24.02.2017 Views

Structural Concrete - Hassoun

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

550 Chapter 15 Design for Torsion<br />

b. Max. s = P h /8 = 91/8 = 11.4 in. Use s = 6 in., as calculated.<br />

c. A vt /s = 50b w /f yt = 50(14)/60,000 = 0.017 in. 2 /in., which is less than the 0.032 in. 2 /in. used. Use<br />

no. 4 closed stirrups spaced at 6 in.<br />

6. Find the distribution of longitudinal bars. Total A l is 1.28 in. 2 Use one-third, or 0.43 in. 2 ,atthe<br />

top, at the bottom, and at middepth.<br />

a. Total top bars = 0.88 + 0.43 = 1.31 in. 2 ; use three no. 6 bars (1.32 in. 2 ).<br />

b. Total bottom bars = 4.0 + 0.43 = 4.43 in. 2 ; use five no. 9 bars (5.0 in. 2 ). Total A l used = (1.32<br />

− 0.88) + (5 − 4) = 1.44 in. 2<br />

c. Use two no. 4 bars at middepth (0.40 in. 2 ). Reinforcement details are shown in Fig. 15.16.<br />

Spacing of longitudinal bars is at 7.5 in. < 12 in. The diameter of no. 4 bars used is 0.5 in.,<br />

which is greater than no. 3 or stirrup spacing, s∕24 = 6 = 0.25in. Add no. 4 longitudinal bars<br />

24<br />

on all corners of closed stirrups in beam web and flange.<br />

SUMMARY<br />

Sections 15.1–15.7<br />

1. Torsional stresses develop in a beam when a moment acts on the beam section parallel to its<br />

surface.<br />

2. In most practical cases, a structural member may be subjected to combined shear and torsional<br />

moments.<br />

3. The design methods for torsion rely generally on two basic theories: the skew bending theory<br />

and the space truss theory. The ACI Code adopted the space truss theory.<br />

Sections 15.8 and 15.9<br />

A summary of the relative equations in U.S. customary units and SI units is given here.<br />

Note that (1.0 √ f ′ c) in psi is equivalent to (0.08 √ f ′ c) in MPa N/mm 2 ,1in.≈25 mm, and<br />

f yt ≤ 400 MPa.<br />

Equation U.S. Customary Units SI Units<br />

15.16 T c = 2φ √ ∑<br />

f c ′ x 2 y T c = 0.17φ √ ∑<br />

f c ′ x 2 y<br />

15.17 φT n ≥ T u Same<br />

15.19 T cr = 4λ √ ( )<br />

A<br />

2<br />

f c<br />

′ cp<br />

T<br />

P cr =(λ √ f c ′ ∕3)(A 2 cp ∕P cp )<br />

cp<br />

15.20 T u = φλ √ ( )<br />

A<br />

2<br />

f c<br />

′ cp<br />

T<br />

P u ≤ φλ( √ f c ′ ∕12)(A 2 cp ∕P cp )<br />

cp 15.21<br />

√ ( ) ( )<br />

V 2 2 [( )<br />

u Tu P<br />

+ h<br />

Vc<br />

≤ φ + 8 √ ]<br />

f<br />

b w d 1.7A 2 c<br />

′ (U.S.)<br />

b<br />

0h<br />

w d<br />

√ ( ) ( )<br />

V 2 2 [( )<br />

u Tu P<br />

+ h<br />

Vc<br />

≤ φ +(2 √ ]<br />

f<br />

b w d 1.7A 2 c ′ ∕3) (SI)<br />

b<br />

0h<br />

w d

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!