Quasi-static angular acceleration

Tim_Gowing

Hi Victor,
You may have seen my related question about angular acceleration over on the Calulix forum.
In reading the Mecway manual, under section 10.16 Centrifugal force, I see the paragraph:
"Angular velocity can be constant or time dependent in nonlinear static analysis. However, since this is
a quasi-static simulation, it can't model the forces induced by angular acceleration."
Gravity or linear acceleration is a vector quantity and available in a static analysis; angular acceleration is not universally accepted as a vector, perhaps a 'pseudovector', does this have any relevence to the statement in the manual?
I'm interested in gaining understanding of quasi-static FE limitations.
Regards, Tim

Victor

Angular acceleration is just not implemented, sorry. There's no major reason other than that not many people have asked for it.

The reason centrifugal force and linear acceleration are available in Quasi-static analysis is because they're putting the whole model in an accelerated reference frame where those fictitious forces are explicitly applied as force rather than acceleration. Angular acceleration would be the same if it was there.

I've heard of a trick to simulate angular acceleration by using two centrifugal forces about perpendicular axes. The internal solver can do that but I don't think CCX can. I can look into this some more if you want.

Another possibility might be a moment load on faces. Mecway converts a face moment to a distributed force that's proportional to radius (see picture). Maybe that's the same force distribution that an angular acceleration would cause? I don't know.

Tim_Gowing

Victor, thank you for describing the situation regarding angular acceleration.
Would the implementation be way down in your enhancements list? next version?
In the meantime I am very interested in the two centrifugal force trick, any further info greatly appreciated.
Tim

Victor

Not sure if/when I'd add it sorry.

Here's a document written by Anthony Falzone which contains the maths. I don't understand it but it seems to work perfectly*. The key equation is:

α = -2 Ω ω

where Ω and ω are two angular velocities about two perpendicular axes. It's important which one is which and from trial and error I found that for acceleration about the Z axis, you need to identify the rotation about the X axis as Outer in the centrifugal force window.

In the attached example:

3 rad/s about the X axis
4 rad/s about the Y axis
α = -24 rad/s^2 from the formula

From the solution, the reaction moment at the support (Solution -> Sum) = 16000 N.m about the Z axis which agrees with the formula's -24 rad/s^2 when you use
τ = I α
I = m L^2 /3 (moment of inertia for a beam about one end)

* EDIT: I notice there's a radial stress and a net reaction force which don't seem right for a pure rotation acceleration at 0 speed. So be careful.

Tim_Gowing

Victor,
Thank you very much for providing the theory and creating the example; I must now work to gain some understanding of what's happening here; your help is much appreciated.
Regards, Tim