architecture structure concepts for architects

Draft LECTURE NOTES

AREN4525

STUCTURAL CONCEPTS AND SYSTEMS

FOR ARCHITECTS

VICTOR E. SAOUMA

SPRING 1997

Dept. of Civil Environmental and Architectural Engineering

University of Colorado, Boulder, CO 80309-0428

July 8, 2003

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In order to invent a structure and to give it ex-act proportions, one must follow both the intu-itive and the mathematical paths.

-Pier Luigi Nervi

Victor Saouma Structural Concepts and Systems for Architects

Draft

Contents

1 INTRODUCTION 1–11.1 Science and Technology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11.2 Structural Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11.3 Structures and their Surroundings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11.4 Architecture & Engineering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–21.5 Architectural Design Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–21.6 Architectural Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–21.7 Structural Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–31.8 Structural Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–31.9 Load Transfer Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–41.10 Structure Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–41.11 Structural Engineering Courses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–121.12 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–13

2 LOADS 2–12.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–12.2 Vertical Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–1

2.2.1 Dead Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22.2.2 Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–2E 2-1 Live Load Reduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–42.2.3 Snow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–5

2.3 Lateral Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–52.3.1 Wind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–5E 2-2 Wind Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–92.3.2 Earthquakes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–11E 2-3 Earthquake Load on a Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–14E 2-4 Earthquake Load on a Tall Building, (Schueller 1996) . . . . . . . . . . . . . . . . 2–16

2.4 Other Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.4.1 Hydrostatic and Earth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–18E 2-5 Hydrostatic Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.4.2 Thermal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–19E 2-6 Thermal Expansion/Stress (Schueller 1996) . . . . . . . . . . . . . . . . . . . . . . 2–19

2.5 Other Important Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–202.5.1 Load Combinations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–202.5.2 Load Placement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–212.5.3 Load Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–212.5.4 Structural Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–212.5.5 Tributary Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–25

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3 STRUCTURAL MATERIALS 3–13.1 Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–1

3.1.1 Structural Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13.1.2 Reinforcing Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–5

3.2 Aluminum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–63.3 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–63.4 Masonry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–73.5 Timber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–73.6 Steel Section Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–83.7 Joists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–17

4 Case Study I: EIFFEL TOWER 4–14.1 Materials, & Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–14.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–34.3 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–44.4 Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–64.5 Internal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–8

5 REVIEW of STATICS 5–15.1 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–1

5.1.1 Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–15.1.2 Equations of Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–35.1.3 Static Determinacy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–35.1.4 Geometric Instability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–45.1.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–5E 5-1 Simply Supported Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–5E 5-2 Three Span Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–6E 5-3 Three Hinged Gable Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–7

5.2 Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–85.2.1 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–85.2.2 Basic Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–95.2.3 Determinacy and Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–95.2.4 Method of Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–10E 5-4 Truss, Method of Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–12

5.3 Shear & Moment Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–145.3.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–14

5.3.1.1 Design Sign Conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–145.3.1.2 Load, Shear, Moment Relations . . . . . . . . . . . . . . . . . . . . . . . 5–155.3.1.3 Moment Envelope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–165.3.1.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–18

E 5-5 Simple Shear and Moment Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 5–18E 5-6 Frame Shear and Moment Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–19E 5-7 Frame Shear and Moment Diagram; Hydrostatic Load . . . . . . . . . . . . . . . . 5–22E 5-8 Shear Moment Diagrams for Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–24E 5-9 Shear Moment Diagrams for Inclined Frame . . . . . . . . . . . . . . . . . . . . . . 5–265.3.2 Formulaes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–28

5.4 Flexure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–375.4.1 Basic Kinematic Assumption; Curvature . . . . . . . . . . . . . . . . . . . . . . . . 5–375.4.2 Stress-Strain Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–385.4.3 Internal Equilibrium; Section Properties . . . . . . . . . . . . . . . . . . . . . . . . 5–39

5.4.3.1 ΣFx = 0; Neutral Axis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–39


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5.4.3.2 ΣM = 0; Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . . 5–395.4.4 Beam Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–40E 5-10 Design Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–415.4.5 Approximate Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–42E 5-11 Approximate Analysis of a Statically Indeterminate beam . . . . . . . . . . . . . . 5–42

6 Case Study II: GEORGE WASHINGTON BRIDGE 6–16.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–16.2 The Case Study . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–4

6.2.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–46.2.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–56.2.3 Cable Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–76.2.4 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–7

7 A BRIEF HISTORY OF STRUCTURAL ARCHITECTURE 7–17.1 Before the Greeks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–17.2 Greeks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–17.3 Romans . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–37.4 The Medieval Period (477-1492) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–47.5 The Renaissance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–5

7.5.1 Leonardo da Vinci 1452-1519 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–67.5.2 Brunelleschi 1377-1446 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–67.5.3 Alberti 1404-1472 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–77.5.4 Palladio 1508-1580 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–87.5.5 Stevin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–87.5.6 Galileo 1564-1642 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–10

7.6 Pre Modern Period, Seventeenth Century . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–127.6.1 Hooke, 1635-1703 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–127.6.2 Newton, 1642-1727 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–137.6.3 Bernoulli Family 1654-1782 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–147.6.4 Euler 1707-1783 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–15

7.7 The pre-Modern Period; Coulomb and Navier . . . . . . . . . . . . . . . . . . . . . . . . . 7–167.8 The Modern Period (1857-Present) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–17

7.8.1 Structures/Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–177.8.2 Eiffel Tower . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–177.8.3 Sullivan 1856-1924 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–177.8.4 Roebling, 1806-1869 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–187.8.5 Maillart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–187.8.6 Nervi, 1891-1979 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–197.8.7 Khan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–197.8.8 et al. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–20

8 Case Study III: MAGAZINI GENERALI 8–18.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–18.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–18.3 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–38.4 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–38.5 Internal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–6


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9 DESIGN PHILOSOPHIES and GUIDELINES 9–19.1 Safety Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–19.2 Working Stress Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–29.3 Ultimate Strength Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–3

9.3.1 † Probabilistic Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–39.3.2 Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–5

9.4 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–7E 9-1 LRFD vs ASD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–7

9.5 Design Guidelines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–7

10 BRACED ROLLED STEEL BEAMS 10–110.1 Nominal Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–210.2 Failure Modes and Classification of Steel Beams . . . . . . . . . . . . . . . . . . . . . . . . 10–210.3 Compact Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–4

10.3.1 Bending Capacity of Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–410.3.2 Design of Compact Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–6

10.4 Partially Compact Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–610.5 Slender Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–710.6 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–7

E 10-1 Z for Rectangular Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–7E 10-2 Beam Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–8

11 REINFORCED CONCRETE BEAMS 11–111.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–1

11.1.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–111.1.2 Modes of Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–211.1.3 Analysis vs Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–211.1.4 Basic Relations and Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–311.1.5 ACI Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–3

11.2 Cracked Section, Ultimate Strength Design Method . . . . . . . . . . . . . . . . . . . . . . 11–411.2.1 Equivalent Stress Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–411.2.2 Balanced Steel Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–511.2.3 Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–611.2.4 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–7E 11-1 Ultimate Strength Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–8E 11-2 Beam Design I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–9E 11-3 Beam Design II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–9

11.3 Continuous Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–1011.4 ACI Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–10

12 PRESTRESSED CONCRETE 12–112.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–1

12.1.1 Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–112.1.2 Prestressing Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–412.1.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–412.1.4 Tendon Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–412.1.5 Equivalent Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–412.1.6 Load Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–4

12.2 Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–5E 12-1 Prestressed Concrete I Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–8

12.3 Case Study: Walnut Lane Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–10


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12.3.1 Cross-Section Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–1012.3.2 Prestressing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–1212.3.3 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–1312.3.4 Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–13

13 Three-Hinges ARCHES 13–113.1 Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1

13.1.1 Uniform Horizontal Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1E 13-1 Design of a Three Hinged Arch . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–3

13.2 Case Study: Salginatobel Bridge (Maillart) . . . . . . . . . . . . . . . . . . . . . . . . . . 13–513.2.1 Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–513.2.2 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–813.2.3 Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–813.2.4 Internal Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–1113.2.5 Internal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–12

13.3 Structural Behavior of Deck-Stiffened Arches . . . . . . . . . . . . . . . . . . . . . . . . . 13–13

14 BUILDING STRUCTURES 14–114.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–1

14.1.1 Beam Column Connections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–114.1.2 Behavior of Simple Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–114.1.3 Eccentricity of Applied Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–2

14.2 Buildings Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–514.2.1 Wall Subsystems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–5

14.2.1.1 Example: Concrete Shear Wall . . . . . . . . . . . . . . . . . . . . . . . . 14–514.2.1.2 Example: Trussed Shear Wall . . . . . . . . . . . . . . . . . . . . . . . . 14–7

14.2.2 Shaft Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–814.2.2.1 Example: Tube Subsystem . . . . . . . . . . . . . . . . . . . . . . . . . . 14–8

14.2.3 Rigid Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–914.3 Approximate Analysis of Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–10

14.3.1 Vertical Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–1114.3.2 Horizontal Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–13

14.3.2.1 Portal Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–13E 14-1 Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads . . . 14–15

14.4 Lateral Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–2714.4.1 Short Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–2714.4.2 Tall Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–2814.4.3 Walls and Lintel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–2814.4.4 Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–2914.4.5 Trussed Frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–3014.4.6 Example of Transverse Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–3214.4.7 Effect of Bracing Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–34


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List of Figures

1.1 Types of Forces in Structural Elements (1D) . . . . . . . . . . . . . . . . . . . . . . . . . . 1–41.2 Basic Aspects of Cable Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–51.3 Basic Aspects of Arches . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–61.4 Types of Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–71.5 Variations in Post and Beams Configurations . . . . . . . . . . . . . . . . . . . . . . . . . 1–81.6 Different Beam Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–91.7 Basic Forms of Frames . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–101.8 Examples of Air Supported Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–111.9 Basic Forms of Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–12

2.1 Approximation of a Series of Closely Spaced Loads . . . . . . . . . . . . . . . . . . . . . . 2–22.2 Snow Map of the United States, ubc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–52.3 Loads on Projected Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–62.4 Wind Map of the United States, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . . . 2–72.5 Effect of Wind Load on Structures(Schueller 1996) . . . . . . . . . . . . . . . . . . . . . . 2–82.6 Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Struc-

tures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–102.7 Vibrations of a Building . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–122.8 Seismic Zones of the United States, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . 2–132.9 Earth and Hydrostatic Loads on Structures . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.10 Load Placement to Maximize Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–212.11 Load Transfer in R/C Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–222.12 Two Way Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–232.13 Load Life of a Structure, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . 2–242.14 Concept of Tributary Areas for Structual Member Loading . . . . . . . . . . . . . . . . . 2–25

3.1 Stress Strain Curves of Concrete and Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–23.2 Standard Rolled Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–23.3 Residual Stresses in Rolled Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–43.4 Residual Stresses in Welded Sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–43.5 Influence of Residual Stress on Average Stress-Strain Curve of a Rolled Section . . . . . . 3–53.6 Concrete microcracking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–73.7 W and C sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–83.8 prefabricated Steel Joists . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–17

4.1 Eiffel Tower (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–14.2 Eiffel Tower Idealization, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . 4–34.3 Eiffel Tower, Dead Load Idealization; (Billington and Mark 1983) . . . . . . . . . . . . . . 4–34.4 Eiffel Tower, Wind Load Idealization; (Billington and Mark 1983) . . . . . . . . . . . . . . 4–44.5 Eiffel Tower, Wind Loads, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . 4–5

Draft0–2 LIST OF FIGURES

4.6 Eiffel Tower, Reactions; (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . 4–54.7 Eiffel Tower, Internal Gravity Forces; (Billington and Mark 1983) . . . . . . . . . . . . . . 4–74.8 Eiffel Tower, Horizontal Reactions; (Billington and Mark 1983) . . . . . . . . . . . . . . . 4–74.9 Eiffel Tower, Internal Wind Forces; (Billington and Mark 1983) . . . . . . . . . . . . . . . 4–8

5.1 Types of Supports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–25.2 Inclined Roller Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–35.3 Examples of Static Determinate and Indeterminate Structures . . . . . . . . . . . . . . . . 5–45.4 Geometric Instability Caused by Concurrent Reactions . . . . . . . . . . . . . . . . . . . . 5–45.5 Bridge Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–95.6 A Statically Indeterminate Truss . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–105.7 X and Y Components of Truss Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–115.8 Sign Convention for Truss Element Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–115.9 Shear and Moment Sign Conventions for Design . . . . . . . . . . . . . . . . . . . . . . . . 5–145.10 Sign Conventions for 3D Frame Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–155.11 Free Body Diagram of an Infinitesimal Beam Segment . . . . . . . . . . . . . . . . . . . . 5–155.12 Shear and Moment Forces at Different Sections of a Loaded Beam . . . . . . . . . . . . . 5–175.13 Slope Relations Between Load Intensity and Shear, or Between Shear and Moment . . . . 5–175.14 Deformation of a Beam un Pure Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–375.15 Elastic Curve from the Moment Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–435.16 Approximate Analysis of Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–44

6.1 Cable Structure Subjected to p(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–26.2 Longitudinal and Plan Elevation of the George Washington Bridge . . . . . . . . . . . . . 6–46.3 Truck Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–56.4 Dead and Live Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–66.5 Location of Cable Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–76.6 Vertical Reactions in Columns Due to Central Span Load . . . . . . . . . . . . . . . . . . 6–76.7 Cable Reactions in Side Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–86.8 Cable Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–96.9 Deck Idealization, Shear and Moment Diagrams . . . . . . . . . . . . . . . . . . . . . . . . 6–10

7.1 Hamurrabi’s Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–27.2 Archimed . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–37.3 Pantheon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–37.4 From Vitruvius Ten Books on Architecture, (Vitruvius 1960) . . . . . . . . . . . . . . . . 7–47.5 Hagia Sophia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–57.6 Florence’s Cathedral Dome . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–77.7 Palladio’s Villa Rotunda . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–97.8 Stevin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–97.9 Galileo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–107.10 Discourses Concerning Two New Sciences, Cover Page . . . . . . . . . . . . . . . . . . . . 7–117.11 “Galileo’s Beam” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–117.12 Experimental Set Up Used by Hooke . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–127.13 Isaac Newton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–137.14 Philosophiae Naturalis Principia Mathematica, Cover Page . . . . . . . . . . . . . . . . . 7–147.15 Leonhard Euler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–157.16 Coulomb . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–167.17 Nervi’s Palazetto Dello Sport . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–19

8.1 Magazzini Generali; Overall Dimensions, (Billington and Mark 1983) . . . . . . . . . . . . 8–28.2 Magazzini Generali; Support System, (Billington and Mark 1983) . . . . . . . . . . . . . . 8–2


DraftLIST OF FIGURES 0–3

8.3 Magazzini Generali; Loads (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . 8–38.4 Magazzini Generali; Beam Reactions, (Billington and Mark 1983) . . . . . . . . . . . . . . 8–38.5 Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983) . . . . . . . 8–48.6 Magazzini Generali; Internal Moment, (Billington and Mark 1983) . . . . . . . . . . . . . 8–48.7 Magazzini Generali; Similarities Between The Frame Shape and its Moment Diagram,

(Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–58.8 Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark

1983) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8–58.9 Magazzini Generali; Effect of Lateral Supports, (Billington and Mark 1983) . . . . . . . . 8–6

9.1 Load Life of a Structure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–29.2 Frequency Distributions of Load Q and Resistance R . . . . . . . . . . . . . . . . . . . . . 9–49.3 Definition of Reliability Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–4

10.1 Lateral Bracing for Steel Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–110.2 Failure of Steel beam; Plastic Hinges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–310.3 Failure of Steel beam; Local Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–310.4 Failure of Steel beam; Lateral Torsional Buckling . . . . . . . . . . . . . . . . . . . . . . . 10–410.5 Stress distribution at different stages of loading . . . . . . . . . . . . . . . . . . . . . . . . 10–410.6 Stress-strain diagram for most structural steels . . . . . . . . . . . . . . . . . . . . . . . . 10–510.7 Nominal Moments for Compact and Partially Compact Sections . . . . . . . . . . . . . . . 10–7

11.1 Failure Modes for R/C Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–211.2 Internal Equilibrium in a R/C Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–311.3 Cracked Section, Limit State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–411.4 Whitney Stress Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–511.5 Reinforcement in Continuous R/C Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–11

12.1 Pretensioned Prestressed Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . . . . . . 12–212.2 Posttensioned Prestressed Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . . . . . . 12–212.3 7 Wire Prestressing Tendon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–312.4 Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978) . . . . 12–512.5 Determination of Equivalent Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–612.6 Load-Deflection Curve and Corresponding Internal Flexural Stresses for a Typical Pre-

stressed Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–612.7 Flexural Stress Distribution for a Beam with Variable Eccentricity; Maximum Moment

Section and Support Section, (Nilson 1978) . . . . . . . . . . . . . . . . . . . . . . . . . . 12–712.8 Walnut Lane Bridge, Plan View . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–1112.9 Walnut Lane Bridge, Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12–12

13.1 Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Linand Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13–2

13.2 Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . 13–213.3 Two Hinged Arch, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . 13–313.4 Arch Rib Stiffened with Girder or Truss, (Lin and Stotesbury 1981) . . . . . . . . . . . . 13–313.5 Salginatobel Bridge; Dimensions, (Billington and Mark 1983) . . . . . . . . . . . . . . . . 13–513.6 Salginatobel Bridge; Idealization, (Billington and Mark 1983) . . . . . . . . . . . . . . . . 13–613.7 Salginatobel Bridge; Hinges, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . . 13–613.8 Salginatobel Bridge; Sections, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . . 13–713.9 Salginatobel Bridge; Dead Load, (Billington and Mark 1983) . . . . . . . . . . . . . . . . 13–813.10Salginatobel Bridge; Truck Load, (Billington and Mark 1983) . . . . . . . . . . . . . . . . 13–913.11Salginatobel Bridge; Total Vertical Load, (Billington and Mark 1983) . . . . . . . . . . . . 13–10


Draft0–4 LIST OF FIGURES

13.12Salginatobel Bridge; Reactions, (Billington and Mark 1983) . . . . . . . . . . . . . . . . . 13–1013.13Salganitobel Bridge; Shear Diagrams, (Billington and Mark 1983) . . . . . . . . . . . . . . 13–1113.14Salginatobel Bridge; Live Load Moment Diagram, (Billington and Mark 1983) . . . . . . . 13–1213.15Structural Behavior of Stiffened Arches, (Billington 1979) . . . . . . . . . . . . . . . . . . 13–14

14.1 Flexible, Rigid, and Semi-Flexible Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–114.2 Deformation of Flexible and Rigid Frames Subjected to Vertical and Horizontal Loads,

(Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–214.3 Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal Frames

Subjected to Vertical and Horizontal Loads . . . . . . . . . . . . . . . . . . . . . . . . . . 14–314.4 Axial and Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–414.5 Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . 14–614.6 Trussed Shear Wall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–814.7 Design Example of a Tubular Structure, (Lin and Stotesbury 1981) . . . . . . . . . . . . . 14–914.8 A Basic Portal Frame, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . 14–1014.9 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments . . . . . . 14–1114.10Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces . . . . 14–1214.11Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments . . . . . 14–1214.12Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear . . . . . . . . 14–1414.13***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment . . . . . 14–1414.14Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force . . . . 14–1514.15Example; Approximate Analysis of a Building . . . . . . . . . . . . . . . . . . . . . . . . . 14–1514.16Approximate Analysis of a Building; Moments Due to Vertical Loads . . . . . . . . . . . . 14–1714.17Approximate Analysis of a Building; Shears Due to Vertical Loads . . . . . . . . . . . . . 14–1814.18Approximate Analysis for Vertical Loads; Spread-Sheet Format . . . . . . . . . . . . . . . 14–2014.19Approximate Analysis for Vertical Loads; Equations in Spread-Sheet . . . . . . . . . . . . 14–2114.20Approximate Analysis of a Building; Moments Due to Lateral Loads . . . . . . . . . . . . 14–2314.21Portal Method; Spread-Sheet Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–2414.22Portal Method; Equations in Spread-Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–2514.23Shear Deformation in a Short Building, (Lin and Stotesbury 1981) . . . . . . . . . . . . . 14–2814.24Flexural Deformation in a Tall Building, (Lin and Stotesbury 1981) . . . . . . . . . . . . 14–2814.25Deflection in a Building Structure Composed of Two Slender Walls and Lintels, (Lin and

Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14–2914.26Portal Method to Estimate Lateral Deformation in Frames, (Lin and Stotesbury 1981) . . 14–3014.27Shear and Flexural Deflection of a Rigid Frame Subsystem, (Lin and Stotesbury 1981) . . 14–3114.28Side-Sway Deflection from Unsymmetrical Vertical Load, (Lin and Stotesbury 1981) . . . 14–3114.29Axial Elongation and Shortening of a Truss Frame, (Lin and Stotesbury 1981) . . . . . . 14–3114.30Transverse Deflection, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . 14–3214.31Frame Rigidly Connected to Shaft, (Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . 14–3414.32Effect of Exterior Column Bracing in Buildings, (Lin and Stotesbury 1981) . . . . . . . . 14–35


Draft

List of Tables

1.1 Structural Engineering Coverage for Architects and Engineers . . . . . . . . . . . . . . . . 1–121.2 tab:secae . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–12

2.1 Unit Weight of Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22.2 Weights of Building Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–32.3 Average Gross Dead Load in Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–32.4 Minimum Uniformly Distributed Live Loads, (UBC 1995) . . . . . . . . . . . . . . . . . . 2–42.5 Wind Velocity Variation above Ground . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–72.6 Ce Coefficients for Wind Load, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . . . . 2–82.7 Wind Pressure Coefficients Cq, (UBC 1995) . . . . . . . . . . . . . . . . . . . . . . . . . . 2–92.8 Importance Factors for Wind and Earthquake Load, (UBC 1995) . . . . . . . . . . . . . . 2–92.9 Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Struc-

tures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–102.10 Z Factors for Different Seismic Zones, ubc . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–132.11 S Site Coefficients for Earthquake Loading, (UBC 1995) . . . . . . . . . . . . . . . . . . . 2–142.12 Partial List of RW for Various Structure Systems, (UBC 1995) . . . . . . . . . . . . . . . 2–152.13 Coefficients of Thermal Expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–19

3.1 Properties of Major Structural Steels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–33.2 Properties of Reinforcing Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–53.3 Joist Series Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–173.4 Joist Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–19

5.1 Equations of Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–35.2 Static Determinacy and Stability of Trusses . . . . . . . . . . . . . . . . . . . . . . . . . . 5–105.3 Section Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–40

9.1 Allowable Stresses for Steel and Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–39.2 Selected β values for Steel and Concrete Structures . . . . . . . . . . . . . . . . . . . . . . 9–59.3 Strength Reduction Factors, Φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–69.4 Approximate Structural Span-Depth Ratios for Horizontal Subsystems and Components

(Lin and Stotesbury 1981) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–8

14.1 Columns Combined Approximate Vertical and Horizontal Loads . . . . . . . . . . . . . . 14–2614.2 Girders Combined Approximate Vertical and Horizontal Loads . . . . . . . . . . . . . . . 14–27

Draft0–2 LIST OF TABLES


Draft

Chapter 1

INTRODUCTION

1.1 Science and Technology

1 “There is a fundamental difference between science and and technology. Engineering or technology isthe making of things that did not previously exist, whereas science is the discovering of things that havelong existed. Technological results are forms that exist only because people want to make them, whereasscientific results are informations of what exists independently of human intentions. Technology dealswith the artificial, science with the natural.” (Billington 1985)

1.2 Structural Engineering

2 Structural engineers are responsible for the detailed analysis and design of:

Architectural structures: Buildings, houses, factories. They must work in close cooperation with anarchitect 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 en-sure the structural safety of those important structures.

1.3 Structures and their Surroundings

3 Structural design is affected by various environmental constraints:

1. Major movements: For example, elevator shafts are usually shear walls good at resisting lateralload (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.

Draft1–2 INTRODUCTION

• 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.

1.4 Architecture & Engineering

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

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

6 The engineer, partly due to his/her education think in reverse, starting with details and withoutsufficient regards for the overall picture. (S)he is a pragmatist who “knows everything about nothing”.

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

8 Engineer’s education is more specialized and in depth than the architect’s. However, engineer mustbe kept aware of overall architectural objective.

9 In the last resort, it is the architect who is the leader of the construction team, and the engineers arehis/her servant.

10 A possible compromise might be an Architectural Engineer.

1.5 Architectural Design Process

11 Architectural design is hierarchical:

Schematic: conceptual overall space-form feasibility of basic schematic options. Collaboration is mostlybetween the owner and the architect.

Preliminary: Establish basic physical properties of major subsystems and key components to provedesign feasibility. Some collaboration with engineers is necessary.

Final design: final in-depth design refinements of all subsystems and components and preparation ofworking documents (“blue-prints”). Engineers play a leading role.

1.6 Architectural Design

12 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 placesevere limitations on the structural system.

Economy: It should be kept in mind that the two largest components of a structure are labors andmaterials. Design cost is comparatively negligible.


Draft1.7 Structural Analysis 1–3

13 Buildings may have different functions:

Residential: housing, which includes low-rise (up tp 2-3 floors), mid-rise (up to 6-8 floors) and highrise buildings.

Commercial: Offices, retail stores, shopping centers, hotels, restaurants.

Industrial: warehouses, manufacturing.

Institutional: Schools, hospitals, prisons, chruch, government buildings.

Special: Towers, stadium, parking, airport, etc.

1.7 Structural Analysis

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

15 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

1.8 Structural Design

16 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 andsizes of reinforcing bars).

17 For new structures, iterative process between analysis and design. A preliminary design is madeusing rules of thumbs (best known to Engineers with design experience) and analyzed. Followingdesign, we check for

Serviceability: deflections, crack widths under the applied load. Compare with acceptable valuesspecified in the design code.

Failure (limit state): and compare the failure load with the applied load times the appropriate factorsof safety.

If the design is found not to be acceptable, then it must be modified and reanalyzed.

18 For existing structures rehabilitation, or verification of an old infrastructure, analysis is the mostimportant component.

19 In summary, analysis is always required.



Figure 1.1: Types of Forces in Structural Elements (1D)

1.9 Load Transfer Mechanisms

20 From Strength of Materials, loads can be transferred through various mechanisms, Fig. 1.1

Axial: cables, truss elements, arches, membrane, shells

Flexural: Beams, frames, grids, plates

Torsional: Grids, 3D frames

Shear: Frames, grids, shear walls.

1.10 Structure Types

21 Structures can be classified as follows:

Tension & Compression Structures: only, no shear, flexure, or torsion. Those are the most effi-cient types of structures.

Cable (tension only): The high strength of steel cables, combined with the efficiency of simpletension, makes cables ideal structural elements to span large distances such as bridges, anddish roofs, Fig. 1.2. A cable structure develops its load carrying capacity by adjusting itsshape so as to provide maximum resistance (form follows function). Care should be exercisedin minimizing large deflections and vibrations.

Arches (mostly compression) is a “reversed cable structure”. In an arch, flexure/shear is mini-mized and most of the load is transfered through axial forces only. Arches are used for largespan roofs and bridges, Fig. 1.3


Draft1.10 Structure Types 1–5

Figure 1.2: Basic Aspects of Cable Systems



Figure 1.3: Basic Aspects of Arches



Trusses have pin connected elements which can transmit axial forces only (tension and com-pression). Elements are connected by either slotted, screwed, or gusset plate connectors.However, due to construction details, there may be secondary stresses caused by relativelyrigid connections. Trusses are used for joists, roofs, bridges, electric tower, Fig. 1.4

Figure 1.4: Types of Trusses

Post and Beams: Essentially a support column on which a “beam” rests, Fig. 1.5, and 1.6.

Beams: Shear, flexure and sometimes axial forces. Recall that σ = McI is applicable only for shallow

beams, i.e. span/depth at least equal to five.

Whereas r/c beams are mostly rectangular or T shaped, steel beams are usually I shaped (if thetop 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 2Dstructures and with respect to two axis in 3D structures), torsion (only in 3D). The frame iscomposed of at least one horizontal member (beam) rigidly connected to vertical ones1. The vertical

1The precursor of the frame structures were the Post and Lintel where the post is vertical member on which the lintelis simply posed.



Figure 1.5: Variations in Post and Beams Configurations



OVERLAPPING SINGLE-STRUTCABLE-SUPPORTED BEAM

CABLE-STAYED BEAM

BRACED BEAM

VIERENDEEL TRUSS TREE-SUPPORTED TRUSS

CABLE-SUPPORTEDMULTI-STRUT

BEAM OR TRUSS

CABLE-SUPPORTED PORTAL FRAMECABLE-SUPPORTED ARCHED FRAME

SUSPENDED CABLESUPPORTED BEAM

CABLE-SUPPORTEDSTRUTED ARCH ORCABLE BEAM/TRUSS

GABLED TRUSS

BOWSTRING TRUSS

Figure 1.6: Different Beam Types



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 rigidconnections between the beams, additional stiffness is introduced by the torsional resistance ofmembers.

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 spanroofs. Note that the plate may be folded circularly rather than longitudinally. Folded plates areused mostly as long span roofs. However, they can also be used as vertical walls to support bothvertical and horizontal loads.

Membranes: 3D structures composed of a flexible 2D surface resisting tension only. They are usuallycable-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 compressiveaxial stresses only, Fig. 1.9.

Figure 1.9: Basic Forms of Shells

Shells are classified in terms of their curvature.

1.11 Structural Engineering Courses

22 Structural engineering education can be approached from either one of two points of views, dependingon the audience, ??.

Architects EngineersApproach Global ElementalEmphasis Structure ComponentAnalysis Approximate, “rules of thumbs” Exact, detailled

preliminary FinalStructures Most Trusses, FramesDesign Approximate Per code

Table 1.1: Structural Engineering Coverage for Architects and Engineers

Table 1.2: tab:secae


Draft1.12 References 1–13

Architects: Start from overall design, and move toward detailed analysis. Emphasis on good under-standing of overall structural behavior. Develop a good understanding of load transfer mechanismfor most types of structures, cables, arches, beams, frames, shells, plates. Approximate analysisfor most of them.

Engineers: Emphasis is on the individual structural elements and not always on the total system.Focus on beams, frames (mostly 2D) and trusses. Very seldom are arches covered. Plates andshells are not even mentioned.

1.12 References

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

Structures for Architect

1. Ambrose, J., Building Structures, second Ed. Wiley, 1993.

2. Billington, D.P. Rober Maillart’s Bridges; The Art of Engineering, Princeton University Pres,1979.

3. †Billington, D.P., The Tower and the Bridge; The new art of structural engineering, PrincetonUniversity Pres,, 1983.

4. †Billington, D.P., Structures and the Urban Environment, Lectures Notes CE 262, Departmentof Civil Engineering, Princeton University, 1978

5. French, S., Determinate Structures; Statics, Strength, Analysis, Design, Delmar, 1996.

6. Gordon, J.E., Structures, or Why Things Do’nt Fall Down, Da Capo paperback, New York,1978.

7. Gordon, J.E., The Science of Structures and Materials, Scientific American Library, 1988.

8. Hawkes, N., Structures, the way things are built, MacMillan, 1990.

9. Levy, M. and Salvadori, M., Why Buildings Fall Down, W.W.Norton, 1992.

10. †Lin, T.Y. and Stotesbury, S.D., Structural Concepts and Systems for Architects and Engi-neers, John Wiley, 1981.

11. †Mainstone, R., Developments in Structural Form, Allen Lane Publishers, 1975.

12. Petroski, H., To Enginer is Human, Vintage Books, 1992.

13. †Salvadori, M. and Heller, R., Structure in Architecture; The Building of Buildings, PrenticeHall, Third Edition, 1986.

14. Salvadori, M. and Levy, M., Structural Design in Architecture, Prentice hall, Second Edition,1981.

15. Salvadori, M., Why Buildings Stand Up; The Strength of Architecture, Norton Paperack, 1990.

16. †Sandaker, B.N. and Eggen, A.P., The Structural Basis of Architecture, Whitney Library ofDesign, 1992.

17. †Schueller, W., The design of Building Structures, Prentice Hall, 1996.

Structures for Engineers

1. † Arbadi, F. Structural Analysis and Behavior, McGraw-Hill, Inc., 1991.

2. Biggs, J.M., Introduction to Structural Engineering; Analysis and Design, Prentice Hall, 1986.



3. Hsieh, Y.Y., Elementary Theory of Structures, Third Edition, Prentice Hall, 1988.

4. Ghali, A., and Neville, A.M., Structural Analysis, Third Edition, Chapman and Hall, 1989

5. White, R. Gergely, P. and Sexmith, R., Structural Engineering; Combined Edition, JohnWiley, 1976.

6. † Nilson, A., and Winter, G. Design of Concrete Structures, Eleventh Edition, McGraw Hill,1991.

7. Galambos, T., Lin, F.J., and Johnston, B.G., Basic Steel Design with LRFD, Prentice Hall,1996.

8. † Salmon C. and Johnson, J. Steel Structures, Third Edition, Harper Collins Publisher, 1990.

9. † Gaylord, E.H., Gaylord, C.N. and Stallmeyer, J.E., Design of Steel Structures, Third Edi-tion, McGraw Hill, 1992.

10. Vitruvius, The Ten Books on Architecture, Dover Publications, 1960.

11. Palladio, A., The Four Books of Architecture, Dover Publication.

Codes

1. ACI-318-89, Building Code Requirements for Reinforced Concrete, American Concrete Insti-tute

2. Load & Resistance Factor Design, Manual of Steel Construction, American Institute of SteelConstruction.

3. Uniform Building Code, International Conference of Building Officials, 5360 South WorkmanRoad; Whittier, CA 90601

4. Minimum Design Loads in Buildings and Other Structures, ANSI A58.1, American NationalStandards Institute, Inc., New York, 1972.


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 settlementof foundations, P-Delta effects (additional moment caused by the product of the vertical force and thelateral 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 ofsafety 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 uniformlydistributed load rather than as discrete loads, Fig. 2.1

Draft2–2 LOADS

P P P P P P P1 2 3 4 5 6 7

TYPICAL SYSTEM OF JOISTS

SUPPORT BEAM

REPETITIVE JOIST LOADS

ACTUAL DISCRETE LOADS ON SUPPORT BEAM

ASSUMED EQUIVALENT UNIFORM LOAD

w LB/FT = TOTAL LOAD / SPAN

SPAN

Figure 2.1: Approximation of a Series of Closely Spaced Loads

2.2.1 Dead Load

7 Dead loads (DL) consist of the weight of the structure itself, and other permanent fixtures (such aswalls, slabs, machinery).

8 For analysis purposes, dead loads can easily be determined from the structure’s dimensions and density,Table 2.1

Material lb/ft3 kN/m3

Aluminum 173 27.2Brick 120 18.9Concrete 145 33.8Steel 490 77.0Wood (pine) 40 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 SteelConstruction.

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

11 Weights for building materials is given in Table 2.2

12 For preliminary design purposes the average dead loads of Table 2.3 can be used:

2.2.2 Live Loads

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

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

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

16 A statistical approach is used to determine a uniformly distributed static load which is equivalent tothe weight of the maximum concentration of occupants. These loads are defined in codes such as theUniform Building Code or the ANSI Code, Table 2.4.

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


Draft2.2 Vertical Loads 2–3

Material lb/ft2

CeilingsChannel suspended system 1Acoustical fiber tile 1

FloorsSteel deck 2-10Concrete-plain 1 in. 12Linoleum 1/4 in. 1Hardwood 4

RoofsCopper or tin 1-55 ply felt and gravel 6Shingles asphalt 3Clay tiles 9-14Sheathing wood 3Insulation 1 in. poured in place 2

PartitionsClay tile 3 in. 17Clay tile 10 in. 40Gypsum Block 5 in. 14Wood studs 2x4 (12-16 in. o.c.) 2Plaster 1 in. cement 10Plaster 1 in. gypsum 5

WallsBricks 4 in. 40Bricks 12 in. 120Hollow concrete block (heavy aggregate)4 in. 308 in. 5512 in. 80Hollow concrete block (light aggregate)4 in. 218 in. 3812 in. 55

Table 2.2: Weights of Building Materials

Material lb/ft2

Timber 40-50Steel 50-80Reinforced concrete 100-150

Table 2.3: Average Gross Dead Load in Buildings


Draft2–4 LOADS

Use or Occupancy lb/ft2

Assembly areas 50Cornices, marquees, residential balconies 60Corridors, stairs 100Garage 50Office buildings 50Residential 40Storage 125-250

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

18 Since there is a small probability that the whole floor in a building be fully loaded, the UBC codespecifies that the occupancy load for members supporting an area A larger than 150 ft2 (i.e. a columnwith a total tributary area, including floors above it, larger than 150 ft2) may be reduced by R where

R = r(A − 150) ≤ 23.1(

1 +DL

LL

)(2.1)

where r = .08 for floors, A is the supported area ( ft2) DL and LL are the dead and live loads per unitarea 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 loadis 60 psf, snow load 20 psf and live load 80 psf. what is the total live load and total load for which acolumn must be designed on the ground floorSolution:

1. The tributary area is 16 × 16 = 256ft2 > 150√

2. The reduction R for the roof is 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 Roof 10 9 8 7 6 5 4 3 2A 256 512 768 1024 1280 1536 1792 2048 2304 2560A − 150 106 362 618 874 1130 1386 1642 1898 2154 2410R′ 8.48 28.96 49.44 69.92 90.40 110.88 131.36 151.84 172.32 192.8R % 8.48 28.96 40.4 40.4 40.4 40.4 40.4 40.4 40.4 40.4LL 20 80 80 80 80 80 80 80 80 80(100 − R) × LL/100 18.3 56.83 47.68 47.68 47.68 47.68 47.68 47.68 47.68 47.68

The resulting design live load for the bottom column has been reduced from

LLBefore = (20) psf(256) ft2︸︷︷︸Roof

+ (9)(80) psf(256) ft2︸︷︷︸9 floors

= 189,440 lbs (2.2)


Draft2.3 Lateral Loads 2–5

toLLReduced = (18.3) psf(256) ft2︸︷︷︸

Roof

+ (9)(47.68) psf(256) ft2︸︷︷︸9 floors

= 114,540 lbs (2.3)

5. The total dead load is DL = (10)(60) psf(256) ft2 k(1,000) lbs = 153.6 Kips, thus the total reduction

in load is from 153.6+189.4 = 343 k to 153.6+114.5 = 268.1 k a reduction of 343−268343 ×100= 22% .

2.2.3 Snow

19 Roof snow load vary greatly depending on geographic location and elevation. They range from20 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.

21 The steeper the roof, the lower the snow retention. For snow loads greater than 20 psf and roof pitchesα more than 20◦ the snow load p may be reduced by

R = (α − 20)( p

40− 0.5

)(psf) (2.4)

2.3 Lateral Loads

2.3.1 Wind

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

23 Wind loads are particularly significant on tall buildings1.1The primary design consideration for very high rise buildings is the excessive drift caused by lateral load (wind and

possibly earthquakes).


Draft2–6 LOADS

LIVE LOAD

DEAD LOAD

LENGTH RIS

E

RUN

WINDLOAD

Figure 2.3: Loads on Projected Dimensions

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

qs =12ρV 2 (2.5)

where the air mass density ρ is the air weight divided by the accleration of gravity g = 32.2 ft/sec2. Atsea level and a temperature of 15oC (59oF), the ai weighs 0.0765 lb/ft3 this would yield a pressure of

qs =12

(0.0765)lb/ft3

(32.2)ft/sec2

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

V

)2

(2.6)

or

qs = 0.00256V 2 (2.7)

where V is the maximum wind velocity (in miles per hour) and qs is in psf. V can be obtained fromwind maps (in the United States 70 ≤ V ≤ 110), Fig. 2.4.

25 During storms, wind velocities may reach values up to or greater than 150 miles per hour, whichcorresponds to a dynamic pressure qs of about 60 psf (as high as the average vertical occupancy load inbuildings).

26 Wind pressure increases with height, Table 2.5.

27 Wind load will cause suction on the leeward sides, Fig. 2.6



Figure 2.4: Wind Map of the United States, (UBC 1995)

Height Zone Wind-Velocity Map Area(in feet) 20 25 30 35 40 45 50<30 15 20 25 25 30 35 4030 to 49 20 25 30 35 40 45 5050 to 99 25 30 40 45 50 55 60100 to 499 30 40 45 55 60 70 75500 to 1199 35 45 55 60 70 80 90>1,200 40 50 60 70 80 90 100

Table 2.5: Wind Velocity Variation above Ground


Draft2–8 LOADS

Figure 2.5: Effect of Wind Load on Structures(Schueller 1996)

28 This magnitude must be modified to account for the shape and surroundings of the building. Thus,the design base pressure (at 33.3 ft from the ground) p (psf) is given by

p = CeCqIqs (2.8)

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 thefact that wind velocity increases with height and that dynamic character of the airflow (i.e thewind pressure is not steady), Table 2.6. l

Ce Exposure1.39-2.34 D Open, flat terrain facing large bodies of water1.06-2.19 C Flat open terrain, extending one-half mile or open from the site in

any full quadrant0.62-1.80 B 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.

I Importance Factor as given by Table 2.8. where



Windward Side Leeward SideGabled Frames (V:H)

Roof Slope <9:12 −0.7 −0.79:12 to 12:12 0.4 −0.7>12:12 0.7 −0.7

Walls 0.8 −0.5Buildings (height < 200 ft)

Vertical Projections height < 40 ft 1.3 −1.3height > 40 ft 1.4 −1.4

Horizontal Projections −0.7 −0.7

Table 2.7: Wind Pressure Coefficients Cq, (UBC 1995)

Imprtance Factor IOccupancy Category Earthquake Wind

I Essential facilities 1.25 1.15II Hazardous facilities 1.25 1.15III Special occupancy structures 1.00 1.00IV Standard occupancy structures 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. communica-tion centers.

II Hazardous Facilities: Structures housing, supporting or containing sufficient quantities oftoxic or explosive substances to be dangerous to the safety of the general public if released.

III Special occupancy structure: Covered structures whose primary occupancy is public as-sembly, 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 aboveJails and detention facilitiesAll structures with occupancy >5,000 persons.Structures and equipment in power generating stations and other public utilitiy facilities notincluded above, and required for continued operation.

IV Standard occupancy structure: All structures having occupancies or functions not listedabove.

29 For the preliminary design of ordinary buildings Ce = 1.0 and Cq = 1.3 may be assumed, yielding

p = (1.3).020256V 2 = .00333V 2 (2.9)

which corresponds to a pressure of 21 psf for a wind speed of 80 mph, Fig. 2.6, Table 2.9.

Example 2-2: Wind Load


Draft2–10 LOADS

ExposureHeight B CAbove Basic Wind Speed (mph)Grade (ft) 70 80 70 800-15 10 13 17 2320 11 14 18 2425 12 15 19 2530 12 16 20 2640 14 18 21 2860 17 22 25 3380 18 24 27 35100 20 26 28 37120 21 28 29 38160 23 30 31 41200 25 33 33 43300 29 37 36 47400 32 41 38 50

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

0 5 10 15 20 25 30 35 40 45 50Approximate Design Wind Pressure (psf)

0

50

100

150

200

250

300

350

400

Hei

ght A

bove

Gra

de (

ft)

Exposure B, 70 mphExposure B, 80 mphExposure C, 70 mphExposure C, 80 mph

Figure 2.6: Approximate Design Wind Pressure p for Ordinary Wind Force Resisting Building Structures



Determine the wind forces on the building shown on below which is built in St Louis and is surroudedby trees.Solution:

1. From Fig. 2.4 the maximum wind velocity is St. Louis is 70 mph, since the building is protectedwe can take Ce = 0.7, I = 1.. The base wind pressure is qs = 0.00256 × (70)2 = 12.54 psf.

2. The slope of the roof is 8:15=6.4:12 which gives Cq = −0.7 for both the windward and leewardsides. The vertical walls have Cq = 0.8 for the winward side and Cq = −0.5 for the leeward one.

3. Thus the applied pressure on the roof is p = 0.7 × (−0.7) × 12.54 = -6.14 psf that is the roof issubjected 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 toresist all possible load combinations.

6. For large structures which may be subjected to large wind loads, testing in a wind tunnel of thestructure itself and its surroundings is often accomplished.

2.3.2 Earthquakes

30 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 nonstruc-tural damage


Draft2–12 LOADS

This is achieved through an appropriate dynamic analysis.

31 For preliminary designs or for small structures an equivalent horizontal static load can be deter-mined.

32 Actual loads depend on the following

1. Intensity of the ground acceleration (including soil/rock properties).

2. Dynamic properties of the building, such as its mode shapes and periods of vibration and itsdamping characteristics.

3. Mass of the building.

33 A critical factor in the dynamic response of a structure is the fundamental period of the structure’svibration (or first mode of vibration). This is the time required for one full cycle of motion, Fig. 2.7. Ifthe earthquake excitation has a frequency close to the one of the building, then resonance may occur.This should be avoided.

Figure 2.7: Vibrations of a Building

34 Earthquake load manifests itself as a horizontal force due to the (primarily) horizontal inertia force(F = ma).

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

V =ZIC

RWW (2.10)

where:

Z: Zone Factor: to be determined from Fig. 2.8 and Table 2.10.



Figure 2.8: Seismic Zones of the United States, (UBC 1995)

Seismic Zone 0 1 2A 2B 3 4Z 0 0.075 0.15 0.2 0.3 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 =1.25ST 2/3

≤ 2.75 (2.11)

T is the fundamental period of vibration of the building in seconds. This can be determined fromeither the free vibration analysis of the building, or estimated from the following empirical formula

T = Ct(hn)3/4 (2.12)

where:hn is the building height above base in ft.

andCt 0.035 steel moment resisting framesCt 0.030 reinforced concrete moment resisting frames and eccentrically braced framesCt 0.020 all other buildings

S: Site Coefficient given by Table 2.11 Note that most of the damages in the 1990? earthquakein San Francisco occurred in the marina where many houses were built on soft soil.

and

C

RW≥ 0.075 (2.13)


Draft2–14 LOADS

Type Description S FactorS1 A soil profile with either rock-like material or stiff/dense soil less

than 200 ft.1.0

S2 Dense or stiff soil exceeding 200 ft 1.2S3 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.1.5

S4 Soil containing more than 40 ft of soft clay 2.0

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

RW is given by Table 2.12.

W Load total structure load.

36 The horizontal force V is distributed over the height of the building in two parts. The first (appliedonly if T ≥ 0.7 sec.) is a concentrated force F1 equal to

Ft = 0.07TV ≤ 0.25V (2.14)

is applied at the top of the building due to whiplash. The balance of the force V − Ft is distributed asa triangular load diminishing to zero at the base.

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

Fx =(V − Ft)hx

h1 + h2 + · · · + hn=

(V − Ft)hx

Σni=1hi

(2.15)

where hi and hx are the height in ft above the base to level i, or x respectively. Note that it is assumedthat all floors have also same width.

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 DLplus 50%LL as the weight of each floor. The building is in zone 3.

Solution:



Structural System RW H (ft)Bearing wall system

Light-framed walls with shear panelsPlywood walls for structures three stories or less 8 65All other light-framed walls 6 65

Shear wallsConcrete 8 240Masonry 8 160

Building frame system using trussing or shear walls)

Steel eccentrically braced ductile frame 10 240Light-framed walls with shear panels

Plywood walls for structures three stories or less 9 65All other light-framed walls 7 65

Shear wallsConcrete 8 240Masonry 8 160

Concentrically braced framesSteel 8 160Concrete (only for zones I and 2) 8 -Heavy timber 8 65

Moment-resisting frame systemSpecial moment-resisting frames (SMRF)

Steel 12 N.L.Concrete 12 N.L.

Concrete intermediate moment-resisting frames (IMRF)only for zones 1 and 2 8 -Ordinary moment-resisting frames (OMRF)

Steel 6 160Concrete (only for zone 1) 5 -

Dual systems (selected cases are for ductile rigid frames only)Shear wallsConcrete with SMRF 12 N.L.Masonry with SMRF 8 160

Steel eccentrically braced ductile frame 6-12 160-N.L.Concentrically braced frame 12 N. L.

Steel with steel SMRF 10 N.L.Steel with steel OMRF 6 160Concrete with concrete SMRF (only for zones 1 and 2) 9 -

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


Draft2–16 LOADS

1. The fundamental period of vibration is

T = Ct(hn)3/4 = (0.030)(24)3/4 = 0.32 sec. (2.16)

2. The C coefficient is

C =1.25ST 2/3

=(1.25)(2.0)(0.32)2/3

= 5.344 > 2.75 (2.17)

use C = 2.75.

3. The other coefficients are: Z =0.3; I=1.25; RW =12

4. CheckC

RW=

2.7512

= 0.23 > 0.075√

(2.18)

5. The total vertical load is

W = 2 ((200 + 0.5(400)) (20) = 16000 lbs (2.19)

6. The total seismic base shear is

V =ZIC

RW=

(0.3)(1.25)(2.75)12

= 0.086W (2.20-a)

= (0.086)(16000) = 1375 lbs (2.20-b)

7. Since T < 0.7 sec. there is no whiplash.

8. The load on each floor is thus given by

F2 =(1375)(24)

12 + 24= 916.7 lbs (2.21-a)

F1 =(1375)(12)

12 + 24= 458.3 lbs (2.21-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 concretestructure in the short direction. The rigid frames are spaced 25 ft apart in the cross section and 20ft in the longitudinal direction. The plan dimension of the building is 175x100 ft, and the structure is25(12)=300 ft high. This office building is located in an urban environment with a wind velocity of 70mph 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

2638 k

1523 k

2(30

0)/3

=20

0’

300/

2=15

0’

25(1

2)=

300’

84000 k

3108 k

5(20)=100’

7(25

)=17

5’

Solution:

1. The total building weight is

W = (0.1926) ksf(100 × 175) ft2 × 25 storeys = 84, 000 k (2.22)

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.√

(2.23)

3. The C coefficient is

C =1.25ST 2/3

=(1.25)(1.5)(2.16)2/3

= 1.12 ≤ 2.75√

(2.24)

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

5. We checkC

RW=

1.1212

= 0.093 ≥ 0.075√

(2.25)

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

V =ZIC

RWW =

(0.4)(1)(1.12)(12)

W = 0.037W (2.26-a)

= (0.037)(84000) = 3108 kip (2.26-b)

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

Ft = 0.07TV = (0.07)(2.16)(3108) = 470 k ¾ (2.27-a)le 0.25V = (0.25)(3108) = 777 k (2.27-b)


Draft2–18 LOADS

Hence the total triangular load is

V − Ft = 3108 − 470 = 2638 k (2.28)

8. let us check if wind load governs. From Table xx we conservatively assume a uniform wind pressureof 29 psf resulting in a total lateral force of

PW = (0.029) psf(175 × 300) ft2 = 1523 k < 3108 k (2.29)

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

2.4 Other Loads

2.4.1 Hydrostatic and Earth

38 Structures below ground must resist lateral earth pressure.

q = Kγh (2.30)

where γ is the soil density, K = 1−sin Φ1+sin Φ is the pressure coefficient, h is the height.

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

40 If the structure is partially submerged, it must also resist hydrostatic pressure of water, Fig. 2.9.

Figure 2.9: Earth and Hydrostatic Loads on Structures

q = γW h (2.31)

where γW = 62.4 lbs/ft3.

Example 2-5: Hydrostatic Load


Draft2.4 Other Loads 2–19

The basement of a building is 12 ft below grade. Ground water is located 9 ft below grade, whatthickness 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

=

(150) lbs/ft3 × h︸︷︷︸concrete

⇒ h = (62.4) lbs/ft3

(150) lbs/ft3(3) ft(12) in/ft = 14.976 in ' 15.0 inch

2.4.2 Thermal

41 If a member is uniformly heated (or cooled) without restraint, then it will expand (or contract).This expansion is given by

∆l = αl∆T (2.32)

where α is the coefficient of thermal expansion, Table 2.13

α (/◦F )Steel 6.5 × 10−6

Concrete 5.5 × 10−6

Table 2.13: Coefficients of Thermal Expansion

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

43 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 masonary (α = 3.6 × 10−6

in./in./oF, E = 2, 400, 000 psi) bearing wall. The structure was built at a temperature of 60oF andis located in the northern part of the United States where the temperature range is between -20o and+120oF.Solution:

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

∆LSummer = α∆TL = (3.6 × 10−6) in/ in/oF (120 − 60)oF (100) ft(12) in/ft = 0.26 in(2.33-a)

∆LWinter = α∆TL = (3.6 × 10−6) in/ in/oF (−20 − 60)oF (100) ft(12) in/ft = -0.35 in(2.33-b)


Draft2–20 LOADS

2. We now assume (conservatively) that the free movement cannot occur (∆L = 0) hence the resultingstress would be equal to σ = Eε = E ∆L

L = E α∆TLL = Eα∆T

σSummer = Eα∆T = (2, 400, 000)lbs

in2 (3.6 × 10−6) in/ in/oF (120 − 60)oF = 518lbs

in2 Tension(2.34-a)

σWinter = Eα∆T = (2, 400, 000)lbs

in2 (3.6 × 10−6) in/ in/oF (−20 − 60)oF = -691lbs

in2 Compression(2.34-b)

(2.34-c)

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

2.5 Other Important Considerations

2.5.1 Load Combinations

44 Live loads specified by codes represent the maximum possible loads.

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

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

47 Denoting D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature;H= soil:

48 For the load and resistance factor design (LRFD) method of concrete structures, the American Con-crete Institute (ACI) Building design code (318) (318 n.d.) requires that the following load combinationsbe considered:

1. 1.4D+1.7L

2. 0.75(1.4D+1.7L+1.7W)

3. 0.9D+1.3W

4. 1.4D +1.7L+1.7H

5. 0.75(1.4D+1.4T+1.7L)

6. 1.4(D+T)

whereas for steel structures, the American Institute of Steel Construction (AISC) code, (of Steel COn-struction 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


Draft2.5 Other Important Considerations 2–21

6. 0.9D+1.3W(or 1.5 E)

49 Analysis can be separately performed for each of the basic loads (L, D, W, etc) and then using theprinciple of superposition the loads can be linearly combined (unless the elastic limit has been reached).

50 Loads are often characterized as Usual, Unusual and Extreme.

2.5.2 Load Placement

51 Only the dead load is static. The live load on the other hand may or may not be applied on a givencomponent 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. 2.10.

Figure 2.10: Load Placement to Maximize Moments

2.5.3 Load Transfer

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

1. The section is part of a beam or girder.

2. The beam or girder is really part of a three dimensional structure in which load is transmittedfrom any point in the structure to the foundation through any one of various structural forms.

53 Load transfer in a structure is accomplished through a “hierarchy” of simple flexural elements whichare then connected to the columns, Fig. 2.11 or by two way slabs as illustrated in Fig. 2.12.

2.5.4 Structural Response

54 Under the action of the various forces and loadings described above, the structure must be able torespond with proper behavior, Fig. 9.1.


Draft2–22 LOADS

Figure 2.11: Load Transfer in R/C Buildings



Figure 2.12: Two Way Actions


Draft2–24 LOADS

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



2.5.5 Tributary Areas

55 For preliminary analyses, the tributary area of a structural component will determine the total appliedload.

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Figure 2.14: Concept of Tributary Areas for Structual Member Loading


Draft2–26 LOADS


Draft

Chapter 3

STRUCTURAL MATERIALS

1 Proper understanding of structural materials is essential to both structural analysis and to structuraldesign.

2 Characteristics of the most commonly used structural materials will be highlighted.

3.1 Steel

3.1.1 Structural Steel

3 Steel is an alloy of iron and carbon. Its properties can be greatly varied by altering the carbon content(always less than 0.5%) or by adding other elements such as silicon, nickle, manganese and copper.

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

5 The yield stress of steel can vary from 40 ksi to 250 ksi. Most commonly used structural steel are A36(σyld = 36 ksi) and A572 (σyld = 50 ksi), Fig. 3.1

6 Structural steel can be rolled into a wide variety of shapes and sizes. Usually the most desirablemembers are those which have a large section moduli (S) in proportion to their area (A), Fig. 3.2.

7 Steel can be bolted, riveted or welded.

8 Sections are designated by the shape of their cross section, their depth and their weight. For exampleW 27× 114 is a W section, 27 in. deep weighing 114 lb/ft.

9 Common sections are:

S sections were the first ones rolled in America and have a slope on their inside flange surfaces of 1 to6.

W or wide flange sections have a much smaller inner slope which facilitates connections and rivetting.W sections constitute about 50% of the tonnage of rolled structural steel.

C are channel sections

MC Miscellaneous channel which can not be classified as a C shape by dimensions.

HP is a bearing pile section.

Draft3–2 STRUCTURAL MATERIALS

Figure 3.1: Stress Strain Curves of Concrete and Steel

Figure 3.2: Standard Rolled Sections


Draft3.1 Steel 3–3

M is a miscellaneous section.

L are angle sections which may have equal or unequal sides.

WT is a T section cut from a W section in two.

10 The section modulus Sx of a W section can be roughly approximated by the following formula

Sx ≈ wd/10 or Ix ≈ Sxd

2≈ wd2/20 (3.1)

and the plastic modulus can be approximated by

Zx ≈ wd/9 (3.2)

11 Properties of structural steel are tabulated in Table 3.1.

ASTMDesig.

Shapes Available Use σy (kksi) σu (kksi)

A36 Shapes and bars Riveted, bolted, welded;Buildings and bridges

36 up through 8 in. (32 above8.)

A500 Cold formed welded andseamless sections;

General structural pur-pose Riveted, welded orbolted;

Grade A: 33; Grade B: 42;Grade C: 46

A501 Hot formed welded and seam-less sections;

Bolted and welded 36

A529 Plates and bars 12

in and lessthick;

Building frames andtrusses; Bolted andwelded

42

A606 Hot and cold rolled sheets; Atmospheric corrosionresistant

45-50

A611 Cold rolled sheet in cutlengths

Cold formed sections Grade C 33; Grade D 40;Grade E 80

A 709 Structural shapes, plates andbars

Bridges Grade 36: 36 (to 4 in.); Grade50: 50; Grade 100: 100 (to2.5in.) and 90 (over 2.5 to 4in.)

Table 3.1: Properties of Major Structural Steels

12 Rolled sections, Fig. 3.3 and welded ones, Fig3.4 have residual stresses. Those originate during therolling or fabrication of a member. The member is hot just after rolling or welding, it cools unevenlybecause of varying exposure. The area that cool first become stiffer, resist contraction, and developcompressive stresses. The remaining regions continue to cool and contract in the plastic condition anddevelop tensile stresses.

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



Compression (-)

Tension (+)(+)

(-)

Maximum compressive

stress, say 12 ksi average

Figure 3.3: Residual Stresses in Rolled Sections

-

+ +

+

_

+

+

-

-

say 12 ksi

20 ksi

Welded H

say 35 ksitension

say 20 ksi

say 40 ksi

Welded box

say 20 ksicompression

Figure 3.4: Residual Stresses in Welded Sections


Draft3.1 Steel 3–5

.

..

Maximumresidualcompressivestress

no residual stressIdeal coupon containing

Members withresidual stress

F

F

y

p

2

1

3

1

2Average copressive strain

Ave

rage

str

ess

P/A

Shaded portion indicates arwhich has achieved a stress

Figure 3.5: Influence of Residual Stress on Average Stress-Strain Curve of a Rolled Section

3.1.2 Reinforcing Steel

14 Steel is also used as reinforcing bars in concrete, Table 3.2. Those bars have a deformation on theirsurface to increase the bond with concrete, and usually have a yield stress of 60 ksi1.

Bar Designation Diameter Area Perimeter Weight(in.) ( in2) in lb/ft

No. 2 2/8=0.250 0.05 0.79 0.167No. 3 3/8=0.375 0.11 1.18 0.376No. 4 4/8=0.500 0.20 1.57 0.668No. 5 5/8=0.625 0.31 1.96 1.043No. 6 6/8=0.750 0.44 2.36 1.5202No. 7 7/8=0.875 0.60 2.75 2.044No. 8 8/8=1.000 0.79 3.14 2.670No. 9 9/8=1.128 1.00 3.54 3.400No. 10 10/8=1.270 1.27 3.99 4.303No. 11 11/8=1.410 1.56 4.43 5.313No. 14 14/8=1.693 2.25 5.32 7.650No. 18 18/8=2.257 4.00 7.09 13.60

Table 3.2: Properties of Reinforcing Bars

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

16 Steel is also used as wire strands and ropes for suspended roofs, cable-stayed bridges, fabric roofs andother structural applications. A strand is a helical arrangement of wires around a central wire. A ropeconsists of multiple strands helically wound around a central plastic core, and a modulus of elasticity of20,000 ksi, and an ultimate strength of 220 ksi.

17 Prestressing Steel cables have an ultimate strength up to 270 ksi.1Stirrups which are used as vertical reinforcement to resist shear usually have a yield stress of only 40 ksi.



3.2 Aluminum

18 Aluminum is used whenever light weight combined with strength is an important factor. Thoseproperties, along with its resistance to corrosion have made it the material of choice for airplanestructures, light roof framing.

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

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

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

22 When aluminum is in contact with other metals in the presence of an electrolyte, galvanic corrosionmay cause damage. Thus, steel and aluminum in a structure must be carefully separated by means ofpainting or a nonconductive material.

3.3 Concrete

23 Concrete is a mixture of Portland cement2, water, and aggregates (usually sand and crushed stone).An ideal mixture is one in which:

1. A minimum amount of cement-water paste is used to fill the interstices between the particles ofaggregates.

2. A minimum amount of water is provided to complete the chemical reaction with cement.

In such a mixture, about 3/4 of the volume is constituted by the aggregates, and the remaining 1/4being the cement paste.

24 Smaller particles up to 1/4 in. in size are called fine aggregates, and the larger ones being coarseaggregates.

25 Contrarily to steel to modulus of elasticity of concrete depends on the strength and is given by

E = 57, 000√

f ′c (3.3)

orE = 33γ1.5

√f ′

c (3.4)

where both f ′c and E are in psi and γ is in lbs/ft3.

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

27 All concrete fail at an ultimate strain of 0.003, Fig. 3.1.

28 Pre-peak nonlinearity is caused by micro-cracking Fig. 3.6.

29 The tensile strength of concrete f ′t is about 10% of the compressive strength.

30 Density of normal weight concrete is 145 lbs/ft3 and 100 lbs/ft3 for lightweight concrete.2Portland cement is a mixture of calcareous and argillaceous materials which are calcined in a kiln and then pulverized.

When mixed with water, cement hardens through a process called hydration.


Draft3.4 Masonry 3–7

εu

f’c

.5f’c

linear

non-linear

Figure 3.6: Concrete microcracking

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

32 When concrete is poured (or rather placed), the free water not needed for the hydration processevaporates over a period of time and the concrete will shrink. This shrinkage is about 0.05% after oneyear (strain). Thus if the concrete is restrained, then cracking will occur3.

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

3.4 Masonry

34 Masonry consists of either natural materials, such as stones, or of manufactured products such asbricks and concrete blocks4, stacked and bonded together with mortar.

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

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

3.5 Timber

37 Timber is one of the earliest construction materials, and one of the few natural materials with goodtensile properties.

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

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

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

41 Members can be laminated together under good quality control, and flexural strengths as high as2,500 psi can be achieved.

3For this reason a minimum amount of reinforcement is always necessary in concrete, and a 2% reinforcement, canreduce the shrinkage by 75%.

4Mud bricks were used by the Babylonians, stones by the Egyptians, and ice blocks by the Eskimos...



3.6 Steel Section Properties

42 Dimensions and properties of rolled sections are tabulated in the following pages, Fig. 3.7.

Figure 3.7: W and C sections

==============


Draft3.6 Steel Section Properties 3–9

Designation A dbf

2tf

hctw

Ix Sx Iy Sy Zx Zy

in2 in in4 in3 in4 in3 in3 in3

W 36x848 249.0 42.45 2.0 12.5 67400 3170 4550 501 3830.0 799.0W 36x798 234.0 41.97 2.1 13.2 62600 2980 4200 467 3570.0 743.0W 36x720 211.0 41.19 2.3 14.5 55300 2690 3680 414 3190.0 656.0W 36x650 190.0 40.47 2.5 16.0 48900 2420 3230 367 2840.0 580.0W 36x588 172.0 39.84 2.7 17.6 43500 2180 2850 328 2550.0 517.0W 36x527 154.0 39.21 3.0 19.6 38300 1950 2490 289 2270.0 454.0W 36x485 142.0 38.74 3.2 21.0 34700 1790 2250 263 2070.0 412.0W 36x439 128.0 38.26 3.5 23.1 31000 1620 1990 235 1860.0 367.0W 36x393 115.0 37.80 3.8 25.8 27500 1450 1750 208 1660.0 325.0W 36x359 105.0 37.40 4.2 28.1 24800 1320 1570 188 1510.0 292.0W 36x328 96.4 37.09 4.5 30.9 22500 1210 1420 171 1380.0 265.0W 36x300 88.3 36.74 5.0 33.3 20300 1110 1300 156 1260.0 241.0W 36x280 82.4 36.52 5.3 35.6 18900 1030 1200 144 1170.0 223.0W 36x260 76.5 36.26 5.7 37.5 17300 953 1090 132 1080.0 204.0W 36x245 72.1 36.08 6.1 39.4 16100 895 1010 123 1010.0 190.0W 36x230 67.6 35.90 6.5 41.4 15000 837 940 114 943.0 176.0W 36x256 75.4 37.43 3.5 33.8 16800 895 528 86 1040.0 137.0W 36x232 68.1 37.12 3.9 37.3 15000 809 468 77 936.0 122.0W 36x210 61.8 36.69 4.5 39.1 13200 719 411 68 833.0 107.0W 36x194 57.0 36.49 4.8 42.4 12100 664 375 62 767.0 97.7W 36x182 53.6 36.33 5.1 44.8 11300 623 347 58 718.0 90.7W 36x170 50.0 36.17 5.5 47.8 10500 580 320 53 668.0 83.8W 36x160 47.0 36.01 5.9 50.0 9750 542 295 49 624.0 77.3W 36x150 44.2 35.85 6.4 52.0 9040 504 270 45 581.0 70.9W 36x135 39.7 35.55 7.6 54.1 7800 439 225 38 509.0 59.7W 33x619 181.0 38.47 2.4 15.2 41800 2170 2870 340 2560.0 537.0W 33x567 166.0 37.91 2.6 16.6 37700 1990 2580 308 2330.0 485.0W 33x515 151.0 37.36 2.8 18.2 33700 1810 2290 276 2110.0 433.0W 33x468 137.0 36.81 3.0 19.7 30100 1630 2030 247 1890.0 387.0W 33x424 124.0 36.34 3.3 21.7 26900 1480 1800 221 1700.0 345.0W 33x387 113.0 35.95 3.6 23.8 24300 1350 1620 200 1550.0 312.0W 33x354 104.0 35.55 3.8 25.8 21900 1230 1460 181 1420.0 282.0W 33x318 93.5 35.16 4.2 28.8 19500 1110 1290 161 1270.0 250.0W 33x291 85.6 34.84 4.6 31.2 17700 1010 1160 146 1150.0 226.0W 33x263 77.4 34.53 5.0 34.5 15800 917 1030 131 1040.0 202.0W 33x241 70.9 34.18 5.7 36.1 14200 829 932 118 939.0 182.0W 33x221 65.0 33.93 6.2 38.7 12800 757 840 106 855.0 164.0W 33x201 59.1 33.68 6.8 41.9 11500 684 749 95 772.0 147.0W 33x169 49.5 33.82 4.7 44.7 9290 549 310 54 629.0 84.4W 33x152 44.7 33.49 5.5 47.2 8160 487 273 47 559.0 73.9W 33x141 41.6 33.30 6.0 49.6 7450 448 246 43 514.0 66.9W 33x130 38.3 33.09 6.7 51.7 6710 406 218 38 467.0 59.5W 33x118 34.7 32.86 7.8 54.5 5900 359 187 33 415.0 51.3W 30x581 170.0 35.39 2.3 13.7 33000 1870 2530 312 2210.0 492.0W 30x526 154.0 34.76 2.5 15.1 29300 1680 2230 278 1990.0 438.0W 30x477 140.0 34.21 2.7 16.6 26100 1530 1970 249 1790.0 390.0W 30x433 127.0 33.66 2.9 18.0 23200 1380 1750 222 1610.0 348.0W 30x391 114.0 33.19 3.2 19.9 20700 1250 1550 198 1430.0 310.0W 30x357 104.0 32.80 3.5 21.8 18600 1140 1390 179 1300.0 279.0W 30x326 95.7 32.40 3.7 23.7 16800 1030 1240 162 1190.0 252.0W 30x292 85.7 32.01 4.1 26.5 14900 928 1100 144 1060.0 223.0W 30x261 76.7 31.61 4.6 29.0 13100 827 959 127 941.0 196.0W 30x235 69.0 31.30 5.0 32.5 11700 746 855 114 845.0 175.0W 30x211 62.0 30.94 5.7 34.9 10300 663 757 100 749.0 154.0W 30x191 56.1 30.68 6.3 38.0 9170 598 673 90 673.0 138.0W 30x173 50.8 30.44 7.0 41.2 8200 539 598 80 605.0 123.0W 30x148 43.5 30.67 4.4 41.5 6680 436 227 43 500.0 68.0W 30x132 38.9 30.31 5.3 43.9 5770 380 196 37 437.0 58.4W 30x124 36.5 30.17 5.7 46.2 5360 355 181 34 408.0 54.0W 30x116 34.2 30.01 6.2 47.8 4930 329 164 31 378.0 49.2W 30x108 31.7 29.83 6.9 49.6 4470 299 146 28 346.0 43.9W 30x 99 29.1 29.65 7.8 51.9 3990 269 128 24 312.0 38.6W 30x 90 26.4 29.53 8.5 57.5 3620 245 115 22 283.0 34.7



Designation A dbf

2tf

hctw

Ix Sx Iy Sy Zx Zy


W 27x539 158.0 32.52 2.2 12.3 25500 1570 2110 277 1880.0 437.0W 27x494 145.0 31.97 2.3 13.4 22900 1440 1890 250 1710.0 394.0W 27x448 131.0 31.42 2.5 14.7 20400 1300 1670 224 1530.0 351.0W 27x407 119.0 30.87 2.7 15.9 18100 1170 1480 200 1380.0 313.0W 27x368 108.0 30.39 3.0 17.6 16100 1060 1310 179 1240.0 279.0W 27x336 98.7 30.00 3.2 19.2 14500 970 1170 161 1130.0 252.0W 27x307 90.2 29.61 3.5 20.9 13100 884 1050 146 1020.0 227.0W 27x281 82.6 29.29 3.7 22.9 11900 811 953 133 933.0 206.0W 27x258 75.7 28.98 4.0 24.7 10800 742 859 120 850.0 187.0W 27x235 69.1 28.66 4.4 26.6 9660 674 768 108 769.0 168.0W 27x217 63.8 28.43 4.7 29.2 8870 624 704 100 708.0 154.0W 27x194 57.0 28.11 5.2 32.3 7820 556 618 88 628.0 136.0W 27x178 52.3 27.81 5.9 33.4 6990 502 555 79 567.0 122.0W 27x161 47.4 27.59 6.5 36.7 6280 455 497 71 512.0 109.0W 27x146 42.9 27.38 7.2 40.0 5630 411 443 64 461.0 97.5W 27x129 37.8 27.63 4.5 39.7 4760 345 184 37 395.0 57.6W 27x114 33.5 27.29 5.4 42.5 4090 299 159 32 343.0 49.3W 27x102 30.0 27.09 6.0 47.0 3620 267 139 28 305.0 43.4W 27x 94 27.7 26.92 6.7 49.4 3270 243 124 25 278.0 38.8W 27x 84 24.8 26.71 7.8 52.7 2850 213 106 21 244.0 33.2W 24x492 144.0 29.65 2.0 10.9 19100 1290 1670 237 1550.0 375.0W 24x450 132.0 29.09 2.1 11.9 17100 1170 1490 214 1410.0 337.0W 24x408 119.0 28.54 2.3 13.1 15100 1060 1320 191 1250.0 300.0W 24x370 108.0 27.99 2.5 14.2 13400 957 1160 170 1120.0 267.0W 24x335 98.4 27.52 2.7 15.6 11900 864 1030 152 1020.0 238.0W 24x306 89.8 27.13 2.9 17.1 10700 789 919 137 922.0 214.0W 24x279 82.0 26.73 3.2 18.6 9600 718 823 124 835.0 193.0W 24x250 73.5 26.34 3.5 20.7 8490 644 724 110 744.0 171.0W 24x229 67.2 26.02 3.8 22.5 7650 588 651 99 676.0 154.0W 24x207 60.7 25.71 4.1 24.8 6820 531 578 89 606.0 137.0W 24x192 56.3 25.47 4.4 26.6 6260 491 530 82 559.0 126.0W 24x176 51.7 25.24 4.8 28.7 5680 450 479 74 511.0 115.0W 24x162 47.7 25.00 5.3 30.6 5170 414 443 68 468.0 105.0W 24x146 43.0 24.74 5.9 33.2 4580 371 391 60 418.0 93.2W 24x131 38.5 24.48 6.7 35.6 4020 329 340 53 370.0 81.5W 24x117 34.4 24.26 7.5 39.2 3540 291 297 46 327.0 71.4W 24x104 30.6 24.06 8.5 43.1 3100 258 259 41 289.0 62.4W 24x103 30.3 24.53 4.6 39.2 3000 245 119 26 280.0 41.5W 24x 94 27.7 24.31 5.2 41.9 2700 222 109 24 254.0 37.5W 24x 84 24.7 24.10 5.9 45.9 2370 196 94 21 224.0 32.6W 24x 76 22.4 23.92 6.6 49.0 2100 176 82 18 200.0 28.6W 24x 68 20.1 23.73 7.7 52.0 1830 154 70 16 177.0 24.5W 24x 62 18.2 23.74 6.0 50.1 1550 131 34 10 153.0 15.7W 24x 55 16.2 23.57 6.9 54.6 1350 114 29 8 134.0 13.3W 21x402 118.0 26.02 2.1 10.8 12200 937 1270 189 1130.0 296.0W 21x364 107.0 25.47 2.3 11.8 10800 846 1120 168 1010.0 263.0W 21x333 97.9 25.00 2.5 12.8 9610 769 994 151 915.0 237.0W 21x300 88.2 24.53 2.7 14.2 8480 692 873 134 816.0 210.0W 21x275 80.8 24.13 2.9 15.4 7620 632 785 122 741.0 189.0W 21x248 72.8 23.74 3.2 17.1 6760 569 694 109 663.0 169.0W 21x223 65.4 23.35 3.5 18.8 5950 510 609 96 589.0 149.0W 21x201 59.2 23.03 3.9 20.6 5310 461 542 86 530.0 133.0W 21x182 53.6 22.72 4.2 22.6 4730 417 483 77 476.0 119.0W 21x166 48.8 22.48 4.6 24.9 4280 380 435 70 432.0 108.0W 21x147 43.2 22.06 5.4 26.1 3630 329 376 60 373.0 92.6W 21x132 38.8 21.83 6.0 28.9 3220 295 333 54 333.0 82.3W 21x122 35.9 21.68 6.5 31.3 2960 273 305 49 307.0 75.6W 21x111 32.7 21.51 7.1 34.1 2670 249 274 44 279.0 68.2W 21x101 29.8 21.36 7.7 37.5 2420 227 248 40 253.0 61.7W 21x 93 27.3 21.62 4.5 32.3 2070 192 93 22 221.0 34.7W 21x 83 24.3 21.43 5.0 36.4 1830 171 81 20 196.0 30.5W 21x 73 21.5 21.24 5.6 41.2 1600 151 71 17 172.0 26.6



Designation A dbf

2tf

hctw

Ix Sx Iy Sy Zx Zy


W 21x 68 20.0 21.13 6.0 43.6 1480 140 65 16 160.0 24.4W 21x 62 18.3 20.99 6.7 46.9 1330 127 58 14 144.0 21.7W 21x 57 16.7 21.06 5.0 46.3 1170 111 31 9 129.0 14.8W 21x 50 14.7 20.83 6.1 49.4 984 94 25 8 110.0 12.2W 21x 44 13.0 20.66 7.2 53.6 843 82 21 6 95.4 10.2W 18x311 91.5 22.32 2.2 10.6 6960 624 795 132 753.0 207.0W 18x283 83.2 21.85 2.4 11.5 6160 564 704 118 676.0 185.0W 18x258 75.9 21.46 2.6 12.5 5510 514 628 107 611.0 166.0W 18x234 68.8 21.06 2.8 13.8 4900 466 558 96 549.0 149.0W 18x211 62.1 20.67 3.0 15.1 4330 419 493 85 490.0 132.0W 18x192 56.4 20.35 3.3 16.7 3870 380 440 77 442.0 119.0W 18x175 51.3 20.04 3.6 18.0 3450 344 391 69 398.0 106.0W 18x158 46.3 19.72 3.9 19.8 3060 310 347 61 356.0 94.8W 18x143 42.1 19.49 4.2 21.9 2750 282 311 56 322.0 85.4W 18x130 38.2 19.25 4.6 23.9 2460 256 278 50 291.0 76.7W 18x119 35.1 18.97 5.3 24.5 2190 231 253 45 261.0 69.1W 18x106 31.1 18.73 6.0 27.2 1910 204 220 39 230.0 60.5W 18x 97 28.5 18.59 6.4 30.0 1750 188 201 36 211.0 55.3W 18x 86 25.3 18.39 7.2 33.4 1530 166 175 32 186.0 48.4W 18x 76 22.3 18.21 8.1 37.8 1330 146 152 28 163.0 42.2W 18x 71 20.8 18.47 4.7 32.4 1170 127 60 16 145.0 24.7W 18x 65 19.1 18.35 5.1 35.7 1070 117 55 14 133.0 22.5W 18x 60 17.6 18.24 5.4 38.7 984 108 50 13 123.0 20.6W 18x 55 16.2 18.11 6.0 41.2 890 98 45 12 112.0 18.5W 18x 50 14.7 17.99 6.6 45.2 800 89 40 11 101.0 16.6W 18x 46 13.5 18.06 5.0 44.6 712 79 22 7 90.7 11.7W 18x 40 11.8 17.90 5.7 51.0 612 68 19 6 78.4 9.9W 18x 35 10.3 17.70 7.1 53.5 510 58 15 5 66.5 8.1W 16x100 29.4 16.97 5.3 24.3 1490 175 186 36 198.0 54.9W 16x 89 26.2 16.75 5.9 27.0 1300 155 163 31 175.0 48.1W 16x 77 22.6 16.52 6.8 31.2 1110 134 138 27 150.0 41.1W 16x 67 19.7 16.33 7.7 35.9 954 117 119 23 130.0 35.5W 16x 57 16.8 16.43 5.0 33.0 758 92 43 12 105.0 18.9W 16x 50 14.7 16.26 5.6 37.4 659 81 37 10 92.0 16.3W 16x 45 13.3 16.13 6.2 41.2 586 73 33 9 82.3 14.5W 16x 40 11.8 16.01 6.9 46.6 518 65 29 8 72.9 12.7W 16x 36 10.6 15.86 8.1 48.1 448 56 24 7 64.0 10.8W 16x 31 9.1 15.88 6.3 51.6 375 47 12 4 54.0 7.0W 16x 26 7.7 15.69 8.0 56.8 301 38 10 3 44.2 5.5W 14x730 215.0 22.42 1.8 3.7 14300 1280 4720 527 1660.0 816.0W 14x665 196.0 21.64 2.0 4.0 12400 1150 4170 472 1480.0 730.0W 14x605 178.0 20.92 2.1 4.4 10800 1040 3680 423 1320.0 652.0W 14x550 162.0 20.24 2.3 4.8 9430 931 3250 378 1180.0 583.0W 14x500 147.0 19.60 2.4 5.2 8210 838 2880 339 1050.0 522.0W 14x455 134.0 19.02 2.6 5.7 7190 756 2560 304 936.0 468.0W 14x426 125.0 18.67 2.8 6.1 6600 707 2360 283 869.0 434.0W 14x398 117.0 18.29 2.9 6.4 6000 656 2170 262 801.0 402.0W 14x370 109.0 17.92 3.1 6.9 5440 607 1990 241 736.0 370.0W 14x342 101.0 17.54 3.3 7.4 4900 559 1810 221 672.0 338.0W 14x311 91.4 17.12 3.6 8.1 4330 506 1610 199 603.0 304.0W 14x283 83.3 16.74 3.9 8.8 3840 459 1440 179 542.0 274.0W 14x257 75.6 16.38 4.2 9.7 3400 415 1290 161 487.0 246.0W 14x233 68.5 16.04 4.6 10.7 3010 375 1150 145 436.0 221.0W 14x211 62.0 15.72 5.1 11.6 2660 338 1030 130 390.0 198.0W 14x193 56.8 15.48 5.5 12.8 2400 310 931 119 355.0 180.0W 14x176 51.8 15.22 6.0 13.7 2140 281 838 107 320.0 163.0W 14x159 46.7 14.98 6.5 15.3 1900 254 748 96 287.0 146.0W 14x145 42.7 14.78 7.1 16.8 1710 232 677 87 260.0 133.0W 14x132 38.8 14.66 7.1 17.7 1530 209 548 74 234.0 113.0W 14x120 35.3 14.48 7.8 19.3 1380 190 495 68 212.0 102.0W 14x109 32.0 14.32 8.5 21.7 1240 173 447 61 192.0 92.7W 14x 99 29.1 14.16 9.3 23.5 1110 157 402 55 173.0 83.6W 14x 90 26.5 14.02 10.2 25.9 999 143 362 50 157.0 75.6W 14x 82 24.1 14.31 5.9 22.4 882 123 148 29 139.0 44.8



Designation A dbf

2tf

hctw

Ix Sx Iy Sy Zx Zy


W 14x 74 21.8 14.17 6.4 25.3 796 112 134 27 126.0 40.6W 14x 68 20.0 14.04 7.0 27.5 723 103 121 24 115.0 36.9W 14x 61 17.9 13.89 7.7 30.4 640 92 107 22 102.0 32.8W 14x 53 15.6 13.92 6.1 30.8 541 78 58 14 87.1 22.0W 14x 48 14.1 13.79 6.7 33.5 485 70 51 13 78.4 19.6W 14x 43 12.6 13.66 7.5 37.4 428 63 45 11 69.6 17.3W 14x 38 11.2 14.10 6.6 39.6 385 55 27 8 61.5 12.1W 14x 34 10.0 13.98 7.4 43.1 340 49 23 7 54.6 10.6W 14x 30 8.9 13.84 8.7 45.4 291 42 20 6 47.3 9.0W 14x 26 7.7 13.91 6.0 48.1 245 35 9 4 40.2 5.5W 14x 22 6.5 13.74 7.5 53.3 199 29 7 3 33.2 4.4W 12x336 98.8 16.82 2.3 5.5 4060 483 1190 177 603.0 274.0W 12x305 89.6 16.32 2.4 6.0 3550 435 1050 159 537.0 244.0W 12x279 81.9 15.85 2.7 6.3 3110 393 937 143 481.0 220.0W 12x252 74.1 15.41 2.9 7.0 2720 353 828 127 428.0 196.0W 12x230 67.7 15.05 3.1 7.6 2420 321 742 115 386.0 177.0W 12x210 61.8 14.71 3.4 8.2 2140 292 664 104 348.0 159.0W 12x190 55.8 14.38 3.7 9.2 1890 263 589 93 311.0 143.0W 12x170 50.0 14.03 4.0 10.1 1650 235 517 82 275.0 126.0W 12x152 44.7 13.71 4.5 11.2 1430 209 454 73 243.0 111.0W 12x136 39.9 13.41 5.0 12.3 1240 186 398 64 214.0 98.0W 12x120 35.3 13.12 5.6 13.7 1070 163 345 56 186.0 85.4W 12x106 31.2 12.89 6.2 15.9 933 145 301 49 164.0 75.1W 12x 96 28.2 12.71 6.8 17.7 833 131 270 44 147.0 67.5W 12x 87 25.6 12.53 7.5 18.9 740 118 241 40 132.0 60.4W 12x 79 23.2 12.38 8.2 20.7 662 107 216 36 119.0 54.3W 12x 72 21.1 12.25 9.0 22.6 597 97 195 32 108.0 49.2W 12x 65 19.1 12.12 9.9 24.9 533 88 174 29 96.8 44.1W 12x 58 17.0 12.19 7.8 27.0 475 78 107 21 86.4 32.5W 12x 53 15.6 12.06 8.7 28.1 425 71 96 19 77.9 29.1W 12x 50 14.7 12.19 6.3 26.2 394 65 56 14 72.4 21.4W 12x 45 13.2 12.06 7.0 29.0 350 58 50 12 64.7 19.0W 12x 40 11.8 11.94 7.8 32.9 310 52 44 11 57.5 16.8W 12x 35 10.3 12.50 6.3 36.2 285 46 24 7 51.2 11.5W 12x 30 8.8 12.34 7.4 41.8 238 39 20 6 43.1 9.6W 12x 26 7.7 12.22 8.5 47.2 204 33 17 5 37.2 8.2W 12x 22 6.5 12.31 4.7 41.8 156 25 5 2 29.3 3.7W 12x 19 5.6 12.16 5.7 46.2 130 21 4 2 24.7 3.0W 12x 16 4.7 11.99 7.5 49.4 103 17 3 1 20.1 2.3W 12x 14 4.2 11.91 8.8 54.3 89 15 2 1 17.4 1.9W 10x112 32.9 11.36 4.2 10.4 716 126 236 45 147.0 69.2W 10x100 29.4 11.10 4.6 11.6 623 112 207 40 130.0 61.0W 10x 88 25.9 10.84 5.2 13.0 534 98 179 35 113.0 53.1W 10x 77 22.6 10.60 5.9 14.8 455 86 154 30 97.6 45.9W 10x 68 20.0 10.40 6.6 16.7 394 76 134 26 85.3 40.1W 10x 60 17.6 10.22 7.4 18.7 341 67 116 23 74.6 35.0W 10x 54 15.8 10.09 8.2 21.2 303 60 103 21 66.6 31.3W 10x 49 14.4 9.98 8.9 23.1 272 55 93 19 60.4 28.3W 10x 45 13.3 10.10 6.5 22.5 248 49 53 13 54.9 20.3W 10x 39 11.5 9.92 7.5 25.0 209 42 45 11 46.8 17.2W 10x 33 9.7 9.73 9.1 27.1 170 35 37 9 38.8 14.0W 10x 30 8.8 10.47 5.7 29.5 170 32 17 6 36.6 8.8W 10x 26 7.6 10.33 6.6 34.0 144 28 14 5 31.3 7.5W 10x 22 6.5 10.17 8.0 36.9 118 23 11 4 26.0 6.1W 10x 19 5.6 10.24 5.1 35.4 96 19 4 2 21.6 3.3W 10x 17 5.0 10.11 6.1 36.9 82 16 4 2 18.7 2.8W 10x 15 4.4 9.99 7.4 38.5 69 14 3 1 16.0 2.3W 10x 12 3.5 9.87 9.4 46.6 54 11 2 1 12.6 1.7



Designation A dbf

2tf

hctw

Ix Sx Iy Sy Zx Zy


W 8x 67 19.7 9.00 4.4 11.1 272 60 89 21 70.2 32.7W 8x 58 17.1 8.75 5.1 12.4 228 52 75 18 59.8 27.9W 8x 48 14.1 8.50 5.9 15.8 184 43 61 15 49.0 22.9W 8x 40 11.7 8.25 7.2 17.6 146 36 49 12 39.8 18.5W 8x 35 10.3 8.12 8.1 20.4 127 31 43 11 34.7 16.1W 8x 31 9.1 8.00 9.2 22.2 110 28 37 9 30.4 14.1W 8x 28 8.2 8.06 7.0 22.2 98 24 22 7 27.2 10.1W 8x 24 7.1 7.93 8.1 25.8 83 21 18 6 23.2 8.6W 8x 21 6.2 8.28 6.6 27.5 75 18 10 4 20.4 5.7W 8x 18 5.3 8.14 8.0 29.9 62 15 8 3 17.0 4.7W 8x 15 4.4 8.11 6.4 28.1 48 12 3 2 13.6 2.7W 8x 13 3.8 7.99 7.8 29.9 40 10 3 1 11.4 2.2W 8x 10 3.0 7.89 9.6 40.5 31 8 2 1 8.9 1.7W 6x 25 7.3 6.38 6.7 15.5 53 17 17 6 18.9 8.6W 6x 20 5.9 6.20 8.2 19.1 41 13 13 4 14.9 6.7W 6x 15 4.4 5.99 11.5 21.6 29 10 9 3 10.8 4.8W 6x 16 4.7 6.28 5.0 19.1 32 10 4 2 11.7 3.4W 6x 12 3.5 6.03 7.1 21.6 22 7 3 2 8.3 2.3W 6x 9 2.7 5.90 9.2 29.2 16 6 2 1 6.2 1.7

W 5x 19 5.5 5.15 5.8 14.0 26 10 9 4 11.6 5.5W 5x 16 4.7 5.01 6.9 15.8 21 9 8 3 9.6 4.6W 4x 13 3.8 4.16 5.9 10.6 11 5 4 2 6.3 2.9

M 14x 18 5.1 14.00 7.4 60.3 148 21 3 1 24.9 2.2M 12x 12 3.5 12.00 6.8 62.5 72 12 1 1 14.3 1.1M 12x 11 3.2 11.97 7.3 63.6 65 11 1 1 13.2 1.0M 12x 10 2.9 11.97 9.1 68.0 62 10 1 1 12.2 1.0M 10x 9 2.7 10.00 6.5 58.4 39 8 1 0 9.2 0.8M 10x 8 2.3 9.95 7.4 59.3 34 7 1 0 8.2 0.7M 10x 8 2.2 9.99 7.8 65.0 33 7 0 0 7.7 0.6M 8x 6 1.9 8.00 6.0 53.8 18 5 0 0 5.4 0.5M 6x 4 1.3 6.00 5.4 47.0 7 2 0 0 2.8 0.3

M 5x 19 5.6 5.00 6.0 11.2 24 10 8 3 11.0 5.0S 24x121 35.6 24.50 3.7 26.4 3160 258 83 21 306.0 36.2S 24x106 31.2 24.50 3.6 34.1 2940 240 77 20 279.0 33.2S 24x100 29.3 24.00 4.2 28.3 2390 199 48 13 240.0 23.9S 24x 90 26.5 24.00 4.1 33.7 2250 187 45 13 222.0 22.3S 24x 80 23.5 24.00 4.0 42.1 2100 175 42 12 204.0 20.7S 20x 96 28.2 20.30 3.9 21.6 1670 165 50 14 198.0 24.9S 20x 86 25.3 20.30 3.8 26.2 1580 155 47 13 183.0 23.0S 20x 75 22.0 20.00 4.0 27.1 1280 128 30 9 153.0 16.7S 20x 66 19.4 20.00 3.9 34.1 1190 119 28 9 140.0 15.3S 18x 70 20.6 18.00 4.5 21.8 926 103 24 8 125.0 14.4S 18x 55 16.1 18.00 4.3 33.6 804 89 21 7 105.0 12.1S 15x 50 14.7 15.00 4.5 23.2 486 65 16 6 77.1 10.0S 15x 43 12.6 15.00 4.4 31.0 447 60 14 5 69.3 9.0S 12x 50 14.7 12.00 4.2 13.9 305 51 16 6 61.2 10.3S 12x 41 12.0 12.00 4.0 20.7 272 45 14 5 53.1 8.9S 12x 35 10.3 12.00 4.7 23.4 229 38 10 4 44.8 6.8S 12x 32 9.4 12.00 4.6 28.6 218 36 9 4 42.0 6.4S 10x 35 10.3 10.00 5.0 13.8 147 29 8 3 35.4 6.2S 10x 25 7.5 10.00 4.7 26.4 124 25 7 3 28.4 5.0S 8x 23 6.8 8.00 4.9 14.5 65 16 4 2 19.3 3.7S 8x 18 5.4 8.00 4.7 23.7 58 14 4 2 16.5 3.2S 7x 20 5.9 7.00 4.9 12.3 42 12 3 2 14.5 3.0S 7x 15 4.5 7.00 4.7 21.9 37 10 3 1 12.1 2.4S 6x 17 5.1 6.00 5.0 9.9 26 9 2 1 10.6 2.4S 6x 12 3.7 6.00 4.6 19.9 22 7 2 1 8.5 1.9S 5x 15 4.3 5.00 5.0 7.5 15 6 2 1 7.4 1.9S 5x 10 2.9 5.00 4.6 17.4 12 5 1 1 5.7 1.4S 4x 10 2.8 4.00 4.8 8.7 7 3 1 1 4.0 1.1S 4x 8 2.3 4.00 4.5 14.7 6 3 1 1 3.5 1.0S 3x 8 2.2 3.00 4.8 5.6 3 2 1 0 2.4 0.8S 3x 6 1.7 3.00 4.5 11.4 3 2 0 0 2.0 0.7



Designation A dbf

2tf

hctw

Ix Sx Iy Sy Zx Zy


C 15.x 50 14.7 15. 0 0 404.0 53.8 11. 3.78 8.20 8.17C 15.x 40 11.8 15. 0 0 349.0 46.5 9.23 3.37 57.20 6.87C 15.x 34 10.0 15. 0 0 315.0 42.0 8.13 3.11 50.40 6.23C 12.x 30 8.8 12. 0 0 162.0 27.0 5.14 2.06 33.60 4.33C 12.x 25 7.3 12. 0 0 144.0 24.1 4.47 1.88 29.20 3.84C 12.x 21 6.1 12. 0 0 129.0 21.5 3.88 1.73 25.40 3.49C 10.x 30 8.8 10. 0 0 103.0 20.7 3.94 1.65 26.60 3.78C 10.x 25 7.3 10. 0 0 91.2 18.2 3.36 1.48 23. 3.19C 10.x 20 5.9 10. 0 0 78.9 15.8 2.81 1.32 19.30 2.71C 10.x 15 4.5 10. 0 0 67.4 13.5 2.28 1.16 15.80 2.35C 9.x 20 5.9 9. 0 0 60.9 13.5 2.42 1.17 16.80 2.47C 9.x 15 4.4 9. 0 0 51.0 11.3 1.93 1.01 13.50 2.05C 9.x 13 3.9 9. 0 0 47.9 10.6 1.76 0.96 12.50 1.95C 8.x 19 5.5 8. 0 0 44.0 11.0 1.98 1.01 13.80 2.17C 8.x 14 4.0 8. 0 0 36.1 9.0 1.53 0.85 10.90 1.73C 8.x 12 3.4 8. 0 0 32.6 8.1 1.32 0.78 9.55 1.58C 7.x 15 4.3 7. 0 0 27.2 7.8 1.38 0.78 9.68 1.64C 7.x 12 3.6 7. 0 0 24.2 6.9 1.17 0.70 8.40 1.43C 7.x 10 2.9 7. 0 0 21.3 6.1 0.97 0.63 7.12 1.26C 6.x 13 3.8 6. 0 0 17.4 5.8 1.05 0.64 7.26 1.36C 6.x 11 3.1 6. 0 0 15.2 5.1 0.87 0.56 6.15 1.15C 6.x 8 2.4 6. 0 0 13.1 4.4 0.69 0.49 5.13 0.99C 5.x 9 2.6 5. 0 0 8.9 3.6 0.63 0.45 4.36 0.92C 5.x 7 2.0 5. 0 0 7.5 3.0 0.48 0.38 3.51 0.76C 4.x 7 2.1 4. 0 0 4.6 2.3 0.43 0.34 2.81 0.70C 4.x 5 1.6 4. 0 0 3.8 1.9 0.32 0.28 2.26 0.57C 3.x 6 1.8 3. 0 0 2.1 1.4 0.31 0.27 1.72 0.54C 3.x 5 1.5 3. 0 0 1.9 1.2 0.25 0.23 1.50 0.47C 3.x 4 1.2 3. 0 0 1.7 1.1 0.20 0.20 1.30 0.40

Designation A wgt Ix Sx Iy Sy Zx Zy

in2 k/ft in4 in3 in4 in3 in3 in3

L 8.0x4.0x1.000 11.00 37.40 69.6 14.1 11.60 3.94 24.30 7.72L 8.0x4.0x0.750 8.44 28.70 54.9 10.9 9.36 3.07 18.90 5.81L 8.0x4.0x0.563 6.43 21.90 42.8 8.4 7.43 2.38 14.50 4.38L 8.0x4.0x0.500 5.75 19.60 38.5 7.5 6.74 2.15 13.00 3.90L 7.0x4.0x0.750 7.69 26.20 37.8 8.4 9.05 3.03 14.80 5.65L 7.0x4.0x0.625 6.48 22.10 32.4 7.1 7.84 2.58 12.60 4.74L 7.0x4.0x0.500 5.25 17.90 26.7 5.8 6.53 2.12 10.30 3.83L 7.0x4.0x0.375 3.98 13.60 20.6 4.4 5.10 1.63 7.87 2.90L 6.0x6.0x1.000 11.00 37.40 35.5 8.6 35.50 8.57 15.50 15.50L 6.0x6.0x0.875 9.73 33.10 31.9 7.6 31.90 7.63 13.80 13.80L 6.0x6.0x0.750 8.44 28.70 28.2 6.7 28.20 6.66 12.00 12.00L 6.0x6.0x0.625 7.11 24.20 24.2 5.7 24.20 5.66 10.20 10.20L 6.0x6.0x0.563 6.43 21.90 22.1 5.1 22.10 5.14 9.26 9.26L 6.0x6.0x0.500 5.75 19.60 19.9 4.6 19.90 4.61 8.31 8.31L 6.0x6.0x0.438 5.06 17.20 17.7 4.1 17.70 4.08 7.34 7.34L 6.0x6.0x0.375 4.36 14.90 15.4 3.5 15.40 3.53 6.35 6.35L 6.0x6.0x0.313 3.65 12.40 13.0 3.0 13.00 2.97 5.35 5.35L 6.0x4.0x0.875 7.98 27.20 27.7 7.2 9.75 3.39 12.70 6.31L 6.0x4.0x0.750 6.94 23.60 24.5 6.3 8.68 2.97 11.20 5.47L 6.0x4.0x0.625 5.86 20.00 21.1 5.3 7.52 2.54 9.51 4.62L 6.0x4.0x0.563 5.31 18.10 19.3 4.8 6.91 2.31 8.66 4.19L 6.0x4.0x0.500 4.75 16.20 17.4 4.3 6.27 2.08 7.78 3.75L 6.0x4.0x0.438 4.18 14.30 15.5 3.8 5.60 1.85 6.88 3.30L 6.0x4.0x0.375 3.61 12.30 13.5 3.3 4.90 1.60 5.97 2.85L 6.0x4.0x0.313 3.03 10.30 11.4 2.8 4.18 1.35 5.03 2.40L 6.0x3.5x0.500 4.50 15.30 16.6 4.2 4.25 1.59 7.50 2.91L 6.0x3.5x0.375 3.42 11.70 12.9 3.2 3.34 1.23 5.76 2.20L 6.0x3.5x0.313 2.87 9.80 10.9 2.7 2.85 1.04 4.85 1.85





L 5.0x5.0x0.875 7.98 27.20 17.8 5.2 17.80 5.17 9.33 9.33L 5.0x5.0x0.750 6.94 23.60 15.7 4.5 15.70 4.53 8.16 8.16L 5.0x5.0x0.625 5.86 20.00 13.6 3.9 13.60 3.86 6.95 6.95L 5.0x5.0x0.500 4.75 16.20 11.3 3.2 11.30 3.16 5.68 5.68L 5.0x5.0x0.438 4.18 14.30 10.0 2.8 10.00 2.79 5.03 5.03L 5.0x5.0x0.375 3.61 12.30 8.7 2.4 8.74 2.42 4.36 4.36L 5.0x5.0x0.313 3.03 10.30 7.4 2.0 7.42 2.04 3.68 3.68L 5.0x3.5x0.750 5.81 19.80 13.9 4.3 5.55 2.22 7.65 4.10L 5.0x3.5x0.625 4.92 16.80 12.0 3.7 4.83 1.90 6.55 3.47L 5.0x3.5x0.500 4.00 13.60 10.0 3.0 4.05 1.56 5.38 2.83L 5.0x3.5x0.438 3.53 12.00 8.9 2.6 3.63 1.39 4.77 2.49L 5.0x3.5x0.375 3.05 10.40 7.8 2.3 3.18 1.21 4.14 2.16L 5.0x3.5x0.313 2.56 8.70 6.6 1.9 2.72 1.02 3.49 1.82L 5.0x3.5x0.250 2.06 7.00 5.4 1.6 2.23 0.83 2.83 1.47L 5.0x3.0x0.625 4.61 15.70 11.4 3.5 3.06 1.39 6.27 2.61L 5.0x3.0x0.500 3.75 12.80 9.4 2.9 2.58 1.15 5.16 2.11L 5.0x3.0x0.438 3.31 11.30 8.4 2.6 2.32 1.02 4.57 1.86L 5.0x3.0x0.375 2.86 9.80 7.4 2.2 2.04 0.89 3.97 1.60L 5.0x3.0x0.313 2.40 8.20 6.3 1.9 1.75 0.75 3.36 1.35L 5.0x3.0x0.250 1.94 6.60 5.1 1.5 1.44 0.61 2.72 1.09L 4.0x4.0x0.750 5.44 18.50 7.7 2.8 7.67 2.81 5.07 5.07L 4.0x4.0x0.625 4.61 15.70 6.7 2.4 6.66 2.40 4.33 4.33L 4.0x4.0x0.500 3.75 12.80 5.6 2.0 5.56 1.97 3.56 3.56L 4.0x4.0x0.438 3.31 11.30 5.0 1.8 4.97 1.75 3.16 3.16L 4.0x4.0x0.375 2.86 9.80 4.4 1.5 4.36 1.52 2.74 2.74L 4.0x4.0x0.313 2.40 8.20 3.7 1.3 3.71 1.29 2.32 2.32L 4.0x4.0x0.250 1.94 6.60 3.0 1.0 3.04 1.05 1.88 1.88L 4.0x3.5x0.500 3.50 11.90 5.3 1.9 3.79 1.52 3.50 2.73L 4.0x3.5x0.438 3.09 10.60 4.8 1.7 3.40 1.35 3.11 2.42L 4.0x3.5x0.375 2.67 9.10 4.2 1.5 2.95 1.16 2.71 2.11L 4.0x3.5x0.313 2.25 7.70 3.6 1.3 2.55 0.99 2.29 1.78L 4.0x3.5x0.250 1.81 6.20 2.9 1.0 2.09 0.81 1.86 1.44L 4.0x3.0x0.500 3.25 11.10 5.1 1.9 2.42 1.12 3.41 2.03L 4.0x3.0x0.438 2.87 9.80 4.5 1.7 2.18 0.99 3.03 1.79





L 4.0x3.0x0.375 2.48 8.50 4.0 1.5 1.92 0.87 2.64 1.56L 4.0x3.0x0.313 2.09 7.20 3.4 1.2 1.65 0.73 2.23 1.31L 4.0x3.0x0.250 1.69 5.80 2.8 1.0 1.36 0.60 1.82 1.06L 3.5x3.5x0.500 3.25 11.10 3.6 1.5 3.64 1.49 2.68 2.68L 3.5x3.5x0.438 2.87 9.80 3.3 1.3 3.26 1.32 2.38 2.38L 3.5x3.5x0.375 2.48 8.50 2.9 1.1 2.87 1.15 2.08 2.08L 3.5x3.5x0.313 2.09 7.20 2.5 1.0 2.45 0.98 1.76 1.76L 3.5x3.5x0.250 1.69 5.80 2.0 0.8 2.01 0.79 1.43 1.43L 3.5x3.0x0.500 3.00 10.20 3.5 1.5 2.33 1.10 2.63 1.98L 3.5x3.0x0.438 2.65 9.10 3.1 1.3 2.09 0.98 2.34 1.76L 3.5x3.0x0.375 2.30 7.90 2.7 1.1 1.85 0.85 2.04 1.53L 3.5x3.0x0.313 1.93 6.60 2.3 1.0 1.58 0.72 1.73 1.30L 3.5x3.0x0.250 1.56 5.40 1.9 0.8 1.30 0.59 1.41 1.05L 3.5x2.5x0.500 2.75 9.40 3.2 1.4 1.36 0.76 2.53 1.40L 3.5x2.5x0.438 2.43 8.30 2.9 1.3 1.23 0.68 2.26 1.24L 3.5x2.5x0.375 2.11 7.20 2.6 1.1 1.09 0.59 1.97 1.07L 3.5x2.5x0.313 1.78 6.10 2.2 0.9 0.94 0.50 1.67 0.91L 3.5x2.5x0.250 1.44 4.90 1.8 0.8 0.78 0.41 1.36 0.74L 3.0x3.0x0.500 2.75 9.40 2.2 1.1 2.22 1.07 1.93 1.93L 3.0x3.0x0.438 2.43 8.30 2.0 1.0 1.99 0.95 1.72 1.72L 3.0x3.0x0.375 2.11 7.20 1.8 0.8 1.76 0.83 1.50 1.50L 3.0x3.0x0.313 1.78 6.10 1.5 0.7 1.51 0.71 1.27 1.27L 3.0x3.0x0.250 1.44 4.90 1.2 0.6 1.24 0.58 1.04 1.04L 3.0x3.0x0.188 1.09 3.71 1.0 0.4 0.96 0.44 0.79 0.79L 3.0x2.5x0.500 2.50 8.50 2.1 1.0 1.30 0.74 1.88 1.35L 3.0x2.5x0.438 2.21 7.60 1.9 0.9 1.18 0.66 1.68 1.20L 3.0x2.5x0.375 1.92 6.60 1.7 0.8 1.04 0.58 1.47 1.05L 3.0x2.5x0.313 1.62 5.60 1.4 0.7 0.90 0.49 1.25 0.89L 3.0x2.5x0.250 1.31 4.50 1.2 0.6 0.74 0.40 1.02 0.72L 3.0x2.5x0.188 1.00 3.39 0.9 0.4 0.58 0.31 0.78 0.55L 3.0x2.0x0.500 2.25 7.70 1.9 1.0 0.67 0.47 1.78 0.89L 3.0x2.0x0.438 2.00 6.80 1.7 0.9 0.61 0.42 1.59 0.79L 3.0x2.0x0.375 1.73 5.90 1.5 0.8 0.54 0.37 1.40 0.68L 3.0x2.0x0.313 1.46 5.00 1.3 0.7 0.47 0.32 1.19 0.58L 3.0x2.0x0.250 1.19 4.10 1.1 0.5 0.39 0.26 0.97 0.47L 3.0x2.0x0.188 0.90 3.07 0.8 0.4 0.31 0.20 0.75 0.36L 2.5x2.5x0.500 2.25 7.70 1.2 0.7 1.23 0.72 1.31 1.31L 2.5x2.5x0.375 1.73 5.90 1.0 0.6 0.98 0.57 1.02 1.02L 2.5x2.5x0.313 1.46 5.00 0.8 0.5 0.85 0.48 0.87 0.87L 2.5x2.5x0.250 1.19 4.10 0.7 0.4 0.70 0.39 0.71 0.71L 2.5x2.5x0.188 0.90 3.07 0.5 0.3 0.55 0.30 0.55 0.55L 2.5x2.0x0.375 1.55 5.30 0.9 0.5 0.51 0.36 0.99 0.66L 2.5x2.0x0.313 1.31 4.50 0.8 0.5 0.45 0.31 0.84 0.56L 2.5x2.0x0.250 1.06 3.62 0.7 0.4 0.37 0.25 0.69 0.46L 2.5x2.0x0.188 0.81 2.75 0.5 0.3 0.29 0.20 0.53 0.35L 2.0x2.0x0.375 1.36 4.70 0.5 0.4 0.48 0.35 0.63 0.63L 2.0x2.0x0.313 1.15 3.92 0.4 0.3 0.42 0.30 0.54 0.54L 2.0x2.0x0.250 0.94 3.19 0.3 0.2 0.35 0.25 0.44 0.44L 2.0x2.0x0.188 0.71 2.44 0.3 0.2 0.27 0.19 0.34 0.34L 2.0x2.0x0.125 0.48 1.65 0.2 0.1 0.19 0.13 0.23 0.23L 1.8x1.8x0.250 0.81 2.77 0.2 0.2 0.23 0.23 0.34 0.34L 1.8x1.8x0.188 0.62 2.12 0.2 0.1 0.18 0.14 0.26 0.26L 1.5x1.5x0.250 0.69 2.34 0.1 0.1 0.14 0.13 0.24 0.24L 1.5x1.5x0.188 0.53 1.80 0.1 0.1 0.11 0.10 0.19 0.19L 1.3x1.3x0.250 0.56 1.92 0.1 0.1 0.08 0.09 0.16 0.16L 1.3x1.3x0.188 0.43 1.48 0.1 0.1 0.06 0.07 0.13 0.13L 1.1x1.1x0.125 0.27 0.90 0.0 0.0 0.03 0.04 0.07 0.07L 1.0x1.0x0.125 0.23 0.80 0.0 0.0 0.02 0.03 0.06 0.06


Draft3.7 Joists 3–17

3.7 Joists

43 Steel joists, Fig. 3.8 look like shallow trusses (warren type) and are designed as simply supported uni-formly loaded beams assuming that they are laterally supported on the top (to prevent lateral torsionalbuckling). The lateral support is often profided by the concrete slab it suppors.

44 The standard open-web joist designation consists of the depth, the series designation and the chordtype. Three series are available for floor/roof construction, Table 3.3

Series Depth (in) Span (ft)K 8-30 8-60LH 18-48 25-96

DLH 52-72 89-120

Table 3.3: Joist Series Characteristics

4"

4"4"

Span

[Design Length = Span – 0.33 FT.]

Figure 3.8: prefabricated Steel Joists

45 Typical joist spacing ranges from 2 to 4 ft, and provides an efficient use of the corrugated steel deckwhich itself supports the concrete slab.

46 For preliminary estimates of the joist depth, a depth to span ratio of 24 can be assumed, therefore

d ≈ L/2 (3.5)

where d is in inches, and L in ft.

47 Table 3.4 list the load carrying capacity of open web, K-series steel joists based on a amximumallowable stress of 30 ksi. For each span, the first line indicates the total safe uniformly distributedload-carrying capacity in pounds per linear foot. Note that the dead load (including the one of the joist)



must be substracted in order to determine the safe live load. The second line indicates the live load(pounds/linear foot) which will produce an approximate delection of L/360.


Draft3.7 Joists 3–19

Joint 8K1 10K1 12K1 12K3 12K5 14K1 14K3 14K4 14K6 16K2 16K3 16K4 16K5 16K6 16K7 16K9Desig.Depth(in.)

8 10 12 12 12 14 14 14 14 16 16 16 16 16 16 16

≈ W 5.1 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(lbs/ft)Span(ft.)8 550

5509 550

55010 550 550

480 55011 532 550

377 54212 444 550 550 550 550

288 455 550 550 55013 377 479 550 550 550

225 363 510 510 51014 324 412 500 550 550 550 550 550 550

179 289 425 463 463 550 550 550 55015 281 358 434 543 550 511 550 550 550

145 234 344 428 434 475 507 507 50716 246 313 380 476 550 448 550 550 550 550 550 550 550 550 550 550

119 192 282 351 396 390 467 467 467 550 550 550 550 550 550 55017 277 336 420 550 395 495 550 550 512 550 550 550 550 550 550

159 234 291 366 324 404 443 443 488 526 526 526 526 526 52618 246 299 374 507 352 441 530 550 456 508 550 550 550 550 550

134 197 245 317 272 339 397 408 409 456 490 490 490 490 49019 221 268 335 454 315 395 475 550 408 455 547 550 550 550 550

113 167 207 269 230 287 336 383 347 386 452 455 455 455 45520 199 241 302 409 284 356 428 525 368 410 493 550 550 550 550

97 142 177 230 197 246 287 347 297 330 386 426 426 426 42621 218 273 370 257 322 388 475 333 371 447 503 548 550 550

123 153 198 170 212 248 299 255 285 333 373 405 406 40622 199 249 337 234 293 353 432 303 337 406 458 498 550 550

106 132 172 147 184 215 259 222 247 289 323 351 385 38523 181 227 308 214 268 322 395 277 308 371 418 455 507 550

93 116 150 128 160 188 226 194 216 252 282 307 339 36324 166 208 282 196 245 295 362 254 283 340 384 418 465 550

81 101 132 113 141 165 199 170 189 221 248 269 298 34625 180 226 272 334 234 260 313 353 384 428 514

100 124 145 175 150 167 195 219 238 263 31126 166 209 251 308 216 240 289 326 355 395 474

88 110 129 56 133 148 173 194 211 233 27627 154 193 233 285 200 223 268 302 329 366 439

79 98 115 139 119 132 155 173 188 208 24628 143 180 216 265 186 207 249 281 306 340 408

70 88 103 124 106 118 138 155 168 186 22029 173 193 232 261 285 317 380

95 106 124 139 151 167 19830 161 180 216 244 266 296 355

86 96 112 126 137 151 17831 151 168 203 228 249 277 332

78 87 101 114 124 137 16132 142 158 190 214 233 259 311

71 79 92 103 112 124 147

Table 3.4: Joist Properties


Draft

Chapter 4

Case Study I: EIFFEL TOWER

Adapted from (Billington and Mark 1983)

4.1 Materials, & Geometry

1 The tower was built out of wrought iron, less expensive than steel,and Eiffel had more expereince withthis material, Fig. 4.1

Figure 4.1: Eiffel Tower (Billington and Mark 1983)

Draft4–2 Case Study I: EIFFEL TOWER

2 The structure is essentially a tower, subjected to gravity andwind load. It is a relatively “light” structure, so dead load is smallcompared to the wind load.

3 To avoid overturning MgravityMwind had to be much higher than 1.

This can be achieved either by:

1. Increase self weight (as in Washington’s monument)

2. Increase the width of the base

3. Design support to resist tension.

4. Post-tension the support.

4 The tower is 984 feet high, and 328 feet wide at the base. Thelarge base was essential to provide adequate stability in the pres-ence of wind load.

5 We can assume that the shape of the tower is parabolic. If wetake the x axis to be along the vertical axis of symmetry and y thehalf width, then we know that at x = 984 the (half) width y = 0and at x = 0 the half width is 328/2 = 164, thus the equation ofthe half width is given by

y = 164(

984 − x

984

)2

︸︷︷︸av(x)2

(4.1)

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

dy

dx=

dy

dv

dv

dx(4.2-a)

d

dxax2 = 2ax (4.2-b)

Thus for our problem

dy

dx= 2(164)

984 − x

984︸︷︷︸dydv

(− 1

984

)︸︷︷︸

dvdx

(4.3-a)

=984 − x

2, 952(4.3-b)

Alsody

dx= tanβ ⇒ β = tan−1 dy

dx(4.4)

where β is the angle measured from the x axis to the tangent to the curve.

3 We now can tabulate the width and slope at various elevations


Draft4.2 Loads 4–3

WidthLocation Height Width/2 Estimated Actual dy

dx βSupport 0 164 328 .333 18.4o

First platform 186 108 216 240 .270 15.1o

second platform 380 62 123 110 .205 11.6o

Intermediate platform 644 20 40 .115 6.6o

Top platform 906 1 2 .0264 1.5o

Top 984 0 0 0.000 0o

4 The tower is supported by four inclined supports, each with a cross section of 800 in2. An idealizationof the tower is shown in Fig. 4.2.

ACTUALCONTINUOUSCONNECTION

ACTUALPOINTS OFCONNECTION

CONTINUOUSCONNECTION

IDEALIZED

Figure 4.2: Eiffel Tower Idealization, (Billington and Mark 1983)

4.2 Loads

5 The total weight of the tower is 18, 800 k.

6 The dead load is not uniformly distributed, and is approximated as follows, Fig. 4.3:

Figure 4.3: Eiffel Tower, Dead Load Idealization; (Billington and Mark 1983)



Location Height Dead WeightGround- second platform 380 ft 15, 500 k

Second platform-intermediate platform 264 ft 2, 200 k

intermediate platform - top 340 ft 1, 100 k

Total 984 ft 18, 800 k

7 From the actual width of the lower two platforms we can estimate the live loads (the intermediate andtop platforms would have negligible LL in comparison):

1st platform: (50) psf(240)2 ft2 kip

(1,000) lbs 2, 880 k

2nd platform: (50) psf(110)2 ft2 k(1,000) lbs 600 k

Total: 3, 480 k

Hence the total vertical load is Pvert = DL + LL = 18, 800 + 3, 480 = 22, 280 k.

8 The wind pressure is known to also have a parabolic distribution (maximum at the top), the crosssectional area over which the wind is acting is also parabolic (maximum at the base). Hence we willsimplify our analysis by considering an equivalent wind force obtained from a constant wind pressure(force/length) and constant cross section Fig. 4.4: The pressure is assumed to be 2.6 k/ft, thus the

Figure 4.4: Eiffel Tower, Wind Load Idealization; (Billington and Mark 1983)

lateral wind force is, Fig. 4.5

Plat = (2.6) k/ft(984) ft = 2,560 k acting at9842

= 492 ft (4.5)

4.3 Reactions

9 Simplifying the three dimensional structure with 4 supports into a two dimensional one with twosupports, the reactions can be easily determined for this statically determinate structure, Fig.4.6.

Gravity Load

Pvert = 22, 280 ? (4.6-a)


Draft4.3 Reactions 4–5

P=2560k

Q=22,280kL/2

TOTALLOADS

��

��

��

��

H

V

M 0

Q

P

0

0

LOADS

REACTIONS

Figure 4.5: Eiffel Tower, Wind Loads, (Billington and Mark 1983)

VERTICALFORCES FORCES

TOTAL

WINDWARDSIDE

LEEWARD

SIDE

+ =

WIND

Figure 4.6: Eiffel Tower, Reactions; (Billington and Mark 1983)



Rgravvert =

22, 2802

= 11,140 k 6 (4.6-b)

Lateral Load

Lateral Moment (we essentially have a cantilivered beam subjected to a uniform load). Themoment at a distance x from the support along the cantilevered beam subjected to a uniformpressure p is given by

Mlat = p(L − x)︸︷︷︸Force

L − x

2︸︷︷︸Moment arm

= p(L − x)2

2(4.7)

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

Mlat = p(L − x)2

2= (2.6) k/ft

(984 − 0)2

2ft2 = 1,260,000 k.ft

¡¢¾ (4.8)

We observe that the shape of the moment diagram is also parabolic, just like the tower itself.This is not accidental, as nearly optimum structures have a shape which closely approximatetheir 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 wehave two supports (one under tension and the other under compression) we use

Rwindvert = ±M

d=

1, 260, 000 k.ft

328 ft= ± 3,850 k 6? (4.9)

Lateral Forces to be resisted by each of the two pairs. By symmetry, the lateral force will beequally divided among the two pairs of supports and will be equal to

Rwindlat =

(2, 560) k

2= 1,280 k ¾ (4.10)

4.4 Internal Forces

10 First, a biref reminder

F

F F

x

y

θ

θ cos θ =Fx

F(4.11-a)

sin θ =Fy

F(4.11-b)

tan θ =Fy

Fx(4.11-c)

11 Internal forces are first determined at the base.

12 Gravity load are first considered, remember those are caused by the dead load and the live load,Fig. 4.7:

cos β =V

N⇒ N =

V

cos β(4.12-a)


Draft4.4 Internal Forces 4–7

INCLINEDINTERNALFORCE: N

KNOWN VERTICALCOMPONENT: VCOMPONENT: H

HORIZONTALCONSEQUENT

β=18.40

V

H

N

β=18.4

FORCE POLYGON

Figure 4.7: Eiffel Tower, Internal Gravity Forces; (Billington and Mark 1983)

N =11, 140 k

cos 18.4o= 11,730 kip (4.12-b)

tanβ =H

V⇒ H = V tanβ (4.12-c)

H = 11, 140 k(tan 18.4o) = 3,700 kip (4.12-d)

The horizontal forces which must be resisted by the foundations, Fig. 4.8.

H H

3700 k 3700 k

Figure 4.8: Eiffel Tower, Horizontal Reactions; (Billington and Mark 1983)

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

14 At the second platform, the total vertical load is Q = 1, 100+2, 200 = 3, 300 k and at that height theangle is 11.6o thus the axial force (per pair of columns) will be

Nvert =3,300 k

2

cos 11.6o= 1, 685 k (4.13-a)

Hvert =3, 300 k

2(tan 11.6o) = 339 k (4.13-b)

Note that this is about seven times smaller than the axial force at the base, which for a given axialstrength, 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.

15 Wind Load: We now have determined at each pair of support the vertical and the horizontal forcescaused by the wind load, the next step is to determine their axial components along the inclined leg,



1,280 k

1,280 sin 18.4

3,850 cos 18.43,850 k

18.4

18.4

Figure 4.9: Eiffel Tower, Internal Wind Forces; (Billington and Mark 1983)

Fig. 4.9:

Nc = −Rwindvert cos β − Rwind

lat sinβ (4.14-a)= −(3, 850) k(cos 18.4o) − (1, 280) k(sin 18.40) (4.14-b)

= -4,050 k Leeward (4.14-c)

Nt = −Rwindvert cos β + Rwind

lat sinβ (4.14-d)= (3, 850) k(cos 18.4o) + (1, 280) k(sin 18.40) (4.14-e)

= 4,050 k Winward (4.14-f)

4.5 Internal Stresses

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

NTotalL = −(11, 730) k︸︷︷︸

gravity

−(4, 050) k︸︷︷︸lateral

= -15,780 k Leeward side (4.15-a)

NTotalW = −(11, 730) k︸︷︷︸

gravity

+(4, 050) k︸︷︷︸lateral

= -7,630 k Winward side (4.15-b)

We observe that even under wind load, the windward side is still under compression.

17 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

σcomp =NT

L

A=

−15, 780 k

1, 600 in2 = -9.9 ksi (4.16)

18 The strength of wrought iron is 45 ksi, hence the safety factor is

Safety Factor =ultimate stressactual stress

=45 ksi

9.9 ksi= 4.5 (4.17)


Draft

Chapter 5

REVIEW of STATICS

To every action there is an equaland opposite reaction.

Newton’s third law of motion

5.1 Reactions

1 In the analysis of structures (hand calculations), it is often easier (but not always necessary) to startby determining the reactions.

2 Once the reactions are determined, internal forces are determined next; finally, internal stresses and/ordeformations (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. 5.1.

Roller: provides a restraint in only one direction in a 2D structure, in 3D structures a roller may providerestraint 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.

5.1.1 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 In a 3D cartesian coordinate system there are a total of 6 independent equations of equilibrium:1This 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 computerimplementation.

2In a dynamic system ΣF = ma where m is the mass and a is the acceleration.

Draft5–2 REVIEW of STATICS

Figure 5.1: Types of Supports

ΣFx = ΣFy = ΣFz = 0ΣMx = ΣMy = ΣMz = 0 (5.1)

8 In a 2D cartesian coordinate system there are a total of 3 independent equations of equilibrium:

ΣFx = ΣFy = ΣMz = 0 (5.2)

9 For reaction calculations, the externally applied load may be reduced to an equivalent force3.

10 Summation of the moments can be taken with respect to any arbitrary point.

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

12 Not all equations are applicable to all structures, Table 5.1

13 The three conventional equations of equilibrium in 2D: ΣFx,ΣFy and ΣMz can be replaced by theindependent moment equations ΣMA

z , ΣMBz , ΣMC

z provided that A, B, and C are not colinear.

14 It is always preferable to check calculations by another equation of equilibrium.

15 Before you write an equation of equilibrium,

1. Arbitrarily decide which is the +ve direction

2. Assume a direction for the unknown quantities

3. The right hand side of the equation should be zero3However for internal forces (shear and moment) we must use the actual load distribution.



Structure Type EquationsBeam, no axial forces ΣFy ΣMz

2D Truss, Frame, Beam ΣFx ΣFy ΣMz

Grid ΣFz ΣMx ΣMy

3D Truss, Frame ΣFx ΣFy ΣFz ΣMx ΣMy ΣMz

Alternate SetBeams, no axial Force ΣMA

z ΣMBz

2 D Truss, Frame, Beam ΣFx ΣMAz ΣMB

z

ΣMAz ΣMB

z ΣMCz

Table 5.1: Equations of Equilibrium

If your reaction is negative, then it will be in a direction opposite from the one assumed.

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

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

5.1.2 Equations of Conditions

18 If a structure has an internal hinge (which may connect two or more substructures), then this willprovide 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 reac-tions.

20 In an inclined roller support with Sx and Sy horizontal and vertical projection, then the reactionR would have, Fig. 5.2.

Rx

Ry=

Sy

Sx

(5.3)

Figure 5.2: Inclined Roller Support

5.1.3 Static Determinacy

21 In statically determinate structures, reactions depend only on the geometry, boundary conditions andloads.



22 If the reactions can not be determined simply from the equations of static equilibrium (and equationsof conditions if present), then the reactions of the structure are said to be statically indeterminate.

23 The degree of static indeterminacy is equal to the difference between the number of reactionsand the number of equations of equilibrium (plus the number of equations of conditions if applicable),Fig. 5.3.

Figure 5.3: Examples of Static Determinate and Indeterminate Structures

24 Failure of one support in a statically determinate system results in the collapse of the structures.Thus a statically indeterminate structure is safer than a statically determinate one.

25 For statically indeterminate structures, reactions depend also on the material properties (e.g. Young’sand/or shear modulus) and element cross sections (e.g. length, area, moment of inertia).

5.1.4 Geometric Instability

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

27 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. 5.4.

Figure 5.4: Geometric Instability Caused by Concurrent Reactions



3. The number of reactions is smaller than the number of equations of equilibrium, that is a mech-anism is present in the structure.

28 Mathematically, this can be shown if the determinant of the equations of equilibrium is equal tozero (or the equations are inter-dependent).

5.1.5 Examples

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

Example 5-1: Simply Supported Beam

Determine the reactions of the simply supported beam shown below.

Solution:The beam has 3 reactions, we have 3 equations of static equilibrium, hence it is statically determinate.

(+- ) ΣFx = 0; ⇒ Rax − 36 k = 0(+ 6) ΣFy = 0; ⇒ Ray + Rdy − 60 k − (4) k/ft(12) ft = 0

(+ ¡¢¾) ΣM cz = 0; ⇒ 12Ray − 6Rdy − (60)(6) = 0

or through matrix inversion (on your calculator) 1 0 0

0 1 10 12 −6

Rax

Ray

Rdy

=

36108360

⇒

Rax

Ray

Rdy

=

36 k

56 k

52 k

Alternatively we could have used another set of equations:

(+ ¡¢¾) ΣMaz = 0; (60)(6) + (48)(12) − (Rdy)(18) = 0 ⇒ Rdy = 52 k 6

(+ ¡¢¾) ΣMdz = 0; (Ray)(18) − (60)(12) − (48)(6) = 0 ⇒ Ray = 56 k 6

Check:(+ 6) ΣFy = 0; ; 56 − 52 − 60 − 48 = 0

√



Example 5-2: 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 ofcondition (ΣMb = 0), thus the structure is statically determinate.

1. Isolating ab:

ΣM ¡¢¾b = 0; (9)(Ray) − (40)(5) = 0 ⇒ Ray = 22.2 k 6(+ ¡¢¾) ΣMa = 0; (40)(4) − (S)(9) = 0 ⇒ S = 17.7 k 6

ΣFx = 0; ⇒ Rax = 30 k ¾

2. Isolating bd:

(+ ¡¢¾) ΣMd = 0; −(17.7)(18) − (40)(15) − (4)(8)(8) − (30)(2) + Rcy(12) = 0⇒ Rcy = 1,236

12 = 103 k 6(+ ¡¢¾) ΣMc = 0; −(17.7)(6) − (40)(3) + (4)(8)(4) + (30)(10) − Rdy(12) = 0

⇒ Rdy = 201.312 = 16.7 k 6



3. CheckΣFy = 0; 6; 22.2 − 40 − 40 + 103 − 32 − 30 + 16.7 = 0

√

Example 5-3: Three Hinged Gable Frame

The three-hinged gable frames spaced at 30 ft. on center. Determine the reactions components on theframe 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 andhorizontal reactions.

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)

ft1

1, 000lbs/k = 20.2 k ?



4. Snow load per frame is

SL = (30) psf(30) ft(30) ft1

1, 000lbs/k = 27. k ?

5. Wind load per frame (ignoring the suction) is

WL = (15) psf(30) ft(35) ft1

1, 000lbs/k = 15.75 k-

6. There are 4 reactions, 3 equations of equilibrium and one equation of condition ⇒ staticallydeterminate.

7. The horizontal reaction H due to a vertical load V at midspan of the roof, is obtained by takingmoment with respect to the hinge

(+ ¡¢¾) ΣMC = 0; 15(V ) − 30(V ) + 35(H) = 0 ⇒ H = 15V35 = .429V

Substituting for roof dead and snow load we obtain

V ADL = V B

DL = 20.2 k 6HA

DL = HBDL = (.429)(20.2) = 8.66 k-

V ASL = V B

SL = 27. k 6HA

SL = HBSL = (.429)(27.) = 11.58 k-

8. The reactions due to wind load are

(+ ¡¢¾) ΣMA = 0; (15.75)(20+152 ) − V B

WL(60) = 0 ⇒ V BWL = 4.60 k 6

(+ ¡¢¾) ΣMC = 0; HBWL(35) − (4.6)(30) = 0 ⇒ HB

WL = 3.95 k ¾(+- ) ΣFx = 0; 15.75 − 3.95 − HA

WL = 0 ⇒ HAWL = 11.80 k ¾

(+ 6) ΣFy = 0; V BWL − V A

WL = 0 ⇒ V AWL = −4.60 k ?

9. Thus supports should be designed for

H = 8.66 k + 11.58 k + 3.95 k = 24.19 k

V = 20.7 k + 27.0 k + 4.60 k = 52.3 k

5.2 Trusses

5.2.1 Assumptions

30 Cables and trusses are 2D or 3D structures composed of an assemblage of simple one dimensionalcomponents which transfer only axial forces along their axis.

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

32 For trusses, it is assumed that

1. Bars are pin-connected


Draft5.2 Trusses 5–9

2. Joints are frictionless hinges4.

3. Loads are applied at the joints only.

33 A truss would typically be composed of triangular elements with the bars on the upper chord undercompression and those along the lower chord under tension. Depending on the orientation of thediagonals, they can be under either tension or compression.

34 In a truss analysis or design, we seek to determine the internal force along each member, Fig. 5.5

Figure 5.5: Bridge Truss

5.2.2 Basic Relations

Sign Convention: Tension positive, compression negative. On a truss the axial forces are indicated asforces acting on the joints.

Stress-Force: σ = PA

Stress-Strain: σ = Eε

Force-Displacement: ε = ∆LL

Equilibrium: ΣF = 0

5.2.3 Determinacy and Stability

35 Trusses are statically determinate when all the bar forces can be determined from the equationsof statics alone. Otherwise the truss is statically indeterminate.

36 A truss may be statically indeterminate with respect to the reactions or externally indeterminateand/or statically indeterminate with respect to the internal forces that is internally indeterminate.

37 a 2D truss is externally indeterminate if there are more than 3 reactions.4In 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 momentsis the dead weight of the element.



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

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

40 If we refer to j as the number of joints, R the number of reactions and m the number of members,then we would have a total of m + R unknowns and 2j (or 3j) equations of statics (2D or 3D at eachjoint). If we do not have enough equations of statics then the problem is indeterminate, if we have toomany equations then the truss is unstable, Table 5.2.

2D 3DStatic Indeterminacy

External R > 3 R > 6Internal m + R > 2j m + R > 3jUnstable m + R < 2j m + R < 3j

Table 5.2: Static Determinacy and Stability of Trusses

41 Fig. 5.6 shows a truss with 4 reactions, thus it is externally indeterminate. This truss has 6 joints(j = 6), 4 reactions (R = 4) and 9 members (m = 9). Thus we have a total of m + R = 9 + 4 = 13unknowns and 2 × j = 2 × 6 = 12 equations of equilibrium, thus the truss is statically indeterminate.

Figure 5.6: A Statically Indeterminate Truss

42 There are two methods of analysis for statically determinate trusses

1. The Method of joints

2. The Method of sections

5.2.4 Method of Joints

43 The method of joints can be summarized as follows

1. Determine if the structure is statically determinate



2. Compute all reactions

3. Sketch a free body diagram showing all joint loads (including reactions)

4. For each joint, and starting with the loaded ones, apply the appropriate equations of equilibrium(ΣFx and ΣFy in 2D; ΣFx, ΣFy and ΣFz in 3D).

5. Because truss elements can only carry axial forces, the resultant force (~F = ~Fx + ~Fy) must bealong the member, Fig. 5.7.

F

l=

Fx

lx=

Fy

ly(5.4)

44 Always keep track of the x and y components of a member force (Fx, Fy), as those might be neededlater on when considering the force equilibrium at another joint to which the member is connected.

Figure 5.7: X and Y Components of Truss Forces

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

46 In truss analysis, there is no sign convention. A member is assumed to be under tension (orcompression). If after analysis, the force is found to be negative, then this would imply that the wrongassumption was made, and that the member should have been under compression (or tension).

47 On a free body diagram, the internal forces are represented by arrow acting on the joints andnot as end forces on the element itself. That is for tension, the arrow is pointing away from the joint,and for compression toward the joint, Fig. 5.8.

Figure 5.8: Sign Convention for Truss Element Forces



Example 5-4: 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

(+ ¡¢¾) ΣME = 0; ⇒ (20 + 12)(3)(24) + (40 + 8)(2)(24) + (40)(24) − RAy(4)(24) = 0

⇒ RAy= 58 k 6

(+ ?) ΣFy = 0; ⇒ 20 + 12 + 40 + 8 + 40 − 58 − REy= 0

⇒ REy= 62 k 6

3. Consider each joint separately:

Node A: Clearly AH is under compression, and AB under tension.

(+ 6) ΣFy = 0; ⇒ FAHy− 58 = 0

FAH = lly

(FAHy)

ly = 32 l =√

322 + 242 = 40⇒ FAH = 40

32 (58) = 72.5 Compression(+- ) ΣFx = 0; ⇒ −FAHx

+ FAB = 0FAB = lx

ly(FAHy

) = 2432 (58) = 43.5 Tension



Node B:

(+- ) ΣFx = 0; ⇒ FBC = 43.5 Tension(+ 6) ΣFy = 0; ⇒ FBH = 20 Tension

Node H:

(+- ) ΣFx = 0; ⇒ FAHx− FHCx

− FHGx= 0

43.5 − 24√242+322 (FHC) − 24√

242+102 (FHG) = 0 (I)(+ 6) ΣFy = 0; ⇒ FAHy

+ FHCy− 12 − FHGy

− 20 = 058 + 32√

242+322 (FHC) − 12 − 10√242+102 (FHG) − 20 = 0 (II)

Solving for I and II we obtain

FHC = −7.5 TensionFHG = 52 Compression

Node E:

ΣFy = 0; ⇒ FEFy= 62 ⇒ FEF = 32√

242+322 (62) = 77.5 k

ΣFx = 0; ⇒ FED = FEFx⇒ FED = 24

32 (FEFy) = 24

32 (62) = 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

5.3 Shear & Moment Diagrams

5.3.1 Theory

5.3.1.1 Design Sign Conventions

48 Before we derive the Shear-Moment relations, let us arbitrarily define a sign convention.

49 The sign convention adopted here, is the one commonly used for design purposes5. With reference toFig. 5.9

+-

+ve Shear

+ve Load + Axial Force

+ve Moment

Figure 5.9: Shear and Moment Sign Conventions for Design

2D:

5Later on, in more advanced analysis courses we will use a different one.


Draft5.3 Shear & Moment Diagrams 5–15

Load Positive along the beam’s local y axis (assuming a right hand side convention), that ispositive upward.

Axial: tension positive.

Flexure A positive moment is one which causes tension in the lower fibers, and compression inthe upper ones. For frame members, a positive moment is one which causes tension along theinner side.

Shear A positive shear force is one which is “up” on a negative face, or “down” on a positive one.Alternatively, a pair of positive shear forces will cause clockwise rotation.

Torsion Counterclockwise positive

3D: Use double arrow vectors (and NOT curved arrows). Forces and moments (including torsions) aredefined with respect to a right hand side coordinate system, Fig. 5.10.

Draft

��*

��*

��*

--

6

6

6

��>

��>

---

x

y

z

Mz

Tx

Mz

Tx

My

My

6

6

Figure 5.10: Sign Conventions for 3D Frame Elements

5.3.1.2 Load, Shear, Moment Relations

50 Let us (re)derive the basic relations between load, shear and moment. Considering an infinitesimallength dx of a beam subjected to a positive load6 w(x), Fig. 5.11. The infinitesimal section must also

Figure 5.11: Free Body Diagram of an Infinitesimal Beam Segment

be in equilibrium.

51 There are no axial forces, thus we only have two equations of equilibrium to satisfy ΣFy = 0 andΣMz = 0.

52 Since dx is infinitesimally small, the small variation in load along it can be neglected, therefore weassume w(x) to be constant along dx.

53 To denote that a small change in shear and moment occurs over the length dx of the element, we addthe differential quantities dVx and dMx to Vx and Mx on the right face.

6In this derivation, as in all other ones we should assume all quantities to be positive.



54 Next considering the first equation of equilibrium

(+ 6) ΣFy = 0 ⇒ Vx + wxdx − (Vx + dVx) = 0

or

dV

dx= w(x) (5.5)

The slope of the shear curve at any point along the axis of a member is given bythe load curve at that point.

55 Similarly

(+ ¡¢¾) ΣMo = 0 ⇒ Mx + Vxdx − wxdxdx

2− (Mx + dMx) = 0

Neglecting the dx2 term, this simplifies to

dM

dx= V (x) (5.6)

The slope of the moment curve at any point along the axis of a member is givenby the shear at that point.

56 Alternative forms of the preceding equations can be obtained by integration

V =∫

w(x)dx (5.7)

∆V21 = Vx2 − Vx1 =∫ x2

x1

w(x)dx (5.8)

The change in shear between 1 and 2, ∆V21, is equal to the area under the loadbetween x1 and x2.

and

M =∫

V (x)dx (5.9)

∆M21 = M2 − M1 =∫ x2

x1

V (x)dx (5.10)

The change in moment between 1 and 2, ∆M21, is equal to the area under theshear curve between x1 and x2.

57 Note that we still need to have V1 and M1 in order to obtain V2 and M2 respectively.

58 Fig. 5.12 and 5.13 further illustrates the variation in internal shear and moment under uniform andconcentrated forces/moment.

5.3.1.3 Moment Envelope

59 For design, we often must consider different load combinations.

60 For each load combination, we should draw the shear, moment diagrams. and then we should use theMoment envelope for design purposes.



Figure 5.12: Shear and Moment Forces at Different Sections of a Loaded Beam

Shear

Moment

Load

Shear

Positive Constant Negative Constant

Positive Constant Negative Constant Negative Increasing Negative Decreasing

Negative Increasing Negative Decreasing

Positive Increasing Positive Decreasing

Positive Increasing Positive Decreasing

Figure 5.13: Slope Relations Between Load Intensity and Shear, or Between Shear and Moment



5.3.1.4 Examples

Example 5-5: Simple Shear and Moment Diagram

Draw the shear and moment diagram for the beam shown below

Solution:The free body diagram is drawn below



Reactions are determined from the equilibrium equations

(+ ¾) ΣFx = 0; ⇒ −RAx+ 6 = 0 ⇒ RAx

= 6 k

(+ ¡¢¾) ΣMA = 0; ⇒ (11)(4) + (8)(10) + (4)(2)(14 + 2) − RFy(18) = 0 ⇒ RFy

= 14 k

(+ 6) 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 isuniform 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 sheardiagram, or ∆MB−A = (13)(4) = 52.

3. etc...

Example 5-6: Frame Shear and Moment Diagram

Draw the shear and moment diagram of the following frame

Solution:



Reactions are determined first

(+ ¾) ΣFx = 0; ⇒ RAx− 4

5(3)︸︷︷︸

load

(15) = 0

⇒ RAx= 36 k

(+ ¡¢¾) ΣMA = 0; ⇒ (3)(30)(302 ) + 3

5 (3)(15)(30 + 9

2

) − 45 (3)(15)12

2 − 39RDy= 0

⇒ RFy= 52.96 k

(+ 6) ΣFy = 0; ⇒ RAy− (3)(30) − 3

5 (3)(15) + 52.96 = 0⇒ RAy

= 64.06 k

Shear:

1. For A−B, the shear is constant, equal to the horizontal reaction at A and negative accordingto 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 transmittedalong 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 slopeequal 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 andshear components. Thus at D the shear is equal to 3

552.96 = 31.78 k and is negative. Basedon 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 − 5

39(3) = −31.78 or Vc = 13.22. The equation forthe 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

(+ ¾) ΣFx = 0 ⇒ 35 (42.37) + 4

5 (13.22) = 25.42 + 10.58 = 36√

(+ 6) ΣFy = 0 ⇒ 45 (42.37) − 3

5 (13.22) = 33.90 − 7.93 = 25.97√

Moment:

1. Along A − B, the moment is zero at A (since we have a hinge), and its slope is equal to theshear, thus at B the moment is equal to (−36)(12) = −432 k.ft

2. Along B − C, the moment is equal to

MB−C = MB +∫ x

0

VB−C(x)dx = −432 +∫ x

0

(64.06 − 3x)dx

= −432 + 64.06x − 3x2

2

which is a parabola. Substituting for x = 30, we obtain at node C: MC = −432+64.06(30)−3 302

2 = 139.8 k.ft

3. If we need to determine the maximum moment along B−C, we know that dMB−C

dx = 0 at thepoint where VB−C = 0, that is VB−C(x) = 64.06 − 3x = 0 ⇒ x = 64.06

3 = 25.0 ft. In otherwords, maximum moment occurs where the shear is zero.

Thus MmaxB−C = −432 + 64.06(25.0) − 3 (25.0)2

2 = −432 + 1, 601.5 − 937.5 = 232 k.ft



4. Finally along C − D, the moment varies quadratically (since we had a linear shear), themoment first increases (positive shear), and then decreases (negative shear). The momentalong C − D is given by

MC−D = MC +∫ x

0VC−D(x)dx = 139.8 +

∫ x

0(13.22 − 3x)dx

= 139.8 + 13.22x − 3x2

2

which is a parabola.Substituting for x = 15, we obtain at node C MC = 139.8 + 13.22(15) − 3 152

2 = 139.8 +198.3 − 337.5 = 0

√

Example 5-7: Frame Shear and Moment Diagram; Hydrostatic Load

The frame shown below is the structural support of a flume. Assuming that the frames are spaced 2ft 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 ascantilevers 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 theflume, 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 theend walls as well as the uniform load on the horizontal members.



End Actions

1. Base force at B is FBx = (.749)62 = 2.246 k

2. Base moment at B is MB = (2.246)63 = 4.493 k.ft

3. End force at B for member B − E are equal and opposite.

4. Reaction at C is RCy = (.749)162 = 5.99 k

Shear forces

1. Base at B the shear force was determined earlier and was equal to 2.246 k. Based on theorientation 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)(32 ) = −7.864 k.ft

3. The maximum moment is equal to Mmax = −7.864 + (.749)(5)(52 ) = 1.50 k.ft

Design: Reinforcement should be placed along the fibers which are under tension, that is on the sideof the negative moment7. The figure below schematically illustrates the location of the flexural8

reinforcement.

Example 5-8: Shear Moment Diagrams for Frame

7That is why in most European countries, the sign convention for design moments is the opposite of the one commonlyused in the U.S.A.; Reinforcement should be placed where the moment is “postive”.

8Shear reinforcement is made of a series of vertical stirrups.



Vba

Hbc

M bc Vbc

Vba

Vbc

30k 10k(10)+(2)(10)

M bc

-200’k(10)(10)+(2)(10)(10)/2

-650’k(-52.5)(12)+(-20)

450’k

(50

)(1

5)-

[(4

)(5

)/2

][(2

)(1

5)/

3)]

450’k

Hbd

Vbd

M bd

82.5k

20k

4k/ft 50k

50k

20k

(50

)-(4

)(1

5)/

2

M bc

12’

A

30k5k/ft

B H

M ba

8’ 10’

D

G

H

V

4k/ft

15’

D

D

AE

V

5k/ft

5’

30k

A

B C

10k

ba

20k

3.5’-22.5k

-52.5k-22.5+(-30)

17.5k

B C

17.5k

17.5-(5)(8)

-20’k

30.6’k

M ba

(17.5)(3.5)/2+(-22.5)(8-3.5)/2

(17.5)(3.5)/2

10k2k/ft

2k/ft

200’k

82.5k

450’k

52.5k 30k

00

650’k

Hbd

VbdM bd

M ba

Vba Vbc

CHECK

0

17.5-5x=0



Example 5-9: Shear Moment Diagrams for Inclined Frame



x

y

F/Fy=z/xF/Fx=z/yFx/Fy=y/x

Fy

Fx

zF

A

B

C

D

E

20k

20’36’

20’

15’

10’10’13’

13’13’

26k26k

2k/ft

125

133

5

4

H

V

a

aVe

80

0k’

2k/ft60k

20k

60k

19.2k

800’k

(20

)(2

0)+

(60

-20

)(2

0)/

219.2k 48.8k

48.8k

0k’20k

1130’k 488’k

+6

0k

+20

k

+25.4k

-26.6k-23.1k

-39.1k800’k

60

-(2

)(2

0)

(60

)(2

0)-

(2)(

20

)(2

0)/

2

26k26k

17.72k

26.6k

778k’

19.2k

0k

800k’

20k 28.8k

11.1k

-0.6-26

800’k1130’k

800+(25.4)(13)

39.1k

28.8k

778k’

0k0k’

488’k(39.1)(12.5)

-23.1-16

24

k 24

k10k10k

18.46k7.38k

20

-10

-10

+25.4k

-26.6k

-23.1k-39.1k

48

8+

(23

.1)(

12

.5)

777k

’

1122k’

-0.58k

1,130-(.58)(13)

25.42-26

777k’

-0.58k

1122k’ 777k’

20

k16

k

12k

17.28k23.1

k

48.9k29.3k

1 2 3 4 5 6

8 B-C 11 C-D

13

1,1

22

-(2

6.6

)(1

3)

9 B-C 12 C-D

107

AB ED

BC CD

(20)(12)/(13)=18.46(19.2)(5)/(13)=7.38(19.2)(12)/(13)=17.72

(26.6)(13)/(12)=28.8(26.6)(5)/(12)=11.1

(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

(28.8)(4)/(5)=23.1

(26)(12)/(13)=24

(20)(15)/13=7.7

14

17.7+7.7

7.69k



5.3.2 Formulaes

Adapted from (of Steel COnstruction 1986)



1) Simple Beam; uniform Load

R = VVx = w

(L2 − x

)at center Mmax = wL2

8Mx = wx

2 (L − x)∆max = 5

384wL4

EI∆x = wx

24EI (L3 − 2Lx2 + x3)2) Simple Beam; Unsym-

metric Triangular Load

R1 = V1 = W3

Max R2 = V2 = 2W3

Vx = W3 − Wx2

L2

at x = .577L Mmax = .1283WLMx = Wx

3L2 (L2 − x2)at x = .5193L ∆max = .01304WL3

EI

∆x = Wx3

180EIL2 (3x4 − 10L2x2 + 7L4)3) Simple Beam; Symmet-

ric Triangular Load

R = V = W2

for x < L2 Vx = W

2L2 (L2 − 4x2)at center Mmax = WL

6

for x < L2 Mx = Wx

(12 − 2

3x2

L2

)for x < L

2 ∆x = Wx480EIL2 (5L2 − 4x2)2

∆max = WL3

60EI



4) Simple Beam; Uniform Load Partially Distributed

Max when a < c R1 = V1 = wb2L (2c + b)

Max when a > c R2 = V2 = wb2L (2a + b)

when a < x < a + b Vx = R1 − w(x − a)when x < a Mx = R1x

when a < x < a + b Mx = R1x − w2 (x − a)2

when a + b < x Mx = R2(L − x)at x = a + R1

w Mmax = R1

(a + R1

2w

)5) Simple Beam; Concen-

trated Load at Center

max R1 = V1 = wa2L (2L − a)

R = V = 2Pat x = L

2 Mmax = PL4

when x < L2 Mx = Px

2

whenx < L2 ∆x = Px

48EI (3L2 − 4x2)at x = L

2 ∆max = PL3

48EI

6) Simple Beam; Concen-trated Load at Any Point

max when a < b R1 = V1 = PbL

max when a > b R2 = V2 = PaL

at x = a Mmax = PabL

when x < a Mx = PbxL

at x = a ∆a = Pa2b2

3EIL

when x < a ∆x = Pbx6EIL (L2 − b2 − x2)

at x =√

a(a+2b)3 & a > b ∆max = Pab(a+2b)

√3a(a+2b)

27EIL



7) Simple Beam; Two Equally Concentrated Symmetric Loads

R = V = PMmax = Pa∆max = Pa

24EI (3L2 − 4a2)when x < a ∆x = Px

6EI (3La − 3a2 − x2)when a < x < L − a ∆x = Pa

6EI (3Lx − 3x2 − a2)8) Simple Beam; Two Equally

Concentrated Unsymmetric Loads

max when a < b R1 = V1 = PL (L − a + b)

max when b < a R2 = V2 = PL (L − b + a)

when a < x < L − b Vx = PL (b − a)

max when b < a M1 = R1amax when a < b M2 = R2b

when x < a Mx = R1xwhen a < x < L − b Mx = R1x − P (x − a)

9) Cantilevered Beam, Uni-form Load

R1 = V1 = 38wL

R2 = V2 = 58wL

Vx = R1 − wx

Mmax = wL2

8at x = 3

8L M1 = 9128wL2

Mx = R1x − wx2

2∆x = wx

48EI (L3 − 3Lx+2x3)at x = .4215L ∆max = wL4

185EI



10) Propped Cantilever, Concentrated Load at Center

R1 = V1 = 5P16

R2 = V2 = 11P16

at x = L Mmax = 3PL16

when x < L2 Mx = 5Px

16

when L2 < x Mx = P

(L2 − 11x

16

)at x = .4472L ∆max = .009317PL3

EI

11) Propped Cantilever;Concentrated Load

R1 = V1 = Pb2

2L3 (a + 2L)R2 = V2 = Pa

2L3 (3L2 − a2)at x = a M1 = R1aat x = L M2 = Pab

2L2 (a + L)at x = a ∆a = Pa2b3

12EIL3 (3L + a)when a < .414L at x = L L2+a2

3L2−a2 ∆max = Pa3EI

(L2−a2)3

(3L2−a2)2

when .414L < a at x = L√

a2L+a ∆max = Pab2

6EI

√a

2L+a2

12) Beam Fixed at BothEnds, Uniform Load

R = V = wL2

Vx = w(

L2 − x

)at x = 0 and x = L Mmax = wL2

12

at x = L2 M = wL2

24

at x = L2 ∆max = wL4

384EI

∆x = wx2

24EI (L − x)2



13) Beam Fixed at Both Ends; Concentrated Load

R = V = P2

at x = L2 Mmax = PL

8

when x < L2 Mx = P

8 (4x − L)at x = L

2 ∆max = PL3

192EI

when x < L2 ∆x = Px2

48EI (3L − 4x)14) Cantilever Beam; Tri-

angular Unsymmetric Load

R = V = 83W

Vx = W x2

L2

at x = L Mmax = WL3

Mx = Wx2

3L2

∆x = W60EIL2 (x5 − 5L2x + 4L5)

at x = 0 ∆max = WL3

15EI

15) Cantilever Beam; Uni-form Load

R = V = wLVx = wx

Mx = wx2

2

at x = L Mmax = wL2

2∆x = w

24EI (x4 − 4L3x + 3L4)at x = 0 ∆max = wL4

8EI



16) Cantilever Beam; Point Load

R = V = Pat x = L Mmax = Pb

when a < x Mx = P (x − a)at x = 0 ∆max = Pb2

6EI (3L − b)at x = a ∆a = Pb3

3EI

when x < a ∆x = Pb2

6EI (3L − 3x − b)when a < x ∆x = P (L−x)2

6EI (3b − L + x)17) Cantilever Beam; Point

Load at Free End

R = V = Pat x = L Mmax = PL

Mx = Px

at x = 0 ∆max = PL3

3EI

∆x = P6EI (2L3 − 3L2x + x3)

18) Cantilever Beam; Con-centrated Force and Moment at Free End

R = V = PMx = P

(L2 − x

)at x = 0 and x = L Mmax = PL

2

at x = 0 ∆max = PL3

12EI

∆x = P (L−x)2

12EI ((L + 2x)



19) Beam Overhanging One Support; Uniform Load Between Supports

R = V = wL2

Vx = w(

L2 − x

)Mx = wx

2 (L − x)at x = L

2 ∆max = 5wL4

384EI∆x = wx

24EI (L3 − 2Lx2 + x3)when L < x < L + a ∆x1 = wL3x1x

24EI

20) Beam Overhanging oneSupport; Concentrated ForceMax when a < b R1 = V1 = Pb

L

Max when b < a R2 = V2 = PaL

at x < aL Mx = PbxL

at x =√

a(a+2b)3 when a > b ∆max = Pab(a+2b)

√3a(a+2b)

27EIL

at x = a ∆a = Pa2b2

3EIL

when x < a ∆x = Pbx6EIL (L2 − b2 − x2)

when a < x ∆x = Pa(L−x)6EIL (2Lx − x2 − a2)

at L < x < L + a ∆x1 = Pabx16EIL (L + a)

21) Continuous Beam,Two Equal Spans; Concentrated Force

R1 = V1 = Pb4L3

[4L2 − a(L + a)

]R2 = V2 + V3 = Pa

2L3

[2L2 + b(L + a)

]R3 = V3 = −Pab

4L3 (L + a)V2 = Pa

4L3

[4L2 + b(L + a)

]Mmax = Pab

4L3

[4L2 − a(L + a)

]at x = L M1 = Pab

4L2 (L + a)



22) Simple Beam, Uniform Load, End Moments

R1 = V1 = wL2 + M1−M2

L

R2 = V2 = wL2 − M1−M2

L

Vx = w(

L2 − x

)+ M1−M2

L

at x = L2 + M1−M2

wL

Mx = wx2 (L − x) +

(M1−M2

L

)x − M1

b =√

L2

4 − (M1+M2

w

)+

(M1−M2

wl

)2

∆x = wx24EI

[x3 − (

2L + 3M1wL − 4M2

wL

)x2

+ 12M1w x + L3 − 8M1L

w − 4M2Lw

]23) Simple Beam; Con-

centrated Force, End Moments

R1 = V1 = P2 + M1−M2

L

R2 = V2 = P2 − M1−M2

L

at x = L2 M3 = PL

4 − M1+M22

when x < L2 Mx =

(P2 + M1−M2

L

)x − M1

when L2 < x Mx = P

2 (L − x) + M1−M2L x − M1

when x < L2 ∆x = Px

48EI

[3L2 − 4x2

− 8(L−x)PL (M1(2L − x) + M2(L + x))

]


Draft5.4 Flexure 5–37

24) Beam Overhanging one Support; Uniform Load

at 0 < x < L Mx = wx2L (L2 − a2 − xL)

at L < x < L + a Mx1 = w2 (a − x1)2

at 0 < x < L ∆x = wx24EIL (L4 − 2L2x2 + Lx3 − 2a2L2 + 2a2x2)

at L < x < L + a ∆x1 = wx124EI (4a2L−L3 + 6a2x1 − 4ax2

1 + x31)

R1 = V1 = w2L (L2 − a2)

R2 = V2 + V3 = w2L (L + a)2

V2 = waV3 = w

2L (L2 + a2)at 0 < x < L Vx = R1 − wx

at L < x < L + a Vx1 = w(a − x1)at x = L

2

(1 − a2

L2

)M1 = w

8L2 (L + a)2(L − a)2

at x = L M2 = wa2

2at 0 < x < L Mx = wx

2L (L2 − a2 − xL)at L < x < L + a Mx1 = w

2 (a − x1)2

at 0 < x < L ∆x = wx24EIL (L4 − 2L2x2 + Lx3 − 2a2L2 + 2a2x2)

at L < x < L + a ∆x1 = wx124EI (4a2L−L3 + 6a2x1 − 4ax2

1 + x31)

5.4 Flexure

5.4.1 Basic Kinematic Assumption; Curvature

61 Fig.5.14 shows portion of an originally straight beam which has been bent to the radius ρ by endcouples M . support conditions, Fig. 5.1. It is assumed that plane cross-sections normal to the length of

Neutral Axis

dx

ρ

E’ F’

E F

O

M M

dθ

X

Y

Z

dA

+ve Curvature, +ve bending

-ve Curvature, -ve Bending

Figure 5.14: Deformation of a Beam un Pure Bending

the unbent beam remain plane after the beam is bent.

62 Except for the neutral surface all other longitudinal fibers either lengthen or shorten, thereby creatinga longitudinal strain εx. Considering a segment EF of length dx at a distance y from the neutral axis,



its original length isEF = dx = ρdθ (5.11)

anddθ =

dx

ρ(5.12)

63 To evaluate this strain, we consider the deformed length E′F ′

E′F ′ = (ρ − y)dθ = ρdθ − ydθ = dx − ydx

ρ(5.13)

The strain is now determined from:

εx =E′F ′ − EF

EF=

dx − y dxρ − dx

dx(5.14)

or after simplification

εx = −y

ρ(5.15)

where y is measured from the axis of rotation (neutral axis). Thus strains are proportional to thedistance from the neutral axis.

64 ρ (Greek letter rho) is the radius of curvature. In some textbook, the curvature κ (Greek letterkappa) is also used where

κ =1ρ

(5.16)

thus,

εx = −κy (5.17)

5.4.2 Stress-Strain Relations

65 So far we considered the kinematic of the beam, yet later on we will need to consider equilibrium interms of the stresses. Hence we need to relate strain to stress.

66 For linear elastic material Hooke’s law states

σx = Eεx (5.18)

where E is Young’s Modulus.

67 Combining Eq. with equation 5.17 we obtain

σx = −Eκy (5.19)



5.4.3 Internal Equilibrium; Section Properties

68 Just as external forces acting on a structure must be in equilibrium, the internal forces must alsosatisfy the equilibrium equations.

69 The internal forces are determined by slicing the beam. The internal forces on the “cut” section mustbe in equilibrium with the external forces.

5.4.3.1 ΣFx = 0; Neutral Axis

70 The first equation we consider is the summation of axial forces.

71 Since there are no external axial forces (unlike a column or a beam-column), the internal axial forcesmust be in equilibrium.

ΣFx = 0 ⇒∫

A

σxdA = 0 (5.20)

where σx was given by Eq. 5.19, substituting we obtain∫A

σxdA = −∫

A

EκydA = 0 (5.21-a)

But since the curvature κ and the modulus of elasticity E are constants, we conclude that

∫A

ydA = 0 (5.22)

or the first moment of the cross section with respect to the z axis is zero. Hence we conclude that theneutral axis passes through the centroid of the cross section.

5.4.3.2 ΣM = 0; Moment of Inertia

72 The second equation of internal equilibrium which must be satisfied is the summation of moments.However contrarily to the summation of axial forces, we now have an external moment to account for,the one from the moment diagram at that particular location where the beam was sliced, hence

ΣMz = 0; ¡¢¾+ve; M︸︷︷︸Ext.

= −∫

A

σxydA︸︷︷︸Int.

(5.23)

where dA is an differential area a distance y from the neutral axis.

73 Substituting Eq. 5.19

M = −∫

A

σxydA

σx = −Eκy

M = κE

∫A

y2dA (5.24)

74 We now pause and define the section moment of inertia with respect to the z axis as

Idef=

∫A

y2dA (5.25)



and section modulus as

Sdef=

I

c(5.26)

75 Section properties for selected sections are shown in Table 5.3.

b

y

x

h

Y

X

A = bhx = b

2

y = h2

Ix = bh3

12

Iy = hb3

12 b’b

h’h

x

y

X

Y

A = bh − b′h′

x = b2

y = h2

Ix = bh3−b′h′312

Iy = hb3−h′b′312

h

b

Y

Xy

a

A = h(a+b)2

y = h(2a+b)3(a+b)

Ix = h3(a2+4ab+b2

36(a+b)

h

b

x

Xy

c

YA = bh

2

x = b+c3

y = h3

Ix = bh3

36

Iy = bh36 (b2 − bc + c2)

X

Y

r

A = πr2 = πd2

4

Ix = Iy = πr4

4 = πd4

64

X

Y

rt A = 2πrt = πdt

Ix = Iy = πr3t = πd3t8

b

b

a a

X

Y

A = πab

Ix = πab3

3

Iy = πba3

4

Table 5.3: Section Properties

5.4.4 Beam Formula

76 We now have the ingredients in place to derive one of the most important equations in structures, thebeam formula. This formula will be extensively used for design of structural components.



77 We merely substitute Eq. 5.25 into 5.24,

M = κE

∫A

y2dA

I =∫

a

y2dA

M

EI = κ = 1ρ (5.27)

which shows that the curvature of the longitudinal axis of a beam is proportional to the bending momentM and inversely proportional to EI which we call flexural rigidity.

78 Finally, inserting Eq. 5.19 above, we obtain

σx = −Eκyκ = M

EI

}σx = −My

I (5.28)

Hence, for a positive y (above neutral axis), and a positive moment, we will have compressive stressesabove the neutral axis.

79 Alternatively, the maximum fiber stresses can be obtained by combining the preceding equation withEquation 5.26

σx = −M

S(5.29)

Example 5-10: Design Example

A 20 ft long, uniformly loaded, beam is simply supported at one end, and rigidly connected at theother. The beam is composed of a steel tube with thickness t = 0.25 in. Select the radius such thatσmax ≤ 18 ksi, and ∆max ≤ L/360.

20’

r 0.25’

1 k/ft

Solution:

1. Steel has E = 29, 000 ksi, and from above Mmax = wL2

8 , ∆max = wL4

185EI , and I = πr3t.

2. The maximum moment will be

Mmax =wL2

8=

(1) k/ft(20)2 ft2

8= 50 k.ft (5.30)

3. We next seek a relation between maximum deflection and radius

∆max = wL4

185EII = πr3t

} ∆ = wL4

185Eπr3t

= (1) k/ft(20)4 ft4(12)3 in3/ ft3

(185)(29,000) ksi(3.14)r3(0.25) in= 65.65

r3

(5.31)



4. Similarly for the stress

σ = MS

S = Ir

I = πr3t

σ = Mπr2t

= (50) k.ft(12) in/ft(3.14)r2(0.25) in

= 764r2

(5.32)

5. We now set those two values equal to their respective maximum

∆max =L

360=

(20) ft(12) in/ft

360= 0.67 in =

65.65r3

⇒ r = 3

√65.650.67

= 4.61 in (5.33-a)

σmax = (18) ksi =764r2

⇒ r =

√76418

= 6.51 in (5.33-b)

5.4.5 Approximate Analysis

80 From Fig. 5.14, and Eq. 5.27 ( MEI = κ = 1

ρ ), we recall that that the moment is directly proportionalto the curvature κ.

81 Thus,

1. A positive and negative moment would correspond to positive and negative curvature respectively(adopting the sign convention shown in Fig. 5.14).

2. A zero moment correspnds to an inflection point in the deflected shape.

82 Hence, for

Statically determinate structure, we can determine the deflected shape from the moment diagram,Fig. 5.15.

Statically indeterminate structure, we can:

1. Plot the deflected shape.2. Identify inflection points, approximate their location.3. Locate those inflection points on the structure, which will then become statically determinate.4. Perform an approximate analysis.

Example 5-11: Approximate Analysis of a Statically Indeterminate beam

Perform an approximate analysis of the following beam, and compare your results with the exactsolution.

16’ 12’ 28’

20k

28’



Figure 5.15: Elastic Curve from the Moment Diagram



Figure 5.16: Approximate Analysis of Beams



Solution:

16’ 12’ 28’

A B C D

28’

28’

Approximate Location of IP20k

20k

28’

22’ 6’

6’22’

1. We have 3 unknowns RA, RC , and RD, all in the vertical directions, and only two applicableequations of equilibrium (since we do not have any force in the x direction), thus the problem isstatically indeterminare.

2. We sketch the anticipated deflected shape, and guess the location of the inflection point.

3. At that location, we place a hinge, and we now have an additional equation of condition at thatlocation (ΣM = 0).

4. If we consider AB, and take the moments with respect to point B:

(+ ¡¢¾) ΣMB = 0; (22)(RA) − (20)(22 − 16) = 0 ⇒ RA = 5.45 k 6(5.34-a)

5. If we now consider the entire beam:

(+ ¡¢¾) ΣMD = 0; (RA)(28 + 28) − (20)(28 + 12) + (RC)(28) = 0(5.45)(56) − (20)(40) + (RC)(28) = 0

⇒ RC = 17.67 k 6(+ 6) ΣFy = 0; RA − 20 + Rc + RD = 0

5.45 − 20 + 17.67 + RD = 0

⇒ RD = −3.12 3.12 k ?



6. Check

(+ ¡¢¾) ΣMA = 0; (20)(16) − (RC)(28) + (RD)(28 + 28) =320 − (17.67)(28) + (3.12)(56) =

320 − 494.76 + 174.72 = 0√

(5.36-a)

7. The moments are determined next

Mmax = RAa = (5.45)(16) = 87.2 (5.37-a)

M1 = RDL = (3.12)(28) = 87.36 (5.37-b)

8. We now compare with the exact solution from Section 5.3.2, solution 21 where:L = 28, a = 16,b = 12, and P = 20

R1 = RA =Pb

4L3

[4L2 − a(L + a)

]=

(20)(12)4(28)3

[4(28)2 − (16)(28 + 16)

]= 6.64 (5.38-a)

R2 = RB =Pa

2L3

[2L2 + b(L + a)

](5.38-b)

=(20)(16)2(28)3

[2(28)2 + 12(28 + 16)

]= 15.28 (5.38-c)

R3 = RD = −Pab

4L3(L + a) (5.38-d)

= − (20)(16)(12)4(28)3

(28 + 16) = −1.92 (5.38-e)

Mmax = R1a = (6.64)(16) = 106.2 (5.38-f)

M1 = R3L = (1.92)(28) = 53.8 (5.38-g)

9. If we tabulate the results we have

Value Approximate Exact % ErrorRA 5.45 6.64 18RC 17.67 15.28 -16RD 3.12 1.92 63M1 87.36 53.8 62Mmax 87.2 106.2 18

10. Whereas the correlation between the approximate and exact results is quite poor, one should notunderestimate the simplicity of this method keeping in mind (an exact analysis of this structurewould have been computationally much more involved). Furthermore, often one only needs a roughorder of magnitude of the moments.


Draft

Chapter 6

Case Study II: GEORGEWASHINGTON BRIDGE

6.1 Theory

1 Whereas the forces in a cable can be determined from statics alone, its configuration must be derivedfrom its deformation. Let us consider a cable with distributed load p(x) per unit horizontal pro-jection of the cable length (thus neglecting the weight of the cable). An infinitesimal portion of thatcable can be assumed to be a straight line, Fig. 6.1 and in the absence of any horizontal load we haveH =constant. Summation of the vertical forces yields

(+ ?) ΣFy = 0 ⇒ −V + wdx + (V + dV ) = 0 (6.1-a)dV + wdx = 0 (6.1-b)

where V is the vertical component of the cable tension at x (Note that if the cable was subjected to itsown weight then we would have wds instead of wdx). Because the cable must be tangent to T , we have

tan θ =V

H(6.2)

Substituting into Eq. 6.1-b yields

d(H tan θ) + wdx = 0 ⇒ − d

dx(H tan θ) = w (6.3)

2 But H is constant (no horizontal load is applied), thus, this last equation can be rewritten as

−Hd

dx(tan θ) = w (6.4)

3 Written in terms of the vertical displacement v, tan θ = dvdx which when substituted in Eq. 6.4 yields

the governing equation for cables−Hv′′ = w (6.5)

4 For a cable subjected to a uniform load w, we can determine its shape by double integration of Eq.6.5

−Hv′ = wx + C1 (6.6-a)

−Hv =wx2

2+ C1x + C2 (6.6-b)

Draft6–2 Case Study II: GEORGE WASHINGTON BRIDGE

H

T+

dT

TH

V

V+

dV

H

θ

θ

q(x)

dx

dyds

ds

L

x

V

q(x)

y

y(x)

dx

h

y’

x’

x

y

L/2

Figure 6.1: Cable Structure Subjected to p(x)


Draft6.1 Theory 6–3

and the constants of integrations C1 and C2 can be obtained from the boundary conditions: v = 0 atx = 0 and at x = L ⇒ C2 = 0 and C1 = −wL

2 . Thus

v =w

2Hx(L − x) (6.7)

This equation gives the shape v(x) in terms of the horizontal force H,

5 Since the maximum sag h occurs at midspan (x = L2 ) we can solve for the horizontal force

H =wL2

8h(6.8)

we note the analogy with the maximum moment in a simply supported uniformly loaded beam M =Hh = wL2

8 . Furthermore, this relation clearly shows that the horizontal force is inversely proportionalto the sag h, as h ↘ H ↗. Finally, we can rewrite this equation as

rdef=

h

L(6.9-a)

wL

H= 8r (6.9-b)

6 Eliminating H from Eq. 6.7 and 6.8 we obtain

v = 4h

(− x2

L2+

x

L

)(6.10)

Thus the cable assumes a parabolic shape (as the moment diagram of the applied load).

7 Whereas the horizontal force H is constant throughout the cable, the tension T is not. The maximumtension occurs at the support where the vertical component is equal to V = wL

2 and the horizontal oneto H, thus

Tmax =√

V 2 + H2 =

√(wL

2

)2

+ H2 = H

√1 +

(wL/2

H

)2

(6.11)

Combining this with Eq. 6.8 we obtain1.

Tmax = H√

1 + 16r2 ≈ H(1 + 8r2) (6.12)

8 Had we assumed a uniform load w per length of cable (rather than horizontal projection), theequation would have been one of a catenary2.

v =H

wcosh

[w

H

(L

2− x

)]+ h (6.13)

The cable between transmission towers is a good example of a catenary.

1Recalling that (a + b)n = an + nan−1b +n(n−1)

2!an−2b2 + · or (1 + b)n = 1 + nb +

n(n−1)b2

2!+

n(n−1)(n−2)b3

3!+ · · ·;

Thus for b2 << 1,√

1 + b = (1 + b)12 ≈ 1 + b

22Derivation of this equation is beyond the scope of this course.



6.2 The Case Study


9 The George Washington bridge, is a suspension bridge spanning the Hudson river from New York Cityto New Jersey. It was completed in 1931 with a central span of 3,500 ft (at the time the world’s longestspan). The bridge was designed by O.H. Amman, who had emigrated from Switzerland. In 1962 thedeck was stiffened with the addition of a lower deck.

6.2.1 Geometry

10 A longitudinal and plan elevation of the bridge is shown in For simplicity we will assume in our

??

377 ft 327 ft

610 ft 650 ft

ELEVATION

N.J. N.Y.HUDSON RIVER

PLAN

3,500 ft

4,760 ft

Figure 6.2: Longitudinal and Plan Elevation of the George Washington Bridge

analysis that the two approaching spans are equal to 650 ft.

11 There are two cables of three feet diameter on each side of the bridge. The centers of each pair are 9ft apart, and the pairs themselves are 106 ft apart. We will assume a span width of 100 ft.

12 The cables are idealized as supported by rollers at the top of the towers, hence the horizontal com-ponents of the forces in each side of the cable must be equal (their vertical components will add up).

13 The cables support the road deck which is hungby suspenders attached at the cables. The cables aremade of 26,474 steel wires, each 0.196 inch in diameter. They are continuous over the tower supportsand are firmly anchored in both banks by huge blocks of concrete, the anchors.


Draft6.2 The Case Study 6–5

14 Because the cables are much longer than they are thick (large IL ), they can be idealized a perfectly

flexible members with no shear/bending resistance but with high axial strength.

15 The towers are 578 ft tall and rest on concrete caissons in the river. Because of our assumptionregarding the roller support for the cables, the towers will be subjected only to axial forces.

6.2.2 Loads

16 The dead load is composed of the weight of the deck and the cables and is estimated at 390 and 400psf respectively for the central and side spans respectively. Assuming an average width of 100 ft, thiswould be equivalent to

DL = (390) psf(100) ftk

(1, 000) lbs= 39 k/ft (6.14)

for the main span and 40 k/ft for the side ones.

17 For highway bridges, design loads are given by the AASHTO (Association of American State HighwayTransportation Officials). The HS-20 truck is often used for the design of bridges on main highways, Fig.6.3. Either the design truck with specified axle loads and spacing must be used or the equivalent uniformload and concentrated load. This loading must be placed such that maximum stresses are produced.

Figure 6.3: Truck Load

18 With two decks, we estimate that there is a total of 12 lanes or

LL = (12)Lanes(.64) k/ ft/Lane = 7.68 k/ft ≈ 8 k/ft (6.15)

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

19 Final DL and LL are, Fig. 6.4: TL = 39 + 8 = 47 k/ft



DEAD LOADS

Lw = 8 k/ft

D D,Sw = 39 k/ft w = 40 k/ftD,Sw = 40 k/ft

Figure 6.4: Dead and Live Loads



6.2.3 Cable Forces

20 The thrust H (which is the horizontal component of the cable force) is determined from Eq. 6.8

H = wL2cs8h

= (47) k/ft(3,500)2 ft2

(8)(327) ft= 220, 000 k

From Eq. 6.12 the maximum tension is

r = hLcs = 327

3,500 = 0.0934Tmax = H

√1 + 16r2

= (2, 200) k√

1 + (16)(0.0934)2

= (2, 200) k(1.0675) = 235,000 k

6.2.4 Reactions

21 Cable reactions are shown in Fig. 6.5.

POINTS WITH REACTIONS TOCABLES

Figure 6.5: Location of Cable Reactions

22 The vertical force in the columns due to the central span (cs) is simply the support reaction, 6.6

A

POINT OF NOMOMENT

B

REACTIONS ATTOP OF TOWER

TOT

L = 3,500 FT

w = 39 + 8 = 47 k/ft

Figure 6.6: Vertical Reactions in Columns Due to Central Span Load

Vcs =12wLcs =

12(47) k/ft(3, 500) ft = 82, 250 k (6.16)



Note that we can check this by determining the vector sum of H and V which should be equal to Tmax:√V 2cs + H2 =

√(82, 250)2 + (220, 000)2 = 235, 000 k

√(6.17)

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

ΣMD = 0; ¡¢¾+;hssH + (wssLss)Lss2

− VssLss = 0 (6.18-a)

= (377) k(220, 000) k + (48) k/ft(650) ft(650) ft

2− 650Vss = 0 (6.18-b)

Vss = 143, 200 k (6.18-c)

24 Hence the total axial force applied on the column is

V = Vcs + Vss = (82, 250) k + (143, 200) k = 225,450 k (6.19)

25 The vertical reaction at the anchor is given by summation of the forces in the y direction, Fig. 6.7:

(+ 6) ΣFy = 0; (wssLss) + Vss + Ranchor = 0 (6.20-a)−(48) k/ft(650) ft + (143, 200) k + Ranchor = 0 (6.20-b)

Ranchor = 112,000 k ? (6.20-c)

(6.20-d)

225,450 k

112,000 k

220,000 k

Figure 6.7: Cable Reactions in Side Span

26 The axial force in the side cable is determined the vector sum of the horizontal and vertical reactions.

T ssanchor =

√R2

anchor + H2 =√

(112, 000)2 + (220, 000)2 = 247, 000 k (6.21-a)

T sstower =

√V 2ss + H2 =

√(143, 200)2 + (220, 000)2 = 262,500 k (6.21-b)



27 The cable stresses are determined last, Fig. 6.8:

Awire =πD2

4=

(3.14)(0.196)2

4= 0.03017 in2 (6.22-a)

Atotal = (4)cables(26, 474)wires/cable(0.03017) in2/wire = 3, 200 in2(6.22-b)

Central Span σ =H

A=

(220, 000) k

(3, 200) in2 = 68.75 ksi (6.22-c)

Side Span Tower σsstower =

T sstowerA

=(262, 500) in2

(3, 200) in2 = 82 ksi (6.22-d)

Side Span Anchor σsstower =

T ssanchor

A=

(247, 000) in2

(3, 200) in2 = 77.2 ksi (6.22-e)

77.2 ksi

81.9 ksi73.4 ksi

68.75 ksi

Figure 6.8: Cable Stresses

28 If the cables were to be anchored to a concrete block, the volume of the block should be at least equalto V = (112,000) k(1,000) lbs/ k

150 lbs/ft3= 747, 000 ft3 or a cube of approximately 91 ft

29 The deck, for all practical purposes can be treated as a continuous beam supported by elastic springswith stiffness K = AL/E (where L is the length of the supporting cable). This is often idealized asa beam on elastic foundations, and the resulting shear and moment diagrams for this idealization areshown in Fig. 6.9.



K=AL/E

Shear

Moment

Figure 6.9: Deck Idealization, Shear and Moment Diagrams


Draft

Chapter 7

A BRIEF HISTORY OFSTRUCTURAL ARCHITECTURE

If I have been able to see a little farther than some others,it was because I stood on the shoulders of giants.

Sir Isaac Newton

1 More than any other engineering discipline, Architecture/Mechanics/Structures is the proud outcomeof a of a long and distinguished history. Our profession, second oldest, would be better appreciated ifwe were to develop a sense of our evolution.

7.1 Before the Greeks

2 Throughout antiquity, structural engineering existing as an art rather than a science. No record existsof any rational consideration, either as to the strength of structural members or as to the behavior ofstructural materials. The builders were guided by rules of thumbs and experience, which were passed fromgeneration to generation, guarded by secrets of the guild, and seldom supplemented by new knowledge.Despite this, structures erected before Galileo are by modern standards quite phenomenal (pyramids,Via Appia, aqueducs, Colisseums, Gothic cathedrals to name a few).

3 The first structural engineer in history seems to have been Imhotep, one of only two commoners to bedeified. He was the builder of the step pyramid of Sakkara about 3,000 B.C., and yielded great influenceover ancient Egypt.

4 Hamurrabi’s code in Babylonia (1750 BC) included among its 282 laws penalties for those “architects”whose houses collapsed, Fig. 7.1.

7.2 Greeks

5 The greek philosopher Pythagoras (born around 582 B.C.) founded his famous school, which wasprimarily a secret religious society, at Crotona in southern Italy. At his school he allowed neithertextbooks nor recording of notes in lectures, on pain of death. He taught until the age of 95, and is

Draft7–2 A BRIEF HISTORY OF STRUCTURAL ARCHITECTURE

228. If a builder build a house for some one and complete it, he shallgive him a fee of two shekels in money for each sar of surface.229 If a builder build a house for some one, and does not construct itproperly, and the house which he built fall in and kill its owner, thenthat builder shall be put to death.230. If it kill the son of the owner the son of that builder shall be put todeath.231. If it kill a slave of the owner, then he shall pay slave for slave tothe owner of the house.232. If it ruin goods, he shall make compensation for all that has beenruined, and inasmuch as he did not construct properly this house whichhe built and it fell, he shall re-erect the house from his own means.233. If a builder build a house for some one, even though he has not yetcompleted it; if then the walls seem toppling, the builder must make thewalls solid from his own means.

Figure 7.1: Hamurrabi’s Code

reported to have coined the term mathematics which means literally the “science of learning” (and alsothe word philosopher meaning “one who loves wisdom”).

6 Aristotle (384-322 B.C.) was Dean of the Lyceum, a college just outside the city gates of Athens,and was a man of universal ability. He is credited with having written in more than 25 different fieldsof knowledge. One of the most influential men of early civilization.

7 A pupil of Aristotle was Alexander the Great (356-323 B.C.) who founded the city of Alexandriain 323. Upon his death, one of his generals Ptolemy I became Pharaoh and established a library.The library of Alexandria was founded with the private library of Aristotle as a nucleus, and laterbecame the largest of the ancient world, containing about 700,000 scrolls. Many of these scrolls weresubsequently brought to the attention of the western world through translations by the arabs.

8 Alexandria was also the seat of the first university (with a reported enrollment of 14,000 students),and its first professor of geometry was Euclid (315-250 B.C.).

9 The greatest of the Greeks was Archimedes (287-212) who was one of the greatest physicist of theancient world and one of its greatest mathematician, Fig. 7.2. He is considered by many as the founderof mechanics because of his treatise “On Equilibrium”. He introduced the concept of center of gravity.He refused to write about “practical stuff” such as machines, catapults, spiral pumps, and others. Itwas one such invention (the lens) which kept the Roman armies at bay outside Syracuse for three years.When the city fell, he was supposed to have had his life spared. But the circumstances of his subsequentdeath are obscure. By some accounts he was killed by an ignorant soldier who disobeyed orders, and byother he was slain because he was too busy solving a mathematical problem to appear in front of theRoman consul and conqueror of Syracuse.


Draft7.3 Romans 7–3

Figure 7.2: Archimed

7.3 Romans

10 Science made much less progress under the Romans than under the Greeks. The Romans apparentlywere more practical, and were not as interested in abstract thinking though they were excellent fightersand builders.

11 As the roman empire expanded, the Romans built great roads (some of them still in use) such as theVia Appia, Cassia, Aurelia; Also they built great bridges (such as the third of a mile bridge over theRhine built by Caesars), and stadium (Colliseum).

12 One of the most notable Roman construction was the Pantheon, Fig. 7.3. It is the best-preserved

Figure 7.3: Pantheon

major edifice of ancient Rome and one of the most significant buildings in architectural history. Inshape it is an immense cylinder concealing eight piers, topped with a dome and fronted by a rectangularcolonnaded porch. The great vaulted dome is 43 m (142 ft) in diameter, and the entire structure islighted through one aperture, called an oculus, in the center of the dome. The Pantheon was erected bythe Roman emperor Hadrian between AD 118 and 128.

13 Marcus Vitruvius Pollio (70?-25 BC) was a Roman architect and engineer. He was an artilleryengineer in the service of the first Roman emperor, Augustus. His Ten Books on Architecture (Vitruvius



1960) is the oldest surviving work on the subject and consists of dissertations on a wide variety ofsubjects relating to architecture, engineering, sanitation, practical hydraulics, acoustic vases, and thelike. Much of the material appears to have been taken from earlier extinct treatises by Greek architects.Vitruvius’s writings have been studied ever since the Renaissance as a thesaurus of the art of classicalRoman architecture, Fig. 7.4.

Figure 7.4: From Vitruvius Ten Books on Architecture, (Vitruvius 1960)

7.4 The Medieval Period (477-1492)

14 This period, also called the Dark Ages, was marked by a general decline of civilization throughoutEurope following the decline and fall of the western Roman Empire.

15 The eastern Roman Empire on the other hand was to continue, and the center of Greek life had bythen been transferred to Constantinople. This city exerted great influence throughout Asia Minor.

16 Hagia Sophia, also Church of the Holy Wisdom, Fig. 7.5, was the most famous Byzantine structurein Constantinople (now Istanbul). Built (532-37) by Emperor Justinian I, its huge size and daringtechnical innovations make it one of the world’s key monuments. The size of its dome though, 112 ft,was nevertheless smaller than the one of the Pantheon in Rome.

17 During that period, the Arabs carried the torch of knowledge, gave birth to algebra, translated someof the great books of the Library of Alexandria.

18 Architecture, was the most important and original art form during the Gothic period, (Anon. xx). Theprincipal structural characteristics of Gothic architecture arose out of medieval masons’ efforts to solvethe problems associated with supporting heavy masonry ceiling vaults over wide spans. The problem wasthat the heavy stonework of the traditional arched barrel vault and the groin vault exerted a tremendousdownward and outward pressure that tended to push the walls upon which the vault rested outward,


Draft7.5 The Renaissance 7–5

Figure 7.5: Hagia Sophia

thus collapsing them. A building’s vertical supporting walls thus had to be made extremely thick andheavy in order to contain the barrel vault’s outward thrust.

Medieval masons solved this difficult problem about 1120 with a number of brilliant innovations.First and foremost they developed a ribbed vault, in which arching and intersecting stone ribs supporta vaulted ceiling surface that is composed of mere thin stone panels. This greatly reduced the weight(and thus the outward thrust) of the ceiling vault, and since the vault’s weight was now carried atdiscrete points (the ribs) rather than along a continuous wall edge, separate widely spaced vertical piersto support the ribs could replace the continuous thick walls. The round arches of the barrel vault werereplaced by pointed (Gothic) arches which distributed thrust in more directions downward from thetopmost point of the arch.

Since the combination of ribs and piers relieved the intervening vertical wall spaces of their supportivefunction, these walls could be built thinner and could even be opened up with large windows or otherglazing. A crucial point was that the outward thrust of the ribbed ceiling vaults was carried across theoutside walls of the nave, first to an attached outer buttress and then to a freestanding pier by meansof a half arch known as a flying buttress. The flying buttress leaned against the upper exterior of thenave (thus counteracting the vault’s outward thrust), crossed over the low side aisles of the nave, andterminated in the freestanding buttress pier, which ultimately absorbed the ceiling vault’s thrust.

These elements enabled Gothic masons to build much larger and taller buildings than their Ro-manesque predecessors and to give their structures more complicated ground plans. The skillful use offlying buttresses made it possible to build extremely tall, thin-walled buildings whose interior structuralsystem of columnar piers and ribs reinforced an impression of soaring verticality.

19 Vilet-Le-Duc classical book, (le Duc 1977) provided an in depth study of Gothic architecture.

7.5 The Renaissance

20 During the Renaissance there was a major revival of interest in science and art.



7.5.1 Leonardo da Vinci 1452-1519

21 Leonardo da Vinci was the most outstanding personality of that period (and of human civilizationfor that matter). He was not only a great artist (Mona Lisa), but also a great scientist and engineer.

22 He did not write books, but much information was found in his notebooks, one the most famous(Codex Leicester) was recently purchased by Bill Gates whose company Corbis made a CD-ROM fromit.

23 He was greatly interested in mechanics, (Timoshenko 1982), and in one of his notes he states

“Mechanics is the Paradise of mathematical science because here we come to the fruits ofmathematics.”

24 He was the first to explore concepts of mechanics, since Archimedes, using a scientific approach.He applied the principle of virtual displacements to analyze various systems of pulleys and levers. Heappears to have developped a correct idea of the thrust produced by an arch.

25 In one of his manuscripts there is a sketch of two members on which a vertical load Q is acting andthe question is asked: What forces are needed at a and b to have equilibrium? (From the dotted lineparallelogram, in the sketch, it can be concluded that he had the right answer).

26 Leonardo also studied the strength of structural materials experimentally. He tried to determine thetensile strength of an iron wire of different length (size effect).

27 He also studied the load carrying capacity of a simply supported uniformly loaded beam and concludedthat “the strength of the beam supported at both ends varies inversely as the length and directly as thewidth” (is this correct? how about the depth of the beam?).

28 For a cantilevered beam he stated

“If a beam 2 braccia long supports 100 libre, a beam 1 braccia long will support 200”

Finally, Leonardo briefly studied the strength of columns and found that

“it varies inversely as its length, but directly as some ratio of its cross section.”

29 Leonardo’s was the first indeed to attempt to apply statics in finding the forces acting in members ofstructures, friction and the first to perform experiments to determine the strength of structural materials.

30 Interestingly, here is Leonardo’s definition of force, (Penvenuto 1991)

“I say that force is a spiritual virtue, an invisible power, which, through accidental exteriorviolence, is caused by motion and placed and infused into bodies which are [thus] removedand deviated from their natural use, giving to such virtue an active life of marvelous power”.

31 Unfortunatly, these important findings, were buried in his notes, and engineers in the fifteenth andsixteenth centuries continued, as in the Roman era, to fix dimensions of structural elements by relyingon experience and judgment.

7.5.2 Brunelleschi 1377-1446

32 Brunelleschi was a Florentine architect and one of the initiators of the Italian Renaissance. Hisrevival of classical forms and his championing of an architecture based on mathematics, proportion, andperspective make him a key artistic figure in the transition from the Middle Ages to the modern era.



33 He was born in Florence in 1377 and received his early training as an artisan in silver and gold. In 1401he entered, and lost, the famous design competition for the bronze doors of the Florence Baptistery. Hethen turned to architecture and in 1418 received the commission to execute the dome of the unfinishedGothic Cathedral of Florence, also called the Duomo. The dome, Fig. 7.6 a great innovation both

Figure 7.6: Florence’s Cathedral Dome

artistically and technically, consists of two octagonal vaults, one inside the other. Its shape was dictatedby its structural needs one of the first examples of architectural functionalism. Brunelleschi made adesign feature of the necessary eight ribs of the vault, carrying them over to the exterior of the dome,where they provide the framework for the dome’s decorative elements, which also include architecturalreliefs, circular windows, and a beautifully proportioned cupola. This was the first time that a domecreated the same strong effect on the exterior as it did on the interior.

34 Completely different from the emotional, elaborate Gothic mode that still prevailed in his time,Brunelleschi’s style emphasized mathematical rigor in its use of straight lines, flat planes, and cubicspaces. This set the tone for many of the later buildings of the Florentine Renaissance.

35 His influence on his contemporaries and immediate followers was very strong and has been felt evenin the 20th century, when modern architects came to revere him as the first great exponent of rationalarchitecture.

7.5.3 Alberti 1404-1472

36 Alberti was an Italian architect and writer, who was the first important art theorist of the Re-naissance and among the first to design buildings in a pure classical style based on a study of ancientRoman architecture.



37 He was born in Genoa, the son of a Florentine noble. He received the best education available inthe 15th century. He was proficient in Greek, mathematics, and the natural sciences. As a poet, aphilosopher, and one of the first organists of his day, Alberti greatly influenced his contemporaries.

38 Alberti’s architectural training began with the study of antique monuments during his first stay inRome. Subsequently he joined the papal court in Florence, where he became intensely involved with thecultural life of the city. Probably at this time he became familiar with the mathematical laws of linearperspective, which Brunelleschi had studied.

39 Alberti took took an active part in the literary life of Florence and championed the literary use ofItalian rather than the use of Latin.

40 In the late 1440s, Alberti began to work as an architect. Although his buildings rank among the bestarchitecture of the Renaissance, he was a theoretical rather than a practical architect. He furnishedthe plans of his buildings but never supervised their construction. His De Re Aedificatoria (1485) wasthe first printed work on architecture of the Renaissance. He also wrote books on sculpture, the family,government, and literature.

7.5.4 Palladio 1508-1580

41 Andrea Palladio was one of the most influential architects in European history. He was born inPadua, and trained as a stonemason, he later joined the poet Trissino who took him to Rome, wherePalladio studied and measured Roman architectural ruins; he also absorbed the treatises of Vitruvius.One outcome of these studies was Palladio’s Antiquities of Rome (1554) (Palladio 19xx), the principalguidebook to Roman ruins for the next two centuries.

42 In and near Vicenza he designed many residences and public buildings. The best known of theseare the Barbarano, Chieregati, Tiene, Porto, and Valmarana palaces and the Villa Capri, or VillaRotunda. Although the historical antecedents of Palladio’s style are the classically Roman-influencedHigh Renaissance works of architects such as the Italian Donato Bramante, Palladio’s own use of classicalmotifs came through his direct, extensive study of Roman architecture. He freely recombined elementsof Roman buildings as suggested by his own building sites and by contemporary needs, Fig. 7.7 Atthe same time he shared the Renaissance concern for harmonious proportion, and his facades have anoteworthy simplicity almost austerity and repose.

43 Palladio was the author of an important scientific treatise on architecture, I Quattro Libri dell’Architettura,(Palladio 19xx) which was widely translated and influenced many later architects. Its precise rules andformulas were widely utilized, especially in England, and were basic to the Palladian style, adoptedby Inigo Jones, Christopher Wren, and other English architects, which preceded and influenced theneoclassical architecture of the Georgian Style.

7.5.5 Stevin

44 Stevin, Fig. 7.8, was a Dutch mathematician and engineer who founded the science of hydrostaticsby showing that the pressure exerted by a liquid upon a given surface depends on the height of the liquidand the area of the surface.

45 Stevin was a bookkeeper in Antwerp, then a clerk in the tax office at Brugge. After this he moved toLeiden where he first attended the Latin school, then he entered the University of Leiden in 1583 (at theage of 35). While quartermaster in the Dutch army, Stevin invented a way of flooding the lowlands inthe path of an invading army by opening selected sluices in dikes. He was an outstanding engineer who



Figure 7.7: Palladio’s Villa Rotunda

Figure 7.8: Stevin



built windmills, locks and ports. He advised the Prince Maurice of Nassau on building fortifications forthe war against Spain.

46 The author of 11 books, Stevin made significant contributions to trigonometry, geography, fortifica-tion, and navigation. Inspired by Archimedes, Stevin wrote important works on mechanics. In his bookDe Beghinselen der Weeghconst in 1586 appears the theorem of the triangle of forces giving impetus tostatics. In 1586 (3 years before Galileo) he reported that different weights fell a given distance in thesame time.

7.5.6 Galileo 1564-1642

47 Galilei Galileo was born in Pisa in 1564. He received his early education in Latin, Greek and logicnear Florence, Fig. 7.9. Just as his father had played an important role in the musical revolution

Figure 7.9: Galileo

from medieval polyphony to harmonic modulation, Galileo came to see Aristotelian physical theology aslimiting scientific inquiry.

48 In 1581 he entered the University of Pisa to study medicine, but he soon turned to philosophy andmathematics, leaving the university without a degree in 1585. For a time he tutored privately and wroteon hydrostatics and natural motions, but he did not publish.

49 In 1589 he became professor of mathematics at Pisa, where he is reported to have shown his studentsthe error of Aristotle’s belief that speed of fall is proportional to weight, by dropping two objects ofdifferent weight simultaneously from the Leaning Tower, thus modern dynamics was born. His contractwas not renewed in 1592, probably because he contradicted Aristotelian professors. The same year, hewas appointed to the chair of mathematics at the University of Padua, where he remained until 1610.

50 In Padua he achieved great fame, and lecture halls capable of containing 2,000 students from allover Europe were used. In 1592 he wrote Della Scienza Meccanica in which various problems of staticswere treated using the principle of virtual displacement. He subsequently became interested inastronomy and built one of the first telescope through which he saw Jupiter and became an ardentproponent of the Copernican theory (which stated that the planets circle the sun as opposed to theAristotelian and Ptolemaic assumptions that it was the sun which was circling Earth). This theorybeing condemned by the church, he received a semiofficial warning to avoid theology and limit himselfto physical reasoning. When he published his books dealing with the two ways of regarding the universe(which clearly favored the Copernican theory) he was called to Rome by the Inquisition, condemned andhad to read his recantation (At the end of his process he murmured the famous e pur se muove).



51 When he was almost seventy years old, his life shattered by the Inquisition, he retired to his villanear Florence and wrote his final book, Discourses Concerning Two New Sciences, (Galilei 1974), Fig.7.10. His first science was the study of the forces that hold objects together -the dialogue taking place

Figure 7.10: Discourses Concerning Two New Sciences, Cover Page

in a shipyard, triggered by observations of craftsmen building the Venetian fleet. His second scienceconcerned local motions - laws governing the trajectory of projectiles. A portion of the book dealingwith the mechanical properties of structural materials and with the strength of beams. Strength ofMaterials as a discipline was born.

52 He observed that if we make structures geometrically similar, then with increase of the dimensionsthey become weaker and weaker, One cannot reason from the small to the large, because many mechanicaldevices succeed on a small scale that cannot exist in great size.

53 It is interesting to note that when Galileo studied the strength of a cantilevered (wooden) beam withan applied load at the end, Fig. 7.11, he failed to properly understand the exact internal stress/strain

Figure 7.11: “Galileo’s Beam”

distribution. He determined that the stress is constant throughout the cross section (whereas as weknow it varies linearly).

54 Galileo’s lifelong struggle to free scientific inquiry from restriction by philosophical and theologicalinterference stands beyond science. Since the full publication of Galileo’s trial documents in the 1870s,



entire responsibility for Galileo’s condemnation has customarily been placed on the Roman Catholicchurch. This conceals the role of the philosophy professors who first persuaded theologians to linkGalileo’s science with heresy. An investigation into the astronomer’s condemnation, calling for its rever-sal, was opened in 1979 by Pope John Paul II. In October 1992 a papal commission acknowledged theVatican’s error.

7.6 Pre Modern Period, Seventeenth Century

7.6.1 Hooke, 1635-1703

55 Hooke was best known for his study of elasticity but also original contributions to many other fieldsof science.

56 Hooke was born on the Isle of Wight and educated at the University of Oxford. He served as assistantto the English physicist Robert Boyle and assisted him in the construction of the air pump. In 1662Hooke was appointed curator of experiments of the Royal Society and served in this position until hisdeath. He was elected a fellow of the Royal Society in 1663 and was appointed Gresham Professor ofGeometry at Oxford in 1665. After the Great Fire of London in 1666, he was appointed surveyor ofLondon, and he designed many buildings.

57 Hooke anticipated some of the most important discoveries and inventions of his time but failed to carrymany of them through to completion. He formulated the theory of planetary motion as a problem inmechanics, and grasped, but did not develop mathematically, the fundamental theory on which Newtonformulated the law of gravitation.

58 His most important contribution was published in 1678 in the paper De Potentia Restitutiva. Itcontained results of his experiments with elastic bodies, and was the first paper in which the elasticproperties of material was discused, Fig. 7.12.

Figure 7.12: Experimental Set Up Used by Hooke

“Take a wire string of 20, or 30, or 40 ft long, and fasten the upper part thereof to a nail,and to the other end fasten a Scale to receive the weights: Then with a pair of compasses takethe distance of the bottom of the scale from the ground or floor underneath, and set down the


Draft7.6 Pre Modern Period, Seventeenth Century 7–13

said distance,then put inweights into the said scale and measure the several stretchings of thesaid string, and set them down. Then compare the several strtchings of the said string, andyou will find that they will always bear the same proportions one to the other that the weightsdo that made them”.

This became Hooke’s Law σ = Eε.

59 Because he was concerned about patent rights to his invention, he did not publish his law when firstdiscovered it in 1660. Instead he published it in the form of an anagram “ceiinosssttuu” in 1676 andthe solution was given in 1678. Ut tensio sic vis (at the time the two symbos u and v were employedinterchangeably to denote either the vowel u or the consonant v), i.e. extension varies directly with force.

7.6.2 Newton, 1642-1727

60 Born on christmas day in the year of Galileo’s death, Newton, Fig. 7.13 was Professor of Mathematics

Figure 7.13: Isaac Newton

at Cambridge university.

61 In 1684 Newton’s solitude was interrupted by a visit from Edmund Halley, the British astronomer andmathematician, who discussed with Newton the problem of orbital motion. Newton had also pursuedthe science of mechanics as an undergraduate, and at that time he had already entertained basic notionsabout universal gravitation. As a result of Halley’s visit, Newton returned to these studies.

62 During the following two and a half years, Newton established the modern science of dynamics byformulating his three laws of motion. Newton applied these laws to Kepler’s laws of orbital motionformulated by the German astronomer Johannes Kepler and derived the law of universal gravitation.Newton is probably best known for discovering universal gravitation, which explains that all bodiesin space and on earth are affected by the force called gravity. He published this theory in his bookPhilosophiae Naturalis Principia Mathematica or simply Principia, in 1687, Fig. 7.14. This bookmarked a turning point in the history of science.

63 The Principia’s appearance also involved Newton in an unpleasant episode with the English philoso-pher and physicist Robert Hooke. In 1687 Hooke claimed that Newton had stolen from him a central



Figure 7.14: Philosophiae Naturalis Principia Mathematica, Cover Page

idea of the book: that bodies attract each other with a force that varies inversely as the square of theirdistance. However, most historians do not accept Hooke’s charge of plagiarism.

64 Newton also engaged in a violent dispute with Leibniz over priority in the invention of calculus.Newton used his position as president of the Royal Society to have a committee of that body investigatethe question, and he secretly wrote the committee’s report, which charged Leibniz with deliberateplagiarism. Newton also compiled the book of evidence that the society published. The effects of thequarrel lingered nearly until his death in 1727.

65 In addition to science, Newton also showed an interest in alchemy, mysticism, and theology. Manypages of his notes and writings particularly from the later years of his career are devoted to these topics.However, historians have found little connection between these interests and Newton’s scientific work.

7.6.3 Bernoulli Family 1654-1782

66 The Bernouilli family originally lived in Antwerp, but because of religious persecution, they leftHolland and settled in Basel. Near the end of the seventeenth century this family produced outstandingmathematicians for more than a hundred years. Jacob and John were brothers. John was the father ofDaniel, and Euler his pupil.

67 Whereas Galileo (and Mariotte) investigated the strength of beams (Strength), Jacob Bernoulli (1654-1705) made calculation of their deflection (Stiffness) and did not contribute to our knowledge of physicalproperties. Jacob Bernouilli is also credited in being the first to to have assumed that a bf plane sectionof a beam remains plane during bending, but assumed rotation to be with respect to the lower fiber(as Galileo did) and this resulted in an erroneous solution (where is the exact location of the axis ofrotation?). He also showed that the curvature at any point along a beam is proportional to the curvatureof the deflection curve.

68 Bernoulli made the first analytical contribution to the problem of elastic flexure of a beam. In 1691he published a logogriph Qrzumubapt dxqopddbbp ... whose secret was revealed in 1694. A letter isreplaced by the next in the Latin alphabet, the second by the letter three away, and the third by theletter six away, so that aaaaa would be encoded as bdgbd. The logogriph reads Portio axis applicatem...and the decoded is that the radius of curvature at any point of an initially straight beam in inverselyproportional to the value of the bending moment at that point.

69 Daniel Bernoulli (1700-1782) first postulated that a force can be decomposed into its equivalent


Draft7.6 Pre Modern Period, Seventeenth Century 7–15

(“Potentiis quibuscunque possunt substitui earundem aequivalentes”. Another hypothesis defined thesum of two “conspiring” forces applied to the same point. According to Bernoulli, this “necessary truth”follows from the metaphysical principle that the whole equalts the sum of its parts, (Penvenuto 1991).

7.6.4 Euler 1707-1783

70 Leonhard Euler was born in Basel and early on caught the attention of John Bernoulli whoseteaching was attracting young mathematicians from all over Europe, Fig. 7.15. He obtained his Master

Figure 7.15: Leonhard Euler

at age 16, and before the age of 20 won a competition from the French Academy of Sciences. At age 20he moved to the Russian Academy of Sciences in St Petersburg along with the two sons of John Bernoulli(Nicholas and Daniel). He was appointed professor of physics in 1730 and professor of mathematics in1733. In 1741 he became professor of mathematics at the Berlin Academy of Sciences at the urging ofthe Prussian king Frederick the Great. Euler returned to St. Petersburg in 1766, remaining there untilhis death.

71 Although hampered from his late 20s by partial loss of vision and in later life by almost total blind-ness, Euler produced a number of important mathematical works and hundreds of mathematical andscientific memoirs. In his Introduction to the Analysis of Infinities, Euler gave the first full analyticaltreatment of algebra, the theory of equations, trigonometry, and analytical geometry. In this work hetreated the series expansion of functions and formulated the rule that only convergent infinite series canproperly be evaluated. He also discussed three-dimensional surfaces and proved that the conic sectionsare represented by the general equation of the second degree in two dimensions. Other works dealt withcalculus, including the calculus of variations, number theory, imaginary numbers, and determinate andindeterminate algebra. Euler, although principally a mathematician, made contributions to astronomy,mechanics, optics, and acoustics.

72 In Russia he wrote a famous book in mechanics in which instead of applying the geometrical methodsused by Newton, he introduced analytical methods.

73 As a mathematician, Euler was interested principally in the geometrical forms of elastic curves. Heapproached problems from the point of view of variational calculus and in the introduction of his bookMethodus inveniendi lineas curva ... he stated

Since the fabric of the universe is most perfect, and is the work of a most wise Creator, nothingwhatsover takes place in the universe in which some relation o maximum an minimum does



not appear. Therefore there is absolutely no doubt that every effect in the universe can beexplained as satisfactorily from final causes, by the aid of the method of maxima and minima,as it can from the effective causes themselves...

74 Euler obtained a near exact expression for the deflection of a cantilever subjected to a point load, andfor the buckling load of a column.

7.7 The pre-Modern Period; Coulomb and Navier

75 Coulomb (1736-1806) was a French military engineer, Fig. 7.16, as was the first to publish the correct

Figure 7.16: Coulomb

analysis of the fiber stresses in flexed beam with rectangular cross section (Sur une Application desRegles de maximis et minimis a quelques problemes de statique relatifs a l’architecture in 1773). He usedHooke’s law, placed the neutral axis in its exact position, developed the equilibrium of forces on thecross section with external forces, and then correcly determined the stresses. He also worked on friction(“Coulomb friction”) and on earth pressure.

76 Coulomb did also research on magnetism, friction, and electricity. In 1777 he invented the torsionbalance for measuring the force of magnetic and electrical attraction. With this invention, Coulombwas able to formulate the principle, now known as Coulomb’s law, governing the interaction betweenelectric charges. In 1779 Coulomb published the treatise Theorie des machines simples (Theory ofSimple Machines), an analysis of friction in machinery. After the war Coulomb came out of retirementand assisted the new government in devising a metric system of weights and measures. The unit ofquantity used to measure electrical charges, the coulomb, was named for him.

77 Navier (1785-1836) Navier was educated at the Ecole Polytechnique and became a professor therein 1831. Whereas the famous memoir of Coulomb (1773) contained the correct solution to numerousimportant problems in mechanics of materials, it took engineers more than forty years to understandthem correctly and to use them in practical application

78 In 1826 he published his Lecons (lecture notes) which is considered the first great textbook in me-chanics for engineering. In it he developed the first general theory of elastic solids as well as the firstsystematic treatment of the theory of structures.

79 It should be noted that no clear division existed between the theory of elasticity and the theoryof structures until about the middle of the nineteenth century (Coulomb and Navier would today beconsidered professional structural engineers).


Draft7.8 The Modern Period (1857-Present) 7–17

80 Three other structural engineers who pioneered the development of the theory of elasticity from thatpoint on were Lame, Clapeyron and de Saint-Venant. Lam’e published the first book on elasticityin 1852, and credited Clapeyron for the theorem of equality between external and internal work. deSaint-Venant was perhaps the greatest elasticians who according to Southwell “... combined with highmathematical ability an essentially practical outlook which gave direction to all his work”. In 1855-6 hepublished his classical work on torsion, flexure, and shear stresses.

7.8 The Modern Period (1857-Present)

7.8.1 Structures/Mechanics

81 From 1857 the evolution of a comprehensive theory of structures proceeded at astonishing rate nowthat the basic and requisite principles had been determined.

82 Great contributors in that period include: Maxwell (first analysis of indeterminate structures), Cul-mann (graphics statics), Mohr (Mohr’s circle, indeterminate analysis), Castigliano (1st and 2nd theo-rems), Cross (moment distribution), Southwell (relaxation method).

7.8.2 Eiffel Tower

83 The Eiffel Tower was designed and built by the French civil engineer Alexandre Gustave Eiffel forthe Paris World’s Fair of 1889. The tower, without its modern broadcasting antennae, is 300 m (984ft) high. The lower section consists of four immense arched legs set on masonry piers. The legs curveinward until they unite in a single tapered tower. Platforms, each with an observation deck, are at threelevels; on the first is also a restaurant.

84 The tower, constructed of about 6300 metric tons (about 7000 tons) of iron, has stairs and elevators.A meteorological station, a radio communications station, and a television transmission antenna, as wellas a suite of rooms that were used by Eiffel, are located near the top of the tower.

7.8.3 Sullivan 1856-1924

85 Sullivan wan an American architect, whose brilliant early designs for steel-frame skyscraper construc-tion led to the emergence of the skyscraper as the distinctive American building type. Through his ownwork, especially his commercial structures, and as the founder of what is now known as the Chicagoschool of architects, he exerted an enormous influence on 20th-century American architecture. His mostfamous pupil was the architect Frank Lloyd Wright, who acknowledged Sullivan as his master.

86 After studying architecture at the Massachusetts Institute of Technology, he spent a year in Parisat the Ecole des Beaux-Arts and in the office of a French architect. Settling in Chicago in 1875, hewas employed as a draftsman, then in 1881 formed a partnership with Dankmar Adler. Together theyproduced more than 100 buildings.

87 Adler secured the clients and handled the engineering and acoustical problems, while Sullivan con-cerned himself with the architectural designs. One of their earliest and most distinguished joint enter-prises was the ten-story Auditorium Building (1886-89) in Chicago. This famous showplace incorporateda hotel, an office building, and a theater renowned for its superb acoustics. The Wainwright Building,also ten stories high, with a metal frame, was completed in 1891 in Saint Louis, Missouri. In 1895 theSullivan-Adler partnership was dissolved, leading to a decline in Sullivan’s practice. The Carson PirieScott (originally Schlesinger and Meyer) Department Store, Chicago, regarded by many as Sullivan’smasterpiece, was completed in 1904.



88 His famous axiom, Form follows function became the touchstone for many in his profession. Sul-livan, however, did not apply it literally. He meant that an architect should consider the purpose of thebuilding as a starting point, not as a rigidly limiting stricture.

89 He also had tremendous respect for the natural world which played an enormous role in forging histheories about architecture (he spent all of his first summers on his grandparents’ farm in Massachusettswhere he developed this love and respect for nature) expressed in his Autobiography of an Idea), 1924).

7.8.4 Roebling, 1806-1869

90 John Augustus Roebling was an American civil engineer, who was one of the pioneers in theconstruction of suspension bridges. He was born in Germany, educated at the Royal Polytechnic Schoolof Berlin and immigrated to the States in 1831.

91 In his first job he was employed by the Pennsylvania Railroad Corp. to survey its route across theAllegheny Mountains between Harrisburg and Pittsburgh. He then demonstrated the practicability ofsteel cables in bridge construction and in 1841 established at Saxonburg the first factory to manufacturesteel-wire rope in the U.S.

92 Roebling utilized steel cables in the construction of numerous suspension bridges and is generallyconsidered one of the pioneers in the field of suspension-bridge construction. He built railroad suspensionbridges over the Ohio and Niagara rivers and completed plans for the Brooklyn Bridge shortly beforehis death. Roebling was the author of Long and Short Span Railway Bridges (1869).

7.8.5 Maillart

From (Billington 1973)

93 Robert Maillart was born on February 6, 1872, in Bern, Switzerland, where his father, a Belgiancitizen, was a banker. He studied civil engineering at the Federal Institute of Technology in Zurich andgraduated in 1894. Ironically, one of his lowest grades was in bridge design, even though he is regardedtoday as one of the half dozen greatest bridge designers of the twentieth century.

94 For eight years following his graduation, he worked with different civil engineering organizations. In1902, he founded his own firm for design and construction; thereafter, his business grew rapidly andexpanded as far as Russia and Spain. In the summer of 1914, he took his wife and three children toRussia. Since the World War prevented their return to Switzerland, Maillart stayed and worked inRussia until 1919, when his business was liquidated by the Revolution. Forced to flee, he returned toSwitzerland penniless and lonely, his wife having died in Russia.

95 Because of these misfortunes Maillart felt unable to take up the construction business again andhenceforth concentrated on design alone. He opened an office in Geneva in 1919 and branches in Bernand Zurich in 1924.

96 During the twenties he began to develop and modify his ideas of bridge design; and from 1930, whenthe Salginatobel and Landquart Bridges were completed, until his death in 1940, he produced overthirty bridge designs of extraordinary originality. Unfortunately, no Swiss municipality would accept hisdesigns for prominent urban locations.

97 In 1936, he was elected an honorary member of the Royal institute of British Architects (R.I.B.A.)although he had never officially acted as architect on any project. The 1941 first edition of Space, Timeand Architecture by art historian Siegfried Giedion introduced Maillart to a wide public in the U.S.A.Finally, Max Bill’s 1949 book, Robert Maillart, with its photographs and commentary on nearly allMaillart’s bridges powerfully presented him as an artist of first rank.



7.8.6 Nervi, 1891-1979

98 Pier Luigi Nervi was an Italian architect and engineer, whose technical innovations, particularlyin the use of reinforced concrete, made possible aesthetically pleasing solutions to difficult structuralproblems.

99 He attended the Civil Engineering School in Bologna and established his own firm in 1920. His firstmajor commission (a stadium in Florence, 1932) features cantilevered beams and a daringly exposedconcrete structure. For airplane hangars he used reinforced concrete to cover enormous spans with alight but strong latticework.

100 Nervi considered himself primarily an engineer and technician, not an architect, and hestrove primarily for strength through form. He maintained that the strong aesthetic appeal of hisbuildings was simply a by-product of their structural correctness. His introduction of a versatile newtype of reinforced concrete layers of fine steel mesh sprayed with cement mortar made possible one of hismasterpieces, the Turin Exposition Hall (1949), in which the approximately 76-m (250-ft) corrugatedlattice roof (only about 5 cm thick) creates an immense interior space as dramatic as a cathedral.

101 The best known and most influential is probably his Palazetto dello Sport (Small Sport Palace, 1960,Rome), Fig. 7.17. Encircled by Y-shaped supports and topped by a shallow scalloped concrete dome,

Figure 7.17: Nervi’s Palazetto Dello Sport

this building has become a paradigm of the 20th-century sports arena.

7.8.7 Khan

Fazlur Khan was born in 1929 in Dacca India, (Anon. xx). After obtaining a B.A. in engineering fromthe University of Dacca in 1950, Khan worked as assistant engineer for the India Highway Department



and taught at the University of Dacca. Qualifying for a scholarship in 1952, he enrolled at the Uni-versity of Illinois, Champaign-Urbana, where he received master’s degrees in both applied mechanicsand structural engineering and a Ph.D. in structural engineering. He returned briefly to Pakistan andwon an important position as executive engineer of the Karachi Development Authority. Frustrated byadministrative demands that kept him from design work, however, he returned to the United States andjoined the prestigious architectural firm of Skidmore, Owings & Merrill in Chicago in 1955, eventuallybecoming a partner (1966).

Among his many designs for skyscrapers are Chicago’s John Hancock Center (1970) and the SearsTower (1973), which are among the world’s tallest buildings, and One Shell Plaza in Houston, Texas.The Sears Tower was his first skyscraper to employ his ”bundled tube” structural system, which consistsof a group of narrow steel cylinders that are clustered together to form a thicker column. The systemwas innovative because it minimized the amount of steel needed for high towers, eliminated the needfor internal wind bracing (since the perimeter columns carried the wind loadings), and permitted freerorganization of the interior space.

His later projects included the strikingly different Haj Terminal of the King Abdul Aziz InternationalAirport, Jiddah, Saudi Arabia (1976-81), and King Abdul Aziz University, also in Jiddah (1977-78).

7.8.8 et al.

102 To name just a few of the most influential Architects/Engineers: Menn, Isler, Candella, Torroja,Johnson, Pei, Calatrava, ...



NEW ARCHITECTUREby

Felix CandelaProfessor - University of Illinois

Unauthorized reprint from (Billington 1973)

The title of my lecture is New Architecture; but I cannot avoid the feeling that I have not too muchto do with this subject. I don’t think I can speak of my work as of any new architecture or even asarchitecture at all, and perhaps the same could be said of Maillart’s work. But, of course, it all dependson what you consider architecture and there is not anymore a general consent about the meaning of thisword,

But, anyway, upon discussing this with David Billington, he told me it could be interesting to knowwhat had been the influence of Maillart on my development, and all of a sudden I realized that he mayhave been one of the strongest influences at the critical moment in my career in which I was trying tobecome a builder of shells.

But let me speak first about my background, because it may be important to know what impelspeople to do things and the circumstances and difficulties that they had to deal with in order to achievetheir purposes.

I was trained as an architect with, you may say, quite a backward kind of curriculum, in Madridduring the thirties. We had only one course in strength of materials, but it was a very good course,dealing mainly with theory of elasticity and following the classical and rigoristic French tradition. Asyou can imagine, most students considered the matter completely useless for their professional practiceand, as it required some knowledge of mathematics, it was very difficult for most of them to pass theexamination. This gave me opportunity to do some private tutoring to my classmates, which was a veryinstructive manner to make some money to pay for my studies. As a result, I became more familiarwith the theoretical bases of the current methods of calculation of indeterminate structures. I discoveredlater that this modest background made me more knowledgeable on the matter than most practicingengineers whose training and interest tend to be directed towards mastering the accepted methods ofanalysis rather than questioning the basic hypotheses.

Anyway, since I never had a high opinion of myself as an artist, I was more interested in the technicalpart of the curriculum and began to read extensively about structures. Among my lectures I foundseveral French and German papers dealing with shells which were beginning to be in vogue at that timein Europe. Examples of such structures built in Germany and France could be found in magazines andTorroja was building the famous roof of the ”Fronton Recoletos” in Madrid, with an unusual shape anda record span.

Shells appeared to be an intriguing challenge for me, and I dreamed about the possibility of buildingsome in the future. But my lack of experience and my youthful faith in the impressive wisdom displayed inlearned magazines led me to believe that the key to shell design was in complete mathematical calculation,and I tried, rather unsucessfully, to understand and follow them and to make some sense of their results.I was not the only one misled and discouraged by the mathematical barrier so cunningly deployed byGerman engineers, a clever move which secured them practically an exclusive on the construction ofbarrel vaults for more than twenty years, hindering the normal employment of such structures duringthe same time.

However, such was my enthusiasm with this mathematical approach that I managed to get a fellowshipto go to Germany, hoping to learn something more from the German professors. But the outbreak ofthe Spanish Civil War saved me from such an ordeal. I could not leave the country and ended in Mexicoafter three years of military service, with no more baggage than my bare hands and no further additionto my academic background. After several years of general practice in Mexico, as draftsman, designerand contractor, I recalled my old fancy with shells and began to collect again papers on the subject.Whatever I learned from then on was to be the hard way, working alone, with no direct help from anyuniversity or engineering office. But I am indebted to many people who did help me through their



writings and Maillart was one of the foremost.I discovered him in Giedion’s Space, Time and Architecture; and then I got Max Bill’s book with

its invaluable collection of Maillart’s essays. I devoured his articles about ”Reinforced Concrete Designand Calculation” (he was very careful to differentiate the meaning of such words and to avoid themore than semantic confusion prevalent nowadays in English-speaking countries)., ”The Engineer andthe Authorities” which expresses his position in front of the establishment and ”Mass and Quality inReinforced Concrete Structures.” Very short papers, indeed, but well provided with opinions, somethingI could rarely find in other engineering articles. I learned later that to express personal opinions isconsidered bad taste among technical writers. Any discussion should be restricted to insignificant details,but never touch fundamental dogmas, in a fashion curiously similar to what could be expected of thecouncils of the Church or the meetings of any Politbureau.

But my attitude with respect to calculations of reinforced concrete structures was becoming un-orthodox, being tired perhaps of performing long and tedious routines whose results were not alwaysmeaningful. Therefore, I found Maillart’s thoughts delightfully sympathetic and encouraging. If a rebelwas able to produce such beautiful and sound structures there could not be anything wrong with be-coming also a rebel, which was besides, my only way to break the mystery surrounding shell analysis.

Thus, I started to follow the bibliographic tread and met, through their writings, with Freudenthal,Johansen, Van der Broek, Kist, Saliger, Kacinczy and so many others who showed me there was morethan a single and infallible manner to approach structural analysis. The discovery of rupture methods,with their emphasis on simple statics and their bearing on the actual properties of construction materialsand their behavior in the plastic range, allowed me to trust in simplified procedures to understand andanalyze the distribution of stresses in shell structures. It also helped me to get out of my naive belief inthe indisputable truth of the printed word and to start reading with a new critical outlook. No longer didI need to believe whatever was in print, no matter how high-sounding the name of the author. I couldmake my own judgements about what methods of stress analysis were better suited for my practice.

Since I was working practically alone, I could not afford nor had time for complex calculations anddid welcome Maillart’s advice that simpler calculations are more reliable than complex ones, especiallyfor somebody who builds his own structures. This was exactly my case and, since most structures I wasbuilding were of modest scale, I could control what was happening, check the results and confirm theaccuracy of my judgement or correct my mistakes. In a way, I was working with full scale models. Iunderstand that this was also true of Maillart who in many cases was the actual builder of his designs.

Following the general trend to mess up issues, there has been a lot of speculation about the engineeras an artist and in some instances, like in the case of Nervi, about the engineer as an architect (as if thetitle of architect could confer, per se, artistic ability to its holder); but few people realize that the onlyway to be an artist in this difficult specialty of building is to be your own contractor. in countries likethis, where the building industry has been thoroughly and irreversibly fragmented and the responsibilitydiluted among so many trades, it may be shocking to think of a contractor as an artist; but it is indeedthe only way to have in your hands the whole set of tools or instruments to perform the forgotten art ofbuilding, to produce ”works of art” which, by the way, was the common expression to designate a bridgein the old French engineering vocabulary.

Implicit in the above statement is the fact that you have to be, besides your own structural designerand calculator and perhaps your own architect, also your own contractor, a very difficult proposition insome countries where such mergings of today’s disparate professions may even be considered unlawful.This means, of course, that the price for being the master of your trade is to accept the whole responsi-bility for the good performance of the structure, and not too many people today would readily endorsesuch an awesome commitment. I am not advocating a return to the past; history is an irreversibleprocess. I am simply stating that the Maillart phenomenom could not happen under today’s situationof the industry.

I like to think, however, that Maillart did not judge himself an artist. As Picasso said of himself ”hewas not looking for beauty; he found it”. His main concerns must have been efficiency and economy ofmeans, since to be able to build one of his bridges he had to win a bidding competition and prove that



he could do it cheaper than anybody else. But an efficient and economical structure has not necessarilyto be ugly. Beauty has no price tag and there is never one single solution to an engineering problem.Therefore, it is always possible to modify the whole or the parts until the ugliness disappears. Thisaversion to ugliness is quite the opposite of the task of the professional artist who has to produce beautyas an obligation or of today’s star-architect who has to be original at any cost in each new project.

Maillart’s works did not need to be beautiful. This word did not even exist in the practical worldof the serious citizens who had to judge his competitive bids. He achieved a beauty without need orpurpose; just for the pure joy of it. The kind of joy that you can feel also in the works of Haydn orVivaldi. They were simply enjoying what they were doing, and so was obviously Maillart.

He did also possess that rare quality, source of artistic creation and of all invention, of being ableto challenge the conventional wisdom and come up with the obvious solution, one, nevertheless, whichnobody could think of before. I can imagine the fits of rage and jealousy of some of his contemporarycolleagues at the sight of one of his bridges (Landquart or Schwandbach), in which the curved routeis supported in a straight arched slab. The problem with this unusual combination - which, of course,looks perfectly logical after the fact - is that it was very difficult, if not impossible, to analyze with themethods available at that time. But Maillart would not take any unnecessary risk and first he tried thesoundness of his approximated calculations in a small example (the Halbkern Bridge) with a span ofonly fifteen meters. This was his testing model which gave him firm ground from which to extrapolateat the next opportunity.

I would like to insist at this moment c-n something that everybody knows but which is easily for-gotten; that all calculations, no matter how sophisticated and complex, can not be more than roughapproximations of the natural phenomenon they try to represent by means of a mathematical model.The complexity, or even elegance, of such a model bears no relation at all with the degree of approxi-mation. There is not such a thing as an exact method of structural analysis and, notwithstanding thepopular belief in the letter of the codes, the accuracy of any calculation is still a question of personaljudgement. This fortunate circumstance allows engineering to reach sometimes the highest category ofart, to the despair of dull and inflexible technicians.

If I find something lacking in this commendable conference in memory of one of the greatest engineersof all times it is that the side of Maillart’s personality as a rebel, with his tireless and successful struggleagainst the establishment of his times, has not been sufficiently stressed.


Draft

Chapter 8

Case Study III: MAGAZINIGENERALI


8.1 Geometry

1 This sotrage house, built by Maillart in Chiasso in 1924, provides a good example of the mariagebetween aesthetic and engineering.

2 The most strking feature of the Magazini Generali is not the structure itself, but rather the shape ofits internal supporting frames, Fig. 8.1.

3 The frame can be idealized as a simply supported beam hung from two cantilever column supports.Whereas the beam itself is a simple structural idealization, the overhang is designed in such a way as tominimize the net moment to be transmitted to the supports (foundations), Fig. 8.2.

8.2 Loads

4 The load applied on the frame is from the weights of the roof slab, and the frame itself. Given thespace between adjacent frames is 14.7 ft, and that the roof load is 98 psf, and that the total frameweight is 13.6 kips, the total uniform load becomes, Fig. 8.3:

qroof = (98) psf(14.7) ft = 1.4 k/ft (8.1-a)

qframe =(13.6) k

(63.6) ft= 0.2 k/ft (8.1-b)

qtotal = 1.4 + 0.2 = 1.6 k/ft (8.1-c)

Draft8–2 Case Study III: MAGAZINI GENERALI

HINGE IDEALIZATION

OF THIN SECTIONS

ACTUAL FRAME

AS A SIMPLE BEAMABSTRACTION OF MID SECTION

63.6 ft

9.2 ft

Figure 8.1: Magazzini Generali; Overall Dimensions, (Billington and Mark 1983)

d d

M =P*d

P

12

2P

M =B*dB 1

M =M -MR B P

B

B

Figure 8.2: Magazzini Generali; Support System, (Billington and Mark 1983)



ROOFq = 1.4 k/ft

FRAME + q = 0.2 k/ft

ROOFq = 1.4 k/ft FRAME

q = 1.6 k/ftTOTAL

+ q = 0.2 k/ft

Figure 8.3: Magazzini Generali; Loads (Billington and Mark 1983)

8.3 Reactions

5 Reactions for the beam are determined first taking advantage of symmetry, Fig. 8.4:

W = (1.6) k/ft(63.6) ft = 102 k (8.2-a)

R =W

2=

1022

= 51 k (8.2-b)

We note that these reactions are provided by the internal shear forces.

63.6 ft

51 k 51 k

TOTALq = 1.6 k/ft

Figure 8.4: Magazzini Generali; Beam Reactions, (Billington and Mark 1983)

8.4 Forces

6 The internal forces are pimarily the shear and moments. Those can be easily determined for a simplysupported uniformly loaded beam. The shear varies linearly from 51 kip to -51 kip with zero at thecenter, and the moment diagram is parabolic with the maximum moment at the center, Fig. 8.5, equalto:

Mmax =qL2

8=

(1.6) k/ft(63.6) ft2

8= 808 k.ft (8.3)

7 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 oneson the lower. Thus the net axial force is zero, however there is a net internal couple, Fig. 8.6.



Mmax

SHE

AR

FO

RC

E

51 K

25 K

25 K

51 K

0L/2

Lx

x

MO

ME

NT

L

0 L/4 L/2 3L/4

Figure 8.5: Magazzini Generali; Shear and Moment Diagrams (Billington and Mark 1983)

A

TOTALq

VA

C

T

d M

Figure 8.6: Magazzini Generali; Internal Moment, (Billington and Mark 1983)


Draft8.4 Forces 8–5

Mext = Cd ⇒ C =Mext

d(8.4-a)

T = C =(808) k.ft

(9.2) ft= ± 88 k (8.4-b)

8 Because the frame shape (and thus d(x)) is approximately parabolic, and the moment is also parabolic,then the axial forces are constants along the entire frame, Fig. 8.7.

MOMENT DIAGRAM

M

CURVE OF DIAGRAM

FRAME :CABLE :

SHAPE OF DIAGRAM

FRAME

d

Figure 8.7: Magazzini Generali; Similarities Between The Frame Shape and its Moment Diagram,(Billington and Mark 1983)

9 The axial force at the end of the beam is not balanced, and the 88 kip compression must be transmittedto the lower chord, Fig. 8.8. Fig. 8.9 This is analogous to the forces transmiited to the support by a

Tied ArchCable Force

Vertical Reaction

Horizontal Component

Axial Force

Compression

88 k

88 kTension88 k

88 k

Figure 8.8: Magazzini Generali; Equilibrium of Forces at the Beam Support, (Billington and Mark 1983)

tied arch.

10 It should be mentioned that when a rigorous computer analysis was performed, it was determinedthat the supports are contributing a compression force of about 8 kips which needs to be superimposedover the central values, Fig. 8.9.



FRAME ACTS AS A

UNIT, UNLIKE THE

ABSTRACTION

16 k

16 k

88 k - 8 k = 80 k

-88 k - 8 k = -96 k

Figure 8.9: Magazzini Generali; Effect of Lateral Supports, (Billington and Mark 1983)

8.5 Internal Stresses

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

σ =P

A=

(88) k

(75) in2 = 1.17 ksi (8.5)

this is much lower than the allowable compressive stress of concrete which is about 1,350 ksi. It shouldbe noted that if the frame was cast along with the roof (monolithic construction), than this stress wouldbe even lower.

12 Since concrete has practically no tensile strength, the tensile force in the lower chord must be resistedby steel. The lower chord has 4 bars with 0.69 in2 and 6 other bars with 0.58 in2, thus we have a totalof

As = 4(0.69) + 6(0.58) = 6.24 in2 (8.6)

Thus the steel stresses will be

σ =P

A=

(88) k

(6.24) in2 = 14.1 ksi (8.7)

which is lower than the allowable steel stress.


Draft

Chapter 9

DESIGN PHILOSOPHIES andGUIDELINES

9.1 Safety Provisions

1 Structures and structural members must always be designed to carry some reserve load above what isexpected under normal use. This is to account for

Variability in Resistance: The actual strengths (resistance) of structural elements will differ fromthose assumed by the designer due to:

1. Variability in the strength of the material (greater variability in concrete strength than insteel strength).

2. Differences between the actual dimensions and those specified (mostly in placement of steelrebars in R/C).

3. Effect of simplifying assumptions made in the derivation of certain formulas.

Variability in Loadings: All loadings are variable. There is a greater variation in the live loads thanin the dead loads. Some types of loadings are very difficult to quantify (wind, earthquakes).

Consequences of Failure: The consequence of a structural component failure must be carefully as-sessed. The collapse of a beam is likely to cause a localized failure. Alternatively the failure of acolumn is likely to trigger the failure of the whole structure. Alternatively, the failure of certaincomponents can be preceded by warnings (such as excessive deformation), whereas other are sud-den and catastrophic. Finally, if no redistribution of load is possible (as would be the case in astatically determinate structure), a higher safety factor must be adopted.

2 The purpose of safety provisions is to limit the probability of failure and yet permit economicalstructures.

3 The following items must be considered in determining safety provisions:

1. Seriousness of a failure, either to humans or goods.

2. Reliability of workmanship and inspection.

3. Expectation of overload and to what magnitude.

Draft9–2 DESIGN PHILOSOPHIES and GUIDELINES

4. Importance of the member in the structure.

5. Chance of warning prior to failure.

4 Two major design philosophies have emerged

1. Working Stress Method

2. Ultimate Strength Method

9.2 Working Stress Method

5 This is the simplest of the two methods, and the one which has been historically used by structuralengineers.

6 Structural elements are designed for their service loads, and are dimensioned such that the stressesdo not exceed some predesignated allowable strength, Fig. 9.1.

Figure 9.1: Load Life of a Structure

7 In R/C this method was the one adopted by the ACI (American Concrete institute) code up to 1971,Working Stress Design Method (WSD).

8 The AISC (American Institute of Steel Construction) code refers to it as the Allowable StressDesign (ASD) and was used until 1986.


Draft9.3 Ultimate Strength Method 9–3

9 In this method:

1. All loads are assumed to have the same average variability.

2. The entire variation of the loads and the strengths is placed on the strength side of the equation.

σ < σall =σyld

F.S.(9.1)

where F.S. is the factor of safety.

10 Major limitations of this method

1. An elastic analysis can not easily account for creep and shrinkage of concrete.

2. For concrete structures, stresses are not linearly proportional to strain beyond 0.45f ′c.

3. Safety factors are not rigorously determined from a probabilistic approach, but are the result ofexperience and judgment.

11 Allowable strengths are given in Table 9.1.

Steel, AISC/ASDTension, Gross Area Ft = 0.6Fy

Tension, Effective Net Area∗ Ft = 0.5Fu

Bending Fb = 0.66Fy

Shear Fv = 0.40Fy

Concrete, ACI/WSDTension 0Compression 0.45f ′

c

∗ Effective net area will be defined in section ??.

Table 9.1: Allowable Stresses for Steel and Concrete

9.3 Ultimate Strength Method

9.3.1 † Probabilistic Preliminaries

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

13 Typical frequency distributions of such random variables are shown in Fig. 9.2.

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

15 Q and R can be combined and the result expressed logarithmically, Fig. 9.3.

X = lnR

Q(9.2)

Failure would occur for negative values of X

16 The probability of failure Pf is equal to the ratio of the shaded area to the total area under thecurve in Fig. 9.3.



Figure 9.2: Frequency Distributions of Load Q and Resistance R

Figure 9.3: Definition of Reliability Index


Draft9.3 Ultimate Strength Method 9–5

17 If X is assumed to follow a Normal Distribution than it has a mean value X =(ln R

Q

)m

and astandard deviation σ.

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

19 For standard distributions and for β = 3.5, it can be shown that the probability of failure is Pf = 19,091

or 1.1 × 10−4. That is 1 in every 10,000 structural members designed with β = 3.5 will fail because ofeither excessive load or understrength sometime in its lifetime.

20 Target values for β are shown in Table 9.2.

Type of Load/Member βAISC

DL + LL; Members 3.0DL + LL; Connections 4.5DL + LL + WL; Members 3.5DL + LL +EL; Members 1.75

ACIDuctile Failure 3-3.5Sudden Failures 3.5-4

Table 9.2: Selected β values for Steel and Concrete Structures

21 Because the strengths and the loads vary independently, it is desirable to have one factor to accountfor variability in resistance, and another one for the variability in loads.

22 These factors are referred to as resistance factor Φ and Load Factor α respectively. The resistancefactor is defined as

Φ =Rm

Rnexp(−0.55βVR) (9.3)

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

9.3.2 Discussion

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

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

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

26 The general form is (LRFD-A4.1)1

ΦRn ≥ ΣαiQi (9.4)

where

Φ is a strength reduction factor, less than 1, and must account for the type of structural element,Table 9.3.

1Throughout the notes we will refer by this symbol the relevant design specification in the AISC code.



Type of Member ΦACI

Axial Tension 0.9Flexure 0.9Axial Compression, spiral reinforcement 0.75Axial Compression, other 0.70Shear and Torsion 0.85Bearing on concrete 0.70

AISCTension, yielding 0.9Tension, fracture 0.75Compression 0.85Beams 0.9Fasteners, Tension 0.75Fasteners, Shear 0.65

Table 9.3: Strength Reduction Factors, Φ

Rn is the nominal resistance (or strength).

ΦRn is the design strength.

αi is the load factor corresponding to Qi and is greater than 1.

ΣαiQi is the required strength based on the factored load:

i is the type of load

27 The various factored load combinations which must be considered are

AISC

1. 1.4D

2. 1.2D+1.6L+0.5(Lr or S)

3. 1.2D+0.5L (or 0.8W)+1.6(Lr or S)

4. 1.2D+0.5L+0.5(Lr or S)+1.3W

5. 1.2D+0.5L(or 0.2 S)+1.5E

6. 0.9D+1.3W(or 1.5 E)

ACI

1. 1.4D+1.7L

2. 0.75(1.4D+1.7L+1.7W)

3. 0.9D+1.3W

4. 1.05D+1.275W

5. 0.9D+1.7H

6. 1.4D +1.7L+1.7H

7. 0.75(1.4D+1.4T+1.7L)


Draft9.4 Example 9–7

8. 1.4(D+T)

where D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature; H= soil.We must select the one with the largest limit state load.

28 Thus, in this method, we must perform numerous analysis, one for each load, of a given structure.For trusses, this is best achieved if we use the matrix method, invert the statics matrix [B], and multiply[B]−1 by each one of the load cases, (Refer to Section ??). For the WSD method, we need not performmore than one analysis in general.

29 Serviceability Limit States must be assessed under service loads (not factored). The most impor-tant ones being

1. Deflections

2. Crack width (for R/C)

3. Stability

9.4 Example

Example 9-1: LRFD vs ASD

To illustrate the differences between the two design approaches, let us consider the design of an axialmember, 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 9.1, the allowable stress is0.6σyld = 0.6 ∗ 36 = 21.6 ksi. Thus the required cross sectional area is

A =18021.6

= 8.33 in2

USD we consider the largest of the two load combinations

ΣαiQi : 1.4D = 1.4(100) = 140 k

1.2D + 1.6L = 1.2(100) + 1.6(80) = 248 k ¾

From Table 9.3 Φ = 0.9, and ΦRn = (0.9)Aσyld. Hence, applying Eq. 9.4 the cross sectional areashould be

A =ΣαiQi

Φσyld

=248

(0.9)(36)= 7.65 in2

Note that whereas in this particular case the USD design required a smaller area, this may not be thecase for different ratios of dead to live loads.

9.5 Design Guidelines

30 To assist in the preliminary design/dimensioning of structures, Table 9.4 provides average, maximumand typical spans for various types of structures.



Average Max Typical SpanFt.

TIMBERPlywood 36 40 3-5Planks 28 32 2-6Joists 22 26 10-25Beams 16 20 15-30Girders 12 16 20-35Gable bents 26 30 30-50Trusses 4 8 30-100I Beams 18 24 15-60Joists 18 25 15-60Plate and I girders 14 20 40-100Trusses 12 18 40-80Gable bents 30 40 50-120Arches span to rise 8 16 80-200Arches span to thickness 40 50 -Simple suspension (span to rise) 10 15 150-300Cable stayed 6 10 150-300

REINFORCED CONCRETESolid slabs 28 32 10-25Slabs with drops and capitals 30 36 20-35Two-way slab on beams 30 36 20-35Waffle slabs 20 24 30-40Joists 22 26 25-45Beams 16 20 15-40Girders 12 16 20-60Gable bents 24 30 40-80Arches span to rise 8 12 60-150Arches span to thickness 30 40 -

Cylindrical thin shell roof (Min. thickness may govern)Longitudinal span to Structural depth 12 15 50-70Transverse span to thickness 50 60 12-30

PRESTRESSED CONCRETESolid slabs 40 44 20-35Slabs with drops 44 48 35-45Two-way slab on beams 44 48 35-45Waffle slabs 28 32 35-70Cored slabs 36 40 30-60Joists 32 36 40-60Beams 24 28 30-80Girders 20 24 40-120

Cylindrical thin shell roof (Min. thickness may govern)Longitudinal span to structural depth 15 20 60-120Transverse span to thickness 60 70 15-35

Table 9.4: Approximate Structural Span-Depth Ratios for Horizontal Subsystems and Components (Linand Stotesbury 1981)


Draft

Chapter 10

BRACED ROLLED STEEL BEAMS

1 This chapter deals with the behavior and design of laterally supported steel beams according to theLRFD provisions.

2 A laterally stable beam is one which is braced laterally in the direction perpendicular to the planeof the web. Thus overall buckling of the compression flange as a column cannot occur prior to its fullparticipation to develop the moment strength of the section.

3 If a beam is not laterally supported, Fig. 10.1, we will have a failure mode governed by lateral torsional

A) COMPOSITE BEAM B) OTHER FRAMING

C) CROSS BRACING

Figure 10.1: Lateral Bracing for Steel Beams

buckling.

4 By the end of this lecture you should be able to select the most efficient section (light weight withadequate strength) for a given bending moment and also be able to determine the flexural strength of agiven beam.

Draft10–2 BRACED ROLLED STEEL BEAMS

10.1 Nominal Strength

5 The strength requirement for beams in load and resistance factor design is stated as

φbMn ≥ Mu (10.1)

where:φb strength reduction factor; for flexure 0.90

Mn nominal moment strengthMu factored service load moment.

6 The equations given in this chapter are valid for flexural members with the following kinds of crosssection and loading:

1. Doubly symmetric (such as W sections) and loaded in plane of symmetry

2. Singly symmetric (channels and angles) loaded in plane of symmetry or through the shear centerparallel to the web1.

10.2 Failure Modes and Classification of Steel Beams

7 A beam is classified as laterally supported depending on Lb which is the distance between lateralsupports (or unbraced length) and Lp.

Lb < Lp (10.2)

Lp =300√Fy, ksi

ry (10.3)

ry =

√Iy

A(10.4)

where ry is the radius of gyration with respect to the (minor) y axis (as opposed to the major x axis).

8 The strength of flexural members is limited by:

Plastic Hinge: at a particular cross section, Fig. 10.2.

local buckling: of a cross-sectional element (e.g. the web or the flange), Fig. 10.3.

Lateral-Torsional buckling: of the entire member, Fig. 10.4.

9 Accordingly, the LRFD manual classifies steel sections as

Compact sections: No local buckling can occur. Strength is based on the plastic moment.

Partially compact sections: Where local buckling may occur

Slender sections: where lateral torsional buckling may occur.

We will cover only the first two cases.

10 Shear should be checked, however with exception of short beams (and no self-respecting architectwill ever conceive such a thing:-), flexure generally controls.

1More about shear centers in Mechanics of Materials II.


Draft10.2 Failure Modes and Classification of Steel Beams 10–3

��

��

��

w

wu

M p

pM =(wL )/82

M=(wL )/82 +

+

-

-

σ

σ

σ

σ

y

y

y

y

Figure 10.2: Failure of Steel beam; Plastic Hinges

COMPACT WEB BUCKLINGFLANGE BUCKLING

b

h

t

h

f

f

c

w

Figure 10.3: Failure of Steel beam; Local Buckling



B

B

A

A

B

A

COMPRESSIONFLANGE

LATERAL DEFLECTION ANDTORSION OF THE COMPRESSIONFLANGE

LATERAL DEFLECTIONAND TORSION OF THE COMPRESSION FLANGE

Figure 10.4: Failure of Steel beam; Lateral Torsional Buckling

10.3 Compact Sections

11 For compact sections, the mode of failure is the formation of a plastic hinge that is the section isfully plastified. Hence we shall first examine the bending behavior of beams under limit load. Then wewill relate this plastic moment to the design of compact sections.

10.3.1 Bending Capacity of Beams

12 The stress distribution on a typical wide-flange shape subjected to increasing bending moment isshown in Fig.10.5. In the service range (that is before we multiplied the load by the appropriate factorsin the LRFD method) the section is elastic. This elastic condition prevails as long as the stress at theextreme fiber has not reached the yield stress Fy. Once the strain ε reaches its yield value εy, increasingstrain induces no increase in stress beyond Fy.

Figure 10.5: Stress distribution at different stages of loading


Draft10.3 Compact Sections 10–5

Figure 10.6: Stress-strain diagram for most structural steels

13 When the yield stress is reached at the extreme fiber, the nominal moment strength Mn, is referredto as the yield moment My and is computed as

Mn = My = SxFy (10.5)

(assuming that bending is occurring with respect to the x − x axis).

14 When across the entire section, the strain is equal or larger than the yield strain (ε ≥ εy = Fy/Es)then the section is fully plastified, and the nominal moment strength Mn is therefore referred to as theplastic moment Mp and is determined from

Mp = Fy

∫A

ydA = FyZ (10.6)

where

Zdef=

∫ydA (10.7)

is the Plastic Section Modulus.

15 The plastic section modulus Z should not be confused with the elastic section modulus S defined,Eq. 5.25 as

S =I

d/2(10.8-a)

Idef=

∫A

y2dA (10.8-b)

16 The section modulus Sx of a W section can be roughly approximated by the following formula

Sx ≈ wd/10 or Ix ≈ Sxd

2≈ wd2/20 (10.9)

and the plastic modulus can be approximated by

Zx ≈ wd/9 (10.10)



10.3.2 Design of Compact Sections

17 A section is compact if the following conditions are met:

1. Flanges are continuously connected to the web

2. Width to thickness ratios, known as the slenderness ratios, of the flange and the web must notexceed the limiting ratios λp defined as follows:

Flange bf

2tf≤ λp λp = 65√

Fy

Web hc

tw≤ λp λp = 640√

Fy

(10.11)

Note that bf

2tfand hc

tware tabulated in Sect. 3.6.

18 The nominal strength Mn for laterally stable compact sections according to LRFD is

Mn = Mp (10.12)

where:Mp plastic moment strength = ZFy

Z plastic section modulusFy specified minimum yield strength

19 Note that section properties, including Z values are tabulated in Section 3.6.

10.4 Partially Compact Section

20 If the width to thickness ratios of the compression elements exceed the λp values mentioned in Eq.10.11 but do not exceed the following λr, the section is partially compact and we can have local buckling.

Flange: λp <bf

2tf≤ λr λp = 65√

Fy

λr = 141√Fy−Fr

Web: λp < hc

tw≤ λr λp = 640√

Fy

λr = 970√Fy

(10.13)

where:Fy specified minimum yield stress in kksibf width of the flangetf thickness of the flangehc unsupported height of the web which is twice the distance from the neutral axis

to the inside face of the compression flange less the fillet or corner radius.tw thickness of the web.Fr residual stress = 10.0 ksi for rolled sections and 16.5 ksi for welded sections.

21 The nominal strength of partially compact sections according to LRFD is, Fig. 10.7

Mn = Mp − (Mp − Mr)(λ − λp

λr − λp) ≤ Mp (10.14)


Draft10.5 Slender Section 10–7

Draft

-

6

�r�p

Mr

Mp

Mn

�

Partially CompactCompact Slender

-� -� -

65pFy

141pFy�Fr

970pFy

640pFy

bf

2tf

hctwWeb

Flanges

Figure 10.7: Nominal Moments for Compact and Partially Compact Sections

where:Mr Residual Moment equal to (Fy − Fr)S

λ bf/2tf for I-shaped member flanges andhc/tw for beam webs.

22 All other quantities are as defined earlier. Note that we use the λ associated with the one beingviolated (or the lower of the two if both are).

10.5 Slender Section

23 If the width to thickness ratio exceeds λr values of flange and web, the element is referred to as slendercompression element. Since the slender sections involve a different treatment, it will not be dealt here.

10.6 Examples

Example 10-1: Z for Rectangular Section

Determine the plastic section modulus for a rectangular section, width b and depth d.



Solution:

1. The internal plastic moment is equal to

M = Fybd

2︸︷︷︸Force

d

2= Fyb

d2

4(10.15)

2. The yield stress, Fy, plastic moment Mp and plastic section modulus Z are related by:

Z =Mp

Z(10.16)

3. Substituting, we get:

Z =Mp

Fy=

Fybd2

4Fy= bd2

4 (10.17)

Note that this is to be contrasted with the elastic section modulus S = bd2

6 .

Example 10-2: Beam Design

Select the lightest W or M section to carry a uniformly distributed dead load of 0.2 kip/ft superim-posed (i.e., in addition to the beam weight) and 0.8 kip/ft live load. The simply supported span is 20ft. The compression flange of the beam is fully supported against lateral movement. Select the sectionsfor the following steels: A36; A572 Grade 50; and A572 Grade 65.Solution:

Case 1: A36 Steel

1. Determine the factored load.

wD = 0.2 k/ft

wL = 0.8 k/ft

wu = 1.2wD + 1.6wL

= 1.2(0.2) + 1.6(0.8) = 1.52 k/ft


Draft10.6 Examples 10–9

2. Compute the factored load moment Mu. For a simply supported beam carrying uniformlydistributed load,

Mu = wuL2/8 = (1.52)(20)2/8 = 76 k.ft

Assuming compact section, since a vast majority of rolled sections satisfy λ ≤ λp for both theflange and the web. The design strength φbMn is

φbMn = φbMp = φbZxFy

The design requirement isφbMn = Mu

or, combing those two equations we have:

φbZxFy = Mu

3. Required Zx is

Zx =Mu

φbFy=

76(12)0.90(36)

= 28.1 in3

From the notes on Structural Materials, we select a W12X22 section which has a Zx = 29.3 in3

Note that Zx is approximated by wd9 = (22)(12)

9 = 29.3.

4. Check compact section limits λp for the flanges from the table

λ = bf

2tf= 4.7

λp = 65√Fy

= 65√36

= 10.8 > λ√

and for the web:λ = hc

tw= 41.8

λp = 640√Fy

= 640√36

= 107√

5. Check the Strength by correcting the factored moment Mu to include the self weight. Selfweight of the beam W12X22 is 22 lb./ft. or 0.022 kip/ft

wD = 0.2 + 0.022 = 0.222 k/ft

wu = 1.2(0.222) + 1.6(0.8) = 1.55 k/ft

Mu = (1.55)(20)2/8 = 77.3 k.ft

Mn = Mp = ZxFy = (29.3) in3(36) ksi

(12) in/ft= 87.9 k.ft

φbMn = 0.90(87.9) = 79.1 k.ft > Mu√

Therefore use W12X22 section.

6. We finally check for the maximum distance between supports.

ry =

√Iy

A=

√5

6.5= 0.88 in (10.18-a)

Lp =300√

Fy

ry (10.18-b)

=300√

360.88 = 43 ft (10.18-c)



Case 2: A572 Grade 65 Steel:

1. same as in case 1

2. same as in case 1

3. Required Zx = Mu

φbFy= 76(12)

0.90(65) = 15.6 in3 Select W12X14: Zx = 17.4 in3 Note that Zx is

approximated by wd9 = (14)(12)

9 = 18.7.

4. Check compact section limits λp :

λ = hc

tw= 54.3

λp = 640√Fy

= 640√65

= 79.4√

λ = bf

2tf= 8.82

λp = 65√Fy

= 65√65

= 8.1 < λ Not Good

In this case the controlling limit state is local buckling of the flange.Since λp < λ < λr, as above, the section is classified as non-compact.

5. Check the strength:Since the section is non-compact, the strength is obtained by interpolation between Mp andMr.For the flanges:

λr = 141√Fy−10

= 141√65−10

= 19.0

Mn = Mp − (Mp − Mr)(λ−λp

λr−λp) ≤ Mp

Mp = ZxFy = (17.4) in3(65) ksi

(12) in/ft= 94.2 k.ft

Mr = Sx(Fy − Fr) = (14.9) in3(65−10) ksi

(12) in/ft= 68.3 k.ft.

Mn = 94.2 − (94.2 − 68.3)(

8.8−8.119.0−8.1

)= 92.5 k.ft

φbMn = 0.90(92.5) = 83.25 k.ft > Mu√

Therefore provide W12X14 section.


Draft

Chapter 11

REINFORCED CONCRETEBEAMS

11.1 Introduction

1 Recalling that concrete has a tensile strength (f ′t) about one tenth its compressive strength (f ′

c),concrete by itself is a very poor material for flexural members.

2 To provide tensile resistance to concrete beams, a reinforcement must be added. Steel is almostuniversally used as reinforcement (longitudinal or as fibers), but in poorer countries other indigenousmaterials have been used (such as bamboos).

3 The following lectures will focus exclusively on the flexural design and analysis of reinforced con-crete rectangular sections. Other concerns, such as shear, torsion, cracking, and deflections are left forsubsequent ones.

4 Design of reinforced concrete structures is governed in most cases by the Building Code Requirementsfor Reinforced Concrete, of the American Concrete Institute (ACI-318). Some of the most relevantprovisions of this code are enclosed in this set of notes.

5 We will focus on determining the amount of flexural (that is longitudinal) reinforcement required at agiven section. For that section, the moment which should be considered for design is the one obtainedfrom the moment envelope at that particular point.

11.1.1 Notation

6 In R/C design, it is customary to use the following notation

Draft11–2 REINFORCED CONCRETE BEAMS

As Area of steelb Widthc Distance from top of compressive fibers to neutral axisd Distance from the top of the compressive fibers to the centroid

of the reinforcementf ′

c Concrete compressive strengthf ′

r Concrete modulus of rupturef ′

s Steel stressf ′

t Concrete tensile strengthfy Steel yield stress (equivalent to Fy in AISC)h Heightρ Steel ratio, As

bd

11.1.2 Modes of Failure

7 A reinforced concrete beam is a composite structure where concrete provides the compression and steelthe tension.

8 Failure is initiated by, Fig. 11.5:

Steel Yielding Concrete Crushing

Figure 11.1: Failure Modes for R/C Beams

1. Yielding of the steel when the steel stress reaches the yield stress (fs = fy). This occurs if wedo not have enough reinforcement that is the section is under-reinforced. This will result inexcessive rotation and deformation prior to failure.

2. Crushing of the concrete, when the concrete strain reaches its ultimate value (εc = εu = 0.003),ACI 318: 10.2.3. This occurs if there is too much reinforcement that is the section is over-reinforced. This is a sudden mode of failure.

9 Ideally in an optimal (i.e. most efficient use of materials) design, a section should be dimensionedsuch that crushing of concrete should occur simultaneously with steel yielding. This would then be abalanced design.

10 However since concrete crushing is a sudden mode of failure with no prior warning, whereas steelyielding is often accompanied by excessive deformation (thus providing ample warning of an imminentfailure), design codes require the section to be moderately under-reinforced.

11.1.3 Analysis vs Design

11 In R/C we always consider one of the following problems:


Draft11.1 Introduction 11–3

Analysis: Given a certain design, determine what is the maximum moment which can be applied.

Design: Given an external moment to be resisted, determine cross sectional dimensions (b and h) aswell as reinforcement (As). Note that in many cases the external dimensions of the beam (b andh) are fixed by the architect.

12 We often consider the maximum moment along a member, and design accordingly.

11.1.4 Basic Relations and Assumptions

13 In developing a design/analysis method for reinforced concrete, the following basic relations will beused, Fig. ??:

C

T

d

εy

T=CM_ext=Cd

Compatibility Equilibrium

Figure 11.2: Internal Equilibrium in a R/C Beam

1. Equilibrium: of forces and moment at the cross section. 1) ΣFx = 0 or Tension in the reinforcement= Compression in concrete; and 2) ΣM = 0 or external moment (that is the one obtained fromthe moment envelope) equal and opposite to the internal one (tension in steel and compression ofthe concrete).

2. Material Stress Strain: We recall that all normal strength concrete have a failure strain εu = .003in compression irrespective of f ′

c.

14 Basic assumptions used:

Compatibility of Displacements: Perfect bond between steel and concrete (no slip). Note that thosetwo materials do also have very close coefficients of thermal expansion under normal temperature.

Plane section remain plane ⇒ strain is proportional to distance from neutral axis.

11.1.5 ACI Code

15 The ACI code is based on limit strength, or ΦMn ≥ Mu thus a similar design philosophy is used asthe one adopted by the LRFD method of the AISC code, ACI-318: 8.1.1; 9.3.1; 9.3.2

16 The required strength is based on (ACI-318: 9.2)



U = 1.4D + 1.7L (11.1)= 0.75(1.4D + 1.7L + 1.7W ) (11.2)

11.2 Cracked Section, Ultimate Strength Design Method

11.2.1 Equivalent Stress Block

17 In determining the limit state moment of a cross section, we consider Fig. 11.3. Whereas the straindistribution is linear (ACI-318 10.2.2), the stress distribution is non-linear because the stress-straincurve of concrete is itself non-linear beyond 0.5f ′

c.

18 Thus we have two alternatives to investigate the moment carrying capacity of the section, ACI-318:10.2.6

1. Use the non-linear stress distribution.

2. Use a simpler equivalent stress distribution.

19 The ACI code follows the second approach. Thus we seek an equivalent stress distribution suchthat:

1. The resultant force is equal

2. The location of the resultant is the same.

We note that this is similar to the approach followed in determining reactions in a beam subjected to adistributed load when the load is replaced by a single force placed at the centroid.

Figure 11.3: Cracked Section, Limit State


Draft11.2 Cracked Section, Ultimate Strength Design Method 11–5

20 It was shown that the depth of the equivalent stress block is a function of f ′c:

β1 = .85 if f ′c ≤ 4, 000

= .85 − (.05)(f ′c − 4, 000) 1

1,000 if 4, 000 < f ′c < 8, 000

(11.3)

Figure 11.4: Whitney Stress Block

11.2.2 Balanced Steel Ratio

21 Next we seek to determine the balanced steel ratio ρb such that failure occurs by simultaneousyielding of the steel fs = fy and crushing of the concrete εc = 0.003, ACI-318: 10.3.2 We willseparately consider the two failure possibilities:

Tension Failure: we stipulate that the steel stress is equal to fy:

ρ = As

bdAsfy = .85f ′

cab = .85f ′cbβ1c

}⇒ c =

ρfy

0.85f ′cβ1

d (11.4)

Compression Failure: where the concrete strain is equal to the ultimate strain; From the straindiagram

εc = 0.003cd = .003

.003+εs

}⇒ c =

.003fs

Es+ .003

d (11.5)

22 Balanced Design is obtained by equating Eq. 11.4 to Eq. 11.5 and by replacing ρ by ρb and fs by fy:

ρfy

0.85f ′cβ1

d = .003fsEs

+.003d

fs = fy

ρ = ρb

ρbfy

.85f ′cβ1

d =.003

fy

Es+ .003

d

When we replace Es by 29, 000 ksi we obtain

ρb = .85β1f ′

c

fy

87, 00087, 000 + fy

(11.6)

This ρb corresponds to the only combination of b, d and As which will result in simultaneous yielding ofthe steel and crushing of the concrete, that is an optimal design.



23 Because we need to have ample warning against failure, hence we prefer to have an under-reinforcedsection. Thus, the ACI code stipulates:

ρ < .75ρb (11.7)

24 In practice, depending on the relative cost of steel/concrete and of labour it is common to select lowervalues of ρ. If ρ < 0.5ρb (thus we will have a deeper section) then we need not check for deflection.

25 A minimum amount of reinforcement must always be used to prevent temperature and shrinkagecracks:

ρmin ≥ 200fy

(11.8)

26 The ACI code adopts the limit state design method

MD = ΦMn > Mu Φ = Φb = 0.9 (11.9)

11.2.3 Analysis

Given As, b, d, f ′c, and fy determine the design moment:

1. ρact = As

bd

2. ρb = (.85)β1f ′

c

fy

8787+fy

3. If ρact < ρb (that is failure is triggered by yielding of the steel, fs = fy)

a = Asfy

.85f ′cb From Equilibrium

MD = ΦAsfy

(d − a

2

)}

MD = Φ Asfy

(d − 0.59

Asfy

f ′cb

)︸︷︷︸

Mn

Combining this last equation with ρ = As

bd yields

MD = Φρfybd2

(1 − .59ρ

fy

f ′c

)(11.10)

4. † If ρact > ρb is not allowed by the code as this would be an over-reinforced section which wouldfail with no prior warning. However, if such a section exists, and we need to determine its momentcarrying capacity, then we have two unknowns:

(a) Steel strain εs (which was equal to εy in the previous case)

(b) Location of the neutral axis c.

We have two equations to solve this problem

Equilibrium: of forces

c =Asfs

.85f ′cbβ1

(11.11)



Strain compatibility: since we know that at failure the maximum compressive strain εc is equalto 0.003. Thus from similar triangles we have

c

d=

.003.003 + εs

(11.12)

Those two equations can be solved by either one of two methods:

(a) Substitute into one single equation

(b) By iteration

Once c and fs = Eεs are determined then

MD = ΦAsfs

(d − β1c

2

)(11.13)

11.2.4 Design

27 We distinguish between two cases. The first one has dimensions as well as steel area unknown, thesecond has the dimensions known (usually specified by the architect or by other constraints), and weseek As.

b, d and As unknowns and MD known:

1. We start by assuming ρ, at most ρ = .75ρb, and if deflection is of a concern (or steel tooexpensive), then we can select ρ = 0.5ρb with ρb determined from Eq. 11.6

ρ = 0.75[.85β1

f ′c

fy

87, 00087, 000 + fy

]

2. From Eq. 11.10

MD = Φ ρfy

(1 − .59ρ

fy

f ′c

)︸︷︷︸

R

bd2

or

R = ρfy

(1 − .59ρ

fy

f ′c

)(11.14)

which does not depend on unknown quantities1. Then solve for bd2:

bd2 =MD

ΦR(11.15)

3. solve for b and d (this will require either an assumption on one of the two, or on their ratio).

4. As = ρbd

1Note analogy with Eq. 10.6 Mp = FyZ for stell beams.



† b, d and Md known, As unknown: In this case there is no assurance that we can have a designwith ρb. If the section is too small, then it will require too much steel resulting in an over-reinforcedsection.

We will again have an iterative approach

1. Since we do not know if the steel will be yielding or not, use fs.2. Assume an initial value for a (a good start is a = d

5 )3. Assume initially that fs = fy

4. Check equilibrium of moments (ΣM = 0)

As =MD

Φfs

(d − a

2

) (11.16)

5. Check equilibrium of forces in the x direction (ΣFx = 0)

a =Asfs

.85f ′cb

(11.17)

6. Check assumption of fs from the strain diagram

εs

d − c=

.003c

⇒ fs = Esd − c

c.003 < fy (11.18)

where c = aβ1

.7. Iterate until convergence is reached.

Example 11-1: Ultimate Strength Capacity

Determine the ultimate Strength of a beam with the following properties: b = 10 in, d = 23 in,As = 2.35 in2, f ′

c = 4, 000 psi and fy = 60 ksi.Solution:

ρact = As

bd = 2.35(10)(23) = .0102

ρb = .85β1f ′

c

fy

8787+fy

= (.85)(.85) 460

8787+60 = .02885 > ρact

√

a = Asfy

.85f ′cb = (2.35)(60)

(.85)(4)(10) = 4.147 in

Mn = (2.35)(60)(23 − 4.1472 ) = 2, 950 k.in

MD = ΦMn = (.9)(2, 950) = 2, 660 k.in

Note that from the strain diagram

c =a

0.85=

4.4140.85

= 4.87 in

Alternative solution

Mn = ρactfybd2(1 − .59ρact

fy

f ′c

)= Asfyd

(1 − .59ρact

fy

f ′c

)= (2.35)(60)(23)

[1 − (.59)60

4 (.01021)]

= 2, 950 k.in = 245 k.ft

MD = ΦMn = (.9)(2, 950) = 2, 660 k.in



Example 11-2: Beam Design I

Design a 15 ft beam to support a dead load of 1.27 k/ft, a live load of 2.44 k/ft using a 3,000 psiconcrete and 40 ksi steel. Neglect beam weightSolution:

wu = 1.4(1.27) + 1.7(2.44) = 5.92 k/ft

MD = (5.92) k/ft(15)2 ft2

8 (12) in/ft = 2, 000 k.in

ρb = .85β1f ′

c

fy

8787+fy

= (.85)(.85) 340

8787+40 = 0.040

ρ = .75ρb = .75(0.04) = 0.030R = ρfy

(1 − .59ρ fy

f ′c

)= (0.03)(40)

[1. − (0.59)(0.03)40

3

]= 0.917 ksi

bd2 = Md

ΦR = 2,000(0.9)(0.917)

k.in in2

k = 2, 423 in3

Assume b = 10 in, this will give d =√

2,42310 = 15.57 in. We thus adopt b = 10 in and d = 16 in .

Finally,As = ρbd = (0.030)(10)(16) = 4.80 in2

we select 3 bars No. 11

Example 11-3: Beam Design II

Select the reinforcement for a cross section with b = 11.5 in; d = 20 in to support a design momentMd = 1, 600 k.in using f ′

c = 3 ksi; and fy = 40 ksi

Solution:

1. Assume a = d5 = 20

5 = 4 in and fs = fy

2. Equilibrium of moments:

As =MD

Φfy(d − a2 )

==1, 600 k.in

(.9)(40) ksi(20 − 42 ) in

= 2.47 in2

3. Check equilibrium of forces:

a =Asfy

.85f ′cb

=(2.47) in2(40) ksi

(.85)(3) ksi(11.5) in= 3.38 in

4. We originally assumed a = 4, at the end of this first iteration a = 3.38, let us iterate again witha = 3.30



5. Equilibrium of moments:

As =MD

Φfy(d − a2 )

==1, 600 k.in

(.9)(40) ksi(20 − 3.32 ) in

= 2.42 in2

6. Check equilibrium of forces:

a =Asfy

.85f ′cb

=(2.42) in2(40) ksi

(.85)(3) ksi(11.5) in= 3.3 in

√

7. we have converged on a.

8. Actual ρ is ρact = 2.42(11.5)(20) = .011

9. ρb is equal to

ρb = .85β1f ′

c

fy

8787 + fy

= (.85)(.85)340

8787 + 40

= .037

10. ρmax = .75ρ = (0.75)(0.037) = .0278 > 0.011√

thus fs = fy and we use As = 2.42 in2

11.3 Continuous Beams

28 Whereas coverage of continuous reinforced concrete beams is beyond the scope of this course, Fig. ??illustrates a typical reinforcement in such a beam.

11.4 ACI Code

Attached is an unauthorized copy of some of the most relevant ACI-318-89 design code provisions.8.1.1 - In design of reinforced concrete structures, members shall be proportioned for adequate

strength in accordance with provisions of this code, using load factors and strength reduction factors Φspecified in Chapter 9.

8.3.1 - All members of frames or continuous construction shall be designed for the maximum effectsof factored loads as determined by the theory of elastic analysis, except as modified according to Section8.4. Simplifying assumptions of Section 8.6 through 8.9 may be used.

8.5.1 - Modulus of elasticity Ec for concrete may be taken as W 1.5c 33

√f ′

c ( psi) for values of Wc

between 90 and 155 lb per cu ft. For normal weight concrete, Ec may be taken as 57, 000√

f ′c.

8.5.2 - Modulus of elasticity Es for non-prestressed reinforcement may be taken as 29,000 psi.9.1.1 - Structures and structural members shall be designed to have design strengths at all sections

at least equal to the required strengths calculated for the factored loads and forces in such combinationsas are stipulated in this code.

9.2 - Required Strength9.2.1 - Required strength U to resist dead load D and live load L shall be at least equal to

U = 1.4D + 1.7L

9.2.2 - If resistance to structural effects of a specified wind load W are included in design, thefollowing combinations of D, L, and W shall be investigated to determine the greatest required strengthU

U = 0.75(1.4D + 1.7L + 1.7W )


Draft11.4 ACI Code 11–11

Ext

erio

r sp

anIn

teri

or s

pan

Inte

rior

spa

n

No.

3 st

irru

p su

ppor

t if

nece

ssar

ySt

irru

psSt

raig

ht b

ar

Ext

erio

r sp

anIn

teri

or s

pan

Inte

rior

spa

n

Cra

cks

Cra

cks

Rei

nfor

cem

ent

Poin

ts o

f de

flec

tion

(a)

Def

lect

ed s

hape

(b)

Mom

ent d

iagr

am u

nder

typi

cal l

oadi

ng

(d)

Str

aigh

t and

ben

t bar

rei

nfor

cem

ent

Inte

rior

col

umn

Ben

t bar

sSt

raig

ht b

otto

m b

ar

Ben

t bar

at n

onco

ntin

uous

end

Top

bar

sB

ent b

ar

Stir

rups

Bot

tom

bar

s

Top

bar

s

Stir

rups

Bot

tom

bar

s

Sect

ion

thro

ugh

beam

Sect

ion

thro

ugh

beam

Stra

ight

bot

tom

bar

Inte

rior

col

umn

Stir

rups

Stra

ight

top

bar

(c)

Str

aigh

t bar

rei

nfor

cem

ent

Figure 11.5: Reinforcement in Continuous R/C Beams



where load combinations shall include both full value and zero value of L to determine the more severecondition, and

U = 0.9D + 1.3W

but for any combination of D, L, and W, required strength U shall not be less than Eq. (9-1).9.3.1 - Design strength provided by a member, its connections to other members, and its cross

sections, in terms of flexure, axial load, shear, and torsion, shall be taken as the nominal strengthcalculated in accordance with requirements and assumptions of this code, multiplied by a strengthreduction factor Φ.

9.3.2 - Strength reduction factor Φ shall be as follows:9.3.2.1 - Flexure, without axial load 0.909.4 - Design strength for reinforcement Designs shall not be based on a yield strength of reinforcement

fy in excess of 80,000 psi, except for prestressing tendons.10.2.2 - Strain in reinforcement and concrete shall be assumed directly proportional to the distance

from the neutral axis, except, for deep flexural members with overall depth to clear span ratios greaterthan 2/5 for continuous spans and 4/5 for simple spans, a non-linear distribution of strain shall beconsidered. See Section 10.7.

10.2.3 - Maximum usable strain at extreme concrete compression fiber shall be assumed equal to0.003.

10.2.4 - Stress in reinforcement below specified yield strength fy for grade of reinforcement usedshall be taken as Es times steel strain. For strains greater than that corresponding to fy, stress inreinforcement shall be considered independent of strain and equal to fy.

10.2.5 - Tensile strength of concrete shall be neglected in flexural calculations of reinforced concrete,except when meeting requirements of Section 18.4.

10.2.6 - Relationship between concrete compressive stress distribution and concrete strain may beassumed to be rectangular, trapezoidal, parabolic, or any other shape that results in prediction ofstrength in substantial agreement with results of comprehensive tests.

10.2.7 - Requirements of Section 10.2.5 may be considered satisfied by an equivalent rectangularconcrete stress distribution defined by the following:

10.2.7.1 - Concrete stress of 0.85f ′c shall be assumed uniformly distributed over an equivalent com-

pression zone bounded by edges of the cross section and a straight line located parallel to the neutralaxis at a distance (a = β1c) from the fiber of maximum compressive strain.

10.2.7.2 - Distance c from fiber of maximum strain to the neutral axis shall be measured in adirection perpendicular to that axis.

10.2.7.3 - Factor β1 shall be taken as 0.85 for concrete strengths f ′c up to and including 4,000 psi.

For strengths above 4,000 psi, β1 shall be reduced continuously at a rate of 0.05 for each 1000 psi ofstrength in excess of 4,000 psi, but β1 shall not be taken less than 0.65.

10.3.2 - Balanced strain conditions exist at a cross section when tension reinforcement reaches thestrain corresponding to its specified yield strength fy just as concrete in compression reaches its assumedultimate strain of 0.003.

10.3.3 - For flexural members, and for members subject to combined flexure and compressive axialload when the design axial load strength (ΦPn) is less than the smaller of (0.10f ′

cAg) or (ΦPb), the ratioof reinforcement p provided shall not exceed 0.75 of the ratio ρb that would produce balanced strainconditions for the section under flexure without axial load. For members with compression reinforcement,the portion of ρb equalized by compression reinforcement need not be reduced by the 0.75 factor.

10.3.4 - Compression reinforcement in conjunction with additional tension reinforcement may beused to increase the strength of flexural members.

10.5.1 - At any section of a flexural member, except as provided in Sections 10.5.2 and 10.5.3, wherepositive reinforcement is required by analysis, the ratio ρ provided shall not be less than that given by

ρmin =200fy


Draft

Chapter 12

PRESTRESSED CONCRETE

12.1 Introduction

1 Beams with longer spans are architecturally more appealing than those with short ones. However, fora reinforced concrete beam to span long distances, it would have to have to be relatively deep (and atsome point the self weight may become too large relative to the live load), or higher grade steel andconcrete must be used.

2 However, if we were to use a steel with fy much higher than ≈ 60 ksi in reinforced concrete (R/C),then to take full advantage of this higher yield stress while maintaining full bond between concrete andsteel, will result in unacceptably wide crack widths. Large crack widths will in turn result in corrosionof the rebars and poor protection against fire.

3 One way to control the concrete cracking and reduce the tensile stresses in a beam is to prestress thebeam by applying an initial state of stress which is opposite to the one which will be induced by theload.

4 For a simply supported beam, we would then seek to apply an initial tensile stress at the top andcompressive stress at the bottom. In prestressed concrete (P/C) this can be achieved through prestressingof a tendon placed below the elastic neutral axis.

5 Main advantages of P/C: Economy, deflection & crack control, durability, fatigue strength, longerspans.

6 There two type of Prestressed Concrete beams:

Pretensioning: Steel is first stressed, concrete is then poured around the stressed bars. When enoughconcrete strength has been reached the steel restraints are released, Fig. 12.1.

Postensioning: Concrete is first poured, then when enough strength has been reached a steel cable ispassed thru a hollow core inside and stressed, Fig. 12.2.

12.1.1 Materials

7 P/C beams usually have higher compressive strength than R/C. Prestressed beams can have f ′c as

high as 8,000 psi.

8 The importance of high yield stress for the steel is illustrated by the following simple example.If we consider the following:

Draft12–2 PRESTRESSED CONCRETE

Anchorage Jacks

Continuoustendon

Casting bed

Tendon

JacksTendon anchorage

Jacks

Supportforce

Hold-downforce

Casting bed

Prestressing bed slab

Precast Concreteelement

Harping hold-downpoint

Verticalbulkhead

Harping hold-uppoint

Figure 12.1: Pretensioned Prestressed Concrete Beam, (Nilson 1978)

Tendon in conduct

BeamJackAnchorage

Wrapped tendon

Slab

AnchorageJack

JackIntermediatediaphragms

Anchorage

Figure 12.2: Posttensioned Prestressed Concrete Beam, (Nilson 1978)



1. An unstressed steel cable of length ls

2. A concrete beam of length lc

3. Prestress the beam with the cable, resulting in a stressed length of concrete and steel equal tol′s = l′c.

4. Due to shrinkage and creep, there will be a change in length

∆lc = (εsh + εcr)lc (12.1)

we want to make sure that this amout of deformation is substantially smaller than the stretch ofthe steel (for prestressing to be effective).

5. Assuming ordinary steel: fs = 30 ksi, Es = 29, 000 ksi, εs = 3029,000 = 1.03 × 10−3 in/ in

6. The total steel elongation is εsls = 1.03 × 10−3ls

7. The creep and shrinkage strains are about εcr + εsh ' .9 × 10−3

8. The residual stres which is left in the steel after creep and shrinkage took place is thus

(1.03 − .90) × 10−3(29 × 103) = 4 ksi (12.2)

Thus the total loss is 30−430 = 87% which is unacceptably too high.

9. Alternatively if initial stress was 150 ksi after losses we would be left with 124 ksi or a 17% loss.

10. Note that the actual loss is (.90 × 10−3)(29 × 103) = 26 ksiin each case

9 Having shown that losses would be too high for low strength steel, we will use

Strands usually composed of 7 wires. Grade 250 or 270 ksi, Fig. 12.3.��

��

��

��

��

��

��

��

��

��

��

��

��

��

Figure 12.3: 7 Wire Prestressing Tendon

Tendon have diameters ranging from 1/2 to 1 3/8 of an inch. Grade 145 or 160 ksi.

Wires come in bundles of 8 to 52.

Note that yield stress is not well defined for steel used in prestressed concrete, usually we take 1% strainas effective yield.

10 Steel relaxation is the reduction in stress at constant strain (as opposed to creep which is reductionof strain at constant stress) occrs. Relaxation occurs indefinitely and produces significant prestress loss.If we denote by fpthe final stress after t hours, fpi the initial stress, and fpy the yield stress, then

fp

fpi= 1 − log t

10

(fpi

fpy− .55

)(12.3)



12.1.2 Prestressing Forces

11 Prestress force “varies” with time, so we must recognize 3 stages:

1. Pj Jacking force. But then due to

(a) friction and anchorage slip in post-tension

(b) elastic shortening in pretension

is reduced to:

2. Pi Initial prestress force; But then due to time dependent losses caused by

(a) relaxation of steel

(b) shrinkage of concrete

(c) creep of concrete

is reduced to:

3. Pe Effective force

12.1.3 Assumptions

12 The following assumptions are made;

1. Materials are both in the elastic range

2. section is uncracked

3. sign convention: +ve tension, −ve compression

4. Subscript 1 refers to the top and 2 to the bottom

5. I, S1 = Ic1

, S2 = Ic2

6. e + ve if downward from concrete neutral axis

12.1.4 Tendon Configuration

13 Through proper arrangement of the tendon (eccentricity at both support and midspan) various internalflexural stress distribution can be obtained, Fig. 12.4.

12.1.5 Equivalent Load

14 An equivalent load for prestressing can be usually determined from the tendon configuration and theprestressing force, Fig. 12.5.

12.1.6 Load Deformation

15 The load-deformation curve for a prestressed concrete beam is illustrated in Fig. 12.6.


Draft12.2 Flexural Stresses 12–5

Q

P Ph/2

h/3

2Q

P Ph/2

h/3

2Q

P P 2h/3

P P h/2

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

W

h

f’

f

f

y

fc

c

c

f =fc t

2fc

2f 2fc

02f =2ft c

c

2fc

0

0 2fc

2f =2ft c

2fc

02fc

f

fc

cc

c

f

f

0

2fc f =f

c

t c

f

fc

c

fcc

c cf f

f

+

+

+

+

+

+ =

=

=

=

=

=0

0

Midspan

Ends

Midspan

Q

Ends

f

Figure 12.4: Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978)

12.2 Flexural Stresses

16 We now identify the following 4 stages:

Initial Stage when the beam is being prestressed (recalling that r2 = IAc

1. the prestressing force, Pi

only

f1 = − Pi

Ac+

Piec1

I= − Pi

Ac

(1 − ec1

r2

)(12.4)

f2 = − Pi

Ac− Piec2

I= − Pi

Ac

(1 +

ec2

r2

)(12.5)

2. Pi and the self weight of the beam M0 (which has to be acconted for the moment the beamcambers due to prestressing)

f1 = − Pi

Ac

(1 − ec1

r2

)− M0

S1(12.6)

f2 = − Pi

Ac

(1 +

ec2

r2

)+

M0

S2(12.7)

Service Load when the prestressing force was reduced from Pi to Pe beacause of the losses, and theactual service (not factored) load is apllied3. Pe and M0



P sin θP sin θ

P cos θ P cos θ

P

P cos θ

P cos θ

P sin θP sin θ

P cos θP sin θ

P

2

P sin θP sin θ

P cos θ

M

P cos θ

P sin θP sin θ

P cos θP sin θ

Pθ

θP

Pe

P

P PP e P e

e

Pθ M

P PNone

P

P

P

2

None

P P

(a)

(b)

(d)

(f)

(g)

(e)

P

(c)

Equivalent load on concrete from tendon Moment from prestressingMember

Figure 12.5: Determination of Equivalent Loads

UltimateSteel yielding

Service load limitincludingtolerable overload

First cracking load

Decompression

Full dead load

TnOverload

Service load range

fcr

or higher

cgs (f=0)Balanced

∆o ∆D ∆L

∆pe

∆pi

Load

Rupture

Deformation ∆(deflection of camber)

∆pi= Initial prestress camber∆pe= Effective prestress camber∆O= Self-weight deflection∆D= Dead load deflection∆L= Live load deflection

Figure 12.6: Load-Deflection Curve and Corresponding Internal Flexural Stresses for a Typical Pre-stressed Concrete Beam, (Nilson 1978)



f1 = −Pe

Ac

(1 − ec1

r2

)− M0

S1(12.8)

f2 = −Pe

Ac

(1 +

ec2

r2

)+

M0

S2(12.9)

4. Pe and M0 + MDL + MLL

f1 = −Pe

Ac

(1 − ec1

r2

)− M0 + MDL + MLL

S1(12.10)

f2 = −Pe

Ac

(1 +

ec2

r2

)+

M0 + MDL + MLL

S2(12.11)

The internal stress distribution at each one of those four stages is illustrated by Fig. 12.7.

S2

Mo+

S2

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

e c 2

c 1

Pi e c

Ic2Stage 1

Stage 2

Stage 4

re c

2 )2( 1 +P i

Ac

Mo+

S2 re c

2 )2( 1 +P i

Ac

P i

Ac

P i

Ac

Pi e c 1

Ic

P i

Ac( 1 -

re c

21 )

re c

2 )2( 1 +P i

Ac

P i

Ac( 1 -

re c

21 ) -

Mo

S1-

Mo

S1( 1 -

re c

21 )

P i

Ac

Ac

P e-

Mo

S( 1 -

re c

21 ) -

Md + Ml

S-

S1

Mt

Ac

P e( 1 -

re c

21 )

Md + Ml

S+ +

Mtr

e c2 )2( 1 +

Ac

P er

e c2 )2( 1 +

Ac

P e Mo+

S2 2

1 1

Figure 12.7: Flexural Stress Distribution for a Beam with Variable Eccentricity; Maximum MomentSection and Support Section, (Nilson 1978)

17 Those (service) flexural stresses must be below those specified by the ACI code (where the subscriptsc, t, i and s refer to compression, tension, initial and service respectively):



fci permitted concrete compression stress at initial stage .60f ′ci

fti permitted concrete tensile stress at initial stage < 3√

f ′ci

fcs permitted concrete compressive stress at service stage .45f ′c

fts permitted concrete tensile stress at initial stage 6√

f ′c or 12

√f ′

c

Note that fts can reach 12√

f ′c only if appropriate deflection analysis is done, because section would

be cracked.

18 Based on the above, we identify two types of prestressing:

Full prestressing (pioneered by Freysinet), no tensile stresses, no crack, but there are some problemswith excessive camber when unloaded.

Partial prestressing (pioneered by Leonhardt, Abeles, Thurliman), cracks are allowed to occur (justas in R/C), and they are easier to control in P/C than in R/C.

19 The ACi code imposes the following limits on the steel stresses in terms of fpu which is the ultimatestrength of the cable: Pj < .80fpuAs and Pi < .70fpuAs. No limits are specified for Pe.

Example 12-1: Prestressed Concrete I Beam

Adapted from (Nilson 1978)The following I Beam has f ′

c = 4, 000 psi, L = 40 ft, DL+LL =0.55 k/ft, concrete density γ = 150lb/ft3 and multiple 7 wire strands with constant eccentricity e = 5.19 in. Pi = 169 k, and the total lossesdue to creep, shinkage, relaxation are 15%.

5"

5"

7"

7"

4"

4"

2"

2"

6"

6"

4"

24"

12"

The section properties for this beam are Ic = 12, 000 in4, Ac = 176 in2, S1 = S2 = 1, 000 in3,r2 = I

A = 68.2 in2.Determine flexural stresses at midspan and at support at initial and final conditions. Solution:

1. Prestressing force, Pi only

f1 = − Pi

Ac

(1 − ec1

r2

)(12.12-a)

= −169, 000176

(1 − (5.19)(12)

68.2

)= −83 psi (12.12-b)

f2 = − Pi

Ac

(1 +

ec2

r2

)(12.12-c)

= −169, 000176

(1 +

(5.19)(12)68.21

)= 1, 837 psi (12.12-d)



2. Pi and the self weight of the beam M0 (which has to be acconted for the moment the beam cambersdue to prestressing)

w0 =(176) in2

(144) in2/ ft2(.150) k/ ft3 = .183 k/ft (12.13-a)

M0 =(.183)(40)2

8= 36.6 k.ft (12.13-b)

The flexural stresses will thus be equal to:

fw01,2 = ∓ M0

S1,2= ∓ (36.6)(12, 000)

1, 000= ∓439 psi (12.14)

f1 = − Pi

Ac

(1 − ec1

r2

)− M0

S1(12.15-a)

= −83 − 439 = −522 psi (12.15-b)

fti = 3√

f ′c = +190

√(12.15-c)

f2 = − Pi

Ac

(1 +

ec2

r2

)+

M0

S2(12.15-d)

= −1, 837 + 439 = −1, 398 psi (12.15-e)

fci = .6f ′c = −2, 400

√(12.15-f)

3. Pe and M0. If we have 15% losses, then the effective force Pe is equal to (1 − 0.15)169 = 144 k

f1 = −Pe

Ac

(1 − ec1

r2

)− M0

S1(12.16-a)

= −144, 000176

(1 − (5.19)(12)

68.2

)− 439 (12.16-b)

= −71 − 439 = −510 psi (12.16-c)

f2 = −Pe

Ac

(1 +

ec2

r2

)+

M0

S2(12.16-d)

= −144, 000176

(1 +

(5.19)(12)68.2

)+ 439 (12.16-e)

= −1, 561 + 439 = −1, 122 psi (12.16-f)

note that −71 and −1, 561 are respectively equal to (0.85)(−83) and (0.85)(−1, 837) respectively.


MDL + MLL =(0.55)(40)2

8= 110 k.ft (12.17)

and corresponding stresses

f1,2 = ∓ (110)(12, 000)1, 000

= ∓1, 320 psi (12.18)

Thus,

f1 = −Pe

Ac

(1 − ec1

r2

)− M0 + MDL + MLL

S1(12.19-a)



= −510 − 1, 320 = −1, 830 psi (12.19-b)

fcs = .45f ′c = −2, 700

√(12.19-c)

f2 = −Pe

Ac

(1 +

ec2

r2

)+

M0 + MDL + MLL

S2(12.19-d)

= −1, 122 + 1, 320 = +198 psi (12.19-e)

fts = 6√

f ′c = +380

√(12.19-f)

5. The stress distribution at each one of the four stages is shown below.

-83

-510

-522

-183

0

-139

8

-112

2

-183

7

+19

8

1234

12.3 Case Study: Walnut Lane Bridge


20 The historical Walnut Lane Bridge (first major prestressed concrete bridge in the USA) is made ofthree spans, two side ones with lengths of 74 ft and a middle one of length 160 feet. Thirteen prestressedcocnrete beams are placed side by side to make up a total width of 44 fet of roadway and two 9.25 feet ofsidewalk. In between the beams, and cast with them, are transverse stiffeners which connect the beamslaterally, Fig. 12.8

12.3.1 Cross-Section Properties

21 The beam cross section is shown in Fig. 12.9 and is simplified

Ac = 2(8.9)(52) + (7)(61.2) = 1, 354 in2 (12.20-a)

I = 2

[(52)(8.9)3

12+ (52)(8.9)

(792

− 8.92

)2]

+(7)(61.2)3

12(12.20-b)

= = 1, 277 × 103 in4 (12.20-c)

c1 = c2 =h

2=

792

= 39.5 in (12.20-d)


Draft12.3 Case Study: Walnut Lane Bridge 12–11

80 ft

CENTERLINEELEVATION OF BEAM HALF

9.25’44 ’9.25’

BEAM CROSS SECTIONS TRANSVERSE DIAPHRAGMS

ROAD

SIDEWALK

CROSS - SECTION OF BRIDGE

CROSS - SECTION OF BEAM

6’-7"3’-3"

7"

10"3"

6 "3 "

7"

1/21/2

30"

52"

10"

7"

TRANSVERSE DIAPHRAGM

SLOTS FOR CABLES

Figure 12.8: Walnut Lane Bridge, Plan View



SIMPLIFIED CROSS - SECTION OF BEAM

8.9"

61.2"

8.9"

22.5"22.5" 7"

52"

6’-7"= 79"

Figure 12.9: Walnut Lane Bridge, Cross Section

S1 = S2 =I

c=

1, 277 × 103

39.5= 32, 329 in3 (12.20-e)

r2 =I

A=

1, 277 × 103

1, 354= 943. in2 (12.20-f)

12.3.2 Prestressing

22 Each beam is prestressed by two middle parabolic cables, and two outer horizontal ones along theflanges. All four have approximately the same eccentricity at midspan of 2.65 ft. or 31.8 inch.

23 Each prestressing cable is made up 64 wires each with a diameter of 0.27 inches. Thus the total areaof prestressing steel is given by:

Awire = π(d/2)2 = 3.14(0.276 in

2)2 = 0.0598 in2 (12.21-a)

Acable = 64(0.0598) in2 = 3.83 in2 (12.21-b)Atotal = 4(3.83) in2 = 15.32 in2 (12.21-c)

24 Whereas the ultimate tensile strength of the steel used is 247 ksi, the cables have been stressed onlyto 131 ksi, thus the initial prestressing force Pi is equal to

Pi = (131) ksi(15.32) in2 = 2, 000 k (12.22)

25 The losses are reported ot be 13%, thus the effective force is

Pe = (1 − 0.13)(2, 000) k = 1, 740 k (12.23)


Draft12.3 Case Study: Walnut Lane Bridge 12–13

12.3.3 Loads

26 The self weight of the beam is q0 = 1.72 k/ft.

27 The concrete (density=.15 k/ ft3) road has a thickness of 0.45 feet. Thus for a 44 foot width, thetotal load over one single beam is

qr,tot =113

(44) ft(0.45) ft(0.15) k/ ft3 = 0.23 k/ft (12.24)

28 Similarly for the sidewalks which are 9.25 feet wide and 0.6 feet thick:

qs,tot =113

(2)(9.25) ft(0.60) ft(0.15) k/ ft3 = 0.13 k/ft (12.25)

We note that the weight can be evenly spread over the 13 beams beacause of the lateral diaphragms.

29 The total dead load isqDL = 0.23 + 0.13 = 0.36 k/ft (12.26)

30 The live load is created by the traffic, and is estimated to be 94 psf, thus over a width of 62.5 feetthis gives a uniform live load of

wLL =113

(0.094) k/ft2(62.5) ft = 0.45 k/ft (12.27)

31 Finally, the combined dead and live load per beam is

wDL+LL = 0.36 + 0.45 = 0.81 k/ft (12.28)

12.3.4 Flexural Stresses

1. Prestressing force, Pi only

f1 = − Pi

Ac

(1 − ec1

r2

)(12.29-a)

= − (2 × 106)1, 354

(1 − (31.8)(39.5)

943.

)= 490. psi (12.29-b)

f2 = − Pi

Ac

(1 +

ec2

r2

)(12.29-c)

= − (2 × 106)1, 354

(1 +

(31.8)(39.5)943.

)= −3, 445. psi (12.29-d)

2. Pi and the self weight of the beam M0 (which has to be acconted for the moment the beam cambersdue to prestressing)

M0 =(1.72)(160)2

8= 5, 504 k.ft (12.30)

The flexural stresses will thus be equal to:

fw01,2 = ∓ M0

S1,2= ∓ (5, 50.4)(12, 000)

943.= ∓2, 043 psi (12.31)



f1 = − Pi

Ac

(1 − ec1

r2

)− M0

S1(12.32-a)

= 490 − 2, 043 = −1, 553 psi (12.32-b)

fti = 3√

f ′c = +190

√(12.32-c)

f2 =Pi

Ac

(1 +

ec2

r2

)+

M0

S2(12.32-d)

= −3, 445 + 2, 043 = −1, 402. psi (12.32-e)

fci = .6f ′c = −2, 400

√(12.32-f)

3. Pe and M0. If we have 13% losses, then the effective force Pe is equal to (1 − 0.13)(2 × 106) =1.74 × 106 lbs

f1 = −Pe

Ac

(1 − ec1

r2

)− M0

S1(12.33-a)

= −1.74 × 106

1, 354

(1 − (31.8)(39.5)

943.

)− 2, 043. = −1, 616 psi (12.33-b)

f2 =Pe

Ac

(1 +

ec2

r2

)+

M0

S2(12.33-c)

= −1.74 × 106

1, 354

(1 +

(31.8)(39.5)943.

)+ 2, 043. = −954. psi (12.33-d)


MDL + MLL =(0.81)(160)2

8= 2, 592 k.ft (12.34)

and corresponding stresses

f1,2 = ∓ (2, 592)(12, 000)32, 329

= ∓962. psi (12.35)

Thus,

f1 = −Pe

Ac

(1 − ec1

r2

)− M0 + MDL + MLL

S1(12.36-a)

= −1, 616 − 962. = −2, 578. psi (12.36-b)

fcs = .45f ′c = −2, 700

√(12.36-c)

f2 =Pe

Ac

(1 +

ec2

r2

)+

M0 + MDL + MLL

S2(12.36-d)

= −954 + 962. = +8. psi (12.36-e)

fts = 6√

f ′c = +380

√(12.36-f)


Draft

Chapter 13

Three-Hinges ARCHES

13.1 Theory

13.1.1 Uniform Horizontal Load

1 In order to optimize dead-load efficiency, long span structures should have their shapes approximate thecoresponding moment diagram, hence an arch, suspended cable, or tendon configuration in a prestressedconcrete beam all are nearly parabolic, Fig. 13.1.

2 Long span structures can be built using flat construction such as girders or trusses. However, for spansin excess of 100 ft, it is often more economical to build a curved structure such as an arch, suspendedcable or thin shells.

3 Since the dawn of history, mankind has tried to span distances using arch construction. Essentiallythis was because an arch required materials to resist compression only (such as stone, masonary, bricks),and labour was not an issue.

4 The basic issues of static in arch design are illustrated in Fig. 13.2 where the vertical load is per unithorizontal projection (such as an external load but not a self-weight). Due to symmetry, the verticalreaction is simply V = wL

2 , and there is no shear across the midspan of the arch (nor a moment). Takingmoment about the crown,

M = Hh − wL

2

(L

2− L

4

)= 0 (13.1)

Solving for H

H =wL2

8h(13.2)

We recall that a similar equation was derived for arches., and H is analogous to the C − T forces in abeam, and h is the overall height of the arch, Since h is much larger than d, H will be much smallerthan C − T in a beam.

5 Since equilibrium requires H to remain constant across thee arch, a parabolic curve would theoreticallyresult in no moment on the arch section.

6 Three-hinged arches are statically determinate structures which shape can acomodate support settle-ments and thermal expansion without secondary internal stresses. They are also easy to analyse throughstatics.

Draft13–2 Three-Hinges ARCHES

M = w L /82

L

w=W/L

BEAMBEAM

W/2 W/2

- C

+ T

- C

+ T

RISE = h

SAG = h

IDEALISTIC ARCHSHAPE GIVEN BYMOMENT DIAGRAM

IDEALISTIC SUSPENSIONSHAPE GIVEN BYMOMENT DIAGRAM

T

C

T

C

NOTE THAT THE "IDEAL" SHAPE FOR AN ARCH OR SUSPENSION

SYSTEM IS EQUIVILENT TO THE DESIGN LOAD MOMENT DIAGRAM

M-ARM smallC-T large

Figure 13.1: Moment Resisting Forces in an Arch or Suspension System as Compared to a Beam, (Linand Stotesbury 1981)

CROWN2

M = VL/2 - wL /8 - H h = 0

BASE2M = wL /8 - H h = 0

w

V = wL/2

H

w

h

L/2

h

V = wL/2

H H

VR R

L

22R = V + H

H = wL /8h2

wL/2

Figure 13.2: Statics of a Three-Hinged Arch, (Lin and Stotesbury 1981)


Draft13.1 Theory 13–3

7 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 compressivestresses.

8 An arch is far more efficient than a beam, and possibly more economical and aesthetic than a truss incarrying loads over long spans.

9 If the arch has only two hinges, Fig. 13.3, or if it has no hinges, then bending moments may existeither at the crown or at the supports or at both places.

APPARENT LINEOF PRESSURE WITHARCH BENDING EXCEPT AT THE BASE

h

h’

V V

M

w

h

H’=wl /8h’<2

wl /8h2

H’H’<H

APPARENT LINE OFPRESSURE WITHARCH BENDING INCLUDING BASE

V V

M

w

h

L

crownM

M

base

baseh’

H’<H

Figure 13.3: Two Hinged Arch, (Lin and Stotesbury 1981)

10 Since H varies inversely to the rise h, it is obvious that one should use as high a rise as possible. Fora combination of aesthetic and practical considerations, a span/rise ratio ranging from 5 to 8 or perhapsas much as 12, is frequently used. However, as the ratio goes higher, we may have buckling problems,and the section would then have a higher section depth, and the arch advantage diminishes.

11 In a parabolic arch subjected to a uniform horizontal load there is no moment. However, in practicean arch is not subjected to uniform horizontal load. First, the depth (and thus the weight) of an archis not usually constant, then due to the inclination of the arch the actual self weight is not constant.Finally, live loads may act on portion of the arch, thus the line of action will not necessarily follow thearch centroid. This last effect can be neglected if the live load is small in comparison with the dead load.

12 Since the greatest total force in the arch is at the support, (R =√

V 2 + H2), whereas at the crownwe simply have H, the crown will require a smaller section than the support.

APPARENT LINEOF PRESSURE WITHARCH BENDING EXCEPT AT THE BASE

h

h’

V V

M

w

h

H’=wl /8h’<2

wl /8h2

H’H’<H

APPARENT LINE OFPRESSURE WITHARCH BENDING INCLUDING BASE

V V

M

w

h

L

crownM

M

base

baseh’

H’<H

Figure 13.4: Arch Rib Stiffened with Girder or Truss, (Lin and Stotesbury 1981)



Example 13-1: Design of a Three Hinged Arch

adapted from (Lin and Stotesbury 1981) A long arch 100 ft high and spanning 510 ft is to bedesigned for a garage and hotel building, using air rights over roads and highways.

510’

100’Garage and hotel Building

It is necessary to determine preliminary dimensions for the size of the arch section. The arches arespaced 60 ft on centers and carry four-story loading totaling 27 k/ft along each arch.Solution:

1. To the initial DL and LL of 27 k/ft we add the arch own weight estimated to be ≈ 25% of theload, thus the total load is

w = (1 + .25)27 = 33.7 ≈ 33 k/ft (13.3)

2. We next determine the various forces:

H =wL2

8h=

(33)(510)2

8(100)= 10, 700 k (13.4-a)

V =wL

2=

(33)(510)2

= 8, 400 k (13.4-b)

R =√

H2 + V 2 =√

(10, 700)2 + (8, 400)2 = 13, 600 k (13.4-c)

3. If we use concrete-filled steel pipe for arch section, and selecting a pipe diameter of 6 ft with athickness 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 (13.5)

4. The concrete area is

Ac = πD2

4= (3.14)

(6)2

4ft2(144) in2/ftsq = 4, 070 in2 (13.6)

5. Assuming that the steel has an allowable stress of 20 ksi and the concrete 2.5 ksi (noting that thestrength of confined concrete can be as high as three times the one of f ′

c), then the load carry-

ing capacity of each component isSteel Ac (113)(20)ksi = 2,260 k

Concrete As (4, 070)(2.5)ksi = 10,180 kTotal 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.


Draft13.2 Case Study: Salginatobel Bridge (Maillart) 13–5

13.2 Case Study: Salginatobel Bridge (Maillart)


13.2.1 Geometry

13 The Salginatobel bridge, perhaps the most famous and influential structure of Maillart is located inhigh up in the Swiss Alps close to Shuders.

14 It is a three hinged pedestrian bridge which crosses a deep valley with a most beautiful shape whichblends perfectly with its surrounding, Fig. 13.5

20 ft 20 ft 20 ft 20 ft 20 ft87.5 ft87.5 ft

295 ft

20 ft

42.6

ft

Figure 13.5: Salginatobel Bridge; Dimensions, (Billington and Mark 1983)

15 The load supporting structure is the arch itself, whereas the bridge deck and the piers are transferingthe vertical load into the arch.

16 The arch cross section is not constant, and can be idealized as in Fig. 13.6

17 The basic shape of the supporting structure is a three hinged arch as shown in Fig. 13.7

18 The arch is parabolic (which as we saw an the optimal shape which minimizes flexure), and the crosssection at the quarter point has an area of

At = 2[(0.62)(12.46) + (0.59)(12.17)] = 29.8 ft2 = 4, 291 in2 (13.7)

19 Each flange has an area of

AF = (0.62)(12.46) = 7.73 ft2 = 1, 113 in2 (13.8)

and the effective depth of the section is d = 12.79 ft, Fig. 13.8.

20 At the crown/hinge the section is rectangular with

Acr = (1.05)(11.48) = 12.05 ft2 = 1, 735 in2 (13.9)



ACTUAL ARCH WITH CENTROID (DOTTED LINE)

IDEALIZATION(ONE DEMENSIONAL)

295 ft

42.6

ft

Figure 13.6: Salginatobel Bridge; Idealization, (Billington and Mark 1983)

CORK PADS

HINGE

ACTUAL SPRINGING HINGE IDEALIZATION

CONCRETE

CONCRETE HINGE

ACTUAL CROWN HINGE IDEALIZATION

CORK PAD

HARD WOOD

Figure 13.7: Salginatobel Bridge; Hinges, (Billington and Mark 1983)



12.46 ft

0.62 ft

0.62 ft

0.59 ft

13.4

1 ft

d=12

.79

ft12

.17

ft

ACTUAL ARCH

SECTIONS295 ft

42.6

ft

Figure 13.8: Salginatobel Bridge; Sections, (Billington and Mark 1983)



13.2.2 Loads

21 The dead load WD is assumed to be linearly distributed (even though it is greater where the arch isdeeper, and the vertical members longer) and is equal to 1,680 kips, Fig. 13.9.

wD =WD

L=

(1, 680) k

(295) ft= 5.7 k/ft (13.10)

L = 295 ft

D

D

W = 1680 k

w = 5.7 k/ft

Figure 13.9: Salginatobel Bridge; Dead Load, (Billington and Mark 1983)

22 For the sake of simplicity we will neglect the snow load (which is actually negligible compared to thedead load).

23 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. 13.10. This placement of the load actually corresponds to one ofthe most critical loading arrangement. The total vertical load is shown in Fig. 13.11

13.2.3 Reactions

24 Reactions are easily determined from equilibrium, Fig. 13.15

VD =1, 680

2= 840 k (13.11-a)

VL =1102

= 55 k (13.11-b)

(+ ¡¢¾) ΣMc = 0 (13.11-c)(840)(147.5) − (840)(73.75) − HD(42.6) = 0 (13.11-d)

⇒ HD = 1, 455 k (13.11-e)

(55)(147.5) − (55)(73.75) − HL(42.6) = 0 (13.11-f)

⇒ HL = 95 k (13.11-g)

RD =√

(840)2 + (1, 455)2 = 1, 680 k (13.11-h)

RL =√

(55)2 + (95)2 = 110 k (13.11-i)



P =

55

kP

= 5

5 k

AR

CH

AB

UT

ME

NT

RO

AD

WA

Y

PLA

N

42.5 ft

295

ft

Figure 13.10: Salginatobel Bridge; Truck Load, (Billington and Mark 1983)



V = 55 kV = 55 kA,L B,LLIVE LOAD

P = 55 k P = 55 k

AB 42

.6 f

t

295 ft

V = 840 k V = 840 kA,D B,D

DQ = 1680 k

DEAD LOAD

AB

Figure 13.11: Salginatobel Bridge; Total Vertical Load, (Billington and Mark 1983)

A

V

H

l/2=147.5 ft

l/4=73.75 ft

d=42.6 ft

CP

Figure 13.12: Salginatobel Bridge; Reactions, (Billington and Mark 1983)



13.2.4 Internal Forces

25 The shear diagrams for the dead, live and combined load is shown in Fig. 13.13.

295 ft

295 ft

295 ft

0L/4 L/2

3L/4

840 k

420 k

420 k

840 k

295 ftx

x

+ 420 k

+ 895 k

+ 475 k

- 475 k

- 895 k

- 420 k

0 L

+

=

L/255 k

-55 k

L0

SHE

AR

FO

RC

E

SHE

AR

FO

RC

E

SHE

AR

FO

RC

E

Figure 13.13: Salganitobel Bridge; Shear Diagrams, (Billington and Mark 1983)

26 At the quarter point the axial force can be expressed as:

N = H cos α + V sin α (13.12)

wheretanα =

2d

L(13.13)

and α = 16.1o at this location. The horizontal force for the dead and live loads was determined previouslyas 1, 455 and 95 kips respectively, and the vertical forces are obtained from the shear diagram, thus

NqrD = (−1, 455) cos 16.1o + (−420) sin 16.1o = −1, 514 k (13.14-a)

N qrL = (−95) cos 16.1o + (−55) sin 16.1o = −106 k (13.14-b)

and at the crown where there is no vertical force (and α = 0)

N crD = (−1, 455) cos 0o + (−420) sin 0o = −1, 455 k (13.15-a)

N crL = (−95) cos 0o + (−55) sin 0o = −95 k (13.15-b)

27 The uniform dead load will not produce a moment on the parabolic arch.

28 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

MVL =

P

2L

4=

11o2

2954

= 4, 050 k.ft (13.16)

2. The second is caused by the horizontal reaction, and the resulting moment is MHL = Hd(x), since

d varies parabolically, and H is constant, that second moment is parabolic with a peak value equalto

MVL = Hdmax = (95)(32.6) = −4, 050 k.ft (13.17)



at the quarter point

MVL = Hd1/4 = (95)

3(32.6)4

= −3, 040 k.ft (13.18)

29 The overall bending moment diagram from the live loads is determined by simply adding those twocomponents, Fig. 13.14.

MO

ME

NT

BE

ND

ING

295 ft

BE

ND

ING

295 ft

MO

ME

NT

BE

ND

ING

L/4L/4

P P

xL/4 L/2 3L/4 L0

MO

ME

NT

+

=+ PL/4 = 4,050 k-ft

-4,050 k-ft

-3,040 k-ft

4,050 ft.k-3,040 ft.k = 1,010 ft.k

Figure 13.14: Salginatobel Bridge; Live Load Moment Diagram, (Billington and Mark 1983)

30 We observe that the actual shape of the arch follows this bending moment diagram for one of themost critical live load case.

31 The maximum moment at midspan is

MmaxL = 4, 050 − 3, 040 = 1, 010 k.ft (13.19)

which would produce internal forces in the upper and lower flanges equal to:

Fint = ±MmaxL

d=

(1, 010) k.ft

(12.8) ft= ±79 k (13.20)

13.2.5 Internal Stresses

32 The axial stresses at the springlines were determined to be −1, 680 and −110 kips for the dead andlive loads respectively.

33 At the support the area of concrete is Ac = 2, 240 in2, thus the axial stresses are

σDsp =

−(1, 680) k

(2, 240) in2 1, 000 = −750 psi (13.21-a)

σLsp =

−(110) k

(2, 240) in2 1, 000 = −49 psi (13.21-b)

σsspTotal = −750 − 49 = −799 psi (13.21-c)


Draft13.3 Structural Behavior of Deck-Stiffened Arches 13–13

34 At the crown, we repeat the same calculations, where the axial force is equal to the horizontalcomponent of the reactions

σDcr =

−(1, 455) k

(1, 735) in2 1, 000 = −839 psi (13.22-a)

σLcr =

−(95) k

(1, 735) in2 1, 000 = −55 psi (13.22-b)

σscrTotal = −839 − 55 = −894 psi (13.22-c)

35 The stresses at the quarter point are determined next. Note that we must include the effect of bothaxial and flexural stresses

σtopqr =

−(1, 514) k

(4, 291) in2 1, 000︸︷︷︸DeadLoad

+−(106) k

(4, 291) in2 1, 000︸︷︷︸LiveLoad︸︷︷︸

AxialStresses

− (79) k

(1, 113) in2︸︷︷︸Flexural

(13.23-a)

= −353 − 25 − 71 −449 psi (13.23-b)

σbotqr =

−(1, 514) k

(4, 291) in2 1, 000︸︷︷︸DeadLoad

+−(106) k

(4, 291) in2 1, 000︸︷︷︸LiveLoad︸︷︷︸

AxialStresses

(79) k

(1, 113) in2︸︷︷︸Flexural

(13.23-c)

= −353 − 25 + 71 −307 psi (13.23-d)

13.3 Structural Behavior of Deck-Stiffened Arches

From (Billington 1979)INCOMPLETE

36 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. 13.15



wL

wL

wL a2

wL a2

wL

wL

wL a

wL

Unstiffened Arch Stiffened Arch

∼∆/10∆

a

0

Figure 13.15: Structural Behavior of Stiffened Arches, (Billington 1979)


Draft

Chapter 14

BUILDING STRUCTURES

14.1 Introduction

14.1.1 Beam Column Connections

1 The connection between the beam and the column can be, Fig. 14.1:

θ b

θc

θ b θcθ b θc=θ b θc=θ b θc=

θ

θ

θ

θ

b b

cc

M=K( - )s s

Flexible Rigid Semi-Flexible

Figure 14.1: Flexible, Rigid, and Semi-Flexible Joints

Flexible that is a hinge which can transfer forces only. In this case we really have cantiliver actiononly. 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 6= 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 appliedmoment at the node), Mcol = Mbeam 6= 0, θcol = θbeam.

Semi-Rigid: The end moments are equal and not equal to zero, but the rotation are different. θbeam 6=θcol, Mcol = Mbeam 6= 0. Furthermore, the difference in rotation is resisted by the spring Mspring =Kspring(θcol − θbeam).

14.1.2 Behavior of Simple Frames

2 For vertical load across the beam rigid connection will reduce the maximum moment in the beam (atthe expense of a negative moment at the ends which will in turn be transferred to the columns).

Draft14–2 BUILDING STRUCTURES

3 The advantages of a rigid connection are greater when the frame is subjected to a lateral load. Underthose conditions, the connection will stiffen the structure and reduce the amount of lateral deflection,Fig. 14.2.

PI PI

PI

PI PI

PI

V V V V

H∆

H∆

θ

PI PI

θ

Figure 14.2: Deformation of Flexible and Rigid Frames Subjected to Vertical and Horizontal Loads, (Linand Stotesbury 1981)

4 Fig. 14.3 illustrates the deformation, shear, moment and axial forces in frames with different boundaryconditions under both vertical and horizontal loads.

14.1.3 Eccentricity of Applied Loads

5 A concentric axial force P and moment M , applied on a support sytem (foundation, columns, pre-stressing) can be replaced by a static equivalent one in which the moment M is eliminated and the forceP applied with an eccentricity

e =M

P(14.1)

6 The induced stresses can be decomposed into uniform (−P/L) (assuming a unit width) and linearlyvarying one (σ = M/S) and the end stresses are

σmin = −P

L− σ (14.2-a)

σmax = −P

L+ σ (14.2-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, andΣM = 0, i.e. Mint = Mext.

7 If we seek the eccentricity ecr for which σmax equals zero, then σ = PL



w

P

w/2

-w/2

w/2

-w/2

w/2

-w/2

w/2

M’

M’

M’/

L

-M’/

L

M’/

L

-M’/

L

a

bh

L

c

d

e

f

g

h

i

j

k

l

p

M

w/2 w/2

M’

-M’/L

p

-M’/L

p

w/2 -w/2

M/h-M/h

p/2p/2

-M/L

w/2 -w/2

0.36

M/h

-0.3

6M/h

-M/L

p/2p/2

-w/2

0.68

M/h

-0.5M’/L

p/2 p/2

M

M

w/2 w/2

w/2 w/2

M M

M/h

w/2 w/2

M’/2 M’/2p/2

M/L

-M/L

0.4M 0.4M

0.64M0.4M/h

w/2 w/2

M’/2p/2

-M’/

L

M’/

L

0.55M

0.45M

M’/4M’/4

M’/4

0.68M/h

p/2

w/2w/2

M’/

2L

-M’/

2L

POST AND BEAM STRUCTURE

SIMPLE BENT FRAME

THREE-HINGE PORTAL

THREE-HINGE PORTAL

TWO-HINGE FRAME

RIGID FRAME

-0.6

8M/h

M’/4

0.45M

M’/2

Deformation Shear Moment AxialFrame TypeW=wL, M=wL/8, M’=Ph

2

Figure 14.3: Deformation, Shear, Moment, and Axial Diagrams for Various Types of Portal FramesSubjected to Vertical and Horizontal Loads



8 The net tensile force due to the eccentric load is

T =12σ

L

2(14.3)

If we want this net tensile force to be equal and opposite to the compressive force, then

T = σL4

σ = PA

}T = P

L

applied at23

L2 from the centroid

(14.4)

Thus the net internal moment is

Mint = 2T23

L

2= 2

P

423

L

2=

PL

6(14.5)

9 To satisfy the equilibrium equation, this internal moment must be equal and opposite to the externalmoment Mext = Pecr hence

PL

6︸︷︷︸Mint

= Pecr︸︷︷︸Mext

(14.6)

or

ecr =L

6(14.7)

in other words to avoid tensile stresses on either side, the resultant force P must be placed within themidle third kernel, Fig. 14.4

L/2L/2 L/3 L/3 L/3

e

L/3 L/3 L/3

e

L/3 L/3L/6

P

L/3 L/3L/6

P

L/3L/3L/2L/2

L/3L/3P PL/6

M/S

+ +

=

+

==

P/A

P

Figure 14.4: 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), thenthere 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 onlyunder the foundation and no undesirable uplift.

3. Buildings: If the eccentricity of the vertical load is within the middle third, all columns will beloaded under compression only.


Draft14.2 Buildings Structures 14–5

14.2 Buildings Structures

11 There are three primary types of building systems:

Wall Subsytem: in which very rigid walls made up of solid masonry, paneled or braced timber, or steeltrusses 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 tubularstructure may be interior (housing elevators, staircases) and/or exterior. Most efficient for veryhigh rise buildings.

Rigid Frame: which consists of linear vertical components (columns) rigidly connected to stiff hor-izontal ones (beams and girders). This is not a very efficient structural form to resist lateral(wind/earthquake) loads.

14.2.1 Wall Subsystems

12 Whereas exterior wall provide enclosure and interior ones separation, both of them can also have astructural role in trnsfering vertical and horizontal loads.

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

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

15 When shear-walls subsytems are used, it is best if the center of orthogonal shear resistance is closeto the centroid of lateral loads as applied. If this is not the case, then there will be torsional designproblems.

14.2.1.1 Example: Concrete Shear Wall

From (Lin and Stotesbury 1981)

16 We consider a reinforced concrete wall 20 ft wide, 1 ft thick, and 120 ft high with a vertical load of400 k acting on it at the base. As a result of wind, we assume a uniform horizontal force of 0.8 kip/ftof vertical height acting on the wall. It is required to compute the flexural stresses and the shearingstresses in the wall to resist the wind load, Fig. 14.5.

1. Maximum shear force and bending moment at the base

Vmax = wL = (0.8) k.ft(120) ft = 96 k (14.8-a)

Mmax =wL2

2=

(0.8) k.ft(120)2 ft2

2= 5, 760 k.ft (14.8-b)

2. The resulting eccentricity is

eActual =M

P=

(5, 760) k.ft

(400) k= 14.4 ft (14.9)

3. The critical eccentricity is

ecr =L

6=

(20) ft

6= 3.3 ft < eActualN.G. (14.10)

thus there will be tension at the base.



��

��

��

+ F

H=96 k;M =5760 k’

120’

1’

- f

- F

+ F

HORIZONTAL

VERTICAL

2/(3d)

7.7’ IN TENSION

+140

+ 600+ 740 PSI

+ f

V

60’

DL

M M

20’

400 kW

w=0.8 k/ft

Figure 14.5: Design of a Shear Wall Subsystem, (Lin and Stotesbury 1981)

4. The moment of inertia of the wall is

I =bh3

12=

(1) ft(20) ft3

12= 667 ft4 (14.11)

5. The maximum flexural stresses will be

σmax = ±Mc

I=

(5, 760) k.ft(10) ft

(667) ft4= ±(86.5) ksf = ±(600) psi (14.12)

6. The average shearing stress is

τ =V

A=

(96) k

(1)(20) ft2= 4.8 ksf = 33.3 psi (14.13)

A concrete with nominal shear reinforcement can carry at least 100 psi in shear, those computedshear streses are permissible.

7. At the base of the wall, the axial stresses will be

σ =−(400) k

(1)(20) ft2= (20) ksf =≈ −140 psi (14.14-a)

8. The maximum stresses will thus be:

σ1 = −140 + 600 = 460 psi (Tension) (14.15-a)

σ2 = −140 − 600 = −740 psi (Compression) (14.15-b)



9. The compressive stress of 740 psi can easily be sustained by concrete, as to the tensile stress of460 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 windload,

σall =43(20) = 26.7 ksi (14.16)

11. The stress distribution is linear, compression at one end, and tension at the other. The length ofthe tension area is given by (similar triangles)

x

460=

20460 + 740

⇒ x =460

460 + 740(20) = 7.7 ft (14.17)

12. The total tensile force inside this triangular stress block is

T =12(460) ksi(7.7 × 12) in (12) in︸︷︷︸

width

= 250 k (14.18)

13. The total amount of steel reinforcement needed is

As =(250) k

(26.7) ksi= 9.4 in2 (14.19)

This amount of reinforcement should be provided at both ends of the wall since the wind oreartquake can act in any direction. In addition, the foundations should be designed to resisttensile uplift forces (possibly using piles).

14.2.1.2 Example: Trussed Shear Wall


17 We consider the same problem previously analysed, but use a trussed shear wall instead of a concreteone, Fig. 14.6.

1. Using the maximum moment of 5, 760 kip-ft (Eq. 14.8-b), we can compute the compression andtension in the columns for a lever arm of 20 ft.

F = ± (5, 760) k.ft

(20) ft= ±288 k (14.20)

2. If we now add the effect of the 400 kip vertical load, the forces would be

C = − (400) k

2− 288 = −488 k (14.21-a)

T = − (400) k

2+ 288 = 88 k (14.21-b)

3. The force in the diagonal which must resist a base shear of 96 kip is (similar triangles)

F

96=

√(20)2 + (24)2

20⇒ F =

√(20)2 + (24)2

20(96) = 154 k (14.22)

4. The design could be modified to have no tensile forces in the columns by increasing the width ofthe base (currently at 20 ft).



20’

120’

24’

1.2

1

~1.6

V

60’

W400 k

+F -FMM

H=96 k

Figure 14.6: Trussed Shear Wall

14.2.2 Shaft Systems

18 Vertical shear-resisting shafts in buildings act as a tubular section and generally have a rectangularcross section. If there is only one shaft, it is generally located in the center and houses the elevators. Ifthere are many shafts, then they should be symmetrically arranged.

19 If the shaft is relatively short and wide, with an aspect ratio under 1 or 2, then the dominant strcutralaction is that of a stiff shear resisting tube. If the aspect ratio is between 3 and 5, then the shear forcesmay not be the controlling criterion, and flexure dominates.

14.2.2.1 Example: Tube Subsystem


20 With reference to Fig. 14.7, the reinforced concrete shaft is 20 ft square, 120 ft high, and with 1 ftthick walls. It is subjected to a lateral force of 0.8 k/ft.

1. Comparing this structure with the one analysed in Sect. 14.2.1.1 the total vertical load acting onthe base is now increased to

V = 4(400) = 1, 600 k (14.23)

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 = Σbd3

12=

(20)(20)3

12− (18)(18)3

12= 4, 600 ft4 (14.24)



~ 20 ’~ 20 ’

N.A.

��

��

��

��

w = 0.8 k/ft

H = 96 k

60 ’

120 ’

20 ’ 20 ’

Figure 14.7: Design Example of a Tubular Structure, (Lin and Stotesbury 1981)

4. The maximum flexural stresses:

σfl = ±MC

I= ± (5, 760) k.ft(20/2) ft

(4, 600) ft4= ±12.5 ksf = ±87 psi (14.25)

5. The average shear stress is

τ =V

A=

(96) k

2(20)(1) ft2= 2.4 ksf = 17 psi (14.26)

6. The vertical load of 1,600 k produces an axial stress of

σax =P

A=

−(1, 600) k

(4(20)(1) ft2= −20 ksf = −140 psi (14.27)

7. The total stresses are thus

σ = σax + σfl (14.28-a)

σ1 = −140 + 87 = −53 psi (14.28-b)

σ2 = −140 − 87 = −227 psi (14.28-c)

thus we do not have any tensile stresses, and those stresses are much better than those obtainedfrom a single shear wall.

14.2.3 Rigid Frames

21 Rigid frames can carry both vertical and horizontal loads, however their analysis is more complexthan for tubes.



22 The rigorous and exact analysis of a rigid frame can only be accomplished through a computer analysis.However, for preliminary design it is often sufficient to perform approximate analyses.

23 There are two approximate methods for the analysis of rigid frames subjected to lateral loads: 1)Portal and 2) Cantilever method.

24 The portal frame method is based on the following major assumptions, Fig. 14.8:

L

P

h

h/2

L/2

PI

P P

h/2

V =P/(2L) V =P/(2L)

H =P/2H =P/2

1

1 2

2

Figure 14.8: A Basic Portal Frame, (Lin and Stotesbury 1981)

1. Each bay of a bent acts as a separate “portal” frame consisting of two adjacent columns and theconnecting 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 exteriorcolumn is half the shear of an interior column.

This method will be discussed in more details in the following section.

14.3 Approximate Analysis of Buildings

25 Despite the widespread availability of computers, approximate methods of analysis are justified by

1. Inherent assumption made regarding the validity of a linear elastic analysis vis a vis of an ultimatefailure design.

2. Ability of structures to redistribute internal forces.

3. Uncertainties in load and material properties

26 Vertical loads are treated separately from the horizontal ones.

27 We use the design sign convention for moments (+ve tension below), and for shear (ccw +ve).

28 Assume girders to be numbered from left to right.

29 In all free body diagrams assume positivee forces/moments, and take algeebraic sums.


Draft14.3 Approximate Analysis of Buildings 14–11

14.3.1 Vertical Loads

30 The girders at each floor are assumed to be continuous beams, and columns are assumed to resist theresulting unbalanced moments from the girders.

31 Basic assumptions

1. Girders at each floor act as continous beams supporting a uniform load.

2. Inflection points are assumed to be at

(a) One tenth the span from both ends of each girder.(b) Mid-height of the columns

3. Axial forces and deformation in the girder are negligibly small.

4. Unbalanced end moments from the girders at each joint is distributed to the columns above andbelow the floor.

32 Based on the first assumption, all beams are statically determinate and have a span, Ls equal to 0.8the original length of the girder, L. (Note that for a rigidly connected member, the inflection point isat 0.211 L, and at the support for a simply supported beam; hence, depending on the nature of theconnection one could consider those values as upper and lower bounds for the approximate location ofthe hinge).

33 End forces are given by

Maximum positive moment at the center of each beam is, Fig. 14.9

0.1L

L

0.1L0.8L

Vrgt

Vlft

Mrgt

Mlft

w

Figure 14.9: Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments

M+ =18wL2

s = w18(0.8)2L2 = 0.08wL2 (14.29)

Maximum negative moment at each end of the girder is given by, Fig. 14.9

M left = M rgt = −w

2(0.1L)2 − w

2(0.8L)(0.1L) = −0.045wL2 (14.30)



Vrgti-1 Vlft

i

Pabove

Pbelow

Figure 14.10: Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces

Girder Shear are obtained from the free body diagram, Fig. 14.10

V lft =wL

2V rgt = −wL

2(14.31)

Column axial force is obtained by summing all the girder shears to the axial force transmitted by thecolumn above it. Fig. 14.10

P dwn = P up + V rgti−1 − V lft

i(14.32)

Column Moment are obtained by considering the free body diagram of columns Fig. 14.11

h/2

h/2

h/2

h/2

Mcolbelow

Mi-1rgt Mi

lft

Mcolabove

Li

h/2

h/2

h/2

h/2

Li-1

Vi-1lft Vi

lft Virgt

Vi-1rgt

Mi-1lft

Mirgt

Figure 14.11: Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments

M top = Mbotabove − M rgt

i−1 + M lfti Mbot = −top (14.33)



Column Shear Points of inflection are at mid-height, with possible exception when the columns onthe first floor are hinged at the base, Fig. 14.11

V =M top

h2

(14.34)

Girder axial forces are assumed to be negligible eventhough the unbalanced column shears above andbelow a floor will be resisted by girders at the floor.

14.3.2 Horizontal Loads

34 We must differentiate between low and high rise buildings.

Low rise buidlings, where the height is at least samller than the hrizontal dimension, the deflectedshape is characterized by shear deformations.

High rise buildings, where the height is several times greater than its least horizontal dimension, thedeflected shape is dominated by overall flexural deformation.

14.3.2.1 Portal Method

35 Low rise buildings under lateral loads, have predominantly shear deformations. Thus, the approximateanalysis of this type of structure is based on

1. Distribution of horizontal shear forces.

2. Location of inflection points.

36 The portal method is based on the following assumptions

1. Inflection points are located at

(a) Mid-height of all columns above the second floor.

(b) Mid-height of floor columns if rigid support, or at the base if hinged.

(c) At the center of each girder.

2. Total horizontal shear at the mid-height of all columns at any floor level will be distributed amongthese columns so that each of the two exterior columns carry half as much horizontal shear as eachinterior columns of the frame.

37 Forces are obtained from

Column Shear is obtained by passing a horizontal section through the mid-height of the columns ateach floor and summing the lateral forces above it, then Fig. 14.12

V ext =∑

F lateral

2No. of baysV int = 2V ext (14.35)



H/2 H/2HHFigure 14.12: Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear

Column Moments at the end of each column is equal to the shear at the column times half the heightof the corresponding column, Fig. 14.12

M top = Vh

2Mbot = −M top (14.36)

Girder Moments is obtained from the columns connected to the girder, Fig. 14.13

h/2

h/2

h/2

h/2

Mcolbelow

Mi-1rgt Mi

lft

Mcolabove

Li-1/2 Li/2h/2

h/2

h/2

h/2

Li/2Li-1/2

Vi-1lft Vi

lft Virgt

Vi-1rgt

Mi-1lft

Mirgt

Figure 14.13: ***Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment

M lfti = Mabove

col − M belowcol + M rgt

i−1 M rgti = −M lft

i(14.37)

Girder Shears Since there is an inflection point at the center of the girder, the girder shear is obtainedby considering the sum of moments about that point, Fig. 14.13

V lft = −2M

LV rgt = V lft (14.38)



Column Axial Forces are obtained by summing girder shears and the axial force from the columnabove, Fig. ??

Vrgti-1 Vlft

i

Pabove

Pbelow

Figure 14.14: Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force

P = P above + P rgt + P lft (14.39)

Example 14-1: Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads

Draw the shear, and moment diagram for the following frame. Solution:

15K

30K

14’

16’

20’ 30’ 24’

5 6

1

12 13 14

11109

87

3 42

0.25K/ft

0.50K/ft

Figure 14.15: Example; Approximate Analysis of a Building

Vertical Loads



1. Top Girder Moments

M lft12 = −0.045w12L

212 = −(0.045)(0.25)(20)2 = − 4.5 k.ft

M cnt12 = 0.08w12L

212 = (0.08)(0.25)(20)2 = 8.0 k.ft

M rgt12 = M lft

12 = − 4.5 k.ft

M lft13 = −0.045w13L

213 = −(0.045)(0.25)(30)2 = − 10.1 k.ft

M cnt13 = 0.08w13L

213 = (0.08)(0.25)(30)2 = 18.0 k.ft

M rgt13 = M lft

13 = − 10.1 k.ft

M lft14 = −0.045w14L

214 = −(0.045)(0.25)(24)2 = − 6.5 k.ft

M cnt14 = 0.08w14L

214 = (0.08)(0.25)(24)2 = 11.5 k.ft

M rgt14 = M lft

14 = − 6.5 k.ft

2. Bottom Girder Moments

M lft9 = −0.045w9L

29 = −(0.045)(0.5)(20)2 = − 9.0 k.ft

M cnt9 = 0.08w9L

29 = (0.08)(0.5)(20)2 = 16.0 k.ft

M rgt9 = M lft

9 = − 9.0 k.ft

M lft10 = −0.045w10L

210 = −(0.045)(0.5)(30)2 = − 20.3 k.ft

M cnt10 = 0.08w10L

210 = (0.08)(0.5)(30)2 = 36.0 k.ft

M rgt10 = M lft

11 = − 20.3 k.ft

M lft11 = −0.045w12L

212 = −(0.045)(0.5)(24)2 = − 13.0 k.ft

M cnt11 = 0.08w12L

212 = (0.08)(0.5)(24)2 = 23.0 k.ft

M rgt11 = M lft

12 = − 13.0 k.ft

3. Top Column Moments

M top5 = +M lft

12 = − 4.5 k.ft

Mbot5 = −M top

5 = 4.5 k.ft

M top6 = −M rgt

12 + M lft13 = −(−4.5) + (−10.1) = − 5.6 k.ft

Mbot6 = −M top

6 = 5.6 k.ft

M top7 = −M rgt

13 + M lft14 = −(−10.1) + (−6.5) = − 3.6 k.ft

Mbot7 = −M top

7 = 3.6 k.ft

M top8 = −M rgt

14 = −(−6.5) = 6.5 k.ft

Mbot8 = −M top

8 = − 6.5 k.ft

4. Bottom Column Moments

M top1 = +Mbot

5 + M lft9 = 4.5 − 9.0 = − 4.5 k.ft

Mbot1 = −M top

1 = 4.5 k.ft

M top2 = +Mbot

6 − M rgt9 + M lft

10 = 5.6 − (−9.0) + (−20.3) = − 5.6 k.ft

Mbot2 = −M top

2 = 5.6 k.ft

M top3 = +Mbot

7 − M rgt10 + M lft

11 = −3.6 − (−20.3) + (−13.0) = 3.6 k.ft

Mbot3 = −M top

3 = − 3.6 k.ft

M top4 = +Mbot

8 − M rgt11 = −6.5 − (−13.0) = 6.5 k.ft

Mbot4 = −M top

4 = − 6.5 k.ft

5. Top Girder ShearV lft

12 = w12L122 = (0.25)(20)

2 = 2.5 k

V rgt12 = −V lft

12 = − 2.5 k

V lft13 = w13L13

2 = (0.25)(30)2 = 3.75 k

V rgt13 = −V lft

13 = − 3.75 k

V lft14 = w14L14

2 = (0.25)(24)2 = 3.0 k

V rgt14 = −V lft

14 = − 3.0 k



0.25K/ft

0.50K/ft 14’

16’

20’ 30’ 24’

5 6

1

12 13 14

11109

87

3 42

-9.0’k

-4.5’k

-10.1’k -10.1’k

-13.0’k

-4.5’k

-4.5’k

+4.5’k

+4.5’k +5.6’k

+5.6’k

-5.6’k

-5.6’k +3.6’k

+3.6’k

-3.6’k

-3.6’k

-6.5’k

-6.5’k

+6.5’k

+6.5’k

-13.0’k

-20.2’k-20.2’k

+8.0’k +18.0’k

+11.5’k

+16.0’k

+32.0’k+23.0’k

-6.5’k-6.5’k

-4.5’k

-9.0’k

Figure 14.16: Approximate Analysis of a Building; Moments Due to Vertical Loads



6. Bottom Girder Shear

V lft9 = w9L9

2 = (0.5)(20)2 = 5.00 k

V rgt9 = −V lft

9 = − 5.00 k

V lft10 = w10L10

2 = (0.5)(30)2 = 7.50 k

V rgt10 = −V lft

10 = − 7.50 k

V lft11 = w11L11

2 = (0.5)(24)2 = 6.00 k

V rgt11 = −V lft

11 = − 6.00 k

7. Column ShearsV5 = Mtop

5H52

= −4.5142

= − 0.64 k

V6 = Mtop6

H62

= −5.6142

= − 0.80 k

V7 = Mtop7

H72

= 3.6142

= 0.52 k

V8 = Mtop8

H82

= 6.5142

= 0.93 k

V1 = Mtop1

H12

= −4.5162

= − 0.56 k

V2 = Mtop2

H22

= −5.6162

= − 0.70 k

V3 = Mtop3

H32

= 3.6162

= 0.46 k

V4 = Mtop4

H42

= 6.5162

= 0.81 k

+2.5K +3.75K

-3.75K

+3.0K

-3.0K

+5.0K

-5.0K

+7.5K

-7.5K

-0.64K -0.80K +0.51K +0.93K

+0.81K+0.45K-0.70K-0.56K

-6.0K

+6.0K

-2.5K

Figure 14.17: Approximate Analysis of a Building; Shears Due to Vertical Loads



8. Top Column Axial Forces

P5 = V lft12 = 2.50 k

P6 = −V rgt12 + V lft

13 = −(−2.50) + 3.75 = 6.25 k

P7 = −V rgt13 + V lft

14 = −(−3.75) + 3.00 = 6.75 k

P8 = −V rgt14 = 3.00 k

9. Bottom Column Axial Forces

P1 = P5 + V lft9 = 2.50 + 5.0 = 7.5 k

P2 = P6 − V rgt10 + V lft

9 = 6.25 − (−5.00) + 7.50 = 18.75 k

P3 = P7 − V rgt11 + V lft

10 = 6.75 − (−7.50) + 6.0 = 20.25 k

P4 = P8 − V rgt11 = 3.00 − (−6.00) = 9.00 k

Horizontal Loads, Portal Method

1. Column ShearsV5 = 15

(2)(3) = 2.5 k

V6 = 2(V5) = (2)(2.5) = 5 k

V7 = 2(V5) = (2)(2.5) = 5 k

V8 = V5 = 2.5 k

V1 = 15+30(2)(3) = 7.5 k

V2 = 2(V1) = (2)(7.5) = 15 k

V3 = 2(V1) = (2)(2.5) = 15 k

V4 = V1 = 7.5 k

2. Top Column Moments

M top5 = V1H5

2 = (2.5)(14)2 = 17.5 k.ft

Mbot5 = −M top

5 = − 17.5 k.ft

M top6 = V6H6

2 = (5)(14)2 = 35.0 k.ft

Mbot6 = −M top

6 = − 35.0 k.ft

M top7 = V up

7 H7

2 = (5)(14)2 = 35.0 k.ft

Mbot7 = −M top

7 = − 35.0 k.ft

M top8 = V up

8 H8

2 = (2.5)(14)2 = 17.5 k.ft

Mbot8 = −M top

8 = − 17.5 k.ft

3. Bottom Column Moments

M top1 = V dwn

1 H12 = (7.5)(16)

2 = 60 k.ft

Mbot1 = −M top

1 = − 60 k.ft

M top2 = V dwn

2 H22 = (15)(16)

2 = 120 k.ft

Mbot2 = −M top

2 = − 120 k.ft

M top3 = V dwn

3 H32 = (15)(16)

2 = 120 k.ft

Mbot3 = −M top

3 = − 120 k.ft

M top4 = V dwn

4 H42 = (7.5)(16)

2 = 60 k.ft

Mbot4 = −M top

4 = − 60 k.ft



Approximate Analysis Vertical Loads APROXVER.XLS Victor E. Saouma

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123456789

101112131415161718192021222324252627282930

A B C D E F G H I J K L M N O P Q

L1 L2 L3Height Span 20 30 24

14 Load 0.25 0.25 0.2516 Load 0.5 0.5 0.5

MOMENTSBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam ColLft Cnt Rgt Lft Cnr Rgt Lft Cnt Rgt

-4.5 8.0 -4.5 -10.1 18.0 -10.1 -6.5 11.5 -6.5-4.5 -5.6 3.6 6.54.5 5.6 -3.6 -6.5

-9.0 16.0 -9.0 -20.3 36.0 -20.3 -13.0 23.0 -13.0-4.5 -5.6 3.6 6.54.5 5.6 -3.6 -6.5

SHEARBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam ColLft Rgt Lft Rgt Lft Rgt

2.50 -2.50 3.75 -3.75 3.00 -3.00-0.64 -0.80 0.52 0.93

5.00 -5.00 7.50 -7.50 6.00 -6.00-0.56 -0.70 0.46 0.81

AXIAL FORCEBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam Col0.00 0.00 0.00

2.50 6.25 6.75 3.000.00 0.00 0.00

7.50 18.75 20.25 9.00

Figure 14.18: Approximate Analysis for Vertical Loads; Spread-Sheet Format



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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

AB

CD

EF

GH

IJ

KL

MN

OP

Q

L1L2

L3H

eig

htSp

an

2030

2414

Loa

d0.

250.

250.

2516

Loa

d0.

50.

50.

5M

OM

ENTS

Bay

1Ba

y 2

Bay

3C

ol

Bea

mC

olu

mn

Bea

mC

olu

mn

Bea

mC

ol

Lft

Cn

tR

gt

Lft

Cn

rR

gt

Lft

Cn

tR

gt

=-0

.045

*D4*

D3^

2=

0.08

*D4*

D3*

D3

=+

D10

=-0

.045

*I4*

I3^

2=

0.08

*I4*

I3*I

3=

+I1

0=

-0.0

45*N

4*N

3^2

=0.

08*N

4*N

3*N

3=

N10

=+

D10

=-F

10+

I10

=-K

10+

N10

=-P

10=

-C11

=-G

11=

-L11

=-Q

11

=-0

.045

*D5*

D3^

2=

0.08

*D5*

D3*

D3

=+

D13

=-0

.045

*I5*

I3^

2=

0.08

*I5*

I3*I

3=

+I1

3=

-0.0

45*N

5*N

3^2

=0.

08*N

5*N

3*N

3=

+N

13

=+

D13

+C

12=

-F13

+I1

3+G

12=

-K13

+N

13+

L12

=-P

13+

Q12

=-C

14=

-G14

=-L

14=

-Q14

SHEA

RBa

y 1

Bay

2Ba

y 3

Co

lBe

am

Co

lum

nBe

am

Co

lum

nBe

am

Co

l

Lft

Rg

tLf

tR

gt

Lft

Rg

t

=+

D3*

D4/

2=

-D20

=+

I3*I

4/2

=-I2

0=

+N

3*N

4/2

=-N

20

=2*

C11

/A4

=2*

G11

/A4

=2*

L11/

A4

=2*

Q11

/A4

=+

D3*

D5/

2=

-D22

=+

I3*I

5/2

=-I2

2=

+N

3*N

5/2

=-N

22

=2*

C14

/A5

=2*

G14

/A5

=2*

L14/

A5

=2*

Q14

/A5

AX

IAL

FO

RCE

Bay

1Ba

y 2

Bay

3C

ol

Bea

mC

olu

mn

Bea

mC

olu

mn

Bea

mC

ol

00

0

=+

D20

=-F

20+

I20

=-K

20+

N20

=-P

20

00

0

=+

C28

+D

22=

+G

28-F

22+

I22

=+

L28-

K22+

N22

=+

Q28

-P22

Figure 14.19: Approximate Analysis for Vertical Loads; Equations in Spread-Sheet



4. Top Girder Moments

M lft12 = M top

5 = 17.5 k.ft

M rgt12 = −M lft

12 = − 17.5 k.ft

M lft13 = M rgt

12 + M top6 = −17.5 + 35 = 17.5 k.ft

M rgt13 = −M lft

13 = − 17.5 k.ft

M lft14 = M rgt

13 + M top7 = −17.5 + 35 = 17.5 k.ft

M rgt14 = −M lft

14 = − 17.5 k.ft

5. Bottom Girder Moments

M lft9 = M top

1 − Mbot5 = 60 − (−17.5) = 77.5 k.ft

M rgt9 = −M lft

9 = − 77.5 k.ft

M lft10 = M rgt

9 + M top2 − Mbot

6 = −77.5 + 120 − (−35) = 77.5 k.ft

M rgt10 = −M lft

10 = − 77.5 k.ft

M lft11 = M rgt

10 + M top3 − Mbot

7 = −77.5 + 120 − (−35) = 77.5 k.ft

M rgt11 = −M lft

11 = − 77.5 k.ft

6. Top Girder ShearV lft

12 = − 2M lft12

L12= − (2)(17.5)

20 = −1.75 k

V rgt12 = +V lft

12 = −1.75 k

V lft13 = − 2M lft

13L13

= − (2)(17.5)30 = −1.17 k

V rgt13 = +V lft

13 = −1.17 k


14L14

= − (2)(17.5)24 = −1.46 k

V rgt14 = +V lft

14 = −1.46 k

7. Bottom Girder Shear

V lft9 = − 2M lft

12L9

= − (2)(77.5)20 = −7.75 k

V rgt9 = +V lft

9 = −7.75 k


10L10

= − (2)(77.5)30 = −5.17 k

V rgt10 = +V lft

10 = −5.17 k


11L11

= − (2)(77.5)24 = −6.46 k

V rgt11 = +V lft

11 = −6.46 k

8. Top Column Axial Forces (+ve tension, -ve compression)

P5 = −V lft12 = −(−1.75) k

P6 = +V rgt12 − V lft

13 = −1.75 − (−1.17) = −0.58 k

P7 = +V rgt13 − V lft

14 = −1.17 − (−1.46) = 0.29 k

P8 = V rgt14 = −1.46 k

9. Bottom Column Axial Forces (+ve tension, -ve compression)

P1 = P5 + V lft9 = 1.75 − (−7.75) = 9.5 k

P2 = P6 + V rgt10 + V lft

9 = −0.58 − 7.75 − (−5.17) = −3.16 k

P3 = P7 + V rgt11 + V lft

10 = 0.29 − 5.17 − (−6.46) = 1.58 k

P4 = P8 + V rgt11 = −1.46 − 6.46 = −7.66 k

Design Parameters On the basis of the two approximate analyses, vertical and lateral load, we nowseek the design parameters for the frame, Table 14.2.



+17.5’K+17.5’K

+17.5’K

-17.5’K

+60’K

+60’K+120’K

-120’K -120’K

+120’K

-60’K

-60’K

-17.5’K

+17.5’K+35’K

-35’K -35’K

+35’K

+17.5’K

+77.5’K +77.5’K +77.5’K

-77.5’K -77.5’K -77.5’K

-17.5K-17.5K -17.5K

14’

16’

20’ 30’ 24’

5 6

1

12 13 14

11109

87

3 42

15K

30K

Figure 14.20: Approximate Analysis of a Building; Moments Due to Lateral Loads



Portal Method PORTAL.XLS Victor E. Saouma





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123456789

101112131415161718192021222324252627282930

A B C D E F G H I J K L M N O P Q R SPORTAL METHOD

# of Bays 3 L1 L2 L320 30 24

MOMENTS# of Storeys 2 Bay 1 Bay 2 Bay 3

Force Shear Col Beam Column Beam Column Beam ColH Lat. Tot Ext Int Lft Rgt Lft Rgt Lft Rgt

17.5 -17.5 17.5 -17.5 17.5 -17.5H1 14 15 15 2.5 5 17.5 35.0 35.0 17.5

-17.5 -35.0 -35.0 -17.577.5 -77.5 77.5 -77.5 77.5 -77.5

H2 16 30 45 7.5 15 60.0 120.0 120.0 60.0-60.0 -120.0 -120.0 -60.0

SHEARBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam ColLft Rgt Lft Rgt Lft Rgt

-1.75 -1.75 -1.17 -1.17 -1.46 -1.462.50 5.00 5.00 2.502.50 5.00 5.00 2.50

-7.75 -7.75 -5.17 -5.17 -6.46 -6.467.50 15.00 15.00 7.507.50 15.00 15.00 7.50

AXIAL FORCEBay 1 Bay 2 Bay 3

Col Beam Column Beam Column Beam Col0.00 0.00 0.00

1.75 -0.58 0.29 -1.460.00 0.00 0.00

9.50 -3.17 1.58 -7.92

Figure 14.21: Portal Method; Spread-Sheet Format



Portal Method PORTAL.XLS Victor E. Saouma

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1

23

4

5

67

8

910

11

1213

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15

16

17

18

1920

21

2223

24

25

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27

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29

30

A B C D E F G H I J K L M N O P Q R S

PORTAL METHOD# of Bays 3 L1 L2 L3

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20 30 24

MOMENTSAAAAAAAA

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# of Storeys 2 Bay 1 Bay 2 Bay 3AAAA

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Force Shear Col Beam Column Beam Column Beam ColAAAA

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H Lat. Tot Ext Int Lft Rgt Lft Rgt Lft Rgt

=+H9 =-I8 =+J8+K9 =-M8 =+N8+O9 =-Q8AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

H1 14 15 =+C9 =+D9/(2*$F$2) =2*E9 =+E9*B9/2 =+F9*B9/2 =+K9 =+H9

=-H9 =-K9 =+K10 =+H10

=+H12-H10 =-I11 =+K12-K10+J11 =-M11 =+O12-O10+N11 =-Q11AAAAAAAAAAA

AAAAAAAAAAA

H2 16 30 =SUM($C$9:C12) =+D12/(2*$F$2) =2*E12 =+E12*B12/2 =+F12*B12/2 =+K12 =+H12

=-H12 =-K12 =+K13 =+H13

SHEARAAAAAAAAAAAA

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Col Beam Column Beam Column Beam ColAAAA

AAAA

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AAAAAAAA

AAAA

Lft Rgt Lft Rgt Lft Rgt

=-2*I8/I$3 =+I18 =-2*M8/M$3 =+M18 =-2*Q8/Q$3 =+Q18AAAA

=+E9 =+F9 =+F9 =+E9

=+H19 =+K19 =+O19 =+S19

=-2*I11/I$3 =+I21 =-2*M11/M$3 =+M21 =-2*Q11/Q$3 =+Q21

=+E12 =+F12 =+F12 =+E12

=+H22 =+K22 =+O22 =+S22

AXIAL FORCE

Bay 1 Bay 2 Bay 3AAAAA

AAAAA

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Col Beam Column Beam Column Beam ColAAAA

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0 0 0

=-I18 =+J18-M18 =+N18-Q18 =+R18

0 0 0

=+H28-I21 =+K28+J21-M21 =+O28+N21-Q21 =+S28+R21AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA





Figure 14.22: Portal Method; Equations in Spread-Sheet



Mem. Vert. Hor. DesignValues

Moment 4.50 60.00 64.501 Axial 7.50 9.50 17.00

Shear 0.56 7.50 8.06Moment 5.60 120.00 125.60

2 Axial 18.75 15.83 34.58Shear 0.70 15.00 15.70Moment 3.60 120.00 123.60






8 Axial 3.00 1.46 4.46Shear 0.93 2.50 3.43

Table 14.1: Columns Combined Approximate Vertical and Horizontal Loads


Draft14.4 Lateral Deflections 14–27

Mem. Vert. Hor. DesignValues

-ve Moment 9.00 77.50 86.509 +ve Moment 16.00 0.00 16.00

Shear 5.00 7.75 12.75-ve Moment 20.20 77.50 97.70

10 +ve Moment 36.00 0.00 36.00Shear 7.50 5.17 12.67-ve Moment 13.0 77.50 90.50




14 +ve Moment 11.50 0.00 11.50Shear 3.00 1.46 4.46

Table 14.2: Girders Combined Approximate Vertical and Horizontal Loads

14.4 Lateral Deflections

38 Even at schematic or preliminary stages of design, it is important to estimate the lateral deflectionsof tall buildings for the following reasons

1. Lateral deflections are often limited by code requirements, for example ∆ < h/500 where h is theheight of the story or of the building. This is important because occupants should not experienceuncomfortable horizontal movements.

2. A building that deflects severly under lateral forces may have damage problems associated withvibration (as with vertical defelctions of beams).

3. Through the evaluation of deflection, one may also get some idea of the relative horizontal loadcarried by the various vertical subsystems in a building (i.e. how much is carried by the shaftcompared to the frames). Since all systems are connected, they must move together and throughtheir stiffness (deformation per unit load) we can determine the contribution of each subsystem.

14.4.1 Short Wall

39 In short structures (as with short beams), shear deflections, Fig. 14.23 dominates. For a concentratedload

∆ ≈ 1.2V h

GA(14.40)

where for concrete and steel G ≈ 25E.



��

��

h

∆

WALL ELEVATION

V

Figure 14.23: Shear Deformation in a Short Building, (Lin and Stotesbury 1981)

14.4.2 Tall Wall

40 Alternatively, in a tall building flexural deformations, Fig. 14.24 are predominant. dominates.

��

��

WALL (OR TUBE) ELEVATION

w h

∆

Figure 14.24: Flexural Deformation in a Tall Building, (Lin and Stotesbury 1981)

∆ ≈ wh4

8EI(14.41)

and the moment of inertia I = bh3

12 for rectangular sections.

14.4.3 Walls and Lintel

41 When two slender walls are connected by (heavy) lintels, the entire subsystem can be made to actas one cantilever supported by the foundations and deflections will be small. However if we have lightlintels, their deformation is larger than those of the walls, Fig. 14.25.



αα

LINTEL BENDING

RELATIONSHIP BETWEEN

WALL AND LINTEL

DEFORMATION

��

��

��

��

w

LLINTELS ∆

α h

a

2 WALLS CONNECTED

BY LINTELS

Figure 14.25: Deflection in a Building Structure Composed of Two Slender Walls and Lintels, (Lin andStotesbury 1981)

42 In this case deflections can be estimated from:

Mmax =wh2

2(14.42-a)

T = C =M

a(14.42-b)

V/Lintel ≈ T

# of Lintels(14.42-c)

αwall = αlintel ≈ V L2

12EI(14.42-d)

(14.42-e)

and

δ ≈ αh (14.43)

14.4.4 Frames

43 Deflection of a rigid frame is essentially caused by shear between stories which produces vertical shearsin the girders. From the portal method we can estimate those deformations, Fig. 14.26.

44 The deformation for the first story at the exterior joint can be approximated from

∆col =VcolE h3

12EIcolE

(14.44-a)

∆gdr =VgdrL

2h

12EIgdr=

2VcolE Lh2

12EIgdr(14.44-b)

∆totE= ∆colE + ∆gdr =

VcolEh2

12E

[h

IcolE

+2L

Igdr

](14.44-c)



OVERALL FRAME ELEVATION

h ST

OR

IES

∆

α θ α

h

L

DUE TO CO

DUE TO GI

DEFORMATION OF ONE BE

V

V

V

h/2

h/2

L/2

col

col

gdrVgd

gdrV

Vcol

V

h/2

h/2

L/2

L/2

EXTERIOR JOINT INTERIOR JOINT

Vgdr (L/2)=V (h)col(L)=Vcol (h)Vgdr

MOMENT EQUILIBRIUM

Figure 14.26: Portal Method to Estimate Lateral Deformation in Frames, (Lin and Stotesbury 1981)

45 For the interior joint:

∆col =VcolI h

3

12EIcolI

(14.45-a)

∆gdr =VgdrL

2h

12EIgdr=

2VcolI Lh2

12EIgdr(14.45-b)

∆totI= ∆colI + ∆gdr =

VcolI h2

12E

[h

IcolI

+L

Igdr

](14.45-c)

and the total displacement will be

∆tot = n∆tot

2(14.46)

where n is the number of stories, and ∆tot is for either the interior or exterior joints.

46 The two major sources of lateral deflection are the bending of column in resisting horizontal shearand girders in resisting vertical shear, Fig. 14.27.

47 A vertical unsymmetric load will cause lateral deflection in frames, Fig. 14.28.

14.4.5 Trussed Frame

48 The cantilever deflection due to column shortening and lengthening (produced by overturning mo-ment) is usually of secondary importance until the building is some 40 stories or higher,Fig. ??.

49 The total deflection ∆ at C is given by

∆ = ΣPPL

AE(14.47)



∆S ∆M ∆S ∆M

MOMENT EFFECT

(OVERALL BENDING)

SHEAR EFFECT

(RACKING)

OVERALL EFFECT

(RACKING + BENDING)

+

SHO

RT

EN

ING

EL

ON

GA

TIO

N

Figure 14.27: Shear and Flexural Deflection of a Rigid Frame Subsystem, (Lin and Stotesbury 1981)

SIDE SWAYP

Figure 14.28: Side-Sway Deflection from Unsymmetrical Vertical Load, (Lin and Stotesbury 1981)

H

H

H

H∆c

1

2

3

4

P 1

C C

T C

T

∆

a

a

h

c

ΣH

δ δ1

Figure 14.29: Axial Elongation and Shortening of a Truss Frame, (Lin and Stotesbury 1981)



where: P is the force in any member due to loading on the whole system, L is the length of the member,A and E the corresponding cross sectional area and modulus of elasticity, P the force in the same memberdue to a unit (1) force applied in the direction of the deflection sought, and at the point in question.

50 Alternatively, we can neglect the web deformation and consider only the axial deformations in thecolums:

∆ ≈ δt + δc

ah (14.48)

δt + δc = 2Th

AE(14.49)

14.4.6 Example of Transverse Deflection

51 Typical plan, elevation and floor section of a building are shown in Fig. 14.30. The lateral resiting

��

��

��

��

��

��

��

��

60’

20’

MOMENT DIAGRAM

TOTAL M58500 K-FT

W=4.8 K/FT

LOAD

12’TYP.

20"

20"

20’

5"

12"2.5’

20’

40’

20" TYP.

13@

12’=

156’

156’

20’

60’

A A

FLOOR PLAN

GIRDER SECTION

COLUMN SECTION

TRANSVERSE ELEVATION OF CORE

TRANSVERSE ELEVATION OF BUILDING

CORECORE SHAFT

Figure 14.30: Transverse Deflection, (Lin and Stotesbury 1981)

elements are the center concrete shaft (20ft×40ft in section and made up of four 12-in walls) and thereinforced-prestressed concrete frames (made up of 12in×30in T beams, Igdr = 3.64 ft4, and 20 inchsquare reinforced concrete columns).

52 We consider a wind load of 4.8 k/ft in the transverse direction and make the following assumptions:

1. Colums are of uniform sectional properties and height for all stories.

2. Shaft walls are of uniform thickness for all stories. We neglect wall openings.

3. The wind load is uniform over the height of the building.

53 The solution proceeds as follows:



1. Determine the flexural deformation of the top of the shaft (we may neglect shear deformationssince the shaft is slender):

∆ =wh4

8EI(14.50-a)

I =b1d

31 − b2d

32

12(14.50-b)

=(41)(21)3 − (39)(19)3

12= 9, 400 ft4 (14.50-c)

E = 3 × 106 psi = 432, 000 ksf (14.50-d)

∆ =(4.8) k.ft(156)4 ft4

8(432, 000) ksf(9, 400) ft4= 0.087 ft (14.50-e)

∆h

=0.087156

=1

1, 800√

(14.50-f)

The ∆h ratio is much less than 1/500 as permitted in most building codes, and s within the usual

index for concrete buildings, which ranges between 1/1,000 and 1/2,500.

If the wall thickness is reduced, and if door openings are considered, the deflection will be corre-spondingly smaller.

The deflection due to moment increases rapidly at the top, the value of 1/1,800 indicates only theaverage drift index for the entire building, whereas the story drift index may be higher, especiallyfor the top floor.

2. We next consider the deflection of the top of the frame. Assuming that each frame takes 1/9 ofthe total wind load and shear, and neglecting column shortening, then:

∆ =VcolE h2

12E

[h

IcolE

+2L

Igdr

](14.51-a)

Icol =bh3

12=

(20/12)(20/12)3

12= 0.64 ft4 (14.51-b)

Igdr = 3.64 ft4 (14.51-c)

V groundcolI

=(4.8) k.ft(156) ft

(2)(9)= 41.7 k/col (14.51-d)

∆ =(41.7) k(12)2

12(432, 000) ksf

[(12) ft

(0.64) ft4+

2(60) ft

(3.46) ft4

](14.51-e)

= 0.00116(18.8 + 34.7) = 0.062 ft (14.51-f)

3. Since the story drift varies with the shear in the story, which decreases linearly to the top, theaverage drift will be 0.062/2 = 0.31 ft per story and the deflection at top of the building isapproximately

∆ = (13)(0.031) = 0.40 ft (14.52)

which indicates a drift ratio of

Drift Ratio for Building =(0.4) ft

(156) ft= 1

400 (14.53-a)

Drift Ratio for Ground Floor =(0.062) ft

(12) ft= 1

194 (14.53-b)



4. Comparing the frame deflection of 0.40 ft with the shaft deflection of 0.087 ft, it is seen that theframe is about five times more flexible than the shaft. Furthermore, the frame would not bestiff enough to carry all the lateral load by itlself. Proportioning the lateral load to the relativestiffnesses, the frame would carry about 1/6 of the load, and the remaining 5/6 will be carried bythe shaft.

Increasing the column size will stiffen the frame, but in order to be really effective, the girderstiffness will also need to be increased, since thegirders contribute about 2/3 of the deflection.Then the frames can be made o carry a larger proportion of the load. Note that the deflectedshapes of the shaft and the frames are quite different, so that the above simple comparison of topdeflections is not an accurate assessment.

Finally, we have not studied the effect of the shaft stiffened by the exterior columns, which arerigidly connected to the shaft walls and will avt with the shaft as a unit, Fig. 14.31. This would

��

��

60’

20’ 20’ 20’

156’

w

CORE

COLUMNS PARTICIPATE

+

+-

-

Figure 14.31: Frame Rigidly Connected to Shaft, (Lin and Stotesbury 1981)

be quite effective as the horizontal floor diaphragms will hold and force them to deflect together.

5. In summary, this appears to be quite an efficient layout, further analysis would refine and optimizeit.

14.4.7 Effect of Bracing Trusses

54 Through strategically located havy trusses at the top and possibly at the middle of a building we canbrace the exterior columns against the core shaft. This will result in a frame-like action in the shaft,



equalize temperature shortening of vertical components, and reduce lateral deflections, Fig. 14.32.

MID - HEIGHT

WITHBRACINGEFFECT

BRACE

CORE BENDINGWITH CANTILEVER

DEFLECTION

HE

IGH

T

BRACING REDUCEOVERALL DEFLECOF BUILDING

TUBE

HAT

FULL

DEFLECTION

CANTILEVER

TENSION COMPRESSION

WIND

TOTAL RESISTANCE ARM ISINCREASED BY COL. ACTION

TIEDOWN

RESISTANCE ARM OFCORE SHAFT ONLY

HAT - TRUSS

TRUSS

T C

TC CORE

Figure 14.32: Effect of Exterior Column Bracing in Buildings, (Lin and Stotesbury 1981)


Draft

Bibliography

318, A. C.: n.d., Building Code Requirements for Reinforced Concrete, (ACI 318-83), American ConcreteInstitute.

Anon.: xx, Envyclopaedia Brittanica, University of Chicago.

Billington, D.: 1973, in D. Billington, R. Mark and J. Abel (eds), The Maillart Papers; Second NationalConference on Civil Engineering: History, Heritage and the Humanities, Department of Civil En-gineering, Princeton University.

Billington, D.: 1979, Robert Maillart’s Bridges; The art of Engineering, Princeton University Press.

Billington, D.: 1985, The Tower and the Bridge, xx.

Billington, D. and Mark, R.: 1983, Structural studies, Technical report, Department of Civil Engineering,Princeton University.

Galilei, G.: 1974, Two New Sciences, Including Centers of Gravity and Forces of Percussion, Universityof Wisconsin Press, Madison, Wisc. S. Drake translation.

le Duc, V.: 1977, Entretiens sur L’Architecture, Pierre Mardaga, Bruxelles, Belgique.

Lin, T. and Stotesbury, S.: 1981, Structural Concepts and Systems for Architects and Engineers, JohnWiley.

Nilson, A.: 1978, Design of Prestressed Concrete, John Wiley and Sons.

of Steel COnstruction, A. I.: 1986, Manual of Steel Construction; Load and Resistance Factor Design,American Institute of Steel Construction.

Palladio, A.: 19xx, The Four Books of Architecture, Dover Publications.

Penvenuto, E.: 1991, An Introduction to the History of Structural Mechanics, Springer-Verlag.

Schueller, W.: 1996, The design of Building Structures, Prentice Hall.

Timoshenko, S.: 1982, History of Strength of Materials, Dover Publications.

UBC: 1995, Uniform building code, Technical report, International COnference of Building Officials.

Vitruvius: 1960, The Ten Books on Architecture, Dover.

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