A helical spring of 10 N/mm rating is mounted on top of another helical spring of 8 N/mm rating. The force required for a total combined deflection of 45 mm through the two springs is

Concept:

Spring on top of one another = Spring in series

 When ‘n’ springs are connected in series

Equivalent stiffness is given by:

\(\frac{1}{{{k_{\left( {eq} \right)}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}} + \ldots + \frac{1}{{{K_n}}}\)

Calculation:

Given:

k₁ = 10 N/mm, k₂ = 8 N/mm, and δ = 45 mm

Equivalent stiffness \({k_{eq}} = \frac{{{k_1}\;{k_2}}}{{{k_1}\; + \;{k_2}}}\)

Equivalent stiffness \({k_{eq}} = \left(\frac{{10\; \times \;8}}{{10\; + \;8}}\right) = \frac{{80}}{{18}} = \frac{{40}}{9}\;N/mm\)

\(Stiffness = \frac{{Load}}{{Deflection}}\)

Load = Stiffness × Deflection

\(\therefore F=\frac{{40}}{9} \times 45 = 200\;N\)