Structural Concrete - Hassoun

3.14 Rectangular Sections with Compression Reinforcement 125
d. Calculate
e. Calculate
φM n = φ
a = A s f y − A ′ s f ′
s
0.85f ′ cb
or
a = β 1 c
[ (As ) (
f y − A ′ s f s
′ d − 1 )
]
2 a + A ′ s f s ′ (d − d ′ )
Example 3.10
Determine the design moment strength of the section shown in Fig. 3.27 using f c ′ = 5ksi,f y = 60 ksi,
A ′ s = 2.37 in.2 (three no. 8 bars), and A s = 7.62 in. 2 (six no. 10 bars).
Figure 3.27
Example 3.10 analysis solution.
Solution
1. Calculate ρ and ρ ′ :
ρ = A s
bd = 7.62
14 × 22.5 = 0.0242 ρ′ = A′ s
bd = 2.37
14 × 22.5 = 0.00753
ρ − ρ ′ = 0.01667
2. Apply Eq. 3.50, assuming β 1 = 0.8 for f c ′ = 5000 psi.
K = 0.85β 1 × f ′ c
f y
× d′
d × 87
87 − f y
= 0.85 × 0.8
( 5
60
(or from Table 3.4, K = 0.456/d = 0.0203):
ρ − ρ ′ = 0.01667 < 0.0203
)( )( 2.5
22.5
87
87 − 60
)
= 0.0203
Therefore, compression steel does not yield, and f s ′ < 60 ksi.
For f c ′ = 5ksi and f y = 60 ksi, find ρ b = 0.0335 and ρ max = 0.02123 (Table 3.2); then check
(ρ − ρ ′ )