Answer:
14778.29 N
Explanation:
Diameter, d=10.1mm= [tex]10.1*10^{-3}m[/tex]
Since stress, [tex]\sigma= \frac {F}{A}[/tex] where F is force, A is area and since specimen is cylindrical,
A= [tex]\pi *(d/2)^{2}[/tex] Therefore, [tex]\sigma= \frac {F}{\pi*(0.5d)^{2}}[/tex]
Also strain [tex]\epsilon= \frac { \triangle L}{L}[/tex] where L is length
Poison’s ratio,v is the ratio of lateral strain to longitudinal strain hence
V= [tex]-\frac {\epsilon_{x}}{\epsilon_{z}}= \frac {\triangle dl_{o}}{d \triangle l}[/tex]
From Hooke’s law, [tex]\sigma=E \epsilon_{z}[/tex]
Conclusively, [tex]E* \epsilon_{z}=E* \frac {- \epsilon_{x}}{v}=\frac {F}{\pi*(0.5d)^{2}}[/tex]
[tex]\frac {4F}{ \pi *d^{2}}=- \frac {E \triangle d}{vd}[/tex]
F=[tex]- \frac {\pi *Ed \triangle d}{4v}[/tex]
F= [tex]- \frac {\pi *207*10^{9} *10.1*10^{-3}* (-2.7*10^{-6})}{4*0.3}[/tex]= 14778.29N
Therefore, required force F is 14778.29 N
