The usual approach to calculate stress is to equate thermal expansion in the unclamped condition to the magnitude of contraction caused by strain produced due to the walls. I have some questions about this approach:
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Wouldn't Young's modulus of the rod change with temperature?
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Moreover, in this method, $L$ is the length in the unclamped condition. After increasing the temperature, the 'original length' of the rod should be $L(1+\alpha \Delta T)$, which should be used in the relation (i) below because the original length of the rod is taken with reference to a particular temperature.
$$\Delta L=\frac{FL}{AY}=L\alpha \Delta T \cdots (i)$$
$$ \frac{F}{A}=Y \alpha\Delta T= \text{stress} \cdots (ii)$$
Am I correct in both? This is really bothering me because I cannot find a discussion of these points in any books, but I strongly feel their validity.
Best Answer
After heating it, the length is $L_0(1+\alpha \Delta T)$ and its length has been increased by $\Delta L=L_0\alpha \Delta T$. To get it back to its original length, we have to compress it so that the second $\Delta L$ is minus the first $\Delta L$: $$\sigma=Y\frac{(L_0\alpha \Delta T)}{L_0(1+\alpha \Delta T)}=Y\frac{\alpha \Delta T}{(1+\alpha \Delta T)}\tag{1}$$But, from the equation for an infinite geometric progression, we have $$\frac{1}{1+\alpha \Delta T}=1-(\alpha \Delta T)+(\alpha \Delta T)^2-(\alpha \Delta T)^3+...$$If we substitute that back into Eqn. 1 and recognize that, in the approximations associated with Hooke's law, we are retaining only linear terms in the strains, we obtain (for the compressional stress): $$\sigma=Y\alpha \Delta T$$