EQUATION:

vc = [phi][1.9(f'c)1/2 + 2500 p] < /=

(44)

2.28[phi](f'c)1/2

where,

vc = maximum shear capacity of an unreinforced web, psi

p = reinforcement ratio of the tension reinforcement at the

support

[phi] = capacity reduction factor equal to 0.85

(3) whenever the nominal shear stress, vu, exceeds the shear

capacity, vc, of the concrete, shear reinforcement must be provided to

carry the excess. Closed ties placed perpendicular to the flexural

reinforcement must be used to furnish the additional shear capacity. Open

stirrups, either single or double leg, are not permitted. The required area

of shear reinforcement is calculated using:

EQUATION:

Av = [(vu - vc) b ss]/[phi]fy

(45)

where,

Av = total area of stirrups, sq in

vu-vc = excess shear stress, psi

ss = spacing of stirrups in the direction parallel to the

longitudinal reinforcement, in

[phi] = capacity reduction factor equal to 0.85

(4) In order to insure the full development of the flexural

reinforcement in a beam, a premature shear failure must be prevented. The

following limitations must be considered in the design of the closed ties:

a) the design shear stress (excess shear stress vu - vc) used in

Equation (45) shall be equal to, or greater than, the shear capacity of

unreinforced concrete, vc, as obtained from Equation (44); b) the nominal

shear stress, vu, must not exceed 10 [phi] (f'c)1/2; c) the area,

Av, of closed ties should not be less than 0.0015bss; d) the required

area, Av, of closed ties shall be determined at the critical section, and

this quantity of reinforcement shall be uniformly distributed throughout the

member; and e) the maximum spacing of closed ties is limited to d/2 when

vu - vc is less than 4 [phi] (f'c)1/2 but is further limited to d/4

when vu - vc is greater than 4 [phi] (f'c)1/2.

e.

Direct Shear.

(1) Direct shear failure of a member is characterized by the rapid

propagation of a vertical crack through the depth of the member. This crack

is usually located at the supports where the maximum shear stresses occur.

Failure of this type is possible even in members reinforced for diagonal

tension.

2.08-74