If (i) The ends do not yield (ii) The ends yield by 0.12 mm.

Take $E_s = 2\times10^5 MPa, \alpha_s = 12\times10^{-6}/\hspace{0.05cm}^\circ{C}$

Given

$E_s = 2\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}MPa,\hspace{0.25cm}\alpha_s = 12\hspace{0.05cm}\times\hspace{0.05cm}10^{-6}/\hspace{0.05cm}^\circ{C}\\ d_s = 30 mm,\hspace{0.25cm}L = 5 m = 5000\hspace{0.05cm}mm\\ t = 95 - 30 = 65^\circ{C}\\ A = \frac{\pi}{4}\hspace{0.05cm}\times\hspace{0.05cm}30^2 = 706.85\hspace{0.05cm}mm^2$

To Find

$\sigma = P =?$

Solution

Case (i) Ends do not yield (No gap)

Therefore, $\delta = 0$

For stresses, $\sigma = E.\alpha.t\\ \hspace{0.25cm}= 2\hspace{0.05cm}\times\hspace{0.05cm}10^5\hspace{0.05cm}\times\hspace{0.05cm}12\hspace{0.05cm}\times\hspace{0.05cm}10^{-6}\hspace{0.05cm}\times\hspace{0.05cm}65\\ \sigma = 156\hspace{0.05cm}N/mm^2$

For pull, $\sigma = \frac{P}{A}\\ P = 10268.6 \hspace{0.05cm}N$

Case(ii) Yield by 0.12 mm ($\delta$ = 0.12 mm)

For stresses, $\sigma = \frac{E(\alpha L t - \delta)}{L} = 151.2\hspace{0.05cm}N/mm^2$

For pull, $\sigma = \frac{P}{A}\\ P = 106875.72\hspace{0.05cm}N$