Page 1
LIMIT STATE OF FLEXURE
1. Characteristic strength of materials
? The term ‘characteristic strength‘ means that value of the strength of material below which not
more than minimum acceptable percentage of test results are expected to fall.
? IS 456:2000 have accepted the minimum acceptable percentage as 5% for reinforced concrete
structures.
? Characteristic strength = Mean strength – K x standard deviation or
fk = fm – K x Sd
where, fk = characteristic strength of the material
fm = mean strength
K = constant = 1.65
Sd = standard deviation for a set of test results.
The value of standard deviation (sd) is given by
2
d
S
n1
?
=
-
?
Where d = deviation of the individual test strength from the average or mean strength of
n samples.
n = number of test results
IS 456:2000 has recommended minimum value of n = 30
2. Partial safety factor for loads
The partial safety for loads, as per IS 456:2000 are given in table below
Page 2
LIMIT STATE OF FLEXURE
1. Characteristic strength of materials
? The term ‘characteristic strength‘ means that value of the strength of material below which not
more than minimum acceptable percentage of test results are expected to fall.
? IS 456:2000 have accepted the minimum acceptable percentage as 5% for reinforced concrete
structures.
? Characteristic strength = Mean strength – K x standard deviation or
fk = fm – K x Sd
where, fk = characteristic strength of the material
fm = mean strength
K = constant = 1.65
Sd = standard deviation for a set of test results.
The value of standard deviation (sd) is given by
2
d
S
n1
?
=
-
?
Where d = deviation of the individual test strength from the average or mean strength of
n samples.
n = number of test results
IS 456:2000 has recommended minimum value of n = 30
2. Partial safety factor for loads
The partial safety for loads, as per IS 456:2000 are given in table below
Load
combination
Limit state of collapse Limit state of Serviceability
DL LL WL/EL DL LL WL/EL
DL+IL 1.5 1.5 - 1.0 1.0 -
DL+WL 1.5 or
0.9*
- 1.5 1.0 - 1.0
DL+IL+WL 1.2 1.2 1.2 1.0 0.8 0.8
(* This value is to be considered when stability against overturning or stress reversal is
critical)
LIMIT STATE OF COLLAPSE IN FLEXURE
? The maximum compressive strain in concrete (at the outermost fibre)
cu
? shall be taken as
0.0035 in bending.
? For design purpose, the compressive strength of concrete in the structure shall be assumed
to be 0.67 times the characteristic strength. The partial factor y, 1.5 shall be applied in
addition to this.
Figure: Stress-strain curve for concrete
? For design purpose of reinforcement, the partial safety factor
m
? equal to 1.15 shall be
applied.
? The maximum strain in the tension reinforcement in the section at failure shall not be
less than :
y
s
f
0.002
1.15E
+
3. Single Reinforced Beam
Page 3
LIMIT STATE OF FLEXURE
1. Characteristic strength of materials
? The term ‘characteristic strength‘ means that value of the strength of material below which not
more than minimum acceptable percentage of test results are expected to fall.
? IS 456:2000 have accepted the minimum acceptable percentage as 5% for reinforced concrete
structures.
? Characteristic strength = Mean strength – K x standard deviation or
fk = fm – K x Sd
where, fk = characteristic strength of the material
fm = mean strength
K = constant = 1.65
Sd = standard deviation for a set of test results.
The value of standard deviation (sd) is given by
2
d
S
n1
?
=
-
?
Where d = deviation of the individual test strength from the average or mean strength of
n samples.
n = number of test results
IS 456:2000 has recommended minimum value of n = 30
2. Partial safety factor for loads
The partial safety for loads, as per IS 456:2000 are given in table below
Load
combination
Limit state of collapse Limit state of Serviceability
DL LL WL/EL DL LL WL/EL
DL+IL 1.5 1.5 - 1.0 1.0 -
DL+WL 1.5 or
0.9*
- 1.5 1.0 - 1.0
DL+IL+WL 1.2 1.2 1.2 1.0 0.8 0.8
(* This value is to be considered when stability against overturning or stress reversal is
critical)
LIMIT STATE OF COLLAPSE IN FLEXURE
? The maximum compressive strain in concrete (at the outermost fibre)
cu
? shall be taken as
0.0035 in bending.
? For design purpose, the compressive strength of concrete in the structure shall be assumed
to be 0.67 times the characteristic strength. The partial factor y, 1.5 shall be applied in
addition to this.
Figure: Stress-strain curve for concrete
? For design purpose of reinforcement, the partial safety factor
m
? equal to 1.15 shall be
applied.
? The maximum strain in the tension reinforcement in the section at failure shall not be
less than :
y
s
f
0.002
1.15E
+
3. Single Reinforced Beam
Table: Limiting depth of neutral axis for different grades of steel
Steel Grade Fe 250 Fe 415 Fe 500
Xu,max/d 0.5313 04791 04791
d= effective width
i. Depth of neutral Axis
=
y st
u
ck
0.87f A
x
0.36f b
ii. Lever Arm= d-xu (d= effective width and xu is depth of neutral axis)
iii. Ultimate Moment of resistance
MuR = 0.36× fck× bxu (d – 0.42xu) ; for all xu
Alternatively, in terms of the steel tensile stress,
MuR = 0.87 fy× Ast (d – 0.42xu) ; for all xu
Case – 1 :
u
x
d
equal to the limiting value
u,max
x
d
: Balanced section
Case – 2:
u
x
d
less than limiting value : under-reinforced section
Case – 3 :
u
x
d
more than limiting value : over-reinforced section.
iv. Computation of Mu
a. xu < xu,max
Page 4
LIMIT STATE OF FLEXURE
1. Characteristic strength of materials
? The term ‘characteristic strength‘ means that value of the strength of material below which not
more than minimum acceptable percentage of test results are expected to fall.
? IS 456:2000 have accepted the minimum acceptable percentage as 5% for reinforced concrete
structures.
? Characteristic strength = Mean strength – K x standard deviation or
fk = fm – K x Sd
where, fk = characteristic strength of the material
fm = mean strength
K = constant = 1.65
Sd = standard deviation for a set of test results.
The value of standard deviation (sd) is given by
2
d
S
n1
?
=
-
?
Where d = deviation of the individual test strength from the average or mean strength of
n samples.
n = number of test results
IS 456:2000 has recommended minimum value of n = 30
2. Partial safety factor for loads
The partial safety for loads, as per IS 456:2000 are given in table below
Load
combination
Limit state of collapse Limit state of Serviceability
DL LL WL/EL DL LL WL/EL
DL+IL 1.5 1.5 - 1.0 1.0 -
DL+WL 1.5 or
0.9*
- 1.5 1.0 - 1.0
DL+IL+WL 1.2 1.2 1.2 1.0 0.8 0.8
(* This value is to be considered when stability against overturning or stress reversal is
critical)
LIMIT STATE OF COLLAPSE IN FLEXURE
? The maximum compressive strain in concrete (at the outermost fibre)
cu
? shall be taken as
0.0035 in bending.
? For design purpose, the compressive strength of concrete in the structure shall be assumed
to be 0.67 times the characteristic strength. The partial factor y, 1.5 shall be applied in
addition to this.
Figure: Stress-strain curve for concrete
? For design purpose of reinforcement, the partial safety factor
m
? equal to 1.15 shall be
applied.
? The maximum strain in the tension reinforcement in the section at failure shall not be
less than :
y
s
f
0.002
1.15E
+
3. Single Reinforced Beam
Table: Limiting depth of neutral axis for different grades of steel
Steel Grade Fe 250 Fe 415 Fe 500
Xu,max/d 0.5313 04791 04791
d= effective width
i. Depth of neutral Axis
=
y st
u
ck
0.87f A
x
0.36f b
ii. Lever Arm= d-xu (d= effective width and xu is depth of neutral axis)
iii. Ultimate Moment of resistance
MuR = 0.36× fck× bxu (d – 0.42xu) ; for all xu
Alternatively, in terms of the steel tensile stress,
MuR = 0.87 fy× Ast (d – 0.42xu) ; for all xu
Case – 1 :
u
x
d
equal to the limiting value
u,max
x
d
: Balanced section
Case – 2:
u
x
d
less than limiting value : under-reinforced section
Case – 3 :
u
x
d
more than limiting value : over-reinforced section.
iv. Computation of Mu
a. xu < xu,max
? In this case the concrete reaches 0.0035, steel has started flowing showing ductility
(Strain >
y
s
0.87f
0.002
E
+ ).
? Therefore, Mu= 0.87 fy Ast (d – 0.42 xu)
b. xu = xu,max
? In this case steel just reaches the value of
y
s
0.87f
0.002
E
+ and concrete also reaches its
maximum value.
?
u,max u,max 2
u,lim ck
xx
M 0.36 1 0.42 f bd
dd
??
=-
??
??
C. xu > xu,max
In this case , concrete reaches the strain of 0.0035, tensile strain of steel is much less
than
y
s
0.87f
0.002
E
??
+
??
??
??
? On the other hand, when steel reaches
y
s
0.87f
0.002
E
+ , the strain of concrete far exceeds
0.0035. Hence, it is not possible. Therefore, such design is avoided and the section
should be redesigned.
? The moment of resistance Mu for such existing beam is calculated by restricting xu to
xu,max only and the corresponding Mu will be as per the case when xu = xu,max .
2. DOUBLY REINFORCED SECTION
? MU2 = 0.87 fy Ast2 (d-d’) = Asc (fsc – fcc) (d – d’)
Where Ast2 = Area of additional tensile reinforcement
Asc = Area of compression reinforcement
fsc = stress in compression reinforcement
Page 5
LIMIT STATE OF FLEXURE
1. Characteristic strength of materials
? The term ‘characteristic strength‘ means that value of the strength of material below which not
more than minimum acceptable percentage of test results are expected to fall.
? IS 456:2000 have accepted the minimum acceptable percentage as 5% for reinforced concrete
structures.
? Characteristic strength = Mean strength – K x standard deviation or
fk = fm – K x Sd
where, fk = characteristic strength of the material
fm = mean strength
K = constant = 1.65
Sd = standard deviation for a set of test results.
The value of standard deviation (sd) is given by
2
d
S
n1
?
=
-
?
Where d = deviation of the individual test strength from the average or mean strength of
n samples.
n = number of test results
IS 456:2000 has recommended minimum value of n = 30
2. Partial safety factor for loads
The partial safety for loads, as per IS 456:2000 are given in table below
Load
combination
Limit state of collapse Limit state of Serviceability
DL LL WL/EL DL LL WL/EL
DL+IL 1.5 1.5 - 1.0 1.0 -
DL+WL 1.5 or
0.9*
- 1.5 1.0 - 1.0
DL+IL+WL 1.2 1.2 1.2 1.0 0.8 0.8
(* This value is to be considered when stability against overturning or stress reversal is
critical)
LIMIT STATE OF COLLAPSE IN FLEXURE
? The maximum compressive strain in concrete (at the outermost fibre)
cu
? shall be taken as
0.0035 in bending.
? For design purpose, the compressive strength of concrete in the structure shall be assumed
to be 0.67 times the characteristic strength. The partial factor y, 1.5 shall be applied in
addition to this.
Figure: Stress-strain curve for concrete
? For design purpose of reinforcement, the partial safety factor
m
? equal to 1.15 shall be
applied.
? The maximum strain in the tension reinforcement in the section at failure shall not be
less than :
y
s
f
0.002
1.15E
+
3. Single Reinforced Beam
Table: Limiting depth of neutral axis for different grades of steel
Steel Grade Fe 250 Fe 415 Fe 500
Xu,max/d 0.5313 04791 04791
d= effective width
i. Depth of neutral Axis
=
y st
u
ck
0.87f A
x
0.36f b
ii. Lever Arm= d-xu (d= effective width and xu is depth of neutral axis)
iii. Ultimate Moment of resistance
MuR = 0.36× fck× bxu (d – 0.42xu) ; for all xu
Alternatively, in terms of the steel tensile stress,
MuR = 0.87 fy× Ast (d – 0.42xu) ; for all xu
Case – 1 :
u
x
d
equal to the limiting value
u,max
x
d
: Balanced section
Case – 2:
u
x
d
less than limiting value : under-reinforced section
Case – 3 :
u
x
d
more than limiting value : over-reinforced section.
iv. Computation of Mu
a. xu < xu,max
? In this case the concrete reaches 0.0035, steel has started flowing showing ductility
(Strain >
y
s
0.87f
0.002
E
+ ).
? Therefore, Mu= 0.87 fy Ast (d – 0.42 xu)
b. xu = xu,max
? In this case steel just reaches the value of
y
s
0.87f
0.002
E
+ and concrete also reaches its
maximum value.
?
u,max u,max 2
u,lim ck
xx
M 0.36 1 0.42 f bd
dd
??
=-
??
??
C. xu > xu,max
In this case , concrete reaches the strain of 0.0035, tensile strain of steel is much less
than
y
s
0.87f
0.002
E
??
+
??
??
??
? On the other hand, when steel reaches
y
s
0.87f
0.002
E
+ , the strain of concrete far exceeds
0.0035. Hence, it is not possible. Therefore, such design is avoided and the section
should be redesigned.
? The moment of resistance Mu for such existing beam is calculated by restricting xu to
xu,max only and the corresponding Mu will be as per the case when xu = xu,max .
2. DOUBLY REINFORCED SECTION
? MU2 = 0.87 fy Ast2 (d-d’) = Asc (fsc – fcc) (d – d’)
Where Ast2 = Area of additional tensile reinforcement
Asc = Area of compression reinforcement
fsc = stress in compression reinforcement
fcc = Compressive stress in concrete at the level of compression reinforcement
Since the addition a reinforcement is balanced by the additional compressive force.
Asc (fsc – fcc) = 0.87 fyAst2
? The strain at level of compression reinforcement is
u
u,max
x
0.0035 1
x
??
-??
??
??
? Total area of reinforcement shall be obtained by
Ast = Ast1 + Ast2
Ast1 = Area of reinforcement for a singly reinforced section for Mu,lim
( )
sc sc cc
st2
y
A f f
A
0.87f
-
=
3. T BEAMS AND L BEAMS
i. Effective width of Flange
a. For Beam casted monolithic with slab
‘effective width of flange bf (Cl. 23.1.2 of Code) are given as follows:
0 w f
f
0 w f
l / 6 b 6D for T Beam
b
l / 12 b 3D forL Beam
+ + - ?
?
=
?
+ + -
?
?
[Eq. 1]
• bw is the breadth of the web,
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