Spoke tension and buckling

[Ben C]

I think it's time to start a new thread. This came from "Maintenance
Manuals".

This is what I've gathered so far, which might be wrong:

1. High spoke tension decreases propensity to buckling because so long
as no spokes go slack, the wheel will hold its shape. Higher tension
means higher loads before it goes slack, where "loads" includes
hitting bumps in the road.
2. High spoke tension increases propensity to buckling. What's the
explanation here?

benc> But does it increase resistance to buckling or not? If not, why
benc> not?

jim beam> michael press seems to have the best handle on this.
jim beam> increasing tension increases propensity to buckle. the only
jim beam> wheel i've ever had spontaneously taco on me was a jobst-tight
jim beam> wheel.

Michael Press explained how in the buckled wheel, tension is thought to
be reduced in all spokes and the wheel is in a lower energy state. It
was thought that probably true and tacoed were both local energy minima
with true being higher energy than tacoed. But I'm not sure anything was
said about whether high tension increases propensity to buckling.

We know that if you tension your MA-2 too high then it buckles when you
stress-relieve it. Then you reduce the tension a bit. In theory hitting
bumps isn't supposed to raise the tension, so you should be able to ride
around safely at just below buckle-tension. But perhaps nasty bumps do
sometimes raise the tension a bit?

If true is a local energy minimum, one question is how steep are the
sides of the energy function at that point, with lower or higher spoke
tension. In other words, how much deformation does it take to snap you
into a buckle, and high vs low tension?

Certainly a wheel doesn't buckle when stress-relieved unless tension is
very high.

What does the graph of buckle-propensity against tension look like? Does
it increase monotonically with tension, or does it go downwards up to
some quite high tension and then suddenly shoot upwards at the end?

 

[[email protected]]

Ben C? writes:

> I think it's time to start a new thread. This came from "Maintenance
> Manuals".

> This is what I've gathered so far, which might be wrong:

> 1. High spoke tension decreases propensity to buckling because so
> long as no spokes go slack, the wheel will hold its shape.
> Higher tension means higher loads before it goes slack, where
> "loads" includes hitting bumps in the road.

You should define "buckling" first. While truing and tensioning a
wheel, rim buckling (becoming untrue in a slight taco shape) is a sign
of excessive tension. Collapsing a wheel from overload occurs when
spoke become sack and no longer furnish lateral stability to the wheel
in conjunction with lateral force.

> 2. High spoke tension increases propensity to buckling. What's the
> explanation here?

You haven't followed the discussion, possibly because there were rude
distractions. The buckling effect has been discussed for some time
now under the previous heading.

>>> But does it increase resistance to buckling or not? If not, why
>>> not?

>> michael press seems to have the best handle on this. increasing
>> tension increases propensity to buckle. the only wheel i've ever
>> had spontaneously taco on me was a jobst-tight wheel.

> Michael Press explained how in the buckled wheel, tension is thought
> to be reduced in all spokes and the wheel is in a lower energy
> state. It was thought that probably true and tacoed were both local
> energy minima with true being higher energy than tacoed. But I'm
> not sure anything was said about whether high tension increases
> propensity to buckling.

I explained that in previous responses. I can't imagine how you
missed that. Press also mentioned the same characteristic in that
there are fairly slack spokes in a forcefully pretzeled wheel. This
is not bucking from truing and tensioning. The two effects not being
parallel.

> We know that if you tension your MA-2 too high then it buckles when
> you stress-relieve it. Then you reduce the tension a bit. In
> theory hitting bumps isn't supposed to raise the tension, so you
> should be able to ride around safely at just below buckle-tension.
> But perhaps nasty bumps do sometimes raise the tension a bit?

Don't confuse that effect with a "tacoed" wheel.

> If true is a local energy minimum, one question is how steep are the
> sides of the energy function at that point, with lower or higher
> spoke tension. In other words, how much deformation does it take to
> snap you into a buckle, and high vs low tension?

You are mixing to different effects.

> Certainly a wheel doesn't buckle when stress-relieved unless tension
> is very high.

How "very"? That is a way of gauging whether a wheel is too tight and
it is dependent on the rim and number of spokes as I discussed at
length here and in "the Bicycle Wheel". It is not a single value for
different numbers of spokes.

> What does the graph of buckle-propensity against tension look like?
> Does it increase monotonically with tension, or does it go downward
> up to some quite high tension and then suddenly shoot upward at the
> end?

It is a single value so there is no curve. If you mean how does the
number of spokes affect it, it is not a linear effect because fewer
spokes act as fewer guy wires to keep the rim straight while at the
same time require more tension for the wheel to carry the same load
(there being fewer spokes in the load affected zone.

Jobst Brandt

 

[jim beam]

[email protected] wrote:
> Ben C? writes:
>
>> I think it's time to start a new thread. This came from "Maintenance
>> Manuals".
>
>> This is what I've gathered so far, which might be wrong:
>
>> 1. High spoke tension decreases propensity to buckling because so
>> long as no spokes go slack, the wheel will hold its shape.
>> Higher tension means higher loads before it goes slack, where
>> "loads" includes hitting bumps in the road.
>
> You should define "buckling" first. While truing and tensioning a
> wheel, rim buckling (becoming untrue in a slight taco shape) is a sign
> of excessive tension. Collapsing a wheel from overload occurs when
> spoke become sack and no longer furnish lateral stability to the wheel
> in conjunction with lateral force.
>
>> 2. High spoke tension increases propensity to buckling. What's the
>> explanation here?
>
> You haven't followed the discussion, possibly because there were rude
> distractions. The buckling effect has been discussed for some time
> now under the previous heading.
>
>>>> But does it increase resistance to buckling or not? If not, why
>>>> not?
>
>>> michael press seems to have the best handle on this. increasing
>>> tension increases propensity to buckle. the only wheel i've ever
>>> had spontaneously taco on me was a jobst-tight wheel.
>
>> Michael Press explained how in the buckled wheel, tension is thought
>> to be reduced in all spokes and the wheel is in a lower energy
>> state. It was thought that probably true and tacoed were both local
>> energy minima with true being higher energy than tacoed. But I'm
>> not sure anything was said about whether high tension increases
>> propensity to buckling.
>
> I explained that in previous responses. I can't imagine how you
> missed that. Press also mentioned the same characteristic in that
> there are fairly slack spokes in a forcefully pretzeled wheel. This
> is not bucking from truing and tensioning. The two effects not being
> parallel.
>
>> We know that if you tension your MA-2 too high then it buckles when
>> you stress-relieve it. Then you reduce the tension a bit. In
>> theory hitting bumps isn't supposed to raise the tension, so you
>> should be able to ride around safely at just below buckle-tension.
>> But perhaps nasty bumps do sometimes raise the tension a bit?
>
> Don't confuse that effect with a "tacoed" wheel.
>
>> If true is a local energy minimum, one question is how steep are the
>> sides of the energy function at that point, with lower or higher
>> spoke tension. In other words, how much deformation does it take to
>> snap you into a buckle, and high vs low tension?
>
> You are mixing to different effects.
>
>> Certainly a wheel doesn't buckle when stress-relieved unless tension
>> is very high.
>
> How "very"? That is a way of gauging whether a wheel is too tight and
> it is dependent on the rim and number of spokes as I discussed at
> length here and in "the Bicycle Wheel". It is not a single value for
> different numbers of spokes.
>
>> What does the graph of buckle-propensity against tension look like?
>> Does it increase monotonically with tension, or does it go downward
>> up to some quite high tension and then suddenly shoot upward at the
>> end?
>
> It is a single value so there is no curve. If you mean how does the
> number of spokes affect it, it is not a linear effect because fewer
> spokes act as fewer guy wires to keep the rim straight while at the
> same time require more tension for the wheel to carry the same load
> (there being fewer spokes in the load affected zone.
>

no mention of the wrong type of rims jobst? it makes all your dodging
and weaving about quantification much more amusing.

 

[Ben C]

On 2007-10-08, [email protected] <[email protected]> wrote:
> Ben C? writes:
[...]
>> Michael Press explained how in the buckled wheel, tension is thought
>> to be reduced in all spokes and the wheel is in a lower energy
>> state. It was thought that probably true and tacoed were both local
>> energy minima with true being higher energy than tacoed. But I'm
>> not sure anything was said about whether high tension increases
>> propensity to buckling.
>
> I explained that in previous responses. I can't imagine how you
> missed that. Press also mentioned the same characteristic in that
> there are fairly slack spokes in a forcefully pretzeled wheel.

Are you saying that it _follows_ from the fact that a buckled wheel has
looser spokes that spokes going slack in use leads to buckling, and
therefore that over-tight spokes does not?

If so, I did miss that part. It's not clear to me that that connection
is so easy to make.

[...]
>> What does the graph of buckle-propensity against tension look like?
>> Does it increase monotonically with tension, or does it go downward
>> up to some quite high tension and then suddenly shoot upward at the
>> end?
>
> It is a single value so there is no curve.

I think you're saying pretzeling (as opposed to buckling) is a single
value at a high tension. But collapsing from slack spokes presumably is
some sort of curve, with propensity to collapse going down as tension is
increased?

> If you mean how does the number of spokes affect it, [...]

No, not asking about number of spokes, just how tension affects
propensity to buckle.

 

[[email protected]]

Ben C? writes:

>>> Michael Press explained how in the buckled wheel, tension is
>>> thought to be reduced in all spokes and the wheel is in a lower
>>> energy state. It was thought that probably true and tacoed were
>>> both local energy minima with true being higher energy than
>>> tacoed. But I'm not sure anything was said about whether high
>>> tension increases propensity to buckling.

>> I explained that in previous responses. I can't imagine how you
>> missed that. Press also mentioned the same characteristic in that
>> there are fairly slack spokes in a forcefully pretzeled wheel.

> Are you saying that it _follows_ from the fact that a buckled wheel
> has looser spokes that spokes going slack in use leads to buckling,
> and therefore that over-tight spokes does not?

> If so, I did miss that part. It's not clear to me that that
> connection is so easy to make.

> [...]
>>> What does the graph of buckle-propensity against tension look like?
>>> Does it increase monotonically with tension, or does it go downward
>>> up to some quite high tension and then suddenly shoot upward at the
>>> end?

>> It is a single value so there is no curve.

> I think you're saying pretzeling (as opposed to buckling) is a
> single value at a high tension. But collapsing from slack spokes
> presumably is some sort of curve, with propensity to collapse going
> down as tension is increased?

>> If you mean how does the number of spokes affect it, [...]

> No, not asking about number of spokes, just how tension affects
> propensity to buckle.

If there are no significant side loads, even wheels with slack spokes
do not collapse (unless greatly overloaded). Lateral collapse occurs
from too loose spokes when leaning the bicycle in the standing
position while pedaling hard, or most commonly after a crossing a
patch of poor traction that induces side slip. If traction recovers
after a slide slip, wheels are loaded laterally and a loosely spoked
wheels more readily fold. I don't call that buckling because buckling
is, in engineering terms, is lateral bowing of a column from
compression and is not a result of plastic deformation (see pole
vault). It is an elastic phenomenon that if carried far enough will
cause yield. That is not what causes most wheel collapse but rather
side loading.

Jobst Brandt

 

[Ben C]

On 2007-10-09, jim beam <[email protected]> wrote:
[...]
> no, excessively compressed rims more readily fold.

Suppose the rim is highly compressed by high spoke tension.

Then I hit a bump. The compression from the bump isn't just added to the
compression from the spoke tension, because as the rim starts to
compress, those spokes start to go slack. So the compression on the rim
stays about the same until the spokes have gone completely slack.

Is this right/wrong/too simplistic?

 

[jim beam]

Ben C wrote:
> On 2007-10-09, jim beam <[email protected]> wrote:
> [...]
>> no, excessively compressed rims more readily fold.
>
> Suppose the rim is highly compressed by high spoke tension.
>
> Then I hit a bump. The compression from the bump isn't just added to the
> compression from the spoke tension, because as the rim starts to
> compress, those spokes start to go slack. So the compression on the rim
> stays about the same until the spokes have gone completely slack.
>
> Is this right/wrong/too simplistic?

the hoop compression is a function of spoke tension. i doubt locally
slackening spokes will reduce the compression of the whole because it's
an equilibrium structure and the bump load has to be taken up by the
whole. as far as local bending is concerned, it's a bending beam, with
a neutral plane, and compressive and tensile regions accordingly.

 

[Peter Cole]

jim beam wrote:
> Ben C wrote:
>> On 2007-10-09, jim beam <[email protected]> wrote:
>> [...]
>>> no, excessively compressed rims more readily fold.
>>
>> Suppose the rim is highly compressed by high spoke tension.
>>
>> Then I hit a bump. The compression from the bump isn't just added to the
>> compression from the spoke tension, because as the rim starts to
>> compress, those spokes start to go slack. So the compression on the rim
>> stays about the same until the spokes have gone completely slack.
>>
>> Is this right/wrong/too simplistic?
>
> the hoop compression is a function of spoke tension. i doubt locally
> slackening spokes will reduce the compression of the whole because it's
> an equilibrium structure and the bump load has to be taken up by the
> whole.

In other words, no, it's not "too simplistic" a bump won't significantly
change the rim compression, which is why you can operate a wheel within
a fairly small safety margin of compression buckling. The limit is more
what happens to wheel shape when you break a spoke.

> as far as local bending is concerned, it's a bending beam, with
> a neutral plane, and compressive and tensile regions accordingly.

A beam which is supported by the spokes, with the consequence that the
stiffness dramatically lowers when the support is removed.

The net is you want to have tight spokes to prevent excessive (damaging)
deflection of the rim under overload.

 

[Ben C]

On 2007-10-09, Peter Cole <[email protected]> wrote:
> jim beam wrote:
>> Ben C wrote:
>>> On 2007-10-09, jim beam <[email protected]> wrote:
>>> [...]
>>>> no, excessively compressed rims more readily fold.
>>>
>>> Suppose the rim is highly compressed by high spoke tension.
>>>
>>> Then I hit a bump. The compression from the bump isn't just added to the
>>> compression from the spoke tension, because as the rim starts to
>>> compress, those spokes start to go slack. So the compression on the rim
>>> stays about the same until the spokes have gone completely slack.
>>>
>>> Is this right/wrong/too simplistic?
[...]
>> as far as local bending is concerned, it's a bending beam, with
>> a neutral plane, and compressive and tensile regions accordingly.
>
> A beam which is supported by the spokes, with the consequence that the
> stiffness dramatically lowers when the support is removed.

Sticking to local bending, _which_ spokes is the region of rim hitting
the bump supported by?

If it's the ones running from the hub to the bump, then those spokes
going slack would reduce the stiffness as you say.

But if it's all the other spokes, which aren't going slack, providing
the support, then the stiffness of the rim (locally) wouldn't change
much at the point where you hit the bump. Or it might even get stiffer
as the spoke goes slack (see below).

If the spokes weren't pretensioned and this was a cart wheel, it would
seem quite intuitive that it's the spokes near the bump that are
propping up the rim at that point. And people keep saying that a
pretensioned wheel behaves the same as a cart wheel.

But does it in every detail? It seems like it should be the other way
round. The effective stiffness of the rim at the bump should _increase_
as those spokes above it go slack, reaching its maximum stiffness when
they are completely slack.

Suppose you have a helical compression coil spring held in compression
with a rubber band around it. The spring holds the band in tension and
the band holds the spring in compression. They're in equilibrium. Both
obey Hooke's Law.

What's the response of the whole thing to compression and tension? If
you compress it, you're squashing the spring, but releasing the band a
bit. If you stretch it, you're stretching the band, and relaxing the
spring a bit.

If K is the spring constant of the spring, and L the spring constant of
the band, I would expect the spring constant in compression (for small
deviations from the equilibrium position) to be K - L and in tension
(for small deviations from the equilibrium position) to be L - K.

(But I would expect the system as a whole not to obey Hooke's Law-- it
should be like a spring that gets a bit stiffer the further you stretch
or compress it as the force contribution from the other element gets
less the further you go).

So, if the stiffness of the rim on its own is K and that of the spoke L,
then squashing the rim in (locally) should react like something with a
stiffness of K - L. When eventually L goes to zero (slack spoke) the
stiffness is K, which is higher than K - L.

Where have I gone wrong here?

> The net is you want to have tight spokes to prevent excessive (damaging)
> deflection of the rim under overload.

 

[[email protected]]

Ben C? writes:

> [...]
> > no, excessively compressed rims more readily fold.

> Suppose the rim is highly compressed by high spoke tension.

I don't understand what you mean by this. If the wheel can withstand
hard rim braking it is not over structurally tensioned although rims
without sockets can fail by cracking (reducing tension in those
spokes) with use.

> Then I hit a bump. The compression from the bump isn't just added
> to the compression from the spoke tension, because as the rim starts
> to compress, those spokes start to go slack. So the compression on
> the rim stays about the same until the spokes have gone completely
> slack.

I think the deflection diagrams in "the Bicycle Wheel" make that
clearer as was mentioned about diagram being worth many words.

> Is this right/wrong/too simplistic?

I think you can test this on your own wheels (if they don't have fewer
than 20 spokes) by having someone pluck spokes while you stand on one
pedal and lean the bicycle with both feet in the pedals as in riding
while standing. You'll notice that the bottom spokes of the wheels
lose tension unequally, the "upperside" ones losing more tension than
the "underside" spokes. You'll need to lean the bicycle more than
what occurs while riding to get any other response. That would be at
an angle that one cannot pedal the bicycle.

Jobst Brandt

 

[Joe Riel]

Ben C <[email protected]> writes:

> Suppose you have a helical compression coil spring held in compression
> with a rubber band around it. The spring holds the band in tension and
> the band holds the spring in compression. They're in equilibrium. Both
> obey Hooke's Law.
>
> What's the response of the whole thing to compression and tension? If
> you compress it, you're squashing the spring, but releasing the band a
> bit. If you stretch it, you're stretching the band, and relaxing the
> spring a bit.
>
> If K is the spring constant of the spring, and L the spring constant of
> the band, I would expect the spring constant in compression (for small
> deviations from the equilibrium position) to be K - L and in tension
> (for small deviations from the equilibrium position) to be L - K.

The spring and band are in parallel. The combined stiffness is K + L,
regardless whether the system is compressed or stretched.

--
Joe Riel

 

[Joe Riel]

Ben C <[email protected]> writes:

> Suppose you have a helical compression coil spring held in compression
> with a rubber band around it. The spring holds the band in tension and
> the band holds the spring in compression. They're in equilibrium. Both
> obey Hooke's Law.
>
> What's the response of the whole thing to compression and tension? If
> you compress it, you're squashing the spring, but releasing the band a
> bit. If you stretch it, you're stretching the band, and relaxing the
> spring a bit.
>
> If K is the spring constant of the spring, and L the spring constant of
> the band, I would expect the spring constant in compression (for small
> deviations from the equilibrium position) to be K - L and in tension
> (for small deviations from the equilibrium position) to be L - K.

The spring and band are in parallel. The combined stiffness is K + L,
regardless whether the system is compressed or stretched.

--
Joe Riel

 

[Peter Cole]

Ben C wrote:
> On 2007-10-09, Peter Cole <[email protected]> wrote:
>> jim beam wrote:
>>> Ben C wrote:
>>>> On 2007-10-09, jim beam <[email protected]> wrote:
>>>> [...]
>>>>> no, excessively compressed rims more readily fold.
>>>> Suppose the rim is highly compressed by high spoke tension.
>>>>
>>>> Then I hit a bump. The compression from the bump isn't just added to the
>>>> compression from the spoke tension, because as the rim starts to
>>>> compress, those spokes start to go slack. So the compression on the rim
>>>> stays about the same until the spokes have gone completely slack.
>>>>
>>>> Is this right/wrong/too simplistic?
> [...]
>>> as far as local bending is concerned, it's a bending beam, with
>>> a neutral plane, and compressive and tensile regions accordingly.
>> A beam which is supported by the spokes, with the consequence that the
>> stiffness dramatically lowers when the support is removed.
>
> Sticking to local bending, _which_ spokes is the region of rim hitting
> the bump supported by?

The spokes that are right there. Think railroad track & ties.


> If it's the ones running from the hub to the bump, then those spokes
> going slack would reduce the stiffness as you say.
>
> But if it's all the other spokes, which aren't going slack, providing
> the support, then the stiffness of the rim (locally) wouldn't change
> much at the point where you hit the bump. Or it might even get stiffer
> as the spoke goes slack (see below).

It's not.


> If the spokes weren't pretensioned and this was a cart wheel, it would
> seem quite intuitive that it's the spokes near the bump that are
> propping up the rim at that point. And people keep saying that a
> pretensioned wheel behaves the same as a cart wheel.

Because it does. And if those spokes were springy, how would the
combination of spoke spring and rim spring combine?


> But does it in every detail? It seems like it should be the other way
> round. The effective stiffness of the rim at the bump should _increase_
> as those spokes above it go slack, reaching its maximum stiffness when
> they are completely slack.

You are confusing stiffness with force. Stiffness is the change in force
with displacement. The force increases, the stiffness does not change
(up to the point where spokes go slack).


> Suppose you have a helical compression coil spring held in compression
> with a rubber band around it. The spring holds the band in tension and
> the band holds the spring in compression. They're in equilibrium. Both
> obey Hooke's Law.
>
> What's the response of the whole thing to compression and tension? If
> you compress it, you're squashing the spring, but releasing the band a
> bit. If you stretch it, you're stretching the band, and relaxing the
> spring a bit.

This is a bit non-intuitive, draw the graphs for each, paying attention
to sign, and you'll see that the slope (force over displacement) of the
combination is greater than either alone. The stiffnesses add.

> If K is the spring constant of the spring, and L the spring constant of
> the band, I would expect the spring constant in compression (for small
> deviations from the equilibrium position) to be K - L and in tension
> (for small deviations from the equilibrium position) to be L - K.

Think about it, when you squash the spring, the band eases, "helping"
you less, making the spring "seem" stiffer. You really have to draw it
out I think to fully grasp it. The spring constants add.

> (But I would expect the system as a whole not to obey Hooke's Law-- it
> should be like a spring that gets a bit stiffer the further you stretch
> or compress it as the force contribution from the other element gets
> less the further you go).
>
> So, if the stiffness of the rim on its own is K and that of the spoke L,
> then squashing the rim in (locally) should react like something with a
> stiffness of K - L. When eventually L goes to zero (slack spoke) the
> stiffness is K, which is higher than K - L.

It obeys Hooke's Law, it must, it's linear, so you just superimpose
forces. Plot combined force against displacement. There is no need to
think about equilibrium or small displacements, assume perfect springs
of arbitrary constants. Don't think about what it "should be".


> Where have I gone wrong here?

Try drawing it out, it's hard to do in your head.

>> The net is you want to have tight spokes to prevent excessive (damaging)
>> deflection of the rim under overload.

 

[Ben C]

On 2007-10-09, Peter Cole <[email protected]> wrote:
> Ben C wrote:
>> On 2007-10-09, Peter Cole <[email protected]> wrote:
>>> jim beam wrote:
>>>> Ben C wrote:
>>>>> On 2007-10-09, jim beam <[email protected]> wrote:
>>>>> [...]
>>>>>> no, excessively compressed rims more readily fold.
>>>>> Suppose the rim is highly compressed by high spoke tension.
>>>>>
>>>>> Then I hit a bump. The compression from the bump isn't just added to the
>>>>> compression from the spoke tension, because as the rim starts to
>>>>> compress, those spokes start to go slack. So the compression on the rim
>>>>> stays about the same until the spokes have gone completely slack.
>>>>>
>>>>> Is this right/wrong/too simplistic?
>> [...]
>>>> as far as local bending is concerned, it's a bending beam, with
>>>> a neutral plane, and compressive and tensile regions accordingly.
>>> A beam which is supported by the spokes, with the consequence that the
>>> stiffness dramatically lowers when the support is removed.
>>
>> Sticking to local bending, _which_ spokes is the region of rim hitting
>> the bump supported by?
>
> The spokes that are right there. Think railroad track & ties.

Yes, that was what I was thinking of.

[...]
>> But does it in every detail? It seems like it should be the other way
>> round. The effective stiffness of the rim at the bump should _increase_
>> as those spokes above it go slack, reaching its maximum stiffness when
>> they are completely slack.
>
> You are confusing stiffness with force. Stiffness is the change in force
> with displacement. The force increases, the stiffness does not change
> (up to the point where spokes go slack).

No, whatever I'm confusing, it's not stiffness and force. See [1] below.

>> Suppose you have a helical compression coil spring held in compression
>> with a rubber band around it. The spring holds the band in tension and
>> the band holds the spring in compression. They're in equilibrium. Both
>> obey Hooke's Law.
>>
>> What's the response of the whole thing to compression and tension? If
>> you compress it, you're squashing the spring, but releasing the band a
>> bit. If you stretch it, you're stretching the band, and relaxing the
>> spring a bit.
>
> This is a bit non-intuitive, draw the graphs for each, paying attention
> to sign, and you'll see that the slope (force over displacement) of the
> combination is greater than either alone. The stiffnesses add.

If you pay attention to the sign, yes. But I will draw some more graphs.

>> If K is the spring constant of the spring, and L the spring constant of
>> the band, I would expect the spring constant in compression (for small
>> deviations from the equilibrium position) to be K - L and in tension
>> (for small deviations from the equilibrium position) to be L - K.
>
> Think about it, when you squash the spring, the band eases, "helping"
> you less, making the spring "seem" stiffer. You really have to draw it
> out I think to fully grasp it.

[1] That's exactly what I thought. Which is why I think the whole
assembly gets more stiff the further you compress (or extend) it, as the
other element helps you less.

Anyway, thanks for your input, I think I can work this out.

 

[Ben C]

On 2007-10-09, [email protected] <[email protected]> wrote:
> Ben C? writes:
>
>> [...]
>> > no, excessively compressed rims more readily fold.
>
>> Suppose the rim is highly compressed by high spoke tension.
>
> I don't understand what you mean by this.

Just that the spokes are pulling the rim inwards compressing it.

[...]
>> Is this right/wrong/too simplistic?
>
> I think you can test this on your own wheels (if they don't have fewer
> than 20 spokes) by having someone pluck spokes while you stand on one
> pedal and lean the bicycle with both feet in the pedals as in riding
> while standing. You'll notice that the bottom spokes of the wheels
> lose tension unequally, the "upperside" ones losing more tension than
> the "underside" spokes.

Interesting, but my question was about something else-- how spoke
tension affects what happens to the rim when you hit a bump.

 

[jim beam]

Peter Cole wrote:
> jim beam wrote:
>> Ben C wrote:
>>> On 2007-10-09, jim beam <[email protected]> wrote:
>>> [...]
>>>> no, excessively compressed rims more readily fold.
>>>
>>> Suppose the rim is highly compressed by high spoke tension.
>>>
>>> Then I hit a bump. The compression from the bump isn't just added to the
>>> compression from the spoke tension, because as the rim starts to
>>> compress, those spokes start to go slack. So the compression on the rim
>>> stays about the same until the spokes have gone completely slack.
>>>
>>> Is this right/wrong/too simplistic?
>>
>> the hoop compression is a function of spoke tension. i doubt locally
>> slackening spokes will reduce the compression of the whole because
>> it's an equilibrium structure and the bump load has to be taken up by
>> the whole.
>
> In other words, no, it's not "too simplistic" a bump won't significantly
> change the rim compression,

as a whole, but local elastic deformation can change local compression
significantly.

> which is why you can operate a wheel within
> a fairly small safety margin of compression buckling.

except that,... see above.


> The limit is more
> what happens to wheel shape when you break a spoke.

no, the limit is spoke tension sufficient to prevent spokes going slack.


>
>> as far as local bending is concerned, it's a bending beam, with a
>> neutral plane, and compressive and tensile regions accordingly.
>
> A beam which is supported by the spokes, with the consequence that the
> stiffness dramatically lowers when the support is removed.
>
> The net is you want to have tight spokes to prevent excessive (damaging)
> deflection of the rim under overload.

except that it has to be balanced against increasing propensity to taco
and rim cracking. spokes only need to be tight enough to not go slack.
no more.

 

[Tom Sherman]

[email protected] aka Carl Fogel wrote:
> ...
> Dear Jobst,
>
> Sigh . . . What a nice post, except for where you call people you
> don't know liars about something that you've never tested....

Not that "Dear Carl" would ever call someone a liar based on NO
evidence. [end sarcasm].

--
Tom Sherman - Holstein-Friesland Bovinia
Beer - It's not just for breakfast anymore!

 

[Ben C]

On 2007-10-09, Joe Riel <[email protected]> wrote:
> Ben C <[email protected]> writes:
>
>> Suppose you have a helical compression coil spring held in compression
>> with a rubber band around it. The spring holds the band in tension and
>> the band holds the spring in compression. They're in equilibrium. Both
>> obey Hooke's Law.
>>
>> What's the response of the whole thing to compression and tension? If
>> you compress it, you're squashing the spring, but releasing the band a
>> bit. If you stretch it, you're stretching the band, and relaxing the
>> spring a bit.
>>
>> If K is the spring constant of the spring, and L the spring constant of
>> the band, I would expect the spring constant in compression (for small
>> deviations from the equilibrium position) to be K - L and in tension
>> (for small deviations from the equilibrium position) to be L - K.
>
> The spring and band are in parallel. The combined stiffness is K + L,
> regardless whether the system is compressed or stretched.

Yes, you're right, and so is Peter Cole.

I have drawn some more graphs and thought about it some more and am
now convinced of this fact.

The difficulty is getting the signs right and consistent so you're
adding the correct two functions together.

It feels OK intuitively now too, although it's not that easy to explain,
and there's probably not much point since it's only myself I need to
explain it to.

Anyway, this means Peter Cole is also right that stiffness is higher
when the spokes aren't slack than when they are (railroad ties etc.).
This means that high tension spokes mean more deformation is possible
before you reach the transition point where the rim responds with K
alone rather than with K+L.

I was on my own tangent there, but now I don't understand why jim beam
says that higher spoke tension (higher than needed to prevent slack
spoke flexing fatigue) achieves nothing, or why he says it makes it more
likely that the rim will dent when you hit a bump.

 

[jim beam]

Ben C wrote:
> On 2007-10-09, Joe Riel <[email protected]> wrote:
>> Ben C <[email protected]> writes:
>>
>>> Suppose you have a helical compression coil spring held in compression
>>> with a rubber band around it. The spring holds the band in tension and
>>> the band holds the spring in compression. They're in equilibrium. Both
>>> obey Hooke's Law.
>>>
>>> What's the response of the whole thing to compression and tension? If
>>> you compress it, you're squashing the spring, but releasing the band a
>>> bit. If you stretch it, you're stretching the band, and relaxing the
>>> spring a bit.
>>>
>>> If K is the spring constant of the spring, and L the spring constant of
>>> the band, I would expect the spring constant in compression (for small
>>> deviations from the equilibrium position) to be K - L and in tension
>>> (for small deviations from the equilibrium position) to be L - K.
>> The spring and band are in parallel. The combined stiffness is K + L,
>> regardless whether the system is compressed or stretched.
>
> Yes, you're right, and so is Peter Cole.
>
> I have drawn some more graphs and thought about it some more and am
> now convinced of this fact.
>
> The difficulty is getting the signs right and consistent so you're
> adding the correct two functions together.
>
> It feels OK intuitively now too, although it's not that easy to explain,
> and there's probably not much point since it's only myself I need to
> explain it to.
>
> Anyway, this means Peter Cole is also right that stiffness is higher
> when the spokes aren't slack than when they are (railroad ties etc.).
> This means that high tension spokes mean more deformation is possible
> before you reach the transition point where the rim responds with K
> alone rather than with K+L.
>
> I was on my own tangent there, but now I don't understand why jim beam
> says that higher spoke tension (higher than needed to prevent slack
> spoke flexing fatigue) achieves nothing, or why he says it makes it more
> likely that the rim will dent when you hit a bump.

simple. look at a stress/strain graph. all you're doing is moving
along the straight line region and approaching the yield point. that is
true of any loaded point in the structure. the only way "increasing"
anything can mitigate a load is if it has an opposite sign to the load.
increasing tension is great for spokes, for exactly that reason. but
it's terrible for rims as loading increase and tension increase are the
same sign!

 

[Ben C]

On 2007-10-10, jim beam <[email protected]> wrote:
[...]
> simple. look at a stress/strain graph. all you're doing is moving
> along the straight line region and approaching the yield point. that is
> true of any loaded point in the structure. the only way "increasing"
> anything can mitigate a load is if it has an opposite sign to the load.
> increasing tension is great for spokes, for exactly that reason. but
> it's terrible for rims as loading increase and tension increase are the
> same sign!

So, the high-tension spokes are compressing the rim, and the bump you
hit compresses it even more? So by pre-loading it, you're bringing it
nearer to yield?

If this is what you're saying, then the part I was having trouble with
was this: how is it that rim compression from spoke tension and from the
bump are added together, since as you hit the bump, the spoke becomes
slacker.

But actually I now think that's exactly the same confusion I had about
the railroad ties/my compression spring wrapped up in a rubber band.

Until the spoke is completely slack, then the rim _is_ being compressed
both by the bump and by the spoke.

The fact that the spoke is there makes the whole assembly stiffer, but
also means that the rim is operating at closer to yield.

So, assuming Hooke's Law for spokes and rim (but assuming that spokes
don't work in compression):

1. Having a spoke that isn't slack supports the rim and makes the whole
assembly stiffer.
2. Making the tension even higher doesn't make it any _more_ stiff.
3. But it does bring the rim closer to yield making it easier to
flat-spot.
4. But it does keep it stiffer for longer-- the stress it can take
before the spokes go completely slack and the assembly loses
stiffness is greater.

There's some critical value of spoke tension such that the spoke goes
slack just as the rim reaches yield stress.

More tension than that critical value and you're definitely reducing rim
strength without any benefit.

Less tension than that and the stress level at which you lose spoke
support is reduced. But a stiffer rim doesn't need as much spoke
support.

And then there's rim fatigue cracking, which may be a problem at a lower
tension than at the "critical value" mentioned above.