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
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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
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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