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Structural Concrete - Hassoun

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3.11 Adequacy of Sections 109<br />

accommodate a concrete strength higher than 5 ksi. The two minimum ratios are equal when f ′ c =<br />

4440 psi. This indicates that<br />

⎧<br />

200<br />

⎪<br />

⎪ f y<br />

ρ min = ⎨3 √ f c<br />

⎪<br />

′<br />

⎪ f y<br />

⎩<br />

for f ′ c < 4500 psi<br />

for f ′ c ≥ 4500 psi<br />

For example, if f y = 60 ksi, ρ min = 0.00333 when f c ′ < 4500 psi, whereas ρ min = 0.00353 when<br />

f c ′ = 5000 psi and 0.00387 when f c ′ = 6000 psi.<br />

In the case of a rectangular section, use b = b w in the preceding expressions. For statically<br />

determinate T-sections with the flange in tension, as in the case of cantilever beams, the value of<br />

A s,min must be equal to or greater than following equation:<br />

( √ )<br />

3 f<br />

′<br />

c<br />

A s,min = (x)(d) ≥ 200xd<br />

f y f y<br />

where<br />

x = 2 b w or b f whichever is smaller<br />

b w = width of web<br />

b f = width of flange<br />

For example if b f = 48 in., b w = 16 in., d = 20 in., f ′ c = 4000 psi, and f y = 60,000 psi, then<br />

(<br />

3 √ )<br />

4000<br />

A s,min =<br />

(32)(20) =2.02 in. 2<br />

60,000<br />

200(32)(20)<br />

= 2.13 in. 2 (controls)<br />

60,000<br />

A s,min = 2.13 in. 2<br />

3.11 ADEQUACY OF SECTIONS<br />

A given section is said to be adequate if the internal moment strength of the section is equal to<br />

or greater than the externally applied factored moment, M u ,orφM n ≥ M u . The procedure can be<br />

summarized as follows:<br />

1. Calculate the external applied factored moment, M u .<br />

M u = 1.2M D + 1.6M L<br />

2. Calculate φM n for the basic singly reinforced section:<br />

a. Check that ρ min ≤ ρ ≤ ρ max .

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