Engineering Basic Mechanics I Statics 工程基础力学Ⅰ 静力学 课件 Chapter 12 Combined Deformation

StaticsStaticsofdeformablebodyChapter12

CombinedDeformation

12.1Introduction12.2Skewbending12.3Combinedaxialloadandbending12.4CombinedbendingandtorsionContents12.1Introduction1.EngineeringexamplesCombinedcompressiveandbendingdeformationCombinedtensileandbendingdeformationcolumnofdrillingmachineCombinedbendingandtorsionaldeformationCombinedcompressiveandbendingdeformationcombineddeformationthedeformationoftenincludesmorethantwobasicdeformations.Forexample:purlinsontheroofframe(skewbending)Fyzq2.Theapproximatestepsforstrengthcalculationofcombineddeformationdecompose+superimposeSimplifyordecomposetheexternalforceintoseveralsimpleforms.Drawtheinternalforcesforeachbasicdeformationandfindthelocationofpossibledangeroussections.Thengivethedistributionofstressesinthedangeroussections.Findthelocationofthedangerpointbasedonthestressdistributioninthedangersection.Usetheprincipleofsuperpositiontodeterminethestressstateatthedangerpoint.Accordingtothestressstateofthedangerpoint，establishthecorrespondingstrengthconditionsandperformstrengthcalculations.12.2SkewbendingPurlinsplaceddiagonallyontheroofframezFjyyjzFyF(b)yzxo(c)oyzxzFTakeacantileverbeamwithrectangularsectionasanexample1.InternalforcesandstressesdecomposeFalongtheprincipalaxisbendingmoments:whereM=F(l-x)isthetotalbendingmomentoftheforceFtothecross-sectionm-m.ommyyFzjz(a)lxxThedirectionsofMyandMz

areshowninFigurea.normalbendingstresscausedbyMz:normalbendingstresscausedbyMy:(b)yzs¢Mz(a)xmyDyMzMm(c)yzs¢¢MyCausedbyCausedby（12-1）Whetherσ'andσ''aretensileorcompressivestressescanbedeterminedbythedeformationofthebarinthespecificproblem.Atthesamepointthesetwostressesareco-linearvectors,andbythesuperpositionprinciplewehave(b)yzs¢Mz(a)xmyDyMzMm(c)yzs¢¢MyCausedbyCausedbyExampleThecantileverbeamshownintheFigureissubjectedtohorizontalforceF1andaperpendicularforceF2attwodifferentsections.F1=800N,F2=1650N,l=2m.Trytofindthemaximumnormalstressinthebeamanditslocationforthefollowingtwocases.(1)Thebeamhasarectangularcrosssectionwithawidthandheightofb=9cm,h=18cm.(2)Thebeamhascircularsectionofdiameterd=13cm.2F1m1md1Fzy2F1m1m1FxObhsolution

RectangularcrosssectioncausedbyF2causedbyF1

Superimposetwostresseszy2F1m1m1FxObh(2)Circularcrosssection：Letthecoordinatesofthedangerpointbe(y,z)causedbyF2causedbyF1

2F1m1md1FSuperimposetwostressesfrom

,wehaveso2.MaximumnormalstressandstrengthconditionsThelinewherethenormalstressiszeroiscalledtheneutralaxisonthesection.Ifthepointontheneutralaxisismarkedas(y0,z0),thenlet

σ=0，Theequationfortheneutralaxiscanbeobtainedas(12-2)yzjp1D2D(a)neutralaxisItcanbeseenthattheneutralaxisisalinepassingthroughthecentroidofthesection.Thepointfurthestfromtheneutralaxishasthehighestnormalstressandisthedangerpoint(D1

andD2).yzjF1D2D(a)neutralaxisThestepsforstrengthcalculation:Determiningthelocationofdangeroussectionsbasedoninternalforces.Calculatethemaximumnormalstressonthedangeroussection,thatis,thenormalstressatthedangerouspoint.Determinationofthedangerpoint:thepointonthedangeroussectionfarthestfromtheneutralaxis.

aFzy(b)2D1DNote:(1)Thelocationofthedangerpointiseasytodetermineforacross-sectionwithangleslikearectangle.(2)Forirregularsections,thepositionoftheneutralaxisshouldbedeterminedfirst,andthenatangentlineparalleltotheneutralaxisshouldbemadeontheperimeterofthesection,andthetangentpointisthedangerpoint.

aFzy(b)2D1DIfthecoordinatesofthedangerpointsare(y1，z1)and(y2，z2),andthetensileandcompressivestrengthsofthematerialareequal,thestrengthconditioncanbeexpressedas

(12-3)Forsectionssuchasrectangles,thestrengthconditioncanbesimplyexpressedas

(12-4)aFzy(b)2D1DNote：（1）Whenthetensileandcompressivestrengthofthematerialaredifferent,themaximumtensilestressandmaximumcompressivestressshouldbecalibratedrespectively.（2）Whenselectingsections,sinceWy,Wzintheintensityconditionsareunknown,theintensityconditionscannotbeusedtodeterminethevaluesofWy,Wzatthesametime.SoweneedtosetavalueofWy/Wz

accordingtoexperienceandthentosolvebytrialcalculation.3.DeflectionofskewbendingThedeflectionofabeaminskewbendingcanalsobecalculatedusingthemethodofsuperposition.FirstlythedeflectionfycausedbyFyandthedeflectionfz

causedbyFzarecalculatedaccordingtotheofplanebending.Andthenthemagnitudeanddirectionofthetotaldeflectionfcanbederivedbyvectorsynthesis.Nowwetakearectangularsectionbeamasanexampletoillustratethemethodoffindingthedeflectionofskewbending.Example

Findthedeflectionofthefreeendofthecantileverbeamshowninthefigure.Byreferringtothetable,wecanget

thetotaldeflectionofthefreeendis

Lettheangleofthetotaldeflectiontothez-axisbe,then

ommyyFzjxzfyyfz

fz12.3CombinedaxialloadandbendingThebarwillbedeformedinacombinationoftension(compression)andbendingundertwotypesofloading:

eccentricaxialloads;

combinedaxialandtransverseloads.FNF1.StrengthchecksTheinternalforcesMy,Mz,andNinthecrosssectionoftherodcorrespondtothepositivestresses.(1)accordingtoMy,MzandN,determinethelocationofthedangeroussectionofthebar;(2)thenbasedontheactualdirectionofMy,MzandN,thelocationofthedangerouspointinthedangeroussectionisdetermined;(3)bysuperimposingthestressescorrespondingtoeachinternalforcecomponentatthedangerpoint,thestressvalueatthedangerpointisobtained.Forrectangularsectionrods,wehavewherethesignisdeterminedbywhethertheactualstressatthedangerpointisintensionorcompression.thedangerouspointonthememberisinauniaxialstressstate,anditsstrengthconditionis

Example

TheshortcolumnshownissubjectedtoloadsFandH.TrytofindtheinternalforcesatthecornerpointsA,B,CandDonthefixedendsection.SolutionInternalforcesinfixedendsection(2)

Geometricparametersofcross-sectionH=5kNF=25kNAyzBCD150100255075(3)CalculatestressesH=5kNF=25kNAyzBCD150100255075AyzBCD150100255075H=5kNF=25kN2.CoreofSectionFigure(a)showsashortcolumnsubjecttoeccentriccompression.Asshowninfigure(b),simplifyFtowardstheaxis:yxzFOpyPzA(a)(b)zxOyzMyM(b)zxOyzMyMInthecrosssectionshown,atpointB(y,z),thestresscomponentscorrespondingtothethreedeformationsareWherethenegativesignindicatesthatthestressiscompressivestress.ByzyzpBythemethodofsuperpositionandtakingintoaccounttherelationshipbetweenIz=Aiz2

andIy=Aiy2,thenormalstressatpointBis

Ifwesee(y0，z0)asthecoordinatesofanypointontheneutralaxis,sincethestressontheneutralaxisisequaltozero,theequationoftheneutralaxiscanbeobtainedasTheneutralaxisineccentriccompressionisastraightlinethatdoesnotpassthroughthesection'scentroid

let

theinterceptsoftheneutralaxisontheyaxisandzaxiscanbeobtainedasByzyzNeutralaxisayazNote:（1）ayandyP,azandzPareofoppositesign.SotheneutralaxisandtheactionpointAoftheexternalforceFareoneachsideofthesectioncentroid.

（2）iftheactionpointofthepressureFgraduallyapproachesthecentroid,theinterceptwillgraduallyincrease.TheneutralaxiswillgraduallymoveawayfromthecentroidofthesectionyxzFopypzA(a)Thereisaareaaroundthecentroid,Whenthepressureisappliedwithinthisarea,theneutralaxiswillnotcrossthecross-section,andtherewillbenotensilestressesinthecross-section.Thisareaiscalledthecoreofsection.yz1ya1234515321za43Determinetheboundaryofthecoreofanarbitrarysection.Theline①canberegardedastheneutralaxis.Letitsinterceptsontheprincipalcentroidalinertiaaxesyandzbeay1,az1.Thepoint1ofexternalforcecorrespondingtothisneutralaxiscanbecalculatedasfollows（*）yz1ya1234515321za43Similarly,otherlines②③...tangenttotheperimeterofthesectioncanberegardedasneutralaxes.Thecoordinatesofcorrespondingpoints2,3...ontheboundaryofthecoreofsectioncanbeobtainedbytheabovemethodinturn.Connectthesepointstoobtainaclosedcurve,whichistheboundaryofthecoreofsection.yz1ya1234515321za4312.4CombinedbendingandtorsionMostofthecommonshaftsinengineeringaresubjectedtobothtorsionalandbendingdeformation,whichiscalledthecombineddeformationofbendingandtorsion.F2zyxF11.Deformationofcircularshaftssubjectedtobendingandtorsion(1)Strengthconditionsdrawthebendingmoment(My

andMz)diagramsandtorque(Mn)diagram.findthetotalbendingmoment

fromthesyntheticdiagramofbendingmomentandthetorque,thepositionofthedangeroussectionofthecircularshaftcanbedetermined.zy1D2DozMyMwM(a)Action

planeThemaximumtorsionalshearstresscorrespondingtothetorqueMninthedangeroussectionoccursattheedgeofthecrosssectionwiththevalue （a）ThenormalstressinbendingcorrespondingtothesyntheticbendingmomentMWismaximumatpointsD1andD2,anditsvalueis（b）1D2Dontntwsws(b)wsws1Dntnt1D(c)wswszy1D2DozMyMwMwM(a)Action

planeThedistributionofshearstressesandnormalstressesarealongthediameterD1D2ofthesectionshowninFigure(b).BothD1

andD2aredangerouspoints.AnelementiscutatpointD1,ThestressesineachsectionoftheelementareshowninFigure(c).zy1D2DozMyMwMwM(a)Action

plane1D2Dontntwsws(b)wsws1Dntnt1D(c)wswsThedangerpointisinaplanestressstate,sostrengthconditionsshouldbeestablishedaccordingtotheoriesofstrength.TheprincipalstressatpointD1is

Accordingtothethirdtheoryofstrength,thestrengthconditionisAftersubstitutingtheprincipalstress,weget

wswswsws1Dntnt1D(c)Accordingtothefourththeoryofstrength,thestrengthconditionisSimilarly,aftersubstitutingtheprincipalstress,wecanget

(2)Strengthconditionsexpressedintheformof“calculatedbendingmoment”

(12-18)Forcircularsections,thereisWn=2W.Thealternativeformulationofthestrengthconditioncanbeobtained.thethirdtheoryofstrength：thefourththeoryofstrength：

(12-19)(12-20)(12-21)

、,

arecalledthe"calculatedbendingmoments"correspondingtothethirdandfourthstrengththeories,respectively.Obviously,afterintroducingtheconceptof"calculatedbendingmoment",Wejustneedtocalculatethevalueofthebendingmomentandtorqueatthedangersection.Andthenthestrengthcheckcanbecarriedoutdirectly.wswswsws1Dntnt1DEquations(12-20)and(12-21)applyonlytocombineddeformationsofbendingandtorsionofcircularorhollowcircularsection.Equations(12-18)and(12-19)aremorewidelyapplicable,Itisapplicableaslongasthedangerouspointisintheplanestressstateshown.(3)CombinedbendingandtorsionaldeformationofacircularshaftwithaxialforcesPropulsionshaftofshipForthestrengthcalculationofthistypeofrod,theformula(12-15)or(12-16)isstillavailable,butσwshouldbereplacedbythesumofthenormalstressinaxialcompression(tension)andthenormalbendingstress.However,fortheCombinedbendingandtorsionaldeformationofacircularshaftwithaxialforces.Theformula(12-15)or(12-16)isunavailable.TeMAB1F2FeMTwswsnt1DExample

AsteelcircularshaftisfittedwithtapewheelsAandB,asshowninFigure(a).BothwheelshavethesamediameterD=1mandweightF=5kN.ThetensioninthetapeonwheelAishorizontalandthetensioninthetapeonwheelBisintheplumbdirection.Theallowablestressforthecircularshaftis[σ]=80Mpa.Trytofindtherequireddiameteroftheshaftaccordingtothethirdtheoryofstrength.

zyxABCD5kN2kN2kN300500500(a)5kNsolution：Simplifytheexternalforcestowardtheaxis.Makethetorquediagram(Mn)andthebendingmomentdiagrams(MyMz).ThesyntheticbendingmomentsofsectionsCandBarezyxABCD2kN2kN3005005005kN7kN1.5kNm12kNnMxxzMyxzyM5kN5kN1.5kNm1.5kNm2.1kNm2.25kNm1.5kNmSyntheticbendingmomentdiagramMwisshowninFigure.ItcanbeprovedthatthecurveofdiagramMwisconcave.Obviously,sectionCisadangeroussection.(3)Accordingtothethirdtheoryofstrength:Bysubstitutingthedata,thereareTherequireddiameterisABCD2kN5kNxwM2.58kNm2.48kNm2kN5kNExample

Figure(a)showsapropulsionshaftofacargoship.ThepowerofthemainengineisknownasN=7277kW,speedn=119r/min,effectivepropulsiveforceT=767kN,theweightofthepropellerbladesF1=180kN,thetotalweightoftheshaftoutreachF2=45kN,diameteroftheshaftd=51.5cm.Andthereisa1=1.9m,a2=1.2m.Thematerialishighqualitycarbonsteelwithayieldstressofσs=250MPa,allowablefactorofsafety[n]=4.CheckthestrengthofsectionAofthepropulsionshaftaccordingtothefourththeoryofstrength.TMAB1F2F2a1aMT(a)d(2)InternalforcesinsectionA

FN=T=767kNyzMnMaFNAsectionTeMAB1F2F2a1aeMT(a)dsolution(1)Torsionalcouplemoment(3)stressesatthepointofdangerNormalstressesduetobendingmomentsandaxialforcesareshowninFigure.Thecompressivestressisgreatestatpointaofthelowersectionedge,withavalueofThemaximumtorsionalshearstressoccursattheedgeofthecircularsectionandthevalueofpointaisBythefourththeoryofStrength,wecanobtainWsNsspointzMnMyaFNAsection(4)CalculatethefactorofsafetyThefactorofsafetyofashaftrepresentsthesafetyreserveoftheshaftinoperationandisequaltothebreakingstressdividedbytheworkingstress.

Thereforetheshaftissafe.PleaseconsiderthefollowingquestionIfthethirdtheoryofstrengthisadopted,canthestre