Correct Answer - Option 2 : 20.16 Nm
Explanation:
Given Data
Wire diameter(d) = 12 mm
Mean coil radius(R) = 84 mm
Angle of helix (\(\alpha\)) = 600
Axial Load (W) = 480 N
Now the twisting moment(T) due to the axial load acting on an open-coiled helical spring is given by
\(T = WR \cos\alpha\)
\(T = 480 \times 84 \times \cos60\) N-mm
T= 20.16 N-m
The axial load acting on an open-coiled helical spring, bending moment(M) is given by
\(M = WR \sin\alpha\)
The Strain energy in open-coiled helical spring due to axial load given by
Strain energy = Strain energy due to bending + Strain energy due to torsion