SA 2 CHAP 12.pdf

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Structural Analysis 7Structural Analysis 7thth Edition in SI UnitsEdition in SI UnitsRussell C. HibbelerRussell C. Hibbeler

Chapter 12:Chapter 12:Displacement Method of Analysis: Moment DistributionDisplacement Method of Analysis: Moment Distribution

General Principles & DefinitionGeneral Principles & Definition

Moment distribution is a method of successiveMoment distribution is a method of successiveapproximations that may be carried out to anyapproximations that may be carried out to anydesired degree of accuracydesired degree of accuracy

The method begins by assuming each joint of aThe method begins by assuming each joint of astructure is fixedstructure is fixed

By unlocking and locking each joint in succession,By unlocking and locking each joint in succession,

the internal moments at the joints arethe internal moments at the joints aredistributeddistributed

& balanced until the joints have rotated to their& balanced until the joints have rotated to theirfinal or nearly final positionsfinal or nearly final positions

2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th Edition

Chapter 12: Displacement Method of Analysis: Moment Distribution

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Member stiffness factorMember stiffness factor

Joint stiffness factorJoint stiffness factor

The total stiffness factor of joint A isThe total stiffness factor of joint A is




L

EIK

4=

10000100050004000 =++== KKT


Distribution Factor (DF)Distribution Factor (DF)

That fraction of the total resisting moment suppliedThat fraction of the total resisting moment suppliedby the member is called the distribution factor (DF)by the member is called the distribution factor (DF)




K

K

DF

K

K

M

MDF

i

ii

i

=

==

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Member relative stiffness factorMember relative stiffness factor

Quite often a continuous beam or a frame will beQuite often a continuous beam or a frame will bemade from the same materialmade from the same material

E will therefore be constantE will therefore be constant




L

IK

R=


CarryCarry--over (CO) factorover (CO) factor

Solving forSolving for and equating these eqn,and equating these eqn,

The moment M at the pin induces a moment of MThe moment M at the pin induces a moment of M

= 0.5M at the wall= 0.5M at the wall In the case of a beam with the far end fixed, theIn the case of a beam with the far end fixed, the

CO factor is +0.5CO factor is +0.5




AA 2

;4

=

=

L

EIM

L

EIM

BAAB

ABBA MM 5.0=

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CarryCarry--over (CO) factorover (CO) factor

The plus sign indicates both moments act in theThe plus sign indicates both moments act in thesame directionsame direction

Consider the beamConsider the beam




6.0)60(4)40(4

)60(4

4.0)60(4)40(4

)40(4

/)10)(60(44

)10)(240(4

/)10)(40(43

)10)(120(4

466

466

=+

=

=+

=

==

==

EE

EDF

EE

EDF

mmmEE

K

mmmEE

K

BC

BA

BC

BA


Note that the above results could also have beenNote that the above results could also have beenobtained if the relative stiffness factor is usedobtained if the relative stiffness factor is used




0)60(4

)60(4

0)40(4

)40(4

=+

=

=+

=

E

EDF

E

EDF

CB

AB

kNmwL

FEM

kNmwLFEM

CB

BC

800012

)(

800012

)(

2

2

==

==

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We begin by assuming joint B is fixed or lockedWe begin by assuming joint B is fixed or locked

The fixed end moment at B then holds span BC inThe fixed end moment at B then holds span BC inthis fixed or locked positionthis fixed or locked position

To correct this, we will apply an equal but oppositeTo correct this, we will apply an equal but oppositemoment of 8000Nm to the joint and allow the jointmoment of 8000Nm to the joint and allow the jointto rotate freelyto rotate freely





As a result, portions of this moment are distributedAs a result, portions of this moment are distributedin spans BC and BA in accordance with the DFs ofin spans BC and BA in accordance with the DFs ofthese spans at the jointthese spans at the joint

Moment in BA is 0.4(8000) = 3200NmMoment in BA is 0.4(8000) = 3200Nm

Moment in BC is 0.6(8000) = 4800NmMoment in BC is 0.6(8000) = 4800Nm

These moment must be carried over sinceThese moment must be carried over since

moments are developed at the far ends of themoments are developed at the far ends of thespanspan




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Using the carryUsing the carry--over factor of +0.5, the results areover factor of +0.5, the results areshownshown

The steps are usually presented in tabular formThe steps are usually presented in tabular form

CO indicates a line where moments are distributedCO indicates a line where moments are distributedthen carried overthen carried over

In this particular case only one cycle of momentIn this particular case only one cycle of momentdistribution is necessarydistribution is necessary

The wall supports at A and CThe wall supports at A and C absorbabsorb thethemoments and no further joints have to bemoments and no further joints have to bebalanced to satisfy joint equilibriumbalanced to satisfy joint equilibrium








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Determine the internal moment at each support of the beam. Themoment of inertia of each span is indicated.

Example 12.2Example 12.2




A moment does not get distributed in the overhanging span AB

So the distribution factor (DF)BA=0

Span BC is based on 4EI/L since the pin rocker is not at the farend of the beam

SolutionSolution




EE

K

EE

K

CD

BC

)10(3203

)10)(240(4

)10(3004

)10)(300(4

66

66

==

==

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SolutionSolution




NmwL

FEM

NmwL

FEM

NmmNFEM

DFDF

EE

EDF

DFDF

CB

BC

BA

DCCD

CB

BABC

200012

)(

200012

)(

4000)2(2000)(overhang,toDue

0;516.0

484.0320300

300

101)(1

2

2

==

==

==

==

=+

=

===

The overhanging span requires the internal moment to the left ofB to be +4000Nm.

Balancing at joint B requires an internal moment of 4000Nm tothe right of B.

-2000Nm is added to BC in order to satisfy this condition.

The distribution & CO operations proceed in the usual manner.

Since the internal moments are known, the moment diagram forthe beam can be constructed.

SolutionSolution




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SolutionSolution




StiffnessStiffness--Factor ModificationsFactor Modifications

The previous e.g. of moment distribution, we haveThe previous e.g. of moment distribution, we haveconsidered each beam span to be constrained by aconsidered each beam span to be constrained by afixed support at its far end when distributing &fixed support at its far end when distributing &carrying over the momentscarrying over the moments

In some cases, it is possible to modify the stiffnessIn some cases, it is possible to modify the stiffnessfactor of a particular beam span & thereby simplifyfactor of a particular beam span & thereby simplifythe process of moment distributionthe process of moment distribution




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Member pin supported at far endMember pin supported at far end

As shown the applied moment M rotates end A byAs shown the applied moment M rotates end A byan amtan amt

To determineTo determine , the shear in the conjugate beam at, the shear in the conjugate beam atAA must be determinedmust be determined




L

EI

MEI

L

V

LLEI

MLVM

A

AB

33'

03

2

2

1)('0'

===

=

=


Member pin supported at far end (contMember pin supported at far end (contd)d)

The stiffness factor in the beam isThe stiffness factor in the beam is

The CO factor is zero, since the pin at B does notThe CO factor is zero, since the pin at B does notsupport a momentsupport a moment

By comparison, if the far end was fixedBy comparison, if the far end was fixed supported,supported,the stiffness factor would have to be modified bythe stiffness factor would have to be modified by to model the case of having the far end pinto model the case of having the far end pinsupportedsupported




L

EIK

3=

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Symmetric beam & loadingSymmetric beam & loading

The bendingThe bending--moment diagram for the beam willmoment diagram for the beam willalso be symmetricalso be symmetric

To develop the appropriate stiffnessTo develop the appropriate stiffness--factorfactormodification consider the beammodification consider the beam

Due to symmetry, the internalDue to symmetry, the internalmoment at B & C are equalmoment at B & C are equal

Assuming this value toAssuming this value to

be M, the conjugatebe M, the conjugatebeam for span BC is shownbeam for span BC is shown





Symmetric beam & loading (contSymmetric beam & loading (contd)d)

Moments for only half the beam can be distributedMoments for only half the beam can be distributedprovided the stiffness factor for the center span isprovided the stiffness factor for the center span iscomputedcomputed




L

EIK

L

EIM

EI

MLV

LL

EI

MLVM

B

BC

2

2

2'

02

)('-0'

=

===

=

+=

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Symmetric beam with asymmetric loadingSymmetric beam with asymmetric loading

Consider the beam as shownConsider the beam as shown

The conjugate beam for its center span BC is shownThe conjugate beam for its center span BC is shown

Due to its asymmetric loading, the internal momentDue to its asymmetric loading, the internal momentat B is equal but opposite to that at Cat B is equal but opposite to that at C





Symmetric beam with asymmetric loadingSymmetric beam with asymmetric loading

Assuming this value to be M, the slopeAssuming this value to be M, the slope at eachat eachend is determined as follows:end is determined as follows:




L

EIK

L

EIM

EI

MLV

LL

EI

MLL

EI

MLV

M

B

B

C

6

6

6'

0622

1

6

5

22

1)('-

0'

=

===

=

+

=

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Determine the internal moments at the supports of the beamshown below. The moment of inertia of the two spans is shown inthe figure.





The beam is roller supported at its far end C.

The stiffness of span BC will be computed on the basis of K =3EI/L

We have:

SolutionSolution




EELEIK

EE

L

EIK

BC

AB

)10(1804

)10)(240(33

)10(1603

)10)(120(44

6

6

66

===

===

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SolutionSolution




1

180

180

5294.0180160

180

4706.0180160

160

0160

160

==

=+

=

=+

=

=+

=

E

EDF

EE

EDF

EE

EDF

E

EDF

CB

BC

BA

AB

NmwL

FEMBC

120008

)4(6000

8)(

22

=

==

The forgoing data are entered into table as shown.

The moment distribution is carried out.

By comparison, the method considerably simplifies thedistribution.

The beams end shears & moment diagrams are shown.

SolutionSolution




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Moment distribution for frames:Moment distribution for frames:No sideswayNo sidesway

Application of the momentApplication of the moment--distribution method fordistribution method forframes having no sidesway follows the sameframes having no sidesway follows the sameprocedure as that given for beamprocedure as that given for beam




Determine the internal moments at the joints of the frame asshown. There is a pin at E and D and a fixed support atA. EI isconstant.





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By inspection, the pin at E will prevent the frame will sidesway.

The stiffness factors of CD and CE can be computed using K =3EI/L since far ends are pinned.

The 60kN load does not contribute a FEM since it is applied atjoint B.

SolutionSolution




455.0545.01

545.0

6/45/4

5/4

0

4

3;

5

3;

6

4;

5

4

==

=

+

=

=

====

BC

BA

AB

CECDBCAB

DF

EIEI

EIDF

DF

EIK

EIK

EIK

EIK

SolutionSolution




1;1

372.0298.0330.01

298.04/35/36/4

5/3

330.04/35/36/4

6/4

==

==

=++

=

=++

=

ECDC

CE

CD

CB

DFDF

DF

EIEIEI

EIDF

EIEIEI

EIDF

kNmwL

FEM

kNmwLFEM

CB

BC

13512

)(

13512

)(

2

2

==

==

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The data are shown in table.

The distribution of moments successively goes to joints B & C.

The final moment are shown on the last line.

Using these data, the moment diagram for the frame isconstructed as shown.

SolutionSolution




Moment distribution for frames: SideswayMoment distribution for frames: Sidesway

To determine sidesway and the internal momentsTo determine sidesway and the internal momentsat the joints using moment distribution, we will useat the joints using moment distribution, we will usethe principle of superpositionthe principle of superposition

The frame shown is first held from sidesway byThe frame shown is first held from sidesway byapplying an artificial joint support at Capplying an artificial joint support at C

Moment distribution is applied & by statics, theMoment distribution is applied & by statics, the

restraining force R is determinedrestraining force R is determined




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The equal but opposite restraining force is thenThe equal but opposite restraining force is thenapplied to the frame The moments in the frameapplied to the frame The moments in the frameare calculatedare calculated





Multistory framesMultistory frames

Multistory frameworks may have severalMultistory frameworks may have severalindependent joints dispindependent joints disp

Consequently, the moment distribution analysisConsequently, the moment distribution analysisusing the above techniques will involve moreusing the above techniques will involve morecomputationcomputation




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The structure shown can have 2 independent jointThe structure shown can have 2 independent jointdisp since the sidesway of the first story isdisp since the sidesway of the first story isindependent of any disp of the second storyindependent of any disp of the second story






These disp are not known initiallyThese disp are not known initially

The analysis must proceed on the basis ofThe analysis must proceed on the basis ofsuperpositionsuperposition

2 restraining forces R2 restraining forces R11 and Rand R22 are appliedare applied

The fixed end moments are determined &The fixed end moments are determined &

distributeddistributed Using the eqn of eqm, the numerical values of RUsing the eqn of eqm, the numerical values of R11

and Rand R22 are then determinedare then determined




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The restraint at the floor of the first story isThe restraint at the floor of the first story isremoved & the floor is given a dispremoved & the floor is given a disp

This disp causes fixed end moment (FEMs) in theThis disp causes fixed end moment (FEMs) in theframe which can be assigned specific numericalframe which can be assigned specific numericalvaluesvalues

By distributing these moments & using the eqn ofBy distributing these moments & using the eqn ofeqm, the associated numerical values of Reqm, the associated numerical values of R11 and Rand R22

can be determinedcan be determined






In a similar manner, the floor of the second story isIn a similar manner, the floor of the second story isthen given a dispthen given a disp

With reference to the restraining forces we requireWith reference to the restraining forces we requireequal but opposite application of Requal but opposite application of R11 and Rand R22 to theto theframe such that:frame such that:




111

222

"'''

"'''

RCRCR

RCRCR

+=

+=

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Simultaneous solution of these eqn yields theSimultaneous solution of these eqn yields thevalues of Cvalues of C and Cand C

These correction factors are then multiplied by theThese correction factors are then multiplied by theinternal joint moments found from momentinternal joint moments found from momentdistributiondistribution

The resultant moments are found by adding theseThe resultant moments are found by adding thesecorrected moments to those obtained for the framecorrected moments to those obtained for the frame




Determine the moments at each joint of the frame shown. EI isconstant.





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First, we consider the frame held from sidesway

The stiffness factor of each span is computed on the basis of4EI/L or using relative stiffness factor I/L

SolutionSolution




kNmFEM

kNmFEM

CB

BC

56.2)5(

)4()1(16)(

24.10)5(

)1()4(16)(

2

2

2

2

==

==

The DFs and the moment distribution are shown in the table.

The eqn of eqm are applied to the free body diagrams of thecolumns in order to determine Ax and DxFrom the free body diagram of the entire frame, the joint restraintR has a magnitude of

SolutionSolution




kNkNkNRFx 92.081.073.1;0 ===

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An equal but opposite value of R = 0.92kN must be applied tothe frame at C and the internal moments computed.

We assume a force R is applied at C causing the frame to deflectas shown.

The joints at B and C are temporarily restrained from rotating.

As a result, the FEM at the ends of the columns are determined.

SolutionSolution




Since both B and C happen to be displaced the same amount andAB and DC have the same E, I and L, the FEM in AB will be thesame as that in DC.

As shown we will arbitrarily assumed this FEM to be

The moment distribution of the FEM is shown below.

SolutionSolution




kNmFEMFEMFEMFEMDCCDBAAB

100)()()()( ====

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From the eqm, the horizontal reactions at A and D are calculated.

For the entire frame, we require:

R=56kN creates the moments tabulated below

Corresponding moments caused by R = 0.92kN can bedetermined by proportion

SolutionSolution




kNRFx

562828';0 ===

( )

kNmMkNmM

kNmMkNmMkNmM

kNmM

DCCD

CBBCBA

AB

63.2;71.3

71.3;79.4;79.4

57.1800.56

92.088.2

==

===

=+=

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SA 2 CHAP 12.pdf