Structural Analysis
Structural Analysis
6th Edition
ISBN: 9781337630931
Author: KASSIMALI, Aslam.
Publisher: Cengage,
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Please solve part A and B Subject is reinforced concrete design show all steps
P01: For the continuous floor system in Fig 1 below, calculate the following: a. be for the beam T-section at section A-A and B-B, b. Design moment for Doubly Reinforced Rectangular beam section (Fig. 4) at section C-C. c. Design moment for Singly reinforced T- beam section (Fig. 2) at section A-A. (See Hint) d. Design moment for Doubly reinforced T- beam section (Fig. 3) at section B-B. (Optional - for further knowledge) Given: a. bw = 1 ft; b. f'c = 4000 psi and fy = 60 ksi. c. Central floor beam spanning in horizontal direction. d. The floor system has a uniform spacing between beams. HINT: For a typical floor system, mid-span sections are subjected to positive bending. Section A-A has a large compression zone; hence compression steel may not be required for additional moment strength in this case. Assuming section will include No. 3 or No. 4 stirrups (compression steel in the section is not shown in the figure which is placed for reinforcement continuity requirements), d= 24 in- 3.5 in= 20.5 in and d= 24 in - 2.5 in = 21.5 in). 7 Slab thickness=5 in. 24 ft Figure 1 Continuous Floor System 10 ft 10 ft
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Transcribed Image Text:P01: For the continuous floor system in Fig 1 below, calculate the following: a. be for the beam T-section at section A-A and B-B, b. Design moment for Doubly Reinforced Rectangular beam section (Fig. 4) at section C-C. c. Design moment for Singly reinforced T- beam section (Fig. 2) at section A-A. (See Hint) d. Design moment for Doubly reinforced T- beam section (Fig. 3) at section B-B. (Optional - for further knowledge) Given: a. bw = 1 ft; b. f'c = 4000 psi and fy = 60 ksi. c. Central floor beam spanning in horizontal direction. d. The floor system has a uniform spacing between beams. HINT: For a typical floor system, mid-span sections are subjected to positive bending. Section A-A has a large compression zone; hence compression steel may not be required for additional moment strength in this case. Assuming section will include No. 3 or No. 4 stirrups (compression steel in the section is not shown in the figure which is placed for reinforcement continuity requirements), d= 24 in- 3.5 in= 20.5 in and d= 24 in - 2.5 in = 21.5 in). 7 Slab thickness=5 in. 24 ft Figure 1 Continuous Floor System 10 ft 10 ft
3 No. 5 bars be= 72 in. 6 No. 7 bars 12 in. I 3.5 in. Figure 2 T-Beam section at A-A section 13 No. 8 bars 2 No. 8 (No. 25) b₂ = 72 in. 6 No. 10 (No. 32) 3 No. 7 bars 12 in. I = 2.5 in. Is 5. in. 12" Figure 3 T-Beam Section at B-B section 2/2 -2.5 in. 3 No. 5 bars 24" Figure 4 Rectangular Beam Section at C-C 24 in. 5 in. 24 in.
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Transcribed Image Text:3 No. 5 bars be= 72 in. 6 No. 7 bars 12 in. I 3.5 in. Figure 2 T-Beam section at A-A section 13 No. 8 bars 2 No. 8 (No. 25) b₂ = 72 in. 6 No. 10 (No. 32) 3 No. 7 bars 12 in. I = 2.5 in. Is 5. in. 12" Figure 3 T-Beam Section at B-B section 2/2 -2.5 in. 3 No. 5 bars 24" Figure 4 Rectangular Beam Section at C-C 24 in. 5 in. 24 in.
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Step 1: Given
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Step 2: Effective flange width for T beam
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Step 3: Doubly beam
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Step 4: Equilibrium equation
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Step 5: Verifying the assumption that the steel is yielding
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Step 6: Design moment capacity of beam
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Solution
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