The document discusses the behavior and analysis of reinforced concrete beams. It describes the three stages a beam undergoes when loaded: uncracked, cracked-elastic, and ultimate strength. The transformed area method is presented for calculating stresses in cracked beams. An example problem demonstrates using this method to find bending stresses in a beam section. The allowable resisting moment is also determined based on specified material stresses.

1. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 1
CE370
REINFORCED CONCRETE-I
Prof. Abdelhamid CHARIF
Flexural Behavior and
Analysis of RC Beams
Behavior stages in a RC beam
• A small transverse load placed on a simply supported RC beam is
gradually increased until beam failure.
• This loading causes positive bending moment (tension, and steel
in bottom) and shear force
• Perfect bond is assumed between steel bars and concrete
• The beam will go through three distinct stages before collapse
occurs. These are:
• The uncracked concrete stage
• The concrete cracked – elastic strength stage
• The ultimate – strength stage.
2

2. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 2
3
Simply supported beam under uniform load
4
Basic Assumptions in Flexure theory of RC Members
1. Plane sections before bending remain plane after bending.
This assumption has been experimentally verified for all normal
beams except deep beams (length to thickness ratio less than 3).
2. The strain in reinforcement equals to the strain in concrete at the
same level (perfect bond, no slip).
1 & 2 imply that strain in reinforcement and concrete is
directly proportional to the distance from N.A.
3. Tensile strength of concrete is neglected in flexural strength
calculations (after cracking).
4. Stresses in concrete and reinforcement are calculated from
strains using material stress-strain curves.
Other assumptions will be made at ultimate state

3. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 3
5
Recall Concrete Stress-Strain Curve
Compression: For a stress less or equal to 50% of fc’ ,
the curve is almost linear.
Tension: Strength very small and curve assumed linear.
6
Recall Steel Stress-Strain Curve
(Large plastic deformations after yielding)







y
s
y
y
s
s
s
s
f
E
f





when
when
grades
all
for
200000MPa
Es 

4. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 4
Uncracked Concrete Stage
• At small loads when the tensile stresses are less than the
modulus of rupture, the entire cross section of the beam resists
bending.
• Standard bending theory can be used.
• Compression develops on top side and tension on bottom (for
positive moments.
7
• Neutral axis passes through
centroid
• Moment of inertia of gross
section is used
8
Uncracked Concrete Stage
Strains and stresses

5. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 5
Concrete-Cracked – Elastic Stage
• As load is increased, the modulus of rupture of the concrete is
exceeded, and cracks begin to develop in bottom of the beam.
• Cracking Moment: The moment at which tensile stress in the
bottom of the beam equals the modulus of rupture (i.e. when
cracks begin to form) is referred to as the cracking moment, Mcr
• As load is further increased, cracks quickly spread closer to the
neutral axis, which then begins to move upward.
• Note: The cracks occur at all places along the beam where the
actual moment is greater than the cracking moment.
9
Concrete-Cracked – Elastic Stage (contd.)
• As concrete in the cracked zone cannot resist tensile stresses –
the steel must do it. Perfect bond is still maintained.
• This stage will continue as long as the compression stress in the
top fibers is less than about one-half of the concrete compressive
strength fc’ and as long as steel stress is less than its yield stress
• In this stage the compressive stresses vary linearly with the
distance from the neutral axis (straight line).
10
ratio
Modular
with
:
point
same
at the
strains
Equal









n
E
E
n
nf
f
E
E
f
f
E
f
E
f
c
s
c
s
c
s
c
s
s
s
c
c
s
c 


6. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 6
• Service or Working Loads: Loads that actually occur when a
structure is in normal service use.
• Under working loads, moments develop which are considerably
larger than the cracking moments. Obviously the tensile side of
the beam will be cracked.
• Under working loads, concrete compression stress is less than
50% of its strength and linear stress-strain variation is assumed.
• Service loads and serviceability limit state correspond thus to
the cracked elastic stage
• To compute concrete and steel stresses in this range, the
transformed-area method is used.
11
Cracked – Elastic Stage and Service Loads
Beam Failure- Ultimate Strength Stage
• As load is increased further, compressive stresses are greater than
0.5fc’, cracks and neutral axis move further upward .
• Concrete compressive stresses begin to change appreciably from
linear to curved, and reinforcing bars yield.
12

7. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 7
Moment Curvature Diagram
• From the measured compressive strain in concrete and tensile
strain in steel, the relationship between the bending moment
and the curvature can be tracked up to failure.
13
d
y
y
s
c 













n
ta
angles
small
Very
NA
from
distance
some
at
fiber
beam
a
in
strain
curvature
or
section
beam
the
of
change
Angle


y
s

c

d

• Bending moments cause
curvatures (rotations).
• Curvature more objective
than deflection as
bending deformation.
14
Typical Moment Curvature Diagram
for a Reinforced Concrete Beam
There are three
different stages
in the curve :
OC – CY - YF

8. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 8
15
Moment Curvature Diagram
1. First stage is for moments less than the cracking moment Mcr .
The entire beam cross section is available to resist bending. The
relationship is quasi-linear.
2. When the moment is increased beyond Mcr, the slope of the
curve decreases as the cracked beam is not as stiff as in the
initial stage. The diagram will still follow a straight line from Mcr
to the point where steel is stressed to its yield point.
15
3. After steel yields, the beam has
very little additional moment
capacity, and only a small
additional load is required to
substantially increase rotations
and deflections. The slope of the
diagram is now almost flat.
Bending stress in an Uncracked Beam
• Steel area is small (usually 2% or less), and its effect on beam
properties is negligible as long as the beam is uncracked
• The bending stress in such a beam can be obtained based on
the gross properties of the beam’s cross section.
• The neutral axis coincides with the centroid
• The stress in the concrete at a distance y from the neutral axis
of the cross section can be determined from the following
flexure formula:
16
section.
cross
gross
the
of
inertia
of
Moment
section
the
of
moment
Bending




g
cr
g
I
M
M
I
My
f

9. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 9
Cracking Moment
17
section.
cross
the
of
inertia
of
moment
Gross
moment
Bending



g
g
I
M
I
My
f
t
g
r
cr
g
t
cr
r
c
r
t
y
I
f
M
I
y
M
f
f
f
f
y
y








:
Thus
9.5.2.3)
section
SBC
/
(ACI
0.7
rupture
of
Modulus
fiber
tension
extreme
to
axis
centroidal
from
distance
Substitute
'
Example 1: Uncracked section
• Assuming concrete is uncracked, compute bending stresses in
extreme fibers of the section below for a bending moment of
34 kN.m. fc’ = 30 MPa.
• To check that the section is uncracked, first determine the
cracking moment of the section.
18
375
450
300
28
3
All dimensions in mm

10. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 10
19
Solution of Example 1
uncracked
is
section
the
kN.m,
34.0
At
kN.m
8
.
38
N.mm
10
8
.
38
225
10
28
.
2
83
.
3
:
)
(
Moment
Cracking
mm
10
28
.
2
450
300
12
1
12
1
MPa
83
.
3
30
7
.
0
7
.
0
:
rupture
of
Modulus
6
9
4
9
3
3
'



















M
M
y
I
f
M
M
bh
I
f
f
cr
t
g
r
cr
cr
g
c
r
19
375
450
300
28
3
Solution of Example 1 (continued)
20
uncracked.
well
is
section
The
MPa
83
.
3
30
7
.
0
7
.
0
rupture
of
Modulus
MPa
35
.
3
10
28
.
2
225
)
10
34
(
225
2
:
fibers
extreme
two
At the
mm
10
28
.
2
450
300
12
1
12
1
:
Stresses
Bending
'
9
6
4
9
3
3























r
c
r
g
g
g
f
f
f
f
I
My
f
mm
h
y
bh
I
I
My
f
An uncracked beam is assumed to be homogeneous with neutral
axis passing through the centroid of the beam section.
Bending stress is given by the standard flexural equation.
375
450
300
28
3

11. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 11
Calculation of elastics stresses
in cracked sections
• When bending moment is sufficiently large to cause tensile stress
in the extreme fibers , greater than the modulus of rupture, it is
assumed that all of the concrete on the tensile side of the beam is
cracked and must be neglected in the flexure calculations.
• Cracking of the beam does not necessarily mean that the beam is
going to fail. The reinforcing bars on the tension side begin to pick
up the tension caused by the applied moment.
• Assumption of perfect bond is made between reinforcing bars
and concrete. Thus the strain in the concrete and in the steel will
be the same at equal distances from the neutral axis.
• Although the strains in the two materials at a particular point are
the same, their stresses can not be the same as they have
different values of elastic modulus.
21
Modular Ratio (n)
22
steel
of
that
times
be
will
concrete
required
of
Area
required?
be
will
two
the
of
areas
what
concrete
or
steel
by
resisted
be
to
is
force
If
with
n
nA
A
n
A
A
A
F
n
A
F
nf
f
F
E
E
n
nf
f
f
E
E
f
E
f
E
f
s
c
s
c
c
s
c
s
c
s
c
s
c
c
s
s
s
s
c
c
s
c















 

Definition: The ratio of the steel modulus to the concrete modulus
is called the modular ratio.
If the modular ratio for a beam is 10, stress in steel will be 10 times
the stress in concrete at the same distance from the neutral axis.
Otherwise: when n = 10, 500 mm2 of steel will carry the same total
force as 5000 mm2 of concrete.
At the same point, steel and concrete strains are equal and the
stresses are proportional to their modulus of elasticity.

12. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 12
Transformed Area Method
• When the steel area As is replaced with an equivalent area of
fictitious concrete (nAs), which supposedly can resist tension,
the area is referred to as the transformed area.
• As the transformed area is of concrete only, it can be analyzed
by the usual methods for elastic homogeneous beams.
23
Tensile concrete is
ignored
Tensile steel area As
replaced by equivalent
concrete area nAs
N.A.
c
f
s
A
n
fs
s
nA
Steps in transformed area method
1. Locate the neutral axis:
– Assume it is located a distance x from the compression
surface of the beam.
– Equate the first moment of the compression area of the beam
cross section about the neutral axis to the first moment of
the tensile area about the neutral axis.
– Solve the resulting quadratic equation.
2. Calculate the moment of inertia of the transformed section
wit respect to the neutral axis.
3. Compute the stresses in the concrete and the steel with the
flexure formula ( f = My/I ).
24

13. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 13
Example 2: Transformed Area Method
• (a) Calculate the bending stresses in the beam shown below
using the transformed area method
• fc’ = 20 MPa, n = 9 and M = 95 kN.m
• (b) Determine the allowable resisting moment of the beam, if
the allowable stresses are : fac = 10 MPa and fas = 140 MPa
25
425
500
300
28
3
All dimensions in mm
26
Solution:
425
500
300
)
1846
(
28
3


s
A
All dimensions in mm
300
425
x
)
425
( x

2
mm
16614

s
nA
4
9
2
3
2
3
2
2
2
2
2
mm
10
571
.
1
168.5)
(425
(16614)
168.5
300
3
1
)
(425
)
(
168.5
300
3
1
:
axis)
neutral
(about
Inertia
of
Moment
mm
5
.
168
is
solution
Positive
84
.
447
78
.
200559
47073
4
)
76
.
110
(
0
47073
76
.
110
0
7060950
16614
150
16614
7060950
150
)
425
(
16614
150
)
425
(
2
)
(300
:
)
is
fiber
top
from
depth
axis
(neutral
axis
neutral
about
moments
first
Taking















































I
x
nA
I
x
x
x
x
x
x
x
x
x
x
nA
x
x
x
s
s

14. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 14
27
MPa
6
.
139
10
571
.
1
5
.
256
)
10
95
(
9
:
mm)
256.5
168.5
-
425
(
level
Steel
MPa
2
.
10
10
571
.
1
5
.
168
)
10
95
(
:
mm)
168.5
(
n
compressio
Extreme
:
Stresses
Bending
9
6
9
6


























I
My
n
f
I
My
n
f
f
y
I
My
f
y
I
My
f
s
s
c
c
kN.m
2
.
95
N.mm
10
2
.
95
5
.
256
9
10
571
.
1
140
)
/
(
mm
256.5
168.5
-
425
:
level
steel
At the
:
steel
in
tension
allowable
to
ing
correspond
moment
Allowable
kN.m
2
.
93
N.mm
10
2
.
93
5
.
168
10
571
.
1
10
:
concrete
in
n
compressio
allowable
to
ing
correspond
moment
Allowable
:
moment
resisting
Allowable
(b)
6
9
6
9





















ny
I
f
y
I
n
f
M
y
y
I
f
M
y
I
f
M
I
y
M
f
as
as
as
ac
ac
allowable
allowable
allowable
allowable
Solution – Cont.
28
Concrete and steel will reach their permissible stresses at these
moments respectively.
Discussion:
Concrete and steel will not reach their maximum allowable stresses
at exactly the same bending moments.
The resisting moment of the section is 93.2 kNm (the smallest)
because if that value is exceeded, concrete becomes overstressed
even though the steel stress is less than its allowable stress.
kN.m
2
.
95
kN.m
2
.
93
:
moments
resisting
Allowable


as
ac
M
M
Solution – Cont.

15. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 15
29
29
d
h
b
s
A
b
h
x
)
( x
d 
s
nA
I
x
d
M
n
f
x
d
y
I
Mx
f
x
y
x
nA
bx
I
x
x
nA
d
nA
bx
x
d
nA
bx
x
d
nA
x
x
x
s
c
s
s
s
s
s
)
(
:
)
(
stress
tension
Steel
:
)
(
stress
n
compressio
concrete
Top
)
(d
)
(
3
:
NA)
(about
inertia
of
Moment
for
Solve
0
2
)
(
2
)
(
2
)
(b
:
)
is
fiber
top
from
depth
axis
(neutral
axis
neutral
about
moments
First
2
3
2
2


























Steps for a General Rectangular Section
Ultimate State and Nominal Flexural Moment
• After concrete compression stresses exceed about 50%
of fc’, they no longer vary linearly. The variation is
rather nonlinear as shown in Figure (b)
30
d
y
f
b
s
A
c
(a) Beam
(b) Actual compression
stress variation
Neutral axis depth
is now noted c

16. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 16
31
The ultimate flexural moment capacity is called “Nominal”
moment and is noted Mn. It is derived using this assumption:
The curved compression diagram can be replaced by a
rectangular one with a constant stress of 0.85 fc’ (Whitney).
The Whitney rectangular stress block of depth “a” is assumed to
have the same area and same centre of gravity as the curved
diagram (generating the same force and the same moment).
d
y
f
b
s
A
c c
a 1


ab
f
C c
'
85
.
0

'
85
.
0 c
f
(a) Beam
(b) Actual block of
compression stress
(c) Whitney block of
compression stress
y
s f
A
T 
Ultimate State and Nominal Flexural Moment
32
Apart from the previous standard assumptions (sections remain
plane after bending, perfect bond between concrete and steel bars,
concrete tensile strength ignored), two other assumptions are added
in the ultimate stage:
• Concrete is assumed to fail (crush) when its compressive strain
at the extreme fiber reaches its ultimate value εcu = 0.003
• The nonlinear compressive stress block for concrete is replaced
by an equivalent rectangular one with constant stress of 0.85 fc’
and a depth a = 1 c
Additional assumptions at ultimate state
 










MPa
f
for
f
MPa
f
for
c
c
c
30
65
.
0
,
008
.
0
09
.
1
Max
30
85
.
0
1

1 shall not be smaller than 0.65

17. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 17
33
Factor 1 to account for skewed shape
of compression stress block
Question: Why does 1 vary (reduced for fc’ > 30 MPa) ?
 










MPa
f
for
f
MPa
f
for
c
c
c
30
65
.
0
,
008
.
0
09
.
1
Max
30
85
.
0
1

34
Variation of coefficient 1
Observe skewed shape of compression
stress-strain curves for higher strength
The shapes of the stress-
strain curves for high
strength concrete are
skewed. This shifts the
centroid of the area
(below the curve) closer
to the origin, and thus
reduces the length of the
equivalent rectangle.
Thus 1 must be smaller
for high strength
concrete.

18. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 18
35
There are three types of flexural failure of a structural
member:
1. Steel reaches its yield strength before concrete reaches
its ultimate strain of 0.003 (Under-reinforced section)
2. Steel reaches yield limit at the same time as concrete
reaches its ultimate strain (Balanced section).
3. Concrete fails before the yield of steel, in the case of
presence of a high percentage of steel in the section
(Over-reinforced section).
Types of ultimate flexural failure
36
Steel reaches its yield strength before concrete reaches its ultimate
strain. Large deformations occur before concrete crushing. Failure
is ductile. After yielding, the steel tensile force remains constant.
To maintain an equal equilibrating concrete compression force, the
neutral axis is shifted up. Cracks will also progress upwards.
Under-reinforced section

19. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 19
37
Steel reaches its yield limit at the same time as concrete
reaches the ultimate strain. Failure is sudden and brittle
Balanced section
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Concrete crushes before the yield of steel, due to the
presence of a high percentage of steel in the section.
Failure is very sudden and brittle
Over-reinforced section

20. 24/2/2013
CE 370 : Prof. Abdelhamid Charif 20
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Moment-Curvature for various steel areas
(Increasing steel area reduces ductility)
Types of flexural failure
• Brittle failures are very dangerous and must be
avoided.
• All codes of practice (ACI, SBC, …) require beams to
be designed as under-reinforced to ensure ductile
behavior.
• In fact SBC and ACI codes require more ductility than
just under reinforced.
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