Hello,
Recently i have done a MSC NASTRAN Free-Free analysis for a FEM consisting of only one beam element (See Figure:1 in attached excel sheet). I could not get the all eigenvalue roots of the SOL 103 modal analysis. In fact I got only 6 modes eigenvalues as shown in Table 1: MSC NASTRAN SOL 103 Summary in the attached excel sheet.
As per theory, a beam element has 6 dof at each nodes. Hence a complete beam should have 6 eigenvalues for elastic modes and 6 rigid body modes (i.e., total of 12 eigenvalues). So my first question is why SOL 103 for this model is giving me only 6 modes even though i requested for 18 eigenvalue roots?
Generally in free-free analysis the first 6 modes should be rigid body modes (i.e., the eigenvalues for these 6 should be zero). But I have noticed that the 6th mode eigenvalue is non Zero. So my second question, Could you please give me a justification for this 6th mode non zero eigenvalue?
I have attached my .bdf file and . F06 file for more details. Also an excel sheet highlighting the SOL 103 summary and model snap from MSC Patran.
Any help will be appreciated,
Thank You,
Paul
Hi Paul
1st question - I believe that SOL103 gets numerical problems and limits its solution set by constraining dofs with too high ratio. .
MAXIMUM RATIO OF MATRIX DIAGONAL TO FACTOR DIAGONAL = 1.0E+15 AT ROW NUMBER 1
2nd question. What you write is absolutely correct. Your mistake lies in forgetting to input the material density on the MAT1 card.
As an exercise - remove the CONM2 elements and you would have no modes at all. The modes you get are because of the mass elements and as these have zero inertia, they cannot provide mass around the beam element length direction.
Add inertia to the CONM2 elements and you will have 6 dofs. Add MAT1 material density and you will still have only 6dofs. Add a few more nodes and you
Have fun
Claes
Hello Claes,
Thank you for your valuable answer.
I have tried your solution given for the second question, still the 6th mode giving me a non zero eigenvalue. The results for all the iterations are tabulated in the attached excel sheet.
Table 2: shows the result for the .bdf SOL 103 run adding only MAT1 density and removing the CONM2 mass.
Table 3: shows the result for the .bdf SOL 103 run adding both MAT1 density and CONM2 mass.
But both the results giving me the Non zero eigenvalue for the 6th mode.
Also the .bdf and F06 files for the two iterations are attached.
Regards,
Paul
Hi Paul
It is a good habit to read the FO6 files and this is where you can pick up tricky modelling problems.
Your F06 writes this
*** USER WARNING MESSAGE 9994 (BULFUN)
MAT1 with MID=1 near line 36
User information: Material has unreasonable or inconsistent values of E, G or NU.
Try to enter data for common steel. In SI E is 2.1e11 and density is 7800 and keep Poisson's ratio at 0,3 as well as stating only G or E.
/C.
Hello Claes,
Sorry for the mistake. Thank you for pointing that out.
This .bdf is generated from a tool, hence it will not represent any standard material property values. I have tried with AL 7050 -T7651 E, G, Nu and Rho values. Now the USER WARNING MESSAGE 9994 has disappeared but eigenvalue problem persists. Attached is the new .bdf and F06 files.
I think the 6th mode eigenvalue is non zero due to the warning message 4698 regarding the FACTOR DIAGONAL RATIO problem.
Regards,
Paul
Hi Paul
Try adding a few more elements. Running a single beam element is somewhat of a special case.
/C
You should use SUPORT,1,123456 or SUPORT ,2,123456 (or other combination) for fine free-free normal modes determination. Without this statement your model is singular, so the numeric resulting rigid body mode frequencies are close to zero, but not equal zero.
Hello Wojciech, You are correct and the issue is resolved. Thanks alot.
Once again Thank you Wojciech and Claes for taking time out to answer my question.
You should have to use your arbitrary column vector is apart from 1 1 1 1 1 1.instead use like 123456 so that you can get your eigen values and vectors
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