Apply outward radial pressure of same magnitude on a inner 2D ring (new)

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Hi I just would like to ask you a question in Comsol simulation, as I tried many times, but still get the same problem over this weekend. The objective was to obtain the relation between the pressure and the diameter of a 2D ring structure.

Firstly, I create the 2D ring as attached. There are two circles (inner and outer) and each circle is made up of four curves which can be generated by selecting 'circle' under geometry.

Then, I select 'pressure' under the boundary load and apply the pressure of the same magnitude of 5Pa to each inner curve. Then, I always get the same simulation issues as attached.

Then, I attempted to address the problem. Initially, I was thinking there could be an error in the direction of pressure and hence I tried to apply a fixed boundary condition to one edge of curve and apply 5 Pa to another edge as attached. Then, from the simulation result as attached, I realized the positive pressure direction is directing radially outward, so there is no problem in the direction of pressure that I have chosen previously. However, I also realized that the reason that the simulation issue occurred is there was no fixing boundary condition as I discussed with you last meeting.

I will very appreciate if you can give me the correct solution for that

Thanks for your time Best wishes Junwei



6 Replies Last Post Oct 21, 2020, 7:23 PM EDT
Edgar J. Kaiser Certified Consultant

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Posted: 1 month ago Oct 19, 2020, 5:30 AM EDT

Junwei,

you can apply prescribed displacements to constrain the rigid body motion that breaks the solver. Apply u0x = 0 to the top outside point and u0y = 0 to the right outside point. Let the other checkboxes unchecked, so displacement remains free in that directions. This should sufficiently constrain the model and it will not affect the results because due to the symmetry of the model no tangential displacments can be expected. Reducing the model to a quarter ring and applying symmetry BCs (under 'More constraints') should do the same.

Cheers Edgar

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Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
Junwei, you can apply prescribed displacements to constrain the rigid body motion that breaks the solver. Apply u0x = 0 to the top outside point and u0y = 0 to the right outside point. Let the other checkboxes unchecked, so displacement remains free in that directions. This should sufficiently constrain the model and it will not affect the results because due to the symmetry of the model no tangential displacments can be expected. Reducing the model to a quarter ring and applying symmetry BCs (under 'More constraints') should do the same. Cheers Edgar

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Posted: 1 month ago Oct 19, 2020, 6:56 AM EDT

Hi Edgar Thanks so much for that response. I tried to do that, but I found it unable to select any point when using the prediscerbed displacement. Any advice on that? Best wishes Junwei

Hi Edgar Thanks so much for that response. I tried to do that, but I found it unable to select any point when using the prediscerbed displacement. Any advice on that? Best wishes Junwei

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Posted: 1 month ago Oct 19, 2020, 7:08 AM EDT

Junwei,

you can apply prescribed displacements to constrain the rigid body motion that breaks the solver. Apply u0x = 0 to the top outside point and u0y = 0 to the right outside point. Let the other checkboxes unchecked, so displacement remains free in that directions. This should sufficiently constrain the model and it will not affect the results because due to the symmetry of the model no tangential displacments can be expected. Reducing the model to a quarter ring and applying symmetry BCs (under 'More constraints') should do the same.

Cheers Edgar Thanks so much for that. Actually, I found its not enough to only apply the prediscribed displacement boundary condition on the top outside point and right outside point, but have to apply that to the bottom outside point and left outside point as well. Thanks for the basic idea and now I solved the problem!

>Junwei, > >you can apply prescribed displacements to constrain the rigid body motion that breaks the solver. >Apply u0x = 0 to the top outside point and u0y = 0 to the right outside point. Let the other checkboxes unchecked, so displacement remains free in that directions. This should sufficiently constrain the model and it will not affect the results because due to the symmetry of the model no tangential displacments can be expected. >Reducing the model to a quarter ring and applying symmetry BCs (under 'More constraints') should do the same. > >Cheers >Edgar Thanks so much for that. Actually, I found its not enough to only apply the prediscribed displacement boundary condition on the top outside point and right outside point, but have to apply that to the bottom outside point and left outside point as well. Thanks for the basic idea and now I solved the problem!

Henrik Sönnerlind COMSOL Employee

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Posted: 1 month ago Oct 19, 2020, 7:21 AM EDT

The easiest way to get sufficient constraints is to add a Rigid Motion Suppression node from Domain Constraints.

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Henrik Sönnerlind
COMSOL
The easiest way to get sufficient constraints is to add a *Rigid Motion Suppression* node from *Domain Constraints*.

Edgar J. Kaiser Certified Consultant

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Posted: 1 month ago Oct 19, 2020, 7:27 AM EDT

@Junwei: you find that under the 'points' menu

@Henrik: I sometimes found that the Rigid Motion Suppression applied unfavourable constraints that affected the results more than needed. But maybe that has improved meanwhile.

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Edgar J. Kaiser
emPhys Physical Technology
www.emphys.com
@Junwei: you find that under the 'points' menu @Henrik: I sometimes found that the Rigid Motion Suppression applied unfavourable constraints that affected the results more than needed. But maybe that has improved meanwhile.

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Posted: 1 month ago Oct 21, 2020, 7:23 PM EDT

Hi Frank, From the images you attached, it seems like you have a symmetric problem here. So how about you consider a quarter of the geometry (you can put an 90 degree angle instead of 360 in your Circle definition)? Then you can add symmetry boundary conditions on the edges that form on the x and y axis. You reduce your problem size and you get the stability.

Cheers, Temesgen

Hi Frank, From the images you attached, it seems like you have a symmetric problem here. So how about you consider a quarter of the geometry (you can put an 90 degree angle instead of 360 in your Circle definition)? Then you can add symmetry boundary conditions on the edges that form on the x and y axis. You reduce your problem size and you get the stability. Cheers, Temesgen

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