Chapter2 FundamentalsoftheMechanicalBehaviorof Materials
QUESTIONS
2.1 Canyou calculatethepercentelongationofmaterials basedonlyontheinformationgiveninFig.2.6?Explain.
RecallthatthepercentelongationisdefinedbyEq.(2.6) onp.35anddependsontheoriginalgagelength(lo)of thespecimen.FromFig.2.6onp.39,onlythenecking strain(trueandengineering)andtruefracturestrain canbedetermined.Thus,wecannotcalculatethepercentelongationofthespecimen;also,notethatthe elongationisafunctionofgagelengthandincreases withgagelength.
2.2 Explainifitispossibleforstress-straincurvesintensionteststoreach0%elongationasthegagelengthis increasedfurther.
Thepercentelongationofthespecimenisafunctionof theinitialandfinalgagelengths.Whenthespecimenis beingpulled,regardlessoftheoriginalgagelength,it willelongateuniformly(andpermanently)untilneckingbegins.Therefore,thespecimenwillalwayshave acertainfiniteelongation.However,notethatasthe specimen’sgagelengthisincreased,thecontribution oflocalizedelongation(thatis,necking)willdecrease, butthetotalelongationwillnotapproachzero.
2.3 Explainwhythedifferencebetweenengineeringstrain andtruestrainbecomeslargerasstrainincreases.Is thisphenomenontrueforbothtensileandcompressive strains?Explain.
Thedifferencebetweentheengineeringandtrue strainsbecomeslargerbecauseofthewaythestrains aredefined,respectively,ascanbeseenbyinspecting Eqs.(2.1)and(2.9).Thisistrueforbothtensileand compressivestrains.
2.4 Usingthesamescaleforstress,thetensiletrue-stresstrue-straincurveishigherthantheengineeringstress-
straincurve.Explainwhetherthisconditionalsoholds foracompressiontest.
Duringacompressiontest,thecross-sectionalarea ofthespecimenincreasesasthespecimenheightdecreases(becauseofvolumeconstancy)astheloadisincreased.Sincetruestressisdefinedasratiooftheload totheinstantaneouscross-sectionalareaofthespecimen,thetruestressincompressionwillbelowerthan theengineeringstressforagivenload,assumingthat frictionbetweentheplatensandthespecimenisnegligible.
2.5 Whichofthetwotests,tensionorcompression,requiresahighercapacitytestingmachinethanthe other?Explain.
Thecompressiontestrequiresahighercapacitymachinebecausethecross-sectionalareaofthespecimen increasesduringthetest,whichistheoppositeofatensiontest.Theincreaseinarearequiresaloadhigher thanthatforthetensiontesttoachievethesamestress level.Furthermore,notethatcompression-testspecimensgenerallyhavealargeroriginalcross-sectional areathanthosefortensiontests,thusrequiringhigher forces.
2.6 Explainhowthemodulusofresilienceofamaterial changes,ifatall,asitisstrained:(a)foranelastic,perfectlyplasticmaterial,and(b)foranelastic,linearly strain-hardeningmaterial.
Recallthatthemodulusofresilienceisgivenby Eq.(2.5)onp.34as S2 y /(2E).(a)Ifthematerialis perfectlyplastic,thentheyieldstrengthdoesnotincreasewithstrain-seeFig.2.7conp.42.Therefore,the modulusofresilienceisunchangedasthematerialis strained.(b)Foralinearstrainhardeningmaterial,the yieldstrengthincreaseswithplasticstrain.Therefore themodulusofresiliencewillincreasewithstrain.
2.7 If youpullandbreakatensile-test specimen rapidly, wherewouldthetemperaturebethehighest?Explain why.
Sincetemperatureriseisduetotheworkinput,the temperaturewillbehighestintheneckedregionbecausethatiswherethestrain,hencetheenergydissipatedperunitvolumeinplasticdeformation,ishighest.
2.8 CommentonthetemperaturedistributionifthespecimeninQuestion2.7ispulledveryslowly.
Ifthespecimenispulledveryslowly,thetemperature generatedwillbedissipatedthroughoutthespecimen andtotheenvironment.Thus,therewillbenoappreciabletemperatureriseanywhere,particularlywith materialswithhighthermalconductivity.
2.9 Inatensiontest,theareaunderthetrue-stress-truestraincurveistheworkdoneperunitvolume(thespecificwork).Also,theareaundertheload-elongation curverepresentstheworkdoneonthespecimen.If youdividethislatterworkbythevolumeofthespecimenbetweenthegagemarks,youwilldeterminethe workdoneperunitvolume(assumingthatalldeformationisconfinedbetweenthegagemarks).Willthis specificworkbethesameastheareaunderthetruestress-true-straincurve?Explain.Willyouranswerbe thesameforanyvalueofstrain?Explain.
Ifwedividetheworkdonebythetotalvolumeofthe specimenbetweenthegagelengths,weobtaintheaveragespecificworkthroughoutthespecimen.However,theareaunderthetruestress-truestraincurve representsthespecificworkdoneatthenecked(and fractured)regioninthespecimenwherethestrainisa maximum.Thus,theanswerswillbedifferent.However,uptotheonsetofnecking(instability),thespecific workcalculatedwillbethesame.Thisisbecausethe strainisuniformthroughoutthespecimenuntilneckingbegins.
2.10 ThenoteatthebottomofTable2.4statesthatastemperatureincreases, C decreasesand m increases.Explainwhy.
Thevalueof C inTable2.4onp.46decreaseswithtemperaturebecauseitisameasureofthestrengthofthe material.Thevalueof m increaseswithtemperature becausethematerialbecomesmorestrain-ratesensitive,duetothefactthatthehigherthestrainrate,the lesstimethematerialhastorecoverandrecrystallize, henceitsstrengthincreases.
2.11 Youaregiventhe K and n valuesoftwodifferentmaterials.Isthisinformationsufficienttodeterminewhich materialistougher?Ifnot,whatadditionalinformationdoyouneed,andwhy?
Althoughthe K and n valuesmaygiveagoodestimateoftoughness,thetruefracturestressandthetrue strainatfracturearerequiredforaccuratecalculation
oftoughness.Themodulusofelasticityandyieldstress wouldprovideinformationabouttheareaunderthe elasticregion;however,thisregionisverysmallandis thususuallynegligiblewithrespecttotherestofthe stress-straincurve.
2.12 ModifythecurvesinFig.2.7toindicatetheeffectsof temperature.Explainyourchanges.
Thesemodificationscanbemadebyloweringtheslope oftheelasticregionandloweringthegeneralheightof thecurves.See,forexample,Fig.2.9onp.43.
2.13 Usingaspecificexample,showwhythedeformation rate,sayinm/s,andthetruestrainratearenotthe same.
Thedeformationrateisthequantity v inEqs.(2.16) and(2.17).Thus,when v isheldconstantduringdeformation(henceaconstantdeformationrate),thetrue strainratewillvary(l increases),whereastheengineeringstrainratewillremainconstant.Hence,thetwo quantitiesarenotthesame.
2.14 Ithasbeenstatedthatthehigherthevalueof m,the morediffusetheneckis,andlikewise,thelowerthe valueof m,themorelocalizedtheneckis.Explainthe reasonforthisbehavior.
AsdiscussedinSection2.2.7,withhigh m values,the materialstretchestoagreaterlengthbeforeitfails;this behaviorisanindicationthatneckingisdelayedwith increasing m.Whenneckingisabouttobegin,the neckingregion’sstrengthwithrespecttotherestofthe specimenincreases,duetostrainhardening.However, thestrainrateintheneckingregionisalsohigherthan intherestofthespecimen,becausethematerialiselongatingfasterthere.Sincethematerialintheneckedregionbecomesstrongerasitisstrainedatahigherrate, theregionexhibitsagreaterresistancetonecking.The increaseinresistancetoneckingthusdependsonthe magnitudeof m.Asthetensiontestprogresses,neckingbecomesmore diffuse,andthespecimenbecomes longerbeforefracture;hence,totalelongationincreases withincreasingvaluesof m.Asexpected,theelongationafternecking(postuniformelongation)alsoincreases withincreasing m.Ithasbeenobservedthatthevalue of m decreaseswithmetalsofincreasingstrength.
2.15 Explainwhymaterialswithhigh m values,suchashot glassandsillyputty,whenstretchedslowly,undergo largeelongationsbeforefailure.Considereventstakingplaceintheneckedregionofthespecimen. TheanswerissimilartoAnswer2.14above.
2.16 Assumethatyouarerunningfour-pointbendingtests onanumberofidenticalspecimensofthesamelength andcross-section,butwithincreasingdistancebetweentheupperpointsofloading(seeFig.2.19b). Whatchanges,ifany,wouldyouexpectinthetestresults?Explain.
As thedistancebetweentheupperpointsofloading inFig. 2.19b increases,themagnitudeofthebending momentdecreases.However,thevolumeofmaterial subjectedtothemaximumbendingmoment(henceto maximumstress)increases.Thus,theprobabilityof failureinthefour-pointtestincreasesasthisdistance increases.
2.17 WouldEq.(2.10)holdtrueintheelasticrange?Explain.
Notethatthisequationisbasedonvolumeconstancy, i.e., Aolo = Al.Weknow,however,thatbecausethe Poisson’sratio ν islessthan0.5intheelasticrange,the volumeisnotconstantinatensiontest;seeEq.(2.47) onp.71.Therefore,theexpressionisnotvalidinthe elasticrange.
2.18 Whyhavedifferenttypesofhardnesstestsbeendeveloped?Howwouldyoumeasurethehardnessofavery largeobject?
Thereareseveralbasicreasons:
1.Theoverallhardnessrangeofthematerials
2.Thehardnessoftheirconstituents;seeChapter3;
3.Thethicknessofthespecimen,suchasbulkversusfoil
4.Thesizeofthespecimenwithrespecttothatofthe indenter
5.Thesurfacefinishofthepartbeingtested.
2.19 Whichhardnesstestsandscaleswouldyouusefor verythinstripsofmaterial,suchasaluminumfoil? Why?
Becausealuminumfoilisverythin,theindentations onthesurfacemustbeverysmallsoasnottoaffect testresults.Suitabletestswouldbeamicrohardness testsuchasKnooporVickersunderverylightloads (seeFig.2.20onp.54).Theaccuracyofthetestcan bevalidatedbyobservinganychangesinthesurface appearanceoppositetotheindentedside.
2.20 Listandexplainthefactorsthatyouwouldconsiderin selectinganappropriatehardnesstestandscalefora particularapplication.
Hardnesstestsmainlyhavethreedifferences: 1.typeofindenter, 2.appliedload,and 3.methodofindentationmeasurement(depthor surfaceareaofindentation,orreboundofindenter).
2.21 InaBrinellhardnesstest,theresultingimpressionis foundtobeanellipse.Givepossibleexplanationsfor thisresult.
Thereareseveralpossiblereasonsforthisphenomenon,butthetwomostlikelyareanisotropyinthe materialandthepresenceofsurfaceresidualstresses inthematerial.
2.22 ReferringtoFig.2.20,thematerialfortestersareeither steel,tungstencarbide,ordiamond.Whyisn’tdiamondusedforallofthetests?
Whilediamondisthehardestmaterialknown,it wouldnot,forexample,bepracticaltomakeanduse a10-mmdiamondindenterbecausethecostswould beprohibitive.Consequently,ahardmaterialsuchas thoselistedaresufficientformosthardnesstests.
2.23 Whatroledoesfrictionplayinahardnesstest?Can highfrictionbetweenamaterialandindenteraffecta hardnesstest?Explain.
Theeffectoffrictionhasbeenfoundtobeminimal.In ahardnesstest,mostoftheindentationoccursthrough plasticdeformation,andthereisverylittleslidingat theindenter-workpieceinterface;seeFig.2.23onp.58.
2.24 Describethedifferencebetweencreepandstressrelaxation,givingtwoexamplesforeachastheyrelateto engineeringapplications.
Creepisthepermanentdeformationofapartthatis underaloadoveraperiodoftime,usuallyoccurringat elevatedtemperatures.Stressrelaxationisthedecrease inthestresslevelinapartunderaconstantstrain.Examplesofcreepinclude:
1.turbinebladesoperatingathightemperatures, and
2.high-temperaturesteamlinesandfurnacecomponents.
Stressrelaxationisobservedwhen,forexample,arubberbandorathermoplasticispulledtoaspecific lengthandheldatthatlengthforaperiodoftime.This phenomenoniscommonlyobservedinrivets,bolts, andguywires,aswellasthermoplasticcomponents.
2.25 ReferringtothetwoimpacttestsshowninFig.2.26, explainhowdifferenttheresultswouldbeifthespecimenswereimpactedfromtheoppositedirections.
NotethatimpactingthespecimensshowninFig.2.26 onp.61fromtheoppositedirectionswouldsubject therootsofthenotchestocompressivestresses,and thustheywouldnotactasstressraisers.Thus,cracks wouldnotpropagateastheywouldwhenundertensilestresses.Consequently,thespecimenswouldbasicallybehaveasiftheywerenotnotched.
2.26 Ifyouremovethelayer ad fromthepartshownin Fig.2.27d,suchasbymachiningorgrinding,which waywillthespecimencurve?(Hint: Assumethat thepartindiagram(d)iscomposedoffourhorizontal springsheldattheends.Thus,fromthetopdown,we havecompression,tension,compression,andtension springs.)
Sincetheinternalforceswillhavetoachieveastateof staticequilibrium,thenewparthastobowdownward (i.e.,itwillholdwater).Suchresidual-stresspatterns
can bemodeledwithasetofhorizontaltension and compressionsprings.Notethatthetoplayerofthematerial ad inFig.2.27d,whichisundercompression,has thetendencytobendthebarupward.Whenthisstress isrelieved(suchasbyremovingalayer),thebarwill compensateforitbybendingdownward.
2.27 Isitpossibletocompletelyremoveresidualstressesina pieceofmaterialbythetechniquedescribedinFig.2.29 ifthematerialiselastic,linearlystrainhardening?Explain.
Byfollowingthesequenceofeventsdepictedin Fig.2.29onp.64,itcanbeseenthatitisnotpossibletocompletelyremovetheresidualstresses.Note thatforanelastic,linearlystrainhardeningmaterial, σc willnevercatchupwith σt
2.28 ReferringtoFig.2.29,woulditbepossibletoeliminateresidualstressesbycompression?Assumethat thepieceofmaterialwillnotbuckleundertheuniaxial compressiveforce.
Yes,bythesamemechanismdescribedinFig.2.29on p.64.
2.29 Listandexplainthedesirablemechanicalproperties for(a)anelevatorcable;(b)abandage;(c)ashoesole; (d)afishhook;(e)anautomotivepiston;(f)aboatpropeller;(g)agas-turbineblade;and(h)astaple.
Thefollowingaresomebasicconsiderations:
(a)Elevatorcable:Thecableshouldnotelongateelasticallytoalargeextentorundergoyieldingasthe loadisincreased.Theserequirementsthuscallfor amaterialwithahighelasticmodulusandyield stress.
(b)Bandage:Thebandagematerialmustbecompliant,thatis,havealowstiffness,buthavehigh strengthinthemembranedirection.Itsinnersurfacemustbepermeableandoutersurfaceresistanttoenvironmentaleffects.
(c)Shoesole:Thesoleshouldbecompliantforcomfort,withahighresilience.Itshouldbetoughso thatitabsorbsshockandshouldhavehighfrictionandwearresistance.
(d)Fishhook:Afishhookneedstohavehigh strengthsothatitdoesn’tdeformpermanently underload,andthusmaintainitsshape.Itshould bestiff(forbettercontrolduringitsuse)and shouldberesistanttheenvironmentitisusedin (suchassaltwater).
(e)Automotivepiston:Thisproductmusthavehigh strengthatelevatedtemperatures,highphysical andthermalshockresistance,andlowmass.
(f)Boatpropeller:Thematerialmustbestiff(to maintainitsshape)andresistanttocorrosion,and alsohaveabrasionresistancebecausethepropellerencounterssandandotherabrasiveparticleswhenoperatedclosetoshore.
(g)Gasturbineblade:Agasturbinebladeoperates athightemperatures(dependingonitslocationin theturbine);thusitshouldhavehigh-temperature strengthandresistancetocreep,aswellastooxidationandcorrosionduetocombustionproducts duringitsuse.
(h)Staple:Thepropertiesshouldbecloselyparallel tothatofapaperclip.Thestapleshouldhave highductilitytoallowittobedeformedwithout fracture,andalsohavelowyieldstresssothatit canbebent(aswellasunbentwhenremovingit) easilywithoutrequiringexcessiveforce.
2.30 Makeasketchshowingthenatureanddistributionof theresidualstressesinFigs.2.28aandbbeforetheparts werecut.Assumethatthesplitpartsarefreefromany stresses.(Hint: Forcethesepartsbacktotheshapethey wereinbeforetheywerecut.)
Asthequestionstates,whenweforcebackthesplit portionsinthespecimeninFig.2.28aonp.63,weinducetensilestressesontheoutersurfacesandcompressiveontheinner.Thustheoriginalpartwould, alongitstotalcrosssection,havearesidualstressdistributionoftension-compression-tension.Usingthe sametechnique,wefindthatthespecimeninFig.2.28b wouldhaveasimilarresidualstressdistributionprior tocutting.
2.31 Itispossibletocalculatetheworkofplasticdeformationbymeasuringthetemperatureriseinaworkpiece, assumingthatthereisnoheatlossandthatthetemperaturedistributionisuniformthroughout?Ifthespecificheatofthematerialdecreaseswithincreasingtemperature,willtheworkofdeformationcalculatedusingthespecificheatatroomtemperaturebehigheror lowerthantheactualworkdone?Explain.
Ifwecalculatetheheatusingaconstantspecificheat valueinEq.(2.62),theworkwillbehigherthanitactuallyis.Thisisbecause,bydefinition,asthespecific heatdecreases,lessworkisrequiredtoraisetheworkpiecetemperaturebyonedegree.Consequently,the calculatedworkwillbehigherthantheactualwork done.
2.32 Explainwhetherornotthevolumeofametalspecimen changeswhenthespecimenissubjectedtoastateof (a)uniaxialcompressivestressand(b)uniaxialtensile stress,allintheelasticrange.
Forcase(a),thequantityinparenthesesinEq.(2.47) onp.71willbenegative,becauseofthecompressive stress.Sincetherestofthetermsarepositive,theproductofthesetermsisnegativeand,hence,therewillbe adecreaseinvolume(Thiscanalsobededucedintuitively.)Forcase(b),itwillbenotedthatthevolume willincrease.
2.33 Itisrelativelyeasytosubjectaspecimentohydrostatic compression,suchasbyusingachamberfilledwitha liquid.Deviseameanswherebythespecimen(say,in
the shapeofacubeoraround disk) canbesubjected tohydrostatictension,oroneapproachingthisstateof stress.(Notethatathin-walled,internallypressurized sphericalshellisnotacorrectanswer,becauseitissubjectedonlytoastateofplanestress.)
Twopossibleanswersarethefollowing:
1.Asolidcubemadeofasoftmetalhasallitssix facesbrazedtolongsquarebars(ofthesamecross sectionasthespecimen);thebarsaremadeofa strongermetal.Thesixarmsarethensubjectedto equaltensionforces,thussubjectingthecubeto equaltensilestresses.
2.Athin,solidrounddisk(suchasacoin)andmade ofasoftmaterialisbrazedbetweentheendsof twosolidroundbarsofthesamediameterasthat ofthedisk.Whensubjectedtolongitudinaltension,thediskwilltendtoshrinkradially.Butbecauseitisthinanditsflatsurfacesarerestrained bythetworodsfrommoving,thediskwillbesubjectedtotensileradialstresses.Thus,astateoftriaxial(thoughnotexactlyhydrostatic)tensionwill existwithinthethindisk.
2.34 ReferringtoFig.2.17,makesketchesofthestateof stressforanelementinthereducedsectionofthetube whenitissubjectedto(a)torsiononly;(b)torsionwhile thetubeisinternallypressurized;and(c)torsionwhile thetubeisexternallypressurized.Assumethatthe tubeisaclosed-endtube.
Thesestatesofstresscanberepresentedsimplybyreferringtothecontentsofthischapteraswellastherelevantmaterialscoveredintextsonmechanicsofsolids.
2.35 Apenny-shapedpieceofsoftmetalisbrazedtothe endsoftwoflat,roundsteelrodsofthesamediameter asthepiece.Theassemblyisthensubjectedtouniaxialtension.Whatisthestateofstresstowhichthesoft metalissubjected?Explain.
Thepenny-shapedsoftmetalpiecewilltendtocontract radiallyduetoPoisson’sratio;however,thesolidrods towhichitattachedwillpreventthisfromhappening. Consequently,thestateofstresswilltendtoapproach thatofhydrostatictension.
2.36 Acirculardiskofsoftmetalisbeingcompressedbetweentwoflat,hardenedcircularsteelpunchesofhavingthesamediameterasthedisk.Assumethatthe diskmaterialisperfectlyplasticandthatthereisno frictionoranytemperatureeffects.Explainthechange, ifany,inthemagnitudeofthepunchforceasthedisk isbeingcompressedplasticallyto,say,afractionofits originalthickness.
Notethatasitiscompressedplastically,thedisk willexpandradially,becauseofvolumeconstancy. Anapproximatelydonut-shapedmaterialwillthenbe pushedradiallyoutward,whichwillthenexertradialcompressivestressesonthediskvolumeunder
thepunches.Thevolumeofmaterialdirectlybetween thepuncheswillnowsubjectedtoatriaxialcompressivestateofstress.Accordingtoyieldcriteria(see Section2.11),thecompressivestressexertedbythe puncheswillthusincrease,eventhoughthematerial isnotstrainhardening.Therefore,thepunchforcewill increaseasdeformationincreases.
2.37 Aperfectlyplasticmetalisyieldingunderthestress state σ1, σ2, σ3,where σ1 >σ2 >σ3.Explainwhat happensif σ1 isincreased.
ConsiderFig.2.32onp.70.Pointsintheinteriorof theyieldlocusareinanelasticstate,whereasthoseon theyieldlocusareinaplasticstate.Pointsoutsidethe yieldlocusarenotadmissible.Therefore,anincrease in σ1 whiletheotherstressesremainunchangedwould requireanincreaseinyieldstress.ThiscanalsobededucedbyinspectingeitherEq.(2.38)orEq.(2.39).
2.38 WhatisthedilatationofamaterialwithaPoisson’sratioof0.5?IsitpossibleforamaterialtohaveaPoisson’sratioof0.7?Givearationaleforyouranswer.
ItcanbeseenfromEq.(2.47)onp.71thatthedilatation ofamaterialwith ν =0.5 isalwayszero,regardlessof thestressstate.Toexaminethecaseof ν =0 7,consider thesituationwherethestressstateishydrostatictension.Equation(2.47)wouldthenpredictcontraction underatensilestress,asituationthatcannotoccur.
2.39 CanamaterialhaveanegativePoisson’sratio?Givea rationaleforyouranswer.
SolidmaterialdonothaveanegativePoisson’sratio, withtheexceptionofsomecompositematerials(see Chapter10),wheretherecanbeanegativePoisson’s ratioinagivendirection.
2.40 Asclearlyaspossible,defineplanestressandplane strain.
Planestressisthesituationwherethestressesinoneof thedirectiononanelementarezero;planestrainisthe situationwherethestrainsinoneofthedirectionare zero.
2.41 Whattestwouldyouusetoevaluatethehardnessofa coatingonametalsurface?Woulditmatterifthecoatingwasharderorsofterthanthesubstrate?Explain.
Theanswerdependsonwhetherthecoatingisrelativelythinorthick.Forarelativelythickcoating,conventionalhardnesstestscanbeconducted,aslongas thedeformedregionundertheindenterislessthan aboutone-tenthofthecoatingthickness.Ifthecoatingthicknessislessthanthisthreshold,thenonemust eitherrelyonnontraditionalhardnesstests,orelse usefairlycomplicatedindentationmodelstoextract thematerialbehavior.Asanexampleoftheformer, atomicforcemicroscopesusingdiamond-tippedpyramidshavebeenusedtomeasurethehardnessofcoatingslessthan100nanometersthick.Asanexampleof
the latter,finite-elementmodelsofacoatedsubstrate being indented byanindenterofaknowngeometry canbedevelopedandthencorrelatedtoexperiments.
2.42 Listtheadvantagesandlimitationsofthestress-strain relationshipsgiveninFig.2.7.
Severalanswersthatareacceptable,andthestudentis encouragedtodevelopasmanyaspossible.Twopossibleanswersare:
1.Thereisatradeoffbetweenmathematicalcomplexityandaccuracyinmodelingmaterialbehavior
2.Somematerialsmaybebettersuitedforcertain constitutivelawsthanothers
2.43 Plotthedataintheinsidefrontcoveronabarchart, showingtherangeofvalues,andcommentontheresults.
Bythestudent.Anexampleofabarchartfortheelastic modulusisshownbelow.
1.Thereisasmallerrangeformetalsthanfornonmetals;
2.Thermoplastics,thermosetsandrubbersareordersofmagnitudelowerthanmetalsandother non-metals;
3.Diamondandceramicscanbesuperiortoothers, butceramicshavealargerangeofvalues.
2.44 Ahardnesstestisconductedonas-receivedmetalasa qualitycheck.Theresultsshowindicatethatthehardnessistoohigh,indicatingthatthematerialmaynot havesufficientductilityfortheintendedapplication. Thesupplierisreluctanttoacceptthereturnofthematerial,insteadclaimingthatthediamondconeusedin theRockwelltestingwaswornandblunt,andhence thetestneededtoberecalibrated.Isthisexplanation plausible?Explain.
RefertoFig.2.20onp.54andnotethatifanindenterisblunt,thenthepenetration, t,underagivenload willbesmallerthanthatusingasharpindenter.This thentranslatesintoahigherhardness.Theexplanation isplausible,butinpractice,hardnesstestsarefairly reliableandmeasurementsareconsistentifthetestingequipmentisproperlycalibratedandroutinelyserviced.
2.45 Explainwhya0.2%offsetisusedtoobtaintheyield strengthinatensiontest.
Thevalueof0.2%issomewhatarbitraryandisused tosetsomestandard.Ayieldstress,representingthe transitionpointfromelastictoplasticdeformation,is difficulttomeasure.Thisisbecausethestress-strain curveisnotlinearlyproportionalaftertheproportionallimit,whichcanbeashighasone-halftheyield strengthinsomemetals.Therefore,atransitionfrom elastictoplasticbehaviorinastress-straincurveisdifficulttodiscern.Theuseofa0.2%offsetisaconvenientwayofconsistentlyinterpretingayieldpoint fromstress-straincurves.
2.46 ReferringtoQuestion2.45,wouldtheoffsetmethodbe necessaryforahighlystrainedhardenedmaterial?Explain.
The0.2%offsetisstilladvisablewheneveritcanbe used,becauseitisastandardizedapproachfordeterminingyieldstress,andthusoneshouldnotarbitrarily abandonstandards.However,ifthematerialishighly coldworked,therewillbeamorenoticeable‘kink’in thestress-straincurve,andthustheyieldstressisfar moreeasilydiscernablethanforthesamematerialin theannealedcondition.
2.47 Explainwhythehardnessofamaterialisrelatedtoa multipleoftheuniaxialcompressivestress,sinceboth involvecompressionofworkpiecematerial.
Typicalcommentsregardingsuchachart are:
Thehardnessisrelatedtoamultipleoftheuniaxial compressivestress,notjusttheuniaxialcompressive stress,because:
1. Thevolumeofmaterialthatisstressed is different -inahardnesstest,thevolumethatisunderstress isnotjustacylinderbeneaththeinventor.
2.Thestressedvolumeisconstrainedbytheelasticmaterialoutsideoftheindentationarea.This oftenrequiresmaterialtodeformlaterallyand countertotheindentationdirection-seeFig.2.21
2.48 Withoutusingthewords“stress”or“strain”,define elasticmodulus.
Thisisactuallyquitechallenging,buthistoricallysig-
PROBLEMS
2.49 Astripofmetalisoriginally1.0mlong.Itisstretched inthreesteps:firsttoalengthof1.5m,thento2.5m, andfinallyto3.0m.Showthatthetotaltruestrainis thesumofthetruestrainsineachstep,thatis,that thestrainsareadditive.Showthat,usingengineering strains,thestrainforeachstepcannotbeaddedtoobtainthetotalstrain.
ThetruestrainisgivenbyEq.(2.9)onp.36as
Therefore,thetruestrainsforthe threestepsare:
1 =ln 1.5 1 0 =0 4055
2 =ln
=0.5108
=0 1823
Thesumofthese truestrainsis =0 4055+0 5108+ 0.1823=1.099.Thetruestrainfromstep1to3is =ln 3 1 = 1 099
Thereforethetrue strains areadditive.Usingthesame approachforengineeringstrainasdefinedbyEq.(2.1), weobtain e1 =0 5, e2 =0 667,and e3 =0 20.Thesum ofthesestrainsis e1 + e2 + e3 =1 367.Theengineering strainfromstep1to3is
e = l lo lo = 3 1 1 = 2
Notethatthisisnotequaltothe sum oftheengineering strainsfortheindividualsteps.ThefollowingMatlab codecanbeusedtodemonstratethatthisisgenerally true,andnotjustaconclusionforthespecificdeformationsstatedintheproblem.
nificant,sinceThomasYoungdidnothavethebenefit oftheconceptofstrainwhenhefirstdefinedmodulusofelasticity.Young’sdefinitionsatisfiestheproject requirement.InYoung’swords:
Themodulusoftheelasticityofanysubstanceisacolumn ofthesamesubstance,capableofproducingapressureonits basewhichistotheweightcausingacertaindegreeofcompressionasthelengthofthesubstanceistothediminution ofthelength.
Therearemanypossibleotherdefinitions,ofcourse.
l0=1; l1=1.5; l2=2.5; l3=3; etot=(l1-l0)/l0+(l2-l0)/l0+(l3-l0)/l0; efin=(l3-l0)/l0; epstot=log(l1/l0)+log(l2/l1)+log(l3/l2); epsfin=log(l3/l0);
2.50 Apaperclipismadeofwire1.00mmindiameter.If theoriginalmaterialfromwhichthewireismadeisa rod15mmindiameter,calculatethelongitudinaland diametricalengineeringandtruestrainsthatthewire hasundergoneduringprocessing.
Assumingvolumeconstancy,wemaywrite
lf lo = do df 2 = 15 1 00 2 =225
Letting lo beunity,thelongitudinalengineering strain is e1 = (225 1)/1=224.Thediametralengineering strainiscalculatedas
ed = 1 15 15 = 0 933
Thelongitudinaltruestrainis given byEq.(2.9)on p.36as
=ln l lo = ln(224)= 5 412
Thediametraltrue strain is d =ln 1 15 = 2 708
Notethelargedifference betweentheengineeringand truestrains,eventhoughbothdescribethesamephenomenon.Notealsothatthesumofthetruestrains (recognizingthattheradialstrainis r =ln 0 5 15 = 2.708)inthethree principal directionsiszero,indicatingvolumeconstancyinplasticdeformation. ThefollowingMatlabcodeisuseful:
d=0.001; d0=0.015; l0=1; lf=(d0/d)ˆ2*l0; e1=(lf-l0)/l0; ed=(d-d0)/d0; epsilon=log(lf/l0); epsilon_d=log(d/d0);
2.51 A materialhasthefollowingproperties: Sut =350 MPa and n =0.20.Calculateitsstrengthcoefficient, K.
NotefromEq.(2.11)onp.36that Sut,true = K n = Knn becauseatnecking = n.FromFig.2.3, Sut = P Ao , where P istheloadatnecking.Thetrueultimate tensile strengthwouldbe
Sut,true = P/A = Sut Ao A
FromEq.(2.10),
ln Ao A = = 0 20
Therefore,
Ao A = exp(0 2)=1 2214
Substitutingintotheexpression for trueultimate strength,
Sut,true =(350 MPa)(1 2214)=427 MPa
Thestrengthcoefficient, K,canthenbefoundas
K = 427 0 20 2 = 589 MPa
2.52 Basedontheinformationgivenin Fig. 2.6,calculatethe ultimatetensilestrengthof304stainlesssteel.
FromFig.2.6onp.39,thetruestressfor304stainless steelatnecking(wheretheslopechanges;seeFig.2.7e) isfoundtobeabout900MPa,whilethetruestrainis about0.4.Wealsoknowthattheratiooftheoriginalto neckedareasofthespecimenisgivenby
ln Ao Aneck =0 40 or
o = e 0 40 =0 670
Thus, Sut =(900)(0.670)= 603 MPa.
2.53 Calculate the ultimatetensilestrength(engineering)of amaterialwhosestrengthcoefficientis300MPaand thatnecksatatruestrainof0.25.
Inthisproblem, K =300 MPaand n =0 25.Following thesameprocedureasinExample2.1onp.41,thetrue ultimatetensilestrengthis σ =(300)(0.25)0 25 =212 MPa and Aneck = Aoe 0 25 =0 779Ao
Consequently, Sut =(212)(0 779)= 165 MPa
2.54 Amaterialhasastrengthcoefficient K =700 MPa.Assumingthatatensile-testspecimenmadefromthismaterialbeginstoneckatatruestrainof0.20,showthat theultimatetensilestrengthofthismaterialis415MPa.
TheapproachisthesameasinExample2.1onp.41. Sincetheneckingstraincorrespondstothemaximum loadandtheneckingstrainforthismaterialisgiven as = n =0.20,wehave,asthetrueultimatetensile strength: Sut,true =(700)(0 20)0 20 =507 MPa.
Thecross-sectionalareaattheonsetofneckingisobtainedfrom
ln Ao Aneck = n =0 20
Consequently, Aneck = Ao exp( 0 20) andthemaximumload, P ,is
P = σA = Sut,trueAo exp( 0 20) =(507)(0 8187)(Ao)= 415 × 106 Ao
Since Sut = P/Ao,wehave Sut =415 MPa. Thisis confirmedwiththefollowingMatlabcode.
K=700e6; n=0.2; Sut_true=K*nˆn; P=Sut_true*exp(-n)
2.55 Acableismadeoffourparallelstrandsofdifferentmaterials,allbehavingaccordingtotheequation σ = K n , where n =0 20.Thematerials,strengthcoefficients andcross-sectionsareasfollows:
MaterialA: K =450 MPa, Ao =7 mm2;
MaterialB: K =600 MPa, Ao =2.5 mm2;
MaterialC: K =300 MPa, Ao =3 mm2;
MaterialD: K =750 MPa, Ao =2 mm2;
(a)Calculatethemaximumtensileforcethatthiscablecanwithstandpriortonecking.
(b)Explainhowyouwouldarriveatananswerifthe n valuesofthethreestrandsweredifferentfrom eachother.
(a) Neckingwilloccurwhen = n =0.20. At thispointthetruestressesineachcableare,from Eq.(2.11)onp.36,
σA =(450)0 20 2 =326 MPa
σB =(600)0 20 2 =435 MPa
σC =(300)0 20 2 =217 MPa
σD =(760)0 20 2 =543 MPa
Theareasatneckingarecalculatedfrom Aneck =
Aoe n (seeExample2.1onp.41):
AA =(7)e 0 2 =5 73 mm2
AB =(2 5)e 0 2 =2 04 mm2
AC =(3)e 0 2 =2 46 mm2
AD =(2)e 0 2 =1 64 mm2
Hencethetotalloadthatthecablecansupportis
P =(326)(5 73)+(435)(2 04) +(217)(2.46)+(543)(1.64) = 4180 N.
ThefollowingMatlabcodeishelpful:
K_A=450e6; K_B=600e6; K_C=300e6; K_D=750e6; A0_A=7/1e6; A0_B=2.5/1e6; A0_C=3/1e6; A0_D=2/1e6; n=0.20;
s_A=K_A*nˆn; s_B=K_B*nˆn; s_C=K_C*nˆn; s_D=K_D*nˆn;
A_A=A0_A*exp(-1*n);
A_B=A0_B*exp(-1*n);
A_C=A0_C*exp(-1*n);
A_D=A0_D*exp(-1*n);
P=s_A*A_A+s_B*A_B+s_C*A_C+s_D*A_D; (b)Ifthe n valuesofthefourstrandsweredifferent,theprocedurewouldconsistofplottingthe load-elongationcurvesofthefourstrandsonthe samechart,thenobtaininggraphicallythemaximumload.Alternately,acomputerprogramcan bewrittentodeterminethemaximumload.
2.56 UsingonlyFig.2.6,calculatethemaximumloadintensiontestingofa304stainless-steelspecimenwithan originaldiameterof6.0mm.
ObservefromFig.2.6onp.39thatneckingbeginsat atruestrainofabout0.4for304stainlesssteel,and that Sut,true isabout900MPa(thisisthelocationof
the‘kink’inthestress-straincurve).Theoriginalcrosssectionalareais Ao = π(0 006 m)2/4=2 827×10 5 m2
Since n =0 4,aproceduresimilartoExample2.1on p.41demonstratesthat
Ao Aneck =exp(0 4)=1 49
Thus
Sut = 900 1 49 =604 MPa
Hencethemaximumloadis
P =(Sut)(Ao) =(604)(2 827 × 10 5)
or P =17.1 kN.ThefollowingMatlabcodeishelpful toinvestigateotherparameters.
n=0.4; Sut_true=900e6; A_0=pi*d0*d0/4; Sut=Sut_true/exp(n); P=Sut*A_0;
2.57 Usingthedatagivenintheinsidefrontcover,calculate thevaluesoftheshearmodulus G forthemetalslisted inthetable.
TheimportantequationisEq.(2.24)whichgivesthe shearmodulusas
G = E 2(1 + ν)
Thefollowingvaluescanbecalculated(mid-rangevalues of ν aretakenasappropriate):
Material E (GPa) ν G (GPa)
Al &alloys69-790.3226-30
Cu &alloys 105-150 0.34 39-56
Pb &alloys140.434.9
Mg &alloys 41-45 0.32 15.5-17.0
Mo &alloys330-3600.32125-136
Ni &alloys 180-214 0.31 69-82
Steels 190-2000.3073-77
Stainless steels 190-200 0.29 74-77
Ti&alloys80-1300.3230-49
W &alloys 350-400 0.27 138-157
Ceramics 70-10000.229-417
Glass 70-80 0.24 28-32
Rubbers 0.01-0.10.50.0033-0.033
Thermoplastics 1.4-3.4 0.36 0.51-1.25
Thermosets 3.5-170.341.3-6.34
2.58 Derive anexpressionforthetoughnessofa material representedbytheequation σ = K ( +0 2)n and whosefracturestrainisdenotedas f
Recallthattoughnessistheareaunderthestress-strain curve,hencethetoughnessforthismaterialwouldbe
2.59 A cylindricalspecimenmadeofabrittlematerial50 mmhighandwithadiameterof25mmissubjectedto acompressiveforcealongitsaxis.Itisfoundthatfracturetakesplaceatanangleof 45◦ underaloadof130 kN.Calculatetheshearstressandthenormalstress, respectively,actingonthefracturesurface.
Assumingthatcompressiontakesplacewithoutfriction,notethattwooftheprincipalstresseswillbezero. Thethirdprincipalstressactingonthisspecimenis normaltothespecimenanditsmagnitudeis
σ3 = 130, 000 (π/4)(0 025)2 = 264 MPa
TheMohr’s circleforthissituationisshownbelow.
2.61
or amodulusofresilienceforsteelof 1.81 MN-m/m3 . For copper,
ModulusofResilience = S2 y 2E = 520 × 106 2 2(124 × 109)
or amodulusofresilienceforcopperof 1.09 MNm/m3 .
Note thatthesevaluesareslightlydifferentthanthe valuesgiveninthetext.Thisisduetothefactthat (a)highlycold-workedmetalssuchasthesehavea muchhigheryieldstressthantheannealedmaterials describedinthetext;and(b)arbitrarypropertyvalues aregiveninthestatementoftheproblem.
Calculatetheworkdoneinfrictionlesscompressionof asolidcylinder40mmhighand15mmindiameter toareductioninheightof50%forthefollowingmaterials:(a)1100-Oaluminum;(b)annealedcopper;(c) annealed304stainlesssteel;and(d)70-30brass,annealed.
TheworkdoneiscalculatedfromEq.(2.59)wherethe specificenergy, u,isobtainedfromEq.(2.57)onp.73. Sincethereductioninheightis50%,thefinalheightis 20mmandtheabsolutevalueofthetruestrainis
=ln 40 20 = 0.6931
K and n areobtainedfromTable 2.3asfollows:
The fractureplaneisorientedatanangle of 45◦,correspondingtoarotationof90◦ ontheMohr’scircle.This correspondstoastressstateonthefractureplaneof σ = τ =264/2=132 MPa.
2.60 Whatisthemodulusofresilienceofahighlycoldworkedpieceofsteelwithahardnessof280HB?Of apieceofhighlycold-workedcopperwithahardness of175HB?
ReferringtoFig.2.22onp.57,thevalueof c inEq.(2.31) isapproximately3.2forhighlycold-workedsteelsand around3.4forcold-workedaluminum.Therefore,approximate c =3 3 forcold-workedcopper.From Eq.(2.31),
Sy,steel = H 3 2 = 280 3 2 =87.5 kg/mm2 =858 MPa
Sy,Cu = H 3 3 = 175 3 3 =53.0 kg/mm2 =520 MPa
Fromthe inside frontcover, Esteel =200 GPaand ECu =124 GPa.ThemodulusofresilienceiscalculatedfromEq.(2.5)onp.34.Forsteel:
ModulusofResilience = S2 y 2E = 858 × 106 2 2(200 × 109)
u is thencalculatedfromEq.(2.57).Forexample, for 1100-O aluminum,where K is180MPaand n is0.20, u iscalculatedas
u = K n+1 n + 1 = (180)(0 6931)1 2 1.2 =96 6 MN/m3
Thevolumeiscalculated as V = πr 2l = π(0 0075)2(0 04)=7 069 × 10 6 m3
Theworkdoneistheproductofthespecificwork, u, andthevolume, V .Therefore,theresultscanbetabulatedasfollows.
u Work Material (MN/m3 ) (Nm) 1100-O Al 96.6682 Cu, annealed 168 1186
304 Stainless,annealed6204389 70-30 brass,annealed 348 2460
The followingMatlabcodecanbeusedtoconfirm results:
h0=0.040; d0=0.015; hf=0.020; epsilon=log(h0/hf); K=180e6; n=0.20; u=K*epsilonˆ(n+1)/(n+1); V=pi*d0ˆ2/4*h0; Work=u*V;
2.62 A tensile-testspecimenismadeofamaterialrepresentedbytheequation σ = K( + n)n.(a)Determine thetruestrainatwhichneckingwillbegin.(b)Show thatitispossibleforanengineeringmaterialtoexhibit thisbehavior.
(a)InSection2.2.4onp.40,itwasnotedthatinstability,hencenecking,requiresthefollowingconditiontobefulfilled:
Consequently,forthismaterialwehave
Thisissolvedas n =0;thusneckingbeginsas soonasthespecimenissubjectedtotension.
(b)Yes,thisbehaviorispossible.Consideratensiontestspecimenthathasbeenstrainedtonecking andthenunloaded.Uponloadingitagainintension,itwillimmediatelybegintoneck.
2.63 Taketwosolidcylindricalspecimensofequaldiameter,butdifferentheights.Assumethatbothspecimens arecompressed(frictionless)bythesamepercentreduction,say50%.Provethatthefinaldiameterswillbe thesame.
Identifytheshortercylindricalspecimenwiththesubscript s andthetalleroneas t,andtheiroriginaldiameteras D.Subscripts f and o indicatefinalandoriginal,respectively.Becausebothspecimensundergothe samepercentreductioninheight,
2.64 Ina disk testperformedonaspecimen50mmindiameterand2.5mmthick,thespecimenfracturesata stressof500MPa.Whatwastheloadonitatfracture?
Equation(2.22)isusedtosolvethisproblem.Noting that σ =500 MPa, d =50 mm=0.05m,and t =2 5 mm=0.0025m,wecanwrite
and fromvolumeconstancy,
and
Because Dto = Dso, notefromtheserelationshipsthat Dtf = Dsf .
ThefollowingMatlabcodeallowsforvariationin parameters forthisproblem.
d=0.050; t=0.0025; sigma=500e6; P=sigma*pi*d*t/2;
2.65 InFig.2.29a,letthetensileandcompressiveresidual stressesbothbe70MPa,andthemodulusofelasticity ofthematerialbe 200 GPawithamodulusofresilience of225kN-m/m3.Iftheoriginallengthindiagram(a) is500mm,whatshouldbethestretchedlengthindiagram(b)sothat,whenunloaded,thestripwillbefree ofresidualstresses?
NotethattheyieldstresscanbeobtainedfromEq.(2.5) onp.34as Mod.ofResilience = MR = S2 y 2E
Thus, Sy = 2(MR)E = 2 (225 × 103)(200 × 109) or Sy =300 MPa.Thestrainrequiredtorelievethe residualstressis: = σc E + Sy E = 70 × 106 200 × 109 + 300 × 106 200 × 109 = 0.00185
Therefore, =ln lf lo = ln lf 0 500 m =0.00185
Therefore, lf =0 50093 m. ThefollowingMatlabcode ishelpful
sigma_c=70e6; E=200e9; MR=225e3; l0=0.5; Sy=(2*MR*E)ˆ(0.5); epsilon=sigma_c/E+Sy/E; lf=l0*exp(epsilon);
2.66 A horizontalrigidbar c-c issubjectingspecimen a totension and specimen b tofrictionlesscompression suchthatthebarremainshorizontal.(Seetheaccompanyingfigure.)Theforce F islocatedatadistanceratio of2:1.Bothspecimenshaveanoriginalcross-sectional areaof0.0001m2 andtheoriginallengthsare a =200 mmand b =115 mm.Thematerialforspecimen a has atrue-stress-true-straincurveof σ =(700 MPa) 0 5 . Plotthetrue-stress-true-straincurvethatthematerial forspecimen b shouldhaveforthebartoremainhorizontal.
mustapproachzero.Thisobservationsuggeststhat specimen b cannothaveatruestress-truestraincurve typicalofmetals,andthatitwillhaveamaximumat somestrain.Thisisseenintheplotof σb shownbelow.
Fromtheequilibriumofverticalforcesand to keepthe barhorizontal,wenotethat 2Fa = Fb.Hence,interms oftruestressesandinstantaneousareas,wehave
Fromvolumeconstancywealsohave,intermsoforiginalandfinaldimensions
and AobLob = A
where Loa =(0 200/0 115)Lob =1 73Lob.Fromthese relationshipswecanshowthat
2.67 Inspect thecurvethatyouobtainedinProblem 2.66. Does atypicalstrain-hardeningmaterialbehaveinthat manner?Explain.
BasedonthediscussionsinSection2.2.3,itisobvious thatordinarymetalswouldnotnormallybehaveinthis manner.However,undercertainconditions,thefollowingcouldexplainsuchbehavior:
• Whenspecimen b isheatedtohigherandhigher temperaturesasdeformationprogresses,withits strengthdecreasingas x isincreasedfurtherafter themaximumvalueofstress.
• Incompressiontestingofbrittlematerials,suchas ceramics,whenthespecimenbeginstofracture.
• Ifthematerialissusceptibletothermalsoftening,thenitcandisplaysuchbehaviorwithasufficientlyhighstrainrate.
Since σa = K 0 5 a where K =700 MPa,wecannow write
Hence, foradeflectionof x,
Thetruestraininspecimenb is givenby
Byinspectingthefigureinthe problemstatement,we notethatwhilespecimen a getslonger,itwillcontinueexertingsomeforce Fa.However,specimen b will eventuallyacquireacross-sectionalareathatwillbecomeinfiniteas x approaches115mm,thusitsstrength
2.68 Showthatyoucantakeabentbarmadeofanelastic, perfectlyplasticmaterialandstraightenitbystretchingitintotheplasticrange.(Hint: Observetheevents showninFig.2.29.)
Theseriesofeventsthattakesplaceinstraightening abentbarbystretchingitcanbevisualizedbystartingwithastressdistributionasinFig.2.29aonp.64, whichwouldrepresenttheunbendingofabentsection.Asweapplytension,wealgebraicallyaddauniformtensilestresstothisstressdistribution.Notethat thechangeinthestressesisthesameasthatdepicted inFig.2.29d,namely,thetensilestressincreasesand reachestheyieldstress, Sy .Thecompressivestress isfirstreducedinmagnitude,thenbecomestensile. Eventually,thewholecrosssectionreachestheconstantyieldstress, Sy .Becausewenowhaveauniform stressdistributionthroughoutitsthickness,thebarbecomesstraightandremainsstraightuponunloading.
2.69 A bar1mlongisbentandthen stressrelieved.The radiusofcurvaturetotheneutralaxisis0.50m.The baris25mmthickandismadeofanelastic,perfectly plasticmaterialwith Sy =500 MPaand E =207 GPa.Calculatethelengthtowhichthisbarshouldbe stretchedsothat,afterunloading,itwillbecomeand remainstraight.
Areviewofbendingtheoryfromasolidmechanics textbookisnecessaryforthisproblem.Inparticular, itshouldberecognizedthatwhenthecurvedbarbecomesstraight,theengineeringstrainitundergoesis givenbytheexpression e = t 2ρ
where t isthethicknessand ρ istheradius totheneutralaxis.Henceinthiscase,
e = (0 025) 2(0 50) =0.025
Since Sy =500 MPaand E =207 GPa,we findthatthe elasticlimitforthismaterialisatanelasticstrainof
e = Sy E = 500 MPa 207 GPa = 0 00242 whichissmallerthan0.025.Therefore,weknowthat thebarmustbeloadedintheplasticrange.FollowingthedescriptioninAnswer2.68above,thestrainrequiredtostraightenthebaristwicetheelasticlimit,or
e =(2)(0 00242)=0 0048
or lf lo lo = 0.0048 → lf =0.005l
or lf =1 0049 m. ThefollowingMatlabcodeishelpful.
l=1; rho=0.5; Sy=500e6; E=207e9; t=0.025; epsilon=t/2/rho; epsilon_y=Sy/E; epsilon_b=2*epsilon_y; lf=l/(1-epsilon_b);
2.70 Assumethatamaterialwithauniaxialyieldstrength Sy yieldsunderastressstateofprincipalstresses σ1, σ2, σ3,where σ1 >σ2 >σ3.Showthatthesuperpositionofahydrostaticstress p onthissystem(suchas placingthespecimeninachamberpressurizedwitha liquid)doesnotaffectyielding.Inotherwords,thematerialwillstillyieldaccordingtoyieldcriteria.
Thissolutionconsidersthedistortion-energycriterion, althoughthesamederivationcouldbeperformedwith
themaximumshearstresscriterionaswell.Equation (2.39)onp.67gives
Nowconsideranewstressstatewheretheprincipal stressesare
whichrepresentsanewloadingwithanadditionalhydrostaticpressure, p.Thedistortion-energycriterion forthisstressstateis
whichcanbesimplifiedas
whichistheoriginalyieldcriterion.Hence,theyield criterionisunaffectedbythesuperpositionofahydrostaticpressure.
2.71 Givetwodifferentandspecificexamplesinwhichthe maximum-shear-stressandthedistortion-energycriteriagivethesameanswer.
Inordertoobtainthesameanswerforthetwoyield criteria,werefertoFig.2.32onp.70forplanestress andnotethecoordinatesatwhichthetwodiagrams meet.Examplesare:simpletension,simplecompression,equalbiaxialtension,andequalbiaxialcompression.Thus,acceptableanswerswouldinclude(a)wire rope,asusedonacranetoliftloads;(b)sphericalpressurevessels,includingballoonsandgasstoragetanks; and(c)shrinkfits.
2.72 Athin-walledsphericalshellwithayieldstrength Sy issubjectedtoaninternalpressure p.Withappropriate equations,showwhetherornotthepressurerequired toyieldthisshelldependsontheparticularyieldcriterionused.
Herewehaveastateofplanestresswithequalbiaxialtension.TheanswertoProblem2.71leadsoneto immediatelyconcludethatboththemaximumshear stressanddistortionenergycriteriawillgivethesame results.Wewillnowdemonstratethismorerigorously. Theprincipalmembranestressesaregivenby
Using themaximumshear-stress criterion, wefindthat σ1 0= Sy
Hence p = 2tSy r
Using thedistortion-energycriterion,wehave (0 0)2 +(σ2 0)2 + (0 σ1)2 = 2S2 y
Since σ1 = σ2,thenthisgives σ1 = σ2 = Sy ,andthe sameexpressionisobtainedforpressure.
2.73 Showthat,accordingtothedistortion-energycriterion, theyieldstrengthinplanestrainis 1.15Sy ,where Sy is theuniaxialyieldstrengthofthematerial.
Aplane-strainconditionisshowninFig.2.35d,where σ1 istheyieldstressofthematerialinplanestrain (Sy ), σ3 iszero,and 2 =0.FromEq.2.43b,wefind that σ2 = σ1/2.SubstitutingtheseintothedistortionenergycriteriongivenbyEq.(2.37),
and
2.74 WhatwouldbetheanswertoProblem2.73ifthemaximumshearstresscriterionwereused?
Because σ2 isanintermediatestressandusing Eq.(2.38)onp.67,theanswerwouldbe
Hence,theyieldstressinplanestrainwillbeequalto theuniaxialyieldstress, Sy
2.75 Aclosed-end,thin-walledcylinderoforiginallength l thickness t,andinternalradius r issubjectedtoan internalpressure p.UsingthegeneralizedHooke’s lawequations,showthechange,ifany,thatoccursin thelengthofthiscylinderwhenitispressurized.Let ν =0 25
Aclosed-end,thin-walledcylinderunderinternalpressureissubjectedtothefollowingprincipalstresses:
wherethesubscript1isthelongitudinaldirection,2 isthehoopdirection,and3isthethicknessdirection. FromHooke’slawgivenbyEq.(2.34)onp.66,
Sinceallthequantitiesarepositive (note thatinorder toproduceatensilemembranestress,thepressureis positiveaswell),thelongitudinalstrainisfiniteand positive.Thusthecylinderbecomeslongerwhenpressurized,asitcanalsobededucedintuitively.
2.76 Around,thin-walledtubeissubjectedtotensioninthe elasticrange.Showthatboththethicknessandthediameterdecreaseastensionincreases.
Thestressstateinthiscaseis σ1, σ2 = σ3 =0.Fromthe generalizedHooke’slawequationsgivenbyEq.(2.34), anddenotingtheaxialdirectionas1,thehoopdirectionas2,andtheradialdirectionas3,wehaveforthe hoopstrain:
Therefore,thediameterisnegativefora tensile (positive)valueof σ1.Fortheradialstrain,thegeneralized Hooke’slawgives
2.77
Therefore,theradialstrainisalsonegative and thewall becomesthinnerforapositivevalueof σ1.
Takealongcylindricalballoonand,withathinfelt-tip pen,markasmallsquareonit.Whatwillbetheshape ofthissquareafteryoublowuptheballoon,(a)alarger square;(b)arectanglewithitslongaxisinthecircumferentialdirection;(c)arectanglewithitslongaxisin thelongitudinaldirection;or(d)anellipse?Perform thisexperiment,and,basedonyourobservations,explaintheresults,usingappropriateequations.Assume thatthematerialtheballoonismadeupofisperfectly elasticandisotropicandthatthissituationrepresents athin-walledclosed-endcylinderunderinternalpressure.
ThisisasimplegraphicwayofillustratingthegeneralizedHooke’slawequations.Aballoonisareadilyavailableandeconomicalmethodofdemonstrating thesestressstates.Itisalsoencouragedtoassignthe studentsthetaskofpredictingtheshapenumerically; anexampleofavaluableexperimentinvolvespartially inflatingtheballoon,drawingthesquare,thenexpandingitfurtherandhavingthestudentspredictthedimensionsofthesquare.
Althoughnotasreadilyavailable,arubbertubecanbe usedtodemonstratetheeffectsoftorsioninasimilar manner.
2.78 Takeacubicpieceofmetalwithasidelength lo and deformitplasticallytotheshapeofarectangularparallelepipedofdimensions l1, l2,and l3.Assumingthat thematerialisrigidandperfectlyplastic,showthat volumeconstancyrequiresthatthefollowingexpressionbesatisfied: 1 + 2 + 3 =0
The initialvolumeandthefinalvolumeare constant, so that lololo = l1l2l3 → l1l2l3 lololo = 1
Takingthenaturallogofbothsides, ln l1l2l3
lololo = ln(1)=0 since ln(AB)=ln(A) +ln(B), ln l1 lo + ln l2 lo + ln l3
o = 0
Fromthedefinitionoftruestraingiven byEq.(2.9)on p.36, ln l1 l0 = 1, etc.,sothat
1 + 2 + 3 =0
2.79 Whatisthediameter of anoriginally40-mm-diameter solidsteelballwhentheballissubjectedtoahydrostaticpressureof2GPa?
FromEq.(2.47)onp.71,andnotingthat,forthiscase, allthreestrainsareequalandallthreestressesare equalto p (accountingforthefactthatapositivepressureisequivalenttoanegativestress), 3 = 1 2ν E ( 3p)
where p isthehydrostatic pressure.Thus,fromtheinsidefrontcover, ν =0 3 and E =200 GPa,sothat = 1 2ν E ( p)= 1 0 6 200 ( 2) or = 0 004.Therefore ln Df Do = 0.004
Solvingfor Df , Df = Doe 0 004 =(40)e 0 004 = 39.84 mm.
ThefollowingMatlabcodeisuseful.
nu=0.3; E=200e9; do=0.040; p=2e9; epsilon=-(1-2*nu)/E*p; df=do*exp(epsilon);
2.80 Determinetheeffectivestressandeffectivestrainin plane-straincompressionaccordingtothedistortionenergycriterion.
ReferringtoFig.2.31d,notethat,forthiscase, σ3 =0 and σ2 = σ1/2,ascanbeseenfromEq.(2.45)onp.70.
Accordingtothedistortion-energycriterionandreferringtoEq.(2.53)onp.72foreffectivestress,wefind that
Note thatforthiscase 3 =0.Sincevolume constancy is maintainedduringplasticdeformation,wealsohave 3 = 1.SubstitutingtheseintoEq.(2.54),theeffectivestrainisfoundtobe
2.81 (a) Calculatetheworkdoneinexpandinga2-mmthick spherical shellfromadiameterof100mmto150 mm,wheretheshellismadeofamaterialforwhich σ =200+50 0 5 MPa.(b)Doesyouranswerdependon theparticularyieldcriterionused?Explain.
(a)Forthiscase,themembranestressesaregivenby σ1 = σ
andthestrainsare
Note thatwehaveabalanced(orequal)biaxial stateof planestress.Thus,thespecificenergy(for aperfectly-plasticmaterial)will,accordingtoeitheryieldcriteria,be
The workdonewillbe,accordingtoEq. (2.59) on p.73, W =(Volume)(u)
Using thepressure-volumemethodofwork,we begin with theformula
W = pdV where V isthevolumeofthesphere.Weintegrate thisequationbetweenthelimits Vo and Vf ,noting that p = 2tSy r
and V = 4πr3 3
so that dV =4πr 2 dr
Also,fromvolumeconstancy, wehave t = r2 o to r2
Combining theseexpressions,weobtain
W =8πSy r 2 o to rf ro dr r = 8πSy r 2 o to ln rf ro
which isthesameexpressionobtainedearlier. To obtainanumericalanswertothisproblem,note that Sy shouldbereplacedwithanaveragevalue
Sy .Alsonotethat 1 = 2 =ln(150/100)=0.4055. Thus, Sy =200+ 50(0.4055)1 5 1 5 =208.6 MPa
Hencetheworkdone is W = 8πSy r 2 o to ln rf ro = 8π(208 6 × 106)(0 1)2(0 001)ln(75/50) =21.26 kN-m
(b)Theyieldcriterionuseddoesnotmatterbecause thisisequalbiaxialtension;seetheanswerto Problem2.71.
2.82 Acylindricalslugthathasadiameterof25mmandis 25mmhighisplacedatthecenterofa50-mm-diameter cavityinarigiddie(seetheaccompanyingfigure).The slugissurroundedbyacompressiblematrix,thepressureofwhichisgivenbytherelation pm =150 ∆V Vom MPa
where m denotesmatrixand Vom istheoriginalvolume ofthecompressible matrix.Boththeslugandthematrixarebeingcompressedbyapistonandwithoutany friction.Theinitialpressureofthematrixiszero,and theslugmaterialhasthetrue-stress-true-straincurve of σ =600 0 4 .
Obtain anexpressionfortheforce F versuspiston travel d up to d = 10 mm.
Thetotalforce, F ,onthepistonwillbe
F = Fw + Fm,
wherethesubscript w denotestheworkpieceand m thematrix.As d increases,thematrixpressureincreases,thussubjectingtheslugtotransversecompressivestressesonitscircumference.Hencetheslug willbesubjectedtotriaxialcompressivestresses,with σ2 = σ3.Usingthemaximumshear-stresscriterionfor simplicity,wehave
+ σ2
where σ1 istherequiredcompressivestressontheslug, σ istheflowstressoftheslugmaterialcorresponding toagivenstrain,andgivenas σ =600 0 4,and σ2 isthe compressivestressduetomatrixpressure.
Theinitialvolumeoftheslugisequalto
s = π 4 (0 025)2(0 025)=1 227 × 10 5 m3
and the volumeofthecavitywhen d =0 is Vco = π 4 (0 050)2(0 025)=4 909 × 10 5
The volume ofthematrixatanyvalueof d isthen Vm = π 4 (0 050)2(0 025 d) 1 227 × 10 5
=3.682 × 10 5 0.001963d
fromwhichweobtain
(3 682 × 10 5) (3 682 × 10 5 0 001963d) 3.682 × 10 5 =53.33d
Notethat when d = 0 01875 m,thevolumeofthe matrixbecomeszero.Therefore,thematrixvolume isstillpositiveintheboundsdefinedbytheproblem (d< 0 010).Thematrixpressure,hence σ2,isnow givenby
2 =150 ∆V V = 150(53.33d)=8000d
Theabsolutevalueof thetruestrainintheslugisgiven by =ln 0 025 0 025 d, withwhichwecandeterminethevalue of σ for any d. Thecross-sectionalareaoftheworkpieceatany d is Aw = (π/4)(0 025)2(0 025) 0 025 d = 1 227 × 10 5 0 025 d
and thatofthematrixis
Am = π 4 (0.050)2 Aw =1.963 × 10 3 Aw
Therequired compressivestressontheslugis
σ1 = σ + σ2 = σ +8000d.
Wemaynowwritethetotalforceonthepistonas F = Aw (σ +8000d)+ Am8000d MN
Thefollowingdatagivessomenumericalresults:
0 4 908 × 10 4 000
0.002 5.335 × 10 4 0.0834 222 150
0.004 5 843 × 10 4 0.174298237
0.006 6.458 × 10 4 0.274 357 325
0.008 7 218 × 10 4 0.386409421
0.010 8 180 × 10 4 0.511 458 532
And thefollowingplotshowsthedesiredresults.
246810
2.83 A specimenintheshapeofacube25 mm oneachside isbeingcompressedwithoutfrictioninadiecavity,as showninFig.2.31d,wherethewidthofthegroove is30mm.Assumethatthelinearlystrain-hardening materialhasthetrue-stress-true-straincurvegivenby σ =70+30 MPa.Calculatethecompressiveforce requiredwhentheheightofthespecimenis3mm,accordingtobothyieldcriteria.
Notethatthevolumeofthespecimenisconstantand canbeexpressedas
(0 025)3 =(h)(x)(x)
where x isthelateraldimensionassumingthespecimenexpandsuniformlyduringcompression.Since h =3 mm,wehave x =72 17 mm.Thus,thespecimentouchesthewallsandhencethisbecomesaplanestrainproblem(seeFig.2.31d).Theabsolutevalueof thetruestrainisgivenbyEq.(2.10)onp.36as
=ln 25 3 = 2 12
Wecannowdeterminethe flow stress, σf ,ofthematerialatthisstrainas
σf =70+30(2.12)=133.6 MPa
Thecross-sectionalareaonwhichtheforceisactingis
Area = (0 025)3 0 003 =5 208 × 10 3 m2
Accordingto the maximumshear-stresscriteriongiven byEq.(2.41)onp.69,wehave σ1 = σf (because σ3 =0),andthus
Force =(133 6)(5 208 × 10 3)= 696 kN
Accordingtothedistortionenergycriterionin Eq.(2.46)onp.71,wehave σ1 =1 15σf ,or
Force =(1.15)(696)= 800 kN.
2.84 Obtainexpressionsforthespecificenergyforamaterialforeachofthestress-straincurvesshownin Fig.2.7,similartothoseshowninSection2.12.
Equation(2.56)onp.72givesthespecificenergyas u = 1 0 σd
1.Foraperfectly-elasticmaterialasshowninFig 2.7a,thisexpressionbecomes
2. Forarigid,perfectly-plasticmaterialasshownin Fig. 2.7b, thisis
3.Foranelastic,perfectlyplasticmaterial,thisis identicaltoanelasticmaterialfor 1 <Sy /E,and for 1 >Sy /E itis
4.Forarigid,linearlystrainhardening material, the specificenergyis
5. Foranelastic,linearstrainhardeningmaterial, the specific energyisidenticaltoanelasticmaterial for 1 <Sy /E andfor 1 >Sy /E itis
2.85 A materialwithayieldstrengthof75 MPa issubjectedtoprincipal(normal)stressesof σ1, σ2 =0,and σ3 = σ1/2.Whatisthevalueof σ1 whenthemetal yieldsaccordingtothevonMisescriterion?
Thedistortion-energycriterion,givenbyEq.(2.39)on p.67,is
Substituting Sy =75 MPaand σ1,
=0 and σ3 = σ1/2,wehave
2(75)
thus, σ1 =56 69 MPa. If Sy =75 MPa and σ1,
2 = σ1/3 and σ3 = σ1/2 isthestressstate,then
2.87 A50-mm-wide,1-mm-thickstripisrolledtoafinal thicknessof0.5mm.Itisnotedthatthestriphasincreasedinwidthto51mm.Whatisthestraininthe rollingdirection?
ThethicknessstrainisgivenbyEq.(2.10)onp.36as t =ln l lo = ln 0.5 mm 1 mm = 0 693
Thewidthstrainis w =ln l lo
Therefore,fromEq. (2.49)onp.71,thestraininthe rolling(orlongitudinal)directionis l =0 0 0198+ 0 693= 0 6732
ThefollowingMatlabcodeisuseful.
Thus, σ1 =64 3 MPa. Therefore, thestressleveltoinitiateyieldingactuallyincreaseswhen σ2 isincreased.
2.86 Asteelplateis100mm × 100mm × 10mmthick.Itis subjectedtobiaxialtension σ1 = σ2 =350 MPa,with thestressinthethicknessdirectionof σ3 =0.Whatis thechangeinvolumeusingHooke’slaw?
Fromtheinsidefrontcover,itisnotedthatforsteelwe canuse E =200 GPaand ν =0 30.Forastressstateof σ1 = σ2 =350 MPaand σ3 =0,Equation(2.48)gives:
∆= 1 2ν E (σx + σy + σz ) = 1 2(0 3)
200 GPa [(350 MPa)+(350 MPa] ==0.0014
Sincethe original volumeis(100)(100)(10)=100,000 mm3,thestressedvolumeis100,140mm3,orthevolumechangeis ∆V =140 mm3
Forcopper,wehave E =125 GPaand ν =0 34.Followingthesamederivation,thedilatationforcopper is0.0018;thestressedvolumeis100,180mm3 andthus thechangeinvolumeis ∆V =180 mm3
ThefollowingMatlabcodeishelpful.Notethatthe valuesof E and ν needtobechangedforeachmaterial considered.
l1=0.1; l2=0.1; t=0.010; s1=350e6; s2=350e6; s3=0; nu=0.30; E=200e9; Delta=(1-2*nu)/E*(s1+s2+s3); V=l1*l2*t; V2=V*(1+Delta); DeltaV=V2-V;
w0=0.050; t0=0.001; tf=0.0005; wf=0.051; et=log(tf/t0); ew=log(wf/w0); el=0-et-ew;
2.88 Analuminumalloyyieldsatastressof50MPainuniaxialtension.Ifthismaterialissubjectedtothestresses σ1 =25 MPa, σ2 =15 MPaand σ3 = 26 MPa,willit yield?Explain.
Accordingtothemaximumshear-stresscriterion,the effectivestressisgivenbyEq.(2.52)onp.72as:
=
1
3 =25 ( 26)=51 MPa
However,accordingtothedistortion-energycriterion, theeffectivestressisgivenbyEq.(2.53)as:
or σ = 46.8 MPa.Therefore,theeffective stressis higherthantheyieldstressforthemaximumshearstresscriterion,andlowerthantheyieldstressforthe distortion-energycriterion.Itisimpossibletostate whetherornotthematerialwillyieldatthisstress state.Anaccuratestatementwouldbethatyielding isimminent,ifitisnotalreadyoccurring.
ThefollowingMatlabcodeishelpful.
Sy=50e6; s1=25e6; s2=15e6; s3=-26e6; s_MSS=s1-s3; se=1/((2)ˆ0.5)* ((s1-s2)ˆ2+(s2-s3)ˆ2+(s3-s1)ˆ2)ˆ0.5;
2.89 A purealuminumcylindricalspecimen25mmin diameter and25mmhighisbeingcompressedbydroppingaweightof1000Nonitfromacertainheight. Afterdeformation,itisfoundthatthetemperaturerise inthespecimenis50◦C.Assumingnoheatlossandno friction,calculatethefinalheightofthespecimen.Use thefollowinginformationforthematerial: K =205 MPa, n =0.4, ρ =7800 kg/m3,and cp =450 J/kg-K. ThisproblemusesthesameapproachasinExample 2.7onp.76.Thevolumeofthespecimenis
V = πd2h 4 = π(0 025)2(0 025) 4 = 1 227 × 10 5 m3
Theexpressionforheat is givenby
Heat = cpρV ∆T =(450)(7800)(1 227 × 10 5)(50) =2150 J.
Since,ideally,fromEq.(2.59),
Heat = Work = Vu = V K n+1 n + 1 = 1 227 × 10 5 (205) 1 4 1 4
Solvingfor , 1 4 = (1 4)(2150) (1.227 × 10 5)(205 × 106) =1 1966
Therefore, = 1 13.Usingabsolutevalues,wehave ln ho hf = ln 25 mm hf = 1 13
Solvingfor hf gives hf =8 02 mm.
Thefollowing Matlab codeconfirmstheanswerandallowsinvestigationofalternatevalues:
do=0.025; ho=0.025; W=1000; DT=50; K=205e6; n=0.4; rho=7800; cp=450; V=pi*doˆ2*ho/4; H=cp*rho*V*DT; eps=((1+n)*H/V/K)ˆ(1/(1+n)); hf=ho/exp(eps);
2.90 Aductilemetalcylinder100mmhighiscompressedto afinalheightof30mmintwostepsbetweenfrictionlessplatens.Afterthefirststepthecylinderis70mm high.Calculateboththeengineeringstrainandthetrue strainforbothsteps,comparethem,andcommenton yourobservations.
Inthefirststep,wenotethat ho =100 mmand h1 =70 mm,sothatfromEq.(2.1)onp.31,
andfromEq.(2.9)on p. 36,
Similarly,forthesecondstep where h1 =70 mmand h2 =30 mm,
Notethatiftheoperationwere conductedinonestep, thefollowingwouldresult:
= 1 204
AswasshowninProblem 2.49, thisindicatesthatthe truestrainsareadditivewhiletheengineeringstrains arenot.
ThefollowingMatlabcodecanbeusedtoconsider otherparameters.
ho=0.1; h1=0.07; h2=0.03; e1=(h1-ho)/ho; eps1=log(h1/ho); e2=(h2-h1)/h1; eps2=log(h2/h1); etot=(h2-ho)/ho; eps_tot=log(h2/ho);
2.91 SupposethecylinderinProblem2.90hasaninitialdiameterof50mmandismadeof1100-Oaluminum.Determinetheloadrequiredforeachstep.
Fromvolumeconstancy,wecalculate
Basedonthesediametersthecross-sectional areaatthe stepsiscalculatedas:
AscalculatedinProblem 2.90, 1 = 0 357 and total = 1.204.Notethatfor1100-Oaluminum, K =180 MPa
and n = 0.20 (seeTable2.2onp.38) so thatEq.(2.11) onp.36yields
σ1 =180(0 357)0 20 =146 5 MPa
σ2 =180(1 204)0 20 =186 8 MPa
Therefore,theloadsarecalculatedas:
P1 = σ1A1 =(146 5) 2 805 × 10 3 =411 kN
P2 =(186.8)(6.545 × 10 3)=1223 kN
ThefollowingMatlabcodeishelpful.
ho=0.1; h1=0.07; h2=0.03; e1=(h1-ho)/ho; eps1=log(h1/ho); e2=(h2-h1)/h1; eps2=log(h2/h1); etot=(h2-ho)/ho; eps_tot=log(h2/ho); do=0.050; K=180e6; n=0.20; d1=do*(ho/h1)ˆ0.5; d2=do*(ho/h2)ˆ0.5; A1=pi/4*d1*d1; A2=pi/4*d2*d2; sigma1=K*(abs(eps1)ˆn); sigma2=K*(abs(eps_tot)ˆn)
P1=sigma1*A1; P2=sigma2*A2;
2.92 Determinethespecificenergyandactualenergyexpendedfortheentireprocessdescribedintheprevious twoproblems.
FromEq.(2.57)onp.73andusing total =1 204, K =180 MPaand n =0 20,wehave u = K n+1 n + 1 = (180)(1 204)1 2 1 2 =187 MPa
2.93 Ametalhasastrainhardening exponentof0.22.Ata truestrainof0.2,thetruestressis80MPa.(a)Determinethestress-strainrelationshipforthismaterial.(b) Determinetheultimatetensilestrengthforthismaterial.
(a)ThissolutionfollowsthesameapproachasinExample2.1onp.41.FromEq.(2.11)onp.36,and recognizingthat n =0 22 and σ =80 MPafor =0 20,
σ = K n → 80= K(0 20)0 22 or K =114 MPa.Therefore,thestress-strainrelationshipforthismaterialis
σ =114 0 22 MPa
(b)Todeterminetheultimatetensilestrengthforthe material,realizethatthestrainatneckingisequal tothestrainhardeningexponent,or = n.Therefore,
σult = K(n)n =114(0 22)0 22 =81 7 MPa
Notethatthisisthetruestresswhereas Sut is basedonengineeringstress.Therefore,theapproachinExample2.1onp.41needstobefollowed.Thecross-sectionalareaattheonsetof neckingisobtainedfrom
ln Ao Aneck = n =0 22
Consequently, Aneck = Aoe 0 22
and the maximumloadis
P = σA = σultAneck.
Hence,
P =(81 7)(Ao)e 0 22 =65 6Ao
Since Sut = P/Ao,wehave Sut = 65 6Ao Ao = 65 6 MPa
ThefollowingMatlabcodeishelpful.
n=0.22; e=0.2; sigma=80e6; K=sigma/(eˆn); sigma_ut=K*nˆn; Sut=sigma_ut*exp(-n);
2.94 The areaofeachfaceofametalcubeis5cm2,andthe metalhasashearyieldstrength k of140MPa.Compressiveloadsof40kNand80kNareappliedtodifferentfaces(sayinthe x-and y-directions).Whatmust bethecompressiveloadappliedtothe z-directionto causeyieldingaccordingtotheTrescayieldcriterion? Assumeafrictionlesscondition.
Sincetheareaofeachfaceis5cm2 = 5 × 10 4 m2,the stressesinthe x-and y-directionsare
σx = 40, 000 5 × 10 4 = 80 MPa
σy = 80, 000 5 × 10 4 = 160 MPa
wherethenegativesignindicatesthat the stresses arecompressive.IftheTrescacriterionisused,then Eq.(2.38)onp.67gives
σmax σmin = Sy =2k =280 MPa
It isstatedthat σ3 iscompressive,andisthereforenegative.Notethatif σ3 iszero,thenthematerialdoesnot yieldbecause σmax σmin =0 ( 160)=160 MPa < 280 MPa.Therefore, σ3 mustbelowerthan σ2,andis calculatedfrom:
or
Thecompressiveloadisthen
ThefollowingMatlabcodeishelpful.
A=5/100/100; k=140e6; Px=-40e3; Py=-80e3; sigma_x=Px/A; sigma_y=Py/A; Sy=2*k; sigma_z=sigma_x-Sy; P=sigma_z*A;
2.95 Atensileforceof9kNisappliedtotheendsofasolid barof7.0mmdiameter.Underload,thediameterreducesto5.00mm.Assuminguniformdeformationand volumeconstancy,(a)determinetheengineeringstress andstrain;(b)determinethetruestressandstrain;(c) Iftheoriginalbarhadbeensubjectedtoatruestressof 345MPaandtheresultingdiameterwas5.60mm,what aretheengineeringstressandstrainforthiscondition?
Firstnotethat,inthiscase, do =7 mm, df =5.00 mm, P =9000 N,andfromvolumeconstancy,
(c) If thefinaldiameteris df =5.60 mm,thenthefinalareais Af = π 4 d2 f = 24 63 mm2.Ifthetrue stress is 345MPa,then
P = σA =(345)(24 63)=8497 ≈ 8500 N
Therefore,theengineeringstressiscalculatedas beforeas
σ = P Ao = 8500 π 4 (0.0056)2 =345 MPa
Similarly,fromvolumeconstancy,
lf lo = d2 o d2 f = 72 5.602 =1
Therefore,the engineering strainis
ThefollowingMatlab code isusefulforparts(a)and (b).
df=0.005; do=0.007; Ao=pi*do*do/4; Af=pi*df*df/4; P=9000; lr=do*do/df/df; s=P/Ao; e=lr-1; sigma=P/Af; eps=log(lr);
2.96 Twoidenticalspecimens20mmindiameterandwith testsections25mmlongaremadeof1112steel.Oneis intheas-receivedconditionandtheotherisannealed. (a)Whatwillbethetruestrainwhenneckingoccurs, andwhatwillbetheelongationofthesesamplesatthat instant?(b)Findtheultimatetensilestrengthforthese materials.
(a)Theengineeringstress iscalculatedfromEq.(2.3) onp.32as:
s = P Ao = 9000 π 4 (0.007)2 =234 MPa
andtheengineeringstrainis calculated from Eq.(2.1)as:
e = l lo lo = lf lo 1 =1 96 1=0 96
(b)Thetrue stressiscalculatedfromEq.(2.8)onp.35 as:
σ = P A = 9000 π 4 (5.00)2 =458 MPa andthetruestrainis calculated fromEq.(2.9)on p.36as:
=ln lf lo = ln1 96=0 673
ThisproblemusesasimilarapproachasforExample2.1onp.41.First,wenotefromTable2.2on p.38thatforcold-rolled1112steel, K =760 MPa and n =0 08.Also,theinitialcross-sectionalareais
Ao = π 4 (0.020)2 =3.142 × 10 4 m2.Forannealed1112 steel, K = 760 MPaand n =0 19.Atnecking, = n,so thatthestrainwillbe =0.08 forthecold-rolledsteel and =0 19 fortheannealedsteel.Forthecold-rolled steel,thefinallengthisgivenbyEq.(2.9)onp.36as = n =ln lf lo
Solving for lf , lf = enlo = e 0 08(25)=27 08 mm
Theelongationis,from Eq.(2.6)onp.35,
Elongation = lf lo lo × 100 = 27.08 25 25 × 100
or 8.32%. Tocalculatetheultimatestrength,we can write, forthecold-rolledsteel,
Sut,true = σut = Knn =760(0 08)0 08 =621 MPa
AsinExample2.1onp.41,wecalculatetheloadat neckingas:
Sothat
Thisexpressionisevaluatedas
Sut = (621)e 0 08 = 573 MPa.
Repeatingthesecalculationsfortheannealedspecimen yields l =30 23 mm, elongation=20.9%,and Sut =458 MPa.
ThefollowingMatlabcodeisuseful,andcanbeused toprovethecalculationsfortheannealedspecimenare correctbyusing n =0.19.
do=0.020; lo=0.025; K=760e6; n=0.08; Ao=pi*do*do/4; lf=exp(n)*lo; elongation=(lf-lo)/lo*100; sigma_ut=K*nˆn; P=sigma_ut*Ao*exp(-n); Sut=sigma_ut*exp(-n);
2.97 Duringtheproductionofapart,ametalwithayield strengthof200MPaissubjectedtoastressstate σ1, σ2 = σ1/3, σ3 =0.SketchtheMohr’scirclediagram forthisstressstate.Determinethestress σ1 necessary tocauseyieldingbythemaximumshearstressandthe vonMisescriteria.
Forthestressstateof σ1, σ1/3,0thefollowingfigure thethree-dimensionalMohr’scircle:
Solving for σ1 gives σ1 =227 MPa.Accordingtothe Tresca criterion, Eq.(2.42)onp.69gives σ1 σ3 = σ1 =0=
or
1 =200 MPa.
2.98 Thefollowingdataaretakenfromastainlesssteel tension-testspecimen:
Also, Ao = 3 5 × 10 5 m2 , Af =1 0 × 10 5 in2 , lo = 50 mm. Plotthetruestress-truestraincurveforthematerial.
ThefollowingarecalculatedfromEqs.(2.6),(2.9), (2.10),and(2.8):
The true stress-true strain curve is then plottedasfollows:
For thevonMisescriterion,Eq.(2.39)onp.67 gives:
2.99 A metalisyieldingplasticallyunderthestress state shown.
(a) Labeltheprincipalaxesaccordingtotheir proper numericalconvention(1,2,3).
(b)WhatistheyieldstrengthusingtheTrescacriterion?
(c)WhatifthevonMisescriterionisused?
(d)Thestressstatecausesmeasuredstrainsof 1 = 0 4 and 2 =0 1,with 3 notbeingmeasured. Whatisthevalueof 3?
(a)The1-directioncorrespondstothe50MPastress, the2directioncorrespondstothe20MPastress, andthe3directioncorrespondstothe-40MPa stress.
(b)Since σ1 ≥ σ2 ≥ σ3,thenfromthefigure σ1 =50 MPa, σ2 =20 MPaand σ3 = 40 MPa.
(c)TheyieldstressusingtheTrescacriterionisgiven byEq.(2.38)onp.67as
min = Sy
Sy =50 MPa ( 40 MPa)= 90 MPa.
(d)IfthevonMisescriterionisused,thenEq.(2.39) gives
canestimatethediameteroftheindentationfromthe expression:
HB = 2P (πD)(D √D2 d2)
fromwhichwefindthat d =3.51 mm for D = 10 mm. Tocalculatethedepthofpenetration,considerthefollowingsketch: 5 mm 3 mm
Because theradiusis5mmandone-halfthe penetration diameteris1.755mm,wecanobtain α as α =sin 1 1 755 5 = 20 5◦
Thedepthofpenetration, t,can be obtainedfrom
2.101 IthasbeenproposedtomodifythevonMisesyieldcriterionas: (σ1 σ2)
or 2S2 y =12, 600 whichissolvedas Sy =79 4 MPa.
(d)Ifthematerialisdeformingplastically,thenfrom Eq.(2.49)onp.71, 1 + 2 + 3 =0.4+0.2+ 3 =0 or 3 = 0 6
ThefollowingMatlabcodeisuseful.
s1=50e6; s2=20e6; s3=-40e6; Sy_T=s1-s3; se=(s1-s2)ˆ2+(s2-s3)ˆ2+(s3-s1)ˆ2; Sy_vM=(se/2)ˆ(0.5);
2.100 EstimatethedepthofpenetrationinaBrinellhardness testusing500kgastheloadwhenthesampleisacoldworkedaluminumwithayieldstrengthof150MPa.
NotefromFig.2.22onp.57thatforcold-workedaluminumwithayieldstressof150MPa,theBrinellhardnessisaround50kg/mm2.FromFig.2.20onp.54,we
where C isaconstantand a isanevenintegerlarger than2.Plotthisyieldcriterionfor a =4 and a =12, alongwiththeTrescaandvonMisescriterion,inplane stress.(Hint: SeeFig.2.32).
Forplanestress,oneofthestresses,say σ3,iszero,and theotherstressesare σA and σB .Theyieldcriterionis then
σA σB )a +(σB )a +(σA)a = C
Foruniaxialtension, σA = Sy and σB =0 sothat C = 2Sa y .Theseequationsaredifficulttosolvebyhand;the followingsolutionwasobtainedusingamathematical programmingpackage: Y Y
B
A Tresca von Mises a=12 a=4
Note thatthesolutionfor a =2 (vonMises)and a =4 are soclosethattheycannotbedistinguishedinthe plot.Whenzoomedintoaportionofthecurve,one wouldseethatthe a =4 curveliesbetweenthevon Misescurveandthe a =12 curve.
2.102 Assumethatyouareaskedtogiveaquiztostudentson thecontentsofthischapter.Preparethreequantitative problemsandthreequalitativequestions,andsupply theanswers.
Bythestudent.Thisisachallenging,open-endedquestionthatrequiresconsiderablefocusandunderstandingonthepartofthestudent,andhasbeenfoundto beaveryvaluablehomeworkproblem.