Structural Analysis Chapter 12.ppt

8/18/2019 Structural Analysis Chapter 12.ppt

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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


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General Principles & DefnitionGeneral Principles & Defnition

! Moment distribution is a method ofMoment distribution is a method ofsuccessi"e appro#imations that may besuccessi"e appro#imations that may be

carried out to any desired de$ree ofcarried out to any desired de$ree of

accuracyaccuracy! %he method be$ins by assumin$ each &oint %he method be$ins by assumin$ each &oint

of a structure is '#edof a structure is '#ed

! (y unloc)in$ and loc)in$ each &oint in(y unloc)in$ and loc)in$ each &oint in

succession* the internal moments at thesuccession* the internal moments at the

&oints are &oints are ““distributeddistributed”” + balanced until the+ balanced until the

&oints ha"e rotated to their 'nal or nearly &oints ha"e rotated to their 'nal or nearly

'nal positions'nal positions © 2009 Pearson Education South Asia Pte LtdStructural Analysis 7th Edition

Chapter 12: Displacement ethod o! Analysis: oment Distri"ution


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! Member sti,ness factorMember sti,ness factor

! -oint sti,ness factor -oint sti,ness factor! %he total sti,ness factor of &oint A is %he total sti,ness factor of &oint A is

© 2009 Pearson Education South Asia Pte Ltd

Structural Analysis 7th Edition


L

EI K

4=

10000100050004000 =++=∑= K K T


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! Distribution actor /D0Distribution actor /D0! %hat fraction of the total resistin$ moment %hat fraction of the total resistin$ moment

supplied by the member is called thesupplied by the member is called the

distribution factor /D0distribution factor /D0




K

K DF

K

K

M

M DF

i

ii

i

∑=

∑==

θ

θ


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! Member relati"e sti,ness factorMember relati"e sti,ness factor! 1uite often a continuous beam or a frame 2ill1uite often a continuous beam or a frame 2ill

be made from the same materialbe made from the same material

! E 2ill therefore be constantE 2ill therefore be constant




L

I K

R =


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! Carry3o"er /C40 factorCarry3o"er /C40 factor

! Sol"in$ forSol"in$ for θθ and e5uatin$ these e5n*and e5uatin$ these e5n*

! %he moment M at the pin induces a moment %he moment M at the pin induces a moment

of Mof M’’ 6 .8M at the 2all6 .8M at the 2all

! In the case of a beam 2ith the far end '#ed*In the case of a beam 2ith the far end '#ed*

the C4 factor is 9.8the C4 factor is 9.8




AA 2

;4

θ θ

=

=

L

EI M

L

EI M

BA AB

AB BA M M 5.0=


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! Carry3o"er /C40 factorCarry3o"er /C40 factor! %he plus si$n indicates both moments act in %he plus si$n 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

46

6

=+

=

=+

=

==

==

E E

E DF

E E

E

DF

mmm E E

K

mmm E E

K

BC

BA

BC

BA


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! :ote that the abo"e results could also ha"e:ote that the abo"e results could also ha"e

been obtained if the relati"e sti,ness factorbeen obtained if the relati"e sti,ness factor

is usedis used




0

)60(4

)60(4

0)40(4

)40(4

=

+∞

=

=+∞

=

E

E DF

E

E DF

CB

AB

kNmwL

FEM

kNmwL FEM

CB

BC

800012

)(

800012

)(

2

2

==

−=−=


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! ;e be$in by assumin$ &oint ( is '#ed or;e be$in by assumin$ &oint ( is '#ed orloc)edloc)ed

! %he '#ed end moment at ( then holds span %he '#ed end moment at ( then holds span

(C in this '#ed or loc)ed position(C in this '#ed or loc)ed position! %o correct this* 2e 2ill apply an e5ual but %o correct this* 2e 2ill apply an e5ual but

opposite moment of


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! As a result* portions of this moment areAs a result* portions of this moment aredistributed in spans (C and (A indistributed in spans (C and (A in

accordance 2ith the Ds of these spans ataccordance 2ith the Ds of these spans at

the &ointthe &oint

! Moment in (A is .=/?:mMoment in (A is .=/?:m

! Moment in (C is .@/


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! Usin$ the carry3o"er factor of 9.8* theUsin$ the carry3o"er factor of 9.8* theresults are sho2nresults are sho2n

! %he steps are usually presented in tabular %he steps are usually presented in tabular

formform! C4 indicates a line 2here moments areC4 indicates a line 2here moments are

distributed then carried o"erdistributed then carried o"er

!In this particular case only one cycle ofIn this particular case only one cycle ofmoment distribution is necessarymoment distribution is necessary

! %he 2all supports at A and C %he 2all supports at A and C ““absorbabsorb”” thethe

moments and no further &oints ha"e to bemoments and no further &oints ha"e to be

balanced to satisfy &oint e5uilibriumbalanced to satisfy &oint e5uilibrium © 2009 Pearson Education South Asia Pte LtdStructural Analysis 7th Edition



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Determine the internal moment at each support of thebeam. %he moment of inertia of each span is indicated.

Example 12.2Example 12.2





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A moment does not $et distributed in the o"erhan$in$span A(

So the distribution factor /D0(A 6

Span (C is based on =EIB since the pin roc)er is not at

the far end of the beam

SolutionSolution




E E K

E E

K

CD

BC

)10(3203

)10)(240(4

)10(3004

)10)(300(4

66

66

==

==


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SolutionSolution




NmwL

FEM

NmwL FEM

Nmm N FEM

DF DF

E E

E DF

DF DF

CB

BC

BA

DC CD

CB

BA BC

200012

)(

200012

)(

4000)2(2000)( overhang,toDue

0 ;516.0

484.0320300

300

101)(1

2

2

==

−=−=

==

==

=

+

=

=−=−=


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! %he o"erhan$in$ span re5uires the internal moment tothe left of ( to be 9=:m.

!(alancin$ at &oint ( re5uires an internal moment of –=:m to the ri$ht of (.

!3?:m is added to (C in order to satisfy this condition.! %he distribution + C4 operations proceed in the usualmanner.

Since the internal moments are )no2n* the moment

dia$ram for the beam can be constructed.

SolutionSolution





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SolutionSolution





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Stiness-Factor o!ifcationsStiness-Factor o!ifcations

! %he pre"ious e.$. of moment distribution* 2e %he pre"ious e.$. of moment distribution* 2eha"e considered each beam span to beha"e considered each beam span to be

constrained by a '#ed support at its far endconstrained by a '#ed support at its far end

2hen distributin$ + carryin$ o"er the2hen distributin$ + carryin$ o"er the

momentsmoments

! In some cases* it is possible to modify theIn some cases* it is possible to modify the

sti,ness factor of a particular beam span +sti,ness factor of a particular beam span +

thereby simplify the process of momentthereby simplify the process of momentdistributiondistribution





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! Member pin supported at far endMember pin supported at far end! As sho2n the applied moment M rotates endAs sho2n the applied moment M rotates end

A by an amtA by an amt θθ

! %o determine %o determine θθ* the shear in the con&u$ate* the shear in the con&u$atebeam at Abeam at A’’ must be determinedmust be determined




θ θ L

EI M

EI

LV

L L EI

M LV M

A

A B

33'

0

3

2

2

1)(' 0'

=⇒==

=

−=∑


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! Member pin supported at far end /contMember pin supported at far end /cont ’’d0d0! %he sti,ness factor in the beam is %he sti,ness factor in the beam is

! %he C4 factor is ero* since the pin at ( does %he C4 factor is ero* since the pin at ( does

not support a momentnot support a moment

!(y comparison* if the far end 2as '#ed(y comparison* if the far end 2as '#edsupported* the sti,ness factor 2ould ha"e tosupported* the sti,ness factor 2ould ha"e to

be modi'ed bybe modi'ed by ¾¾ to model the case of ha"in$to model the case of ha"in$

the far end pin supportedthe far end pin supported




L

EI K

3=


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! Symmetric beam + loadin$Symmetric beam + loadin$! %he bendin$3moment dia$ram for the beam %he bendin$3moment dia$ram for the beam

2ill also be symmetric2ill also be symmetric

! %o de"elop the appropriate sti,ness3factor %o de"elop the appropriate sti,ness3factormodi'cation consider the beammodi'cation consider the beam

! Due to symmetry* the internalDue to symmetry* the internal

moment at ( + C are e5ualmoment at ( + C are e5ual

! Assumin$ this "alue toAssumin$ this "alue to

be M* the con&u$atebe M* the con&u$ate

beam for span (C is sho2nbeam for span (C is sho2n





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! Symmetric beam + loadin$ /contSymmetric beam + loadin$ /cont’’d0d0

! Moments for only half the beam can beMoments for only half the beam can bedistributed pro"ided the sti,ness factor fordistributed pro"ided the sti,ness factor for

the center span is computedthe center span is computed




θ

θ θ

L

EI K

L

EI

M EI

ML

V

L L

EI

M LV M

B

BC

2

22

'

02

)('- 0'

=

=⇒==

=

+=∑


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! Symmetric beam 2ith asymmetric loadin$Symmetric beam 2ith asymmetric loadin$! Consider the beam as sho2nConsider the beam as sho2n

! %he con&u$ate beam for its center span (C is %he con&u$ate beam for its center span (C is

sho2nsho2n! Due to its asymmetric loadin$* the internalDue to its asymmetric loadin$* the internal

moment at ( is e5ual but opposite to that atmoment at ( is e5ual but opposite to that at

CC





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! Symmetric beam 2ith asymmetric loadin$Symmetric beam 2ith asymmetric loadin$! Assumin$ this "alue to be M* the slopeAssumin$ this "alue to be M* the slope θθ atat

each end is determined as follo2seach end is determined as follo2s




L

EI K

L

EI M

EI

MLV

L L

EI

M L L

EI

M LV

M

B

B

C

6

6

6

'

0622

1

6

5

22

1)('-

0'

=

=⇒==

=

−

+

=∑

θ θ


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Determine the internal moments at the supports of thebeam sho2n belo2. %he moment of inertia of the t2ospans is sho2n in the '$ure.

Example 12."Example 12."





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! %he beam is roller supported at its far end C.! %he sti,ness of span (C 2ill be computed on the basis of 6 >EIB

!;e ha"e

SolutionSolution




E

E

L

EI

K

E E

L

EI K

BC

AB

)10(1804

)10)(240(33

)10(1603

)10)(120(44

66

66

===

===


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SolutionSolution




1180180

5294.0180160

180

4706.0

180160

160

0160

160

==

=+

=

=+

=

=+∞

=

E E DF

E E

E DF

E E

E DF

E

E DF

CB

BC

BA

AB

NmwL

FEM BC 120008

)4(6000

8)(

22

−=−

=−=


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%he for$oin$ data are entered into table as sho2n.%he moment distribution is carried out.

(y comparison* the method considerably simpli'es thedistribution.

%he beam’s end shears + moment dia$rams are sho2n.

SolutionSolution





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oment !istri#ution $or $rames:oment !istri#ution $or $rames:

%o si!esa'%o si!esa'

! Application of the moment3distributionApplication of the moment3distributionmethod for frames ha"in$ no sides2aymethod for frames ha"in$ no sides2ay

follo2s the same procedure as that $i"en forfollo2s the same procedure as that $i"en for

beambeam





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Determine the internal moments at the &oints of the frameas sho2n. %here is a pin at E and D and a '#ed support atA. EI is constant.

Example 12.(Example 12.(





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(y inspection* the pin at E 2ill pre"ent the frame 2illsides2ay.

%he sti,ness factors of CD and CE can be computed usin$ 6 >EIB since far ends are pinned.

%he @): load does not contribute a EM since it isapplied at &oint (.

SolutionSolution




455.0545.01

545.06/45/4

5/4

0

4

3 ;

5

3 ;

6

4 ;

5

4

=−=

=+

=

=

====

BC

BA

AB

CE CD BC AB

DF

EI EI EI DF

DF

EI K

EI K

EI K

EI K


32/47

SolutionSolution




1 ;1

372.0298.0330.01

298.04/35/36/4

5/3

330.04/35/36/4

6/4

==

=−−=

=

++

=

=++

=

EC DC

CE

CD

CB

DF DF

DF

EI EI EI

EI DF

EI EI EI

EI DF

kNmwL

FEM

kNm

wL

FEM

CB

BC

13512

)(

13512)(

2

2

==

−=−

=


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! %he data are sho2n in table.! %he distribution of moments successi"ely $oes to &oints( + C.

! %he 'nal moment are sho2n on the last line.

!Usin$ these data* the moment dia$ram for the frame isconstructed as sho2n.

SolutionSolution





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Si!esa'Si!esa'

! %o determine sides2ay and the internal %o determine sides2ay and the internalmoments at the &oints usin$ momentmoments at the &oints usin$ moment

distribution* 2e 2ill use the principle ofdistribution* 2e 2ill use the principle of

superpositionsuperposition

! %he frame sho2n is 'rst held from sides2ay %he frame sho2n is 'rst held from sides2ay

by applyin$ an arti'cial &oint support at Cby applyin$ an arti'cial &oint support at C

! Moment distribution is applied + by statics*Moment distribution is applied + by statics*

the restrainin$ force R is determinedthe restrainin$ force R is determined





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Si!esa'Si!esa'

! %he e5ual but opposite restrainin$ force is %he e5ual but opposite restrainin$ force isthen applied to the frame %he moments inthen applied to the frame %he moments in

the frame are calculatedthe frame are calculated





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Si!esa'Si!esa'

! Multistory framesMultistory frames! Multistory frame2or)s may ha"e se"eralMultistory frame2or)s may ha"e se"eral

independent &oints dispindependent &oints disp

! Conse5uently* the moment distributionConse5uently* the moment distributionanalysis usin$ the abo"e techni5ues 2illanalysis usin$ the abo"e techni5ues 2ill

in"ol"e more computationin"ol"e more computation




!i i# i $ $


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Si!esa'Si!esa'

! Multistory framesMultistory frames! %he structure sho2n can ha"e ? independent %he structure sho2n can ha"e ? independent

&oint disp since the sides2ay of the 'rst story &oint disp since the sides2ay of the 'rst story

is independent of any disp of the secondis independent of any disp of the second

storystory

© 2009 Pearson Education South Asia Pte LtdStructural Analysis 7th Edition


!i i# i $ $


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Si!esa'Si!esa'

! Multistory framesMultistory frames! %hese disp are not )no2n initially %hese disp are not )no2n initially

! %he analysis must proceed on the basis of %he analysis must proceed on the basis of

superpositionsuperposition! ? restrainin$ forces R? restrainin$ forces REE and Rand R?? are appliedare applied

! %he '#ed end moments are determined + %he '#ed end moments are determined +

distributeddistributed

! Usin$ the e5n of e5m* the numerical "aluesUsin$ the e5n of e5m* the numerical "alues

of Rof REE and Rand R?? are then determinedare then determined



!i i# i $ $


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Si!esa'Si!esa'

! Multistory framesMultistory frames! %he restraint at the Foor of the 'rst story is %he restraint at the Foor of the 'rst story is

remo"ed + the Foor is $i"en a dispremo"ed + the Foor is $i"en a disp

! %his disp causes '#ed end moment /EMs0 in %his disp causes '#ed end moment /EMs0 inthe frame 2hich can be assi$ned speci'cthe frame 2hich can be assi$ned speci'c

numerical "aluesnumerical "alues

! (y distributin$ these moments + usin$ the(y distributin$ these moments + usin$ the

e5n of e5m* the associated numerical "aluese5n of e5m* the associated numerical "aluesof Rof REE’’ and Rand R??’’ can be determinedcan be determined



!i i# i $ $ t !i t i# ti $ $


40/47


Si!esa'Si!esa'

! Multistory framesMultistory frames! In a similar manner* the Foor of the secondIn a similar manner* the Foor of the second

story is then $i"en a dispstory is then $i"en a disp

! ;ith reference to the restrainin$ forces 2e;ith reference to the restrainin$ forces 2ere5uire e5ual but opposite application of Rre5uire e5ual but opposite application of REE

and Rand R?? to the frame such thatto the frame such that



111

222

'''

'''

RC RC R

RC RC R

−+=+−=

!i i# i $ $ t !i t i# ti $ $


41/47


Si!esa'Si!esa'

! Multistory framesMultistory frames! Simultaneous solution of these e5n yields theSimultaneous solution of these e5n yields the

"alues of C"alues of C’’ and Cand C””

! %hese correction factors are then multiplied %hese correction factors are then multipliedby the internal &oint moments found fromby the internal &oint moments found from

moment distributionmoment distribution

! %he resultant moments are found by addin$ %he resultant moments are found by addin$

these corrected moments to those obtainedthese corrected moments to those obtainedfor the framefor the frame




42/47

Determine the moments at each &oint of the frame sho2n.EI is constant.

Example 12.)Example 12.)




43/47

irst* 2e consider the frame held from sides2ay

%he sti,ness factor of each span is computed on the basisof =EIB or usin$ relati"e sti,ness factor IB

SolutionSolution



kNm FEM

kNm FEM

CB

BC

56.2)5(

)4()1(16)(

24.10)5(

)1()4(16)(

2

2

2

2

==

−=−=


44/47

%he Ds and the moment distribution are sho2n in thetable.

%he e5n of e5m are applied to the free body dia$rams ofthe columns in order to determine A# and D#

rom the free body dia$ram of the entire frame* the &ointrestraint R has a ma$nitude of

SolutionSolution



kN kN kN R F x 92.081.073.1 ;0 =−==∑


45/47

!An e5ual but opposite "alue of R 6 .G?): must beapplied to the frame at C and the internal momentscomputed.

!;e assume a force R’ is applied at C causin$ the frameto deFect as sho2n.

! %he &oints at ( and C are temporarily restrained fromrotatin$.

!As a result* the EM at the ends of the columns aredetermined.

SolutionSolution




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!Since both ( and C happen to be displaced the sameamount and A( and DC ha"e the same E* I and B* the EMin A( 2ill be the same as that in DC.

!As sho2n 2e 2ill arbitrarily assumed this EM to be

%he moment distribution of the EM is sho2n belo2.

SolutionSolution



kNm FEM FEM FEM FEM DC CD BA AB 100)()()()( −====


47/47

!rom the e5m* the horiontal reactions at A and D arecalculated.

!or the entire frame* 2e re5uire

!R’68@): creates the moments tabulated belo2

!Correspondin$ moments caused by R 6 .G?): can bedetermined by proportion

SolutionSolution


kN R F x

562828' ;0 =+==∑

( )

kNm M kNm M

kNm M kNm M kNm M

kNm M

DC CD

CB BC BA

AB

63.2 ;71.3

71.3 ;79.4 ;79.4

57.180

0.56

92.088.2

−=−=

=−==

=−+=

Structural Analysis Chapter 12.ppt

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Transcript of Structural Analysis Chapter 12.ppt