Shear stress damage due to thermal gradient

  • Thread starter Rob B
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I'm trying to use my rudimentary understanding of material physics to understand a simple problem, and am getting stuck - I hope you can help!

My idealized case involves a sheet of infinite extent in length and width direction, to which a linear thermal gradient in the depth dimension is applied, i.e. deltaT/deltaz=constant. We could express this as T=T1 at z=0 and T=T2 at z=d. The material has a known thermal expansion coefficient alpha and elastic Young's modulus M, so we can calculate the stress induced by the temperature difference - assuming zero stress at the mid-plane, the tensile/compressive stress at the surfaces will be +/- (T2-T1)/2*alpha*M . Now I'd like to know if the material will undergo plastic deformation or brittle fracture as a result of the temperature gradient. I assume that this is much more likely with a steep temperature gradient (such as would be induced by thermal shock) than with a mild gradient - that is if the same temperature difference is imposed across a sheet of thickness 1 mm and a sheet of thickness 1 m, the former would be more likely to fail. But I'm struggling to come up with the formulas that would allow me to calculate shear stress levels that I could compare to the moduli of such failures - where does the thickness come in? It must be simple (I think) - please help!
 

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Baluncore
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Welcome to PF.
Now I'd like to know if the material will undergo plastic deformation or brittle fracture as a result of the temperature gradient.
That will be a characteristic of the material ductility, not so much of the gradient.

An unrestrained flat sheet, heated on one side, will take the form of a spherical shell. That is contradicted by the assumption of an infinite sheet.
The opposed faces will have different shell circumferences due to thermal expansion, that will imply a ratio of the inner and outer surface radii. The thickness of the sheet is the difference in those radii.
 
  • #3
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Thanks Baluncore. I agree that that's what will happen with a sheet that can change its shape - my formulation of the problem was just a simplification of a real problem, in which the sheet cannot change its shape, that is, it will remain planar (the sheet is in actuality just the surface of a larger object, which has a temperature gradient that penetrates only part-way into the object; the remainder of the object is rigid, preventing the curling action you describe.
I also agree that the potential for plastic deformation depends on the material ductility - but for an object of a given ductility (characterized by some modulus?), I imagine that deformation will happen beyond a certain temperature gradient. I'm trying to figure out how to relate the material ductility parameter to the maximum temperature gradient that can be tolerated. Does that make sense?
 
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It could be easily solved using thermo-mechanical FEA (Finite Element Analysis). It’s even possible to define various damage criteria and see how the structure actually breaks.
 
  • #5
Baluncore
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Does that make sense?
Yes.
The topic title refers to “shear stress damage”. A metal plate will not de-laminate due to thermal gradient shear. The failure will be a three point star fracture or a pentagonal / hexagonal tessellation of tension cracks in the cold face.
Study the stress/strain diagram for the material to find the elastic limit, yield point and material elongation before tension cracks will form.
 
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Right, which gets to the heart of my question: If I have the elastic limit, how do I know at which linear temperature gradient that limit is exceeded? I'm sorry if I'm missing the obvious - this is well outside of my core area of expertise ...
 
  • #7
Baluncore
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The first mode of failure would be for the plate to bow towards the heat. But if the plate could be held flat, then the compression on one side should be opposed by tension in the other. That is independent of plate thickness and the magnitude of the thermal gradient, it is due only to the temperature difference of the opposite faces.

Thermal expansion will change the length, but because the plate is constrained, the internal force will increase, until it reaches the yield point. The yield point occurs at a particular proportional length extension for a material. You can calculate back to the temperature difference required for the crack to occur. If the material can stretch 2% before yield, then a temperature difference giving a 2% differential length will be sufficient to open a crack on the cold face.

Alternatively, if the heat softens the material, the hot side of the plate may thicken to accommodate the increased volume. When it cools, tension will appear on the hot side, with compression on the cold.
 

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