Often, in papers presenting updated Lagrangian simulation methods for solid dynamics, the following procedure for updating the (Cauchy) stress tensor is presented:

First, the Cauchy stress tensor is split into a hydrostatic and a deviatoric part:

$$\sigma^{ij} = -p\delta^{ij} + S^{ij}$$

The pressure is found using an equation of state. Often the following isothermal approach is used: $$p = c_0^2(\rho - \rho_0)$$

$c_0^2$ being the adiabatic sound speed, $\rho$ the density and $\rho_0$ the reference density. Then, it is stated that Hooke's law is assumed and the deviatoric part of the stress tensor evolves as follows:

$$\frac{dS^{ij}}{dt} = 2\mu(\dot{\epsilon}^{ij} - \frac{1}{3}\delta^{ij}\dot{\epsilon}^{kk}) + S^{ij}\Omega^{jk} + \Omega^{ik}S^{kj}$$

where $\mu$ is the shear modulus, $$\Omega^{ij} = \frac{1}{2} (\frac{\delta v^i}{\delta x^j} - \frac{\delta v^j}{\delta x^i})$$ is the spin tensor and $$\dot{\epsilon}^{ij} = \frac{1}{2} (\frac{\delta v^i}{\delta x^j} + \frac{\delta v^j}{\delta x^i})$$ is the rate of deformation tensor. Now, since:

$$\dot{\sigma} = \overset{\Delta J}{\sigma} + \sigma\cdot\Omega + \Omega\cdot\sigma$$

where $\overset{\Delta J}{\sigma}$ is the Jaumann rate it holds that:

$$ \overset{\Delta J}{\sigma}_{ij} = 2\mu(\dot{\epsilon}^{ij} - \frac{1}{3}\delta^{ij}\dot{\epsilon}^{kk})$$

Now on to my questions:

  1. How does one come up with the above equation for the Jaumman Rate? Or, particularly, how does the assumption of Hooke's law yield that equation for the Jaumann rate?

  2. Is that equation for the Jaumann rate also valid for other objective stress rates? For example for the Truesdell rate, giving a stress update as follows:

$$\frac{dS}{dt} = 2\mu(\dot{\epsilon} - \frac{1}{3}\mathbf{1}{\rm Tr}(\dot{\epsilon})) - {\rm Tr}(\dot{\epsilon})S + \dot{\epsilon}\cdot S + S\dot{\epsilon}^T$$

($\rm Tr(.)$ being the trace)


1 Answer 1


You probably know that while the Cauchy stresses are objective, its stress rate (material derivative) is not. If $Q(t)$ is an orthogonal tensor representing a change of frame, the stress is the new frame is

$$ T^* = Q T Q^T $$

However, if you take material derivatives on both sides, you have,

$$ \dot{T^*} = \dot{Q} T Q^T + Q \dot{T} Q^T + Q T \dot{Q}^T \quad \quad (1) $$

Terms 1 and 3 above are "extra" -> they imply that naively taking the material derivative of the stress $T$ won't get you an objective rate.

A similar problem arises because the "spin" part of the velocity gradient tensor $L$, (I'll call it $W$ rather than $\Omega$) is not objective. i.e., $L = W + D$, but while $D$ is objective, $W$ is not. Starting from first principles, it is not hard to show that (I'll add this bit if you want)

$$ W^* = Q W Q^T + \dot{Q} Q^T \quad \quad (2) $$

Also, because $Q$ is orthogonal ($Q^T=Q^{-1}$) you have the lemma

$$ \dot{Q} Q^T = -Q \dot{Q}^T $$

We can actually use this second problem to construct a solution to the first. We'll eliminate $\dot{Q}, \dot{Q}^T$ from (1) in turn. Right multiply (2) by Q and rearrange; you have

$$ \dot{Q} = W^* Q - Q W $$ Transpose both sides and you have an expression for $\dot{Q}^T$. $$ \dot{Q}^T = -Q^T W^* + W Q^T $$ The last result uses the fact that the spin tensor $W$ is skew-symmetric.

Let's substitute these expressions for $\dot{Q},\dot{Q}^T$ back in (1).

$$ \dot{T^*} = (W^* Q - Q W) T Q^T + Q \dot{T} Q^T + Q T (-Q^T W^* + W Q^T) \\ \quad = W^* Q T Q^T - Q W T Q^T + Q \dot{T} Q^T - Q T Q^T W^* + Q T W Q^T) \\ \quad = W^* T^* - Q W T Q^T + Q \dot{T} Q^T - T^* W^* + Q T W Q^T \\ \Rightarrow \quad \text {(on rearranging)} \\ \quad \dot{T^*} + T^* W^* - W^* T^* = Q (\dot{T} + T W - W T) Q^T \\ $$

Notice that the right had side has the form $Q R Q^T$, and the left side is precisely the bracketed quantity in the starred frame of reference. In other words, we have constructed a rate $J = \dot{T} + T W - W T$ such that $\dot{J^*} = Q J Q^T $, which is objective by definition. You can now easily specialize your equations to elasticity.

The important thing to remember is that the Jaumann rate is merely one of an infinite number of objective stress rates. In general, Lie derivatives of objective tensor fields are objective.

  • $\begingroup$ Note @Matthias: your expression relating the Jaumann rate and $\dot{\sigma}$ has a sign problem. $\endgroup$ Commented Jul 15, 2014 at 17:36
  • $\begingroup$ Thank you for your extensive answer. Your derivation of the Jaumann Rate is clear and easy to follow. However, I still struggle to "easily" specialize the equations to elasticity. Could you maybe comment on my questions 1 & 2 concretely? It would be much appreciated. (Can't upvote because of lack of reputation) $\endgroup$
    – Matthias
    Commented Jul 16, 2014 at 10:39
  • $\begingroup$ @Matthias: You can see the following page (Scroll down to 3.9.3): solidmechanics.org/text/Chapter3_9/Chapter3_9.htm Note that there is an approximation involved (namely, that stresses are much smaller than the elastic moduli) $\endgroup$ Commented Jul 18, 2014 at 14:37
  • $\begingroup$ Also, always be careful to use work conjugate quantities in your stress-strain relations. $\endgroup$ Commented Jul 18, 2014 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.