This question pertains to the general coefficient of thermal expansion (CTE) tensor, with no assumption about isotropy. The CTE tensor is $\alpha_{ij}$. Is it possible for $\alpha_{ij}$ to be non-zero for $i \neq j$? Essentially this is asking if there is a such thing as a shear component of thermal strain?
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$\begingroup$ You might be able to get more responses if you defined exactly what a CTE tensor is because I don't think that that's a very widely used concept. I and, I think, most people are only familiar with the thermal coefficient of expansion being represented by a scalar. $\endgroup$– user93237Commented Nov 9, 2018 at 2:18
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$\begingroup$ I did not define it because I do not know if CTE "tensor" is an actual thing. But if the CTE is a scalar then the material is isotropic. So for a material that is not isotropic, a single scalar should not be sufficient to represent the CTE? $\endgroup$– roulette01Commented Nov 9, 2018 at 2:27
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$\begingroup$ physics.stackexchange.com/questions/404627/… See the answer in this link for somewhat of a definition. $\endgroup$– roulette01Commented Nov 9, 2018 at 2:31
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1 Answer
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If it's a 2nd order symmetric tensor (not necessarily isotropic), there are going to be 3 orthogonal principal directions. But, for a set of orthogonal axes that don't coincide with the principal directions, there are going to be off-diagonal elements. So, in short, there certainly can be shear components to the thermal strain.
answered Nov 9, 2018 at 3:43
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$\begingroup$ Is there any difference between the terminology "orthogonal principal directions," "orthogonal axes," and "principal directions" for an ORTHOTROPIC material? I recall that the 3 orthogonal axes of an orthotropic material ARE the principal directions at which there's no shear? $\endgroup$ Commented Nov 9, 2018 at 16:41
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$\begingroup$ That is my understanding too. $\endgroup$ Commented Nov 9, 2018 at 17:38