Determine the diameter of wire - close coiled helical spring:
A close coiled helical spring has to absorb 100 N-m of energy while compressed to 10 cm. The coil diameter is 10 times the wire diameter. The number of coils is 12. Determine the diameter of wire, mean radius & the maximum shear stress.
Take G = 80 GPa.
Solution
We know
U = (½) W Δ
∴ 100 × 10^{3} = (1 /2 )× W × 100
∴ W = 2000 N
D = 10 d or R = 5 d
G = 80 GPa = 80 × 10^{3} MPa = 80 × 10^{3} N/mm^{2}
Δ= 64 W R^{3} n/ Gd ^{4}
∴ 100 =6 4× 2000 × (5d ) ^{3} × 12 / (80 × 10^{3} ) d ^{4}
or d = 24 mm
R = 5d = 5 × 24 = 120 mm
τ _{max} = 16 W R / π d ^{3}
= 16 × 2000 × 120/ π (24)^{3} = 88.4 N/mm^{2}