《作业讲评》PPT课件.pptVIP

第九次作业 中文,4-1： 铝的弹性模量为70GPa,泊松比为0.34，在83MPa的静水压时，此单位晶胞的体积是多少？ 由E=3K(1-2) 得 K=E/3(1-2) =70Gpa/3(1-2*0.34)=72.9Gpa V/V=/K=83Mpa/72.9GPa=1.14 V=4.04963*10-30*(1-1.14)=66.310-30 (m3),4-3 直径为12.83mm的试棒，标距长度为50mm，轴向受200kN的作用力后拉长0.456mm，且直径变成12.79mm， (a) 此试棒的体积模量是多少？ (b) 剪切模量是多少？ 解：=F/S=F/(d2/4)=1.56GPa =L/L=0.456/50=0.912% 正弹性模量：E=/=1.56Gpa/0.912%=172.9Gpa 泊松比：=-eY/eX =-(12.79-12.83)/12.83/0.912%=0.342 (a) 体积模量：K= E/3(1-2)=172.9/3(1-2*0.342) =182Gpa (b) 剪切模量： 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 150mm long is to be subjected to a tensile stress of 50 Mpa; at this stress level the resulting deformation will be totally elastic. (a) If the elongation must be less than 0.072mm, which of the metals in Table7.1 are suitable candidates? Why ? = l / l0=0.072mm/150mm=0.00048 =E ，E=/ =50MPa/0.00048=104GPa 要使l 104MPa, 因此 in Table7.1, the metals of Tungsten, steel, nickel, titanium and copper are suitable candidates.,(b) If, in addition, the maximum permissible diameter decrease is 2.310-3 mm,which of the metals in Table 7.1 may be used ? Why? y= d /d0=0.0023mm/15mm=0.000153 v =- y/ x=0.000153/ 0.00048=0.319 要使d 0.0023mm, 则v 0.319, 因此 in Table7.1, the metals of Tungsten, steel and nickel may be used.,7.24 A cylindrical rod 380 mm long, having a diameter of 10.0 mm, is to be subjected to a 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Steel alloy and Brass alloy才有可能。 另外： = l /l0=0.9mm/380mm=0.00237 =E ，E=/ =312MPa/0.00237=131MPa, 因此，l 大于131MPa, 因此Steel alloy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 subjected 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：0-0.0015， 塑性变形：0.0015,(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? E=slope=/=(2-1)/(2-1)=(300-0)MPa/(0.0013-0)=231GPa =F/A0=F/(a*b) =33400N /(19*3.2mm2)=549.3MPa 图中可知,在该应力时的总应变为总= 0.005, 最大弹性为: 弹= 0.0015 去除应力后弹性应变回复,故长度为: l0 *(1+ 总 - 弹 )= 460 *(1+ 0.005 0.0015 ) =461.61 mm,8.24 (a) Show, for a tensile test, that if there is no change in specimen volume during the deformation process (i.e., A0 l0 =Ad ld). CW%=( A0 -Ad )/A0 100=(1-Ad /A0)*100 A0 l0 =Ad ld , Ad /A0= l0 / ld = l0 /( l0 + l)=1/ ( l0 + l)/ l0=1/1+ 所以CW%=( A0 -Ad )/A0 100=(1-Ad /A0)100=1-1/(1+)100 =/(1+)100,即上式。 (b) Using the result of part a, compute the percent cold work experienced by naval brass (the stressstrain behavior of which is shown in Figure 7.12) when a stress of 400 MPa is applied. =0.12 CW%=/(1+)100=0.12/(1+0.12)100%=10.7%,4-6.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，并说明原因。 a. 拉伸强度 伸长率 冲击强度（悬臂梁） 弹性模量 MPa % Nm MPa103 (1) 62.1 110 19.04 2.415 51.8 0 0.41 6.90 27.6 72 4.08 0.828 69.0 0 1.09 6.90 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,思考题,（1） PC;（2) 酚醛;（3）HDPE;（4）环氧树脂;（5） PTFE,4-14.有哪些途径可以提高材料的刚性？ 复合材料、提高材料刚性、结晶、交联、提高分子量、热处理,7.22 Cite the primary differences between elastic, anelastic, and plastic deformation behaviors. 分别从概念、原子论角度、施加应力后的应变、材料的差别（或对应的材料）等几个方面阐述。,8.18 Describe in your own words the three strengthening mechanisms discussed in this chapter (i.e., grain size reduction, solid solution strengthening, and strain hardening). Be sure to explain how dislocations are involved in each of the st