Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.
Plotting Neutral Axis Shift for Asymmetric Material Model
Posted 17.11.2022, 12:40 GMT-5 Results & Visualization, Structural Mechanics, Material Models Version 6.0 2 Replies
Please login with a confirmed email address before reporting spam
Hello,
I have a question about whether anyone has tried plotting the shift in the neutral axis along a cantilever beam, while using an asymmetric material model. I have geometric and material nonlinearity included in the COMSOL model and I'm importing my material model based on uniaxial data. For reference, I am comparing with an analytical model I am developing that includes tension-compression asymmetry.
I have tried using a contour plot of the stress (in my case I am using the second Piola-Kirchoff stress (solid.SXX)) and I set the levels to zero in order to find where the stress is equal to zero. This would give the position of the neutral axis. However, since I am not really clear on how COMSOL is averaging the stresses on the backend, I am not sure if this approach is correct. And I feel that it is also highly mesh dependent. I am currently using an extremely fine mapped mesh instead of the default free triangular one.
I am finding that since there may be stress singularities especially at the tip of the beam where the force is being applied, the results I am getting may not be accurate. The image attached shows that for a very large tip force, the neutral axis shift drops to zero towards the end of the beam. I'm assuming because there are also artifical stresses that equal to zero at the top and bottom of the beam where the force is applied, and the model is lumping those stress levels together. From my analytical model, it should be a constant value after a certain level of stress.
Has anyone else tried tracking the neutral axis shift before in COMSOL using a different method or have had similar issues in the approach I am using?
I appreciate any insight, and can give more information if necessary.
Attachments:
Please login with a confirmed email address before reporting spam
I think your idea if using the contour value for zero solid.SXX is a good approach. I would not be too worried about the exact stress evaluation/interpolation, as long as you have a reasonable mesh. No need to make it extremely fine.
Now to your problem. By looking at the plots, my guess is the following:
Your load at the beam end may be pointing in the space-fixed negative Y-direction. At larger displacements, that will be not only a bending load, but also partially a tensile load, due to the rotation of the end of the beam. That could explain why your neutral axis drops. We know from beam theory that the rotation increases quadratically along the beam, so the effect should be much stronger as we approach the tip. At the same time, the bending moment is smaller there (varies linearly inwards).
Under these assumptions, I think a first order estimate for the neutral axis location is proportional to x^2/(L-x), where x is the coordinate along the beam (length L). That looks a somewhat like your results.
Finally, if you want as little as possible influence of the load application, distribute it parabolically like (h/2-y)*(h/2+y), where h is the beam height.
-------------------Henrik Sönnerlind
COMSOL
Please login with a confirmed email address before reporting spam
Thank you very much, Henrik.
I could tell from my results that the force at the end was influencing the location of the neutral axis, but I didn't quite understand why. But, your explanation helps me a lot. And correct, I am applying a boundary load in the negative Y direction at the end of the beam. I'll have to refresh my understanding of beam theory since there's a lot of assumptions I make in my analytical model ignoring the tensile load due to the applied force, which is probably not neglected in COMSOL.
I will try your suggestion of the distributed load and see how it works out.
Thank you again,
Brianne Hargrove