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Tire-making: bead stress, tire width, math, woe........



 
 
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  #1  
Old August 22nd 11, 05:47 PM posted to rec.bicycles.tech
DougC
external usenet poster
 
Posts: 1,276
Default Tire-making: bead stress, tire width, math, woe........

I'm wondering if it is possible that a rough estimate of necessary bead
strength could be determined for a given size tire. There is info online
about figuring wall strength of hoses and cylinders containing pressure,
but they don't deal with the situation of a tire--where the wall is
interrupted.

I've cut up a couple cheap cruiser tires and seen that for them (about
2.1" wide, with a max pressure of 40 PSI) that each of the beads is a
piece of cable with a total breaking strength of about 300 lbs. I could
just match that (even cheap tire beads very rarely fail at their rated
pressures) but it would be nice to know a rule of thumb when making
different-size tires.

From what I have seen of tires I have on hand, the tire's overall
diameter has very little if anything to do with bead loads. I have a
pair of 1.5" wide Kenda Kwest 100 psi tires in 406mm and 559mm, and both
have the same width casings, and both beads measure right about the same
thickness (.118").

{-I am just measuring on the outside of the tire bead, rubber and all,
but anyway}

I also have a 2.3" 559 Big Apple (60 psi), and it is about 6.5" across
the casing, and the beads of it measure right about .150". The BA beads
feel quite stiffer than the Kwests as well.

So then-

1) assuming they have the same cross-section, a larger-diameter tire
(26") does not seen to need a thicker bead than a smaller-diameter tire
(20")

2) the overall tire pressure does necessitate a stronger bead as the
tire pressure increases, but-

3) the stress on the bead increases with the tire's cross-section more
than it does for the pressure (the BA's pressure is only 60% of the
Kwests, but the BA bead is still considerably thicker)

4) I am also wondering now what difference in rim width would make, as
you can get road 26" rims that are ~25mm wide, as well as cruiser 26"
rims that are 80mm wide. On the 80mm rim, not only does the tire's
internal volume increase, but the portion borne by the beads increases
as well.

Is this a problem that can even be estimated roughly, or would it
require 3-d modeling to figure out? It would seem to be fairly simple,
as the tire casing always expands into a circle (the cross-section of
the tire, that is...).

----------

Also when I went looking for such info online, I ran across a lot of
reports of people trying to use non-tubeless tires on tubeless rims.

It's pretty surprising (to me) how common it is for people to say that
the tubeless setups ride much better, but also how common the problem is
of a tire blowing off a tubeless rim and the bead being permanently
damaged from it (ruining the tire). Usually this seems to be with tires
that are not intended for tubeless use; I've already read that a lot of
tires not specified as tubeless are not warrantied for this purpose.

I've never seen these IRL as the first one came out right as I got rid
of the last MTB I had. I've already read a lot of accounts of it, but if
there's any websites that have a lot of pictures and explanations of the
different rims it'd be interesting to see.
Ads
  #2  
Old August 22nd 11, 07:18 PM posted to rec.bicycles.tech
thirty-six
external usenet poster
 
Posts: 10,049
Default Tire-making: bead stress, tire width, math, woe........

On Aug 22, 5:47*pm, DougC wrote:
I'm wondering if it is possible that a rough estimate of necessary bead
strength could be determined for a given size tire. There is info online
about figuring wall strength of hoses and cylinders containing pressure,
but they don't deal with the situation of a tire--where the wall is
interrupted.

I've cut up a couple cheap cruiser tires and seen that for them (about
2.1" wide, with a max pressure of 40 PSI) that each of the beads is a
piece of cable with a total breaking strength of about 300 lbs. I could
just match that (even cheap tire beads very rarely fail at their rated
pressures) but it would be nice to know a rule of thumb when making
different-size tires.

*From what I have seen of tires I have on hand, the tire's overall
diameter has very little if anything to do with bead loads. I have a
pair of 1.5" wide Kenda Kwest 100 psi tires in 406mm and 559mm, and both
have the same width casings, and both beads measure right about the same
thickness (.118").

{-I am just measuring on the outside of the tire bead, rubber and all,
but anyway}

I also have a 2.3" 559 Big Apple (60 psi), and it is about 6.5" across
the casing, and the beads of it measure right about .150". The BA beads
feel quite stiffer than the Kwests as well.

So then-

1) assuming they have the same cross-section, a larger-diameter tire
(26") does not seen to need a thicker bead than a smaller-diameter tire
(20")

2) the overall tire pressure does necessitate a stronger bead as the
tire pressure increases, but-

3) the stress on the bead increases with the tire's cross-section more
than it does for the pressure (the BA's pressure is only 60% of the
Kwests, but the BA bead is still considerably thicker)

4) I am also wondering now what difference in rim width would make, as
you can get road 26" rims that are ~25mm wide, as well as cruiser 26"
rims that are 80mm wide. On the 80mm rim, not only does the tire's
internal volume increase, but the portion borne by the beads increases
as well.

Is this a problem that can even be estimated roughly, or would it
require 3-d modeling to figure out? It would seem to be fairly simple,
as the tire casing always expands into a circle (the cross-section of
the tire, that is...).

* ----------

Also when I went looking for such info online, I ran across a lot of
reports of people trying to use non-tubeless tires on tubeless rims.

It's pretty surprising (to me) how common it is for people to say that
the tubeless setups ride much better, but also how common the problem is
of a tire blowing off a tubeless rim and the bead being permanently
damaged from it (ruining the tire). Usually this seems to be with tires
that are not intended for tubeless use; I've already read that a lot of
tires not specified as tubeless are not warrantied for this purpose.

I've never seen these IRL as the first one came out right as I got rid
of the last MTB I had. I've already read a lot of accounts of it, but if
there's any websites that have a lot of pictures and explanations of the
different rims it'd be interesting to see.


The wire stiffens the connection with the rim so that the tyre stays
in place. The wire's strength is of no particular importance, it's
the resistance to bending which is key. This applies so long as the
gap between the rim's walls is around 0.7 of the tyre section. The
tyre's carcass width of fabric from wire to wire will be just over 3/4
of the circumference of the section size. Steel generally presents
the best material for making the wire because of how slim the bead
ends up. This makes fitting and removing easier for the not so wide
rims. The stiffness of a required bead is related to riders'
expectations as to running pressure, loading, road conditions and
riding style. A conservative approach is best unless you can
realistically mark your tyres as racing tyres. Too flexible a wire
will have the tyre rolling off should it puncture.
  #3  
Old August 22nd 11, 07:30 PM posted to rec.bicycles.tech
DirtRoadie
external usenet poster
 
Posts: 2,915
Default Tire-making: bead stress, tire width, math, woe........

On Aug 22, 10:47*am, DougC wrote:
I'm wondering if it is possible that a rough estimate of necessary bead
strength could be determined for a given size tire. There is info online
about figuring wall strength of hoses and cylinders containing pressure,
but they don't deal with the situation of a tire--where the wall is
interrupted.

I've cut up a couple cheap cruiser tires and seen that for them (about
2.1" wide, with a max pressure of 40 PSI) that each of the beads is a
piece of cable with a total breaking strength of about 300 lbs. I could
just match that (even cheap tire beads very rarely fail at their rated
pressures) but it would be nice to know a rule of thumb when making
different-size tires.

*From what I have seen of tires I have on hand, the tire's overall
diameter has very little if anything to do with bead loads. I have a
pair of 1.5" wide Kenda Kwest 100 psi tires in 406mm and 559mm, and both
have the same width casings, and both beads measure right about the same
thickness (.118").

{-I am just measuring on the outside of the tire bead, rubber and all,
but anyway}

I also have a 2.3" 559 Big Apple (60 psi), and it is about 6.5" across
the casing, and the beads of it measure right about .150". The BA beads
feel quite stiffer than the Kwests as well.

So then-

1) assuming they have the same cross-section, a larger-diameter tire
(26") does not seen to need a thicker bead than a smaller-diameter tire
(20")

2) the overall tire pressure does necessitate a stronger bead as the
tire pressure increases, but-

3) the stress on the bead increases with the tire's cross-section more
than it does for the pressure (the BA's pressure is only 60% of the
Kwests, but the BA bead is still considerably thicker)

4) I am also wondering now what difference in rim width would make, as
you can get road 26" rims that are ~25mm wide, as well as cruiser 26"
rims that are 80mm wide. On the 80mm rim, not only does the tire's
internal volume increase, but the portion borne by the beads increases
as well.

Is this a problem that can even be estimated roughly, or would it
require 3-d modeling to figure out? It would seem to be fairly simple,
as the tire casing always expands into a circle (the cross-section of
the tire, that is...).

* ----------

Also when I went looking for such info online, I ran across a lot of
reports of people trying to use non-tubeless tires on tubeless rims.

It's pretty surprising (to me) how common it is for people to say that
the tubeless setups ride much better, but also how common the problem is
of a tire blowing off a tubeless rim and the bead being permanently
damaged from it (ruining the tire). Usually this seems to be with tires
that are not intended for tubeless use; I've already read that a lot of
tires not specified as tubeless are not warrantied for this purpose.

I've never seen these IRL as the first one came out right as I got rid
of the last MTB I had. I've already read a lot of accounts of it, but if
there's any websites that have a lot of pictures and explanations of the
different rims it'd be interesting to see.


Just for clarification, when you say "bead strength" or "stress on the
bead," you seem to be referring to the tensile force on the wire or
whatever material provides the reinforcement of the bead. Is that
correct? There is also the issue of the security of the interface
between rim and tire at the bead. Essentially, with a secure enough
bead "hook," the rim would serve in place of the tire bead. By way of
further example, a tubular tire requires nothing but the tire itself
(no extra circumferential reinforcement) to withstand applicable
pressures.

DR
  #4  
Old August 22nd 11, 09:20 PM posted to rec.bicycles.tech
DougC
external usenet poster
 
Posts: 1,276
Default Tire-making: bead stress, tire width, math, woe........

On 8/22/2011 1:30 PM, DirtRoadie wrote:


Just for clarification, when you say "bead strength" or "stress on the
bead," you seem to be referring to the tensile force on the wire or
whatever material provides the reinforcement of the bead. Is that
correct?


Yes that's correct.
What I'd like to be able to predict is, given a tire's width
(cross-section area) and knowing an inflation pressure it should reach,
how much tensile strength do the beads need to have.

There is also the issue of the security of the interface
between rim and tire at the bead. Essentially, with a secure enough
bead "hook," the rim would serve in place of the tire bead.


Not with any modern tire. The only tires that were retained by a
hook-bead rim were the (cotton) cord-bead tires used previous to 1920 or
so at the latest.

One image, from 1913-
http://www.norcom2000.com/users/dcim..._rims_k25d.jpg

I have not found anything written that indicated that the tiny ridges on
the edges of modern clincher bicycle tires plays any part in keeping
them on the rim.

By way of
further example, a tubular tire requires nothing but the tire itself
(no extra circumferential reinforcement) to withstand applicable
pressures.


The physics of inflation pressure a tubular tire are considerably
different than a clincher.



  #5  
Old August 22nd 11, 09:24 PM posted to rec.bicycles.tech
thirty-six
external usenet poster
 
Posts: 10,049
Default Tire-making: bead stress, tire width, math, woe........

On Aug 22, 7:30*pm, DirtRoadie wrote:
On Aug 22, 10:47*am, DougC wrote:



I'm wondering if it is possible that a rough estimate of necessary bead
strength could be determined for a given size tire. There is info online
about figuring wall strength of hoses and cylinders containing pressure,
but they don't deal with the situation of a tire--where the wall is
interrupted.


I've cut up a couple cheap cruiser tires and seen that for them (about
2.1" wide, with a max pressure of 40 PSI) that each of the beads is a
piece of cable with a total breaking strength of about 300 lbs. I could
just match that (even cheap tire beads very rarely fail at their rated
pressures) but it would be nice to know a rule of thumb when making
different-size tires.


*From what I have seen of tires I have on hand, the tire's overall
diameter has very little if anything to do with bead loads. I have a
pair of 1.5" wide Kenda Kwest 100 psi tires in 406mm and 559mm, and both
have the same width casings, and both beads measure right about the same
thickness (.118").


{-I am just measuring on the outside of the tire bead, rubber and all,
but anyway}


I also have a 2.3" 559 Big Apple (60 psi), and it is about 6.5" across
the casing, and the beads of it measure right about .150". The BA beads
feel quite stiffer than the Kwests as well.


So then-


1) assuming they have the same cross-section, a larger-diameter tire
(26") does not seen to need a thicker bead than a smaller-diameter tire
(20")


2) the overall tire pressure does necessitate a stronger bead as the
tire pressure increases, but-


3) the stress on the bead increases with the tire's cross-section more
than it does for the pressure (the BA's pressure is only 60% of the
Kwests, but the BA bead is still considerably thicker)


4) I am also wondering now what difference in rim width would make, as
you can get road 26" rims that are ~25mm wide, as well as cruiser 26"
rims that are 80mm wide. On the 80mm rim, not only does the tire's
internal volume increase, but the portion borne by the beads increases
as well.


Is this a problem that can even be estimated roughly, or would it
require 3-d modeling to figure out? It would seem to be fairly simple,
as the tire casing always expands into a circle (the cross-section of
the tire, that is...).


* ----------


Also when I went looking for such info online, I ran across a lot of
reports of people trying to use non-tubeless tires on tubeless rims.


It's pretty surprising (to me) how common it is for people to say that
the tubeless setups ride much better, but also how common the problem is
of a tire blowing off a tubeless rim and the bead being permanently
damaged from it (ruining the tire). Usually this seems to be with tires
that are not intended for tubeless use; I've already read that a lot of
tires not specified as tubeless are not warrantied for this purpose.


I've never seen these IRL as the first one came out right as I got rid
of the last MTB I had. I've already read a lot of accounts of it, but if
there's any websites that have a lot of pictures and explanations of the
different rims it'd be interesting to see.


Just for clarification, when you say "bead strength" or "stress on the
bead," you seem to be referring to the tensile force on the wire or
whatever material provides the reinforcement of the bead. Is that
correct? There is also the issue of the security of the interface
between rim and tire at the bead. Essentially, with a secure enough
bead "hook," the rim would serve in place of the tire bead. *By way of
further example, a tubular tire requires nothing but the tire itself
(no extra circumferential reinforcement) to withstand applicable
pressures.

DR


Usually sewn together although there was an East European tubular tyre
which relied on the gluing of the base tape and the tyre pressure upon
the rim to hold it together. I know people liked them, cheap, so
there must have been good experiences. Proper wires are the best way
to secure a wired-on tyre. The gist of it is in the name.
  #6  
Old August 22nd 11, 09:28 PM posted to rec.bicycles.tech
DougC
external usenet poster
 
Posts: 1,276
Default Tire-making: bead stress, tire width, math, woe........

On 8/22/2011 1:18 PM, thirty-six wrote:


The wire stiffens the connection with the rim so that the tyre stays
in place. The wire's strength is of no particular importance, it's
the resistance to bending which is key. ...


I don't know if I believe that.

It would mean that you could take a clincher tire and cut both beads
completely through--and then mount & inflate it and still have it stay
on the rim, with just as much pressure as with the beads uncut.

Obviously the beads are placed under great tension in use, since for
~100 years steel was the only material used and in the last several
decades the only other material used has been kevlar (which is also a
high tensile strength material).

  #7  
Old August 22nd 11, 09:28 PM posted to rec.bicycles.tech
thirty-six
external usenet poster
 
Posts: 10,049
Default Tire-making: bead stress, tire width, math, woe........

On Aug 22, 9:20*pm, DougC wrote:
On 8/22/2011 1:30 PM, DirtRoadie wrote:



Just for clarification, when you say "bead strength" or "stress on the
bead," you seem to be referring to the tensile force on the wire or
whatever material provides the reinforcement of the bead. Is that
correct?


Yes that's correct.
What I'd like to be able to predict is, given a tire's width
(cross-section area) and knowing an inflation pressure it should reach,
how much tensile strength do the beads need to have.


None. See my first reply.

* There is also the issue of the security of the interface

between rim and tire at the bead. Essentially, with a secure enough
bead "hook," the rim would serve in place of the tire bead.


Not with any modern tire. The only tires that were retained by a
hook-bead rim were the (cotton) cord-bead tires used previous to 1920 or
so at the latest.

One image, from 1913-http://www.norcom2000.com/users/dcimper/assorted/inanities/recumbent/...

I have not found anything written that indicated that the tiny ridges on
the edges of modern clincher bicycle tires plays any part in keeping
them on the rim.


Works with high pressures and flexible wires. Still has the problem
that upon deflation the tyre folds up.

* By way of

further example, a tubular tire requires nothing but the tire itself
(no extra circumferential reinforcement) to withstand applicable
pressures.


The physics of inflation pressure a tubular tire are considerably
different than a clincher.


Then you're looking at it wrong.

  #8  
Old August 22nd 11, 09:32 PM posted to rec.bicycles.tech
thirty-six
external usenet poster
 
Posts: 10,049
Default Tire-making: bead stress, tire width, math, woe........

On Aug 22, 9:28*pm, DougC wrote:
On 8/22/2011 1:18 PM, thirty-six wrote:



The wire stiffens the connection with the rim so that the tyre stays
in place. *The wire's strength is of no particular importance, it's
the resistance to bending which is key. *...


I don't know if I believe that.

It would mean that you could take a clincher tire and cut both beads
completely through--and then mount & inflate it and still have it stay
on the rim, with just as much pressure as with the beads uncut.


You would be breaking the beam and so the hold of the tyre is
compromised around the cut. Test it statically. Do not ride it at
speed.

Obviously the beads are placed under great tension in use, since for
~100 years steel was the only material used and in the last several
decades the only other material used has been kevlar (which is also a
high tensile strength material).


The steel wire is used as a beam for the tyre's unfortunate edge. The
tubular tyre does not of course suffer from this liability as the
edges are joined together making the tyre essentially edgeless.

  #9  
Old August 22nd 11, 09:55 PM posted to rec.bicycles.tech
DougC
external usenet poster
 
Posts: 1,276
Default Tire-making: bead stress, tire width, math, woe........

On 8/22/2011 3:32 PM, thirty-six wrote:
On Aug 22, 9:28 pm, wrote:
On 8/22/2011 1:18 PM, thirty-six wrote:



The wire stiffens the connection with the rim so that the tyre stays
in place. The wire's strength is of no particular importance, it's
the resistance to bending which is key. ...


I don't know if I believe that.

It would mean that you could take a clincher tire and cut both beads
completely through--and then mount& inflate it and still have it stay
on the rim, with just as much pressure as with the beads uncut.


You would be breaking the beam and so the hold of the tyre is
compromised around the cut. Test it statically. Do not ride it at
speed.

Obviously the beads are placed under great tension in use, since for
~100 years steel was the only material used and in the last several
decades the only other material used has been kevlar (which is also a
high tensile strength material).


The steel wire is used as a beam for the tyre's unfortunate edge. The
tubular tyre does not of course suffer from this liability as the
edges are joined together making the tyre essentially edgeless.


You need to sort the terminology you're using.

Stiffness = resistance to bending.
Tensile strength = resistance to a longitudinal pulling force.

Bicycle tire beads do not need to be stiff at all, steel is stiff but
[foldable] kevlar is not, and both obviously work.
  #10  
Old August 22nd 11, 10:05 PM posted to rec.bicycles.tech
thirty-six
external usenet poster
 
Posts: 10,049
Default Tire-making: bead stress, tire width, math, woe........

On Aug 22, 9:55*pm, DougC wrote:
On 8/22/2011 3:32 PM, thirty-six wrote:



On Aug 22, 9:28 pm, *wrote:
On 8/22/2011 1:18 PM, thirty-six wrote:


The wire stiffens the connection with the rim so that the tyre stays
in place. *The wire's strength is of no particular importance, it's
the resistance to bending which is key. *...


I don't know if I believe that.


It would mean that you could take a clincher tire and cut both beads
completely through--and then mount& *inflate it and still have it stay
on the rim, with just as much pressure as with the beads uncut.


You would be breaking the beam and so the hold of the tyre is
compromised around the cut. * Test it statically. *Do not ride it at
speed.


Obviously the beads are placed under great tension in use, since for
~100 years steel was the only material used and in the last several
decades the only other material used has been kevlar (which is also a
high tensile strength material).


The steel wire is used as a beam for the tyre's unfortunate edge. *The
tubular tyre does not of course suffer from this liability as the
edges are joined together making the tyre essentially edgeless.


You need to sort the terminology you're using.

Stiffness = resistance to bending.
Tensile strength = resistance to a longitudinal pulling force.

Bicycle tire beads do not need to be stiff at all, steel is stiff but
[foldable] kevlar is not, and both obviously work.


I've explained why.
 




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