Thermal stresses in the stress tensor

[hunt_mat]

TL;DR

How to I include thermal stresses in the stress tensor

Suppose I have a mechanical stress tensor $\sigma$. Say I have the stress tensor for viscous flow:
$$\boldsymbol{\sigma}=-p\mathbf{I}+\frac{1}{2}(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})$$
If the thermal flux is given by $\boldsymbol{\sigma}_{th}=\alpha T\mathbf{I}$, so I have a total flux as:
$$\boldsymbol{\sigma}=-p\mathbf{I}+\frac{1}{2}(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})+\alpha T\mathbf{I}$$
Is this correct?

 

[Chestermiller]

TL;DR Summary: How to I include thermal stresses in the stress tensor

Suppose I have a mechanical stress tensor $\sigma$. Say I have the stress tensor for viscous flow:
$$\boldsymbol{\sigma}=-p\mathbf{I}+\frac{1}{2}(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})$$
If the thermal flux is given by $\boldsymbol{\sigma}_{th}=\alpha T\mathbf{I}$, so I have a total flux as:
$$\boldsymbol{\sigma}=-p\mathbf{I}+\frac{1}{2}(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})+\alpha T\mathbf{I}$$
Is this correct?

Your original equation is for an incompressible fluid.

 

[Andy Resnick]

I don't think so, at least the reference I found below indicates something different in Eq. 4:

https://www.sciencedirect.com/scien...be calculated,is the Kronecker delta function.

 

[Chestermiller]

The correct equation for a compressible viscous fluid without thermal expansion is $$\boldsymbol{\sigma}=-(p+\frac{2\mu}{3 }\nabla \centerdot \mathbf u)\mathbf{I}+\mu(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})$$
With thermal expansion, this becomes $$\boldsymbol{\sigma}=-(p+\frac{2\mu}{3 }(\nabla \centerdot \mathbf u-\alpha \frac{D T}{Dt}))\mathbf{I}+\mu(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})$$The thermal expansion term is usually considered negligible in determining the stress.

 

[hunt_mat]

The correct equation for a compressible viscous fluid without thermal expansion is $$\boldsymbol{\sigma}=-(p+\frac{2\mu}{3 }\nabla \centerdot \mathbf u)\mathbf{I}+\mu(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})$$
With thermal expansion, this becomes $$\boldsymbol{\sigma}=-(p+\frac{2\mu}{3 }\nabla \centerdot \mathbf u-\alpha \frac{\partial T}{\partial t})\mathbf{I}+\mu(\nabla\mathbf{u}+(\nabla\mathbf{u})^{T})$$The thermal expansion term is usually considered negligible in determining the stress.

Hi, thanks for your reply. I'm not interested in expansion, but thermal stresses within a material. I want temperature to be coupled to Navier's equations. I would include the $\partial_{t}T$ term as part of the stress tensor to fully couple the derivative?

I'm thinking of sintering with this application, and how thermal expansion affects everything.