Structural Concrete - Hassoun

Review Problems on Concrete Building Components 951
)
A s f y
M u = φA s f y
(d −
2 ∗ 0.85f c ′ b e
1080 =(0.90)A s (60)
(
)
A s f y
18 −
1.7 (3) (45)
A s = 1.13 in. 2 , use 2 no. 7 bars (A s = 1.20 in. 2 )
Check that ρ w = A s
b w d = 1.57
(10)(18) = 0.0087 ≥ ρ min = 0.00333
Check for φ ∶ ε t ∶ a = 0.82 in., c = 0.82
0.85 = 0.96 in., d t = d = 22.5in.
0.003(d − c)
ε t = =
c
Beam details are shown in Figure 23.5.
0.003(18 − 0.96)
0.96
= 0.053 > 0.005,φ= 0.9
45″
5″
18″
3″
10″
2 no. 7
Figure 23.5
Detail of the beam.
b. Design for shear reinforcement
Calculate the factored shear from external loading:
w u = 1.2w DL + 1.6w LL = 1.2(2)+1.6(0.5) =3.2K∕ft
V u = w u L = (3.2)(15) = 24K
2 2
Calculate V u at distance d from face of support: (ACI Code, Section 9.4.3)
( ) ( )
d 18
V ud = V u − w u = 24 = 3.2 = 19.20K
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 (10)(18) =14.80 K
φV c
2 = 7.40 K
V c1 = 4 √ √
(f c ′ ) b w d = 4 3000 (10)(18) =39.4K
V c2 = 8 √ √
f c ′ b w d = 8 3000(10)(18) =78.8 K