Civil Engineering Reference
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where:
A
t
= area of one leg of closed stirrups resisting torsion, in
2
A
o
= 0.85A
oh
, in as defined in Step 4
s = spacing of stirrups, in.
(Note that for simplicity the angle of compression diagonals in truss analogy is assumed to be 45°)
(6) Calculate the required area of stirrups for shear:
V
u
V
s
=
φ
−
V
c
A
v
V
s
f
y
d
s
=
where:
A
v
= Area of shear reinforcement within spacing s (two legs)
(7) Calculate required combined stirrups for shear and torsion (for one leg):
A
t
A
v
2s
s
=
Select the size and spacing of the combined stirrups to satisfy the following conditions:
a. The minimum area of stirrups A
v
+2A
t
(two legs) for the selected concrete with compressive
strength = 4000 psi is 50b
w
s/f
y
b. The maximum spacing of transverse torsion reinforcement s is the smaller of p
h
/8 or 12 in.
(8) Calculate the required additional longitudinal reinforcement A˜ for torsion
5
f
c
A
cp
f
y
ʹ
A
t
s
A
t
s
⎛
⎜
⎞
⎟
A
=
p
h
≥
−
p
h
(ACI 11.5.3.7)
A
t
/s in the above equation must be taken as the actual amount calculated in Step 5 but not less than 25b
w
/f
y
(ACI 11.5.5.3). The additional longitudinal reinforcement must be distributed around the perimeter of the
closed stirrups with maximum spacing of 12 in.
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