2
$\begingroup$

I have a rectangular metal plate with a hole in it (with diameter of 300 mm). The plate has a temperature gradient going from one of its short sides to the other, I can measure the temperature anywhere in the plate.

I want to compute how much the hole is deforming from its round shape.

I know that I can calculate the expansion of the hole at uniform temperature with $\frac{ΔL}{L_0}=αΔT$ . So I was wondering if it is correct to measure a bunch temperature points at the edge of the hole and just apply that formula to each of them independently. But since there's a temperature gradient I'm assuming that there will be mechanical stresses between hot and cold zones working against the expansion, is this the case?

Is there a way to compute this by hand?

In the picture $T1>T2$. enter image description here

$\endgroup$
  • $\begingroup$ A sketch might help here. $\endgroup$ – grfrazee Apr 20 '19 at 22:03
  • $\begingroup$ No, you can’t use that formula. That formula is for uniaxial strain. I don’t think this problem has an analytical solution either, so you’ll need to use something like finite element method on the uncoupled thermoelasticity equations. $\endgroup$ – Paul Apr 20 '19 at 22:54
  • $\begingroup$ @Paul FEM provides indeed more accurate answer, but i guess if the plate is constrained though that's is not the case here, we can apply Lamé equations for plane stress, of course this would be just an approximation. $\endgroup$ – Sam Farjamirad Apr 21 '19 at 6:04
-1
$\begingroup$

Assuming the plate is insulated on the sides and around the hole, only open at two ends, and assuming is settled into a stationary heat profile, yes you can corollate your temperature contour line with the expansion in the plate.

It takes some time for the expansion of the plate and deformation of the circle into a kind of egg shape to settle. Then stresses redistribute and relax.

Reminds me of a jobsite I had where the stock pile of steel beams heaped on top of each other would groan at sunrise or when the clouds opened up because of the squeezing stresses due to expansion, but shortly thereafter things became quiet down again.

So at the stationary state, if we assume the warping of surface limited to small angles, $$\frac{\Delta L_{I,j,k}}{L_{i,j,k}} =\alpha\cdot\Delta T $$

| improve this answer | |
$\endgroup$
  • $\begingroup$ The strain wouldn’t be uniaxial. How can you justify applying the OP’s formula? $\endgroup$ – Paul Apr 21 '19 at 0:00
  • $\begingroup$ It is no , I edited my answer to clarify that expansion will be three dimensional. $\endgroup$ – kamran Apr 21 '19 at 0:25
  • $\begingroup$ The expression you post is the relative change in length $\Delta L / L$ for the entire bar after it has experienced an overall temperature change $\Delta T$. The system at hand has a thermal gradient at any point $z$ along the bar. These are not the same cases. $\endgroup$ – Jeffrey J Weimer May 22 '19 at 2:11
  • $\begingroup$ An expression for the temperature can be written as a function of a range variable from the left to right. The change in length also written in these terms can be integrated to find the overall displacement. remember Thermal expansion happens in all direction not just L. Its pretty tough to get this all into a hand calc I reckon. $\endgroup$ – ShadowMan May 22 '19 at 7:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.