Draft - Ahmed Mansour and Design of Reinfor… · ... 3{8 E 3-1 Shear Design . ... 9.2.6 Biaxial Bending . . . ... 7.6 Approximate Analysis of Frames Subjected to Lateral Loads; Column

DraftDRAFT

Lecture Notes in:

Mechanics and Design of

REINFORCED CONCRETE

CVEN4555

c©VICTOR E. SAOUMA,

Fall 2002

Dept. of Civil Environmental and Architectural Engineering

University of Colorado, Boulder, CO 80309-0428

January 13, 2003

Draft0–2

Victor Saouma Mechanics and Design of Reinforced Concrete

Draft

Contents

1 INTRODUCTION 1–11.1 Material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–1

1.1.1 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–11.1.1.1 Mix Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–1

1.1.1.1.1 Constituents . . . . . . . . . . . . . . . . . . . . . . . . 1–11.1.1.1.2 Preliminary Considerations . . . . . . . . . . . . . . . . 1–51.1.1.1.3 Mix procedure . . . . . . . . . . . . . . . . . . . . . . . 1–51.1.1.1.4 Mix Design Example . . . . . . . . . . . . . . . . . . . 1–8

1.1.1.2 Mechanical Properties . . . . . . . . . . . . . . . . . . . . . . . . 1–91.1.2 Reinforcing Steel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–13

1.2 Design Philosophy, USD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–141.3 Analysis vs Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–151.4 Basic Relations and Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–161.5 ACI Code . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–16

2 FLEXURE 2–12.1 Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–1

E 2-1 Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22.2 Section Cracked, Stresses Elastic . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–3

2.2.1 Basic Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–32.2.2 Working Stress Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–4E 2-2 Cracked Elastic Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–5E 2-3 Working Stress Design Method; Analysis . . . . . . . . . . . . . . . . . . . 2–6E 2-4 Working Stress Design Method; Design . . . . . . . . . . . . . . . . . . . 2–7

2.3 Cracked Section, Ultimate Strength Design Method . . . . . . . . . . . . . . . . . 2–82.3.1 Whitney Stress Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–82.3.2 Balanced Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–102.3.3 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–112.3.4 Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–11

2.4 Practical Design Considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–122.4.1 Minimum Depth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–122.4.2 Beam Sizes, Bar Spacing, Concrete Cover . . . . . . . . . . . . . . . . . . 2–132.4.3 Design Aids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–13

2.5 USD Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–15E 2-5 Ultimate Strength; Review . . . . . . . . . . . . . . . . . . . . . . . . . . 2–15E 2-6 Ultimate Strength; Design I . . . . . . . . . . . . . . . . . . . . . . . . . . 2–16E 2-7 Ultimate Strength; Design II . . . . . . . . . . . . . . . . . . . . . . . . . 2–17

Draft0–2 CONTENTS

E 2-8 Exact Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–172.6 T Beams, (ACI 8.10) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–20

2.6.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–212.6.2 Design, (balanced) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22E 2-9 T Beam; Moment Capacity I . . . . . . . . . . . . . . . . . . . . . . . . . 2–22E 2-10 T Beam; Moment Capacity II . . . . . . . . . . . . . . . . . . . . . . . . . 2–23E 2-11 T Beam; Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–24

2.7 Doubly Reinforced Rectangular Beams . . . . . . . . . . . . . . . . . . . . . . . . 2–262.7.1 Tests for fs and f ′

s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–272.7.2 Moment Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–29E 2-12 Doubly Reinforced Concrete beam; Review . . . . . . . . . . . . . . . . . 2–30E 2-13 Doubly Reinforced Concrete beam; Design . . . . . . . . . . . . . . . . . . 2–32

2.8 Moment-Curvature Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–332.9 Bond & Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–35

2.9.1 Moment Capacity Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 2–39

3 SHEAR 3–13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13.2 Shear Strength of Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . 3–23.3 Shear Strength of Cracked Sections . . . . . . . . . . . . . . . . . . . . . . . . . . 3–53.4 ACI Code Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–63.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–8

E 3-1 Shear Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–83.6 Shear Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–9

E 3-2 Shear Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–113.7 Brackets and Corbels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–123.8 Deep Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–12

4 CONTINUOUS BEAMS 4–14.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–14.2 Methods of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–2

4.2.1 Detailed Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–24.2.2 ACI Approximate Method . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–2

4.3 Effective Span Design Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–44.4 Moment Redistribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–4

4.4.1 Elastic-Perfectly Plastic Section . . . . . . . . . . . . . . . . . . . . . . . . 4–44.4.2 Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–6E 4-1 Moment Redistribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–6

4.5 Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–7

5 ONE WAY SLABS 5–15.1 Types of Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–15.2 One Way Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–45.3 Design of a One Way Continuous Slab . . . . . . . . . . . . . . . . . . . . . . . . 5–5


DraftCONTENTS 0–3

6 SERVICEABILITY 6–16.1 Control of Cracking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–1

E 6-1 Crack Width . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–36.2 Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–3

6.2.1 Short Term Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–46.2.2 Long Term Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–5E 6-2 Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–7

7 APPROXIMATE FRAME ANALYSIS 7–17.1 Vertical Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–17.2 Horizontal Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–4

7.2.1 Portal Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–4E 7-1 Approximate Analysis of a Frame subjected to Vertical and Horizontal

Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–6

8 COLUMNS 8–1

9 COLUMNS 9–19.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–1

9.1.1 Types of Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–19.1.2 Possible Arrangement of Bars . . . . . . . . . . . . . . . . . . . . . . . . . 9–2

9.2 Short Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–29.2.1 Concentric Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–29.2.2 Eccentric Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–2

9.2.2.1 Balanced Condition . . . . . . . . . . . . . . . . . . . . . . . . . 9–39.2.2.2 Tension Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–59.2.2.3 Compression Failure . . . . . . . . . . . . . . . . . . . . . . . . . 9–6

9.2.3 ACI Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–79.2.4 Interaction Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–79.2.5 Design Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–7E 9-1 R/C Column, c known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–7E 9-2 R/C Column, e known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–9E 9-3 R/C Column, Using Design Charts . . . . . . . . . . . . . . . . . . . . . . 9–139.2.6 Biaxial Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–14E 9-4 Biaxially Loaded Column . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–17

9.3 Long Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–189.3.1 Euler Elastic Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–189.3.2 Effective Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–199.3.3 Moment Magnification Factor; ACI Provisions . . . . . . . . . . . . . . . 9–21E 9-5 Long R/C Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–24E 9-6 Design of Slender Column . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–25

10 PRESTRESSED CONCRETE 10–110.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1

10.1.1 Materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–110.1.2 Prestressing Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–410.1.3 Assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–410.1.4 Tendon Configuration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–4


Draft0–4 CONTENTS

10.1.5 Equivalent Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–410.1.6 Load Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–4

10.2 Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–6E 10-1 Prestressed Concrete I Beam . . . . . . . . . . . . . . . . . . . . . . . . . 10–8

10.3 Case Study: Walnut Lane Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1010.3.1 Cross-Section Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1210.3.2 Prestressing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1210.3.3 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1310.3.4 Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–13


Draft

List of Figures

1.1 Schematic Representation of Aggregate Gradation . . . . . . . . . . . . . . . . . 1–21.2 MicroCracks in Concrete under Compression . . . . . . . . . . . . . . . . . . . . 1–101.3 Concrete Stress Strain Curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–111.4 Modulus of Rupture Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–111.5 Split Cylinder (Brazilian) Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–111.6 Biaxial Strength of Concrete . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–121.7 Time Dependent Strains in Concrete . . . . . . . . . . . . . . . . . . . . . . . . . 1–13

2.1 Strain Diagram Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . . 2–12.2 Transformed Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22.3 Stress Diagram Cracked Elastic Section . . . . . . . . . . . . . . . . . . . . . . . 2–32.4 Desired Stress Distribution; WSD Method . . . . . . . . . . . . . . . . . . . . . . 2–42.5 Cracked Section, Limit State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–82.6 Whitney Stress Block . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–102.7 Bar Spacing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–152.8 T Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–202.9 T Beam as Rectangular Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–202.10 T Beam Strain and Stress Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 2–212.11 Decomposition of Steel Reinforcement for T Beams . . . . . . . . . . . . . . . . . 2–212.12 Doubly Reinforced Beams; Strain and Stress Diagrams . . . . . . . . . . . . . . . 2–262.13 Different Possibilities for Doubly Reinforced Concrete Beams . . . . . . . . . . . 2–272.14 Strain Diagram, Doubly Reinforced Beam; is As Yielding? . . . . . . . . . . . . . 2–272.15 Strain Diagram, Doubly Reinforced Beam; is A′

s Yielding? . . . . . . . . . . . . . 2–282.16 Summary of Conditions for top and Bottom Steel Yielding . . . . . . . . . . . . . 2–292.17 Bending of a Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–342.18 Moment-Curvature Relation for a Beam . . . . . . . . . . . . . . . . . . . . . . . 2–352.19 Bond and Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–362.20 Actual Bond Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–372.21 Splitting Along Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–372.22 Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–382.23 Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–382.24 Hooks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–402.25 Bar cutoff requirements of the ACI code . . . . . . . . . . . . . . . . . . . . . . . 2–412.26 Standard cutoff or bend points for bars in approximately equal spans with uni-

formly distributed load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–422.27 Moment Capacity Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–43

3.1 Principal Stresses in Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–1

Draft0–2 LIST OF FIGURES

3.2 Types of Shear Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13.3 Shear Strength of Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . 3–23.4 Mohr’s Circle for Shear Strength of Uncracked Section . . . . . . . . . . . . . . . 3–33.5 Shear Strength of Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . 3–43.6 Free Body Diagram of a R/C Section with a Flexural Shear Crack . . . . . . . . 3–53.7 Equilibrium of Shear Forces in Cracked Section . . . . . . . . . . . . . . . . . . . 3–63.8 Summary of ACI Code Requirements for Shear . . . . . . . . . . . . . . . . . . . 3–73.9 Corbel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–93.10 Shear Friction Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–103.11 Shear Friction Across Inclined Reinforcement . . . . . . . . . . . . . . . . . . . . 3–10

4.1 Continuous R/C Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–14.2 Load Positioning on Continuous Beams . . . . . . . . . . . . . . . . . . . . . . . 4–14.3 ACI Approximate Moment Coefficients . . . . . . . . . . . . . . . . . . . . . . . . 4–34.4 Design Negative Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–44.5 Moment Diagram of a Rigidly Connected Uniformly Loaded Beam . . . . . . . . 4–54.6 Moment Curvature of an Elastic-Plastic Section . . . . . . . . . . . . . . . . . . . 4–54.7 Plastic Moments in Uniformly Loaded Rigidly Connected Beam . . . . . . . . . . 4–54.8 Plastic Redistribution in Concrete Sections . . . . . . . . . . . . . . . . . . . . . 4–64.9 Block Diagram for R/C Building Design . . . . . . . . . . . . . . . . . . . . . . . 4–8

5.1 Types of Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–15.2 One vs Two way slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–25.3 Load Distribution in Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–25.4 Load Transfer in R/C Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–3

6.1 Crack Width Equation Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . 6–26.2 Uncracked Transformed and Cracked Transformed X Sections . . . . . . . . . . . 6–46.3 Time Dependent Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–56.4 Time Dependent Strain Distribution . . . . . . . . . . . . . . . . . . . . . . . . . 6–66.5 Short and long Term Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–6

7.1 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments . 7–27.2 Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces7–37.3 Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments 7–37.4 Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear . . . 7–57.5 Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment . . 7–57.6 Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force7–67.7 Example; Approximate Analysis of a Building . . . . . . . . . . . . . . . . . . . . 7–77.8 Approximate Analysis of a Building; Moments Due to Vertical Loads . . . . . . . 7–97.9 Approximate Analysis of a Building; Shears Due to Vertical Loads . . . . . . . . 7–107.10 Approximate Analysis for Vertical Loads; Spread-Sheet Format . . . . . . . . . . 7–127.11 Approximate Analysis for Vertical Loads; Equations in Spread-Sheet . . . . . . . 7–137.12 Approximate Analysis of a Building; Moments Due to Lateral Loads . . . . . . . 7–147.13 Portal Method; Spread-Sheet Format . . . . . . . . . . . . . . . . . . . . . . . . . 7–167.14 Portal Method; Equations in Spread-Sheet . . . . . . . . . . . . . . . . . . . . . . 7–17

9.1 Types of columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–19.2 Tied vs Spiral Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–1


DraftLIST OF FIGURES 0–3

9.3 Possible Bar arrangements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–29.4 Sources of Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–39.5 Load Moment Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 9–39.6 Strain and Stress Diagram of a R/C Column . . . . . . . . . . . . . . . . . . . . 9–49.7 Column Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–89.8 Failure Surface of a Biaxially Loaded Column . . . . . . . . . . . . . . . . . . . . 9–149.9 Load Contour at Plane of Constant Pn, and Nondimensionalized Corresponding

plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–159.10 Biaxial Bending Interaction Relations in terms of β . . . . . . . . . . . . . . . . . 9–169.11 Bilinear Approximation for Load Contour Design of Biaxially Loaded Columns . 9–169.12 Euler Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–189.13 Column Failures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–199.14 Critical lengths of columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–209.15 Effective length Factors Ψ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–219.16 Standard Alignment Chart (ACI) . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–229.17 Minimum Column Eccentricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–229.18 P-M Magnification Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . 9–23

10.1 Pretensioned Prestressed Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . 10–210.2 Posttensioned Prestressed Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . 10–210.3 7 Wire Prestressing Tendon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–310.4 Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978)10–510.5 Determination of Equivalent Loads . . . . . . . . . . . . . . . . . . . . . . . . . . 10–510.6 Load-Deflection Curve and Corresponding Internal Flexural Stresses for a Typi-

cal Prestressed Concrete Beam, (Nilson 1978) . . . . . . . . . . . . . . . . . . . . 10–610.7 Flexural Stress Distribution for a Beam with Variable Eccentricity; Maximum

Moment Section and Support Section, (Nilson 1978) . . . . . . . . . . . . . . . . 10–710.8 Walnut Lane Bridge, Plan View . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1110.9 Walnut Lane Bridge, Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . 10–12


Draft0–4 LIST OF FIGURES


Draft

List of Tables

1.1 ASTM Sieve Designation’s Nominal Sizes Used for Concrete Aggregates . . . . . 1–31.2 ASTM C33 Grading Limits for Coarse Concrete Aggregates . . . . . . . . . . . . 1–31.3 ASTM C33 Grading Limits for Fine Concrete Aggregates . . . . . . . . . . . . . 1–31.4 Example of Fineness Modulus Determination for Fine Aggregate . . . . . . . . . 1–51.5 Recommended Slumps (inches) for Various Types of Construction . . . . . . . . 1–61.6 Recommended Average Total Air Content as % of Different Nominal Maximum

Sizes of Aggregates and Levels of Exposure . . . . . . . . . . . . . . . . . . . . . 1–61.7 Approximate Mixing Water Requirements, lb/yd3 of Concrete For Different

Slumps and Nominal Maximum Sizes of Aggregates . . . . . . . . . . . . . . . . . 1–71.8 Relationship Between Water/Cement Ratio and Compressive Strength . . . . . . 1–71.9 Volume of Dry-Rodded Coarse Aggregate per Unit Volume of Concrete for Dif-

ferent Fineness Moduli of Sand . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–81.10 Creep Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–131.11 Properties of Reinforcing Bars . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–141.12 Strength Reduction Factors, Φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1–14

2.1 Total areas for various numbers of reinforcing bars (inch2) . . . . . . . . . . . . . 2–142.2 Minimum Width (inches) according to ACI Code . . . . . . . . . . . . . . . . . . 2–14

4.1 Building Structural Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–7

5.1 Recommended Minimum Slab and Beam Depths . . . . . . . . . . . . . . . . . . 5–4

7.1 Columns Combined Approximate Vertical and Horizontal Loads . . . . . . . . . 7–187.2 Girders Combined Approximate Vertical and Horizontal Loads . . . . . . . . . . 7–19

Draft0–2 LIST OF TABLES


Draft

Chapter 1

INTRODUCTION

1.1 Material

1.1.1 Concrete

This section is adapted from Concrete by Mindess and Young, Prentice Hall, 1981

1.1.1.1 Mix Design

1.1.1.1.1 Constituents

1 Concrete is a mixture of Portland cement, water, and aggregates (usually sand and crushedstone).

2 Portland cement is a mixture of calcareous and argillaceous materials which are calcined ina kiln and then pulverized. When mixed with water, cement hardens through a process calledhydration.

3 Ideal mixture is one in which:

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

2. A minimum amount of water is provided to complete the chemical reaction with cement.Strictly speaking, a water/cement ratio of about 0.25 is needed to complete this reaction,but then the concrete will have a very low “workability”.

In such a mixture, about 3/4 of the volume is constituted by the aggregates, and the remaining1/4 being the cement paste.

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

5 Portland Cement has the following ASTM designation

I Normal

II Moderate sulfate resistant, moderate heat of hydration

III High early strength (but releases too much heat)

Draft1–2 INTRODUCTION

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Figure 1.1: Schematic Representation of Aggregate Gradation

IV Low heat Portland cement, minimizes thermal cracking but must control initial temper-ature

V Sulfate resistant (marine environment)

6 Aggregate usually occupy 70% to 80% of the volume of concrete. They are granular materialderived, for the most part, from natural rock, crushed stone, natural gravels and sands.

7 ASTM C33 (Standard Specifications for Concrete Aggregates) governs the types of rock whichcan produce aggregates.

8 The shape can be rounded, irregular, angular, flaky, or elongated.

9 The surface texture can be glassy, smooth, granular, rough, crystalline or honeycombed.

10 The particle size distribution or grading of aggregates is very important as it determinesthe amount of paste for a workable concrete, Fig. 1.1. Since cement is the most expensivecomponent, proper gradation is of paramount importance.

11 The grading of an aggregate supply is determined by a sieve analysis. A representativesample of the aggregate is passed through a stack of sieves aranged in order of decreasing sizeopening of the sieve.

12 We divide aggregates in two categories

Coarse aggregate fraction is that retained on the No. 4 sieve, Table 1.1.

Fine aggregate fraction is that passing the No. 4 sieve.

13 ASTM C33 sets grading limits for coarse and fine aggregates, Table 1.2 and 1.3 respectively.

14 If a concrete does not comply with these limits, than there will be a need for more paste,and there will be the possibility of aggregate segregation.

15 Since aggregates contain some porosity, water can be absorbed. Also water can be retainedon the surface of the particle as a film or moisture. Hence, it is necessary to quantify themoisture content of the aggregates in order to make adjustments to the water. Because dryaggregates will remove water from the paste, then the w/c is effectively reduced. On the otherhand moist aggregates may effectively increase the w/c ratio.


Draft1.1 Material 1–3

ASTM SizeDesign. mm in.

Coarse Aggregate3 in. 75 3

21/2 in. 63 2.52 in. 50 2

11/2 in. 37.5 1.51 in. 25 1

3/4 in. 19 0.751/2 in. 12.5 0.503/8 in. 9.5 0.375

Fine AggregateNo. 4 4.75 0.187No. 8 2.36 0.0937

No. 16 1.18 0.0469No. 30 0.60 (600 µm) 0.0234No. 50 300 µm 0.0124

No. 100 150 µm 0.0059

Table 1.1: ASTM Sieve Designation’s Nominal Sizes Used for Concrete Aggregates

Sieve Size % Passing Each Sieve(Nominal Maximum Size)

11/2 in. 1 in. 3/4 in. 1/2 in.11/2 in. 95-100 100 - -1 in. - 95-100 100 -3/4 in. 35-70 - 90-100 1001/2 in. - 25-60 - 90-1003/8 in. 10-30 - 20-55 40-70No. 4 0-5 0-10 0-10 0-15No. 8 - 0-5 0-5 0-5

Table 1.2: ASTM C33 Grading Limits for Coarse Concrete Aggregates

Sieve Size % Passing3/4 in. 100No. 4 95-100No. 8 80-100No. 16 50-85No. 30 25-60No. 50 10-30No. 100 2-10

Table 1.3: ASTM C33 Grading Limits for Fine Concrete Aggregates



16 Moisture states are defined as

Oven-dry (OD): all moisture is removed from the aggregate.

Air-dry (AD): all moisture is removed from the surface, but internal pores are partially full.

Saturated-surface-dry (SSD): All pores are filled with water, but no film of water on thesurface.

Wet: All pores are completely filled with a film of water on the surface.

17 Based on the above, we can determine

Absorption capacity (AC): is the maximum amount of water the aggregate can absorb

AC =WSSD − WOD

WOD× 100% (1.1)

most normal -weight aggregates (fine and coarse) have an absorption capacity in the rangeof 1% to 2%.

Surface Moisture (SM): is the water in excess of the SSD state

SM =WWet − WSSD

WSSD× 100% (1.2)

18 The fineness modulus is a parameter which describe the grading curve and it can be usedto check the uniformity of the grading. It is usually computed for fine aggregates on the basisof

F.M. =∑

cumulative percent retained on standard sieves100

(1.3)

where the standard sieves used are No. 100, No. 50, No. 30, No. 16, No. 8, and No. 4, and3/8 in, 3/4 in, 11/2 in and larger.

19 The fineness modulus for fine aggregate should lie between 2.3 and 3.1 A small numberindicates a fine grading, whereas a large number indicates a coarse material.

20 Table 1.4 illustrates the determination of the fineness modulus.

21 Fineness modulus of fine aggregate is required for mix proportioning since sand gradationhas the largest effect on workability. A fine sand (low fineness modulus) has much higher pasterequirements for good workability.

22 The fineness modulus of coarse aggregate is not used for mix design purposes.

23 no-fines concrete has little cohesiveness in the fresh state and can not be compacted to avoid-free condition. Hence, it will have a low strength, high permeability. Its only advantage islow density, and high thermal insulation which can be used if structural requirements are nothigh.



Sieve Weight Amount Cumulative CumulativeSize Retained Retained Amount Amount

(g) (wt. %) Retained (%) Passing (%)No. 4 9 2 2 98No. 8 46 9 11 89No. 16 97 19 30 70No. 30 99 20 50 50No. 50 120 24 74 26No. 100 91 18 92 8

Sample Weight 500 g.∑

= 259Fineness modulus=259/100=2.59

Table 1.4: Example of Fineness Modulus Determination for Fine Aggregate

1.1.1.1.2 Preliminary Considerations

24 There are two fundamental aspects to mix design to keep in mind:

1. Water/Cement ratio: where the strength is inversely proportional to the water to cementratio, approximately expressed as:

f ′c =

A

B1.5w/c(1.4)

For f ′c in psi, A is usually taken as 14,000 and B depends on the type of cement, but may

be taken to be about 4. It should be noted that w/c controls not only the strength, butalso the porosity and hence the durability.

2. Aggregate Grading: In order to minimize the amount of cement paste, we must maximizethe volume of aggregates. This can be achieved through proper packing of the granularmaterial. The “ideal” grading curve (with minimum voids) is closely approximated bythe Fuller curve

Pt =(

d

D

)q

(1.5)

where Pt is the fraction of total solids finer than size d, and D is the maximum particlesize, q is generally taken as 1/2, hence the parabolic grading.

1.1.1.1.3 Mix procedure

25 Before starting the mix design process, the following material properties should be deter-mined:

1. Sieve analysis of both fine and coarse aggregates

2. Unit weight of the coarse aggregate

3. Bulk specific gravities



4. absorption capacities of the aggregates

1. Slump1 must be selected for the particular job to account for the anticipated methodof handling and placing concrete, Table 1.5 As a general rule, adopt the lowest possible

Type of Construction Max MinFoundation walls and footings 3 1Plain footings, caissons 3 1Beams and reinforced walls 4 1Building columns 4 1Pavement and slabs 3 1Mass concrete 3 1

Table 1.5: Recommended Slumps (inches) for Various Types of Construction

slump.

2. Maximum aggregate size: in general the largest possible size should be adopted.However, it should be noted that:

(a) For reinforced concrete, the maximum size may not exceed one-fifth of the mini-mum dimensions between the forms, or three-fourths of the minimum clear spacingbetween bars, or between steel and forms.

(b) For slabs on grade, the maximum size may not exceed one-third the slab depth.

In general maximum aggregate size is 3/4 in or 1 in.

3. Water and Air content Air content will affect workability (some time it is better toincrease air content rather than increasing w/c which will decrease strength). Air contentcan be increased through the addition of admixtures. Table 1.6 tabulates recommendedvalues of air content (obtained through such admixtures) for different conditions (forinstance under severe freezing/thawing air content should be high).

Recommended water requirements are given by Table 1.7.

Sizes of AggregatesExposure 3/8 in. 1/2 in. 3/4 in. 1 in. 11/2 in.

Mild 4.5 4.0 3.5 3.5 3.0Moderate 6.0 5.5 5.0 4.5 4.4Extreme 7.5 7.0 6.0 6.05 5.5

Table 1.6: Recommended Average Total Air Content as % of Different Nominal Maximum Sizesof Aggregates and Levels of Exposure

1The slump test (ASTM C143) is a measure of the shear resistance of concrete to flowing under its own weight.It is a good indicator of the concrete “workability”. A hollow mold in the form of a frustum of a cone is filledwith concrete in three layers of equal volume. Each layer is rodded 25 times. The mold is then lifted vertically,and the slump is measured by determining the difference between the height of the mold and the height of theconcrete over the original center of the base of the specimen.



Slump Sizes of Aggregatesin. 3/8 in. 1/2 in. 3/4 in. 1 in. 11/2 in.

Non-Air-Entrained Concrete1-2 350 335 315 300 2753-4 385 365 340 325 3006-7 410 385 360 340 315

Air-Entrained Concrete1-2 305 295 280 270 2503-4 340 325 305 295 2756-7 365 345 325 310 290

Table 1.7: Approximate Mixing Water Requirements, lb/yd3 of Concrete For Different Slumpsand Nominal Maximum Sizes of Aggregates

4. Water/cement ratio: this is governed by both strength and durability. Table 1.8provides some guidance in terms of strength.

28 days w/c Ratio by Weightf ′

c Non-air-entrained Air-entrained6,000 0.41 -5,000 0.48 0.404,000 0.57 0.483,000 0.68 0.592,000 0.82 0.74

Table 1.8: Relationship Between Water/Cement Ratio and Compressive Strength

For durability, if there is a severe exposure (freeze/thaw, exposure to sea-water, sulfates),then there are severe restrictions on the W/C ratio (usually to be kept just under 0.5)

5. Cement Content: Once the water content and the w/c ratio are determined, the amountof cement per unit volume of concrete is determined simply by dividing the estimatedwater requirement by the w/c ratio.

6. Coarse Aggregate Content: Volume of coarse aggregate required per cubic yard ofconcrete depends on its maximum size and the fineness modulus of the fine aggregate,Table 1.9. The oven dry (OD) volume of coarse aggregate in ft3 required per cubic yardis simply equal to the value from Table 1.9 multiplied by 27. This volume can then beconverted to an OD weight by multiplying it by the dry-rodded2 weight per cubic foot ofcoarse aggregate.

7. The fine aggregate content can be estimated by subtracting the volume of cement,water, air and coarse aggregate from the total volume. The weight of the fine aggregatecan then be obtained by multiplying this volume by the density of the fine aggregate.

2Dry Rodded Volume (DRV) is the normal volume of space a material occupies.



Agg. Size Sand Fineness Moduliin 2.40 2.60 2.80 3.003/8 0.50 0.48 0.46 0.441/2 0.59 0.57 0.55 0.533/4 0.66 0.64 0.62 0.601 0.71 0.69 0.67 0.65

11/2 0.76 0.74 0.72 0.70

Table 1.9: Volume of Dry-Rodded Coarse Aggregate per Unit Volume of Concrete for DifferentFineness Moduli of Sand

8. Adjustment for moisture in the aggregates: is necessary. If aggregates are airdry, they will absorb some water (thus effectively lowering the w/c), or if aggregates aretoo wet they will release water (increasing the w/c and the workability but reducing thestrength).

1.1.1.1.4 Mix Design Example

Concrete is required for an exterior column to be located above ground in an area wheresubstantial freezing and thawing may occur. The concrete is required to have an average 28-day compressive strength of 5,000 psi. For the conditions of placement, the slump should bebetween 1 and 2 in, the maximum aggregate size should not exceed 3/4 in. and the propertiesof the materials are as follows:

Cement: Type I specific gravity = 3.15

Coarse Aggregates: Bulk specific gravity (SSD) = 2.70; absorption capacity= 1.0%; Totalmoisture content = 2.5%; Dry-rodded unit weight = 100 lb/ft3

Fine Aggregates: Bulk specific gravity (SSD) = 2.65; absorption capacity = 1.3 %; Totalmoisture content=5.5%; fineness modulus = 2.70

The sieve analyses of both the coarse and fine aggregates fall within the specified limits. Withthis information, the mix design can proceed:

1. Choice of slump is consistent with Table 1.5.

2. Maximum aggregate size (3/4 in) is governed by reinforcing details.

3. Estimation of mixing water: Because water will be exposed to freeze and thaw, it mustbe air-entrained. From Table 1.6 the air content recommended for extreme exposure is6.0%, and from Table 1.7 the water requirement is 280 lb/yd3

4. From Table 1.8, the water to cement ratio estimate is 0.4

5. Cement content, based on steps 4 and 5 is 280/0.4=700 lb/yd3

6. Coarse aggregate content, interpolating from Table 1.9 for the fineness modulus ofthe fine aggregate of 2.70, the volume of dry-rodded coarse aggregate per unit volume ofconcrete is 0.63. Therefore, the coarse aggregate will occupy 0.63 × 27 = 17.01 ft3/yd3.



The OD weight of the coarse aggregate is 17.01 ft3/yd3, × 100 lbs/ft3=1,701 lb. The SSDweight is 1,701 × 1.01=1,718 lb.

7. Fine aggregate content Knowing the weights and specific gravities of the water, cement,and coarse aggregate, and knowing the air volume, we can calculate the volume per yd3

occupied by the different ingredients.

Water 280/62.4 = 4.49 ft3

Cement 700/(3.15)(62.4) = 3.56 ft3

Coarse Aggregate (SSD) 1,718/(2.70)(62.4) = 1.62 ft3

Air (0.06)(27) = 1.62 ft3

19.87 ft3

Hence, the fine aggregate must occupy a volume of 27.0 − 19.87 = 7.13 ft3. The requiredSSD weight of the fine aggregate is 7.13 ft3 (2.65)(62.4)lb/ft3 =1,179 lbs lb.

8. Adjustment for moisture in the aggregate. Since the aggregate will be neither SSD orOD in the field, it is necessary to adjust the aggregate weights for the amount of watercontained in the aggregate. Only surface water need be considered; absorbed water doesnot become part of the mix water. For the given moisture contents, the adjusted aggre-gate weights become:

Coarse aggregate (wet)=1,718(1.025-0.01) = 1,744 lb/yd3 of dry coarseFine aggregate (wet)=1,179(1.055-0.013) = 1,229 lb/yd3 of dry fine

Surface moisture contributed by the coarse aggregate is 2.5-1.0 = 1.5%; by the fine ag-gregate: 5.5-1.3 = 4.2%; Hence we need to decrease water to280-1,718(0.015)-1,179(0.042) = 205 lb/yd3.

Thus, the estimated batch weight per yd3 are

Water 205 lbCement 700 lbWet coarse aggregate 1,744 lbWet fine aggregate 1,229 lb

3,878 lb/yd3

3,87827 143.6 lb/ft3

1.1.1.2 Mechanical Properties

26 Contrarily to steel to modulus of elasticity of concrete depends on the strength and is givenby

E = 57, 000√

f ′c (1.6)

or

E = 33γ1.5√

f ′c (1.7)

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



27 Normal weight and lightweight concrete have γ equal to 150 and 90-120 lb/ft3 respectively.

28 Poisson’s ratio ν = 0.15.

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

30 Stress-strain curve depends on

1. Properties of aggregates

2. Properties of cement

3. Water/cement ratio

4. Strength

5. Age of concrete

6. Rate of loading, as rate↗, strength ↗

31 Non-linear part of stress-strain curve is caused by micro-cracking around the aggregates, Fig.1.2

~ 0.5 f’c

εu

f’c

Linear

Non-Linear

Figure 1.2: MicroCracks in Concrete under Compression

32 Irrespective of f ′c, maximum strain under compression is ≈ 0.003, Fig. 1.3

33 Full strength of concrete is achieved in about 28 days

f ′ct =

t

4.0 + .85tf ′

c,28 (1.8)

or

t (days) 1 2 4 7 10 15%f ′

c,28 20 35 54 70 80 90

34 Concrete always gain strength in time, but a decreasing rate



εu =

f’c

f’c

0.003 ε

σ

/ 2

Figure 1.3: Concrete Stress Strain Curve

35 The tensile strength of concrete f ′t is very difficult to measure experimentally. Accepted

values

f ′t ≈ 0.07 − 0.11f ′

c (1.9-a)≈ 3 − 5

√f ′

c (1.9-b)

36 Rather than the tensile strength, it is common to measure the modulus of rupture f ′r, Fig.

1.4

��

��

��

Figure 1.4: Modulus of Rupture Test

σσ

Figure 1.5: Split Cylinder (Brazilian) Test

f ′r ≈ 7.5

√f ′

c (1.10)



f’c

f’c

~ 20% increase in strength

f’

f’

t

t

σ

σ

σ

σ

11

2

2

1

2

Figure 1.6: Biaxial Strength of Concrete

37 Using split cylinder (or brazilian test), Fig. 1.5 f ′t ≈ 6−8

√f ′

c. For this test, a nearly uniformtensile stress

σ =2P

πdt(1.11)

where P is the applied compressive load at failure, d and t are diameter and thickness of thespecimen respectively.

38 In most cases, concrete is subjected to uniaxial stresses, but it is possible to have biaxial(shells, shear walls) or triaxial (beam/column connections) states of stress.

39 Biaxial strength curve is shown in Fig. 1.6

40 Concrete has also some time-dependent properties

Shrinkage: when exposed to air (dry), water tends to evaporate from the concrete surface, ⇒shrinkage. It depends on the w/c and relative humidity. εsh ≈ 0.0002−0.0007. Shrinkagecan cause cracking if the structure is restrained, and may cause large secondary stresses.

If a simply supported beam is fully restrained against longitudinal deformation, then

σsh = Eεsh (1.12-a)

= 57, 000√

3, 000(0.0002) = 624 psi >3, 000

10︸︷︷︸f ′

t

(1.12-b)

if the concrete is restrained, then cracking will occur3.

Creep: can be viewed as the “squeezing” out of water due to long term stresses (analogous toconsolidation in clay), Fig. 1.7.

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



ε

no load constant load

creepElastic recovery

Residual

Creep recovery

no load

Figure 1.7: Time Dependent Strains in Concrete

Creep coefficient, Table 1.10

Cu =εct

εci≈ 2 − 3 (1.13-a)

Ct =t0.6

10 + t0.6Cu (1.13-b)

f ′c 3,000 4,000 6,000 8,000

Cu 3.1 2.9 2.4 2.0

Table 1.10: Creep Coefficients

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

1.1.2 Reinforcing Steel

42 Steel is used as reinforcing bars in concrete, Table 1.11.

43 Bars have a deformation on their surface to increase the bond with concrete, and usuallyhave a yield stress of 60 ksi.

44 Maximum allowable fy is 80 ksi.

45 Stirrups, used as vertical reinforcement to resist shear, usually have a yield stress of only 40ksi

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

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



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 1.11: Properties of Reinforcing Bars

48 Welded wire fabric is often used to reinforce slabs and shells. It has both longitudinal andtransverse cold-drawn steel. They are designated by A×A−WB×B, such as 6×6−W1.4×1.4where spacing of the wire is 6 inch, and a cross section of 0.014 in2.

1.2 Design Philosophy, USD

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

ΦRn ≥ ΣαiQi (1.14)

where

Φ is a strength reduction factor, less than 1, and must account for the type of structuralelement, Table 1.12 (ACI 9.3.2)

Type of Member ΦAxial Tension 0.9Flexure 0.9Axial Compression, spiral reinforcement 0.75Axial Compression, other 0.70Shear and Torsion 0.85Bearing on concrete 0.70

Table 1.12: Strength Reduction Factors, Φ

Rn is the nominal resistance (or strength).


Draft1.3 Analysis vs Design 1–15

Ru = Rd = 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

ΦMn ≥ Mu (1.15-a)ΦVn ≥ Vu (1.15-b)ΦPn ≥ Pu (1.15-c)

50 Note that the subscript d and u are equivalent.

51 The various factored load combinations which must be considered (ACI: 9.2) are

1. 1.4D+1.7L

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

3. 0.9D+1.3W

4. 1.05D+1.275W

5. 0.9D+1.7H

6. 1.4D +1.7L+1.7H

7. 0.75(1.4D+1.4T+1.7L)

8. 1.4(D+T)

where D= dead; L= live; Lr= roof live; W= wind; E= earthquake; S= snow; T= temperature;H= soil. We must select the one with the largest limit state load.

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

1. Deflections

2. Crack width (for R/C)

3. Stability

1.3 Analysis vs Design

53 In R/C we always consider one of the following problems:

Analysis: Given a certain design, determine what is the maximum moment which can beapplied.

Design: Given an external moment to be resisted, determine cross sectional dimensions (b andh) as well as reinforcement (As). Note that in many cases the external dimensions of thebeam (b and h) are fixed by the architect.

54 We often consider the maximum moment along a member, and design accordingly.



1.4 Basic Relations and Assumptions

55 In developing a design/analysis method for reinforced concrete, the following basic relationswill be used:

1. Equilibrium: of forces and moment at the cross section. 1) ΣFx = 0 or Tension in thereinforcement = Compression in concrete; and 2) ΣM = 0 or external moment (that is theone obtained from the moment envelope) equal and opposite to the internal one (tensionin steel and compression of the concrete).

2. Material Stress Strain: We recall that all normal strength concrete have a failure strainεu = .003 in compression irrespective of f ′

c.

56 Basic assumptions used:

Compatibility of Displacements: Perfect bond between steel and concrete (no slip). Notethat those two materials do also have very close coefficients of thermal expansion undernormal temperature.

Plane section remain plane ⇒ strain is proportional to distance from neutral axis.

Neglect tensile strength in all cases.

1.5 ACI Code

Attached is an unauthorized copy of some of the most relevant ACI-318-89 design code provi-sions.

8.1.1 - In design of reinforced concrete structures, members shall be proportioned for ad-equate strength in accordance with provisions of this code, using load factors and strengthreduction factors Φ specified in Chapter 9.

8.3.1 - All members of frames or continuous construction shall be designed for the maximumeffects of factored loads as determined by the theory of elastic analysis, except as modifiedaccording to Section 8.4. Simplifying assumptions of Section 8.6 through 8.9 may be used.

8.5.1 - Modulus of elasticity Ec for concrete may be taken as W 1.5c 33

√f ′

c ( psi) for valuesof Wc between 90 and 155 lb per cu ft. For normal weight concrete, Ec may be taken as57, 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 insuch combinations as 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 (1.16)

9.2.2 - If resistance to structural effects of a specified wind load W are included in design,the following combinations of D, L, and W shall be investigated to determine the greatestrequired strength U

U = 0.75(1.4D + 1.7L + 1.7W ) (1.17)


Draft1.5 ACI Code 1–17

where load combinations shall include both full value and zero value of L to determine the moresevere condition, and

U = 0.9D + 1.3W (1.18)

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 nominalstrength calculated in accordance with requirements and assumptions of this code, multipliedby a strength reduction 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 toclear span ratios greater than 2/5 for continuous spans and 4/5 for simple spans, a non-lineardistribution of strain shall be considered. See Section 10.7.

10.2.3 - Maximum usable strain at extreme concrete compression fiber shall be assumedequal to 0.003.

10.2.4 - Stress in reinforcement below specified yield strength fy for grade of reinforcementused shall be taken as Es times steel strain. For strains greater than that corresponding to fy,stress in reinforcement shall be considered independent of strain and equal to fy.

10.2.5 - Tensile strength of concrete shall be neglected in flexural calculations of reinforcedconcrete, except when meeting requirements of Section 18.4.

10.2.6 - Relationship between concrete compressive stress distribution and concrete strainmay be assumed to be rectangular, trapezoidal, parabolic, or any other shape that results inprediction of strength in substantial agreement with results of comprehensive tests.

10.2.7 - Requirements of Section 10.2.5 may be considered satisfied by an equivalent rect-angular concrete stress distribution defined by the following:

10.2.7.1 - Concrete stress of 0.85f ′c shall be assumed uniformly distributed over an equiva-

lent compression zone bounded by edges of the cross section and a straight line located parallelto the neutral axis 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 measuredin a direction 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 foreach 1000 psi of strength in excess of 4,000 psi, but β1 shall not be taken less than 0.65.

10.3.2 - Balanced strain conditions exist at a cross section when tension reinforcementreaches the strain corresponding to its specified yield strength fy just as concrete in compressionreaches its assumed ultimate strain of 0.003.

10.3.3 - For flexural members, and for members subject to combined flexure and compres-sive axial load when the design axial load strength (ΦPn) is less than the smaller of (0.10f ′

cAg)or (ΦPb), the ratio of reinforcement p provided shall not exceed 0.75 of the ratio ρb that wouldproduce balanced strain conditions for the section under flexure without axial load. For mem-bers with compression reinforcement, the portion of ρb equalized by compression reinforcementneed not be reduced by the 0.75 factor.

10.3.4 - Compression reinforcement in conjunction with additional tension reinforcementmay be used to increase the strength of flexural members.



10.5.1 - At any section of a flexural member, except as provided in Sections 10.5.2 and10.5.3, where positive reinforcement is required by analysis, the ratio ρ provided shall not beless than that given by

ρmin =200fy

(1.19)


Draft

Chapter 2

FLEXURE

1 This is probably the longest chapter in the notes, we shall cover in great details flexuraldesign/analysis of R/C beams starting with uncracked section to failure conditions.

1. Uncracked elastic (uneconomical)

2. cracked elastic (service stage)

3. Ultimate (failure)

2.1 Uncracked Section

h d

b

A

ε

ε

c

s

s

Figure 2.1: Strain Diagram Uncracked Section

2 Assuming perfect bond between steel and concrete, we have εs = εc, Fig. 2.1

εs = εc ⇒ fs

Es=

fc

Ec⇒ fs =

Es

Ecfc ⇒ fs = nfc (2.1)

where n is the modular ratio n = EsEc

3 Tensile force in steel Ts = Asfs = Asnfc

4 Replace steel by an equivalent area of concrete, Fig. 2.2.

Draft2–2 FLEXURE

S(n-1)A

2S(n-1)A

2��

��

Figure 2.2: Transformed Section

5 Homogeneous section & under bending

fc =Mc

I⇒ fs = nfc (2.2)

6 Make sure that σ+max < f ′

t

Example 2-1: Uncracked Section

Given f ′c = 4,000 psi; f ′

t = 475 psi; fy = 60,000 psi; M = 45 ft-k = 540,000 in-lb; As = 2.35in2

Determine f+max, f−

max, and fs

s

2

t

bA = 2.35 in

25"23"

10"

y

y

Solution:

n =29, 000

57√

4, 000= 8 ⇒ (n − 1)As = (8 − 1)(2.35) = 16.45 in2 (2.3-a)

yb =(10)(25)(25

2 ) + (16.45)(2)(25)(10) + 16.45

(2.3-b)

yb = 11.8 in (2.3-c)yt = 25 − 11.8 = 13.2 in (2.3-d)

I =(10)(25)3

12+ (25)(10)(13.2 − 12.5)2 + (16.45)(23 − 13.2)2 (2.3-e)

= 14, 722 in2 (2.3-f)

fcc =Mc

I=

(540, 000) lb.in(13.2)in(14, 722) in4 = 484 psi (2.3-g)


Draft2.2 Section Cracked, Stresses Elastic 2–3

fct =Mc

I=

(540, 000) lb.in(25 − 13.2) in(14, 722) in4

= 433 psi < 475 psi√

(2.3-h)

fs = nMc

I= (8)

(540, 000)(23 − 13.2) in(14, 722)

= 2, 876 psi (2.3-i)

2.2 Section Cracked, Stresses Elastic

7 This is important not only as an acceptable alternative ACI design method, but also for thelater evaluation of crack width under service loads.

2.2.1 Basic Relations

8 If fct > fr, fcc <≈ .5f ′c and fs < fy we will assume that the crack goes all the way to the

N.A and we will use the transformed section, Fig. 2.3

S(n-1)A

2S(n-1)A

2

C

T

f

��

��

��

��

��

b

dkd

c

kd/3

(1-k/3)d=jd

Figure 2.3: Stress Diagram Cracked Elastic Section

9 To locate N.A, tension force = compressive force (by def. NA) (Note, for linear stress distri-bution and with ΣFx = 0;σ = by ⇒ ∫

bydA = 0, thus b∫

ydA = 0 and∫

ydA = yA = 0, bydefinition, gives the location of the neutral axis)

10 Note, N.A. location depends only on geometry & n(

EsEc

)11 Tensile and compressive forces are equal to C = bkd

2 fc & T = Asfs and neutral axis isdetermined by equating the moment of the tension area to the moment of the compressionarea

b(kd)(

kd

2

)= nAs(d − kd) 2nd degree equation (2.4-a)

M = Tjd = Asfsjd ⇒ fs =M

Asjd(2.4-b)

M = Cjd =bkd

2fcjd =

bd2

2kjfc ⇒ fc = M

12bd2kj

(2.4-c)

where j = (1 − k/3).


Draft2–4 FLEXURE

2.2.2 Working Stress Method

12 Referred to as Alternate Design Method (ACI Code Appendix A); Based on WorkingStress Design method.

13 Places a limit on stresses and uses service loads (ACI A.3).

fcc ≤ .45f ′c

fst ≤ 20 ksi for grade 40 or 50 steelfst ≤ 24 ksi for grade 60 steel

(2.5)

14 Location of neutral axis depends on whether we are analysing or designing a section.

Review: We seek to locate the N.A by taking the first moments:

ρ = Asbd

b(kd) (kd)2 = nAs(d − kd)

⇒ k =

√2ρn + (ρn)2 − ρn (2.6)

Design: Objective is to have fc & fs preset & determine As, Fig. 2.4, and we thus seek theoptimal value of k in such a way that concrete and steel reach their respective limitssimultaneously.

cf

f s

dkd

kd/3C

T

s

cε

ε

(1-k/3)d=jd

Figure 2.4: Desired Stress Distribution; WSD Method

εcεs

= kdd−kd

εc = fc

Ec

εs = fs

Es

fc

Ec

Esfs

= k1−k

n = EsEc

r = fs

fc

k = n

n+r (2.7)

15 Balanced design in terms of ρ: What is the value of ρ such that steel and concrete will bothreach their maximum allowable stress values simultaneously

C = bkd2 fc

T = Asfs

C = T

ρ = Asbd

fc

2 bkd = ρbfsbdk = n

n+r

}ρb = n

2r(n+r) (2.8)

16 Governing equations



Review Start by determining ρ,

• If ρ < ρb steel reaches max. allowable value before concrete, and

M = Asfsjd (2.9)

• If ρ > ρb concrete reaches max. allowable value before steel and

M = fcbkd

2jd (2.10)

or

M =12fcjkbd2 = Rbd2 (2.11)

where

k =√

2ρn + (ρn)2 − ρn

Design We define

Rdef=

12fckj (2.12)

where k = nn+r , solve for bd2 from

bd2 =M

R(2.13)

assume b and solve for d. Finally we can determine As from

As = ρbbd (2.14)

17 Summary

Review Designb, d, As

√M

√M? b, d, As?ρ = As

bd k = nn+r

j = 1 − k3

k =√

2ρn + (ρn)2 − ρn r = fs

fc

r = fs

fcR = 1

2fckj

ρb = n2r(n+r) ρb = n

2r(n+r)

ρ < ρb M = Asfsjd bd2 = MR

ρ > ρb M = 12fcbkd2j As = ρbbd or As = M

fsjd

Example 2-2: Cracked Elastic Section


Draft2–6 FLEXURE

Same problem as example 2.1 f ′c = 4,000 psi; f ′

t = 475 psi; fy = 60,000 psi; As = 2.35 in2

however, M is doubled to M = 90 k.ft (instead of 45). Determine concrete and steel stressesSolution:

Based on previous example, fct would be 866 psi À fr and the solution is thus no longervalid.

The neutral axis is obtained from

ρ =As

bd=

2.35(10)(23)

= 0.0102 (2.15-a)

ρn = (0.010)(8) = 0.08174 (2.15-b)k =

√2ρn + (ρn)2 − ρn (2.15-c)

=√

2(0.08174) + (0.08174)2 − (0.08174) = 0.33 (2.15-d)kd = (.33)(23) = 7.6 in (2.15-e)

jd =(

1 − 0.333

)(23) = 20.47 in (2.15-f)

fs =M

Asjd(2.15-g)

=(90)(1, 000)(12)(2.35)(20.47)

= 22, 400 psi (2.15-h)

fc =2M

bjkd2(2.15-i)

=(2)(90)(12, 000)(10) (20.47)︸︷︷︸

jd

(7.6)︸︷︷︸kd

= 1, 390 psi (2.15-j)

I =(10)(7.6)3

12+ (10)(7.6)

(7.62

)2

+ 8(2.35)(23 − 7.6)2 = 5, 922 in4 (2.15-k)

Uncracked Cracked Cracked/uncrackedM k.ft 45 90 2

N.A in 13.2 7.6fcc psi 485 1,390 (< .5f ′

c) 2.9I in4 14,710 5,910 .4 (δα1

I )fs psi 2,880 22,400 (≈ 7 )δ in 1 ≈ 4 4

Example 2-3: Working Stress Design Method; Analysis

Same problem as example 2.1 f ′c = 4,000 psi; f ′

t = 475 psi; fy = 60,000 psi; As = 2.35 in2.Determine Moment capacity.Solution:



ρ =As

bd=

2.35(10)(23)

= .0102 (2.16-a)

fs = 24 ksi (2.16-b)fc = (.45)(4, 000) = 1, 800 psi (2.16-c)k =

√2ρn + (ρn)2 − ρn =

√2(.0102)8 + (.0102)2 − (8)(.0102) = .331 (2.16-d)

j = 1 − k

3= .889 (2.16-e)

N.A. @ (.331)(23) = 7.61 in (2.16-f)

ρb =n

2r(n + r)=

8(2)(13.33)(8 + 13.33)

= .014 > ρ ⇒ Steel reaches elastic limit(2.16-g)

M = Asfsjd = (2.35)(24)(.889)(23) = 1, 154 k.in = 96 k.ft (2.16-h)

Note, had we used the alternate equation for moment (wrong) we would have overestimatedthe design moment:

M = =12fcbkd2j (2.17-a)

=12(1.8)(10)(0.33)(0.89)(23)2 = 1, 397 k.in > 1, 154 k.in (2.17-b)

If we define αc = fc/1, 800 and αs = fs/24, 000, then as the load increases both αc and αs

increase, but at different rates, one of them αs reaches 1 before the other.

Load

1 α αs c

Example 2-4: Working Stress Design Method; Design

Design a beam to carry LL = 1.9 k/ft, DL = 1.0 k/ft with f ′c = 4, 000 psi, fy = 60, 000 psi,

L = 32 ft.Solution:

fc = (.45)(4, 000) = 1, 800 psi (2.18-a)


Draft2–8 FLEXURE

fs = 24, 000 psi (2.18-b)

n =Es

Ec=

29, 00057√

4, 000= 8 (2.18-c)

r =fs

fc=

241.8

= 13.33 (2.18-d)

k =n

n + r=

88 + 13.33

= .375 (2.18-e)

j = 1 − d

3= 1 − .375

3= .875 (2.18-f)

ρb =n

2r(n + r)=

82(13.33)(8 + 13.33)

= .01405 (2.18-g)

R =12fckj =

12(1, 800)(.375)(.875) = 295 psi (2.18-h)

Estimate beam weight at .5 k/ft, thus

M = [(1.9) + (1.0 + .5)](32)2

8= 435 k.ft (2.19-a)

bd2 =M

R=

435 k.ft in2(12, 000) lb.in(295) lbs ft k

= 17, 700 in3 (2.19-b)

Take b = 18 in & d = 31.4 in ⇒ h = 36 inCheck beam weight (18)(36)

145 (.15) in2 ft2in2

kft3

= .675 k/ft√

As = (.01405)(18)(31.4) = 7.94 in2 ⇒ use 8# 9 bars in 2 layers ⇒ As = 8.00 in2

2.3 Cracked Section, Ultimate Strength Design Method

2.3.1 Whitney Stress Block

b

dh

A

a= cc

ε

C= f’abC= f’cb γ c

c f’γ

α

Actual

cβ

ε

σ

c

s

1β

sfsf

c

βa/2 = c

Figure 2.5: Cracked Section, Limit State

Figure


Draft2.3 Cracked Section, Ultimate Strength Design Method 2–9

18 At failure we have, linear cross strain distribution (ACI 10.2.2) (except for deep beams),non-linear stress strain curve for the concrete, thus a non-linear stress distribution.

19 Two options:

1. Analytical expression of σ ⇒ exact integration

2. Replace exact stress diagram with a simpler and equivalent one, (ACI 10.2.6)

Second approach adopted by most codes.

20 For the equivalent stress distribution, all we need to know is C & its location, thus α and β.We adopt a rectangular stress, with depth a = β1c, and stress equal to γf ′

c (ACI 10.2.7.1)

C = αf ′cbc = γf ′

cab (2.20-a)

α =fav

f ′c

(2.20-b)

a = β1c (2.20-c)

Thusγ =

α

β1(2.21)

But the location of the resultant forces must be the same, hence

β1 = 2β (2.22)

21 From Experiments

f ′c ( psi) <4,000 5,000 6,000 7,000 8,000

α .72 .68 .64 .60 .56β .425 .400 .375 .350 .325β1 = 2β .85 .80 .75 .70 .65γ = α/β1 0.85 0.85 0.85 0.86 0.86

22 Thus we have, (ACI-318 10.2.7.3):

β1 = .85 if f ′c ≤ 4, 000

= .85 − (.05)(f ′c − 4, 000) 1

1,000 if 4, 000 < f ′c < 8, 000

(2.23)

23 Failure can occur by either

yielding of steel: εs = εy; Progressive

crushing of concrete: εc = .003; Sudden; (ACI 10.3.2).


Draft2–10 FLEXURE

C=0.85f’ abc

h d

b

Aε

s

s

0.85 f’c

c

d

T

ε =0.003u

a= cβ1

Figure 2.6: Whitney Stress Block

2.3.2 Balanced Design

Tension Failure:fs = fy

Asfs = .85f ′cab = .85f ′

cbβ1c

ρ = Asbd

c = ρfy

.85f ′cβ1

d (2.24)

Compression Failure:

εc = .003 (2.25-a)

εs =fs

Es(2.25-b)

c

d=

.003.003 + εs

⇒ c = .003fsEs

+.003d (2.25-c)

Balanced Design:

24 Balanced design occurs if we have simultaneous yielding of the steel and crushing of theconcrete. Hence, we simply equate the previous two equations

ρfy

.85f ′cβ1

d = .003fsEs

+.003d

ρ = ρb

}ρbf2d

.85f ′cβ1

= .003fs

E−s+.003

d

Es = 29, 000 ksi

}ρb = .85β1

f ′c

fy

87,00087,000+fy

(ACI 8.4.3)(2.26)

25 To ensure failure by yielding,

ρ < .75ρb (2.27)

26 ACI strength requirements

U = 1.4D + 1.7L (ACI 9.2.1)U = 0.75(1.4D + 1.7L + 1.7W ) (ACI 9.2.2)

Md = Mu = φMn (ACI 9.1.1)φ = .90 (ACI 9.3.2.2)

(2.28)


Draft2.3 Cracked Section, Ultimate Strength Design Method 2–11

27 Also we need to specify a minimum reinforcement ratio

ρmin ≥ 200fy

(ACI 10.5.1) (2.29)

to account for temperature & shrinkage

28 Note, that ρ need not be as high as 0.75ρb. If steel is relatively expensive, or deflection is ofconcern, can use lower ρ.

29 As a rule of thumb, if ρ < 0.5ρb, there is no need to check for deflection.

2.3.3 Review

30 Given, b, d, As, f′c, fy, determine the moment capacity M .

ρact = Asbd

ρb = (.85)β1f ′

cfy

8787+fy

(2.30)

• ρact < ρb: Failure by yielding and

a = Asfy

.85f ′cb ΣFx = 0

Md = φAsfy(d − a2 ) ΣM = 0

(2.31)

• ρact > ρb is not allowed by code, in this case we have an extra unknown fs.

31 We now have one more unknown fs, and we will need an additional equation (from straindiagram).

c = Asfs

.85f ′cbβ1

ΣFx = 0cd = .003

.003+εsFrom strain diagram

Md = φAsfs(d − β1c2 ) ΣM = 0

(2.32)

We can solve by iteration, or substitution and solution of a quadratic equation.

2.3.4 Design

32 We consider two cases:

I b d and As, unknown; Md known; Since design failure is triggered by fs = fy

ΣFx = 0 a = Asfy

0.85f ′cb

ρ = Asbd

}a = ρfy

0.85f ′c

Md = Asfy

(d − a

2

)}

Md = Φ ρfy

(1 − .59ρ

fy

f ′c

)︸︷︷︸

R

bd2(2.33-a)

where ρ is specified by the designer; or

R = ρfy

(1 − .59ρ

fy

f ′c

)(2.34)


Draft2–12 FLEXURE

which does not depend on unknown quantities. Then solve for bd2:

bd2 =Md

ΦR(2.35)

Solve for b and d (this will require either an assumption on one of the two, or on theirratio).

As = ρbd

II b & d known & Md known ⇒ there is no assurance that we can have a design with ρb

If the section is too small, then it will require too much steel resulting in an over-reinforcedsection.

Iterative approach

(a) Since we do not know if the steel will be yielding or not, use fs.

(b) Assume an initial value for a (a good start is a = d5)

(c) Assume initially that fs = fy

(d) Check equilibrium of moments (ΣM = 0)

As =Md

Φfs

(d − a

2

) (2.36)

(e) Check equilibrium of forces in the x direction (ΣFx = 0)

a =Asfs

.85f ′cb

(2.37)

(f) Check assumption of fs by either comparing ρ with ρb, or from the strain diagram

εs

d − c=

.003c

⇒ fs = Esd − c

c.003 < fy (2.38)

where c = aβ1

.

(g) Iterate until convergence is reached.

2.4 Practical Design Considerations

2.4.1 Minimum Depth

33 ACI 9.5.2.1 stipulates that the minimum thickness of beams should be

Simply One end Both ends Cantileversupported continuous continuous

Solid Oneway slab L/20 L/24 L/28 L/10Beams orribbed One way slab L/16 L/18.5 L/21 L/8

where L is in inches, and members are not supporting partitions.

34 Smaller values can be taken if deflections are computed.


Draft2.4 Practical Design Considerations 2–13

2.4.2 Beam Sizes, Bar Spacing, Concrete Cover

35 Beam sizes should be dimensioned as

1. Use whole inches for overall dimensions, except for slabs use 12 inch increment.

2. Ideally, the overall depth to width ratio should be between 1.5 to 2.0 (most economical).

3. For T beams, flange thickness should be about 20% of overall depth.

36 Reinforcing bars

1. Minimum spacing between bars, and minimum covers are needed to

(a) Prevent Honeycombing of concrete (air pockets)(b) Concrete (usually up to 3/4 in MSA) must pass through the reinforcement(c) Protect reinforcement against corrosion and fire

2. Use at least 2 bars for flexural reinforcement

3. Use bars #11 or smaller for beams.

4. Use no more than two bar sizes and no more than 2 standard sizes apart (i.e #7 and #9acceptable; #7 and #8 or #7 and #10 not).

5. Use no more than 5 or 6 bars in one layer.

6. Place longest bars in the layer nearest to face of beam.

7. Clear distance between parallel bars not less that db (to avoid splitting cracks) nor 1 in.(to allow concrete to pass through).

8. Clear distance between longitudinal bars in columns not less that 1.5db or 1.5 in.

9. Minimum cover of 1.5 in.

10. Summaries in Fig. 2.7 and Table 2.1, 2.2.

2.4.3 Design Aids

37 Basic equations developed in this section can be easily graphed.

Review Given b d and known steel ratio ρ and material strength, φMn can be readily obtainedfrom φMn = φRbd2

Design in this case

1. Set Md = φRbd2

2. From tabulated values, select ρmax and ρmin often 0.5ρb is a good economical choice.3. Select R from tabulated values of R in terms of fy, f ′

c and ρ. Solve for bd2.4. Select b and d to meet requirements. Usually depth is about 2 to 3 times the width.5. Using tabulated values select the size and number of bars giving preference to larger

bar sizes to reduce placement cost (careful about crack width!).6. Check from tables that the selected beam width will provide room for the bars chosen

with adequate cover and spacing.


Draft2–14 FLEXURE

Bar Nominal Number of BarsSize Diam. 1 2 3 4 5 6 7 8 9 10#3 0.375 0.11 0.22 0.33 0.44 0.55 0.66 0.77 0.88 0.99 1.10#4 0.500 0.20 0.40 0.60 0.80 1.00 1.20 1.40 1.60 1.80 2.00#5 0.625 0.31 0.62 0.93 1.24 1.55 1.86 2.17 2.48 2.79 3.10#6 0.750 0.44 0.88 1.32 1.76 2.20 2.64 3.08 3.52 3.96 4.40#7 0.875 0.60 1.20 1.80 2.40 3.00 3.60 4.20 4.80 5.40 6.00#8 1.000 0.79 1.58 2.37 3.16 3.95 4.74 5.53 6.32 7.11 7.90#9 1.128 1.00 2.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00

#10 1.270 1.27 2.54 3.81 5.08 6.35 7.62 8.89 10.16 11.43 12.70#11 1.410 1.56 3.12 4.68 6.24 7.80 9.36 10.92 12.48 14.04 15.60#14 1.693 2.25 4.50 6.75 9.00 11.25 13.50 15.75 18.00 20.25 22.50#18 2.257 4.00 8.00 12.00 16.00 20.00 24.00 28.00 32.00 36.00 40.00

Table 2.1: Total areas for various numbers of reinforcing bars (inch2)

Bar Number of bars in single layer of reinf.Size 2 3 4 5 6 7 8#4 6.8 8.3 9.8 11.3 12.8 14.3 15.8#5 6.9 8.5 10.2 11.8 13.4 15.1 16.7#6 7.0 8.8 10.5 12.3 14.0 15.8 17.5#7 7.2 9.1 11.0 12.8 14.7 16.6 18.5#8 7.3 9.3 11.3 13.3 15.3 17.3 19.3#9 7.6 9.9 12.1 14.4 16.6 18.9 21.2

#10 7.8 10.3 12.9 15.4 18.0 20.5 23.0#11 8.1 10.9 13.7 16.6 19.4 22.2 25.0#14 8.9 12.3 15.7 19.1 22.5 25.9 29.3#18 10.6 15.1 19.6 24.1 28.6 33.2 37.7

Table 2.2: Minimum Width (inches) according to ACI Code


Draft2.5 USD Examples 2–15

Figure 2.7: Bar Spacing

2.5 USD Examples

Example 2-5: Ultimate Strength; Review

Determine the ultimate moment capacity of example 2.1 f ′c = 4,000 psi; f ′

t = 475 psi; fy =60,000 psi; As = 2.35 in2

s

2

t

bA = 2.35 in

25"23"

10"

y

y

Solution:

ρact =As

bd=

2.35(10)(23)

= .0102 (2.39-a)

ρb = .85β1f ′

c

fy

8787 + fy

= (.85)(.85)460

8787 + 60

= .0285 > ρact√

(2.39-b)

a =Asfy

.85f ′cb

=(2.35)(60)

(.85)(4)(10)= 4.15 in (2.39-c)

Mn = Asfy

(d − a

2

)= (2.35)(60)

(23 − 4.15

2

)= 2, 950 k.in (2.39-d)

Md = φMn = 0.9(2, 950) = 2, 660 k.in (2.39-e)

Note:


Draft2–16 FLEXURE

1. From equilibrium, ΣFx = 0 ⇒ c = Asfy

.85β1bf ′c

= (2.35)(60)(.85)(.85)(4)(10) = 4.87 in

2. Comparing with previous analysis

uncracked cracked ultimatec 13.2 7.61 4.87M 45 90 245

1.7 = 144

3. Alternative solution:

Mn = ρactfybd2(1 − .59ρact

fy

f ′c

) (2.40-a)

= Asfyd(1 − 59ρactfy

f ′c

) (2.40-b)

= (2.35)(60)(23)[1 − (.59)604

(.0102)] = 2, 950 k.in = 245 k.ft (2.40-c)

Md = φMn = (.9)(2, 950) = 2, 660 k.in (2.40-d)

Example 2-6: Ultimate Strength; Design I

Design a R/C beam with L = 15 ft; DL = 1.27 k/ft; LL = 2.44 k/ft; f ′c = 3,000 psi; fy =

40 ksi; Neglect beam own’s weight; Select ρ = 0.75ρb

Solution:

wu = 1.4(1.27) + 1.7(2.44) = 5.92 k/ft Factored load (2.41-a)

Md =wuL2

8=

(5.92)(15)2

8= 166.5 k.ft(12) in/ft = 1, 998 k.in (2.41-b)

ρ = 0.75ρb = (0.75)(0.85)β1f ′

c

fy

8787 + fy

(2.41-c)

= (0.75)(.85)2340

8787 + 40

= .0278 (2.41-d)

R = ρfy

(1 − .59ρ

fy

f ′c

)(2.41-e)

= (.0278)(40)(

1 − (0.59)(.0278)403

)= 0.869 psi (2.41-f)

bd2 =Md

φR=

1, 998(0.9)(0.869)

= 2, 555 in3 (2.41-g)

Take b = 10 in, d = 16 in ⇒ As = (.0278)(10)(16) = 4.45 in2 ⇒ use 3 # 11


Draft2.5 USD Examples 2–17

Example 2-7: Ultimate Strength; Design II

Design a R/C beam for b = 11.5 in; d = 20 in; f ′c = 3 ksi; fy = 40 ksi; Md = 1, 600 k.in

Solution:

Assume a = d5 = 20

5 = 4 in

As =Md

φfy(d − a2 )

=(1, 600)

(.9)(40)(20 − 42)

= 2.47 in2 (2.42)

check assumption,

a =Asfy

(.85)f ′cb

=(2.47)(40)

(.85)(3)(11.5)= 3.38 in (2.43)

Thus take a = 3.3 in.

As =(1, 600)

(.9)(40)(20 − 3.32 )

= 2.42 in2 (2.44-a)

⇒ a =(2.42)(40)

(.85)(3)(11.5)= 3.3 in

√(2.44-b)

ρact =2.42

(11.5)(20)= .011 (2.44-c)

ρb = (.85)(.85)340

8787 + 40

= .037 (2.44-d)

ρmax = .75ρb = .0278 > ρact√

(2.44-e)

Example 2-8: Exact Analysis

As an Engineer questioning the validity of the ACI equation for the ultimate flexural capacityof R/C beams, you determined experimentally the following stress strain curve for concrete:

σ =2 f ′

cεmax

ε

1 +(

εεmax

)2 (2.45)

where f ′c corresponds to εmax.

1. Determine the exact balanced steel ratio for a R/C beam with b = 10”, d = 23”, f ′c =

4, 000 psi, fy = 60 ksi, εmax = 0.003.

(a) Determine the equation for the exact stress distribution on the section.

(b) Determine the total compressive force C, and its location, in terms of the locationof the neutral axis c.


Draft2–18 FLEXURE

(c) Apply equilibrium

2. Using the ACI equations, determine the:

(a) Ultimate moment capacity.

(b) Balanced steel ratio.

3. For the two approaches, compare:

(a) Balanced steel area.

(b) Location of the neutral axis.

(c) Centroid of resultant compressive force.

(d) Ultimate moment capacity.

Solution:

1. Stress-Strain:

σ =24,000

.003 ε

1 +(

ε.003

)2 =2.667 × 106ε

1 + 1.11 × 105ε2(2.46)

2. Assume crushing at failure, hence strain distribution will be given by

ε =0.003

cy (2.47)

3. Combine those two equation:

σ =8, 000y

c

1 +(y

c

)2 (2.48)

4. The total compressive force is given by

F =∫ c

0dF = b

∫ c

0σdy = b

∫ c

0

8, 000yc

1 +(y

c

)2 dy = b8, 000

c

∫ c

0

y

1 +(

1c

)2 dy (2.49)

= 8, 000b

c

1

2(

1c

)2 ln[1 +

(y

c

)2]∣∣∣∣∣

c

0

= 8, 000b

c

c2

2ln

[1 +

(y

c

)2]∣∣∣∣c

0

(2.50)

= 4, 000bc ln(2) = 2, 773bc (2.51)

5. Equilibrium requires that T = C

2, 773bc = Asfy (2.52)

From the strain diagram:

.003c

=εy + .003

d⇒ c =

(.003)dεy + .003

(2.53)

c =(.003)(23)60

29,000 + .003= 13.6 in (2.54)


Draft2.6 T Beams, (ACI 8.10) 2–19

6. Combining Eq. 2.52 with Eq. 2.54

As =(2, 773)(10)(13.6)

60, 000= 6.28 in2 (2.55)

7. To determine the moment, we must first determine the centroid of the compressive forcemeasured from the neutral axis

y =

∫ydA

A=

b

∫yσdy

2, 773bc=

b

∫ c

0

8, 000y2

c

1 + (y/c)2

2, 773bcdy =

8, 000b

2, 773bc2

∫ c

0

y2

1 +(

1c

)2y2

dy (2.56)

=2.885

c2

∫ c

0

y2

1 +(

1c

)2y2

dy =2.885

(13.61)2

[y(1c

)2 − 1(1c

)2

∫ c

0

dy

1 +(

1c

)2y2

](2.57)

= .01557

yc2 − c2

1√

1c2

tan−1 y

√1c2

∣∣∣∣∣∣c

0

(2.58)

= .01557[c3 − c3 tan−1(1)

]= (.01557)(13.61)3(1 − tan−1(1)) = 8.43 in (2.59)

8. Next we solve for the moment

M = Asfy(d − c + y) = (6.28)(60)(23 − 13.61 − 8.43) = 6, 713 k.in (2.60)

9. Using the ACI Code

ρb = .85β1f ′

c

fy

8787 + 60

= (.85)2460

87147

= .0285 (2.61)

As = ρbbd = (.0285)(10)(23) = 6.55 in2 (2.62)

a =Asfy

.85f ′cb=

(6.55)(60)(.85)(4)(10)

= 11.57 in (2.63)

M = Asfy

(d − a

2

)= (6.55)(60)

(23 − 11.57

2

)= 6, 765 k.in (2.64)

c =a

β1=

11.57.85

= 13.61 (2.65)

10. We summarize

Exact ACIAs 6.28 6.55

c 13.6 13.6y′ 5.18 5.78M 6,713 6,765


Draft2–20 FLEXURE

��

��

fh

b

wb

eb

Figure 2.8: T Beams

2.6 T Beams, (ACI 8.10)

38 Equivalent width for uniform stress, Fig. 2.8 must satisfy the following requirements (ACI8.10.2):

1. 12(b − bw) ≤ 8hf

2. b < 4bw for isolated T beams only

3. hf > bw2

4. b < L4

39 Two possibilities:

1. Neutral axis within the flanges (c < hf ) ⇒ rectangular section of width b, Fig. 2.9.

2. Neutral axis in the web (c > hf ) ⇒ T beam.

��

h d

As

b

fh

Figure 2.9: T Beam as Rectangular Section

40 For T beams, we have a large concrete area, start by assuming that failure will occur by steelyielding, Fig. 2.10.

41 The approach consists in decomposing As into 2 components Fig. 2.11.



As

s

0.85 f’cu

s y

f

w

b

h

dh

b

ε

ε =0.003

T=A f

d

c C=0.85f’ ac

β1a= c

��

��

Figure 2.10: T Beam Strain and Stress Diagram

1. Asf → resists compression force in (b − bw)hf

2. (As − Asf ) → resists compression force in bwc

2.6.1 Review

42 Given, b, d, hf , As, f′c, fy, determine the moment capacity M , Fig. 2.11.

b c(b−b )hbw

w w

fh

b

f

��

��

��

��

s A sfA −

= +

c

s

��

��

A sfA

Figure 2.11: Decomposition of Steel Reinforcement for T Beams

43 The moment is obtained from

Flanges:Asf = .85f ′

c(b−bw)hf

fyΣF = 0

Mn1 = Asffy(d − hf

2 ) ΣM = 0(2.66)

Web:

a =(As − Asf )fy

.85f ′cbw

ΣF = 0 (2.67-a)

Mn2 = (As − Asf )fy(d − a

2) ΣM = 0 (2.67-b)

Total moment:Mn = Mn1 + Mn2 (2.68)


Draft2–22 FLEXURE

2.6.2 Design, (balanced)

44 Let us derive an expression for ρb and use it for design

c

d=

εu

εu + εyStrain Compatibility (2.69-a)

Asfy = .85f ′cβ1cbw + .85f ′

c(b − bw)hf︸︷︷︸Asf fy

ΣF = 0 (2.69-b)

thus,

Asfy = .85f ′cβ1cbw + Asffy

ρwdef= As

bwd

ρfdef= Asf

bwd

ρw =

ρb︷︸︸︷.85

f ′c

fyβ1

εu

εu + εy+ρf (2.70)

Hence,

ρwb = ρb + ρf (2.71)ρw,max = .75(ρb + ρf ) (2.72)

Example 2-9: T Beam; Moment Capacity I

For the following beam: As= 8 # 11 ( 12.48 in2); f ′c=3,000 psi; fy = 50,000 psi. Determine

Mn

C=0.85f’ abc

��

��

εs

0.85 f’c

c

d

ε =0.003u

a= cβ1

T=A fs y

14"

30"

7"

36"

Solution:

1. Check requirements for isolated T sections

(a) bw = 30 in should not exceed 4bw = 4(14) = 56 in√

(b) hf ≥ bu2 ⇒ 7 ≥ 14

2

√

2. Assume Rectangular section

a =Asfy

.85f ′cb

=(12.48)(50)

(0.85)(3)(30)= 8.16 in > hf (2.73)



3. For a T section

Asf =.85f ′

chf (b − bw)fy

(2.74-a)

=(.85)(3)(7)(30 − 14)

50= 5.71 in2 (2.74-b)

ρf =Asf

bwd=

5.71(14)(36)

= .0113 (2.74-c)

Asw = As − Asf = 12.48 − 5.71 = 6.77 in2 (2.74-d)

ρw =Asw

bwd=

12.48(14)(36)

= .025 (2.74-e)

ρb = .85β1f ′

c

fy

8787 + fy

(2.74-f)

= (.85)(.85)350

8787 + 50

= .0275 (2.74-g)

4. Maximum permissible ratio

ρmax = .75(ρb + ρf ) (2.75-a)= .75(.0275 + .0113) = .029 > ρw

√(2.75-b)

5. The design moment is then obtained from

Mn1 = (5.71)(50)(

36 − 72

)= 9, 280 k.in (2.76-a)

a =(As − Asf )fy

.85f ′cbw

(2.76-b)

=(6.77)(50)

(.85)(3)(14)= 9.48 in (2.76-c)

Mn2 = (6.77)(50)(36 − 9.482

) = 10, 580 k.in (2.76-d)

Md = (.9)(9, 280 + 10, 580) = 17, 890 k.in → 17, 900 k.in (2.76-e)

Example 2-10: T Beam; Moment Capacity II

Determine the moment capacity of the following section, assume flange dimensions to satisfyACI requirements; As = 6#10 = 7.59 in2; f ′

c = 3 ksi; fy=60 ksi.


Draft2–24 FLEXURE

C=0.85f’ abc

��

��

εs

0.85 f’c

c

d

ε =0.003u

a= cβ1

T=A fs y

10"

28"

6"

26"

Solution:

Assume rectangular beam

ρ =7.59

(28)(26)= .0104 (2.77-a)

ρb = (.85)(.85)(

360

) (87

87 + 60

)= .0214 > ρ ⇒ fs = fy (2.77-b)

a =(As − Asf )fy

.85f ′cbw

(2.77-c)

=(7.59)(60)

(.85)(3)(28)= 6.37 in > 6 in ⇒ T beam (2.77-d)

Asf =(.85)(3)(18)(6)

60= 4.59 in2 (2.77-e)

Asw = 7.59 − 4.59 = 3.00 in2 (2.77-f)

ρw =7.59

(26)(10)= .0292 (2.77-g)

ρf =4.59

(26)(10)= .0177 (2.77-h)

ρmax = .75(.0214 + .0177) = .0294 > .0292 ⇒ Ductile failure (2.77-i)Mn1 = (4.59)(60)(26 − 3) = 6, 330 k.in (2.77-j)

As − Asf = 7.59 − 4.59 = 3. in2 (2.77-k)

a =(3)(60)

(.85)(3)(10)= 7.07 in (2.77-l)

Mn2 = (3.00)(60)(26 − 7.072

) = 4, 050 k.in (2.77-m)

Md = (.9)(6, 330 + 4, 050) = 9, 350 k.in (2.77-n)

Example 2-11: T Beam; Design

given L = 24 ft; fy= 60 ksi; f ′c = 3 ksi; Md = 6, 400 k.in; Design a R/C T beam.



3"

11"

20"

47"

Solution:

1. Determine effective flange width:

12(b − bw) ≤ 8hf

16hf + bw = (16)(3) + 11 = 59 inL4 = 24

4 12 = 72 inCenter Line spacing = 47 in

b = 47 in (2.78-a)

2. Assume a = 3 in

As =Md

φfy(d − a2 )

=6, 400

(0.9)(60)(20 − 32)

= 6.40 in2 (2.79-a)

a =Asfy

(.85)f ′cb

=(6.4)(60)

(.85)(3)(47)= 3.20 in > hf (2.79-b)

3. Thus a T beam analysis is required.

Asf =.85f ′

c(b − bw)hf

fy=

(.85)(3)(47 − 11)(3)60

= 4.58 in2 (2.80-a)

Md1 = φAsffy(d − hf

2) = (.90)(4.58)(60)(20 − 3

2) = 4, 570 k.in (2.80-b)

Md2 = Md − Md1 = 6, 400 − 4, 570 = 1, 830 k.in (2.80-c)(2.80-d)

4. Now, this is similar to the design of a rectangular section. Assume a = d5 = 20

5 = 4. in

As − Asf =1, 830

(.90)(60)(20 − 4

2

) = 1.88 in2 (2.81)

5. check

a =(1.88)(60)

(.85)(3)(11)= 4.02 in ≈ 4.00 (2.82-a)

As = 4.58 + 1.88 = 6.46 in2 (2.82-b)

ρw =6.46

(11)(20)= .0294 (2.82-c)


Draft2–26 FLEXURE

ρf =4.58

(11)(20)= .0208 (2.82-d)

ρb = (.85)(.85)(

360

) (87

87 + 60

)= .0214 (2.82-e)

ρmax = (.0214 + .0208) = .042 > ρw√

(2.82-f)

6. Note that 6.46 in2 (T beam) is close to As = 6.40 in2 if rectangular section was assumed.

2.7 Doubly Reinforced Rectangular Beams

45 Negative steel reinforcement is needed to

1. Increase internal moment resistance capacity (not very efficient)

2. Support stirrups

3. Reverse moments (moving load)

4. Provide ductility (earthquake)

5. Reduce creep (long term deflections)

46 Approach will again be based on a strain compatibility analysis & equilibrium equation, Fig.2.12.

A s

a= cβ1

A fs s

a= cβ1

εs

sε ’

h d

b

ε =0.003u

sA’ d’

=

0.85 f’c

(A − A’ )fs s s

c

d

0.85 f’cA’ f’ss

+d−d’

A fs s

A’ f’s s

Figure 2.12: Doubly Reinforced Beams; Strain and Stress Diagrams

47 If ρ ≤ ρmax = .75ρb we can disregard compression steel

48 As for T beams, we decompose the tension steel into two components

1. A′s to resist the force in the top steel (assuming both yield)

2. As − A′s to resist compression in the concrete.

and we define

ρ′ =A′

s

bd(2.83)


Draft2.7 Doubly Reinforced Rectangular Beams 2–27

2.7.1 Tests for fs and f ′s

49 Different possibilities: Fig. 2.13

sA’ yield?sA’ yield?

sA yield?

f = fs y

Yes No

f = fs

s

sf’ = fs

s

sf’ < f

f < f f < f

f’ < f

Yes YesNo No

I II IVIII

Not Allowed by ACI

f’ = fs

yyy

y y y y

Figure 2.13: Different Possibilities for Doubly Reinforced Concrete Beams

Test 1 fs = fy?

Assuming εs = εy, and f ′s 6= fy, we have from the strain diagram, Fig. 2.14

u

ss

s

’

h

b

A’d’

ε =0.003

d

yε = ε

ε

As

Figure 2.14: Strain Diagram, Doubly Reinforced Beam; is As Yielding?

c =εu

εu + εyd (2.84-a)

ε′s = εu − d′

d(εu + εy) (2.84-b)

f ′s = Esε

′s (2.84-c)

From equilibrium:ρbdfy = ρ′bdf ′

s + .85f ′cβ1bc (2.85)

Combining:

ρb = ρ1 = ρ′f ′

s

fy+ .85

f ′c

fyβ1

εu

εu + εy︸︷︷︸ρb

(2.86)


Draft2–28 FLEXURE

thus

ρb = ρ1 = ρ′f ′

s

fy+ ρb (2.87)

ρmax = 0.75ρb + ρ′f ′

s

fy(2.88)

Note that 0.75 premultiplies only one term as in the other failure is ipso facto by yielding.We also note the similarity with ρmax of T Beams (where 0.75 premultiplied both terms).

Test 2 fs = fy is f ′s = fy?

We set ε′s = εy, and from the strain diagram

As

u

ss

s

’

y

h

b

A’d’

ε =0.003

ε > ε

ε = ε

d

y

Figure 2.15: Strain Diagram, Doubly Reinforced Beam; is A′s Yielding?

c =εu

εu − εyd′ (2.89)

from equilibriumρbdfy = ρ′bdfy + .85f ′

cβ1cb (2.90)

combining

ρmin ≡ ρ2 = ρ′ + .85β1f ′

c

fy

d′

d

8787 − fy

(2.91)

which corresponds to the minimum amount of steel to ensure yielding of compression steelat failure. Thus, if ρ < ρmin then f ′

s < fy. Note that some times ρmin can be larger thanρ.

Test 3 fs < fy, is f ′s = fy (not allowed by ACI)?

From strain diagram:

c =εu

εu − εyd′ (2.92-a)

εs =d − c

c − d′εy (2.92-b)



From equilibrium

ρbdfs = ρ′bdfy + .85f ′cβ1bc (2.93-a)

combining

ρ = ρ3 =c − d′

d − c

[ρ′ + .85β1

f ′c

fy

c

d

](2.94)

Summary of the tests are shown in Fig. 2.16

IVIII III

ρρ ρ ρ

Test 2 Test 1

s

Test 3

f’ = fs

f = fs

f’ = fy sy y ys y yf’ < fs

f’ < f

min

f < f

Figure 2.16: Summary of Conditions for top and Bottom Steel Yielding

2.7.2 Moment Equations

Case I fs = fy and f ′s = fy, usually occur if we have small bottom and top reinforcement

ratios.

Asfy = A′sfy + .85f ′

cab (2.95-a)

a =(As − A′

s)fy

.85f ′cb

(2.95-b)

M In = .85f ′

cab(d − a

2

)+ A′

sfy(d − d′) (2.96)

Case II We have fs = fy and f ′s < fy (small bottom and large top reinforcement ratios, most

common case)

ε′s = εuc − d′

c(2.97-a)

f ′s = Esε

′s (2.97-b)

Asfy = A′sf

′s + .85f ′

cbβ1c (2.97-c)

solve for c and f ′s by iteration.

Alternatively, those equations can be combined yielding

(0.85f ′cb)a

2 + (0.003EsA′s − Asfy)a − (0.003EsA

′sβ1d

′) = 0 (2.98)


Draft2–30 FLEXURE

Using a = β1c

M IIn = .85f ′

cab(d − a

2

)+ A′

sf′s(d − d′) (2.99)

Case III fs < fy and f ′s = fy (large bottom and small top reinforcement ratios, rare)

εs = εud − c

c(2.100-a)

fs = Esεs (2.100-b)Asfs = A′

sfy + .85f ′cab (2.100-c)

a = β1c (2.100-d)

solve for a

M IIIn = .85f ′

cab(d − a

2

)+ A′

sfy(d − d′) (2.101)

Case IV (not allowed by ACI) fs < fy and f ′s < fy (large bottom and top reinforcement

ratios, rare)

ε′s = εuc − d′

c(2.102-a)

εs = εud − c

c(2.102-b)

Asfs = A′sf

′s + .85f ′

cab (2.102-c)a = β1c (2.102-d)

solve for a

M IVn = .85f ′

cab(d − a

2

)+ A′

sfs(d − d′) (2.103)

50 Note that in most beams of normal size and proportions, it will be found that f ′s < fy when

fs = fy. We nevertheless use A′s in order to ensure ductility, stiffness and support for the

stirrups.

Example 2-12: Doubly Reinforced Concrete beam; Review

Given, f ′c = 4, 000 psi, fy= 60,000 psi, A′

s = 3 (1.56) = 4.68 in2, As= 4 (1.56) = 6.24 in2,determine the moment carrying capacity of the following beam.



βA’ f’

a= c

0.85 f’

d

c

ε

β

s

ε =0.003

A = 4 # 1116"

3"

27.3"

A’ = 3 # 11

s

s c

=

u A’ f’

1

c

sss(A − A’ )A fs y

s y

1

s

A fs s

s

+

0.85 f’

a= c

d−d’

’εs

Solution:

1. Determine ρ:

ρb = (.85)β1f ′

c

fy

8787 + fy

= (.85)(.85)460

8787 + 60

= .0285 (2.104-a)

ρ =6.24

(16)(27.3)= .0143 (2.104-b)

ρ′ =4.68

(16)(27.3)= .0107 (2.104-c)

2. Check for ρmin

ρmin = ρ′ + .85f ′

c

fyβ1

d′

d

εu

εu − εy(2.105-a)

= .0107 + (.85)460

(.85)3

27.3.003

.003 − 6029,000

= .0278 > ρ (2.105-b)

Henceρ < ρmin < ρb

.0143 < .0278 < .0285(2.106)

and thus fs = fy and f ′s < fy and we have case II

3. We have two equations: strain compatibility (nonlinear equation) and summation of forces(linear equation), and two unknowns c and f ′

s

f ′s = Esεu

c − d′

c= (29, 000)(.003)

c − 3c

(2.107-a)

= 87c − 3

c(2.107-b)

Asfy = A′sf

′s + .85f ′

cbβ1c (2.107-c)(6.24)(60) = (4.68)f ′

s + (.85)(4)(16)(.85)c (2.107-d)374.4 = 4.68f ′

s + 46.24c (2.107-e)f ′

s = −9.9c + 80.2 (2.107-f)

Note that if we were to plott those two equations,


Draft2–32 FLEXURE

2 3 4 5 6

-100

-75

-50

-25

25

50

We note that f ′s increases with c from the strain diagram, but f ′

s decreases with c fromequilibrium. If c increases, force in concrete increases too and force in steel decreases.Graphically the solution is around 4.9.

4. Combining those two equations1

c2 + .7085c − 26.42 = 0 (2.108)

we obtain c = 4.80 in a = 0.85(4.8) = 4.078 in, and f ′s = (.003)(29, 000)4.80−3

4.80 = 32.6 ksi

5. Substituting into the moment equation

Mn = .85f ′cab

(d − a

2

)+ A′

sf′s(d − d′) (2.109-a)

= (.85)(4)(4.078)(16)(

27.3 − 4.0782

)+ (4.68)(32.62)(27.3 − 3) (2.109-b)

= 9, 313 k.in (2.109-c)Md = 0.9(9, 313) = 8, 382 k.in = 699 k.ft (2.109-d)

6. Check

ρmax = .75ρb +f ′

s

fyρ′ (2.110-a)

= (.75)(.0285) +32.660

(.0107) = .027︸︷︷︸ρ

√(2.110-b)

Example 2-13: Doubly Reinforced Concrete beam; Design

Given Md = 505 k.ft, f ′c = 4 ksi, fy = 60 ksi, b = 12 in, h = 24.5 in, d = 21 in, and

d′ = 2.5 in, determine the reinforcement As and possibly A′s.

Solution:

1In this problem, unfortunately an iterative method diverges if we were to start with a = d5.


Draft2.8 Moment-Curvature Relations 2–33

1. Check if singly or doubly reinforced section:

Md = (505)(12) = 6, 060 k.in (2.111-a)

ρb = .85β1f ′

c

fy

8787 + fy

= (.85)(.85)460

8787 + 60

= .0285 (2.111-b)

ρmax = .75ρb = (.75)(.0285) = .0213 (2.111-c)Amax

s = (.0213)(12)(21) = 5.37 in2 (2.111-d)

a =Asfy

.85f ′cb

=(5.37)(60)

(.85)(4)(12)= 7.89 in (2.111-e)

Mmax = (0.9)Asfy

(d − a

2

)= (.9)(5.37)(60)

(21 − 7.89

2

)(2.111-f)

= 4, 943 k.in < 6, 060 k.in (2.111-g)

Thus compression steel is required.

2. Assuming that f ′s = fy

Md2 = 6, 060 − 4, 943 = 1, 117 k.in (2.112-a)

A′s =

Md2

φfy(d − d′)=

1, 117(0.9)(60)(21 − 2.5)

= 1.12 in2 (2.112-b)

⇒ A′s = 1.12 in2 (2.112-c)

As = 5.37 + 1.12 = 6.49 in2 (2.112-d)

3. Check that f ′s = fy

ρ′ =1.12

(12)(21)= .00444 (2.113-a)

ρ =6.49

(12)(21)= .0257 (2.113-b)

ρmin = ρ′ + .85β1f ′

c

fy

d′

d

εu

εu − εy(2.113-c)

= .00444 + (.85)(.85)460

2.521.0

8787 − 60

= .0229 < ρ(.0257)√

(2.113-d)

Note that if it turned out that f ′s < fy, then we will need to make an assumption on A′

s (suchas A′

s = As2 , as we will have three equations (2 of equilibrium and one of strain compatibility)

and four unknowns (As, A′s, f ′

s and c).

2.8 Moment-Curvature Relations

51 In ordinary reinforced concrete design, we need not be concerned by the moment curva-ture relation of a flexural member. Yet this relation is important to properly understand (insubsequent chapters)


Draft2–34 FLEXURE

1. Redistribution of moments (reducing negative moments, increasing positive ones).

2. Short and long term deflections with the shifting of the neutral axis under service load.

3. Ductility in seismic design, i.e. the ability of a section to exhibit enough flexibility duringseismic excitation, and thus absorbs enough energy.

52 Fig.2.17 shows portion of an originally straight beam which has been bent to the radius ρ

f y

f’c

εel

ψ

y

ε

y

ε

CONCRETE

STEEL

εε

ψ

C

a bMM

O

ρ

B

Ac d y

D

Figure 2.17: Bending of a Beam

by end couples M , thus the segment is subjected to pure bending. It is assumed that planecross-sections normal to the length of the unbent beam remain plane after the beam is bent.Therefore, considering the cross-sections AB and CD a unit distance apart, the similar sectorsOab and bcd give

ε =y

ρ(2.114)

where y is measured from the axis of rotation (neutral axis), ρ the radius of curvature.

53 Furthermore, we define the curvature Ψ as

Ψ =ε

y=

εc

c(2.115)

54 Next we seek to derive the moment curvature for a beam. This will clearly depend on thelocation of the neutral axis, and we identify the following key stages, Fig. ??:

Uncracked Elastic is the first stage

Mcr =frIut

c2(2.116-a)

Ψcr =εr

c2=

fr

Ecc2(2.116-b)

Cracked Elastic: when the section cracks, the stiffness is immediately reduced and the cur-vature increases (the moment does not change). We will then have a linear response up


Draft2.9 Bond & Development Length 2–35

ψcr

ε = εcr

ε < εel

1c ψ

inel

ε < ε < ε el u

ε < εy

S εS

T=A ES

< A fS yS ε

ST=A E

S

1f = f

c

Mcr

Mel

E Iut

E Ict

ψ

Cracking

Proportional

Limit of Concrete

Failure

Mn

h

f

ε = ε

c

c

ε = εcs

d

Z

d

el

1

2

r

d

jd

1c = kd

c = d−kdS

ψel

��

��

��

��

��

��

��

Figure 2.18: Moment-Curvature Relation for a Beam

to the limit of elasticity of the concrete where, Eq. 2.4-c

Mel =bd2

2kjfc (2.117-a)

εel =fc,el

Ec(2.117-b)

Ψ1el =

εel

c1(2.117-c)

we should check that fs = MAsjd ≤ fy.

Cracked Section, Inelastic Material: For this case

1. Select top face concrete strain εel ≤ ε ≤ εu

2. Assume the neutral axis depth to be at a distance c1 below top fibers.3. From the strain diagram and similarity of triangles, determine εs = εcs.4. Determine the tensile steel stress fs = Esεs ≤ fy, and T + Asfs.5. Check equilibrium of forces (C = T ), this requires computing C (area under the

nonlinear stress curve). If equilibrium is not satified, adjust location of neutral axisupward or downward until equilibrium is reached. The internal lever arm z is thendetermined.

6. Solve for

M iinel = Cizi (2.118-a)

Ψiinel =

εi1

ci1

(2.118-b)

7. Plot the results.

2.9 Bond & Development Length

55 Considering the equilibrium of forces acting on an infinitesimal portion of a rebar, Fig. 2.19,and defining U as the force per unit length, we have


Draft2–36 FLEXURE

��

��

��

��

jd

dx

T T+dT

C + dC

T+dTV+dVV

dx

T

C

M + M∆M

��

Figure 2.19: Bond and Development Length

Udx = dT ⇒ U =dT

dx(2.119)

56 The tensile force is a function of the moment

M = Tjd (2.120-a)

dT =dM

jd(2.120-b)

57 But the shear is related to the moment

V =dM

dx(2.121)

Combing those equations together, we obtain

U =V

jd(2.122)

58 We define u as the bond stress, and is equal to

u =U

Σ0(2.123)

where Σ0 is the sum of all the bars perimeters.

59 If plain bar → weak adhesion → slip → need end anchorage → no bond → u = 0 → dT = 0 →steel stress is constant over entire length → T = Mmax

jd → total steel elongation >than if bondpresent → large deflection and large crack width.

60 Actual stress distribution along steel bar is quite complex, Fig. 2.20.

61 If bond stress is too large ⇒ splitting along reinforcement, Fig. 2.21.

62 Failure will initiate at points of high shear(large dM

dx

).

63 It frequently starts at diagonal cracks ⇒ dowel action increases the tendancy of splitting ⇒shear and bond failures are often interrelated.

64 Based on tests with one single bar, ultimate average bond force/inch of length of bar isUn ≈ 35

√f ′

c.



��

��

u stresses on rebar

u stresses on concreteMM

Bond stress u

Steel tension slope = dTdx

Figure 2.20: Actual Bond Distribution

Figure 2.21: Splitting Along Reinforcement


Draft2–38 FLEXURE

65 If we have several bars in one layer spaced 6 in or less, then the ultimate bond capacity is80% of the single bar case.

66 In terms of bond stress, Fig. 2.22

un =35

√f ′

c

Σ0(2.124)

s bT = A fy

��

T = 0s Ld

Figure 2.22: Development Length

67 Putting it differently, the minimum length necessary to develop through bond a force Asfy

is, Fig. ??.Ld = Abfy

Un

Un = unΣ0

un = 35√

f ′c

Σ0

ld =

0.028Abfy√f ′

c

(2.125)

A f

L

Us y

d

Figure 2.23: Development Length

68 For small bar spacing, we have to decrease the bond stress

ld =1

0.80.028Abfy√

f ′c

=0.035Abfy√

f ′c

(2.126)

69 If actual development length l is smaller than ld, then we must provide anchorage in orderto avoid a bond failure.

70 Note:

1. un is independent of diameter



2. For a given fs

T = Abfs

= fsπd2

b4

ld = Abfs

Un

ld = fsπd2

b4Un

(2.127)

ld increases with the square of db ⇒ small bar diameters require shorter developmentlength.

71 Top bars, with more than 12 inch of concrete below them, will have a reduced bond stress(due to rise of water during vibration). This reduction in bond results in an increase of ld by40%

72 ACI 12.2.2 may be obtained from above but rather than use φ we increase ld by 15% forsafety.

ldb = .04Abfy√f ′

c

#11 or smaller; and deformed wire

= .085 fy√f ′

c

#14

= .125 fy√f ′

c

#18

> 12 in. in all cases

(2.128)

Consult ACI 12.5 code for hooks geometry, and corrections to this basic equation.

73 Check ACI code for modifications related to top reinforcement, lightweight aggregate, highstrength reinforcement, excess reinforcement, and spiral confinement.

ld = λdλddldb (2.129)

74 If not enough development length can be provided ⇒ provide hooks, Fig. 2.24 at

1. 90 degrees: bar must extend by 12db

2. 180 degrees: see code.

wherelhb = 1200 db√

f ′c

ldh = λdlhb

(2.130)

and λd is given in the ACI code.

2.9.1 Moment Capacity Diagram

75 Ideally, the steel should be everywhere as nearly fully stressed as possible. Since the steelforce is proportional to the moment, then the steel area is nearly proportional to the momentdiagram.

76 Requirements include, Fig. 2.25:


Draft2–40 FLEXURE

db

db

As in part (b)

12db

ldh

ldh

Criticalsection

Criticalsection

(a)

(b)

4db

6db

5db

Nos. 3 through 8

Nos. 9, 10, 11

Nos. 14 and 18

4db or 2 1/2in. min.

Figure 2.24: Hooks

1. At least As3 in simple beams and As

4 for continuous beams should be extended at least 6in. into support.

2. If negative bars are cut, they must extend at least ld beyond face of support.

3. Negative bars must extend d or 12db beyond theoretical cutoff point defined by momentdiagram.

4. At least one third of top reinforcement at support must extend at least ld beyond the-oretical cutoff point of other bars, and d, 12db or ln

16 beyond the inflection point of thenegative moment diagram.

77 Determination of cutoff points can be rather tedious, for nearly equal spans uniformly loaded,in which no more than about one half the tensile steel is to be cut off or bent, locations shownin Fig. 2.26 are satisfactory (note that left support is assumed simply supported).

78 Fig. 2.27 is an illustration of the moment capacity diagram for a beam.



Moment Capacityof bars O

Inflection pointfor (+As)

Inflection pointfor (-As)

Theoreticalnegativemoment

Theoreticalpositivemoment

Bars N

Bars OBars L

Bars M

ld

ld

ld

ld

d or 12 db

d or 12 db

6" for at least1/4 of (+AS)(1/3 for simple spans)

Greatest of d, 12 db, ln/16for at least 1/3 of (-AS)

C L o

f sp

an

Moment capacityof bars M

Fac

e o

f su

pp

ort

Figure 2.25: Bar cutoff requirements of the ACI code


Draft2–42 FLEXURE

L1

L1

L1

L1

L2

L2

L2

L2

L2

L2

L2

L2

L1

L1L1

4

4

3

3

3

3

3

3

8

4

8

4

8

47

6"

6"

6"

6"

L1

L1

L2

L2

0"

0"

0"

0"6"

6"

Figure 2.26: Standard cutoff or bend points for bars in approximately equal spans with uni-formly distributed load



B

BA

A C

C

5 bars4 bars2 bars

Ld

Ld

Ld

d or 12’’

AA

BB

CC

Mcap of 2 bars

Mcap of 4 bars

Mcap of 5 bars

Md=φMn

Figure 2.27: Moment Capacity Diagram


Draft2–44 FLEXURE


Draft

Chapter 3

SHEAR

3.1 Introduction

1 Beams are subjected to both flexural and shear stresses. Resulting principal stresses (or stresstrajectory) are shown in Fig. 3.1.

Compression trajectoriesTension trajectories

τ α

2

21

1σ

σσ

σ

σσ τ

ττ

ττ

τττ

τ τ

σο

45

9045α

τ

σ

Figure 3.1: Principal Stresses in Beam

2 Due to flexure, vertical flexural cracks develop from the bottom fibers.

3 As a result of the tensile principal stresses, two types of shear cracks may develop, Fig. 3.2:

Flexural Shear Cracks Flexural CracksWeb Shear CracksFlexural Cracks

Small MLarge VSmall V

Large MSmall MLarge V

Figure 3.2: Types of Shear Cracks

Web shear cracks: Large V, small M. They initiate in the web & spread up & down at ≈ 45o.

Draft3–2 SHEAR

Flexural shear cracks: Large V, large M. They initiate as an extension of a pre-existingflexural crack, initially vertical, then curve.

4 Shear failure is sudden ⇒ φ = 0.85

5 Some of the important parameters controlling shear failure:

1. Shear span ratio MV d

2. Steel ratio ρ = Asbd

3. f ′t = 4

√f ′

c note that f ′r = 7.5

√f ′

c

6 We shall first examine the shear strength of uncracked sections, then the one of crackedsections (with shear reinforcement).

3.2 Shear Strength of Uncracked Section

7 Question: What is the maximum shear force which can be applied before a flexural crackdevelop into a flexural shear crack?

1. Apply M → flexural crack

2. Apply V → flexural shear crack

8 Note that all shear resistance is provided by the concrete. As with flexural reinforcement,steel is ineffective as long as the section is uncracked.

C

T

jd +

Flexure

cf

Shear

σ

τ

v c

Shear

v n

Figure 3.3: Shear Strength of Uncracked Section

9 Solution strategy:

1. Determine the flexural compressive stress fc in terms of M

2. Determine shear stress v in terms of V


Draft3.2 Shear Strength of Uncracked Section 3–3

3. Compute the principal stresses

4. Equate principal tensile stress to the tensile strength

10 Using a semi-analytical approach

1. Assume that fc is directly proportional to steel stress

fc = α fs

n

Mn = Asfsjd ⇒ fs = MnAsjd

fc = α Mn

nAsjd

ρ = Asbd

}fc =

αMn

nρjbd2= F1

Mn

ρnbd2(3.1)

2. Shear stressvn = F2

Vn

bd(3.2)

3. From Mohr’s circle, the tensile principal stress is

R

1

τ

σf

vn

vn

cf

Figure 3.4: Mohr’s Circle for Shear Strength of Uncracked Section

f1 =fc

2+

√(fc

2

)2

+ v2n (3.3)

4. Set f1 equal to the tensile strength

f1 = f ′t ⇒ f1

Vn

bd= f ′

t

Vn

bd(3.4-a)

Vn

bd=

f ′t

f1

Vn

bd(3.4-b)

=f ′

tf1bdVn

(3.4-c)

Combining Eq. 3.1, 3.2, and 3.3


Draft3–4 SHEAR

Vn

bd=

f ′t

F1

2Ec

Es︸︷︷︸C′

1

MnρVnd +

F1

2Ec

Es︸︷︷︸C′

1

MnρVnd

2

+ F 22︸︷︷︸

C′2

1/2(3.5)

5. set f ′t = 4

√f ′

cVn

bd√

f ′c

=1

C1

√f ′

c

ρMnVnd +

√(C1

√f ′

c

ρMnVnd

)2

+ C2

(3.6)

6. Let the variables be Vn

bd√

f ′c

& Mn

√f ′

c

ρVnd

7. This is how far we can go analytically. To determine the exact factors associated withthis equation, one has to undertake a series of tests.

8. From 440 tests, Fig. 3.5 it is found that

cf’

nV

cf’nM

bd

1.9

2.0

3.5

nV d

Figure 3.5: Shear Strength of Uncracked Section

Vn

bd√

f ′c

= 1.9 + 2, 500ρVnd

Mn

√f ′

c

≤ 3.5 (3.7)

or if we set vc = Vnbd , then

vc = 1.9√

f ′c + 2, 500

ρVnd

Mn≤ 3.5

√f ′

c (ACI 11.3.2.1) (3.8)

9. Note that vc is in terms of VndM or inverse of shear span (M

V )


Draft3.3 Shear Strength of Cracked Sections 3–5

10. This equation is usually found acceptable for predicting the flexure shear cracking loadfor shear span/depth ratio Mn

Vnd of 2.5 to 6 & is found to be very conservative for lowervalues

11. Increasing ρ has a beneficial effort as a larger amount of steel results in narrower & smallerflexural tension cracks before formation of diagonal cracks ⇒ larger area of uncrackedconcrete can resist the shear.

12. Use of Vu & Mu instead of Vn = Vuφ & Mn = Mu

φ

3.3 Shear Strength of Cracked Sections

11 If the shear stress exceeds 1.9√

f ′c + 2, 500ρVnd

Md, then the flexural crack will extend into a

flexural shear crack, Fig. 3.6. and if

C

z

V

A f

V

p

T=A ssf

Va

d

s

c

vv

Figure 3.6: Free Body Diagram of a R/C Section with a Flexural Shear Crack

1. No shear reinforcements ⇒ failure

2. Stirrups are present ⇒ stirrups will carry part of shear force

12 If the section is cracked, Fig. 3.7

Vext = Vc + ΣnAvfv + Vd + Vay︸︷︷︸Vint

(3.9)

whereVc Shear resisted by uncracked sectionn # of stirrup traversing the crack n = p

sAv Area of shear reinforcementfv Shear reinforcement stressVd Dowel force in steelVa Aggregate interlock

13 We must determine the internal (resisting) shear forces at failure where fv = fy


Draft3–6 SHEAR

Vs

Vd

Vcz

Vay

Vint

ΣV int

Incl

ined

cra

ckin

g

Yie

ld o

f st

irup

sFa

ilure

Flex

ural

cra

ckin

g

Vext

Figure 3.7: Equilibrium of Shear Forces in Cracked Section

1. Due to yielding → large separation between 2 sides of cracks ⇒ Va → 0

2. Neglect Vd

3. Vext = Vn = Vc︸︷︷︸unknown

+nAvfy

4. We will assume that at failure the shear force provided by concrete is equal to the onewhich caused the diagonal crack to form ⇒ va = 1.9

√f ′

c + 2, 500ρVndMd

. Thus, Vc = vabwd

5. Finally, if we assume p = d (implying a crack at 45◦)

Vn = Vc + Avfyd

s︸︷︷︸Vs

(ACI 11.1.1) (3.10)

3.4 ACI Code Requirements

14 The ACI code requirements (∮

11) are summarized by Fig. 3.8:

1. Design for Vu (factored shear) rather than Vn = Vuφ (ACI 11.1.1), ⇒ plot Vu diagram.

2. Determine φVc (nominal shear carried by the concrete) where

Vc = 2√

f ′cbwd (ACI 11.3.1.1)

orVc = [1.9

√f ′

c + 2, 500ρwVudMu

]bwd ≤ 3.5√

f ′cbwd (ACI 11.3.2.1)

whereVudMu

< 1

(3.11)

3. If Vu <0.5 φVc no shear reinforcement is needed (ACI 11.5.5.1)


Draft3.4 ACI Code Requirements 3–7

4. If 0.5φVc < Vu ≤ φVc use minimum shear reinforcement; select Av (usually #3 bars) anddetermine

s = Avfy

50bw(ACI 11.5.5.3)

s < d2 (ACI 11.5.4.1)

s < 24 in (ACI 11.5.4.1)(3.12)

5. If Vu > φVc ⇒ provide stirrup such that

Vu

φ= Vn = Vc + Vs = Vc +

Avfyd

s(ACI 11.17) (3.13)

or

s =AvfydVuφ − Vc

=φAvfy

(vu − φvc)b(3.14)

6. If Vu − Vc > 4√

f ′cbwd, then s < d

4 and s < 12 in, (ACI 11.5.4.3).

7. Upper limit:Vu − Vc < 8

√f ′

cbwd (ACI 11.5.6.8) (3.15)

8. fy ≤ 60 ksi, (ACI 11.5.2)

9. Critical section is at d from support (reduces design shear force), (ACI 11.1.3.1)

f ’c

f ’c

f ’c

f ’c

f ’cf ’c

2

6

10

4

8

d

not allowable

no s

tirup

ssmax=d/4 or 12" smax=d/2 or 24"

min

. stir

ups

s=

Av

f y50

b w

s= =Avfy d Avfy

Vu

Vu

(vuφ

φ

φ

−Vc−φvc)b

Steel

Concrete

distance from support

Vb wd

Figure 3.8: Summary of ACI Code Requirements for Shear


Draft3–8 SHEAR

3.5 Examples

Example 3-1: Shear Design

b = 12 in.; d=22 in.; wu= 8.8 k/ft; L= 20 ft.; As= 3# 11; f ′c= 4 ksi; fy= 40 ksi;

Design vertical stirrupSolution:

1. At support: Vu = 8.8 (20)2 = 88 k and vu = 88

(12)(22) = .333 ksi

2. At d from support Vu = 88 − 2212(8.8) = 71.9 k and vu = 71.9

(12)(22) = .272 ksi

3. vc = 2√

f ′c = 2

√4, 000 = 126 psi; φvc = (0.85)(126) = 107.1 psi

4. φvc

2 = 53.6 psi

5. 4√

f ′c = 2(126) = 252 psi;

6. vu − vc = 272 − 126 = 146 psi < 4√

f ′c

√

19"min. reinforcement no reinforcement

38.6"

x

dpsi

333

272

107.1

53.6

φv c

φv c

2

Vu

7. vu − φvc = 0 ⇒ 333(10)(12)x = 107.1 ⇒ x = 38.6 in = 3.2 ft from mid-span

8. vu − φvc

2 = 0 ⇒ 333(10)(12)x = 53.6 ⇒ x = 19.3 in = 1.6 ft

9. Selecting #3 bars, Av = 2(.11) = .22 in2

smax = Avfy

50bw= (.22)(40,000)

(50)(12) = 14.66 ind2 = 22

2 = 11 in

}smax = 11 in (3.16)


Draft3.6 Shear Friction 3–9

10. at support

s =AvfydVuφ − Vc

=φAvfy

(vu − φvc)b(3.17-a)

=(.85)(.22)(40, 000)(272 − 107.1)(12)

(3.17-b)

= 3.78 in (3.17-c)(3.17-d)

3.6 Shear Friction

15 Previous design procedure was applicable to diagonal tension cracks (where tension wasinduced by shear), for those cases where we do have large pure shear, Fig. 3.9, use shearfriction concept.

#7 Vu

weldNuc

close sriru(usually #3)

assumed crack+ shear plane

remainder of A v f

A n part of A v f

An=Nuc

φfy

assumed crack

Avf

Vu

Figure 3.9: Corbel

16 The crack for which shear-friction reinforcement is required may not have been caused byshear. However once the crack has occurred a shear transfer mechanism must be provided for,Fig. 3.10. The shear friction theory is based on the assumption that a crack will occur andthen reinforcement across it will resist relative displacement along the crack.

17 If we assume separation to be sufficient⇒ steel will yield

Vn = µAvffy (3.18)

18 If the shear reinforcement is inclined with respect to the crack, Fig. 3.11

19 Component of tensile force in reinforcement gives rise to compression force at interface C ⇒µc vertical force due to friction;

Vn = T cos αf + µCC = T sinαf

}Vn = T (cos αf − µ sinαf )T = Avffy

}Vn = Avffy(cos αf + µ sin αf(3.19-a)


Draft3–10 SHEAR

Vn Vn

Vn

Vn

Vn

crack

Shear−transferreinforcement

crac

k se

para

tion

due

to s

lip

Avf

f yA f A fvf vfy y

µA vff y

22

Figure 3.10: Shear Friction Mechanism

Tcos αf

Tsin αf

assumed crack

applied shear=Vn

Avffy

Tαf

µC

C=Tsin α f

Figure 3.11: Shear Friction Across Inclined Reinforcement


Draft3.6 Shear Friction 3–11

20 Note: Vu = φVn and φ = 0.85

21 The preceding equation can be rewritten as

Avf =Vu

φµfy(3.20)

Avf =Vu

φfy(cos αf + µ sin αf )ACI − 11.27 (3.21)

22 ACI-11.7.4.3 specifies µ as such thatconcrete cast monolithically µ = 1.4λconcrete against hardened concrete µ = 1.0λconcrete against steel µ = 0.7λ

where λ = 1.0 for normal weight concrete and λ = .75 for lightweight concrete. and

Vn <

{0.2f ′

cAc

800Ac(3.22)

and Ac( in2) is the area of concrete resisting shear.

Example 3-2: Shear Friction

Design reinforcement needed at the bearing region of a precast beam 14” wide & 28” deepsupported on a 4” bearing pad. Vu = 105k, horizontal force due to restraint, shrinkage, creepis 0.3 Vu

15

20

possible crackA vf

Vu

N uc

4"24" Vuc

15

Nuc

3#6

2#6

Solution:

1. Assume all the shear Vu will be acting parallel to crack (small angle 20◦)

2. Assume all Vu is parallel to crack ⇒ required Avf = Vuφfyµ = 105

(0.85)(60)(1.4) = 1.47 in2

3. As = Nacφfy

= (0.3)(105)(0.85)(60) = 0.62 in2 for horizontal force

⇒ As = Avf + An = 1.47 + 0.62 in2 = 2.09 in2 ⇒ use 5# 6 (As = 2.20 in2)

4. Note: ACI 11-9-3-4 Nuc > 0.2 Vu for Corbels;


Draft3–12 SHEAR

3.7 Brackets and Corbels

To be Edited

23 Nu might be due to shrinkage, prestressing · · ·

24 Design based on truss analogy

25 A.C.I. provisions (Chapter 11)

1. For ad < 1

2 , use shear friction theory

2. For ad > 1, use ordinary beam theory

3. For 12 ≤ a

d ≤ 1

Vn = [6.5 − 5.1√

Nuc

Vu](1 − 0.5

α

d)

{1 + [64 + 160

√(Nu

Vu)3]ρ

}√f ′

cbwd (3.23)

where ρ = As?? ; and ρ ≤ 0.13 f ′

cfy

; NuVu

not to be taken < 0.20 in calculating vu; Nu = (+ve)compression, and (-ve) tension; Ah < As also Ah ≥ 0.50As distributed uniformly; thru23d adjacent to As; ρ = As

bd ≥ .04 f ′c

fy.

3.8 Deep Beams


Draft

Chapter 4

CONTINUOUS BEAMS

4.1 Continuity

1 R/C bldgs constructions commonly have floor slabs, beams, girders and columns continuouslyplaced to form a monolithic system

Figure 4.1: Continuous R/C Structures

2 In a continuous system, load must be placed in such a way to maximize desired effect (M+vemax

M−vemax Vmax, Fig. 4.2

Max +ve M @ AB_CD_EF

Max -ve M @ B

Min -ve @ B

Max -ve @ C

Min -ve @ C

Max -ve @ D

Min -ve @ D

A B C D E F G

Figure 4.2: Load Positioning on Continuous Beams

3 Given the moment diagram for various load cases, a designer should draw the momentenveloppe and design for the maximum negative and positive moments (eventhough they maynot be caused by the same load case).

Draft4–2 CONTINUOUS BEAMS

4.2 Methods of Analysis

4 Two approaches:

1. Detailed analysis

(a) Moment distribution

(b) Computer analysis (such as RISA, SAP, etc...)

2. Approximate (but conservative) based on ACI 8.3.3 moment coefficients

4.2.1 Detailed Analysis

5 Refer to CVEN3525/3535/4525

4.2.2 ACI Approximate Method

6 This method, Fig. 4.3 can be used if:

1. 2 or more spans

2. Spans are approximately equals, and the larger of adjacent ones not greater than theshorter by more than 20%

3. Loads are uniformably distributed

4. LL < 3DL

5. Prismatic members

Positive MomentEnd Spans

Continuous end unrestrained 111wuL2

n

Continuous end integral with support 114wuL2

n

Interior spans 116wuL2

n

Negative MomentNegative moment at exterior face of first of first interior support

Two spans 19wuL2

n

> Two spans 110wuL2

n

Negative moment at other faces of interior support 111wuL2

n

...................................... ......Shear

Shear in end member at face of first interior support 1.15wuLn2

Shear at face of all other supports wuLn2

where wu is the factored load, and Ln is the clear span.

7 These moment coefficients take into account some inelastic action (stress redistribution).

8 They are conservative compared to an exact analysis.


Draft4.2 Methods of Analysis 4–3

Figure 4.3: ACI Approximate Moment Coefficients



��

��

V

aL

VaL

2

2

Column width aLL2

VaL6

VaL

VaL

aL

VaL

VaL

VaL3

3

2

6

62

Adjusted Moment Curve

Moment curve based on prismatic member

C beam

C beam

C beam

C spanC column

C column

L

L

L

L

LL

��

��

Figure 4.4: Design Negative Moment

4.3 Effective Span Design Moment

9 Negative moments should be the one at the face of the columns, Fig. 4.4.

10 We recall that the change in moment is equal to the area under shear diagram.

M = Mcl − V aL

2(4.1)

but V and M vary in some unknown way between center line of column and edge, thus we canreduce M by V b/3 where b is the width of the column. Thus

M−ved ≈ M−ve

max − V b

3(4.2)

11 This can substantially reduce high M−ve.

4.4 Moment Redistribution

4.4.1 Elastic-Perfectly Plastic Section

12 Let us consider a uniformly loaded rigidly connected beam, Fig. 4.5

13 The beam has an elastic plastic moment curvature relation, Fig. 4.6


Draft4.4 Moment Redistribution 4–5

��

��

��

��

W

W L2

12

W L2

24

+

− −W L2

12

Figure 4.5: Moment Diagram of a Rigidly Connected Uniformly Loaded Beam

Φu

M

Φy

Mp

Curvature

X

Figure 4.6: Moment Curvature of an Elastic-Plastic Section

14 |M−ve| > |M+ve| as w ↗, M−ve → Mp first ⇒

wL2

12= Mp ⇒ w =

12Mp

L2

15 Thus we will have a plastic hinge at the support however this is not synonymous with collapse.

16 Collapse or failure occurs when we have a mechanism or 3 adjacent hinges (plastic or other-wise). This can be easily determined from statics, Fig. 4.7

∆

w

16 M

2L

p

2L

12 Mp

pM

pM

pM

Figure 4.7: Plastic Moments in Uniformly Loaded Rigidly Connected Beam

2Mp = wuL2

8

wu = 16Mp

L2

17 Thus capacity was increased 33% after first plastic hinge occurred.

18 This is accompanied by large rotation of the plastic hinges at the supports, and when com-pared with the linear elastic solution M−ve ↘ and M+ve ↗19 The section must be designed to accomodate this rotation.



4.4.2 Concrete

20 Concrete is brittle hence by itself no appreciable plastic deformation can occur, however inR/C, Fig. 4.8

ε = εA f

cf

d−kd

kd

ysys

A f

cf

y

.003

moment redistributionStrain caused by

ceε

yε

cε.003

Unit rotation

Steel yielding

First crack

ε > ε

u

crθ

crM

yM

M

s

d−c

c

ys

uθy

θ

yθ

uθ

uθ

yθ

Figure 4.8: Plastic Redistribution in Concrete Sections

21 If certain rotation capacity exists (i.e., if ρ−ρ′ is low) M is controlled by yielding of the steelwhile the concrete strain is still low compared to 0.003 ⇒ reserve rotation capacity θu − θy isthen available for a redistribution of moment to occur before ε → 0.003

22 M−ve moment at support of continuous flexural members calculated by elastic theory canbe decreased by no more than

∆M = 20(1 − ρ − ρ′

ρb)% ACI 8.4.1 (4.3)

where ρb = 0.85β1f ′

cfy

( 8787+fy

) provided that

1. Moments are exactly determined (i.e., not ACI coefficients)

2. ρ or ρ − ρ′ < 0.5ρb

23 M+ve must be increased accordingly.

24 This capacity to redistribute moments (reduce M−ve and increase M+ve) is a characterisitcof ductile members.

25 Earthquake resistant structures must have a certain ductility to absorb the lateral oscillatingload ⇒ large amount of reinforcement at the joints.


Draft4.5 Buildings 4–7

Example 4-1: Moment Redistribution

Determine the moment redistribution for the following singly reinforced beam with ρ = 0.5ρb

W L2

12W L

2

12

W L2

24

+

− − W L2

12

W L

+

− −W L2

120.9 0.9

20

2

Solution:

From above, amout of redistribution

∆M = 20(1 − ρ−ρ′ρb

)%

= 20(1 − 0.5) = 10%M−ve = 0.9wL2

12

M+ve = 1.2wL2

24 = wL2

20

4.5 Buildings

26 Building types, Table 4.1

Structural System Number of StoriesFrame Up to 15Shear Wall-Frame up to 40Single framed tube up to 40Tube in Tube up to 80

Table 4.1: Building Structural Systems

27 We analyse separately for vertical and horizontal loads.

Vertical loads: DL and LL. This is typically done for a floor, through a grid analysis. Noneed to model the entire structure. We can use

ACI Approximate equations

Exact (Moment distribution, computer)

Lateral laod: WL, EL. This requires the analysis of a 2D or 3D frame. Two approaches:

Approximate method: Portal method, or cantilever method.

Exact Moment distribution, computer.

28 Recommended analysis/design procedures



1. Use ACI approximate equations for the design of the slab. Then, there is no need to worryabout optimal placement of load to maximize positive or negative moments, or momentredistribution.

2. Once the slab is designed, use exact method for beams, girders. Reduce negative moments.

3. Tabulate maximum +ve and -ve moments for each beam.

4. Determine the column loads, tabulate.

5. Can use approximate or “exact” method of analysis for frames. Tabulate results.

6. Add maximum positive and negative moments due to vertical and lateral loads.

7. Design accordingly.

29 A block diagram for the various steps is shown in Fig. 4.9

N

E

S

W

L L LL

hf

h h h

DL LLw0 w0 w0 w0PW PW PWWL WL

b b

b

wu wu wu wu

M M M MV V

V

Col

Fou

VR R

E-W SLAB N-S BEAM E-W GIRDER N-S GIRDER

L Spanhf Slab thicknessh Beam/girder depthM FlexureV ShearR ReationPW Partition wallWL Wind loadW0 Self weightWa Total factored loadCol ColumnFou Foundation R/C Bldg Design

Figure 4.9: Block Diagram for R/C Building Design


Draft

Chapter 5

ONE WAY SLABS

5.1 Types of Slabs

1 Types of slabs, Fig. 5.1

Bea

m

Bea

m

Beam

Beam

Bea

m

Bea

m

one−way slab two−way slab

one−way slab Flat plate slabFlat slab

Grid slab

Figure 5.1: Types of Slabs

2 Two types of slabs, Fig. 5.2

1. One way slab: long span/short span > 2. Load is transmitted along the short span.

2. Two Way slab: Long span/short span <2. Load is transmitted along two orthogonaldirections.

3 If Ls > 2 than most of the load (≈ 95%) is carried in the short directions, Fig. 5.3

4 Load transfer in one way slabs is accomplished hierarchically through an interaction of slab,beam, girder, column and foundations, Fig. 5.4

Draft5–2 ONE WAY SLABS

StripStrip

B B

BB

1’−0" 1’−0"

S S

Beam

1

Beam

2

Beam

2

Beam 1

Beam 1

Beam

1

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

L

Figure 5.2: One vs Two way slabs

AA

B

B

ρB

ρAP

��

��

��

��

��

��

Figure 5.3: Load Distribution in Slabs


Draft5.1 Types of Slabs 5–3

Figure 5.4: Load Transfer in R/C Buildings



Simply One end Both ends Cantileversupported continuous continuous

Solid Oneway slab L/20 L/24 L/28 L/10Beams orribbed One way slab L/16 L/18.5 L/21 L/8

Table 5.1: Recommended Minimum Slab and Beam Depths

5.2 One Way Slabs

5 Preliminary considerations for one way slabs:

1. Load on slabs ksf.

2. Design an imaginary 12 in strip.

3. The area of reinforcement is As/ft of width or

As

ft= Ab

(12 in

bar spacing in inches

)(5.1)

where Ab is the area of one bar. or

Bar spacing in inches =12Ab

As(5.2)

4. Slab thickness t is usually assumed, and we design reinforcement. ACI 9.5.2.1 recom-mended minimum thickness of beams/slabs are given by Table 5.1. where L is in inches,and members are not supporting partitions. If a slab is so dimensioned

(a) Deflection need not be checked(b) Usually, neither flexure, nor shear controls

5. In reinforcement design, a good initial guess for ad is 0.15.

6. Slab thickness are rounded to the neares 1/4 inch for slabs less than 6 inch, and 1/2 forthicker ones.

7. ACI Sect. 7.7.1 gives minimum cover for corrosion control

(a) Concrete not exposed to weather or in contact with ground, No. 11 or smaller 3/4inch.

(b) Concrete exposed to weather or in contact with ground:i. No. 5 bars and smaller, 1.5 inch.ii. No. 6 and larger, 2. inch.

8. Transverse reinforcement (shrinkage, temperature) must be provided

As

bh=

0.002 Grade 40 and 50 barsACI 7.12.2.1

0.0018 Grade 60 and welded wire fabric(5.3)


Draft5.3 Design of a One Way Continuous Slab 5–5

9. Shear does not usually control & no minimum reinforcement is needed (vc = 2√

f ′c)

10. Principal reinforcement shall not be spaced at more than 3 times the slab thickness nor18 in (ACI 7.6.5).

11. Usually No. 4 and larger bars are used for flexural reinforcement, as No. 3 may bebent out of position by workers walking on it. This is more critical for top than bottomreinforcement.

12. Sometimes, No.3 is used for bottom, and No. 4 for top.

13. Shrinkage/temperature reinforcement shall not be spaced at more than 5 times the slabthickness nor 18 in (ACI 7.12.2.2).

5.3 Design of a One Way Continuous Slab

Design an 8 span floor slab. Each span is 15 ft long, f ′c = 3, 750 psi, fy= 60 ksi, wl=100 psf,

floor cover is 0.5 psf, mechanical equipment 4 psf, and ceiling 2 psf. Interior supporting beamshave a width of 14 inch, and exterior ones 16 inches. First span is measured from exterior ofexterior beam to center of first interior beam.

Thickness: of the floor is based on ACI recommendation:

le = (15)(12) − 162

− 142

= 165 in (5.4-a)

li = (15)(12) − 2142

= 166 in (5.4-b)

hemin =

l

24=

16524

= 6.88 in (5.4-c)

himin =

l

28=

16628

= 5.93 in (5.4-d)

We round h up to h = 7.25 in. Assuming 3/4 in. cover and No. 4 bars

d = 7.25 −(

0.75 +0.52

)= 6.25 in (5.5)

Factored Loads Slab

wd =7.2512

(150) = 90.6 psf of floor surface (5.6)

Total dead load 90.6 + 0.5 + 4 + 2 = 97.1 psf

Factored loadwu = 1.4(97.1) + 1.7(100) = 306 psf (5.7)

The load per foot of strip is 306 lbs/ft

Since wl < 3wd we can use the ACI 8.3.3 coefficients to compute the moments.

Net spans

1. First interior span ln = (15)(12) − 162 − 14

2 = 165 in = 13.75 ft



2. Second interior span ln = (15)(12) − 14 = 166 in = 13.83 ft

3. Average span ln = 12(165 + 166) 1

12 = 13.79 ft

Flexural Design

ai = 0.15d = 0.15(6.25) = 0.9375 in (5.8-a)

As =Mu

φfy(d − a2 )

=12Mu

0.9(60)(6.25 − a2 )

=0.222

6.25 − a2

Mu (5.8-b)

a =Asfy

0.85f ′cb

=60

(0.85)(3.75)(12)As = 1.569As (5.8-c)

Amins = 0.0018bh = 0.0018(12)(7.25) = 0.157 in2/ft (5.8-d)

For maximum spacing, ACI specifies 3h = 3(7.25) = 21.75 in but no more than 18 in,⇒ smax = 18 in.

Support Midspan Support Midspan Support Midspanln, ft 13.75 13.75 13.79 13.83 13.83 13.83wul2n 57.85 57.85 58.19 58.53 58.53M Coeff. 1/24 1/14 1/10 1/11 1/16 1/11 1/16Mu ft-kip/ft 2.41 4.13 5.82

√5.82 5.29 3.66 5.32 3.66

a 0.937 0.937 0.937 0.937 0.937As 0.092 0.159 0.223 0.141 0.204a 0.145 0.249 0.351 0.221 0.320As 0.087 0.150 0.213 0.132 0.194a 0.136 0.235 0.334 0.207 0.304As 0.087 0.150 0.212

√0.132 0.194

√Amin

s 0.157√

0.157√

0.157 0.157√

0.157Reinf. #4@15 #4 @15 #4@12 #4@15 #4@12Aprov

s 0.16 0.16 0.20 0.16 0.20

Shear Since we have unequal spans we must check at

1. Exterior face of the first interior support

Vu = 1.15wuln2

=(1.15)(306)(157)

2= 2, 302 lb/ft of width (5.9)

2. Typical interior span

Vu = 1.0wuln2

=(1.0)(306)(166)

2= 2, 117 lb/ft of width (5.10)

The shear resistance is

φVc = (0.85)2√

f ′cbwd = (0.85)(2)√

3, 750(12)(6.25) = 7, 808 lb/ft (5.11)

hence the slab is adequate for shear.


Draft5.3 Design of a One Way Continuous Slab 5–7

Shrinkage and Temperature Reinforcement must be provided perpendicular to the spanof the slab

As = 0.0018bh = 0.0018(12)(7.25) = 0.157 in2/ft (5.12)

and maximum spacing is 18 in. Therefore, we can provide # 4 bars at 15 in. as shrinkageand temperature reinforcement. They should be placed on top of the lower layer of steel.

Note that in this problem a 6.5 in. thickness was acceptablee for the six interior spans, but a7.25 in. thickness was required for the end spans.

If the entire floor were made of 6. in. thick slab instead of 7.25 in. about 45 cubic yards ofconcrete could have been saved (for a total floor width of about 90 ft) per flor or 180 kips ofdead load per floor. This would represent a considerable saving in say a 20 story building.

In this case, it would be advisable to use 6., and check for delfections in the end spans.




Draft

Chapter 6

SERVICEABILITY

1 So far we have focused on the ultimate structural behaviour (failure), Vu & Mu, i.e the strengthof a member.

2 It is important to also control the behaviour of structural elements under service load (unfac-tored)

1. Cracking

2. Deflection

6.1 Control of Cracking

3 As σy ↗, εy ↗⇒ larger crack width is associated with large fy. This is why the ACI codeplaces a limitation on max fy = 80ksi. (ACI 9.4)

4 The concern is not the # of crack (we can not control it) but rather the crack width.

5 Crack width should be minimized because:

1. Appearance

2. Corrosion of steel

3. Redistributions of internal stresses

4. Effect on deflection

6 The controlling parameters are:

1. Surface of the reinforcing bar

(a) Round & smooth ⇒ few wide cracks (bad)

(b) Irregular & deformed ⇒ many small cracks (better)

2. Steel stress

3. Concrete cover

Draft6–2 SERVICEABILITY

4. Distribution of steel over the tensile zone of concrete ←

7 Based on purely experimental research, the following emperical relation was determined, Fig.6.1:

w = .076βfs3√

dcA Gergely & Lutz Eq. (6.1)

wherew width in 1/1,000 infs Steel service stress ksi (if not computed can be assumed as 0.6 fy)dc Thickness of concrete cover measured from tension face to center of bar

closest to this face, in.β h2

h1

A Area of concrete surrounding one bar = Total effective tensile area# of bars in2

2yy

h h

d

1 2

cw

Neutral Axis

Steel Centroid

��

��

Figure 6.1: Crack Width Equation Parameters

8 ACI

1. Expresses the crack width indirectly by z where

z =w

.076β= fs

3√

dcA (ACI 10.6.4) (6.2)

and assumes β = h2h1

= 1.2 ⇒ w = .091z

Interior beams z ≤ 175 (w = .016 in)Exterior beams z ≤ 145 (w = .013 in)

Note that to reduce z (beneficial) we must

reduce A or increase the number of bars.

2. Only deformed bars can be used

3. Bars should be well distributed in tension zone

4. fy < 80 ksi

5. In lieu of an accurate evaluation, fs = 0.6fy.

9 Maximum acceptable crack width (ACI Committee 224).


Draft6.2 Deflections 6–3

Exposure wmax (in.)dry air, or protective membrane .016humidity, moist air, soil .012deicing chemicals .007seawater, salt .006water retaining structures .004

Example 6-1: Crack Width

f ′c = 3,000 ksi; fy = 40 ksi; As = 4 # 8; LL = 2.44 k/ft; DL = 1.27 k/ft; L = 15 ft.;

Determine z and crack width

��

��

11.5"

20"

22.5

"

2.5"

7.85

"12

.15"

14.6

5"

��

Solution:

1. w = .076βfs3√

dcA

2. Ec = 57√

3, 000 = 3, 120 ksi

3. n = 29×103

3,120 = 9.29

4. Taking first moment (Eq. 2.6) b(kd)2

2 − nAs(d − kd) = 0 ⇒ k = .393 ⇒ j = 1 − k3 =

.869 ⇒ kd = 7.85 in

5. M = wL2

8 and fs = MAsjd ⇒ fs = (1.27+2.44)(15)2(12)

8(3.14)(.869)(20) = 22.9 ksiNote that ACI allows 0.6fy = (0.6)(40) = 24 ksi conservative

6. β = 22.5−7.8520−7.85 = 14.65

12.15 = 1.206 (note ACI stipulates 1.2)

7. A = (2.5)(2)(11.5)4 = 14.38 in2

8. w = (.076)(1.206)(22.9) 3√

(2.5)(14.38) 11,000 = .00696 in .

9. or z = f3s

√dcA = (22.9) 3

√(2.5)(14.38) = 75.64

6.2 Deflections

10 ACI Code Sect. 9.5

11 Every structural design must satisfy requirements of strength, stiffness & stability



12 With the increased usuage of: a) high strength material (resulting in smaller cross section)& b) use of refined design methods, we can no longer rely on the factor of safety to “take care”of deflection, ⇒ we but must detemine it

13 Deflection should be controlled because of:

1. Visually unacceptable

2. Possible ponding of water

3. Cracking in partition walls

4. Functional difficulties (windows, doors, etc · · ·)5. Machine misalignment

6. Vibration

14 Deflection are computed for service loads only

15 Both long term & short term deflection should be considered.

16 As a rule of thumb, deflections seldom control if ρ < 0.5ρb

6.2.1 Short Term Deflection

17 In general δ = f(w,L)EI , i.e., uniform load over simply supported beam in 5wL4

384EI

18 f(w, l) and E are known, but how do we determine I? (uncracked transformed or cracked),Fig. 6.2

B

Μ

Μ

Ε Ι

Α

Ε Ι Ε Ι

Μ

Ε Ι

Α Α BB B B Α ∆Α ∆

2

c

cr

c cr

2

c ut

1

c e2e1

1

Figure 6.2: Uncracked Transformed and Cracked Transformed X Sections

19 It would be too complicated to have I = I(M)

20 ACI recommends to use a weighted average expression for I → Ieff

Ieff =(

Mcr

Ma

)3

Ig +

[1 −

(Mcr

Ma

)3]

Icr (ACI 9.5.2.3) (6.3)



whereIeff ≤ Ig

Mcr = f ′r

Ig

yb

f ′r = 7.5

√f ′

c

and Ma is the maximum (service) moment at stage in which deflection is computed

21 For continuous beams average

Ieff = 0.70Im + 0.15(Ie1 + Ie2)

22 For beams with one end continuous Ieff = 0.85Im + 15(Icon) where Im, Ie are the moment ofinertia at the middle and the end respectively.

23 Note that Ig may be substituted for Iut

24 Deflection evaluation is a nonlinear problem, as w ↗ M ↗ McrMa

↘ Ieff ↘ and for a continu-ous beam

∆ =5

384wL4

EI

{w ↗ ∆ ↗I ↘ ∆ ↗

6.2.2 Long Term Deflection

δ t

δ inst.

t

Figure 6.3: Time Dependent Deflection

25 Creep coefficient:Cc = εf

εi

Etc = σ

ε = σεi(1+Cc)

= Ec1+Cc

⇒ Creep tends to reduce the elastic modulus of concrete, Fig. 6.4

26 From Strain diagram:

1. Steel strain remains unchanged

2. As concrete undergoes creep, the N.A. moves down ⇒ larger area of concrete is undercompression but since C = T ⇒ stress in concrete is slightly reduced



εs

Asfs

fci

fct

ε tεib

d

kd

As

φ i

φ t

Crackedelasticneutral axis

Figure 6.4: Time Dependent Strain Distribution

3. But since C is now lower and we still satisfy Mext = Mint both stresses in steel & concretemust increase with time

27 According to ACI section 9.5.2.5:

1. Additional long term deflection δt

δt = δi × λ (6.4)

whereλ = ξ

1+50ρ′

ρ′

= A′s

bd

and

Time (months) 3 6 12 ≥ 60ξ 1.0 1.2 1.4 2.0

Thus compressive reinforcement can substantially reduce long term deflections

δtotal = δinitial(1 + λ) (6.5)

LL short

DL sustained

A B C

��

��

Figure 6.5: Short and long Term Deflections



28 Short and long term deflections, Fig. 6.5

A → δi,sust

B → δi,sust + δt,sust

C → δsust + δi,short

Note that we are usually interested in the live load deflection (C-B), thus

δi, short = δi, sust + short︸︷︷︸Ieff(DL+LL)

− δi, sust︸︷︷︸Ieff(DL)

(6.6)

29 ACI max. deflections (ACI 9.5.2.6)

Flat roof not supporting nonstructural elements likely to be damaged δi,sh < L180

Floors not supporting nonstructural elements likely to be damaged δi,sh < L360

Roofs or floors supporting nonstructural elements likely to be damaged δt,sust + δi,sh < L480

Floors not supporting nonstructural elements not likely to be damaged δt,sus + δi,sh < L240

Example 6-2: Deflections

b = 11.5 in.; h = 22.5 in,; d = 20 in.; As = 4 # 8; f ′c = 3,000 psi; fy = 40 ksi; DL = 1.27

k/ft; LL = 2.44 k/ft; L = 15 ft.

1. Determine the short term deflection

2. Find the creep portion of the sustained load deflection & immediate live load deflections

Solution:

1. δi, short = δi,short︸︷︷︸2.44

+sust︸︷︷︸1.27

− δi, sust︸︷︷︸1.27

2. Moment of inertias:

Ieff =(

McrMa

)3Ig +

[1 −

(McrMa

)3]

Ict

Ig = bh3

12 = (11.5)(22.5)3

12 = 10, 916 in4

3. To find Ict, need to locate N.A @ service



11.5"

7.85"

12.15"

20"

b(kd)2

2 − nAs(d − kd) = 0 ⇒ k = .393 ⇒ kd = 7.85 in

Ict = (11.5)(7.85)3

12 + (11.5)(7.85)(

7.852

)2 + (9.29)︸︷︷︸n

(3.14)︸︷︷︸As

(12.152) = 6, 130 in4

f ′r = 7.5

√3, 000 = 410.8 psi

Mcr = f ′rIg

yb

= (410.8)(10,916)11.25 = 33.2 k.ft = 399 k.in

4. Ma for sustained load

M susta =

(1.27)(15)2(12)8

= 428.6 k.in = 35.72 k.ft

5. Ma for sustained and short load

M sust+shorta =

(1.27 + 2.44)(15)2(12)8

= 1, 252 k.in = 104 k.ft

6. Moment of inertias

Ieff, sust + short =(

33.2104.3

)3 (10, 916) +[1 − (

33.2104.3

)3](6, 130) = 6, 209 in4

Ieff, sust =(

33.235.7

)3 (10, 916) +[1 − (

33.235.7

)3](6, 130) = 9, 993 in4

7. DeflectionsE = 57

√3, 000 = 3, 120 ksi

δ = 5384

wL4

EI

δi, short + sust = 5384

(1.27+2.44)[(15)(12)]4

(3,120)(6,209) = .218 in

δi, sust = 5384

(1.27)[(15)(12)]4

(3,120)(9,993) = .046 in

δi = .218 − .046 = .172 in

8. δcreep = λδi, sust

λ =2.

1 + 0= 2. ⇒ δcreep = (2)(.046) = .092 in


Draft

Chapter 7

APPROXIMATE FRAMEANALYSIS

1 Despite the widespread availability of computers, approximate methods of analysis are justi-fied by

1. Inherent assumption made regarding the validity of a linear elastic analysis vis a vis ofan ultimate failure design.

2. Ability of structures to redistribute internal forces.

3. Uncertainties in load and material properties

2 Vertical loads are treated separately from the horizontal ones.

3 We use the design sign convention for moments (+ve tension below), and for shear (ccw +ve).

4 Assume girders to be numbered from left to right.

5 In all free body diagrams assume positivee forces/moments, and take algeebraic sums.

7.1 Vertical Loads

6 The girders at each floor are assumed to be continuous beams, and columns are assumed toresist the resulting unbalanced moments from the girders.

7 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 columnsabove and below the floor.

Draft7–2 APPROXIMATE FRAME ANALYSIS

8 Based on the first assumption, all beams are statically determinate and have a span, Ls

equal to 0.8 the original length of the girder, L. (Note that for a rigidly connected member, theinflection point is at 0.211 L, and at the support for a simply supported beam; hence, dependingon the nature of the connection one could consider those values as upper and lower bounds forthe approximate location of the hinge).

9 End forces are given by

Maximum positive moment at the center of each beam is, Fig. 7.1

0.1LL

0.1L0.8L

Vrgt

Vlft

MM rgt

lft

w

��

��

��

��

��

��

Figure 7.1: Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments

M+ =18wL2

s = w18(0.8)2L2 = 0.08wL2 (7.1)

Maximum negative moment at each end of the girder is given by, Fig. 7.1

M left = M rgt = −w

2(0.1L)2 − w

2(0.8L)(0.1L) = −0.045wL2 (7.2)

Girder Shear are obtained from the free body diagram, Fig. 7.2

V lft =wL

2V rgt = −wL

2(7.3)

Column axial force is obtained by summing all the girder shears to the axial force transmit-ted by the column above it. Fig. 7.2

P dwn = P up + V rgti−1 − V lft

i(7.4)


Draft7.1 Vertical Loads 7–3

Vrgti−1 Vlft

i

Pabove

Pbelow

Figure 7.2: Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces

Mcolbelow

Mi−1rgt M i

lftM col

above

Li

h/2

h/2

h/2

h/2

Li−1

Vi−1lft V i

lft Vi

rgtVi−1

rgt

Mi−1lft

Mi

rgt

Figure 7.3: Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments



Column Moment are obtained by considering the free body diagram of columns Fig. 7.3

M top = M botabove − M rgt

i−1 + M lfti M bot = −top (7.5)

Column Shear Points of inflection are at mid-height, with possible exception when the columnson the first floor are hinged at the base, Fig. 7.3

V =M top

h2

(7.6)

Girder axial forces are assumed to be negligible eventhough the unbalanced column shearsabove and below a floor will be resisted by girders at the floor.

7.2 Horizontal Loads

10 We must differentiate between low and high rise buildings.

Low rise buidlings, where the height is at least samller than the hrizontal dimension, thedeflected shape is characterized by shear deformations.

High rise buildings, where the height is several times greater than its least horizontal di-mension, the deflected shape is dominated by overall flexural deformation.

7.2.1 Portal Method

11 Low rise buildings under lateral loads, have predominantly shear deformations. Thus, theapproximate analysis of this type of structure is based on

1. Distribution of horizontal shear forces.

2. Location of inflection points.

12 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 dis-tributed among these columns so that each of the two exterior columns carry half asmuch horizontal shear as each interior columns of the frame.

13 Forces are obtained from


Draft7.2 Horizontal Loads 7–5

H/2 H/2HH

Figure 7.4: Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear

Column Shear is obtained by passing a horizontal section through the mid-height of thecolumns at each floor and summing the lateral forces above it, then Fig. 7.4

V ext =∑

F lateral

2No. of baysV int = 2V ext (7.7)

Column Moments at the end of each column is equal to the shear at the column times halfthe height of the corresponding column, Fig. 7.4

M top = Vh

2M bot = −M top (7.8)

Girder Moments is obtained from the columns connected to the girder, Fig. 7.5

M colbelow

M i−1rgt

M ilft

M col

above

Li−1/2 Li /2h/2

h/2

h/2

h/2

Li /2Li−1/2

Vi−1lft V

ilft V i

rgtV

i−1

rgt

M i−1lft

M irgt

Figure 7.5: 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(7.9)

Girder Shears Since there is an inflection point at the center of the girder, the girder shearis obtained by considering the sum of moments about that point, Fig. 7.5

V lft = −2M

LV rgt = V lft (7.10)

Column Axial Forces are obtained by summing girder shears and the axial force from thecolumn above, Fig. ??

Vrgti−1 Vlft

i

Pabove

Pbelow

Figure 7.6: Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force

P = P above + P rgt + P lft (7.11)

Example 7-1: Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads

Draw the shear, and moment diagram for the following frame. Solution:

Vertical Loads



��

14

161 2 3 4

5 6 7 8

9 10

12 14

11

20’ 30’ 24’

0.25 k/ft

0.5 k/ft13

30 k

15 k

Figure 7.7: Example; Approximate Analysis of a Building

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.ftM lft

13 = −0.045w13L213 = −(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.ftM lft

14 = −0.045w14L214 = −(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.ftM lft

10 = −0.045w10L210 = −(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.ftM lft

11 = −0.045w12L212 = −(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.ftM bot

5 = −M top5 = 4.5 k.ft

M top6 = −M rgt

12 + M lft13 = −(−4.5) + (−10.1) = − 5.6 k.ft

M bot6 = −M top

6 = 5.6 k.ftM top

7 = −M rgt13 + M lft

14 = −(−10.1) + (−6.5) = − 3.6 k.ftM bot

7 = −M top7 = 3.6 k.ft

M top8 = −M rgt

14 = −(−6.5) = 6.5 k.ftM bot

8 = −M top8 = − 6.5 k.ft



4. Bottom Column Moments

M top1 = +M bot

5 + M lft9 = 4.5 − 9.0 = − 4.5 k.ft

M bot1 = −M top

1 = 4.5 k.ftM top

2 = +M bot6 − M rgt

9 + M lft10 = 5.6 − (−9.0) + (−20.3) = − 5.6 k.ft

M bot2 = −M top

2 = 5.6 k.ftM top


10 + M lft11 = −3.6 − (−20.3) + (−13.0) = 3.6 k.ft

M bot3 = −M top

3 = − 3.6 k.ftM top


11 = −6.5 − (−13.0) = 6.5 k.ftM bot

4 = −M top4 = − 6.5 k.ft

5. Top Girder Shear

V lft12 = w12L12

2 = (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

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

8. Top Column Axial Forces

P5 = V lft12 = 2.50 k

P6 = −V rgt12 + V lft

13 = −(−2.50) + 3.75 = 6.25 kP7 = −V rgt

13 + V lft14 = −(−3.75) + 3.00 = 6.75 k

P8 = −V rgt14 = 3.00 kVictor Saouma Mechanics and Design of Reinforced Concrete


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 7.8: Approximate Analysis of a Building; Moments Due to Vertical Loads



+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 7.9: Approximate Analysis of a Building; Shears Due to Vertical Loads



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 kP3 = P7 − V rgt

11 + V lft10 = 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 kV7 = 2(V5) = (2)(2.5) = 5 kV8 = V5 = 2.5 kV1 = 15+30

(2)(3) = 7.5 k

V2 = 2(V1) = (2)(7.5) = 15 kV3 = 2(V1) = (2)(2.5) = 15 kV4 = V1 = 7.5 k

2. Top Column Moments

M top5 = V1H5

2 = (2.5)(14)2 = 17.5 k.ft

M bot5 = −M top

5 = − 17.5 k.ft

M top6 = V6H6

2 = (5)(14)2 = 35.0 k.ft

M bot6 = −M top

6 = − 35.0 k.ft

M top7 = V up

7 H7

2 = (5)(14)2 = 35.0 k.ft

M bot7 = −M top

7 = − 35.0 k.ft

M top8 = V up

8 H8

2 = (2.5)(14)2 = 17.5 k.ft

M bot8 = −M top

8 = − 17.5 k.ft

3. Bottom Column Moments

M top1 = V dwn

1 H1

2 = (7.5)(16)2 = 60 k.ft

M bot1 = −M top

1 = − 60 k.ft

M top2 = V dwn

2 H2

2 = (15)(16)2 = 120 k.ft

M bot2 = −M top

2 = − 120 k.ft

M top3 = V dwn

3 H3

2 = (15)(16)2 = 120 k.ft

M bot3 = −M top

3 = − 120 k.ft

M top4 = V dwn

4 H4

2 = (7.5)(16)2 = 60 k.ft

M bot4 = −M top

4 = − 60 k.ft

4. Top Girder Moments

M lft12 = M top

5 = 17.5 k.ftM rgt

12 = −M lft12 = − 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.ftM lft

14 = M rgt13 + M top

7 = −17.5 + 35 = 17.5 k.ftM rgt

14 = −M lft14 = − 17.5 k.ft



Approximate Analysis Vertical Loads APROXVER.XLS Victor E. Saouma

AAAAAAAAAAAAAAAAAAAAAAAAAAAA




AAAAAAA



















AAAAAAA





AAAAAAAAAAAAAAAAAAAAAAAA








AAAAAAAAAAAA









AAAAAAAAAAAAAAAAAA


















AAAAAAAAAAAAAA









AAAAAAAAAAAAAAAAAAAAA










AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA




AAAAAAAA



















AAAAAAAA













AAAAAAAAAAAA









AAAAAAAAAAAAAAAAAA


















AAAAAAAAAAAAAA























AAAAAAA

























AAAAAAA
















AAAAAAAAAAAAAA























AAAAAAA

























AAAAAAA
















AAAAAAAAAAAAAA























AAAAAAA



















AAAAAAA













AAAAAAAAAAAAAA























AAAAAAAA



















AAAAAAAA













AAAAAAAAAAAA









AAAAAAAAAAAAAAAAAA










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 7.10: Approximate Analysis for Vertical Loads; Spread-Sheet Format



Ap

pro

xim

ate

An

aly

sis V

ert

ica

l Lo

ad

sA

PRO

XV

ER.X

LSV

icto

r E. S

ao

um

a

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAA

AAA

AAA

AAA

AAA

AAA

AAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAA

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 7.11: Approximate Analysis for Vertical Loads; Equations in Spread-Sheet



5. Bottom Girder Moments

M lft9 = M top

1 − M bot5 = 60 − (−17.5) = 77.5 k.ft

M rgt9 = −M lft

9 = − 77.5 k.ftM lft

10 = M rgt9 + M top

2 − M bot6 = −77.5 + 120 − (−35) = 77.5 k.ft

M rgt10 = −M lft

10 = − 77.5 k.ftM lft

11 = M rgt10 + M top

3 − M bot7 = −77.5 + 120 − (−35) = 77.5 k.ft

M rgt11 = −M lft

11 = − 77.5 k.ft

+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 7.12: Approximate Analysis of a Building; Moments Due to Lateral Loads



6. Top Girder Shear

V lft12 = −2M lft

12L12

= − (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

V lft14 = −2M lft

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

V lft10 = −2M lft

10L10

= − (2)(77.5)30 = −5.17 k

V rgt10 = +V lft

10 = −5.17 k

V lft11 = −2M lft

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 kP7 = +V rgt

13 − V lft14 = −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 kP3 = P7 + V rgt

11 + V lft10 = 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 now seek the design parameters for the frame, Table 7.2.



Portal Method PORTAL.XLS Victor E. Saouma





AAAAAAA























AAAAAAA









































AAAAAAA























AAAAAAA









































AAAAAA









AAAAAAAAAAAAAAAAAA














AAAAAA









































AAAAAAA























AAAAAAA









































AAAAAAA























AAAAAAA























AAAAAAAA























AAAAAAAA



















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 7.13: Portal Method; Spread-Sheet Format


Draft7.2 Horizontal Loads 7–17Portal Method PORTAL.XLS Victor E. Saouma

AAAAAAAAAAAAAAAAAAAA











AAAAAAAAAAAAAAA







AAAAAAAAAAAAAAA




AAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAA












AAAAAAAAAA







AAAAAAAAAAAAAAA












AAAAAAAAAAAAAAA







AAAAAAAAAAAAAAA




AAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAA












AAAAAAAAAA







AAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAA












AAAAAAAAAA







AAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAA












AAAAAAAAAAAAAAA







AAAAAAAAAAAAAAA




AAAAAAAAAA












AAAAAAAAAA







AAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAAAAAA












AAAAAAAAAAAAAAA







AAAAAAAAAAAAAAA




AAAAAAAAAA












AAAAAAAAAA







AAAAAAAAAAAAAAA












AAAAAAAAAAAAAAA







AAAAAAAAAAAAAAA




AAAAAAAAAA












AAAAAAAAAA







AAAAAAAAAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAA

AAAAAAAAAAA

AAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAAAAAA

AAAAAAAA

AAAAAA

AAAAAAAAAAA



AAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA














AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAA


AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AA

AAAAAAAAAAAAAA

1

23

4

5

67

8

910

11

1213

14

15

16

17

18

1920

21

2223

24

25

26

27

28

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

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAAAAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

20 30 24

MOMENTSAAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAAAAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

AAAAAAAA

# of Storeys 2 Bay 1 Bay 2 Bay 3AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

Force Shear Col Beam Column Beam Column Beam ColAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAA

AAAAAAAA

AAAA

AAAA

AAAA

AAAA

AAAA

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

AAAAAAAAAAAA

Bay 1 Bay 2 Bay 3AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAAAAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

Col Beam Column Beam Column Beam ColAAAA

AAAA

AAAA

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

AAAAA

AAAAA

AAAAA

AAAAAAAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

AAAAA

Col Beam Column Beam Column Beam ColAAAA

AAAA

AAAA

AAAAAAAA

AAAA

0 0 0

=-I18 =+J18-M18 =+N18-Q18 =+R18

0 0 0

=+H28-I21 =+K28+J21-M21 =+O28+N21-Q21 =+S28+R21AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA





Figure 7.14: 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 7.1: Columns Combined Approximate Vertical and Horizontal Loads



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 7.2: Girders Combined Approximate Vertical and Horizontal Loads




Draft

Chapter 8

COLUMNS

Draft

Chapter 9

COLUMNS

9.1 Introduction

1 Columns resist a combination of axial P and flexural load M , (or M = Pe for eccentricallyapplied load).

9.1.1 Types of Columns

Types of columns, Fig. 9.1

Tied column

Spiral column

Composite colu

Pipe column

tie steel

main longitudinal steel reinforcement

Figure 9.1: Types of columns

2 Lateral reinforcement, Fig. 9.2

P

Spiral

TiedX X

δ

Figure 9.2: Tied vs Spiral Reinforcement

1. Restrains longitudinal steel from outward buckling

2. Restrains Poisson’s expansion of concrete

3. Acts as shear reinforcement for horizontal (wind & earthquake) load

4. Provide ductility

Draft9–2 COLUMNS

very important to resist earthquake load.

9.1.2 Possible Arrangement of Bars

3 Bar arrangements, Fig. 9.3

4 bars 6 bars 8 bars

10 bars

12 bars

14 bars

16 bars

Wall column

Corner column

Figure 9.3: Possible Bar arrangements

9.2 Short Columns

9.2.1 Concentric Loading

4 No moment applied,

Elastic Behaviour

P = fcAc + fsAs

= fc(Ac + nAs)

Ultimate Strength

Pd = φPn

Pn = .85f ′cAc + fyAs

note:

1. 0.85 is obtained also from test data

2. Matches with beam theory using rect. stress block

3. Provides an adequate factor of safety

9.2.2 Eccentric Columns

5 Sources of flexure, Fig. 9.4

1. Unsymmetric moments ML 6= MR

2. Uncertainty of loads (must assume a minimum eccentricity)


Draft9.2 Short Columns 9–3

ML M

R

Pe

e= −

nP

nP

RM

LM

RM

LM

Figure 9.4: Sources of Bending

3. Unsymmetrical reinforcement

6 Types of Failure, Fig. 9.5

1. Large eccentricity of load ⇒ failure by yielding of steel

2. Small eccentricity of load ⇒ failure by crushing of concrete

3. Balanced condition

M n

Pn

e small

e =

0; a

= h

; c =

c ~ h; e=

ec

Compressionfailure range

Radial lines show

Balanced Failure

Tension failure range

Load path forgivin e

e large

c

e

c

cb

constant e=M n

Pn

8

8 M0

P0

cuε

ε

εy

cu

εy

εsu

ε cu

>

ε cu

Figure 9.5: Load Moment Interaction Diagram

7 Assumptions A′s = As; ρ = As

bd = A′s

bd ; f ′s = fy

9.2.2.1 Balanced Condition

8 There is one specific eccentricity eb = MP such that failure will be triggered by simultaneous

1. Steel yielding

2. Concrete crushing


Draft9–4 COLUMNS

9 From the strain diagram (and compatibility of concrete and steel strains), Fig. 9.6

s’ε

bs s

A A’

h/2d

a

d’

y

y

’sA f

sA f

c

sε

c

ssA fc

nP

c0.85f’

e’e

s’sA f

sε

ssA f

Figure 9.6: Strain and Stress Diagram of a R/C Column

εc = .003 (9.1-a)

εy =fy

Es(9.1-b)

c =εu

εu + εyd =

.003fy

Es+ .003

d (9.1-c)

Furthermore,

ε′sc − d′

=εc

c(9.2-a)

⇒ ε′s =c − d′

cεc (9.2-b)

thus the compression steel will be yielding (i.e. ε′s = εy) for εc = .003 and d′ = 2 in if c > 6 in

10 Equilibrium (neglecting ceter steel for now):

Pn = .85f ′cab + A′

sf′y − Asfs

fs = fy

As = A′s

a = Pn.85f ′

cb

a = β1cb

cb = .003fyEs

+.003d

Pn,b = .85β1f

′cbd

.003fy

Es+ .003

(9.3)



or

Pnb = .85β1f′cbd

87, 000fy + 87, 000

(9.4)

11 To obtain Mnb we take moment about centroid of tension steel As of internal forces, thismust be equal and opposite to the externally applied moment, Fig. 9.6.

Mnb = Pnbeb︸︷︷︸Mext

= .85f ′cab(d − a

2) + A′

sfy(d − d′)︸︷︷︸Mint

(9.5)

12 Note: Internal moments due to Asfy and A′sfy cancel each other for symmetric columns.

9.2.2.2 Tension Failure

Case I, e is known and e > eb

In this case a and Pn are unknowns, and for failure to be triggered by fy in As we musthave e > eb.

Can still assume Asfy = A′sfy

ΣFy = 0 ⇒ Pn = .85f ′cab ⇒ a =

Pn

.85f ′cb

(9.6-a)

ΣM = 0 ⇒ Pne′ = Pn(d − a

2) + A′

sf′y(d − d′) (9.6-b)

Two approaches

1. Solve iteratively for those two equations

(a) Assume a (a < h2 )

(b) From strain compatibility solve for fsc, center steel stress if applicable.(c) ΣFy = 0 ⇒ solve for Pn

(d) ΣM = 0 with respect to tensile reinforcement,⇒ solve for Pn

(e) If no convergence among the two Pn, iterate by solving for a from ΣFy = 0

2. Combine them into a quadratic equation in Pn

Pn = .85f ′cbd

−ρ − (

e′

d− 1) +

√(1 − e′

d

)2

+ 2ρ

[µ

(1 − d′

d

)+

e′

d

] (9.7)

whereρ = As

bd = A′s

bd

µ = fy

.85f ′c


Draft9–6 COLUMNS

Case II c is known and c < cb; Pn is unknown

In this case, we only have two unknown, Pn and f ′s.

a = β1c (9.8-a)

fsdef= fy (9.8-b)

f ′s = εuEs

c − d′

c≤ fy (9.8-c)

C = 0.85f ′cab (9.8-d)

Pn = C + A′sf

′s − Asfy (9.8-e)

Mn = Ch − a

2+ A′

sf′s

(h

2− d′

)+ Asfs

(d − h

2

)(9.8-f)

e =Mn

Pn(9.8-g)

Note this approach is favoured when determining the interaction diagram.

9.2.2.3 Compression Failure

Case I e is known and e < eb; Pn, a and fs are unknown

Compression failure occurs if e′ < e′b ⇒ εu = .003, assume f ′s = fy, and fs < fy

From geometry

c =εu

fs

Es+ εu

d

⇒ fs = Esεud − c

c

= Esεu

d − aβ1

aβ1

(9.9-a)

Pn = .85f ′cab + A′

sfy − Asfs (9.9-b)

Pne′ = .85f ′cab(d − a

2) + A′

sfy(d − d′) (9.9-c)

this would yield a cubic equation in Pn, which can be solved analytically or by iteration.

1. Assume a (a ' h)

2. Solve for ΣM = 0 with respect to tensile reinforcement & solve for Pn

3. From strain compatibility solve for fs

4. Check that ΣFy = 0 & solve for a

5. If ai+1 6= ai go to step 2

Case II: c is known and c > cb; fs, f ′s, and Pn are unknown



In this case

a = β1c (9.10-a)

fs = εcEsd − c

c≤ fy (9.10-b)

f ′s = εcEs

c − d′

c≤ fy (9.10-c)

C = 0.85f ′cab (9.10-d)

Pn = C + Asfs + A′sf

′s (9.10-e)

Mn = Ch − a

2+ A′

sf′s

(h

2− d′

)+ Asfs

(d − h

2

)(9.10-f)

9.2.3 ACI Provisions

1. Governing equations

ρmin = 1% ACI 10.9.1ρmax = 8%

ρs = 0.45(Ag

Ac− 1) f ′

cfy

ACI 10.5φ = 0.7 for tied columnsφ = 0.75 for spiral columns

(9.11)

whereρs minimum ratio of spiral reinforcementAg gross area of sectionAc area of core

2. A minimum of 4 bars for tied circular and rect

3. A minimum of 6 bars for spirals (ACI10.9.2)

4. φ increases linearly to 0.9 as φPn decreases from 0.10f ′cAg or φP0, whichever is smaller,

to zero (ACI 9.3.2).

5. Maximum strength is 0.8φP0 for tied columns (φ = 0.7) and 0.85φP0 for spirally reinforcedcolumns (φ = 0.75).

9.2.4 Interaction Diagrams

13 Each column is characterized by its own interaction diagram, Fig. 9.7

9.2.5 Design Charts

14 To assist in the design of R.C. columns, design charts have been generated by ACI in termof non dimensionalized parameters χ = Pn

bhf ′cvs Mn

bh2f ′c

= χe

h for various ρt where ρt = As+A′s

bh and

µ = fy

.85f ′c

Example 9-1: R/C Column, c known


Draft9–8 COLUMNS

0 φM n M n M

0.10f’ cA g

eb

Pd−M d

Pn−M n (M n’ Pn)

(Mnb’

Pnb )e min

φPn(max)

Pn(max)

P0

P

A

Tied: Pn(max)

n(max)

= 0.80 P

= 0.85 P0

0Spir. reinf: P

1e

Compressioncontrol region

Tensioncontrol region

Balanced failure

e~h; e = infty

e=0;

a=

h; c

= in

fty

Figure 9.7: Column Interaction Diagram

A 12 by 20 in. column is reinforced with four No. 4 bars of area 1.0 in2 each, at eachcorner. f ′

c = 3.5 ksi, fy = 50 ksi, d′ = 2.5 in. Determne: 1) Pb and Mb; 2) The load and momentfor c = 5 in; 3) load and moment for c = 18 in.Solution:

Balanced Conditions is derived by revisiting the fundamental equations, rather than meresubstitution into previously derived equation.

d = h − d′ = 20 − 2.5 = 17.5 in (9.12-a)

cb =.003

fy

Es+ .003

d =.003

5029,000 + .003

17.5 = 11.1 in (9.12-b)

a = β1cb = (0.85)(11.1) = 9.44 in (9.12-c)

fsdef= fy = 50 ksi (9.12-d)

f ′s = Es

c − d′

cεc = (29, 000)(

11.1 − 2.511.1

(0.003) = 67.4 ksi > fy ⇒ f ′s = 50 ksi(9.12-e)

C = 0.85f ′cab = (0.85)(3.5)(9.44)(12) = 337 k (9.12-f)

Pnb = C + Asfs − A′sfs = 337 + (2.0)(50) + (2.0)(−50) = 337 k (9.12-g)

Mnb = Pnbe′ = .85f ′

cab(d − a

2) + A′

sfy(d − d′) (9.12-h)

= 337(

17.5 − 9.442

)+ (2.0)(50)(17.5 − 2.5) = 5, 807 k.in = 484 k.ft (9.12-i)

eb =5, 807337

= 17.23 in (9.12-j)

Tension failure, c = 5 in

fsdef= fy = 50 ksi (9.13-a)



f ′s = εcEs

c − d′

c≤ fy (9.13-b)

= (0.003)(29, 000)5.0 − 2.5

5.0= 43.5 ksi (9.13-c)

a = β1c = 0.85(5.0) = 4.25 in (9.13-d)C = 0.85f ′

cab (9.13-e)= (0.85)(3.5)(4.25)(12) = 152 k (9.13-f)

Pn = C + A′sf

′s − Asfy (9.13-g)

= 152 + (2.0)(43.5) − (2.0)(50) = 139 k (9.13-h)

Mn = Ch − a

2+ A′

sf′s

(h

2− d′

)+ Asfs

(d − h

2

)about section centroid (9.13-i)

= (152)(

20 − 4.252

)+ (2.0)(43.5)

(202

− 2.5)

+ (2.0)(50)(

17.5 − 202

)(9.13-j)

= 2, 598 k.in = 217 k.ft (9.13-k)

e =2, 598139

= 18.69 in (9.13-l)

Compression failure, c = 18 in

a = β1c = 0.85(18) = 15.3 in (9.14-a)

fs = εcEsd − c

c≤ fy (9.14-b)

= (0.003)(29, 000)17.5 − 18.0

18.0= −2.42 ksi As is under compression (9.14-c)

f ′s = εcEs

c − d′

c≤ fy (9.14-d)

= (0.003)(29, 000)18.0 − 2.5

18.0= 75 ksi > fy ⇒ f ′

s = 50 ksi (9.14-e)

C = 0.85f ′cab = (0.85)(3.5)(15.3)(12) = 546 k (9.14-f)

Pn = C + A′sf

′s − Asfs (9.14-g)

= 546 + (2.0)(50) − (−2.42)(2) = 650 k (9.14-h)

Mn = Ch − a

2+ A′

sf′s

(h

2− d′

)+ Asfs

(d − h

2

)about section centroid (9.14-i)

= (546)(

20 − 15.32

)+ (2.0)(50)

(202

− 2.5)

+ (2.0)(−2.42)(

17.5 − 202

)(9.14-j)

= 2, 000 k.in = 167 k.ft (9.14-k)

e =2, 000650

= 3.07 in (9.14-l)

Example 9-2: R/C Column, e known


Draft9–10 COLUMNS

For the following column, determine eb, Pb, Mb; Pn and Mn for e = 0.1h and e = h.f ′

c = 3, 000 psi and fy = 40, 000 psi. The area of each bar is 1.56 in2.

3" 3"

24"

20"

12"

y

Cc

.003

cε

Balanced Condition:

εy =fy

Es=

4029, 000

= .001379 (9.15-a)

cb =εu

εu + εyd =

.003.003 + .001379

.003 = 14.4 in (9.15-b)

a = β1cb = (.85)(14.4) = 12.2 in (9.15-c)Cc = .85f ′

cab = (.85)(3)(12.2)(20) = 624 k (9.15-d)

εsc =c − h/2

cεu =

14.4 − 1214.4

.003 = .0005 (9.15-e)

fsc = (29, 000)(.0005) = 15 ksi center bars (9.15-f)Cs = (.0005)(29, 000)(2)(1.56) = 46.8 k (9.15-g)

Pnb = 624 + 46.8 = 670.8 k (9.15-h)

Note that the compression steel is yielding because d′ > 2” and c > 6” (as previouslyproven)

Taking moment about centroid of section

Mnb = Pnbe (9.16-a)

= .85f ′cab

(h

2− a

2

)+ Asfy

(h

2− d′

)+ A′

sfy

(h

2− d′

)(9.16-b)

= (.85)(3)(12.2)(20)(

12 − 12.22

)+ 4(1.56)(9)(40)

+4(1.56)(40 − .85 × 3)(12 − 3) (9.16-c)= 3, 671 + 2, 246 + 2, 246 (9.16-d)= 8, 164 k.in; 680 k.ft (9.16-e)

eb =8, 164670.8

= 12.2 in (9.16-f)



e= .1 h e = (.1)(24) = 2.4 in < eb ⇒ failure by compression. Pn, a and fs are unknown.Available equations: 1) ΣF = 0; 2) ΣM = 0; and 3) strain diagram; Solve by iterations.

c=23.5"

A fA’ f

a=20" a/2

A f

h−c−

d’

ε εεyscs

3" 3"

24"

20"

12"

y

P

C

n

ssc c

c

s s

ysc

9"

e’=11.4"

.85f’

h/2=12"

e=2.

4"

.003

ε

1. Assume a = 20 in

c =a

β1=

20.85

= 23.5 in (9.17)

2. For center steel (from geometry)εsc

c − h2

=.003

c(9.18-a)

⇒ εsc =c − h

2

c.003 (9.18-b)

fsc = Esεsc (9.18-c)

= Esc − h

2

c.003 (9.18-d)

= 29, 00023.5 − 12

23.5.003 = 42.5 ksi > fy ⇒ fsc = fy (9.18-e)


Draft9–12 COLUMNS

3. Take moment about centroid of tensile steel bar

Pne′ = 0.85f ′cab(d − a

2) + A′

sfy(h − 2d′) + Ascfy(h

2− d′) (9.19-a)

Pn(9 + 2.4) = (.85)(3)(20)(20)(21 − 202

) + 4(1.56)(40)(24 − 6) + 2(1.56)(40)(9)(9.19-b)

= 11, 220 + 4, 493 + 259.7 (9.19-c)⇒ Pn = 1, 476 k (9.19-d)

4. Get εs in tension bar (from strain diagram)

εs

h − d′ − 23.5=

.003c

(9.20-a)

⇒ εs =.00323.5

(24 − 3 − 23.5) (9.20-b)

= −.000319 (9.20-c)fs = Eεs = (29, 000)(0.000319) = 9.25 ksi (9.20-d)

5. Take ΣF = 0 to check assumption of a

Pn = 0.85f ′cab + A′

sfy + Ascfsc + Asfy (9.21-a)1, 476 k = (.85)(3)(a)(20) + (4)(1.56)(40) + (2)(1.56)(40) + (4)(1.56)(9.25)(9.21-b)

1, 476 = 51a + 432.1 (9.21-c)⇒ a = 20.4 in

√(9.21-d)

Pn = 1, 476 k (9.21-e)

Mn = (1, 476)(2.4) = 3, 542 k.in = 295 k.ft (9.21-f)

e=h 1. In this case e = 24 in > eb ⇒ failure by tension. Pn and a are unknown. We havetwo equations: 1) ΣF = 0, and 2) ΣM = 0.

2. Assume a = 7.9 in ⇒ c = aβ1

= 7.9.85 = 9.3 in

3. Steel stress at centroid

c

.003=

12 − c

εsc(9.22-a)

⇒ εsc =12 − 9.3

9.3.003 = .00087 (9.22-b)

⇒ fsc = (29, 000)(0.00087) = 25.3 ksi (9.22-c)

4. Iterate

ΣF = 0 ⇒ Pn = (.85)f ′cab + Ascfsc (9.23-a)

= (.85)(3)(7.9)(20) − 2(1.56)(25.3) (9.23-b)= 403 − 79 = 324 k (9.23-c)

ΣM = 0 ⇒ Pn(e + h/2 − d′) = .85f ′cab(d − a

2) + A′

sfy(d − d′)

−Ascfsc(d − d′

2) (9.23-d)



Pn(24 + 9) = (.85)(3)(7.9)(20)(21 − 7.92

) + 4(1.56)(40)(21 − 3)

+2(1.56)(25.3)(9) (9.23-e)Pn(33) = 6, 870 + 4, 493 − 710 = 10, 653 k.in (9.23-f)⇒ Pn = 323 k

√(9.23-g)

5. Determine Mn

Mn = Pne = (323)(24) = 7, 752 k.in = 646 k.ft (9.24)

Example 9-3: R/C Column, Using Design Charts

Design the reinforcement for a column with h = 20 in, b = 12 in, d′ = 2.5 in, f ′c = 4, 000 psi,

fy = 60, 000 psi, to support PDL = 56 k, PLL = 72 k, MDL = 88 k.ft, MLL = 75 k.ft,Solution:

1. Ultimate loads

Pu = (1.4)(56) + (1.7)(72) = 201 k ⇒ Pn =2010.7

= 287 k (9.25-a)

Mu = (1.4)(88) + (1.7)(75) = 251 k.ft ⇒ Mn =2510.7

= 358 k.ft (9.25-b)

2. Chart parameters

e

h=

(358)(12)(287)(20)

= 0.75 (9.26-a)

γ =h − 2d′

h

20 − (2)(2.5)20

= 0.75 ⇒ interpolate between A3 and A4(9.26-b)

κ =Pn

bhf ′c

=287

(12)(20)(4)= 0.3 (9.26-c)

κe

h= (0.3)(0.75) = 0.225 (9.26-d)

3. Interpolating between A.3 and A.4 ⇒ ρtµ = 0.4

4. Reinforcement

ρt = Atbh

µ = fy

.85f ′c

}At =

(0.4)(b)(h)(.85)(f ′c)

fy= (0.4)(12)(20)(.85)(4)

1(60)

= 5.45 in2(9.27-a)

⇒ use 4 # 9 & 2 # 8, ⇒ At = 5.57 in2


Draft9–14 COLUMNS

9.2.6 Biaxial Bending

15 Often columns are subjected to biaxial moments (such as corner columns)

16 An exact approach entails the trial and eror determination of an inclined neutral axis, thisis an exact method but too cumbersome to use in practice.

17 Hence, we seek an approximate solution, the most widely used method is the load contourmethod or Bresler-Parme method.

18 The failure surface of a biaxialy loaded column is shown in Fig. 9.8, and the general nondi-

��

��

Pn

M 0 x

M 0 y

M n y

M n x

Figure 9.8: Failure Surface of a Biaxially Loaded Column

mensional equation for the moment contour at a constant Pn may be expressed as(Mnx

M0x

)α1

+(

Mny

M0y

)α1

= 1.0

whereMnx = Pney

Mny = Pnex

M0x = Mnx capacity at axial load Pn when Mny (or ex) is zeroM0y = Mny capacity at axial load Pn when Mnx (or ey) is zero

and α1 and α2 are exponent which depend on geometry and strength.

19 Bresler suggested that we set α1 = α2 = α. For practical purposes, a value of α = 1.5 forrectangular columns, and between 1.5 and 2.0 for square sections has proven acceptable.

20 An improvement of Bresler equation was devised by Parme. The main assumption is that atany load Pn, Fig. 9.9

Mny

Mnx=

M0y

M0x

orMnx = βM0x; Mny = βM0y



1.0

1.0

β

β

AA

BB

CC

M ny /M0yMny

M0y

M0y

Mnx /M0x

Mnx

M0x

M0x

βM0x

βM0y

45 o

Figure 9.9: Load Contour at Plane of Constant Pn, and Nondimensionalized Correspondingplots

21 Thus, β is the portion of the uniaxial moment strength permitted to act simultaneously onthe column section. It depends on the cross section, strength, and layout.

22 The usual range is between 0.55 and 0.70, with a recommended value of 0.65 for design.

23 Hence, once β is selected, we can substitute in Bresler’s equation(β M0x

M0x

)α+

(β

M0y

M0y

)α= 1.0

βα = 12

α log β = log 0.5α = log 0.5

log β

thus, (Mnx

M0x

)log 0.5/logβ

+(

Mny

M0y

)log 0.5/logβ

= 1.0 (9.28)

24 Effect of β is shown in Fig. 9.10.

25 Gouwens proposed to replace the above curves, by a bilinear model, Fig. 9.11

Review of a section

Mny

M0y+

Mnx

M0x

(1 − β

β

)= 1 If

Mny

M0y≥ Mnx

M0x(9.29)

Mnx

M0x+

Mny

M0y

(1 − β

β

)= 1 If

Mny

M0y≤ Mnx

M0x(9.30)

Design of a column

Mny + Mnx

(M0y

M0x

) (1 − β

β

)= M0y If

Mny

Mnx≥ M0y

M0x(9.31-a)

Mnx + Mny

(M0x

M0y

) (1 − β

β

)= M0x If

Mny

Mnx≤ M0y

M0x(9.31-b)


Draft9–16 COLUMNS

0.0 0.2 0.4 0.6 0.8 1.0

Mnx/M0x

0.0

0.2

0.4

0.6

0.8

1.0

Nny

/M0y

Biaxial Bending Interaction Diagram

beta=0.50

0.55

0.60

0.65

0.70

0.75

0.800.85

0.90

Figure 9.10: Biaxial Bending Interaction Relations in terms of β

��

��

Pn

M 0 x

M 0 y

M n y

M n x

45o

1−β

1−β

(1−β/ )

(1− β/ )

β

β

β

ββ

β

B

C

A

Mny

/M0y

M ny /M 0y

M ny /M 0y

M nx /M 0x

M nx /M 0x

(Mnx

/M 0x)α+(M ny /M 0y)α=1

M nx /M 0x

1.0

1.0

+

+

=1

=1

Figure 9.11: Bilinear Approximation for Load Contour Design of Biaxially Loaded Columns



26 Note, circular or square columns with symmetric reinforcement should always be consideredfirst for biaxially loaded columns.

Example 9-4: Biaxially Loaded Column

Determine the adequacy of a 16 in. square tied column with 8 # 9 bars. d′ = 2.5 in, andthere are 3 bars on each side. The section is to carry factored loads of Pu = 144 k, Mux = 120 k.ftand Muy = 54 k.ft, f ′

c = 3 ksi and fy = 40 ksi. P0 = 952 k, M0x = M0y = 207 k.ft (we have asymmetrical reinforcement).Solution:

ey = MuxPu

= (120)(12)144 = 10.0 in

ex = Muy

Pu= (54)(12)

144 = 4.5 in

The interaction diagram for e = 10 in, e = 4.5 in and e = 0 will give Pn equal to 254, 486, and952 kips respectively.

The required load Pn = 1440.7 = 205 k, the corresponding moments are M0x = M0y = 207 k.ft

from the interaction diagram. Using β = 0.65

Required Mnx

M0x=

1200.7207 = 0.828

Required Mny

M0y=

540.7207 = 0.373

We shall use both solutions

Bresler-Parme which is exact solution

log(0.5)log β = log 0.5

log 0.65 = 1.609(MnxM0x

)log 0.5/logβ+

(Mny

M0y

)log 0.5/logβ=

(0.828)1.609 + (0.373)1.609 = 0.943√

Note that we could have first solved for MnxM0x

, and then determined Mny

M0yfrom Fig. 9.10.

This would have given Mny

M0y≈ 0.45 which is greater than the actual value, hence the design

is safe.

Gouwens which is an approximate solution

MnxM0x

+ Mny

M0y

(1−β

β

)≤ 1

0.828 + 0.337(

1−0.650.65

)= 0.828 + 0.1815 = 1.0095

which indicates a slight overstress.

We note that the approximate method is on the conservative side.


Draft9–18 COLUMNS

9.3 Long Columns

9.3.1 Euler Elastic Buckling

27 Column buckling theory originated with Leonhard Euler in 1744. An initially straight mem-ber is concentrically loaded, and all fibers remain elastic until buckling occur.

28 For buckling to occur, it must be assumed that the column is slightly bent as shown in Fig.9.12. Note, in reality no column is either perfectly straight, and in all cases a minor imperfection

P Px

x

L

y

x and y are

principal axes

Slightly bent position

Figure 9.12: Euler Column

is present.

29 At any location x along the column, the imperfection in the column compounded by theconcentric load P , gives rise to a moment

Mz = −Py (9.32)

Note that the value of yis irrelevant.

30 Recalling thatd2y

dx2=

Mz

EI(9.33)

upon substitution, we obtain the following differential equation

d2y

dx2− P

EI= 0 (9.34)

31 Letting k2 = PEI , the solution to this second-order linear differential equation is

y = −A sin kx − B cos kx (9.35)

32 The two constants are determined by applying the essential boundary conditions

1. y = 0 at x = 0, thus B = 0

2. y = 0 at x = L, thusA sin kL = 0 (9.36)


Draft9.3 Long Columns 9–19

This last equation can e satisfied if: 1) A = 0, that is there is no deflection; 2) kL = 0, that isno applied load; or 3)

kL = nπ (9.37)

Thus buckling will occur if PEI =

(nπL

)2 or

P =n2π2EI

L2

33 The fundamental buckling mode, i.e. a single curvature deflection, will occur for n = 1; ThusEuler critical load for a pinned column is

Pcr =π2EI

L2(9.38)

The corresponding critical stress is

σcr =π2E(Lr

)2(9.39)

where I = Ar.

34 Note that buckling will take place with respect to the weakest of the two axis.

9.3.2 Effective Length

35 Large kLr column buckling, small kL

r column crushing, Fig. 9.13.

tan−1

E t

tan−1E

fp

f

ε

Pn

Crushing Buckling

(kl/r) lim(kl/r)

Pfail

Pcr

Figure 9.13: Column Failures

36 Recall from strength of material slenderness ratio

λ =Le

r

where Le is the effective length and is equal to Le = kL and r the radius of gyration (r =√

IA).

37 Le is the distance between two adjacent (fictitious or actual) inflection points, Fig. 9.13


Draft9–20 COLUMNS

i.p.

i.p.

i.p.

i.p.i.p.

i.p.

kl=l l

P

cr crcr

cr

crcr cr

cr

crcrcr

cr

PPP

P PP

PP

P PP

l/4

l/4

kl=2l

<kl

<l

2l

l

i.p.

i.p.

i.p.

i.p.l l

l

kl=21kl=1

l<kl< 8

k=1 1/2<k<1

l<kl< 8

k=1k=2

k=1/2

Figure 9.14: Critical lengths of columns



38 k is known for some simple highly idealized cases, but for most cases k depends on ΨA +ΨB

(relative stiffnesses of columns to connected beams), Fig. 9.15

Ψ =Σ(EI

L )of columnsΣ(EI

L )of floor members(9.40)

and k is then determined from the chart shown in Fig. 9.16.

P P PEI

EI

EI

EI

ll

l

l n

n

n

n

(

(

((

(

(

((

1

2

2

1

MB

MB

MB

MA

MA M A

AA

Aψ Aψ A ψ A

ψ Bψ B ψ BB

B B

Single curvature Double curvature

Unbraced

∆

Braced

Figure 9.15: Effective length Factors Ψ

9.3.3 Moment Magnification Factor; ACI Provisions

39 The critical stress in a column is given by(P

A

)cr

=π2E(kLr

)2

40 Code recommends some minimum eccentricity to account for imperfectly placed load, Fig.9.17

41 For an eccentrically placed load

Mmax = M01

1 − P1−Pcr︸︷︷︸

Moment magnification factor

(9.41)

42 The moment magnification factor reflects the amount by which the beam moment M0 ismagnified by the presence of an axial load, Fig. 9.18

43 The previous equation assumes the presence of hinges at each end (Euler column). In themost general case we will have

Mmax = M0Cm

1 − PPcr

(9.42-a)


Draft9–22 COLUMNS

Sidesway Inhibited

Ga

0.

0.1

0.2

0.3

0.4

0.5

0.60.70.8

1.

2.

3.

5.

10.50.∞

K

0.5

0.6

0.7

0.8

0.9

1.0Gb

0.

0.1

0.2

0.3

0.4

0.5

0.60.70.8

1.

2.

3.

5.

10.50.∞

Sidesway Uninhibited

Ga

0

1.

2.

3.

4.

5.

6.7.8.9.10.

20.30.

50.100.∞

K

1.

1.5

2.

3.

4.

5.

10.20.∞

Gb

0

1.

2.

3.

4.

5.

6.7.8.9.10.

20.30.

50.100.∞

Figure 9.16: Standard Alignment Chart (ACI)

eP

Figure 9.17: Minimum Column Eccentricity



M0

P ∆

kl/r

M

M

M0

P cr

P n

P

M0

Mn

M

P

φ P 0

P0(max)

Pu

M 2

M 2

M c

M c =δ

0 M

C

B

Pu e Pu ∆

em

in

Figure 9.18: P-M Magnification Interaction Diagram

Cm = .6 + .4M1

M2≥ .4 (9.42-b)

whereM1 is numerically smaller than M2 (not algebracially)M1M2

> 0 if single curvatureM1M2

< 0 if double curvatureCm < 1 if members are braced against sideswayCm=1 if members are not braced against sidesway

44 ACI CodeLu unsupported length ACI 10.11.1k ≤ 1.0 braced columns ACI 10.11.2k ≥ 1.0 unbraced columns ACI 10.11.2r = .3h rectangular x section ACI 10.11.3r = .25d circular cross sectionkLu

r < 34 − 12M1M2

braced, neglect slenderness ACI 10.11.4kLu

r < 22 unbraced, neglect slenderness

45 From conventional elastic analysis get Pn&Mn

Mc = δM2 (9.43)

δ =Cm

1 − PnφPcr

≥ 1.0 (9.44)

Pcr =π2EI

(kLu)210.11.5 (9.45)

Cm = .6 + .4M1

M2(9.46)

EI =EcIg

5 + EsIs

1 + βd(9.47)

or EI =EcIg

2.5

1 + βd(9.48)

βd =MD

MD + ML=

1.4PDL

1.4PDL + 1.7PLL(9.49)

βd is the ratio of maximum design load moment to maximum design total load moment (always


Draft9–24 COLUMNS

+ve) as β ↗ EI ↘⇒ dead load has a detrimental effect (creep)

Example 9-5: Long R/C Column

A 15 ft long, 14” circular column is connected to 40 ft long 14” by 22” beams. The columnis on the last floor, below it the column is circular and has a 16” diameter. No sidesway.

Given, Pn = 500 k, 14 × 22 has ρ = .015, f ′c = 5, 000 psi, fy = 40, 000 psi

Solution:

Lu = 15 ft − 2212

= 13.17 ft (9.50-a)

r = .25d = (.25)(14 in) = 3.5 in (9.50-b)Ec = 57, 000

√f ′

c = 57, 000√

5, 000 = 4, 030 ksi (9.50-c)

Ig =πd4

64=

π(14)4

64= 1, 886 in4 (9.50-d)

EIcol =EcIg2.5

1+βd

βd = 0

}EIcol = (4, 030)(1, 886)

12.5

= 3, 040, 000 k in2 (9.51-a)

(EI

L

)c

=3, 040, 000(15)(12)

= 16, 890 k.in (9.52-a)

Ibeam = Icr ' Ig

2=

(14)(22)3

1212

= 6, 210 in4 (9.52-b)(EI

L

)beam

=(4, 030)(6, 210)

(12)(40)= 52, 140 k.in (9.52-c)

ΨA =Σ(EI/L)col

Σ(EI/L)beam=

(16, 890)2(52, 140)

= .162 (9.52-d)

bottom column I = π(16)4

64 = 3, 217 in4

EI =(4, 030)(3, 217)

2.5= 5, 186, 000 (9.53-a)(

EI

L

)col

=5, 186, 000(15)(12)

= 28, 800 k.in (9.53-b)

ΨB =16, 890 + 28, 800

2(52, 140)= .438 (9.53-c)

From ACI commentary ΨA = .162, ΨB = .438,⇒ k ' .62 and

kLu

r=

(.65)(13.16)(12)3.5

= 29.3 (9.54-a)

34 − 12M1

M2= 34 − 12 = 22 assuming M1 = M2 (9.54-b)

kL

r> 22 ⇒ consider column instability (9.54-c)



Pcr =π2EI

(kl)2=

π2(3, 040, 000)[(.65)(13.16)(12)]2

= 2, 848 k (9.54-d)

CM = .6 + .4M1

M2= 1 (9.54-e)

δ =1

1 − PuφPcr

=1

1 − 500(.75)(2,848)

= 1.3 (9.54-f)

Example 9-6: Design of Slender Column

Given: frame not braced, design AB as square column. PD = 46 k, MD = 92 k.ft, PL = 94 k,ML = 230 k.ft, f ′

c = 4 ksi, fy = 60 ksi

Lu =18’Ll =43.3in

3

A

B

��

��

��

��

Solution:

Pu = 1.4 × 46 + 1.7 × 94 = 224 k (9.55-a)Mu = 1.4 × 92 + 1.7 × 230 = 520 k.ft (9.55-b)

βd =MDL

MDL + MLL=

(1.4)92520

= .24 (9.55-c)

Assume a 22 × 22 inch column and ρt = .03

8.5

2.5

22"

22"


Draft9–26 COLUMNS

If =224

12= 19, 500 in4 (9.56-a)

Is = (2)(.015)(22)2(8.5)2 = 1, 050 in4 (9.56-b)Ec = 57, 000

√4, 000 = 3.6 × 106 psi (9.56-c)

Es = 29 × 106 ksi (9.56-d)

EI =EcIg

5 + EsIs

1 + βd

=(3.6×106)(19,500)

5 + (29 × 106)(1, 050)1 + .24

= 3.59 × 1010 (9.56-e)

EIc

L=

3.59 × 1010

12 × 18= 1.66 × 108 (9.56-f)

EIb

L= (3.6 × 106)(43.3) = 1.56 × 108 (9.56-g)

AtA&B Ψ =2(1.66 × 108)1.56 × 108

= 2.13 from ACI commentary k = 1.65 (9.56-h)

if kLr = 22 neglect slenderness

r = (.3)(22) = 6.6 in (9.57-a)

⇒ kL

r=

(1.65)(18)(12)6.6

= 54 > 22 (9.57-b)

Pcr =π2EI

(kL)2=

π2(3.59 × 1010)[(1.65)(18)(12)]2

= 279 × 106 lbs (9.57-c)

Pu = 2.24 × 105lb (9.57-d)Cm = 1.0(unbraced) (9.57-e)

Moment Magnification δ =1

1 − PuφPcr

=1

1 − (2.24×105)(.7)(2.79×106)

(9.57-f)

= 1.13 (9.57-g)

⇒ Moment for which the column is to be designed (1.13) (520) = 587 k.ft and Pu = 224


Draft

Chapter 10

PRESTRESSED CONCRETE

10.1 Introduction

1 Beams with longer spans are architecturally more appealing than those with short ones.However, for a reinforced concrete beam to span long distances, it would have to have to berelatively deep (and at some point the self weight may become too large relative to the liveload), or higher grade steel and concrete must be used.

2 However, if we were to use a steel with fy much higher than ≈ 60 ksi in reinforced concrete(R/C), then to take full advantage of this higher yield stress while maintaining full bond betweenconcrete and steel, will result in unacceptably wide crack widths. Large crack widths will inturn result in corrosion of the rebars and poor protection against fire.

3 One way to control the concrete cracking and reduce the tensile stresses in a beam is toprestress the beam by applying an initial state of stress which is opposite to the one which willbe induced by the load.

4 For a simply supported beam, we would then seek to apply an initial tensile stress at thetop and compressive stress at the bottom. In prestressed concrete (P/C) this can be achievedthrough prestressing of a tendon placed below the elastic neutral axis.

5 Main advantages of P/C: Economy, deflection & crack control, durability, fatigue strength,longer spans.

6 There two type of Prestressed Concrete beams:

Pretensioning: Steel is first stressed, concrete is then poured around the stressed bars. Whenenough concrete strength has been reached the steel restraints are released, Fig. 10.1.

Postensioning: Concrete is first poured, then when enough strength has been reached a steelcable is passed thru a hollow core inside and stressed, Fig. 10.2.

10.1.1 Materials

7 P/C beams usually have higher compressive strength than R/C. Prestressed beams can havef ′

c as high as 8,000 psi.

8 The importance of high yield stress for the steel is illustrated by the following simple example.

Draft10–2 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 10.1: Pretensioned Prestressed Concrete Beam, (Nilson 1978)

Tendon in conduct

BeamJackAnchorage

Wrapped tendon

Slab

AnchorageJack

JackIntermediatediaphragms

Anchorage

Figure 10.2: Posttensioned Prestressed Concrete Beam, (Nilson 1978)


Draft10.1 Introduction 10–3

If we consider the following:

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 steelequal to L′

s = L′c.

4. Due to shrinkage and creep, there will be a change in length

∆Lc = (εsh + εcr)Lc (10.1)

we want to make sure that this amout of deformation is substantially smaller than thestretch of the 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 stress which is left in the steel after creep and shrinkage took place is thus

(1.03 − .90) × 10−3(29 × 103) = 4 ksi (10.2)

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 a17% loss.

10. Note that the actual loss is (.90 × 10−3)(29 × 103) = 26 ksi in 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. 10.3.

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Figure 10.3: 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 take1% strain as effective yield.

10 Steel relaxation is the reduction in stress at constant strain (as opposed to creep whichis reduction of strain at constant stress) occurs. Relaxation occurs indefinitely and producessignificant 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

)(10.3)



10.1.2 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

10.1.3 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

, (section modulus)

6. e + ve if downward from concrete neutral axis

10.1.4 Tendon Configuration

13 Through proper arrangement of the tendon (eccentricity at both support and midspan)various internal flexural stress distribution can be obtained, Fig. 10.4.

10.1.5 Equivalent Load

14 An equivalent load for prestressing can be usually determined from the tendon configurationand the prestressing force, Fig. 10.5.

10.1.6 Load Deformation

15 The load-deformation curve for a prestressed concrete beam is illustrated in Fig. 10.6.


Draft10.1 Introduction 10–5

yf

f

f’

h

W

f

ct

2f =2f

0

c2f

2f

c2f

tc f =

f

c

c c

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

c

+ + + +

f

ff

cc c

c f cc

+

f

Ends

Q

Midspan

Ends

Midspan

00= = = = = =

+

f

c2f

0

c2f

ct

2f =2f c2f

0 0

c2ff

f

ct

c

f =f

c2f

0

ff

ccc c f

h/3

h/2

PP

Q 2Q

h/2P

P

h/3

h/2

PP

2h/3P

P

2Q

Figure 10.4: Alternative Schemes for Prestressing a Rectangular Concrete Beam, (Nilson 1978)

P sin θ

θ

θP sin

M

P cos

P cos

2

2

P eP e

θ

θ

θ

θ

P cos

P

θ

P cos

θP cos

Moment from prestressingEquivalent load on concrete from tendonMember

None

None

P cos

θP cos

θP sin

θP sinθ P cos

P θ P sin

PeP

Pθ

θ

P

P

P

P

P

(b)

(a)

(d)

(e)

P

P

(c)

P

(g)

(f)

P

P

P

Pe

θP sinM

θP sin P sin θ

θP sinθP sin

Figure 10.5: Determination of Equivalent Loads



Steel yieldingService load limitincludingtolerable overload

First cracking load

Decompression

Full dead load

TnOverload

Service load range

fcror 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 10.6: Load-Deflection Curve and Corresponding Internal Flexural Stresses for a TypicalPrestressed Concrete Beam, (Nilson 1978)

10.2 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

)(10.4)

f2 = − Pi

Ac− Piec2

I= − Pi

Ac

(1 +

ec2

r2

)(10.5)

2. Pi and the self weight of the beam M0 (which has to be acconted for the momentthe beam cambers due to prestressing)

f1 = − Pi

Ac

(1 − ec1

r2

)− M0

S1(10.6)

f2 = − Pi

Ac

(1 +

ec2

r2

)+

M0

S2(10.7)

Service Load when the prestressing force was reduced from Pi to Pe beacause of the losses,and the actual service (not factored) load is apllied3. Pe and M0


Draft10.2 Flexural Stresses 10–7

f1 = −Pe

Ac

(1 − ec1

r2

)− M0

S1(10.8)

f2 = −Pe

Ac

(1 +

ec2

r2

)+

M0

S2(10.9)

4. Pe and M0 + MDL + MLL

f1 = −Pe

Ac

(1 − ec1

r2

)− M0 + MDL + MLL

S1(10.10)

f2 = −Pe

Ac

(1 +

ec2

r2

)+

M0 + MDL + MLL

S2(10.11)

The internal stress distribution at each one of those four stages is illustrated by Fig. 10.7.

2)

2e cr

2S

+M

o

Ac

P i( 1 +

2)

2e cr

Stage 4

Stage 2

( 1 +

2e cr

)12

e cr( 1 −

A

c

P i

Ic1

Pi e c

Ac

P i

Ac

P i

Ac

P i

Stage 1

2

Ic

Pi e c

1c

2c

e

)

e crM

t+

+S

Md +

Ml

)12

e cr( 1 −

P

e

Ac

Mt1

S−

2

11

22

S+

Mo

P e

Ac

( 1 +2

)2

e crP

e

Ac

( 1 +2

)

S 1S M

o−

1S M

o−

)12

e cr( 1 −

A

c

P i

Ac

P i( 1 +

2

( 1 −

Md +

Ml

−

)12

e cr( 1 −

S M

o−

P e

Ac

Ac

P i

)12

e cr

+M

o2S

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

2S

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

��

Figure 10.7: Flexural Stress Distribution for a Beam with Variable Eccentricity; MaximumMoment Section and Support Section, (Nilson 1978)



17 Those (service) flexural stresses must be below those specified by the ACI code (where thesubscripts c, 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 someproblems with excessive camber when unloaded.

Partial prestressing (pioneered by Leonhardt, Abeles, Thurliman), cracks are allowed tooccur (just as 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 theultimate strength of the cable: Pj < .80fpuAs and Pi < .70fpuAs. No limits are specified forPe.

Example 10-1: 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 densityγ = 150 lb/ft3 and multiple 7 wire strands with constant eccentricity e = 5.19 in. Pi = 169 k,and the total losses due 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

)(10.12-a)


Draft10.2 Flexural Stresses 10–9

= −169, 000176

(1 − (5.19)(12)

68.2

)= −83 psi (10.12-b)

f2 = − Pi

Ac

(1 +

ec2

r2

)(10.12-c)

= −169, 000176

(1 +

(5.19)(12)68.21

)= −1, 837 psi (10.12-d)

2. Pi and the self weight of the beam M0 (which has to be acconted for the moment thebeam cambers due to prestressing)

w0 =(176) in2

(144) in2/ ft2(.150) k/ ft3 = .183 k/ft (10.13-a)

M0 =(.183)(40)2

8= 36.6 k.ft (10.13-b)

The flexural stresses will thus be equal to:

fw01,2 = ∓ M0

S1,2= ∓(36.6)(12, 000)

1, 000= ∓439 psi (10.14)

f1 = − Pi

Ac

(1 − ec1

r2

)− M0

S1(10.15-a)

= −83 − 439 = −522 psi (10.15-b)

fti = 3√

f ′c = +190

√(10.15-c)

f2 = − Pi

Ac

(1 +

ec2

r2

)+

M0

S2(10.15-d)

= −1, 837 + 439 = −1, 398 psi (10.15-e)

fci = .6f ′c = −2, 400

√(10.15-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(10.16-a)

= −144, 000176

(1 − (5.19)(12)

68.2

)− 439 (10.16-b)

= −71 − 439 = −510 psi (10.16-c)

f2 = −Pe

Ac

(1 +

ec2

r2

)+

M0

S2(10.16-d)

= −144, 000176

(1 +

(5.19)(12)68.2

)+ 439 (10.16-e)

= −1, 561 + 439 = −1, 122 psi (10.16-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 (10.17)

and corresponding stresses

f1,2 = ∓(110)(12, 000)1, 000

= ∓1, 320 psi (10.18)

Thus,

f1 = −Pe

Ac

(1 − ec1

r2

)− M0 + MDL + MLL

S1(10.19-a)

= −510 − 1, 320 = −1, 830 psi (10.19-b)

fcs = .45f ′c = −2, 700

√(10.19-c)

f2 = −Pe

Ac

(1 +

ec2

r2

)+

M0 + MDL + MLL

S2(10.19-d)

= −1, 122 + 1, 320 = +198 psi (10.19-e)

fts = 6√

f ′c = +380

√(10.19-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

10.3 Case Study: Walnut Lane Bridge

Adapted from (Billington and Mark 1983)

20 The historical Walnut Lane Bridge (first major prestressed concrete bridge in the USA) ismade of three spans, two side ones with lengths of 74 ft and a middle one of length 160 feet.Thirteen prestressed cocnrete beams are placed side by side to make up a total width of 44fet of roadway and two 9.25 feet of sidewalk. In between the beams, and cast with them, aretransverse stiffeners which connect the beams laterally, Fig. 10.8


Draft10.3 Case Study: Walnut Lane Bridge 10–11

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 10.8: Walnut Lane Bridge, Plan View



10.3.1 Cross-Section Properties

21 The beam cross section is shown in Fig. 10.9 and is simplified

SIMPLIFIED CROSS - SECTION OF BEAM

8.9"

61.2"

8.9"

22.5"22.5" 7"

52"

6’-7"= 79"

Figure 10.9: Walnut Lane Bridge, Cross Section

Ac = 2(8.9)(52) + (7)(61.2) = 1, 354 in2 (10.20-a)

I = 2

[(52)(8.9)3

12+ (52)(8.9)

(792

− 8.92

)2]

+(7)(61.2)3

12(10.20-b)

= = 1, 277 × 103 in4 (10.20-c)

c1 = c2 =h

2=

792

= 39.5 in (10.20-d)

S1 = S2 =I

c=

1, 277 × 103

39.5= 32, 329 in3 (10.20-e)

r2 =I

A=

1, 277 × 103

1, 354= 943. in2 (10.20-f)

10.3.2 Prestressing

22 Each beam is prestressed by two middle parabolic cables, and two outer horizontal onesalong the flanges. All four have approximately the same eccentricity at midspan of 2.65 ft. or31.8 inch.

23 Each prestressing cable is made up 64 wires each with a diameter of 0.27 inches. Thus thetotal area of prestressing steel is given by:

Awire = π(d/2)2 = 3.14(0.276 in

2)2 = 0.0598 in2 (10.21-a)



Acable = 64(0.0598) in2 = 3.83 in2 (10.21-b)Atotal = 4(3.83) in2 = 15.32 in2 (10.21-c)

24 Whereas the ultimate tensile strength of the steel used is 247 ksi, the cables have beenstressed only to 131 ksi, thus the initial prestressing force Pi is equal to

Pi = (131) ksi(15.32) in2 = 2, 000 k (10.22)

25 The losses are reported ot be 13%, thus the effective force is

Pe = (1 − 0.13)(2, 000) k = 1, 740 k (10.23)

10.3.3 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,the total load over one single beam is

qr,tot =113

(44) ft(0.45) ft(0.15) k/ ft3 = 0.23 k/ft (10.24)

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 (10.25)

We note that the weight can be evenly spread over the 13 beams beacause of the lateraldiaphragms.

29 The total dead load isqDL = 0.23 + 0.13 = 0.36 k/ft (10.26)

30 The live load is created by the traffic, and is estimated to be 94 psf, thus over a width of62.5 feet this gives a uniform live load of

wLL =113

(0.094) k/ft2(62.5) ft = 0.45 k/ft (10.27)

31 Finally, the combined dead and live load per beam is

wDL+LL = 0.36 + 0.45 = 0.81 k/ft (10.28)

10.3.4 Flexural Stresses

1. Prestressing force, Pi only

f1 = − Pi

Ac

(1 − ec1

r2

)(10.29-a)

= −(2 × 106)1, 354

(1 − (31.8)(39.5)

943.

)= 490. psi (10.29-b)

f2 = − Pi

Ac

(1 +

ec2

r2

)(10.29-c)

= −(2 × 106)1, 354

(1 +

(31.8)(39.5)943.

)= −3, 445. psi (10.29-d)



2. Pi and the self weight of the beam M0 (which has to be acconted for the moment thebeam cambers due to prestressing)

M0 =(1.72)(160)2

8= 5, 504 k.ft (10.30)

The flexural stresses will thus be equal to:

fw01,2 = ∓ M0

S1,2= ∓(5, 50.4)(12, 000)

943.= ∓2, 043 psi (10.31)

f1 = − Pi

Ac

(1 − ec1

r2

)− M0

S1(10.32-a)

= 490 − 2, 043 = −1, 553 psi (10.32-b)

fti = 3√

f ′c = +190

√(10.32-c)

f2 =Pi

Ac

(1 +

ec2

r2

)+

M0

S2(10.32-d)

= −3, 445 + 2, 043 = −1, 402. psi (10.32-e)

fci = .6f ′c = −2, 400

√(10.32-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(10.33-a)

= −1.74 × 106

1, 354

(1 − (31.8)(39.5)

943.

)− 2, 043. = −1, 616 psi (10.33-b)

f2 =Pe

Ac

(1 +

ec2

r2

)+

M0

S2(10.33-c)

= −1.74 × 106

1, 354

(1 +

(31.8)(39.5)943.

)+ 2, 043. = −954. psi (10.33-d)


MDL + MLL =(0.81)(160)2

8= 2, 592 k.ft (10.34)

and corresponding stresses

f1,2 = ∓(2, 592)(12, 000)32, 329

= ∓962. psi (10.35)

Thus,

f1 = −Pe

Ac

(1 − ec1

r2

)− M0 + MDL + MLL

S1(10.36-a)

= −1, 616 − 962. = −2, 578. psi (10.36-b)

fcs = .45f ′c = −2, 700

√(10.36-c)

f2 =Pe

Ac

(1 +

ec2

r2

)+

M0 + MDL + MLL

S2(10.36-d)

= −954 + 962. = +8. psi (10.36-e)

fts = 6√

f ′c = +380

√(10.36-f)




Draft

Bibliography

Billington, D. and Mark, R.: 1983, Structural studies, Technical report, Department of CivilEngineering, Princeton University.

Nilson, A.: 1978, Design of Prestressed Concrete, John Wiley and Sons.

Documents

Draft - Ahmed Mansour and Design of Reinfor… · ... 3{8 E 3-1 Shear Design . ... 9.2.6 Biaxial Bending . . . ... 7.6 Approximate Analysis of Frames Subjected to Lateral Loads; Column