24.02.2017 Views

Structural Concrete - Hassoun

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

3.9 Singly Reinforced Rectangular Section in Bending 103<br />

For members subjected to flexure, the relationship between the steel ratio, ρ, was given in<br />

Eq. 3.24:<br />

or<br />

ε t + 0.003 = 0.003 + f y∕E s<br />

ρ∕ρ b<br />

(3.24)<br />

ρ<br />

ρ b<br />

= 0.003 + f y∕E s<br />

0.003 + ε t<br />

(3.29)<br />

For f y = 60 ksi and E s = 29,000 ksi, f y /E s may be assumed to be 0.00207.<br />

ρ<br />

ρ b<br />

= 0.00507<br />

0.003 + ε t<br />

(3.30)<br />

The limit for tension to control is ε t ≥ 0.005 according to ACI. For ε t = 0.005, Eq. 3.30 becomes<br />

ρ<br />

= 0.005<br />

ρ b 0.008 = 5 = 0.625 (3.30a)<br />

8<br />

or ρ ≤ 0.63375 ρ b for tension-controlled sections if ε t = 0.00507 = f y /E s . Both values can be used<br />

for practical analysis and design. The small increase in ρ will slightly increase the moment capacity<br />

of the section. For example, if f ′ c = 4ksiandf y = 60 ksi, ρ b = 0.0285 and ρ ≤ 0.01806 for tension<br />

to control (as in the case of flexural members). The φ factor in this case is 0.9. This value is less<br />

than ρ max = 0.75ρ b = 0.0214 allowed by the ACI Code for flexural members when φ = 0.9 can<br />

be used.<br />

Design of beams and other flexural members can be simplified using the limit of ε t = 0.005.<br />

ρ<br />

= 0.003 + f y∕E s<br />

ρ b 0.008<br />

In this case, ρ = ρ max = upper limit for tension-controlled sections.<br />

( )<br />

0.003 + fy ∕E s<br />

ρ max =<br />

ρ<br />

0.008 b<br />

(3.31)<br />

(3.31a)<br />

Note that when ρ used ≤ ρ max , tension controls and φ = 0.9. When ρ>ρ max , the section will be in<br />

the transition region with φ

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

Saved successfully!

Ooh no, something went wrong!