Determine the axial extension along a load:

An open coiled helical spring is created having 10 turns of a mean radius of 60 mm. The wire diameter is 10 mm and coils make an angle of 38^o along a plane perpendicular to the axis of the coil. Determine

 (1) the axial extension along a load of 200 N,

 (2) the angle of rotation of free end.

E = 200 GPa, G = 80 GPa.

Solution

n = 10, R = 60 mm, d = 10 mm, α = 38, E = 200 GPa, G = 80 GPa

(a) Axial Extension

 Δ= 64 W R³ sec α/d⁴ ( cos2 α/G +2 sin ² α /E)

=64 (200) (60)³ (10) sec 30 / 10⁴ ((cos² 30/80 × 10³⁾) + (2 sin ² 30 /200 × 10³ ))  = 36.7 mm

(a) Angle of Rotation at Free End

φ= 64 W R³ sin α/d⁴ ((1/G -  2/E))

=64 (200) (60)³ (10) sec 30/10⁴ ( 1/ (80 × 10³)  - 2/ (200 × 10³)) = 0.0567 radian