ultraman6970

I think just like you Captain Blight. Well the wheel has only one point of contact to start with, the ground (im not taking in consideration the hub ok?). The question is pretty good but in which situation the wheel will stand forces in the top and in the ground at the same time? In my opinion this does not happen in the real world as the op mentioned. As a note, from the Op question im under the impression he meant forces instead of stress? My asnwer is based on this ok?

Now imagine the situation (are you talking about this, right?)

If you put the wheel perfectly perpendicular to the ground and you seat over the wheel you are putting all the weight over it, lets say 200 pounds of weight (90 kilos to the metric people), no matter how many spokes u have in there anyways but well, you seat over it, then some spokes above the hub will compress and the ones under the hub will compress, with compress i meant lose tension (some body said this already) because thats what the rim will do, rims work compressed in the contact points and traction in the left and right of the wheel in this case. In this case the amount of force applied in the upper rim and in the lower rim should be equal. You apply 200 pounds going down perpendicular to the ground, then from the ground up you will get the same amount of force going up, this is not new anyways, just basic physics. In this particular case u have two contact points, one at the top and one at the bottom and all the math should be done in both contact points but the amount of force should be equal or almost equal in both contact points (haven't done the math ok?)

Now if you put the wheel in the fork, this change the scenery because the force now is divided in different components, one perpendicular to the ground, one goes toward the fork and the last one goes toward the downtube.

For example u just put 200 pounds of weight in the handlebar (imagine the op seated in the handlebars), the weight/force/vector will have more components because now some of the force will go down and some of the force will go throwout the fork to the hub, so the force I imagine will have 3 components, one perpendicular to the ground and the second one should be something like (F*sin@) being the @ the angle of the fork or headtube (haven't done this in centuries so the sen could be cos and im not doing any drawings to know how the forces go, neither how is the right composition of those vectors ok?), and the third one should be like F*cos@ that goes towards the downtube. So now the force that u get into the hub is not the full 200 pounds but a fraction of it. One part goes perpendicular to the ground, one goes to the downtube and the last one to the hub towards the fork. So the answer to the op second question is that the forces/stress the wheel stands in this case is just a fraction of the total weight applied.

Good luck

ultraman6970

Why would it happen under load <-- I think is because the rim is compressed under load, the spokes lose tension then unwind a little bit then to get in the right position. Is the whole spoke the one that unwind, not the head of the spoke.

I have some bladed wheels and as rydaddy mentioned, if you do not let the spoke to twist while truing the spoke wont ping at all, in a matter of fact haven't heard mines to ping ever after a good truing work. When i got them i had to release all the spokes and true back from zero.

DannoXYZ 

Back to the original question.

Load on a tensioned wheel is similar to a Bernoulli buoyancy equation where the downward load is balanced by an equal upward force. In the case of the wheel, adding 100kg of load to the wheel will unload/untension the spokes in the loaded zone by 100kgf tension regardless if the load is applied at the hubs or at the top of the wheel.

In the case of applying that load to the hub, only the bottom spokes will be unloaded. Say the two directly below unloads by 67kgf (33.3 each) and the two adjacent ones by 33kgf (16.5 each). At that amount of sink & rim-compression, all the other spokes will have increased in tension by a sum of 100kgf, all forces balance out, and the system will be come static.

In the case of applying that same load to the rim at the top of the wheel, you will still unload 100kgf of total spoke-tension. However, you will now unload 8-spokes instead of just 4-spokes (4 on top and 4 on the bottom). In which case, the actual amount of spoke untensioning will be only one-half of what it would be if the load was applied to the hubs. So the top four spokes will loose 50kgf of tension and the bottom 4 spokes will lose 50kgf. The remaining spokes will still increase tension by a sum total of 100kgf.

If you measure rim-deflection due to load, you'll find that the loading at the hub creates a flat spot on the bottom. Loading at the top of the wheel will result in a oval shape with flattening at the top and bottom. BUT, the actual amount of total compression will be the same. So loading on the rim at the top will compress the top by 1/2 and the bottom by 1/2 of the amount of the wheel that's loaded at the hub.

dasding

yeah, that was the original query: how strong a wheel is with weight put on top of it while it is standing on the ground, not attached to anything. kind of an academic question, really. i've started a whole wheel war here

dasding

wow thanks for all of this! next activity tonight is googling "Bernoulli buoyancy equation"!!

vredstein

The Buddhist in me refuses to consider the question as it does not lead to the cessation of suffering.

rydaddy

So, if I am translating Danno's excellent post correctly, it will take 2x as much load at the top of the wheel (vs at the hub) before any spokes go slack.

ratdog

That is because like a wheel, suffering does not have a beginning or end but is a continuous cycle.

SCROUDS

I'm recommending that we have mythbusters test this... with C4.