# Thermal stress in a bar.

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Suppose a bar of length $L$, Young Modulus $Y$, temperature coefficient $\alpha$ is at temperature $T_o$ is just held between two walls. Now temperature increases, to make a change in temperature $\triangle T$ (measured from initial $T_o$). My book says thermal stress developed is $Y\alpha\triangle T$.

My question:

As the rod is rigidly held, walls will not let rod to changes its length further from $L$, for this it applies force $F$. Its original length at $T_o+\triangle T$ would have been $L(1+\alpha\triangle T)$, if walls were not present, hence it causes $\triangle L=l\alpha\triangle T$.

So $Y=\dfrac{F_{T_f}\cdot L_{T_f}}{A\cdot\triangle L}\tag*{1}$ where $T_f$ signifies their respective values at final temperature. Putting these it gives to me themal stress as, $$\sigma=\dfrac{Y\alpha\triangle T}{1+\alpha\triangle T}$$.

If I were to put $L_{T_f}=L$ then this gives what my book says, but we should apply $\text{eq. 1}$ and put everything at the same temperature, so why putting its length as $L$ mandatory. Please help.

asked Apr 19
edited Apr 19

## 1 Answer

0 votes

Young's Modules is defined as,

$Y = \frac{\sigma}{\epsilon}$

where $\sigma$ is the strain defined as $\sigma = F/A$ and $\epsilon$ is the stress defined as $\epsilon = \frac{\Delta L}{L_0}$.

In this case we require the stress, thus re-arranging the first equation gives,

$\sigma = \epsilon Y$

After plugging in the definition of $\epsilon$ we then arrive at,

$\sigma = \frac{\Delta L}{L_0}Y$

Now as you already worked out, $\Delta L = L_0\alpha\Delta T$. Therefore after plugging this in, we arrive at the required result:

$\sigma = \frac{L_0\alpha\Delta T}{L_0}Y = \alpha\Delta T Y$

answered Apr 30 by (1,486 points)