Springs in Strength Of Materials : Practice MCQ Questions & Answers with Explanations

Concept: Deflection of Helical Springs of Circular Wire: $\delta = \frac{{8W{D^3}n}}{{Gd^4}}$ 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 Calculation: It is given that D2 = 2D1 d2 = 2d1 ${\delta 1} = \frac{{8\;W\;D1^3n}}{{G\;{d^41}}}$ ${\delta 2} = \frac{{8\;W\;D2^3n}}{{G\;{d^42}}}$ $\delta = \frac{{8W{D^3}n}}{{Gd^4}} \Rightarrow \delta \propto \frac{D^3}{d^4}$ (rest other parameters are same) $\frac{{{\delta 2}}}{{{\delta 1}}} = (\frac{{{D2}}}{{{D1}}})^3 \times(\frac{d1}{d2})^4= \frac{{{2^3}}}{{{2^4}}} = \frac{1}{2}$ ${\delta 2} = \frac{1}{2}{\delta 1}$ Therefore If d and D is doubled, δ will be halved.