Thermal Expansion in a rigidly fixed rod

Question

I was reading this topic, and this is what I Found:

Consider a rod of length $L$ which is fixed between to rigid end separated at a distance $L$. Now, if the temperature of the rod is increased by $Δθ$, then the strain produced in the rod will be:

Thermal strain,$$ε = \frac{ΔL}{L}=\frac{\text{Final length - Original Length}}{\text{Original Length}}=\alpha \Delta\theta,$$ where $\alpha=\frac{\Delta L}{L} \times \frac{1}{\Delta \Theta}.$

Now my doubt is that why are we taking Original Length as L and not as L = $L(1+\alpha\Delta\theta)$? Reason being that when we are considering a rigidly fixed rod , we can visualise that the wall/support has caused the rod to shorten or compress from its changed length of L(1+αΔθ) back to L, and so the value of strain $\epsilon$ should have been = $\frac{\Delta L}{L(1+\alpha\Delta\theta)}.$ So please explain me where I am wrong.

Answer 1

In addition to nice @Gandalf61 answer there's some caveat in what you propose. Original definition of thermal expansion is $$\tag 1 \varepsilon_0 = \frac{\Delta L}{L}$$,

while you propose it to be: $$\tag 2 \varepsilon_1 = \frac{\Delta L}{L+\Delta L}$$

In (1) equation it measures expansion percentage (ratio) based on length before expansion, while your (2) equation measures percentage after expansion.

It would be not so bad, in a first glance, but... in (1) thermal strain is always proportional to rod $\Delta L$, that is $$\tag 3 \varepsilon_0 \propto \Delta L$$

While in (2) it does not, i.e. $$\tag 4 \varepsilon_1 \not \propto \Delta L$$

in case $\Delta L \gg L$, then $~\varepsilon_1 \approx 1$, and proportionality is lost. What's the point of thermal expansion percentage if it stops responding to great prolongations ?

Answer 2

The original length of the rod is its length before it is heated. You can think of its as being the length of the rod when $\Delta \theta$ is zero. We are told that this length is $L$.

So the original length of the rod is $L$, its final length if it were free to expand would be $L(1+\alpha \Delta \theta)$, the difference in length would be $\Delta L = L \alpha \Delta \theta$, and the strain $\epsilon$ would be $\alpha \Delta \theta$.

Answer 3

This passage is only considering the effect of temperature. The term "strain" only refers to the change in size of the material, not the forces acting on it. My guess is that, shortly after this passage in your reading, the stress put upon the rod by the supports to compress the rod back to its original length will be considered.