When a closely-coiled helical spring of mean diameter (D) is subjected to an axial load (W), the stiffness of the spring is given by 

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ISRO URSC Technical Assistant Mechanical 24 March 2019 Official Paper
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  1. Gd4/D3n
  2. Gd4/4D3n
  3. Gd4/2D3n
  4. Gd4/8D3n

Answer (Detailed Solution Below)

Option 4 : Gd4/8D3n
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Explanation:

Deflection of Helical Springs of Circular Wire:

\(\delta = \frac{{8W{D^3}n}}{{G{d^4}}} = \frac{{8W{C^3}n}}{{Gd}}\)

Stiffness of the spring or spring rate

\(k = \frac{W}{\delta } = \frac{W{G{d^4}}}{{8W{D^3}n}}\)

\(k = \frac{{G{d^4}}}{{8{D^3}n}} = \frac{{G{d^4}}}{{8{{\left( {2R} \right)}^3}n}} = \frac{{G{d^4}}}{{64{R^3}n}}\)

Where D = Mean diameter of the spring coil, d = Diameter of the spring wire, n = Number of active coils, G = Modulus of rigidity for the spring material, W = Axial load on the spring

C = Spring index = D/d

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