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Structural Concrete - Hassoun

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548 Chapter 15 Design for Torsion<br />

2 no. 6<br />

4 no. 9<br />

3 no. 6 6 no. 4 bars<br />

2 no. 4<br />

5 no. 9<br />

Solution<br />

Figure 15.16 Example 15.5.<br />

1. Design forces are V u = 53 K and T u = 240 K ⋅ in. = 20 K⋅ft.<br />

2. .a. Shear reinforcement is needed when V u >φV c /2.<br />

φV c = φ2λ √ √<br />

f c ′ b w d = 0.75(2)(1) 4000(14)(18) =23.9K<br />

V u > φV c<br />

2 = 11.95K<br />

Shear reinforcement is required.<br />

b. Check if torsional reinforcement is needed. Assuming that flange is contributing to resist torsion,<br />

the effective flange length is h w = 15 in. < 4 × 6 = 24 in.<br />

x 0 = 14in. and y 0 = 21in.<br />

A cp =(14 × 21)(web)+(15 × 6)(flange) =384in. 2<br />

P cp = 2(21 + 29) =100in.<br />

T a (Eq. 15.20) = 0.75(1)√ 4000(384) 2<br />

100<br />

= 70K ⋅ in.<br />

T u > T a<br />

Torsional reinforcement is required.

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