Originally Posted by
desmodue
Your thinking is highly flawed in that you assume that the axle and skewer are both limited to 2 axis, in reality what happens is the skewer being much thinner than the axle will flex instead of compressing. Simply put, its a freaking piece of crap aluminum hub not stress rupture machine. For all the idiots that think a steel axle compresses, here's a simple test" get a .500" thick piece of plate steel. Drill a 9/32' hole in it 1/2' from the edge. Measure the plate thickness with a caliper. Put a 1/4' x 20 tpi grade 8 bolt in it with washers and a nut tighten the nut and bolt as hard as you dare without twisting it off. Measure the plate thickness again. Not that the dimension remans the same. That's all I have to say about that.
dear Gormless Gob****e:
Let's see how this works out. A QR skewer is about 4.8 mm in diameter. That's about 18 X 10^-6 m^2
The axle is about 10mm in diameter with about a 5mm hole in the center. That's about 60 X 10^-6 m^2.
So, the cross section of the axle is about 3.3 times the cross section of the skewer, not the 8 times (or 9 times) that you proposed.
So, how much tension should it be reasonable for a QR skewer to experience? I don't know the mechanical properties of the QR skewer , but I don't know the mechanical properties of a 1.8 mm diameter spoke, either. What I do know, is that a 1.8 mm spoke and a QR skewer probably have about the same yield strength. So, it is quite obviious that a QR skewer can handle the stress that a 1.8 mm spoke can handle. A 1.8 mm spoke can easily handle in excess of 1300 N. A 4.8 mm QR skewer, therefeore, can easily handle about 9000 N as the area of the skewer is about 7 times the area of the spoke.
So, if you take a 130 mm axle and apply 9000 N compressive force, how much does the axle compress? That is clearly explained by the well known stress/strain relationship.
Stress/strain = Young's modulus, which for various steels is about 200 X 10^9 N/m^2
So, how much does a steel axle compress if a compressive tension of 9000 N is applied?
strain = stress/Young's modulus X axle length
strain (i.e. compression) = (9X 10^3 N / 60 X 10^-6 m^2) / (200 X 10 ^ 9 N/m^2) X (130 X 10 ^-3 m)
If you calculate that out, you come up with a compression of .0975 mm -- a little less than .004".
It was suggested above by Kontact that the compression would be (1/24)/16". About .003"
Seems like it's reasonable for a QR skewer to cmpress an axle by at least that much.
I did these calculations with a slide rule and I'm assuming that a QR and spoke have about the same yield strength and I'm assuming that the very high tension that can be exerted with a QR eccentric is about 9000 N. If my calculations are off, or my assumptions are unwarranted, please explain. Else, go back to serving coffee.