Is deviatoric strain associated with thermal effects? Does temperature have any effects on deviatoric strain for a linearly elastic isotropic material?
 A: We can write constitutive equation for linear elastic isotropic material as :
$$ \epsilon_{ij} = \frac{1}{E}\big [ (1 + \nu) \sigma_{ij} - \nu \delta_{ij} \sigma_{kk} \big ] + \alpha \Delta T \delta_{ij}$$
[ nomenclature: $\epsilon$ is strain, $E$ is elastic modulus, $\nu$ is Poisson's ratio, $\alpha$ is thermal conductivity, $T$ is temperature]
Multiplying above equation by $\delta_{ij}$
$$\epsilon_{kk} = \frac{1}{E} (1 -2 \nu) \sigma_{kk}  + \alpha \Delta T \delta_{kk} $$
Multiplying above equation by $\frac{1}{3} \delta_{ij}$
$$\frac{1}{3}\delta_{ij}\epsilon_{kk} = \frac{1}{3E} (1 -2 \nu) \sigma_{kk}\delta_{ij}  + \frac{1}{3}\alpha \Delta T\delta_{ij}\delta_{kk} $$
Subtracting third equation from the first one,
$$ \epsilon_{ij} -\frac{1}{3}\delta_{ij}\epsilon_{kk} = \frac{1+\nu}{E}\big [ \sigma_{ij} - \frac{1}{3} \delta_{ij}\sigma_{kk} \big ]  $$
This cancels the temperature terms. 
$$\therefore \epsilon_{dev} = \frac{1+\nu}{E} \sigma_{dev}$$ [$<>_{dev}$ isdeviatoric part. ] So, there is no temperature dependence.
