# Damping for Dynamic Problem using FEM

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!

• Do you have a reference for the piece of text that you show? Also, I think that $\eta=1$ might be small or big... we don't know right now, for that we need to compare it with your other parameters in the simulation. – nicoguaro Oct 13 '16 at 0:52
• As the quoted text implies, the extra term means that the stress tensor contains a viscous term relating stress to the rate of strain (actually here strictly the rate of compression). In some some senses this means the material is now behaving like a cross between a solid and a liquid. – origimbo Oct 13 '16 at 1:18
• This is just a toy rate-dependent model (probably created by the author) and cannot be used for damping (where damping is a decrease in strain energy over time under uniaxial tension or compression). To see why, just plot the response of the model at various strain rates. You will see differences between the plots if you use large enough values of the strain rate. – Biswajit Banerjee Oct 13 '16 at 1:21
• @nicoguaro I did not add the reference because I was afraid that the moderators would qualify this as FEniCS question. Other parmeters in the simulation were: E, nu, rho = 30000.0, 0.3, 5000.0 (where rho = mass density); lambda = E*nu/((1.0+nu)*(1.0-2.0*nu)); mu = E/(2.0*(1.0+nu)); The text is taken from this book. fenicsproject.org/pub/book/book/fenics-book-2011-06-14.pdf – CRG Oct 13 '16 at 1:48
• @origimbo You said that the material is behaving as a cross between solid and liquid. So conclusively, is this a reasonable form of damping in your opinion or not? – CRG Oct 13 '16 at 1:54

## 1 Answer

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.