Cannondale's tests of disks and QRs

well, as i understand the argument, it goes like this:

resolved "ejection" force is postulated to be 1825N, and iso pullout
spec is alledged to be only 500N, therefore the sky will fall tomorrow.

what's being ignored is that a conservative estimate for pullout force
based on just part of one face of a serrated axle [what all disk brake
hub manufacturers afaik use] is in excess of 5000N. and that's /not/
including any effect of serrated skewer nut facings either. if that's
not ignoring a critical part of the analysis, i don't know what is.

I'm just trying to resolve a force of 5000N through small ridges on an alloy
fork, coupled with what is known about the properties of the alloy to which
it is clamped and not getting deformation of the fork dropout faces.

Look at almost any bike fork and you will see marks from the skewer nuts on
the outside and marks from the axle on the inside. That is even before
placing braking stresses on it.

It doesn't take a rocket scientist to work out that if you are putting a
system like that where you are relying on ductile metals to hold something
in place under strong vibrational loads, then it can potentially deform and
work loose.

I think jim's arguement can be stated in "If it is properly tightened then
it can't work loose". And the observation being that the wheel was properly
tightened but did work loose.

The point in question could be the timescale between being verifiably
correctly fastened and the skewer allegedly working loose.

I am not terribly much in favour of the static models produced so far. It
doesn't take a great leap of imagination to determine realistic scenarios
where there is no loading on the handlebars but a maximal braking force on
the front wheel, potentially a pull-out force before the brakes even come
on. A more realistic test would be the following:

Hold the forks rigid and put the brakes on. monitor slippage with respect to
the torsional force on the wheel. Do we get slippage before the wheel
collapses? At what force do we get slippage?

With that data we have to ask how representative of real life are these
forces.


The moral of the story is that when riding any reasonable definition of
rough terrain one should always reseat the q/r (undo and retighten). Ideally
I'd like to see manufacturers moving towards the better style of axle
attachment for off-road machines.

Having said that, I cant remember the last time I checked the q/r on my
bikes but I don't ride o/r very often, and when I do it is hardly severe
stuff.

...d

On Tue, 28 Sep 2004 20:40:35 -0700, jim beam wrote:

what's being ignored is that a conservative estimate for pullout force
based on just part of one face of a serrated axle [what all disk brake
hub manufacturers afaik use] is in excess of 5000N. and that's /not/
including any effect of serrated skewer nut facings either. if that's
not ignoring a critical part of the analysis, i don't know what is.

Not including effect of serrations, so just friction?

Well, steel-on-steel coefficient of friction is in teh order of 0.1 to
0.3, so ignoring the fact that you claim a conservative estimate, and
aiming for an optimistic estimate, that assumes a normal force of
5000/0.3, or 16.7E3 N. Suppose a typical skewer is 4mm diameter,
that's a CSA of 12.5mm2.

Thus we require teh skewer to be working at 1326 N/mm2, or about 5
times the yield stress of mild steel. Admittedly, it's possible to
get steel with those sort of yields, but I'd question whether skewers
are made of them.

When we return to your claim of 'conservative' estimate, I think I'd
need to allow for a low coefficient of friction, say 0.1, requiring a
stress in teh skewer of about 4000 N/mm2, which I think exceeds any
sensible steel.

All of which ignores the fact that this force has to be exerted by teh
person closing the QR - a quick work calculation, based on teh
assumption that teh skewer shortens by say 2mm while the lever moves
through 90 degrees and has ane effective length 50mm, so 90mm stroke,
so 45:1, so it would (using teh conservative friction figure) require
a closing force of aboiut 1100 N, or 110kg. Do you really claim that
QRs should be adjusted such that it takes a 110kg force on teh lever
to close them?

I'd be interested to see your conservative estimate that gets 5000N
pull-out.

regards, Ian SMith
--
|\ /| no .sig
|o o|
|/ \|

On Wed, 29 Sep 2004 02:10:12 +0100, Tony Raven
wrote:


The only thing I can't recall seing come loose is a
Nord-Lock washer set, and I bet even they will in the right (or wrong)
circumstances.

That still works because serrations on the top of the top washer lock it
to the nut rotation. i.e. it uses serrations to work just like a
Shimano skewer.

Sure. But the serrations are much deeper, and the torque is typically
much greater.

All this is speculation. Why have the people with the wherewithal to
actually test the hypothesis (coincidentaly the people who make money
selling the product) not done so?

Guy
--
May contain traces of irony. Contents liable to settle after posting.
http://www.chapmancentral.co.uk

88% of helmet statistics are made up, 65% of them at Washington University

Ian Smith wrote:
On Tue, 28 Sep 2004 20:40:35 -0700, jim beam wrote:


what's being ignored is that a conservative estimate for pullout force
based on just part of one face of a serrated axle [what all disk brake
hub manufacturers afaik use] is in excess of 5000N. and that's /not/
including any effect of serrated skewer nut facings either. if that's
not ignoring a critical part of the analysis, i don't know what is.


Not including effect of serrations, so just friction?

no, _including_ effect of serrations - which are evident on all
forks/hubs that i've seen for disk brakes.


Well, steel-on-steel coefficient of friction is in teh order of 0.1 to
0.3, so ignoring the fact that you claim a conservative estimate, and
aiming for an optimistic estimate, that assumes a normal force of
5000/0.3, or 16.7E3 N. Suppose a typical skewer is 4mm diameter,
that's a CSA of 12.5mm2.

Thus we require teh skewer to be working at 1326 N/mm2, or about 5
times the yield stress of mild steel. Admittedly, it's possible to
get steel with those sort of yields, but I'd question whether skewers
are made of them.

When we return to your claim of 'conservative' estimate, I think I'd
need to allow for a low coefficient of friction, say 0.1, requiring a
stress in teh skewer of about 4000 N/mm2, which I think exceeds any
sensible steel.

All of which ignores the fact that this force has to be exerted by teh
person closing the QR - a quick work calculation, based on teh
assumption that teh skewer shortens by say 2mm while the lever moves
through 90 degrees and has ane effective length 50mm, so 90mm stroke,
so 45:1, so it would (using teh conservative friction figure) require
a closing force of aboiut 1100 N, or 110kg. Do you really claim that
QRs should be adjusted such that it takes a 110kg force on teh lever
to close them?

I'd be interested to see your conservative estimate that gets 5000N
pull-out.

regards, Ian SMith

here's my original post.

http://groups.google.com/groups?selm...&output=gplain

in message , Just zis Guy,
you know? ') wrote:

On Wed, 29 Sep 2004 02:10:12 +0100, Tony Raven
wrote:


The only thing I can't recall seing come loose is a
Nord-Lock washer set, and I bet even they will in the right (or
wrong) circumstances.

That still works because serrations on the top of the top washer lock
it
to the nut rotation. i.e. it uses serrations to work just like a
Shimano skewer.

Sure. But the serrations are much deeper, and the torque is typically
much greater.

All this is speculation. Why have the people with the wherewithal to
actually test the hypothesis (coincidentaly the people who make money
selling the product) not done so?

If you refer to Cannondale in this, may I again point out to you that in
Cannondales 2005 lineup you will find at most two models which combine
disk brakes with front quick-releases?

Cannondale did the test. Whatever they thought of the results which they
submitted back to the consumer protection people, they clearly decided
that building bikes which combined disk brakes with front quick
releases was not a business they really wished to be in.

--
(Simon Brooke) http://www.jasmine.org.uk/~simon/

;; no eternal reward will forgive us now for wasting the dawn.
;; Jim Morrison

Slacker wrote:
[disc brake ejection]
Because you're sucking valuable brandwidth.

Yes! It must be saved for the endless rbt US politics flamewar. God
forbid anyone talk about bicycles...
--
David Damerell flcl?

On Wed, 29 Sep 2004 06:01:30 -0700, jim beam wrote:
Ian Smith wrote:

Not including effect of serrations, so just friction?

no, _including_ effect of serrations - which are evident on all
forks/hubs that i've seen for disk brakes.

I'd be interested to see your conservative estimate that gets 5000N
pull-out.

here's my original post.

http://groups.google.com/groups?selm...&output=gplain


You maintain that fork legs are always soft and axles always embed in
them. Possibly you have in mind that fork dropouts are always
aluminium, but that's not true. Having maintained that the drop-outs
are soft enough that embedment occurs, you then decide that they
require 200 N/mm2 to shear (maintaining that this is a conservative
figure), but assuming a Tresca yield criterion, this requires a
material with a uniaxial yield stress in excess of 400 N/mm2, which is
high for aluminium, and moderate for steel (well above mild steel).

You go from a 94mm2 gross contact area, to a 25mm2 area of sheared
material. I'm not sure how you'd predict the area of material sheared
by a given embedment of a given serration pattern, but I'll try and
work out how you came up with that. Suppose a 3mm seration spacing,
implies 31mm length of contact in teh area. Given your 25mm2, that in
turn suggests a shearing plane width of 0.8mm at each serration, which
if we assume a 45 degree plane, suggests teh serrations must bite into
teh dropout by about 0.6mm. Is that compatible with your assumptions?

It seems excessive to me to assume that every time you tighten teh QR,
you have to embed teh serations 0.6mm into teh material of teh fork
dropout. Can you justify that? I think after a few weeks use, fork
ends would be so chewed to bits you'd need new ones.

regards, Ian SMith
--
|\ /| no .sig
|o o|
|/ \|

Ian Smith wrote:

You maintain that fork legs are always soft and axles always embed in
them. Possibly you have in mind that fork dropouts are always
aluminium, but that's not true. Having maintained that the drop-outs
are soft enough that embedment occurs, you then decide that they
require 200 N/mm2 to shear (maintaining that this is a conservative
figure), but assuming a Tresca yield criterion, this requires a
material with a uniaxial yield stress in excess of 400 N/mm2, which is
high for aluminium, and moderate for steel (well above mild steel).

You go from a 94mm2 gross contact area, to a 25mm2 area of sheared
material. I'm not sure how you'd predict the area of material sheared
by a given embedment of a given serration pattern, but I'll try and
work out how you came up with that. Suppose a 3mm seration spacing,
implies 31mm length of contact in teh area. Given your 25mm2, that in
turn suggests a shearing plane width of 0.8mm at each serration, which
if we assume a 45 degree plane, suggests teh serrations must bite into
teh dropout by about 0.6mm. Is that compatible with your assumptions?

It seems excessive to me to assume that every time you tighten teh QR,
you have to embed teh serations 0.6mm into teh material of teh fork
dropout. Can you justify that? I think after a few weeks use, fork
ends would be so chewed to bits you'd need new ones.

The word "teh" appears seven times. What does "teh" mean?

--
Tom Sherman

Ian Smith wrote:
On Wed, 29 Sep 2004 06:01:30 -0700, jim beam wrote:

Ian Smith wrote:

Not including effect of serrations, so just friction?


no, _including_ effect of serrations - which are evident on all
forks/hubs that i've seen for disk brakes.


I'd be interested to see your conservative estimate that gets 5000N
pull-out.


here's my original post.

http://groups.google.com/groups?selm...&output=gplain



You maintain that fork legs are always soft and axles always embed in
them.

suspension forks, yes.

Possibly you have in mind that fork dropouts are always
aluminium, but that's not true.

correct, they're frequently magnesium alloy. i never stated aluminum,
so please don't make assumptions for me.

Having maintained that the drop-outs
are soft enough that embedment occurs, you then decide that they
require 200 N/mm2 to shear (maintaining that this is a conservative
figure), but assuming a Tresca yield criterion,

Tresca is a fudge. use von Mises instead.

this requires a
material with a uniaxial yield stress in excess of 400 N/mm2, which is
high for aluminium, and moderate for steel (well above mild steel).

eh? you'll have to explain that leap of logic for me.


You go from a 94mm2 gross contact area, to a 25mm2 area of sheared
material. I'm not sure how you'd predict the area of material sheared
by a given embedment of a given serration pattern, but I'll try and
work out how you came up with that. Suppose a 3mm seration spacing,
implies 31mm length of contact in teh area. Given your 25mm2, that in
turn suggests a shearing plane width of 0.8mm at each serration, which
if we assume a 45 degree plane, suggests teh serrations must bite into
teh dropout by about 0.6mm. Is that compatible with your assumptions?

no. i have no idea where you're getting your shear analysis from. this
is not cupping of a tensile sample, it's simple planar shear. if you
want to get into an analysis based on serration orientation,
differential embedding depths, etc., be my guest, but don't make
assumptions like your others here; get yourself access to a mtb fork and
make some measurements. your comment below seems to indicate that you
don't have direct personal experience.


It seems excessive to me to assume that every time you tighten teh QR,
you have to embed teh serations 0.6mm into teh material of teh fork
dropout. Can you justify that? I think after a few weeks use, fork
ends would be so chewed to bits you'd need new ones.

that guess makes no allowance for serration patterns to be persistent,
which is what we see in practice. once serrations are made, fork & axle
mesh in the same place time after time.

http://home.comcast.net/~carlfogel/d...d/Img_3199.jpg

this is my fork. that pic was taken somewhere north of the 2000 mile
mark. persistent serrations are clearly evident. and the wheel gets
taken out of this fork each and every ride, so it's got to be roughly
100 insertions you see there. not exactly "chewed to bits" is it?


regards, Ian SMith

Tom Sherman wrote in message ...
Ian Smith wrote:


It seems excessive to me to assume that every time you tighten teh QR,
you have to embed teh serations 0.6mm into teh material of teh fork
dropout. Can you justify that? I think after a few weeks use, fork
ends would be so chewed to bits you'd need new ones.

The word "teh" appears seven times. What does "teh" mean?

http://www.winternet.com/~mikelr/flame31.html

twinks