[QUOTE=Poolside]Radial loads from the weight of the bike and hitting bumps is carried by the stepped diameter on the drive hub and its mating diameter in the whel hub.. That much is easy and obvious.[/quote]
You're starting down a slippery slope by saying that ANY of the loads are carried by anything other than clamping force. What's the basis for this?

[QUOTE]And the tire friction cannot come within what, a tenth of the force needed to slide the mating surfaces?[/QUOTE]
I was bored while riding down to Road Atlanta last week, started to approximate this one in my head and came out with an alarmingly low Fs. Refiguring it here with numbers I found on the internet* I get something like this:
Total clamping force 120kN (from boltscience.com article)
Cf of wheel and flange: .35 (general assumption from some online example)
So friction of wheel against hub is around 42000N or 4281 kg. Bottoming out the shock might represent 500 kg maybe? I don't know how to accurately predict impact loading, but provided the rim doesn't strike something solid I don't see how the radial loading is going to get much past about 700kg. That's a rough guess based on enough impact to make a 40 SQIN patch contact the ground at 40 PSI. Somewhere around 1600 pound load (I know this doesn't account for sidewall and tread stiffness, feel free to contribute) which is a bit more than 700kg.

The torque strength of that joint relies on the contact area being some distance from the axle. In this case the center of the contact area is slightly beyond the bolt hole so I'll call it 1.4 inches. 4281 kg is 9418 pounds, times the 1.4 inches gives a torque resistance of 1098 foot pounds.

* for that guaranteed accuracy confidence :)