Structural Concrete - Hassoun

11.7 Strength of Columns for Tension Failure 369
3. Taking moments about A s (Eq. 11.12),
P n = 1 [ (
C
e ′ c d − a )
]
+ C
2 s (d − d ′ )
Note that for the plastic centroid at the center of the section, d ′′ = 8.5 in.
4. Equating Eqs. I and II,
or
5. From Eq. I:
e ′ = e + d ′′ = 20 + 8.5 = 28.5in.
[ (
47.6a 19.5 − a 2
P n = 1
28.5
P n = 32.56a − 0.835a 2 + 135.0
)
+ 226.4 × 17
P n = 47.6a − 13.6 = 32.56a − 0.835a 2 + 135.0
a 2 + 18a − 178.0 = 0
a = 7.1in.
P n = 47.6 × 7.1 − 13.6 = 324.4K
M n = P n e = 324.4 × 20 = 540.67 K ⋅ ft
12
6. Check if compression steel has yielded:
c =
a
0.85 = 7.1
0.85 = 8.35 in. ε y = 60
29,000 = 0.00207
ε ′ 8.35 − 2.5
s = 0.003 = 0.0021 >ε
8.35
y
Compression steel yields. Check strain in tension steel:
( )
19.5 − 8.35
ε s =
× 0.003 = 0.004 >ε
8.35
y
If compression steel does not yield, use f s ′ as calculated from f
′
s = ε′ s E s and revise the calculations.
7. Calculate φ: Since ε t = 0.004, the section is in the transition region.
φ = 0.65 +(ε t − 0.002)(250∕3) =0.817
φP n = 0.817(324.4) =264.9 K
φM n = 0.817(540.67) =441.7K⋅ ft
8. Because e = 20 in. > e b = 15 in. (Example 11.2), there is a tension failure condition.
9. The same results can be obtained using the values of A, B,andC given earlier.
where
Aa 2 + Ba + C = 0
A = 0.425f c ′ b = 0.425(4)(14) =23.8
B = 2A(e ′ − d) =2(23.8)(28.5 − 19.5) =428.4
C = 4(60 − 0.85 × 4)(28.5 − 19.5 + 2.5)−4(60)(28.5)
=−4236.4
Solve for a to get a = 7.1 in. and P n = 324.4 K.
]
(II)