Riks in ABAQUS: Is this the forth stopping criteria for Riks?

[KyleSong]

Hi all friends here:

I am a new researcher and employ ABAQUS to analyse steel bar buckling through Riks algorithm, and concerning the stopping criteria, I want to ask a question. Since when we edit "step", we can specify as much as three stopping criterion: Maximum load proportionality factor, Maximum displacement, and the maximum number of increments. But, since we can also define a displacement along certain direction when we impose displacement control to the model, I want to know whether the displacement I defined is another stopping criteria of Riks? If not, what is the significance of defining a displacement? 

Best Regards,

Kyle

 

[KyleSong]

Dear TGS4:

Thanks very much for your fast reply!

Yes, I understand what you mean by now! and yes, not too many people consider the imperfection in actual design, and this is definitely a problem for designing.

And from your information provided, the challenge of introducing geometrical imperfection is to define appropriate scaling value that can make the model produce expected critical buckling load, is that right? Did you apply all the nodes of the body as the imperfection to Riks or just use the imperfection nodes at two ends after eigenvalue buckling analysis ?

And can you help me to check if the progressing of the diagram in attachment is a typical shape of bifurcation buckling?


Best Regards,

Kyle

 

[TGS4]

KyleSong - perhaps you should provide some details to your problem. Most of your specific questions make little sense without that context.

And when you provide plots, labels would be very helpful - I have no idea what you are actually plotting in your figure.

 

[KyleSong]

sorry TGS4!

My problem is: acting compressive load to a steel bar, which has round section shape, diameter is 12mm, two ends are pinned boundary conditions, I used displacement control to impose load to the bar. At this stage, I try to model its non-linear buckling behaviour, as well as load-displacement diagram. In the diagram I attached the same as the previous one, the X-axis means the displacement of midpoint of the bar in mm Unit, the load is the total load of the top section, in KN Unit. The Euler buckling load is approximately the value shown on the diagram, but after that point, there should be a horizontal line, but for this diagram it dropped down, and it doesn't make sense (see the screenshot of the correct diagram in the second attachment)

Sorry about that, I made such a mistake to rashly post the incomplete information, Please forgive me!



Best Regards,
Kyle

 

[KyleSong]

sorry TGS4!

My problem is: acting compressive load to a steel bar, which has round section shape, diameter is 12mm, two ends are pinned boundary conditions, I used displacement control to impose load to the bar. At this stage, I try to model its non-linear buckling behaviour, as well as load-displacement diagram. In the diagram I attached in last reply, the X-axis means the displacement of midpoint of the bar in mm Unit, the load is the total load of the top section, in KN Unit. The Euler buckling load is approximately the value shown on the diagram, but after that point, there should be a horizontal line, but for this diagram it dropped down, and it doesn't make sense (see the screenshot of the correct diagram in the second attachment)

Sorry about that, I made such a mistake to rashly post the incomplete information, Please forgive me, and thanks for your kind help!



Best Regards,
Kyle

 

[Mustaine3]

The Euler buckling load is approximately the value shown on the diagram, but after that point, there should be a horizontal line, but for this diagram it dropped down, and it doesn't make sense (see the screenshot of the correct diagram in the second attachment)

Why do you think it makes no sense? The postbuckling behavior depends heavily on the structure. If you have a real collapse of your structure, then the load that can be taken will be much lower after the buckling. So the curve does not always have to up after buckling.

 

[KyleSong]

Yeah, I agree with you largely, but based on the attached diagram I got from bar buckling introduction in Wikipedia, for a bar during elastic state, it should shows a bifurcation buckling.

But you said a collapse of my structure, actually it won't collapse at all, I use Riks to analyse Bifurcation buckling is because only this algorithm can analyse non-linear plastic buckling. So, that is the reason I believe this result doesn't make sense...

Thanks for your time to help me analyse this, I will keep trying based on your guidance, thanks again!

Sincerely yours,
Kyle

 

[TGS4]

You will have unexpected behaviour with displacement controlled buckling as compared to load controlled buckling. Nevertheless, as a load (X) vs displacement (Y) plot, your graph is not unusual.

 

[KyleSong]

Dear TGS4 and other friendly mates here:

I am coming back, I got another tough issue that sounds very silly, but it still makes me struggle a lot, I beg for your help again！

I am modelling a steel bar that under compression, and want to find its non-linear buckling behaviour, and I use C3D8 element, and want to apply pin-end boundary condition at one end, and axial symmetry boundary condition at another end, but this "pin" is really hard to get!

Since for solid element, there is no rotational DOF for its nodes, but the element can still rotate based on the difference of translational displacement. At the beginning, I fixed all the lateral DOF of all the top surface, and since I was asked to use displacement control, I have to set a displacement value for axial DOF, and after that, I find the result is exactly a fixed end boundary condition!

And some other kind researchers here told me "only axial restraints on all the nodes will fix rotation" I think that makes sense. but another difficulty will be risen up, since I want to use displacement control to the solid bar, I have to set a axial displacement for all the nodes, and then, it won't rotate at that situation. what should I do to get a pin end and can still use displacement control?

Best Regards and Sincerely yours,
Kyle