材料科学与工程基础作业讲评-6

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1、第九次作业第九次作业中文中文4-1：铝的弹性模量为70GPa,泊松比为0.34，在83MPa的静水压时，此单位晶胞的体积是多少？　　由E=3K(1-2ν)得K=E/[3(1-2ν)]=70Gpa/[3(1-2*0.34)]=72.9Gpa△V/V=σ/K=83Mpa/72.9GPa=1.14‰V=4.04963*10-30*(1-1.14‰)=66.310-30(m3)4-3　直径为12.83mm的试棒，标距长度为50mm，轴向受200kN的作用力后拉长0.456mm，且直径变成12.79mm，(a)此试棒的体积模量是多少？(b)剪切模量是多少？解：σ=F/S=F

2、/(πd2/4)=1.56GPaε=ΔL/L=0.456/50=0.912%正弹性模量：E=σ/ε=1.56Gpa/0.912%=172.9Gpa泊松比：ν=-e/eYX=[-(12.79-12.83)/12.83]/0.912%=0.342(a)体积模量：K=E/[3(1-2ν)]=172.9/[3(1-2*0.342)]=182Gpa(b)剪切模量：G=E/(2(1+ν))=172.9/[2*(1+0.342)]=64Gpa英文书英文书7.20Acylindricalmetalspecimen15.0mmindiameterand150mmlongistobes

3、ubjectedtoatensilestressof50Mpa;atthisstressleveltheresultingdeformationwillbetotallyelastic.(a)Iftheelongationmustbelessthan0.072mm,whichofthemetalsinTable7.1aresuitablecandidates?Why?=l/l=0.072mm/150mm=0.000480=E，E=/=50MPa/0.00048=104GPa要使l<0.072mm,则E>104MPa,因此inTable7.1,theme

4、talsofTungsten,steel,nickel,titaniumandcopperaresuitablecandidates.(b(b)If,inaddition,themaximumpermissiblediameterdecreaseis2.3×10-3mm,whichofthemetalsinTable7.1maybeused?Why?=d/d=0.0023mm/15mm=0.000153y0v=-/=0.000153/0.00048=0.319yx要使d<0.0023mm,则v<0.319,因此inTable7.1,themetalsofT

5、ungsten,steelandnickelmaybeused.7.24Acylindricalrod380mmlong,havingadiameterof10.0mm,istobesubjectedtoatensileload.Iftherodistoexperienceneitherplasticdeformationnoranelongationofmorethan0.9mmwhentheappliedloadis24,500N,whichofthefourmetalsoralloyslistedbelowarepossiblecandidates?=F/

6、A=F/A=F/(=F/(dd22/4)=24500N/(3.14/4)=24500N/(3.14**1010220000mmmm22/4)=312MPa,/4)=312MPa,因此从屈服强度来看，只有因此从屈服强度来看，只有SteelalloyandSteelalloyandBrassalloyBrassalloy才有可能。才有可能。另外：另外：==l/ll/l=0.9mm/380mm=0.00237=0.9mm/380mm=0.0023700==EE，，EE==//=312MPa/0.00237=131MPa,=312MPa/0.00

7、237=131MPa,因此，因此，ll<0.9mm,<0.9mm,EE必须　必须　>>大于大于131MPa,131MPa,因因此此SteelalloySteelalloy合适。合适。7.47Asteelspecimenhavingarectangularcrosssectionofdimensions19mm×3.2mm(0.75in×0.125in.)hasthestress–strainbehaviorshowninFigure7.33.Ifthisspecimenissubjectedtoatensileforceof33,400N(7,500lbf)