Elementary Theory of Structures

8/4/2019 Elementary Theory of Structures

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PRINCIPLESOFVIRTUALWORK

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= . In obtaining the external strain energy, we first determine the internal moment asa function of the position x and the apply the strain energy in bending formula. In the case of

the cantilever beam shown in Fig 1.1, the internal moment is = . So that in applyingthe strain energy in bending formula, we obtain

= 2 =

() 20 =

16

Now equating the internal strain energy to the external strain energy and solving for the

unknown displacement, we obtain;

= 1

2= 1

6

= 3Even though this principle has a direct method of approach, it is limited to only a limited set

of problems. It should be realised that only one load may be applied to the structure at a time

as applying more loads would generate as many displacements as the applied load, whilst its

only possible to write just one work equation for the beam. It should further be noted that

only displacement under the applied force can be obtained, as the external work depends on

both the force and its corresponding displacement. A way out of this is by the use of the

principle of virtual work and the method of least work.

1.3PRINCIPLESOFVIRTUALWORKIf a series of external loads P are applied to a deformable structure of any given shape, it will

produce internal u loads at points throughout the structure. For the system to be in

equilibrium the internal loads must equate to the external loads by the equations of

equilibrium. The outcomes of these loadings will be the external displacements which will

occur at the point of the external loads P and internal displacements which will occur at the

position of the internal loadings. Displacements do not have to be elastic and may not be

related to the loads; but the displacements will have to be related by the compatibility of

displacement. That is to say that, a change in the external displacement should yield a

proportional change in the internal displacement; not necessarily by the same amount, but

should be proportionate. Thus if the external displacements are known, the internal

P

Figure 0-1.1

P


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displacement can then be easily found. Thus in summary the principle of work and energy

states that;

=

The principle of virtual work will now be developed from the concept above. In developing

the principle of virtual work, a structure of arbitrary shape is considered. It may be necessary

to determining the displacement at a point say X on the body caused by a series of forces ,, . Notice should be taken of the fact that the loads do not cause any movementin the support but can strain the material beyond its elastic limits. Since there is no external

load applied at the point X, in the direction of the displacement , the displacement can be

found by the introduction of a virtual load on the structure such that the virtual loadacts inthe same direction as the displacement. For ease of calculation the virtual force is assumedto have a unit value; = 1. The term virtual is being used to describe the introduced force

because it does not exist actually as part of the real loading. The virtual force creates an

internal virtual load in a representativeelement or fibre of the body. It is required that

and be related by the equation of equilibrium. The moment the virtual loads are applied, thebody is subjected to the loads, , . Point X would be displaced by an amount which will cause the element to deform an amount. As a consequence, the externalvirtual force and internal virtual force translates along and respectively and thus

performs external work of 1 on the body and on the element in the body. Noting thatthe external virtual work is equal to the internal virtual work done on all the elements of the

body, the virtual work equation can be written as

1 =

Workof

Internalloads

Workof

Externalloads

Figure1.0-1=1

A


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Where

= 1 = =

= = By making = 1, it can be seen that the solution for follows directly, as = .In the same regards, if the rotational displacement or slope of the tangent at a point on the

body is to be determined, a virtual couple moment with a unit magnitude is applied at thepoint. As a result, the couple moment will cause a virtual load in one of the elements ofthe body. Given that the real loads deform the element by an amount, the rotation can befound from the virtual work equation;

1 = Where

= 1 = = =

=

1.4METHODOFVIRTUALWORKASAPPLIEDTOBEAMSANDFRAMESThe principle of virtual work can be formulated for beam and frame deflection, this is

illustrated by considering the beam shown if Fig 1.3, in which the displacement at the point A

is to be determined. To determine the displacement , a virtual unit load is acting in the

direction of the displacement is applied to the beam at the point A. the internal virtual

moment is then determined by the method o section at an arbitrary location , from the leftsupport. When the real load acts on the beam it is displaced by . If the load causes an elasticmaterial response, then the element deforms or rotates = (/). M is the internalmoment at caused by the real loads. As a consequence the external virtual work done by theunit load is 1 and the internal virtual work done by the moment m is = (/).Summing the effect on all the elements along the beam requires integration. This

becomes;

1 =


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Where

1 = =

= = = =

Similarly to determine the tangent rotation or slope angle at a point on the beams elastic

curve, a unit couple moment is applied at the point, and the corresponding internal moments

have to be determined. Since the work of the unit couple is 1., then

1 =

When using the above formulas, it is worth noting that the definite integral on the right hand

side represents the amount of virtual strain that is stored up in the beam. Where concentrated

forces or couple moments act on the beam, or the distributed moment is not continuous, a

single integration cannot be performed for the whole span. Separate coordinates will haveto be chosen within regions that have no discontinuity of loadings. It is important to note that,

it is not necessary for each of the to have the same origin; however, the selected fordetermining the real moment M in a particular region must be the same as selected fordetermining the virtual moment and within the same region.

1.5PROCEDUREFORANALYSISThe following procedure may be used to determine the slope and displacement at appoint on

the elastic curve of a beam using the method of virtual work.

1. Apply a unit load on the beam at the point and in the direction of the desireddisplacement, if it is the slope that is to be determined, place a unit couple moment at

that point

2. Establish appropriate coordinates valid within the regions of the beam where there isno discontinuity of real or virtual loading

3. Calculate the internal moment and as a function of each coordinate, when thevirtual load is in place and the all the real loads have been removed.

4. Assume and acts in the positive direction.5. With the same coordinates as those established by and , determine the internal

moments M caused only by the real loads

6. Apply the virtual work equation to determine the desired displacement or rotation7. If the algebraic sum of all the integral for the entire beam or frame is positive then

slope and rotation is in the same direction as the virtual unit load or unit couple


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moment respectively. If a negative value is realised, the direction of the displacement

and the slope is in the opposite direction to that of the virtual load or virtual couple

moment.

1.6SOLVEDEXAMPLESThe following examples would be used to illustrate the application of the principle of virtual

work in the determination of the slope and displacement in a beam or frame.

EXAMPLE 1

Determine the displacement at point B of the cantilever beam shown in Fig1.4a

SOLUTION

The question requires that the displacement at be determined, from the loading of thecantilever, the value of the concentrated load at is zero. Thus to find the displacement at,a virtual load of unit magnitude is applied at the point, Fig 1.4b. From inspection it can beseen that the loading on the beam is continuous, as such a single coordinate can be used inthe determination of the virtual moment. The origin of the coordinate is taken from, thusignoring the reaction s at.

Thus the internal moment for the virtual force is = 1

The internal moment for the real load is formulated using the same coordinate, whichgives;

= 2

= 20 2

Figure 1.4a

A B

20kN/m

7m

1kN

Figure 1.4b

7m

A B


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= 20

2 = 10

The two moments are then substituted into the virtual work equation. Thus the displacement

at becomes

1 =

= (1)(10

)

1 = (10)

1 = 10

4

1 = 10(7)

4

= 6.010

EXAMPLE 2

Determine the slope at point B of the cantilever beam shown in Fig1.5a

SOLUTION

The question requires that the slope at be determined, this is achieved by applying a unitvirtual couple at the point. For this question because the loading is not inform throughout,two x coordinates are require to determine the total virtual strain energy in the beam.

Coordinate will account for the strain energy in the beam segment, and the coordinate will account for the strain energy in the beam segment BA. Using the method of sectionthe internal moment at each of the segment is calculated for.

Figure 1.5a

C

8kN

4m

A B

4m

CB

1 kNm

A


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The internal moment for the beam segment CB is calculated as;

= 0The internal moment for the beam segment BA is calculated as;

= 1Using the same coordinates the internal moment for the real load is calculated for


= 81The internal moment for the beam segment BA is calculated as;

= 8(4 + 2)

The two moments are then substituted into the virtual work equation. Thus the displacement

at becomes1 =

= ()(1) + ()(2)

= 160

EXAMPLE 3

Determine the displacement at point D of the cantilever beam shown in Fig1.6a. Take =200, = 300(10).

Figure 1.4b

4

8kN

4

8kN

Figure 1.6a

CB

25kN

A

4 4 D5

25


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SOLUTION

In determining the displacement at D, the beam is loaded with a virtual load at D, Fig 1.6b.

By inspection three coordinates

, ,are required to determine the total virtual

strain energy in the beam. The three coordinated covers regions where there are no

discontinuities in either the real loading or the virtual loading. Using the method of sections

the internal moments for the virtual loading is calculated below.

The internal moment for the beam segment DC is calculated as;

= 1The internal moment for the beam segment DB is calculated as;

= 0.75 + 5

The internal moment for the beam segment AB is calculated as;

= 0.65

Using the same coordinates the internal moment for the real load is calculated for


= 0

Figure 1.6b

1kN

0.65kN 1.65kN

1kN

5

1kN

+ 5

1.75kN

1kN

30kN

25

11.87kN 18.13kN


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The internal moment for the beam segment BA is calculated as;

= 18.13

The internal moment for the beam segment AB is calculated as;

= 25 + 11.87The two moments; virtual moments and real moments, are then substituted into the virtual

work equation. Thus the displacement at becomes

1 =

= (1)(0) +

(0.75 +5)(18.13)

+ (0.65)(25+11.87)

=1309.22

18.13kN

30kN

1kN


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CONJUGATEBEAMMETHOD

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From Fig 1.0 the change in slope between A and C is given by;

= The slope at C is thus found as

= It can be seen that,

=

= 1

Therefore

=

It can be seen that deflection at C is;

= =

C

WW

BA

L

x

Fig2.1(a)

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x

B

A

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C2

C BA

Fig2.1(c)


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CONJUGATEBEAMMETHOD

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= ... = ( )

Lets look at a case of a similar beam loaded with the

diagram and having the span equal to

the above considered beam. The reaction at the point A would be given by,

=

= And the shear force at C would be given by,

= =

It can be observed that the slope at C, in the beam earlier considered is equal to the shearforce in the latter beam considered which was loaded with the

diagram. And that the

deflection at C in the earlier beam is equal to the bending moment in the latter beam with the

diagram as its load. The latter beam considered is the conjugate beam. The aboutdiscussion 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 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

equally zero. These conditions exist in a beam with a free end, as such; a fixed support in an


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CONJUGATEBEAMMETHOD

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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 its only a fixed support that

resists both rotation and translation. In this case a free end/support in an original beam willgive 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

2.3ILLUSTRATIONSThe 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 bears 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 () of thebeam 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).

The other dissimilarity between the original beam and the conjugate beam is in their supportreactions. The support reactions in the original beam are not always the same as that in the


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CONJUGATEBEAMMETHOD

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ORIGINALBEAM CONJUGATEBEAMSINO

1

0 = 0Simplysupportedor

Rollerend

0 = 0Simplysupportedor

Rollerend

2

0 = 0Hingedend

0 = 0Hingedend

3 = 0 = 0

Fixedend

= 0 = 0Freeend

4

0 0Freeend

0 0Fixedend

0

= 0

0

= 0

5

Interiorsupport Interiorhinge

Interiorsupport

6

0 0

Interiorhinge

0 0

Table2.1


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CONJUGATEBEAMMETHOD

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8

7

1

6

4

3

2

ORIGINALBEAM CONJUGATEBEAMSINO

5

Table2.2


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CONJUGATEBEAMMETHOD

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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 thedeflection 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 theother, 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), hencedeflection 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 bothrotation 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.

2.4PROCEDUREIn solving problems related to the conjugate beam method, the following step should be

followed,

1. Solve for the reactions of the original beam; that is determining the values of thereaction for the original beam.

2.

Draw the bending moment diagram for the original beam


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CONJUGATEBEAMMETHOD

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3. Draw the conjugate beam from the original beam, having in mind that the length ofthe 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 diagram. The load is assumed to be

uniformly distributed over the conjugate beam and is directed upwards when the

diagram is positive and directed downwards when the 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 theconjugate beam, this will correspond to the slope and deflection in the original beam.

2.5SOLVEDEXAMPLESEXAMPLE 2.1 Determine the slope at A, B, C (, , ) and the deflection at C in the

beam shown in Fig 2.2(a)

SOLUTION

Applying the steps outlined above;

1. Determining the reaction of the original beam. = 0

+ = 35 = 0( 6) (353) = 06 = 105

= 17.5 = 17.5

Fig2.2(a)

CBA

3m 3m

35


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CONJUGATEBEAMMETHOD

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2. Drawing the bending moment diagram for the original beam. The moment diagram isdrawn 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 theoriginal 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 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 and should beequal to the area of the

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 1

2 52.5 6 =

157.5

Summing forces in the y-direction,

= 0 + = 157.5 (1) = 0 6 = 157.5 3

6 = 472.5

Fig2.2(b)

Fig2.2(d)

Fig2.2(c) 3m 3m

52.5

3m 3m

52.5


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CONJUGATEBEAMMETHOD

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= 78.75 + = 157.5

= 157.5 78.75 = 78.75 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 theconjugate beam is calculated for. These values correspond to the slope and deflection ofthe original beam at their respective places.

Thus the slope at A and B are given as,

Slope at A,

= . = = 78.75 ()Slope at B,

= . = = 78.75 ( )

Slope at C,

= . = 78.75 78.75 = 0

Deflection at C,

The centre of gravity of the load is at the mid-span

52.5

3m 3m

Fig2.2(e)


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CONJUGATEBEAMMETHOD

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= = 78.75 3 78.75 1.5

= 118.13 ()

EXAMPLE 2.2

Determine the slope at A, B, C (, , ) and the deflection at C in the beam shown inFig 2.

SOLUTION

Applying the steps outlined above;

1. Determining the reaction of the original beam.

= 0

+ = 40 = 0( 8) (404) = 08 = 160 = 20 = 20

2. Drawing the bending moment diagram for the original beam. The moment diagram isdrawn by calculating for the moment values at critical section of the beam. This is shownin the Fig 2.3(b) below.


be simply supported as from Table 2.1 and 2.2. The conjugate beam bean is illustrated inFig 2.3(c)

Fig2.3(a)

Fig2.3(b)

2II C

BA

4m 4m

40

80


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CONJUGATEBEAMMETHOD

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4. The conjugate beam is now loaded with the diagram of the original beam, Fig 2.3(d)



moment about B, the reaction of the conjugate beam is calculated for.

The load between A and C is 1

2 80

4 =160

The centre of gravity of the load is

4 + 13 4 = 5.33The load between A and C is 1

2 40 4 =

80

The centre of gravity of the load is 23 4 = 2.67


= 0 + = 160 +

80 =

240

= 0 8 = 160 5.33 +

80 2.67

8 = 1066.40

= 133.30

80 40

Fig2.3(d)

4m 4m

Fig2.3(c)


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CONJUGATEBEAMMETHOD

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+ = 240 = 240

133.3

= 106.7 6. Since the original beam is simply supported, it allows for rotation at the support but does




7. Using the equilibrium condition the shear and the moment at various sections on theconjugate 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,

= . = = 133.30 ()Slope at B,

= . = = 106.7 ( )Slope at C,

= . =133.30

106.7 =

26.6

Deflection at C,

= = 133.30 4 160

43

= 319.87 ()

EXAMPLE 2.3

Determine the rotations at A, B, C, E

(, , , , )and the deflection at C, and E

( , )in the beam shown in Fig 2.4(a).

80 40

Fig2.3(e)


24/112

CONJUGATEBEAMMETHOD

24|P a g e

SOLUTION

1. The first step is to determine the reaction of the original beam. = 0

+ = 60

= 0( 12) (603) = 012 = 180 = 15 = 60

2. The next is to draw the bending moment diagram for the original beam. The momentdiagram is drawn by calculating for the moment values at critical section of the beam.

This is shown in the Fig 2.4(b) below.



Fig 2.4(c)


Fig2.4(b)

135

Fig2.4(a)

ED I

2I 2II C

BA

60

3m3m 3m 3m

I2I 2II

Fig2.4(c)

3m3m 3m 3m


25/112

CONJUGATEBEAMMETHOD

25|P a g e


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 1

2 135 3 =

202.5

The centre of gravity of the load is 23 3 = 2

The load between C and E is

= 270

The centre of gravity of the load is = 2.53 + 3 = 5.53The load between E and B is 1

2 45 3 =

67.5

The centre of gravity of the load is 13 3 + 9 = 10


= 0 + = 202.5 +270 +

67.5 =

540

= 0 8 = 202.5 2 +

270 5.53+

67.5 10

12 = 2573.1

=214.43

Fig2.4(d)

45

22.5

135

67.5


26/112

CONJUGATEBEAMMETHOD

26|P a g e

+ = 240 = 540

214.43

= 325.57 6. Since the original beam is simply supported, it allows for rotation at the support but doesnot permit translation in the y-direction and as so can develop slopes at the support but





Thus the slopes at A, B, C, D and E are given as,

Slope at A,

= . = = 325.57 ()Slope at B,

= . = = 214.43 ( )Slope at C,

= . = 325.57 202.5 =

123.07 ()

Slope at E,

= . = 325.57 202.5

270

= 146.93 ( )Deflection at C,

Fig2.4(e)

45

22.5

135

67.5


27/112

CONJUGATEBEAMMETHOD

27|P a g e

= = 325.57 3202.5

33

= 774.21 ()Deflection at E,

= = 325.57 9

202.5

33 +6

270 3.47

= 575.10 ()

EXAMPLE 2.4

Find 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 ofthe original beam.

= 0 + =

= 0( 3) ( 4) = 03 = 4

= 43 = 3 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.


1EI3EI

C

BA

3L L

P

Fig2.5(a)

Fig2.5(b)


28/112

CONJUGATEBEAMMETHOD

28|P a g e



Fig 2.5(c)



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

3 3 =

2


+ 13 3 = The load between B and C is

12 =2The centre of gravity of the load is 1

3 =3

Summing forces in the y-direction, = 0 3 =

2

3 = 2

C

1EI3EI BA

Fig2.5(c)

C

1EI3EI BA

Fig2.5(d)

3


29/112

CONJUGATEBEAMMETHOD

29|P a g e

=

6 6. Since the original beam is simply supported, it allows for rotation at the support but does


cannot develop deflections at the support. The conjugate beam can thus develop shear atthe support but cannot develop moments at the support, Fig 2.5(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 C,

= . = =

6 +

2 +

2 =5

6 ()

Deflection at C,

= =

6 4+2 2+

2

23

= 23 ()

EXAMPLE 2.5

Determine the slope and deflection of the cantilevered beam with the load at the free end in

Fig 2.6(a).

C1EI3EI BA

Fig2.5(e)

3

Fig2.6(a)

A B

25

7m


30/112

CONJUGATEBEAMMETHOD

30|P a g e

SOLUTION


= 0 = 25

= 0 (257) = 0 = 175

2. The next is to draw the bending moment diagram for the original beam. The momentdiagram is drawn by calculating for the moment values at critical section of the beam. Thisis shown in the Fig 2.6(b) below.



Fig 2.6(c)


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

and

should be

BA

175 Fig2.6(b)

BA

Fig2.6(c)

7m

BA

175 Fig2.6(d)


31/112

CONJUGATEBEAMMETHOD

31|P a g e

equal to the area of the 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

12 175 7 = 12252 The centre of gravity of the load is 2

3 7 =143

Summing forces in the y-direction, = 0 = 1225

2

6. Since the original beam is fixed at one end and free at the other end, it will not allow forrotation 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,

= . = = 12252 ()Deflection at B,

= = 12252 143 =

85753 ()

EXAMPLE 2.6

Determine the slope and deflection of the cantilevered beam with the load at the mid-spanend in Fig 2.7(a).

BA

175 Fig2.6(e)


32/112

CONJUGATEBEAMMETHOD

32|P a g e

SOLUTION


= 0

= 20

= 0 (203) = 0

= 602. 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.




Fig 2.7(c)

4. The conjugate beam is now loaded with the diagram of the original beam, Fig 2.7(d)C

Fig2.7(a)

BA

20

3m 3m

BA

60 Fig2.7(b)

BA

6m

Fig2.7(c)

BA

60 Fig2.7(d)


33/112

CONJUGATEBEAMMETHOD

33|P a g e


diagram. By summing up the forces in the y-direction and takingmoment about B, the reaction of the conjugate beam is calculated for.

The load between A and C is 12

60 7 =

30


3 + 23 3 = 5Summing forces in the y-direction,

= 0 = 30

6. Since the original beam is fixed at one end and free at the other end, it will not allow forrotation about the fixed end as well as translation in the y-direction at the fixed support,








Slope at B,


= = 60

5 = 300

()

, BA

60 Fig2.7(e)


34/112

CONJUGATEBEAMMETHOD

34|P a g e

EXAMPLE

Determine the slope and deflection of the cantilevered beam with the load at the mid-span

end in Fig 2.8(a).

SOLUTION


= 0 = 25 5 = 125 = 0

205

2 = 0

= 6252 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.



Fig 2.8(c)

25

Fig2.8(a)

5m

BA

6252

Fig2.8(b)

Fig2.8(c)

5m


35/112

CONJUGATEBEAMMETHOD

35|P a g e




moment about B, the reaction of the conjugate beam is calculated for. The load between A and B is1

3 6252 5 =

520.83

The centre of gravity of the load is13 5 =

53


= 0 = 12252 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,








,

BA

6252

BA

6252

Fig2.8(d)

Fig2.8(e)


36/112

CONJUGATEBEAMMETHOD

36|P a g e

Slope at B,


= = 12252 143 =

85753 ()


37/112

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38/112

SLOPEDEFLECTIONMETHOD

38|P a g e

dimensioned member each node can have at most three linear displacements are three

rotational displacements; and in two dimensions, each node, each node can have at most two

linear displacements and one rotational displacement. Also nodal displacement may be

restricted by the supports or assumptions based on behaviour of the structure.

3.3SLOPEDEFLECTIONEQUATION

As stated earlier, there are two methods that are used to ensure the safety of a structure; the

force method and the load-displacement method. The force method of analysis is limited to

structures that are not highly indeterminate. This because much work is required to set up the

compatibility equation and also each equation written involves the entire unknown, making it

difficult to solve the resulting set of equations. By comparism, the slope-deflection method is

that not very involving. As would be seen later on, the slope deflection method requires less

work both to write the necessary equation for the solution of a problem and to solve these

equations for the unknown displacement and associated internal loads.

The method was originally developed by Heinrich Manderla and Otto Mohr for studying

secondary stresses in trusses. G. A. Maney refined the method in 1915 and applied it to the

analysis of indeterminate beams and frames.

The method is so named because it relates the unknown slope and deflection to the applied load

on the structure. To develop the general form of the formula, the span AB of the continuous

beam in Fig3.1 would be considered, with an arbitrary loading and a constant flexural rigidity,

EI. The internal moments of the beam,

and

will be related in terms of the three

degree of freedom that the beam has, namely; its angular displacement and , and its lineardisplacement which would be caused by a relative settlement between the supports. Moments

and angular displacement that acts clockwise on the span are going to be considered positive

and downward linear displacement too are to be considered positive.

The principle of super positioning is going to be used to determine the slope deflection

equation. This is going to be achieved by considering separately the moment developed at each

support due to each of the displacements, , and

, and the loads.

A B

P

L

EIisconstant

Figure 3.1


39/112


39|P a g e

3.3.1AngularDisplacements

Assuming the node at A to rotate through an angle as the node at the other end remainsfixed and unyielding. And the moment caused by the displacement is being calculated by the

conjugate beam method. It can be seen that the shear at acts downward on the beam since theangular displacement at A is clockwise, Fig 3.2. Because the deflection at each ends of theactual beam is zero, the sum of moment about those points in the conjugate beam should also

be zero. These leaves as with;

= 012

3

12

23 = 0

= 01

2

3 1

2

2

3= 0

Evaluating the above expression will yield the following load-displacement relationships.

= 4 = 2

In the same way, if the nodes at B is rotated through an angle whilst the node at is heldfixed, and the moment caused by the angular displacement calculated by the conjugate beam

method. It would yield a load-displacement relationship of the magnitudes;

= 4 = 2

`

LRealBeam

(a)

LRealBeam

(b)

ConjugateBeam

`

= (c)Figure 3.2


40/112


40|P a g e

3.3.2RelativeLinearDisplacement

Also, if the node at B is displaced compared to A such that the beam rotates in a clockwise

manner about A; whilst preventing the rotation of both of the nodes, then equal and opposite

shear reactions will be developed in the beam. From the conjugate beam analysis, the moment

is the beam can be related to the displacement of the member at B. Because of the fixity at thenodes of the beam, the nodes of it conjugate beam would be free, meaning no moment in its

conjugate beam. But because of the displacement of the real beam at B, the moment of the

conjugate beam taken about the node at B should be equal to the displacement of the real beam

at B. Thus summing up the moment about B in the conjugate beam method produces;

= 012

23

3

12

13

23 = 0

Where

= = = 6

3.3.3FixedEndMoments

The previous section looked at the moments that are developed in the beams as a result of the

displacements that occur in the beam. As it would be noted the displacements are brought

about by the loads that are acting on the beam. To come out with the slope deflection formula,

the loading of the span is going to be converted to moments that are acting at the nodes of the

beam. These will the n be used to develop the load-displacement relation as before. This can be

achieved simply by finding reaction moments that each load develops at the nodes.

Considering the fixed ended beam in Fig3.3a, subject to a concentrated load at the mid part of

its span, the conjugate counterpart of which shown if Fig3.3b. Since the slope at each end of

the beam is required to be zero, summing up forced in the conjugate counterpart and equating

to zero would yield,

= 012

4 2

12

= 0

=

8

The reaction moment as so calculated is referred to as the fixed end moment. It should be noted

that due to the sign convention, it is negative at the nodes A and positive at the nodes B. The

fixed end moments for various span loading and end condition can be found at the end

A B

P

M M

V V2

2

4


41/112


41|P a g e

of the course material. Taking it that the fixed end moment have been calculated specific

purposes, e.g. in Fig 3.3a, then we have;

= =

3.3.4SlopeDeflectionEquationIf the end moments due to the displacement and the loadings are summed up for each of the

nodes, the end moments can be summarised as;

= 2 2 + 3

And

= 2 2 + 3 +

Since the above equation are similar the results can be expressed as a single equation, referring

to the end of the span at the near end as N, and the end of the span at the far end as F. The

generalised equation now becomes;

= 2 2 + 3 Where

= Internal moment in the near end of the spam; moment is positive when acting on thespan.

, = Modulus of elasticity of material and span stiffness = , = Near-end and far-end slopes or angular displacements of the span at the support;

angles are measured in radian and are positive clockwise.

= Span rotation due to displacement, thats = ; this is measured in radians andpositive clockwise.

= Fixed end moment at the near-end support; refer to table at the back of coursematerial.

The slope deflection equation developed can be used to analyse any type of beam or frame

irrespective of the end or support condition. This is achieved by substituting into the equation

the variable as the case may be for each beam or frame.

3.4PROCEDUREFORANALYSIS

The procedure of analysis for the slope deflection equation can be summarised into these basic

steps;

1. Determine the degree of freedom at the various nodes of the beam or frame

Figure3.3(a)

b


42/112


42|P a g e

2. Write the slope deflection equation for each of the nodes at the ends of each of thespans of the beam or frame

3. Write the equilibrium equation of the unknown degree of freedom for the structure4. Solve for the unknown degree of freedoms from the equilibrium equations5.

Solve for the end span moment using the calculated degree of freedom

6. Solve for the reactions at the various supports.

3.5SOLVEDEXAMPLES

The following example will be used to illustrate the application of the slope deflection equation

in the analysis of beams and frames.

Example 1

Draw the shear force and moment diagrams for the beam shown in Fig.3.3a

`

SOLUTION

The beam in question has only one degree of freedom at the support B. The two degree of

freedom is the slope at B.

Thus the slope deflection equation would be written in terms of the slope at B. The fixed end

moment on the span BC is calculated for. The fixed end moment for the node B on the span BC

is calculated as;

=

12

= 20 4

12

= 26.67

The fixed end moment for the node B on the span BC is calculated as;

Figure 3.3a

A

B6 4

20kN/m

C

A

B6 4

C

Figure 3.3b


43/112


43|P a g e

=

12

= 20 4

12

= 26.67The slope deflection equation for the span can then be written. It should be noted that the

slopes at the support A and C are zero, since they are all fixed. It should also be noted that

there are no displacements along the length of the beam. The general formula for slope

deflection is;

= 2 2 + 3 Thus the slope deflection formula for the node A on the span AB becomes;

= 2 2 + 3

= 2 6 20 + 30 0

= 3

The slope deflection formula for the node B on the span AB becomes;

= 2 2 + 3 + = 2 6 2 + 0 30 + 0

= 3 2

Thus the slope deflection formula for the node B on the span BC becomes;

= 2

2 + 3

= 2 4 2 + 0 30 26.67

= 2 2 26.67

The slope deflection formula for the node C on the span BC becomes;

= 2

2 + 3

+

= 2 4 0 + 30 + 26.67


44/112


44|P a g e

= 2 + 26.67

The equilibrium equations for the beam are;

0And

+ = 0Solving the equilibrium equations will yield us;

3 2 + 2 2 26.67 = 0

23 + = 26.67

53 = 26.67

= 16.0

= 16.0

Substituting the values into the slope deflection equations gives;

= 3 = 3

16.0 = 5.33

= 3 2 = 3 2

16.0 = 10.67

= 2 2 26.67 = 2 2

16.0 26.67 = 10.67

= 2 + 26.67 =

2

16.0 + 26.67 = 34.67

The reactions at the support are determined using the equilibrium equations. The reactions for

the span AB are;

+ =

5.33 + 10.676 = 2.67

The reactions for the span BC are

+

2 =

10.67 + 34.67

6 +

20 4

2 = 64

The reactions are as follows;


45/112


45|P a g e

= 2.67 = 66.67 = 64

Example 2

Draw the shear force and moment diagrams for the beam shown in Fig.3.4a

SOLUTION

The beam in question has only three degree of freedom, being the slope at A, B and C

Thus the slope deflection equation would be written in terms of the slope at A, B and C. The

fixed end moment on the span BC is calculated for. The fixed end moment for the node B on

the span BC is calculated as;

=

12

= 8 512 = 16.67


=

12

= 8 5

12

= 16.67

Figure 3.4a

A

B

4 5

8kN/m

C

Figure 3.4b

A

B

4 5C


46/112


46|P a g e

The slope deflection equation for the span can then be written. It should be noted that the

slopes at the support A and C are zero, since they are all fixed. It should also be noted that

there are no displacements along the length of the beam. The general formula for slope

deflection is;


= 2 2 + 3

= 2 4 2 + 30 0

= 2 2 + 1 The slope deflection formula for the node B on the span AB becomes;

= 2 2 + 3 +

= 2 4 2 + 30 + 0 = 2 2 + 2

Thus the slope deflection formula for the node B on the span BC becomes;

= 2 2 + 3

= 2 5 2 + 30 16.67

= 25 2 + 16.67 3

The slope deflection formula for the node C on the span BC becomes;

= 2 2 + 3 +

= 2 5 2 + 30 + 16.67 = 25 2 + + 16.67 4


= = 0


47/112


47|P a g e

And


2 2 + + 25 2 + 16.67 = 0

2 2 +2 +

25 2 +

25 = 16.67

2 +95 2 +

25 = 16.67 5

Making in 1the subject

2 2 + = 0

2 2 + 2 = 0

= 2

=

2

Substituting into 5

2 2 +

2215 2 +

25 = 16.67

3120 +25 = 16.67 6

Solving simultaneously 4 and 6 yields

= 18.52

Substitute the value of the slope at B into the equation 1 to determine the slope at A.

2 2 + = 0

2 = 18.52

= 9.26


48/112


48|P a g e

Substitute the value of the slope at B into the equation 4 to determine the slope at C

25 2 + + 16.67 = 0

25 2 +

18.25 = 16.67

2 + 18.25 =41.68

2 = 60.20

= 30.10


= 2 2 + = 2 2

9.26 +

18.52 = 0

= 2 2 + = 2 2

18.52

9.26 = 13.89

= 25 2 + 16.67 = 25 2

18.52 +

30.10 16.67 = 13.89

= 25 2 + + 16.67 = 25 2 30.10 + 18.52 + 16.67 = 0The reactions at the support are determined using the equilibrium equations. The reactions for

the span AB are;

+ =

0 + 13.894 = 3.47

The reactions for the span BC are

+ + 2 = 10.67 +05 + 8 42 = 16The reactions are as follows;

= 3.47 = 19.47 = 16

Example 3


49/112


49|P a g e

Determine the moments for the beam shown in Fig.3.5a.

SOLUTION

The beam in question has only one degree of freedom. The o degree of freedom is the slope at

B.

Thus the slope deflection equation would be written in terms of the slope at B. The span AB is

the only span to be considered, as the moment , can be calculated from statics.The slope deflection equation for the span can then be written. It should be noted that the

slopes at the support A is zero, since it is fixed. The general formula for slope deflection is;


= 2 2 + 3

= 2 5 20 + 30 0

= 25 The slope deflection formula for the node B on the span AB becomes;

= 2 2 + 3 +

= 2 3 2 + 0 30 + 0 = 2

3 2


Figure 3.5a

A

B5 3

C

A

B

5 3 CFigure 3.5b

20kN/m

20kN/m


50/112


50|P a g e

0And


23 2 + 20 3 = 0

23 2 = 60

= 45


= 25 = 25

45 = 18

= 23 2 = 23 2

45 = 60

3.6ANALYSISOFFRAMESWITHOUTSIDESWAY

Properly restrained frame will neither sway to the left or the right. Also frames that are

symmetrically loaded will not sway eve if they are unrestrained. For the above mentioned

cases, there term in the slope deflection equation is zero. This is due to the fact that bendingdoes not lead to linear displacement of joints.

This will be illustrated with the following examples.

EXAMPLE 4

Determine the moment at each joint of the frame in Fig 3.6a

4m

3kN/m


51/112


51|P a g e

SOLUTION

For this problem, three spans are going to be considered; span AB, BC and CD. Because the

frame is fixed at A and D, the slope at those points too will be zero. The slope deflection

equation for the various spans will thus be calculated for.

Since the frames loading is on only span BC, the fixed end moment will also be calculated for

that span only. The fixed end moments for that span are;

=

12

= 3 3

12 = 2.25


=

12

= 3 3

12 = 2.25

The slope deflection equations for the span AB are:

= 2 2 + 3

= 2 4 0 + 30 0

= 2

= 2 2 + 3 +

= 2 4 2 + 0 30 + 0

= 2 2


52/112


52|P a g e

The slope deflection equations for the span BC are:

= 2 2 + 3

= 2 3 2 + 30 2.25

= 23 2 + 2.25

= 2 2 + 3 +

= 2 3 2 + 30 + 2.25

= 23 2 + + 2.25

The slope deflection equations for the span CD are:

= 2 2 + 3

= 2 4 0 + 30 0

= 2 = 2 2 + 3

+

= 2 4 2 + 0 30 + 0

= 2 2The equilibrium equations for the beam are;

0And

+ = 0 + = 0

The equilibrium equations are now solve for to determine the values of the slopes,

+ = 0


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2 2 + 23 2 + 2.25 = 0

1.83 + 0.67 = 2.25

+ = 0

23 2 + + 2.25 + 2 2 = 0

1.83 + 0.67 = 2.25Solving the two equations simultaneously yields;

= = 1.44

The moments at the various joints are;

= 2 = 2

1.44 = 0.72

= = 1.44 = 1.44

= 23 2 1.44 +

1.44 2.25 = 1.44

= 23 2 1.44 + 1.44 2.25 = 1.44

= = 1.44 = 1.44

= 2 = 2

1.44 = 0.72


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55/112

MOMENTDISTRIBUTIONMETHOD

71|P a g e

4.3.3DISTRIBUTEDMOMENTS: The unbalanced moment at a joint is the differencebetween the fixed end moment and the actual moment that causes a joint to rotate. The

rotation produces a twist at the ends of the members resulting in a change in their moments.

The member resists the twist caused by the unbalanced moment by developing a resisting

moment. This continues until an equilibrium condition is attained - when the resistingmoment is equal to the resisting moment, and the sum of the two moments equals zero. The

resisting moment developed in the member is referred to as the distributed moment.

4.3.5CARRYOVERMOMENTS: The distributed moments in the ends of the membercauses moments in the other ends of the member, which is assumed to be fixed; these are

referred to as carry over moments.

4.3.6CARRYOVERFACTOR: the carry over factor is the ratio between the momentsproduced at a joint at one end of a beam to the moment produced at the other joint at the other

end of the beam. When moment is applied to one end of a beam, the effect of the beam is feltat the other end of the beam. The ratio of the magnitude of the beam felt at the point of

application to that felt at the far end of the beam is what we are referring to as the carry over

factor. The carry over factor is dependent on the type of joint or support. Two type of beam

connect ions will be considered:

1. Where the beam is fixed at one end and simply supported at the other end.2. Where the beam is fixed at both ends of the beam.

4.4CARRYOVERFACTORFORBEAMSFIXEDATONEENDANDSIMPLYSUPPORTEDATTHEOTHEREND.Considering the beam AB in Fig 4.1 fixed at A and simply supported at B, with a clockwise

moment applied at the fixed support B.

Given that;

=

= (. . , ) =

Xx

B

A

R

R

Figure 4.1


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Because the beam is not subject to any external loads, the reaction of the beam to the moment

applied should be equal and act in the opposite direction to each other.

Taking moment about the point A,

(. ) = 0(. ) = + ()Considering a point X, which is x distance away from the support A. Taking moments about

the point X would yield;

= (. )From the course module BT 254, we learnt that the equation of a deflected beam is given by;

=

Therefore

= (. )

The slope was found by integrating the differentiating the equation. Thusdifferentiating the above equation would yield;

= (. ) 2 +

where is the constant of integration, we know that the slope and deflection at a fixedsupport is zero. Therefore when = 0, = 0 and = 0.

= (. ) 2 ()

Integrating the equation above on more time gives

. = .

2 6 +

where is the constant of integration, we know that the slope and deflection at a fixedsupport is zero. Therefore when = 0, = 0 and = 0,and

. = .

2 6 ()

When

= , = 0, therefore substituting it into equation

()


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0 = .

2 6

6

= .

2

. = 3Substituting the value of. in equation ()3 = + 2 = But =

2 = =

12.1

Therefore the carry over factor is one-half.

It can be realised from equation (ii) that

= (. ) 2

The slope at B is found by substituting into the above equation =

. = (. ) 2 . = 3

. = (. ) 3.

2

. = . 2 =

4 =

2

= 4

The negative sign is an indication that the tangent at B makes an angle with the beam AB in

the anticlockwise direction.

=4

= 4.

. 2


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4.5 CARRY OVER FACTOR FOR A BEAM SIMPLY SUPPORTED ON BOTHSIDES.Considering the beam AB in Fig 4.2 fixed at A and simply supported at B, with a clockwise

moment applied at the fixed support B.

Given that;

= = (. . , )

=

Because the beam has a simply support at both ends, it cannot develop fixed end moments.

But the reactions should be equal and opposite in other to maintain statical equilibrium

Taking moment about the point A,

(. ) = 0

(. ) = ()Considering a point X, which is x distance away from the support A. Taking moments about

the point X would yield;

= (. )From the course module BT 254, we learnt that the equation of a deflected beam is given by;

=

Therefore

= (. )

The slope was found by integrating the differentiating the equation. Thusdifferentiating the above equation would yield;

=

2 + ()

where is the constant of integration. Integrating equation one more time produces,

Xx

B

A

R

R

Figure 4.2


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. =

6 + +

where is the constant of integration. We know that the slope and deflection at a fixedsupport is zero. Therefore when

= 0,

= 0and

= 0,and

. =

6 + ()

When = , = 0, therefore substituting it into equation ()

0 =

6 +

=

6=

6

Substituting the value of in equation ()

= 2 +

6 =

2 +

6

=

2 +6

The slope at B is found by substituting into the above equation =

. = 2 + 6 = 2 + 6 = 3

= 3

The negative sign is an indication that the tangent at B makes an angle with the beam AB in

the anticlockwise direction.

=

3

= 3. .3

4.6STIFFNESSFACTORThe stiffness factor is the moment needed to rotate the end of the beam through a unit angle

without causing the far end to move or translate while still moving the beam end. From the

above discussion, we realised that the moment on a beam with one fixed end and the other

simply supported is;


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= 4.

Therefore the stiffness factor for such a beam (given =1) is

= 4 .4

In the same manner the moment on the beam with both end simply supported was found as;

= 3.

The stiffness factor for such a beam (given =1) is

=3

.5

4.7DISTRIBUTIONFACTORThis is the fraction of the total moment that would be supplied by each beam that is

connected at the joints rigidly. Whenever moment is applied to a fixed joint, it develops a

reacting moment to resist the applied moment to keep the system in equilibrium. The sum of

the moment that is developed is as a result of the moments developed in the beams that join

up into that joint. If the applied moment M causes the joint to rotate through an angle , theeach member

of the joint will rotate by the same amount. Given that the stiffness factor for

the member is, then the moment contributed by that member is = . Forequilibrium to be established, it is required that = + = + . Therefore thedistribution factor for the member is

= =

If the common term, which is the angle of rotation, is cancelled out it would be seen the

distribution factor for that member is the stiffness factor for that member divided by the sum

of the entire stiffness factor for that joint. It can be expressed generally as;

= .6

4.8PROCEDUREFORANALYSISThe following methods can be employed generally to determine the moments in beam spans

using the moment distribution method.

1.

Identify the joints on the beam and calculate for the stiffness for each joint sat the spans.


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2. Using the stiffness factors, the distribution factors can then be determined from therelation in equation 6. It is worth noting that for supports that are fixed = 0 and forsimple supports such rollers and pins = 1.

3. The fixed end moment for each loaded span is determined using the table given at the endof the course material. Fixed end moments (FEMs) that are positive acts clockwise on thespan and those that are negative acts counter clockwise on the span.

4. Assuming that the joints at which moments in the connecting spans must be determinedare initially locked, determine the moment that is required to achieve equilibrium of

moment at that joint. That is the difference between the moments at both sides of the joint

is determined.

5. The joint that is assumed to be initially locked is unlocked and the unbalancing moment isdistributed to the connecting span at the joints.

6. The distributed moment is then carried over to the far end of the respective spans bymultiplying by the carry over factor which was determined in equation 1.

7. Step 4 to 6 is repeated successively till the unbalancing moment at the entire joint iscancelled out or becomes very small. Then the cycle is stopped by not carrying over the

last distributed moment.

8. The fixed end moment, fixed end moment, distributed moment and the carry overmoment for each side of the support is summed out. Given that the arithmetic is the

correct, the moment at each sides of the same support will be the same or almost the

same. The values of the moment become the moment value for that particular support.

9. The reaction at the various supports are thus calculated for by summing the values of theloadings on the span and dividing by two; summing the moment at the ends of the same

span, dividing it by the span. When these two values are summed together they give thevalue of the reaction (shear) at that support.

4.9ANALYSISOFSIMPLEINDETERMINATEBEAMThe following examples are going to be used to illustrate the use of the conjugate beam in the

analyses of simple indeterminate beams.

EXAMPLE 4.1

Determine the bending moment and shear force and draw their respective diagram in the

beam shown Fig 4.3a below.

6m 4m 4m

20kN2kN m

Figure 4.3a

A B C


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SOLUTION

These are the information provided in the question.

() = 6() = 8 = 2/ = 20

The stiffness factor for BA and BC is given by,

=4 =

46 =

23

=4 =

48 =

12

Therefore the distribution factor for the members BA and BC are

23

23 +

12

= 47 13

12 +

23

= 37

The fixing moment for the span AB at A

=

12 = 2 6

12

= 6.0The fixing moment for the span AB at B

=

12

= 2 612

= 6.0The fixing moment for the span BC at B

= 8

= 208

8

= 20


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The fixing moment for the span BC at B

= 8

= 2088 = 20

From the above calculation the fixing moment at B in the span AB is 6.0 kNm and the

moment at B on the span BC is 20 kNm. This presents as with a difference of 14 kNm, the

difference in the moment at the various sides of the support is what is referred to as the

unbalanced moment. This unbalanced moment is to be distributing to the two sides of the

support based on the distribution factor for the two sides of the support. Thus the portion of

the unbalanced moment that will be distributed to the span AB side of the support B is

47 14 = 8

The portion of the unbalanced moment that will be distributed to the BC side of the support B

is

37 14 = 6

The fixing moment at the two end of the beam will not be distributed because it is a fixed

support; the distribution factor for fixed end support is zero.

The distributed moments are then carried over to the far ends of the spans. That is the

distributed moment on the span AB at B is carried over to A. The carrying over is done by

sending half of the distributed moment from B to A. Thus the carry over moment from B to A

is half of the 8 kNm which 4 kNm. The same treatment is given to the span BC. The

distributed moment at B in the span BC is carried over to the support C. The carried over

moment from B to C is half the distributed moment at B which is 6 kNm. The treatment is

carried on until distributed moment at the common is or near to zero. The final moment is

achieved by algebraically summing the various moments at each of the supports. The moment

distribution calculation is best done in a tabular form. This is illustrated in the table below.

Distribution factors

Initial moment

Distributed moment

Carrying over moment

Distributed moment

Final moment

47

37

-6 6 -20 20

8 6

4 3

0 0 0 0

-2 14 -14 23


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The mid-span moments for the various spans are treated as though the spans are simply

supported. Thus the mid-span moment for span AB is given by

=8

= 2 6

8

= 9.0And the mid-span moment for span BC is given by

= 4

= 2084 = 40

To determine the reactions at the various supports, assume

= = =

Taking moment about B and equating the same to zero,

( 6) + + (2 3 6) = 06 14 + 2 + 36 = 0

6 = 24 = 4

Taking moment about B and equating the same to zero,

( 8) + + (204) = 0

8 14 + 23 + 80 = 08 = 89 = 11.13

Summing all forces and equating to zero

+ + = (2 6) + 204 + + 11.13 = 32

= 16.87


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EXAMPLE 4.2

Determine the bending moment and shear force diagram in the beam shown Fig 4.4a below.

SOLUTION

These are the information provided in t

Documents

Elementary Theory of Structures