Question

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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Answer :

8.24


Stiffness of helical spring , K = Gd4/64R³n

                                     where, d = spring wire diameter

                                                  R= mean coil radius

                                                  n= number of turns

                                                  K d4

                                                  K'/K = (d'/d)4

                                                  K' = (1.02d/d)4K

                                                  K' = 1.08243 K

          % increase in stiffness =  (K'-K)*100%/K = 8.234%

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