Structural Concrete - Hassoun

3.14 Rectangular Sections with Compression Reinforcement 123
3.14.2 When Compression Steel Does Not Yield
As was explained earlier, if
ρ − ρ ′ < 0.85β 1 × f ′ c
f y
× d′
d × 87
87 − f y
= K (3.50)
then compression steel does not yield. This indicates that if ρ − ρ ′ < K, the tension steel will yield
before concrete can reach its maximum strain of 0.003, and the strain in compression steel, ε ′ s, will
not reach ε y at failure (Fig. 3.25). Yielding of compression steel will also depend on its position
relative to the extreme compressive fibers d ′ . A higher ratio of d ′ /c will decrease the strain in the
compressive steel, ε ′ s, as it places compression steel A ′ s nearer to the neutral axis.
If compression steel does not yield, a general solution can be performed by analysis based on
statics. Also, a solution can be made as follows: Referring to 3.23 and 3.24,
( )
( ) ( )
ε ′ c − d
′
s = 0.003
f s ′ = E
c
s ε ′ c − d
′ c − d
′
s = 29,000(0.003) = 87
c
c
Let C c = 0.85f ′ cβ 1 cb:
Because T = A s f y = C c + C s , then
Rearranging terms yields
C s = A ′ s( f ′
s − 0.85f ′ c)=A ′ s
A s f y =(0.85f ′ cβ 1 cb)+A ′ s
[
87
[
87
(
c − d
′
c
(
c − d
′
c
) ]
− 0.85f c
′
) ]
− 0.85f c
′
(0.85f ′ cβ 1 b)c 2 +[(87A ′ s)−(0.85f ′ cA ′ s)−A s f y ]c − 87A ′ sd ′ = 0
This is similar to A 1 c 2 + A 2 c + A 3 = 0, where
A 1 = 0.85f ′ cβ 1 b
A 2 = A ′ s(87 − 0.85f ′ c)−A s f y
A 3 =−87A ′ sd ′
Solve for c:
c = 1
[
]
−A
2A 2 ±
√A 2 2 − 4A 1A 3
1
Once c is determined, then calculate f s ′ , a, C c ,andC s :
[ ]
f s ′ c − d
′
= 87 a = β 1 c C c = 0.85f cab ′ C s = A ′ s( f s ′ − 0.85f c)
′
c
φM n = φ
[C c
(d − 1 )
]
2 a + C s (d − d ′ )
(3.51)
(3.52)
When compression steel does not yield, f s ′ < f y , and the maximum total tensile steel reinforcement
needed for a rectangular section is
Max A s = ρ w,max bd + A ′ f s
′
s = bd
f y
(ρ w,max + ρ′ f ′
s
f y
)
(3.53)