Dear all,
I'm having a problem with Elmer solver:
In my real case, I have a clamp beam with a static load at the other side. The beam is an I-type with triangular holes.
I'm having results from Catia with the GPS module and I wanted to reproduce this case with Elmer.
Unfortunately, the results are completely different from my reference for the Von Misses as well as for the displacement.
As I couldn't find what was wrong, I decided to simplify my case to a simple geometry for which analytical results exists.
So I have a circular beam, not hollowed, clamp at one side and loaded perpendicularly at the other side.
It's possible to compute analytically the stresses and displacement for that case.
I could reproduce similar value from Catia+GPS, but once again, Elmer is giving completely different results.
I tried also an other solver (z88Aurora) which is also "wrong" (and with different results from Elmer).
So I'm really wondering what is wrong in my case. I was thinking of a unit issue but I did my analyses in m and mm and
results are coherent between the 2 cases.
Please find attached the files (mesh + sif) in meter and millimeter.
The Beam is 1m long (L) and 25cm Radius (R), load (F) is 1000N. Steel: Young modulus (E): 210e9Pa
I=pi*D^4/64
Stress=F*L*R/I
Displ=F*L^3/(3*E*I)
Stress Displ
Analytical 81.46e6[N/m^2] 5.2e-3[m]
Catia 74.13e6[N/m^2] 4.9e-3[m] (acceptable)
Elmer [m] 1.49e5[N/m^2] -8.3e-6[m]
Elmer [mm] 1.54e-1[N/mm^2] -8.09e-3[mm] (similar to Elmer [m])
I'm not completely sure about this last if it's necessary to convert the load from N=(kg*m/s^2) to kg*mm/s^2 but if I don't do,
I have different results. I'm also not sure the output is in N/mm^2 maybe it's kg*mm/(s^2*mm^2) Then the results are different.
Could somebody have a look to my sif (and meshes) and tell me what's wrong?
Thanks a lot
PS:I did also a Dynamic study (clamped) and modes are similar in Elmer and Catia
Static clamped beam: Elmer and Analytical solutions differ
Static clamped beam: Elmer and Analytical solutions differ
- Attachments
-
- beam_mm.zip
- Mesh and sif in [mm]
- (75.04 KiB) Downloaded 306 times
-
- beam_m.zip
- Mesh and sif in [m]
- (75.17 KiB) Downloaded 302 times
Re: Static clamped beam: Elmer and Analytical solutions differ
Didn't check your sif, but a quick look at your mesh.elements file reveals you are using linear tetrahedral elements (504). You'll want to up them to quadratic elements to avoid the so-called "locking."
I think the command to increase the element order is something like:
I think the command to increase the element order is something like:
Code: Select all
ElmerGrid 2 2 [mesh folder name] -increase
Re: Static clamped beam: Elmer and Analytical solutions differ
Hinguyent,
thanks for the info, I'll try that but do you really think it can affect that much the solution (10^3)?
And what is the locking?
Thanks
thanks for the info, I'll try that but do you really think it can affect that much the solution (10^3)?
And what is the locking?
Thanks
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Re: Static clamped beam: Elmer and Analytical solutions differ
Hi
Or you can equally well set in the Solver section
to envoke the quadratic elements using the hierarchical p-basis.
Google for "shear locking".
-Peter
Or you can equally well set in the Solver section
Code: Select all
Element = p:2
Google for "shear locking".
-Peter
Re: Static clamped beam: Elmer and Analytical solutions differ
Dear Peter,
thanks for the explanation. I read about it and understand. Indeed I did a run with quad elements but it didn't improve my results. As expected, it will not cover the "e3" gap between the analytical results and Elmer results. I think I get something like 1.56e5 instead of 1.49e5 with linear elements. Still far away from 8.14e7 of analytical results.
I tried with quad element mesh or with "Element = p:2": same results.
Any other ideas?
Thanks,
Max
thanks for the explanation. I read about it and understand. Indeed I did a run with quad elements but it didn't improve my results. As expected, it will not cover the "e3" gap between the analytical results and Elmer results. I think I get something like 1.56e5 instead of 1.49e5 with linear elements. Still far away from 8.14e7 of analytical results.
I tried with quad element mesh or with "Element = p:2": same results.
Any other ideas?
Thanks,
Max
-
- Site Admin
- Posts: 4832
- Joined: 22 Aug 2009, 11:57
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- Location: Espoo, Finland
- Contact:
Re: Static clamped beam: Elmer and Analytical solutions differ
Hi
Had a look on the .sif file. My suspect would be the interpretation of "Force 3". To clarify, it is a distributed force hence the SI unit for it is [F]/[A]=N/m^2.
-Peter
Had a look on the .sif file. My suspect would be the interpretation of "Force 3". To clarify, it is a distributed force hence the SI unit for it is [F]/[A]=N/m^2.
-Peter
Re: Static clamped beam: Elmer and Analytical solutions differ
Hello Peter,
thanks for the post.
I'm not sure to clearly understand the definition you wrote (I know what a distributed force is).
When you say "...it is [F]/[A]...", I understand that the value I put in the BC editor is 1000 (N) distributed on the A defined (this is the behaviour in Catia (distributed force, defined in N)).
If I put a value of 1000 is it?
- 1000N/m^2 -> 1.9635e-3N (surface is 19.36cm^2)
which means that I have to put 500000N/m^2 to be equivalent.
Apparently it is.
I redid completely my computation and I realized I made a mistake (2 in fact):
- Analytically and in Catia I did R=0.25m while it is R=0.025 in Elmer...
So here are the correct table:
_ Stress Displ
Analytical 8.149e7[N/m^2] 5.2e-3[m]=5.2[mm]
Catia 7.428e7[N/m^2] 4.9e-3[m]=4.9[mm] (acceptable)
- The Force definition 1000N vs 1000N/m^2
Elmer [m] 1.84e5[N/m^2] 7.9e-6[m]=7.9e-3[mm] With F=1000 [N/m^2]
Elmer [m] 9.38e7[N/m^2] 4.0e-3[m]=4.0[mm] With F=509000[N/m^2]
So, it's now coherent.
I should have payed more attention to the dimension and read the manual to know it was in N/m^2.
Maybe it's good to put in the BC editor that the distributed force is per unit of surface.
Thanks a lot for your help, I'm very happy with the tool and the support.
Greetings
thanks for the post.
I'm not sure to clearly understand the definition you wrote (I know what a distributed force is).
When you say "...it is [F]/[A]...", I understand that the value I put in the BC editor is 1000 (N) distributed on the A defined (this is the behaviour in Catia (distributed force, defined in N)).
If I put a value of 1000 is it?
- 1000N/m^2 -> 1.9635e-3N (surface is 19.36cm^2)
which means that I have to put 500000N/m^2 to be equivalent.
Apparently it is.
I redid completely my computation and I realized I made a mistake (2 in fact):
- Analytically and in Catia I did R=0.25m while it is R=0.025 in Elmer...
So here are the correct table:
_ Stress Displ
Analytical 8.149e7[N/m^2] 5.2e-3[m]=5.2[mm]
Catia 7.428e7[N/m^2] 4.9e-3[m]=4.9[mm] (acceptable)
- The Force definition 1000N vs 1000N/m^2
Elmer [m] 1.84e5[N/m^2] 7.9e-6[m]=7.9e-3[mm] With F=1000 [N/m^2]
Elmer [m] 9.38e7[N/m^2] 4.0e-3[m]=4.0[mm] With F=509000[N/m^2]
So, it's now coherent.
I should have payed more attention to the dimension and read the manual to know it was in N/m^2.
Maybe it's good to put in the BC editor that the distributed force is per unit of surface.
Thanks a lot for your help, I'm very happy with the tool and the support.
Greetings