steinandre
Marine/Ocean
I have heard that if you use all tetrahedons in your mesh the construction will become much stiffer than in real life.... Is this true??
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Tetrahedons 3
- Thread starter steinandre
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Andrea Mino
Mechanical
It's true, if you compare a bricks and tetrahedons model with the same DOFS.
If you use parabolic tetrahedons model with elements much smaller than in a bricks model the response is different.
The use of tetrahedons are very usefull because you can use automatic mesh to generate the model. So you can try to generate models with different elements' size to analalize the mesh influence on stiffness.
Bye,
Minoand
If you use parabolic tetrahedons model with elements much smaller than in a bricks model the response is different.
The use of tetrahedons are very usefull because you can use automatic mesh to generate the model. So you can try to generate models with different elements' size to analalize the mesh influence on stiffness.
Bye,
Minoand
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steinandre
Marine/Ocean
First thanks for all the help! 
The reason I ask is tha I'm using DesignSpace 6 to test a underwater construction. But in DesignSpace there are no posibilities to change the elements in the mesh. It uses tet's as a default.....
The reason I ask is tha I'm using DesignSpace 6 to test a underwater construction. But in DesignSpace there are no posibilities to change the elements in the mesh. It uses tet's as a default.....
steinandre, you may better use ANSYS in such a case, because DesignSpace is designed, to my know, only to help the designer have a common feeling about s structure, so it is not too accurate.
To your fisrt question, i think, it is a general case, that the FEM-Models are stiffner then the real models (because of discretitations), i mean, that this is not only the tetrahedron case (think about shear locking phenomen). Normally, the higher the polinomial order, the less stiffner the model is. So the TET10 (parabolic) will approximate the displacement better then the TET4 (linear) and so does Quad9 to Quad4 (Shell). But, the calculation becomes more expensive also. Any comment?
cheers
To your fisrt question, i think, it is a general case, that the FEM-Models are stiffner then the real models (because of discretitations), i mean, that this is not only the tetrahedron case (think about shear locking phenomen). Normally, the higher the polinomial order, the less stiffner the model is. So the TET10 (parabolic) will approximate the displacement better then the TET4 (linear) and so does Quad9 to Quad4 (Shell). But, the calculation becomes more expensive also. Any comment?
cheers
All,
I would have hope that someone would have pointed to the reason why constant strain tetrahedral elements are "stiffer" than with bricks or in reality (which was the original question). I do not have ready access to the equations of the proof. If some one could post this it would benefit this thread greatly. I will look around for something as well. Best regards,
Matthew Ian Loew
I would have hope that someone would have pointed to the reason why constant strain tetrahedral elements are "stiffer" than with bricks or in reality (which was the original question). I do not have ready access to the equations of the proof. If some one could post this it would benefit this thread greatly. I will look around for something as well. Best regards,
Matthew Ian Loew
MLoew, i think that the reason why constant shear tetrahedral elements are "stiffer" than with bricks or in reality is that, because the bricks have more DOFs compare to tetrahedral and the "reality" have more (infinite!) DOFs then both FEM-Modelling technique, right? Rule of thumbs : The more DOFs the softer a model is (think about h-methodes). You can proove this by taking a look to the aproximation curve of a structure's FEM-solution to it's exact calculation : it aproximates the theoritical solution from the bottom up, i.e it is stiffer then the theoretical results.
cheers
cheers
Minoand is right about coparing DOFs. If the element math is right then both the constant strain tet and the higher order elements will converge to the "theoretically" correct answer - but it will take a lot more constant strain tets to get there.
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Zuardy's statement may be misleading (his intent may be correct, but as written it does not capture the reality). The issue of overly-stiff behavior in 1st order tetrahedrals is not just due to lower dofs (in fact that is not the problem with the elements). An equivalent-dof mesh of 1st order tets will still yield a far stiffer behavior than the hexa/pentahedrals will. This is due to formulational issues with the tetrahedrals. These are formulated as constant-strain elements, so they do not have within each element a strain gradient (as opposed to the hexahedrals, which do).
For both types of elements, a finer mesh will yield a more accurate stiffness response.
Second-order tetrahedrals are not constant-strain, so they do not have the problem which first-order tetrahedrals exhibit. As such, they converge very quickly (even with equivalent dofs to the first-order tets).
I recommend against using first-order tets for anything other than "filling-in" a mesh primarily composed of hexahedrals and some pentrahedrals.
Brad
For both types of elements, a finer mesh will yield a more accurate stiffness response.
Second-order tetrahedrals are not constant-strain, so they do not have the problem which first-order tetrahedrals exhibit. As such, they converge very quickly (even with equivalent dofs to the first-order tets).
I recommend against using first-order tets for anything other than "filling-in" a mesh primarily composed of hexahedrals and some pentrahedrals.
Brad
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