Stress Relieving Spokes in Small Wheels---Analysis

Discussion in 'Cycling Equipment' started by Joe Riel, Mar 13, 2003.

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  1. Joe Riel

    Joe Riel Guest

    While replacing a spoke on my Moulton---it has small diameter wheels, the spokes are approximatly
    half the length of a 700C wheel---I wondered whether the force required to properly stress relieve
    it, see Jobst Brandt's book "The Bicycle Wheel," was dependent on the spoke length. Intuitively it
    seemed as though I should squeeze substantially harder to achieve the same tension increase.

    Let

    L = nominal length of spoke F = sideways force applied at center of spoke T = increase in spoke
    tension D = sideways displacement of the spoke K = series compliance of the structure dL = strain
    of structure

    From statics, and using a small angle approximation

    (1) F = 4*T*D/L Brandt's book [revised ed.] has an error

    From geometry

    (2) dL = L - 2*sqrt(L^2/4 - D^2)

    Assuming D << L, this is well approximated by

    (3) dL/L = 2*D^2/L^2

    From elasticity, with K being the sum of compliances of the wheel structure and the spoke

    (4) dL/L = K*T

    Combining (1), (3) and (4) gives

    (5) T = 1/2*K^(1/3)*F^(2/3)

    L does not appear in (5), that is, the increase in tension of a spoke due to deflecting it
    horizontally by a fixed force is independent of its length. There is a relatively small (1/3 power)
    dependency on K, the compliance of the structure, which depends on a number of factors (number of
    spokes, spoke thickness, rim shape).

    Joe Riel
     
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  2. Joe Riel

    Joe Riel Guest

    Joe Riel <[email protected]> writes:

    Two minor corrections:

    dL is the displacement (not the strain, that is dL/L).

    The exponent of K is -1/3, not +1/3, i.e.,

    T = 1/2*F^(2/3)/K^(1/3)

    Joe
     
  3. Joe Riel

    Joe Riel Guest

    Having derived a relation between spoke tension and applied normal force (i.e. the force you
    generate by squeezing adjacent spokes),

    (1) T = 1/2*(F^2/K)^(1/3)

    we would like to estimate the compliance, K and then use it to compute the maximum allowable force.
    I apologize if my choice of "K" for compliance flies in the face of tradition, i.e. using it for
    stiffness. If there is a more usual term, please let me know [I'm not a mechanical engineer].

    Consider the force loop that occurs when spokes are stress-relieved. Two (roughly) parallel spokes
    are tensioned. The compliances in the loop can be expressed as

    (2) Kl = Ks/2 + Km
    (3) Kw = Ks/2 | Km

    where

    Kl is the compliance of the loop Ks is the compliance of a single spoke Kw is the radial
    compliance of the wheel Km is the radial compliance of the wheel minus the two spokes
    | is the parallel operator, a|b = 1/(1/a+1/b)

    Eliminating Km and solving for Kl gives

    (4) Kl = (Ks/2)^2/(Ks/2 - Kw)

    From the "Equations" chapter in [1] we have

    (5) Ks = (dL/L)/F
    = (1.24mm/320mm)/180kgf
    = 21.5 ppm/kgf

    From the radial load finite-element table in [1]

    (6) Kw = (dL/L)/F
    = (.15mm/320mm)/50kgf
    = 9.4 ppm/kgf

    Inserting (5) and (6) into (4) we get

    (7) Kl = 38 ppm/kgf

    We can plug this into (1), but must divide the spoke tension in half because half the tension goes
    into each spoke. Thus

    (8) Tspoke = 1/2*1/2*(F^2/Kl)^(1/3)
    = (F^2/Kl/4^3)^(1/3)
    = 7.4*(F^2*kgf)^(1/3)

    In the "Spoke Strength" chapter of [1], the heavy gauge spokes broke at a tension of 320kgf, the
    lighter gauge spokes at 270kgf. In the "Spoke Elongation" section of the "Equations" chapter, ibid,
    the force in a fully tensioned spokes is given as 180kgf. The increase in tension due to stress
    relieving must then be limited to 160kgf and 90kgf, for the heavier and lighter gauge spokes,
    respectively. Solving for the maximum permissible applied force we get

    (9) Fsqueeze(max) = 80kgf = 37 lbf [heavy gauge (14) spoke]
    = 40kgf = 19 lbf [light gauge (15?) spoke]

    I've never tried to measure my grip strength, but doubt that I could achieve the higher value.
    Probably just as well, when I stress relieve spokes I use leather gloves and squeeze about as
    hard as I can.

    [10] Jobst Brandt, "The Bicycle Wheel", revised edition, 1983, Avocet, Inc.

    Joe Riel
     
  4. Joe Riel

    Joe Riel Guest

    Joe Riel <[email protected]> writes:

    > (9) Fsqueeze(max) = 80kgf = 37 lbf [heavy gauge (14) spoke]
    > = 40kgf = 19 lbf [light gauge (15?) spoke]

    Oops, an obvious error here, That should be

    (9) Fsqueeze(max) = 80kgf = 180 lbf [heavy gauge (14) spoke]
    = 40kgf = 90 lbf [light gauge (15?) spoke]

    Joe
     
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