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

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


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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


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 . . . . . . .

3.7

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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18–1 . 18–1 . 18–1 . 18–2 . 18–2 . 18–2 . 18–4 . 18–4 . 18–4 . 18–5 . 18–6 . 18–7 . 18–11

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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21–1 . 21–1 . 21–4 . 21–5 . 21–5 . 21–5 . 21–9 . 21–11 . 21–13 . 21–16 . 21–17 . 21–18

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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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25–1 . 25–1 . 25–3 . 25–3 . 25–3 . 25–4 . 25–5 . 25–6 . 25–7 . 25–9 . 25–10 . 25–11

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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26–1 . 26–1 . 26–1 . 26–2 . 26–2 . 26–3 . 26–3 . 26–5 . 26–5 . 26–7 . 26–8 . 26–8 . 26–8 . 26–8 . 26–10

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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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. 26–12 . 26–12 . 26–13 . 26–13 . 26–13 . 26–14 . 26–15 . 26–16 . 26–16 . 26–17 . 26–18 . 26–18 . 26–23 . 26–25 . 26–29 . 26–31 . 26–31 . 26–31 . 26–32 . 26–33 . 26–34 . 26–34 . 26–35 . 26–36 . 26–37 . 26–37 . 26–38 . 26–39 . 26–40 . 26–40 . 26–41 . 26–42 . 26–43 . 26–44 . 26–45 . 26–49

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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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28–1 . 28–1 . 28–1 . 28–1 . 28–2 . 28–4 . 28–10 29–1

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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


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


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


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


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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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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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


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


Draft

Part I

STRUCTURAL ENGINEERING I; CVEN 3525

Draft

Draft Chapter 1

INTRODUCTION 1.1 1


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


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


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


1.9 Structure Types

1–5

Draft

Figure 1.3: Basic Aspects of Arches

Victor Saouma


1–6

INTRODUCTION

Draft

Figure 1.4: Types of Trusses

Victor Saouma


1.9 Structure Types

1–7

Draft

Figure 1.5: Variations in Post and Beams Configurations

Victor Saouma


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


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


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


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


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


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


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


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


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


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


2.3 Lateral Loads

2–7

Draft

Height Zone (in feet) 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


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 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


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


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


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)


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


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


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)


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


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


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


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


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


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 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’ 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


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


3–20

STRUCTURAL MATERIALS

4"

4"

Span

[Design Length = Span – 0.33 FT.]

4"

Draft

Figure 3.9: prefabricated Steel Joists

Victor Saouma


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


3–22

Draft

STRUCTURAL MATERIALS

Victor Saouma


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


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


4–4


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


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


4–6


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


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


4–8


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


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


4–10


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


4.7 Arches

Draft 4.7 31

4–11

Arches

See Section ??.

Victor Saouma


4–12

Draft


Victor Saouma


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


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


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


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


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


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


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


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


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


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


=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


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


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


5–16

Draft

TRUSSES

Victor Saouma


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)


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


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 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


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


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


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)



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)


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



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


8–8

DESIGN I

Draft 8.2

Magazini Generali by Maillart


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)


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


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


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


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


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


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


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


8–16

DESIGN I

Draft 8.3.1.2.1 44

Example: Tube Subsystem


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


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


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)


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


8.5 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


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



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


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



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


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


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


8–26

Draft

DESIGN I

Victor Saouma


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


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


9–4


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



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


9–6


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



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


9–8


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)


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


9–10


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



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)


9–12


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



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)


9–14


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



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


9–16


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



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


9–18

Draft


{ 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[x0,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


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


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


9–22


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


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


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


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


6

12

1 2 3 4 5 6

13

1 2 3


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


9–24


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


16

1 2 3 4 5

LOAD CASE:

3 6


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


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


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


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


10–4


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


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


10–6

Draft


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


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


10–8


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


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


10–10


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


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


10–12

Draft


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


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


10–14


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


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


10–16


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


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


10–18


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


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


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’ 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)


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

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


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


18–8


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


18.4 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


18–10


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)


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


18–12

Draft


Victor Saouma


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


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


19–4


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


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


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


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


19–8


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


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


19–10

Draft


Victor Saouma


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


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’ 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


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


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


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


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)


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


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


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


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


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


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


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


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


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


28–12

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


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

Draft

INFLUENCE LINES (unedited)

Victor Saouma


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

30–2

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


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


30–4


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


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)


30–6


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



Draft

30–7

Victor Saouma


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-----------------------------------------------------------------------------

30–8

Draft


Victor Saouma


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


30–9

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


30–10


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



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


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


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.

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