written 3.0 years ago by | modified 24 months ago by |

**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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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.

written 3.0 years ago by | modified 24 months ago by |

**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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written 3.0 years ago by | • modified 3.0 years ago |

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

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