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Override and Contribution (boundaries) - Solid Mechanics

Posted Feb 10, 2015, 11:34 a.m. EST MEMS & Nanotechnology, RF & Microwave Engineering, MEMS & Piezoelectric Devices, Modeling Tools & Definitions, Parameters, Variables, & Functions Version 5.0 5 Replies

Manuel Soeiro
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Hi! I just started my first project on COMSOL! pretty excited!
I have a problem, that I can't seem to solve.

With Solid mechanics physics, I can't make the default Free 1 with an added boundary mass to contribute in the same boundary. It puts the mass to override by default with no option on changing it.

The only solution I can think of is to add a Free 2 boundary, which will overrite the default Free 1 with a All boundaries selection, and then contribute with the added mass.

So my question is: is it impossible to do it with the default Free 1?
5 Replies Last Post Jan 28, 2016, 1:54 a.m. EST
Henrik Sönnerlind COMSOL Employee

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Posted: 9 years ago Feb 11, 2015, 3:45 a.m. EST
Hi,

The override and contribute rules are rather intricate, but I will try to give a quick explanation:

1, The default 'Free' node is actually not a boundary condition at all. It does not do anything, it just acts as a container for all boundaries which do not have any other boundary condition. So it is more of a display convenience.

2. When you add any condition (load, constraint, added mass, spring, ...) to a boundary it will be considered as overriding the default 'Free' node.

3. Manually adding another 'Free' node in the model tree will remove (by overriding) almost all preceding conditions on the selected boundaries. This applies to load and constraints, for example, but *NOT* Added Mass. The reason for this behavior is that added mass is considered more as a 'material property' than as a 'boundary condition' even though it could be argued that it is just another force contribution.

In practice, you will very seldom have to worry about the override rules. The interaction between several boundary conditions on the same geometrical object almost always does what you intuitively expect them to do.

Regards,
Henrik
Hi, The override and contribute rules are rather intricate, but I will try to give a quick explanation: 1, The default 'Free' node is actually not a boundary condition at all. It does not do anything, it just acts as a container for all boundaries which do not have any other boundary condition. So it is more of a display convenience. 2. When you add any condition (load, constraint, added mass, spring, ...) to a boundary it will be considered as overriding the default 'Free' node. 3. Manually adding another 'Free' node in the model tree will remove (by overriding) almost all preceding conditions on the selected boundaries. This applies to load and constraints, for example, but *NOT* Added Mass. The reason for this behavior is that added mass is considered more as a 'material property' than as a 'boundary condition' even though it could be argued that it is just another force contribution. In practice, you will very seldom have to worry about the override rules. The interaction between several boundary conditions on the same geometrical object almost always does what you intuitively expect them to do. Regards, Henrik
Manuel Soeiro
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Posted: 9 years ago Feb 11, 2015, 5:29 a.m. EST
Hi again,

I understood now. For my problem at hand I will add manually a Free node then.
Thank you for you time.

My best regards
Hi again, I understood now. For my problem at hand I will add manually a Free node then. Thank you for you time. My best regards
Henrik Sönnerlind COMSOL Employee

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Posted: 9 years ago Feb 11, 2015, 5:49 a.m. EST
Hi,

Could you explain what you want to achieve with the 'Free' node? I have essentially never seen a case where it is needed to add one.

Regards,
Henrik
Hi, Could you explain what you want to achieve with the 'Free' node? I have essentially never seen a case where it is needed to add one. Regards, Henrik
Manuel Soeiro
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Posted: 9 years ago Feb 12, 2015, 8:56 a.m. EST
Well, I have a cantilever with a tip mass at one end. And I'm actually transferring an old COMSOL component (piezoelectric devices) to the new multiphysics piezoelectric effect component, (which includes the solid mechanics).
The block in question has the added mass contributing with the free node.
Maybe it doesn't matter if it's contributing or overriding, since it's the node to define no loads or or constraints, however I wanted to transfer exactly like it is on the old component version.

Thank you for the quick responses and all the help given.

Best regards,
Manuel Soeiro
Well, I have a cantilever with a tip mass at one end. And I'm actually transferring an old COMSOL component (piezoelectric devices) to the new multiphysics piezoelectric effect component, (which includes the solid mechanics). The block in question has the added mass contributing with the free node. Maybe it doesn't matter if it's contributing or overriding, since it's the node to define no loads or or constraints, however I wanted to transfer exactly like it is on the old component version. Thank you for the quick responses and all the help given. Best regards, Manuel Soeiro
Yasaman Kaiasat
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Posted: 8 years ago Jan 28, 2016, 1:54 a.m. EST
hi Henry,

I am simulating a unit cell of a periodic structure. I have two ports to measure , S11,S12,S21 ,and S22. As I have a Chiral structure I would like to investigate the rotation of the polarization of the incident field in the reflected spectrum. This means that I have to introduce the third port located exactly where port 1 is with different polarization to measure S13. I successfully did it but, when I make the polarization of port 1 and port 3 identical I expect to have S11 = S13 which surprisingly is not the case.
I appreciate your advise,


Kind Regards,
Yasaman
hi Henry, I am simulating a unit cell of a periodic structure. I have two ports to measure , S11,S12,S21 ,and S22. As I have a Chiral structure I would like to investigate the rotation of the polarization of the incident field in the reflected spectrum. This means that I have to introduce the third port located exactly where port 1 is with different polarization to measure S13. I successfully did it but, when I make the polarization of port 1 and port 3 identical I expect to have S11 = S13 which surprisingly is not the case. I appreciate your advise, Kind Regards, Yasaman

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