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Week9
4.7MomentDistributionMethodforMultiRedundantBeams
Thissectiondealswithcontinuousbeamsandproppedcantilevers.AnAmericanengineer,Professor
HardyCross,developedaverysimple,elegantandpracticalmethodofanalysisforsuchstructurescalledMomentDistribution.Thistechniqueisoneofdevelopingsuccessiveapproximationsandis
basedonseveralbasicconceptsofstructuralbehaviour.
4.7.1Bending(Rotational)Stiffness
Afundamentalrelationshipwhichexistsintheelasticbehaviourofstructuresandstructural
elementsisthatbetweenanappliedforcesystemandthedisplacementswhichareinducedbythat
system,i.e.
Force=StiffnessxDisplacement
P=k
Where
Pistheappliedforce,
kisthestiffness,
isthedisplacement.
Adefinitionofstiffnesscanbederivedfromthisequationbyrearrangingitsuchthat:
k=P/
when =1.0 (i.e.unitdisplacement)thestiffnessis:theforcenecessarytomaintainaUNIT
displacement,allotherdisplacementsbeingequaltozero.
Thedisplacementcanbeasheardisplacement,anaxialdisplacement,abending(rotational)
displacementoratorsionaldisplacement,eachinturnproducingtheshear,bendingortorsional
stiffness.
Whenconsideringbeamelementsincontinuousstructuresusingthemomentdistributionmethod
ofanalysis,thebendingstiffnessistheprincipalcharacteristicwhichinfluencesbehaviour.
ConsiderthebeamelementABshowninFigure4.67whichissubjecttoaUNITrotationatendAand
isfixedatendBasindicated.
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Figure4.67
Theforce(MA)necessarytomaintainthisdisplacementcanbeshowntobeequalto(4EI)/L.From
thedefinitionofstiffnessgivenpreviously,thebendingstiffnessofthebeamisequalto(Force/1.0),
thereforek=(4EI)/L.Thisisknownastheabsolutebendingstiffnessoftheelement.Sincemost
elementsincontinuousstructuresaremadefromthesamematerial,thevalueofYoung'sModulus
(E)isconstantthroughoutand4Einthestiffnesstermisalsoaconstant.Thisconstantisnormally
ignored,togivek=I/Lwhichisknownastherelativebendingstiffnessoftheelement.ItisthisvalueofstiffnesswhichisnormallyusedinthemethodofMomentDistribution.
ItisevidentfromFigure4.67thatwhenthebeamelementdeformsduetotheappliedrotationat
endA,anadditionalmoment(MB)isalsotransferredbytheelementtotheremoteendifithaszero
slope(i.e,isfixed).ThemomentMBisknownasthecarryovermoment.
4.7.2 CarryOverMoment
Usingthe
same
analysis
asthat
todetermine
MA,
itcan
be
shown
that
MB
=(2EI)/L,
i.e.
(1/2
xMA),
It
canthereforebestatedthatifamomentisappliedtooneendofabeamthenamomentofthe
samesenseandequaltohalfofitsvaluewillbetransferredtotheremoteendprovidedthatitis
fixed.
Iftheremoteendispinned,thenthebeamislessstiffandthereisnocarryovermoment.
4.7.3 PinnedEnd
ConsiderthebeamshowninFigure4.68inwhichaunitrotationisimposedatendAasbeforebut
theremoteendBispinned.
Figure4.68
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Theforce(MA)necessarytomaintainthisdisplacementcanbeshown(e.g.usingMcCaulay's
Method)tobeequalto (3EI)/L,whichrepresentsthereducedabsolutestiffnessofapinended
beam.Itcanthereforebestatedthatthestiffnessofapinendedbeamisequalto4
3xthestiffness
ofafixedendbeam.Inadditionitcanbeshownthatthereisnocarryovermomenttotheremote
end.ThesetwocasesaresummarisedinFigure4.69.
Figure4.69
4.7.4FreeandFixedBending;Moments
WhenabeamisfreetorotateatbothendsasshowninFigures4.70(a)and(b)suchthatnobending
momentcandevelopatthesupports,thenthebendingmomentdiagramresultingfromtheapplied
loadsonthebeamisknownastheFreeBendingMomentDiagram.
Figure4.70 FreeBendingMomentDiagrams
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Whenabeamisfixedattheends(encastre)suchthatitcannotrotate,i.e.zeroslopeatthe
supports,asshowninFigure4.71,thenbendingmomentsareinducedatthesupportsandarecalled
FixedEndMoments.Thebendingmomentdiagramassociatedonlywiththefixedendmomentsis
calledtheFixedBendingMomentDiagram.
Figure4.71 FixedBendingMomentDiagram
Usingtheprincipleofsuperposition,thisbeamcanbeconsideredintwopartsinordertoevaluate
thesupportreactionsandtheFinalbendingmomentdiagram:
i. The,fixedreactions(momentsandforces)atthesupportswithoutappliedloadonthebeam
Figure4.72
ii. Thefreereactionsatthesupportsandthebendingmomentsthroughoutthelengthdueto
theappliedload,assumingthesupportstobepinned
Figure4.73
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Combining(i)+(ii)givesthefinalbendingmomentdiagramasshowninFigure4.74:
Figure4.74
Note:
ThevaluesofMAandMBforthemostcommonlyappliedloadcasesaregivenAppendix2.Theseare
standardFixedEndMomentsrelatingtosinglespanencastrebeamsandareusedextensivelyin
structuralanalysis.
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4.7.5 Example4.19:SinglespanEncastreBeam
Determinethesupportreactionsanddrawthebendingmomentdiagramfortheencastrebeam
loadedasshowninFigure4.75.
Figure4.75
Solution:
Considerthebeamintwoparts.
(i)FixedSupportReactions
ThevaluesofthefixedendmomentsaregiveninAppendix2.
Considertherotationalequilibriumofthebeam:
Considertheverticalequilibriumofthebeam:
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(ii)Considertherotationalequilibriumofthebeam:
Considertheverticalequilibriumofthebeam:
BendingMomentunderthepointload=(+13.33x2.0)=+26.67kNm
(Thisinducestensioninthebottomofthebeam)
Thefinalverticalsupportreactionsaregivenby(i)+(ii):
Checktheverticalequilibrium:
Totalverticalforce=+ 14.81+5.19=?
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M=?
Fixedbendingmomentdiagram
Freebendingmomentdiagram
Finalbendingmomentdiagram
Figure4.76
Notethesimilaritybetweentheshapeofthebendingmomentdiagramandthefinaldeflected
shapeasshowninFigure4.77.
Figure4.77Deflectedshapeindicatingtensionzonesandthesimilaritytotheshapeofthebending
momentdiagram.
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4.7.6ProppedCantilevers
Thefixedendmomentforproppedcantilevers(i.e.oneendfixedandtheotherendsimply
supported)canbederivedfromthestandardvaluesgivenforencastrebeamsasfollows.
ConsidertheproppedcantilevershowninFigure4.78,whichsupportsauniformlydistributedloadasindicated.
Figure4.78
Thestructurecanbeconsideredtobethesuperpositionofanencastrebeamwiththeadditionofan
equalandoppositemomenttoMBappliedatBtoensurethatthefinalmomentatthissupportis
equaltozero,asindicatedinFigure4.79.
Figure4.79.
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4.7.7Example4.20:ProppedCantilever
Determinethesupportreactionsanddrawthebendingmomentdiagramfortheproppedcontilever
showninFigure4.80.
Figure4.80.
Solution
FixedEndmomentforProppedCantilever:
Considerthebeamfixedatbothsupports.
ThevaluesofthefixedendmomentsforencastrebeamsaregiveninAppendix2.
Themoment MBmustbecancelledoutbyapplyinganequalandoppositemomentatBwhichin
turnproducesacarryovermomentequalto (0.5xMB)atsupportA.
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Considertherotationalequilibriumofthebeam:
Considertheverticalequilibriumofthebeam:
Considertherotationalequilibriumofthebeam:
Considertheverticalequilibriumofthebeam:
Thefinalverticalsupportreactionsaregivenby(i)+(ii):
Checkthe
vertical
equilibrium:
Total
vertical
force
=+50.0
+30.0
=+80
kN
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Figure4.81
Notethesimilaritybetweentheshapeofthebendingmomentdiagramandthefinaldeflected
shapeasshowninFigure4.82.
Figure4.82
Deflectedshapeindicatingtensionzonesandthesimilaritytotheshapeofthebendingmoment
diagram
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Thepositionofthemaximumbendingmomentcanbedeterminedbyfindingthepointofzeroshear
forceasshowninFigure4.83.
Figure4.83
4.7.8DistributionFactors
Considerauniformtwospancontinuousbeam,asshowninFigure4.84.
Figure4.84
IfanexternalmomentMisappliedtothisstructureatsupportBitwillproducearotationofthe
beamatthesupport;partofthismomentisabsorbedbyeachofthetwospansBAandBC,as
indicatedinFigure4.85.
Figure4.85
Theproportionofeachmomentinducedineachspanisdirectlyproportionaltotherelative
stiffnesses,e.g.
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Totalstiffnessofthebeamatthesupport
ThemomentabsorbedbybeamBA
ThemomentabsorbedbybeamBC
Theratio isknownastheDistributionFactorforthememberatthejointwherethemomentisapplied.
AsindicatedinSection 4.7.2,whenamoment(M)isappliedtooneendofabeaminwhichthe
otherendisfixed,acarryovermomentequalto50%ofMisinducedattheremotefixedendand
consequentlymomentsequalto1/2M1and1/2M2,willdevelopatsupportsAandCrespectively,as
showninFigure4.86.
Figure4.86
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4.7.9 ApplicationoftheMethod
AlloftheconceptsoutlinedinSections 4.7.1to4.7.8areusedwhenanalysingindeterminate
structuresusingthemethodofmomentdistribution.Considerthetwoseparatebeamspans
indicatedinFigure4.87.
Figure4.87
SincethebeamsarenotconnectedatthesupportBtheybehaveindependentlyassimplysupported
beamswithseparatereactionsandbendingmomentdiagrams,asshowninFigure4.88.
Figure4.88
WhenthebeamsarecontinuousoversupportBasshowninFigure4.89(a),acontinuitymoment
developsforthecontinuousstructureasshowninFigures4.89(b)and(c).Notethesimilarityofthe
bendingmomentdiagramformemberABtotheproppedcantileverinFigure4.81.Bothmembers
ABandBDaresimilartoproppedcantileversinthisstructure.
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Figure4.89
Momentdistributionenablestheevaluationofthecontinuitymoments.Themethodisideallysuited
totabularrepresentationandisillustratedinExample4.21.
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4.7.10 Example4.21:ThreespanContinuousBeam
Anonuniform,threespanbeamABCDEFisfixedatsupportAandpinnedatsupportF,asillustrated
inFigure4.90.Determinethesupportreactionsandsketchthebendingmomentdiagramforthe
appliedloadingindicated.
Figure4.90
Solution:
Step1
Thefirststepistoassumethatallsupportsarefixedagainstrotationandevaluatethe'fixedend
moments'.
ThevaluesofthefixedendmomentsforencastrebeamsaregiveninAppendix2.
SpanAC
MAC=?
MCA=?
SpanCD
MCD=?
MDC=?
SpanDF*
MDF=?
MFD=?
*SincesupportFispinned,thefixedendmomentsare(MDF 0.5MFD)atDandzeroatF(seeFigure
4.79):(MDF 0.5MFD)=[ 46.89 (0.5x46.89)]= 70.34kNm.
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Step2
Thesecondstepistoevaluatethememberandtotalstiffnessateachinternaljoint/supportand
determinethedistributionfactorsateachsupport.Notethattheappliedforcesystemisnot
requiredtodothis.
SupportC
SupportD
Stiffness of DC= kDC= ?
ktotal= (? + 0.225) I= 0.475 I
Stiffness of DF= kDF*=
4
3(1.5 I / 5.0)= 0.225 I
*Note: The remote end F is pinned and k=3/4 (I/L) Figure 4.69
Distribution factor (DF) for DC=
?. ?DFs= 1.0
Distribution factor (DF) for DF= ?
Thestructureandthedistributionfactorscanberepresentedintabularform,asshowninFigure
4.91.
Figure4.91
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Thedistributionfactorforfixedsupportsisequaltozerosinceanymomentisresistedbyanequal
andoppositemomentwithinthesupportandnobalancingisrequired.Inthecaseofpinned
supportsthedistributionfactorisequalto1.0since100%ofanyappliedmoment,e.g.bya
cantileveroverhang,mustbebalancedandacarryoverof1/2xthebalancingmomenttransferred
totheremoteendattheinternalsupport.
Step3
Thefixedendmomentsarenowenteredintothetableattheappropriatelocations,takingcareto
ensurethatthesignsarecorrect.
Step4
Whenthestructureisrestrainedagainstrotationthereisnormallyaresultantmomentatatypical
internalsupport.Forexample,considerthemomentsC:
Theoutofbalancemomentisequaltothealgebraicdifferencebetweenthetwo:
Iftheimposedfixityatonesupport(allothersremainingfixed),e.g.supportC,isreleased,thebeam
willrotatesufficientlytoinduceabalancingmomentsuchthatequilibriumisachievedandthe
momentsMCAandMCDareequalandopposite.TheapplicationofthebalancingmomentisdistributedbetweenCAandCDinproportiontothedistributionfactorscalculatedpreviously.
MomentappliedtoCA=+(48.89x0.4)=+19.56kNm
MomentappliedtoCD=+(48.89x0.6)=+29.33kNm
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AsindicatedinSection4.7.2,whenamomentisappliedtooneendofabeamwhilsttheremoteend
isfixed,acarryovermomentequalto(1/2xappliedmoment)andofthesamesignisinducedatthe
remoteend.Thisisenteredintothetableasshown.
Step5
Theprocedureoutlineaboveisthencarriedoutforeachrestrainedsupportinturn.Thereader
shouldconfirmthevaluesgiveninthetableforsupportD.
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Ifthetotalmomentsateachinternalsupportarenowcalculatedtheyare:
MCA=(+4.44+19.56)=+24.0kNm
Thedifference=?kN.m
MCD=( 53.33+29.33+0.62)=?kNm i.e.thevalueofthecarryovermoment
MDC=(+53.33+14.67+1.24)=?kNm
Thedifference=?
MCD=( 70.34+1.10)= 69.24kNm
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Itisevidentthatafteroneiterationofeachsupportmomentthetruevaluesarenearerto23.8kNm
and69.0kNmforCandDrespectively.Theexistingoutofbalancemomentswhichstillexist,0.62
kNm,canbedistributedinthesamemannerasduringthefirstiteration.Thisprocessiscarriedout
untilthedesiredlevelofaccuracyhasbeenachieved,normallyafterthreeorfouriterations.
Aslightmodificationtocarryingoutthedistributionprocesswhichstillresultsinthesameanswersis
tocarryoutthebalancingoperationforallsupportssimultaneouslyandthecarryoveroperation
likewise.Thisisquickerandrequireslesswork.Thereadershouldcompleteafurtherthree/four
iterationstothesolutiongivenaboveandcomparetheresultswiththoseshowninFigure4.92.
*Thefinalcarryover,tothefixedsupportonly,meansthatthisvalueisoneiterationmoreaccurate
thantheinternaljoints.
Figure4.92
ThecontinuitymomentsareshowninFigure4.93.
Figure4.93
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Thesupportreactionsandthebendingmomentdiagramsforeachspancanbecalculatedusing
superpositionasbeforebyconsideringeachspanseparately.
i. FixedSupportReactions
ConsiderspanAC:
Considertheverticalequilibriumofthebeam:
ConsiderspanCD:
Considertheverticalequilibriumofthebeam:
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ConsiderspanDF
Considertheverticalequilibriumofthebeam:
Fixedverticalreactions
Thetotalverticalreactionateachsupportduetothecontinuitymomentsisequaltothealgebraicsumofthecontributionsfromeachbeamatthesupport.
ii. Freebendingmoments
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ConsiderspanAC:
Considertheverticalequilibriumofthebeam:
ConsiderspanCD:
Considertheverticalequilibriumofthebeam:
Considerspan
DF
Considertheverticalequilibriumofthebeam:
Verticalreactionforfreebendingmomentcondition
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Thefinalverticalsupportreactionsaregivenby(i)+(ii):
Checktheverticalequilibrium:Totalverticalforce=+2.58+41.81+109.33+36.28
=+190kN(=totalappliedload)
ThefinalbendingmomentdiagramisshowninFigure4.94.
Figure4.94
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4.7.11Problems:MomentDistribution ContinuousBeams
AseriesofcontinuousbeamsareindicatedinProblems4.28to4.32inwhichtherelativeEIvalues
andtheappliedloadingaregiven.Ineachcase:
i. determinethesupportreactions,ii. sketchtheshearforcediagram and
iii. sketchthebendingmomentdiagram
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