材料科学与工程基础作业讲评（课堂PPT）

4-12热处理（退火）的实质是什么？它对材料的拉伸强度、硬度、尺寸稳定性、冲击强度和断裂伸长率有什么影响？结构弛豫，非晶至结晶转变,第十次作业中文,1,4-20玻璃的理论强度超过7000MPa。一块平板玻璃在60MPa弯曲张力下破坏。我们假定裂纹尖端为氧离子尺寸（即裂纹尖端曲率半径为氧离子半径，RO2-=0.14nm），问对应这种低应力断裂，相应的裂纹深度为多大？max=01+2（a/）0.5=2f（a/）0.57000=601+2（a/0.14）0.5a=468nm，2a=936.5nm括号中1省略则a=476nm,2,4-21某钢材的屈服强度为1100MPa,抗拉强度为1200MPa，断裂韧性（KIC）为90MPam1/2。（a）在一钢板上有2mm的边裂，在他产生屈服之前是否会先断裂？（b）在屈服发生之前，不产生断裂的可容许断裂缝的最大深度是多少？（假设几何因子Y等于1.1，试样的拉应力与边裂纹垂直）=KIC/Y（a）0.5=1032Mpa，先断裂a=（KIC/Y）2/=1.5mm11001.7612001.5,3,7.29Acylindricalspecimenofaluminumhavingadiameterof12.8mmandagaugelengthof50.800mmispulledintension.Usetheloadelongationcharacteristicstabulatedbelowtocompleteproblemsathroughf.(a)Plotthedataasengineeringstressversusengineeringstrain.(b)Computethemodulusofelasticity.(c)Determinetheyieldstrengthatastrainoffsetof0.002.(d)Determinethetensilestrengthofthisalloy.(e)Whatistheapproximateductility,inpercentelongation?(f)Computethemodulusofresilience.,4,5,6,(a)A0=d02/4=3.14*12.82mm2/4=128.6mm2=F/A0=l/l0,l0=50.8mm(b)E=slope=/=(2-1)/(2-1)=(57.0-0)MPa/(0.001-0)=57.0GPa117.4MPa(d)369.4MPa(e)%EL=(lf-l0)/l0*100=(59.18250.8)/50.8*100=16.5(f)Ur=1/2*y*y=0.5*117.4*106*0.002J/m3=117400J/m3,7,7.40Forsomemetalalloy,atruestressof415MPaproducesaplastictruestrainof0.475.Howmuchwillaspecimenofthismaterialelongatewhenatruestressof325MPaisappliediftheoriginallengthis300mm?Assumeavalueof0.25forthestrain-hardeningexponentn.T=KTn,K=T/Tn=415MPa/(0.4750.25)=500MPaT=(T/K)1/n=(325MPa/500MPa)1/0.25=0.179T=ln(li/l0),li=l0*eT=300mm*e0.179=300mm*1.196=358.8mm,8,9.17Someaircraftcomponentisfabricatedfromanaluminumalloythathasaplanestrainfracturetoughnessof35MPa.m1/2.Ithasbeendeterminedthatfractureresultsatastressof250MPawhenthemaximum(orcritical)internalcracklengthis2.0mm.Forthissamecomponentandalloy,willfractureoccuratastresslevelof325Mpawhenthemaximuminternalcracklengthis1.0mm?Whyorwhynot?KIC=Y(a)1/2,Y=KIC/(a)1/2=(35Mpam1/2)/(250Mpa)(3.14*0.002/2m)1/2=2.50另外受力：=KIC/(a)1/2Y=(35Mpam1/2)/(3.14*0.001/2m)1/2*2.50=353.3MPa,因此不断裂。,9,9.18Supposethatawingcomponentonanaircraftisfabricatedfromanaluminumalloythathasaplanestrainfracturetoughnessof40Mpam1/2.Ithasbeendeterminedthatfractureresultsatastressof365MPawhenthemaximuminternalcracklengthis2.5mm.Forthissamecomponentandalloy,computethestresslevelatwhichfracturewilloccurforacriticalinternalcracklengthof4.0mm.,KIC=Y(a)1/2,Y=KIC/(a)1/2=(40Mpam1/2)/(365Mpa)(3.14*0.0025/2m)1/2=1.75另外受力：=KIC/(a)1/2Y=(40Mpam1/2)/(3.14*0.004/2m)1/21.75=288.4MPa,10,9.32A12.5mmdiametercylindricalrodfabricatedfroma2014-T6alloy(Figure9.46)issubjectedtoarepeatedtension-compressionloadcyclingalongitsaxis.Computethemaximumandminimumloadsthatwillbeappliedtoyieldafatiguelifeof1.0107cycles.Assumethatthestressplottedontheverticalaxisisstressamplitude,anddataweretakenforameanstressof50MPa.,11,12,Atafatiguelifeof1.0107cycles,thestressamplitudeaisabout175(160)MPa.a=(max-min)/2,英文书P257m=(max+min)/2,英文书P258Somax=m+a,min=m-a,Fmax/A=m+aFmax=(m+a)A=27598N(25758)Fmin=(m-a)A=-15332N(-13492),13,思考题4-7一条长212cm的铜线，直径为0.76mm。当外加载荷为8.7kg时开始产生塑性变形(a)此作用力是多少牛顿？(b)外加载荷为15.2kg时，此线的应变为0.011，则去除载荷后，铜线的长度为多少？此铜线的屈服强度是多少？解：（a）F1=mg=8.7*9.8=85.3N（b）=F2/S=F2/(d2/4)=15.2*9.8/(*0.762/4)=329(MPa)=/E=329MPa/110.3Gpa=3L=L0（1+）=212*（1+0.011-0.003）=213.7cm（3）=F1/S=85.3N/(*0.762/4)=187.9Mpa,14,4-9从拉伸试验如何获得常用的力学性能数据？拉伸强度、屈服强度、断裂强度、断裂伸长率、弹性模量等，公式可计算。,15,4-13有哪些方法可以改善材料的韧性，试举例说明。晶粒细化（晶格类型）；成分，高分子共混橡胶，金属种杂质；热处理；高分子中的银纹。,16,4-19某钢板的屈服强度为690MPa，KIC值为70MPam1/2，如果可容许最大裂缝是2.5mm，且不许发生塑性变形，则此钢的设计极性强度是多少？KIC=c(a)1/2,c=KIC(a)-1/2=70Mpa.m1/2(3.14*1.25mm)-1/2=1117MPa,17,9.6Brieflyexplain(a)whytheremaybesignificantscatterinthefracturestrengthforsomegivenceramicmaterial,and(b)whyfracturestrengthincreaseswithdecreasingspecimensize.陶瓷材料的结构，晶相、玻璃相和气相;裂纹等缺陷。,18,9.28BrieflyexplainwhyBCCandHCPmetalalloysmayexperienceaductile-to-brittletransitionwithdecreasingtemperature,whereasFCCalloysdonotexperiencesuchatransition.FCC韧性，甚至在低温。P254跟裂纹有关。Cracksinductilematerialsaresaidtobestable；Forbrittlefracture,cracksareunstable,andthefracturesurfaceisrelativelyflatandperpendiculartothedirectionoftheappliedtensileload.,19,4-24按照粘附摩擦的机理，说明为什么极性高聚物与金属材料表面间的摩擦系数较大，而非极性高聚物则较小。摩擦系数：=S/PmS为剪切强度；Pm为抗压强度，极性聚合物作用力大，S大。,第十一次中文,20,4-29试从金属、陶瓷和高聚物材料的结构差别解释它们在热容、热膨胀系数和热导率等性能方面的差别。原子键合方式不同；结构不同，包括结晶。,21,热容高分子无机非金属金属热运动基团单元大小不同热膨胀系数金属无机非金属高分子导热的微观机制不同,22,英文书17.6(a)BrieflyexplainwhyCvriseswithincreasingtemperatureattemperaturesnear0K.(b)BrieflyexplainwhyCvbecomesvirtuallyindependentoftemperatureattemperaturesfarremovedfrom0K.(a)TheCviszeroat0K,butitrisesrapidlywithtemperature;thiscorrespondstoanincreasedabilityofthelatticewavestoenhancetheiraverageenergywithascendingtemperature.(b)AbovewhatiscalledtheDebyetemperatureD,Cvlevelsoffandbecomesessentiallyindependentoftemperatureatavalueofapproximately3R,Rbeingthegasconstant.Thuseventhoughthetotalenergyofthematerialisincreasingwithtemperature,thequantityofenergyrequiredtoproduceaone-degreetemperaturechangeisconstant.,23,Solution:l/l0=lT,l=(l/l0)/T=0.0002/0.1/80=2.510-5(1/C),17.10A0.1m(3.9in.)rodofametalelongates0.2mm(0.0079in.)onheatingfrom20to100(68to212).Determinethevalueofthelinearcoefficientofthermalexpansionforthismaterial.,24,17.16(a)Calculatetheheatfluxthroughasheetofsteel10mmthickifthetemperaturesatthetwofacesare300and100C;assumesteady-stateheatflow.(b)Whatistheheatlossperhouriftheareaofthesheetis0.25m2?(c)Whatwillbetheheatlossperhourifsodalimeglassinsteadofsteelisused?(d)Calculatetheheatlossperhourifsteelisusedandthethicknessisincreasedto20mm.(a):heatflux(任一点):q=(T1-T2)/d=51.9(W/m.K)200K/0.001m=1.038106W/m2(b):Q=qA3600s/1h=9.34108J/h(c):Q=qgA3600s/1h=3.06107J/h=1.7(W/m.K)(c):Q=4.67108J/h,25,17.23Foreachofthefollowingpairsofmaterials,decidewhichhasthelargerthermalconductivity.Justifyyourchoices.(a)Puresilver;sterlingsilver(92.5wt%Ag7.5wt%Cu).(b)Fusedsilica;polycrystallinesilica.(c)Linearpolyethylene(Mn=450,000g/mol);lightlybranchedpolyethylene(Mn=650,000g/mol).(d)Atacticpolypropylene(Mw=106g/mol);isotacticpolypropylene(Mw=5105g/mol).（a）纯银合金的杂质会妨碍自由电子的运动，减少传导作用（b）熔融SiO2熔融SiO2有对流现象（c）LLPE结晶度更大（d）isotacticPPisotacticPP结晶,26,思考题：,4-27为什么非晶态高聚物在玻璃化转变前后热膨胀系数不同？玻璃态和高弹态；高弹态分子是解缠的。,27,4-30何为半分解温度？它与高分子化学键之间有什么关系？半分解温度为高聚物在真空中加热30min后质量损失一半所需要的温度K=Ae-E/RT，键能越大，半分解温度越高杂环或环状结构、元素高分子,28,4