Potential good news for Mt. Washington access. - Page 26 - Cycling Mob: Road Cycling, Mountain Biking and More

01-18-2005, 08:52 AM

Joe Riel <[Only registered and activated users can see links. ]> writes:

> Here is an approximation. Assume that we can model this by computing
> the transverse bending of a circular plate of radius a with a load F
> uniformly distributed around a concentric circle of radius b. The
> diameter of the plate is set to the width of the rim, the diameter of
> the loading circle to that of the spoke hole. From [1,article 314
> (vi)] the displacement, w, is
>
> w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
>
> where
>
> D = 2/3*E*(T/2)^3/(1-sigma^2)
> T = material thickness = 1mm
> sigma = Poisson ratio = 1/3
> E = elasticity of aluminum = 75kN/mm^2
>
> a = half rim width = 25mm/2
> b = radius of spoke hole = 3.2mm
>
> This gives
>
> w/F = 8.3um/kgf
>
> So a change in the spoke tension causes a deflection in the nipple
> hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
> 0.2% of the spoke length. For a 100degC rise the diameter of the rim
> increased by 0.25%. So it appears as though much of the compliance is
> in the bottom of the rim. The proper technique is to combine this
> compliance with the previous computation---but its late and I'm tired.
>
>
> Reference
> ---------
> [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"

I assumed the entire load was supported by one wall (the inner
surface) of the rim. With sockets distributing the load to both
walls, the compliance should be about half that computed.

Joe Riel

01-18-2005, 08:52 AM

Joe Riel <[Only registered and activated users can see links. ]> writes:

> Here is an approximation. Assume that we can model this by computing
> the transverse bending of a circular plate of radius a with a load F
> uniformly distributed around a concentric circle of radius b. The
> diameter of the plate is set to the width of the rim, the diameter of
> the loading circle to that of the spoke hole. From [1,article 314
> (vi)] the displacement, w, is
>
> w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
>
> where
>
> D = 2/3*E*(T/2)^3/(1-sigma^2)
> T = material thickness = 1mm
> sigma = Poisson ratio = 1/3
> E = elasticity of aluminum = 75kN/mm^2
>
> a = half rim width = 25mm/2
> b = radius of spoke hole = 3.2mm
>
> This gives
>
> w/F = 8.3um/kgf
>
> So a change in the spoke tension causes a deflection in the nipple
> hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
> 0.2% of the spoke length. For a 100degC rise the diameter of the rim
> increased by 0.25%. So it appears as though much of the compliance is
> in the bottom of the rim. The proper technique is to combine this
> compliance with the previous computation---but its late and I'm tired.
>
>
> Reference
> ---------
> [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"

I assumed the entire load was supported by one wall (the inner
surface) of the rim. With sockets distributing the load to both
walls, the compliance should be about half that computed.

Joe Riel

01-18-2005, 08:52 AM

Joe Riel <[Only registered and activated users can see links. ]> writes:

> Here is an approximation. Assume that we can model this by computing
> the transverse bending of a circular plate of radius a with a load F
> uniformly distributed around a concentric circle of radius b. The
> diameter of the plate is set to the width of the rim, the diameter of
> the loading circle to that of the spoke hole. From [1,article 314
> (vi)] the displacement, w, is
>
> w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))
>
> where
>
> D = 2/3*E*(T/2)^3/(1-sigma^2)
> T = material thickness = 1mm
> sigma = Poisson ratio = 1/3
> E = elasticity of aluminum = 75kN/mm^2
>
> a = half rim width = 25mm/2
> b = radius of spoke hole = 3.2mm
>
> This gives
>
> w/F = 8.3um/kgf
>
> So a change in the spoke tension causes a deflection in the nipple
> hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
> 0.2% of the spoke length. For a 100degC rise the diameter of the rim
> increased by 0.25%. So it appears as though much of the compliance is
> in the bottom of the rim. The proper technique is to combine this
> compliance with the previous computation---but its late and I'm tired.
>
>
> Reference
> ---------
> [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"

I assumed the entire load was supported by one wall (the inner
surface) of the rim. With sockets distributing the load to both
walls, the compliance should be about half that computed.

Joe Riel

01-18-2005, 09:44 AM

Joe Riel writes:

>> Here is an approximation. Assume that we can model this by
>> computing the transverse bending of a circular plate of radius a
>> with a load F uniformly distributed around a concentric circle of
>> radius b. The diameter of the plate is set to the width of the
>> rim, the diameter of the loading circle to that of the spoke hole.
>> From [1,article 314 (vi)] the displacement, w, is

>> w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))

>> where

>> D = 2/3*E*(T/2)^3/(1-sigma^2)
>> T = material thickness = 1mm
>> sigma = Poisson ratio = 1/3
>> E = elasticity of aluminum = 75kN/mm^2

>> a = half rim width = 25mm/2
>> b = radius of spoke hole = 3.2mm

>> This gives

>> w/F = 8.3um/kgf

>> So a change in the spoke tension causes a deflection in the nipple
>> hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
>> 0.2% of the spoke length. For a 100degC rise the diameter of the
>> rim increased by 0.25%. So it appears as though much of the
>> compliance is in the bottom of the rim. The proper technique is to
>> combine this compliance with the previous computation---but its
>> late and I'm tired.

>> Reference
>> ---------
>> [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"

> I assumed the entire load was supported by one wall (the inner
> surface) of the rim. With sockets distributing the load to both
> walls, the compliance should be about half that computed.

The rims I used in deflection analysis were socketed and support spoke
load on both walls and distribute the load in such a way that there is
no tent pole effect around spokes. If there were, it would be
permanent, considering the concentration of the deformation. It would
take the rim to yield and remain as a permanent shape. Besides, if
this were the case, rims would crack there fairly soon... as some do,
but for other reasons.

Rims can readily be modeled as straight beams between spokes, rigidly
connected at their ends with no effect on the deflection results. The
arch in rim segments is not a buckling element in that axis.

Jobst Brandt
[Only registered and activated users can see links. ]

01-18-2005, 09:44 AM

Joe Riel writes:

>> Here is an approximation. Assume that we can model this by
>> computing the transverse bending of a circular plate of radius a
>> with a load F uniformly distributed around a concentric circle of
>> radius b. The diameter of the plate is set to the width of the
>> rim, the diameter of the loading circle to that of the spoke hole.
>> From [1,article 314 (vi)] the displacement, w, is

>> w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))

>> where

>> D = 2/3*E*(T/2)^3/(1-sigma^2)
>> T = material thickness = 1mm
>> sigma = Poisson ratio = 1/3
>> E = elasticity of aluminum = 75kN/mm^2

>> a = half rim width = 25mm/2
>> b = radius of spoke hole = 3.2mm

>> This gives

>> w/F = 8.3um/kgf

>> So a change in the spoke tension causes a deflection in the nipple
>> hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
>> 0.2% of the spoke length. For a 100degC rise the diameter of the
>> rim increased by 0.25%. So it appears as though much of the
>> compliance is in the bottom of the rim. The proper technique is to
>> combine this compliance with the previous computation---but its
>> late and I'm tired.

>> Reference
>> ---------
>> [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"

> I assumed the entire load was supported by one wall (the inner
> surface) of the rim. With sockets distributing the load to both
> walls, the compliance should be about half that computed.

The rims I used in deflection analysis were socketed and support spoke
load on both walls and distribute the load in such a way that there is
no tent pole effect around spokes. If there were, it would be
permanent, considering the concentration of the deformation. It would
take the rim to yield and remain as a permanent shape. Besides, if
this were the case, rims would crack there fairly soon... as some do,
but for other reasons.

Rims can readily be modeled as straight beams between spokes, rigidly
connected at their ends with no effect on the deflection results. The
arch in rim segments is not a buckling element in that axis.

Jobst Brandt
[Only registered and activated users can see links. ]

01-18-2005, 09:44 AM

Joe Riel writes:

>> Here is an approximation. Assume that we can model this by
>> computing the transverse bending of a circular plate of radius a
>> with a load F uniformly distributed around a concentric circle of
>> radius b. The diameter of the plate is set to the width of the
>> rim, the diameter of the loading circle to that of the spoke hole.
>> From [1,article 314 (vi)] the displacement, w, is

>> w = F/8/Pi/D*((a^4-b^4)/2/a^2 - (a^2+b^2)*ln(a/b))

>> where

>> D = 2/3*E*(T/2)^3/(1-sigma^2)
>> T = material thickness = 1mm
>> sigma = Poisson ratio = 1/3
>> E = elasticity of aluminum = 75kN/mm^2

>> a = half rim width = 25mm/2
>> b = radius of spoke hole = 3.2mm

>> This gives

>> w/F = 8.3um/kgf

>> So a change in the spoke tension causes a deflection in the nipple
>> hole of approximately (75kgF)(8.3um/kgF) = 0.6mm. That represents
>> 0.2% of the spoke length. For a 100degC rise the diameter of the
>> rim increased by 0.25%. So it appears as though much of the
>> compliance is in the bottom of the rim. The proper technique is to
>> combine this compliance with the previous computation---but its
>> late and I'm tired.

>> Reference
>> ---------
>> [1] A. E. H. Love "A Treatise on the Mathematical Theory of Elasticity"

> I assumed the entire load was supported by one wall (the inner
> surface) of the rim. With sockets distributing the load to both
> walls, the compliance should be about half that computed.

The rims I used in deflection analysis were socketed and support spoke
load on both walls and distribute the load in such a way that there is
no tent pole effect around spokes. If there were, it would be
permanent, considering the concentration of the deformation. It would
take the rim to yield and remain as a permanent shape. Besides, if
this were the case, rims would crack there fairly soon... as some do,
but for other reasons.

Rims can readily be modeled as straight beams between spokes, rigidly
connected at their ends with no effect on the deflection results. The
arch in rim segments is not a buckling element in that axis.

Jobst Brandt
[Only registered and activated users can see links. ]