ xf (x)dx Continuous systems *N (30.2) E(x) = µx = i=1 xf (x) Discrete systems

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ELEMENTS of STRUCTURAL RELIABILITY

Draft

Median: of a sorted series (xi−1 < xi < xi+1 ) is deﬁned as: 3 x N +1 2 xmed = 1 2 (x N + x N +1 ) 2

2

σ2

N odd N even = or

(30.3)

1 ΣN (xi − µ)2 N − 1 i=1

(30.4)

1 N Σ (xi − µ)2 (30.5)Note Variance: is an indication of the “width” of the cluster: σ 2 = N i=1 that if N is less than 10, it is more appropriate to use the second equation, otherwise use the ﬁrst one. Standard Deviation: is deﬁned as the square root of the Variance σ=

√ σ2

(30.6)

Coeﬃcient of Variation: is the standard deviation normalized with respect to the mean: V=

σ µ

(30.7)

When insuﬃcient data are available to accurately compute the moments, the coeﬃcient of variation is often estimated on the basis of experience. Covariance: Pairs of random variables may be correlated or independent. If correlated, then the likelihood of y depends on the likelihood of x. Thus, covariance σxy measures the combined eﬀect of how two variables vary together. σxy =

N 1 ) (xi − µx )(yi − µy ) N i=1

(30.8)

Correlation Coeﬃcient: ρxy is a nondimentional measure of the degree of correlation ρxy =

σxy σx σy

(30.9)

A correlation coeﬃcient of 1.0 or −1.0 indicates a perfect linear correlation. A positive value indicates that the variables either increase or decrease together, a negative one indicates that one value increases while the other decreases. A zero value indicates that there is no linear correlation between the variables. Skewness: characterizes the degree of asymmetry of a distribution around its mean. It is deﬁned in a non-dimensional value. A positive one signiﬁes a distribution with an asymmetric tail extending out toward more positive x 3 xi − µ 1 N (30.10) Skew = Σi=1 N σ Kurtosis: is a nondimensional quantity which measures the “ﬂatness” or “peakedness” of a distribution. It is normalized with respect to the curvature of a normal distribution. Hence a negative value would result from a distribution resembling a loaf of bread, while a positive one would be induced by a sharp peak: 4 xi − µ 1 N Kurt = Σi=1 −3 (30.11) N σ the −3 term makes the value zero for a normal distribution. 6 The expected value (or mean), standard deviation and coeﬃcient of variation are interdependent: knowing any two, we can determine the third.

Victor Saouma

Structural Engineering

30.3 Distributions of Random Variables

30–3

Draft 30.3 7

Distribution of variables can be mathematically represented.

30.3.1 8

Uniform Distribution

Uniform distribution implies that any value between xmin and xmax is equaly likely to occur.

30.3.2 9

Distributions of Random Variables

Normal Distribution

The general normal−3.0 (or −2.5 Gauss) distribution given0.0by, 0.5 Fig. 30.1: −2.0 −1.5 −1.0 is−0.5 1.0 1.5

2.0

2.5

3.0

0.40 0.35 0.30

2σ

PDF

0.25 0.20 0.15 0.10 0.05

0.5

1.0

1.5

2.0

2.5

3.0

0.5

1.0

1.5

2.0

2.5

3.0

CDF

0.00 −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0.0 µ 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0.0 x

Figure 30.1: Normalized Gauss Distribution, and Cumulative Distribution Function

1 x−µ 2 1 e− 2 [ σ ] φ(x) = √ 2πσ

10

(30.12)

A normal distribution N (µ, σ 2 ) can be normalized by deﬁning

y=

x−µ σ

(30.13)

and y would have a distribution N (0, 1): y2 1 φ(y) = √ e− 2 2π

(30.14)

The normal distribution has been found to be an excellent approximation to a large class of distributions, and has some very desirable mathematical properties:

11

Victor Saouma

Structural Engineering

30–4

ELEMENTS of STRUCTURAL RELIABILITY

Draft

1. f (x) is symmetric with respect to the mean µ. 2. f (x) is a “bell curve” with inﬂection points at x = µ ± σ.

3. f (x) is a valid probability distribution function as: ∞ f (x) = 1

(30.15)

−∞

4. The probability that xmin < x < xmax is given by:

xmax

P (xmin < x < xmax ) =

f (x)dx

(30.16)

xmin

5. Cumulative distribution functions (cdf) of the normal distribution deﬁned as: s 1 x−µ 2 1 Φ(s) = √ e− 2 [ σ ] dx 2πσ −∞

(30.17)

and is expressed in terms of the error function (erf). 6. The cdf of normalized normal distribution function is given by: s 1 x2 e− 2 dx Φ(s) = √ 2π −∞

(30.18)

and is usually tabulated in books.

30.3.3

Lognormal Distribution

A random variable is lognormally distributed if the natural logarithm of the variable is normally distributed.

12

13

It provides a reasonable shape when the coeﬃcient of variation is large.

30.3.4

Beta Distribution

Beta distributions are very ﬂexible and can assume a variety of shapes including the normal and uniform distributions as special cases.

14

15

On the other hand, the beta distribution requires four parameters.

16

Beta distributions are selected when a particular shape for the probability density function is desired.

30.3.5

BiNormal distribution

30.4

Reliability Index

30.4.1

Performance Function Identiﬁcation

Designating F the capacity to demand ratio C/D (or performance function), in general F is a function of one or more variables xi which describe the geometry, material, loads, and boundary conditions

17

F = F (xi )

(30.19)

and thus F is in turn a random variable with its own probability distribution function, Fig. 30.2. A performance function evaluation typically require a structural analysis, this may range from a simple calculation to a detailled ﬁnite element study.

18

Victor Saouma

Structural Engineering

30.4 Reliability Index

30–5

Draft 30.4.2

Deﬁnitions

Reliability indices, β are used as a relative measure of the reliability or conﬁdence in the ability of a structure to perform its function in a satisfactory manner. In other words they are a measure of the performance function.

19

Probabilistic methods are used to systematically evaluate uncertainties in parameters that aﬀect structural performance, and there is a relation between the reliability index and risk.

20

21 Reliability index is deﬁned in terms of the performance function capacity C, and the applied load or demand D. It is assumed that both C and D are random variables. 22 The safety margin is deﬁned as Y = C − D. Failure would occur if Y < 0 Next, C and D can be combined and the result expressed logarithmically, Fig. 30.2.

X = ln

C D

(30.20)

Failure would occur for negative values of X The probability of failure Pf is equal to the ratio of the shaded area to the total area under the curve in Fig. 30.2.

23

10

140

Demand 8

σ=300; µ=50 σ=150; µ=30

PDF

100

Capacity

80

60

pdf ln(C/D)

120 6

βσ ln(C/D) 4

2

Failure

40 0 −1.00

20

0 0.0

−0.50

0.00

0.50 ln(C/D) µ

1.00

1.50

2.00

Failure 100.0

200.0 300.0 Deﬁnition 400.0 500.0 Figure 30.2: of Reliability Index Tensile Stress [psi]

C If X is assumed to follow a Normal Distribution than it has a mean value X = ln D and a m standard deviation σ.

24

25

We deﬁne the safety index (or reliability index) as the distance between mean performance value µln C/D β = (30.21) σln C/D σ2

and the limit state normalized with respect to the standard deviation

26

=

ln µ C/D − ln 2C/D 2 σ ln 1 + µ C/D C/D

(30.22)

C The limit state is deﬁned as a ratio of capacity to demand approaching unity (i.e. ln D → 0).

Alternatively, the reliability index can be expressed in terms of the individual means of C and D, and their respective coeﬃcients of variations

27

ln µµC β= 2 D 2 VC + VD Victor Saouma

(30.23)

Structural Engineering

30–6

ELEMENTS of STRUCTURAL RELIABILITY

Draft

1 For standard distributions and for β = 3.5, it can be shown that the probability of failure is Pf = 9,091 −4 or 1.1 × 10 . That is 1 in every 10,000 structural members designed with β = 3.5 will fail because of either excessive load or understrength sometime in its lifetime. 28

30.4.3

Mean and Standard Deviation of a Performance Function

In general, some of the analysis parameters are constants, and others are variables. The variables will be characterized by a mean and a standard deviation, and a distribution function.

29

The objective is to determine the mean and standard deviation of the performance function deﬁned in terms of C/D.

30

31

Those two parameters, in turn, will later be required to compute the reliability index.

30.4.3.1

Direct Integration

32 Given a function random variable x, the mean value of the function is obtained by integrating the function over the probability distribution function of the random variable ∞ µ[F (x)] = g(x)f (x)dx (30.24)

−∞

33

For more than one variable, ∞ ∞ µ[F (x)] = ··· −∞

−∞

∞ −∞

F (x1 , x2 , · · · , xn )F (x1 , x2 , · · · , xn )dx1 dx2 · · · dxn

(30.25)

34 Note that in practice, the function F (x) is very rarely available for practical problems, and hence this method is seldom used.

30.4.3.2 35

Monte Carlo Simulation

The performance function is evaluated for many possible values of the random variables.

Assuming that all variables have a normal distribution, then this is done through the following algorithm

36

1. initialize random number generators 2. Perform n analysis, for each one: (a) For each variable, determine a random number for the given distribution (b) Transform the random number (c) Analyse (d) Determine the performance function, and store the results 3. From all the analyses, determine the mean and the standard deviation, compute the reliability index. 4. Count the number of analyses, nf which performance function indicate failure, the likelihood of structural failure will be p(f ) = nf /n. A sample program (all subroutines are taken from (Press, Flannery, Teukolvsky and Vetterling 1988) which generates n normally distributed data points, and then analyze the results, determines mean and standard deviation, and sort them (for histogram plotting), is shown below:

37

Victor Saouma

Structural Engineering

30.4 Reliability Index

Draft

30–7

Victor Saouma

Structural Engineering

program nice parameter(ns=100000) real x(ns), mean, sd write(*,*)’enter mean, standard-deviation and n ’ read(*,*)mean,sd,n c initialize tmp=gasdev(-1) do i=1,n c obtain gausian distribution and adjust for mean and standard deviation x(i)=gasdev(1)*sd+mean end do c compute mean and standard deviation call moment(x,n,ave,adev,sdev,var,skew,curt) c sort and print results call sort(n,x) write(*,*)(x(i),i=1,n) write(*,*)’ average standard deviation’,ave,sdev stop end c----------------------------------------------------------------------------function gasdev(idum) c*** Function to produce a normally distributed variable with zero c mean and unit ariance. data iset/0/ if (iset.eq.0) then 1 v1=2.*ran1(idum)-1. v2=2.*ran1(idum)-1. r=v1**2+v2**2 if(r.ge.1..or.r.eq.0.)go to 1 fac=sqrt(-2.*log(r)/r) gset=v1*fac gasdev=v2*fac iset=1 else gasdev=gset iset=0 endif return end c----------------------------------------------------------------------------function ran1(idum) c *** Returns a uniform random variable between 0.0 and 1.0, c set idum to any negative value to initialize, or reinitialize the sequence dimension r(97) parameter (m1=259200,ia1=7141,ic1=54773,rm1=3.8580247e-6) parameter (m2=134456,ia2=8121,ic2=28411,rm2=7.4373773e-6) parameter (m3=243000,ia3=4561,ic3=51349) data iff /0/ if (idum.lt.0.or.iff.eq.0) then iff=1 ix1=mod(ic1-idum,m1) ix1=mod(ia1*ix1+ic1,m1) ix2=mod(ix1,m2) ix1=mod(ia1*ix1+ic1,m1) ix3=mod(ix1,m3) do 11 j=1,97 ix1=mod(ia1*ix1+ic1,m1) ix2=mod(ia2*ix2+ic2,m2) r(j)=(float(ix1)+float(ix2)*rm2)*rm1 11 continue idum=1 endif ix1=mod(ia1*ix1+ic1,m1) ix2=mod(ia2*ix2+ic2,m2) ix3=mod(ia3*ix3+ic3,m3) j=1+(97*ix3)/m3 if(j.gt.97.or.j.lt.1)pause ran1=r(j) r(j)=(float(ix1)+float(ix2)*rm2)*rm1 return end c-----------------------------------------------------------------------------

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ELEMENTS of STRUCTURAL RELIABILITY

Victor Saouma

Structural Engineering

subroutine moment(data,n,ave,adev,sdev,var,skew,curt) c given array of data of length N, returns the mean AVE, average c deviation ADEV, standard deviation SDEV, variance VAR, skewness SKEW, c and kurtosis CURT dimension data(n) if(n.le.1)pause ’n must be at least 2’ s=0. do 11 j=1,n s=s+data(j) 11 continue ave=s/n adev=0. var=0. skew=0. curt=0. do 12 j=1,n s=data(j)-ave adev=adev+abs(s) p=s*s var=var+p p=p*s skew=skew+p p=p*s curt=curt+p 12 continue adev=adev/n var=var/(n-1) sdev=sqrt(var) if(var.ne.0.)then skew=skew/(n*sdev**3) curt=curt/(n*var**2)-3. else pause ’no skew or kurtosis when zero variance’ endif return end c-----------------------------------------------------------------------------subroutine sort(n,ra) dimension ra(n) l=n/2+1 ir=n 10 continue if(l.gt.1)then l=l-1 rra=ra(l) else rra=ra(ir) ra(ir)=ra(1) ir=ir-1 if(ir.eq.1)then ra(1)=rra return endif endif i=l j=l+l 20 if(j.le.ir)then if(j.lt.ir)then if(ra(j).lt.ra(j+1))j=j+1 endif if(rra.lt.ra(j))then ra(i)=ra(j) i=j j=j+j else j=ir+1 endif go to 20 endif ra(i)=rra go to 10 end

30.4 Reliability Index

30–9

Draft

As an example, using the above program, and assuming a normal distribution with a mean of 100. and a standard deviation of 20, results shown in Table 30.1 are obtained.

38

n 10 100 1,000 10,000 100,000

µ 100.33 100.49 100.123 99.964 100.029

σ 7.17 9.73 9.73 9.88 9.97

xmin 91.05 77.01 72.54 66.20 53.65

xmax 111.29 120.92 131.43 134.31 145.39

Table 30.1: Tabulated Results of a Simple Monte-Carlo Simulation. Note that for monte carlo simulations to be eﬀective, hundreds (and occasionally thousands) of analyses must be performed.

39

30.4.3.3

Point Estimate Method

As noted above, even for the simplest problem a successful Monte-Carlo simulation can be computationally very expensive. This is even more though when the evaluation of the performance function can only be achieved through a ﬁnite element analysis.

40

Whereas in the Monte Carlo simulations numerous random analyses had to be performed, in this method we select only few points such that they match the mean, standard deviation, and coeﬃcient of skewness of the distribution1 , (Harr 1987, Wolﬀ 1985).

41

42

This is accomplished by limiting ourselves to all possible combinations of µi ± σi .

43

For each analysis we determine SFF, as well as its logarithm.

Mean and standard deviation of the logarithmic values are then determined from the 2n analyses, and then β is the ratio of the mean to the standard deviation.

44

30.4.3.4

Taylor’s Series-Finite Diﬀerence Estimation

In the previous method, we have cut down the number of deterministic analyses to 2n , in the following method, we reduce it even further to 2n + 1, (Anonymous 1992, Anonymous 1993, Bryant, Brokaw and Mlakar 1993).

45

This simpliﬁed approach starts with the ﬁrst order Taylor series expansion of Eq. 30.19 about the mean and limited to linear terms, (Benjamin and Cornell 1970).

46

(30.26)

µF = F (µi ) where µi is the mean for all random variables. 47

For independent random variables, the variance can be approximated by

1 Such

V ar(F ) = σF2

=

∂F ∂xi

≈

) ∂F ∂xi

2 σi

Fi+ − Fi− 2σi

(30.27-a) (30.27-b)

a simpliﬁcation is behind the development of the Whitney stress block in the ACI code; a complex stress distribution is replaced by a simpler one (rectangular), such that both have the same resultant force and location.

Victor Saouma

Structural Engineering

30–10

ELEMENTS of STRUCTURAL RELIABILITY

Draft

Fi+ Fi−

= F (µ1 , · · · , µi + σi , · · · , µn ) = F (µ1 , · · · , µi − σi , · · · , µn )

(30.27-c) (30.27-d)

where σi are the standard deviations of the variables. Hence, σF =

) F+ − F− i

i

(30.28)

2

Finally, the reliability index is given by β=

48

ln µF σF

(30.29)

The procedure can be summarized as follows: 1. Perform an initial analysis in which all variables are set equal to their mean value. This analysis provides the mean µ. 2. Perform 2n analysis, in which all variables are set equal to their mean values, except variable i, which assumes a value equal to µi + σi , and then µi − σi . 3. For each pair of analysis in which variable xi is modiﬁed, determine (a) The standard deviation, which will provide an indication of the sensitivity of the results to variation of this particular variable. + − Fi −Fi (b) 2 + − Fi −Fi terms. from the 2n 4. The standard deviation is then determined by simply adding all the 2 analyses. by the mean. 5. Having determined the mean and standard deviation of C/D, the corresponding values for the logarithm of C/D, µln C/D and σln C/D , and β are determined from Eq. 30.22.

30.4.4

Overall System Reliability

Reliability indices for a number of components, or a number of modes of performance may be used to estimate the overall reliability of a structure.

49

50

We consider two extreme cases:

Series System: Such a system will perform poorly if any one component perform unsatisfactorily. If a system has n components in series, the probability of unsatisafcatorily performance of the ith component is pi and its reliability Ri = 1 − pi , then the reliability of the system, or probability that all components will perform satisfactorily is R = =

R1 R2 · · · Ri · · · Rn (1 − pi )(1 − p2 ) · · · (1 − pi ) · · · (1 − pn )

(30.30-a)

Parallel Systems: Such a system will only perform unsatisfactorily if all component perform unsatisfactorily. Hence the reliability is R = 1 − p1 p2 · · · pi · · · pn

30.4.5

(30.31)

Target Reliability Factors

Reliability indices are a relative measure of the current condition and provide a qualitative estimate of the structural performance.

51

Structures with relatively high reliable indices will be expected to perform well. If the value is too low, then the structure may be classiﬁed as a hazard.

52

53

Target values for β are shown in Table 30.2, and in Fig. 30.3

Victor Saouma

Structural Engineering

30.4 Reliability Index

30–11

Draft

Probability of Failure in terms of β

7

10

6

Good

10

5

1

Poor

2

10

Unsatisfactory

3

10

Above Average

Below Average

4

10

Hazardous

1/(Probability of Failure)

10

10

0

10

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

β

Figure 30.3: Probability of Failure in terms of β

Victor Saouma

Structural Engineering

30–12

Draft

ELEMENTS of STRUCTURAL RELIABILITY Expected Performance High Good Above Average Below Average Poor Unsatisfactory Hazardous

β 5 4 3 2.5 2.0 1.5 1.0

Failures 3/10 million 3/100,000 1/1,000 6/1,000 2.3/100 7/100 16/100

Table 30.2: Selected β values for Steel and Concrete Structures

30.5

Reliability Analysis

The steps involved in a reliability analysis are: 1. Model Develop a model which describes the performance mode and limit state to be evaluated. 2. Limit state determine the limit states in terms of ratios of capacity to demand. 3. Parameters Identify material properties and structure geometric parameters which enter into the analysis/ Separate those which may be considered constants from those which should be treated as random variables. 4. Variables Assign means, standard deviation, and coeﬃcients of correlations to random variables based on available or obtainable information. These may be derived from distributions, calculated directly from data, or assigned judgmentally. 5. Performance function Formulate the performance function C/D, in terms of constants and variables that describe the performance mode of interest. 6. Means and standard deviations of performance functions Determine the means and standard deviations of the performance function (C/D) for the performance modes of interest. 7. Calculate reliability index β Calculate β using any of the methods described above.

Victor Saouma

Structural Engineering

Draft Bibliography 318, A. C.: n.d., Building Code Requirements for Reinforced Concrete, (ACI 318-83), American Concrete Institute. Anonymous: 1992, Reliability assessment of navigation structures, ETL 1110-2-532, Department of the Army, US Army Corps of Engineers, Washington, D.C. Anonymous: 1993, Reliability assessment of navigation structures; stability of existing gravity structures, ETL 1110-2-321, Department of the Army, US Army Corps of Engineers, Washington, D.C. Arbabi, F.: 1991, Structural Analysis and Behavior, McGraw-Hill, Inc. Benjamin, J. and Cornell, C.: 1970, Probability, Statistics, and Decision for Civil Engineers, McGraw Hill, New York. Billington, D.: 1979, Robert Maillart’s Bridges; The art of Engineering, Princeton University Press. Billington, D. and Mark, R.: 1983, Structural studies, Technical report, Department of Civil Engineering, Princeton University. Bryant, L., Brokaw, J. and Mlakar, P.: 1993, Reliability modeling of concrete overstressing, Technical report, JAYCOR, Vicksburg Mississippi. Report Submitted to U.S. Army Engineer Waterways Experiment Station. Chajes, A.: 1983, Prentice-Hall. Gerstle, K.: 1974, Basic Structural Analysis, XX. Out of Print. Harr, M.: 1987, Reliability Based Design in Civil Engineering, McGraw Hill, New York. Kinney, J.: 1957, Indeterminate Structural Analysis, Addison-Wesley. Out of Print. Lin, T. and Stotesbury, S.: 1981, Structural Concepts and Systems for Architects and Engineers, John Wiley. of Steel COnstruction, A. I.: 1986, Manual of Steel Construction; Load and Resistance Factor Design, American Institute of Steel Construction. Penvenuto, E.: 1991, An Introduction to the History of Structural Mechanics, Springer-Verlag. Press, W., Flannery, B., Teukolvsky, S. and Vetterling, W.: 1988, Numerical Recipies in Fortran, The Art of Scientiﬁc Computing, Cambridge University Press. Salmon, C. and Johnson, J.: 1990, Steel Structures; Design and Behavior, Harper Colins. Third Edition. Schueller, W.: 1996, The design of Building Structures, Prentice Hall. UBC: 1995, Uniform building code, Technical report, International COnference of Building Oﬃcials. White, R., Gergely, P. and Sexmith, R.: 1976, Structural Engineering, John Wiley & Sons. Out of Print. Wolﬀ, T.: 1985, Analysis and Design of Embankment Dam Slopes: A Probabilistic Approach, PhD thesis, Purdue University.