Question
Download Solution PDFIn a closed helical spring subjected to an axial load, other quantities remain the same, if the wire diameter is doubled and the mean radius of the coil is also doubled, then the stiffness of the spring when compared to the original one will become _____.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFStiffness 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 diameter of the spring wire; n is number of active coils and G is modulus of rigidity for the spring material.
Here d’ = 2d; D’ = 2D
\(k' = \frac{{Gd{'^4}}}{{8D{'^3}n}} = \frac{{16}}{8}\frac{{G{d^4}}}{{8{D^3}n}} = 2\frac{{G{d^4}}}{{8{D^3}n}} = 2k\)
Last updated on Jul 15, 2025
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