Structural Concrete - Hassoun

962 Chapter 23 Review Problems on Concrete Building Components
Section A
45′
Section B
45′
2 no. 3
7′
2 no. 3
7′
25′
25′
2 no. 3
12′
2 no. 8
12′
Figure 23.12 Detail of the beam at sections A and B.
8. Detail the sections (Figure 23.12).
9. Design for shear reinforcement.
Calculate the factored shear from external loading:
V u = 21.84 K
V u = w u L
2
w u = 1.37 K∕ft
→ 21.84 = w u (32)
2
Calculate V ud at distance d from face of support: (ACI Code, Section 9.4.3)
( )
( )
d 22.5
V ud = V u − w u = 21.84 − 1.37 = 19.3 K
12
12
Calculate φV c and φV c
: (ACI Code, Section 22.5.5.1)
2
(
φV c = φ 2λ √ )
√
f c
′ b w d = 0.75(2)(1) 4000 (12)(22.5) =25.61 K
φV c
2 = 12.81 K
Since φV c
2 ≤ V ud ≤ φV c , use minimum shear reinforcement
Calculate maximum spacing. (ACI Code, Section 9.7.6.2)
Select the smallest of:
s 1 = 24 in.
s 2 = d 2 = 22.5 = 11.25 in.
2
s 3 = A v f y (2 × 2.11)(60000)
= = 22 in.
50b w (50)(12)
Therefore use = 11 in
Using similar triangles (Figure 23.13)
21.84
192 = 12.81
x 1
x 1 = 112.59 in.
x 2 = 192 − 112.59 = 79.41 in.