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Strain or stress can be caused by different sources. I categorized theses sources as mechanical, thermal and electrical loads and formulated the total stress as follows:

$$ \epsilon_{total} = \epsilon_{mechanical} + \epsilon_{thermal} + \epsilon_{electrical} $$

I know that the thermo-mechanical stress-strain relationship can be expressed as follows:

$$\epsilon_{x} = \frac{1}{E}\{\sigma_{x} - v(\sigma_{y}+\sigma_{z})\}+\alpha\Delta T\\ \epsilon_{y} = \frac{1}{E}\{\sigma_{y} - v(\sigma_{z}+\sigma_{x})\}+\alpha\Delta T\\ \epsilon_{z} = \frac{1}{E}\{\sigma_{z} - v(\sigma_{x}+\sigma_{y})\}+\alpha\Delta T $$

However, I'm wondering how the equations look like for a conducting material with high density current. In other words, I want to know what is $\epsilon_{electrical}$ in the following model:

$$\epsilon_{x} = \frac{1}{E}\{\sigma_{x} - v(\sigma_{y}+\sigma_{z})\}+\alpha\Delta T + \epsilon_{electrical-x}\\ \epsilon_{y} = \frac{1}{E}\{\sigma_{y} - v(\sigma_{z}+\sigma_{x})\}+\alpha\Delta T + \epsilon_{electrical-y}\\ \epsilon_{z} = \frac{1}{E}\{\sigma_{z} - v(\sigma_{x}+\sigma_{y})\}+\alpha\Delta T +\epsilon_{electrical-z} $$

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Can I express it in term of Lorentz body forces? – Ali Abbasinasab Aug 13 '13 at 6:37
Are you speaking of piezoelectricity? – WetSavannaAnimal aka Rod Vance Aug 13 '13 at 7:09
@WetSavannaAnimalakaRodVance it's actually something like the opposite of piezoelectricity. That is the high density electric current make the electrons to collide the material atoms or ions and force them to move. Therefore, after a while the accumulation of atoms in anode and the depletion of atom in cathode produce a mechanical stress in the wire. – Ali Abbasinasab Aug 14 '13 at 1:13

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