Structural Concrete - Hassoun

5.11 Examples Using SI Units 219
Table 5.4
Shear Reinforcement Formulas
U.S. Customary Units
V u = design shear
SI Units
V u = design shear
Maximum
(
design V u at distance d from face of support.
V c = 2.0λ √ )
(
f c
′ b w d V c = 0.17λ √ )
f c
′ b w d
[
V c = 1.9λ √ ( )]
[
f c ′ V
+ 2500ρ u d
w b
M w d V c = 0.16λ √ ( )]
f c ′ V
+ 17.2ρ u d
w b
u M w d
u
V u d
V u d
ρ w = A s
b w d
V c ≤
(
3.5λ √ f ′ c
≤ 1.0
M)
u
ρ w = A s
b w d
b w d
V c ≤
(
0.29λ √ f ′ c
≤ 1.0
M u )
b w d
V u = φV c + φV s
V u = φV c + φV s
Vertical stirrups
φV s = V u −φ
V c
S = A v f yt d/V s
Minimum A v = 50b w S ≤
f yt
0.75 √ ( )
bw
f c
′ S
f yt
Maximum S = A v f yt
50b w
≥
φV s = V u − φ
V c
S = A v f yt d/V s
Minimum A v = 0.35b w S ≤
f yt
0.0062 √ ( )
bw
f c
′ S
f y
A v f yt
0.75 √ Maximum S = A v f yt
≥
f c ′ b 0.35b w w
For vertical web reinforcement
Maximum S = 1 d ≤ 24 in.
2 Maximum S = 1 d ≤ 600 mm
2
If V s ≤ 4.0 √ (
f c
′ bw d ) If V s ≤ 0.33 √ (
f c
′ bw d )
Maximum S = d∕4 = 12 in.
Maximum S = d∕4 = 300 mm
If V s > 4.0 √ (
f c
′ bw d ) If V s > 0.33 √ (
f c
′ bw d )
V s ≤ 8 √ (
f c
′ bw d ) V s ≤ 0.67 √ (
f c
′ bw d )
Otherwise increase the dimensions of the
section.
Series of bent bars or inclined stirrups
V
A v =
s S
f yt d (sin α + cos α)
For α = 45 ∘ , S = 1.4A υ f y d∕V s
V
A v =
s S
f yt d (sin α + cos α)
For α = 45 ∘ , S = 1.4A υ f y d∕V s
For single bent bar or group of bars, parallel and bent in one position
A v =
V s
A
f yt sin α
v =
V s
f yt sin α
For α
(
= 45 ∘ , A y = 1.4V s ∕f yt
For α = 45 ∘ , A v = 1.4V s ∕f yt
V s ≤ 3 √ )
(
f c
′ b w d
V s ≤ 0.25 √ )
f c
′ b w d
A v f y
0.062 √ f ′ c b w