Structural Concrete - Hassoun

3.9 Singly Reinforced Rectangular Section in Bending 101
For a balanced or an underreinforced section, T = A s f y . Then
(
M n = A s f y d − 1 )
2 a
(3.19)
To get the usable design moment φ M n , the previously calculated M n must be reduced by the capacity
reduction factor, φ,
(
φM n = φA s f y d − a ) (
= φA
2 s f y d −
A )
s f y
1.7f cb
′ (3.19a)
Equation 3.19a can be written in terms of the steel percentage ρ:
Equation 3.20 can be written as
ρ = A s
A
bd s = ρbd
(
φM n = φf y ρbd d − ρbdf ) (
y
1.7f cb
′ = φρf y bd 2 1 − ρf )
y
1.7f c
′
(3.20)
φM n = R u bd 2 (3.21)
where
(
R u = φρf y 1 − ρf )
y
1.7f c
′ (3.22)
The ratio of the equivalent compressive stress block depth, a, to the effective depth of the section,
d, can be found from Eq. 3.15:
0.85f ′ cab = ρbdf y
a
d =
ρf y
0.85f ′ c
(3.23)
3.9.2 Upper Limit of Steel Percentage
The upper limit or the maximum steel percentage, ρ max , that can be used in a singly reinforced
concrete section in bending is based on the net tensile strain in the tension steel, the balanced steel
ratio, and the grade of steel used. The relationship between the steel percentage, ρ, in the section
and the net tensile strain, ε t , is as follows:
( )
0.003 + fy ∕E s
ε t =
− 0.003 (3.24)
ρ∕ρ b
For f y = 60 ksi, and assuming f y /E s = 0.002,
( )
0.005
ε t = − 0.003 (3.25)
ρ∕ρ b
These expressions are obtained by referring to Fig. 3.12. For a balanced section,
Similarly, for any steel ratio, ρ,
c b = a b
β 1
=
c =
A sb f y
0.85f cbβ ′ = ρ b f y d
1 0.85f cβ ′ 1
ρf yd
0.85f ′ cβ 1
and
c
c b
= ρ ρ b