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If the wire diameter of a compressive helical spring is increased by 2%, the change in spring stiffness (in %) is _____ (correct to two decimal places).

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Concept:

Stiffness of the spring is given by:

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

Where D is mean diameter of the spring coil, d is the diameter of the spring wire; n is the number of active coils and G is the modulus of rigidity for the spring material.

Calculation:

d2 = 1.02d1

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

\(\frac{{{k_2}}}{{{k_1}}} = {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^4} = {\left( {1.02} \right)^4} = 1.082\)

k2 = 1.082 k1

\({\rm{\Delta }}k = \frac{{{k_2} - {k_1}}}{{{k_1}}} = 0.082 = 8.2\% \)

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