材料科学与工程基础作业讲评-6课件.ppt

第九次作业中文;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：σ=F/S=F/(πd2/4)=1.56GPa ε=ΔL/L=0.456/50=0.912% 正弹性模量：E=σ/ε=1.56Gpa/0.912%=172.9Gpa 泊松比：ν=-eY/eX =[-(12.79-12.83)/12.83]/0.912%=0.342 (a) 体积模量：K= E/[3(1-2ν)]=172.9/[3(1-2*0.342)] =182Gpa (b) 剪切模量： G=E/(2(1+ν))=172.9/[2*(1+0.342)]=64Gpa ;英文书;(b) If, in addition, the maximum permissible diameter decrease is 2.3×10-3 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 0.9mm, 　E　必须　大于131MPa, 因此Steel alloy合适。;7.47 A steel specimen having a rectangular cross section of dimensions 19 mm×3.2 mm (0.75in×0.125in.) has the stress–strain 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