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I have a volume which is deforming (using explicit time-integration scheme) uniformly with velocity gradient $L$ and stress tensor $\sigma$. I would like to determine work done by the volume deformation during one timestep $\Delta t$, knowing both current and previous values of $L$, $\sigma$ and volume $V$.

I've seen somewhere the formula $\Delta W={\mathrm tr}(L\sigma)V\Delta t$, but I don't know if it is correct and how to derive it. $L\sigma$ should be energy density, but why are its non-diagonal terms discarded?

Remark: the integration scheme actually is leap-frog, but I ignored mid-step/on-step business for now and supposed everything are on-step values. The formula above would correctly compute the increment mid-step, reading $\Delta W(t-\Delta t/2)={\mathrm tr}\left(L(t-\Delta t/2)\frac{\sigma(t-\Delta t)+\sigma(t)}{2}\right)\frac{V(t-\Delta t)+V(t)}{2}\Delta t$

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up vote 2 down vote accepted

The key idea here is the concept of "power-conjugate" stress and strain-rate measures. For the Cauchy stress $\sigma$, the stress power is given by:

$$ \dot W/V = \sigma:D $$ where $D$ is the rate of deformation tensor defined as the symmetric part of the velocity gradient.

$$D = sym(L) = \frac{1}{2}(L + L^T)$$

The quantity $\sigma:D$ gives the stress power per unit volume. Therefore, using the explicit time integration scheme:

$$ \frac{\Delta W}{\Delta t} = (\sigma:D)V $$

The tensor contraction can be re-written as $\sigma:D = tr(\sigma^TD)$. This is most easily observed if you work it out in index notation:

$$ \begin{align} \sigma:D &= \sigma_{ij}D_{ij} \\ &=(\sigma^T)_{ji}D_{ij} \\ &=(\sigma^T)_{ji}D_{ik}\delta_{kj} \\ &= (\sigma^TD)_{jk}\delta_{kj} \\ &=tr(\sigma^TD) \end{align} $$ where $\delta_{kj}$ is the Kronecker delta.

Due to the symmetry of the stress tensor, you don't really have to compute $D$ explicitly because $\sigma:D = \sigma:L$.

In short, the off-diagonal terms of the stress tensor do factor into the new energy, but it's just not obvious due to the way the formula is evaluating the stress power.

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