Civil Engineering Reference
In-Depth Information
Stirrups Specified
#4 deformed bars at 0.00 inches on center
Stirrup angle: 45.000 degrees
Design Forces Specified
Tu = 360000.00 lb-in
Vu = 60000.00 lb
Compute the Section Properties
Acp = 16.00(26.00) = 416.00 sq.in.
Pcp = 2(16.00) + 2(26.00) = 84.00 in.
Aoh = [16.00 — 2(1.75)][26.00 — 2(1.75)] = 281.25 sq.in.
Ao = 0.85(281.25) = 239.06 sq.in.
Ph = 2[16.00 — 2(1.75)] + 2[26.00 — 2(1.75)] = 70.00 in.
Section complies with ACI 11.6.3.1
Torsional Force when Stirrups are Required
Ts = (0.75)sqrt(4000.00)(416.00)(416.00)/(84.00) = 97723.41
lb-in. (8.14 k-ft)
Check Specified Stirrup Spacing
Av/s = 0.02301 in.
At/s = 0.01673 in.
*Required s = 6.94 in.
Maximum s = 8.75 in.
Specified stirrup spacing is adequate
Compute Required Additional Longitudinal Reinforcement
Al Req'd = (0.02)(70.00)(1.00
∧
2) = 1.17 sq.in.
Al min = 1.02 sq.in.
Specified Additional Al is Adequate
PROBLEMS
For Problems
15.1
to
15.3
, determine the equilibrium torsional capacity of the sections if no torsional reinforcing is
used.
4000 psi and
f
y
60,000 psi.
Problem 15.1
(
Ans.
6.57 ft-k)
c
f
21"
24"
4 #8
3"
15"
*This value for required
s
varies a little from the solution given for Example 15.1 due to the different number of
decimal places used for specifying the stirrup bar area.
