当前位置:首页 > 临时分类 > 材料科学与工程基础英文版第五版课后习题
Homework11.1Whatarematerials?Listeightcommonlyencounteredengineeringmaterials.Answer1.1:Materialsaresubstancesofwhichsomethingiscomposedormade.Steels,aluminumalloys,concrete,wood,glass,plastics,ceramicsandelectronicmaterials.1.2Whatarethemainclassesofengineeringmaterials?Answer1.2:Metallic,polymeric,ceramic,composite,andelectronicmaterialsarethefivemainclasses.1.3Whataresomeoftheimportantpropertiesofeachofthefivemainclassesofengineeringmaterials?Answer1.3:MetallicMaterials•manyarerelativelystrongandductileatroomtemperature•somehavegoodstrengthathightemperature•mosthaverelativelyhighelectricalandthermalconductivitiesPolymericMaterials•generallyarepoorelectricalandthermalconductors•mosthavelowtomediumstrengths•mosthavelowdensities•mostarerelativelyeasytoprocessintofinalshape•somearetransparentCeramicMaterials•generallyhavehighhardnessandaremechanicallybrittle•somehaveusefulhightemperaturestrength•mosthavepoorelectricalandthermalconductivitiesCompositeMaterials•haveawiderangeofstrengthfromlowtoveryhigh•somehaveveryhighstrength-to-weightratios(e.g.carbon-fiberepoxymaterials)•somehavemediumstrengthandareabletobecastorformedintoavarietyofsha(e.g.fiberglass-polyestermaterials)•somehaveuseablestrengthsatverylowcost(e.g.woodandconcrete)ElectronicMaterials•abletodetect,amplifyandtransmitelectricalsignalsinacomplexmanner•arelightweight,compactandenergyefficient1.8Whatarenanomaterials?Whataresomeproposedadvantagesofusingnanomaterialsovertheirconventionalcounterparts?Answer1.8:Aredefinedasmaterialswithacharacteristiclengthscalesmallerthan100nm.Thelengthscalecouldbeparticlediameter,grainsizeinamaterial,layerthicknessinasensor,etc.Thesematerialshavepropertiesdifferentthanthatatbulkscaleoratthemolecularscale.Thesematerialshaveoftenenhancedpropertiesandcharacteristicsbecauseoftheirnano-featuresincomparisontotheirmicro-featuredcounterparts.Thestructural,chemical,electronic,andthermalproperties(amongothercharacteristics)areoftenenhancedatthenano-scale.Homework2Chapter3,Problem4Whatarethethreemostcommonmetalcrystalstructures?Listfivemetalsthathaveeachofthesecrystalstructures.Chapter3,Solution4Thethreemostcommoncrystalstructuresfoundinmetalsare:body-centeredcubic(BCC),face-centeredcubic(FCC),andhexagonalclose-packed(HCP).Examplesofmetalshavingthesestructuresincludethefollowing.BCC:iron,vanadium,tungsten,niobium,andchromium.FCC:copper,aluminum,lead,nickel,andsilver.HCP:magnesium,titanium,zinc,beryllium,andcadmium.Chapter3,Problem5ForaBCCunitcell,(a)howmanyatomsarethereinsidetheunitcell,(b)whatisthecoordinationnumberfortheatoms,(c)whatistherelationshipbetweenthelengthofthesideaoftheBCCunitcellandtheradiusofitsatoms,and(d)APF=0.68or68%Chapter3,Solution5(a)ABCCcrystalstructurehastwoatomsineachunitcell.(b)ABCCcrystalstructurehasacoordinationnumberofeight.(c)InaBCCunitcell,onecompleteatomandtwoatomeighthstoucheachotheralongthecubediagonal.Thisgeometrytranslatesintotherelationship34.aRChapter3,Problem6ForanFCCunitcell,(a)howmanyatomsarethereinsidetheunitcell,(b)Whatisthecoordinationnumberfortheatoms,(c)24Ra,and(d)whatistheatomicpackingfactor?Chapter3,Solution6(a)EachunitcelloftheFCCcrystalstructurecontainsfouratoms.(b)TheFCCcrystalstructurehasacoordinationnumberoftwelve.(d)Bydefinition,theatomicpackingfactorisgivenas:volumeofatomsinFCCunitcellAtomicpackingfactorvolumeoftheFCCunitcellThesevolumes,associatedwiththefour-atomFCCunitcell,are33416433atomsVRRand3unitcellVawherearepresentsthelatticeconstant.Substituting42Ra,33unitcell6422RVaTheatomicpackingfactorthenbecomes,3316122APF(FCCunitcell)3632RR=0.74Chapter3,Problem7ForanHCPunitcell(considertheprimitivecell),(a)howmanyatomsarethereinsidetheunitcell,(b)Whatisthecoordinationnumberfortheatoms,(c)whatistheatomicpackingfactor,(d)whatistheidealc/aratioforHCPmetals,and(e)repeatathroughcconsideringthe“larger”cell.Chapter3,Solution7Theprimitivecellhas(a)twoatoms/unitcell;(b)ThecoordinationnumberassociatedwiththeHCPcrystalstructureistwelve.(c)theAPFis0.74or74%;(d)Theidealc/aratioforHCPmetalsis1.633;(e)allanswersremainthesameexceptfor(a)wherethenewansweris6.Homework3Chapter3,Problem25Lithiumat20CisBCCandhasalatticeconstantof0.35092nm.Calculateavaluefortheatomicradiusofalithiumatominnanometers.Chapter3,Solution25ForthelithiumBCCstructure,whichhasalatticeconstantofa=0.35092nm,theatomicradiusis,33(0.35092nm)44Ra0.152nmChapter3,Problem27PalladiumisFCCandhasanatomicradiusof0.137nm.Calculateavalueforitslatticeconstantainnanometers.Chapter3,Solution27LettingarepresenttheFCCunitcelledgelengthandRthepalladiumatomicradius,Chapter3,Problem31DrawthefollowingdirectionsinaBCCunitcellandlistthepositioncoordinatesoftheatomswhosecentersareintersectedbythedirectionvector:(a)[100](b)[110](c)[111]Chapter3,Solution31(a)PositionCoordinates:(b)PositionCoordinates:(c)PositionCoordinates:(0,0,0),(1,0,0)(0,0,0),(1,1,0)(0,0,0),(1,1,1)Chapter3,Problem32Drawdirectionvectorsinunitcellsforthefollowingcubicdirections:(a)(b)(c)(d)Chapter3,Solution324424or(0.137nm)22aRaR0.387nm111
本文标题:材料科学与工程基础英文版第五版课后习题
链接地址:https://www.777doc.com/doc-2382832 .html