Adding integrated material qois (dissipation for HyperVisco)

[ralberd]

I quickly added this, but perhaps there are better ways to get these quantities out.

• Added a uniaxial material point test comparing the computed dissipated energy to the difference between the internal work and free energy.
• If you plot the energy vs time (in the test), the energy at the end of the unloading cycle is almost all dissipated. Does this make physical sense?

[btalamini]

This looks great. Thanks for putting it all together! I found one potential bug; see my comments. It's easy to fix.

[btalamini]

I think this is a bug. The default values for namedtuple are applied from the right, so I think this will now default the value for compute_material_qoi to 0.0 instead of density. I think you should put compute_material_qoi before density?

[cmhamel]

Don't merge just yet... i"m working on adding a uniaxial strain solution this morning

[cmhamel]

We're good on the analytic solution see attached file for the viscous logarithmic strain vs. time. This is using a relaxation time of 10.0s and total loading time of 100.0s. Dots are the optimism solution at a material point and the line is the analytic solution.

[cmhamel]

Also good in terms of the elastic logarithmic strain for this analytic problem

[cmhamel]

Looking good on dissipation as well

[cmhamel]

Here the total deformation graident we will write as
which eminates from the deformation gradient

$$\mathbf{F} = \begin{bmatrix} \exp\left(\dot{\varepsilon}t\right) & 0 & 0 \\\ 0 & 1 & 0 \\\ 0 & 0 & 1 \end{bmatrix}$$

The kinematics are given by $\mathbf{F}^e\mathbf{F}^v$ where we assumed that $\det\mathbf{F}^v = 1$ are put in terms of principal values $\det\mathbf{F}^v = \lambda^v_1\lambda^v_2\lambda^v_3 = 1$ where due to symmetry we can assumed that $\lambda^v_2 = \lambda^v_3$ to show that we can write $\mathbf{F}^v$ for this simple motion as follows

$$\mathbf{F}^v = \begin{bmatrix} \lambda^v & 0 & 0 \\\ 0 & \frac{1}{\sqrt{\lambda^v}} & 0 \\\ 0 & 0 & \frac{1}{\sqrt{\lambda^v}} \end{bmatrix}$$

The rate can be written for convenience as

$$\dot{\mathbf{F}}^v = \begin{bmatrix} \dot{\lambda}^v & 0 & 0 \\\ 0 & -\frac{\dot{\lambda}^v}{2\left(\lambda^v\right)^{3/2}} & 0 \\\ 0 & 0 & -\frac{\dot{\lambda}^v}{2\left(\lambda^v\right)^{3/2}} \end{bmatrix}$$

Now calculate $\mathbf{F}^e$ from the above

$$\mathbf{F}^e = \begin{bmatrix} \frac{\exp\left(\dot{\varepsilon}t\right)}{\lambda^v} & 0 & 0 \\\ 0 & \sqrt{\lambda^v} & 0 \\\ 0 & 0 & \sqrt{\lambda^v} \end{bmatrix}$$

Since this motion is rotation free, we can readily write $\mathbf{E}^e = \ln\mathbf{U}^e = \ln\mathbf{F}^e$ where

$$\mathbf{E}^e = \begin{bmatrix} \dot{\varepsilon}t - \ln\lambda^v & 0 & 0 \\\ 0 & \frac{1}{2}\ln\lambda^v & 0 \\\ 0 & 0 & \frac{1}{2}\ln\lambda^v \end{bmatrix}$$

with $\tr\mathbf{E} = \ln\left(\dot{\lambda}t\right)$ so that

$$\dev\mathbf{E}^e = \begin{bmatrix} \frac{2}{3}\dot{\varepsilon}t - \ln\lambda^v & 0 & 0 \\\ 0 & \frac{1}{2}\ln\lambda^v - \frac{1}{3}\dot{\varepsilon}t & 0 \\\ 0 & 0 & \frac{1}{2}\ln\lambda^v - \frac{1}{3}\dot{\varepsilon}t \end{bmatrix}$$

Need to assemble and solve $\dot{\mathbf{F}}^v = \mathbf{D}^v\mathbf{F}^v$ with $\mathbf{D}^v = \frac{1}{\tau}\dev\mathbf{E}^e$

This gives the following equation

$$\dot{\lambda}^v = \left[\frac{2}{3\tau}\dot{\varepsilon}t - \frac{1}{\tau}\ln\lambda^v\right]\lambda^v$$ $$\frac{\dot{\lambda}^v}{\lambda^v} + \frac{1}{\tau}\ln\lambda^v = \frac{2}{3\tau}\dot{\varepsilon}t$$

Set $\varepsilon^v = \ln\lambda^v$ to rewrite the previous equation as

$$\dot{\varepsilon}^v + \frac{1}{\tau}\varepsilon^v = \frac{2}{3\tau}\dot{\varepsilon}t$$

Choose an integrating factor as follows

$$I\left(t\right) = \exp\left(t/\tau\right)$$ $$\exp\left(t/\tau\right)\dot{\varepsilon}^v + \frac{1}{\tau}\exp\left(t/\tau\right)\varepsilon^v = \exp\left(t/\tau\right)\frac{2}{3\tau}\dot{\varepsilon}t$$ $$\varepsilon^v = \frac{2\dot{\varepsilon}}{3\tau}\exp\left(-t/\tau\right)\left[\int_0^t t\exp\left(t/\tau\right) dt\right]$$ $$\varepsilon^v = \frac{2\dot{\varepsilon}}{3\tau}\exp\left(-t/\tau\right)\left[\tau\exp\left(t/\tau\right)\left(t - \tau\right) - \tau\left(0 - \tau\right)\right]$$ $$\varepsilon^v = \frac{2\dot{\varepsilon}}{3}\left[\left(t - \tau\right) + \tau\exp\left(-t/\tau\right)\right]$$ $$\varepsilon^v = \frac{2\dot{\varepsilon}t}{3} - \frac{2\dot{\varepsilon}\tau}{3}\left[1 - \exp\left(-t/\tau\right)\right]$$

If loading stops at $t = t_L$ then $\dot{\varepsilon} = 0$ and we then get the following equation instead

$$\varepsilon^v = \frac{2}{3\tau}\exp\left(-t/\tau\right)\left[\int_0^{t_L} \exp\left(t/\tau\right)\dot{\varepsilon}t dt + \int_{t_L}^t \exp\left(t/\tau\right)\dot{\varepsilon}t_L dt\right]$$ $$\varepsilon^v = \frac{2}{3\tau}\exp\left(-t/\tau\right)\left[\dot{\varepsilon}\int_0^{t_L} t\exp\left(t/\tau\right) dt + \dot{\varepsilon}t_L\int_{t_L}^t \exp\left(t/\tau\right) dt\right]$$ $$\varepsilon^v = \frac{2}{3\tau}\exp\left(-t/\tau\right)\left[\dot{\varepsilon}\tau\exp\left(t_L/\tau\right)\left(t_L - \tau\right) - \dot{\varepsilon}\tau\left(0 - \tau\right) + \dot{\varepsilon}t_L\tau\exp\left(t/\tau\right) - \dot{\varepsilon}t_L\tau\exp\left(t_L/\tau\right)\right]$$

[cmhamel]

@ralberd let me know when you're done with this guy. I'll give it a review and smash the merge button.