Modeling a thin plate elastic leaf spring in Elmer

[foadsf]

I had another attempt trying to use the third tutorial on this PDF. You may see the .geo and the .sif file in the attachments. Opening the results in ParaView



I get a symmetric deflection, which is an expected, and a maximum deflection of approximately 1.8mm which also is not that far away from the theory. One thing that I do not understand, and that might have been the core cause of trouble today, is that ParaView is confusing the X direction and the Z direction somehow. it truly shows Z as perpendicular to the spring's plane but considers X as the deflection in that direction. Am I making a mistake here?!

[raback]

Hi

Paraview just brutally assumes that it is a normal displacement vector. In fact it is not. It is a vector whose 1st component is deflection in the normal direction (i.e. z) and the 2nd component is its derivative in x-direction and 3rd component is its derivative in y-direction.

-Peter

[foadsf]

Thanks, Peter. So I assume my 2D simulation is not far wrong.

The next step is to run the same simulation except this the time the force is planar. for example 10N in X or Y direction (Gmsh .geo coordinate system). However, I could not find any option other than Pressure. Any idea how that should be done?

[foadsf]

As a follow up I tried to project the second example on this PDF tutorial, on the 3D version of my geometry. However, the max vertical displacement is six orders of magnitude wrong.



I am not sure where my mistakes are.

[kevinarden]

coherence is under the geometry commands, or you can just put coherence as the last line in your geo file

coherence.png (36.55 KiB) Viewed 2883 times

I will start looking at your other posts.

[foadsf]

I think the main reason why the 3D model doesn't work, is that I do not understand what a "Force" boundary condition is. If I want a total force of 10N to be applied in the middle part of the geometry, should I divide it by the surface area? Then it is actually pressure 10 / (pi * 5e-3^2) = 127324Pa.

[foadsf]

I used the

Code: Select all

Element = p:2

in the solver section as Peter has suggested on several forum posts. Though it horribly slows things down. And I had to have a cheaper preconditioner:

Code: Select all

Linear System Preconditioning = ILU0 

to resolve the error:

Code: Select all

ERROR:: CRS_IncompleteLU: Number of nonzeros larger than HUGE(Integer)
ERROR:: CRS_IncompleteLU: Try some cheaper preconditioner! 

however, the result is still three orders of magnitude smaller than what is expected.

[raback]

Hi

Any real valued keyword in Elmer has a feature that allows it to be divided by the area. This may be handy because many keywords on boundary conditions are per unit area. Like the force density for Smitc is "Pressure". You can always say,

Code: Select all

Boundary Condition i
  ...
  Keyword = 1.23
  Keyword Normalize By Area = Logical True

which makes Elmer to integrate over the area for the 1st time and use then that area to divide the expression for the keyword. So "Pressure Normalize By Area = Logical True" might allow you to skip the cumbersome computation of area by hand.

-Peter

[mika]

the force is planar. for example 10N in X or Y direction (Gmsh .geo coordinate system). However, I could not find any option other than Pressure. Any idea how that should be done?

The plate solver (SMITC) doesn't allow in-plane (membrane) stress resultants, while the shell solver (or the old facet shell solver) takes into account the membrane forces. The shell solver knows keywords "Resultant Force i = ..." in BC sections, or nodal forces might be used.

In addition, 3D modelling of thin structures by standard, low-order finite elements is a bad approach due to finite element locking, which means that the displacements usually are far from the right values (the computed values are too small without refining the mesh considerably). The 3D model should use high-order finite elements in order to avoid this trouble. Here plate and shell finite elements should give better results as they employ special tricks to treat the locking problem.

-- Mika

[kevinarden]

There are 3 cantilevers carrying the 10N load, however they are not free to rotate at the end. The deflection of a guided cantilever (no end rotation) is much less. However, the end rotation of the beam is not free nor zero. Taking the average deflection of free or fixed rotation the result is 0.00225 making the Elmer solution of 0.00173 likely accurate.

result.png (97.67 KiB) Viewed 2863 times

deflection.png (63.23 KiB) Viewed 2863 times