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My name is Ann [436]
3 years ago
7

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 ba

r 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.
Physics
1 answer:
amm18123 years ago
3 0

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

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