Question

A copper wire $4m$ long has diameter of $1mm$, if a load of $10kg$ wt is attached at other end. What extension is produced, if Poisson's ratio is $0.26$? How much lateral compression is produced in it? $(Y_{cu}=12.5\times {10}^{10}N/m^2)$.

Answer 1

Given : $L=4m$ $D=1mm={10}^{-3}m$

Area $A=\frac{\pi D^2}{4}=\frac{\pi \times ({10}^{-3}{)}^2}{4}=0.785\times {10}^{-6}{m}^2$

Weight $F=mg=10\times 10=100N$

Using $Y_{cu}=\frac{FL}{A\delta L}$

Or $12.5\times {10}^{10}=\frac{100\times 4}{0.785\times {10}^{-6}\times \delta L}$

$⟹\delta L=0.4\times {10}^{-2}m$

Poisson's ratio $\sigma =\frac{\delta D/D}{\delta L/L}$

Or $0.26=\frac{\delta D/{10}^{-3}}{0.4\times {10}^{-2}/4}$

$⟹\delta D=-0.26\times {10}^{-6}m$