Edge as a rigid domain in contact mechanics

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In 2d, one can use a real 2d rigid domain or an edge (just a line) to represent a rigid domain. Which is better?


5 Replies Last Post 2024年4月2日 GMT+2 08:42
Henrik Sönnerlind COMSOL Employee

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Posted: 5 months ago 2024年3月20日 GMT+1 13:54

A rigid domain is a material model that you select for a domain. If you want to achive the same thing using a rigid connector, you would have to select all boundaries around the domain (geometrically lines in 2D). In that case, you would still solve for all DOFs inside the domain, unless you remove it from the selection of the physics interface.

In almost all cases, using a rigid domain is better. You will, for example, not have an empty hole in your plots.

If you just want to make a certain boundary rigid, then it is a rigid connector that you should use. But this is a different representation of your physics than if you are using a rigid domain.

Either representation can be used for contact problems, depending on what you want to achieve.

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Henrik Sönnerlind
COMSOL
A rigid domain is a material model that you select for a domain. If you want to achive the same thing using a rigid connector, you would have to select all boundaries around the domain (geometrically lines in 2D). In that case, you would still solve for all DOFs inside the domain, unless you remove it from the selection of the physics interface. In almost all cases, using a rigid domain is better. You will, for example, not have an empty hole in your plots. If you just want to make a certain boundary rigid, then it is a rigid connector that you should use. But this is a different representation of your physics than if you are using a rigid domain. Either representation can be used for contact problems, depending on what you want to achieve.

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Posted: 5 months ago 2024年3月26日 GMT+1 06:32

Thank you very much Dr Sönnerlind. In the attached figure the long straight line representing the ground is what I refer to as the edge or line and what I wish to be rigid. Then I tried to add your rigid connector but I am not allowed to choose that straight line as the rigid connector, yet instead I can only choose the bottom line of the square. If the bottom line of the square isn't rigid, may I not use a rigid connector any more?

Thank you very much Dr Sönnerlind. In the attached figure the long straight line representing the ground is what I refer to as the edge or line and what I wish to be rigid. Then I tried to add your rigid connector but I am not allowed to choose that straight line as the rigid connector, yet instead I can only choose the bottom line of the square. If the bottom line of the square isn't rigid, may I not use a rigid connector any more?


Henrik Sönnerlind COMSOL Employee

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Posted: 5 months ago 2024年3月27日 GMT+1 13:36

As I understand it, you want to simulate contact against a plane that is not really a part of your structural mechanics problem.

A quick (but not elegant) solution is to add a thin strip of material also below the long line, and proceed as for any contact problem.

But you can also just draw the line and mesh it (1 element is enough if it is straight), without having any physics attached. You can then use that line as source side in a contact pair. Note however that when the line does not have an adjacent domain, it does not have a natural 'outwards normal'. So for the contact search to work, the line must be drawn with the correct orientation (you want to have the solid on the left side of the line).

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Henrik Sönnerlind
COMSOL
As I understand it, you want to simulate contact against a plane that is not really a part of your structural mechanics problem. A quick (but not elegant) solution is to add a thin strip of material also below the long line, and proceed as for any contact problem. But you can also just draw the line and mesh it (1 element is enough if it is straight), without having any physics attached. You can then use that line as source side in a contact pair. Note however that when the line does not have an adjacent domain, it does not have a natural 'outwards normal'. So for the contact search to work, the line must be drawn with the correct orientation (you want to have the solid on the left side of the line).

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Posted: 5 months ago 2024年3月31日 GMT+2 07:37

But you can also just draw the line and mesh it (1 element is enough if it is straight), without having any physics attached. You can then use that line as source side in a contact pair. Note however that when the line does not have an adjacent domain, it does not have a natural 'outwards normal'. So for the contact search to work, the line must be drawn with the correct orientation (you want to have the solid on the left side of the line).

Thank you very much Dr Sönnerlind. By drawing a line, you don't mean using a rigid connector any more do you? Then I think a rigid connector can be used only when the bottom line of the square is rigid.

>But you can also just draw the line and mesh it (1 element is enough if it is straight), without having any physics attached. You can then use that line as source side in a contact pair. Note however that when the line does not have an adjacent domain, it does not have a natural 'outwards normal'. So for the contact search to work, the line must be drawn with the correct orientation (you want to have the solid on the left side of the line). Thank you very much Dr Sönnerlind. By drawing a line, you don't mean using a rigid connector any more do you? Then I think a rigid connector can be used only when the bottom line of the square is rigid.

Henrik Sönnerlind COMSOL Employee

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Posted: 5 months ago 2024年4月2日 GMT+2 08:42

That's correct. If there is no physics at all attached to the line, it will not move.

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Henrik Sönnerlind
COMSOL
That's correct. If there is no physics at all attached to the line, it will not move.

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