TheStructureofMetals
QualitativeProblems
1.25.Explainyourunderstandingofwhythestudyofthecrystalstructureofmetalsisimportant.
Thestudyofcrystalstructureisimportantforanumberofreasons.Basically,thecrystalstructure influencesamaterial’sperformancefrombothadesignandmanufacturingstandpoint.Forexample, thenumberofslipsystemsinacrystalhasadirectbearingontheabilityofametaltoundergoplastic deformationwithoutfracture.Similarly,thecrystalstructurehasabearingonstrength,ductility andcorrosionresistance.Metalswithface-centeredcubicstructure,forexample,tendtobeductile whereashexagonalclose-packedmetalstendtobebrittle.Thecrystalstructureandsizeofatom determinesthelargestinterstitialsites,whichhasabearingontheabilityofthatmaterialtoform alloys,andwithwhichmaterials,asinterstitialsorsubstitutionals.
1.26.Whatisthesignificanceofthefactthatsomemetalsundergoallotropism?
Allotropism(alsocalledpolymorphism)meansthatametalcanchangefromonecrystalstructure toanother.Sincepropertiesvarywithcrystalstructures,allotropismisusefulandessentialinheat treatingofmetalstoachievedesiredproperties(Chapter4).Amajorapplicationishardeningofsteel, whichinvolvesthechangeinironfromthefccstructuretothebccstructure(seeFig.1.3).Byheating thesteeltothefccstructureandquenching,itdevelopsintomartensite,whichisaveryhard,hence strong,structure.
1.27.Isitpossiblefortwopiecesofthesamemetaltohavedifferentrecrystallizationtemperatures?Isit possibleforrecrystallizationtotakeplaceinsomeregionsofapartbeforeitdoesinotherregionsof thesamepart?Explain.
Twopiecesofthesamemetalcanhavedifferentrecrystallizationtemperaturesifthepieceshave beencoldworkedtodifferentamounts.Thepiecethatwascoldworkedtoagreaterextent(higher strains),willhavemoreinternalenergy(storedenergy)todrivetherecrystallizationprocess,hence itsrecrystallizationtemperaturewillbelower.Recrystallizationmayalsooccurinsomeregionsofthe partbeforeothersifithasbeenunevenlystrained(sincevaryingamountsofcoldworkhavedifferent recrystallizationtemperatures),oriftheparthasdifferentthicknessesinvarioussections.Thethinner sectionswillheatuptotherecrystallizationtemperaturefaster.
1.28.Describeyourunderstandingofwhydifferentcrystalstructuresexhibitdifferentstrengthsandductilities.
Differentcrystalstructureshavedifferentslipsystems,whichconsistofaslipplane(theclosest packedplane)andaslipdirection(theclose-packeddirection).Thefccstructurehas12slipsystems,bcchas48,andhcphas3.Theductilityofametaldependsonhowmanyoftheslipsystemscan beoperative.Ingeneral,fccandbccstructurespossesshigherductilitythanhcpstructures,because theyhavemoreslipsystems.Theshearstrengthofametaldecreasesfordecreasing b/a ratio(b is inverselyproportionaltoatomicdensityintheslipplaneand a istheplanespacing),andthe b/a ratio dependsontheslipsystemofthechemicalstructure(seeSection1.4).
1.29.Acold-workedpieceofmetalhasbeenrecrystallized.Whentested,itisfoundtobeanisotropic. Explaintheprobablereason.
Theanisotropyoftheworkpieceislikelyduetopreferredorientationremainingfromtherecrystallizationprocess.Copperisanexampleofametalthathasaverystrongpreferredorientationafter annealing.Also,ithasbeenshownthatbelowacriticalamountofplasticdeformation,typically5%, norecrystallizationoccurs.
1.30.Whatmaterialsandstructurescanyouthinkof(otherthanmetals)thatexhibitanisotropicbehavior?
Thisisanopen-endedproblemandthestudentsshouldbeencouragedtodeveloptheirownanswers. However,someexamplesofanisotropicmaterialsarewood,polymersthathavebeencoldworked, bone,anywovenmaterial(suchascloth)andcompositematerials.
1.31.Twopartshavebeenmadeofthesamematerial,butonewasformedbycoldworkingandtheother byhotworking.Explainthedifferencesyoumightobservebetweenthetwo.
Therearealargenumberofdifferencesthatwillbeseenbetweenthetwomaterials,including:
1.Thecoldworkedmaterialwillhaveahigherstrengththanthehotworkedmaterial,andthiswill bemorepronouncedformaterialswithhighstrainhardeningexponents.
2.Sincehardness(seeSection2.6.2)isrelatedtostrength,thecoldworkedmaterialwillalsohave ahigherhardness.
3.Thecoldworkedmaterialwillhavesmallergrainsandthegrainswillbeelongated.
4.Thehotworkedmaterialwillprobablyhavefewerdislocations,andtheywillbemoreevenly distributed.
5.Thecoldworkedmaterialcanhaveasuperiorsurfacefinishwheninanas-formedcondition. Also,itcanhavebettertolerances.
6.Acoldworkedmaterialwillhavealowerrecrystallizationtemperaturethanahotworkedmaterial.
1.32.Explaintheimportanceofhomologoustemperature.
Thehomologoustemperatureisdefinedastheratioofametal’scurrenttemperaturetoitsmelting temperatureonanabsolutescale(KelvinorRankine,notCelsiusorFahrenheit).Thisisimportantfor determiningwhetherornotthemetalwillencounterrecrystallizationorgraingrowth.Thehomologoustemperatureismoreimportantthanactualtemperature,becauserecrystallizationoccursatvery differenttemperaturesfordifferentmetals,butthehomologoustemperatureeffectsarefairlyconsistent.Thehomologoustemperaturethereforeallowsonetodistinguishbetweencold,warm,andhot working,asdiscussedinTable1.2.
1.33.Doyouthinkitmightbeimportanttoknowwhetherarawmaterialtobeusedinamanufacturing processhasanisotropicproperties?Whataboutanisotropyinthefinishedproduct?Explain.
Anisotropyisimportantincold-workingprocesses,especiallysheet-metalformingwherethematerial’spropertiesshouldpreferablybeuniformintheplaneofthesheetandstrongerinthethickness direction.AsshowninSection16.7,thesecharacteristicsallowfordeepdrawingofparts(likebeveragecans)withoutearing,tearing,orcrackingintheformingoperationsinvolved.Inafinished part,anisotropyisimportantsothatthestrongestdirectionofthepartcanbedesignedtosupportthe largestloadinservice.Also,theefficiencyoftransformerscanbeimprovedbyusingasheetsteelwith anisotropythatcanreduce magnetichysteresis losses.Hysteresisiswellknowninferromagneticmaterials.Whenanexternalmagneticfieldisappliedtoaferromagnet,theferromagnetabsorbssome oftheexternalfield.Whensheetsteelishighlyanisotropic,itcontainssmallgrainsandacrystallographicorientationthatisfarmoreuniformthanforisotropicmaterials,andthisorientationwill reducemagnetichysteresislosses.
1.34.Whatisthedifferencebetweenaninterstitialatomandasubstitutionalatom?
ThedifferencecanbeseeninFig.1.8.Asubstitutionalatomreplacesanatomintherepeatinglattice withoutdistortion.Aninterstitialdoesnotfitinthenormallattice;itcanfitinthegapbetween atoms,orelseitdistortsthelattice.Examplesofsubstitutionallyarecopper-nickel,gold-silver,and molybdenum-tungsten.Interstitialscanbeself-interstitials,butcommonotherexamplesarecarbon, lithium,sodium,andnitrogen.
1.35.Explainwhythestrengthofapolycrystallinemetalatroomtemperaturedecreasesasitsgrainsize increases.
Strengthincreasesasmoreentanglementsofdislocationsoccurwithgrainboundaries(Section1.4.2). Metalswithlargergrainshavelessgrain-boundaryareaperunitvolume,andhencewillnotbeas abletogenerateasmanyentanglementsatgrainboundaries,thusthestrengthwillbelower.
1.36.Describethetechniqueyouwouldusetoreducetheorange-peeleffectonthesurfaceofworkpieces. Orangepeelissurfacerougheninginducedbyplasticstrain.Thereareanumberofwaysofreducing theorangepeeleffect,including:
• Performingallformingoperationswithoutalubricant,orelseaverythinlubricantfilm(smaller thanthedesiredroughness)andverysmoothtooling.Thegoalistohavethesurfaceroughness ofthetoolingimpartedontotheworkpiece.
• Largegrainsexacerbateorangepeel,sotheuseofsmallgrainedmaterialswouldreduceorange peel.
• Ifdeformationprocessescanbedesignedsothatthesurfacesseenodeformation,thenthere wouldbenoorangepeel.Forexample,upsettingbeneathflatdiescanleadtoareductionin thicknesswithverylittlesurfacestrainsbeneaththeplaten(seeFig.14.3).
• Finishingoperationscanremoveorangepeeleffects.
1.37.Whatisthesignificanceofthefactthatsuchmetalsasleadandtinhavearecrystallizationtemperaturethatisaboutroomtemperature?
Recrystallizationaroundroomtemperaturepreventsthesemetalsfromworkhardeningwhencold worked.Thischaracteristicpreventstheirstrengtheningandhardening,thusrequiringarecrystallizationcycletorestoretheirductility.Thisbehaviorisalsousefulinexperimentalverificationof analyticalresultsconcerningforceandenergyrequirementsinmetalworkingprocesses(seePartIII ofthetext).
1.38.Itwasstatedinthischapterthattwinningusuallyoccursinhcpmaterials,butFig.1.6bshowstwinning inarectangulararrayofatoms.Canyouexplainthediscrepancy?
ThehcpunitcellshowninFig.1.6ahasahexagononthetopandbottomsurfaces.However,an intersectingplanethatisverticalinthisfigurewouldintersectatomsinarectangulararrayasdepicted inFig.1.6b.Thus,twinningoccursinhcpmaterials,butnotinthehexagonal(closepacked)plane suchasinthetopoftheunitcell.
1.39.Ithasbeennotedthatthemoreametalhasbeencoldworked,thelessitstrainhardens.Explain why.
Thisphenomenoncanbeobservedinstress-straincurves,suchasthoseshowninFigs.2.2and2.5. Recallthatthemaineffectsofcoldworkingarethatgrainsbecomeelongatedandthattheaverage grainsizebecomessmaller(asgrainsbreakdown)withstrain.Strainhardeningoccurswhendislocationsinterferewitheachotherandwithgrainboundaries.Whenametalisannealed,thegrains arelarge,andasmallstrainresultsingrainsmovingrelativelyeasilyatfirst,buttheyincreasingly interferewitheachotherasstrainincreases.Thisexplainsthatthereisstrainhardeningforannealed materialsatlowstrain.Tounderstandwhythereislessstrainhardeningathigherlevelsofcold work,considertheextremecaseofaveryhighlycold-workedmaterial,withverysmallgrainsand verymanydislocationsthatalreadyinterferewitheachother.Forthishighlycold-workedmaterial, thestresscannotbeincreasedmuchmorewithstrain,becausethedislocationshavenowhereelseto go-theyalreadyinterferewitheachotherandarepinnedatgrainboundaries.
1.40.Isitpossibletocoldworkametalattemperaturesabovetheboilingpointofwater?Explain.
Themetallurgicaldistinctionbetweencoldandhotworkingisassociatedwiththe homologoustemperature.Coldworkingisassociatedwithplasticdeformationofametalwhenitisbelowone-thirdof itsmeltingtemperatureonanabsolutescale.Attheboilingpointofwater,thetemperatureis100◦C, or373Kelvin.Ifthisvalueisone-thirdthemeltingtemperature,thenametalwouldhavetohavea meltingtemperatureof1119K,or846◦C.AscanbeseeninTable3.1,therearemanysuchmetals.
1.41.CommentonyourobservationsregardingFig.1.14.
Thisisanopen-endedproblemwithmanypotentialanswers.Studentsmaychoosetoaddressthis problembyfocusingontheshapeofindividualcurvesortheirrelationtoeachother.Theinstructor maywishtofocusthestudentsonacurveortwo,oraskifthefigurewouldgivethesametrendsfor amaterialthatisquicklyheated,heldatthattemperatureforafewseconds,andthenquenched,or alternativelyforonethatismaintainedatthetemperaturesforverylongtimes.
1.42.Isitpossibleforametaltobecompletelyisotropic?Explain.
Thisanswercanbeansweredonlyifisotropyisdefinedwithinlimits.Forexample:
• Asinglecrystalofametalhasaninherentanunavoidableanisotropy.Thus,atalengthscalethat isontheorderofamaterial’sgrainsize,ametalwillalwaysbeanisotropic.
• Ametalwithelongatedgrainswillhavealowerstrengthandhardnessinonedirectionthanin others,andthisisunavoidable.
• However,ametalthatcontainsalargenumberofsmallandequiaxedgrainswillhavethefirst twoeffectsessentiallymadeverysmall;themetalmaybeisotropicwithinmeasurementlimits.
• Annealingcanleadtoequiaxedgrains,anddependingonthemeasurementlimits,thiscanessentiallyresultinanisotropicmetal.
• Ametalwithaverysmallgrainsize(i.e.,ametalglass)canhavenoapparentcrystalstructure orslipsystems,andcanbeessentiallyisotropic.
1.43.ReferringtoFig.1.1,assumeyoucanmakeaballbearingfromasinglecrystal.Whatadvantages anddisadvantageswouldsuchabearinghave?
Thisisachallengingproblem,andisonethatcanbereintroducedattheendofChapters1-4.Students shouldbeencouragedtoprovideanswersthatarebasedontheirunderstandingandexperience;these areintendedtostartadiscussion.
Theadvantagesofasinglecrystalbearinginclude:
• Corrosiongenerallystartsatagrainboundary,wherelooselypackedatomsaremoresusceptible tochemicalattack;therefore,thebearingcouldbemorecorrosionresistant.
• Fatiguecracksmaybemoredifficulttoform,butwithsomematerialsaremoredifficulttopropogatethroughagrainthanalongagrainboundary.
• Thestrengthcouldbehigherthanapolycrystallinemetal.
• Distortionofthebearingatelevatedtemperaturesmaynotbeanissue.
Disadvantagesinclude:
• Theductilityofthebearingmaybesuspect,sinceplasticdeformationassociatedwithdislocation motionarenotpresent.
• Becauseoftheanisotropyofthematerial,producingaperfectspheremaybeverydifficult.
• Costmaybeaseriousconcern.
1.44.ReferringtoFig.1.10,explainwhyedgedislocationscannotcrossgrainboundariesusingappropriate sketches.
Figure1.10showsthatdislocationscanmoveinsidealatticeinclose-packeddirections.Atgrain boundaries,itshouldberecognizedthatthedirectionofthelatticechangesoneachside;thereisthen nonaturaldirectionforadislocationtotravel.Thisisshownbelowforasimplecubicarrangement. Notethatthegrainontheleft(blue)andthegrainontheright(red)arerotated15◦ fromeachother. Thegrainboundaryinbetweenhasonlyoccasionaldirectcontactbetweenatoms,andhasalarge gapwheredislocationswouldhaveadifficulttimemovingandtranslatingacross.Notethatgrain boundariesoftenhaveimpurities,whicharenotshown.
QuantitativeProblems
1.45.Howmanyatomsareinasinglerepeatingcellofanbcccrystalstructure?Howmanyinarepeating cellofanhcpstructure?
Foranfccstructure,refertoFig.1.4a.Theatomsateachcorneraresharedbyeightunitcells,and thereareeightoftheseatoms.Therefore,thecornerscontributeonetotalatom.Theatomsonthe facesareeachsharedbytwocells,andtherearesixoftheseatoms.Therefore,theatomsonthefaces contributeatotalofthreeatomstotheunitcell.Therefore,thetotalnumberofatomsinanfccunit cellisfouratoms.
Forthehcp,refertoFig.1.5.Theatomsontheperipheryofthetopandbottomareeachsharedby sixcells,andthereare12oftheseatoms(ontopandbottom),foracontributionoftwoatoms.The atomsinthecenterofthehexagonaresharedbytwocells,andtherearetwooftheseatoms,foranet contributionofoneatom.Therearealsothreeatomsfullycontainedintheunitcell.Therefore,there aresixatomsinanhcpunitcell.
1.46.Theatomicweightofgoldis196.97,meaningthat 6 023 × 1023 atomsweigh196.97grams.The densityofgoldis19,320kg/m3,andpuregoldformsfcccrystals.Estimatethediameterofagold atom.
Considerthefaceofthefccunitcell,whichconsistsofarighttrianglewithsidelength a andhypoteneuseof 4r.FromthePythagoreantheorem, a =4r/√2.Therefore,thevolumeoftheunitcell is
Eachfccunitcellhasfouratoms(seeProb.1.34),andeachatomhasamassof
Sothatthedensityinsideafccunitcellis
Solvingfor r yields r =1 444 × 10 10 m,or1.44 ˚ A.Notethattheacceptedvalueis1.6 ˚ A;thedifference isattributabletoanumberoffactors,includingimpuritiesincrystalstructureandaconcentrationof massinthenucleusoftheatom.
1.47.PlotthedatagiveninTable1.1intermsofgrains/mm2 vs.grains/mm3,anddiscussyourobservations.
Theplotisshownbelow.Itcanbeseenthatthegrainspercubicmillimeterincreasesfasterthanthe grainspersquaremillimeter.Thisrelationshipistobeexpectedsincethevolumeofanequiaxedgrain dependsonthediametercubed,whereasitsareadependsonthediametersquared.
01 x 106
Grains/mm3
1.48.Astripofmetalisreducedfrom40mminthicknessto20mmbycoldworking;asimilarstripis reducedfrom50mmto30mm.Whichofthesecold-workedstripswillrecrystallizeatalowertemperature?Why?
Themetalthatisreducedto20mmbycoldworkingwillrecrystallizeatalowertemperature.The morethecoldworkthelowerthetemperaturerequiredforrecrystallization.Thisisbecausethe numberofdislocationsandenergystoredinthematerialincreaseswithcoldwork.Thus,when recrystallizingamorehighlycoldworkedmaterial,thisenergycanberecoveredandlessenergy needstobeimpartedtothematerial.
1.49.Theballofaballpointpenis0.5mmindiameterandhasanASTMgrainsizeof12.Howmanygrains arethereintheball?
FromTable1.1,ametalwithanASTMgrainsizeof10hasabout520,000grains/mm3.Thevolumeof theballis
Multiplyingthevolumebythegrainspercubicmillimetergivesthenumberofgrainsinthepaper clipasabout2.18million.
1.50.Howmanygrainsareonthesurfaceoftheheadofapin?Assumethattheheadofapinisspherical witha1-mmdiameterandhasanASTMgrainsizeof10.
Notethatthesurfaceareaofasphereisgivenby A =4πr2.Therefore,a1mmdiameterheadhasa surfaceareaof
FromEq.(1.2),thenumberofgrainsperareais
N =211 =2048
Thisisthenumberofgrainsper0.0645mm2 ofactualarea;thereforethenumberofgrainsonthe surfaceis
1.51.TheunitcellsshowninFigs.1.3through1.5canberepresentedbytennisballsarrangedinvarious configurationsinabox.Insuchanarrangement,the atomicpackingfactor (APF)isdefinedastheratio ofthesumofthevolumesoftheatomstothevolumeoftheunitcell.ShowthattheAPFis0.68for thebccstructureand0.74forthefccstructure.
NotethatthebccunitcellinFig.1.3ahas2atomsinsideofit;oneinsidetheunitcellandeightatoms thathaveone-eighthoftheirvolumeinsidetheunitcell.Thereforethevolumeofatomsinsidethe cellis 8πr3/3,sincethevolumeofasphereis 4πr3/3.Notethatthediagonalofafaceofaunitcellhas alengthof a√2,whichcanbeeasilydeterminedfromthePythagoreantheorem.Usingthatdiagonal andtheheightof a resultsinthedeterminationofthediagonalofthecubeas a√3.Sincethereare fourradiiacrossthatdiagonal,itcanbededucedthat
Thevolumeoftheunitcellis
Therefore,theatomicpackingfactoris
Forthefcccell,therearefouratomsinthecell,sothevolumeofatomsinsidethefccunitcellis 16πr3/3.Onafaceofthefcccell,itcanbeshownfromthePythagoreantheoremthatthehypotenuse is a√2.Also,therearefourradiiacrossthediameter,sothat
Therefore,thevolumeoftheunitcellis
sothattheatomicpackingfactoris
1.52.Showthatthelatticeconstant a inFig.1.4aisrelatedtotheatomicradiusbytheformula α =2√2R, where R istheradiusoftheatomasdepictedbythetennis-ballmodel.
ForafacecenteredcubicunitcellasshowninFig.1.4a,thePythagoreantheoremyields
Therefore,
1.53.Showthat,forthefccunitcell,theradius r ofthelargestholeisgivenby r =0.414R.Determinethe sizeofthelargestholefortheironatomsinthefccstructure.
Thelargestholeisshowninthesketchbelow.Notethatthisholeoccursinotherlocations,infactin threeotherlocationsofthissketch.
ForafacecenteredcubicunitcellasshowninFig.1.4a,thePythagoreantheoremyields
Therefore,
Also,thesidedimension a is
Therefore,substitutingfor a,
1.54.Atechniciandeterminesthatthegrainsizeofacertainetchedspecimenis8.Uponfurtherchecking, itisfoundthatthemagnificationusedwas 150×,insteadofthe 100× thatisrequiredbytheASTM standards.Determinethecorrectgrainsize.
Ifthegrainsizeis8,thenthereare2048grainspersquaremillimeter(seeTable1.1).However,the magnificationwastoolarge,meaningthattoosmallofanareawasexamined.Foramagnificationof 100×,theareaisreducedbyafactorof1/1.82=0.309.Therefore,therereallyare632grainspermm2 , whichcorrespondstoagrainsizebetween6and7.
1.55.Ifthediameterofthealuminumatomis0.28nm,howmanyatomsarethereinagrainofASTMgrain size10?
Ifthegrainsizeis8,thereare65,000grainspercubicmillimeterofaluminum-seeTable1.1.Each grainhasavolumeof 1/65, 000=1 538 × 10 5 mm3.Notethatforanfccmaterialtherearefouratoms perunitcell(seesolutiontoProb.1.43),withatotalvolumeof 16πR3/3,andthatthediagonal, a,of theunitcellisgivenby
Hence,
Notethataslongasalltheatomsintheunitcellhavethesamesize,theatomicpackingfactorsdo notdependontheatomicradius.Therefore,thevolumeofthegrainwhichistakenupbyatomsis (4 88 × 10 4)(0 74)=3 61 × 10 4 mm3.(Recallthat1mm=106 nm.)Ifthediameterofanaluminum atomis0.5nm,thenitsradiusis0.25nmor 0 25 × 10 6 mm.Thevolumeofanaluminumatomis then
Dividingthevolumeofaluminuminthegrainbythevolumeofanaluminumatomyieldsthetotal numberofatomsinthegrainas (1.538 × 10 5)/(6.54 × 10 20)=2.35 × 1014 . 1.56.Thefollowingdataareobtainedintensiontestsofbrass:
DoesthematerialfollowtheHall-Petcheffect?Ifso,whatisthevalueof k?
First,itisobviousfromthistablethatthematerialbecomesstrongerasthegrainsizedecreases,which istheexpectedresult.However,itisnotclearwhetherEq.(1.1)isapplicable.Itispossibletoplotthe yieldstressasafunctionofgraindiameter,butitisbettertoplotitasafunctionof d 1/2,asfollows:
Theleast-squarescurvefitforastraightlineis
Y =35 22+458d 1/2
withan R factorof0.990.Thissuggeststhatalinearcurvefitisproper,anditcanbeconcludedthat thematerialdoesfollowtheHall-Petcheffect,withavalueof k =458 MPa-õm.
1.57.Doeswaterhaveahomologoustemperature?Whatisthehighesttemperaturewherewater(ice)can getcoldworked?
Allmaterialswithameltingpointhaveahomologoustemperature,definedastheratiooftheworking temperaturetothemeltingtemperatureonanabsolutescale.Ifwaterhadstrengtheningmechanisms likemetals,itcouldbecoldworkeduptoone-thirdofitsmeltingtemperature.Sincewatermeltsat 0◦Cor273K,itcouldbecoldworkedupto273/3=91K=-182 ◦C.However,waterisacomplicated materialwithmanypossiblecrystalstructures,anddoesnotingeneraldisplaycoldworkinglike metals.
1.58.Theatomicradiusofironis0.125nm,whilethatofacarbonatomis0.070nm.Canacarbonatom fitinsideasteelbccstructurewithoutdistortingtheneighboringatoms?
ConsiderthebcccellshowninFig.1.3a.Thediagonalis 4R.Thelengthofasideofthebccunitcellis a,sothat a 2 + a 2 + a 2 =(4R)2 or a =2 31R.Sincetherearetworadiionanedge,thesizeoftheopeningis 0 31R.Foriron,this wouldbe(0.31)(0.125)=0.0387nm,whichistoosmallforacarbonatomtofitwithoutdistortion.
1.59.Estimatetheatomicradiusforthefollowingmaterialsanddata:(a)Aluminum(atomicweight=26.98 gramspermole,density=2700kg/m3)(b)silver(atomicweight=107.87grams/mole,density=10,500 kg/m3);(c)titanium(atomicweight=47.87gramspermole,density=4506kg/m3).
RefertoProblem1.46.Aluminumandsilverarefccmaterials;titaniumcanbefccorhcp.Theradii foraluminumisfoundas r =1 44 ˚ A;1.45 ˚ Aforsilver,and1.46forTitanium.ThefollowingMatlab codecanbeusedtoevaluateothermaterials:
M=47.87/(6.023e23)/1000; rho=4506; r=(4*M/rho/22.63)ˆ0.3333;
1.60.Asimplecubicstructureinvolvesatomslocatedatthecubecornersthatareincontactwitheachother alongthecubeedges.Makeasketchofasimplecubicstructure,andcalculateitsatomicpacking factor.
Thesketchisasfollows:
Theedgehasalengthoftwotimestheatomradius,sothatthevolumeofthecubeis 8a3.Thereisone atominthestructure,whichwouldtakeavolumeof πa3.Therefore,theatomicpackingfactoris
APF = πa3 8a3 =39%.
1.61.EstimatetheASTMgrainsizenumberfora300mm(12in.)siliconwaferusedtoproducecomputer chips.
Thenotionofagrainisnotusefulforsiliconwafers,butitprovidesaninterestingdemonstrationof theASTMgrainsizenumber.Ifacrosssectionofa300-mmdiameterisconsidered,itwouldhave anareaof A = πd2/4= π(300)2/4=70, 680 mm2.0.0645mmwouldrepresent 9.12 × 10 7 grains. Therefore,
or n = 19
1.62.PurecopperandpuretitaniumfollowtheHall-Petchequation.Forcopper, Syi =24 MPaand k =0.12 MPa-m1/2.Fortitanium, Syi =80 MPaand k =0.40 MPa-m1/2.(a)Plottheyieldstrengthofthese metalsasafunctionofgrainsizeforASTMgrainsizesof-3to11.(b)Explainwhichmaterialwould seegreaterstrengtheningfromareductioningrainsize,asincoldworking.
NotefromEq.(1.2)thatforanASTMgrainsizeof-3,therewouldbe N =2 4 =0.0625 grainsin 0.0645mm2.Hence,eachgrainwouldhaveanareaofaround1.03mm2.Using A = πd2/4 would giveanestimateofagrainofdiameter1.15mm.Foragrainsizeof11,thegraindiameterissimilarly estimatedas d =0 008955 mm =8 955 × 10 6 m.TheHall-Petchequationisgivenby
y = Syi + kd 1/2
Thedesiredplotisasfollows:
Ascanbeseen,titaniumstrengthensmorewithadecreaseingrainsize(higherASTMgrainsize number).
Synthesis,Design,andProjects
1.63.Bystretchingathinstripofpolishedmetal,asinatension-testingmachine,demonstrateandcommentonwhathappenstoitsreflectivityasthestripisbeingstretched.
Thepolishedsurfaceisinitiallysmooth,whichallowslighttobereflecteduniformlyacrossthesurface.Asthemetalisstretched,thereflectivesurfaceofthepolishedsheetmetalwillbegintobecome dull.Theslipandtwinbandsdevelopedatthesurfacecauseroughening(seeFig.1.7),whichtends toscatterthereflectedlight.
1.64.Drawsomeanalogiestomechanicalfibering—forexample,layersofthindoughsprinkledwithflour ormeltedbutterbetweeneachlayer.
Awidevarietyofacceptableanswersarepossiblebasedonthestudent’sexperienceandcreativity. Someexamplesofmechanicalfiberinginclude:(a)foodproductssuchaslasagna,wherelayersof noodlesboundsauce,orpastrieswithmanythinlayers,suchasbaklava;(b)logcabins,wheretree trunksareorientedtoconstructwallsandthensealedwithamatrix;and(c)straw-reinforcedmud.
1.65.Drawsomeanalogiestothephenomenonofhotshortness.
Someanalogiestohotshortnessinclude:(a)abrickwallwithdeterioratingmortarbetweenthe bricks,(b)time-releasedmedicine,whereaslowlysolublematrixsurroundsdosesofquicklysolublemedicine,and(c)anOreocookieatroomtemperaturecomparedtoafrozencookie.
1.66.Obtainanumberofsmallballsmadeofplastic,wood,marble,ormetal,andarrangethemwithyour handsorgluethemtogethertorepresentthecrystalstructuresshowninFigs.1.3–1.5.Commenton yourobservations.
Bythestudent.Therearemanypossiblecomments,includingtherelativedensitiesofthethreecrystal structures(hcpisclearlydensest).Also,theingeniousandsimplesolid-ballmodelsarestrikingwhen performingsuchdemonstrations.
1.67.Takeadeckofplayingcards,placearubberbandaroundit,andthenslipthecardsagainsteachother torepresentFigs.1.6aand1.7.Ifyourepeatthesameexperimentwithmoreandmorerubberbands aroundthesamedeck,whatareyouaccomplishingasfarasthebehaviorofthedeckisconcerned?
Bythestudent.Withanincreasednumberofrubberbands,youarephysicallyincreasingthefriction forcebetweeneachcard.Thisisanalogoustoincreasingthemagnitudeoftheshearstressrequired tocauseslip.Furthermore,thegreaterthenumberofrubberbands,thehighertheshearorelastic modulusofthematerial(seeSection2.4).Thisproblemcanbetakentoaveryeffectiveextremeby usingsmallC-clampstohighlycompressthecards;theresultisanobjectthatactslikeonesolid,with muchhigherstiffnessthantheloosecards.
1.68.Giveexamplesinwhichanisotropyisscaledependent.Forexample,awireropecancontainannealedwiresthatareisotropiconamicroscopicscale,buttheropeasawholeisanisotropic.
Allmaterialsmaybehaveinananisotropicmannerwhenconsideredatatomicscales,butwhentaken asacontinuum,manymaterialsareisotropic.Otherexamplesinclude:
• Clothing,whichoverallappearstobeisotropic,butclearlyhasanisotropydefinedbythedirectionofthethreadsinthecloth.Thisanisotropicbehaviorcanbeverifiedbypullingsmallpatches oftheclothindifferentdirections.
• Woodhasdirectionality(orthotropic)butitcanbeignoredformanyapplications.
• Humanskin:itappearsisotropicatlargelengthscales,butmicroscopicallyitconsistsofcells withvaryingstrengthswithinthecell.
1.69.ThemovementofanedgedislocationwasdescribedinSection1.4.1bymeansofananalogyinvolvingahumpinacarpetonthefloorandhowthewholecarpetcaneventuallybemovedbymovingthe humpforward.Recallthattheentanglementofdislocationswasdescribedintermsoftwohumpsat differentangles.Useapieceofclothplacedonaflattabletodemonstratethesephenomena. Bythestudent.Thiscanbeclearlydemonstrated,especiallywithacloththatiscompliant(flexible) buthashighfrictionwithaflatsurface.Twomethodsofensuringthisisthecaseare(a)touseacotton material(asfoundinT-shirts)andwettingitbeforeconductingtheexperiments,or(b)sprayingthe bottomsideofthefabricwithtemporaryadhesives,asfoundinmostartsandofficesupplystores. Theexperiments(singleandtwolumps)canthenbeconductedandobservationsmade.
1.70.Ifyouwantedtostrengthenamaterial,wouldyouwishtohaveitconsistofonegrain,orwouldyou wantittohavegrainsthatcontaintheminimumnumberofatoms? Explain. Mostlikely,thedesirewouldbetohavegrainswiththeminimumnumberofatoms,assuggested byEq.(1.1)(TheHall-Petchequation).Thisistherationalefortheuseofmetallicglasses,withno identifiablegrainstructure.
MechanicalBehavior,Testing,and ManufacturingPropertiesofMaterials
QualitativeProblems
2.29. Onthesamescaleforstress,thetensiletruestress-truestraincurveishigherthantheengineering stress-engineeringstraincurve.Explainwhetherthisconditionalsoholdsforacompressiontest. Duringacompressiontest,thecross-sectionalareaofthespecimenincreasesastheloadisincreased. Sincetruestressisdefinedasloaddividedbytheinstantaneouscross-sectionalareaofthespecimen, thetruestressincompressionwillbelowerthantheengineeringstressforagivenload,providing thatfrictionalforces(betweentheplatensandthespecimen)arenegligible.
2.30. Explainwhyitisdifficulttobreakasheetofpaperintension,buteasytocutitwithscissors.
Thereareafewreasonsforthis.First,intension,theentirecrosssectionofthepaperisloaded, whereasincuttingwithscissorsonlyasmallarea(theoreticallyazeroarealine)seesastress.Also, thedesignofscissorsincorporatesnaturalleverage-theshearingtakesplaceclosertothepivotthan theforcethatisapplied.
2.31. Whatarethesimilaritiesanddifferencesbetweendeformationandstrain?
Thesimilaritiesarethattheyarebothameasureofachangeinshape;strainisadeformationnormalizedbyinitiallength,andthereforeisdimensionless.
2.32. CanamaterialhaveanegativePoisson’sratio?Givearationaleforyouranswer.
SolidmaterialdonothaveanegativePoisson’sratio,withtheexceptionofsomecompositematerials (seeChapter10),wheretherecanbeanegativePoisson’sratioinagivendirection.
Therationaleishardertoexpress.Itshouldmakesensetostudentsthatamaterial,whenstretched, shouldbecomenarrowerinthetransversedirections.IfPoisson’sratiowerezero,thentherewould benolateraldeflection.IfthePoisson’sratiowerenegative,itwouldexpandlaterallywhenstretched longitudinally.Alsoconsidercompression-ifcompressedaxially,thematerialwouldneedtothin laterally.Thisshouldn’tmakesensetostudents.
Thiscanbeproventoviolatethesecondlawofthermodynamicsbycalculatingallcomponentsof strainenergyinthiscase,butthisisanadvancedproof.
2.33. ReferringtoTable2.2,explainwhytherecanbesomuchvariationinthestrengthandelongationin aclassofmetalalloys.
Consideraluminumanditsalloys,forexample,wheretheyieldstrengthhasarangeof35to550MPa, andanelongationof45to4%.Anumberofreasonscanexplainthislargerange,asdescribedinthis chapter,including:
• Theloweststrengthisassociatedwithpurealuminum,thehighestwithhighlyalloyedaluminum.Thisdemonstratesthedramaticeffectofalloyingelementsonmechanicalproperties, andisthemostimportantfactor.
• Ifamaterialiscoldworked,itcanstrainharden,leadingtoanincreaseinmechanicalproperties.
• Heattreatingorannealingcanincreaseordecreasethemechanicalproperties,respectively.
• Residualstressesorthepresenceofinternalflawscanhavealargeeffectonthestrengthand elongationofamaterial.
2.34. ReferringtoTable2.2,explainwhythestiffnessofdiamondhassomuchvariation.
AsshowninTable2.2,thestiffnessofdiamondvariesfrom820to1050GPa.Thiscanbeexplainedby anumberoffactors:
• Singlecrystaldiamondwillhavedifferentstiffnessesindifferentdirections,whichisassociated withthecrystaldirection.
• Diamondcanbesinglecrystalorpolycrystalline.Apolycrystallinematerialwillhavemore isotropicproperties,butnottheextremevaluesofstiffness.
• Intension,thepresenceofinternalflawsorcrackscanleadtoalowerelasticmodulus,although thisisgenerallylessimportantthantheotherfactors.
2.35. Ithasbeenstatedthatthehigherthevalueof m,themorediffusetheneckis,andlikewise,thelower thevalueof m,themorelocalizedtheneckis.Explainthereasonforthisbehavior.
AsdiscussedinSection2.2.7,withhigh m values,thematerialstretchestoagreaterlengthbefore itfails;thisbehaviorisanindicationthatneckingisdelayedwithincreasing m.Whenneckingis abouttobegin,theneckingregion’sstrengthwithrespecttotherestofthespecimenincreases,due tostrainhardening.However,thestrainrateintheneckingregionisalsohigherthanintherestof thespecimen,becausethematerialiselongatingfasterthere.Sincethematerialintheneckedregion becomesstrongerasitisstrainedatahigherrate,theregionexhibitsagreaterresistancetonecking. Theincreaseinresistancetoneckingthusdependsonthemagnitudeof m.
2.36. Explainwhymaterialswithhigh m values,suchashotglassandtaffy,whenstretchedslowly,undergolargeelongationsbeforefailure.Considereventstakingplaceintheneckedregionofthespecimen.
TheanswerissimilartoAnswer2.35above.Theuseofglassandtaffyasexamples,andtheavailabilityofvideosofdeformingglassandtaffyontheInternet,clearlydemonstratestheimportanceof strainhardeninginachievinglargestrainsbeforefracture.Whenneckingisabouttobegin,theneckingregion’sstrengthwithrespecttotherestofthespecimenincreases,duetostrainhardening.With enoughstrainhardening,thedeformedsectionresistsfurtherstrain,forcingthematerialtostrain elsewhere.Thisleadstomoreuniformdeformationandapronouncedneck.
2.37. Explainifitispossibleforstress-straincurvesintensionteststoreach0%elongationasthegage lengthisincreasedfurther.
RefertoFigure2.1toseethegagelength, lo,andtheelongation, (le lo).Itwouldappearthatif thegagelengthbecomesinfinite,thentheengineeringstrainwouldapproachzero.However,thisis notlikelyfromapracticalstandpoint,sincetestingequipmentgenerallydoesnothavethecapacity forverylongpieces,recognizingthattheinitialspecimensizeneedstobesmallerthanthelargest displacementbythemachinejaws.However,gagelengthclearlyinfluencesthemeasuredelongation.
2.38. Withasimplesketch,explainwhetheritisnecessarytousetheoffsetmethodtodeterminetheyield strength, Sy ,ofamaterialthathasbeenhighlycoldworked.
AscanbeseenbyreviewingFig.2.3,ahighlycold-workedmetalwillhaveadistinctchangeinslope onitsstress-straincurveoccurringattheyieldpoint,sothattheoffsetmethodisnotnecessary(see alsoFig.2.5).
2.39. Explainwhythedifferencebetweenengineeringstrainandtruestrainbecomeslargerasstrainincreases.Doesthisdifferenceoccurforbothtensileandcompressivestrains?Explain.
Theanswerliesinthefactthatthedefinitionsofengineeringstrainandtruestrainaredifferent,the latterbeingbasedontheactualorinstantaneousdimensions,ascanbeseeninEqs.(2.2)and(2.7), respectively.Inbothcasesoftensionandcompression,thedifferenceincreasesasstrainincreases. ThisisshownquantitativelyinProblem2.74.
2.40. Consideranelastomer,suchasarubberband.Thismaterialcanundergoalargeelasticdeformation beforefailure,butafterfractureitrecoverscompletelytoitsoriginalshape.Isthismaterialbrittleor ductile?Explain.
Thisisaninterestingquestionandonethatcanbeansweredinanumberofways.Fromastress analysisstandpoint,thelargeelasticdeformationswouldleadtobluntingofstressconcentrations, andthematerialwouldbeconsideredductile.However,inmanufacturing,ductilityimpliesanability toachieveapermanentchangeinshape;inthiscase,arubberbandisextremelybrittle,asthereis essentiallynopermanentdeformationwhentherubberbandfracturesinatensiontest,forexample.
2.41. Ifamaterial(suchasaluminum)doesnothaveanendurancelimit,howthenwouldyouestimateits fatiguelife?
Materialswithoutendurancelimitshavetheirfatiguelifedefinedasacertainnumberofcyclesto failureatagivenstresslevel.Forengineeringpurposes,thisdefinitionallowsforanestimateofthe expectedlifetimeofapart.Thepartisthenusuallytakenoutofservicebeforeitslifetimeisreached. Analternativeapproachistousenondestructivetesttechniques(Section36.10)toperiodicallymeasuretheaccumulateddamageinapart,andthenusefracturemechanicsapproachestoestimatethe remaininglife.
2.42. Whatrole,ifany,doesfrictionplayinahardnesstest?Explain.
Theeffectoffrictionhasbeenfoundtobeminimal.Inahardnesstest,mostoftheindentationoccurs throughplasticdeformation,andthereisverylittleslidingattheindenter-workpieceinterface;see Fig.2.14.
2.43. Whichhardnesstestsandscaleswouldyouuseforverythinstripsofmetal,suchasaluminumfoil? Explain.
Ahardnesstestthatproducessmallindentationswouldhavetobeused;also,sincealuminumfoilis relativelysoft,averylightloadwouldberequired.Twoscalesthatsatisfytheserequirementsarethe Knoopmicrohardness(HK)andtheVickershardness(HV)atverylightloads(seeFig.2.13).Anarea ofcurrentresearchistheuseofatomicforcemicroscopyandnanoindenterstoobtainthehardnessof verythinmaterialsandcoatings.TheshapeoftheindenterusedisnotexactlythesameasinFig.2.13, andtheloadsareinthemicro-tomilli-Newtonrange.
2.44. ConsiderthecircumstancewhereaVickershardnesstestisconductedonamaterial.Sketchthe resultingindentationshapeifthereisaresidualstressonthesurface. Therearemanypossibleshapes.Considerthesimplesketchesbelow,wheretheVickersindentation foramaterialwithoutresidualstresses(thebaseline)isshowninred.Thegreenexamplewouldbe
particulartothecasewhereauniaxialresidualstressisalignedwiththeindenterandisconstant throughthethickness.Theblueexampleismoretypical,andshowsabiaxialresidualcompressive residualstress.IftheVickersinventorisnotalignedwiththestress,thentheshapewillbemoreofa rhomboidthansquare.
2.45. Whichofthetwotests,tensionorcompression,wouldrequireahighercapacityoftestingmachine, andwhy?
Thecompressiontestrequiresahighercapacitymachinesincethecross-sectionalareaofthespecimen increasesasthetestprogresses.Theincreaseinarearequiresaloadhigherthanthatforthetension testtoachievethesamestresslevel.Also,thereisfrictionbetweentheflatdies(platens)andthe workpiecesurfacesinacompressiontest(seeSections2.3and14.2)whichresultsinhigherpressures thanintension;thishigherpressurethenrequireslargerforcesforthesamecross-sectionalarea.In addition,thereismoreredundantworkincompressiontestingthanintensiontesting,sothematerial willworkhardenmore(unlessthetestisconductedatelevatedtemperatures).
2.46. InaBrinellhardnesstest,theresultingimpressionisfoundtobeanellipse.Givepossibleexplanationsforthisresult.
TwopossibleexplanationsforanellipticalimpressionafteraBrinelltestare:(a)Anobviousreasonis thepossiblepresenceofasymmetricresidualstressesinthesurfacelayersofthematerialbeforethe test.(b)Thematerialitselfmaybehighlyanisotropic,suchasafiber-reinforcedcompositematerial, orduetoseverecoldworking.
2.47. Listandexplainbrieflytheconditionsthatinducebrittlefractureinanotherwiseductilemetal.
Brittlefracturecanbeinducedbyhighdeformationrates,lowertemperatures(particularlythosewith bccstructure),thepresenceofstressconcentration(notchesandcracks),stateofstress,radiationdamage,corrosion(includinghydrogenembrittlement).Ineachcase,thestressneededtocauseyielding israisedabovethestressneededtocausefailure,orthestressneededforacracktopropagateisbelow theyieldstressofthematerial(aswithstressconcentration).
2.48. Listthefactorsthatyouwouldconsiderinselectingahardnesstest.Explainwhy.
Hardnesstestsmainlyhavethreedifferences:(a)typeofindenter,(b)appliedload,and(c)method ofindentationmeasurement,i.e.,depthorsurfaceareaofindentation,orreboundofindenter.The hardnesstestselectedwoulddependontheestimatedhardnessoftheworkpiece,itssizeandthickness,andifaveragehardnessorthehardnessofindividualmicrostructuralcomponentsisdesired. Forinstance,thescleroscope,whichisportable,iscapableofmeasuringthehardnessoflargepieces thatcannotbeusedformeasurementbyothertechniques.
TheBrinellhardnesstestleavesafairlylargeindentation,thusprovidingagoodmeasureofaverage hardness,whiletheKnooptestleavesasmallindentationthatallowsfordeterminationofthehardnessoftheindividualphasesinatwo-phasealloy.ThesmallindentationoftheKnooptestalsoallows ittobeusefulinmeasuringthehardnessofverythinlayersorplatedlayersonparts.Notethatthe depthofindentationshouldbesmallrelativetopartthickness,andthatanychangeintheappearance ofthebottomsurfacethepartwillmakethetestresultsinvalid.
Figure2.15isausefulguidefordeterminingwhichhardnesstestisvalidforaclassofmaterial.Note thatoftennumeroushardnesstestsaresuitableforamaterial.Inthesecases,thebesthardnesstestis theonethathasoneormoreofthefollowingcharacteristics:
• Thebesthardnesstestisoftenonethatcanbeperformedquickly;thus,itmaybedesirableto alsoselectahardnesstestbasedonavailableequipment.
• Hardnesstestsareoftenspecifiedbycustomersaspartofaqualitycontrolrequirement.Whateverformofhardnesstestisspecifiedbythecustomeristheappropriateonetouse.
• Ahardnesstestthatismostcommonlyusedinaplantmaybethebestchoicesincetechnicians willbemostfamiliarwiththetestprotocolandtheequipmentismostlikelytobeingoodcalibration.
• Experimentalerrorcanbeminimizedbyselectingahardnesstestthatgivesthelargestpenetrationorindentationsize.
2.49. Listtwosituationswhereamaterial’stoughnessisimportantfromadesignstandpoint. Toughnessisimportantforapplicationswherefractureistobeavoidedorwhereenergyistobe absorbed.Forexample,metalsthatareusedinfatigueapplicationsaregenerallytough(although strengthisalsoimportant).Bumpersonvehiclesorusedasdevicestostopmotionaretough,sothat thematerialcandeformandabsorbenergy.Studentsshouldbeencouragedtoproducemanymore examples.
2.50. OnthebasisofFig.2.5,canyoucalculateifametaltension-testspecimenisrapidlypulledand broken,wherewouldthetemperaturebehighest,andwhy?
Sincetemperatureriseisduetoworkinput,itisobviousthatthetemperaturewillbehighestinthe neckedregionbecausethatiswherethestrainishighestand,hence,theenergydissipatedperunit volumeinplasticdeformationishighest.
2.51. CommentonthetemperaturedistributionifthespecimeninQuestion2.50ispulledveryslowly. Regardlessofthespeed,thereisworkputintothespecimen.However,thisenergylikelywillnot causeanappreciableorevennoticeableincreaseintemperature,sincethereismoretimeforheatto beconductedinthespecimen.
2.52. CommentonyourobservationsregardingthecontentsofTable2.2.
Bythestudent.Therearealargenumberofacceptableanswerstothisproblem.Astudentmay comparethevaluesinthetablebetweendifferentmaterialsormaterialclasses.Alternatively,the studentsmaycommentonthesizeoftherangeinproperties.Studentsshouldbeencouragedto developwellthought-outanswerstothisquestion.
2.53. Willthedisktestbeapplicabletoaductilematerial?Whyorwhynot?
Withaductilematerial,apointloadonadiskresultsinthecirculardiskbeingflattenedattheplatens andattainingellipticalshapeoftheoriginallyroundspecimen.Theflatteningconvertsthepointload toadistributedload,completelychangingthestressstateinthepiece.Therefore,Eq.(2.10)isnot valid,andtheusefulnessofthetestiscompromised.
2.54. RefertoTable2.4,andnotethetruestrainencounteredbyamaterialindifferentmanufacturing processes.Explainwhysometypicalstrainsarelargeandothersaresmall. Someoftheanswerswillbeclearerasthestudentprogressesthroughlaterchapters.However,it canbeseenthatmuchlargerstrainsareoftenencounteredinextrusionascomparedtoforging.The
mainreasonisthatwithlargeextrusionratios,thematerialcanhaveapronouncedimprovementin itsductilityandavoidanceofchevroncracking.Inadifferentoperationsuchasforging,theexterior ofthepartwillcrackasitisnotunderhydrostaticcompressionasinextrusion.
2.55. RefertoTable2.4,andsketchtheoriginalanddeformedshapeofa25mmspecimensubjectedto thelargesttypicalstrainforeachprocess.Whatareyourobservationsregardingstrainsthatcanbe achieved?
Asexamples,theleftfigureshowsacylinderthatisupsetforgedwithatotaltruestrainof-0.3 (midrangeofthevaluesinTable2.4).Thatis,theinitialheightis lo =25 mm,thefinalheightis 18.5mm.Theinitialdiameteris25mm,sothatthefinaldiameter(tomaintainthepartvolume)is 29.06mm.Thesketch(atscale)is:
Asamoreextremecase,considerextrusion,wherethe1mmheightisextended(duetoamid-range strainof3.5)to827mm.Thediameterforvolumeconstancywouldchangefrom25mmto4.34mm. Thefollowingsketchisat50%reductioninsize:
Thisdemonstratesthatverylargestrainscanbeachievedinmetalforming,butmoremoderatestrains arecommon.
2.56. Ifatensiontestoncarbonsteelisconductedatroomtemperature,andthenwithabathofboiling water,wouldyouexpectthestrengthtobedifferent?Explain.
Thestrengthwouldnotchange.Themeltingpointofsteelisaround1500◦C,oraround1770K.Onethirdofthistemperatureisaround860◦C.Boilingwaterisnothahighenoughtemperaturetoexceed ahomologoustemperatureofone-third.
2.57. Whathardnesstestissuitablefordeterminingthehardnessofathinceramiccoatingonapieceof metal?
Forathinceramiccoating,itisstillimportantthatthehardnessofthecoatingandnotthesubstrate bemeasured.Mostceramicshavelimitedductility(Section8.3),sothatKnooporVickerstestsare suitable,althoughtheMohstestcanalsobeusedtoobtainaqualitativevalue.Becauseoftheincreasingimportanceofcoatings,specialmicrohardnesstestshavebeendevelopedfortheirhardness measurement.
2.58. Wireropeconsistsofmanywiresthatbendandunbendastheropeisrunoverasheave.Awire-rope failureisinvestigated,anditisfoundthatsomeofthewires,whenexaminedunderascanningelectronmicroscope,displaycup-and-conefailuresurfaces,whileothersdisplaytransgranularfracture surfaces.Commentontheseobservations.
Therearealargenumberofpotentialreasonsforthisbehavior.However,alikelyexplanationisthat whenthewireropewasinuse,itwasrunoverasheaveordrumrepeatedly.Asaresult,someofthe wiresintheropefailedduetofatigue,sotheydisplaybrittlefracturesurfaces.Atsomepoint,enough wireshavefailedsothattheremainingwiresfailduetostaticoverload,anddisplaycup-and-cone failuresurfacesasaresult.
2.59. AstatisticalsamplingofRockwellChardnesstestsareconductedonamaterial,anditisdetermined thatthematerialisdefectivebecauseofinsufficienthardness.Thesupplierclaimsthatthetestsare flawedbecausethediamondconeindenterwasprobablydull.Isthisavalidclaim?Explain.
RefertoFig.2.13andnotethatifanindenterisblunt,thenthepenetration, t,underagivenloadwillbe smallerthanthatusingasharpindenter.Thisthentranslatesintoahigherhardness.Theexplanation isplausible,butinpractice,hardnesstestsarefairlyreliableandmeasurementsareconsistentifthe testingequipmentisproperlycalibratedandroutinelyserviced.
2.60. InaBrinellhardnesstest,theresultingimpressionisfoundtobeelliptical.Givepossibleexplanations forthisresult.
TwopossibleexplanationsforanellipticalimpressionafteraBrinelltestare:(a)Anobviousreasonis thepossiblepresenceofasymmetricresidualstressesinthesurfacelayersofthematerialbeforethe test.(b)Thematerialitselfmaybehighlyanisotropic,suchasafiber-reinforcedcompositematerial, orduetoseverecoldworking.
2.61. Inthemachiningofanextrudedaluminumblocktoproduceasmartphonecase,itisseenthatthere issignificantwarpageaftermachining.Explainwhy.Whatwouldyoudotoreducethiswarpage? Thewarpageislikelyduetoresidualstressinthealuminum.Therefore,strategiesshouldbefollowed toreduceresidualstressesortheireffects.Thesecaninclude:
• Thealuminumcanbeannealedbeforemachining.
• Thealuminumcanbestretchedtoasmallplasticstrain,say10%,toeliminateresidualstresses.
• Thestiffnessofthephonecasecanbeincreasedbyplacingmorematerialatcornersorevenin walls.
• Theresidualstresscanbemeasuredusinganeutronradiationsource,andthenplanningamachiningpaththatallowstheparttorelaxtothedesiredshape.
Notethatthefirstthreeapproachesare much easierthanthelastone.
2.62. Somecoatingsareextremelythin–someasthinasafewnanometers.ExplainwhyeventheKnoop testisnotabletogivereliableresultsforsuchcoatings.Recentinvestigationshaveattemptedtouse highlypolisheddiamonds(withatipradiusaround5nm)toindentsuchcoatingsinatomicforce microscopes.Whatconcernswouldyouhaveregardingtheappropriatenessofthetestresults?
Withacoatingofthicknessof5nm,thestressedvolumehastobeapproximatelyone-tenththisdepth, whichbeginstoapproachthesizeofindividualatoms.Thus,aknoopindentorwouldneedtohavea tipradiusthatwasatomicallysharpinordertogetresults.Evenwithhighlypolisheddiamondtipsin atomicforcemicroscopes,thisscaleproblemisunavoidable.However,thereareadditionalconcerns inthatthediamondindentermaynotbesymmetric,therearelargeadhesiveforcesatthesmallscales, therearecomplicatedelasticandviscoelasticrecoveryatsmalllengthscales,theremayberesidual stressesatthesurface,andthestressedvolumemayormaynotcontainadislocation(whereaswith Knooptests,thereisalwaysanumberofdislocations).
2.63. Selectanappropriatehardnesstestforeachofthefollowingmaterials,andjustifyyouranswer:
1.Cubicboronnitride
2.Lead
3.Cold-drawn0.5%Csteel
4.Diamond
5.Caramelcandy
6.Granite
Figure2.15isausefulguideforselectinghardnesstests.
1.Cubicboronnitrideisveryhard,andusefuldatacanbeobtainedonlyfromtheKnoopand Mohstests.TheMohsscaleisqualitativeanddoesnotgivenumericalvaluesforhardness,so theKnooptestispreferable.
2.Lead.AsshowninFig.2.15,leadissosoftthatonlytheBrinellandVickerstestsyielduseful data.Recognizingthatleadisverysoft,thelightestloadsinthesetestsshouldbeused.Consider theexpectedresultsinthistestifatypicalvalueofhardnessis4HBor4HV.FortheBrinelltest, Fig.2.13suggeststhattheexpectedindentationfora500kgloadis:
Notethatthisdimensionisalmostthesameasthediameteroftheindentor,andmakesthe usefulnessofthetesthighlyquestionable.FortheVickerstest,theexpectedindentationtest, usingthelowestallowableloadof1kg,is:
Thisismuchmorereasonable,suggestingthattheVickerstestisthebestalternativeforlead.
3.Cold-drawn0.5%steel.FromFig.2.15,allofthehardnesstestsaresuitableforthismaterial.As discussedinProblem2.40,thebestchoiceforthismaterialwilldependonanumberoffactors.
4.Diamond.Thehardnessofdiamondisdifficulttoobtain.Thehardnessofdiamondisreally determinedbyextrapolatingthehardnessontheMohscurvetoanotherscaleinFig.2.15.The hardnessofdiamondisusuallyquotedas8000to10,000HK.
5.Caramel(candy).Thiswouldbeaninterestingexperimenttoperform,buttheresultwillbe thatnoneofthehardnesstestscanbeusedforthismaterialbecauseitisfartoosoft.Also, thehardnessofcaramelisstronglytemperature-dependentandthatitcreeps,sothathardness measurementmaybemeaningless.
6.Granite.Thehardnessofgranitevariesaccordingtothesource,butitisapproximatelyaround apatiteontheMohsscale.Thus,varioushardnesstestscangivevaluableinformationongranite. Note,however,thatininspectinggranitesurfaces,onecanseevariousregionswithinwhich therewouldbehardnessvariations.Theparticularhardnesstestselectedwilldependonvarious factors,asdiscussedinpart(c)above.
