Chapter 2 Beams (Part 3)
(b) Doubly reinforced section
If depth and width are restricted.
If this beam has to support a BM more than Mr of the balanced section.
Ther are 2 options.
• Provide an overreinforced section, or
• Provide a doubly Reinforcementsection
Over reinforced section has many disadvantage like brittle failure so always a doubly reinforced
section is a better option.
Properties
• Steel is provided on both side of NA
• Permissible stress for compression steel = 1.5 mC
where C' = stress in concrete around compression steel.
• Equivalent area of steel in terms of concrete for compression steel = 1.5 mAsc For tension steel
equivalent area = mAst.
(i) Actual Depth of Neutral Axis, (Xa)
Equating moment of area on both sides of NA
( ) sc ( a c ) st ( a )
2a
1.5m –1 A X d mA d X
2
BX +  = 
Here, Xa = Actual depth of Neutral axis
(ii) Critical Depth of Neutral Axis, (Xc)
d .
t mc
X mc
c +
=
(iii) Moment of Resistance (Mr)
Mr = C1 (LA1) + C2 LA2
Mr = BXa 2
Ca (1.5m –1)
3
d Xa + ÷ø
ö
çè
æ  Asc . C' (d–dc)
C' can be calcualted using similar triangles
For a balanced section
Xa = Xc
Ca = cbc s
Mr = Q.B.d2 + (1.5 m – 1) Asc. C' (d – dc )
Mr = Mr1 + Mr2
Where, Mr1 = QBd2 = moment of resistance of singly reinforced balanced section
The above doubly reinforced Beam can be assumed to be madeup of two Beams A and B, as shown in
the figure.
Moment of resistance from tension side
For balanced section
ta = st
Design Steps
1. Given values : B, D.
2. If load is given BM =
8
wl2
design moment = BM
3. Find moment of resistance of the balanced section (singly reinforced)
Mr(bal) = Q.B.d2 =Mrl
4. If BM > Mr (balance) then only, a dubly reinforced section is required.
5. Area of steel for singly reinforced balanced section from BeamA
Mr1 = C1 × LA1 = T1 × LA1
6. Area of steel for remaining BM (Part  II)
Mr2 = (BM – Mr1)
Mr2 = (1.5 m –1) Asc C' (d – dc)
= A (d d
7. From eq. (ii)
or equating moment of area on both sides of N.A. in Beams–B.
(1.5 m – 1)Asc (Xa – dc) = mAst2 (d – Xa)
Limitation in Working Stress Method of Design
1. The shrinkage and the creep of concrete have a considerable effect of stresses due to service loads and
these can not be estimated easily (Shrinkage usually means the drying shrinkage, i.e. decrease in all the 3
dimension due to drying).
Shrinkage strain of concrete, e 0.03% cs »
Elastic strain under service loads = 0.035% which is of the same order as that shrinkage. So shrinkage has
considerable effect. Creep occurs (in case of steel) only when stress > 0.5 fy creep is a time dependent
strain under sustained stress. Creep strain has a considerable effect when it is comparable to eci and it is
so. We define creep coefficient, cc = 1.6
e
e
ci
cc » (approx).
eci
Its value varies from 1.1 to 2.2 depending upon the age of loading, the greater value corresponding to
lesser age of loading.
2. Owing to the presence of cracks, the elastic theory for the analysis of stresses due to shear can not be
extended to R.C. beams and the design for shear must be based on the behaviour at ultimate state. The
actual margin of safety is not equal to the 'factory of sately' because stressstrain relationship is not linear
upto collapse. So' limit state method is a bettter method of design.
Chapter 2 (Part 4) Beams
(II) Limit State Method
According to this method a structure is so designed that the probability of reaching the limit state is
acceptably low during the design life of the structure.
Limit state is the state in which the structure becomes unfit for use.
(a) Limit States
It can be classified into two categories
1. Limit state of serviceability
2. Limit state of collapse.
1. Limit Stateof Serviceability
These relate to the satisfactory performance of the structure under service loads.
These includes the limit states of deflection, cracking vibration etc.
2. Limit State of Collapse
These provide adequate margin of safety for normal overloads. There include limit states of overturning
sliding, failure of one or more critical sections, failure due to buckling etc.
We consider all the relevant limit states during design.
3. Difference b/w Serviceability & Collapse
If a structure raches to the limit state of serviceability. it recovers when the loads are removed but at the
limit state of collapse, the structure cannot recovery.
Flexure
Flexural members are those members subjected to bending.
In working stress method, the Stressstrain relation of the concrete and steel are assumed to the lines within the
range of permissible stresses.
In limit state method, the design stress strain curves concrete and steel are assumed as follows:
• Design value of strength
For concrete
Here, mc g Partial factor of safety for concrete
= 1.5
fd = desgin value of strength
For steel
Apart from other common assumption in flexutre we take here an important assumption i.e. On the
tension side, tension is borne entirely by steel. Though in real, this is not true as there is a certain area below
Neutral axis which carries tensile stresses (of small magnitude) and remains uncracked.Force carried is small
and its resultant is also small. So it is ignored and it is the justification of the assumption.
(b) Assumption in LSM
1. Plane section before bending remain plane after bending.
2. Maximum strain in concrete at outermost compression fibre = 0.0035
3. Tensile strength of concrete is ignored
4. Stress in reinforcement are derived from stressstrain curve. For design purpose partial factor of
safety = 1.15 shall be applied
Permissible stress in steel = 0.87 fy
5. Maximum stress is tension reinforcement at failure shall not be less than
A. Analysis
(i) Analysis of stress block parameter
The above stress diagram can be divided into two portions rectangular portion ABED and parabolic
portion BCE.
Compressive force C1 = Width of crosssection x Area of rectangular portion
Now depth of CG of rectangular portion (y1)
Now Compressive force C2 = Width of crosssection × Area of parabolic portion
= 0.1714 fck BXu
Now, depth of CG of parabolic portion (y2)
Now, Total compressive force
C = C1 +C2 = (0.1928 + 0.1714) fck BXu
C = 0.36fck BXu
Now depth of CG of stress diagram from top
= 0.42 Xu
(ii) Limiting depth of N.A. (Xu(lim))
The limiting values of depth of neutral axis Xu(lim) for different grades of steel can be obtained from the
strain diagram.
From similar triangles
The value of Xu(lim) for three grades of steel are given in table.
fy (N/mm2) Xulimit/d
250 0.53
415 0.48
500 0.46
(iii) Actual depth of N.A.
The actual depth of neutral axis can be obtained by considering the equilibrium of the normal forces, that
is
Resultant force of compression = Average stress × area = 0.36 fck BXu
Resultant force of tension = 0.87fy Ast
Force of compression should be equal to force of tension
0.36 fck BXu = 0.87 fy Ast
(iv) Ultimate moment of resistance (Mu(lim))
Since the maximum depth of neutral axis is limited, the maximum value of the moment of resistance is also
limited.
Mu(lim) with respect of concrete
= 0.36 fck BXu(lim) × lever arm
= 0.36 fck BXu(lim) × lever arm
Lever arm = d – 0.42 Xu(lim)
Mu(lim) = 0.36 fck BXu(lim) × (d–0.42 Xu(lim))
Mu(lim) with respect to steel
= 0.87 fy Ast × Lever arm
= 0.87 fy Ast × (d – 0.42 Xu(lim))
For a rectangular beam section, the limiting value of Mu(lim) depends on the type of concrete mix and the
grade of steel. The values of limiting moment of resistance Mu(lim) with respect to concrete for different
grades of concrete and steel are given in table.
In general moment of resistance of balance section = QBd2
where Q = moment of resistance coefficient
= 0.148 fck for Fe 250
= 0.138 fck for Fe 415
= 0.133 fck for Fe 500
Chapter 2 (Part 5) Beams
Design of Rectangular Beam
The design of a section consists of determination of (i) crosssectional dimensions B and d, and (ii) area of
steel, so as to develop a given moment of resistance. Though the objective of a designer would be to design a
balanced section so that the ultimate stresses in both the material are developed simultaneouly, such a design
may not be the most economical. It should be noted that a balanced design gives smallest concrete section and
maximum area of reinforcement. Since the cost of steel is very high in comparison to that of concrete, a
balanced design may not be economical. Also, for practical consideration, sometimes, it may be necessary to
fix some uniform crosssectional dimensions. In such a case, the design may result in a singly reinforced balanced
section, underreinforcement section or doubly reinforced section. However, if the section has to be
singly reinforced, an underreinforced section is always more desirable in the limit state design.
Design to Determine Crossectional Dimensions and Reinforcement
This is the most usual case of design in which the ultimate moment of resistance (Mu) to be developed by the
section in given, and it is required to determine B, d and Ast. The design is done in the following steps..
1. Determine the limiting depth of N.A.
Xu(lim) = d
0.0055 0.87f / E
0.0035
y s +
2. Choose some suitable ratio of d and B. The value of d/B in the range of 1.5 to 3 is usually taken.
3. Find d from the relation
Mu(lim) = 0.36 fck BXu(lim) × (d–0.42 Xu(lim))
= QBd2
4. Knowing B and d, determine area of reinforcement from the relation
Mu(lim) = 0.87 fy Ast (d – 0.42 Xu (lim))
Design to Determine Area of Reinforcement if B and d are known
In this case Mu, B and d are given. To start with, we have to ascertain whether it will result in a singly
reinforced section or a doubly reinforced section. The design is done in the following steps:
1. Determine the limiting moment of resistance
Mu(lim) = 0.36 fck BXu(lim) × (d – 0.42 Xu(lim))
= QBd2
2. (a) If Mu = Mu(lim), it will result in a balanced section, and area of steel can be determined as dis
cussed earlier.
(b) If Mu > Mu (lim), it will result in a doubly reinforced section which will be discussed later.
(c) If Mu < Mu (lim), it will result, in an under reinforced section, and the design is done in the
followingt steps.
3. If Mu < Mu (lim), determine actual Xu from relationship:
Mu = 0.36 fck B Xu (d – 0.42 Xu)
This will result in a quadratic equation in terms of Xu which can be easily solved
4. Determine area of steel from the relation
Mu = 0.87 fy Ast (d – 0.42 Xu)
Alternatively, steps 3 and 4 can be avoided and Ast can be fond from equation given below which is
applicable for underreinforced section.
Mu = 0.87 fy Ast d
The gives a quadratic equation in terms of Ast, the solutions of which works out as under:
(b) Doubly rienforced section
A doubly reinforced concrete section is reinforced in both compression and tension regions. The section
of the beam or slab my be a rectangle, T and L section. The necessity of using steel in the compression
region arises due to two main reasons:
(a) When depth of the section is restricted, the strength available from a singly reinforced is inadequate.
(b) At a support of a continuous beam or slab where bending moment changes sign.
(i) Stress Block and Actual Depth of N.A
Figure shows a doubly reinforced section having compression reinforced section having compression
reinforcement fibre. Figure (b) shows the strain diagram while figure (c) the stress block.
Let,
Ast = total reinforcement at tension face.
Asc = Reinforcement in compression side.
Xu = Depth of N.A
C1 = Compressive force in concrete
= 0.36 fck BXu
C2 = Compressive force in compression steel
= fsc Asc
fsc = Design stress in compression reinforcement read off from the stress strain curve corresponding to
the strain Îsc in compression reinforcement.
Îsc = Strain in compression reinforcement using similar triangles,
sc Î =
( )
u
u u
X
0.0035 X  d
... (i)
Total compressive force is given by
C = C1 + C2
or C = 0.36 fck B Xu + fsc Asc ... (ii)
(neglecting the loss of concrete area occupied by compressive steel)
Total tensile force is given by
T = 0.87 fy Ast ... (iii)
In order to locate the N.A. equate the total compresive force to the total tensile force:
0.36 fck B Xu + fsc Asc = 0.87 fy Ast ... (iv)
From the above relation, Xu can be found. However for the solution of the above equation, an iterative
procedure will have to be adopted, since fsc depends upon, which in turn depends upon Xu. If the loss of
compressive area, occupied by the compressive steel is taken into account equations (ii) and (iv) are
modified as under.
Cu = 0.36 fck B Xu + fsc Asc – 0.446 fck Asc
= 0.36 fck B Xu + (fsc – 0.446 fck) Asc
and
0.36 fck B Xu + (fsc – 0.446 fck) Asc = 0.87 fy Ast
Note :
Normally, the term 0.446 fck Asc is very small and can be neglected witout causing and appreciable error.
Iterative procedure for computation of Xu
1. Compute the depth of N.A. of a balanced section given by strain compatibility
Xu(lim) =
y s 0.0055 0.87f / E
0.0035
+
2. For the given doubly reinforced section, assume Xu equal to Xu(lim).
3. Compute the value of sc Î from equation (i).
4. Compute value of fsc from the stress strain curve of steel corresponding to this value of sc Î .
5. Substitute the value of fsc in equation (iv) and compute the modified value of sc Î
6. Repeat steps 3 to 5 till convergence for the value of Xu is achieved.
(ii) Ultimate moment of resistance
Ultimate moment of resistance is given by
Mu = 0.36fckBXu (d–0.42Xu) + (fsc –0.446 fck) Asc (d–dc)
where
fsc = stress in compression steel and it is calculated by strain at the location of compression steel (fsc)
Design Steps
A doubly reinforced beam can be assumed to be made up of two beams A and B as shown in Figure. In
beam A which is singly reinforced beam, the tension steel Ast1 is required to balance the force of compression
C1 in concrete. In beam B, which is imaginary, the tension steel Ast2 is required to balance the force of compression
C2 in compression steel.
Ast Ast1
1. Determine the limiting moment of resistance Mu(lim) for the given crosssection using the equation for a
singly reinforced beam A.
ie.
Mu(lim) = 0.87 fy Ast1 (d – 0.42 Xu(lim))
or Mu(lim) = 0.36 fck B Xu(lim) (d – 0.42 Xu(lim))
After calculating moment of resistance Mu(lim)
st1 A can be calculated by equating force of compression to force of tension i.e.
0.87fy Ast1 = 0.36 fck B Xu(lim)
Where
st1 A = Area of tension steel corresponding to a balanced singly reinforced beam.
2. If the factored moment M exceeds Mu(lim), a doubly reinforced section is required to be designed for the
additional moment (M–Mu(lim)). This moment is resisted by an internal couple consisting of compression
force C2 in the compression steel and tension force T2 in an additional tension steel in beams B, that is
M – Mu(lim) = (fsc – 0.446 fck) Asc (d–dc)
» fsc Asc (d – dc)
Since 0.446 fck < < fsc
Asc =
u u(lim)
sc c
M –M
f (d–d )
Where
Asc = Area of compression reinforcement
3. The additional area of tension steel st2 A is obtained by considering the equilibrium of force of compression
C2 in compression steel and force of tension T2 in the additional tension steel, i.e.,
fsc Asc – 0.446 fck Asc = 0.87 fy st2 A
or fsc Asc » 0.87 fy st2 A
Ast 2 =
y
sc sc
0.87 f
f A
4. The total tension Steel At is given by
At = st1 A + st2 A
TBeam
1. Effective width of flange
Discussed in WSM
2. Limiting depth of neutral axis
Xu(lim) = d
0.0055 0.87f / E
0.0035
y s +
• Singly reinforced TBeam
Case1: When NA is in flange area
i.e., Xu < Df
(a) for Xu
f
ck f
y st
u D
0.36f B
0.87f A
X = <
(b) Ultimate moment of resistance
Mu = 0.36 fck Bf Xu (d – 0.42Xu)
Mu = 0.87 fy Ast (d – 0.42 Xu)
Case–2 : When NA is in web area (Xu > Df)
Case (a) when Df < u X
7
3
i.e. depth of flange is less than the depth of rectangular portion of stress diagram.
1. For actual depth of neutral ais
0.36fckbwXu + 0.446fck (Bf – bw) Df = 0.87 fy Ast
2. Ultimate moment of resistance
Mu = 0.36fckbwXu(d–0.42 Xu) + 0.446 fck (Bf–bw)Df
÷ø
ö
çè
æ 
2
d Df
Mu = 0.87fy st1 A (d–0.42 Xu) + 0.87 fy st2 A ÷ø
ö
çè
æ 
2
d Df
y
ck w u
st 0.87 f
A 0.36f b X 1
=
( )
y
ck f w f
st 0.87f
A 0.45f B b D 2

=
• Special Case (2) : When Xu > Df
and Df > u X
7
3
i.e. depth of flange is more than depth of rectangular portion of stress diagram.
(B D ) f f (B b ) f w
yf
As per IS : 456 – 2000
(Bf – bw) Df portion of flange is coverted into (Bf–bw)yf section for which stress is taken constant throughout
the section is 0.45 fck.
As per IS : 456–2000
yf = 0.15 Xu + 0.65Df < Df
1. For actual depth of neutral axis
0.36fckbwXu+ 0.446fck (Br–bw)yf = 0.87fy st2 A
or 0.36fckbwXu+ 0.446fck (Br–bw)yf = 0.87fyAst
Chapter 2 (Part 4) Beams
(II) Limit State Method
According to this method a structure is so designed that the probability of reaching the limit state is
acceptably low during the design life of the structure.
Limit state is the state in which the structure becomes unfit for use.
(a) Limit States
It can be classified into two categories
1. Limit state of serviceability
2. Limit state of collapse.
1. Limit Stateof Serviceability
These relate to the satisfactory performance of the structure under service loads.
These includes the limit states of deflection, cracking vibration etc.
2. Limit State of Collapse
These provide adequate margin of safety for normal overloads. There include limit states of overturning
sliding, failure of one or more critical sections, failure due to buckling etc.
We consider all the relevant limit states during design.
3. Difference b/w Serviceability & Collapse
If a structure raches to the limit state of serviceability. it recovers when the loads are removed but at the
limit state of collapse, the structure cannot recovery.
Flexure
Flexural members are those members subjected to bending.
In working stress method, the Stressstrain relation of the concrete and steel are assumed to the lines within the
range of permissible stresses.
In limit state method, the design stress strain curves concrete and steel are assumed as follows:
• Design value of strength
For concrete
Here, mc g Partial factor of safety for concrete
= 1.5
fd = desgin value of strength
For steel
Apart from other common assumption in flexutre we take here an important assumption i.e. On the
tension side, tension is borne entirely by steel. Though in real, this is not true as there is a certain area below
Neutral axis which carries tensile stresses (of small magnitude) and remains uncracked.Force carried is small
and its resultant is also small. So it is ignored and it is the justification of the assumption.
(b) Assumption in LSM
1. Plane section before bending remain plane after bending.
2. Maximum strain in concrete at outermost compression fibre = 0.0035
3. Tensile strength of concrete is ignored
4. Stress in reinforcement are derived from stressstrain curve. For design purpose partial factor of
safety = 1.15 shall be applied
Permissible stress in steel = 0.87 fy
5. Maximum stress is tension reinforcement at failure shall not be less than
A. Analysis
(i) Analysis of stress block parameter
The above stress diagram can be divided into two portions rectangular portion ABED and parabolic
portion BCE.
Compressive force C1 = Width of crosssection x Area of rectangular portion
Now depth of CG of rectangular portion (y1)
Now Compressive force C2 = Width of crosssection × Area of parabolic portion
= 0.1714 fck BXu
Now, depth of CG of parabolic portion (y2)
Now, Total compressive force
C = C1 +C2 = (0.1928 + 0.1714) fck BXu
C = 0.36fck BXu
Now depth of CG of stress diagram from top
= 0.42 Xu
(ii) Limiting depth of N.A. (Xu(lim))
The limiting values of depth of neutral axis Xu(lim) for different grades of steel can be obtained from the
strain diagram.
From similar triangles
The value of Xu(lim) for three grades of steel are given in table.
fy (N/mm2) Xulimit/d
250 0.53
415 0.48
500 0.46
(iii) Actual depth of N.A.
The actual depth of neutral axis can be obtained by considering the equilibrium of the normal forces, that
is
Resultant force of compression = Average stress × area = 0.36 fck BXu
Resultant force of tension = 0.87fy Ast
Force of compression should be equal to force of tension
0.36 fck BXu = 0.87 fy Ast
(iv) Ultimate moment of resistance (Mu(lim))
Since the maximum depth of neutral axis is limited, the maximum value of the moment of resistance is also
limited.
Mu(lim) with respect of concrete
= 0.36 fck BXu(lim) × lever arm
= 0.36 fck BXu(lim) × lever arm
Lever arm = d – 0.42 Xu(lim)
Mu(lim) = 0.36 fck BXu(lim) × (d–0.42 Xu(lim))
Mu(lim) with respect to steel
= 0.87 fy Ast × Lever arm
= 0.87 fy Ast × (d – 0.42 Xu(lim))
For a rectangular beam section, the limiting value of Mu(lim) depends on the type of concrete mix and the
grade of steel. The values of limiting moment of resistance Mu(lim) with respect to concrete for different
grades of concrete and steel are given in table.
In general moment of resistance of balance section = QBd2
where Q = moment of resistance coefficient
= 0.148 fck for Fe 250
= 0.138 fck for Fe 415
= 0.133 fck for Fe 500
Chapter 2 (Part 5) Beams
Design of Rectangular Beam
The design of a section consists of determination of (i) crosssectional dimensions B and d, and (ii) area of
steel, so as to develop a given moment of resistance. Though the objective of a designer would be to design a
balanced section so that the ultimate stresses in both the material are developed simultaneouly, such a design
may not be the most economical. It should be noted that a balanced design gives smallest concrete section and
maximum area of reinforcement. Since the cost of steel is very high in comparison to that of concrete, a
balanced design may not be economical. Also, for practical consideration, sometimes, it may be necessary to
fix some uniform crosssectional dimensions. In such a case, the design may result in a singly reinforced balanced
section, underreinforcement section or doubly reinforced section. However, if the section has to be
singly reinforced, an underreinforced section is always more desirable in the limit state design.
Design to Determine Crossectional Dimensions and Reinforcement
This is the most usual case of design in which the ultimate moment of resistance (Mu) to be developed by the
section in given, and it is required to determine B, d and Ast. The design is done in the following steps..
1. Determine the limiting depth of N.A.
Xu(lim) = d
0.0055 0.87f / E
0.0035
y s +
2. Choose some suitable ratio of d and B. The value of d/B in the range of 1.5 to 3 is usually taken.
3. Find d from the relation
Mu(lim) = 0.36 fck BXu(lim) × (d–0.42 Xu(lim))
= QBd2
4. Knowing B and d, determine area of reinforcement from the relation
Mu(lim) = 0.87 fy Ast (d – 0.42 Xu (lim))
Design to Determine Area of Reinforcement if B and d are known
In this case Mu, B and d are given. To start with, we have to ascertain whether it will result in a singly
reinforced section or a doubly reinforced section. The design is done in the following steps:
1. Determine the limiting moment of resistance
Mu(lim) = 0.36 fck BXu(lim) × (d – 0.42 Xu(lim))
= QBd2
2. (a) If Mu = Mu(lim), it will result in a balanced section, and area of steel can be determined as dis
cussed earlier.
(b) If Mu > Mu (lim), it will result in a doubly reinforced section which will be discussed later.
(c) If Mu < Mu (lim), it will result, in an under reinforced section, and the design is done in the
followingt steps.
3. If Mu < Mu (lim), determine actual Xu from relationship:
Mu = 0.36 fck B Xu (d – 0.42 Xu)
This will result in a quadratic equation in terms of Xu which can be easily solved
4. Determine area of steel from the relation
Mu = 0.87 fy Ast (d – 0.42 Xu)
Alternatively, steps 3 and 4 can be avoided and Ast can be fond from equation given below which is
applicable for underreinforced section.
Mu = 0.87 fy Ast d
The gives a quadratic equation in terms of Ast, the solutions of which works out as under:
(b) Doubly rienforced section
A doubly reinforced concrete section is reinforced in both compression and tension regions. The section
of the beam or slab my be a rectangle, T and L section. The necessity of using steel in the compression
region arises due to two main reasons:
(a) When depth of the section is restricted, the strength available from a singly reinforced is inadequate.
(b) At a support of a continuous beam or slab where bending moment changes sign.
(i) Stress Block and Actual Depth of N.A
Figure shows a doubly reinforced section having compression reinforced section having compression
reinforcement fibre. Figure (b) shows the strain diagram while figure (c) the stress block.
Let,
Ast = total reinforcement at tension face.
Asc = Reinforcement in compression side.
Xu = Depth of N.A
C1 = Compressive force in concrete
= 0.36 fck BXu
C2 = Compressive force in compression steel
= fsc Asc
fsc = Design stress in compression reinforcement read off from the stress strain curve corresponding to
the strain Îsc in compression reinforcement.
Îsc = Strain in compression reinforcement using similar triangles,
sc Î =
( )
u
u u
X
0.0035 X  d
... (i)
Total compressive force is given by
C = C1 + C2
or C = 0.36 fck B Xu + fsc Asc ... (ii)
(neglecting the loss of concrete area occupied by compressive steel)
Total tensile force is given by
T = 0.87 fy Ast ... (iii)
In order to locate the N.A. equate the total compresive force to the total tensile force:
0.36 fck B Xu + fsc Asc = 0.87 fy Ast ... (iv)
From the above relation, Xu can be found. However for the solution of the above equation, an iterative
procedure will have to be adopted, since fsc depends upon, which in turn depends upon Xu. If the loss of
compressive area, occupied by the compressive steel is taken into account equations (ii) and (iv) are
modified as under.
Cu = 0.36 fck B Xu + fsc Asc – 0.446 fck Asc
= 0.36 fck B Xu + (fsc – 0.446 fck) Asc
and
0.36 fck B Xu + (fsc – 0.446 fck) Asc = 0.87 fy Ast
Note :
Normally, the term 0.446 fck Asc is very small and can be neglected witout causing and appreciable error.
Iterative procedure for computation of Xu
1. Compute the depth of N.A. of a balanced section given by strain compatibility
Xu(lim) =
y s 0.0055 0.87f / E
0.0035
+
2. For the given doubly reinforced section, assume Xu equal to Xu(lim).
3. Compute the value of sc Î from equation (i).
4. Compute value of fsc from the stress strain curve of steel corresponding to this value of sc Î .
5. Substitute the value of fsc in equation (iv) and compute the modified value of sc Î
6. Repeat steps 3 to 5 till convergence for the value of Xu is achieved.
(ii) Ultimate moment of resistance
Ultimate moment of resistance is given by
Mu = 0.36fckBXu (d–0.42Xu) + (fsc –0.446 fck) Asc (d–dc)
where
fsc = stress in compression steel and it is calculated by strain at the location of compression steel (fsc)
Design Steps
A doubly reinforced beam can be assumed to be made up of two beams A and B as shown in Figure. In
beam A which is singly reinforced beam, the tension steel Ast1 is required to balance the force of compression
C1 in concrete. In beam B, which is imaginary, the tension steel Ast2 is required to balance the force of compression
C2 in compression steel.
Ast Ast1
1. Determine the limiting moment of resistance Mu(lim) for the given crosssection using the equation for a
singly reinforced beam A.
ie.
Mu(lim) = 0.87 fy Ast1 (d – 0.42 Xu(lim))
or Mu(lim) = 0.36 fck B Xu(lim) (d – 0.42 Xu(lim))
After calculating moment of resistance Mu(lim)
st1 A can be calculated by equating force of compression to force of tension i.e.
0.87fy Ast1 = 0.36 fck B Xu(lim)
Where
st1 A = Area of tension steel corresponding to a balanced singly reinforced beam.
2. If the factored moment M exceeds Mu(lim), a doubly reinforced section is required to be designed for the
additional moment (M–Mu(lim)). This moment is resisted by an internal couple consisting of compression
force C2 in the compression steel and tension force T2 in an additional tension steel in beams B, that is
M – Mu(lim) = (fsc – 0.446 fck) Asc (d–dc)
» fsc Asc (d – dc)
Since 0.446 fck < < fsc
Asc =
u u(lim)
sc c
M –M
f (d–d )
Where
Asc = Area of compression reinforcement
3. The additional area of tension steel st2 A is obtained by considering the equilibrium of force of compression
C2 in compression steel and force of tension T2 in the additional tension steel, i.e.,
fsc Asc – 0.446 fck Asc = 0.87 fy st2 A
or fsc Asc » 0.87 fy st2 A
Ast 2 =
y
sc sc
0.87 f
f A
4. The total tension Steel At is given by
At = st1 A + st2 A
TBeam
1. Effective width of flange
Discussed in WSM
2. Limiting depth of neutral axis
Xu(lim) = d
0.0055 0.87f / E
0.0035
y s +
• Singly reinforced TBeam
Case1: When NA is in flange area
i.e., Xu < Df
(a) for Xu
f
ck f
y st
u D
0.36f B
0.87f A
X = <
(b) Ultimate moment of resistance
Mu = 0.36 fck Bf Xu (d – 0.42Xu)
Mu = 0.87 fy Ast (d – 0.42 Xu)
Case–2 : When NA is in web area (Xu > Df)
Case (a) when Df < u X
7
3
i.e. depth of flange is less than the depth of rectangular portion of stress diagram.
1. For actual depth of neutral ais
0.36fckbwXu + 0.446fck (Bf – bw) Df = 0.87 fy Ast
2. Ultimate moment of resistance
Mu = 0.36fckbwXu(d–0.42 Xu) + 0.446 fck (Bf–bw)Df
÷ø
ö
çè
æ 
2
d Df
Mu = 0.87fy st1 A (d–0.42 Xu) + 0.87 fy st2 A ÷ø
ö
çè
æ 
2
d Df
y
ck w u
st 0.87 f
A 0.36f b X 1
=
( )
y
ck f w f
st 0.87f
A 0.45f B b D 2

=
• Special Case (2) : When Xu > Df
and Df > u X
7
3
i.e. depth of flange is more than depth of rectangular portion of stress diagram.
(B D ) f f (B b ) f w
yf
As per IS : 456 – 2000
(Bf – bw) Df portion of flange is coverted into (Bf–bw)yf section for which stress is taken constant throughout
the section is 0.45 fck.
As per IS : 456–2000
yf = 0.15 Xu + 0.65Df < Df
1. For actual depth of neutral axis
0.36fckbwXu+ 0.446fck (Br–bw)yf = 0.87fy st2 A
or 0.36fckbwXu+ 0.446fck (Br–bw)yf = 0.87fyAst
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2 videos122 docs55 tests


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