Why do my spokes keep breaking? - Bike wheel science.

[PapaZita]

I was contemplating this on my cycle home, and have a thought experiment to offer which may clarify things.

I was thinking along similar lines, but came to a slightly different conclusion! In your step 2, why is the load not shared equally between the two spokes? (P+W/2) and (P-W/2)?

I skipped the idea of infinitely stiff spokes, and just considered elastic ones. The change in length of a spoke is proportional to the change in load. The rim stays round (because it's still infinitely stiff), so as the hub moves in the direction of the load, one spoke grows in length by exactly the same amount as the other contracts. The change in tension in each spoke must be equal and opposite.

This suggests to me that, with an infinitely stiff rim, the load would be borne equally by the upper and lower spokes. Yet, we do know (e.g. from measurement) that it really is the lower spokes that do most of the work in real wheels. Can we conclude that the difference is due to the fact that rims are actually not very stiff?

Intuitively, I think it makes sense that deformation of the elastic rim (radially) occurs mostly in the region where it contacts the road. The lower spokes must be the ones getting shorter, and therefore, must be the ones doing the work. You wouldn't expect a 'flat spot' on top of the wheel.

 

[roubaixtuesday]

why is the load not shared equally between the two spokes? (P+W/2) and (P-W/2)?

Statics is linear, so you can add any two valid solutions.

 

[roubaixtuesday]

I was thinking along similar lines, but came to a slightly different conclusion! In your step 2, why is the load not shared equally between the two spokes? (P+W/2) and (P-W/2)?

I skipped the idea of infinitely stiff spokes, and just considered elastic ones. The change in length of a spoke is proportional to the change in load. The rim stays round (because it's still infinitely stiff), so as the hub moves in the direction of the load, one spoke grows in length by exactly the same amount as the other contracts. The change in tension in each spoke must be equal and opposite.

This suggests to me that, with an infinitely stiff rim, the load would be borne equally by the upper and lower spokes. Yet, we do know (e.g. from measurement) that it really is the lower spokes that do most of the work in real wheels. Can we conclude that the difference is due to the fact that rims are actually not very stiff?

Intuitively, I think it makes sense that deformation of the elastic rim (radially) occurs mostly in the region where it contacts the road. The lower spokes must be the ones getting shorter, and therefore, must be the ones doing the work. You wouldn't expect a 'flat spot' on top of the wheel.

Having thought some more, I think you're right. In my version, E ends up as W/2.

 

[PapaZita]

Statics is linear, so you can add any two valid solutions.

So, let’s change step 1 to a lower spoke bearing the force W in compression. Now, in step 2, we get an upper spoke with tension P, and a lower spoke with tension P-W. (Of course, real wire spokes don’t do compression, but I think it’s fine for the purposes of this analysis, particularly when W < P).

What does this mean? Is your method invalid? It’s been a long time since I studied it. I know it’s OK to superpose different load cases on the same structure. Are we allowed to change the structure rather than the load? Are there two (or more) equally valid solutions?

 

[silva]

I think a spoke has everything to do with tension not pressure.
Could the simplest viewpoint then not be alike a spoke is a rope that is knotted to a ring and in a center of that ring to others?

 

[SkipdiverJohn]

This suggests to me that, with an infinitely stiff rim, the load would be borne equally by the upper and lower spokes. Yet, we do know (e.g. from measurement) that it really is the lower spokes that do most of the work in real wheels. Can we conclude that the difference is due to the fact that rims are actually not very stiff?.

When a bike is just coasting; i.e. the wheels are rotating and in a straight line but not being either driven by the rider pedalling or being slowed down by the application of the brakes, I cannot see how any spokes other than the top ones are actually doing any work. Spokes cannot be compressed, because if you push the edge of the rim towards the hub, all that happens is the tension comes off the spoke and the nipple becomes loose! A loose spoke rattling around in its hole cannot be carrying any load! When a bike is being accelerated or braked though, there are forces involved that are trying to twist the wheel rim relative to the hub, and if cornering forces are present, then those will be trying to push the hub out of the wheel sideways along the axis of the wheel axle.

 

[Ajax Bay]

Morning - sunny to start with early doors down here. Lovely.

3. We now allow the spokes to be elastic. Both spokes must deform by the same amount as they form a straight line, and the rim is infinitely stiff. The tension in the lower spoke must reduce, as its length reduces. We will call this elasticity reduction E. The tension in the upper spoke must also reduce by the same amount, to keep the overall hub forces in balance.
The upper spoke now has a tension of (P+W-E)
The lower spoke has a tension of (P-E)
This leads to our second conclusion: allowing for elasticity, tension in upper spokes will increase, and tension in lower spokes will decrease. The amount of each depends on the elasticity of the spokes.

Able to stay with you till the last point. Your conclusion seems plausible until E=W when its first element fails: tensions in the upper spokes do not increase. For me E=W so upper spoke tension (in the two-spoke wheel) stays at P and lower spoke tension drops to P-W (and I now see that @PapaZita sort of suggested this yesterday). And this does not depend on how elastic the spokes are, provided they are elastic (your model in scenario 3), so I think that "The amount of each [tension increase/decrease] depends on the elasticity of the spokes." is false.
You may wish to consider a scenario where the top spoke is thinner (and more elastic) and the bottom spoke is thicker (and less elastic). Can the (unloaded) tensions be the same if the hub is to stay central? Notwithstanding that, how do the tensions (two spoke wheel, stiffer than a stiff one rim) vary when a load is applies to the hub (balanced by an opposite normal reaction force at the (bottom of the) rim.

I cannot see how any spokes other than the top ones are actually doing any work.

Depends what you mean by 'work'. When static, don't think any spokes do any 'work' (in the physics sense of 'work'). Work = force through a distance so the work done by a rotating spoke is an integral over a cycle of its elongation (for a 45kg load this is about 0.01mm) times the (?average?) tension. All the 'work' is elastic (but not perfectly elastic) so much of the energy consumed is elastically returned but some energy is lost (fast rolling wheels etc etc).
Take a static wheel with no load on it (any orientation). All the tensions will be (more or less, depends on rim) the same. Now place the wheel normally on the ground and apply a load, balanced by the normal force from the ground (wheel still static). The top spoke tensions scarcely change; the bottom spoke tensions drop by (in sum) the force of the load applied; the side spokes' tensions increase slightly to maintain the essentially circular integrity of the rim. So if you mean by "doing any work" applying force to the hub then the upper spokes apply more force than the lower spokes but actually the side spokes apply even more, balanced of course by the opposite spoke.

 

[PapaZita]

I cannot see how any spokes other than the top ones are actually doing any work. Spokes cannot be compressed, because if you push the edge of the rim towards the hub, all that happens is the tension comes off the spoke and the nipple becomes loose! A loose spoke rattling around in its hole cannot be carrying any load!

It’s certainly true that spokes can’t be compressed, and if a spoke ever becomes so loose that it’s rattling around then it can’t contribute to the structure of the wheel.

But, remember that spokes start with an awful lot of tension, even before you put any load on the wheel. Something like the equivalent of a 100 kg weight pulling on each one. This is why they ping if you pluck them. And, compression is just negative tension, and as @roubaixtuesday said above, the system is linear, so we can just add up the different loads.

So, suppose you put a load on the wheel, e.g. by sitting on the bike, that could only be supported by compressing a bottom spoke. Say that load was the equivalent of 25 kg of weight compressing the spoke. But the spoke started with 100 kg of tension. If we add up the loads we have the original 100, minus the 25 applied load (because adding compression is the same as subtracting tension). The spoke now has a tension equivalent to a 75 kg weight hanging off it. It’s still in tension, just less than before.

If the load applied is so great that it’s bigger than the initial tension in the spoke, either because the load is huge, or the spoke wasn’t very tight to begin with, that’s when the spoke goes slack and wobbles around in its hole. That’s a situation that we want to avoid.

The initial tension in the wheel is what allows spokes to appear to be working ‘in compression’. Of course, some spokes may support loads by increasing in tension too. The question is, which mechanism is more prevalent, and which spokes are doing it?

I wonder if an analogy helps? Suppose we’re having a tug of war competition, both leaning back, in equilibrium, with lots of tension in the rope. We all know that you can’t push on a rope, but I can make you fall over backwards, just by momentarily reducing the amount I’m pulling.

 

[Ajax Bay]

From a John Forrester article (adapted):
It explains how the spokes carry the cyclist's weight. When we build a wheel, all the spokes are stretched in tension as we tighten them, until (with a true rim in a well-built wheel) all the spokes are pulling almost evenly. Now if we load the wheel by applying weight through the hub axle (the normal way) the bottom part of the rim where the tire touches the ground gets forced inward. This inward movement shortens the distance between hub and rim and thereby reduces the tension force in the bottom few spokes. These bottom few spokes then pull the hub downward less strongly than they did before the load was applied; the reduction in downward force is just equal to the weight applied to the hub, so the hub (and all the wheel except the bottom two inches or so of the rim) drops the few thousandths of an inch by which the rim is deflected, and then remains at that level. (Of course the tire deflects also; its effect is a change in level many times that of the wheel deflection.)
When you are on your bicycle your weight is supported by the pull of the spokes on the hubs; it's just that the downward pull (by the lower spokes) is less than it would have been had you not been on the bike.

 

[roubaixtuesday]

Morning - sunny to start with early doors down here. Lovely.

Able to stay with you till the last point. Your conclusion seems plausible until E=W when its first element fails: tensions in the upper spokes do not increase. For me E=W so upper spoke tension (in the two-spoke wheel) stays at P and lower spoke tension drops to P-W (and I now see that @PapaZita sort of suggested this yesterday). And this does not depend on how elastic the spokes are, provided they are elastic (your model in scenario 3), so I think that "The amount of each [tension increase/decrease] depends on the elasticity of the spokes." is false.
You may wish to consider a scenario where the top spoke is thinner (and more elastic) and the bottom spoke is thicker (and less elastic). Can the (unloaded) tensions be the same if the hub is to stay central? Notwithstanding that, how do the tensions (two spoke wheel, stiffer than a stiff one rim) vary when a load is applies to the hub (balanced by an opposite normal reaction force at the (bottom of the) rim.

Depends what you mean by 'work'. When static, don't think any spokes do any 'work' (in the physics sense of 'work'). Work = force through a distance so the work done by a rotating spoke is an integral over a cycle of its elongation (for a 45kg load this is about 0.01mm) times the (?average?) tension. All the 'work' is elastic (but not perfectly elastic) so much of the energy consumed is elastically returned but some energy is lost (fast rolling wheels etc etc).
Take a static wheel with no load on it (any orientation). All the tensions will be (more or less, depends on rim) the same. Now place the wheel normally on the ground and apply a load, balanced by the normal force from the ground (wheel still static). The top spoke tensions scarcely change; the bottom spoke tensions drop by (in sum) the force of the load applied; the side spokes' tensions increase slightly to maintain the essentially circular integrity of the rim. So if you mean by "doing any work" applying force to the hub then the upper spokes apply more force than the lower spokes but actually the side spokes apply even more, balanced of course by the opposite spoke.

"For me E=W"

Why?

 

[roubaixtuesday]

Expanding on the last, it is clearly possible from a statics perspective for the change in tension to be entirely in the upper spokes.

It's only deformation in the system which defines where the load is taken.

Thinking about the two spoke system, for simplicity, if all the change in tension were in the lower spoke, it would reduce in length. The upper spoke would then increase in length and would therefore increase in tension as a result.

So I think it's not possible to have the entire change in load either on the upper, or on the lower spokes. It must be shared.

 

[roubaixtuesday]

There was talk somewhere above of a reference where these tensions were measured. Can anyone post a link please?

 

[Yellow Saddle]

There was talk somewhere above of a reference where these tensions were measured. Can anyone post a link please?

You guys haven't bothered to read the thread properly and are now trying to re-hash the debate from the ground up.

There was no mention of tension measured, just an experiment to show a variance in tension or not. It is in the thread. Read it.

 

[PapaZita]

From a John Forrester article (adapted): #snip

I like this way of looking at it. The deflection of the rim is directly related to the change in tension of the spokes, and it makes intuitive sense, to me, that the deflection is going to occur at the bottom, near to where the force from the road is applied.

What I’m still missing is how to show, with engineering and maths, that this is what will happen. I’ve a feeling I once knew how to approach such an analysis, but I fear it’s beyond me now.

 

[Mr Celine]

Spokes cannot be compressed, because if you push the edge of the rim towards the hub, all that happens is the tension comes off the spoke and the nipple becomes loose! A loose spoke rattling around in its hole cannot be carrying any load!

So, go and build a radially spoked wheel on a very stiff rim but only screw the nipples in far enough to just contact the rim, so that there is no tension at all on any spokes. If you put that wheel on a bike then you are correct in so far as the bottom spokes cannot take any load and the hub must be hanging from the top spokes, which are able to take a load in tension.

But that is not how a spoked wheel works in practice and you wouldn't want to ride a bike with such a wheel. All the spokes in a properly built wheel are under a tensile force which greatly exceeds any compressive force that they undergo. The spokes at the bottom of the wheel are carrying a compressive load but are still under tension.

For an example of the opposite scenario consider a simply supported pre-stressed concrete beam. The entire beam is always in compression, but the lower half is carrying a tensile load.