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Electronics Engineer needs help with Belleville calculations
I have an alternator with a splined dog bolted to an M10x1.25 threaded shaft. Originally it used a shouldered nut torqued to 40 Nm, but it was prone to loosening so the factory added a Belleville washer and changed the nut from 10 mm A/F to 15 mm A/F.
When I removed this alternator I found that nut was loose, so I plan to upgrade to the new fixing system.
The Belleville washer specified is DiN2093 20x10.2x1.0x1.0 which I believe is flat at 1.0 mm deflection and is 1.0 mm thick.
The axial load on the "bolt" would be,
F = T / c / D
= 40 (N.m) / 0.2 / 0.010 (m)
= 20,000 N
where c = 0.2, the coefficient of friction of a steel thread
(ref: [URL]https://www.engineersedge.com/torque.htm[/URL])
I understand that the load on a "bolt" should be shared with the Belleville washer, in order for the washer to immediately compensate for the relaxation of the load, due to creep or thermal expansion, so I would expect that the Belleville washer should be sized to flatten at 20,000 N.
The calculator I found:
[ATTACH=CONFIG]2426[/ATTACH]
determines the flattening load of the washer above as 3,300 N when deflected 1.0 mm, using E = 210 GPa, which I understand is that of spring steel, as used in a DIN2093 washer.
So, my question is, assuming the calculations are correct, how useful can this washer be if the axial load must reduce from 20,000 N to 3,300 N before it will begin to move?
