17th August 2010 - 05:54 PM
I want to know how to calculate the stresses (hoop stresses primarily I imagine) on a retaining sleeve used to hold magnets onto the rotor of a high speed motor.
Suppose there are six magnets bonded to a hexagonal shaft symmetrically but the adhesive alone is insufficient to keep the magnets from flying off the shaft during operation. The outer surface of the magnets are curved so that they are uniformly in contact with the sleeve (which is a thin walled cylinder). There may or may not be small gaps between the magnets.
I can calculate the hoop stresses caused by the rotating of the sleeve by itself but I don't know how to determine the added effect the "centrifugal" force of the magnets will have.
This is my first post on this forum. Thanks for your help.
29th August 2010 - 07:30 AM
Acceleration = velocity squared times radius, gravity is like 32 ft a second.. divide acceleration by gravity to get G force or centrifugal force. However aren't the magnets also being subjected to force against the coils while in operation? And magnets are generally very light and being so small I doubt anything will break.. but then again this might just be homework eh?
sorry bout the late reply.
3rd September 2010 - 02:40 PM
Thanks for your reply.
I can calculate the centrifugal force but I donít know how this translates into a hoop stress in the sleeve. I know how to calculate hoop stresses due to uniform pressure like for a pressure vessel but I donít know how to do it for several discreet, symmetric forces that you would get in this situation. (Iím considering just taking the centrifugal force and dividing it by the surface area of the sleeve and just calling it pressure but I suspect this wonít be quite accurate).
This is for extremely high rotational speeds where even the relatively small mass of the magnets plays a very important role and often breaks the adhesive holding the magnets to the rotor.
Yes, there will be a significant force on the magnets due to the magnetic field generated by the coils but I am considering that separately.
16th January 2011 - 06:49 PM
At each small peripheral length unit of the sleeve, the sleeve's tension provides a centripetal force equal to the tension times the small angle unit (a diagram may help), where the small length divided by the small angle give the radius.
So you can relate the centrifugal force and the angle in the sleeve (like 60į) to the tension, and then you have the section needed at the sleeve.
Did you hear about Vacuumschmelze? They have the Koerflex300 alloy with some 3500MPa yield strength (no, not 350MPa) which is an acceptable permanent magnet. This allows it to spin like mad, which more than compensates its limited magnetic strength, as it allows huge speeds hence radii.