A Mechanics Mode for the Compressive Response of Fiber Reinforced Composites
Lecture Note on Mechanics and Design of Reinforced Concrete_Victor E.saouma_(2002)
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Transcript of Lecture Note on Mechanics and Design of Reinforced Concrete_Victor E.saouma_(2002)
DraftDRAFT
Lecture Notes in:
Mechanics and Design of
REINFORCED CONCRETE
CVEN4555
c©VICTOR E. SAOUMA,
Fall 2001
Dept. of Civil Environmental and Architectural Engineering
University of Colorado, Boulder, CO 80309-0428
May 18, 2002
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Victor Saouma Mechanics and Design of Reinforced Concrete
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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
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2.6 T Beams, (ACI 8.10) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–172.6.1 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–192.6.2 Design, (balanced) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–19E 2-8 T Beam; Moment Capacity I . . . . . . . . . . . . . . . . . . . . . . . . . 2–20E 2-9 T Beam; Moment Capacity II . . . . . . . . . . . . . . . . . . . . . . . . . 2–21E 2-10 T Beam; Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–22
2.7 Doubly Reinforced Rectangular Beams . . . . . . . . . . . . . . . . . . . . . . . . 2–232.7.1 Tests for fs and f ′
s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–242.7.2 Moment Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–26E 2-11 Doubly Reinforced Concrete beam; Review . . . . . . . . . . . . . . . . . 2–28E 2-12 Doubly Reinforced Concrete beam; Design . . . . . . . . . . . . . . . . . . 2–30
2.8 Bond & Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–312.8.1 Moment Capacity Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 2–35
3 SHEAR 3–13.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13.2 Shear Strength of Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . 3–23.3 Shear Strength of Cracked Sections . . . . . . . . . . . . . . . . . . . . . . . . . . 3–43.4 ACI Code Requirements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–63.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–7
E 3-1 Shear Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–73.6 Shear Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–8
E 3-2 Shear Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–113.7 Brackets and Corbels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–113.8 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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–5E 4-1 Moment Redistribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–6
4.5 Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4–7
5 SERVICEABILITY 5–15.1 Control of Cracking . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–1
E 5-1 Crack Width . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–35.2 Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–4
5.2.1 Short Term Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–45.2.2 Long Term Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–5E 5-2 Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–7
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6 APPROXIMATE FRAME ANALYSIS 6–16.1 Vertical Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–16.2 Horizontal Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–4
6.2.1 Portal Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–4E 6-1 Approximate Analysis of a Frame subjected to Vertical and Horizontal
Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6–6
7 ONE WAY SLABS 7–17.1 Types of Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–17.2 One Way Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–47.3 Design of a One Way Continuous Slab . . . . . . . . . . . . . . . . . . . . . . . . 7–5
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 . . . . . . . . . . . . . . . . . . . . . . . . . 9–1
9.2 Short Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–39.2.1 Concentric Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–39.2.2 Eccentric Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–3
9.2.2.1 Balanced Condition . . . . . . . . . . . . . . . . . . . . . . . . . 9–49.2.2.2 Tension Failure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–59.2.2.3 Compression Failure . . . . . . . . . . . . . . . . . . . . . . . . . 9–6
9.2.3 ACI Provisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–79.2.4 Interaction Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–89.2.5 Design Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–8E 9-1 R/C Column, c known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–8E 9-2 R/C Column, e known . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–10E 9-3 R/C Column, Using Design Charts . . . . . . . . . . . . . . . . . . . . . . 9–149.2.6 Biaxial Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–15E 9-4 Biaxially Loaded Column . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–18
9.3 Long Columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–199.3.1 Euler Elastic Buckling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–199.3.2 Effective Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–209.3.3 Moment Magnification Factor; ACI Provisions . . . . . . . . . . . . . . . 9–22E 9-5 Long R/C Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–25E 9-6 Design of Slender Column . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–26
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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–410.1.5 Equivalent Load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–410.1.6 Load Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–4
10.2 Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–6
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E 10-1 Prestressed Concrete I Beam . . . . . . . . . . . . . . . . . . . . . . . . . 10–810.3 Case Study: Walnut Lane Bridge . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–10
10.3.1 Cross-Section Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1210.3.2 Prestressing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1210.3.3 Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1310.3.4 Flexural Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–13
11 FOOTINGS 11–1
12 DEEP BEAMS 12–1
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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 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.9 T Beam as Rectangular Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.10 T Beam Strain and Stress Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 2–182.11 Decomposition of Steel Reinforcement for T Beams . . . . . . . . . . . . . . . . . 2–192.12 Doubly Reinforced Beams; Strain and Stress Diagrams . . . . . . . . . . . . . . . 2–242.13 Different Possibilities for Doubly Reinforced Concrete Beams . . . . . . . . . . . 2–242.14 Strain Diagram, Doubly Reinforced Beam; is As Yielding? . . . . . . . . . . . . . 2–252.15 Strain Diagram, Doubly Reinforced Beam; is A′
s Yielding? . . . . . . . . . . . . . 2–262.16 Summary of Conditions for top and Bottom Steel Yielding . . . . . . . . . . . . . 2–272.17 Bond and Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–312.18 Actual Bond Distribution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–322.19 Splitting Along Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–322.20 Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–332.21 Development Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–332.22 Hooks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–352.23 Bar cutoff requirements of the ACI code . . . . . . . . . . . . . . . . . . . . . . . 2–362.24 Standard cutoff or bend points for bars in approximately equal spans with uni-
formly distributed load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–372.25 Moment Capacity Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2–38
3.1 Principal Stresses in Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13.2 Types of Shear Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–13.3 Shear Strength of Uncracked Section . . . . . . . . . . . . . . . . . . . . . . . . . 3–2
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3.4 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 . . . . . . . . . . . . . . . . . . . 3–53.8 Summary of ACI Code Requirements for Shear . . . . . . . . . . . . . . . . . . . 3–73.9 Corbel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–93.10 Shear Friction Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3–93.11 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 . . . . . . . . 4–44.6 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 Crack Width Equation Parameters . . . . . . . . . . . . . . . . . . . . . . . . . . 5–25.2 Uncracked Transformed and Cracked Transformed X Sections . . . . . . . . . . . 5–45.3 Time Dependent Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–55.4 Time Dependent Strain Distribution . . . . . . . . . . . . . . . . . . . . . . . . . 5–65.5 Short and long Term Deflections . . . . . . . . . . . . . . . . . . . . . . . . . . . 5–6
6.1 Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments . 6–26.2 Approximate Analysis of Frames Subjected to Vertical Loads; Column Axial Forces6–36.3 Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments 6–36.4 Approximate Analysis of Frames Subjected to Lateral Loads; Column Shear . . . 6–56.5 Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment . . 6–56.6 Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force6–66.7 Example; Approximate Analysis of a Building . . . . . . . . . . . . . . . . . . . . 6–76.8 Approximate Analysis of a Building; Moments Due to Vertical Loads . . . . . . . 6–96.9 Approximate Analysis of a Building; Shears Due to Vertical Loads . . . . . . . . 6–106.10 Approximate Analysis for Vertical Loads; Spread-Sheet Format . . . . . . . . . . 6–126.11 Approximate Analysis for Vertical Loads; Equations in Spread-Sheet . . . . . . . 6–136.12 Approximate Analysis of a Building; Moments Due to Lateral Loads . . . . . . . 6–146.13 Portal Method; Spread-Sheet Format . . . . . . . . . . . . . . . . . . . . . . . . . 6–166.14 Portal Method; Equations in Spread-Sheet . . . . . . . . . . . . . . . . . . . . . . 6–17
7.1 Types of Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–17.2 One vs Two way slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–27.3 Load Distribution in Slabs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–27.4 Load Transfer in R/C Buildings . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7–3
9.1 Types of columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–19.2 Tied vs Spiral Reinforcement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–29.3 Possible Bar arrangements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–29.4 Sources of Bending . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–3
Victor Saouma Mechanics and Design of Reinforced Concrete
DraftLIST OF FIGURES 0–3
9.5 Load Moment Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . 9–49.6 Strain and Stress Diagram of a R/C Column . . . . . . . . . . . . . . . . . . . . 9–59.7 Column Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–89.8 Failure Surface of a Biaxially Loaded Column . . . . . . . . . . . . . . . . . . . . 9–159.9 Load Contour at Plane of Constant Pn, and Nondimensionalized Corresponding
plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–169.10 Biaxial Bending Interaction Relations in terms of β . . . . . . . . . . . . . . . . . 9–179.11 Bilinear Approximation for Load Contour Design of Biaxially Loaded Columns . 9–179.12 Euler Column . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–199.13 Column Failures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–209.14 Critical lengths of columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–219.15 Effective length Factors Ψ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–229.16 Standard Alignment Chart (ACI) . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–239.17 Minimum Column Eccentricity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9–239.18 P-M Magnification Interaction Diagram . . . . . . . . . . . . . . . . . . . . . . . 9–24
10.1 Pretensioned Prestressed Concrete Beam, (?) . . . . . . . . . . . . . . . . . . . . 10–210.2 Posttensioned Prestressed Concrete Beam, (?) . . . . . . . . . . . . . . . . . . . . 10–210.3 7 Wire Prestressing Tendon . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–310.4 Alternative Schemes for Prestressing a Rectangular Concrete Beam, (?) . . . . . 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, (?) . . . . . . . . . . . . . . . . . . . . . . . . . . 10–610.7 Flexural Stress Distribution for a Beam with Variable Eccentricity; Maximum
Moment Section and Support Section, (?) . . . . . . . . . . . . . . . . . . . . . . 10–710.8 Walnut Lane Bridge, Plan View . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10–1110.9 Walnut Lane Bridge, Cross Section . . . . . . . . . . . . . . . . . . . . . . . . . . 10–12
11.1 xxx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–111.2 xxx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–211.3 xxx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–211.4 xxx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–311.5 xxx . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11–3
Victor Saouma Mechanics and Design of Reinforced Concrete
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 % For 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
6.1 Columns Combined Approximate Vertical and Horizontal Loads . . . . . . . . . 6–186.2 Girders Combined Approximate Vertical and Horizontal Loads . . . . . . . . . . 6–19
7.1 Recommended Minimum Slab and Beam Depths . . . . . . . . . . . . . . . . . . 7–4
Draft0–2 LIST OF TABLES
Victor Saouma Mechanics and Design of Reinforced Concrete
DraftLIST OF TABLES 0–3
Tentative ScheduleFall 1994
1 Aug. 28 Intro; MaterialAug. 30 Concrete mix design
2 Sep. 4 Elastic UncrackedSep. 6 WSD; USD singly reinforced
3 Sep. 11 USD singly, examplesSep. 13 T Beams
4 Sep. 18 T Beams, Doubly ReinfSep. 20 Doubly Reinf Development length
5 Sep. 25 ShearSep. 27 Shear
6 Oct. 2 TP LabOct. 4 Fall Break
7 Oct. 9 Crack widthOct. 11 EXAM I
8 Oct. 16 DeflectionOct. 18 Crack Width-Defelction
9 Oct. 23 Deflection, Continuous SystemsOct. 25 Continuous Systems; One way slabs
10 Oct. 30 Columns; IntroNov. 1 Columns
11 Nov. 6 LABNov. 8 Columns
12 Nov. 13 Biaxial bendingNov. 15 Long column
13 Nov. 20 LabNov. 22 Thanksgiving
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft0–4 LIST OF TABLES
Victor Saouma Mechanics and Design of Reinforced Concrete
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
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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.
Victor Saouma Mechanics and Design of Reinforced Concrete
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 0.0937No. 16 1.18 0.0469No. 30 0.60 (600 µm) 0.0234No. 50 300 µm 0.0124No. 100 150 µm 0.0059
Table 1.1: ASTM Sieve Designation’s Nominal Sizes Used for Concrete Aggregates
Sieve Size % (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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1–4 INTRODUCTION
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 sieves
100(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 frsh 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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1.1 Material 1–5
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∑
= 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 in 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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1–6 INTRODUCTION
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 3.0
Moderate 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 % For Different Nominal MaximumSizes of 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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1.1 Material 1–7
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-entrained
6,000 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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1–8 INTRODUCTION
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 0.6511/2 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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1.1 Material 1–9
The OD weight of the coarse aggregate is 17.01 × 100 = 1, 701 lb. The SSD weight is1, 701× 1.01 = 1, 718 lb.
7. Fine aggregate contentKnowing 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× 2.65× 62.4 = 1, 179 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
Fine aggregate (wet)=1,179(1.055-0.013) = 1,229 lb/yd3
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 the additional water required is then280-1,718(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/yd33,87827 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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1–10 INTRODUCTION
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
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1.1 Material 1–11
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 (1.8-a)≈ 3− 5
√f ′c (1.8-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 (1.9)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1–12 INTRODUCTION
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 =2Pπdt
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
= 57, 000√3, 000(0.0002) = 624 psi >
3, 00010︸ ︷︷ ︸f ′t
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.
Creep coefficient, Table 1.10Cu = εct
εci≈ 2− 3
Ct = t0.6
10+t0.6Cu
3For this reason a minimum amount of reinforcement is always necessary in concrete, and a 2% reinforcement,can reduce the shrinkage by 75%.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1.1 Material 1–13
ε
no load constant load
creepElastic recovery
Residual
Creep recovery
no load
Figure 1.7: Time Dependent Strains in Concrete
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. F47 Prestressing Steel cables have an ultimate strength up to 270 ksi.
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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1–14 INTRODUCTION
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
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 (1.10)
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).
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:
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1.3 Analysis vs Design 1–15
i is the type of load
ΦMn ≥ Mu
ΦVn ≥ VuΦPn ≥ Pu
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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1–16 INTRODUCTION
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 values
of 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 29,000psi.
9.1.1 - Structures and structural members shall be designed to have design strengths at allsections 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
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 )
Victor Saouma Mechanics and Design of Reinforced Concrete
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
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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft1–18 INTRODUCTION
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
Victor Saouma Mechanics and Design of Reinforced Concrete
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 ⇒ fsEs
=fcEc
⇒ fs =EsEc
fc ⇒ 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
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���������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������
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, 00057√4, 000
= 8⇒ (n− 1)As = (8− 1)(2.35) = 16.45 in2 (2.3-a)
yb =(10)(25)(252 ) + (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)
Victor Saouma Mechanics and Design of Reinforced Concrete
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
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��������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������
��������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������������
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, by
definition, 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 ⇒ fsM
Asjd(2.4-b)
M = Cjd =bkd
2fcjd =
bd2
2kjfc ⇒ fc =
M12bd
2kj(2.4-c)
where j = (1− k/3).
Victor Saouma Mechanics and Design of Reinforced Concrete
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-a)
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
fs
dkd
kd/3C
T
s
cε
ε
(1-k/3)d=jd
Figure 2.4: Desired Stress Distribution; WSD Method
εcεs
= kdd−kd
εc = fcEc
εs = fsEs
fcEc
Esfs
= k1−k
n = EsEc
r = fsfc
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 = AsfsC = T
ρ = Asbd
fc2 bkd = ρbfsbd
k = nn+r
}ρb = n
2r(n+r) (2.8-a)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.2 Section Cracked, Stresses Elastic 2–5
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 =12fcjkbd
2 = Rbd2 (2.11)
where
k =√2ρn+ (ρn)2 − ρn
Design We define
R =12fckj (2.12)
solve for bd2 frombd2 =
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 = fsfc
R = 12fckj
ρb = n2r(n+r) ρb = n
2r(n+r)
ρ < ρb M = Asfsjd bd2 = MR
ρ > ρb M = 12fcbkd
2j As = ρbbd or As = Mfsjd
Example 2-2: Cracked Elastic Section
Victor Saouma Mechanics and Design of Reinforced Concrete
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).Solution:
Based on previous example, fct would be 866 psi >> fr and the solution is thus no longervalid.
The neutral axis is obtained from
ρ =Asbd
=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− 0.33
3
)(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 =2Mbjkd2
(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 2N.A in 13.2 7.6fcc psi 485 1,390 (< .5f ′
c) 2.9I in4 14,710 5,910 .4 (δα1I )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:
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.2 Section Cracked, Stresses Elastic 2–7
ρ =Asbd
=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 = =12fcbkd
2j (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 αsincrease, 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)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–8 FLEXURE
fs = 24, 000 psi (2.18-b)
n =EsEc
=29, 00057√4, 000
= 8 (2.18-c)
r =fsfc=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)
R =12fckj =
12(1, 800)(.375)(.875) = 295 psi (2.18-g)
ρb =n
2r(n+ r)=
82(13.33)(8 + 13.33)
= .01405 (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) in
2 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
a/2 = cβ
c
fs fs
h d
b
As
c
σ
ε
βc
Actual
αc =
βa
1
γ f’c
cγa
C= f’cb C= f’ab
ε
Figure 2.5: Cracked Section, Limit State
Figure
Victor Saouma Mechanics and Design of Reinforced Concrete
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 we need to know is C & its location, thus α andβ 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 =favf ′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).
Victor Saouma Mechanics and Design of Reinforced Concrete
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 = fyAsfs = .85f ′
cab = .85f ′cbβ1c
ρ = Asbd
c = ρfy
.85f ′cβ1d (2.24-a)
Compression Failure:
εc = .003 (2.25-a)
εs =fsEs
(2.25-b)
c
d=
.003.003 + εs
⇒ c = .003fsEs+.003
d (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+.003
d
ρ = ρb
}ρbf2d.85f ′cβ1
= .003fs
E−s+.003
d
Es = 29, 000 ksi
}ρb = .85β1
f ′cfy
87,00087,000+fy
(ACI 8.4.3)(2.26-a)
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)
Victor Saouma Mechanics and Design of Reinforced Concrete
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 = 0Md = φAsfy(d− a
2 ) 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;
ΣFx = 0 a = Asfy0.85f ′cb
ρ = Asbd
}a = ρfy
0.85f ′cMd = Asfy
(d− a
2
)}
Md = Φ ρfy
(1− .59ρ
fyf ′c
)︸ ︷︷ ︸
R
bd2(2.33-a)
or
R = ρfy
(1− .59ρ
fyf ′c
)(2.34)
Victor Saouma Mechanics and Design of Reinforced Concrete
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
εsd− 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.
Victor Saouma Mechanics and Design of Reinforced Concrete
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 largerbar sizes to reduce placement cost (careful about crack width!).
6. Check from tables that the selected beam width will provide room for the bars chosenwith adequate cover and spacing.
Victor Saouma Mechanics and Design of Reinforced Concrete
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
Victor Saouma Mechanics and Design of Reinforced Concrete
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 =Asbd
=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:
Victor Saouma Mechanics and Design of Reinforced Concrete
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
fyf ′c
) (2.40-a)
= Asfyd(1− 59ρactfyf ′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 = 0.75ρbSolution:
wu = 1.4(1.27) + 1.7(2.44) = 5.92 k/ft Factored load (2.41-a)
Md =wuL
2
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
fyf ′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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.6 T Beams, (ACI 8.10) 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 =
205 = 4 in
As =Md
φfy(d− a2 )=
(1, 600)(.9)(40)(20− 4
2)= 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)
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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–18 FLEXURE
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fh
b
wb
eb
Figure 2.8: T Beams
��������������������������������������������������������������������
h d
As
b
fh
Figure 2.9: T Beam as Rectangular Section
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As
s
0.85 f’cu
β1
s y
f
w
b
h
dh
b
ε
ε =0.003a= c
T=A f
d
c C=0.85f’ ac
Figure 2.10: T Beam Strain and Stress Diagram
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.6 T Beams, (ACI 8.10) 2–19
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.
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 )hb w
w w
fh
b
f
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s AsfA −
= +
c
s
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AsfA
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(2.45)
Web:a = (As−Asf )fy
.85f ′cbwΣF = 0
Mn2 = (As −Asf )fy(d− a2 ) ΣM = 0
(2.46)
Total moment:Mn = Mn1 +Mn2 (2.47)
2.6.2 Design, (balanced)
44 Let us derive an expression for ρb and use it for design
cd = εu
εu+εyStrain Compatibility
Asfy = .85f ′cβ1cbw + .85f ′
c(b− bw)hf︸ ︷︷ ︸Asffy
ΣF = 0 (2.48)
thus,
Asfy = .85f ′cβ1cbw +Asffy
ρw = Asbwd
ρf = Asf
bwd
ρw =
ρb︷ ︸︸ ︷.85
f ′c
fyβ1
εuεu + εy
+ρf (2.49)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–20 FLEXURE
Hence,
ρwb = ρb + ρf (2.50)ρw,max = .75(ρb + ρf ) (2.51)
Example 2-8: 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
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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 (2.52)
3. For a T section
Asf =.85f ′
chf (b− bw)fy
(2.53-a)
=(.85)(3)(7)(30− 14)
50= 5.71 in2 (2.53-b)
ρf =Asfbwd
=5.71
(14)(36)= .0113 (2.53-c)
Asw = As −Asw = 12.48− 5.71 = 6.77 in2 (2.53-d)
ρw =Aswbwd
=12.48(14)(36)
= .025 (2.53-e)
ρb = .85β1f ′c
fy
8787 + fy
(2.53-f)
= (.85)(.85)350
8787 + 50
= .0275 (2.53-g)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.6 T Beams, (ACI 8.10) 2–21
4. Maximum permissible ratio
ρmax = .75(ρb + ρf ) (2.54-a)= .75(.0275 + .0113) = .029 > ρw
√(2.54-b)
5. The design moment is then obtained from
Mn1 = (5.71)(50)(36− 7
2
)= 9, 280 k.in (2.55-a)
a =(As −Asf )fy
.85f ′cbw
(2.55-b)
=(6.77)(50)(.85)(3)(14)
= 9.48 in (2.55-c)
Mn2 = (6.77)(50)(36− 9.482) = 10, 580 k.in (2.55-d)
Md = (.9)(9, 280 + 10, 580) = 17, 890 k.in→ 17, 900 k.in (2.55-e)
Example 2-9: 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.
C=0.85f’ abc
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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 (2.56-a)
ρb = (.85)(.85)(360
)(87
87 + 60
)= .0214 > ρ ⇒ fs = fy (2.56-b)
a =(As −Asf )fy
.85f ′cbw
(2.56-c)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–22 FLEXURE
=(7.59)(60)(.85)(3)(28)
= 6.37 in > 6 in⇒ T beam (2.56-d)
Asf =(.85)(3)(18)(6)
60= 4.59 in2 (2.56-e)
Asw = 7.59− 4.59 = 3.00 in2 (2.56-f)
ρw =7.59
(26)(10)= .0292 (2.56-g)
ρf =4.59
(26)(10)= .0177 (2.56-h)
ρmax = .75(.0214 + .0177) = .0294 > .0292⇒ Ductile failure (2.56-i)Mn1 = (4.59)(60)(26− 3) = 6, 330 k.in (2.56-j)
As −Asf = 7.59− 4.59 = 3. in2 (2.56-k)
a =(3)(60)
(.85)(3)(10)= 7.07 in (2.56-l)
Mn2 = (3.00)(60)(26− 7.072) = 4, 050 k.in (2.56-m)
Md = (.9)(6, 330 + 4, 050) = 9, 350 k.in (2.56-n)
Example 2-10: 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) ≤ 8hf16hf + bw = (16)(3) + 11 = 59 inL4 =
244 12 = 72 in
Center Line spacing = 47 in
b = 47 in (2.57-a)
2. Assume a = 3 in
As =Md
φfy(d− a2 )=
6, 4000.9)(60)(20− 3
2)= 6.40 in2 (2.58-a)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.7 Doubly Reinforced Rectangular Beams 2–23
a =Asfy(.85)f ′
cb=
(6.4)(60)(.85)(3)(47)
= 3.20 in > hf (2.58-b)
3. Thus a T beam analysis is required.
Asf =.85f ′
c(b− bw)hffy
=(.85)(3)(47− 11)(3)
60= 4.58 in2 (2.59-a)
Md1 = φAsffy(d− hf2) = (.90)(4.58)(60)(20− 3
2) = 4, 570 k.in (2.59-b)
Md2 = Md −Md1 = 6, 400− 4, 570 = 1, 830 k.in (2.59-c)(2.59-d)
4. Now, this is similar to the design of a rectangular section. Assume a = d5 =
205 = 4. in
As −Asf =1, 830
(.90)(60)(20− 4
2
) = 1.88 in2 (2.60)
5. check
a =1.88)(60)(.85)(3)(11)
= 4.02 in ≈ 4.00 (2.61-a)
As = 4.58 + 1.88 = 6.46 in2 (2.61-b)
ρw =6.46
(11)(20)= .0294 (2.61-c)
ρf =4.58
(11)(20)= .0208 (2.61-d)
ρb = (.85)(.85)(360
)(87
87 + 60
)= .0214 (2.61-e)
ρmax = .75(.0214 + .0208) = .0316 > ρw√
(2.61-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)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–24 FLEXURE
As
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
46 Approach will again be based on a strain compatibility analysis & equilibrium equation, Fig.2.12.
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(2.62)
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 �= fy, we have from the strain diagram, Fig. ??
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.7 Doubly Reinforced Rectangular Beams 2–25
As
u
ss
s
’
y
h
b
A’d’
ε =0.003
d
ε = ε
ε
Figure 2.14: Strain Diagram, Doubly Reinforced Beam; is As Yielding?
ε′s = εu − d′
d(εu + εy) (2.63-a)
f ′s = Esε
′s (2.63-b)
c =εu
εu + εyd (2.63-c)
(2.63-d)
From equilibrium:ρbdfy = ρ′bdf ′
s + .85f ′cβ1bc (2.64)
Combining:
ρb = ρ1 = ρ′f ′s
fy+ .85
f ′c
fyβ1
εuεu + εy︸ ︷︷ ︸ρb
(2.65)
thus
ρb = ρ1 = ρ′f ′s
fy+ ρb (2.66)
ρmax = 0.75ρb + ρ′f ′s
fy(2.67)
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
c =εu
εu − εyd′ (2.68)
from equilibriumρbdfy = ρ′bdfy + .85f ′
cβ1cb (2.69)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–26 FLEXURE
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?
combining
ρmin ≡ ρ2 = ρ′ + .85β1f ′c
fy
d′
d
8787− fy
(2.70)
which corresponds to the minimum amount of steel to ensure yielding of compression steelat failure. Thus, if ρ < ρmin then f ′
s < fy.
Test 3 fs < fy, is f ′s = fy?
From strain diagram:
c =εu
εu − εyd′ (2.71-a)
εs =d− c
c− d′εy (2.71-b)
From equilibrium
ρbdfs = ρ′bdfy + .85f ′cβ1bc (2.72-a)
combining
ρ = ρ3 =c− d′
d− c
[ρ′ + .85β1
f ′c
fy
c
d
](2.73)
Summary of the tests are shown in Fig. 2.16
2.7.2 Moment Equations
Case I fs = fy and f ′s = fy, (small bottom and top reinforcement ratios)
Asfy = A′sfy + .85f ′
cab (2.74-a)
a =(As −A′
s)fy.85f ′
cb(2.74-b)
M In = .85f ′
cab(d− a
2
)+A′
sfy(d− d′) (2.75)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.7 Doubly Reinforced Rectangular Beams 2–27
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
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(2.76-a)
f ′s = Esε
′s (2.76-b)
Asfy = A′sf
′s + .85f ′
cbβ1c (2.76-c)
solve for c and f ′s by iteration.
Using a = β1c
M IIn = .85f ′
cab(d− a
2
)+A′
sf′s(d− d′) (2.77)
Case III fs < fy and f ′s = fy (large bottom and small top reinforcement ratios, rare)
εs = εud− c
c(2.78-a)
fs = Esεs (2.78-b)Asfs = A′
sfy + .85f ′cab (2.78-c)
a = β1c (2.78-d)
solve for a
M IIIn = .85f ′
cab(d− a
2
)+A′
sfy(d− d′) (2.79)
Case IV fs < fy and f ′s < fy (large bottom and top reinforcement ratios, rare)
ε′s = εuc− d′
c(2.80-a)
εs = εud− c
c(2.80-b)
Asfs = A′sf
′s + .85f ′
cab (2.80-c)a = β1c (2.80-d)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–28 FLEXURE
solve for a
M IVn = .85f ′
cab(d− a
2
)+A′
sfs(d− d′) (2.81)
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 2-11: 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.
sε ’
u
=
cs
s
A’ = 3 # 11
27.3"
3"
16"A = 4 # 11
ε =0.003
s
β
ε
c
d
0.85 f’
a= c
A’ f’β
A’ f’
d−d’
a= c
0.85 f’
+
s
ssA f
s
1
ys
ysA f(A − A’ )s s s
c
1
Solution:
1. Determine ρ:
ρb = (.85)β1f ′c
fy
8787 + fy
= (.85)(.85)460
8787 + 60
= .0285 (2.82-a)
ρ =6.24
(16)(27.3)= .0143 (2.82-b)
ρ′ =4.68
(16)(27.3)= .0107 (2.82-c)
2. Check for ρmin
ρmin = ρ′ + .85f ′c
fyβ1
d′
d
εuεu − εy
(2.83-a)
= .0107 + (.85)460(.85)
327.3
.003.003− 60
29,000
= .0278 > ρ (2.83-b)
Henceρ < ρmin < ρb
.0143 < .0278 < .0285(2.84)
and thus fs = fy and f ′s < fy and we have case II
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.7 Doubly Reinforced Rectangular Beams 2–29
3. We have two equations (strain compatibility and summation of forces) and two unknownsc and f ′
s
f ′s = Esεu
c− d′
c= (29, 000)(.003)
c− 3c
(2.85-a)
= 87c− 3
c(2.85-b)
Asfy = A′sf
′s + .85f ′
cbβ1c (2.85-c)(6.24)(60) = (4.68)f ′
s + (.85)(4)(16)(.85)c (2.85-d)374.4 = 4.68f ′
s + 46.24c (2.85-e)f ′s = −9.9c+ 80.2 (2.85-f)
Note that if we were to plott those two equations,
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 from
equilibrium. Graphically the solution is around 4.9.
4. Combining those two equations1
c2 + .7085c− 26.42 = 0 (2.86)
we obtain c = 4.80 in a = 0.85(4.8) = 4.078 in, and f ′s = (.003)(29, 000)4.80−34.80 = 32.6 ksi
5. Substituting into the moment equation
Mn = .85f ′cab
(d− a
2
)+A′
sf′s(d− d′) (2.87-a)
= (.85)(4)(4.078)(16)(27.3− 4.078
2
)+ (4.68)(32.62)(27.3− 3) (2.87-b)
= 9, 313 k.in (2.87-c)Md = 0.9(9, 313) = 8, 382 k.in = 699 k.ft (2.87-d)
6. Check
ρmax = .75ρb +f ′s
fyρ′ (2.88-a)
= (.75)(.0285) +32.660
(.0107) = .027︸︷︷︸ρ
√(2.88-b)
1In this problem, unfortunately an iterative method diverges if we were to start with a = d5.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–30 FLEXURE
Example 2-12: 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:
1. Check if T or rectangular:
Md = (505)(12) = 6, 060 k.in (2.89-a)
ρb = .85β1f ′c
fy
8787 + fy
= (.85)(.85)460
8787 + 60
= .0285 (2.89-b)
ρmax = .75ρb = (.75)(.0285) = .0213 (2.89-c)Amaxs = (.0213)(12)(21) = 5.37 in2 (2.89-d)
a =Asfy.85f ′
cb=
(5.37)(60)(.85)(4)(12)
= 7.89 in (2.89-e)
Mmax = (0.9)Asfy(d− a
2
)= (.9)(5.37)(60)
(21− 7.89
2
)= 4, 943 k.in < 6, 060 k.in(2.89-f)
Thus compression steel is required.
2. Assuming that f ′s = fy
Md2 = 6, 060− 4, 943 = 1, 117 k.in (2.90-a)
A′s =
Md2
φfy(d− d′)=
1, 117(0.9)(60)(21− 2.5)
= 1.12 in2 (2.90-b)
⇒ A′s = 1.12 in2 (2.90-c)
As = 5.37 + 1.12 = 6.49 in2 (2.90-d)
3. Check that f ′s = fy
ρ′ =1.12
(12)(21)= .00444 (2.91-a)
ρ =6.49
(12)(21)= .0257 (2.91-b)
ρmin = ρ′ + .85β1f ′c
fy
d′
d
εuεu − εy
(2.91-c)
= .00444 + (.85)(.85)460
2.521.0
8787− 60
= .0229 < ρ(.0257)√
(2.91-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).
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.8 Bond & Development Length 2–31
2.8 Bond & Development Length
51 Considering the equilibrium of forces acting on an infinitesimal portion of a rebar, Fig. 2.17,and defining U as the force per unit length, we have
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M ∆M + M
C
T
dx
V V+dVT+dT
C + dC
ZT+dTT
dx
Figure 2.17: Bond and Development Length
Udx = dT ⇒ U =dT
dx(2.92)
52 The tensile force is a function of the moment
M = Tjd (2.93-a)
dT =dM
jd(2.93-b)
53 But the shear is related to the moment
V =dM
dx(2.94)
Combing those equations together, we obtain
U =V
jd(2.95)
54 We define u as the bond stress, and is equal to
u =U
Σ0(2.96)
where Σ0 is the sum of all the bars perimeters.
55 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.
56 Actual stress distribution along steel bar is quite complex, Fig. 2.18.
57 If bond stress is too large ⇒ splitting along reinforcement, Fig. 2.19.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–32 FLEXURE
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u stresses on rebar
u stresses on concreteMM
Bond stress u
Steel tension slope = dTdx
Figure 2.18: Actual Bond Distribution
Figure 2.19: Splitting Along Reinforcement
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.8 Bond & Development Length 2–33
58 Failure will initiate at points of high shear(large dM
dx
).
59 It frequently starts at diagonal cracks ⇒ dowel action increases the tendancy of splitting ⇒shear and bond failures are often interrelated.
60 Based on tests with one single bar, ultimate average bond force/inch of length of bar isUn ≈ 35
√f ′c.
61 If we have several bars in one layer spaced 6 in or less, then the ultimate bond capacity is80% of the single bar case.
62 In terms of bond stress, Fig. 2.20
un =35√
f ′c
Σ0(2.97)
s bT = A fy
��������������������������������������������������������������������
T = 0s Ld
Figure 2.20: Development Length
63 Putting it differently, the minimum length necessary to develop through bond a force Asfyis, Fig. ??.
Ld = AbfyUn
Un = unΣ0
un = 35√f ′c
Σ0
ld =
0.028Abfy√f ′c
(2.98)
A f
L
Us y
d
Figure 2.21: Development Length
64 For small bar spacing, we have to decrease the bond stress
ld =10.8
0.028Abfy√f ′c
=0.035Abfy√
f ′c
(2.99-a)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–34 FLEXURE
65 If actual development length l is smaller than ld, then we must provide anchorage in orderto avoid a bond failure.
66 Note:
1. Un is independent of diameter
2. For a given fs
T = Abfs
= fsπd2b4
ld = AbfsUn
ld =
fsπd2b4Un
(2.100-a)
ld increases with the square of db ⇒ small bar diameters require shorter developmentlength.
67 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%
68 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
(2.101)
Consult ACI 12.5 code for hooks geometry, and corrections to this basic equation.
69 Check ACI code for modifications related to top reinforcement, lightweight aggregate, highstrength reinforcement, excess reinforcement, and spiral confinement.
ld = λdλddldb (2.102)
70 If not enough development length can be provided ⇒ provide hooks, Fig. 2.22 at
1. 90 degrees: bar must extend by 12db
2. 180 degrees: see code.
wherelhb = 1200 db√
f ′cldh = λdlhb
(2.103)
and λd is given in the ACI code.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.8 Bond & Development Length 2–35
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 2.22: Hooks
2.8.1 Moment Capacity Diagram
71 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.
72 Requirements include, Fig. 2.23:
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.
73 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. 2.24 are satisfactory (note that left support is assumed simply supported).
74 Fig. 2.25 is an illustration of the moment capacity diagram for a beam.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–36 FLEXURE
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 2.23: Bar cutoff requirements of the ACI code
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2.8 Bond & Development Length 2–37
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 2.24: Standard cutoff or bend points for bars in approximately equal spans with uni-formly distributed load
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft2–38 FLEXURE
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 2.25: Moment Capacity Diagram
Victor Saouma Mechanics and Design of Reinforced Concrete
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.
σ
τ
α45 90
45ο
τ
ττ
τσ σ
σ
σ σ
σ 1
1 2
2
α
Tension trajectoriesCompression trajectories
σ
τττ
τ ττ
τ
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:
Large VSmall M
Large VLarge M
Flexural CracksWeb Shear Cracks Web Shear Cracks Flexural Cracks
Figure 3.2: Types of Shear Cracks
Web shear cracks: Large V, small M. They initiate in the web & spread up & down at ≈ 45o.
Flexural shear cracks: Large V, large M. They initiate as an extension of a pre-existingflexural crack, initially vertical, then curve.
Draft3–2 SHEAR
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 ′e 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 Q: What is the maximum shear force which can be applied before a flexural crack developinto 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.
+
Flexure Shear
C
T
jd
v c
σ
τ
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
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 fsn
Mn = Asfsjd ⇒ fs = MnAsjd
fc = α Mn
nAsjd
ρ = Asbd
}fc =
αMn
nρjbd2= F1
Mn
ρnbd2(3.1)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3.2 Shear Strength of Uncracked Section 3–3
2. Shear stressvn = F2
Vnbd
(3.2)
3. From Mohr’s circle, the tensile principal stress is
f σ
τ
1fc
nv
Figure 3.4: Mohr’s Circle for Shear Strength of Uncracked Section
f1 =fc2+
√(fc2
)2+ v2n (3.3)
4. Set f1 equal to the tensile strength
f1 = f ′t ⇒ f1
Vnbd
= f ′t
Vnbd
(3.4-a)
Vnbd
=f ′t
f1
Vnbd
(3.4-b)
=f ′t
f1bdVn
(3.4-c)
Combining Eq. 3.1, 3.2, and 3.3
Vnbd
=f ′t
F12
EcEs︸ ︷︷ ︸C′
1
MnρVnd
+
F12
EcEs︸ ︷︷ ︸C′
1
MnρVnd
2
+ F 22︸︷︷︸C′
2
1/2
(3.5)
5. set f ′t = 4
√f ′c
Vn
bd√
f ′c
=1
C1
√f ′cρ
MnVnd
+
√(C1
√f ′c
ρMnVnd
)2+ C2
(3.6)
6. Let the variables be Vn
bd√f ′c& Mn
√f ′c
ρVnd
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3–4 SHEAR
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 (MV )
10. This equation is usually found acceptable for predicting the flexure shear cracking loadfor shear span/depth ratio Mn
Vndof 2.5 to 6 & is found to be very conservative for lower
values
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
1. No shear reinforcements ⇒ failure
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3.3 Shear Strength of Cracked Sections 3–5
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
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
Vs
Vd
Vcz
Vay
Vint
ΣV int
Incl
ined
cra
ckin
g
Yie
ld o
f st
irup
s
Failu
re
Flex
ural
cra
ckin
g Vext
Figure 3.7: Equilibrium of Shear Forces in Cracked Section
13 We must determine the internal (resisting) shear forces at failure where fv = fy
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3–6 SHEAR
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, 500ρ
VndMd. 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, 500ρw
VudMu]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)
4. If 0.5φVc < Vu ≤ φVc use minimum shear reinforcement; select Av (usually #3 bars) anddetermine
s = Avfy50bw
(ACI 11.5.5.3)s < d
2 (ACI 11.5.4.1)s < 24 in (ACI 11.5.4.1)
(3.12)
5. If Vu > φVc ⇒ provide stirrup such that
Vuφ
= 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)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3.5 Examples 3–7
f ’cf
’ c
f ’ c
f ’c
f ’cf ’c
2
6
104
8
d
not allowable
no s
tirup
ss max=d/4 or 12" s max=d/2 or 24"
min
. stir
ups
s=
Av
f y50
bw
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
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. vc = 4√
f ′c = 2(126) = 252 psi; φvc
2 = 53.6 psi
5. vu − vc = 272− 126 = 146 psi < 4√
f ′c
√
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3–8 SHEAR
19"min. reinforcement no reinforcement
38.6"
x
dpsi
333
272
107.1
53.6
φv c
φv c
2
uV
6. vu − φvc = 0⇒ 333(10)(12)x = 107.1⇒ x = 38.6 in = 3.2 ft from mid-span
7. vu − φvc
2 = 0⇒ 333(10)(12)x = 53.6⇒ x = 19.3 in = 1.6 ft
8. Selecting #3 bars, Av = 2(.11) = .22 in2
smax =Avfy50bw
= (.22)(40,000)(50)(12) = 14.66 in
d2 =
222 = 11 in
}smax = 11 in (3.16)
9. 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 shear frictionconcept.
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, Fig.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3.6 Shear Friction 3–9
#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
Vn Vn
Vn
Vn
Vn
crack
Shear−transferreinforcement
crac
k se
para
tion
due
to s
lip
Avf
fyA f A fvf vfy y
µAvffy
22
Figure 3.10: Shear Friction Mechanism
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3–10 SHEAR
3.10. The shear friction theory is based on the assumption that a crack will occur and thenreinforcement 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
Tcos αf
Tsin αf
assumed crack
applied shear=Vn
Avffy
Tαf
µC
C=Tsin α f
Figure 3.11: Shear Friction Across Inclined Reinforcement
19 Component of tensile force in reinforcement gives rise to compression force at interface C ⇒µc vertical force due to friction;
Vn = T cosαf + µCC = T sinαf
}Vn = T (cosαf − µ sinαf )T = Avffy
}Vn = Avffy(cosαf + µ sinαf(3.19-a)
20 Note: Vu = φVn and φ = 0.85
21 The preceding equation can be rewritten as
Avf =Vu
φµfy(3.20)
Avf =Vu
φfy(cosαf + µ sinαf )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 = 0.7λwhere λ = 1.0 for normal weight concrete and λ = .75 for lightweight concrete. and
Vn <
{0.2f ′
cAc800Ac
(3.22)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3.7 Brackets and Corbels 3–11
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;
3.7 Brackets and Corbels
23 Nu might be due to shrinkage, prestressing · · ·
24 Design based on truss analogy
25 A.C.I. provisions (Chapter 11)
1. For ad < 12 , use shear friction theory
2. For ad > 1, use ordinary beam theory
3. For 12 ≤ ad ≤ 1
Vn = [6.5− 5.1√
NucVu
](1− 0.5α
d)
{1 + [64 + 160
√(NuVu)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′cfy.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft3–12 SHEAR
3.8 Deep Beams
Victor Saouma Mechanics and Design of Reinforced Concrete
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
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 SpansContinuous end unrestrained 1
11wuL2n
Continuous end integral with support 114wuL
2n
Interior spans 116wuL
2n
Negative MomentNegative moment at exterior face of first of first interior supportTwo spans 1
9wuL2n
> Two spans 110wuL
2n
Negative moment at other faces of interior support 111wuL
2n
...................................... ......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 cleas span.
7 These moment coefficients take into account some inelastic action.
8 They are conservative compared to an exact analysis.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft4.2 Methods of Analysis 4–3
Figure 4.3: ACI Approximate Moment Coefficients
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft4–4 CONTINUOUS BEAMS
V
aL2
Column width aLL2
VaL6
VaL
VaL
VaL
aL
VaL
VaL3
3
6
2
62
Adjusted Moment Curve
Moment curve based on prismatic member
C beam
C beam
C beam
C spanC column
C column
L
L
L
LL
L
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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 which is, Fig. 4.4
M−ved ≈ M−ve
max −V b
3(4.1)
10 This can substantially reduce high M−ve.
4.4 Moment Redistribution
4.4.1 Elastic-Perfectly Plastic Section
11 Let us consider a uniformly loaded rigidly connected beam, Fig. 4.5
���
���
������
������
W
W L2
12
W L2
24
+
− −W L2
12
Figure 4.5: Moment Diagram of a Rigidly Connected Uniformly Loaded Beam
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft4.4 Moment Redistribution 4–5
Φu
M
Φy
Mp
Curvature
X
Figure 4.6: Moment Curvature of an Elastic-Plastic Section
12 The beam has an elastic plastic moment curvature relation, Fig. 4.6
13 |M−ve| > |M+ve| as w ↗,M−ve → Mp first ⇒wL2
12= Mp ⇒ w =
12Mp
L2
14 Thus we will have a plastic hinge at the support however this is not synonymous with collapse.
15 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
Mp
Mp
Mp
Figure 4.7: Plastic Moments in Uniformly Loaded Rigidly Connected Beam
2Mp = wuL2
8
wu = 16Mp
L2
16 Thus capacity was increased 33% after first plastic hinge occurred.
17 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 ↗18 The section must be designed to accomodate this rotation.
4.4.2 Concrete
19 Concrete is brittle hence by itself no appreciable plastic deformation can occur, however inR/C, Fig. 4.8
20 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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft4–6 CONTINUOUS BEAMS
θy
θu
θu
θy
φ
f c
A fs y
s y
kd
d−kd
f c
A fε = ε
ys y
c
d−c
s
M
My
Mcr
θcr
u
ε > ε
First crack
Steel yielding
Unit rotation
.003
εc
εy
ε εce ce
Strain caused bymoment redistribution
Figure 4.8: Plastic Redistribution in Concrete Sections
21 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 (4.2)
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
22 M+ve must be increased accordingly.
23 This capacity to redistribute moments (reduce M−ve and increase M+ve) is a characterisitcof ductile members.
24 Earthquake resistant structures must have a certain ductility to absorb the lateral oscillatingload ⇒ large amount of reinforcement at the joints.
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft4.5 Buildings 4–7
Solution:
From above, amout of redistribution
∆M = 20(1− ρ−ρ′ρb)%
= 20(1− 0.5) = 10%M−ve = 0.9wL
2
12
M+ve = 1.2wL2
24 = wL2
20
4.5 Buildings
25 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
26 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.
27 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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft4–8 CONTINUOUS BEAMS
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.
28 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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft
Chapter 5
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
5.1 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
Draft5–2 SERVICEABILITY
4. Distribution of steel over the tensile zone of concrete ←
7 Based on purely experimental research, the following emperical relation was determined, Fig.5.1:
w = .076βfs3√
dcA Gergely & Lutz Eq. (5.1)
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
Steel Centroid
Neutral Axis
y
w
2y
h h
d
1 2
c
���������������
���������������
Figure 5.1: 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) (5.2)
and assumes β = h2h1= 1.2⇒ w = .091z
Interior beams z ≤ 175 (w = .016 in)Exterior beams z ≤ 145 (w = .013 in)
2. Only deformed bars can be used
3. Bars should be well distributed in tension zone
4. fy < 80ksi
5. In lieu of an accurate evaluation, fs = 0.6fy.
9 Maximum acceptable crack width (ACI Committee 224).
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft5.1 Control of Cracking 5–3
Exposure wmax (in.)dry air, or protective membrane .016humidity, moist air, soil .012deicing chemicals .007seawater, salt .006water retaining structures .004
Example 5-1: 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
"
Solution:
1. w = .076βfs 3√dcA
2. Ec = 57√3, 000 = 3, 120 ksi
3. n = 29×1033,120 = 9.29
4. b(kd)2
2 − nAs(d− kd) = 0⇒ k = .393⇒ j = 1− k3 = .869⇒ kd = 7.85 in
5. fs = MAsjd
⇒ fs =(1.27+2.44)(15)2(12)8(3.14)(.869)(20) = 22.9 ksi
Note 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) 1
1,000 = .00696 in .
9. or z = f3s√dcA = (22.9) 3
√(2.5)(14.38) = 75.64
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft5–4 SERVICEABILITY
5.2 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
5.2.1 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?, Fig. 5.2
Α B
Ε Ι Ε Ι
Μ
Ε Ι Ε Ι
Μ
Μ
∆ ∆BΑBBB ΑΑΑ
c
cr
e 2ut c cr
1
c c
2
1
2
e 1
Figure 5.2: Uncracked Transformed and Cracked Transformed X Sections
19 It would be too complicated to have I = I(M)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft5.2 Deflections 5–5
20 ACI recommends to use a weighted average expression for I → Ieff
Ieff =(Mcr
Ma
)3Ig +
[1−
(Mcr
Ma
)3]Icr (ACI 9.5.2.3) (5.3)
whereIeff ≤ IgMcr = f ′
rIgyb
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
∆ =5384
wL4
EI
{w ↗ ∆↗I ↘ ∆↗
5.2.2 Long Term Deflection
δ t
δ inst.
t
Figure 5.3: Time Dependent Deflection
25 Creep coefficient:Cc = εf
εiEtc = σ
ε =σ
εi(1+Cc)= Ec
1+Cc
⇒ Creep tends to reduce the elastic modulus of concrete, Fig. 5.4
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft5–6 SERVICEABILITY
εs
Asfs
fci
fct
ε tεib
d
kd
As
φ i
φ t
Crackedelasticneutral axis
Figure 5.4: Time Dependent Strain Distribution
3. But since C is now lower and we still satisfyMext =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 (5.4)
whereλ = ξ
1+50ρ′
ρ′= A′
sbd
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) (5.5)
LL short
DL sustained
A B C
����������������
����������������
Figure 5.5: Short and long Term Deflections
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft5.2 Deflections 5–7
28 Short and long term deflections, Fig. 5.5
A → δi,sustB → δi,sust + δt,sustC → δsust + δi,short
and
δi, short = δi, sust + short︸ ︷︷ ︸Ieff(DL+LL)
− δi, sust︸ ︷︷ ︸Ieff(DL)
(5.6)
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 5-2: 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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft5–8 SERVICEABILITY
11.5"
7.85"
12.15"
20"
b(kd)2
2 − nAs(d− kd) = 0⇒ k = .393⇒ kd = 7.85 inIct = (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 ′rIgyb
= (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 inδi = .218− .046 = .172 in
8. δcreep = λδi, sust
λ =2.
1 + 0= 2. ⇒ δcreep = (2)(.046) = .092 in
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft
Chapter 6
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.
6.1 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.
Draft6–2 APPROXIMATE FRAME ANALYSIS
8 Based on the first assumption, all beams are statically determinate and have a span, Lsequal 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. 6.1
0.1LL
0.1L0.8L
Vrgt
Vlft
MM rgt
lft
w
����������
����������
�����
�����
����������������
������������
Figure 6.1: Approximate Analysis of Frames Subjected to Vertical Loads; Girder Moments
M+ =18wL2s = w
18(0.8)2L2 = 0.08wL2 (6.1)
Maximum negative moment at each end of the girder is given by, Fig. 6.1
M left =M rgt = −w
2(0.1L)2 − w
2(0.8L)(0.1L) = −0.045wL2 (6.2)
Girder Shear are obtained from the free body diagram, Fig. 6.2
V lft =wL
2V rgt = −wL
2(6.3)
Column axial force is obtained by summing all the girder shears to the axial force transmit-ted by the column above it. Fig. 6.2
P dwn = P up + V rgti−1 − V lfti(6.4)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6.1 Vertical Loads 6–3
Vrgti−1 Vlft
i
Pabove
Pbelow
Figure 6.2: 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 6.3: Approximate Analysis of Frames Subjected to Vertical Loads; Column Moments
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6–4 APPROXIMATE FRAME ANALYSIS
Column Moment are obtained by considering the free body diagram of columns Fig. 6.3
M top = M botabove −M rgt
i−1 +M lfti M bot = −top (6.5)
Column Shear Points of inflection are at mid-height, with possible exception when the columnson the first floor are hinged at the base, Fig. 6.3
V =M top
h2
(6.6)
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.
6.2 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.
6.2.1 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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6.2 Horizontal Loads 6–5
H/2 H/2HH
Figure 6.4: 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. 6.4
V ext =∑
F lateral
2No. of baysV int = 2V ext (6.7)
Column Moments at the end of each column is equal to the shear at the column times halfthe height of the corresponding column, Fig. 6.4
M top = Vh
2M bot = −M top (6.8)
Girder Moments is obtained from the columns connected to the girder, Fig. 6.5
Mcolbelow
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 Vi
rgtV
i−1
rgt
M i−1lft
M irgt
Figure 6.5: Approximate Analysis of Frames Subjected to Lateral Loads; Girder Moment
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6–6 APPROXIMATE FRAME ANALYSIS
M lfti =Mabove
col −M belowcol +M rgt
i−1 M rgti = −M lft
i(6.9)
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. 6.5
V lft = −2ML
V rgt = V lft (6.10)
Column Axial Forces are obtained by summing girder shears and the axial force from thecolumn above, Fig. ??
Vrgti−1 Vlft
i
Pabove
Pbelow
Figure 6.6: Approximate Analysis of Frames Subjected to Lateral Loads; Column Axial Force
P = P above + P rgt + P lft (6.11)
Example 6-1: Approximate Analysis of a Frame subjected to Vertical and Horizontal Loads
Draw the shear, and moment diagram for the following frame. Solution:
Vertical Loads
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6.2 Horizontal Loads 6–7
���� ���� ���� ����
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 6.7: Example; Approximate Analysis of a Building
1. Top Girder Moments
M lft12 = −0.045w12L212 = −(0.045)(0.25)(20)2 = − 4.5 k.ft
M cnt12 = 0.08w12L212 = (0.08)(0.25)(20)2 = 8.0 k.ft
M rgt12 = M lft
12 = − 4.5 k.ftM lft13 = −0.045w13L213 = −(0.045)(0.25)(30)2 = − 10.1 k.ft
M cnt13 = 0.08w13L213 = (0.08)(0.25)(30)2 = 18.0 k.ft
M rgt13 = M lft
13 = − 10.1 k.ftM lft14 = −0.045w14L214 = −(0.045)(0.25)(24)2 = − 6.5 k.ft
M cnt14 = 0.08w14L214 = (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 = −0.045w9L29 = −(0.045)(0.5)(20)2 = − 9.0 k.ft
M cnt9 = 0.08w9L29 = (0.08)(0.5)(20)2 = 16.0 k.ft
M rgt9 = M lft
9 = − 9.0 k.ftM lft10 = −0.045w10L210 = −(0.045)(0.5)(30)2 = − 20.3 k.ft
M cnt10 = 0.08w10L210 = (0.08)(0.5)(30)2 = 36.0 k.ft
M rgt10 = M lft
11 = − 20.3 k.ftM lft11 = −0.045w12L212 = −(0.045)(0.5)(24)2 = − 13.0 k.ft
M cnt11 = 0.08w12L212 = (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 bot5 = −M top
5 = 4.5 k.ftM top6 = −M rgt
12 +M lft13 = −(−4.5) + (−10.1) = − 5.6 k.ft
M bot6 = −M top
6 = 5.6 k.ftM top7 = −M rgt
13 +M lft14 = −(−10.1) + (−6.5) = − 3.6 k.ft
M bot7 = −M top
7 = 3.6 k.ftM top8 = −M rgt
14 = −(−6.5) = 6.5 k.ftM bot8 = −M top
8 = − 6.5 k.ft
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6–8 APPROXIMATE FRAME ANALYSIS
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 top2 = +M bot
6 −M rgt9 +M lft
10 = 5.6− (−9.0) + (−20.3) = − 5.6 k.ftM bot2 = −M top
2 = 5.6 k.ftM top3 = +M bot
7 −M rgt10 +M lft
11 = −3.6− (−20.3) + (−13.0) = 3.6 k.ftM bot3 = −M top
3 = − 3.6 k.ftM top4 = +M bot
8 −M rgt11 = −6.5− (−13.0) = 6.5 k.ft
M bot4 = −M top
4 = − 6.5 k.ft
5. Top Girder Shear
V lft12 = w12L122 = (0.25)(20)
2 = 2.5 kV rgt12 = −V lft12 = − 2.5 kV lft13 = w13L13
2 = (0.25)(30)2 = 3.75 k
V rgt13 = −V lft13 = − 3.75 kV lft14 = w14L14
2 = (0.25)(24)2 = 3.0 k
V rgt14 = −V lft14 = − 3.0 k
6. Bottom Girder Shear
V lft9 = w9L92 = (0.5)(20)
2 = 5.00 kV rgt9 = −V lft9 = − 5.00 kV lft10 = w10L10
2 = (0.5)(30)2 = 7.50 k
V rgt10 = −V lft10 = − 7.50 kV lft11 = w11L11
2 = (0.5)(24)2 = 6.00 k
V rgt11 = −V lft11 = − 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 kP6 = −V rgt12 + V lft13 = −(−2.50) + 3.75 = 6.25 kP7 = −V rgt13 + V lft14 = −(−3.75) + 3.00 = 6.75 kP8 = −V rgt14 = 3.00 kVictor Saouma Mechanics and Design of Reinforced Concrete
Draft6.2 Horizontal Loads 6–9
0.25K/ft
0.50K/ft 14’14’
16’16’
20’20’ 30’30’ 24’24’
5 6
1
12 13 14
11109
87
3 42
-9.0’-9.0’k
-4.5’-4.5’k
-10.1’-10.1’k -10.1’-10.1’k
-13.0’-13.0’k
-4.5’-4.5’k
-4.5’-4.5’k
+4.5’+4.5’k
+4.5’+4.5’k +5.6’+5.6’k
+5.6’+5.6’k
-5.6’-5.6’k
-5.6’-5.6’k +3.6’+3.6’k
+3.6’+3.6’k
-3.6’-3.6’k
-3.6’-3.6’k
-6.5’-6.5’k
-6.5’-6.5’k
+6.5’+6.5’k
+6.5’+6.5’k
-13.0’-13.0’k
-20.2’-20.2’k-20.2’-20.2’k
+8.0’+8.0’k +18.0’+18.0’k
+11.5’+11.5’k
+16.0’+16.0’k
+32.0’+32.0’k+23.0’+23.0’k
-6.5’-6.5’k-6.5’-6.5’k
-4.5’-4.5’k
-9.0’-9.0’k
Figure 6.8: Approximate Analysis of a Building; Moments Due to Vertical Loads
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6–10 APPROXIMATE FRAME ANALYSIS
+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 6.9: Approximate Analysis of a Building; Shears Due to Vertical Loads
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6.2 Horizontal Loads 6–11
9. Bottom Column Axial Forces
P1 = P5 + V lft9 = 2.50 + 5.0 = 7.5 kP2 = P6 − V rgt10 + V lft9 = 6.25− (−5.00) + 7.50 = 18.75 kP3 = P7 − V rgt11 + V lft10 = 6.75− (−7.50) + 6.0 = 20.25 kP4 = P8 − V rgt11 = 3.00− (−6.00) = 9.00 k
Horizontal Loads, Portal Method
1. Column ShearsV5 = 15
(2)(3) = 2.5 kV6 = 2(V5) = (2)(2.5) = 5 kV7 = 2(V5) = (2)(2.5) = 5 kV8 = V5 = 2.5 kV1 = 15+30
(2)(3) = 7.5 kV2 = 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.ftM top6 = V6H6
2 = (5)(14)2 = 35.0 k.ft
M bot6 = −M top
6 = − 35.0 k.ftM top7 = V up
7 H7
2 = (5)(14)2 = 35.0 k.ft
M bot7 = −M top
7 = − 35.0 k.ftM 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 rgt12 = −M lft
12 = − 17.5 k.ftM lft13 = M rgt
12 +M top6 = −17.5 + 35 = 17.5 k.ft
M rgt13 = −M lft
13 = − 17.5 k.ftM lft14 = M rgt
13 +M top7 = −17.5 + 35 = 17.5 k.ft
M rgt14 = −M lft
14 = − 17.5 k.ft
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6–12 APPROXIMATE FRAME ANALYSIS
Approximate Analysis Vertical Loads APROXVER.XLS Victor E. Saouma
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123456789
101112131415161718192021222324252627282930
A B C D E F G H I J K L M N O P Q
L1 L2 L3Height Span 20 30 24
14 Load 0.25 0.25 0.2516 Load 0.5 0.5 0.5
MOMENTSBay 1 Bay 2 Bay 3
Col Beam Column Beam Column Beam ColLft Cnt Rgt Lft Cnr Rgt Lft Cnt Rgt
-4.5 8.0 -4.5 -10.1 18.0 -10.1 -6.5 11.5 -6.5-4.5 -5.6 3.6 6.54.5 5.6 -3.6 -6.5
-9.0 16.0 -9.0 -20.3 36.0 -20.3 -13.0 23.0 -13.0-4.5 -5.6 3.6 6.54.5 5.6 -3.6 -6.5
SHEARBay 1 Bay 2 Bay 3
Col Beam Column Beam Column Beam ColLft Rgt Lft Rgt Lft Rgt
2.50 -2.50 3.75 -3.75 3.00 -3.00-0.64 -0.80 0.52 0.93
5.00 -5.00 7.50 -7.50 6.00 -6.00-0.56 -0.70 0.46 0.81
AXIAL FORCEBay 1 Bay 2 Bay 3
Col Beam Column Beam Column Beam Col0.00 0.00 0.00
2.50 6.25 6.75 3.000.00 0.00 0.00
7.50 18.75 20.25 9.00
Figure 6.10: Approximate Analysis for Vertical Loads; Spread-Sheet Format
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6.2 Horizontal Loads 6–13
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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
AB
CD
EF
GH
IJ
KL
MN
OP
Q
L1L2
L3H
eig
htSp
an
2030
2414
Loa
d0.
250.
250.
2516
Loa
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50.
5M
OM
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Bay
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*I4*
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*D5*
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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 6.11: Approximate Analysis for Vertical Loads; Equations in Spread-Sheet
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6–14 APPROXIMATE FRAME ANALYSIS
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 lft10 = M rgt
9 +M top2 −M bot
6 = −77.5 + 120− (−35) = 77.5 k.ftM rgt10 = −M lft
10 = − 77.5 k.ftM lft11 = M rgt
10 +M top3 −M bot
7 = −77.5 + 120− (−35) = 77.5 k.ftM 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 6.12: Approximate Analysis of a Building; Moments Due to Lateral Loads
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6.2 Horizontal Loads 6–15
6. Top Girder Shear
V lft12 = −2M lft12
L12= − (2)(17.5)
20 = −1.75 kV rgt12 = +V lft12 = −1.75 kV lft13 = −2M lft
13L13
= − (2)(17.5)30 = −1.17 k
V rgt13 = +V lft13 = −1.17 kV lft14 = −2M lft
14L14
= − (2)(17.5)24 = −1.46 k
V rgt14 = +V lft14 = −1.46 k
7. Bottom Girder Shear
V lft9 = −2M lft12L9
= − (2)(77.5)20 = −7.75 k
V rgt9 = +V lft9 = −7.75 kV lft10 = −2M lft
10L10
= − (2)(77.5)30 = −5.17 k
V rgt10 = +V lft10 = −5.17 kV lft11 = −2M lft
11L11
= − (2)(77.5)24 = −6.46 k
V rgt11 = +V lft11 = −6.46 k
8. Top Column Axial Forces (+ve tension, -ve compression)
P5 = −V lft12 = −(−1.75) kP6 = +V rgt12 − V lft13 = −1.75− (−1.17) = −0.58 kP7 = +V rgt13 − V lft14 = −1.17− (−1.46) = 0.29 kP8 = V rgt14 = −1.46 k
9. Bottom Column Axial Forces (+ve tension, -ve compression)
P1 = P5 + V lft9 = 1.75− (−7.75) = 9.5 kP2 = P6 + V rgt10 + V lft9 = −0.58− 7.75− (−5.17) = −3.16 kP3 = P7 + V rgt11 + V lft10 = 0.29− 5.17− (−6.46) = 1.58 kP4 = 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 6.2.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6–16 APPROXIMATE FRAME ANALYSIS
Portal Method PORTAL.XLS Victor E. Saouma
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123456789
101112131415161718192021222324252627282930
A B C D E F G H I J K L M N O P Q R SPORTAL METHOD
# of Bays 3 L1 L2 L320 30 24
MOMENTS# of Storeys 2 Bay 1 Bay 2 Bay 3
Force Shear Col Beam Column Beam Column Beam ColH Lat. Tot Ext Int Lft Rgt Lft Rgt Lft Rgt
17.5 -17.5 17.5 -17.5 17.5 -17.5H1 14 15 15 2.5 5 17.5 35.0 35.0 17.5
-17.5 -35.0 -35.0 -17.577.5 -77.5 77.5 -77.5 77.5 -77.5
H2 16 30 45 7.5 15 60.0 120.0 120.0 60.0-60.0 -120.0 -120.0 -60.0
SHEARBay 1 Bay 2 Bay 3
Col Beam Column Beam Column Beam ColLft Rgt Lft Rgt Lft Rgt
-1.75 -1.75 -1.17 -1.17 -1.46 -1.462.50 5.00 5.00 2.502.50 5.00 5.00 2.50
-7.75 -7.75 -5.17 -5.17 -6.46 -6.467.50 15.00 15.00 7.507.50 15.00 15.00 7.50
AXIAL FORCEBay 1 Bay 2 Bay 3
Col Beam Column Beam Column Beam Col0.00 0.00 0.00
1.75 -0.58 0.29 -1.460.00 0.00 0.00
9.50 -3.17 1.58 -7.92
Figure 6.13: Portal Method; Spread-Sheet Format
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6.2 Horizontal Loads 6–17Portal Method PORTAL.XLS Victor E. Saouma
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AA
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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
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20 30 24
MOMENTSAAAAAAAA
AAAAAAAA
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# of Storeys 2 Bay 1 Bay 2 Bay 3AAAA
AAAA
AAAA
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Force Shear Col Beam Column Beam Column Beam ColAAAA
AAAA
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H Lat. Tot Ext Int Lft Rgt Lft Rgt Lft Rgt
=+H9 =-I8 =+J8+K9 =-M8 =+N8+O9 =-Q8AAAAAAAA
AAAAAAAA
AAAAAAAA
AAAAAAAA
AAAAAAAA
H1 14 15 =+C9 =+D9/(2*$F$2) =2*E9 =+E9*B9/2 =+F9*B9/2 =+K9 =+H9
=-H9 =-K9 =+K10 =+H10
=+H12-H10 =-I11 =+K12-K10+J11 =-M11 =+O12-O10+N11 =-Q11AAAAAAAAAAA
AAAAAAAAAAA
H2 16 30 =SUM($C$9:C12) =+D12/(2*$F$2) =2*E12 =+E12*B12/2 =+F12*B12/2 =+K12 =+H12
=-H12 =-K12 =+K13 =+H13
SHEARAAAAAAAAAAAA
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Bay 1 Bay 2 Bay 3AAAAA
AAAAA
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AAAAA
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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
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Col Beam Column Beam Column Beam ColAAAA
AAAA
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AAAA
0 0 0
=-I18 =+J18-M18 =+N18-Q18 =+R18
0 0 0
=+H28-I21 =+K28+J21-M21 =+O28+N21-Q21 =+S28+R21AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
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Figure 6.14: Portal Method; Equations in Spread-Sheet
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6–18 APPROXIMATE FRAME ANALYSIS
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
3 Axial 20.25 14.25 34.50Shear 0.45 15.00 15.45Moment 6.50 60.00 66.50
4 Axial 9.00 7.92 16.92Shear 0.81 7.50 8.31Moment 4.50 17.50 22.00
5 Axial 2.50 1.75 4.25Shear 0.64 2.50 3.14Moment 5.60 35.00 40.60
6 Axial 6.25 2.92 9.17Shear 0.80 5.00 5.80Moment 3.60 35.00 38.60
7 Axial 6.75 2.63 9.38Shear 0.51 5.00 5.51Moment 6.50 17.50 24.00
8 Axial 3.00 1.46 4.46Shear 0.93 2.50 3.43
Table 6.1: Columns Combined Approximate Vertical and Horizontal Loads
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6.2 Horizontal Loads 6–19
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
11 +ve Moment 23.00 0.00 23.00Shear 6.00 6.46 12.46-ve Moment 4.50 17.50 22.00
12 +ve Moment 8.00 0.00 8.00Shear 2.50 1.75 4.25-ve Moment 10.10 17.50 27.60
13 +ve Moment 18.00 0.00 18.00Shear 3.75 1.17 4.92-ve Moment 6.50 17.50 24.00
14 +ve Moment 11.50 0.00 11.50Shear 3.00 1.46 4.46
Table 6.2: Girders Combined Approximate Vertical and Horizontal Loads
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft6–20 APPROXIMATE FRAME ANALYSIS
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft
Chapter 7
ONE WAY SLABS
7.1 Types of Slabs
1 Types of slabs, Fig. 7.1
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 7.1: Types of Slabs
2 Two types of slabs, Fig. 7.2
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 (≈ ρ5%) is carried in the short directions, Fig. 7.3
4 Load transfer in one way slabs is accomplished hierarchically through an interaction of slab,beam, girder, column and foundations, Fig. 7.4
Draft7–2 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
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B
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Figure 7.3: Load Distribution in Slabs
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft7.1 Types of Slabs 7–3
Figure 7.4: Load Transfer in R/C Buildings
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft7–4 ONE WAY SLABS
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 7.1: Recommended Minimum Slab and Beam Depths
7.2 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
Asft
= Ab
(12 in
bar spacing in inches
)(7.1)
where Ab is the area of one bar. or
Bar spacing in inches =12AbAs
(7.2)
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 7.1. 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
Asbh
=
0.002 Grade 40 and 50 bars
ACI 7.12.2.10.0018 Grade 60 and welded wire fabric
(7.3)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft7.3 Design of a One Way Continuous Slab 7–5
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).
7.3 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 (7.4-a)
li = (15)(12)− 2142= 166 in (7.4-b)
hemin =l
24=16524
= 6.88 in (7.4-c)
himin =l
28=16628
= 5.93 in (7.4-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 (7.5)
Factored Loads Slab
wd =7.2512
(150) = 90.6 psf of floor surface (7.6)
Total dead load 90.6 + 0.5 + 4 + 2 = 97.1 psf
Factored loadwu = 1.4(97.1) + 1.7(100) = 306 psf (7.7)
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft7–6 ONE WAY SLABS
2. Second interior span ln = (15)(12)− 14 = 166 in = 13.83 ft
3. Average span ln = 12(165 + 166)
112 = 13.79 ft
Flexural Design
ai = 0.15d = 0.15(6.25) = 0.9375 in (7.8-a)
As =Mu
φfy(d− a2 )=
12Mu
0.9(60)(6.25− a2 )=
0.2226.25− a
2
Mu (7.8-b)
a =Asfy0.85f ′
cb=
60(0.85)(3.75)(12)
As = 1.569As (7.8-c)
Amins = 0.0018bh = 0.0018(12)(7.25) = 0.157 in2/ft (7.8-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 13.83wul
2n 57.85 57.85 58.19 58.53 58.53
M 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
√Amins 0.157
√0.157
√0.157 0.157
√0.157
Reinf. #4@15 #4 @15 #4@12 #4@15 #4@12Aprovs 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 (7.9)
2. Typical interior span
Vu = 1.0wuln2=(1.0)(306)(166)
2= 2, 117 lb/ft of width (7.10)
The shear resistance is
φVc = (0.85)2√
f ′cbwd = (0.85)(2)√3, 750(12)(6.25) = 7, 808lb/ft (7.11)
hence the slab is adequate for shear.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft7.3 Design of a One Way Continuous Slab 7–7
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 (7.12)
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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft7–8 ONE WAY SLABS
Victor Saouma Mechanics and Design of Reinforced Concrete
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
1. Restrains longitudinal steel from outward buckling
2. Restrains Poisson’s expansion of concrete
3. Acts as shear reinforcement for horizontal (wind & earthquake) load
4. Provide ductility
very important to resist earthquake load.
9.1.2 Possible Arrangement of Bars
3 Bar arrangements, Fig. 9.3
Draft9–2 COLUMNS
P
Spiral
TiedX X
δ
Figure 9.2: Tied vs Spiral Reinforcement
4 bars 6 bars 8 bars
10 bars
12 bars
14 bars
16 bars
Wall column
Corner column
Figure 9.3: Possible Bar arrangements
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.2 Short Columns 9–3
9.2 Short Columns
9.2.1 Concentric Loading
4 No moment applied,Elastic Behaviour
P = fcAc + fsAs= fc(Ac + nAs)
Ultimate Strength
Pd = φPnPn = .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
ML M
R
Pe
ML
MR
ML
MR
Pn
Pn
−e=
Figure 9.4: Sources of Bending
1. Unsymmetric moments ML �= MR
2. Uncertainty of loads (must assume a minimum eccentricity)
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
7 Assumptions A′s = As; ρ = As
bd =A′
sbd = f ′
s = fy
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–4 COLUMNS
M n
Pn
e small
e =
0; a
= h
; c =
c ~ h; e=
c e
Compressionfailure range
Radial lines show
Balanced Failure
Tension failure range
Load path forgivin e
e large
e b
constant e=M n
Pn
8
8
M0
P0
y
cu
yε
ε
εsu
ε
cuε
cu
ε
>
ε cu
Figure 9.5: Load Moment Interaction Diagram
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
9 From the strain diagram (and compatibility of concrete and steel strains), Fig. 9.6
εc = .003 (9.1-a)
εy =fyEs
(9.1-b)
c =εu
εu + εyd =
.003fyEs+ .003
d (9.1-c)
(9.1-d)
Furthermore,
ε′sc− d′
=εcc
(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:
Pn = .85f ′cab+A′
sf′y −Asfs
fs = fyAs = A′
s
a = Pn.85f ′cb
a = β1cbcb = .003
fyEs+.003
d
Pn,b = .85β1f ′
cbd.003
fyEs+ .003
(9.3)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.2 Short Columns 9–5
A fs s
εs
A fs’ s
dd’
h/2
A’Ass
b
ε’s
ae
e’
0.85f’c
Pnc
A fs s
c
εsc
A fs
A fs’
y
y
Figure 9.6: Strain and Stress Diagram of a R/C Column
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)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–6 COLUMNS
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′
sbd
µ = fy.85f ′c
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 = εcEs
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
fsEs+ εu
d
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.2 Short Columns 9–7
⇒ fs = Esεud− c
c
= Esεud− a
β1
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 �= 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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–8 COLUMNS
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
0 φMn Mn M
0.10f’ c A g
eb
P d −M d
Pn−M n (M n’ Pn
)
(Mnb’
Pnb )e min
φPn(max)
Pn(max)
P0
P
A
Tied: P
n(max)
n(max)
= 0.85 P
= 0.80 P0
0Spir. reinf: P
1e
Compressioncontrol region
Tensioncontrol region
Balanced failure
Figure 9.7: Column Interaction Diagram
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 Mnbh2f ′c
= χe
h for various ρt where ρt =As+A′
sbh and
µ = fy.85f ′c
Example 9-1: R/C Column, c known
A 12 by 20 in. column is reinforced with four No. 4 bars of area 1.0 in2 each, at 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:
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.2 Short Columns 9–9
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
fyEs+ .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− 9.44
2
)+ (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− 2.55.0
= 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− 4.25
2
)+ (2.0)(43.5)
(202
− 2.5)+ (2.0)(50)
(17.5− 20
2
)(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)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–10 COLUMNS
= (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− 2.518.0
= 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− 15.3
2
)+ (2.0)(50)
(202
− 2.5)+ (2.0)(−2.42)
(17.5− 20
2
)(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
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 =fyEs
=40
29, 000= .001379 (9.15-a)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.2 Short Columns 9–11
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 =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 co0mpression 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− 12.2
2
)+ 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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–12 COLUMNS
A f
a/2
c=23.5"
a=20"
A’ fA f
h−c−
d’
ε εεscs y
3" 3"
24"
20"
12"
y
C
P
ysc c
s
sc
s
n
s
c
h/2=12"e’=11.4
9"
.003
e=2.
4"
.85f’
ε
1. Assume a = 20 inc =
a
β1=20.85
= 23.5 in (9.17)
2. For center steel (from geometry)
εsc
c− h2
=.003c
(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− 1223.5
.003 = 42.5 ksi > fy ⇒ fsc = fy (9.18-e)
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)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.2 Short Columns 9–13
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
εsh− 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.2. Assume a = 7.9 in⇒ c = a
β1= 7.9.85 = 9.3 in
3. Steel stress at centroidc
.003=
12− c
εsc(9.22-a)
⇒ εsc =12− 9.39.3
.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)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–14 COLUMNS
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 =Pnbhf ′
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.2 Short Columns 9–15
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Pn
M 0 x
M 0 y
M n y
M n x
Figure 9.8: Failure Surface of a Biaxially Loaded Column
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-mensional equation for the moment contour at a constant Pn may be expressed as(
Mnx
M0x
)α1
+(Mny
M0y
)α1
= 1.0
whereMnx = PneyMny = PnexM0x = Mnx capacity at axial load Pn when Mny (or ex) is zeroM0y = Mny capacity at axial load Pn when Mnx (or ey) is zeroand 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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–16 COLUMNS
1.0
1.0
β
β
AA
BB
CC
M ny /M0yMny
M0y
M0y
Mnx /M0x
MnxM0x
M0x
βM0x
βM 0y
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(βM0xM0x
)α+(βM0y
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-a)
Mnx
M0x+
Mny
M0y
(1− β
β
)= 1 If
Mny
M0y≤ Mnx
M0x(9.29-b)
Design of a column
Mny +Mnx
(M0y
M0x
)(1− β
β
)= M0y If
Mny
Mnx≥ M0y
M0x(9.30-a)
Mnx +Mny
(M0x
M0y
)(1− β
β
)= M0x If
Mny
Mnx≤ M0y
M0x(9.30-b)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.2 Short Columns 9–17
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
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Pn
M 0 x
M 0 y
M n y
M n x
45 o
1−β
1−β
(1− β/ )
(1− β/ )
β
β
β
ββ
β
B
C
A
Mny
/M0y
M ny /M 0y
M ny /M0y
M nx /M 0x
M nx /M 0x
(Mnx
/M 0x )α+(Mny /M0y)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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–18 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′ = 2.5in, and thereare 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.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−ββ
)≤ 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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.3 Long Columns 9–19
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 (9.31)
Note that the value of yis irrelevant.
30 Recalling thatd2y
dx2=
Mz
EI(9.32)
upon substitution, we obtain the following differential equation
d2y
dx2− P
EI= 0 (9.33)
31 Letting k2 = PEI , the solution to this second-order linear differential equation is
y = −A sin kx−B cos kx (9.34)
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 (9.35)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–20 COLUMNS
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 (9.36)
Thus buckling will occur if PEI =
(nπL
)2 orP =
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(9.37)
The corresponding critical stress is
σcr =π2E(Lr
)2 (9.38)
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 kLr 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 =Ler
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.3 Long Columns 9–21
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–22 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(EIL )of columns
Σ(EIL )of floor members(9.39)
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
)240 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
(9.40)
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
(9.41-a)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.3 Long Columns 9–23
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–24 COLUMNS
M 0
P∆
kl/r
M
M
M 0
Pcr
P n
P
M 0 M n 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 (9.41-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 sectionkLur < 34− 12M1
M2braced, neglect slenderness ACI 10.11.4
kLr < 22 unbraced, neglect slenderness
45 From conventional elastic analysis get Pn&Mn
Mc = δM2 (9.42)
δ =Cm
1− PnφPcr
≥ 1.0 (9.43)
Pn =π2EI
(kLu)210.11.5 (9.44)
Cm = .6 + 4M1
M2(9.45)
EI =EcIg5 + EsIs
1 + βd(9.46)
or EI =EcIg2.5
1 + βd(9.47)
βd =MD
MD +ML(9.48)
βd is the ratio of maximum design load moment to maximum design total load moment (always
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.3 Long Columns 9–25
+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.
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 (9.49-a)
r = .25d = (.25)(14 in) = 3.5 in (9.49-b)Ec = 57, 000
√f ′c = 57, 000
√5, 000 = 4, 030 ksi (9.49-c)
Ig =πd4
64=
π(14)4
64= 1, 886 in4 (9.49-d)
EIcol =EcIg2.51+βd
βd = 0
}EIcol = (4, 030)(1, 886)
12.5
= 3, 040, 000 ksi (9.50-a)
(EI
L
)c
=3, 040, 000(15)(12)
= 16, 890 k.in (9.51-a)
Ibeam = Icr � Ig2=(14)(22)3
1212= 6, 210 in4 (9.51-b)(
EI
L
)beam
=(4, 030)(6, 210)
(12)(40)= 52, 140 k.in (9.51-c)
ΨA =Σ(EI/L)colΣ(EI/L)beam
=2(16, 890)2(52, 140)
= .324 (9.51-d)
bottom column I = π(16)4
64 = 3, 217 in4
EI =(4, 030)(3, 217)
2.5= 5, 186, 000 (9.52-a)(
EI
L
)col
=5, 186, 000(15)(12)
= 28, 800 k.in (9.52-b)
ΨB =16, 890 + 28, 800
2(52, 140)= .438 (9.52-c)
From ACI commentary ΨA = .324,ΨB = .438,⇒ k � .65 and
kLur
=(.65)(13.16)(12)
3.5= 29.3 (9.53-a)
34− 12M1
M2= 34− 12 = 22 assuming M1 = M2 (9.53-b)
kL
r> 22⇒ consider column instability (9.53-c)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–26 COLUMNS
Pcr =π2EI
(kl)2=
π2(3, 040, 000)[(.65)(13.16)(12)]2
= 2, 848 k (9.53-d)
CM = .6 + .4M1
M2= 1 (9.53-e)
δ =1
1− PuφPcr
=1
1− 500(.75)(2,848)
= 1.3 (9.53-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
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Solution:
Pu = 1.4× 46 + 1.7× 94 = 224 k (9.54-a)Mu = 1.4× 92 + 1.7× 230 = 520 k.ft (9.54-b)
βd =(1.4)92520
= .24 (9.54-c)
Assume a 22× 22 inch column and ρt = .03
8.5
2.5
22"
22"
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9.3 Long Columns 9–27
If =224
12= 19, 500 in4 (9.55-a)
Is = (2)(.015)(22)2(8.5)2 = 1, 050 in4 (9.55-b)Ec = 57, 000
√4, 000 = 3.6× 106 psi (9.55-c)
Es = 29× 106 ksi (9.55-d)
EI =EcIg5 + EsIs
1 + βd
=(3.6×106)(19,500)
5 + (29× 106)(1, 050)1 + .24
= 3.59× 1010 (9.55-e)
EIcL
=3.59× 1010
12× 18= 1.66× 108 (9.55-f)
EIbL
= (3.6× 106)(43.3) = 1.56× 108 (9.55-g)
AtA&B Ψ =2(1.66× 108)1.56× 108
= 2.13 from ACI commentary k = 1.65 (9.55-h)
if kLr = 22 neglect slenderness
r = (.3)(22) = 6.6 ∈ (9.56-a)
⇒ kL
r=
(1.65)(18)(12)6.6
= 54 > 22 (9.56-b)
Pcr =π2EI
(kL)2=
π2(3.59× 1010)[(1.65)(18)(12)]2
= 279× 106 lbs (9.56-c)
Pu = 2.24× 105lb (9.56-d)Cm = 1.0(unbraced) (9.56-e)
Moment Magnification δ =1
1− PuφPcr
=1
1− (2.24×105)(.7)(2.79×106)
(9.56-f)
= 1.13 (9.56-g)
⇒ Moment for which the column is to be designed (1.13) (520) = 587 k.ft and Pu = 224
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft9–28 COLUMNS
Victor Saouma Mechanics and Design of Reinforced Concrete
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, (?)
Tendon in conduct
BeamJackAnchorage
Wrapped tendon
Slab
AnchorageJack
JackIntermediatediaphragms
Anchorage
Figure 10.2: Posttensioned Prestressed Concrete Beam, (?)
Victor Saouma Mechanics and Design of Reinforced Concrete
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.
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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
fpfpi
= 1− log t10
(fpifpy
− .55)
(10.3)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10–4 PRESTRESSED CONCRETE
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 = I
c2, (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.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10.1 Introduction 10–5
Q
P Ph/2
h/3
2Q
P Ph/2
h/3
2Q
P P 2h/3
P P h/2
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h
f’
f
f
y
fc
c
c
f =fc t
2fc
2f 2fc
02f =2ft c
c
2fc
0
0 2fc
2f =2ft c
2fc
02fc
f
fc
cc
c
f
f
0
2fc f =f
c
t c
f
fc
c
fcc
c cf f
f
+
+
+
+
+
+ =
=
=
=
=
=0
0
Midspan
Ends
Midspan
Q
Ends
f
Figure 10.4: Alternative Schemes for Prestressing a Rectangular Concrete Beam, (?)
θ
θP
P e P PP e P e
ePθM
P PNone
PP
2
None
P P
(f)
(g)
P
(c)
Moment from prestressingMember
P
P
P
2
P cos
P cos
P sinP sin
MθP sin
P sin θθ
P cosP sin
P cosθ
Equivalent load on concrete from tendon
P
θ P sin
θ
θθ
θθ
P cos θ P cos
θ
θP sin
P cos
P sinθP
P cosθ θP sinθ
θ
P sin
(e)
(d)
(a)
(b)
Figure 10.5: Determination of Equivalent Loads
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10–6 PRESTRESSED CONCRETE
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, (?)
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 prestressingforce, Pi only
f1 = − PiAc
+Piec1
I= − Pi
Ac
(1− ec1
r2
)(10.4)
f2 = − PiAc
− Piec2I
= − PiAc
(1 +
ec2r2
)(10.5)
2. Pi and the self weight of the beam M0 (which has to be acconted for the moment thebeam cambers due to prestressing)
f1 = − PiAc
(1− ec1
r2
)− M0
S1(10.6)
f2 = − PiAc
(1 +
ec2r2
)+
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10.2 Flexural Stresses 10–7
f1 = −PeAc
(1− ec1
r2
)− M0
S1(10.8)
f2 = −PeAc
(1 +
ec2r2
)+
M0
S2(10.9)
4. Pe and M0 +MDL +MLL
f1 = −PeAc
(1− ec1
r2
)− M0 +MDL +MLL
S1(10.10)
f2 = −PeAc
(1 +
ec2r2
)+
M0 +MDL +MLL
S2(10.11)
The internal stress distribution at each one of those four stages is illustrated by Fig. 10.7.
S2
Mo+
S2
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e c 2
c 1
Pi e c
Ic2Stage 1
Stage 2
Stage 4
re c
2 )2( 1 +P i
Ac
Mo+
S2 re c
2 )2( 1 +P i
Ac
P i
Ac
P i
Ac
Pi e c 1
Ic
P i
Ac( 1 -
re c
21 )
re c
2 )2( 1 +P i
Ac
P i
Ac( 1 -
re c
21 ) -
Mo
S1-
Mo
S1( 1 -
re c
21 )
P i
Ac
Ac
P e-
Mo
S( 1 -
re c
21 ) -
Md + Ml
S-
S1
Mt
Ac
P e( 1 -
re c
21 )
Md + Ml
S+ +
Mtr
e c2 )2( 1 +
Ac
P er
e c2 )2( 1 +
Ac
P e Mo+
S2 2
1 1
Figure 10.7: Flexural Stress Distribution for a Beam with Variable Eccentricity; MaximumMoment Section and Support Section, (?)
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):
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10–8 PRESTRESSED CONCRETE
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 (?)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 = − PiAc
(1− ec1
r2
)(10.12-a)
= −169, 000176
(1− (5.19)(12)
68.2
)= −83 psi (10.12-b)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10.2 Flexural Stresses 10–9
f2 = − PiAc
(1 +
ec2r2
)(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 = − PiAc
(1− ec1
r2
)− M0
S1(10.15-a)
= −83− 439 = −522 psi (10.15-b)
fti = 3√
f ′c = +190
√(10.15-c)
f2 = − PiAc
(1 +
ec2r2
)+
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 = −PeAc
(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 = −PeAc
(1 +
ec2r2
)+
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.
4. Pe and M0 +MDL +MLL
MDL +MLL =(0.55)(40)2
8= 110 k.ft (10.17)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10–10 PRESTRESSED CONCRETE
and corresponding stresses
f1,2 = ∓(110)(12, 000)1, 000
= ∓1, 320 psi (10.18)
Thus,
f1 = −PeAc
(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 = −PeAc
(1 +
ec2r2
)+
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.-8
3-5
10-5
22
-183
0
-139
8
-112
2
-183
7
+19
8
1234
10.3 Case Study: Walnut Lane Bridge
Adapted from (?)
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
Victor Saouma Mechanics and Design of Reinforced Concrete
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
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10–12 PRESTRESSED CONCRETE
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
10.3.1 Cross-Section Properties
21 The beam cross section is shown in Fig. 10.9 and is simplified
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)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10.3 Case Study: Walnut Lane Bridge 10–13
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 = − PiAc
(1− ec1
r2
)(10.29-a)
= −(2× 106)1, 354
(1− (31.8)(39.5)
943.
)= 490. psi (10.29-b)
f2 = − PiAc
(1 +
ec2r2
)(10.29-c)
= −(2× 106)1, 354
(1 +
(31.8)(39.5)943.
)= −3, 445. psi (10.29-d)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10–14 PRESTRESSED CONCRETE
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 = − PiAc
(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 =PiAc
(1 +
ec2r2
)+
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 andM0. If we have 13% losses, then the effective force Pe is equal to (1−0.13)(2×106) =1.74× 106 lbs
f1 = −PeAc
(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 =PeAc
(1 +
ec2r2
)+
M0
S2(10.33-c)
= −1.74× 106
1, 354
(1 +
(31.8)(39.5)943.
)+ 2, 043. = −954. psi (10.33-d)
4. Pe and M0 +MDL +MLL
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 = −PeAc
(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 =PeAc
(1 +
ec2r2
)+
M0 +MDL +MLL
S2(10.36-d)
= −954 + 962. = +8. psi (10.36-e)
fts = 6√
f ′c = +380
√(10.36-f)
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10.3 Case Study: Walnut Lane Bridge 10–15
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft10–16 PRESTRESSED CONCRETE
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft
Chapter 11
FOOTINGS
Read Text 12.1 to 12.4, ACI - Ch. 15, 11.12 31-1/5Unedited
property line
Figure 11.1: xxx
Draft11–2 FOOTINGS
PPP
a
Figure 11.2: xxx
45
Figure 11.3: xxx
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft 11–3
Two-way
f’c b0d
Cl
Cs
Beam
Vc= 2+ f’c b0d( (4βc
Vc
4
2
0 0.5 1.0
dd/2
d/2
a
b c
d
e
fg
h
Figure 11.4: xxx
A1
A2
Figure 11.5: xxx
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft11–4 FOOTINGS
9.5"
9.5"
Column18"x18"
19"2.42’
4.00’
a b
cd
e h
f g
9’-6"
9’-6
"
9’-6"
9’-6
"
9’-0" long
11-#8 each way
18"
18"
3"clear
18"
8-#8 dowels3’-6" long
2’-0
"
9’-6"
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft 11–5
8-#7-6’-0"
18"x24"Column
13-#7-6’-0"
16-#7-6’-3"Bottom
24" sq. col.
6’-6"
11-# 9
19’-
6"to
p
2’-6" 2’-6"
3" clear 3’-5"
4’-6"18’-0"9"
Grade
6’-0"
3" clear
3’-0"23’-3"
Dowels sameas col. bars
1’-6"16’-3"
2’-0"
3’-6"418,000 lb
173,000 lb
385,000 lb
9.30’
1,149,00023.25 = 49,400lb./ft.
ShearDiagram
0.05’ 3,630,000 in.-lb.
3,460,000 in.-lb.
21,400,000 in.-lb.
Moment diagram
REWRITE AND TAKE EXAMPLES FROM NILSON
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft11–6 FOOTINGS
Footings
figure
Isolated spread Wall footingfooting
Combined footing
Bearing capacity of soil
qa = 2, 000 psf 12, 000 psf Safety factor� of 2.5 3.0soft clay
Wall Footings
Similar to the design of a cantilever slab.
Example
DL = 10 k/ft (including thewall weight)
LL + 5 k/ft
Masonry wall
Reinforced & concrete
f′c = 3, 000 psi
fy = 40, 000 psi
qa = 4, 000 psf
Design the footing
Assume hf = 10 in.wf = 10
12 × 150 pct = 125 psf
Effective bearing capacity qe = 4, 000 − 125 = 3, 875 psf
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft 11–7
Consider a unit length of 12′′
← no need to apply keoverload fact or XXee safetyfactor is already including in qa.
Required b = 153.875 = 3.87 ftuse b = 4 ft.
Determine thickness hf → controlled by shear
Wu = 10 (1.4) + 5 (1.7) = 22.5 k/ft
qu = 22.54 = 5.625 ksf
= 5, 625 psf
Vu = Vc = 2√
f ′c bw d
Vu = φ Vc
(4−12 − d)(5625) = 0.85 (2√5, 000)(12)(12)in“ft′′(d)
1.5− d = 2.38 d
d = 0.44 ft = 5.32 in.
hf = 5.32 + 3 (cover) + 0.5 (radius 5/bar)↑ ↑ACI − 7.7.1 Assume # 8
= 8.82 in.
Use hf = 9 in.
Determine flexural Steel
critical section from masonary walls
critical sectionfor concrete columnsor walls
ACI −15.4.2
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft11–8 FOOTINGS
For Masonry Wall,
Mu = 12 qu
(b2 − a
4
)2= 1
2 (5.625)(2 − 1
4
)2= 8.61 ft− kips
Mn = Muφ = 8.61
0.9 = 9.57 ft− kips
Mn = φ (b)(d)2 fy
(1− 0.59 φfy
f′c
)
9.57 × 12 = φ (12)(5.32)2 (40)(1− 0.59 φ(40)
3
)93142 − 118 φ + 1 = 0
φ = 0.0091
As = (0.0291)(12)(5.32) = 0.58 in2
Use # 6 bars
S = AbAs/12
= 12As/Ab
= 0.44 × 120.58 = 9.1
′′
Smax = 3hf = 27 in. or 18 in− ak
Use # 6 @ 9′′
Development length Aab = 0.04 Ab δy√f′c
= 0.04(0.44)(40,000)√3,000
= 12.85 in.
Ad = 12.85 × 0.8 = 10.3 in. < 2/in.↑ ↑
bar spacing from criticalsection to end
Column footings
figure
Failure Modes
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft 11–9
1. Shear Failure - Punching Shear - two-way actionBeam Shear - one-way action
2. Flexure Failure - Bending in each direction
Shear Strength
Punching Shear
ACI − 11.12.2
Vc =(2 + 4
βc
) √f ′c b0 d
βc = ba for b > a
b0 = α[(a+ d) + (b+ d)]
or Vc =(αs db0
+ 2) √
f ′c b0 d
αs = 40 int. cols.
30 edge cols.
20 corner cols.
or Vc = 4√
f ′c b0 d
Shear strength is larger under two-way action than under one-way action because of tri-axialstress.
As βe → very large, Ve = 2√
f ′c b d ← for beams
Beam Shear
Ve = 2√
f ′c bd
← critical section
Vu ≤ φ Vc
Moment Strength (ACI: Ch. 15)
figure
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft11–10 FOOTINGS
Reinf. in Band Width = 2β+1
Total Reinf. inshort direction
β = :2:1
Example
DL = 235 k, LL = 115 k↓
18′′square
hf
f′c = 3, 000 psi
fy = 40, 000 psiqc = 5, 500 psf
7ft. (max.)
Assume hf = 2′
Wf = (2)(150) = 300 psf
qe = 5, 500 − 300 = 5, 200 psf
Required A = (235+115)1,0005,200 (7) = 9.6
′
Use A = 9′ − 8
′′
Determine hf
Pu = 235 (1.4) + 115 (1.7) = 525 kips
qu = PuA = 525 × 1,000
(7)(9.67) = 7.756 psf
Punching Shear
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft 11–11
Vu = 7756[(7)(267) − (
1812 + d
)2]φ Vc = φ (2 + 4
βc)√
f ′c b0 d or φ 4
√f ′c b0
βc = 1φ Vc = (0.85)(4)
√3, 000 (4)(18 + 12d)(12)
or φ Vc =(αsd
b0+ 2
) √f ′c b0 d︸ ︷︷ ︸
will not be critical
φ Vc = Vu
10[(7)(9.67) − (1812 + d)2
]= (18 + 12d)(12d)
154 d2 + 246 d − 677 = 0d = 1.4 ft = 17.3 in.
Beam Shear
Vu = 7, 756(4−672 − 9
12 − d)(7)
φ Vc = φ 2√
f ′ b d
= (0.85)(2)√3, 000 (7 × 12)(12d)
0.528(4.085− d) = d
d = 1.5 ft = 18 in.
use d = 18 in
hf = 18 + 3 (cover) + 0.5 (upper layer) + 1.0 (bottom layer)
= 22.5 in
Select hf = 23 in
Bending:
figure
Long direction:
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft11–12 FOOTINGS
Mu = 7.76 (7) (4.085)2
2 = 453 ft.− kips
Mn = Muφ = 504 ft.− kips ⇒ As = 8.5 in2
14−#7
Short direction:
Mu = 7.76 (9− 67) (2.75)2
2 = 284 ft.− kips
Mn = 2840.9 = 315 ft− kips = 5.5 in2
9−#7
Min Reinf. ACI - 7.12
φg = 0.002
For one short direction, min As = 0.002 (9.67)(12)(23)= 5.3 in2
Reinf. as Band WidthTotal Reinf. = 2
1+β = 21+ 9.67
7
= 0.84
�9 bars
8−#7︸ ︷︷ ︸figure
Check development length
min As = 0.005 Ag�gross areaof columnACI − 15.8.2.1
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft
Chapter 12
DEEP BEAMS
DEEP BEAMS
Examples of occurance
tanks
folded plates
Shear wall & diaphragms
figure
All of these are plane stress problems, and stresses can not be found by classical elastictheory because plane sections do not remain plane after bending (Bernoville-Novier Hypothesisis not valid any more).
need 20 analysis i) elasticityii) finite differenceiii) finite element
Deep beam design
figure
• Main differences between deep & shallow beams:
i) internal stress distribution before cracking
ii) location & orientation of cracks
iii) strength
iv) optimum reinforcement pattern
• Usually deep beams are shear critical & not flexure critical by reducing the span, momen-tum reduced but shear remains constant
Draft12–2 DEEP BEAMS
• Capacity to redistribute stresses is much higher in deep beams
• plane section do not remain plane ⇒ r = My
I + y = V QIb are wrong
XX XX& we have a warping of the x sections tend to relieve compressive stresse @ topmight get more than one neutral ones
stress trajectary
if no vertical compression with vertical compression
tension is much reducedβ < 45◦
• cracks will form at almost vertical directions & thus web reinforcement are not veryefficient ⇒ horz. steel more efficient main steel mostly distributed @ bottoms 13 depth
• After cracking, stress redistributions occurs
crack stopped by high −ve stresses
behaviour becomes similar to that of a tied crack.
J becomes irrelevant what cracking there is, as long as we have this mechanism.
before cracking fs = mAsJd
after cracking (arch mechanism)
• deflection for tied arch is larger than for beam (because stress is larger): but we need towatch out for bond failure, are hooks or to be the steel out & find against the plate.
• better to use horizontal hook XXXXX instead of vertical hook because might have afailure along weak face
figure
• FAILURE by
1. yielding of main longitudinal steel ← preferred use ρ < < ρbal.
2. crushing in high moment regionon the basis of test for beam beams εcu > > > .003 � .008mostly because simultaneous actions in compression XXXXin any case beams would be underreinforced.
Victor Saouma Mechanics and Design of Reinforced Concrete
Draft 12–3
3. crushing of the concrete over the reactions XXXXXXXXXX4. bond splitting failure
compare for normal beams = vu−φ vc
φ fy dbw
s = Avfy(vu−vc)bw
or Avs = vu−vc
fybw
for very deep beams :n1 smallArch is most effectiveas :nd ↗ effect of horzonital steel ↘
Lnd
Av ≥ .0015 bw S
Avh ≥ .0025 bw S2
S ≤ d5 + ≤ 18
′′
S2 ≤ d3 d ≤ 18
′′A.C.I. 11.9
SHEAR WALLS ACI 10.10
of very high better to incorporate shear walls.
Introduction
• Basic relations used:a) Equilibrium i) Tension = compression Σ Fx = 0
ii) Mint = Mext Σ M = 0
b) Material Stress Strain
c) Compatibility of displacements (no slip)
• Basic assumptions used:a) Perfect bond between steel & concrete εs = εcb) Plain section remain plane ⇒ strain is
proportional to distance from N.A.c) Neglect shrinkage & creep (for strength).
• Design: we are going to consider a reinforced concrete beam subjected to an increasingload with:a) sections uncrackedb) sections cracked, elastic → u.s.o.c) sections cracked, inelastic → U.s.o.
Victor Saouma Mechanics and Design of Reinforced Concrete