Structural Concrete - Hassoun

15.9 Summary of ACI Code Procedures 549
3. Design for shear:
a. V u = φV c + φV s
53 = 23.9 + 0.75V s
V s = 38.8K
b. Maximum V s = 8 √ f cb ′ w d = 127.5K > V s
c. A v
s = V s
f y d = 38.8
60 × 18 = 0.036in.2 ∕in. (two legs)
A v
2s = 0.036 = 0.018in. 2 ∕in.
2
4. Design for torsion: T u = 240 K ⋅ in.
a. Determine section properties assuming a concrete cover of 1.5 in. and no. 4 stirrups:
Web x 1 = b − 3.5in. = 14 − 3.5 = 10.5in.
Flange x 1 = 15in.(stirrups extend to the web)
y 1 = h − 3.5 = 21 − 3.5 = 17.5in.
y 1 = 6 − 3.5 = 2.5in.
A 0h =(15 × 2.5)+(10.5 × 17.5) =221in. 2 A 0 = 0.85A 0h = 188in. 2
P h = 2(15 + 2.5)+2(10.5 + 17.5) =91in. θ = 45 ∘ cot θ = 1.0
b. Check the adequacy of the section using Eq. 15.21: V u = 53 K, φV c = 23.9 K, V c = 31.9 K, and
T u = 240 K ⋅ in.
√ ( ) 2 [ ] 2
53,000 240,000 × 91
Left-handside =
+
= 434psi
14 × 18 1.7(184) 2
[ 31,900 √
]
Right-handside = 0.75
14 × 18 + 8 4000 = 475psi
c. Determine the torsional closed stirrups, A t /s, from Eq. 15.25:
A t
s =
T n
2A 0 f yt
=
240
0.75 × 2 × 188 × 60 = 0.014in.2 ∕in. (for one leg)
d. Calculate the additional longitudinal reinforcement from Eq. 15.28 (for f y = 60 ksi and
cot θ = 1.0):
( )
At
A l = P
s h = 0.014(91) =1.28in. 2
A l,min (from Eq. 15.30) is
[
5 √ ]
4000 (384)
A l =
−(0.014 × 91) =0.75in. 2
60,000
The contribution of the flange may be neglected with slight a difference in results and less
labor cost.
5. Determine the total area of the closed stirrups.
a. For one leg, A vt /s = A t /s + A v /2s.
Required A vt = 0.014 + 0.018 = 0.032in. 2 ∕in. (per leg)
Choose no. 4 closed stirrups, area = 0.2 in. 2
Use 6 in.
Spacing of stirrups = 0.2
0.032 = 6.25in.