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#31
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MA3 rim failure, where to now
Just zis Guy, you know? wrote:
I have nothing against a simplistic description of a complex system, as long as it makes reasonable sense, but describing a hub as "standing" on spokes which patently cannot support a net compressive load does not provide any kind of illumination. It does if you understand that the load-supporting is considered as a change relative to the unloaded state. Since all sides seem to agree on the actual tensions and changes this seems like a rather sterile debate on semantics, but I personally find the 'standing on the lower spokes' description a more useful and succinct description than the alternative. James |
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#32
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MA3 rim failure, where to now
Ted Bennett wrote:
If it hangs from the top, then the tension in the top spoke would increase with load. But it doesn't; the tension in the lower spoke decreases. A simple test, plucking a few spokes, may help convince you. That alone doesn't convince me because rim deformation may be responsible for the decrease in tension in the lower spokes (for all I know). The tension in the rest of the spokes may be increasing for all I can tell by plucking the spokes because the load could be spread over so many spokes (not just those right at the top) that the change in pitch is not enough to notice by ear. That's not to say that I'm convinced the rim hangs from the top, just that you need better arguments. ~PB |
#33
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MA3 rim failure, where to now
"James Annan" wrote in message
... I have nothing against a simplistic description of a complex system, as long as it makes reasonable sense, but describing a hub as "standing" on spokes which patently cannot support a net compressive load does not provide any kind of illumination. It does if you understand that the load-supporting is considered as a change relative to the unloaded state. But once you take into account the fact that without those spokes which do nothing the wheel collapses, and without the spokes which do all the work it stays up, and the mechanism by which the substantial change occurs is simply the deformation of the rim, the word "stand" ceases to have any meaning or use. I refer the hon. gentleman to my earlier answer: the hub is supported by the spoke nipples pushing on the rim ;-P -- Guy === WARNING: may contain traces of irony. Contents may settle after posting. http://www.chapmancentral.com |
#35
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MA3 rim failure, where to now
In article , bikerider@-
nospam-thanks-rogers.com says... So the spokes are compressed. This is understood by analyzing the changes in loading rather than absolute loads. But it is confusing to many non-technical people. The spokes at the bottom of a loaded bicycle wheel are "compressed" relative to their unloaded condition, but are not "in compression". The spokes remain in tension--the net axial force acting on the spokes is tensile, not compressive. Rick |
#36
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MA3 rim failure, where to now
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#37
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MA3 rim failure, where to now
David Damerell writes:
I think you (and many other people) are missing the fact that Simon (and Guy) are disputing the terminology only. No-one is claiming anything other than that there is a tension change in the bottom spokes. Because this is not a semantic difference but a technical one, the book shows an aluminum a common die-cast moped wheel that looks as though it might have wire spokes. http://mopedarmy.com/photos/brand/6/1681/ In such wheels, knowing that they are not tensioned, evokes the response that "of course, this wheel stands on its bottom spokes." However, by selective cooling in the die cast process, these spokes can be tensioned, and the answer becomes unclear. Visually it appears to be between a wire spoked wheel and a wooden wagon wheel. Does prestress of a spoke change its function and that of the wheel? Jobst Brandt |
#38
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MA3 rim failure, where to now
On Wed, 10 Sep, Just zis Guy, you know? wrote:
"Ian Smith" wrote in message ... So, the spokes that have the same stress state whether loaded or unloaded are doing all teh work, and the spokes in which the stress changes dramatically are doing nothing at all. As is readily understood by considering the limiting case where tension = 0 in the bottom-most spoke, quite right. I think you need to think about cause and effect a little. You are saying the effect of a large stress change is caused by nothing happening. Meanwhile, no stress change is caused by the spokes carrying teh weight applied to the wheel. Curious. regards, Ian SMith -- |\ /| no .sig |o o| |/ \| |
#39
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MA3 rim failure, where to now
On Wed, 10 Sep 2003, Simon Brooke wrote:
Some stuff that's so badly wrong I have trouble believing he actually meant it. However, just in case he believes what he wrote, I'll go through it slowly... Ian Smith writes: So, the spokes that have the same stress state whether loaded or unloaded are doing all teh work, Yes, they are. Take a bicycle wheel. Cut all the spokes below the hub. Does the hub move? At real levels of load, yes, the wheel collapses due to flexural failure of the rim in the lower portion. The hub hits the floor. So, your argument actually falls at teh very first hurdle, however, let's suppose you want to talk about useless wheels that can't support much load at all, and thus the rim manages to withstand the flexural loads. The argument is still rubbish. You're now not talking about a bicycle wheel but a totally different structure that carries load in a different way. Stated more formally: teh wheel is statically indeterminate, so altering the stiffness of components fundamentally alters the loads in teh other components. If you remove spokes, you effectively set their stiffness to zero - as big a change as can be made. Statical indeterminancy is a tricky concept to grasp, but there's some discussion and simple examples on my web page. As an example of why it;s so badly wrong, let's test your method on another structure. What your method seems to be is to imagine removing a component, work out if the structure still works, and if it does, decide the component did not contribute to the working even when it was there. Let's apply this theory to teh Eiffel Tower. Suppose we remove the northmost leg. Does the tower fall over? No - it has three remaining legs, each with a large footprint, so it continues to stand (albeit somewhat less stable). So, your argument deduces that the northernmost leg does not contribute to the tower standing up. Now, since it's symmetrical in 4 parts, exactly the same argument can be applied to each leg - anything that works for the north leg must work for the east leg, for example. Thus, we reach the conclusion, applying your argument, that the Eiffel Tower stands up without any contribution from any of teh legs - it simply floats there, hanging (presumably) from sky-hooks. This is clearly not the case, so (evidently) your thought experiment actually doesn't tell you anything useful about how a structurally indeterminate structure stands up. and the spokes in which the stress changes dramatically are doing nothing at all. The change is when they _cease_ to do work, not when they _start_ to do work. Eh? You appear to be saying taht something can resist some load without any change in its state of stress. That's a fairly novel concept, and on teh assumption that you're not intent on creating a whole new understanding of structural mechanics, it might be helpful to explain what you mean in more detail. Consider a tug of war. Two teams heave on a rope, and the hankerchief stays over the line, because each team is heaving equally hard. Now suppose the North team go off and get a beer. Their end of the rope goes slack, and the handkerchief moves. Is this because the North team are doing more work? That is your argument. Sorry, no. My argument is that the north team pulls less hard, and teh handkerchief moves. Which team caused teh handkerchief to move? You're saying the south team did, but they are doing nothing different, so it's not sensible (by cause and effect) to say they caused teh change. The North team, on teh other hand, have changed what they are doing, and as soon as they changed what they were doing, the effect changed. Your argument says that teh team that didn't change somehow made an effect happen, but the team that did change had no effect. Suppose two people have two buttons. Both press button A, and nothing happens. Both press button A, nothing happens. Both press button A, nothing happens. Both press button A, nothing happens. Person 1 presses button A, and person 2 presses button B and this time a bell rings. Your argument is that the person 1, by pressing button A, must have made the bell ring (even though every previous time, pressing button A caused nothing to happen). Frankly, that's just barmy - by occam, the most likely explanation is that pressing button B causes a bell to ring, and person 2 caused the bell. Why don't you do the analysis? At teh very least, you could read Jobst Brandt's book. regards, Ian SMith -- |\ /| no .sig |o o| |/ \| |
#40
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MA3 rim failure, where to now
On Wed, 10 Sep 2003 08:35:05 GMT, Simon Brooke wrote:
If they 'shorten' when they get to the bottom, do they 'shorten' at every revolution? If so, the wheel must logically get smaller and smaller until, eventually, it disappears with a *pop* of collapsing credibility. Unless, of course, they 'unshorten' again somewhere else in the revolution. What do we call 'unshortening', children? Oh, that's right, 'enlongation'. And if they shorten when they get to the bottom, where do they enlongate? That's right, not at the bottom. Very good. Actually, they elongate at teh bottom - alongside teh contact patch. In fact, the greatest elongation is in the lower half of the wheel. Jobst Brandt describes this in his book, and it's clearly visible in my own analysis at http://www.astounding.org.uk/ian/wheel/. The elongation at this pont (in the bottom half of teh wheel, remember) is about twice any elongation occurring in teh upper half. If you're going to be facile, getting your facts right would make you look less silly. regards, Ian SMith -- |\ /| no .sig |o o| |/ \| |
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