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)
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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