Structural Concrete - Hassoun

194 Chapter 5 Shear and Diagonal Tension
1. ACI Code, Section 22.5.5.1, adopted this equation for the nominal shear force to be resisted
by concrete for members subjected to shear and flexure only by
[
V c = 1.9λ √ ]
f c ′ V
+ 2500ρ u d
w b
M w d ≤ 3.5λ √ f cb ′ w d (5.6)
u
where ρ w = A s /b w , d and b w are the web width in a T-section or the width of a rectangular
section, and V u and M u are the factored shearing force and bending moment occurring
simultaneously on the considered section.
The value of V u d/M u must not exceed 1.0 in Eq. 5.6. If M u is large in Eq. 5.6, the second
term becomes small and v c approaches 1.9λ √ f c.IfM ′ u is small, the second term becomes
large and the upper limit of 3.5λ √ f c ′ controls. As an alternative to Eq. 5.6, the ACI Code,
Section 22.5.5.1, permits evaluating the shear strength of concrete as follows:
{ √
2λ f
′
V c = c b w d (lb.in.)
0.17λ √ (5.7)
f cb ′ w d (SI)
2. For members subjected to axial compression force N u (ACI Code, Section 22.5.6.1) V c shall
be calculated by:
(
V c = 1.9λ √ )
f c ′ V
+ 2500ρ u d
w b
M w d
m
( ) 4h − d
M m = M u − N u (5.8)
8
where
ρ w = A s /(b w d)
h = overall depth
and V u d/M u may be greater than 1.0 but V c must not exceed
√
V c = 3.5λ √ f ′ cb w d
where A g is the gross area in square inches.
1 + N u
500A g
(5.9)
Shear failure near a middle support.