Structural Concrete - Hassoun

5.11 Examples Using SI Units 217
M u = 14 × 48 + 2.5
12 × (48)2 = 912 K ⋅ in.
2
24, 000 912 × 1000 × 0.1042
v u =
− = 160 psi
0.75 × 10 × 14.5 0.75 × 10 ×(14.5) 2
Similarly, at 6 ft from the support (2 ft from the free end), E67
7. The shear stress resisted by concrete is
2λ √ √
f c ′ =(2)(1) 4000 = 126.5psi
The minimum shear stress to be resisted by shear reinforcement is
v s = 196 − 126.6 = 69.5psi
(V u and consequently v s have already been increased by the ratio 1/φ in Eq. 5.28).
8. Choose no. 3 stirrups with two legs:
A v = 2 × 0.11 = 0.22 in. 2
S (required) = A v f yt 0.22 × 60, 000
= = 19 in.
v s b w 69.5 × 10
S max (for d∕2at fixed end)=9.75 in. to S max = 4.75 in. at free end
S max (for minimum A v )= A v f yt 0.22 × 60, 000
= = 26.4in.
50b w 50 × 10
9. Check for maximum spacing (d/4): v us ≤ 4 √ f c ′ ,
4 √ √
f c ′ =(4) 4000 = 253 > 69.5in.
10. Distribution of stirrups (distances from free end):
There is 4 in. left to the face of the support.
1 stirrup at 2 in. = 2in.
10 stirrups at 4.5in. = 45 in.
3 stirrups at 7 in. = 21 in.
3 stirrups at 8 in. = 24 in.
Total = 92 in.
5.11 EXAMPLES USING SI UNITS
The general design requirements for shear reinforcement according to the ACI Code are summarized
in Table 5.4, which gives the necessary design equations in both U.S. customary and SI units.
The following example shows the design of shear reinforcement using SI units.