Structural Concrete - Hassoun

15.9 Summary of ACI Code Procedures 545
2 no. 4 3 no. 5
2 no. 4
5 no. 9
2 no. 10
3 no. 9
Figure 15.15 Example 15.3.
b. Torsional reinforcement is needed when
T u >φλ √ f ′ c
( )
A
2
cp
= T
P a
cp
where
A cp = x 0 y 0 = 16(23) =368in. 2
P cp = 2(x 0 + y 0 )=2(16 + 23) =78in.
T u = 0.75(1)√ 4000(368) 2
78
= 82.36K ⋅ in.
T u = 360K ⋅ in. >82.36K ⋅ in. (Eq. 15.20)
Torsional reinforcement is needed. Note that if T u is less than 82.36 K ⋅ in., torsional reinforcement
is not required, but shear reinforcement may be required.
3. Design for shear:
a. V u = φV c + φV s , φV c = 48 = 31.1 + 0.75 V s , V s = 22.5 K.
b. Maximum V s = 8 √ f ′ cbd = 8 √ 4000(16)(20.5) =166K > V s .
c. A v /s = V s /f y d = 22.5/(60 × 20.5) = 0.018 in. 2 /in. (two legs). A v /2s = 0.018/2 = 0.009 in. 2
/in. (one leg).
4. Design for torsion:
a. Design T u = 360 K ⋅ in. Determine sectional properties, assuming 1.5 in. concrete cover and no.
4 stirrups:
x 1 = widthtocenterofstirrups = 16 − 2(1.5 + 0.25) =12.5in.
y 1 = depthtocenterofstirrups = 23 − 2(1.5 + 0.25) =19.5in.
Practically, x 1 can be assumed to be b − 3.5 in. and y 1 = h − 3.5 in.
For θ = 45 ∘ and cot θ = 1.0.
A 0h = x 1 y 1 =(12.5 × 19.5) =244in. 2
A 0 = 0.85A 0h = 207.2in. 2
P h = 2(x 1 + y 1 )=2(12.5 + 19.5) =64in.