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metals of Tungsten, steel, nickel, titanium and copper are suitable candidates.
(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?
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:
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?
=182Gpa (b) 剪切模量：
G=E/(2(1+ν))=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.
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
(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?
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)]
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.
(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 <0.072mm, 则E >104MPa, 因此 in Table7.1, the
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 chaHale Waihona Puke Baiduge in specimen volume during the deformation
(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?
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:
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?
=182Gpa (b) 剪切模量：
G=E/(2(1+ν))=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.
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
(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?
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)]
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.
(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 <0.072mm, 则E >104MPa, 因此 in Table7.1, the
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 chaHale Waihona Puke Baiduge in specimen volume during the deformation