Question

A cylindrical bar of steel 10.1 mm (0.3976 in.) in diameter is to be deformed elastically by application of a force along the bar axis. Determine the force that will produce an elastic reduction of 2.7 ×10-3 mm (1.063 ×10-4 in.) in the diameter. For steel, values for the elastic modulus (E) and Poisson's ratio (ν) are, respectively, 207 GPa and 0.30.

Answer 1

Answer:

14778.29 N

Explanation:

Diameter, d=10.1mm= $10.1*10^{-3}m$

Since stress, $\sigma= \frac {F}{A}$ where F is force, A is area and since specimen is cylindrical,  

A= $\pi *(d/2)^{2}$  Therefore, $\sigma= \frac {F}{\pi*(0.5d)^{2}}$

Also strain $\epsilon= \frac { \triangle L}{L}$ where L is length

Poison’s ratio,v is the ratio of lateral strain to longitudinal strain hence

V= $-\frac {\epsilon_{x}}{\epsilon_{z}}= \frac {\triangle dl_{o}}{d \triangle l}$

From Hooke’s law, $\sigma=E \epsilon_{z}$

Conclusively, $E* \epsilon_{z}=E* \frac {- \epsilon_{x}}{v}=\frac {F}{\pi*(0.5d)^{2}}$

$\frac {4F}{ \pi *d^{2}}=- \frac {E \triangle d}{vd}$

F=$- \frac {\pi *Ed \triangle d}{4v}$

F= $- \frac {\pi *207*10^{9} *10.1*10^{-3}* (-2.7*10^{-6})}{4*0.3}$= 14778.29N

Therefore, required force F is 14778.29 N