Compression Reinforcement
What happens if the beam section is fixed and
K
M
bd f
ck
=
2
> allowable limit
Is there any way in which the section can be reinforced so that it
still fails in a ductile manner?
Consider the effect of adding extra steel in the compression
region. Effectively this has the same effect as increasing the area
of the compression block. Thus providing compression
reinforcement provides a means of designing a stronger ductile
section than is possible with tension steel only.
If
K
M
bd f
ck
=
2
> allowable limit then compression steel is
required. The standard procedure adopted by most codes, and
the associated reasoning, is as follows;
First consider the case when
K
is the maximum allowable value
K
M
bd f
M bd
ck
ck
= =
⇒ =
'
.
'.
2
2
0166
0166
f
This
M’
is the largest moment which the single reinforced
section can resist if it is to fail in a ductile manner. The area of
tension steel
A
s
required to resist this moment is easily found
using the standard formulae.
{ }
min
1
maxmin
2
87.0
'
13.125.05.0
'
zf
M
A
dKdz
fbd
M
K
y
s
cu
=⇒
−+=
=
Consider the effect of adding additional tensile steel
A
s2
and at
the same time adding a matching amount of steel
A’
in the
compression region such that;
A f
A
f
s
s y
s
''=
2
γ
In this case the additional forces are selfequilibrating,
equilibrium of axial forces is satisfied without having to re
calculate
x
. In most simple cases the compression steel is placed
close to the extreme compression fibre, which at failure has a
strain of 0.0035, and hence has a strain greater than yield strain,
so that typically
A’=A
s2
.
The addition of this extra steel increases the ultimate moment
capacity of the beam by
(
)
(
)
∆M
f
A d d A f d d
ult
y
s
s s s
= − =
γ
2
'''
−'
Therefore to design a section for a moment M greater than M’ it
is necessary to provide compression reinforcement
A’
where,
(
)
(
)
A
M M
f d d
s
s
'
'
''
=
−
−
and tension reinforcement
A
s
where,
(
)
(
)
A A A
M
f z
M M
f
d d
s s s
y
y
s
= + = +
−
−
1 2
087
'
.
'
'
min
γ
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