Structural Concrete - Hassoun

334 Chapter 9 One-Way Slabs
No. 5 @
No. 4 @
Figure 9.7
Example 9.4: Reinforcement details.
4. Moments and steel reinforcement required at other sections using d = 3.9 in. are as follows:
Location
Moment
Coefficient
A − 1
24
B + 1
14
C − 1
10
D − 1
11
E + 1
16
M u
(K ⋅ in.)
R u
= M u
/bd 2
(psi)
ρ(%)
A s
(in. 2 )
Bars and
Spacings
22.7 Small 0.50 0.23 No. 4 at 10 in.
38.9 213 0.65 0.30 No. 5 at 12 in.
54.5 300 0.90 0.44 No. 5 at 8 in.
49.6 271 0.80 0.38 No. 5 at 8 in.
34.1 187 0.55 0.26 No. 4 at 8 in.
The arrangement of bars is shown in Fig. 9.7.
5. Maximum shear occurs at the exterior face of the second support, section C.
V u (at C) = 1.15UL n
2
= 1.15(0.3755)(11)
2
= 2.375 K∕ft of width
φV c = φ2λ √ f c ′ bd = 0.75(2)(1)(√ 3000)(12)(3.9)
= 3.84 K
1000
This result is acceptable. Note that the provision of minimum area of shear reinforcement when
V u exceeds 1 φV 2 c does not apply to slabs (ACI Code, Section 9.6.3.1).
Example 9.5
Determine the uniform factored load on an intermediate beam supporting the slabs of Example 9.4. Also
calculate the axial load on an interior column; refer to the general plan of Fig. 9.5. Assume the beam
span = 24 ft.
Solution
1. The uniform factored load per foot length on an intermediate beam is equal to the factored uniform
load on slab multiplied by S, the short dimension of the slab. Therefore,
U(beam) =U(slab)×S = 0.3755 × 12 = 4.5K∕ft