UNIT 2
CONJUGATE BEAM METHOD
2.1 INTRODUCTIONIn the course module BT 254, we were introduced to the concept of Moment Area Method as a means of calculating the slope and the deflection of beams at various sections. The concept was based on the theorems by Castigliano, which stated that; the change in slope between two
points on a straight member under flexure is equal to the area of theMEI
diagram between
those two points. That is, the slope between the point A and B on beam is described mathematically as;
θA−θB=ABMD
EIThus, in finding the magnitude of one of the slope, say slope at A, the value of the slope at B should be known before hand or should be a point on the beam with a slope value of zero.
2.2 DERIVATION OF THEOREMThe conjugate beam theorem is derived from the moment area theorems. It has the advantage of being able to calculate for the slope in a member even when they are no points on the beam where the slopes are zero. To illustrate the theorem, we will consider the simply
supported beam which is uniformly loaded as shown in Fig. 2.0(a) below. The MEI
diagram
and the elastic curve for the beam is show in Fig. 2.0(b) and (c) respectively.
OBJECTIVES
At the end of this unit, students are supposed to;
1. Understand Mohr’s first and second conjugate beam theorem.
2. Be able to convert real beams into conjugate beams
3. Be analyse beams and frames using the conjugate beam method4.
CONJUGATE BEAM METHOD
From Fig 1.0 the change in slope between A and C is given by;
θA−θC=Area ofMEI
diagram between A∧C
The slope at C is thus found as
θC=θA−Area ofMEI
diagram between A∧C
It can be seen that,
θA=BB1
AB
¿ 1L
moment of area ofMEI
diagrambetween A∧B about B
Therefore
θC=Moment of M / EI diagramabout BL
−AreaofMEI
diagram between A∧C
It can be seen that deflection at C is;
12 | P a g e
Fig 2.1(a)
x
L
A B
W W
C
Fig 2.1(b)RA RB
Fig 2.1(c)
A BCC2
C1
θA
θB
x
CONJUGATE BEAM METHOD
¿CC2=CC 1−C2C1
¿ xC θ A−D efelction of C w . r .t .tangent at A
¿ xC × Moment of ¿¿
Let’s look at a case of a similar beam loaded with the MEI
diagram and having the span equal
to the above considered beam. The reaction at the point A would be given by,
R'A=
Moment of the load about BL
¿ Moment of M /EI d iagram between A∧B about BL
And the shear force at C would be given by,
V C=R 'A−load between A∧C
¿ Moment of M /EI diagramabout BL
−AreaofMEI
diagr am between A∧C
It can be observed that the slope at C,θC in the beam earlier considered is equal to the shear
force in the latter beam considered which was loaded with the MEI
diagram. And that the
deflection at C in the earlier beam is equal to the bending moment in the latter beam with the MEI
diagram as its load. The latter beam considered is the conjugate beam. The about
discussion is the result of two theorems;
Theorem I: The rotation/slope at a point in a beam is equal to the shear force in the conjugate beam
Theorem II: The deflection in a beam is equal to the bending moment in the conjugate beam.
The generalizations above are made on the premise that the beam is simply supported. Students are being advised, however, to make suitable changes considering the boundary conditions for different beams to see if the above theorem would hold. The conjugate beam can thus be defined as;
An imaginary beam with a span which is equal to the span of the original beam
and is loaded with the MEI
diagram of the original beam, in a way that the shear
force and the bending moments at a section of the conjugate beam represents slope and the deflection in the original beam.
It is worth noting that in the original beam with a fixed end, the slope and the deflection is zero. Thus in the corresponding conjugate beam, the shear and the bending moment would be
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CONJUGATE BEAM METHOD
equally zero. These conditions exist in a beam with a free end, as such; a fixed support in an original beam connotes a free end/support in a conjugate beam. The opposite holds in both cases. As in an original beam with a free end, the slope and the deflection at end is note equal to zero. Similarly, in the conjugate beam, the shear and the bending moment at the support would not be zero. This indicates a support which is fixed; since it’s only a fixed support that resists both rotation and translation. In this case a free end/support in an original beam will give e fixed end/support in the conjugate beam.
There are various end conditions and there corresponding end conditions in the conjugate beam. These are illustrated in Table 2.1 below.
Also original beams and their corresponding conjugate beams are illustrated in Table 2.2.
SING CONVENTION: The sign convention that has been adopted for the purpose of this chapter is as followed.
1. Sagging moment are positive2. Left side upwards force or right side downwards force (↑↓) gives positive shear,
Thus positive shear gives clockwise rotation and positive moment gives downward deflection
2.3 ILLUSTRATIONSThe conjugate beam as seen from the above discussion is an image of an/the original beam. Using the illustration in Fig 2.1, the conjugate beam illustrated in Fig 2.1(b) is an image of the original beam illustrated in Fig 2.1(a). As an image of the original beam, the conjugate beam and the original beam bear’s certain similarities. The similarity between the original beam and the conjugate beam is with their span. That is the span of the original beam is the same as the span of the conjugate beam.
Despite the fact that the conjugate beam is an image of the original beam, and as such are supposed to have similar features, there are certain dissimilarities about the two. The dissimilarities lie in their reaction and their loadings. The loading in the case of the original beam is the loads that are imposed on the beam. That is the various loads the beam is supposed to carry/resist; the point loads, distributed loads and the concentrated moments. The loads in the case of the conjugate beam is the value of the moment at any point on the beam, induced by the loadings on the original beam, divided by the flexural rigidity (EI ) of the beam at that section. In other words, the load of the conjugate beam is given by the moment diagram of the original beam for the section divided by the flexural rigidity of that section of the beam. This in most instances is defined by a varyingly distributed load as seen from the illustration in Fig 2.1(b).
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CONJUGATE BEAM METHOD
The other dissimilarity between the original beam and the conjugate beam is in their support reactions. The support reactions in the original beam are not always the same as that in the
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F ≠ 0∧discontinuousM ≠ 0
Interior hinge
θ ≠ 0∧discontinuous∆ ≠ 0
6
Interior support
Interior hingeInterior support
5
θ ≠ 0∧continuous∆=0
F ≠ 0∧continuousM=0
Fixed end
F ≠ 0 M ≠ 0
Free end
θ ≠ 0∆ ≠ 0
4
Free end
F=0 M=0
Fixed end
θ=0 ∆=0 3
Hinged end
F ≠ 0 M=0
Hinged end
θ ≠ 0∆=0
2
Simply supported or Roller end
F ≠ 0 M=0
Simply supported or Roller end
θ ≠ 0∆=0
1
SI NO CONJUGATE BEAMORIGINAL BEAM
Table 2.1
CONJUGATE BEAM METHOD
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5
SI NO CONJUGATE BEAMORIGINAL BEAM
2
3
4
6
1
7
8
Table 2.2
CONJUGATE BEAM METHOD
conjugate beam. As would be realised from Table 2.1 and 2.2, the reaction in the conjugate beam do sometimes differ from that of the original beam. Again the condition at the support of the original beam differs from the conditions at the support of the conjugate beam. Whereas the condition that might exist at the support of an original beam would be either a translation or movement in any of the two perpendicular plane which is termed as the deflection in that particular plane and a rotation about a given point the same when measured in the conjugate beam is reckoned as the shear and the moment respectively. That is to say that whenever the slope of the original beam is measures or calculated for, the same passes as the shear in the conjugate beam. And whenever the deflection of the original beam is measured it passes as the moment in the conjugate beam. At this moment it is worth saying that a slope in the original beam is equal to a shear in the conjugate beam. And a deflection in the original beam is a moment in the conjugate beam.
To throw a little more light on this lets consider Table 2.1. The table has two main columns, one corresponding to original beam configurations and their respective end conditions and the other, conjugate beam configuration and their end conditions for their respective original beams in the first column. It is worth noting these facts, considering loadings in the y-plane only;
a. A simple support does not allow for translation in that plane (y-plane), hence deflection in the y-plane is zero. But then the beam can rotate about that reaction since a simple support does not have the capacity to resist rotation. In which case there is going to be a slope between the beam at its original position and the beam in its deflected position. This can be summarised that at a simple support deflection is zero but slope is not equal to zero.
b. A free end of a beam does allow for both translation in the y-plane and rotation about that point. Hence at the free end of a beam there exists a slope as much as there is a rotation. Thus in summary, at the free end of a beam slope is not equal to zero and deflection is not equal to zero
c. The fixed end of a beam or a beam with fixed supports has the capacity to resist both rotation and translation within any plane. Since the beam cannot translate at that support, deflection at that support is zero. Because the beam does not allow for rotation at that support as well, the deflected beam would lie parallel the profile of the beam before deflection, in which case slope would be equal to zero. It can thus be said of the fixed support that it has a zero slope and a zero deflection.
2.4 PROCEDURE In solving problems related to the conjugate beam method, the following step should be followed,
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CONJUGATE BEAM METHOD
1. Solve for the reactions of the original beam; that is determining the values of the reaction for the original beam.
2. Draw the bending moment diagram for the original beam3. Draw the conjugate beam from the original beam, having in mind that the length of
the conjugate beams is the same as that of the original beam, that where a support allows for a slope in the original beam, the conjugate beam must develop a shear, and where the support in an original beam allows for deflection the conjugate beam must develop a moment.
4. The conjugate beam is then loaded with the MEI
diagram. The load is assumed to be
uniformly distributed over the conjugate beam and is directed upwards when the MEI
diagram is positive and directed downwards when the MEI
diagram is negative.
5. Calculate the reaction of the conjugate beam using the equation of equilibrium.6. On the conjugate beam show where the unknown moment and shear forces would be
acting corresponding to the section where the slope and deflection would be calculated at.
7. Using the equation of equilibrium, determine the shear and the moments in the conjugate beam, this will correspond to the slope and deflection in the original beam.
2.5 SOLVED EXAMPLESEXAMPLE 2.1 Determine the slope at A, B, C (θA ,θB , θC ) and the deflection at C ∆Cin the
beam shown in Fig 2.2(a)
SOLUTIONApplying the steps outlined above;
1. Determining the reaction of the original beam.
∑ f y=0
RA+RB=35
∑ M A=0
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RARB
Fig 2.2(a)
35 kN
3m3m
A BC
CONJUGATE BEAM METHOD
( RB× 6 )−(35 ×3 )=0
6 RB=105
RB=17.5 kN
∴RA=17.5 kN
2. Drawing the bending moment diagram for the original beam. The moment diagram is drawn by calculating for the moment values at critical section of the beam. This is shown in the Fig 2.2(b) below.
3. The conjugate beam for the original beam is now drawn having the same span as the original beam. Since the original beam is simply supported, the conjugate beam will also be simply supported as from Table 2.1 and 2.2. The conjugate beam bean is illustrated in Fig 2.2(c)
4. The conjugate beam is now loaded with the MEI
diagram of the original beam, Fig 2.2(d).
5. Using the equation of equilibrium, the reaction of the conjugate beam is calculated for.
Under the conditions of equilibrium, the sum of the two reaction RA 'and RB'
should be
equal to the area of the MEI
diagram. By summing up the forces in the y-direction and
taking moment about B, the reaction of the conjugate beam is calculated for.The load between A and B is
12
×52.5EI
× 6=157.5EI
Summing forces in the y-direction,
∑ f y=0
RA '+RB'
=157.5EI
…(1)
∑ M B=0
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Fig 2.2(b)
Fig 2.2(d)
Fig 2.2(c)
3m3mRA ' RB'
3m3mRA ' RB'
52.5EI
52.5 kNm
CONJUGATE BEAM METHOD
RA '×6=157.5
EI×3
6 RA '=472.5
EI
∴RA '=78.75
EIunits
But R A'+RB'
=157.5EI
RB'=157.5
EI−78.75
EI
∴RB'=78.75
EIunits
6. Since the original beam is simply supported, it allows for rotation at the support but does not permit translation in the y-direction and as so can develop slopes at the support but cannot develop deflections at the support. The conjugate beam can thus develop shear at the support but cannot develop moments at the support, Fig 2.2(e).
7. Using the equilibrium condition the shear and the moment at various sections on the conjugate beam is calculated for. These values correspond to the slope and deflection of the original beam at their respective places.
Thus the slope at A and B are given as,
Slope at A,
θA=S . F at A=V A=78.75
EIrad (clockwise )
Slope at B,
θB=S . F at B=−V B=−78.75
EIrad (anti−clockwise )
Slope at C,
θC=S . F at C=78.75EI
−78.75EI
=0
Deflection at C,
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3m3mV A V B
52.5EI
Fig 2.2(e)
CONJUGATE BEAM METHOD
The centre of gravity of the load is at the mid-span
∆C=moment at C=(78.75EI
×3)−( 78.75EI
×1.5)=118.13EI
(downward)
EXAMPLE 2.2
Determine the slope at A, B, C (θA ,θB , θC ) and the deflection at C ∆Cin the beam shown in
Fig 2.
SOLUTIONApplying the steps outlined above;
1. Determining the reaction of the original beam.
∑ f y=0
RA+RB=40
∑ M A=0
( RB× 8 )−(40 × 4 )=0
8 RB=160
RB=20 kN
∴RA=20 kN
2. Drawing the bending moment diagram for the original beam. The moment diagram is drawn by calculating for the moment values at critical section of the beam. This is shown in the Fig 2.3(b) below.
3. The conjugate beam for the original beam is now drawn having the same span as the original beam. Since the original beam is simply supported, the conjugate beam will also be simply supported as from Table 2.1 and 2.2. The conjugate beam bean is illustrated in Fig 2.3(c)
21 | P a g e
RA RBFig 2.3(a)
Fig 2.3(b)
40 kN
4m4m
A BCI 2I
80 kNm
4m4mRA ' RB'
CONJUGATE BEAM METHOD
4. The conjugate beam is now loaded with the MEI
diagram of the original beam, Fig 2.3(d)
5. Using the equation of equilibrium, the reaction of the conjugate beam is calculated for.
Under the conditions of equilibrium, the sum of the two reaction RA 'and RB'
should be
equal to the area of the MEI
diagram. By summing up the forces in the y-direction and
taking moment about B, the reaction of the conjugate beam is calculated for.
The load between A and C is 12
×80EI
×4=160EI
The centre of gravity of the load is
4+ 13
× 4=5.33¿B
The load between A and C is 12
×40EI
×4= 80EI
The centre of gravity of the load is 23
× 4=2.67¿B
Summing forces in the y-direction,
∑ f y=0
RA '+RB'
=160EI
+ 80EI
=240EI
∑ M B=0
RA '×8=[(160
EI×5.33)+( 80
EI×2.67)]
8 RA '=1066.40
EI
∴RA '=133.30
EIunits
22 | P a g e
RB'
RA '
40EI
80EI
Fig 2.3(d)
Fig 2.3(c)
CONJUGATE BEAM METHOD
But R A'+RB'
=240EI
RB'=240
EI−133.3
EI
∴RB'=106.7
EIunits
6. Since the original beam is simply supported, it allows for rotation at the support but does not permit translation in the y-direction and as so can develop slopes at the support but cannot develop deflections at the support. The conjugate beam can thus develop shear at the support but cannot develop moments at the support, Fig 2.3(e).
7. Using the equilibrium condition the shear and the moment at various sections on the conjugate beam is calculated for. These values correspond to the slope and deflection of the original beam at their respective places.
Thus the slopes at A, B and C are given as,
Slope at A,
θA=S . F at A=RA '=133.30
EIrad (clockwise)
Slope at B,
θB=S . F at B=−RB'=−106.7
EIrad (anti−clockwise )
Slope at C,
θC=S . F at C=133.30EI
−106.7EI
=26.6EI
rad
Deflection at C,
∆C=moment at C=(133.30EI
× 4)−(160EI
×43 )=319.87
EI(downward )
EXAMPLE 2.3
Determine the rotations at A, B, C, E (θA ,θB , θC , ,θE ) and the deflection at C, and E (∆C , ∆E ¿
in the beam shown in Fig 2.4(a).
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V BV A
40EI
80EI
Fig 2.3(e)
60 kN
CONJUGATE BEAM METHOD
SOLUTION
1. The first step is to determine the reaction of the original beam.
∑ f y=0
RA+RB=60
∑ M A=0
( RB× 12 )− (60 × 3 )=0
12 RB=180
RB=15 kN
∴RA=60 kN
2. The next is to draw the bending moment diagram for the original beam. The moment diagram is drawn by calculating for the moment values at critical section of the beam. This is shown in the Fig 2.4(b) below.
3. The conjugate beam for the original beam is now drawn having the same span as the original beam. Since the original beam is simply supported, the conjugate beam will also be simply supported as from Table 2.1 and 2.2. The conjugate beam bean is illustrated in Fig 2.4(c)
4. The conjugate beam is now loaded with the MEI
diagram of the original beam, Fig 2.4(d)
24 | P a g e
Fig 2.4(b)
135 kN
67.5EI
135EI
22.5EI
45EI
RBRAFig 2.4(a)
EDI 2I2I ICBA
3m 3m 3m 3m
I 2I2I I 3m 3m 3m 3mRA '
RB'
Fig 2.4(c)
CONJUGATE BEAM METHOD
5. Using the equation of equilibrium, the reaction of the conjugate beam is calculated for.
Under the conditions of equilibrium, the sum of the two reaction RA 'and RB'
should be
equal to the area of the MEI
diagram. By summing up the forces in the y-direction and
taking moment about B, the reaction of the conjugate beam is calculated for.
The load between A and C is 12
×135EI
× 3=202.5EI
The centre of gravity of the load is 23
×3=2¿ A
The load between C and E is
¿ 270EI
The centre of gravity of the load is ¿2.53+3=5.53¿ A
The load between E and B is 12
×45EI
×3=67.5EI
The centre of gravity of the load is 13
×3+9=10 fr om A
Summing forces in the y-direction,
∑ f y=0
RA '+RB'
=202.5EI
+ 270EI
+ 67.5EI
=540EI
∑ M B=0
RB'×8=[( 202.5
EI×2)+( 270
EI×5.53)+( 67.5
EI×10)]
12 RB'=2573.1
EI
∴RB'=214.43
EIunits
But R A'+RB'
=240EI
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CONJUGATE BEAM METHOD
RA '=540
EI−214.43
EI
∴RA '=325.57
EIunits
6. Since the original beam is simply supported, it allows for rotation at the support but does not permit translation in the y-direction and as so can develop slopes at the support but cannot develop deflections at the support. The conjugate beam can thus develop shear at the support but cannot develop moments at the support, Fig 2.4(e).
7. Using the equilibrium condition the shear and the moment at various sections on the conjugate beam is calculated for. These values correspond to the slope and deflection of the original beam at their respective places.
Thus the slopes at A, B, C, D and E are given as,
Slope at A,
θA=S . F at A=RA '=325.57
EIrad (clockwise )
Slope at B,
θB=S . F at B=−RB'=−214.43
EIrad (anti−clockwise)
Slope at C,
θC=S . F at C=325.57EI
−202.5EI
=123.07EI
rad (clockwise )
Slope at E,
θC=S . F at C=325.57EI
−202.5EI
−270EI
=−146.93EI
rad (anti−clockwise)
Deflection at C,
∆C=moment at C=(325.57EI
×3)−( 202.5EI
×33 )=774.21
EI(downward)
Deflection at E,
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67.5EI
135EI
22.5EI
45EI
V A V BFig 2.4(e)
CONJUGATE BEAM METHOD
∆E=moment at E=( 325.57EI
× 9)−( 202.5EI
×33+6)−( 270
EI×3.47)=575.10
EI(downward )
EXAMPLE 2.4Find the rotation and deflection at the free end of the overhanging beam shown in Fig 2.5(a)
SOLUTION
1. The first step in solving a problem of this nature is to determine the support reactions of the original beam.
∑ f y=0
RA+RB=P
∑ M A=0
( RB× 3 L )−( P × 4 L )=0
3 L RB=4 PL
RB=4 P3
kN
∴RA=−P
3kN
2. The next is to draw the bending moment diagram for the original beam. The moment diagram is drawn by calculating for the moment values at critical section of the beam. This is shown in the Fig 2.5(b) below.
3. The conjugate beam for the original beam is now drawn having the same span as the original beam. Since the original beam is simply supported, the conjugate beam will also be simply supported as from Table 2.1 and 2.2. The conjugate beam bean is illustrated in Fig 2.5(c)
27 | P a g e
PL
P
L 3L
A BC
3EI 1EI
Fig 2.5(a)
Fig 2.5(b)
Fig 2.5(c)
A B3EI 1EIC
RA '
CONJUGATE BEAM METHOD
4. The conjugate beam is now loaded with the MEI
diagram of the original beam, Fig 2.5(d)
5. Using the equation of equilibrium, the reaction of the conjugate beam is calculated for.
Under the conditions of equilibrium, the sum of the two reaction RA 'and RB'
should be
equal to the area of the MEI
diagram. By summing up the forces in the y-direction and
taking moment about B, the reaction of the conjugate beam is calculated for.
The load between A and B is
12
×PL
3 EI× 3 L=P L2
2 EIThe centre of gravity of the load is
L+ 13
×3 L=L¿B
The load between B and C is
12
×PLEI
× L= P L2
2 EIThe centre of gravity of the load is
13
× L=L3
¿ B
Summing forces in the y-direction,
∑ M C=0
−RA '×3 L= P L2
2 EI× L
−3 RA'L=P L3
2 EI
∴RA '=−P L2
6 EIunits
6. Since the original beam is simply supported, it allows for rotation at the support but does not permit translation in the y-direction and as so can develop slopes at the support but cannot develop deflections at the support. The conjugate beam can thus develop shear at the support but cannot develop moments at the support, Fig 2.5(e).
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PLEI
PL3 EI
Fig 2.5(d)
A B3EI 1EIC
V A
PLEI
PL3 EI
Fig 2.5(e)
V B
M B
A B3EI 1EIC
CONJUGATE BEAM METHOD
7. Using the equilibrium condition the shear and the moment at various sections on the conjugate beam is calculated for. These values correspond to the slope and deflection of the original beam at their respective places.
Thus the slopes at A, B and C are given as,
Slope at C,
θC=S . F at C=RA'=−P L2
6 EI+ P L2
2 EI+ P L2
2 EI=5 P L2
6 EIrad (clockwise )
Deflection at C,
∆C=moment at C=−(P L2
6 EI× 4 L)+( P L2
2 EI×2 L)+(P L2
2 EI×
2 L3 )=2P L3
3 EI(downward)
EXAMPLE 2.5
Determine the slope and deflection of the cantilevered beam with the load at the free end in Fig 2.6(a).
SOLUTION
1. The first step in solving a problem of this nature is to determine the support reactions of the original beam.
∑ f y=0
RA=25 kN
∑ M A=0
M A−(25× 7 )=0
M A=175 kNm
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Fig 2.6(a)
A B
25
7m
CONJUGATE BEAM METHOD
2. The next is to draw the bending moment diagram for the original beam. The moment diagram is drawn by calculating for the moment values at critical section of the beam. This is shown in the Fig 2.6(b) below.
3. The conjugate beam for the original beam is now drawn having the same span as the original beam. Since the original beam is simply supported, the conjugate beam will also be simply supported as from Table 2.1 and 2.2. The conjugate beam bean is illustrated in Fig 2.6(c)
4. The conjugate beam is now loaded with the MEI
diagram of the original beam, Fig 2.6(d)
5. Using the equation of equilibrium, the reaction of the conjugate beam is calculated for.
Under the conditions of equilibrium, the sum of the two reaction RA 'and RB'
should be
equal to the area of the MEI
diagram. By summing up the forces in the y-direction and
taking moment about B, the reaction of the conjugate beam is calculated for.
The load between A and B is 12
×175EI
× 7=−12252 EI
The centre of gravity of the load is 23
×7=143
¿B
Summing forces in the y-direction,
∑ f y=0
30 | P a g e
Fig 2.6(b)175 kNm
A B
7m
Fig 2.6(c)
A B
Fig 2.6(d)175 kNm
A B
CONJUGATE BEAM METHOD
RB'=1225
2 EI6. Since the original beam is fixed at one end and free at the other end, it will not allow for
rotation about the fixed end as well as translation in the y-direction at the fixed support, therefore it can neither develop slope and deflection at the fixed end. Conversely it will allow for rotation and translation in the y-direction at the free end and as so can develop slope deflection at the free end. The conjugate beam can thus develop shear and moments at the free end but not the fixed support, Fig 2.6(e).
7. Using the equilibrium condition the shear and the moment at various sections on the conjugate beam is calculated for. These values correspond to the slope and deflection of the original beam at their respective places.
Thus the slopes at B are given as,
Slope at B,
θB=S . F at B=RB'=
−12252 EI
rad (clockwise)
Deflection at B,
∆B=moment at B=−( 12252 EI
×143 )=−8575
3 EI(downward)
EXAMPLE 2.6
Determine the slope and deflection of the cantilevered beam with the load at the mid-span end in Fig 2.7(a).
SOLUTION
1. The first step in solving a problem of this nature is to determine the support reactions of the original beam.
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V B'
M B'
BA
175 kNm Fig 2.6(e)
3m3m
20 kN
A B
Fig 2.7(a)
CONJUGATE BEAM METHOD
∑ f y=0
RA=20 kN
∑ M A=0
M A−(20× 3 )=0
M A=60 kN
2. The next is to draw the bending moment diagram for the original beam. The moment diagram is drawn by calculating for the moment values at critical section of the beam. This is shown in the Fig 2.7(b) below.
3. The conjugate beam for the original beam is now drawn having the same span as the original beam. Since the original beam is simply supported, the conjugate beam will also be simply supported as from Table 2.1 and 2.2. The conjugate beam bean is illustrated in Fig 2.7(c)
4. The conjugate beam is now loaded with the MEI
diagram of the original beam, Fig 2.7(d)
5. Using the equation of equilibrium, the reaction of the conjugate beam is calculated for.
Under the conditions of equilibrium, the sum of the two reaction RA 'and RB'
should be
equal to the area of the MEI
diagram. By summing up the forces in the y-direction and
taking moment about B, the reaction of the conjugate beam is calculated for.
The load between A and C is 12
×60EI
×7=−30EI
The centre of gravity of the load is
32 | P a g e
C
Fig 2.7(b)60 kNm
A B
6mA B
Fig 2.7(c)
Fig 2.7(d)60 kNm
A B
CONJUGATE BEAM METHOD
3+ 23
× 3=5 m ¿B
Summing forces in the y-direction,
∑ f y=0
RB'=
30EI
6. Since the original beam is fixed at one end and free at the other end, it will not allow for rotation about the fixed end as well as translation in the y-direction at the fixed support, therefore it can neither develop slope and deflection at the fixed end. Conversely it will allow for rotation and translation in the y-direction at the free end and as so can develop slope deflection at the free end. The conjugate beam can thus develop shear and moments at the free end but not the fixed support, Fig 2.7(e).
7. Using the equilibrium condition the shear and the moment at various sections on the conjugate beam is calculated for. These values correspond to the slope and deflection of the original beam at their respective places.
Thus the slopes at B are given as,
Slope at B,
θB=S . F at B=RB'=
−60EI
rad (clockwise)
Deflection at B,
∆B=moment at B=−( 60EI
×5)=−300EI
(downward )
EXAMPLE
Determine the slope and deflection of the cantilevered beam with the load at the mid-span end in Fig 2.8(a).
SOLUTION
33 | P a g e
V B'
M B,
Fig 2.7(e)60 kNm
A B
25 kNm
Fig 2.8(a)
5m
CONJUGATE BEAM METHOD
1. The first step in solving a problem of this nature is to determine the support reactions of the original beam.
∑ f y=0
RA=25 ×5=125kN ∑ M A=0
M A−( 20 ×52
2 )=0
M A=625
2kNm
2. The next is to draw the bending moment diagram for the original beam. The moment diagram is drawn by calculating for the moment values at critical section of the beam. This is shown in the Fig 2.8(b) below.
3. The conjugate beam for the original beam is now drawn having the same span as the original beam. Since the original beam is simply supported, the conjugate beam will also be simply supported as from Table 2.1 and 2.2. The conjugate beam bean is illustrated in Fig 2.8(c)
4. The conjugate beam is now loaded with the MEI
diagram of the original beam, Fig 2.8(d)
34 | P a g e
6252
kNm
A B
6252
kNm
A B
Fig 2.8(b)
5m
Fig 2.8(c)
Fig 2.8(d)
CONJUGATE BEAM METHOD
5. Using the equation of equilibrium, the reaction of the conjugate beam is calculated for.
Under the conditions of equilibrium, the sum of the two reaction RA 'and RB'
should be
equal to the area of the MEI
diagram. By summing up the forces in the y-direction and
taking moment about B, the reaction of the conjugate beam is calculated for.
The load between A and B is 13
×6252 EI
× 5=−520.83EI
The centre of gravity of the load is 13
×5=53¿ B
Summing forces in the y-direction,
∑ f y=0
RB'=1225
2 EI6. Since the original beam is fixed at one end and free at the other end, it will not allow for
rotation about the fixed end as well as translation in the y-direction at the fixed support, therefore it can neither develop slope and deflection at the fixed end. Conversely it will allow for rotation and translation in the y-direction at the free end and as so can develop slope deflection at the free end. The conjugate beam can thus develop shear and moments at the free end but not the fixed support, Fig 2.8(e).
7. Using the equilibrium condition the shear and the moment at various sections on the conjugate beam is calculated for. These values correspond to the slope and deflection of the original beam at their respective places.
Thus the slopes at B are given as,
Slope at B,
θB=S . F at B=RB'=
−12252 EI
rad (clockwise)
Deflection at B,
∆B=moment at B=−( 12252 EI
×143 )=−8575
3 EI(d ownward)
35 | P a g e
V B'
M B,
6252
kNm
A B
Fig 2.8(e)
CONJUGATE BEAM METHOD
36 | P a g e
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