Conjugate Beam

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.


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


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


It can be seen that deflection at C is;

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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


¿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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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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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


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5

SI NO CONJUGATE BEAMORIGINAL BEAM

2

3

4

6

1

7

8

Table 2.2


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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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


( 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


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)


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.


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RA RBFig 2.3(a)

Fig 2.3(b)

40 kN

4m4m

A BCI 2I

80 kNm

4m4mRA ' RB'



diagram of the original beam, Fig 2.3(d)



should be



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


×40EI

×4= 80EI

The centre of gravity of the load is 23

× 4=2.67¿B


∑ 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

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RB'

RA '

40EI

80EI

Fig 2.3(d)

Fig 2.3(c)


But R A'+RB'

=240EI

RB'=240

EI−133.3

EI

∴RB'=106.7

EIunits



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,


−106.7EI

=26.6EI

rad

Deflection at C,


× 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


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.




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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)




should be





×135EI

× 3=202.5EI


×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


×3+9=10 fr om A


∑ 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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RA '=540

EI−214.43

EI

∴RA '=325.57

EIunits



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,


−202.5EI

=123.07EI

rad (clockwise )

Slope at E,


−202.5EI

−270EI

=−146.93EI

rad (anti−clockwise)

Deflection at C,


×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)


∆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



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PL

P

L 3L

A BC

3EI 1EI

Fig 2.5(a)

Fig 2.5(b)

Fig 2.5(c)

A B3EI 1EIC

RA '






should be




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


∑ M C=0

−RA '×3 L= P L2

2 EI× L

−3 RA'L=P L3

2 EI

∴RA '=−P L2

6 EIunits


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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



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


∑ 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








should be




The load between A and B is 12

×175EI

× 7=−12252 EI


×7=143

¿B


∑ f y=0

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Fig 2.6(b)175 kNm

A B

7m

Fig 2.6(c)

A B

Fig 2.6(d)175 kNm

A B


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).


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


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V B'

M B'

BA

175 kNm Fig 2.6(e)

3m3m

20 kN

A B

Fig 2.7(a)


∑ f y=0

RA=20 kN

∑ M A=0

M A−(20× 3 )=0

M A=60 kN







should be





×60EI

×7=−30EI

The centre of gravity of the load is

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C

Fig 2.7(b)60 kNm

A B

6mA B

Fig 2.7(c)

Fig 2.7(d)60 kNm

A B


3+ 23

× 3=5 m ¿B


∑ 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).



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

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V B'

M B,

Fig 2.7(e)60 kNm

A B

25 kNm

Fig 2.8(a)

5m



∑ f y=0

RA=25 ×5=125kN ∑ M A=0

M A−( 20 ×52

2 )=0

M A=625

2kNm





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6252

kNm

A B

6252

kNm

A B

Fig 2.8(b)

5m

Fig 2.8(c)

Fig 2.8(d)




should be




The load between A and B is 13

×6252 EI

× 5=−520.83EI


×5=53¿ B


∑ 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).



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)

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V B'

M B,

6252

kNm

A B

Fig 2.8(e)


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Conjugate Beam

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