Consider the wheel of a bicycle 🚲. enter image description here

Now from what I understand, all the spokes that are above the upper half of the wheel happen to be in tension as the hub of the wheel hangs on these.

What I don’t understand, however, is the loading (tensile or none) experienced by the spokes that are in the lower half of the wheel.

So, is all the weight of the bicycle 🚲 supported by the rim of the wheel, as that is where the Normal Force acts at vertically upwards ? Are rims then designed to undertake compressive loading ?

And if so, then why ? Why not instead use tubes or rods instead of spokes which can handle both tensile and compressive loading ?

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    $\begingroup$ The modeling for tire and wheel stress are horrendous. It was the last piece of the modelling puzzle for the auto industry. It wasn't until about the '90s that anyone had a handle on this as far as tires dynamics are concerned. Spoke wheels are a bit easier just by themselves. You have to pretension the spokes enough that they never go slack, even when skidding around a corner or braking. When weight is added to the axle, the equal tension becomes unequal, but must never go to zero. Spoke strechiness, spoke pattern, and the rim's resistance to egging determine how the spokes get loaded $\endgroup$
    – Phil Sweet
    Jan 5 '18 at 15:51
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    $\begingroup$ Yes, all of the rim should always be in compression. On the bleeding edge of time-trial bike racing, the spoke blades can handle complex stresses, not just tension. You see bladed carbon wheels where the spokes act like beams and can take compression as well. This is mostly done to improve areodynamics. $\endgroup$
    – Phil Sweet
    Jan 5 '18 at 15:57

I remember several years ago watching a documentary about building Wembley Stadium, that had a good segment on wooden vs wire wheels: https://youtu.be/aqOzcKdvFDI?t=2m32s

To summarise: The spokes are pretensioned, to the point that under load, or even impact, the lower spokes never go slack.

Further to this, the spokes pulling in on the rim mean that the rim itself is in pulled in all the way around, which makes it significantly stronger. If you squash a circle you will end up with an elipse shape. The tensioned spokes resist this by pulling back on the part of the rim that is now further away from the axle than usual.

Circle vs elipse with equal circumference

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    $\begingroup$ I was going to mention wagon wheels, but went with the time trial example instead. Wagon wheel spokes, of course, work like beams. But they had rim brakes and didn't transmit driving torque. Side force and rolling resistance were their main challenges. The steel 'tire' kept the whole thing is compression via thermal contraction. $\endgroup$
    – Phil Sweet
    Jan 5 '18 at 16:14
  • $\begingroup$ Good point re: driving torque - I wonder if that's part of the cause of the "quintessential fixie" look, with spokes rear (where the drive is), and beams front? (unrtd.co/wp-content/uploads/2017/01/…) $\endgroup$ Jan 5 '18 at 17:48

Let's assume we are looking at a wheel of a bicycle turning clockwise and we refer to positions on this wheel by the clock hours: 12,1,2,3...

The upper spokes gradually get loaded in tension as the wheel turns and the leftmost spoke passes the 9 position and moves to 10. It gets loaded by a factor of sin(a), a being the angle clockwise from horizontal or 9 position. Gradually this tension increases till at vertical, or 12 it is maximum. So the spokes above horizon contribute to supporting the load proportional to their angle. As mentioned in other answers there is also residual load of pretention in spokes.

Now let's assume that you hit a bump or a small object on your way. The rim flexes and wants to deflect inside into the circle. This deflection will create bending moment in the rim just like a horizontal beam and will be resisted by the well known equation

m= sigma.I/c

with sigma being allowable stress of rim, I it's bending second moment of area and c half of thickness of rim, approximately.


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