# Stress Strain Analysis In Wall

I have a 20mm thick and 10m long wall. On one side of this wall (inner) is 80 degree celcius water, the outer side is air at an ambient temp of 20 degrees celcius. My coeff of thermal expansion is 15e-5 and the elastic modulus is 1500 MPa (hypothetical wall is hard plastic). The wall is unrestricted.

I want to do a stress and strain analysis over a period of 1 hour of heating by sectioning this wall into 5 sections. In doing so, the following is true:

0mm = 80 degrees C
4mm = 70 degrees C
8mm = 56 degrees C
12mm = 37 degrees C
16mm = 28 degrees C
20mm = 25 degrees C

Now, I know the basic equations for thermal stress and strain (https://www.engineeringtoolbox.com/stress-restricting-thermal-expansion-d_1756.html), however I am struggling to find how to incorporate the impact of section 1 on section 2 and etc. I am expecting the outer side to be in tension stress (+ve) and the inside to be in compression stress (-ve).\

Note: Apologies if the numbers don't make theoretical sense, I am not using the real values.

There will be shear deformation between the layers deforming the transitional interface into a stepped cross-section with rounded corners.

If there is no constraint there is no compression on the ambient side. the wall will curve out like a thermocouple with an angle. $$\alpha$$

$$\alpha=\frac{dL_{inside}-dL_{outside}}{2m}$$ And radius $$r$$

$$r=\frac{2m}{sin \alpha}$$ In reality, there will be a gradual gradient of temperature distribution into the thickness of the wall and there will be no stepwise deformation. Deformation will be even and the wall will warp nicely.

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Unrestricted thermal expansion would basically extend every part according to its temperature. In your case however, there is a bit of restriction in form of wall sections being connected into one sheet. Effect of the temperature would create bending stress if the plate was restricted in flat shape, so the unrestricted plate will bend.

## Simplification: linear temperature profile across thickness

If the temperature change profile across the wall was linear (with mean wall temperature change $$\Delta T_m$$ and thickness $$t$$, coordinate $$z$$ starts in the middle of thickness):

$$\Delta T(x) = \Delta T_m + k\cdot z$$

this would lead to strains in all directions:

$$\epsilon_x(z) = \epsilon_y(z) = \epsilon_z(z) = \alpha\cdot \Delta T(z) = \alpha\cdot \Delta T_m + \alpha\cdot k\cdot z$$

The part with mean wall temperature change, $$\alpha\cdot \Delta T_m$$, is not very interesting, because it just elongates the plate in all directions, but does not change its shape. The other part however will try to bend the plate into a spherical surface in case of very small deformations.

When you bend a beam or plate around one axis using constant bending moment, the longitudinal strain is linear across the thickness:

$$\epsilon(x) = \frac{z}{R_b}$$

where $$R_b$$ is radius of the resulting shape (at the middle of plate thickness). If you equal this strain with a bending part of thermal strain, you can derive a radius resulting just from bending:

$$R_b = \frac{1}{\alpha\cdot k}$$

Now the full radius would be of course affected also by mean wall temperature change, which will elongate it by $$\Delta R$$:

$$\Delta R = \alpha\cdot \Delta T_m \cdot R_b$$

So the full expression for the sphere radius $$R$$ would be: $$R = \frac{1 + \alpha\cdot \Delta T_m}{\alpha\cdot k}$$

In a case similar to yours, where you have 20 mm thickness and temperature starting at 20 °C and rising to 25 up to 80 °C across the thickness ($$\Delta T_m = 32.5 K$$) with $$\alpha = 15e-5K^{-1}$$, you should end up with sphere radius $$R = 2.436 m$$. This could be quite right for a plate of small dimensions, but 10x10 m plate would of course not deform so much. So the resulting deformation would probably be different with some residual stress in the wall. If for example the length of the wall is 10 m, but the height is just 1 m, it would bend mostly around the vertical axis, so it would roll into a cylinder. Bending around the other major axis would not be possible, so most of the residual stress would be the bending stress associated with that axis.

If the temperature profile is not linear across the thickness, there will be additional residual stress.