Structural Concrete - Hassoun

15.9 Summary of ACI Code Procedures 547
b. Maximum s = P h ∕8 = 64 = 8in. or 12 in., whichever is smaller. The value of s used is 7.0 in.
8
< 8in.
c. Minimum A vt /s = 50 b w /f yt = 50(16)/60,000 = 0.0133 in. 2 /in. This is less than 0.028 in. 2 /in.
provided.
6. To find the distribution of longitudinal bars, note that total A l = 1.21 in. 2 Use one-third at the
top, or 1.21/3 = 0.4 in. 2 , to be added to the compression steel A ′ s . Use one-third, or 0.4 in.2 ,atthe
bottom, to be added to the tension steel, and one-third, or 0.4 in. 2 , at middepth.
a. The total area of top bars is 0.4 (two no. 4) + 0.4 = 0.8 in. 2 use three no. 5 bars (A s = 0.91 in. 2 ).
b. The total area of bottom bars is 5 (five no. 9) + 0.4 = 5.4 in. 2 ; use three no. 9 and two no. 10
bars at the corners (total A s = 5.53 in. 2 ).
c. At middepth, use two no. 4 bars (A s = 0.4 in. 2 ). Reinforcement details are shown in Fig. 15.15.
Spacing of longitudinal bars is equal to 9 in., which is less than the maximum required of 12 in.
The diameter of no. 4 bars used is greater than the minimum of no. 3 or stirrup spacing, or s/24
= 0.21 in.
Example 15.4 Compatibility Torsion
Repeat Example 15.3 if the factored torsional torque is a compatibility torsion.
Solution
Referring to the solution of Example 15.3:
1. Design forces are V u = 48 K and compatibility torsion is 360 K ⋅ in.
2. Steps (a) and (b) are the same as in Example 15.3. Web reinforcement is required.
3. Step (c) is the same.
4. Design for torsion:
Because this is a compatibility torsion of 360 K ⋅ in., the design T u is the smaller of 360 K ⋅ in. or
φT cr given in Eq. 15.19.
φT cr = φ4λ √ ( )
A
2
f c
′ cp
= 0.75(4)(1)√ 4000(368) 2
= 329.4K⋅ in. (Eq. 15.19)
P cp 78
Because φT cr < 360 K ⋅ in., use T u = 329.4 K ⋅ in. Repeat all the steps of Example 15.3 using
T u = 329.3 K ⋅ in. to determine that the section is adequate.
A t
s = 0.018in.2 ∕in. (one leg)
A l = 0.018(64) =1.152in. 2
Use 1.2 in. 2 > min. A l .
5. Required A vt = 0.018/2 + 0.018 = 0.027 in. 2 /in. (one leg).
s = 0.2
0.027 = 7.4in.
Use 7 in. Choose bars, stirrups, and spacing similar to Example 15.3.
Example 15.5 L-Section with Equilibrium Torsion
The edge beam of a building floor system is shown in Fig. 15.16. The section at a distance d from the
face of the support is subjected to V u = 53 K and an equilibrium torque T u = 240 K ⋅ in. Design the
necessary web reinforcement using f c ′ = 4ksiandf y = 60 ksi for all steel bars and stirrups.