Civil Engineering Reference
In-Depth Information
b
0.4
s
cu
Resultant compressive
load in concrete
1
2
d
1
C
d
1
3
4
d
1
T
0.87
s
Y
F
IGURE
12.11
Approximation of
stress distribution
in concrete
A
s
Resultant tensile
load in steel
(a)
(b)
M
u
is then given by
C
4
d
1
=
0
.
40
σ
cu
b
2
d
1
4
d
1
M
u
=
which gives
0
.
15
σ
cu
b
(
d
1
)
2
M
u
=
(12.23)
or
T
4
d
1
=
0
.
87
σ
Y
A
s
4
d
1
M
u
=
from which
M
u
=
0
.
65
σ
Y
A
s
d
1
(12.24)
whichever is the lesser. For applied bending moments less than
M
u
a rectangular stress
block may be assumed for the concrete in which the stress is 0
.
4
σ
cu
but in which the
depth of the neutral axis must be calculated. For beam sections in which the applied
bending moment is greater than
M
u
, compressive reinforcement is required.
E
XAMPLE
12.7
A reinforced concrete beam having an effective depth of 600mm
and a breadth of 250mm is subjected to a bending moment of 350 kNm. If the 28-day
cube strength of the concrete is 30N/mm
2
and the yield stress in tension of steel is
400N/mm
2
, determine the required area of reinforcement.
First it is necessary to check whether or not the applied moment exceeds the ultimate
moment of resistance provided by the concrete. Hence, using Eq. (12.23)
600
2
10
−
6
M
u
=
0
.
15
×
30
×
250
×
×
=
405 kNm
Since this is greater than the applied moment, the beam section does not require
compression reinforcement.
We now assume the stress distribution shown in Fig. 12.12 in which the neutral axis
of the section is at a depth
n
below the upper surface of the section. Thus, taking
