工程基础力学Ⅰ 静力学 课件 Chapter 5 Friction、Chapter 6 Fundamental Concepts of Bar Deformation

StaticsStaticsofrigidChapter5Friction§5.1Slidingfriction§5.2Frictionangleandself-lockingphenomenon§5.3Equilibriumproblemswithfriction§5.4RollingresistanceMaincontents1.Concept:Twoobjectswithroughsurfaceincontactwitheachother,whenthereisarelativeslidingorrelativeslidingtrendbetweenitscontactsurface,thereisresistancetoeachotherthathindersrelativesliding,thatisslidingfriction.5.1Slidingfriction2.Status:

(1)Static:(equilibriumconditions)

(2)Critical:;fs—thecoefficientofstaticfriction(3)sliding:f'—thecoefficientofkineticfriction5.2Frictionangleandself-lockingphenomenon1.FrictionangleIfthemagnitudeofthestaticforceoffrictionincreasesfromzerotothemaximumvalue,themaximumanglewhichthetotalreactionofaroughsupportmakeswiththenormaltothesurfaceiscalledtheangleoffriction.Calculation:5.2Frictionangleandself-lockingphenomenon2.Self-lockingphenomenonAbodycanbeatrestbythemutualactionoftheforceoffrictionandthenormalpressure(i.e.,fullReactionForce),itwillnotslide(nomatterhowgreatanappliedforceis),thisphenomenoniscalledself-locking.Conditionsofself-locking:Whenthesystemisalwaysinequilibrium(i.e.,itisself-locking).5.3EquilibriumproblemswithfrictionThefrictionforceisusuallyunknown,whilethedirectionoffrictionforceisoppositetothedirectionoftherelativesliding

tendency.Tosolvetheequilibriumproblemwiththefrictionforce,itisnecessarytolistthesupplementaryequationsFs≤fFN,andthenumberofsupplementaryequationsisthesameasthenumberofthefrictionforces.Example1haαADGFThehomogeneouswoodenboxweighsG=5kNandhasastaticfrictionfactorfs=0.4withtheground,h=2a=2mand.(1)AskwhethertheboxisinequilibriumwhenthetensionF=1kNatD?(2)Findthemaximumtensionforcethatwillkeeptheboxinbalance.5.3EquilibriumproblemswithfrictionTakethewoodenboxastheobjectofstudy,forceanalysisasshowninthefigure.hdaαADGFfFNFEquilibriumequationsBecauseFf<Fmax,sothewoodenboxdoesnotslide.Andbecaused>0,thewoodenboxwillnottoppleover.Solution：Tomaintainthebalanceofthewoodenbox,itisnecessaryto(1)Cannotslide,Ff<Fmax=fsFN.(2)DonottopplearoundpointA,d>0.SolvingtheequationyieldsThemaximumfrictionbetweenthewoodenboxandthegroundis5.3EquilibriumproblemswithfrictionEquilibriumequationsTheconditionsunderwhichthewoodenboxslidesare

Ff=Fmax=fsFNhdaαADGFfFNFThesolutionisTheconditionthatthewoodenboxtopplesoveraroundpointAisd=0,then2.Tofindthemaximumtensileforceatequilibrium,thatis,tofindtheminimumforce

Fintheslidingcriticalandoverturningcritical.F=FF=1443NSinceFF<Fs,themaximumtensiontokeepthewoodenboxinbalanceis5.4Rollingresistance(a)(b)(c)(d)Whenthewheelisequilibrious,theequilibriumequationsThecouplemomentMf

calledrollingresistancecouplemoment5.4RollingresistanceMf

rollingresistance(1)therollingresistancecouplemomentMfincreaseswiththeincreaseoftheappliedforce;(2)

therollingresistancecouplemomentreachesitsmaximumvalue,calledthemaximumrollingresistantcouplemomentMfmax;IfFbecomelarger,thewheelwillroll,andtherollingresistantcouplemomentisapproximatelyequaltoMmaxduringtherollingprocess,andthereis5.4RollingresistanceMf

rollingresistanceRollingresistancelaw:ThemaximumrollingresistancecouplemomentMmaxisindependentoftherollerradiusandproportionaltothemagnitudeofthenormalforceFNonthesupportsurface,

discalled

thecoefficientofrollingresistance:(1)Itsdimensionisthatofalength.Generally,theunitismmorcm;(2)Itdependsontherigidityandtemperatureofthematerialsoftherollerandofthesupportingsurface;(3)Thephysicalmeaningofdisshowninthefigure.δ=d(1)

TheEnd

StaticsStaticsofdeformablebodyChapter6

FundamentalConceptsofBarDeformation6.1Tasks6.2Simplificationofengineeringmembers6.3Internalforceandstress6.4Displacementandstrain6.5BasicformsofbardeformationContents6.1Tasksstructuralmember：Thepartsofengineeringstructures(variousmachines,instruments,andbuildingstructures,etc.)arecollectivelyreferredtoasmembers.Somebasicrequirementsmetbythemembers：1.sufficientstrengthItisrequiredthatthemembersarenotfracturedorpermanentlydeformed(plasticdeformation)whensubjectedtoload.Theabilityofthemembertoresistdamage.Forexample:craneslingsarenotallowedtobreak,etc2.sufficientrigidityTheelasticdeformationofthememberundertheexternalloadcannotexceedtheallowedvalue.Theabilityofamembertoresistelasticdeformationunderload.Forexample:lathespindle,housefloor,etc.3.adequatestabilityItisrequiredthatthebalanceofthememberisstableunderworkingconditions.Theabilityofthemembertomaintainitsoriginalformofbalance.Forexample:jackingrods,pillarsinmines,etc.Theabilityofthemembertomeettherequirementsofstrength,stiffnessandstabilityiscalledtheload-bearingcapacityofthemember.Bystudyingthestrength,stiffnessandstabilityofmembers,andthemechanicalpropertiesofmaterials,thebasictheoryandcalculationmethodsareprovidedfortheselectionofappropriatematerials,thedeterminationofreasonablecross-sectionalshapesanddimensionsofmembersunderthepremiseofensuringsafety,reliabilityandeconomicsavings.Maintasks：Ⅰ.Basicformsofmembers1.Bar

abarisamemberwhoselengthismuchgreaterthanitstransversedimensions(heightandwidth).bar6.2Simplificationofengineeringmembersbody2.block

Memberswithequivalentdimensionsinthelength,widthandheightdirectionsarecalledblocks.Ⅰ.Basicformsofmembersplateshell3.plate

(shell)

Aplate(shell)isamemberwhosethicknessismuchsmallerthanthedimensionsofothertwodirections.Ifthemiddlesurfaceisflat,thememberiscalledaplate.Ifthemiddlesurfaceiscurved,itiscalledashell.Ⅰ.BasicformsofmembersTheassumptionofcontinuity：thememberiscontinuouslyandvoid-freethroughoutitsgeometricvolume.(Physicalquantitiescanbeexpressedasacontinuousfunctionofcoordinates)Theassumptionofhomogeneity:themechanicalpropertiesareidenticalatallpointsinthemember.(anysmallsegmentoftheobjectcanberemovedforanalysisandtheconclusionsobtainedcanbeappliedtothewholeobject)Theassumptionofisotropic：materialshavethesamemechanicalpropertiesalonganydirection.Ⅱ.BasicassumptionsDeformationofdeformablebodyunderloadcanoccurintwodifferentnatureofdeformation：

1.elasticdeformation

2.plasticdeformationElasticrange：Undertheactionofexternalload,theexternalforcerangewithonlyelasticdeformationandnoplasticdeformation(negligible)istheelasticrange.

Smalldeformation：Thedeformationofthememberisextremelysmallcomparedtoitsoriginaldimensions.Ⅲ.Smalldeformationconstraints1.Internalforce——Interactionforcesbetweenapartofamemberanditsadjacentparts

Additionalinternalforce

InternalforcesaredistributedoverthecrosssectionInternalforceispaired2.Sectionmethod：Themethodforshowinganddeterminingtheinternalforcesisthesectionmethod.Thecross-sectionalmethodisbasedontheprinciplethat"anobjectinequilibriumshouldalsohaveallitspartsinequilibrium".6.3InternalforceandstressTheprocessoffindingtheinternalforcebythesectionmethodcanbesummarizedasfollows：(1)CuttingImaginecuttingthememberalongthatsectionanddividingitintotwosegments.(2)Substituteforce

Selectanyonesegmentofthememberandtheotherisremoved.Theactionoftheremovedsegmentontheselectedsegmentisreplacedbythecorrespondinginternalforces.(3)EquilibriumSolvetheinternalforceproblemsusingequilibriumequations3.stressConceptTheinternalforcesatapointiscalledthestressatthatpoint.Lettheinternalforceactingonasmallareabe,thentheaveragesetoftheinternalforceonis（isaveragestress）ThickrodThinrodquestionCADPDCADPDCtsPWhenPointC（infinitelysmall）AlimitvalueWecallPthetotalstressatpointC.Theexpressionis：Theunitofstressis[force]/[length]2(N/m2)

,whichis.Pisavector，Fortheconvenienceandneedofthestudy,wealwaysdecomposeitintotwocomponentsperpendiculartothesectionandparalleltothesection.Theformeriscalledthenormalstressandisdenotedbyσ.Thelatteriscalledshearstressandisdenotedbyτ.CtsPThedifferencebetween"stress"and"pressure"Although"stress"hasthesame