Structural Concrete - Hassoun

Review Problems on Concrete Building Components 955
A s2 =
M u2
φf y (d − d ′ ) = (160.27)(12)
= 2.04 in.2
(0.9)(60)(20 − 2.5)
A s total = A s1 + A s2 = 3.25 + 2.04 = 5.29 in. 2 (6 no. 9 bars)
(
f s ′ = 87000 cmax − d ′ ) ( ) 7.5 − 2.5
= 87000
= 58, 000 psi < f
c max 2.5
y = 60, 000 psi
A ′ s f ′
s = A s2 f y
A ′ s = A s2 f y
f ′
s
= (2.04)(60)
58
Beam reinforcement details are shown in Figure 23.8.
= 2.11 in. 2 (3 no. 8 bars, A s = 2.35 in. 2 )
3 no. 8
20″
6 no. 9
2.5″
12″
4.5″
Figure 23.8
Cross sectional details.
b. Design for shear reinforcement
Calculate the factored shear at support from external loading:
w = 1.2w DL + 1.6w LL = 1.2(3)+1.6(1) =5.2 K∕ft
V u = w u L = (5.2)(25) = 65 K
2 2
Calculate V u at distance d from face of support: (ACI Code, Section 9.4.3)
( ) ( )
d 20
V ud = V u − w u = 65 − 5.2 = 56.33 K
12
12
φV
Calculate φV c , c
, V 2 c1 , V c2 : (ACI Code, Section 22.5.5.1)
(
φV c = φ 2λ √ )
√
f c
′ b w d = 0.75(2)(1) 3000 (12)(20) =19.72 K
φV c
= 9.86 K
2
V c1 = 4 √ √
f c ′ b w d = 4 3000(12)(20) =52.58 K
V c2 = 8 √ √
f c ′ b w d = 8 3000(12)(20) =105.16 K