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Symmetry boundary condition in eigenfrequency study
Posted May 4, 2015, 10:10 a.m. EDT Acoustics & Vibrations, Geometry, Parameters, Variables, & Functions Version 4.4 1 Reply
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Hi,
I want to use the symmetry boundary condition for a heavy model in 3D to reduce computation time. Before to use it, i just want to validate the efficiency of the symmetry boundary condition with a really simple model: a rectangular box.
SO i solve the problem in the full case (the symmetry is not used and the geometry is full), and after a made an other simulation with the half of the geometry and the symmetry boundary condition .
When i compare the results for the eigenfrequencies, don't find the same things. In fact when i use the model with the half geometry and symmetry, i find the sames eigenfrequencies with the same model without the symmetry boundary condition. All this indicate that the symmetry boundary condition is not working.
So can you explain to me what i'm doing wrong?
Thanks a lot.
Find here in this post the (simple) models.
JP
I want to use the symmetry boundary condition for a heavy model in 3D to reduce computation time. Before to use it, i just want to validate the efficiency of the symmetry boundary condition with a really simple model: a rectangular box.
SO i solve the problem in the full case (the symmetry is not used and the geometry is full), and after a made an other simulation with the half of the geometry and the symmetry boundary condition .
When i compare the results for the eigenfrequencies, don't find the same things. In fact when i use the model with the half geometry and symmetry, i find the sames eigenfrequencies with the same model without the symmetry boundary condition. All this indicate that the symmetry boundary condition is not working.
So can you explain to me what i'm doing wrong?
Thanks a lot.
Find here in this post the (simple) models.
JP
1 Reply Last Post May 5, 2015, 12:19 p.m. EDT
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Posted:
10 years ago
Hi,
You are actually not doing anything wrong. The 'Symmetry' boundary condition is mathematically the same as the 'Sound hard' boundary condition. The latter is the default boundary condition, so it is applied when you remove the 'Symmetry' node.
The reason for having this duplicate, is that it is easier to see what a model is intended to do with a specific symmetry condition.
With a half model and a symmetry condition, you will however only get the symmetric modes, and not the antisymmetric ones. You can use the 'Sound soft' boundary condition to compute the antisymmetric modes.
Regards,
Henrik
You are actually not doing anything wrong. The 'Symmetry' boundary condition is mathematically the same as the 'Sound hard' boundary condition. The latter is the default boundary condition, so it is applied when you remove the 'Symmetry' node.
The reason for having this duplicate, is that it is easier to see what a model is intended to do with a specific symmetry condition.
With a half model and a symmetry condition, you will however only get the symmetric modes, and not the antisymmetric ones. You can use the 'Sound soft' boundary condition to compute the antisymmetric modes.
Regards,
Henrik