Avehicle 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

answer:

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explanation: