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.

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