Buckling analysis help in FEMAP

[Srki1982]

Hi,

I want to check if I have done a correct buckling analysis of plate. I receive a very high eigenvalue result and it's just hard to believe that I need to apply such much higher load in order to buckle the plate.
Basically, the model is just a cylinder formed plate with 5 mm of thickness and it is fixed on one end. The load (plate press force) is applied on other end as 1500 kg (14715 N), see picture.

Is my model correct? If not, how do I do a correct buckling analysis in this case?

In the Results, I receive 2 outputs. Am I correct that NX NASTRAN Case 1 is just a result of static analysis as von Mises stress because I receive the same stress if I do a static analysis? In the result output Eigenvalue1 488.7516, the von Mises stress is 274.3 MPa. What is the difference between the stress result in Case 1 and Eigenvalue output?

All the best!

/Srdjan

 

[BlasMolero]

Dear Srdjan,
The first output set is used by NX NASTRAN to compute the differential stiffness matrix automatically generated for each element that supports differential stiffness, then the Case Control Section must contain at least two subcases, this is equivalent to a linear static analysis (SOL101).

What you need to take attention is the Eigenvalue output: the "488.75" means the BUCKLING LOAD FACTOR (BLF), the rest of results (displacements, stress, etc..) are USELESS. The BLF is the factor of safety of your structure againts linear buckling failure.

If you apply an unitary load force, then the resulting Eigenvalue will be the total Buckling Load that will support your structure, OK?.

Best regards,
Blas.

~~~~~~~~~~~~~~~~~~~~~~
Blas Molero Hidalgo
Ingeniero Industrial
Director

IBERISA
48011 BILBAO (SPAIN)
WEB:

http://www.iberisa.com

Blog de FEMAP & NX Nastran:

http://iberisa.wordpress.com/

 

[Srki1982]

Dear Blas,

Thank you for a fast reply!
So basicly, in this case, I would need to apply 488 times higher force in order to buckle the plate. Am I correct?

 

[BlasMolero]

Dear Srdjan,
Yes, this is the result of a LINEAR BUCKLING analysis, remember, this is the value for the bifurcation point.

Please note the following assumptions and limitations apply to linear buckling analysis (SOL105) with FEMAP & NX NASTRAN:
• The deflections must be small.
• The element stresses must be elastic.
• The differential stiffness is supported for the following elements: CONROD, CROD, CTUBE, CBAR, CBEAM, CBEND, CQUAD4, CQUAD8, CQUADR, CTRIA3, CTRIA6, CTRIAR, CSHEAR, CTRAX3, CTRAX6, CQUADX4, CQUADX8, CHEXA, CPENTA, CPYRAM, and CTETRA.
• A minimum of five grid points per half sine wave (buckled shape) is recommended.
• The distribution of the internal element forces due to the applied loads remains constant.
• Offsets SHOULD NOT BE USED in beam, plate, or shell elements (except CQUADR/CTRIAR) for buckling analysis.
• For 3-D buckling problems, the use of PARAM,K6ROT is recommended for CQUAD4 and CTRIA3 elements. A value of 100 is recommended.
• For structures that exhibit nonlinear material or large deflection deformations, the linear buckling load obtained from Solution 105 may be different than the actual buckling load. For structures with significant nonlinearities, it is recommended that you perform a nonlinear buckling analysis using Solution 106.

Best regards,
Blas.

~~~~~~~~~~~~~~~~~~~~~~
Blas Molero Hidalgo
Ingeniero Industrial
Director

IBERISA
48011 BILBAO (SPAIN)
WEB:

http://www.iberisa.com

Blog de FEMAP & NX Nastran:

http://iberisa.wordpress.com/

 

[Srki1982]

Thank you for a very detailed information

 

[Wvandenbos]

Adding to the comments of blas, be aware that everything is linear and tolerances might have a large influence on your result, also (material) non-linearities can result in much lower buckling results.
The cilinder seems to be loaded with a complete symmetric pressure. A slight offset will influence your results considerably.

success

W. van den Bos
SDC Verifier

 

[rb1957]

you could try a hand calc ... do an euler calc of the cyclinder (without holes), fixed both ends.