class moment dis week 9-12

8/2/2019 Class Moment Dis Week 9-12

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Moment Distribution Method -----------------------------------------------------------------------------------------------------------------------------------

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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:


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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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:


ConsiderspanCD:



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ConsiderspanDF


Fixedverticalreactions

Thetotalverticalreactionateachsupportduetothecontinuitymomentsisequaltothealgebraicsumofthecontributionsfromeachbeamatthesupport.

ii. Freebendingmoments


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ConsiderspanAC:


ConsiderspanCD:


Considerspan

DF


Verticalreactionforfreebendingmomentcondition


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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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class moment dis week 9-12

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