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4-124-12热处理（退火）的实质是什么？它对热处理（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？击强度和断裂伸长率有什么影响？结构弛豫，非晶至结晶转变结构弛豫，非晶至结晶转变第十次作业第十次作业中文中文4-204-20玻璃的理论强度超过玻璃的理论强度超过7000MPa7000MPa。一块平板玻。一块平板玻璃在璃在60MPa60MPa弯曲张力下破坏。我们假定裂纹尖端弯曲张力下破坏。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为氧离子尺寸（即裂纹尖端曲率半径为氧离子半径，径，RROO2-2-=0.14nm=0.14nm），问对应这种低应力断裂，相），问对应这种低应力断裂，相应的裂纹深度为多大？应的裂纹深度为多大？σσmaxmax=σ=σ00[1+2[1+2（（a/ρa/ρ））0.50.5]]=2σ=2σff（（a/ρa/ρ））0.50.57000=60[1+27000=60[1+2（（a/0.14a/0.14））0.50.5]]a=468nma=468nm，，2a=936.5nm2a=936.5nm括号中括号中11省略则省略则a=476nma=476nm4-214-21某钢材的屈服强度为某钢材的屈服强度为11001100MPa,MPa,抗拉强度为抗拉强度为1200MP1200MPaa，断裂韧性（，断裂韧性（KKICIC）为）为90MPa·m90MPa·m1/21/2。（。（aa）在一钢板）在一钢板上有上有2mm2mm的的边裂边裂，在他产生屈服之前是否会先断裂？，在他产生屈服之前是否会先断裂？（（bb）在屈服发生之前，不产生断裂的可容许断裂缝的）在屈服发生之前，不产生断裂的可容许断裂缝的最大最大深度深度是多少？（假设几何因子是多少？（假设几何因子YY等于等于1.11.1，试样的，试样的拉应力与边裂纹垂直拉应力与边裂纹垂直））σ=Kσ=KICIC/Y/Y（（πaπa））0.50.5=1032Mpa=1032Mpa，先断裂，先断裂a=a=（（KKICIC/Yσ/Yσ））22/π=1.5mm/π=1.5mm11001.7611001.7612001.512001.57.297.29Acylindricalspecimenofaluminumhavingadiameterof12.8Acylindricalspecimenofaluminumhavingadiameterof12.8mmandagaugelengthof50.800mmispulledintension.Usethemmandagaugelengthof50.800mmispulledintension.Usetheload–elongationcharacteristicstabulatedbelowtocompleteproblemsload–elongationcharacteristicstabulatedbelowtocompleteproblemsathroughf.athroughf.(a)(a)PlotthedataasengineeringstressversusPlotthedataasengineeringstressversusengineeringstrain.engineeringstrain.(b)(b)Computethemodulusofelasticity.Computethemodulusofelasticity.(c)(c)DeterminetheyieldstrengthatastrainDeterminetheyieldstrengthatastrainoffsetof0.002.offsetof0.002.(d)(d)DeterminethetensilestrengthofthisDeterminethetensilestrengthofthisalloy.alloy.(e)(e)Whatistheapproximateductility,inpercentelongation?Whatistheapproximateductility,inpercentelongation?(f)(f)Computethemodulusofresilience.Computethemodulusofresilience.05010015020025030035040000.050.10.150.2StrainStress(MPa)(a)(a)AA00==dd0022/4=3.14*12.8/4=3.14*12.822mmmm22/4/4=128.6mm=128.6mm22=F/A=F/A00==l/ll/l00,l,l00=50.8mm=50.8mm(b)(b)EE=slope==slope=//=(=(22--11)/()/(22--11))=(57.0-0)MPa/(0.001-=(57.0-0)MPa/(0.001-0)=57.0GPa0)=57.0GPa©©117.4MPa117.4MPa(d)369.4MPa(d)369.4MPa(e)(e)%EL=[(%EL=[(llff--ll00)/)/ll00]*100=[(59.182]*100=[(59.182–50.8)/50.8–50.8)/50.8]*100]*100=16.5=16.5(f)(f)UUrr=1/2*=1/2*yy**yy=0.5*117.4*10=0.5*117.4*1066*0.002J/m*0.002J/m33=117400=117400J/mJ/m337.40Forsomemetalalloy,atruestressof415MPaproducesaplastictruestrainof0.475.Howmuchwillaspecimenofthismaterialelongatewhenatruestressof325MPaisappliediftheoriginallengthis300mm?Assumeavalueof0.25forthestrain-hardeningexponentn.T=KTn,K=T/Tn=415MPa/(0.4750.25)=500MPaT=(T/K)1/n=(325MPa/500MPa)...