Forces acting on a Wheel/Rim?

[bigpat]

You cannot disregard lateral loads, or else it's a meaningless exercise. Remember the cars can pull 3G plus cornering load, and this lateral rigidity has a big bearing on design. Remember the cornering forces try to load the rim in bending too.

The inflation loading significant too. At 20 psi, you have a lot of energy loading the rim, nearly 4000 lbs of stored force.

I would also consider rotational (centrifugal ) load, as reducing material thickness of the rim and spokes makes a meaning difference when spinning at 2000- 3000 rpm on a straight at Monza.

The biggest loading would be torsion cause by braking torque, often at 4g's.

Also, the nut does not transfer torque to the wheel, the drive pegs do that.

Finally, have to allow for the strength of the magnesium alloy at over 100 degrees average temperature, due to the brakes. The more pure magnesium alloys lose considerable stiffness as temperature increases....

[Jersey Tom]

In series where tire inflation pressures get to the 90+ psi range the stress imposed on the wheel may not be trivial.

That's mental... who are doing that?

The publicly available, published minimum RF inflation for NASCAR at say Las Vegas is 58 psi cold. No tire blankets so they start off at ambient... take your best guess for tire running temperature... do an ideal gas estimation for where hot pressure ends up.. big number!

Then start multiplying that by how many square inches of surface area you have on the rim... also a big number.

[Greg Locock]

You might want to look at SAE tests J328a and J175 for typical fatigue and impact tests on wheels. In my experience on road cars, designing the spider (ie from just inside the wheel stud to the rim) for fatigue is the most difficult for steel wheels, and designing the area from the outboard bead seat through the radius to the flange area for impact is the most difficult on an alloy wheel.