Two closely-coiled helical springs 'A' and 'B' of the same material, same number of turns and made from same wire are subjected to an axial load W. The mean diameter of spring 'A' is double the mean diameter of spring 'B'. The ratio of deflections in spring 'B' to spring 'A' will be

Concept:

Deflection of closely coiled helical spring subjected to axial force is given by,

\(δ=\frac{8WD^3n}{Gd^4}\) .... (1) 

Further, the Stiffness of spring is, \(S=\frac{Gd^4}{8D^3n}\) ..... (2)

Where, δ = deflection of spring, W = axial load, D = mean diameter, d = coil diameter, n = the number of turns, and G = modulus of rigidity.

From Equation (1), it can be seen that \(δ \propto D^3 \) 

Calculation:

Given that, in two springs A & B, \(D_A = 2 \times D_B\)

Now,  \(\frac{\delta_A}{\delta_B} = \frac{D_A^3}{D_B^3} \)

\(\frac{\delta_A}{\delta_B} = 2^3 = 8 \)

Therefore, \(\frac{\delta_B}{\delta_A} = \frac{1}{8} \)