Structural Concrete - Hassoun

122 Chapter 3 Flexural Analysis of Reinforced Concrete Beams
Solution
1. Check if compression steel yields:
A s = 6.0in. 2
A ′ s = 1.2in.2
A s − A ′ s = 4.8in.2 ρ − ρ ′ = 0.01778
For compression steel to yield,
ρ = A s
bd = 6.0
12 × 22.5 = 0.02222
ρ ′ = A′ s
bd = 1.2
12 × 22.5 = 0.00444
ρ − ρ ′ ≥ 0.85β 1
f ′ c
f y
× d′
d × 87
87 − f y
= K
Let β 1 be 0.85 because f c ′ = 4000 psi; therefore,
( )( )( )
K =(0.85) 2 4 2.5 87
= 0.0175
60 22.5 87 − 60
ρ − ρ ′ = 0.01778 > 0.0175
Therefore, compression steel yields.
2. Check that ρ − ρ ′ ≤ ρ max (Eq. 3.45): For f c ′ = 4ksiand f y = 60 ksi, ρ b = 0.0285 and ρ max = 0.01806
(Table 3.2). Then ρ − ρ ′ = 0.01778 0.005
or
c
= 0.353 < 0.375
d (OK)
6. The maximum total tension steel for this section, max A s , is equal to
Max A s = bd(ρ max + ρ ′ )=12 × 22.5(0.01806 + 0.00444)
= 6.08 in. 2 A s = 6.0in. 2 (used in the section)