Structural Concrete - Hassoun

372 Chapter 11 Members in Compression and Bending
2. From the equations of equilibrium,
where
P n = C c + C s − T (Eq. 11.10)
C c = 0.85f c ′ ab = 0.85 × 4 × 14a = 47.6a
C s = A ′ s (f y − 0.85f c ′ )=4(60 − 0.85 × 4) =226.4K
Assume compression steel yields. (This assumption will be checked later.)
3. Taking moments about A s ,
T = A s f s = 4f s
f s < f y
P n = 47.6a + 226.4 − 4f s
P n = 1 [ (
C
e ′ c d − a )
]
+ C
2 s (d − d ′ )
(I)
(Eq. 11.11)
The plastic centroid is at the center of the section and d ′′ = 8.5 in.
e ′ = e + d ′′ = 10 + 8.5 = 18.5 in.
[ (
47.6a 19.5 − a 2
P n = 1
18.5
4. Assume c = 13.45 in., which exceeds c b (11.54 in.).
Substitute a = 11.43 in Eq. II:
)
+ 226.4(19.5 − 2.5)
= 50.17a − 1.29a 2 + 208 (II)
a = 0.85 × 13.45 = 11.43 in.
P n1 = 50.17 × 11.43 − 1.29(11.43) 2 + 208 = 612.9K
5. Calculate f s from the strain diagram when c = 13.45 in.
( )
19.5 − 13.45
f s =
87 = 39.13 ksi ε
13.45
s = ε t = f s
= 0.00135
E s
6. Substitute a = 11.43 in. and f s = 39.13 ksi in Eq. I to calculate P n2 :
P n2 = 47.6(11.43)+226.4 − 4(39.13) =613.9K
which is very close to the calculated P n1 of 612.9 K (less than 1% difference).
( ) 10
M n = P n e = 612.9 = 510.8 K⋅ ft
12
7. Check if compression steel yields. From the strain diagram,
ε ′ 13.45 − 2.5
s = (0.003) =0.00244 >ε
13.45
y = 0.00207
Compression steel yields, as assumed.
8. P n = 612.9 K is greater than P b = 453.4 K, and e = 10 in. < e b = 15 in., both calculated in the
previous example, indicating that compression controls, as assumed. Note that it may take a few
trials to get P n1 close to P n2 .
]