15 Replies Latest reply on Mar 25, 2014 1:23 AM by Jared Conway

    Why is there no symmetry in this linear buckling simulation?

    Philip Eddy Eriksen

      I have checked the geometry, the restraints and the material several times. I don`t see why there is a difference in the symmetry of the local bucklingresult from the linear bucklingsimulation. Can somebody please explain me why? The block with the load has a roller/slider on each sidesurface.

       

      Thanks!

       

      bilde-4.JPG

        • Re: Why is there no symmetry in this linear buckling simulation?
          J. Mather

          Can you attach the file here?

           

          Also, to get a good screen capture

          go to the Windows globe icon in lower left corner of screen

          start typing Snipping Tool

          and start the Windows Snipping Tool

          you can use this to capture the screen

           

          or

          hit Print Screen on the keyboard and then start MS Paint and Ctrl V to paste into Paint.

          Save the file.

           

          My first thought (without seeing the file) is the difference in symmetry is not significant.

          • Re: Why is there no symmetry in this linear buckling simulation?
            Jared Conway

            If the load was purely symmetric would it buckle?

            • Re: Why is there no symmetry in this linear buckling simulation?
              Shaun Densberger

              I'm a little confused on what your question is, so please bear with me. Are you asking why the solution varies when you use symmetry compared to when you don't use symmetry? If so, then remember that a linear buckling analysis is an eigenvalue problem (like a modal analysis). If you use symmetry you must keep in mind that the symmetric constraint will remove all non-symmetric modes of buckling (or vibration in the modal analysis) from the solution.

                • Re: Why is there no symmetry in this linear buckling simulation?
                  Philip Eddy Eriksen

                  The symmetryproblem concerns the visual bucklingresult (see picture above). Why is the bucklingdeformation different on each side of the transversal webstiffener?

                   

                  Another thing:

                  I have tried an non-linear static study using arc-lenght control to find the buckling load. Is it possible to find the critical load exactly when the initial buckling of the web occurs?

                   

                   

                  I`m sorry for the uploading of the file. I tried the "pack and go", so I hope this new file will work.

                    • Re: Why is there no symmetry in this linear buckling simulation?
                      Jared Conway

                      have you gone through the examples in the training manual on arc length? it should do exactly what you're describing.

                       

                      as for the assymmetry of your results, i'm asking a more physical question. if everything was perfectly symmetric, would the system buckle? it would be very hard right? in the simualtion, the prediction for buckling somewaht comes from the fact that numerically it isn't perfectly symmetric and to me it is expected that what you see won't be perfectly symmetric. If your real world is symmetric perfectly, the only way to guarantee that is to add symmetry. but heed the warning from shaun.

                      • Re: Why is there no symmetry in this linear buckling simulation?
                        Shaun Densberger

                        The symmetryproblem concerns the visual bucklingresult (see picture above). Why is the bucklingdeformation different on each side of the transversal webstiffener?

                         

                        If you look at Eularian buckling, you'll see that the solutions to the classic problems are sinusoidal in nature. That being said, you might not be seeing the buckling mode shape that you would expect because:

                         

                        1. The geometry, loads, and constraints are such that a non-intuitive mode is the weakest. This mode seems to be characterized by a twisting of the column.
                        2. There are multiply buckling modes that are very close in value to one another, and due to numerical variations, the one you're seeing happens to be the "first". The "second" mode of buckling could be 0.001% greater than the "first" (as calculated by the solver) and have the buckling mode shape that you'd expect. Have you looked at the other modes of buckling? Remember, this is a numerical solution (as opposed to an analytical or exact solution) to an approximated system.

                         

                        To illustrate item (2) better, imagine you're doing a modal analysis on a cube that doesn't have any constraints applied, and you've asked the solver for the first six modes of vibration. The solver would return six non-zero values; something like:

                         

                        1. 2.49E-03 Hz
                        2. 2.56E-03 Hz
                        3. 2.60E-03 Hz
                        4. 3.46E-03 Hz
                        5. 3.83E-03 Hz
                        6. 4.24E-03 Hz

                         

                        Now, intuitively, we know all the modes should be zero and equal in value. However, we get non-zero and un-equal values because we're using numerical methods to solve our system. The fact that first mode (in this case) is characterized by rigid body rotation about the z-axis is completely arbitrary; it could have easily be translation in the y-direction (or any of the other four remaining degrees of freedom). The same is true for a buckling analysis; if there are several buckling modes that are almost identical in terms of the buckling load factor, then numerical variability will determine the order of the buckling modes. This becomes an issue if you only look at the first mode of buckling.

                         

                        Another thing:

                        I have tried an non-linear static study using arc-lenght control to find the buckling load. Is it possible to find the critical load exactly when the initial buckling of the web occurs?

                         

                        Yes, you should be able to do this. As Jared stated, you should go through some of the arc-length (and more over, general non-linear) examples before you tackle your model. I also recommend that you use an enforced displacement (instead of an applied load) for numerical stability.

                          • Re: Why is there no symmetry in this linear buckling simulation?
                            Philip Eddy Eriksen

                            Thanks for the nice feedback!

                             

                            I have checked the training manual, but i dont find anything on the arc lenght control (ALC) for non linear (NL) simulation. Can it have something to do that I use a educational version of the program? 

                             

                            For point nr. 2; I have looked at other modes of buckling in the linear buckling study. The bucklingform is a litle different, but the buckling load factor (BLF) just increases in value.

                             

                            For the linear buckling(B)  the BLF seems way to high. How can I get a more valid result using linear B. ? Is a load eccentricity the way to go here? If so - how do i set this, and how big should the eccentricity of the force be?

                             

                            I have tried the NL and ALC, but i think it is a litle bit to compicated for my experience. I`m very close on the NL study though. The bucklingdeformation seem more correct in comparison to the LAB-result. But I get so many load factors (LF), and I don`t know which LF that is the most correct. If I go with the load factor that matches my LAB-result, the bucklingdeformation is almost equal to nothing. If I choose a plot step wich the bucklingdeformation is more close to the LAB-result, then the LF is again way to high

                             

                            My purpose with this is to find the bucklingform and the design buckling load for the web, localy. The slenderness of the web is very high, so I can`t use the yieldstrenght as a reference to when the web will buckle. I assume that the buckling will occure before the steel reaches it`s yieldingpoint, which is most likely because of the slendernessratio .

                             

                            If I must go with the Euler buckling load in the linear B.study, does it excist some kind of a safety factor that i can use that gives me the approx. design buckling load?

                             

                            Thanks!

                              • Re: Why is there no symmetry in this linear buckling simulation?
                                Shaun Densberger

                                I have checked the training manual, but i dont find anything on the arc lenght control (ALC) for non linear (NL) simulation. Can it have something to do that I use a educational version of the program?

                                 

                                Maybe, but I'm not sure; Jared might know. What exactly do you want to learn about it?

                                 

                                For the linear buckling(B)  the BLF seems way to high. How can I get a more valid result using linear B. ? Is a load eccentricity the way to go here? If so - how do i set this, and how big should the eccentricity of the force be?

                                 

                                It very well could be. If you look at my response to your other post, you'll see that LBA give higher BLF values. The eccentricity of your load does have an effect on the BLF from a LBA, so if you know this then go ahead and add it. Please keep in mind the two points I made in the other post (especially the second one) when you decide how to use these results.

                                 

                                I have tried the NL and ALC, but i think it is a litle bit to compicated for my experience. I`m very close on the NL study though. The bucklingdeformation seem more correct in comparison to the LAB-result. But I get so many load factors (LF), and I don`t know which LF that is the most correct. If I go with the load factor that matches my LAB-result, the bucklingdeformation is almost equal to nothing. If I choose a plot step wich the bucklingdeformation is more close to the LAB-result, then the LF is again way to high

                                 

                                Are you using displacement control for your NL analysis? If so,then you shouldn't need to use ALC (it's typically needed when you're using force control). What do you mean by, "I get so many load factors"? Are you talking about the LBA?

                                 

                                My purpose with this is to find the bucklingform and the design buckling load for the web, localy. The slenderness of the web is very high, so I can`t use the yieldstrenght as a reference to when the web will buckle. I assume that the buckling will occure before the steel reaches it`s yieldingpoint, which is most likely because of the slendernessratio .

                                 

                                This is going to be difficult to do (and risky) if you base everything off of FEA. I'd get your model working such that it matches what yuo've tested in the lab, and then use FEA to test different designs and look at the trends. When you have a design that you like, go and do the physical test and see how well it compares.

                                 

                                If I must go with the Euler buckling load in the linear B.study, does it excist some kind of a safety factor that i can use that gives me the approx. design buckling load?

                                 

                                Not that I'm aware of, but maybe there is a standard out there that have more information.

                                • Re: Why is there no symmetry in this linear buckling simulation?
                                  Jared Conway

                                  what training manual did you check? arc length is in the nonlinear training available from your reseller. i don't think there is an example in the tutorials though if you're looking in the software.

                          • Re: Why is there no symmetry in this linear buckling simulation?
                            Roland Schwarz

                            Is your mesh perfectly symmetric?  Doubtful.