Structural Concrete - Hassoun

11.8 Strength of Columns for Compression Failure 371
4 no. 9
Figure 11.10
Example 11.4: Compression controls.
where e ′ = e + d − h/2 when A s = A ′ s or e ′ = e + d ′′ in general, C c = 0.85f cab, ′ C s = A ′ s(f s ′ −
0.85f c), ′ andT = A s f s .
4. Assume a value for c such that c > c b (calculated in step 1). Calculate a = β 1 c. Assume
f s ′ = f y .
5. Calculate f s based on the assumed c:
( )
dt − c
f s = ε s E s = 87 ksi ≤ f
c
y
6. Substitute the preceding values in Eq. 11.10 to calculate P n1 and in Eq. 11.11 to calculate P n2 .
If P n1 is close to P n2 , then choose the smaller or average of P n1 and P n2 .IfP n1 is not close to
P n2 , assume a new c or a and repeat the calculations starting from step 4 until P n1 is close to
P n2 . (1% is quite reasonable.)
7. Check that compression steel yields by calculating ε ′ s = 0.003[(c − d ′ )∕c] and comparing it
with ε y = f y /E s . When ε ′ s ≥ ε y , compression steel yields; otherwise, f s ′ = ε ′ sE s or, directly,
( )
f s ′ c − d
′
= 87 ≤ f
c y ksi
8. Check that e < e b or P n > P b for compression failure. Example 11.4 illustrates the procedure.
9. The net tensile strain, ε t , in the section is normally less than 0.002 for compression-controlled
sections (Fig. 11.4). Consequently, the strength reduction factor (φ) = 0.65 (or 0.70 for spiral
columns).
Example 11.4
Determine the nominal compressive strength, P n , for the section given in Example 11.2 if e = 10 in. (See
Fig. 11.10.)
Solution
(
1. Because e = 10 in. < 2
3)
d = 13 in., assume compression failure. This assumption will be checked
later. Calculate the distance to the neutral axis for a balanced section, c b :
c b = 87 87
d
87 + f t = (19.5) =11.54 in.
y 87 + 60