A coil-spring of stiffness k is cut exactly at the middle and the two springs thus made are arranged in parallel to take up together a compressive load. The equivalent stiffness of the two springs is

Concept:

Stiffness of a coil spring is given by,

\(K = \frac{{G{d^4}}}{{64{R^3}n}}\)

G = Rigidity modulus, D = mean coil diameter, R = mean radius, d = wire diameter, n = number of coils

Stiffness is inversely proportional to number of coils

When ‘n’ springs are connected in series

Equivalent stiffness is given by,

\(\frac{1}{{k\left( {eq} \right)}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \ldots + \frac{1}{{{k_n}}}\) 

When ‘n’ spring are connected in parallel

k_eq = k₁ + k₂ + k₃ + ….. + k_n

Calculation:

When the spring is cut into two equal parts.

Stiffness of each part becomes 2 times since n become n/2

Let the original stiffness be k

New stiffness = 2k each

Connected in parallel = 2k + 2k = 4k

Note:

If spring is divided in ‘n’ equal parts and those parts are connected in parallel.

New stiffness = n²k

The parts connected in series