Jaumann deviatoric stress rate

Question

Background about terms in this question: Hookes law and objective stress rates

From my understading, the Jaumann rate of deviatoric stress is written as: $$dS/dt = \overset{\bigtriangleup}{{S}} = {\dot{S}} +{S} \cdot {w} -{w} \cdot {S}$$

Yet when I see it in practice it is written as:

$$\mathrm{d}{{S}}^{ij}/ \mathrm{d}t = 2\mu\left({\dot{{\epsilon}}}^{ij} - \frac{1}{3}{\delta}^{ij}{\dot{{\epsilon}}}^{ij} \right)+{{S}}^{ik}{{\Omega}}^{jk}+{{\Omega}}^{ik}{{S}}^{kj}$$

To me that says:

$$dS/dt = \overset{\bigtriangleup}{{S}} = {\dot{S}} +{S} \cdot {w'} +{w} \cdot {S}$$

My question is - where did the spin tensor transpose and plus sign come from?

I thought that this might have something to do with Oldroyd and convective stress rates but that uses the tensor of velocity gradients rather than the spin tensor.

Edit - Clarifying where the deviatoric strain rate term came from.

For an isotropic, elastic solid the stress tensor is given by:

$${\sigma}^{ij} = 2\mu{\epsilon}^{ij} + \lambda \delta^{ij}({\epsilon}^{kk})$$

Then the deviatoric stress can be written as:

$${S}^{ij} = 2\mu{\epsilon}'^{ij} + \lambda \delta^{ij}({\epsilon'}^{kk})$$

Given that the deviatoric strain is traceless, the deviatoric stress rate can be written as:

$${\dot{S}}^{ij} = 2\mu\dot{{\epsilon'}}^{ij}$$

The deviatoric strain rate can be rewritten in terms of the strain rate giving:

$${\dot{S}}^{ij} = 2\mu\left[{\dot{\epsilon}}^{ij} - \frac{1}{3}\delta^{ij}{\dot{\epsilon}}^{ij}\right]$$

Answer 1

It seems that you confused the Jaumann derivative $\overset{o}{{S}}$ (in your notation $\overset{\bigtriangleup}{{S}}$) with the time derivative ${\dot{S}}$

$$\frac{dS}{dt} = {\dot{S}} = \overset{o}{{S}} -{S} \cdot {w} +{w} \cdot {S}$$

See how it is derived in "http://www.continuummechanics.org/cm/corotationalderivative.html". Using the argument that $w^T = -w$ we get

$$\frac{dS}{dt} = {\dot{S}} = \overset{o}{{S}} +{S} \cdot {w^T} +{w} \cdot {S}$$

Moreover, when you "clarified" where does the deviatoric strain come from it is not quite clear to me all the steps you followed. Because actually is not the time derivative that equals the term $\left({\dot{{\epsilon}}}^{ij} - \frac{1}{3}{\delta}^{ij}{\dot{{\epsilon}}}^{ij} \right)$ but the Jaumann derivative. For this you can take a look at Hookes law and objective stress rates

More concretely, the derivation of the formula for the Jaumann stress

$$\overset{o}{{S}^{ij}} = 2\mu\left[{\overset{o}{\epsilon}^{ij}} - \frac{1}{3}\delta^{ij}{\overset{o}{\epsilon}^{ij}}\right]$$

is derived in Hooke's law wikipedia and in "An Introduction to Continuous Mechanics, Klaus Hackl, Mehdi Goodarzi". However, here I show a summary of the derivation of the Jaumann stress.

The definition of deviatoric stress is just

$$\sigma = \begin{pmatrix} \frac{\sigma_{11}}{3} & 0 & 0\\ 0 & \frac{\sigma_{22}}{3} & 0\\ 0 & 0 & \frac{\sigma_{33}}{3}\\ \end{pmatrix} + \begin{pmatrix} \sigma_{11} - \frac{\sigma_{11}}{3} & \sigma_{12} & \sigma_{13}\\ \sigma_{21} & \sigma_{22} - \frac{\sigma_{22}}{3} & \sigma_{23}\\ \sigma_{31} & \sigma_{32} & \sigma_{33}- \frac{\sigma_{33}}{3}\\ \end{pmatrix}$$

the second term of the previous equation. If we apply Hooke's law to it (which relates stress $\sigma$ and strain $\varepsilon$ linearly)

$${{S}} = 2\mu\left({{\epsilon} - \frac{1}{3}\mathrm{tr}(\varepsilon)}\right)$$

With the Einstein notation

$${{S}}^{ij} = 2\mu\left[{{\epsilon}}^{ij} - \frac{1}{3}\delta^{ij}\epsilon^k{}_k\right]$$

where $\epsilon$ is the so called strain tensor.

Finally, we can compute the Jaumann derivative to get what we wanted.

Answer 2

TL;DR The expression in the paper is correct. It's just written weirdly.

Take a look at where the "=" is in your index equation. You'll notice that your $\dot{\mathbf{S}}$ and your spin terms are on opposite sides, so you should expect it to look a little goofy. Here's the algebra (in direct notation since I hate indices):

Let $\mathbf{\epsilon}' = \mathbf{\epsilon} - \frac{1}{3}\mathrm{tr}(\epsilon) \mathbf{1}$

$$ \dot{\mathbf{S}} = 2\mu\,\dot\epsilon' + \mathbf{SW}^T + \mathbf{WS} $$

$$ \dot{\mathbf{S}} - \mathbf{SW}^T - \mathbf{WS} = 2\mu\,\dot\epsilon' $$

Recall that $\mathbf{W} = -\mathbf{W}^T$. If you didn't know this, you should read an introductory textbook on continuum mechanics. It should clarify why this is the case.

Thus, the second line can be rewritten as

$$ \dot{\mathbf{S}} + \mathbf{SW} - \mathbf{WS} = 2\mu\,\dot\epsilon' $$

and you get your Jaumann deviatoric stress rate back.