Find out ratio of strain-energy:
A circular rod of diameter d and length l is fixed at one end and a torque T applied at the other. Now, if any close coiled helical spring is built of the same rod & applied same moment T around the axis, find out ratio of strain-energy stored.
E = 200 GPa; G = 80 GPa.
Solution
Circular Rod
J = 2I
θ= Tl / GJ
U_{1} = T ^{2} l / 2 GJ = T l / 4 G I ------------- (1)
Helical Spring
φ= M l/ E I = T l/ E I
U_{ 2} = (½) M φ = T ^{2} l / 2 E I -------------. (2)
U_{1} /U_{2} = E/ 2 G = 200 / (2 × 80) = 1.25