Structural Concrete - Hassoun

3.14 Rectangular Sections with Compression Reinforcement 119
Therefore,
(
ρ − ρ ′ 1 − 0.85 f c
′ )
( )
0.003 + fy ∕E s
≤ ρ
f max = ρ b
y 0.008
(3.46)
Although Eq. 3.46 is more accurate than Eq. 3.45, it is quite practical to use both equations to
check the condition for maximum steel ratio in rectangular sections when compression steel yields.
For example, if f ′ c = 3ksiandf y = 60 ksi, Eq. 3.46 becomes ρ −0.9575 ρ ′ ≤ 0.016; if f ′ c = 4ksi
and f y = 60 ksi, then ρ − 0.9433 ρ ′ ≤ 0.02138.
The maximum total tensile steel ratio, ρ, that can be used in a rectangular section when compression
steel yields is as follows:
Maxρ = ρ max + ρ ′ (3.47)
where ρ max is maximum tensile steel ratio for the basic singly reinforced tension-controlled concrete
section. This means that maximum total tensile steel area that can be used in a rectangular section
when compression steel yield is as follows:
Max A s = bd(ρ max + ρ ′ )
(3.47a)
In the preceding equations, it is assumed that compression steel yields. To investigate this
condition, refer to the strain diagram in Fig. 3.24. If compression steel yields, then
ε ′ s ≥ ε y = f y
E s
From the two triangles above the neutral axis, substitute E s = 29,000 ksi and let f y be in ksi.
Then
c
d = 0.003
= 87
′ 0.003 − f y ∕E s 87 − f y
( )
87
c = d ′ (3.48)
87 − f y
From Eq. 3.37,
but
A s1
f y = 0.85f ′ cab
A s1
= A s − A ′ s and ρ 1 = ρ − ρ ′
Figure 3.24
Strain diagram in doubly reinforced section.