Determine number of coils and length of the spring:
A close coiled helical spring ought to extend by 120 mm under an axial force of 1200 N. If mean coil radius is equal to 40 mm and maximum shear stress is 300 MPa, determine the wire diameter, number of coils and length of the spring. Take G = 80 GPa.
Solution
Δ = 120 mm, W = 1200 N, R = 40 mm, τ_{max } = 300 MPa ,
G = 80 GPa = 80 × 103 N/mm^{2}, d = ?, n = ?, l = ?
Using Eq. (3) for maximum shearing stress
τ = 16W R / π d^{3}
or 300 = 16 × 1200 × R / π d ^{3}
∴ R/ d ^{3}^{ } = 0.049 or R = 0.049 d ^{3} ----------- (1)
40/0.049 or d = 9.35 mm --------- (2)
From Eq. (7)
Δ= 64 W R^{3} n / Gd ^{4}
∴ 120 =6 4× 1200 × R^{3} n/80 × 10^{3} × d ^{4}
or R^{3} n/ d ^{4} = 125
Using values of R and d,
40^{3} × n / 9.354 = 125
∴ n = 14.9 ; 15
l = 2 π R n
= 2 π (40) (15)
= 3770 mm