Damping for Dynamic Problem using FEM

Question

I came across this form of damping implemented in an elastodynamics problem.

The stress tensor without the damping would look like:

$ \sigma = 2 \mu \epsilon + (\lambda \, \text{tr} (\epsilon)) I $

where, $\mu$ and $\lambda$ are Lame' parameters dependent on the Young's Modulus and the Poisson's ratio. Trace is represented by '$\text{tr}$', and $\epsilon = (\nabla u + {(\nabla u)}^T)$ is the strain tensor. Here, $I$ is the second order identity tensor.

I have the following questions:

1. What is this type of damping called?
2. Does $\eta$ control the damping %? For eg. should $\eta = 0.1$ mean 10% damping? I don't think it works like that though.
3. Any other comments on this type of damping?

I ran some wave propagation simulations based on this type of damping and observed the following in Paraview for $\eta=0$ and $\eta=1$. I don't see substantial difference in the displacement values. It is 17% for the highlighted point. Any observations/comments from experts will be useful. Please click on the picture to enlarge it.

Thank you!

Answer 1

The constitutive model is $$ \boldsymbol{\sigma} = 2\mu \boldsymbol{\varepsilon} + \lambda \text{tr}(\boldsymbol{\varepsilon})\,\mathbf{I} + \eta\, \text{tr}(\dot{\boldsymbol{\varepsilon}})\,\mathbf{I} \,. $$ Consider a hydrostatic creep test where $\boldsymbol{\sigma}$ is kept fixed. We want to find the evolution of $\boldsymbol{\varepsilon}$ under that condition.

We have $$ \text{tr}[ \eta\, \text{tr}(\dot{\boldsymbol{\varepsilon}})\,\mathbf{I}] = \text{tr}[\boldsymbol{\sigma} - 2\mu \boldsymbol{\varepsilon} - \lambda \text{tr}(\boldsymbol{\varepsilon})\,\mathbf{I}] $$ or $$ 3\eta\, \text{tr}(\dot{\boldsymbol{\varepsilon}}) = \text{tr}(\boldsymbol{\sigma}) - 2\mu\,\text{tr}(\boldsymbol{\varepsilon}) - 3\lambda\, \text{tr}(\boldsymbol{\varepsilon})\,. $$ Define $\varepsilon_v := \text{tr}(\boldsymbol{\varepsilon})$ and $\sigma_m := \text{tr}(\boldsymbol{\sigma})$. Then, $$ \dot{\varepsilon_v} = \frac{\sigma_m}{\eta} - \frac{2\mu + 3\lambda}{3\eta}\,\varepsilon_v = \frac{\sigma_m}{\eta} - \frac{K}{\eta} \,\varepsilon_v \,. $$ Solving this ODE gives $$ \varepsilon_v(t) = \frac{\sigma_m}{K} + C \exp\left(-\frac{K t}{\eta}\right) $$ where $C$ is determined from the initial conditions. For zero initial strain (is this physically possible in this case?), $$ \varepsilon_v(t) = \frac{\sigma_m}{K}\left[1 - \exp\left(-\frac{K t}{\eta}\right)\right] \,. $$ This is a creep model. But the effect of the value of $\eta$ will not be observed except at small values of $t$ and large values ($O(K)$) of $\eta$. This why this is not a good model for damping.

However, you can interpret the parameter $\eta$ as a damping parameter. To see why you have to repeat the above exercise with $$ \sigma_m = \sigma_m \exp(-i \omega t) $$ and look at the imaginary part of the strain.

For the deviatoric stresses/strains, the model has no viscoelastic effect.