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.

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Chapter2: Loads On Structures
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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).
Slab thickness= 5 in.
24 ft
Figure 1 Continuous Floor System
1
10 ft
10 ft
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). Slab thickness= 5 in. 24 ft Figure 1 Continuous Floor System 1 10 ft 10 ft
3 No. 5 bars
be= 72 in.
6 No. 7 bars
12 in.
Figure 2 T-Beam section at A-A section
I 3.5 in.
b₂ = 72 in.
13 No. 8 bars
2 No. 8 (No. 25)
3 No. 7 bars
●
12 in.
6 No. 10 (No. 32) -
I=2.5 in.
Ism.1
5 in.
Figure 3 T-Beam Section at B-B section
12"
2.5 in.
3 No. 5 bars
24"
Figure 4 Rectangular Beam Section at C-C
24 in.
5 in.
24 in.
Transcribed Image Text:3 No. 5 bars be= 72 in. 6 No. 7 bars 12 in. Figure 2 T-Beam section at A-A section I 3.5 in. b₂ = 72 in. 13 No. 8 bars 2 No. 8 (No. 25) 3 No. 7 bars ● 12 in. 6 No. 10 (No. 32) - I=2.5 in. Ism.1 5 in. Figure 3 T-Beam Section at B-B section 12" 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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