[ItBites]

Tbuttrick,

Youre guessing wrong. Very wrong. Dangerously wrong. The applied preload in force due to torquing fasteners is inversely proportional to the mean pitch diameter of the threads. The equation according to the widely accepted Shigley is F = T/.2D The .2 varies with thread lube, etc..., but the T is torque applied (in inch-pounds), the D is the mean thread pitch diameter (in inches), and the F is the amount of preload or 'clamping' force in the bolted joint (in pounds).

Now, most cars have say qty 5 x 1/2 inch studs, each getting say 90 ft-lbs (1080 in-lb) of torque. If we use the .2 constant for now, you can see that each stud produces 10800 pounds of clamping force and with qty 5, you get a total of 54000 pounds of clamping force to keep the wheel on the spindle.

On the other hand if you have one set of threads in your knock-off that are (I dont have them to measure the diameter, but just say they are 1.25" diamteter) 1.25" diameter and you torque them to 70 ft-lb (840 in-lb), then you only get 3360 lb of clamping force to keep the wheel on the spindle. To get the same 54000 lb that a 5 lug pattern can produce you would need to torque a single 1.25 inch diameter thread to 1125 ft-lb.

This is somewhat modified by using anti-sieze on the knock-offs and not using ift on the 5-lug setup (these are installed dry). This changes the .2 constant as I mentioned earlier and actually reduces the applied clamp force from the 5-lug set-up and increases the clamping force from the knockoff. You can consult your Shigley for a discussion on this factor, but it does not go above .3, or below .1. If we use an absolute worst of .3 in the 5-lug, and a best of .1 in the knock-off, the knock-off torque to produce the same clamp force to hold the wheel in place is still 375 ft-lb.

Once you understand that the bigger diameter threads you have, the less clamping force you produce for a given applied torque, you see why the torques on a large knock-off nut are very high.