i)Find The ratio of extension of copper wire to the steel wire.

ii) For the same extension find the diameter of the copper wire to the steel wire.

Take: $E_S =2X10^5 MPa, E_{CU} = 1.2X10^5 MPa.$

Given

$E_s = 2\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}N/mm^2,\hspace{05cm}E_{cu} = 1.2\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}N/mm^2$

To Find

(i) For $d_{st} = d_{cu},\hspace{03cm}\frac{\delta_{cu}}{\delta_{st}} = ?\\ (ii)For \delta_{cu} = \delta_{st}\hspace{02cm}\frac{d_{cu}}{d_{st}} = ?$

Solution

(i)$\frac{\delta_{cu}}{\delta_{st}} = \frac{(\frac{PL}{AE})_{cu}}{(\frac{PL}{AE})_{st}} = \frac{E_{st}}{E_{cu}} = \frac{2\hspace{0.05cm}\times\hspace{0.05cm}10^5}{1.2\hspace{0.05cm}\times\hspace{0.05cm}10^5}\\ \frac{\delta_{cu}}{\delta_{st}} = 1.66$

(ii)$\delta_{cu} = \delta_{st}\\ (\frac{PL}{AE})_{cu} = (\frac{PL}{AE})_{st}\\ \frac{P_{cu}L{cu}}{A_{cu}E_{cu}} = \frac{P_{st}L_{st}}{A_{st}E_{st}}\\ \frac{1}{\frac{\pi}{4}\hspace{0.05cm}\times\hspace{0.05cm}d_{cu}^2\hspace{0.05cm}\times\hspace{0.05cm}1.2\hspace{0.05cm}\times\hspace{0.05cm}10^5} = \frac{1}{\frac{\pi}{4}\hspace{0.05cm}\times\hspace{0.05cm}d_{st}^2\hspace{0.05cm}\times\hspace{0.05cm}2\hspace{0.05cm}\times\hspace{0.05cm}10^5}\\ \frac{d_{cu}}{d_{st}} = \sqrt{1.67} = 1.29 mm$