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I know, based on research, that the load of a car is transmitted to the wheels by properly torquing the lug nuts. I'd like to verify this through science. Alas, it's been a decade since I last completed a physics problem of this nature, so I could use some help.

Diagram of wheel fitment from www.8thcivic.com

In the above diagram, you can see the general idea of how the pieces fit together. Aftermarket wheels will usually have a center bore that is larger than the hub on the car so that it can fit multiple cars. This is overcome by a hub centric ring (the piece in the middle of the diagram), which is used to make installing the wheel on center easier. It is not there to bear any load (answers should confirm or correct this statement).

How is the load (weight) of the car distributed to the wheels?

Let's throw out some stats roughly based on the car I drive:

Car's weight: 3,000 lbs (assume even weight distribution)
Number of wheels: 4
Number of studs per rotor: 5
Lug nut torque specification: 90 ft/lbs
Hub diameter: 56 mm
Wheel center bore: 65 mm

Any stats not provided should be asked about in comments or can be reasonably assumed by the answerer.

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4 Answers 4

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Assuming a properly centered wheel, most of the force is transmitted through the clamped faces of the wheel and hub, by friction. Also torsionally through the bolts, but much less so. Proper lug nut tightness is critical for that friction. The hub centric lips are there to center the wheel, not to carry load.

https://link.springer.com/article/10.1007/s11668-017-0297-0#:~:text=The%20mechanics%20of%20securing%20the,its%20mating%20component%2C%20and%20the

https://www.maximummotorsports.com/tech_wheels_spacers.aspx

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It doesn't have anything to do with the torque on the lug nuts. The torque spec is there to make sure the nuts are tight enough not to loosen on their own, but not so tight as to damage the nut, stud or wheel.

To a first approximation (given four wheels, five studs per wheel, and a weight of 3000lbs), each stud will have to support 150 lbs of vertical weight. This will be true even if you just slide the wheels on the lugs and don't put any nuts on at all (assuming you can get the wheels to stay on when you put the weight on them). Tightening the nuts will add to the horizontal tension on the lugs, but doesn't have any effect on the vertical forces that support the car.

The question gets a lot more complicated if you consider a moving car. Acceleration, braking, turning, and hitting bumps will all put additional loads on the lugs. The standing weight is just a fraction of their design load.

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  • $\begingroup$ Can you show this to be the case? I would expect the friction between the wheel and the rotor to be significant, considering how tight lug nuts must be torqued at. Various cars have different torque specs, so I would expect there to be more to those specs beyond just "make sure they don't come loose". $\endgroup$
    – Ellesedil
    Commented Dec 12, 2014 at 19:12
  • $\begingroup$ I guess I don't understand what you're asking. The torque on the lug nuts shouldn't have anything to do with friction between wheel and rotor. And yes, I believe the torque specs are essentially "make sure they don't come loose" balanced against "don't damage the stud, nut or wheel." Different cars have different torque specs because the wheels are different sizes, the studs are different diameters, and the gauge of the wheel's metal is different. $\endgroup$
    – David Rose
    Commented Dec 12, 2014 at 20:36
  • $\begingroup$ If you have two surfaces that are pressed together, friction is involved. If you apply pressure on one of these surfaces (the wheel), the friction should increase. In terms of the wheel/car relationship, you now have a force that is reacting against gravity. Torquing the lug nuts will increase the pressure being applied on the wheel against the rotor. I don't think you can dismiss that by saying "all of the weight is supported by the studs." $\endgroup$
    – Ellesedil
    Commented Dec 12, 2014 at 20:44
  • $\begingroup$ To be more clear, consider the following statement: "The wheel is held in place by friction between the back of the wheel and the hub face." Is this true, or even applicable in any significant way? $\endgroup$
    – Ellesedil
    Commented Dec 12, 2014 at 20:48
  • $\begingroup$ Maybe we have a terminology problem. Let's say the "wheel" is the metal part (rim?) that comes off the car and goes in your spare tire well. Let's say the "rotor" is the part that stays on the car, and which has the lug nuts permanently mounted to it. Is that consistent with your definition? If so, then tightening the lug nuts increases the friction between the nut, the lug, and the wheel. The car-side of the lug nut has a conical shape that matches the inverse conical shape of the hole in the wheel. That forces the wheel to center itself around the lug when you tighten the nut. $\endgroup$
    – David Rose
    Commented Dec 12, 2014 at 20:54
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I’m afraid you all only have it half correct.

I by accident have proven my theory of why manufacturers all have rims fitting the centre of axles and hubs neat. These centres share the load of the vehicle by at least 80%, while the lug nuts centre the rim/wheel and fasten the rim to distribute the the weight evenly across the studs. The studs are engieer chosen to withstand forces of friction, flex and shock while carry loads. If you change to a thicker rim using no centre bearing load, only carrying the load on studs, the studs can an will fail.

The Centric Ring is used to put load up of after market wheels back onto the centres, contrary to what was explained above.

Proof? Just look at every car that leaves the factories.

Heavy vehicles operate differently using much more complex science and engineering.

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The tapered seats of the wheels and nuts / bolts not only centre the wheels, but with the friction between wheel and hub take up the weight of the vehicle. Consider the old VW Beetle wheel, which did not have a central tight fitting hole and boss at all. enter image description here

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