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1、第九次作业中文4-14-1：铝的弹性模量为铝的弹性模量为铝的弹性模量为铝的弹性模量为70GPa,70GPa,泊松比为泊松比为泊松比为泊松比为0.340.34，在，在，在，在83MPa83MPa的静水压时，此单位晶胞的体积是多少？的静水压时，此单位晶胞的体积是多少？的静水压时，此单位晶胞的体积是多少？的静水压时，此单位晶胞的体积是多少？由由由由E=3K(1-2)E=3K(1-2)得得得得 K=E/3(1-2)K=E/3(1-2)=70Gpa/3(1-2*0.34)=72.9Gpa=70Gpa/3(1-2*0.34)=72.9Gpa V/V=/K=83Mpa/72.9GPa=1.14V/V=/K=
2、83Mpa/72.9GPa=1.14 V=4.0496V=4.04963 3*10*10-30-30*(1-1.14)=*(1-1.14)=66.366.3 1010-30-30(m(m3 3)4-34-3直径为直径为直径为直径为12.83mm12.83mm的试棒，标距长度为的试棒，标距长度为的试棒，标距长度为的试棒，标距长度为50mm50mm，轴向受轴向受轴向受轴向受200kN200kN的作用力后拉长的作用力后拉长的作用力后拉长的作用力后拉长0.456mm0.456mm，且直径变成且直径变成且直径变成且直径变成12.79mm12.79mm，(a)(a)此试棒的体积模量是多少？此试棒的体积模量
3、是多少？此试棒的体积模量是多少？此试棒的体积模量是多少？(b)(b)剪切模量是多少？剪切模量是多少？剪切模量是多少？剪切模量是多少？解：解：解：解：=F/S=F/(d=F/S=F/(d2 2/4)=1.56GPa/4)=1.56GPa =L/L=0.456/50=0.912%=L/L=0.456/50=0.912%正弹性模量：正弹性模量：正弹性模量：正弹性模量：E=/=1.56Gpa/0.912%=172.9Gpa E=/=1.56Gpa/0.912%=172.9Gpa 泊松比：泊松比：泊松比：泊松比：=-=-e eY Y/e eX X =-(12.79-12.83)/=-(12.79-12.
4、83)/12.8312.83/0.912%=0.912%=0.3420.342(a)(a)体积模量：体积模量：体积模量：体积模量：K=E/3(1-2)=172.9/3(1-2*0.342)K=E/3(1-2)=172.9/3(1-2*0.342)=182Gpa =182Gpa (b)(b)剪切模量：剪切模量：剪切模量：剪切模量：G=E/(2(1+)=172.9/2*(1+0.342)=64Gpa G=E/(2(1+)=172.9/2*(1+0.342)=64Gpa 英文书7.20 A cylindrical metal specimen 15.0mm in diameter and 7.20
5、A cylindrical metal specimen 15.0mm in diameter and 150mm long is to be subjected to a tensile stress of 50 150mm long is to be subjected to a tensile stress of 50 MpaMpa;at this stress level the resulting deformation will;at this stress level the resulting deformation will be totally elastic.be tot
6、ally elastic.(a)(a)If If the the elongation elongation must must be be less less than than 0.072mm,0.072mm,which which of the metals in Table7.1 are suitable candidates?Why?of the metals in Table7.1 are suitable candidates?Why?=l/ll/l0 0=0.072mm/150mm=0.00048=0.072mm/150mm=0.00048 =E E ，E E=/=50MPa/
7、0.00048=104GPa=50MPa/0.00048=104GPa要要要要使使使使 l l 0.072mm,104MPa,104MPa,因因因因此此此此 in in Table7.1,Table7.1,the the metals metals of of Tungsten,Tungsten,steel,steel,nickel,nickel,titanium titanium and and copper copper are suitable candidates.are suitable candidates.(b)If,If,in in addition,addition,the
8、the maximum maximum permissible permissible diameter diameter decrease decrease is is 2.32.31010-3-3 mm,which mm,which of of the the metals in Table 7.1 may be used?Why?metals in Table 7.1 may be used?Why?y y=d/dd/d0 0=0.0023mm/15mm=0.000153=0.0023mm/15mm=0.000153v v=-=-y y/x x=0.000153/0.00048=0.31
9、9=0.000153/0.00048=0.319要要要要使使使使 d d 0.0023mm,0.0023mm,则则则则v v 0.319,0.319,因因因因此此此此 in in Table7.1,Table7.1,the the metals metals of of Tungsten,Tungsten,steel steel and and nickel nickel may may be be used.used.7.24 A cylindrical rod 380 mm long,having a diameter of 10.0 mm,is to be subjected to a
10、tensile load.If the rod is to experience neither plastic deformation nor an elongation of more than 0.9 mm when the applied load is 24,500 N,which of the four metals or alloys listed below are possible candidates?=F/A0=F/(d02/4)=24500N/(3.14*102 mm2/4)=312 MPa,因此从屈服强度来看，只有因此从屈服强度来看，只有Steel alloy and
11、 Brass alloy才有可能。才有可能。另外：另外：=l/l0=0.9mm/380mm=0.00237 =E ，E=/=312MPa/0.00237=131MPa,因此，因此，l 大于大于131MPa,因因此此Steel alloy合适。合适。7.47 A steel specimen having a rectangular cross section of dimensions 19 mm3.2 mm(0.75in0.125in.)has the stressstrain behavior shown in Figure 7.33.If this specimen is subject
12、ed to a tensile force of 33,400 N(7,500lbf),then(a)Determine the elastic and plastic strain values.(b)If its original length is 460 mm(18 in.),what will be its final length after the load in part a is applied and then released?(a)Determine the elastic and plastic strain values.弹性变形应变数值大约：弹性变形应变数值大约：
13、0-0.0015，塑性变形：塑性变形：0.0015(b)If its original length is 460 mm(18 in.),what(b)If its original length is 460 mm(18 in.),what will be its final length after the load in part a is will be its final length after the load in part a is applied and then released?applied and then released?E E=slope=slope=/=(=
14、(2 2-1 1)/()/(2 2-1 1)=(300-)=(300-0)MPa/(0.0013-0)=231GPa0)MPa/(0.0013-0)=231GPa =F/A=F/A0 0=F/(=F/(a a*b b)=33400N)=33400N/(19*3.2mm/(19*3.2mm2 2)=549.3MPa)=549.3MPa图中可知图中可知图中可知图中可知,在该应力时的总应变为在该应力时的总应变为在该应力时的总应变为在该应力时的总应变为 总总总总=0.005,=0.005,最大弹性为最大弹性为最大弹性为最大弹性为:弹弹弹弹=0.0015=0.0015去除应力后弹性应变回复去除应力后弹性
15、应变回复去除应力后弹性应变回复去除应力后弹性应变回复,故长度为故长度为故长度为故长度为:l l0 0 *(*(1+1+总总总总 -弹弹弹弹 )=)=460 460*(1+0.005*(1+0.005 0.0015 0.0015 )=461.61 mm =461.61 mm8.24(a)Show,for a tensile test,that8.24(a)Show,for a tensile test,thatif there is no change in specimen volume during the deformation if there is no change in speci
16、men volume during the deformation process(i.e.,process(i.e.,A A0 0 l l0 0 =A Ad d l ld d).).CW%=(CW%=(A A0 0-A Ad d )/)/A A0 0 100=(1-100=(1-A Ad d/A A0 0)*100)*100A A0 0 l l0 0=A=Ad d l ld d ,A ,Ad d/A/A0 0=l=l0 0/l/ld d=l=l0 0/(l/(l0 0 +l)=l)=1/1/(l(l0 0 +l)/l)/l l0 0=1/1+=1/1+所以所以所以所以CW%=(CW%=(A
17、A0 0-A Ad d )/)/A A0 0 100=(1-100=(1-A Ad d/A A0 0)100=1-1/(1+100=1-1/(1+)100100=/(1+/(1+)100,100,即上式。即上式。即上式。即上式。(b)Using the result of part a,compute the percent cold work(b)Using the result of part a,compute the percent cold work experienced by naval brass(the stressstrain behavior of which exper
18、ienced by naval brass(the stressstrain behavior of which is shown in Figure 7.12)when a stress of 400 MPa is applied.is shown in Figure 7.12)when a stress of 400 MPa is applied.=0.12=0.12CW%=CW%=/(1+/(1+)100=0.12/(1+0.12)100=0.12/(1+0.12)100%=10.7%100%=10.7%4-6.已知温度为已知温度为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
19、，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。并说明原因。a.拉伸强度拉伸强度 伸长率伸长率 冲击强度（悬臂梁）冲击强度（悬臂梁）弹性模量弹性模量 MPa%Nm MPa103(1)62.1 110 19.04 2.415(2)51.8 0 0.41 6.90 (3)27.6 72 4.08 0.828 (4)69.0 0 1.09 6.90 (5)17.3 200 5.44 0.414 名称：环氧树脂、聚四氟乙烯、聚乙烯、酚醛树脂、聚碳酸酯名称：环氧树脂、聚四氟乙烯、聚乙烯、酚醛树脂、聚碳酸酯思考题思考题（1）PC;（2)酚醛酚醛;（3）HDPE;（4）环氧树脂）环氧树脂
20、;（5）PTFE4-14.有哪些途径可以提高材料的刚性？有哪些途径可以提高材料的刚性？复复合合材材料料、提提高高材材料料刚刚性性、结结晶晶、交交联联、提提高高分分子子量量、热处理热处理 7.22 Cite the primary differences between elastic,anelastic,and plastic deformation behaviors.分分别别从从概概念念、原原子子论论角角度度、施施加加应应力力后后的的应应变变、材材料料的的差差别别（或或对对应应的的材材料料）等等几几个方面阐述。个方面阐述。8.18 Describe in your own words th
21、e three strengthening 8.18 Describe in your own words the three strengthening mechanisms discussed in this chapter(i.e.,grain size mechanisms discussed in this chapter(i.e.,grain size reduction,solid solution strengthening,and strain reduction,solid solution strengthening,and strain hardening).Be su
22、re to explain how dislocations are hardening).Be sure to explain how dislocations are involved in each of the strengthening techniques.involved in each of the strengthening techniques.Grain size reduction:Grain size reduction:晶粒尺寸减少，位错时滑移晶粒尺寸减少，位错时滑移晶粒尺寸减少，位错时滑移晶粒尺寸减少，位错时滑移减少方向的改变；原子位置不连续减少。减少方向的改变；
23、原子位置不连续减少。减少方向的改变；原子位置不连续减少。减少方向的改变；原子位置不连续减少。Solid solution strengtheningSolid solution strengthening：加入不同种的原子加入不同种的原子加入不同种的原子加入不同种的原子相成形成固溶体或合金，这些加入的原子限制相成形成固溶体或合金，这些加入的原子限制相成形成固溶体或合金，这些加入的原子限制相成形成固溶体或合金，这些加入的原子限制位错移动。位错移动。位错移动。位错移动。Strain hardeningStrain hardening：先施加应力产生位错。而位错先施加应力产生位错。而位错先施加应力产生位错。而位错先施加应力产生位错。而位错之间的作用是排斥的，结果是存在的一个位错之间的作用是排斥的，结果是存在的一个位错之间的作用是排斥的，结果是存在的一个位错之间的作用是排斥的，结果是存在的一个位错限制另一位错的移动。限制另一位错的移动。限制另一位错的移动。限制另一位错的移动。