Rcc Beams

ANALYSIS AND DESIGN OF STRUCTURES
RCC BEAMS
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AMIE(I) STUDY CIRCLE(REGD.)
A FOCUSSED APPROACH
RCC Beams
A beam is reinforced by placing steel bars in the tensile zone of the concrete beam so that the
compressive bending stress is carried by concrete and tensile bending stress is carried entirely
by steel reinforcing bars.
MODULAR RATIO
Consider composite section consisting of two bars, one of concrete and other of steel, well
bonded together with the help of end plugs, and subjected to a compressive load P.
Let Es = Modulus of elasticity of steel
Ec = Modulus of elasticity of concrete.
Modular ratio will be
s
c
E
m
E

It means that load carried by steel is m limes the load carried by concrete.
Singly Reinforced Beam
NEUTRAL AXIS OF BEAM SECTION
(a) Section (b) Strain diagram (c) Stress diagram
Case 1: Stresses in Concrete and steel known
Let c = compressive stress in the extreme fibre of concrete
t = tensile stress in steel reinforcement b = breadth of beam
d = depth to the centre of reinforcement (known as the effective depth)
n = kd = depth of N.A. below the top of the beam
k = neutral axis depth factor = n/d
Ast = area of tensile reinforcement.

RCC BEAMS
A FOCUSSED APPROACH
Since there is no resultant force across the section,
Total compression = total tension
Neglecting the tensile stress in concrete, we get
1
. . .
2
stc b kd t A
.
. 2
stA c
k
b d t


mc
k
mc t


(k is neutral axis depth factor)
Also, percentage steel p will be
100
.
stA
p x
b d

or it can also be written as
2
50
( )
mc
p
t mc t


Case 2: Dimensions of the beam completely known
Let us now take the case when the dimensions of the section, including the reinforcement are
known, but the stresses in concrete and steel are not known.
Equating the moment of area in compression to the moment of the equivalent area in tension
about the N.A., we get,
. . ( )
2
st
kd
b kd mA d kd 
MOMENT OF RESISTANCE
The total compressive force acting at the centre of gravity of the compressive forces is equal
to the total tensile force acting at the centre of gravity of the steel reinforcement. Hence the
moment of resistance of the reinforced concrete beam section is equal to the moment of the
couple consisting of the compressive force and the tensile force.
The total compressive force
1 1
. . . . . .
2 2
cbcC c k b d kd b 
acts at a distance of kd/3 from the top of the section.
Similarly, the total tensile force
. .st st stT t A A  and acts at a distance d from the top of the section.

RCC BEAMS
A FOCUSSED APPROACH
Lever Arm
If jd is the lever arm (i.e. the distance between the compressive force and the tensile force),
we have
1
3 3
kd k
a jd d d
 
     
 
Now, moment of resistance Mr is given by
Mr = force x lever arm = 2 21 1
. . .( )
2 2
c kd b jd cjk bd Rbd
 
  
 
where
1
2
R cjk
Also . . .r st st stM t A jd A jd 

. .
r
st
st
M M
A
t jd jd
 
VALUES OF DESIGN CONSTANTS
Refer following table.
Concrete
Grade of concrete M15 M20 M25 M30
Modular ratio (m) 18.67 13.33 10.98 9.33
cbc (N/mm2
) 5 7 8.5 10
Steel
Grade of steel Fe 250 Fe 415
cbc (N/mm2
) 140 230
BALANCED SECTION
In a beam section, if the area of steel reinforcement Ast is of such magnitude that the
permissible stresses c (cbc) and t (st) in concrete and steel respectively, are developed
simultaneously, the section is known as the balanced section, critical section or economical
section. For such a balanced section, the moment of resistance obtained from the
compressive force will be equal to the moment of resistance obtained from the tensile force.

RCC BEAMS
A FOCUSSED APPROACH
Percent area of reinforcement in balanced section
1 1
. 100 . . 100
2 2
cbc
st
c
p k x k x
t


 
UNDER REINFORCED SECTION
An under reinforced section is the one in which the percentage steel provided is less than that
given as 100
.
stA
p x
b d
 and therefore full strength of concrete in compression is not
developed.
The actual neutral axis of such a section will fall above the critical neutral axis of a balanced
section.
Given figure shows the under-reinforced section in which the actual N.A. is above the critical
N.A.
In a under-reinforced concrete, the concrete is not fully stressed to its permissible value when
stress in steel reaches its maximum value t = st.
The moment of resistance of an under-reinforced section is, therefore, computed on the basis
of the tensile force in steel:

RCC BEAMS
A FOCUSSED APPROACH
. . .r st st stM tA jd A jd 
OVER REINFORCED SECTION
In an over-reinforced section, the reinforcement provided is more than critical one and
therefore the actual N.A. of such a section falls below the critical N.A. of a balanced section.
Thus, in a over-reinforced section, steel reinforcement is not fully stressed to its permissible
value and the moment of resistance is determined on the basis of compressive force
developed in concrete:
2 21 1 1
. . . . . . . . .
2 2 2
r cbcM c kd b jd c j k bd j k bd  
Or it may also be written as
1 1
. . .
2 3 2 3
r cbc
n n
M c n d b n d b
   
      
   
where n is depth of neutral axis.
TYPES OF PROBLEMS IN SINGLY-REINFORCED BEAMS
There may be three types of problems in singly reinforced concrete beams :
1. Determination of the moment of resistance of the given section
2. Determination of stresses in the given section subjected to a given bending moment.
3. Design of section to resist a given bending moment.
Example
For a balanced rectangular section (b x d) of a singly reinforced beam, determine (i) depth of
neutral axis (ii) moment of resistance and (iii) percentage of steel using M 15 concrete and
using Fe 415 steel. If b = 200 mm and d = 300 mm, determine the numerical values of n, M,
and Ast.

RCC BEAMS
A FOCUSSED APPROACH
Solution
Given data
c = cbc = 5 N/mm2
; m = 19; t = st = 230 N/mm2
Depth of neutral section
For balanced section
19 5
0.292
19 5 230
mc x
k
mc t x
  
 
 Depth of neutral axis = n = kd = 0.292 x d
Moment of resistance
2 21 1
. 5 0.903 0.292 0.659
2 2
rM cjk bd x x x bd   N-mm
Here, we calculated j as
0.292
1 1 0.903
3 3
k
j
 
     
 
Percentage of steel
2 2
50 50 19(5)
0.318%
( ) 230(19 5 230)
mc x
p
t mc t x
  
 
Numerical values of n, M, and Ast
n = 0.292 x 300 = 87.6 mm
Mr = 0.659(200)(300)2
= 11.86 x 106
N-mm = 11.86 kN-m
20.318 200 300
. 190.8
100 100
st
bd x x
A p mm  
Example
Determine the moment of resistance of a singly reinforced beam 160 mm wide and 300 mm
deep to the centre of reinforcement, if the stresses in steel and concrete are not to exceed 140
N/mm2
and 5 N/mm2
. The reinforcement consists of 4 bars of 16 mm diameter. Take m = 18.
If the above beam is used over an effective span of 5 m, find the maximum load the beam can
carry, inclusive of its own weight.

RCC BEAMS
A FOCUSSED APPROACH
Solution
Given data
Ast (/4) x 162
= 804 mm2
; c = cbc = 5 N/mm2
; t = st = 140 N/mm2
; m = 18
Neutral axis
Equating the moment of inertia in compression to the moment of the equivalent area in
tension about neutral axis (NA), we get
( )
2
st
n
b x n x mA d n 

2
160
18 804(300 )
2
x n
x n 
Solving this quadratic equation
n = 159.5 mm
Depth of critical neutral axis
If nc is the depth of critical neutral axis, we have
. .c
mc
n k d d
mc t
 


18 5
300 117.4
(18 5) 140
c
x
n x mm
x
 

Stress diagram

RCC BEAMS
A FOCUSSED APPROACH
Since the depth of actual neutral axis is more than the critical one, the section is over
reinforced. Thus, concrete reaches its maximum stress earlier to steel. Hence the moment of
resistance is found on the basis of compressive force developed in concrete.
Lever arm =
159.5
300 246.8
3 3
n
d mm   

1 1
. . 5 159.5 160(246.8) 15.75
2 3 2
r
n
M c n b d x x x kN m
 
     
 
Maximum bending moment and maximum load
It will be
=
2
25
8 8
wl wx
kN m 
Equating this to the moment of resistance of the beam, we get
15.75 8
5.04 /
25
x
w kN m 
Problem
A singly reinforced concrete beam with an effective span of 4 m has a rectangular cross
sections having a width of 250 mm and an overall depth of 450 mm. The beam is reinforced
with 3 bars of 10 mm diameter Fe 415 HYSD bars at an effective depth of 400 mm. The self
weight of the beam together with dead load on the beam is 4 kN/m. Calculate the maximum
permissible live load on the beam assuming M15 grade concrete.
Answer: 5.96 kN/m
Hint: st = 230 N/mm2
Example
The cross-section of a simply supported reinforced beam is 200 mm wide and 300 mm deep to
the centre of the reinforcement which consists of 3 bars of 16 mm dia. Determine from the
first principles the depth of NA and the maximum stress in concrete when steel is stressed to
120 N/mm2
. Take m = 19.
Solution
Given data
b = 200 mm; d = 300 mm; Ast = 3 x (/4) x 162
= 603.2 mm2
; m = 19;

RCC BEAMS
A FOCUSSED APPROACH
Neutral axis
Let the depth of N.A. be n. Equating the moment of the compressive area to the moment of
equivalent area of steel, about N.A., we get
. . ( )
2
st
n
b n mA d n 
or 2
100 19(603.2)(300 )n n 
or 2
114.6 34382 0n n  
 n = 136.8 mm
Stress diagram
Maximum stress in concrete
Stress in steel = t = 120 N/mm2
(Given)

120
6.316
19
t
m
 
From stress diagram
/c t m
n d n


6.316
136.8 300 136.8
c


 c = 5.29 N/mm2
Example
A rectangular, singly reinforced beam, 300 mm wide and 500 mm effective depth is used as a
simply supported beam over an effective span of 6 m. The reinforcement consists of 4 bars of

RCC BEAMS
A FOCUSSED APPROACH
20 mm diameter. If the beam carries a load of 12 kN/m, inclusive of the self-weight,
determine the stresses developed in concrete and steel. Take m = 19.
Solution
Neutral axis and lever arm
Equating the moments of two areas about NA, we get
( )
2
st
n
b x n x mA d n 

2
300 19 1256.6(500 )
2
n
x x n 
 n = 213.5 mm
Lever arm =
213.5
500 428.8
3 3
n
a d mm    
Maximum bending moment
It will be
2 2
612(6)
54 54 10
8 8
wl
kN m x N mm    
Let c be the compressive stress in concrete.
61 1
. . 300 213.5 428.8 13.732 10
2 2
rM c n b x a c x x x x c N mm   
Compressive stress in concrete
Equating moment of resistance to external bending moment,
6 6
13.732 10 54 10x c x
 c = 254
3.93 /
13.732
N mm
Stresses in steel
Method 1
If t is the corresponding stress in steel, we get
/c t m
n d n


 219 3.93
( ) (500 213.5) 100.2 /
213.5
mc x
t d n N mm
n
    

RCC BEAMS
A FOCUSSED APPROACH
Method 2
. . 1256.6 428.8r stM A t a xt x 
 6
1256.6 428.8 54 10x t x
 t = 100.2 N/mm2
Problem
A reinforced concrete beam has section 300 mm wide and 600 mm deep. The reinforcement
consists of mild steel bars of 25 mm diameter with a cover of 50 mm to the centre of
reinforcement. If it is subjected to a bending moment of 120 kN-m, determine the stresses
developed in steel and concrete. Take m = 15.
Answer: 6.57 N/mm2
; 106 N/mm2
Example
A reinforced concrete beam 200 mm x 400 mm effective depth is used over an effective span
of 5 m. It is subjected to a uniformly distributed load of 5 kN/m inclusive of Its own weight.
Find the necessary steel reinforcement at the centre of the span. Take allowable stresses in
steel and concrete as 130 N/mm2
and 4 N/mm2
respectively, and m=16.
Solution
Given data
c = cbc = 4 N/mm2
; t = st = 130 N/mm2
; m = 16
2
65 25
15.625 15.625 10
8 8
wl x
M kN m x N m     
2 6
15.625 10rM Rbd x 
 2
0.488rM bd (1)
For balanced section
16 4
0.33
16 4 130
mc x
k
mc t x
  
 
1 1 0.11 0.89
3
k
j     

1 1
(4)(0.89)(0.33) 0.587
2 2
R cjk  
 Mr = 0.587bd2
(2)

RCC BEAMS
A FOCUSSED APPROACH
Neutral axis
The moment of resistance of the given beam has to be less than the moment of resistance of
the critical section. Hence the steel reinforcement will be corresponding to an under
reinforced beam. For such a section, the stress in steel will reach the maximum value of 130
N/mm2
.
Let n = depth of N.A. of the actual section.
Then the corresponding stress in concrete is given by
/
c n
t m d n



130
.
16 400
t n n
c x
m d n n
 
 
(3)
Now
1
. .
2 3
r
n
M c n b d
 
  
 
 6 1 130
15.625 10 200 400
2 16 400 3
n n
x x n x
n
   
    
   
 2 (1200 )
57700
400
n
n
n



This is a cubic equation. Solving by trial and error
n = 121.8 mm
Stress in concrete
From (3)
2130 121.8
3.56 /
16 400 121.8
c x N mm 

Area of steel

1
.
2
stcnb A t
 21 1 3.56 121.8 200
. 333
2 2 130
st
cnb x x
A mm
t
  
Problem
A reinforced concrete beam 200 mm x 400 mm effective depth is used over an effective span
of 5 m. It is subjected to a uniformly distributed load of 7 kN/m inclusive of its own weight

RCC BEAMS
A FOCUSSED APPROACH
Find the necessary steel reinforcement at the centre of the span. Take allowable stresses in
steel and concrete as 130 N/mm2
and 4 N/mm2
respectively and m = 16.
Solution
Given data
c = cbc = 4 N/mm2
; t = st = 130 N/mm2
; m = 16
2
67 25
21.88 21.88 10
8 8
wL x
M kN m x N mm     

6
2 2 2
21.88 10
0.684
200 (400)
rM M x
R
bd bd x
   
 Mr = 0.684bd2
(1)
Neutral axis
Since the moment of resistance of the given beam is greater than the moment of resistance of
the critical section, the steel reinforcement will be corresponding to an over-reinforced beam.
For such a section, the stress in concrete will reach the maximum value of 4 N/mm2
1
2 3
r
n
M cnb d
 
  
 
 6 1
21.88 10 4 200 400
2 3
n
x x n x
 
  
 
 n = 157.5 mm
Stress in steel
/t m d n
c n


 2400 157.5
16 4 98.5 /
157.5
d n
t mc x x N mm
n
  
   
 
Area of steel
total tension = total compression

1
. .
2
stt A cn b
 21 1 4 157.5 200
. . 639.6
2 2 98.5
st
cnb x x
A mm
t
  

RCC BEAMS
A FOCUSSED APPROACH
 In this example of an over-reinforced section, the stress in steel is only 98.5 N/mm2
which
is very much less than the permissible value of 130 N/mm2
. Thus the strength of steel is not
fully utilized. Such a design is undesirable and uneconomical. In circumstances where the
dimensions of the beam are limited, and where the section has to develop greater moment of
resistance than that of the balanced section, it is always desirable to design the section as
doubly reinforced. In doubly reinforced section, steel reinforcement is placed in the
compression zone also, which increases its moment of resistance. Hence a doubly reinforced
section is always preferred over an over reinforced section.
Example
Design a reinforced concrete beam subjected to a bending moment of 20 kN-m. Use M 20
concrete, and Fe 415 reinforcement. Keep the width of the beam equal to half the effective
depth.
Solution
Choose data
For M 20 concrete, c = cbc = 7 N/mm2
; m = 13
For Fe 415 steel, st = 230 N/mm2
.
Balanced section
13 7
0.283
13 7 230
cbc
cbc st
m x
k
m x

 
  
 
0.283
1 1 0.906
3 3
k
j     
1 1
. . (7)(0.906)(0.283) 0.898
2 2
R c j k  
Now
3
2 2 3 30.898
. . 0.449
2 2 2
r
d Rd
M Rbd R d d d N mm     
Depth of beam
Given BM = M = 20 kN-m = 20 x 106
N-mm
 Mr = M
3 6
0.449 20 10d x
 d = 354 mm
 b = d/2 = 177 mm

RCC BEAMS
A FOCUSSED APPROACH
Area of steel
6
220 10
271.1
230 0.906 354
st
st
M x
A mm
jd x x
  
Problem
Design a R.C. beam 10 carry a load of 6 kN/m inclusive of its own weight on an effective
span of 6 m. Keep the breath to be 2/3rd of effective depth. The permissible stresses in
concrete and steel are not to exceed 5 N/mm2
and 140 N/mm2
respectively. Take m = 18.
Answer: b = 240 mm; d = 360 mm; Ast = 616 mm2
Doubly Reinforced Beam
The moment of resistance of a balanced section is equal to Rbd2
. If a beam of specified
dimensions (b x d) is required to resist a moment much greater than Rbd2
, there are two
alternatives:
 to use an over-reinforced section, or
 to use doubly reinforced section
An over reinforced section is always uneconomical since the increase in the moment of
resistance is not in proportion to the increase in the area of tensile reinforcement since the
concrete, having reached maximum allowable stress, cannot take more additional load
without adding compression steel.
The other alternative is to provide reinforcement in the compression side of the beam and
thus to increase the moment of resistance of the beam beyond the value Rbd2
for a singly
reinforced balanced section. Sections, reinforced with steel in compression and tension are
known as doubly reinforced sections.
A doubly reinforced section is generally provided under the following conditions;
 When the depth and breadth of the beam are restricted and it has to resist greater
bending moment than a singly reinforced beam of that section would do.
 When the beam is continuous over several supports, the section of the beam at the
supports is usually designed as a doubly reinforced section.
 When the member is subjected to eccentric loading.
 When the bending moment in the member reverses according to the loading
conditions e.g., the wall of an underground R.C.C. storage reservoir, brackets etc.
 When the member is subjected to shocks, impact or accidental lateral thrust.
NEUTRAL AXIS
Following figure shows a doubly reinforced section.

RCC BEAMS
A FOCUSSED APPROACH
Let b = breadth of the beam.
d = effective depth of the beam
dc = depth of the centre of the compressive steel=e.d
e = compressive steel depth factor =dc/d
c = maximum stress in concrete
t = maximum stress in steel
c' = stress in concrete surrounding compressive steel
tc = stress in compressive steel = mc .c' = 1.5 mc'
Ast = area of tensile steel
Asc = area of compressive steel
mc = modular ratio for compression zone = 1.5 m
If maximum stresses in concrete and tensile steel are known the N.A. can be located from
stress diagram exactly in the same manner as in singly reinforced section:
/ 1
c n kd k
t m d n d kd k
  
  
 cbc
cbc st
mmc
k
mc t m

 
 
 
where cbc and st are the permissible stresses in concrete and steel respectively. Thus the
presence of steel in compression zone does not affect the depth of N.A.
Also
( )
' . cn d
c c
n


Again neglecting the concrete in the tensile zone and equating the moment of the
compressive area about N.A. to the moment of the tensile area about N.A. we get
2
( ) ( ) ( )
2
c sc c sc c st
bn
m A n d A n d mA d n     

RCC BEAMS
A FOCUSSED APPROACH
MOMENT OF RESISTANCE
The moment of resistance of a doubly reinforced section can be determined by taking the
moment of the compressive force about the centre of the tensile reinforcement.
Let, M = total bending moment to be resisted by the section = moment of resistance of the
section = M1 + M2
M1 = moment that can be developed by the balanced section, without compressive
reinforcement = moment of the compressive force in concrete about the centre
of tensile reinforcement.
M2 = moment, in excess of M1, that is to be provided by the compression steel =
moment of the compressive force in steel, about the centre of the tensile
reinforcement.
Ast = total tensile reinforcement = Ast1 + Ast2
Ast1 =tensile steel for the balanced section, corresponding to the moment M1
Ast2 = additional tensile steel necessary to develop the remaining moment M2
Taking the moments of the compressive forces about the centre of the tensile steel, we get
( 1) . '( )
2 3
c sc c
c n
M bn d m A c d d
 
     
 
. . ( 1) . '( )
2 3
c sc
c kd
M b kd d m A c d ed
 
     
 
or 21
. ( 1) . ' (1 )
2
c scM jck bd m A c d e   
or 
1
2
2
' ( 1) (1 )c sc
M
M
M Rbd c d m A e   

or M = M1 + M2
Now, 1 1 1
1
.
st
st
M M M
A
t xlever arm tjd jd
  
and 2 2
2
( )
st
c
M M
A
t xlever arm t d d
 

where t may be equal to or less than st.
The additional tensile area Ast2 is actually required to balance the compressive steel. Hence
Ast2 can also alternatively be determined by equating the moment of the equivalent concrete
area of the compressive steel to the moment of equivalent concrete area of the tensile steel
Ast2 about N.A.
Thus, 2( 1) ( ) ( )c sc c stm A n d mA d n   

RCC BEAMS
A FOCUSSED APPROACH
Example
A beam section, 100 mm wide and 560 mm deep is reinforced with 4 bars of 25 mm diameter
in the tensile zone and 4 bars of 12 mm diameter in the compression zone. The cover to the
centre of both the reinforcement is 40 mm. Determine the moment of resistance of the section,
if M 20 concrete and HYSD bars are used.
Solution
Given data
cbc = 7 N/mm2
; st = 230 N/mm2
; m = 13;mc = 1.5m=1.5 x 13 = 19.5
Area of compressive and tension steel
2 2
4. (12) 452.4
4
scA mm

 
2 2
4. (25) 1963.5
4
stA mm

 
Neutral axis
Let us determine the position of the neutral axis by equating the moments of area of
compressive concrete and the area of compressive steel expressed in equivalent concrete to
the moment of the area of the concrete equivalent to the area of steel in tension, about the
N.A.
 2300
(19.5 1) 452.4( 40) 13 1963.5(520 )
2
n x n x n    
Solving n = 208.7 mm
Critical neutral axis
Critical neutral axis is given by

RCC BEAMS
A FOCUSSED APPROACH
13 7
. (520) 147.4
13 7 230
cbc
c
cbc st
m x
n d mm
m x

 
  
 
Stresses
Since the actual NA is below the critical NA, the stress in concrete will reach its maximum
permissible value first.
 c = cbc = 7 N/mm2
2208.7 40
' 7 5.658 /
208.7
cn d
c c x N mm
n
 
  
7 208.7
300(208.7) x 520 (19.5 1)(452.4)(5.658(520 40)
2 3
rM
 
     
 
= 121.44 x 106
N-mm = 121.44 kN-m
Problem
A beam 250 mm x 500 mm in section is reinforced with 3 bars of 14 mm diameter at top and
5 bars of 20 mm diameter at the bottom, each at an effective cover of 40 mm. Determine the
moment of resistance of the beam section. Take the permissible stress in steel and concrete as
126 N/mm2
and 5.2 N /mm2
respectively, and m = 18.
Answer: 73.5 kN-m
Example
A doubly reinforced concrete beam is 400 mm wide and 600 mm deep to the centre of tensile
reinforcement. The compression reinforcement consists of 4 bars of 16 mm diameter, and is
placed with its centre at a depth of 40 mm from the top. The tensile reinforcement consists of
4 bars of 20 mm diameter. The section is subjected to a bending moment of 100 kN-m.
Determine the stresses in concrete and steel. Take m = 16.
Solution
Area of compressive and tensile steels
2 2
4 (16) 805
4
scA mm
 
  
 
2 2
4 (20) 1250
4
stA mm
 
  
 

RCC BEAMS
A FOCUSSED APPROACH
Stress diagram
Neutral axis
Taking moment of equivalent areas about NA
2
( 1) ( ) ( )
2
c sc c st
bn
m A n d mA d n    
or 2400
(1.5 16 1)(805)( 40) 16(1250)(600 )
2
n x n n    
Solving n = 173.8 mm
Stress in compression steel
Let the maximum stress in concrete be c. The stress c' in concrete surrounding the
compression steel is given by
173.8 40
' 0.77
173.8
cn d
c c c c
n
 
  
Now ( 1) . '( )
2 3
r c sc c
c n
M bn d m A c d d
 
     
 
 6 173.8
100 10 400 173.8. 600 (1.5 16 1)(805) 0.77 (600 40)
2 3
c
x x x x c
 
     
 
 c = 3.73 N/mm2
Now c' = 0.77c = 0.77 x 3.73 = 2.87 N/mm2
 Stress in compression steel will be = 1.5mc' = 1.5 x 16 x 2.87 = 68.9 N/mm2
Stress in tension steel
2( ) 3.73 16(600 173.8)
146.3 /
173.8
cm d n x
t N mm
n
 
  

RCC BEAMS
A FOCUSSED APPROACH
Problem
A doubly reinforced concrete beam is 250 mm wide and 500 mm deep from the compression
edge to the centre of tensile steel. The area of compression and tensile steel are both 1300
mm2
each. The centre of compression steel is 50 mm from the compression edge. If the beam
is subjected to a total bending moment of 70 kN-m, determine the stress in concrete, and
tension and compression steel. Take m = 18.
Answer: c = 3.56 N/mm2
; t = 123.8 N/mm2
; tc = 68.3 N/mm2
Example
A doubly reinforced rectangular beam is 240 mm wide and 500 mm deep. If the limiting
stresses in concrete and steel are 5 N/mm2
and 230 N/mm2
respectively determine the steel
areas for bending moment of 80 kN-m. Assume that steel is buried on both faces with its
centre 40 mm from either face. Take m = 19.
Solution
Given data
c = cbc = 5 N/mm2
; t = st = 230 N/mm2
Neutral axis
19 5
. 460 134.5
19 5 230
c
mc x
n n d x mm
mc t x
   
 
Area of compression steel
Max stress in concrete = c = 5 N/mm2
2134.5 40
' . 5 3.513 /
134.5
cn d
c c x N mm
n
 
  
Now . . ( 1) . '( )
2 3
r c sc c
c n
M b n d m A c d d
 
     
 

RCC BEAMS
A FOCUSSED APPROACH
or 6 5 134.5
80 10 240 134.5 460 (1.5 19 1) 3.513(460 40)
2 3
scx x x x A x
 
     
 
 Asc = 1146 mm2
Area of tension steel
 ( 1) . ' .
2
c sc st
c
bn m A c A t  
or
5
240 134.5 (1.5 19 1)(1146)(3.513) (230)
2
stx x x A  
 Ast = 832 mm2
 It is found here that Asc > Ast. Such a design is uneconomical because the HYSD bars (Asc)
provided in compression zone are highly under-stressed. For such a circumstance, it is better
to re-design the beam.
Problem
A doubly reinforced rectangular beam is 300 mm wide and 450 mm deep and is subjected to
a bending moment of 90 kN-m. If the limiting stresses in concrete and steel are 5 N/mm2
and
N/mm2
. determine the steel areas. Assume that steel is buried on both faces, with its centre
mm from either face. Take m = 18.
Answer: Ast = 1770 mm2
; Asc = 1300 mm2

RCC BEAMS
A FOCUSSED APPROACH
ASSIGNMENT
Q.1. (AMIE W13, S15, 10 marks): A RCC beam is 30 cm wide and 70 cm deep. The centres of the steel are 5
cm from the respective edges. Determine the area of steel in tension and compression zone for a bending
moment of 13 x 105
kg-cm. The beam is doubly reinforced. Assume the limiting stress in concrete and steel are
50 kg/cm2
and 1400 kg/cm2
, respectively. Given modular ratio = m = 18.
Answer: Asc = 540.87 mm2
; Ast = 1626.61 mm2
Q.2. (AMIE W15, 5 marks): Briefly describe the major steps for design of axially loaded tension member.
Q.3. (AMIE W15, 14 marks): A simply-supported RCC beam, 300 mm wide and 600 mm deep, has four bars
of 20 mm diameter as tension reinforcement, the centre of the bars being 50 mm from the bottom of the beam.
Determine the uniformly distributed load the beam can carry over an effective span of 6 m. Take the permissible
stresses in concrete and steel as 5 N/mm2
and 140 N/mm2
, respectively. Given that m = 18.
Answer: 14.63 kN/m
Q.4. (AMIE S16, 6 marks): Define effective cover, development length of rebar and balanced beam section.
Q.5. (AMIE S16, 14 marks): A simply supported RCC beam is 400 mm and 800 mm. Assume effective cover
of 50 mm and calculate the minimum reinforcement required and that required for a balanced section. Also,
write the procedure to calculate shear force requirement. Use M20 grade concrete and Fe 415 steel.
Answer: Ast,bal = 1320 mm2
;
Q.6. (AMIE S18, 20 marks): Design a rectangular beam 300 mm in width and 500 mm in depth to carry a load
of 25 kN/m for a span of 5 m. Check if compression reinforcement is required. M20 concrete and Fe 415 steel
bars are to be used.
Q.7. (AMIE W18, 20 marks): A singly reinforced concrete beam with an effective span of 4 m has a
rectangular cross section of 250 mm x 550 mm. It is reinforced with 3 bars of 10 mm dia Fe 415 bars at an
effective depth of 500 mm. The self weight of the beam together with dead load is 4 kN/m. Calculate the
maximum permissible live load on the beam. Assume M20 grade concrete.
Answer: 5041.54 N/m2
Important Note
Candidates intend to appear in Section B examination, are permitted to use reference material
for different engineering discipline as indicated below in the Examination Hall. In case the
Examination Centre is not in a position to provide the required reference materials on request,
candidates may use their own copy, subject to prior verification by the Officer in-Charge/
Invigilator to the extent that no extraneous matter is written inside pages of the concerned
reference material:
 IS 800-1984
 IS 456-2000
 IS 1893
 IS 13920
 IS 4326
 IS 13827

RCC BEAMS
A FOCUSSED APPROACH
 IS 13935
 SP 6(1) and SP (16)
 IRC - 6- 1966 Section III, Loads and Stresses Steel Tables
(For online support such as eBooks, video lectures, audio lectures, unsolved papers, online
objective questions, test series and course updates, visit www.amiestudycircle.com)

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