structural analysis 7 th edition in si units russell c. hibbeler chapter 12: displacement method of...

Structural Analysis 7Structural Analysis 7thth Edition in SI Units Edition 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 Moment distribution is a method of successive approximations that may be successive approximations that may be carried out to any desired degree of carried out to any desired degree of accuracyaccuracy

• The method begins by assuming each joint The method begins by assuming each joint of a structure is fixedof a structure is fixed

• By unlocking and locking each joint in By unlocking and locking each joint in succession, the internal moments at the succession, the internal moments at the joints are “distributed” & balanced until the joints are “distributed” & balanced until the joints have rotated to their final or nearly joints have rotated to their final or nearly final positionsfinal positions

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th EditionChapter 12: Displacement Method of Analysis: Moment Distribution


• Member stiffness factorMember stiffness factor

• Joint stiffness factorJoint stiffness factor• The total stiffness factor of joint A is The 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 That fraction of the total resisting moment

supplied by the member is called the supplied by the member is called the distribution factor (DF)distribution factor (DF)



K

KDF

K

K

M

MDF

i

iii


• Member relative stiffness factorMember relative stiffness factor• Quite often a continuous beam or a frame will Quite often a continuous beam or a frame will

be made from the same materialbe made from the same material• E will therefore be constantE will therefore be constant



L

IKR


• Carry-over (CO) factorCarry-over (CO) factor

• Solving for Solving for and equating these eqn, and equating these eqn,

• The moment M at the pin induces a moment The moment M at the pin induces a moment of M’ = 0.5M at the wallof M’ = 0.5M at the wall

• In the case of a beam with the far end fixed, In the case of a beam with the far end fixed, the CO factor is +0.5the CO factor is +0.5



AA 2

; 4

L

EIM

L

EIM BAAB

ABBA MM 5.0


• Carry-over (CO) factorCarry-over (CO) factor• The plus sign indicates both moments act in The plus sign indicates both moments act in

the same directionthe same 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 Note that the above results could also have been obtained if the relative stiffness factor been obtained if the relative stiffness factor is usedis used



0)60(4

)60(4

0)40(4

)40(4

E

EDF

E

EDF

CB

AB

kNmwL

FEM

kNmwL

FEM

CB

BC

800012

)(

800012

)(

2

2


• We begin by assuming joint B is fixed or We begin by assuming joint B is fixed or lockedlocked

• The fixed end moment at B then holds span The fixed end moment at B then holds span BC in this fixed or locked positionBC in this fixed or locked position

• To correct this, we will apply an equal but To correct this, we will apply an equal but opposite moment of 8000Nm to the joint opposite moment of 8000Nm to the joint and allow the joint to rotate freelyand allow the joint to rotate freely




• As a result, portions of this moment are As a result, portions of this moment are distributed in spans BC and BA in distributed in spans BC and BA in accordance with the DFs of these spans at accordance with the DFs of these spans at the jointthe 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 since These moment must be carried over since

moments are developed at the far ends of moments are developed at the far ends of the spanthe span




• Using the carry-over factor of +0.5, the Using the carry-over factor of +0.5, the results are shown results are shown

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

• CO indicates a line where moments are CO indicates a line where moments are distributed then carried overdistributed then carried over

• In this particular case only one cycle of In this particular case only one cycle of moment distribution is necessarymoment distribution is necessary

• The wall supports at A and C “absorb” the The wall supports at A and C “absorb” the moments and no further joints have to be moments and no further joints have to be balanced to satisfy joint equilibriumbalanced to satisfy joint equilibrium© 2009 Pearson Education South Asia Pte Ltd


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

Example 12.2Example 12.2



A moment does not get distributed in the overhanging span ABSo the distribution factor (DF)BA =0

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

SolutionSolution



EE

K

EE

K

CD

BC

)10(3203

)10)(240(4

)10(3004

)10)(300(4

66

66

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 of B to be +4000Nm.•Balancing at joint B requires an internal moment of –4000Nm to the 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 for the beam can be constructed.

SolutionSolution



SolutionSolution



Stiffness-Factor ModificationsStiffness-Factor Modifications

• The previous e.g. of moment distribution, we The previous e.g. of moment distribution, we have considered each beam span to be have considered each beam span to be constrained by a fixed support at its far end constrained by a fixed support at its far end when distributing & carrying over the when distributing & carrying over the momentsmoments

• In some cases, it is possible to modify the In some cases, it is possible to modify the stiffness factor of a particular beam span & stiffness factor of a particular beam span & thereby simplify the process of moment thereby simplify the process of moment distributiondistribution




• Member pin supported at far endMember pin supported at far end• As shown the applied moment M rotates end As shown the applied moment M rotates end

A by an amt A by an amt • To determine To determine , the shear in the conjugate , the shear in the conjugate

beam at A’ must be determinedbeam at A’ must be determined



L

EIM

EI

LV

LLEI

MLVM

A

AB

33'

03

2

2

1)(' 0'


• Member pin supported at far end (cont’d)Member pin supported at far end (cont’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 The CO factor is zero, since the pin at B does not support a momentnot support a moment

• By comparison, if the far end was fixed By comparison, if the far end was fixed supported, the stiffness factor would have to supported, the stiffness factor would have to be modified by ¾ to model the case of having be modified by ¾ to model the case of having the far end pin supportedthe far end pin supported



L

EIK

3


• Symmetric beam & loadingSymmetric beam & loading• The bending-moment diagram for the beam The bending-moment diagram for the beam

will also be symmetricwill also be symmetric• To develop the appropriate stiffness-factor To develop the appropriate stiffness-factor

modification consider the beam modification consider the beam • Due to symmetry, the internal Due to symmetry, the internal

moment at B & C are equalmoment at B & C are equal• Assuming this value to Assuming this value to

be M, the conjugate be M, the conjugate beam for span BC is shownbeam for span BC is shown




• Symmetric beam & loading (cont’d)Symmetric beam & loading (cont’d)

• Moments for only half the beam can be Moments for only half the beam can be distributed provided the stiffness factor for distributed provided the stiffness factor for the center span is computedthe center span is computed



L

EIK

L

EIM

EI

MLV

LL

EI

MLVM

B

BC

2

2

2'

02

)(' - 0'


• 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 The conjugate beam for its center span BC is

shownshown• Due to its asymmetric loading, the internal Due to its asymmetric loading, the internal

moment at B is equal but opposite to that at moment at B is equal but opposite to that at CC




• Symmetric beam with asymmetric loadingSymmetric beam with asymmetric loading• Assuming this value to be M, the slope Assuming this value to be M, the slope at at

each end is determined as follows:each 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'

Determine the internal moments at the supports of the beam shown below. The moment of inertia of the two spans is shown in the 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



EE

L

EIK

EE

L

EIK

BC

AB

)10(1804

)10)(240(33

)10(1603

)10)(120(44

66

66

SolutionSolution



1180

180

5294.0180160

180

4706.0180160

160

0160

160

E

EDF

EE

EDF

EE

EDF

E

EDF

CB

BC

BA

AB

NmwL

FEM BC 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 the distribution.The beam’s end shears & moment diagrams are shown.

SolutionSolution



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

• Application of the moment-distribution Application of the moment-distribution method for frames having no sidesway method for frames having no sidesway follows the same procedure as that given for follows the same procedure as that given for beambeam



Determine the internal moments at the joints of the frame as shown. There is a pin at E and D and a fixed support at A. EI is constant.




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 at joint B.

SolutionSolution



455.0545.01

545.06/45/4

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

kNmwL

FEM

CB

BC

13512

)(

13512

)(

2

2

•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 is constructed as shown.

SolutionSolution



Moment distribution for frames: Moment distribution for frames: SideswaySidesway

• To determine sidesway and the internal To determine sidesway and the internal moments at the joints using moment moments at the joints using moment distribution, we will use the principle of distribution, we will use the principle of superpositionsuperposition

• The frame shown is first held from sidesway The frame shown is first held from sidesway by applying an artificial joint support at Cby applying an artificial joint support at C

• Moment distribution is applied & by statics, Moment distribution is applied & by statics, the restraining force R is determinedthe restraining force R is determined




• The equal but opposite restraining force is The equal but opposite restraining force is then applied to the frame The moments in then applied to the frame The moments in the frame are calculatedthe frame are calculated




• Multistory framesMultistory frames• Multistory frameworks may have several Multistory frameworks may have several

independent joints dispindependent joints disp• Consequently, the moment distribution Consequently, the moment distribution

analysis using the above techniques will analysis using the above techniques will involve more computationinvolve more computation




• Multistory framesMultistory frames• The structure shown can have 2 independent The structure shown can have 2 independent

joint disp since the sidesway of the first story joint disp since the sidesway of the first story is independent of any disp of the second is independent of any disp of the second storystory




• Multistory framesMultistory frames• These disp are not known initiallyThese disp are not known initially• The analysis must proceed on the basis of The analysis must proceed on the basis of

superpositionsuperposition

• 2 restraining forces R2 restraining forces R11 and R and R22 are applied are applied

• The fixed end moments are determined & The fixed end moments are determined & distributeddistributed

• Using the eqn of eqm, the numerical values Using the eqn of eqm, the numerical values of Rof R11 and R and R22 are then determined are then determined




• Multistory framesMultistory frames• The restraint at the floor of the first story is The restraint at the floor of the first story is

removed & the floor is given a dispremoved & the floor is given a disp• This disp causes fixed end moment (FEMs) in This disp causes fixed end moment (FEMs) in

the frame which can be assigned specific the frame which can be assigned specific numerical valuesnumerical values

• By distributing these moments & using the By distributing these moments & using the eqn of eqm, the associated numerical values eqn of eqm, the associated numerical values of Rof R11’ and R’ and R22’ can be determined’ can be determined




• Multistory framesMultistory frames• In a similar manner, the floor of the second In a similar manner, the floor of the second

story is then given a dispstory is then given a disp• With reference to the restraining forces we With reference to the restraining forces we

require equal but opposite application of Rrequire equal but opposite application of R11 and Rand R22 to the frame such that: to the frame such that:



111

222

"'''

"'''

RCRCR

RCRCR


• Multistory framesMultistory frames• Simultaneous solution of these eqn yields the Simultaneous solution of these eqn yields the

values of C’ and C”values of C’ and C”• These correction factors are then multiplied These correction factors are then multiplied

by the internal joint moments found from by the internal joint moments found from moment distribution moment distribution

• The resultant moments are found by adding The resultant moments are found by adding these corrected moments to those obtained these corrected moments to those obtained for the framefor the frame



Determine the moments at each joint of the frame shown. EI is constant.




First, we consider the frame held from sidesway

The stiffness factor of each span is computed on the basis of 4EI/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 the columns in order to determine Ax and Dx

From the free body diagram of the entire frame, the joint restraint R has a magnitude of

SolutionSolution



kNkNkNRFx 92.081.073.1 ;0

•An equal but opposite value of R = 0.92kN must be applied to the frame at C and the internal moments computed.•We assume a force R’ is applied at C causing the frame to deflect as 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 and AB and DC have the same E, I and L, the FEM in AB will be the same as that in DC.•As shown we will arbitrarily assumed this FEM to be

The moment distribution of the FEM is shown below.

SolutionSolution



kNmFEMFEMFEMFEM DCCDBAAB 100)()()()(

•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 be determined 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

structural analysis 7 th edition in si units russell c. hibbeler chapter 12: displacement method of...

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