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 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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