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1、材料科学与工程基础作业讲评-6英文书7.20Acylindricalmetalspecimen15.0mmindiameterand150mmlongistobesubjectedtoatensilestressof50Mpa;atthisstressleveltheresultingdeformationwillbetotallyelastic.(a)Iftheelongationmustbelessthan0.072mm,whichofthemetalsinTable7.1aresuitablecandidates?Why?=l/l0=0.072mm/15
2、0mm=0.00048=E,E=/=50MPa/0.00048=104GPa要使l<0.072mm,则E>104MPa,因此inTable7.1,themetalsofTungsten,steel,nickel,titaniumandcopperaresuitablecandidates.(b)If,inaddition,themaximumpermissiblediameterdecreaseis2.3×10-3mm,whichofthemetalsinTable7.1maybeused?Why?y=d/d0=0.0023mm/15mm=0.000
3、153v=-y/x=0.000153/0.00048=0.319要使d<0.0023mm,则v<0.319,因此inTable7.1,themetalsofTungsten,steelandnickelmaybeused.7.24Acylindricalrod380mmlong,havingadiameterof10.0mm,istobesubjectedtoatensileload.Iftherodistoexperienceneitherplasticdeformationnoranelongationofmorethan0.9mmwhentheappl
4、iedloadis24,500N,whichofthefourmetalsoralloyslistedbelowarepossiblecandidates?=F/A0=F/(d02/4)=24500N/(3.14*102mm2/4)=312MPa,因此从屈服强度来看,只有SteelalloyandBrassalloy才有可能。另外:=l/l0=0.9mm/380mm=0.00237=E,E=/=312MPa/0.00237=131MPa,因此,l<0.9mm,E必须 >大于131MPa,因此Steelalloy合适。7.47Asteelspeci
5、menhavingarectangularcrosssectionofdimensions19mm×3.2mm(0.75in×0.125in.)hasthestress–strainbehaviorshowninFigure7.33.Ifthisspecimenissubjectedtoatensileforceof33,400N(7,500lbf),then(a)Determinetheelasticandplasticstrainvalues.(b)Ifitsoriginallengthis460mm(18in.),whatwillbeitsfinalleng
6、thaftertheloadinpartaisappliedandthenreleased?(a)Determinetheelasticandplasticstrainvalues.弹性变形应变数值大约:0-0.0015,塑性变形:>0.0015(b)Ifitsoriginallengthis460mm(18in.),whatwillbeitsfinallengthaftertheloadinpartaisappliedandthenreleased?E=slope=/=(2-1)/(2-1)=(300-0)MPa/(0.0013-0)=231GP
7、a=F/A0=F/(a*b)=33400N/(19*3.2mm2)=549.3MPa图中可知,在该应力时的总应变为总=0.005,最大弹性为:弹=0.0015去除应力后弹性应变回复,故长度为:l0*(1+总-弹)=460*(1+0.005–0.0015)=461.61mm8.24(a)Show,foratensiletest,thatifthereisnochangeinspecimenvolumeduringthedeformationprocess(i.e.,A0l0=Adld).CW%=(A0-Ad)/A0100=(1-Ad/A0)*100A0l0
8、=Adld