Applying a Force in a different angle

[jeff]

Hi everyone,
is it possible to apply a force in a different direction then x-/y- or z-direction? I would like to apply a force on a screw in a certain angle but I could not find how I can do it.
Hope somoene can help me and thanks a lot in advance.
Best regards
Jeff

[maas]

Hi Jeff,

I assume the screw is a rigid body? You can apply a force in a particular direction by decomposing the force in its vector components and then applying three forces, one for each direction. For instance, in 2D, since you know the angle, you can a force F in direction a, by applying two forces: Fx = F*cos(a), Fy = F*sin(a).

Cheers,

Steve

[jeff]

Hi Steve,
thanks a lot for your help. I understand the decomposing of the force. But how do I enter it in FEBio. If I add a load, I can only enter a number for the load and not a cos(a) or sin(a).
Best regards,
Jeff

[maas]

Hi Jeff,

We have some support for defining equations, but not for all features yet, and I don't think you can use equations in rigid body forces (if that's what you are using?). I can definitely look into adding that. In the meantime however, you'll have to do all your calculations beforehand, and enter the final numbers in the value fields for the three components of the force. Or is there something more to what you want to do that I am missing?

Cheers,

Steve

[jeff]

Hi Steve,
thank you for your support!
I have entered all the values but now I cannot run the simulation. Am I missing something?
I just want a simple simulation of one vertebra and a force in a certain angle on a pedicle screw (see the link).

https://www.dropbox.com/sh/ksb3i9uykbegr48/AACMZv-tICJunkYATiV9MiQ_a?dl=0

[maas]

Hi Jeff,

The main problem is that the two rigid parts (the vertebra and the screws) are not attached to each other. When you apply the force to one of the screws it just flies off to infinity. You need to attach the screws to the vertebra. Given that both are rigid bodies, I think the easiest way to do this is to make them one rigid body. (Assign the same rigid body material to both the vertebra and the screws). Also, constraints on rigid bodies have to be applied via "rigid constraints". The "fixed displacement" that is applied to the vertebra will not have any effect.

By the way, I had some problems with this model. It seems that some parts had the same ID, which made it impossible to select the screws. Do you have this problem as well? I was able to fix that issue so if you want the fixed model, please reach out to me via email (steve dot maas at utah dot edu).

Cheers,

Steve

[jeff]

Hi Steve,
sorry for the late reply and thanks a lot for your help!
Is there another way to attach the screws to the vertebra? As I understand your suggestion, the material properties of the vertebra and the screws will be the same, but they are actually not. Furthermore I do not understand what you mean by "Also, constraints on rigid bodies have to be applied via "rigid constraints". The "fixed displacement" that is applied to the vertebra will not have any effect". How do I have to change it, so that it has an effect?

For me, I could select the screws and did not have any problem.

Best regards,
Jeff

[maas]

Hi Jeff,

Since both objects are rigid and you are doing a static analysis, none of rigid body material parameters have any effect here. (The density only has an effect in dynamic analysis, and the elastic properties are not used by anything anymore. They used to be used in the auto-penalty calculation of some contact formulations, but I don't think that is the case anymore. The center-of-mass parameters are only relevant when you allow rotations.)

In FEBio, the kinematics of rigid bodies is defined in terms of its six degrees of freedom (three displacement and three rotational). It are these degrees of freedom that you need to fix. When a node is assigned to a rigid body, it does not participate in the usual boundary conditions, since its position can also be described in terms of the rigid body kinematics. So if you assign a fixed degree of freedom to a rigid node, then this will not have any effect. I hope that explains it better.

Cheers,

Steve