1
Chapter II
Beams with Compression Reinforcement
（
L散瑵r攠㈱
）
Compression reinforcement helps to:
1)
reduce long

term deflections
2)
increase deformation capacity
(
deformation at the limit state,
at the
boundary
between stages III and IV)
On the other hand, compression
reinforcement is not effective in increasing flexural strength.
In
this lecture
, we study how compression reinforcement affects
the
deformation capacity and the
flexural
strength
of
a
section.
k d
k k d
C = A'
s
x f
'
s
s
'
s
d'
A
s
A'
s
d
b
h
k f
'
cu
dk d
s
c
jd
T=A
s
f
y
C
u
u
2
u
3
c
We define the limit state again
by fixing the maximum compr
essive unit strain in the concrete at
0.003. The distribution of strain is again assumed to be linear.
U
nit strain in the compression
reinforcement
is therefore computed as
:
d
d
k
d
k
u
u
cu
s
defining
c
=d’/d,
c
u
u
cu
s
k
k
or, rearranging terms,
c
s
cu
cu
u
k
The expression we used before to relate strains in the reinforcement in tension and the maximum
compressive strain still holds:
s
cu
cu
u
k
2
But, now,
the
exp
ression for equilibrium of forces requires a new term describing the force
carried by the steel in compression
(C
s
)
:
s
c
C
C
T
T
is tensile force.
If the steel in tension yields:
T = As fy.
C
c
is the compressive force resisted by the con
crete:
s
u
c
1
c
A
d
k
b
f
85
.
0
k
C
We subtract the area A’
s
because it is occupied by steel, not concrete.
C
s
is the compressive force resisted by steel:
s
s
s
f
A
C
Replacing
:
c
1
s
s
u
c
1
y
s
f
85
.
0
k
f
A
d
k
b
f
85
.
0
k
f
A
Rearranging terms and
discarding the term
k
1
x0.85xf’
c
(usually
much
smaller than f’
s
)
c
1
s
y
u
f
85
.
0
k
f
f
k
’ = A’
s
/(bd)
is the ratio of compressive reinforcement
(not
e
that
we divide by d,
not
d’)
This expression tells us that as
the ratio of
compression reinforcement
increases,
neutral axis
depth
decreases
. Because cu
rvature is inversely proportion
al to neutral axis depth (
=
c
/(k
u
d)),
we conclude that the limiting curvature increases with increasing amount of compression
reinforcement.
Displacement
s
and signs
warning us of potential fai
lure also increase for beams
with larger amounts of compression reinforcement.
The expression we have obtained also
indicates that the ratio of
tension
reinforcement associated with balanced conditions (
k
u
=0.6)
increases as the ratio of compres
sion
reinfo
rcement
is increased
.
We also conclude that,
because in beams
k
u
is always
larger than zero, the
unit
stress in the
compression reinforcement
must
decrease as
’ increases. To estimate the
unit
stress in the
compression reinforcement
(f’
s
)
we
equate
the expressions:
c
1
s
y
u
f
85
.
0
k
f
f
k
and
c
s
cu
cu
u
k
and
obtain
f’
s
as a
function of
’
s
:
c
1
c
s
cu
cu
y
s
f
85
.
0
k
f
1
f
V
alues of
f’
s
and
’
s
satisfying
this
relationship
are plotted
on the right
(solid curves) for the following
parameters:
f’
c
= 3000psi
0
20
40
60
80
.000
.001
.002
.003
'
s
f'
s
[ksi]
¾
½
¼
¾
½
¼
1
=1%
=2%
f'c = 3,000 psi
fy = 60ksi
d'/d = 0.1
'=
x
1
3
f
y
= 60ksi
c
= 0.1
= 1% (thin lines), 2% (thick lines)
As we have done before,
we
superimpose
th
e
curves
we have obtained and
f’
s
= E
s
’
s
< f
y
.
The
intersection points
give us estimates of
f’
s
and
’
s
that satisfy all the expressions in our
formulation
. Our solutions show that as we increase
’,
f’
s
and
’
s
decrease
, and that
if we are to
limit
to 2%
(to avoid congestion)
,
it is rather difficult t
o reach yielding in compression
reinforcement.
Having estimates of f’
s
and
’
s
allows us to
compute
k
u
and check our assumption about the stress
in the tension reinforcement
(f
s
= f
y
; which is the case if k
u
<3/5
and f
y
= 60ksi
)
.
It also allows us
to esti
mate m
oment capacity as:
d
d
C
d
k
k
1
C
M
s
u
1
2
1
c
n
Observe that we are computing moments about the tension reinforcement and we include a term
related to stresses in concrete and another related to stresses in steel.
The result obtained with this
expressi
on is usually close to 0.9A
s
f
y
.
II

3.1
Example
Compute nominal moment capacity
and limiting curvature
for the following parameters:
= 1.5%
f’
c
= 5000psi
f
y
= 60ksi
c
= 0.1
’ = 0, 0.5%, 1%
Solution
For the case with
’ = 0
:
26
.
0
ksi
5
85
.
0
8
.
0
ksi
60
015
.
0
f
85
.
0
k
f
k
c
1
y
u
(<3/5 => f
s
= f
y
)
4
u
cu
u
10
6
.
4
in
25
26
.
0
003
.
0
d
k
1/in,
9
.
0
26
.
0
8
.
0
1
k
k
1
j
2
1
u
1
2
1
in
kip
8100
in
25
9
.
0
ksi
60
in
25
in
16
%
5
.
1
jd
f
d
b
M
y
n
For the case with
’ = 0.5%
:
ksi
5
85
.
0
8
.
0
1
.
0
003
.
0
003
.
0
ksi
60
015
.
0
005
.
0
1
f
to
leads
f
85
.
0
k
f
f
s
s
c
s
cu
cu
c
1
s
y
2.5
”
A
s
A'
s
25”
16”
2
7.
5
”
4
w
e compare this equation with the unit stress
–
unit
strain relationship assumed for steel and obtain
f’
s
=
44
ksi and
’
s
= 0.001
5
.
With this estimate we compute:
2
.
0
1
.
0
%
15
.
0
%
3
.
0
%
3
.
0
k
c
s
cu
cu
u
(
<3/5
)
4
u
cu
u
10
6
in
25
2
.
0
003
.
0
d
k
1/in,
d
d
f
A
k
k
1
d
k
b
f
85
.
0
k
M
s
s
u
1
2
1
2
u
c
1
n
in
kip
8200
in
kip
1980
in
kip
6250
M
in
5
.
2
25
ksi
44
in
25
in
16
%
5
.
0
2
.
0
8
.
0
1
in
25
2
.
0
in
16
ksi
5
85
.
0
8
.
0
M
n
2
1
2
n
For the case with
’ =
1
%
:
ksi
5
85
.
0
8
.
0
1
.
0
003
.
0
003
.
0
ksi
60
015
.
0
01
.
0
1
f
to
leads
f
85
.
0
k
f
f
s
s
c
s
cu
cu
c
1
s
y
we compare this equation with the unit stress
–
unit
strain relationship assumed for steel and obtain f’
s
=
34ksi and
’
s
= 0.0012.
Now
17
.
0
1
.
0
%
12
.
0
%
3
.
0
%
3
.
0
k
c
s
cu
cu
u
(<3/5
)
4
u
cu
u
10
7
in
25
17
.
0
003
.
0
d
k
1/in,
in
kip
8400
in
kip
3060
in
kip
5380
M
in
5
.
2
25
ksi
34
in
25
in
16
%
1
17
.
0
8
.
0
1
in
25
17
.
0
in
16
ksi
5
85
.
0
8
.
0
M
d
d
f
A
k
k
1
d
k
b
f
85
.
0
k
M
n
2
1
2
n
s
s
u
1
2
1
2
u
c
1
n
O
ur results
and other moment

curvature pairs are
plotted in the Figure on
the right. Observe that the relative changes in strength
associated with
addition of compression reinforcement
a
re much sm
aller that the changes in limiting curvature.
II

3.2
Exercise
Repeat the previous example using f’
c
= 3000psi.
0.0
20.0
40.0
60.0
80.0
.000
.001
.002
.003
'
s
f'
s
[ksi]
0
4000
8000
12000
0.0000
0.0002
0.0004
0.0006
0.0008
Curvature
[1/in.]
Moment [kipin.]
' = 0%
' = 0.5%
' = 1.0%
Remember
Compression reinforcement can be used effectively to increase the
deformation
capacity of a
beam
subjected
to bending but it is not effective in increasing its strength.
0.0
20.0
40.0
60.0
80.0
.000
.001
.002
.003
'
s
f'
s
[ksi]
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