Structural Concrete - Hassoun

380 Chapter 11 Members in Compression and Bending
5. Determine φ: For a balanced section, ε t = ε y = 0.002, φ = 0.65,
φP b = 0.65P b = 472 K and φM b = 0.65M b = 618.8K⋅ ft
Example 11.8
Repeat the previous example when e = 6.0 in.
Solution
1. Because e = 6in.< e b = 15.735 in., this is a compression failure condition. Assume c = 16.16 in.
(by trial) and a = 0.85(16.16) = 13.74 in. (Fig. 11.13b).
2. Calculate the forces in concrete and steel bars:
C c = 0.85(4)(13.74)(22) =1027.75 K
In a similar approach to the balanced case, f s1 = 60 ksi and C s1 = 359.41.
f s2 = 50.66 ksi C s2 = 120.0 K
f s3 = 27.78 ksi
f s4 = 4.9ksi
f s5 = 18 ksi
3. Calculate P n = C c + ΣC s − ΣT = 1458.7 K.
C s3 = 61.92 K
C s4 = 3.81 K
M n = P n e = 729.35 K ⋅ ft
T = 6.35(18) =114.2 K
(e = 6in.)
4. Check P n by taking moments about A s ,
P n = 1 [ (
C
e ′ c d − a )
+ C
2 s1 (d − d ′ )+C s2 (d − d ′ − s)
+ C s3 (d − d ′ − 2s)+C s4 (d − d ′ − 3s)]
e ′ = e + d − h = 6 + 19.5 − 22∕2 = 14.5in.
2
5. Calculate φ:
s = distance between side bars
= 4.25 in. (s = constant in this example.)
[ (
1027.75 19.5 − 13.74
2
P n = 1
14.5
+ 120(17 − 4.25)+61.92(17 − 8.5)
+ 3.81(17 − 12.75)] = 1459 K
d t = d = 19.5in.
)
+ 359.41(17)
c = 16.16 in.
ε t (at tension steel level) = 0.003(d t − c)
c
0.003(19.5 − 16.16)
= = 0.00062
16.16