Stress Equation and Implicit part

Tobi · July 12, 2016, 09:39

Dear all,

at the moment I am focusing at the stress equation of the solidDisplacementFoam (again) but now I want to write the equations down for future. There are some question marks that came up, if we calculate the stress equation. Starting with the general stuff:

$\frac{\partial^2 \textbf{D}}{\partial t^2} = \nabla \bullet \boldsymbol{\sigma}$

Where D is the displacement vector and sigma the Cauchy stress tensor defined as:

$\boldsymbol{\sigma} = 2\mu \boldsymbol \epsilon + \lambda \mathrm{tr} (\boldsymbol\epsilon) \textbf{I}$

And epsilon is the strain tensor defined as:

$\boldsymbol \epsilon = \frac{1}{2}(\nabla \otimes \textbf{D} + (\nabla \otimes \textbf{D})^T)$

It is clear that we have to solve that equation:

$\frac{\partial^2 \textbf{D}}{\partial t^2} = \nabla \bullet [ 2 \mu (\frac{1}{2} (\nabla \otimes \textbf{D}) + (\nabla \otimes \textbf{D})^T) + \lambda \mathrm{tr}(\nabla \otimes \textbf{D}) \textbf{I}]$

The rhs is given by divSigmaExp in the code. After we have the field divSigmaExp, we subtract a part (I think for numerical stabilisation, thanks to Alex):

$\nabla \bullet [(2 \mu + \lambda ) \nabla \otimes \textbf{D}]$

And here is the problem. Till now I do not get the point why we subtract this kind of term. Even this term is treated different in the numerical point of view (compact stress switch???). I can not figure out why it is (2mu + lambda) grad (D). As far as I derived it, we should end up with mu only:

$\nabla \bullet \boldsymbol \sigma = \nabla \bullet (\mu \nabla \otimes \textbf{D}) + \nabla \bullet (\mu (\nabla \otimes \textbf{D})^T) + \lambda \mathrm{tr}(\nabla \otimes \textbf{D}) \textbf{I}$

As we can see, the first part is implicit that can be removed from the explicit part but it would be:

$\nabla \bullet (\mu \nabla \otimes \textbf{D})$

and not

$\nabla \bullet ([2\mu + \lambda]\nabla \otimes \textbf{D})$

The only way that I found is that we add and subtract the term

$\nabla \bullet ([\mu + \lambda] \nabla \otimes \textbf{D}) - \nabla \bullet ([\mu + \lambda] \nabla \otimes \textbf{D})$

Then I would get the term that I can not derive till now but then the negative term is missing in my equations somehow.

Any suggestion would be appreciated.
Thanks in advance,

Tobi

Tobi · July 13, 2016, 07:25

Hi all,

my last assumption was correct. The resource is here: https://www.researchgate.net/publica...ear_Elasticity

July 12, 2016, 09:39	Stress Equation and Implicit part	#1
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Bad Wörishofen Posts: 2,711 Blog Entries: 6 Rep Power: 52	Dear all, at the moment I am focusing at the stress equation of the solidDisplacementFoam (again) but now I want to write the equations down for future. There are some question marks that came up, if we calculate the stress equation. Starting with the general stuff: $\frac{\partial^2 \textbf{D}}{\partial t^2} = \nabla \bullet \boldsymbol{\sigma}$ Where D is the displacement vector and sigma the Cauchy stress tensor defined as: $\boldsymbol{\sigma} = 2\mu \boldsymbol \epsilon + \lambda \mathrm{tr} (\boldsymbol\epsilon) \textbf{I}$ And epsilon is the strain tensor defined as: $\boldsymbol \epsilon = \frac{1}{2}(\nabla \otimes \textbf{D} + (\nabla \otimes \textbf{D})^T)$ It is clear that we have to solve that equation: $\frac{\partial^2 \textbf{D}}{\partial t^2} = \nabla \bullet [ 2 \mu (\frac{1}{2} (\nabla \otimes \textbf{D}) + (\nabla \otimes \textbf{D})^T) + \lambda \mathrm{tr}(\nabla \otimes \textbf{D}) \textbf{I}]$ The rhs is given by divSigmaExp in the code. After we have the field divSigmaExp, we subtract a part (I think for numerical stabilisation, thanks to Alex): $\nabla \bullet [(2 \mu + \lambda ) \nabla \otimes \textbf{D}]$ And here is the problem. Till now I do not get the point why we subtract this kind of term. Even this term is treated different in the numerical point of view (compact stress switch???). I can not figure out why it is (2mu + lambda) grad (D). As far as I derived it, we should end up with mu only: $\nabla \bullet \boldsymbol \sigma = \nabla \bullet (\mu \nabla \otimes \textbf{D}) + \nabla \bullet (\mu (\nabla \otimes \textbf{D})^T) + \lambda \mathrm{tr}(\nabla \otimes \textbf{D}) \textbf{I}$ As we can see, the first part is implicit that can be removed from the explicit part but it would be: $\nabla \bullet (\mu \nabla \otimes \textbf{D})$ and not $\nabla \bullet ([2\mu + \lambda]\nabla \otimes \textbf{D})$ The only way that I found is that we add and subtract the term $\nabla \bullet ([\mu + \lambda] \nabla \otimes \textbf{D}) - \nabla \bullet ([\mu + \lambda] \nabla \otimes \textbf{D})$ Then I would get the term that I can not derive till now but then the negative term is missing in my equations somehow. Any suggestion would be appreciated. Thanks in advance, Tobi __________________ Keep foaming, Tobias Holzmann

July 13, 2016, 07:25		#2
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Bad Wörishofen Posts: 2,711 Blog Entries: 6 Rep Power: 52	Hi all, my last assumption was correct. The resource is here: https://www.researchgate.net/publica...ear_Elasticity __________________ Keep foaming, Tobias Holzmann

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