Understanding How Tensioned Members in a Bicycle Wheel Support Loads

[nov0798]

Hi gang. I need some help with a technical issue.

I am an avid mountain biker, and one of the forums I post on has an ongoing discussion of whether or not the bottom spokes in a bicycle wheel support the load, or the upper spokes support the load.

In a typical wheel the spokes are tensioned, sometimes up to 350 lbs of tension per spoke in high end wheels, with up to 36 spokes per wheel. All the spokes attach at a center hub/flange. One group believes that the load is supported by the top spokes, while the other group believes that the load is supported by the bottom spokes.

Numerous tests have been done using a tensiometer on the spokes showing that the bottom spokes tension reduces when a riders weight is applied, however the top spokes show no change in tension. It would leave me to believe which I do, that the spokes showing the change in tension are the ones carrying the load. In addition, it is my belief that the spokes act as a rigid column as long as they remain under tension, at which point they carry the load.

One of the issues we are having is that some cannot understand how a nipple on the rim side of the spoke (tensioning end), since it is not affixed to a solid structure could remain in place under a compressive load (rider weight).

So bottom line is I need some folks smarter than me in the physics would to easily explain how this works. I would love to actually build a model in which I tension a single spoke, and then stand on it. the issue I am having is how would I create a fixture that would allow me to do this since neither end is really supported?

Any help would be GREATLY appreciated.

Thanks
Brian

 

[sophiecentaur]

Hi and welcome.
If you imagine replacing the spokes with rubber bands, that would give you a way of figuring out the problem. All the bands would be under tension but the hub would end up (sagging) a bit nearer the ground. Hence, the tension in the higher bands /spokes would be a bit higher and the lower bands / spokes would be a bit less tense.

I hope this releases some of the tension on your members.

 

[nov0798]

Thanks, but the issue is that the top spoked don't change tension. Here is a post from the other forum.

This morning I measured the spoke tension of the three top and three bottom spokes in my 2008 Mavic Crossride rear wheel. The tension measurements confirm the theory (as well as Jobst Bradt’s FEM model results) that the bottom spokes carry the rider’s weight. The tension in the top spokes did not change when I sat my 205 pound body on the bike seat.
2008 Mavic Crossride rear Mountain Bike wheel with (24) 1.4 x 3.0 bladed stainless steel spokes laced in a 2 X pattern.
The three Bottom Spokes (Park TM-1 scale reading/(approximate Kgf):
Spoke #1 in front of spoke 2: Unloaded=13/ (58Kgf) and loaded under 205 pounds=11/ (<53Kgf).
Spoke #2-drive side at BDC: Unloaded=20/ (114Kgf) and loaded=18/ (92Kgf).
Spoke #3 behind spoke 2: Unloaded=10/ (<<53Kgf) and loaded=7/ (<<<53Kgf).

The Three Top Spokes:
Spoke #4 behind spoke 5: Unloaded=17.5/ (87.5Kgf) and loaded=17.5/ (87.5Kgf).
Spoke #5-non-drive side spoke at TDC: Unloaded=12/ (53Kgf) and loaded=12/ (53Kgf).
Spoke #6 in front of spoke 5: Unloaded=18/ (92Kgf) and loaded=18/ (92Kgf).
Note; Park does not have a direct conversion for a 3.0mm x 1.4mm steel bladed spoke so I used a 2.9 x 1.4 spoke listed on their updated conversion chart that I downloaded from Park’s home page.

Any idea how to build an experiment that shows that a tensioned structure (spoke) can support a load?

 

[AlephZero]

For a spoked structure with a stiff outside ring, the load in the spokes is distributed as Sophiecentaur describes. This is a common situation in rotating machinery where the bearings are spoked structures connected to the outer casing. Whether the spokes are radial or angled does not change the basic pattern of the loads.

I expect that two things are causing of the different results for a bike wheel. First, the wheel rim and the tire are far from being rigid structures (as anybody who has tried to build a bike wheel knows, it is easy to build a wheel, but harder to build a circular wheel!). Second, the reaction force from the road is only applied at one point around the circumference, and then distributed into the metal rim through the tire.

One of the issues we are having is that some cannot understand how a nipple on the rim side of the spoke (tensioning end), since it is not affixed to a solid structure could remain in place under a compressive load (rider weight).

That is the main reason why the spokes are pre-tensioned. If the pre-tension is say 350lbf and a load of 100lbf is all transmitted through one spoke, the tension in that spoke reduces to 250lbf, but it is still in tension. If it was pre-tensioned to less than 100lbf, it would "come loose at the rim" and the rim would deform so the rest of the load went through the adjacent spokes.

If all the spokes have the same tension, the resultant load on the wheel hub is zero (ignoring the fact that the spokes are trying to pull the hub apart and increase its size). If you reduce the tension in the bottom spoke by 100lbf, that meas there is 100lbf less force UP on the hub from the spokes, which balances the 100lbf force DOWN through the frame and the wheel bearings.

The tension measurements confirm the theory (as well as Jobst Bradt’s FEM model results) that the bottom spokes carry the rider’s weight.

Do you have a reference to the FEM model and results?

 

[Rap]

I think you have to say the top spokes are doing the supporting. Just because their tension does not change does not mean they are not doing the supporting.

Look at a simple example, suppose you have a tennis ball with two rubber bands. You pull up on the top one with one hand, down on the bottom one with the other hand so that the tennis ball stays still. Equal tension in both bands. If you clip the top band, the tennis ball falls, but if you clip the bottom band, it stays. The top band is what keeps it up.

Now have a friend pull down on the tennis ball a little bit. Less tension in the bottom band, more in the top band. Now suppose you don't pull as hard on the top band, you drop your top hand down until the top tension is the same as it was before. There will be even less tension in the bottom band. Still, if you clip the top band, the ball falls, but if you clip the bottom band, it stays. The top band is still holding up the ball.

Saying that you drop your hand a little means that for the case of the wheel, the wheel itself is springy, it wants to goes out of round when someone sits on the bike. The spokes are a lot less springy than the wheel, and it turns out that this means that the top spokes keep almost the same tension, the bottom ones give a little, and the wheel stays almost round.

When you say "support" from the bottom, you mean "push up" and a spoke never pushes, it always pulls. If it tries to push, it will bend. The top spoke pulls up with about the same tension always, and is supporting, the bottom spokes pull down, but not as much when you sit on the bike.

 

[K^2]

Top spokes support the weight plus the tension of the lower spokes. That's all there is to it.

If you measurements aren't showing a change in tension in the top spokes, something is wrong with the measurement. In order to reduce tension in lower spokes, the axis shifts slightly down. That reduces the tension in lower spokes and increases tension in top spokes. Since displacement is minuscule, the Young modulus in the spokes cannot change significantly. (Second-order effect.) That means the tension in lower spokes should go down by the SAME amount as tension goes up in the top spokes.

 

[sophiecentaur]

The non-rigid rim makes a difference and transfers force to all spokes. But, resolving the forces on the hub. There must be a net upwards force in order to support the weight of the bike. The hub will move down a finite amount. This must imply length changes - hence changed tensions.

 

[sophiecentaur]

On second thoughts, even with a rigid rim, the forces on all spokes will change when the hub is loaded, as soon as one spoke stretches, force is transferred to others.

 

[Rap]

If you measurements aren't showing a change in tension in the top spokes, something is wrong with the measurement.

This is only true if the wheel rim is rigid, but its not, its flexible, much more flexible than the tensioned spokes.

 

[sophiecentaur]

@Rap
It's time for some basics. If the bicycle is not sinking to the ground then the sum of all forces on the hub must be upwards (to balance the weight). Therefore there must be more up forces than down forces - however you care to look at it. If the up forces are greater than the down forces then the top spokes must have more tension in them - or the lower spokes are in compression. There is nowhere else for the forces to come from, is there?
This is a classical example of a Mechanics problem which has a 'best' approach to the solution. Considering forces the hub clinches it and doesn't involve bothering with the rim at all.

The only time your idea could be considered would be if the arrangement were so 'saggy' that the loaded wheel ended up with more spokes pulling up than pulling down so that the up force was shared between more spokes. I don't care to do the sums but I am pretty sure that even that condition wouldn't make things behave as you claim. But how can your rim change shape significantly, as you claim, without the spokes changing length? You'e got Triangles in the structure. The strongest shape known to Mankind - next to the sphere (tongue in cheek).

 

[Rap]

@Rap
It's time for some basics. If the bicycle is not sinking to the ground then the sum of all forces on the hub must be upwards (to balance the weight). Therefore there must be more up forces than down forces - however you care to look at it. If the up forces are greater than the down forces then the top spokes must have more tension in them - or the lower spokes are in compression.

Yes, the top spokes must have more tension in them than the down spokes when the weight is on. But that does not say anything about whether the top spokes must have more tension in them than when the weight is off. That's the problem at hand, why is the tension in the top spokes the same (or very nearly the same) whether the weight is on or off?

What I am saying is that if the wheel rim is absolutely rigid, then yes, when the weight goes on, the hub must drop a little from the center of the still-circular wheel, and so the top spokes will stretch and the tension in the top spokes will increase from the zero-weight case. The bottom spokes will shrink and the tension in the bottom spokes will decrease from the zero weight case. But if the rim is springy, then all bets are off. The hub will still drop a little from the zero weight case, but the rim will squish down slightly too, into an ellipse. It could be that when the weight goes on, the top spokes don't stretch much at all and the bottom spokes are the only ones that change length. What I am saying is that if the rim is very springy compared to the spokes, that's just what will happen.

 

[sophiecentaur]

Perhaps you need to specify you model more accurately. I was assuming a normal bicycle.

 

[Rap]

Perhaps you need to specify you model more accurately. I was assuming a normal bicycle.

I will try to make a mathematical model, if that would help. But what is a normal bicycle? I think if you pull on a normal bicycle rim with no spokes it will deform a lot more than if you pull on a lone spoke with the same amount of force.

 

[AlephZero]

The point is that the rim IS springy, compared with the spokes.

First, the spokes are acting in tension which is the most efficient use of the materal to provide stiffness. If the rim goes non-circular, it is bending, which is much more flexible.

Second, the rim is in compression, because of the tension in the spokes. This reduces the effective bending stiffeness of the rim from its "no-load" condition, because the tension is tending to make the rim buckle.

If you take a "normal" bike wheel rim on its own, with no spokes, it only takes a very small load to squash it slightly into an ellipse shape. The amount of local squashing required to take say 100 lbf of pre-tension out of one spoke is only a fraction of a millimeter. So when the ground "pushes" against the outside of the rim at just one point, the rim deforms in that local region to reduce the tension in a small number of spokes, and the forces on the other side of the wheel are not changed by any significant amount.

If the rim is very stiff, the situation is different, and the load path from force on the ground goes right round the rim (which stays circular), and the tension in the spokes on the opposite side does increase.

If you want to see some "standard" theory about this, google for "beams on elastic foundations". A typical example in civil engineering is a beam laid on the ground (e.g the concrete foundation of a house wall) with a point load acting on it.

This is almost the same configuration as the bike wheel, if you imagine the wheel rim to be "unwrapped" into a straight beam. The distributed stiffness (or flexibility) of the ground corresponds to the stiffness of the spokes, and the point load corresponds to the force of the ground on the bike wheel. There are standard formulas to give the length over which the point load is spread into the ground, depending on the relativie stiffness of the beam itself and the ground it is resting on.

The analogy is not perfect, but it is near enough to see the basics of what is happening.

 

[sophiecentaur]

This is turning ridiculous. If the rim is "flexible" because it will bend when not part of a wheel then you can also say that the spokes are flexible if you take one on its own and bend it with your hands. We are considering a structure as a structure and not as components.
The rim has a huge cross section compared with the spokes and is in compression. If you want to flex the rim (when part of a wheel) then you must also stretch the spokes differentially. The compression force is spoke tension times the sine of the (small) angle between the spokes - much less and the spoke CSA is a small fraction of the rim's. That means that the stress is much much smaller on the rim than the spokes. The rim needs also to handle other forces such as uneven road surfaces. You can permanently knacker a spoke with thumb and forefinger but, unlike the rims, they are not designed for that kind of abuse.
Are you saying that the small arc between spoke attatchments 'straightens out' when the ground force acts on it? The tyre makes sure that there is no high local stress on the rim from ground contact, by spreading this force over more than one segment of arc. You would need to draw a picture to show what form this "deformation" of the rim you are describing. Does it not act in the same way as an arch, in which all the forces are directed tangentially?

But, in any case, you still haven't defined your configuration or the model you are discussing. Is the wheel loaded by a bicycle? Do the spokes / rim / hub have any mass? In a "normal bicycle", the significant forces are the stresses in the spokes and rim and the only other forces of relevance will be the weight of the bike and the normal reaction from the ground. Is that the situation you are discussing? That is "normal" isn't it? Else, if you take an isolated wheel, it could be horizontal and that would be another matter.

You also haven't answered my reasoning that, considering forces on the hub, there must be more force from the wheel up than down, to balance the weight of bike plus load. More force means more tension. What's wrong with that? It answers the OP perfectly well.

Btw, I just read something in the OP which made me laugh. There is no way the bottom spokes can support the weight of anything because that would imply they are acting in compression. How could they do that? There is nothing at all to prevent a spoke from pushing right through the rim and going into the inner tube. They may have less tension in them but never compression.

 

[AlephZero]

This is turning ridiculous. If the rim is "flexible" because it will bend when not part of a wheel then you can also say that the spokes are flexible if you take one on its own and bend it with your hands.

The flexibility of the spokes in bending is irrelevant. The reason they are pre-tensioned is to make sure they don't bend.

We are considering a structure as a structure and not as components.

Agreed. That is why you need to consider the relative stiffness of the components, to understand correctly where the load paths through the structure are. There are two load paths topologically "in parallel" between the road and the wheel hub. One is along the spoke at the bottom of the wheel (by reducing the pre-tension in the spoke). The other one is around the rim and into other spokes. Which load path carries the most load depends on the relative stiffness of the rim and the spokes.

The compression force is spoke tension times the sine of the (small) angle between the spokes

No, if you draw a free body diagram for one spoke, you will see it is the spoke tension divided by the sine of a small angle. (There are some factors of 2 involved as well, but let's not quibble about those).
An easier way to find the compression in the rim, and see that it is greater than the tension in a single spoke, is to consider the free body diagram for half of the rim.

The tyre makes sure that there is no high local stress on the rim from ground contact, by spreading this force over more than one segment of arc.

I agree the tyre will spread the load over a circumferential distance comparable with the contact area between tyre and road, but that is usually small for bike tires. The tire will only spread the load around a large arc if it is stiffer than the rim, which doesn't seem very plausible to me.

But, in any case, you still haven't defined your configuration or the model you are discussing. Is the wheel loaded by a bicycle? Do the spokes / rim / hub have any mass? In a "normal bicycle", the significant forces are the stresses in the spokes and rim and the only other forces of relevance will be the weight of the bike and the normal reaction from the ground. Is that the situation you are discussing? That is "normal" isn't it? Else, if you take an isolated wheel, it could be horizontal and that would be another matter.

Now you are being ridiculous. The OP was talking about a load equal to the weight of the rider. To keep it simple, let's assume the bike is not moving.

You also haven't answered my reasoning that, considering forces on the hub, there must be more force from the wheel up than down, to balance the weight of bike plus load.

Agreed, but...

More force means more tension. What's wrong with that?

... more force can also mean less tension in the spokes at the bottom of the wheel, and the same tension in the spokes at the top. That is what the OP measured, and I don't have any problem understanding why the measurents were correct.

 

[Rap]

You also haven't answered my reasoning that, considering forces on the hub, there must be more force from the wheel up than down, to balance the weight of bike plus load. More force means more tension. What's wrong with that? It answers the OP perfectly well.

It does not. Please read my post #11. To reiterate mathematically - Let's suppose for a wheel with a massless hub, the upward force on the hub is F due to the tension in the upper spokes. The downward force due to the tension in the lower spokes is the same F, by symmetry. The total force on the hub is F-F=zero. The total force on the massless rim is the downward force due to the upper spokes, again F, and the upward force due to the bottom spokes, which is again F, and whose sum is likewise zero. Now press down on the hub with a weight W. The upward force on the weighted hub will be F1, due to the new tensions in the upper spokes. The downward force will be F2+W, due to the weight and the new tensions in the lower spokes. Their total will be zero, so that F1=F2+W. The force on the rim will be F1 downward, due to the tension in the upper spokes, F2 upward due to the tensions in the lower spokes, and W upwards applied by the road at the junction between the road and the tire. The total force on the rim is zero, so, again F1=F2+W. What you are in effect saying is that F1 is greater than F2, which is obviously true, because F1=F2+W. But the original problem is asking why is F1 equal to F? This is not so obviously true.

It certainly will not be true if the rim is rigid. If L is the distance from the top of the rim to the hub in the unloaded case, then L is also the distance from the hub to the bottom of the rim in the unloaded case. When the hub is loaded, it will drop from the center of the circular rim. Let's say L1 is now the distance from the top of the rim to the hub and L2 is the distance from the hub to the bottom of the rim. For a rigid rim, we must have L1+L2=2L. Since the hub drops, L1 gets larger, L2 gets smaller. As a rough first model, we can express the spring equations for the spokes by F=K(L-L0), F1=K(L1-L0), and F2=K(L2-L0) where K is the effective spring constant of all the upper (or lower) spokes taken together and L0 is the untensioned length of the spoke. Since L1+L2=2L, we can see that F1+F2=2F. Since F1=F2+W, we see that F1=F+W/2 and F2=F-W/2. In this case the tension in the upper spokes is larger than F, the tension in the lower spokes is smaller than F.

If the rim is not rigid, then the restriction that L1+L2=2L no longer holds. We would expect that L1+L2 is smaller than 2L in this case. It COULD BE the case that L1=L and L2 is smaller than L. In that case, using the spring equations and F1=F2-W, we get F1=F and F2=F-W. In terms of lengths, L1=L and L2=L-W/K.

Ok, it COULD BE but what is the actual case? You can use the principle of minimum potential energy (See http://en.wikipedia.org/wiki/Minimum_total_potential_energy_principle) which says that the potential energy stored in the rim due to deformation of the rim plus the potential energy stored in the spokes due to the deformation of the spokes plus the increase in potential energy as the weight W goes down from the center of the wheel must be at a minimum. That is, L1 and L2 will adjust so that that potential energy sum is at a minimum. We can ignore the potential energy in the rim if it is much springier than the spokes (much smaller spring constant). The potential energy change in the upper spokes is like (K/2)(L1-L)^2 and in the lower spokes its (K/2)(L2-L)^2, and the increase in potential energy of the hub is -W (L2-L). If you minimize the sum of these three terms with respect to L1 and L2 you get L1=L and L2=L-W/K.

Btw, I just read something in the OP which made me laugh. There is no way the bottom spokes can support the weight of anything because that would imply they are acting in compression. How could they do that? There is nothing at all to prevent a spoke from pushing right through the rim and going into the inner tube. They may have less tension in them but never compression.

Exactly, as I said in #5.

 

[sophiecentaur]

The OP asks what happened to the load on the top spokes. The answer is that the loads are shared by changes in the tensions of all spokes. The bottom line is that the measurements were not complete or accurate enough. There can be no argument with the fact that, if you measure the tension is all the spokes and add them up, vectorially, the resultant will be to produce an upward force equal to the load.

@AZ
You speak sense, mostly but I don't agree with all of your post. Btw my "normal bicycle' comment was hoovering up some statements made in subsequent posts. A bit of irony, perhaps? We are now back on a bike like yours and mine.

Yes, of course you are right about the 1/sin. I just had to draw it out!

In order to simplify the problem it must be ok to replace the circular rim with a polygonal one. Once the tyre is in place, a wheel of either construction would behave the same, I'm sure. This takes away the problem of 'flattening' the bottom segment/s of the rim, which I think is a bit of red herring.
I even reckon you could replace this rigid polygon with a set of struts, joined to each other and to the spokes by flexible links (giving a totally floppy rim) - for the purpose of this discussion, at least - but no good for a practical bicycle because, no doubt, the spokes would need to have even more tension and twisting would also be a problem.

I don't think any great stiffness is needed for the tyre, btw, will there not be a general pressure on the rim, the same way that there's a general pressure on the ground footprint?

You might reconsider you last statement, though: "More force can mean less tension"
How? OMG- I just read another post of yours which agrees with me. 'Letting go' isn't pushing.

I think the real explanation for the measurements that are quoted earlier is that the situation is not symmetrical (up/down). The lower few spokes have measurably less load on them but the extra vertical load will be spread around by changes in all the other spokes.

Imagine a wheel with just four spokes connected at the corners of a infinitely stiff square rim, with two spokes vertical. After the necessary sag, the two almost horizontal spokes will increase greatly in tension and the vertical component of that tension will take away a portion of that needed by the top spoke. So the reduction in load on the bottom spoke is shared by all three spokes - not just the top spoke. Increasing the number of spokes can continue until the rim is circular enough to put a tyre round it. Spokes below the horizontal will have less than static tension and spokes above will have more. If the rim is very rigid, the tensions in the more horizontal spokes could, I think, be large. The detailed way in which the tensions vary would depend on the stiffness of rim and spokes.

A real wheel is not, of course, strung from the centre and the hub has a very finite diameter and the spokes are attached tangentially, almost, to allow for propulsion and braking (hub brakes, that is). This complicates things a lot but one thing at a time, I think.

 

[sophiecentaur]

Rap
What is the point of a "massless hub"?
The whole bike rests on the hub so the two add together, surely.

How can you possibly object to the fact that the hub is in equilibrium and that upwards forces must balance downward forces? The vector sum of all spoke tensions just has to equal the weight on the hub. What else could it equal? You wanted "maths" and you can't get more precise than that.

 

[Studiot]

Spoked cycle wheels are very difficult to analyse, structurally.

This is in part due to the fact that the spokes are not radial, but connected from the rim to the edge of a flange circle at the hub. In particular the tension in the spoke is not normal to the rim tangent, but has both radial and tangential components.
Secondly spokes are fitted in crossing pairs which have a friction contact point about one quarter the radius out from the flange. Substantial force is therefore transferred by this contact.

 

[sophiecentaur]

Spoked cycle wheels are very difficult to analyse, structurally.

This is in part due to the fact that the spokes are not radial, but connected from the rim to the edge of a flange circle at the hub. In particular the tension in the spoke is not normal to the rim tangent, but has both radial and tangential components.
Secondly spokes are fitted in crossing pairs which have a friction contact point about one quarter the radius out from the flange. Substantial force is therefore transferred by this contact.

Very difficult.
And it's interesting that you have introduced the friction between spokes too! That must imply that the spokes actually deflect!

Until my previous post, the thread had been assuming radial spokes, for simplicity but the torque when under power must be pretty significant., the hub 'dragging' the rim behind it as it goes round. If the spokes weren't strung like they are, they would just tend to be 'wound round' the spindle when the pedals were turned and be subjected to very high tension.. So, the ideal front wheel could be strung differently from the ideal back wheel??
I would imagine that a lot of wheel design used to rely on suck it and see. There must have been some pretty rubbish designs on the market before people sorted out the basics.

 

[Rap]

Rap
What is the point of a "massless hub"?
The whole bike rests on the hub so the two add together, surely.

Just to keep the analysis simple. The weight of the hub (less than a pound) is negligible compared to the weight applied to the hub when the 205 pound guy sits on the bike, which, for a two-wheel bike is about half that.

How can you possibly object to the fact that the hub is in equilibrium and that upwards forces must balance downward forces? The vector sum of all spoke tensions just has to equal the weight on the hub. What else could it equal? You wanted "maths" and you can't get more precise than that.

I don't object to it - its obviously true. Please read #17. When there is no weight on the hub, the upward force is F, the downward force is F, and vector equilibrium says that F-F=0. That expresses the equilibrium for the first case. When there is weight W on the hub, the upward force is F1, the downward force is F2+W and I explicitly use the equilibrium equation F1=F2+W (or, equivalently, F1-(F2+W)=0) all over the place.

Do you understand the difference between saying

A) F1 > F2

and

B) F1 > F

I am saying sure, F1>F2 because by equilibrium F1=F2+W. But that is not the same as saying F1 > F. In fact, for a springy rim, F1=F is very close to true.

 

[AlephZero]

In order to simplify the problem it must be ok to replace the circular rim with a polygonal one.

You are probably right, but the two are different. A straight section of rim is very stiff in the direction along its length. between the ends of two adjacent spokes. A curved section is much less stiff in that direction.

BTW finite element models can add more confusion to this situation, depending how the particular FE system models curved beams in a "linear" analysis.

You might reconsider you last statement, though: "More force can mean less tension"
How? OMG- I just read another post of yours which agrees with me. 'Letting go' isn't pushing.

I was thinking about the change of forces relative to the equilibrium configuration (pre-stressed spokes, and no external forces on the wheel.). In that sense, the spokes can "push" on the hub. (I spend most of my "real life" doing dynamics not statics, and thinking about perturbations about steady state solutions.)

The detailed way in which the tensions vary would depend on the stiffness of rim and spokes.

YESSSSS! That is exactly the point!

A real wheel is not, of course, strung from the centre and the hub has a very finite diameter and the spokes are attached tangentially, almost, to allow for propulsion and braking (hub brakes, that is). This complicates things a lot but one thing at a time, I think.

It has no significant effect on how radial forces are transferred from the rim to the hub. A pair of "tangential" spokes which are mirror images of each other, attached to either side of the hub, act the same way as one "radial" spoke.

Of course it makes a huge difference to the torsional stiffness of the wheel, as you said, but I don't think the OP was asking about that.

 

[Alex James]

It's a just a matter of practice. Try starting at the supports. You know the direction of the reaction there.



____________________
http://www.kindsofcats.org/abyssinian-cats-colors/

 

[rcgldr]

Assuming the rim isn't rigid, then what the test results indicate is that almost all the deformation of the rim occurs near the point of contact with the road, which make sense. The hub also deforms, but having a much smaller radius, most of the deformation occcurs at the rim. This would correspond to the measured results that the lower spokes experience a reduction in tension, due to the rim deformation, while the upper spokes experience no change, since the upper portion of the rim does not deform.

Regardless if the rim is rigid or not, the hub will be displaced downwards slightly due the load. If most of this corresponds with deformation of the rim below the hub, then most of the reaction will be reduced tension in the lower spokes. If the rim is relatively rigid compared to the spokes, then the load would be more evenly distributed between the lower (less tension) and upper spokes (more tension).