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Structural Concrete - Hassoun

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378 Chapter 11 Members in Compression and Bending<br />

(a)<br />

(b)<br />

Figure 11.12 Side bars in rectangular sections: (a) six side bars and (b) two side bars<br />

(may be neglected).<br />

Example 11.7<br />

Determine the balanced load, P b moment, M b ,ande b for the section shown in Fig. 11.13. Use f c ′ = 4ksi<br />

and f y = 60 ksi.<br />

Solution<br />

The balanced section is similar to Example 11.2. Given: b = h = 22 in., d = 19.5 in., d ′ = 2.5in., A s =<br />

A ′ s = 6.35 in.2 (five no. 10 bars), and six no. 10 side bars (three on each side).<br />

1. Calculate the distance to the neutral axis:<br />

( ) (<br />

87<br />

c b = d<br />

87 + f t =<br />

y<br />

a b = 0.85(11.54) =9.81 in.<br />

87<br />

87 + 60<br />

)<br />

19.5 = 11.54 in.<br />

2. Calculate the forces in concrete and steel bars; refer to Fig. 11.13a. In the compression zone,<br />

C c = 0.85f c ′ ab = 0.85(4)(9.81)(22) =733.8K.<br />

( )<br />

f s ′ c − d = 87 ′ ( )<br />

11.54 − 2.5<br />

= 87<br />

= 68.15 ksi > 60 ksi<br />

c<br />

11.54<br />

Then f s ′ = 60 ksi. C s1 = A ′ s (f y − 0.85f c ′ )=6.35(60 − 0.85 × 4) =359.4 K<br />

( )<br />

11.54 − 2.5 − 4.25<br />

f s2 = 87<br />

= 36.11 ksi<br />

11.54<br />

C s2 = 2(1.27)(36.11 − 0.85 × 4) =83.1K<br />

Similarly, f s3 = 4.07 ksi and C s3 = 2(1.27)(4.07 − 0.85 × 4) = 1.7 K.<br />

In the tension zone,<br />

ε s4 = 964.50 × 10 −6<br />

T 1 = 2(1.27)(28) =71 K<br />

f s4 = 28 ksi<br />

T 2 = A s f y = 6.35(60) =381 K

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