Question
Download Solution PDFIn a closed helical spring subjected to an axial load, other quantities remaining the same, if the wire diameter is doubled and mean radius of the coil is also doubled, then stiffness of spring when compared to original one will become _____.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
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 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
\(k' = \frac{{Gd{'^4}}}{{8D{'^3}n}} = \frac{{16}}{8}\frac{{G{d^4}}}{{8{D^3}n}} = 2\frac{{G{d^4}}}{{8{D^3}n}} = 2k\)
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