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27 January, 09:25

A helical compression spring is made with oil-tempered wire with wire diameter of 0.2 in, mean coil diameter of 2 in, a total of 12 coils, a free length of 5 in, with squared ends. (a) Find the solid length. (b) Find the force necessary to deflect the spring to its solid length. (c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

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  1. 27 January, 11:14
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    a. Solid length Ls = 2.6 in

    b. Force necessary for deflection Fs = 67.2Ibf

    Factor of safety FOS = 2.04

    Explanation:

    Given details

    Oil-tempered wire,

    d = 0.2 in,

    D = 2 in,

    n = 12 coils,

    Lo = 5 in

    (a) Find the solid length

    Ls = d (n + 1)

    = 0.2 (12 + 1) = 2.6 in Ans

    (b) Find the force necessary to deflect the spring to its solid length.

    N = n - 2 = 12 - 2 = 10 coils

    Take G = 11.2 Mpsi

    K = (d^4*G) / (8D^3N)

    K = (0.2^4*11.2) / (8*2^3*10) = 28Ibf/in

    Fs = k*Ys = k (Lo - Ls)

    = 28 (5 - 2.6) = 67.2 lbf Ans.

    c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.

    For C = D/d = 2/0.2 = 10

    Kb = (4C + 2) / (4C - 3)

    = (4*10 + 2) / (4*10 - 3) = 1.135

    Tau ts = Kb { (8FD) / (Πd^3) }

    = 1.135 { (8*67.2*2) / (Π*2^3) }

    = 48.56 * 10^6 psi

    Let m = 0.187,

    A = 147 kpsi. inm^3

    Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi

    Ssy = 0.50 Sut

    = 0.50 (198.6) = 99.3 kpsi

    FOS = Ssy/ts

    = 99.3/48.56 = 2.04 Ans.
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