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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
Drill four holes(two62inchand 25inch)
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