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Structural Concrete - Hassoun

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172 Chapter 4 Flexural Design of Reinforced <strong>Concrete</strong> Beams<br />

42"<br />

14" 14"<br />

18.5"<br />

in.<br />

A 2<br />

s A s1<br />

14" 14"<br />

A st = 3.57 in 2<br />

Figure 4.9 Analysis of Example 4.8.<br />

3. Find the portion of the design moment taken by the overhanging portions of the flange (Fig. 4.9).<br />

First calculate the area of steel required to develop a tension force balancing the compressive force<br />

in the projecting portions of the flange:<br />

A sf = 0.85f c ′(b − b w )t 0.85 × 3 ×(42 − 14)×3<br />

= = 3.57in 2 .<br />

f y 60<br />

Here, φ M n = M u1 + M u2 , that is, the sum of the design moment of the web and the design<br />

moment of the flanges:<br />

(<br />

M u2 = φA sf f y d − 1 )<br />

2 t (<br />

= 0.9 × 3.57 × 60 18.5 − 3 )<br />

= 3277K ⋅ in.<br />

2<br />

4. Calculate the design moment of the web (as a singly reinforced rectangular section):<br />

M u1 = M u − M u2 = 5080 − 3277 = 1803K ⋅ in.<br />

R u =<br />

M u1 1, 803, 000<br />

=<br />

(b w d 2 ) 14 ×(18.5) = 376psi<br />

2<br />

From Eq. 4.2 or the tables in Appendix A, for R u = 376 psi, ρ 1 = 0.0077:<br />

A s1 = ρ 1 b w d = 0.0077(14)(18.5) =1.99in. 2<br />

Total A s = A sf + A s1 = 3.57 + 1.99 = 5.56in. 2 (uses ix no.9 bars in two rows)<br />

Total h 5= 18.5 + 3.5 = 22 in. Calculate A s,max for T-sections using Eq. 3.72:<br />

MaxA s = 7.02in. 2 > 5.56in. 2<br />

5. Check ε t : a = 1.99 ×60/(0.85 ×3 ×14) = 3.34 in., c = 3.93 in., and d t = 19.5 in. Then ε t = 0.003<br />

(d t − c)/c = 0.0119 > 0.005, tension-controlled section (φ = 0.9).<br />

Example 4.9<br />

In a slab–beam system, the flange width was determined to be 48 in., the web width was b w = 16 in., and<br />

the slab thickness was t = 4 in. (Fig. 4.10). Design a T-section to resist an external factored moment of<br />

M u = 812 K ⋅ ft. Use f c ′ = 3ksi and f y = 60 ksi.<br />

Solution<br />

1. Because the effective depth is not given, let a = t and calculate A sft for the whole flange:<br />

A sft = 0.85f c ′ bt = 0.85(3)(48)(4) = 8.16in. 2<br />

f y 60

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