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A steel rod of 30mm dia and 5m long is connected to two grips and the rod is maintained at 95deg C.Determine the stresses and pull exerted when temperature falls to 30 deg C.

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

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

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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)

enter image description here

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)

enter image description here

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

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

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