机械工程导论(双语)课件四

mechanicalengineeringAnintroductionto1Lecturer:LiuJu-rongCHAPTER4ForcesinStructuresOVERVIEW1FORCESANDRESULTANTS23MOMENTOFAFORCE

2EQUILIBRIUMOFFORCESANDMOMENTS4Vocabulary3Mechanics力学neteffect净效应leverarm杆臂Quadrant象限Bracket支架Protractor量角器Eyebolt吊环螺栓pivotpoint支点Particle质点RigidBody刚体FreeBodyDiagram自由体受力图boldfacenotation粗体字的标识1OVERVIEW4Mechanicalengineersusemathematicsandphysicallawstodesignhardwarebetterandfasterthanwouldbepossibleotherwise.Byapplyingtheprincipleofforcebalance,forinstance,anengineercanoftenanalyzeadesigntoareasonablelevelofaccuracybeforeanyhardwareisbuilt.Engineersreducethetimeandexpenseassociatedwithconstructingandtestingprototypesbyfirstrefiningtheirdesignsonpaper.Computer-aidedengineeringtoolsfurtherincreasethelevelofsophisticationthatisavailableforsuchanalyses.1OVERVIEW---Theabilitytobeavailableinthischapter5Afterstudy,whatshouldyoubeabletohave?Describeaforceintermsofitsrectangularandpolarcomponents.Calculatetheresultantofasystemofforcesbyusingthevectoralgebraandpolygonmethods.Calculatethemomentofaforceaboutapointusingtheperpendicularleverarmandmomentcomponentmethods.Understandtherequirementsforequilibrium,andcalculateunknownforces.Explainthedifferencebetweenlaminarandturbulentflowingfluids.CalculateanddescribethedimensionlessReynoldsandMachnumbers.Discussthefluidforcesknownasbuoyancy,drag,andlift,andcalculatethemincertainapplication.1OVERVIEW6FINGURE4.1Heavyconstructionequipmentisdesignedtosupportthelargeforcesdevelopedduringoperation.Source:ReprintedwithpermissionofmechanicalDynamics,Incorporated,andbyCaterpillarIncorporated.2FORCESANDRESULTANTS---RectangularandPolarForms

Whatisvectornotation?

Rectangularcomponents7Inthistextbook,wewilluseboldfacenotation--F--todenoteforcevectors.Acommonmethodtodescribeaforceisintermsofitshorizontalandverticalcomponents.TheprojectionofFinthehorizontaldirection(thexRectangularaxis)iscalledFx,andtheverticalprojection(yaxis)iscalledFy.Infact,thepairofnumbers(Fx,Fy)isjustthecoordinatesoftheforcevector'stip.2FORCESANDRESULTANTS---RectangularandPolarForms

Unitvectors8theunitvectorsiandjareusedtoindicatethedirectionsinwhichFxandFyact.Vectoripointsalongthepositivexdirection,andjisavectorpointinginthepositiveydirection.F=Fxi+FyjFINGURE4.2Representingaforcevectorintermsofitsrectangularcomponents(Fx,Fy),anditspolarcomponents(F,θ).2FORCESANDRESULTANTS---RectangularandPolarForms

polarcomponentsVectormagnitude9Thelatterviewpointisbasedonpolarcoordinates.AsalsoshowninFigure,Factsattheangleθ,whichismeasuredrelativetothehorizontalaxis.Themagnitudeorlengthoftheforcevectorisascalarquantity,anditisdenotedbyF=|F|,wherethe|·|notationdesignatesthevector'sabsolutevalue.VectordirectionInsteadofspecifyingFxandFy,wecannowviewVectordirectiontheforcevectorFintermsofthequantitiesFandθ.Fx=FcosθandFy=Fsinθ

2FORCESANDRESULTANTS---RectangularandPolarForms10FINGURE4.3Determiningtheangleofactionforaforcethat(a)liesinthefirstquadrantand(b)liesinthesecondquadrant.2FORCESANDRESULTANTS---Resultants

Forcesystem11Aforcesystemisacollectionofseveralforcesthatsimultaneouslyactonanobject.WithNindividualforcesdenotedbyFi(i=1,2,…,N),theyaresummedaccordingtobyusingtherulesofvectoralgebra.Resultant2FORCESANDRESULTANTS---Resultants

vectoralgebramethod12Inthistechnique,eachforceFiisbrokendownintoitshorizontalandverticalcomponents,whichwelabelasFxiandFyifortheit’sforce.Theresultant'shorizontalcomponentRxisfoundbyLikewise,weseparatelysumtheverticalcomponentsthroughTheresultantforceisthenexpressedasR=Rxi+Ryj.IfweareinterestedinthemagnitudeRanddirectionθofR,weapplytheexpressions2FORCESANDRESULTANTS---Resultants

VectorpolygonMethod----head–to–tailrule13Alternatively,theresultantofaforcesystemcanbefoundbysketchingapolygontorepresentadditionoftheFivectors.Themagnitudeanddirectionoftheresultantarethendeterminedbyapplyingrulesoftrigonometrytothepolygon'sgeometry.ReferringtothemountingpostofFigure4.4,thevectorpolygonforthosethreeforcesisdrawnbyaddingtheindividualFi'sinachainHead-to-tail(头尾相接)ruleaccordingtothehead-to-tailrule.2FORCESANDRESULTANTS---Resultants14FINGURE4.4Amountingpostandbracketthatareloadedbythreeforces.FINGURE4.5ThebracketRextendsfromthestarttotheendofthechainformedbyaddingF1,F2andF3together3MOMENTOFAFORCE---perpendicularleverarm15Thetermtorquecanalsobeusedtodescribetheeffectaforceactingoveraleverarm,butmechanicalengineersgenerallyreservetorquetodescribemomentsthatcauserotationofashaftinamotor,engine,orgearbox.

ThemagnitudeofamomentisfoundfromitsdefinitionMo=Fdanddistheperpendicularleverarmdistancefromtheforce'slineofactiontopointO.

momentmagnitude

leverarm

torque3MOMENTOFAFORCE---perpendicularleverarm16theunitforMoistheproductofforceanddistance.Infact,Fcouldbeappliedtothebracketatanypointalongitslineofaction,andthemomentproducedaboutOwouldremainunchangedbecausedwouldlikewisenotchange.MomentunitsLineofaction3MOMENTOFAFORCE---momentcomponents17Wefirstchoosethefollowingsignconvention:Amomentthatisdirectedclockwiseispositive,andacounterclockwisemomentisnegative.Thesignconventionisjustabookkeepingtoolforcombiningthevariousclockwiseandcounterclockwisemomentcomponents.Signconvection3MOMENTOFAFORCE---momentcomponents18FINGURE4.11Calculatingmomentsbasedoncomponents.(a)BothFxandFycreateclockwisemomentsaboutpointO.(b)Fxexertsaclockwisemoment,butFyexertsacounterclockwisemoment.3MOMENTOFAFORCE---momentcomponents19Regardlessofwhichmethodyouusetocalculateamoment,whenreportinganansweryoushouldstate(1)thenumericalmagnitudeofthemoment,(2)theunits,and(3)thedirection.(CWorCCW)Inthegeneralcaseofthemomentcomponentsmethod,wewriteMo=±Fx∆y±Fy∆x20thephysicaldimensionsoftheobjectareunimportantincalculatingforces.thelength,width,andbreadthofanobjectareimportantfortheproblemathand.particlerigidbody4EQUILIBRIUMOFFORCESANDMOMENTS--ParticlesandRigidBodies21Aparticleisinequilibriumiftheforcesactingonitbalancewithzeroresultant.Becauseforcescombineasvectors,theresultantmustbezerointwoperpendiculardirections,whichwelabelxandy:Forarigidbodytobeinequilibrium,itisnecessarythat(1)theresultantofallforcesiszero,and

(2)thenetmomentisalsozero.ForcebalanceMomentbalance4EQUILIBRIUMOFFORCESANDMOMENTS--ParticlesandRigidBodies∑Fxi=0and∑Fyi=0∑Fxi=0and∑Fyi=0∑Moi=0Itisnotpossibletoobtainmoreindependentequationsofequilibriumbyresolvingmomentsaboutanalternativepointorbysummingforcesindifferentdirections.22Freebodydiagrams(abbreviatedFBD)aresketchesusedtoanalyzetheforcesandmomentsthatactonstructuresandmachines,andtheirconstructionisanimportantskill.TheFBDisusedtoidentifythemechanicalsystemthatisbeingexaminedandtorepresentalloftheknownandunknownforcesthatarepresent.FBD4EQUILIBRIUMOFFORCESANDMOMENTS–Freebodydiagrams23thephysicaldimensionsoftheobjectareunimportantincalculatingforces.thelength,width,andbreadthofanobjectareimportantfortheproblemathand.particlerigidbody4EQUILIBRIUMOFFORCESANDMOMENTS–FreeBodyDiagrams4EQUILIBRIUMOFFORCESANDMOMENTS–FreeBodyDiagrams24ThreemainstepsarefollowedwhenaFBDisdrawn:3.Inthefinalstep,allforcesandmomentsaredrawnandlabeled.2.Thecoordinatesystemisdrawnnexttoindicatethepositivesignconventionsforforcesandmoments.1.Selectanobjectthatwillbeanalyzedbyusingtheequilibriumequations.Imaginethatadottedlineisdrawnaroundtheobject,andnotehowthelinewouldcutthroughandexposevariousforces.4EQUILIBRIUMOFFORCESANDMOMENTS–FreeBodyDiagrams25SOLUTION(a)ThefreebodydiagramofthebuckleisshowninFigure4.15(b).Thexycoordinatesystemisalsodrawntoindicateoursignconventionforthepositivehorizontalandverticaldirections.Threeforcesactonthebuckle:thetwogiven300-1bforcesandtheunknownforceintheanchorstrap.Forthebuckletobeinequilibrium,thesethreeforcesmustbalance.AlthoughboththemagnitudeTanddirection0oftheforceinstrapABareunknown,bothquantitiesareshownonthefreebodydiagramforcompleteness.EXAMPLEDuringcrashtestingofanautomobile,thelapandshoulderseatbeltseachbecometensionedto300lb,asshowninFigure4.15(a).TreatingthebuckleBasaparticle,(a)drawafreebodydiagram,(b)determinethetensionTintheanchorstrapAB,and(c)determinetheangleatwhichTacts.4EQUILIBRIUMOFFORCESANDMOMENTS–FreeBodyDiagrams26FINGURE4.15EquilibriumanalysisoftheseatbeltlatchinExample4.5.4EQUILIBRIUMOFFORCESANDMOMENTS–FreeBodyDiagrams27(b)Wesumthethreeforcesbyusingthevectorpolygonapproach,asshowninFigure4.15(c).Thepolygon'sstartandendpointsarethesamebecausethethreeforcesactingtogetherhavezeroresultant;thatis,thedistancebetweenthepolygon'sstartandendpointsiszero.Thetensionisdeterminedbyapplyingthelawofcosines(equationsforobliquetrianglesarereviewedinAppendixB)totheside-angle-sidetriangleinFigure4.15(c):T2=(300lb)2+(300lb)2-2(300lb)(300lb)cos120ofromwhichwecalculateT=519.6lb.(c)Theanchorstrap'sangleisfoundfromthelawofsines:

andθ=30o.SUMMARY28Inthischapter,weintroducedtheengineeringscienceconceptsofforces,resultants,moments,andequilibrium.Weexaminedthosequantitiesinthecontextofforcesactingonmachines