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Structural Concrete - Hassoun

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Review Problems on <strong>Concrete</strong> Building Components 957<br />

x 2 = 22.75 in.<br />

1st stirrup at face of support<br />

2nd stirrup at s/2 = 5/2 = 2.5 in.<br />

21 stirrups at 5 in. → 107.5 in.<br />

2 stirrups at 10 in. → 127.5 in.<br />

x 3 = 150 − 22.75 − 22.75 = 104.50 in.<br />

c. Check development length<br />

Calculate the development length in compression l dc (ACI Code, Section 25.4.9)<br />

⎧0.02f ⎪ y<br />

⎪<br />

l dc = largest of<br />

λ √ 0.02 (60, 000)<br />

d b =<br />

f c<br />

′ (1) √ (1) =21.9 in. = 22 in.<br />

3000<br />

⎨<br />

⎪(0.0003f y )d b =(0.0003 x 60, 000)(1) =18 in.<br />

⎪8in.<br />

⎩<br />

Therefore l dc = 22 in.<br />

Calculate the development length for tension bar l d .<br />

Check if conditions for spacing and cover are met to select an equation. (ACI Code, Section<br />

25.4.2.2)<br />

d b = 1in.<br />

Since clear cover = 2.5 in.>d b<br />

and clear spacing = 12 − 6<br />

2<br />

then conditions are met to use l d = Ψ t Ψ e f y<br />

20λ √ f ′ c<br />

− 1.128 = 1.9in. ≥ d b<br />

Determine the multiplication factors: (ACI Code, Section 25.4.2.4)<br />

d b<br />

Ψ t = 1.0 (bottom bars)<br />

Ψ e = 1.0 (no coating)<br />

Ψ t Ψ e < 1.7 OK<br />

λ = 1.0 (normal − weight concrete)<br />

√<br />

f<br />

′<br />

c =<br />

√<br />

3000 = 54.8psi< 100 psi<br />

Calculate l d : (ACI Code, Section 25.4.2.2)<br />

l d = Ψ t Ψ e f y<br />

20λ √ d b =<br />

f c<br />

′<br />

(1)(1)(60, 000)<br />

20(1) √ (1.128) =61.8in. = 62 in. ≥ 12 in.<br />

3000<br />

Example 23.4<br />

A simply supported one-story building frame is shown in Figure 23.10. The building is 16 ft high and<br />

has a 33 ft x 11 ft slab.<br />

a. Design the slab.<br />

b. Design the beam along points A and C.

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