Why Solid Elements like Tetrahedron has only 3 dofs per node, why not has 6dofs? what is the physical significance of that?

Why Solid Elements like Tetrahedron has only 3 dofs per node, why not has 6dofs? what is the physical significance of that?

Hi,

You can see in the SolidWorks Simulation. After solving a problem, In the simulation tree, right click and select "details". there it gives details of elements, nodes and dofs etc. from that we can calculate dofs/node by dividing total no. dofs / total no. nodes.

Is'nt this due to the fact, that this is the degrees of freddom left after Fixtures (restrainment)?

Where exactly due you right click to get this info?

In my opinion a tetrahydron will always have 6 DOFs, but restrained (fixed) the level of total will be less.

If you divide the DOFs with numbers of elements or nodes, it makes no sense - doesnt yield a whole number.

It simply notes how many DOFs there are on the model with restrainsBut you are right!!! A model with no restrains yields nodes * 3 = DOF !?!

I dont know why, but 16087 x 3 = 48261From Solidworks help file - it says 6

Maybe i will ask my VAR, what this is about

The tets solidworks simulation uses has 3DOF.

From the solidworks help files:

*For structural problems, each node in a solid element has three degrees of freedom that represent the translations in three orthogonal directions. The software uses the X, Y, and Z directions of the global Cartesian coordinate system in formulating the problem.**For thermal problems, each node has one degree of freedom which is the temperature.*The "degrees of freedom" quote that you have copied from the help files has nothing to do with FEM, it is talking about sketches.

You only need the 3 dofs to solve the PDEs you're typically trying to solve. With a bunch of completely reasonable and fairly universal assumptions this is perfectly fine to do. Such as the elements don't rotate too much relative to each other (in which case you tend to start looking into non-linear solution methods instead). In the case of stress it's sigma=E.e, where e is easily described without rotation.

Compare the above to the beam equation, which to be solved needs rotational information, and hence rotational DOFs end up falling into the descritised solution.

There are solid elements derived that have rotational degrees of freedom. I beleive they're used mostly for situations where those reasonable assumptions are broken and it becomes more efficient to have more DOFs per element than smaller elements and more time steps in the simulation. I believe, i've never used/seen/formulated them.

Hello,

Thanks for coming back.

Do you have any material to read and understand about how it is decided the dofs / node in solid elements?

I found this with some quick googling, it seems to condense the main stuff.

There's plenty of documentation inside the solidworks help files, but it's a bit hidden in my opinion, and also not the easiest to follow.

This is easily the best book on the subject I've ever used. I've read quite a lot of different books over the years, but this one introduces everything in baby steps, with really nice explanations of stuff that no one else bothers with. Examples of this are when assumptions are made, it actually goes into detail, with worked out examples, of what happens if you make different assumptions. I've never seen anyone bother to do that in other books before, which always just leaves me feeling like my picture of all this is incomplete.

https://www.amazon.com/Introduction-Finite-Element-Method-Ottosen/dp/0134738772

Pricey though...

Hi,

in FEA, a node in solid (volume) mesh have 3 DOF (3 translations), a node in shell (surface mesh) have 6 DOF ( 3 translations + 3 rotations)

volume : 2018 SOLIDWORKS Help - Solid Mesh

Where do you see this fact?

It should be 3 rotational and 3 transitional - 6 in all, typically.