Structural Concrete - Hassoun

5.8 Design Summary 207
Choose no. 3 U-stirrups and calculate S 1 based on V s :
S 1 = A v f yt d
= 0.22(60)(21.5) = 4.1in. say, 4in.
V s 70
Calculate maximum spacings: S 2 = d∕4 = 21.5∕4 = 5.3in., say, 5.0 in.; S 3 = 22 in.; and S 4 =
12 in. Hence S max = 5-in. controls.
Because S = 4in.< S max = 5 in., then use no. 3 stirrups spaced at 4 in.
e. Assume V u = 128 K >φV c , so shear reinforcement is required.
Calculate V s = (V u –φV c )/φ = (128–24.5)/0.75 = 138 K.
Because V s > V c2
= 130.2 K, the section is not adequate. Increase one or both dimensions of
the beam section.
Notes: Table 5.2 and Fig. 5.10 can be used to calculate the spacing S for (c) and (d).
1. For (c), V s = 39.3 K, from Fig. 5.10 (or Table 5.2 for no. 3 U-stirrups), S∕d = 0.34 and S 1 = 7.3
in., which is less than d∕2 = 10.5 in. Note that S max based on V s is d/2 and not d/4. Also, from
Table 5.1, S 3 = A v f yt ∕50b w = 22 in.
2. For (d), V s = 70 K, S∕d = 0.19 and S 1 = 4.1 in. So V s = 70 > 52.8K, and S max = d∕4 is
required.
Example 5.2
A 17-ft-span simply supported beam has a clear span of 16 ft and carries uniformly distributed dead and
live loads of 4.5 and 3.75 K/ft, respectively. The dimensions of the beam section and steel reinforcement
are shown in Fig. 5.12. Check the section for shear and design the necessary shear reinforcement. Given
f c ′ = 3 ksi normal-weight concrete and f yt = 60 ksi.
Solution
Given: b w (web) =14 in. and d = 22.5in.
1. Calculate factored shear from external loading:
Factoreduniformload = 1.2(4.5)+1.6(3.75) =11.4K∕ft
V u (at face of support) = 11.4(16) = 91.2K
2
Design V u (at distance d from the face of the support) = 91.2 − 22.5(11.4)/12 = 69.83 K.
2. Calculate φV c :
φV c = φ(2λ √ f c ′ )b w d = 0.75(2)(1)(√ 3000)(14)(22.5)
= 25.88 K
1000
1
2 φV c = 12.94 K
Calculate V c1
=(4 √ f c ′ )b w d =(4 √ 3000)(14)(22.5)∕1000 = 69 K. Calculate V c2
=(8 √ f c ′ )
b w d = 138 K.
3. Design V u = 69.83 K >φV c = 25.88 K; therefore, shear reinforcement must be provided. The
distance x ′ at which no shear reinforcement is needed (at 1 φV 2 c )is
( )
x ′ 91.2 − 12.94
=
(8) =6.86 ft = 82 in.
91.2
(from the triangles of the shear diagram, Fig. 5.12).