# Heating a hollow disc maintaining same temperature at centre? [duplicate]

• If a hollow disc made of brass is heated at the same time the circular hole is kept at a constant temperature, would the circular hole still show a change in shape?

• Also, does the position of the circular hole in the disc dictate how much it expands / contracts ?

Ps. Any mathematical proof would be really appreciated. • By hollow you mean it has a hole? You would produce a heat gradient. The temperature would increase stress anyway and this would affect the hole. The stress distribution/strength depends on the position of the hole. Mar 17, 2017 at 14:37
• Yep by hollow I mean a disc with a hole just like the one in the image. But will the hole also expand if the disc is heated and the hole maintained at a constant temperature ? Mar 17, 2017 at 14:39
• Well, stress is a long range phenomenon, meaning that heating the disk at one point changes the stress in the entire disk, regardless of the local temperature. So my guess is, yes, it will deformed, depending on its position. Mar 17, 2017 at 14:44
• everything will expand Mar 17, 2017 at 14:44
• How can I mathematically show that it does expand and by how much if i have say coefficient of linear expansion given ? Mar 17, 2017 at 14:48

I think in detail the calculation will be sort of tricky, but a general picture of what is going to happen is possible.Let's start with constant temperature and let's assume to neighboring atomic circles (for simplicity I assume that the atoms forming the disc sit on circles). The inner circle has circumference $U_\mathrm{i}=2\pi R_\mathrm{i}$, the outer $U_\mathrm{o}=2\pi R_\mathrm{o}= 2 \pi (R_\mathrm{i}+\Delta R)$. If the expand it changes, according to linear expansion, so $\tilde U_\mathrm{i}=2\pi R_\mathrm{i}(1+\alpha)$ and $\tilde U_\mathrm{o}=2\pi R_\mathrm{o}(1+\alpha)= 2 \pi (1+\alpha)(R_\mathrm{i}+\Delta R)$, from which we see that $\Delta \tilde R=(1+\alpha)\Delta R$. Hence all expanded equally and everything is in equilibrium.

Now this is different if the outer ring is heated while the inner stays on its temperature. If The outer circle would increase by ($1+\alpha$) while the inner stayed constant, the $\delta R$ would increase by even more than $(1+\alpha)$, while the real temperature gradient would suggest something smaller. so this cannot be in equilibrium. The forces from the inner to the outer ring will shrink the outer slightly and increase the inner as well.

To make a full calculation you need to consider strain and stress differences according to temperature differences. So you'd have to deal with the elasticity as well. In the radial symmetric case this might be possible. If your hole is off centre, I would suggest simulation software.

• I came across a numerical where the radius of the inner hole is (R), kept at 20degC, while the disc itself heated to 120degC. then the answer for the increase in radius amounts to 100xR, where (x=coeff of linear expansion). So there is no change in formula whether we keep the inner hole temperature constant or not. Do you think this is correct considering that the hole is in centre of the circle ? Mar 17, 2017 at 16:51
• No, I think that is incorrect. Mar 20, 2017 at 7:36

Your question is not very clear. However, judging by your comments I think you mean that the inner edge of the annulus is held at one temperature $T_1$ while the outer edge is held at a higher temperature $T_2$.

There will be a temperature gradient along the radius of the disk. We cannot assume that this temperature gradient will be linear. To find it we must solve the Laplace Equation for heat transfer with circular symmetry.

Unlike the case in which the annulus is heated uniformly, the temperature gradient will set up thermal stresses in the annulus. Each infinitessimally thin concentric ring will be pulled outwards and inwards by adjacent rings, the difference being balanced by elastic tension in that ring.

This is not an easy calculation, though not impossible. It could certainly be solved numerically.

Without any calculation we can say that the inner radius will increase, because it is pulled outwards by adjacent expanding material, although it will not increase as much as when the annulus is heated uniformly. The material expands most where the temperature greatest, which is nearest the outer rim. This means that the hole will be distorted out of its circular shape if it is not concentric with the outer rim.

• I came across a numerical where the radius of the inner hole is (R), kept at 20degC, while the disc itself heated to 120degC. then the answer for the increase in radius amounts to 100xR, where (x=coeff of linear expansion). So there is no change in formula whether we keep the inner hole temperature constant or not. Do you think this is correct considering that the hole is in centre of the circle ? Mar 19, 2017 at 11:38
• Do you mean that the inner edge is thermally insulated from the rest of the annulus but bonded to it? This is similar to having a disk made out of 2 materials with different expansion coefficients, bonded together. The answer you are giving is the same as in the link - the hole expands exactly the same as the material which was removed from the centre. ... It is still very unclear what the conditions of your problem are. If your question is based on a problem in a book, please can you post the problem exactly as it is written. Mar 19, 2017 at 19:54