毕业设计数控车床外文

毕业设计数控车床外文五、外文资料翻译StressandStrainIntroductiontoMechanicsofMaterialsMechanicsofmaterialsisabranchofappliedmechanicsthatdealswiththebehaviorofsolidbodiessubjectedtovarioustypesofloading.Itisafieldofstudythatisknownbyavarietyofnames,including“strengthofmaterials”and“mechanicsofdeformablebodies”.Thesolidbodiesconsideredinthisbookincludeaxially-loadedbars,shafts,beams,andcolumns,aswellasstructuresthatareassembliesofthesecomponents.Usuallytheobjectiveofouranalysiswillbethedeterminationofthestresses,strains,anddeformationsproducedbytheloads;ifthesequantitiescanbefoundforallvaluesofloaduptothefailureload,thenwewillhaveobtainedacompletepictureofthemechanicsbehaviorofthebody.Theoreticalanalysesandexperimentalresultshaveequallyimportantrolesinthestudyofmechanicsofmaterials.Onmanyoccasionwewillmakelogicalderivationstoobtainformulasandequationsforpredictingmechanicsbehavior,butatthesametimewemustrecognizethattheseformulascannotbeusedinarealisticwayunlesscertainpropertiesofthebeenmadeinthelaboratory.Also,manyproblemsofimportanceinengineeringcannotbehandledefficientlybytheoreticalmeans,andexperimentalmeasurementsbecomeapracticalnecessity.Thehistoricaldevelopmentofmechanicsofmaterialsisafascinatingblendofboththeoryandexperiment,withexperimentspointingthewaytousefulresultsinsomeinstancesandwiththeorydoingsoinothers①.SuchfamousmenasLeonardodaVinci(1452-1519)andGalileoGalilei(1564-1642)madeexperimentstoadequatetodeterminethestrengthofwires,bars,andbeams,althoughtheydidnotdevelopanyadequatetheories(bytoday’sstandards)toexplaintheirtestresults.Bycontrast,thefamousmathematicianLeonhardEuler(1707-1783)developedthemathematicaltheoryanyofcolumnsandcalculatedthecriticalloadofacolumnin1744,longbeforeanyexperimentalevidenceexistedtoshowthesignificanceofhisresults②.Thus,Euler’stheoreticalresultsremainedunusedformanyyears,althoughtodaytheyformthebasisofcolumntheory.Theimportanceofcombiningtheoreticalderivationswithexperimentallydeterminedpropertiesofmaterialswillbeevidenttheoreticalderivationswithexperimentallydeterminedpropertiesofmaterialswillbeevidentasweproceedwithourstudyofthesubject③.Inthissectionwewillbeginbydiscussingsomefundamentalconcepts,suchasstressandstrain,andthenwewillinvestigatebathebehavingofsimplestructuralelementssubjectedtotension,compression,andshear.StressTheconceptsofstressandstraincanbeillustratedinelementarywaybyconsideringtheextensionofaprismaticbar[seeFig.1.4(a)].Aprismaticbarisonethathascrosssectionthroughoutitslengthandastraightaxis.InthisillustrationthebarisassumedtobeloadedatitsendsbyaxisforcesPthatproduceauniformstretching,ortension,ofthebar.Bymakinganartificialcut(sectionmm)throughthebaratrightanglestoitsaxis,wecanisolatepartofthebarasafreebody[Fig.1.4(b)].Attheright-handendtheforcePisapplied,andattheotherendthereareforcesrepresentingtheactionoftheremovedportionofthebaruponthepartthatremain.Theseforceswillbecontinuouslydistributedoverthecrosssection,analogoustothecontinuousdistributionofhydrostaticpressureoverasubmergedsurface.Theintensityofforce,thatis,theperunitarea,iscalledthestressandiscommonlydenotedbytheGreekletterб.Assumingthatthestresshasauniformdistributionoverthecrosssection[seeFig.1.4(b)],wecanreadilyseethatitsresultantisequaltotheintensityбtimesthecross-sectionalareaAofthebar.Furthermore,fromtheequilibriumofthebodyshowinFig.1.4(b),Fig.1.4PrismaticbarintensionwecanalsoseethatthisresultantmustbeequalinmagnitudeandoppositeindirectiontotheforceP.Hence,weobtainб=P/A(1.3)astheequationfortheuniformstressinaprismaticbar.Thisequationshowsthatstresshasunitsofforcedividedbyarea--------forexample,Newtonspersquaremillimeter(N/mm²)orpoundsofpersquareinch(psi).WhenthebarisbeingstretchedbytheforcesP,asshowninthefigure,theresultingstressisatensilestress;iftheforcearereversedindirection,causingthebattobecompressed,theyarecalledcompressivestress.AnecessaryconditionforEq.(1.3)tobevalidisthatthestressбmustbeuniformoverthecrosssectionofthebat.Thisconditionwillberealizediftheaxialforcepactsthroughthecentroidofthecrosssection,ascanbedemonstratedbystatics.WhentheloadPdosesnotactatthuscentroid,bendingofthebarwillresult,andamorecomplicatedanalysisisnecessary.Throughoutthisbook,however,itisassumedthatallaxialforcesareappliedatthecentroidofthecrosssectionunlessspecificallystatedtothecontrary④.Also,unlessstatedotherwise,itisgenerallyassumedthattheweightoftheobjectitselfisneglected,aswasdonewhendiscussingthisbarinFig.1.4.3.StrainThetotalelongationofabarcarryingforcewillbedenotedbytheGreekletterб[seeFig.1.4(a)],andtheelongationperunitlength,orstrain,isthendeterminedbytheequationε=δ/L(1.4)WhereListhetotallengthofthebar.Nowthatthestrainεisanondimensionalquantity.ItcanbeobtainedaccuratelyformEq.(1.4)aslongasthestrainisuniformthroughoutthelengthofthebar.Ifthebarisintension,thestrainisatensilestrain,representinganelongationorastretchingofthematerial;ifthebarisincompression,thestrainisacompressivestrain,whichmeansthatadjacentcrosssectionofthebarmoveclosertooneanother.(SelectedfromStephenP.TimoshenkoandJamesM.Gere,MechanicsofMaterials,VanNostrandReinholdCompanyLtd.,1978.)ShearForceandBendingMomentinBeamsLetusnowconsider,asanexample,acantileverbeamacteduponbyaninclinedloadPatitsfreeend[Fig.1.5(a)].Ifwecutthroughthebeamatacrosssectionmnandisolatetheleft-handpartofthebeamasfreebody[Fig.1.5(b)],weseethattheactionoftheremovedpartofthebeam(thatis,theright-handpart)upontheleft-handpartmustastoholdtheleft-handinequilibrium.Thedistributionofstressesoverthecrosssectionmnisnotknownatthisstageinourstudy,butweedoknowthattheresultantofthesestressesmustbesuchastoequilibratetheloadP.ItisconvenienttoresolvetotheresultantintoanaxialforceNactingnormaltothecrosssectionandpassingthroughthecentriodofthecrosssection,ashearforceVactingparalleltothecrosssection,andabendingmomentMactingintheplaneofthebeam.Theaxialforce,shearforce,andbendingmomentactingatacrosssectionofabeamareknownasstressresultants.Forastaticallydeterminatebeam,thestressresultantscanbedeterminedfromequationsofequilibrium.Thus,forthecantileverbeampicturedinFig.1.5,wemaywriterthreeequationsofstacticsforthefree-bodydiagramshowninthesecondpartofthefigure.Fromsummationsofforcesinthehorizontalandverticaldirectionswefind,respectively,N=PcosβV=Psinβand,fromasummationofmomentsaboutanaxisthroughthecentroidofcrosssectionmn,weobtainM=Pxsinβwherexisthedistancefromthefreeendtosectionmn.Thus,throughtheuseofafree-bodydiagramandequationsofstaticequilibrium,weareabletocalculatethestressresultantswithoutdifficulty.ThestressinthebeamduetotheaxialforceNactingalonehavebeendiscussedinthetextofUnit.2;NowwewillseehowtoobtainthestressesassociatedwithbendingmomentMandtheshearforceV.ThestressresultantsN,VandMwillbeassumedtobepositivewhenthetheyactinthedirectionsshowninFig.1.5(b).Thissignconventionisonlyuseful,however,whenwearediscussingtheequilibriumoftheleft-handpartofthebeamisconsidered,wewillfindthatthestressresultantshavethesamemagnitudesbutoppositedirections[seeFig.1.5(c)].Therefore,wemustrecognizethatthealgebraicsignofastressresultantdoesnotdependuponitsdirectioninspace,suchastotheleftortotheright,butratheritdependsuponitsdirectionwithrespecttothematerialagainst,whichitacts.Toillustratethisfact,thesignconventionsforN,VandMarerepeatedinFig.1.6,wherethestressresultantsareshownactingonanelementofthebeam.Weseethatapositiveaxialforceisdirectedawayfromthesurfaceuponwhichisacts(tension),apositiveshearforceactsclockwiseaboutthesurfaceuponwhichitacts,andapositivebendingmomentisonethatcompressestheupperpartofthebeam.ExampleAsimplebeamABcarriestwoloads,aconcentratedforcePandacoupleMo,actingasshowninFig.1.7(a).Findtheshearforceandbendingmomentinthebeamatcrosssectionslocatedasfollows:(a)asmalldistancetotheleftofthemiddleofthebeamand(b)asmalldistancetotherightofthemiddleofthebeam.SolutionThefirststepintheanalysisofthisbeamistofindthereactionsRAandRB.TakingmomentsaboutendsAandBgivestwoequationsofequilibrium,fromwhichwefindRA=3P/4–Mo/LRB=P/4+mo/LNext,thebeamiscutatacrosssectionjusttotheleftofthemiddle,andafree-bodydiagramisdrawnofeitherhalfofthebeam.Inthisexamplewechoosetheleft-handhalfofthebean,andthecorrespondingdiagramisshowninFig.1.7(b).TheforcepandthereactionRAappearinthisdiagram,asalsodotheunknownshearforceVandbendingmomentM,bothofwhichareshownintheirpositivedirections.ThecoupleModoesnotappearinthefigurebecausethebeamiscuttotheleftofthepointwhereMoisapplied.AsummationofforcesintheverticaldirectiongivesV=R–P=-P/4-M0/LWhichshownthattheshearfor