Wheel deflection

On Tue, 28 Oct 2008 12:50:22 -0700 (PDT), Chalo
wrote:

jim beam wrote:

Jobst Brandt wrote:

Time of loading has no effect on metals

untrue. *many materials react differently depending on loading rate. *for
example, this is why you have nail guns - they succeed at a high rate on
driving nails into concrete whereas trying to do the same job at a lower
rate always fails. *

Concrete is the issue. Whack a nail into wood, or push it in with an
arbor press, and it's about the same (although the whacked nail might
hold better due to local heating of the wood resins).

[snip]

Dear Chalo,

Usually I follow you because your posts are clear to even the meanest
intelligence.

But this is a concrete example (sorry, couldn't resist it) of how even
you can overestimate the dregs of your readership.

Can you explain concrete versus wood a little more from the nail gun's
point of view?

Cheers,

Carl Fogel

Frank Krygowski wrote:

Does anybody know if a rim would react differently to a sudden
load versus a gradual load? Â*What I mean is if you were to apply
a gradual load on a wheel would that rim be able to handle a
higher load before permanantly deforming than if you were to
apply the same load very quickly?

With suddenly applied loads, there's an inertia effect that
generally makes things worse.

Let's not leave it at such a vague description. Â*The question is
load, regardless of its origin, inertial or dead weight. Â*The rate
of stress application has no effect on metals, at least up to
acoustic frequencies.

That depends on exactly what Steve meant by "load."

A load is a load is a load. A mass can be a load statically and
dynamically but that is a load. Confusing inertial loads with stress
confuses the issue at hand. Steve's question has a semantic problem.
Does load mean force or mass in his question?

The classic example is gently placing a load on a spring,
vs. suddenly releasing the same load onto the same spring. Â*Even
if the load is (barely) touching the spring just before it's
released, the spring momentarily deflects twice as far as with the
gently placed load, and the peak force on the spring is twice as
great.

Oops! What do you mean by load? Getting force mass and load
scrambled does not lead to a clarification of the matter.

Those are two different loads. Â*The question was about load
application and possibly duration causing plastic deformation.

If Steve could tell us more about exactly what he means, we could
determine whether the sudden application of an object onto a spring is
properly analogous. I believe it's likely, but he's given no details.

I don't think your adoption of his terms helps either.

If the load were applied by a mechanism that didn't involve mass
and inertia, I don't believe that effect would be present. Â*But
bike wheel loads do involve mass and inertia.

Bicycle wheels do not involve mass AND inertia. Â*Loads are mainly
rider and bicycle mass and their inertia. Â*The mass of the wheel
and tire is insignificant in that respect. Â*An example of that
would be to toss a wheel, with inflated tire, into the air and
watch it bounce undamaged on pavement. Â*To damage a wheel in that
manner would take a hefty toss, especially with a fat (2+ tire)
inch.

Bicycle wheels support mass. That mass has inertia. It makes a
difference.

You said "But bike wheel loads do involve mass and inertia." which is
a jumble of misstatement. You may have meant that bicycle wheels bear
inertial loads of the bicycle but they themselves have insignificant
inertial effects on their own stress and deformation.

Example: A 200 pound bike+rider sitting still on level ground causes
certain stresses in the various wheel components. That same bike
+rider dropping one foot down onto level ground causes much more
stress. If Steve was envisioning anything similar, the fact that
wheel masses are low has no bearing.

There you go again mixing the bike+rider with wheel inertial effects.
That is not an example of "But bike wheel loads do involve mass and
inertia." You must mean "Bicycle loads involve inertia." Tossing the
word wheel in there doesn't clarify what is meant.

So, Steve, what do you have in mind?

So, Frank, what do you have in mind?

Jobst Brandt

On Oct 28, 1:58*pm, wrote:
Frank Krygowski wrote:
[content-free definitional posturing]
[more content-free definitional posturing]

I wonder why is is that an insistence on rigid adherence to
(arbitrarily decided and often inconsistent) terminology so often
correlates with the same person's complete failure to communicate,
whether in the role of producer or receiver of information.

The typical failure insists that accurate diction is essential to
accurate communication; yet when the process of communication has
progressed so that any outside observer can see what information is in
play, the failure goes on long after the fact to insist that words be
used *his* way -- though by any measure the communication of
information is long complete. One is forced to conclude from the fact
of his actions, in contradiction to his stated beliefs, the failure
has no pragmatic interest in the transmission of accurate information.

-pm

On 2008-10-28, wrote:
Frank Krygowski wrote:

Does anybody know if a rim would react differently to a sudden
load versus a gradual load? Â*What I mean is if you were to apply
a gradual load on a wheel would that rim be able to handle a
higher load before permanantly deforming than if you were to
apply the same load very quickly?

With suddenly applied loads, there's an inertia effect that
generally makes things worse.

Let's not leave it at such a vague description. Â*The question is
load, regardless of its origin, inertial or dead weight. Â*The rate
of stress application has no effect on metals, at least up to
acoustic frequencies.

That depends on exactly what Steve meant by "load."

A load is a load is a load.

If you ride the same bike over the same bump faster you will have more
load.

It is helpful to think about collisions. In a collision momentum and
energy are conserved (although the energy won't all end up as kinetic
energy unless it's an "elastic" collision).

If you hang a rim from the ceiling and strike it with a toffee
hammer some amount of momentum is transferred from the hammer to the
wheel.

If you now put a tyre on that rim and hit it with the same hammer in the
same way, roughly the same amount of momentum is transferred (actually
more because the tyre weighs quite a lot, but let's pretend it doesn't).

But the momentum is transferred more slowly. Rate of change of momentum
is force (Newton's second law), so the force on the rim is less if it
has a tyre on. Both the average force and the peak force during the
collision are reduced, so the rim doesn't yield (get dented).

It's force that yields a rim. A big rate of change of force (applying
the force suddenly) doesn't make much difference to how much the rim
yields. But a big rate of change of momentum makes all the difference
because a big rate of change of momentum _is_ a big force.

The OP talks about "sudden load" vs "gradual load". But is he really
talking literally about rate of change of force? I don't know, but
certainly riding over a kerb fast or without a tyre is more likely to
dent the rim than riding over it slowly, because the peak rate of change
of _momentum_ is higher.

Rate of change of momentum = force. Force yields things. Rate of change
of force is rate of change of rate of change of momentum. So that's just
the rate at which you yield your rim. Since we aren't talking about
hammering nails into concrete or making spoons, the rate of change of
force probably doesn't make much difference to how much the rim yields
in this case.

[...]
The classic example is gently placing a load on a spring,
vs. suddenly releasing the same load onto the same spring. Â*Even
if the load is (barely) touching the spring just before it's
released, the spring momentarily deflects twice as far as with the
gently placed load, and the peak force on the spring is twice as
great.

This example is slightly different. If I put the mass on the spring and
let go the mass falls under the influence of gravity and I introduce
energy into the system, so it oscillates, with the result that at the
bottom of the oscillation, there is a higher load on the spring. If I
lower the mass onto the spring gently, then I am acting as a damper,
absorbing some of that energy. That's the significant effect, not the
different rate of change of force.

On Tue, 28 Oct 2008 17:58:52 -0500, Ben C wrote:

On 2008-10-28,
wrote:
Frank Krygowski wrote:

Does anybody know if a rim would react differently to a sudden load
versus a gradual load? ÂÂ*What I mean is if you were to apply a
gradual load on a wheel would that rim be able to handle a higher
load before permanantly deforming than if you were to apply the
same load very quickly?

With suddenly applied loads, there's an inertia effect that
generally makes things worse.

Let's not leave it at such a vague description. ÂÂ*The question is
load, regardless of its origin, inertial or dead weight. ÂÂ*The rate
of stress application has no effect on metals, at least up to
acoustic frequencies.

That depends on exactly what Steve meant by "load."

A load is a load is a load.

If you ride the same bike over the same bump faster you will have more
load.

It is helpful to think about collisions. In a collision momentum and
energy are conserved (although the energy won't all end up as kinetic
energy unless it's an "elastic" collision).

If you hang a rim from the ceiling and strike it with a toffee hammer
some amount of momentum is transferred from the hammer to the wheel.

If you now put a tyre on that rim and hit it with the same hammer in the
same way, roughly the same amount of momentum is transferred (actually
more because the tyre weighs quite a lot, but let's pretend it doesn't).

But the momentum is transferred more slowly. Rate of change of momentum
is force (Newton's second law), so the force on the rim is less if it
has a tyre on. Both the average force and the peak force during the
collision are reduced, so the rim doesn't yield (get dented).

It's force that yields a rim. A big rate of change of force (applying
the force suddenly) doesn't make much difference to how much the rim
yields. But a big rate of change of momentum makes all the difference
because a big rate of change of momentum _is_ a big force.

The OP talks about "sudden load" vs "gradual load". But is he really
talking literally about rate of change of force? I don't know, but
certainly riding over a kerb fast or without a tyre is more likely to
dent the rim than riding over it slowly, because the peak rate of change
of _momentum_ is higher.

Rate of change of momentum = force. Force yields things. Rate of change
of force is rate of change of rate of change of momentum. So that's just
the rate at which you yield your rim. Since we aren't talking about
hammering nails into concrete or making spoons, the rate of change of
force probably doesn't make much difference to how much the rim yields
in this case.

in this case. but it's not a blanket statement - different materials have
different deformation characteristics depending on rate. remember silly
putty?




[...]
The classic example is gently placing a load on a spring, vs.
suddenly releasing the same load onto the same spring. ÂÂ*Even if the
load is (barely) touching the spring just before it's released, the
spring momentarily deflects twice as far as with the gently placed
load, and the peak force on the spring is twice as great.

This example is slightly different. If I put the mass on the spring and
let go the mass falls under the influence of gravity and I introduce
energy into the system, so it oscillates, with the result that at the
bottom of the oscillation, there is a higher load on the spring. If I
lower the mass onto the spring gently, then I am acting as a damper,
absorbing some of that energy. That's the significant effect, not the
different rate of change of force.

On Tue, 28 Oct 2008 14:27:02 -0700, pm wrote:

On Oct 28, 1:58Â*pm, wrote:
Frank Krygowski wrote:
[content-free definitional posturing]
[more content-free definitional posturing]

I wonder why is is that an insistence on rigid adherence to (arbitrarily
decided and often inconsistent) terminology so often correlates with the
same person's complete failure to communicate, whether in the role of
producer or receiver of information.

The typical failure insists that accurate diction is essential to
accurate communication; yet when the process of communication has
progressed so that any outside observer can see what information is in
play, the failure goes on long after the fact to insist that words be
used *his* way -- though by any measure the communication of information
is long complete. One is forced to conclude from the fact of his
actions, in contradiction to his stated beliefs, the failure has no
pragmatic interest in the transmission of accurate information.

-pm


potm!

On Tue, 28 Oct 2008 20:58:32 +0000, jobst.brandt wrote:

Frank Krygowski wrote:

Does anybody know if a rim would react differently to a sudden load
versus a gradual load? Â*What I mean is if you were to apply a
gradual load on a wheel would that rim be able to handle a higher
load before permanantly deforming than if you were to apply the same
load very quickly?

With suddenly applied loads, there's an inertia effect that generally
makes things worse.

Let's not leave it at such a vague description. Â*The question is load,
regardless of its origin, inertial or dead weight. Â*The rate of stress
application has no effect on metals, at least up to acoustic
frequencies.

That depends on exactly what Steve meant by "load."

A load is a load is a load.

with the jobstian exception of course being when that load is connected to
a rubber tire, in which case it mysteriously disappears. unless you're
prepared to admit time and thus rates into your simplistic little purview,
in which case load is not load, etc.



A mass can be a load statically and
dynamically but that is a load. Confusing inertial loads with stress
confuses the issue at hand. Steve's question has a semantic problem.
Does load mean force or mass in his question?

given your obvious confusion, would it make a difference to you?





The classic example is gently placing a load on a spring, vs.
suddenly releasing the same load onto the same spring. Â*Even if the
load is (barely) touching the spring just before it's released, the
spring momentarily deflects twice as far as with the gently placed
load, and the peak force on the spring is twice as great.

Oops! What do you mean by load? Getting force mass and load scrambled
does not lead to a clarification of the matter.

Those are two different loads. Â*The question was about load
application and possibly duration causing plastic deformation.

If Steve could tell us more about exactly what he means, we could
determine whether the sudden application of an object onto a spring is
properly analogous. I believe it's likely, but he's given no details.

I don't think your adoption of his terms helps either.

If the load were applied by a mechanism that didn't involve mass and
inertia, I don't believe that effect would be present. Â*But bike
wheel loads do involve mass and inertia.

Bicycle wheels do not involve mass AND inertia. Â*Loads are mainly
rider and bicycle mass and their inertia. Â*The mass of the wheel and
tire is insignificant in that respect. Â*An example of that would be to
toss a wheel, with inflated tire, into the air and watch it bounce
undamaged on pavement. Â*To damage a wheel in that manner would take a
hefty toss, especially with a fat (2+ tire) inch.

Bicycle wheels support mass. That mass has inertia. It makes a
difference.

You said "But bike wheel loads do involve mass and inertia." which is a
jumble of misstatement. You may have meant that bicycle wheels bear
inertial loads of the bicycle but they themselves have insignificant
inertial effects on their own stress and deformation.

Example: A 200 pound bike+rider sitting still on level ground causes
certain stresses in the various wheel components. That same bike
+rider dropping one foot down onto level ground causes much more
stress. If Steve was envisioning anything similar, the fact that wheel
masses are low has no bearing.

There you go again mixing the bike+rider with wheel inertial effects.
That is not an example of "But bike wheel loads do involve mass and
inertia." You must mean "Bicycle loads involve inertia." Tossing the
word wheel in there doesn't clarify what is meant.

So, Steve, what do you have in mind?

So, Frank, what do you have in mind?

Jobst Brandt

Carl Fogel wrote:

Chalo wrote:

jim beam wrote:

Jobst Brandt wrote:

Time of loading has no effect on metals

untrue. *many materials react differently depending on loading rate. *for
example, this is why you have nail guns - they succeed at a high rate on
driving nails into concrete whereas trying to do the same job at a lower
rate always fails. *

Concrete is the issue. *Whack a nail into wood, or push it in with an
arbor press, and it's about the same (although the whacked nail might
hold better due to local heating of the wood resins).

[snip]

DearChalo,

Usually I follow you because your posts are clear to even the meanest
intelligence.

But this is a concrete example (sorry, couldn't resist it) of how even
you can overestimate the dregs of your readership.

Can you explain concrete versus wood a little more from the nail gun's
point of view?

Concrete is very strong in compression, but subject to being
disintegrated by shock. The shock created by a powder-fired fastener
accomplishes what a similarly large but steady axial load on the same
fastener can't, pulverizing a small zone around the nail and allowing
it to penetrate. Without the shock, a nail pushed sufficiently hard
would bend or break before penetrating the concrete enough to fasten
to it.

Wood is a resilient material and does not need to be shocked to allow
a fastener to penetrate. Thus a nail can be shot, pounded, or simply
shoved into place with comparable results in whichever case.

Chalo

On 2008-10-29, jim beam wrote:
On Tue, 28 Oct 2008 17:58:52 -0500, Ben C wrote:
[...]
Rate of change of momentum = force. Force yields things. Rate of change
of force is rate of change of rate of change of momentum. So that's just
the rate at which you yield your rim. Since we aren't talking about
hammering nails into concrete or making spoons, the rate of change of
force probably doesn't make much difference to how much the rim yields
in this case.

in this case. but it's not a blanket statement - different materials have
different deformation characteristics depending on rate. remember silly
putty?

Yes, I loved that stuff. I didn't know that stainless steel was more
ductile at high rates.

Although presumably you draw it with more or less constant force
(reducing it as the wire gets thinner?), but at a high rate of
displacement.

On Wed, 29 Oct 2008 03:38:21 -0500, Ben C wrote:

On 2008-10-29, jim beam wrote:
On Tue, 28 Oct 2008 17:58:52 -0500, Ben C wrote:
[...]
Rate of change of momentum = force. Force yields things. Rate of
change of force is rate of change of rate of change of momentum. So
that's just the rate at which you yield your rim. Since we aren't
talking about hammering nails into concrete or making spoons, the rate
of change of force probably doesn't make much difference to how much
the rim yields in this case.

in this case. but it's not a blanket statement - different materials
have different deformation characteristics depending on rate. remember
silly putty?

Yes, I loved that stuff. I didn't know that stainless steel was more
ductile at high rates.

indeed it is. large deep drawn stainless is often explosive formed to get
the job done. a traditional cold drawing process can be long, tedious,
and not achieve the required deformation despite multiple phases of work.
this is not limited to stainless - many structural materials can be formed
in this way



Although presumably you draw it with more or less constant force
(reducing it as the wire gets thinner?), but at a high rate of
displacement.

wire is a different story - wire drawing is done at a fraction of the rate
above. and it's easy to reduce progressively [and anneal between
operations] to achieve the required total deformation.