材料科学与工程基础作业讲评-6

10九月2026材料科学与工程基础作业讲评-64-3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

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=0.000153v=-

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.Iftherodistoexperienceneitherplasticdeformationnoranelongationofmorethan0.9mmwhentheappliedloadis24,500N,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，

l<0.9mm,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.,A0

l0

=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)62.111019.042.41551.800.416.9027.6724.080.82869.001.096.9017.32005.440.414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