Structural Concrete - Hassoun

19.8 Design for Shear 771
3. The minimum shear reinforcement, A v , required by the ACI Code is
A v, min
= 0.75 √ f ′ c
(
bw s
f yt
)
≥ 50b ws
f yt
(Eq. 19.58)
When the effective prestress, f pe , is greater than or equal to 0.4f pu , the minimum A v is
A v,min = A ps
80 × f pu
f yt
× s
d p
×
√
d p
b w
(Eq. 19.59)
The effective depth, d p , need not be taken less than 0.8h. Generally, Eq. 19.59 requires greater
minimum shear reinforcement than Eq. 19.58.
Example 19.7
For the beam of Example 19.4, determine the nominal shear strength and the necessary shear reinforcement.
Check the sections at h/2 and 10 ft from the end of the beam. Use f y = 60 ksi for the shear
reinforcement, and a live load = 1.33 K/ft. using normal-weight concrete.
Solution
1. For the section at h/2:
h
2 = 40 = 20 in. = 1.67 ft from the end
2
2. The factored uniform load on the beam is
V u
W u = 1.2(0.388 + 0.9)+1.6 × 1.33 = 3.68 K∕ft
at a distance 1 h = 3.68(24 − 1.67) =82.2K
2
Using the simplified ACI method (Eq. 19.50), determine M u at section h/2:
M u =(3.68 × 21)×1.67 − 3.68 (1.67)2 = 142.4K⋅ ft = 1708 K ⋅ in.
2
The value of d p at section h/2 from the end (Fig. 19.6a) is
16 − 1.67
d p = 33.7(at midspan)−
16
V u d p 82.2 × 20.7
= = 0.966 ≤ 1.0
M u 1708
× 14.5 = 20.7in.
as required by the ACI Code.
(
V c = 0.6λ √ f c ′ + 700 V u d )
p
b
M w d
u
√
=(0.6 × 1 × 5000 + 700 × 0.996)6 × 20.7 = 91, 800 lb = 91.8K
Minimum V c = 2λ √ √
f c ′ b w d p = 2 × 1 × 5000 × 6 × 20.7 = 17.6K
Maximum V c = 5λ √ f ′ c b w d p = 43.9K
The maximum V c of 43.9 K controls.