Cite three reasons why ferrous alloys are（一）引用三个原因铁合金

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CHAPTER11

APPLICATIONSANDPROCESSINGOFMETALALLOYS

PROBLEMSOLUTIONS

FerrousAlloys

11.1(a)Listthefourclassificationsofsteels.(b)Foreach,brieflydescribethepropertiesandtypicalapplications.

Solution

Thisquestionasksthatwelistfourclassificationsofsteels,and,foreach,todescribepropertiesandcitetypicalapplications.

LowCarbonSteels

Properties:nonresponsivetoheattreatments;relativelysoftandweak;machinableandweldable.

Typicalapplications:automobilebodies,structuralshapes,pipelines,buildings,bridges,andtincans.

MediumCarbonSteels

Properties:heattreatable,relativelylargecombinationsofmechanicalcharacteristics.

Typicalapplications:railwaywheelsandtracks,gears,crankshafts,andmachineparts.

HighCarbonSteels

Properties:hard,strong,andrelativelybrittle.

Typicalapplications:chisels,hammers,knives,andhacksawblades.

HighAlloySteels(StainlessandTool)

Properties:hardandwearresistant;resistanttocorrosioninalargevarietyofenvironments.

Typicalapplications:cuttingtools,drills,cutlery,foodprocessing,andsurgicaltools.

11.2(a)Citethreereasonswhyferrousalloysareusedsoextensively.(b)Citethreecharacteristicsofferrousalloysthatlimittheirutilization.

Solution

(a)Ferrousalloysareusedextensivelybecause:

(1)Ironoresexistinabundantquantities.

(2)Economicalextraction,refining,andfabricationtechniquesareavailable.

(3)Thealloysmaybetailoredtohaveawiderangeofproperties.

(b)Disadvantagesofferrousalloysare:

(1)Theyaresusceptibletocorrosion.

(2)Theyhavearelativelyhighdensity.

(3)Theyhaverelativelylowelectricalconductivities.

11.3Whatisthefunctionofalloyingelementsintoolsteels?

Solution

Thealloyingelementsintoolsteels(e.g.,Cr,V,W,andMo)combinewiththecarbontoformveryhardandwear-resistantcarbidecompounds.

11.4ComputethevolumepercentofgraphiteVGrina3.5wt%Ccastiron,assumingthatallthecarbonexistsasthegraphitephase.Assumedensitiesof7.9and2.3g/cm3forferriteandgraphite,respectively.

Solution

Weareaskedtocomputethevolumepercentgraphiteina3.5wt%Ccastiron.Itfirstbecomesnecessarytocomputemassfractionsusingtheleverrule.Fromtheiron-carbonphasediagram(Figure11.2),thetie-lineintheandgraphitephasefieldextendsfromessentially0wt%Cto100wt%C.Thus,fora3.5wt%Ccastiron

ConversionfromweightfractiontovolumefractionofgraphiteispossibleusingEquation9.6aas

=0.111or11.1vol%

11.5Onthebasisofmicrostructure,brieflyexplainwhygrayironisbrittleandweakintension.

Solution

Grayironisweakandbrittleintensionbecausethetipsofthegraphiteflakesactaspointsofstressconcentration.

11.6Comparegrayandmalleablecastironswithrespectto(a)compositionandheattreatment,(b)microstructure,and(c)mechanicalcharacteristics.

Solution

Thisquestionasksustocomparevariousaspectsofgrayandmalleablecastirons.

(a)Withrespecttocompositionandheattreatment:

Grayiron--2.5to4.0wt%Cand1.0to3.0wt%Si.Formostgrayironsthereisnoheattreatmentaftersolidification.

Malleableiron--2.5to4.0wt%Candlessthan1.0wt%Si.Whiteironisheatedinanonoxidizingatmosphereandatatemperaturebetween800and900Cforanextendedtimeperiod.

(b)Withrespecttomicrostructure:

Grayiron--Graphiteflakesareembeddedinaferriteorpearlitematrix.

Malleableiron--Graphiteclustersareembeddedinaferriteorpearlitematrix.

(c)Withrespecttomechanicalcharacteristics:

Grayiron--Relativelyweakandbrittleintension;goodcapacityfordampingvibrations.

Malleableiron--Moderatestrengthandductility.

11.7Comparewhiteandnodularcastironswithrespectto(a)compositionandheattreatment,(b)microstructure,and(c)mechanicalcharacteristics.

Solution

Thisquestionasksustocomparewhiteandnodularcastirons.

(a)Withregardtocompositionandheattreatment:

Whiteiron--2.5to4.0wt%Candlessthan1.0wt%Si.Noheattreatment;however,coolingisrapidduringsolidification.

Nodularcastiron--2.5to4.0wt%C,1.0to3.0wt%Si,andasmallamountofMgorCe.Aheattreatmentatabout700Cmaybenecessarytoproduceaferriticmatrix.

(b)Withregardtomicrostructure:

Whiteiron--Thereareregionsofcementiteinterspersedwithinpearlite.

Nodularcastiron--Nodulesofgraphiteareembeddedinaferriteorpearlitematrix.

(c)Withrespecttomechanicalcharacteristics:

Whiteiron--Extremelyhardandbrittle.

Nodularcastiron--Moderatestrengthandductility.

11.8Isitpossibletoproducemalleablecastironinpieceshavinglargecross-sectionaldimensions?Whyorwhynot?

Solution

Itisnotpossibletoproducemalleableironinpieceshavinglargecross-sectionaldimensions.Whitecastironistheprecursorofmalleableiron,andarapidcoolingrateisnecessaryfortheformationofwhiteiron,whichmaynotbeaccomplishedatinteriorregionsofthickcross-sections.

NonferrousAlloys

11.9Whatistheprincipaldifferencebetweenwroughtandcastalloys?

Solution

Theprincipaldifferencebetweenwroughtandcastalloysisasfollows:wroughtalloysareductileenoughsoastobehotorcoldworkedduringfabrication,whereascastalloysarebrittletothedegreethatshapingbydeformationisnotpossibleandtheymustbefabricatedbycasting.

11.10Whymustrivetsofa2017aluminumalloyberefrigeratedbeforetheyareused?

Solution

Rivetsofa2017aluminumalloymustberefrigeratedbeforetheyareusedbecause,afterbeingsolutionheattreated,theyprecipitationhardenatroomtemperature.Onceprecipitationhardened,theyaretoostrongandbrittletobedriven.

11.11Whatisthechiefdifferencebetweenheat-treatableandnon-heat-treatablealloys?

Solution

Thechiefdifferencebetweenheat-treatableandnonheat-treatablealloysisthatheat-treatablealloysmaybestrengthenedbyaheattreatmentwhereinaprecipitatephaseisformed(precipitationhardening)oramartensitictransformationoccurs.Nonheat-treatablealloysarenotamenabletostrengtheningbysuchtreatments.

11.12Givethedistinctivefeatures,limitations,andapplicationsofthefollowingalloygroups:titaniumalloys,refractorymetals,superalloys,andnoblemetals.

Solution

TitaniumAlloys

Distinctivefeatures:relativelylowdensity,highmeltingtemperatures,andhighstrengthsarepossible.

Limitation:becauseofchemicalreactivitywithothermaterialsatelevatedtemperatures,thesealloysareexpensivetorefine.

Applications:aircraftstructures,spacevehicles,andinchemicalandpetroleumindustries.

RefractoryMetals

Distinctivefeatures:extremelyhighmeltingtemperatures;largeelasticmoduli,hardnesses,andstrengths.

Limitation:someexperiencerapidoxidationatelevatedtemperatures.

Applications:extrusiondies,structuralpartsinspacevehicles,incandescentlightfilaments,x-raytubes,andweldingelectrodes.

Superalloys

Distinctivefeatures:abletowithstandhightemperaturesandoxidizingatmospheresforlongtimeperiods.

Applications:aircraftturbines,nuclearreactors,andpetrochemicalequipment.

NobleMetals

Distinctivefeatures:highlyresistanttooxidation,especiallyatelevatedtemperatures;softandductile.

Limitation:expensive.

Applications:jewelry,dentalrestorationmaterials,coins,catalysts,andthermocouples.

FormingOperations

11.13Citeadvantagesanddisadvantagesofhotworkingandcoldworking.

Solution

Theadvantagesofcoldworkingare:

(1)Ahighqualitysurfacefinish.

(2)Themechanicalpropertiesmaybevaried.

(3)Closedimensionaltolerances.

Thedisadvantagesofcoldworkingare:

(1)Highdeformationenergyrequirements.

(2)Largedeformationsmustbeaccomplishedinsteps,whichmaybeexpensive.

(3)Alossofductility.

Theadvantagesofhotworkingare:

(1)Largedeformationsarepossible,whichmayberepeated.

(2)Deformationenergyrequirementsarerelativelylow.

Thedisadvantagesofhotworkingare:

(1)Apoorsurfacefinish.

(2)Avarietyofmechanicalpropertiesisnotpossible.

11.14(a)Citeadvantagesofformingmetalsbyextrusionasopposedtorolling.(b)Citesomedisadvantages.

Solution

(a)Theadvantagesofextrusionasopposedtorollingareasfollows:

(1)Pieceshavingmorecomplicatedcross-sectionalgeometriesmaybeformed.

(2)Seamlesstubingmaybeproduced.

(b)Thedisadvantagesofextrusionoverrollingareasfollows:

(1)Nonuniformdeformationoverthecross-section.

(2)Avariationinpropertiesmayresultoveracross-sectionofanextrudedpiece.

Casting

11.15Listfoursituationsinwhichcastingisthepreferredfabricationtechnique.

Solution

Foursituationsinwhichcastingisthepreferredfabricationtechniqueare:

(1)Forlargepiecesand/orcomplicatedshapes.

(2)Whenmechanicalstrengthisnotanimportantconsideration.

(3)Foralloyshavinglowductilities.

(4)Whenitisthemosteconomicalfabricationtechnique.

11.16Comparesand,die,investment,lostfoam,andcontinuouscastingtechniques.

Solution

Forsandcasting,sandisthemoldmaterial,atwo-piecemoldisused,ordinarilythesurfacefinishisnotanimportantconsideration,thesandmaybereused(butthemoldmaynot),castingratesarelow,andlargepiecesareusuallycast.

Fordiecasting,apermanentmoldisused,castingratesarehigh,themoltenmetalisforcedintothemoldunderpressure,atwo-piecemoldisused,andsmallpiecesarenormallycast.

Forinvestmentcasting,asingle-piecemoldisused,whichisnotreusable;itresultsinhighdimensionalaccuracy,goodreproductionofdetail,andafinesurfacefinish;andcastingratesarelow.

Forlostfoamcasting,thepatternispolystyrenefoam,whereasthemoldmaterialissand.Complexgeometriesandtighttolerancesarepossible.Castingratesarehigherthanforinvestment,andtherearefewenvironmentalwastes.

Forcontinuouscasting,attheconclusionoftheextractionprocess,themoltenmetaliscastintoacontinuousstrandhavingeitherarectangularorcircularcross-section;theseshapesaredesirableforsubsequentsecondarymetal-formingoperations.Thechemicalcompositionandmechanicalpropertiesarerelativelyuniformthroughoutthecross-section.

MiscellaneousTechniques

11.17Ifitisassumedthat,forsteelalloys,theaveragecoolingrateoftheheat-affectedzoneinthevicinityofaweldis10°C/s,comparethemicrostructuresandassociatedpropertiesthatwillresultfor1080(eutectoid)and4340alloysintheirHAZs.

Solution

Thisproblemasksthatwespecifyandcomparethemicrostructuresandmechanicalpropertiesintheheat-affectedweldzonesfor1080and4340alloysassumingthattheaveragecoolingrateis10C/s.Figure10.27showsthecontinuouscoolingtransformationdiagramforaniron-carbonalloyofeutectoidcomposition(1080),and,inaddition,coolingcurvesthatdelineatechangesinmicrostructure.Foracoolingrateof10C/s(whichislessthan35C/s)theresultingmicrostructurewillbetotallypearlite--probablyareasonablyfinepearlite.Ontheotherhand,inFigure10.28isshowntheCCTdiagramfora4340steel.Fromthisdiagramitmaybenotedthatacoolingrateof10C/sproducesatotallymartensiticstructure.Pearliteissofterandmoreductilethanmartensite,and,therefore,ismostlikelymoredesirable.

11.18Describeoneproblemthatmightexistwithasteelweldthatwascooledveryrapidly.

Solution

Ifasteelweldiscooledveryrapidly,martensitemayform,whichisverybrittle.Insomesituations,cracksmayformintheweldregionasitcools.

AnnealingProcesses

11.19Inyourownwordsdescribethefollowingheattreatmentproceduresforsteelsand,foreach,theintendedfinalmicrostructure:fullannealing,normalizing,quenching,andtempering.

Solution

Fullannealing--Heattoabout50CabovetheA3line,Figure11.10(iftheconcentrationofcarbonislessthantheeutectoid)orabovetheA1line(iftheconcentrationofcarbonisgreaterthantheeutectoid)untilthealloycomestoequilibrium;thenfurnacecooltoroomtemperature.Thefinalmicrostructureiscoarsepearlite.

Normalizing--Heattoatleast55CabovetheA3lineFigure11.10(iftheconcentrationofcarbonislessthantheeutectoid)orabovetheAcmline(iftheconcentrationofcarbonisgreaterthantheeutectoid)untilthealloycompletelytransformstoaustenite,thencoolinair.Thefinalmicrostructureisfinepearlite.

Quenching--Heattoatemperaturewithintheaustenitephaseregionandallowthespecimentofullyaustenitize,thenquenchtoroomtemperatureinoilorwater.Thefinalmicrostructureismartensite.

Tempering--Heataquenched(martensitic)specimen,toatemperaturebetween450and650C,forthetimenecessarytoachievethedesiredhardness.Thefinalmicrostructureistemperedmartensite.

11.20Citethreesourcesofinternalresidualstressesinmetalcomponents.Whataretwopossibleadverseconsequencesofthesestresses?

Solution

Threesourcesofresidualstressesinmetalcomponentsareplasticdeformationprocesses,nonuniformcoolingofapiecethatwascooledfromanelevatedtemperature,andaphasetransformationinwhichparentandproductphaseshavedifferentdensities.

Twoadverseconsequencesofthesestressesaredistortion(orwarpage)andfracture.

11.21Givetheapproximateminimumtemperatureatwhichitispossibletoaustenitizeeachofthefollowingiron–carbonalloysduringanormalizingheattreatment:(a)0.20wt%C,(b)0.76wt%C,and(c)0.95wt%C.

Solution

(a)For0.20wt%C,heattoatleast905C(1660F)sincetheA3temperatureis850C(1560F).

(b)For0.76wt%C,heattoatleast782C(1440F)sincetheA3temperatureis727C(1340F).

(c)For0.95wt%C,heattoatleast840C(1545F)sincetheAcmtemperatureis785C(1445F).

11.22Givetheapproximatetemperatureatwhichitisdesirabletoheateachofthefollowingiron–carbonalloysduringafullannealheattreatment:(a)0.25wt%C,(b)0.45wt%C,(c)0.85wt%C,and(d)1.10wt%C.

Solution

(a)For0.25wt%C,heattoabout880C(1510F)sincetheA3temperatureis830C(1420F).

(b)For0.45wt%C,heattoabout830C(1525F)sincetheA3temperatureis780C(1435F).

(c)For0.85wt%C,heattoabout777C(1430F)sincetheA1temperatureis727C(1340F).

(d)For1.10wt%C,heattoabout777C(1430F)sincetheA1temperatureis727C(1340F).

11.23Whatisthepurposeofaspheroidizingheattreatment?Onwhatclassesofalloysisitnormallyused?

Solution

Thepurposeofaspheroidizingheattreatmentistoproduceaverysoftandductilesteelalloyhavingaspheroiditicmicrostructure.Itisnormallyusedonmedium-andhigh-carbonsteels,which,byvirtueofcarboncontent,arerelativelyhardandstrong.

HeatTreatmentofSteels

11.24Brieflyexplainthedifferencebetweenhardnessandhardenability.

Solution

Hardnessisameasureofamaterial'sresistancetolocalizedsurfacedeformation,whereashardenabilityisameasureofthedepthtowhichaferrousalloymaybehardenedbytheformationofmartensite.Hardenabilityisdeterminedfromhardnesstests.

11.25Whatinfluencedoesthepresenceofalloyingelements(otherthancarbon)haveontheshapeofahardenabilitycurve?Brieflyexplainthiseffect.

Solution

Thepresenceofalloyingelements(otherthancarbon)causesamuchmoregradualdecreaseinhardnesswithpositionfromthequenchedendforahardenabilitycurve.Thereasonforthiseffectisthatalloyingelementsretardtheformationofpearliticandbainiticstructureswhicharenotashardasmartensite.

11.26Howwouldyouexpectadecreaseintheaustenitegrainsizetoaffectthehardenabilityofasteelalloy?Why?

Solution

Adecreaseofaustenitegrainsizewilldecreasethehardenability.Pearlitenormallynucleatesatgrainboundaries,andthesmallerthegrainsize,thegreaterthegrainboundaryarea,and,consequently,theeasieritisforpearlitetoform.

11.27Nametwothermalpropertiesofaliquidmediumthatwillinfluenceitsquenchingeffectiveness.

Solution

Thetwothermalpropertiesofaliquidmediumthatinfluenceitsquenchingeffectivenessarethermalconductivityandheatcapacity.

11.28Constructradialhardnessprofilesforthefollowing:

(a)A50-mm(2-in.)diametercylindricalspecimenofan8640steelalloythathasbeenquenchedinmoderatelyagitatedoil

Solution

InthemannerofExampleProblem11.1,theequivalentdistancesandhardnessestabulatedbelowweredeterminedfromFigures11.14and11.17b.

Radial Equivalent HRC

Position Distance,mm Hardness

Surface 7 54

3/4R 11 50

Midradius 14 45

Center 16 44

Theresultinghardnessprofileisplottedbelow.

(b)A75-mm(3-in.)diametercylindricalspecimenofa5140steelalloythathasbeenquenchedinmoderatelyagitatedoil

Solution

InthemannerofExampleProblem11.1,theequivalentdistancesandhardnessestabulatedbelowweredeterminedfromFigures11.14and11.17b.

Radial Equivalent HRC

Position Distance,mm Hardness

Surface 13 41

3/4R 17.5 37

Midradius 22 33

Center 25 32

Theresultinghardnessprofileisplottedbelow.

(c)A65-mm(2-in.)diametercylindricalspecimenofan8620steelalloythathasbeenquenchedinmoderatelyagitatedwater

Solution

InthemannerofExampleProblem11.1,theequivalentdistancesandhardnessestabulatedbelowweredeterminedfromFigures11.15and11.17a.

Radial Equivalent HRC

Position Distance,mm Hardness

Surface 2.5 42

3/4R 7 31

Midradius 11 25

Center 13 24

Theresultinghardnessprofileisplottedbelow.

(d)A70-mm(2-in.)diametercylindricalspecimenofa1040steelalloythathasbeenquenchedinmoderatelyagitatedwater.

Solution

InthemannerofExampleProblem11.1,theequivalentdistancesandhardnessestabulatedbelowweredeterminedfromFigures11.14and11.17a.

Radial Equivalent HRC

Position Distance,mm Hardness

Surface 3 48

3/4R 8 30

Midradius 13 23

Center 15 22

Theresultinghardnessprofileisplottedbelow.

11.29Comparetheeffectivenessofquenchinginmoderatelyagitatedwaterandoilbygraphing,onasingleplot,radialhardnessprofilesfor65-mm(2-in.)diametercylindricalspecimensofan8630steelthathavebeenquenchedinbothmedia.

Solution

Weareaskedtocomparetheeffectivenessofquenchinginmoderatelyagitatedwaterandoilbygraphing,onasingleplot,hardnessprofilesfora65mm(2-1/2in.)diametercylindricalspecimenofan8630steelthathasbeenquenchedinbothmedia.

Formoderatelyagitatedwater,theequivalentdistancesandhardnessesfortheseveralradialpositions[Figures11.17aand11.15]aretabulatedbelow.

Radial Equivalent HRC

Position Distance,mm Hardness

Surface 2.5 52

3/4R 7 43

Midradius 11 36

Center 13 33

Whileformoderatelyagitatedoil,theequivalentdistancesandhardnessesfortheseveralradialpositions[Figures11.17band11.15]aretabulatedbelow.

Radial Equivalent HRC

Position Distance,mm Hardness

Surface 10 37

3/4R 15 32

Midradius 18 29

Center 20 28

Thesedataareplottedhere.

PrecipitationHardening

11.30Compareprecipitationhardening(Section11.9)andthehardeningofsteelbyquenchingandtempering(Sections10.5,10.6,and10.8)withregardto

(a)Thetotalheattreatmentprocedure

(b)Themicrostructuresthatdevelop

(c)Howthemechanicalpropertieschangeduringtheseveralheattreatmentstages

Solution

(a)Withregardtothetotalheattreatmentprocedure,thestepsforthehardeningofsteelareasfollows:

(1)Austenitizeabovetheuppercriticaltemperature.

(2)Quenchtoarelativelylowtemperature.

(3)Temperatatemperaturebelowtheeutectoid.

(4)Cooltoroomtemperature.

Withregardtoprecipitationhardening,thestepsareasfollows:

(1)Solutionheattreatbyheatingintothesolidsolutionphaseregion.

(2)Quenchtoarelativelylowtemperature.

(3)Precipitationhardenbyheatingtoatemperaturethatiswithinthesolidtwo-phaseregion.

(4)Cooltoroomtemperature.

(b)Forthehardeningofsteel,themicrostructuresthatformatthevariousheattreatingstagesinpart(a)are:

(1)Austenite

(2)Martensite

(3)Temperedmartensite

(4)Temperedmartensite

Forprecipitationhardening,themicrostructuresthatformatthevariousheattreatingstagesinpart(a)are:

(1)Singlephase

(2)Singlephase--supersaturated

(3)Smallplate-likeparticlesofanewphasewithinamatrixoftheoriginalphase.

(4)Sameas(3)

(c)Forthehardeningofsteel,themechanicalcharacteristicsforthevariousstepsinpart(a)areasfollows:

(1)Notimportant

(2)Thesteelbecomeshardandbrittleuponquenching.

(3)Duringtempering,thealloysoftensslightlyandbecomesmoreductile.

(4)Nosignificantchangesuponcoolingtoormaintainingatroomtemperature.

Forprecipitationhardening,themechanicalcharacteristicsforthevariousstepsinpart(a)areasfollows:

(1)Notimportant

(2)Thealloyisrelativelysoft.

(3)Thealloyhardenswithincreasingtime(initially),andbecomesmorebrittle;itmaysoftenwithoveraging.

(4)Thealloymaycontinuetohardenoroverageatroomtemperature.

11.31Whatistheprincipaldifferencebetweennaturalandartificialagingprocesses?

Solution

Forprecipitationhardening,naturalagingisallowingtheprecipitationprocesstooccurattheambienttemperature;artificialagingiscarriedoutatanelevatedtemperature.

DesignProblems

FerrousAlloys

NonferrousAlloys

11.D1Belowisalistofmetalsandalloys:

Plaincarbonsteel

Magnesium

Brass

Zinc

Graycastiron

Toolsteel

Platinum

Aluminum

Stainlesssteel

Tungsten

Titaniumalloy

Selectfromthislisttheonemetaloralloythatisbestsuitedforeachofthefollowingapplications,andciteatleastonereasonforyourchoice:

(a)Theblockofaninternalcombustionengine

(b)Condensingheatexchangerforsteam

(c)Jetengineturbofanblades

(d)Drillbit

(e)Cryogenic(i.e.,verylowtemperature)container

(f)Asapyrotechnic(i.e.,inflaresandfireworks)

(g)High-temperaturefurnaceelementstobeusedinoxidizingatmospheres

Solution

(a)Graycastironwouldbethebestchoiceforanengineblockbecauseitisrelativelyeasytocast,iswearresistant,hasgoodvibrationdampingcharacteristics,andisrelativelyinexpensive.

(b)Stainlesssteelwouldbethebestchoiceforaheatexchangertocondensesteambecauseitiscorrosionresistanttothesteamandcondensate.

(c)Titaniumalloysarethebestchoiceforhigh-speedaircraft

jetengineturbofanbladesbecausetheyarelightweight,strong,andeasilyfabricatedveryresistanttocorrosion.However,onedrawbackistheircost.

(d)Atoolsteelwouldbethebestchoiceforadrillbitbecauseitisveryhardretainsitshardnessathightemperatureandiswearresistant,and,thus,willretainasharpcuttingedge.

(e)Foracryogenic(low-temperature)container,analuminumalloywouldbethebestchoice;aluminumalloyshaveanFCCcrystalstructure,andtherefore,areductileatverylowtemperatures.

(f)Asapyrotechnicinflaresandfireworks,magnesiumisthebestchoicebecauseitigniteseasilyandburnsreadilyinairwithaverybrightflame.

(g)Platinumisthebestchoiceforhigh-temperaturefurnaceelementstobeusedinoxidizingatmospheresbecauseitisveryductile,hasarelativelyveryhighmeltingtemperature,andishighlyresistanttooxidation.

11.D2Agroupofnewmaterialsarethemetallicglasses(oramorphousmetals).Writeanessayaboutthesematerialsinwhichyouaddressthefollowingissues:(1)compositionsofsomeofthecommonmetallicglasses,(2)characteristicsofthesematerialsthatmakethemtechnologicallyattractive,(3)characteristicsthatlimittheirutilization,(4)currentandpotentialuses,and(5)atleastonetechniquethatisusedtoproducemetallicglasses.

Solution

(a)Compositionally,themetallicglassmaterialsarerathercomplex;severalcompositionsareasfollows:Fe80B20,Fe72Cr8P13C7,Fe67Co18B14Si,Pd77.5Cu6.0Si16.5,andFe40Ni38Mo4B18.

(b)Thesematerialsareexceptionallystrongandtough,extremelycorrosionresistant,andareeasilymagnetized.

(c)Principaldrawbacksforthesematerialsare1)complicatedandexoticfabricationtechniquesarerequired;and2)inasmuchasveryrapidcoolingratesarerequired,atleastonedimensionofthematerialmustbesmall--i.e.,theyarenormallyproducedinribbonform.

(d)Potentialusesincludetransformercores,magneticamplifiers,headsformagnetictapeplayers,reinforcementsforpressurevesselsandtires,shieldsforelectromagneticinterference,securitytapesforlibrarybooks.

(e)Productiontechniquesincludecentrifugemeltspinning,planar-flowcasting,rapidpressureapplication,arcmeltspinning.

11.D3Ofthefollowingalloys,picktheone(s)thatmaybestrengthenedbyheattreatment,coldwork,orboth:R50500titanium,AZ31Bmagnesium,6061aluminum,C51000phosphorbronze,lead,6150steel,304stainlesssteel,andC17200berylliumcopper.

Solution

Thisquestionprovidesuswithalistofseveralmetalalloys,andthenasksustopickthosethatmaybestrengthenedbyheattreatment,bycoldwork,orboth.Thosealloysthatmaybeheattreatedareeitherthosenotedas"heattreatable"(Tables11.6through11.9),orasmartensiticstainlesssteels(Table11.4).Alloysthatmaybestrengthenedbycoldworkingmustnotbeexceptionallybrittle,and,furthermore,musthaverecrystallizationtemperaturesaboveroomtemperature(whichimmediatelyeliminateslead).Thealloysthatfallwithinthethreeclassificationsareasfollows:

HeatTreatable ColdWorkable Both

6150steel 6150steel 615