Is deviatoric strain associated with thermal effects?

Question

Does temperature have any effects on deviatoric strain for a linearly elastic isotropic material?

Answer 1

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.