Structural Concrete - Hassoun

206 Chapter 5 Shear and Diagonal Tension
Example 5.1
A simply supported beam has a rectangular section with b = 12 in., d = 21.5 in., and h = 24 in. and is
reinforced with four no. 8 bars. Check if the section is adequate for each of the following factored shear
forces. If it is not adequate, design the necessary shear reinforcement in the form of U-stirrups. Use
f c ′ = 4ksiandf yt = 60 ksi. Assume normal-weight concrete.
(a) V u = 12 K, (b) V u = 24 K, (c) V u = 54 K, (d) V u = 77 K, (e) V u = 128 K
Solution
In general, b w = b = 12 in., d = 21.5 in., and
φV c = φ(2λ √ √
f c ′ )bd = 0.75(2)(1)( 4000)(12)(21.5) =24.5K
1
2 φV c = 12.25 K
V c1
=(4 √ f ′ c )bd = (4√ 4000)(12)(21.5)
1000
V c2
=(8 √ f ′ c )bd = 130.6K
= 65.3K
a. Assume V u = 12 K < 1 φV 2 c = 12.25 K, the section is adequate, and shear reinforcement is not
required.
b. Assume V u = 24 K > 1 φV 2 c , but it is less than φV c = 24.5 K. Therefore, V s = 0 and minimum
shear reinforcement is required. Choose a no. 3 U-stirrup (two legs) at maximum spacing. Let
A v = 2(0.11) = 0.22 in 2 . Maximum spacing is the least of
S 2 = d 2 = 21.5 = 10.75 in. say, 10.5in. (controls)
2
S 3 = A v f yt 0.22(60, 000)
= = 22 in. (or use Table 5.1)
50b w 50(12)
S 4 = 24 in.
Use no.3 U-stirrups spaced at 10.5 in.
c. Assume V u = 54 K >φV c . Shear reinforcement is needed. Calculation may be organized in steps:
Calculate V s = (V u –φ V c )/φ = (54–24.5)/0.75 = 39.3 K.
Check if V s ≤ V c1
=(4 √ f ′ c)b w d = 65.3 K. Because V s < 65.3K,thenS max = d∕2, and the d/4
condition does not apply.
Choose no. 3 U-stirrups and calculate the required spacing based on V s :
S 1 = A v f yt d
= 0.22(60)(21.5) = 7.26 in. say, 7in.
V s 39.3
Calculate maximum spacing: S 2 = 10.5 in., S 3 = 22 in., and S 4 = 24 in. and maximum S =
10.5 in. [calculated in (b)].
Because S = 7in.< S max = 10.5 in., then use no. 3 U-stirrups spaced at 7 in.
d. Assume V u = 77 K >φV c , so stirrups must be provided.
Calculate V s = (V u –φV c )/φ = (77–24.5)/0.75 = 70 K.
Check if V s ≤ V c1
=(4 √ f c ′ )b w d = 65.3 K. Because V s > 65.3K, then S max = d∕4 = 12 in.
must be used.
Check if V s ≤ V c1
=(8 √ f c ′ )b w d = 130.6 K. Because V c1
< V s < V c2
, then stirrups can be used
without increasing the section.