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24 July, 12:26

A vehicle has individual wheel suspension in the form of helical springs. The free length of the spring lf = 360 mm and the solid length ls = 160 mm at a compressive force of 5000 N. The shear modulus G = 80 GPa. Use D/d = 9 and calculate the shear stress for pure torsion of the spring wire. The spring ends are squared and ground. Find Na, p, d, D, and  max.

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  1. 24 July, 12:33
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    d = 68.44, D = 615.96 mm, N (a) = 0.3378 coils, p = 11065.67 mm τ = 28.43 GPa

    Explanation:

    Given that l (f) = 360 mm, l (s) = 160 mm, P = 5000 N, G = 80 GPa, D/d = 9

    l (f) = l (s) + δ

    360 = 160 + δ

    δ = 200

    δ = 8. P. D³. N (a) / G. d⁴

    We know that D/d = 9 and also that N (a) can be written as (l (s) / d) - 2, hence

    S = (8. P/G). (D/d). (D/d). (D/d). (1/d).{ (160/d) - 2}

    Substitute the values and solve for d

    200 = (8 x 5000/80) x 9 x 9 x 9 x (1/d) x ((160/d) - 2)

    200 = (40500/d). (160/d - 2)

    200 = 6480000/d² - 81000/d

    Multiplying the above equation by (d²/200) and rearranging

    d² + 405d - 32400 = 0

    Solve simultaneously to get

    d = 68.44 and - 473.44

    Use the positive value which is d = 68.44 mm

    We know that D = 9 x d

    D = 9 x 68.44 = 615.96 mm

    N (a) = (160/d) - 2

    = (160/68.44) - 2 = 0.3378 coils

    p = l (f) / N (a)

    = 360/0.3378 = 11065.67 mm

    Tau = 8WD/πd³ x P

    Where W, is the Wahl correction factor which is given by

    W = (4C - 1/4C - 4) + (0.615/C)

    = 35/32 + 0.615/9

    = 1.162

    τ = 8 x (1.162) x 615.96/π x (68.44) ³ x (5000)

    = 0.0056855 x 5000 = 28.427 N/mm

    = 28.43 GPa
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