A close-coiled helical spring of 10 active turns is made of 8 mm diameter steel wire. The mean coil diameter is 10 cm. If G = 80 GPa for the material of spring, the extension of spring under a tensile load of 200 N will be nearly

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ESE Mechanical 2014 Official Paper - 2

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Answer 1

Concept:

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

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

Calculation:

Given:

n = 10, D = 10 cm = 100 mm, d = 8 mm, G = 80 GPa = 80 × 10³ MPa, P = 200 N

Deflection (extension) of a spring:

\(\Delta = \frac{{8P{D^3}n}}{{G{d^4}}}\)

\(\Delta = \frac{{8\: × \:200\: × \:{{100}^3}\: × \:10}}{{80\: ×\: {{10}^3}\: ×\: {8^4}}}\) = 48.828 mm

Δ ≃ 49 mm