J
Joe Riel
Guest
Joe Riel <[email protected]> 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
> 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