CONJUGATE BEAM METHOD
By
Murtaza zulfiqar
OBJECTIVES
At the end of this unit, students are supposed to;
• Understand what is conjugate beam ?
• Be able to understand what is conjugate beam method
• Will know Advantages of this method over others methods
• Be able to derive Conjugate beam method derivation/proof
• Be analyse beams and frames using the conjugate beam
method
• examples
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Conjugate beam method
The conjugate-beam method is an engineering
method to derive the slope and displacement of a
beam. The conjugate-beam method was developed
by H. Müller-Breslau in 1865
• Many credit Heinrich Müller-Breslau (1851-1925)
with the development of this method, while others,
say the method was developed by Christian Otto
Mohr (1835-1918).
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Theorm of conjugate beam method
• Therefore, the two theorems related to the
conjugate beam method are:
• Theorem 1 : The slope at a point in the real beam
is equal to the shear at the corresponding point in
the conjugate beam.
• Theorem 2 : The displacement of a point in the
real beam is equal to the moment at the
corresponding point in the conjugate beam
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• The basis for the method comes from the similarity of Eq.
1 and Eq 2 to Eq 3 and Eq 4. To show this similarity, these
equations are shown below.
Integrated, the equations look like this.
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Procedure for analysis of beams through CBM
• Procedure for analysis
• 1. Construct the conjugate beam with the M/EI
loading. Remember when the M/EI diagram is
positive the loading is upward and
• when the M/EI diagram is negative the loading is
downward.
• 2. Use the equations of equilibrium to solve for the
• reactions of the conjugate beam.
• This may be difficult if the moment diagram is
complex.
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Procedure
• 3. Solve for the shear and moment at the point or points
• where the slope and displacement are desired.
• If the values are positive, the slope is counterclockwise and the
• displacement is upward.
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• The following procedure provides a method that
may be used to determine
the displacement and slope at a point on the elastic
curve of a beam using the conjugate-beam method.
• Conjugate beam
• Draw the conjugate beam for the real beam. This
beam has the same length as the real beam and has
corresponding supports as listed above.
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• In general, if the real support allows a slope, the
conjugate support must develop shear; and if the
real support allows a displacement, the conjugate
support must develop a moment.
• The conjugate beam is loaded with the real beam's
M/EI diagram. This loading is assumed to be
distributed over the conjugate beam and is directed
upward when M/EI is positive and downward when
M/EI is negative. In other words, the loading always
acts away from the beam.[
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• Equilibrium[
• Using the equations of statics, determine the
reactions at the conjugate beams supports.
• Section the conjugate beam at the point where the
slope θ and displacement Δ of the real beam are to
be determined. At the section show the unknown
shear V' and M' equal to θ and Δ, respectively, for
the real beam. In particular, if these values are
positive, and slope is counterclockwise and the
displacement is upward
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EXAMPLE
&
SOLUTIONS
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Example
Draw the shear and moment diagrams for the beam
shown in Fig. The support at B settles 1.5 in. Take E =
29(103) ksi, I = 750 in4.
20 k
1.5 in
A
C
B
12 ft
12 ft
24 ft
Actual Beam
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Solution
Principle of Superposition
20 k
1.5 in
A
Actual Beam
B
12 ft
12 ft
C
24 ft
• The beam is first degree statically indeterminate.
• The centre support B is chosen as redundant, so that the
roller at B is removed.
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• By is assumed to act downward on the beam.
20 k
1.5 in
A
C
Actual Beam
B
=
20 k
B
A
C
Primary
Structure
C
Redundant By
applied
ΔB
+
By
A
B
Δ’BB=ByfBB
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Compatibility Equation
• With reference to point B, using units of ft, we require
 
1.5
  B  By f BB
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(1)
• Use conjugate beam method to compute ΔB and fBB since
the moment diagrams consists straight line segments.
• For ΔB
20 k
B
A
C
20 k
A
C
15 k
5k
12 ft
36 ft
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Compatibility Equation
20 k
A
C
15 k
5k
12 ft
1080
EI
2520
EI
8 ft
36 ft
180
EI
3240
EI
16 ft
24 ft
1800
EI
conjugate beam
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Compatibility Equation
1080
EI
180
EI
3240
EI
  MB'  0
2520
EI
8 ft
16 ft
24 ft
120
EI
1800
EI
1440
EI
MB’
1440
1800
8 
24   0
 M B' 
EI
EI
31680 31680
M B'  


EI
EI
VB’
8 ft
16 ft
1800
EI
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Compatibility Equation
• For fBB
1k
B
A
C
1k
A
C
0.5 k
0.5 k
24 ft
24 ft
288
EI
12
EI
conjugate beam
144
EI
24 ft
24 ft
144
EI
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Compatibility Equation
• For fBB
144
EI
288
EI
conjugate beam
24 ft
24 ft
  MB'  0
144
144
8  24   0
 mB ' 
EI
EI
2304 2304
mB '  


EI
EI
12
EI
12
EI
144
EI
144
EI
mB’
vB’
24 ft
8 ft
144
EI
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OR
Pa  3 2
2 
B 
L a 

1 2EI  4

31680

EI
PL3
f BB 
4 8EI
2304

EI
Compatibility Equation
• Substituting these results into eq. (1), we have
 
1.5
  B  By f BB
12
(1)
1.5 31680
 2304

 By 

12
EI
 EI 
• Expressing the units of E and I in terms of k and ft, we
have
  




 1.5 
2
4
3
2
2
2
4
4




ft
29
10
k
in
12
in
ft
750
in
ft
12
in 4


 12 
 31680 By 2304
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By  5.56k
Equilibrium Equations
• The negative sign indicates that By acts upward on the
beam.
20 k
A
Ay
C
12 ft
12 ft
By=5.56 k
24 ft
Cy
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Equilibrium Equations
  MA  0
 2012  5.5624  Cy 48  0
Cy  2.22 k
   Fy  0
Ay  20  5.56  2.22  0
Ay  12.22 k
20 k
A
Ay=12.22 k
C
12 ft
12 ft
By=5.56 k
24 ft
Cy=2.22 k
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• Using these results, shear and moment diagrams are
20 k
A
Ay=12.22 k
C
12 ft
12 ft
By=5.56 k
Cy=2.22 k
24 ft
V (k)
12.22
x (ft)
-2.22
-7.78
-20
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• Using these results, shear and moment diagrams are
20 k
A
Ay=12.22 k
C
12 ft
12 ft
By=5.56 k
24 ft
Cy=2.22 k
M (k.ft)
146.7
53.3
x (ft)
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Advantages of this method over others methods
• Generally, the is a more direct and effective method than the
consistency method in determining reactions and deflections of
beams.
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THE END
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•References:
•
Rc hibler , 2009 edition
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Be calm Guys , feel free to ask any questions ….
We are here ,to satisfy U …..

But one by one :P
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