algorithmic modeling for Rhino
Hey,
I'm trying to observe and compare kangaroo to ANSYS (FE method) in a small example in my work.
I'm well known to the compatible unit systems (m - N - Pa) (mm - N - Mpa) etc.
I presume spring stiffness can be chosen as A*E/L if I want to model a beam. (L = total length, A = cross section, E = 200GPa steel)
However the model will "explode" during the simulation if I change the stiffness higher than 6000 +/-.
Is this a kangaroo problem or is the structure too stiff above that number, and thus a design issue? But since the default setting is 1000 I presume the units are allready defined in some sort? I haven't found anything in the manual or in previous posts. So sorry I'm repeating solved problems.
kind regards
Tags:
Hi Carlo,
Great to hear you are doing some testing and comparisons with ANSYS - this is something I've been wanting to do for some time.
You are quite right to use A*E/L as the stiffness.
The simulation 'exploding' is an issue with the explicit integration methods used. It can be avoided by decreasing the timestep (try dividing it by 10 until it becomes stable, then work slowly back up from there to find the best value). This will of course mean that it takes more iterations to reach equilbrium. To keep the simulation still running at a reasonable speed you can also increase the SubIterations, so that it only outputs the result once every 100 iterations, say.
There are also some other integration methods available in the Kangaroo settings, including RK4 and a 6th order one from Yoshida, but I find that their increased calculation time tends to outweigh their advantage in stability over the default velocity verlet in many situations.
I'd be very interested to see your results, and it would be great to finally clear up the issue of correct units and real material stiffness values, and give people an idea of how accurate they can expect the results to be.
This will also help me assess how worthwhile and necessary it is to pursue alternative methods beyond explicit integration.
Hey, thanks for the info the model works now.
I'm just defining one imaginary 2D truss and comparing the displcaments.
This on the simple model, but also an actuated version to check if the kinematics are also approximated in a proper way.
I'll send the data in a PM if you want to take a look at it.
greetz Carlo
Hey Carlo,
I'm intersted in seeing your tests too. When you'll have a chance, please share your results!
Thanks to you and Daniel
Lio
Hi Carlo.
I'm very intersted in your work, I also find very useful your method to define stiffness values.
Did you make the kinematic test?
Is Kangoroo approximation acceptable?
Thanks
Andrea
Hi Carlo - I would also be interested in hearing about the results of your work.
thanks
Ash
Nice! I've been using kangaroo this year in the bar position and pattern design stage for a big stadium project at my school. I've used SAP2000 for the final calculations.
The results and process was really nice because kangaroo give us kick feedback about how the structure could work; with this data in our hand modifications and redesign was really straight forward.
For final steps because the nature of the project (we design 2 big supports attached to the structure and floor with steel wires) we needed to use multi-step calculus.
My experience in this workflow is that Kangaroo in impressively good and quick for first design stages but when things comes harder, you always will need a good FEA app.
You can see that I implemented a simple interface to check compression-traction bar state.
Best.
really interesting to see your feedback here on the accuracy of kangaroo for real loads. I've found it great for early stage concept work too.
Hi Angel,
would it be possible to upload the image in a higher resolution.
I am trying to learn from your script but im finding it difficult to recognize certain components.
Thanks a lot.
Hi Carlo,
Thanks for posting this topic about units of the stiffness has been very helpful. Just one question the L in your formula es the complete length of the truss? or the individual elements / divisions?
Thank you for your help,
Bernardo
Hi Bernardo,
The L would be the length of an individual element/spring
Thank you Daniel
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