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
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1、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
2、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
3、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
4、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
5、(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
6、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 throug
7、h 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 s
8、trainn 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 res
9、ilience.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
10、.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 prod
11、uces 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)1/0.2
12、5=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 toughn
13、ess 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
14、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
15、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 tha
16、t 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 de
17、termined 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
18、,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*0.0025/2
19、m)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
20、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
21、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.
22、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0.76mm0.7
23、6mm。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
24、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
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