Dear friends here:
Thanks for your reply to my previous questions, and I learnt really a lot from this forum, thanks everyone who helped me, May all the things goes well on you!
Now, I found a serious problem when I using buckling in linear perturbation. I know the critical load equals to the load I set in load module multiplied by the corresponding eigenvalue, so I set the load as 1 to avoid further calculation, but after I get the result, I feel surprised at two tings.
Firstly, the eigenvalue is much more bigger than the critical value calculated from Euler critical load formula, I need this value to estimate the load added to the following steps, but now I was totally stuck here...
Secondly, I tried to changed the load value in load module from 1 to 0.1, and hope to see the difference that the eigenvalue for the case has load as 0.1 should be ten times of the case has 1 as load, but to my surprise, the results are just the same! that doesn't make any sense...
Could anyone help me with this issue, I feel really frustrating about this.
Best Regards,
Kyle
Please check your basic information about the buckling and determine which type of fixation for your model in order to select the proper condition. Also, check your model carefully. check also the B.C. for your model. After that I think the problem will disappear.
Yes, Mr. Saleh, you are right, the final critical buckling load for the same model which have determined stiffness should be always be the same no matter what value I set as the load in load module of ABAQUS, but my fact is that I changed the load value but the eigenvalue always keep the same, that means the final critical load is always change for the same model, that doesn't make sense.
Since my model is just a cylindrical rod, very simple model, and use one end pin boundary condition, the other side symmetrical boundary condition, so I really cannot find out why this will happen..
I need your help Mr. Saleh!
Please check your basic information about the buckling and determine which type of fixation for your model in order to select the proper condition. Also, check your model carefully. check also the B.C. for your model. After that I think the problem will disappear.
Perhaps stating the obvious, apologies if so: 1. Check if the first buckling mode value is equal to pi()^2*EI/L^2 (I believe that is the value of the first buckling mode for a pinned column.) 2. Make sure that you have allowed the column to move in the direction of the applied load, i.e., you should not have a translational restraint in the same direction as the applied force. 3. Also, make sure the column can bend into the buckling mode (i.e., rotations are permitted at the end node).
Do also note that using a symmetry boundary condition means you can only capture symmetric buckling modes - you will miss all non-symmetric and anti-symmetric ones. Therefore, using symmetry for a buckling problem is not a good idea unless you are certain you want only the symmetric modes.
All of the above discussion assumes you are trying to capture the buckling due to bending of the member. Make sure you are using a beam element and not a rod/truss element.
Can I just apply displacement control, and then sum the reaction force at the end where I apply displacement, and with that sum load value times the eigenvalue to get the critical buckling load?
Thanks very much!
No you can't use enforced displacement. You should be able to keep displacement in the axial direction free, You can fix lateral translations if need be. Fix torsional rotation if you are using beam elements. Free both rotational restraints. Ask a professor or call the Abaqus guys. This is an easy problem. Ask someone to look at your model. Not worth overcomplicating the problem.
Thanks for your reply Mr.Ranga!
I finally accurately get the result, thanks for all your help!
But another question for you: since I use C3D8 element (solid) to model a bar that is under compression, and since for solid element, no rotational DOF for its nodes, so when I add boundary condition, even if I only suppress two lateral DOF, and leave the axial DOF free ( it will deform in that direction), the bar will get a "fixed end" boundary condition, right? at least I find in the animation recording, it deformed as a fixed end?
My question is: how I add boundary condition to make it works as a pin end? It sounds like a very simple question, but it really kills me, I tried did not suppress the lateral DOF at the end, but the bar at certain time will freely have rigid body motion, which is not the way we want it to be like a pin end...
Please help me, I feel so discouraged by this small point, it is so small that I even do not have the courage to ask my supervisor, I am a first year PhD candidature, still on my way, all the help are highly appreciated !
Best Regards and sincerely,
Mengli
Hi!
Even very simple BC's are not as simple to model as one would think. You can see many types of modelling BC's on pinned columns if you dig around a bit, such as adding additional stiff end plates, through linear constraints, etc. I would suggest trying at first to apply the BC's in just 2 nodes at the ends of the column and take it from there. However, if you got the correct eigenvalue and the buckling mode looks ok, then I guess your model is good. I mean you can see in the buckling mode if the column you have modelled looks pinned or fixed at the ends.
OK, rotational restraints are irrelevant to solid element. And simple questions are often quite hard! You can apply lateral restraint to all nodes at the load end if you want. That will not fix rotations, only axial restraints on multiple nodes will fix rotations. But the only buckling modes you will get are symmetric bending modes since you are also using symmetry. If your column is upright and you have 9 nodes on the top plane and 9 nodes at the symmetry plane, you can restrain all the top 9 nodes in the two lateral directions and apply axial load there. At the symmetry plane you can restrain all nodes in all three translational directions.
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