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# Re: [femm] r-component of force in problems with axisymmetric geometry

- To: femm@yahoogroups.com
- Subject: Re: [femm] r-component of force in problems with axisymmetric geometry
- From: David Meeker <dmeeker@ieee.org>
- Date: Tue, 02 Jul 2002 23:43:00 -0400
- References: <NDBBLMCJKLJMKGPONOMCOECLCFAA.JA-Rabchuk@wiu.edu>
- User-agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.0.0) Gecko/20020530

James Rabchuk wrote:

> `Hi,
> I've just begun using FEMM, and have found it to be very useful and
> accurate. What a great tool! Not only so, but the archive of discussions on
> the program have been very useful as well. Thanks so much.
> `

I hope that the program proves to be useful to you.

> `
> I have a question regarding the calculation of forces on objects lying along
> the axis of symmetry using the stress tensor integral. It's clear from the
> Roters' example provided on the website that to find the z-component of the
> force on such an object, I need to create a contour with beginning and
> ending points on the axis (r = 0) line which "encloses" the object, like so:
> -------|
> | ^
> -| | | z I hope this picture is preserved!
> -| | -> r
> -------|
> `

Yes, this is right. You could imagine the surface of revolution made by this countour as a pill shape that encloses your object.

> `
> Doing so has given me very believable results for the force, if I use a
> small enough mesh, though the magnitude of the force is quite small (on the
> order of 10^-4 Newtons). However, just as a check, since I am looking for an
> equilibrium solution, I wanted to see what the program would say about any
> forces acting in the r direction. Of course, I expected that they should
> cancel out because of the symmetry of the problem (allowing for possible
> numerical error). Surprisingly, the program predicted a quite large,
> positive force in the r-direction (on the order of 10^-1 N), which varied
> smoothly with vertical position of the object.
> As I was doing this calculation using lua scripting, I just changed the
> _expression_ found in the Roters' example to
> fr,p,fz=lineintegral(3)
> where fr and fz should pick out the real parts of the r and z components of
> the 1x force, as I understand it.
> Can anybody explain to me what is going on? Even if I were to explain away
> the positive force in the r-direction as being equivalent for every value of
> theta, and therefore cancelling out as expected, I expect and know that
> there should in fact be a restoring force pushing the object back toward the
> center line, so that the sign of the force should be negative in this case.
> Not only so, but the magnitude of the force is just too big.
> `

The result that you get from the lua function for radial force is garbage. As you say, the net radial force for an axisymmetric problem should be zero. If you take the line integral "by hand," the program doesn't report a radial force component (because it is supposed to be zero). However, when you take the integral using the lua function, it reports an essentially meaningless result for the radial force, rather than zero. I'll go back and make sure that it reports a zero force for the radial direction for the lineintegral(3) in axisymmetric problems--thanks for pointing this out.

> `
> I did notice when I graphed B dot n along an upward vertical line, the
> program automatically gave the sign of the dot product to be positive, when
> the field should have a negative r-component, as verified by the plot of B
> dot t for a horizontal contour moving away from the axis. Does the normal
> always point 90 degrees ccw from the direction of the line segment?
> `

Yes, the normal always points ccw from the direction of the line segment. Therefore, an upward vertical line would have the normal pointed in the -r direction, i.e. along the vector (-1,0). If the field has negative r component along your line, the dot product with (-1,0) would yield a positive-valued result.

> `
> I should also mention that in this case, the stress tensor proved much MORE
> useful than did the coenergy method that is usually recommended by the
> program author. I guess that it is because in my problem, the object which
> is moving is so small compared to the entire solution domain. I could never
> resolve a clear trend in coenergy vs. position.
> `

It's hard to say quite what's going on without seeing the problem. Generally, however, it's good to make the elements in the region that is in motion small enough so that each step traverses a length longer than the element size--this has given OK results for me in the past.
Dave.
--
David Meeker

- Follow-Ups:
  - Re: [femm] r-component of force in problems with axisymmetric geometry
    - From: Xia Chen <xchen4@umbc.edu>

- References:
  - r-component of force in problems with axisymmetric geometry
    - From: "James Rabchuk" <ja-rabchuk@wiu.edu>

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