材料科学和工程基础作业讲评PPT课件市公开课一等奖百校联赛获奖课件.pptx

n n4-12热处理（退火）实质是什么？它对材料拉伸强度、硬度、尺寸稳定性、冲击强度和断裂伸长率有什么影响？n n 结构弛豫结构弛豫 ，非晶至结晶转变，非晶至结晶转变第十次作业汉字第1页n n4-204-20玻玻玻玻璃璃璃璃理理理理论论论论强强强强度度度度超超超超出出出出7000MPa7000MPa。一一一一块块块块平平平平板板板板玻玻玻玻璃璃璃璃在在在在60MPa60MPa弯弯弯弯曲曲曲曲张张张张力力力力下下下下破破破破坏坏坏坏。我我我我们们们们假假假假定定定定裂裂裂裂纹纹纹纹尖尖尖尖端端端端为为为为氧氧氧氧离离离离子子子子尺尺尺尺寸寸寸寸（即即即即裂裂裂裂纹纹纹纹尖尖尖尖端端端端曲曲曲曲率率率率半半半半径径径径为为为为氧氧氧氧离离离离子子子子半半半半径径径径，R ROO2-2-=0.14nm=0.14nm），问问问问对对对对应应应应这这这这种种种种低低低低应应应应力力力力断断断断裂裂裂裂，对对对对应应应应裂纹深度为多大？裂纹深度为多大？裂纹深度为多大？裂纹深度为多大？n n maxmax=0 01+21+2（a/a/）0.50.5 n n =2=2=2=2f f f f（a/a/a/a/）0.50.50.50.5n n7000=601+27000=601+27000=601+27000=601+2（a/0.14a/0.14a/0.14a/0.14）0.50.50.50.5 n na=468nma=468nma=468nma=468nm，2a=936.5nm2a=936.5nm2a=936.5nm2a=936.5nmn n括号中括号中括号中括号中1 1 1 1省略则省略则省略则省略则a=476nma=476nma=476nma=476nm第2页n n4-214-21某某某某钢钢钢钢材材材材屈屈屈屈服服服服强强强强度度度度为为为为11001100MPa,MPa,抗抗抗抗拉拉拉拉强强强强度度度度为为为为1200MPa1200MPa，断断断断裂裂裂裂韧韧韧韧性性性性（K KICIC）为为为为90MPam90MPam1/21/2。（a a）在在在在一一一一钢钢钢钢板板板板上上上上有有有有2mm2mm边边边边裂裂裂裂，在在在在他他他他产产产产生生生生屈屈屈屈服服服服之之之之前前前前是是是是否否否否会会会会先先先先断断断断裂裂裂裂？（b b）在在在在屈屈屈屈服服服服发发发发生生生生之之之之前前前前，不不不不产产产产生生生生断断断断裂裂裂裂可可可可允允允允许许许许断断断断裂裂裂裂缝缝缝缝最最最最大大大大深深深深度度度度是是是是多多多多少少少少？（假假假假设设设设几几几几何何何何因因因因子子子子Y Y等等等等于于于于1.11.1，试试试试样样样样拉拉拉拉应应应应力力力力与与与与边边边边裂裂裂裂纹纹纹纹垂垂垂垂直直直直）n n=K=K=K=KICICICIC/Y/Y/Y/Y（aaaa）0.5 0.5 0.5 0.5 =1032Mpa=1032Mpa=1032Mpa=1032Mpa，先断裂，先断裂，先断裂，先断裂n na=a=a=a=（K K K KICICICIC/Y/Y/Y/Y）2 2 2 2/=1.5mm/=1.5mm/=1.5mm/=1.5mm 1100 1.76 1100 1.76 1200 1.5 1200 1.5第3页n n7.297.29 A A cylindrical cylindrical specimen specimen of of aluminum aluminum having having a a diameter diameter of of 12.8 12.8 mm mm and and a a gauge gauge length length of of 50.800 50.800 mm mm is is pulled pulled in in tension.tension.Use Use the the loadelongation loadelongation characteristics characteristics tabulated tabulated below to complete problems a through f.below to complete problems a through f.n n(a)(a)Plot the data as engineering stress versusPlot the data as engineering stress versusn n engineering strain.engineering strain.n n(b)(b)Compute the modulus of elasticity.Compute the modulus of elasticity.n n(c)(c)Determine the yield strength at a strainDetermine the yield strength at a strainn n offset of 0.002.offset of 0.002.n n(d)(d)Determine the tensile strength of this Determine the tensile strength of this n n alloy.alloy.n n(e)(e)What What is is the the approximate approximate ductility,ductility,in in percent percent elongation?elongation?n n(f)(f)Compute the modulus of resilience.Compute the modulus of resilience.第4页第5页第6页(a)(a)A A0 0=d d0 02 2/4=3.14*12.8/4=3.14*12.82 2 mm mm2 2/4/4 =128.6mm =128.6mm2 2 =F/A=F/A0 0 =l/ll/l0 0,l,l0 0 =50.8mm=50.8mm(b)(b)E E=slope=slope=/=(=(2 2-1 1)/()/(2 2-1 1)=(57.0-0)MPa/(0.001-0)=57.0GPa =(57.0-0)MPa/(0.001-0)=57.0GPa 117.4MPa117.4MPa(d)369.4MPa(d)369.4MPa(e)(e)%EL=(%EL=(l l f f-l l 0 0)/)/l l 0 0*100=(59.182*100=(59.182 50.8)/50.8 50.8)/50.8 *100 *100 =16.5 =16.5(f)(f)U U r r=1/2*=1/2*y y*y y=0.5*117.4*10=0.5*117.4*106 6*0.002J/m*0.002J/m3 3=117400 =117400 J/mJ/m3 3第7页7.40 For some metal alloy,a true stress of 415 MPa produces a plastic true strain of 0.475.How much will a specimen of this material elongate when a true stress of 325 MPa is applied if the original length is 300 mm?Assume a value of 0.25 for the strain-hardening exponent n.n n T=K Tn ,K=T/Tn =415MPa/(0.4750.25)=500MPan n T=(T/K)1/n=(325MPa/500MPa)1/0.25=0.179n n T=ln(l i/l 0),l i=l 0*e T T =300mm*e0.179=300mm*1.196=358.8mm第8页9.17 Some aircraft component is fabricated from an aluminum 9.17 Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa.malloy that has a plane strain fracture toughness of 35 MPa.m1/21/2.It has been determined that fracture results at a stress of 250 It has been determined that fracture results at a stress of 250 MPa when the maximum(or critical)MPa when the maximum(or critical)internal crackinternal crack length is length is 2.0 mm.For this same component and alloy,will fracture occur 2.0 mm.For this same component and alloy,will fracture occur at a stress level of 325 Mpa when the maximum internal crack at a stress level of 325 Mpa when the maximum internal crack length is 1.0 mm?Why or why not?length is 1.0 mm?Why or why not?KKICIC=Y=Y (a)a)1/21/2,Y=K Y=KICIC/(a)a)1/21/2=(35 Mpa m=(35 Mpa m 1/21/2)/(250 Mpa)/(250 Mpa)(3.14*0.002/2m)(3.14*0.002/2m)1/21/2=2.50=2.50另外受力：另外受力：另外受力：另外受力：=K=KICIC/(/(a)a)1/21/2 Y=(35 Mpa m Y=(35 Mpa m 1/2 1/2)/)/(3.14*0.001/2m)(3.14*0.001/2m)1/21/2*2.50=353.3MPa,*2.50=353.3MPa,所以不停裂。所以不停裂。所以不停裂。所以不停裂。第9页9.18 Suppose that a wing component on an aircraft is 9.18 Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 Mpa mfracture toughness of 40 Mpa m1/21/2.It has been .It has been determined that fracture results at a stress of 365 MPa determined that fracture results at a stress of 365 MPa when the maximum internal crack length is 2.5 mm.For when the maximum internal crack length is 2.5 mm.For this same component and alloy,compute the stress level this same component and alloy,compute the stress level at which fracture will occur for a critical internal crack at which fracture will occur for a critical internal crack length of 4.0 mm.length of 4.0 mm.KKICIC=Y=Y (a)a)1/21/2,Y=KY=KICIC/(a)a)1/21/2=(40 Mpa m=(40 Mpa m 1/21/2)/)/(365Mpa)(3.14*0.0025/2m)(365Mpa)(3.14*0.0025/2m)1/21/2=1.75=1.75另外受力：另外受力：=K=KICIC/(/(a)a)1/21/2 Y=(40 Mpa mY=(40 Mpa m 1/2 1/2)/)/(3.14*0.004/2m)(3.14*0.004/2m)1/21/2 1.75=288.4MPa,1.75=288.4MPa,第10页9.32 A 12.5mm diameter cylindrical rod fabricated 9.32 A 12.5mm diameter cylindrical rod fabricated from a-T6 alloy(Figure 9.46)is subjected to a from a-T6 alloy(Figure 9.46)is subjected to a repeated tension-compression load cycling repeated tension-compression load cycling along its axis.Compute the maximum and along its axis.Compute the maximum and minimum loads that will be applied to yield a minimum loads that will be applied to yield a fatigue life of 1.0fatigue life of 1.0 10107 7cycles.Assume that the cycles.Assume that the stress plotted on the vertical axis is stress stress plotted on the vertical axis is stress amplitude,and data were taken for a mean amplitude,and data were taken for a mean stress of 50 MPa.stress of 50 MPa.第11页第12页At a fatigue life of 1.0107 cycles,the stress amplitude a is about 175(160)MPa.a=(max-min)/2,英文书英文书P257 m=(max+min)/2,英文书英文书P258 So max=m+a,min=m-a ,F max/A=m+aF max=(m+a)A=27598N(25758)F min=(m-a)A=-15332N(-13492)第13页思索题思索题思索题思索题4-74-7 一条长一条长一条长一条长212cm212cm铜线，直径为铜线，直径为铜线，直径为铜线，直径为0.76mm0.76mm。当外加载荷为。当外加载荷为。当外加载荷为。当外加载荷为8.7kg8.7kg时开时开时开时开始产生塑性变形始产生塑性变形始产生塑性变形始产生塑性变形(a)(a)此作用力是多少牛顿？此作用力是多少牛顿？此作用力是多少牛顿？此作用力是多少牛顿？(b)(b)外加载荷为外加载荷为外加载荷为外加载荷为15.2kg15.2kg时，此线应变为时，此线应变为时，此线应变为时，此线应变为0.0110.011，则去除载荷后，铜线长度为多少，则去除载荷后，铜线长度为多少，则去除载荷后，铜线长度为多少，则去除载荷后，铜线长度为多少？此铜线屈服强度是多少？此铜线屈服强度是多少？此铜线屈服强度是多少？此铜线屈服强度是多少？解：解：解：解：（a a a a）F F F F1 1 1 1=mg=8.7*9.8=85.3N=mg=8.7*9.8=85.3N=mg=8.7*9.8=85.3N=mg=8.7*9.8=85.3N （b b b b）=F=F=F=F2 2 2 2/S=F/S=F/S=F/S=F2 2 2 2/(d/(d/(d/(d2 2 2 2/4)/4)/4)/4)=15.2*9.8 =15.2*9.8 =15.2*9.8 =15.2*9.8 /(*0.76/(*0.76/(*0.76/(*0.762 2 2 2/4)=329(MPa)/4)=329(MPa)/4)=329(MPa)/4)=329(MPa)=/E=329MPa/110.3Gpa=3 =/E=329MPa/110.3Gpa=3 =/E=329MPa/110.3Gpa=3 =/E=329MPa/110.3Gpa=3 L=L L=L L=L L=L0 0 0 0（1+1+1+1+）=212*=212*=212*=212*（1+0.011-0.0031+0.011-0.0031+0.011-0.0031+0.011-0.003）=213.7cm=213.7cm=213.7cm=213.7cm （3 3 3 3）=F=F=F=F1 1 1 1/S=85.3N/(*0.76/S=85.3N/(*0.76/S=85.3N/(*0.76/S=85.3N/(*0.762 2 2 2/4)/4)/4)/4)=187.9Mpa =187.9Mpa =187.9Mpa =187.9Mpa第14页4-9 从拉伸试验怎样取得惯用力学性能数据？从拉伸试验怎样取得惯用力学性能数据？拉伸强度、屈服强度、断裂强度、断裂伸长拉伸强度、屈服强度、断裂强度、断裂伸长率、弹性模量等，公式可计算。率、弹性模量等，公式可计算。第15页4-13 有有哪哪些些方方法法能能够够改改进进材材料料韧韧性性，试试举举例例说说明。明。晶晶粒粒细细化化（晶晶格格类类型型）；成成份份，高高分分子子共共混混橡橡胶，金属种杂质；热处理；高分子中银纹。胶，金属种杂质；热处理；高分子中银纹。第16页4-19某钢板屈服强度为某钢板屈服强度为690MPa，KIC值为值为70MPam1/2，假如可允许最大裂缝是，假如可允许最大裂缝是2.5mm，且不许发生塑性变形，则此钢设计极性强度是且不许发生塑性变形，则此钢设计极性强度是多少？多少？KIC=c(a)1/2,c=KIC(a)-1/2=70Mpa.m1/2(3.14*1.25mm)-1/2 =1117MPa第17页9.6 Briefly explain(a)why there may be significant scatter in the fracture strength for some given ceramic material,and(b)why fracture strength increases with decreasing specimen size.陶瓷材料结构，晶相、玻璃相和气相陶瓷材料结构，晶相、玻璃相和气相;裂纹裂纹等缺点。等缺点。第18页9.28 Briefly explain why BCC and HCP metal alloys may experience a ductile-to-brittle transition with decreasing temperature,whereas FCC alloys do not experience such a transition.FCC 韧性，甚至在低温。韧性，甚至在低温。P254跟裂纹相关。跟裂纹相关。Cracks in ductile materials are said to be stable；For brittle fracture,cracks are unstable,and the fracture surface is relatively flat and perpendicular to the direction of the applied tensile load.第19页4-24 按照粘附摩擦机理，说明为何极性高聚物与金属材料表面间摩擦系数较大，而按照粘附摩擦机理，说明为何极性高聚物与金属材料表面间摩擦系数较大，而非极性高聚物则较小。非极性高聚物则较小。摩擦系数：摩擦系数：=S/Pm S为剪切强度；Pm为抗压强度，极性聚合物作用力大，S大。第十一次第十一次汉字汉字第20页4-29试从金属、陶瓷和高聚物材料结构差异解释试从金属、陶瓷和高聚物材料结构差异解释它们在热容、热膨胀系数和热导率等性能方它们在热容、热膨胀系数和热导率等性能方面差异。面差异。原子键合方式不一样；结构不一样，包含结晶。第21页n n热容热容热容热容 高分子高分子高分子高分子 无机非金属无机非金属无机非金属无机非金属 金属金属金属金属n n 热运动基团单元大小不一样热运动基团单元大小不一样热运动基团单元大小不一样热运动基团单元大小不一样n n热膨胀系数热膨胀系数热膨胀系数热膨胀系数 金属金属金属金属无机非金属无机非金属无机非金属无机非金属 高分子高分子高分子高分子n n 导热微观机制不一样导热微观机制不一样导热微观机制不一样导热微观机制不一样第22页英文书英文书17.6 (a)Briefly explain why 17.6 (a)Briefly explain why Cv Cv rises with increasing rises with increasing temperature at temperatures near 0 K.temperature at temperatures near 0 K.(b)Briefly explain why(b)Briefly explain why Cv Cv becomes virtually independent of becomes virtually independent of temperature at temperatures far removed from 0 K.temperature at temperatures far removed from 0 K.(a)The(a)The Cv Cv is zero at 0 K,but it rises rapidly with temperature;is zero at 0 K,but it rises rapidly with temperature;this corresponds to an increased ability of the lattice waves to this corresponds to an increased ability of the lattice waves to enhance their average energy with ascending temperature.enhance their average energy with ascending temperature.(b)Above what is called the(b)Above what is called the Debye temperature Debye temperature D,D,Cv Cv levels off levels off and becomes essentially independent of temperature at a and becomes essentially independent of temperature at a value of approximately 3value of approximately 3R R,R R being the gas constant.Thus being the gas constant.Thus even though the total energy of the material is increasing even though the total energy of the material is increasing with temperature,the quantity of energy required to produce with temperature,the quantity of energy required to produce a one-degree temperature change is constant.a one-degree temperature change is constant.第23页Solution:l/l 0=l T,l=(l/l 0)/T=0.0002/0.1/80 =2.510-5(1/C),17.10 A 0.1 m(3.9 in.)rod of a metal elongates 0.2 mm(0.0079 in.)on heating from 20 to 100(68 to 212).Determine the value of the linear coefficient of thermal expansion for this material.第24页17.16(a)17.16(a)Calculate the heat flux through a sheet of steel 10mm Calculate the heat flux through a sheet of steel 10mm thick if the temperatures at the two faces are 300 and 100Cthick if the temperatures at the two faces are 300 and 100C;assume steady-state heat flow.assume steady-state heat flow.(b)(b)What is the heat loss per What is the heat loss per hour if the area of the sheet is 0.25 mhour if the area of the sheet is 0.25 m2 2?(c)(c)What will be the What will be the heat loss per hour if sodalime glass instead of steel is used?heat loss per hour if sodalime glass instead of steel is used?(d)(d)Calculate the heat loss per hour if steel is used and the Calculate the heat loss per hour if steel is used and the thickness is increased to 20 mm.thickness is increased to 20 mm.(a):heat flux(a):heat flux(任一点任一点任一点任一点):):q=q=(T(T1 1-T-T2 2)/)/d d =51.9(=51.9(W/m.K)W/m.K)200200K K/0.001/0.001mm =1.03810=1.038106 6W/mW/m2 2(b):Q=q(b):Q=q A A 3600s/1h=9.3410 3600s/1h=9.34108 8J/J/h h(c):Q=q(c):Q=qg g A A 3600s/1h=3.0610 3600s/1h=3.06107 7J/J/h h =1.7(1.7(W/m.K)W/m.K)(c):Q=4.67(c):Q=4.6710108 8J/J/h h第25页17.23 17.23 For each of the following pairs of materials,decide which has For each of the following pairs of materials,decide which has the larger thermal conductivity.Justify your choices.the larger thermal conductivity.Justify your choices.(a)Pure silver;sterling silver(92.5 wt%Ag 7.5 wt%Cu).(a)Pure silver;sterling silver(92.5 wt%Ag 7.5 wt%Cu).(b)Fused silica;polycrystalline silica.(b)Fused silica;polycrystalline silica.(c)Linear polyethylene(c)Linear polyethylene(Mn=Mn=450,000g/mol);lightly branched 450,000g/mol);lightly branched polyethylene(polyethylene(Mn=Mn=650,000g/mol).650,000g/mol).(d)Atactic polypropylene(d)Atactic polypropylene(Mw=Mw=10106 6g/mol);isotactic polypropylene g/mol);isotactic polypropylene(Mw=Mw=5 510105 5g/mol).g/mol).（a a a a）纯银纯银纯银纯银 合金杂质会妨碍自由电子运动，降低传导作用合金杂质会妨碍自由电子运动，降低传导作用合金杂质会妨碍自由电子运动，降低传导作用合金杂质会妨碍自由电子运动，降低传导作用 （b b b b）熔融熔融熔融熔融SiOSiOSiOSiO2 2 2 2 熔融熔融熔融熔融SiOSiOSiOSiO2 2 2 2有有有有对流现象对流现象对流现象对流现象（c c c c）LLPELLPELLPELLPE 结晶度更大结晶度更大结晶度更大结晶度更大（d d d d）isotactic PPisotactic PPisotactic PPisotactic PP isotactic PP isotactic PP isotactic PP isotactic PP 结晶结晶结晶结晶第26页思索题：思索题：n n4-27为何非晶态高聚物在玻璃化转变前后热为何非晶态高聚物在玻璃化转变前后热膨胀系数不一样？膨胀系数不一样？n n玻璃态和高弹态；玻璃态和高弹态；n n高弹态分子是解缠。高弹态分子是解缠。第27页n n4-30何何为为半半分分解解温温度度？它它与与高高分分子子化化学学键键之间有什么关系？之间有什么关系？n n半半分分解解温温度度为为高高聚聚物物在在真真空空中中加加热热30min后质量损失二分之一所需要温度后质量损失二分之一所需要温度n nK=Ae-E/RT，n n键能越大，半分解温度越高键能越大，半分解温度越高n n杂环或环状结构、元素高分子杂环或环状结构、元素高分子第28页n n4-354-354-354-35增加高分子材料阻燃性普通有哪些方法？增加高分子材料阻燃性普通有哪些方法？增加高分子材料阻燃性普通有哪些方法？增加高分子材料阻燃性普通有哪些方法？结构和组成结构和组成结构和组成结构和组成 提升热稳定性提升热稳定性提升热稳定性提升热稳定性 引入卤族、磷、氮等元素引入卤族、磷、氮等元素引入卤族、磷、氮等元素引入卤族、磷、氮等元素 阻燃剂和无机填料阻燃剂和无机填料阻燃剂和无机填料阻燃剂和无机填料 吸收热量吸收热量吸收热量吸收热量 降低温度降低温度降低温度降低温度 隔离氧隔离氧隔离氧隔离氧第29页n n17.1 Estimate the energy required to raise the temperature of 2 kg(4.42 lbm)of the following materials from 20 to 100C(68 to 212F):aluminum,steel,sodalime glass,and high density polyethylene.n n热容相关。n nE=cmt铝铝钢钢钠钙玻璃钠钙玻璃HDPEHDPEC(J.KgC(J.Kg-1-1.K.K-1 1)90090048648684084021002100E(KJ)E(KJ)14414477.877.8134134336336第30页 17.3(a)Determine the room temperature heatcapacities at constant pressure for the followingmaterials:aluminum,silver,tungsten,and 70Cu-30Zn brass.(b)How do these values compare with one another?How do you explain this?TABLE B.8铝铝银银钨钨黄铜黄铜C(J.KgC(J.Kg-1-1.K.K-1 1)900900235235142142375375第31页17.15 Explain why,on a cold day,the metal door handle of an automobile feels colder to the touch than a plastic steering wheel,even though both are at the same temperature.17.20 Briefly explain why metals are typically better thermal conductors than ceramic materials.第32页