Structural Concrete - Hassoun

542 Chapter 15 Design for Torsion
15.8.8 Minimum Torsional Reinforcement
Where torsional reinforcement is required, the minimum torsional reinforcement may be computed
as follows (ACI Code, Section 9.6.4.2):
1. Minimum transverse closed stirrups for combined shear and torsion (see Section 8.6):
A v + 2A t ≥ 0.75 √ ( )
bw
f c
′ s
f yt
where
A v = area of two legs of closed stirrup determined from shear
A t = area of one leg of closed stirrup determined from torsion
s = spacing of stirrups
f yt = yield strength of closed stirrups ≤60 ksi
But shall not be less than 50b ws
f yt
(15.29)
Spacing of stirrups, s, should not exceed P h /8 or 12 in., whichever is smaller. This spacing
is needed to control cracking width.
2. Minimum total area of longitudinal torsional reinforcement:
A l,min =
( √ )
5 f
′
c A cp
−
f y
(
At
s
)
P h
( fyt
f y
)
(15.30)
where f y is the yield strength of longitudinal torsional reinforcement. Also A t /s shall not be
taken less than 25 b w /f yt .
The minimum A l in Eq. 15.30 was determined to provide a minimum ratio of the volume
of torsional reinforcement to the volume of concrete of about 1% for reinforced concrete
subjected to pure torsion.
15.9 SUMMARY OF ACI CODE PROCEDURES
The design procedure for combined shear and torsion can be summarized as follows:
1. Calculate the factored shearing force, V u , and the factored torsional moment, T u , from the
applied forces on the structural member. Critical values for shear and torsion are at a section
distance d from the face of the support.
2. a. Shear reinforcement is needed when V u >φ V c ∕2, where V c = 2λ √ f cb ′ w d
b. Torsional reinforcement is needed when
T u >φλ √ ( ) Acp
f c
′ 2
(Eq. 15.20)
P cp
If web reinforcement is needed, proceed as follows.