Structural Concrete - Hassoun

4.4 Rectangular Sections with Compression Reinforcement 167
ρ − ρ ′ = A s1
bd = 3.576
(12)(16.5) = 0.01806 ≤ K
Therefore, compression steel does not yield: f s ′ < f y .
4. Calculate f s ′ ∶ f s ′ = 87[(c − d′ )∕c] ≤ f y . Determine c from A s1 : A s1 = 3.576 in. 2 ,
a =
5. Calculate A ′ s from M u2 = φA′ s f ′
s (d − d′ ):
f ′
A s1 f y 3.576 × 60
0.85f c ′ =
b 0.85 × 4 × 12 = 5.26in.
c = a = 5.26
β 1 0.85 = 6.19in.
( ) 6.19 − 2.5
s = 87 × = 51.8ksi < 60ksi
6.19
902 = 0.9A ′ s (51.8)(16.5 − 2.5)
Thus, A ′ s = 1.38in.2 , or calculate A ′ s from A′ s = A s2 (f y ∕f s ′)=1.38in.2
(two no. 8 bars). Note
that the condition [ρ − ρ ′ (f s ′ ∕f y )]=(ρ − ρ′ ) ≤ ρ max is already met:
(ρ − ρ ′ f s
′ )
= 1
f y bd (A s − A 3.576
s2 )=
12 × 16.5 = 0.01806
as assumed in the solution.
6. These calculations using ρ max and R u are based on a strain of 0.005 at the centroid of the tension
steel:
( )
dt − c
ε t (at bottom row) = 0.003
as expected.
d t = 20 − 2.5 = 17.5in. ε t =
c
( 17.5 − 6.19
6.19
)
0.003 = 0.00548 > 0.005
Solution 2
Assuming two rows of tension bars and a strain at the lower row, ε t = 0.005, the solution will be as
follows:
1. Calculate d t = 20 −2.5 = 17.5 in. From the strain diagram,
c
= 0.003 = 0.003
d t 0.003 + ε t 0.008 = 0.375
c = 0.375(17.5) =6.5625 in.
2. The compression force in the concrete = C 1 = 0.85 f ′ c ab,
a = 0.85c = 5.578 in.
C 1 = 0.85(4)(5.578)(12) =227.6K = T 1 (as singly reinforced)
A s1 = C 1
= T 1
= 227.6 = 3.793in. 2
f y f y 60
d = 20 − 3.5 = 16.5in.
(
M u1 = φA s1 f y d − a )
(
= 0.9(3.793)(60) 16.5 − 5.578 )
= 2808K ⋅ in.
2
2
= 234K ⋅ ft