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

U₁ = 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)

U₁ /U₂  =   E/ 2 G =   200 / (2 × 80) = 1.25