A closely coiled helical spring of 20 cm mean diameter is having 25 coils of 2 cm diameter rod. The modulus of rigidity of the material is 10⁷ N/cm². What is the stiffness for the spring in N/cm?

Question

Question

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NPCIL SA/ST ME GJ held on 07/11/2019 Shift-1

Answer 1

Concept:

The extension of spring is given as, \(\Delta = \frac{{8P{D^3}n}}{{G{d^4}}}\)

The stiffness for the spring is given as, \(k=\frac{Gd^4}{8D^3n}\)

where, P = load, D = diameter of coil, d = diameter of wire, n = no. of turns, G = modulus of rigidity, k = stiffness

Calculation:

Given:

n = 25, D = 20 cm, d = 2 cm, G = 107 N/cm2,

\(k=\frac{10^7 \times2^4}{8\times20^3\times25}=100~N/cm\)