材料科学与工程基础作业讲评-6

第九次作业中文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/(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=-eY/eX=[-(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7.20Acylindricalmetalspecimen15.0mmindiameterand150mmlongistobesubjectedtoatensilestressof50Mpa;atthisstressleveltheresultingdeformationwillbetotallyelastic.(a)Iftheelongationmustbelessthan0.072mm,whichofthemetalsinTable7.1aresuitablecandidates?Why?=l/l0=0.072mm/150mm=0.00048=E，E=/=50MPa/0.00048=104GPa要使l0.072mm,则E104MPa,因此inTable7.1,themetalsofTungsten,steel,nickel,titaniumandcopperaresuitablecandidates.(b)If,inaddition,themaximumpermissiblediameterdecreaseis2.3×10-3mm,whichofthemetalsinTable7.1maybeused?Why?y=d/d0=0.0023mm/15mm=0.000153v=-y/x=0.000153/0.00048=0.319要使d0.0023mm,则v0.319,因此inTable7.1,themetalsofTungsten,steelandnickelmaybeused.7.24Acylindricalrod380mmlong,havingadiameterof10.0mm,istobesubjectedtoatensileload.Iftherodistoexperienceneitherplasticdeformationnoranelongationofmorethan0.9mmwhentheappliedloadis24,500N,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，l0.9mm,E必须大于131MPa,因此Steelalloy合适。7.47Asteelspecimenhavingarectangularcrosssectionofdimensions19mm×3.2mm(0.75in×0.125in.)hasthestress–strainbehaviorshowninFigure7.33.Ifthisspecimenissubjectedtoatensileforceof33,400N(7,500lbf),then(a)Determinetheelasticandplasticstrainvalues.(b)Ifitsoriginallengthis460mm(18in.),whatwillbeitsfinallengthaftertheloadinpartaisappliedandthenreleased?(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)=231GPa=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=Adld,Ad/A0=l0/ld=l0/(l0+l)=1/[(l0+l)/l0]=1/[1+]所以CW%=(A0-Ad)/A0100=(1-Ad/A0)100=[1-1/(1+)]100=[/(1+)]100,即上式。(b)Usingtheresultofparta,computethepercentcoldworkexperiencedbynavalbrass(thestress–strainbehaviorofwhichisshowninFigure7.12)whenastressof400MPaisapplied.=0.12CW%=[/(1+)]100=[0.12/(1+0.12)]100%=10.7%4-6.已知温度为25℃时五种高聚物的性能，用下面列的名称来识别是哪种高聚物，并说明原因。a.拉伸强度伸长率冲击强度（悬臂梁）弹性模量MPa%N·mMPa×103(1)62.111019.042.415(2)51.800.416.90(3)27.6724.080.828(4)69.001.096.90(5)17.32005.440.414名称：环氧树脂、聚四氟乙烯、聚乙烯、酚醛树脂、聚碳酸酯思考题（1）PC;（2)酚醛;（3）HDPE;（4）环氧树脂;（5）PTFE4-14.有哪些途径可以提高材料的刚性？复合材料、提高材料刚性、结晶、交联、提高分子量、热处理7.22Citetheprimarydifferencesbetweenelastic,anelastic,andplasticdeformationbehaviors.分别从概念、原子论角度、施加应力后的应变、材料的差别（或对应的材料）等几个方面阐述。8.18Describeinyourownwordsthethreestrengtheningmechanismsdiscussedinthischapter(i.e.,grainsizereduction,solidsolutionstrengthening,andstrainhardening).Besuretoexplainhowdislocationsareinvolvedineachofthestrengtheningtechniques.Grainsizereduction:晶粒尺寸减少，位错时滑移减少方向的改变；原子位置不连续减少。Solidsolutionstrengthening：加入不同种的原子相成形成固溶体或合金，这些加入的原子限制位错移动。Strainhardening：先施加应力产生位错。而位错之间的作用是排斥的，结果是存在的一个位错限制另一位错的移动。