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Topic Title: DC machine - is the torque produced on the armature wires or steelwork?
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Created On: 14 February 2013 05:10 PM
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 14 February 2013 05:10 PM
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Avatar for aroscoe.

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Calling machines experts!

I am brushing up on my electromagnetic theory etc. in preparation for teaching some undergrads. I've been going through the standard "DC Motor" threory and come across a conundrum I can't resolve. It has been causing some good debates in the coffee area today but so far, I didn't find any convincing arguments.

The question is, on a DC motor armature (rotor), is the force (torque) produced on the WIRES, or on the STEEL (core), or a mixture of them both? According to standard texts, there are two ways to work out the force.

1) You can work out the B field due to the field winding, and then do F=BIl on the armature turns, ignoring the MMF that THEY produce which slightly adjusts the B field due to the field winding. You then convert F to a torque by using the armature radius.

2) You can work out the change in field energy using the changing mutual inductance between the armature coil and the field winding, and then do T=dW/dtheta where W is the stored field energy. I'm ignoring the reluctance torques since I'm assuming a nice round armature core.

If I do the above methods carefully, they actually give the SAME torque result which is reassuring (both numerically and also algebraically if you collapse all the equations together right back to equivalent symbols (I, u0, dimensions etc)).

My conundrum is that method 1) clearly suggests that the force (torque) is on the wires, and that they would need to be glued to the rotor very well or they will rip themselves off.

However, method 2) suggests that it is much more a magetic field interaction between the steelwork parts, and that if I put a permanent magnet in, instead of a wound rotor, it would still have a force and torque, and there would be no armature wires - the force/torque would have to be between the steel/magnet structures.

So, which is it, how are the forces actually distributed in a DC machine armature? Is there some force on both, or all on the wires, or all on the steel? The textbooks all seem to gloss over this ....


Dr. Andrew Roscoe

 14 February 2013 07:08 PM
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My limited understanding sugests that it is a combination, the magnetic field is produced by the current in the wires and this field is concentrated by the steel.
If you replaced the steel with plastic the motor would still work albeit not as well.
If the steel works to concentrate the field then it follows that most of the force must be on the steel.

Interesting question
 14 February 2013 10:24 PM
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A DC contactor or a large industrial magnet wouldn't be as effective without the magnetic circuit, The ferrous metal would concentrate the useful magnetic path. Likewise, I think that the same argument applies to any ferrous magnetic circuit, encompassing a DC coil.

 15 February 2013 09:25 AM
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Agree with both those comments, but it goes deeper. If I built an electromagnet and used it to lift up metal objects, I am pretty convinced that the force would be on the metal and not the coil wires - in that case the wires are wrapped around the outside of the electromagnet core (which contains 99.9% of the flux) and so they are OUTSIDE the main B field and so logically there is (almost) zero force on them by F=BiL - all the force would be between the steel objects due to field energy etc. I wouldn't need to glue the wires particularly well to the magnet core.

The problem is this thing with the DC motor where the wires are right in the B field. If you swap out the ferrous armature core for plastic, or some other thing with ur=1, then the force must be on the wires, logically. Still, both calculation methods for Torque, using F=BiL and using field energy, give the same answer (which is reassuring). It is a rubbish motor, though, with very small torques, because the path reluctance is so high. But, if I put in an armature core with ur inceasing up to several thousand (for steel), the machine gets much more effective, as everyone agrees. while both calculation results always still give the same answer for torque (the overall machine reluctance path is the same in both calculations), the majority of people I asked so far have a "gut feel" that the force gets distributed between the wires and the armature steel. But is this true? Is there any textbook reference which describes this process?

I might need to build something or do an experiment to get to the bottom of this!

For reference, the two calculation methods are given below:

Dr. Andrew Roscoe

 15 March 2013 02:52 PM
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Avatar for aroscoe.

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I finally resolved this in my head, while stopped at some traffic lights on my pedal bike.

Force is all on the wires if the steel rotor is cylindrical. Because, even if it is ferromagnetic and has N-S poles "created" by the current in the armature wires, a rotation of the rotor steel (if it could rotate freely without the winding rotating) would not result in the poles rotating - they would stay orthogonal to the winding. Therefore, a "virtual work" examination tells you that there is no torque on the rotor steelwork. It must all be on the wires.

If the rotor is not perfectly cylindrical then it will have a reluctance torque, though.

Dr. Andrew Roscoe

 18 March 2013 06:22 PM
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The Books you want on this subject you may or could obtain from the library is by Charles S. Siskind. (Direct - current Machinery)

Some on DC motors below if you scroll down

 29 April 2013 08:37 PM
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I finally stumbled on a proper and more full explanation, just the other day, in the 1967 book by Eric Laithwaite "The Engineer in Wonderland". It is buried in the Appendix to chapter 2, where he describes a hefty physical setup built specifically to measure the balance of forces on wires and steel rotor structures embedded in fields.

If the rotor is smooth (cylindrical) and the wires (turns) are stuck on the outside the rotor structure (in the "uniform" airgap), then the force is ALL on the wires (like my previous post suggested ought to be the case, which makes sense).

However, if you cut slots in the rotor and embed the wires (turns, coils) into the slots so that they don't protrude into the airgap, then almost ALL the force is transmitted via the rotor steelwork. The wires (turns, coils) experience almost zero force. In practice, most machines look like this. In a squirrel cage motor, this means that while the currents flow (almost entirely) in the conductor bars/ends/cage, the force itself is almost entirely produced on the rotor steelwork.

If you make a kind of in between version where perhaps the rotor steelwork has more complex non-cylindrical structure, then the balance of forces is harder to determine, and in some cases the force on the steel rotor can even be opposite to what you would expect, but is balanced by a larger force in the "right" direction on the wire(s), leading to a net force in the direction you would expect with F=BIL. I guess this really boils down to "salient pole" type reluctance effects which modify the simpler clear-cut cases of cylindrical rotors, or cylindrical rotors with slots cut in them and the wires fully embedded in the slots.

See Also:
