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 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 i and d₁= 24 in – 2.5 in = 21.5 in). 7 Slab thickness = 5 in. 24 ft Figure 1 Continuous Floor System 10 ft 10 ft
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 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 i and d₁= 24 in – 2.5 in = 21.5 in). 7 Slab thickness = 5 in. 24 ft Figure 1 Continuous Floor System 10 ft 10 ft
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Reinforced Concrete Design
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