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)

• A. 8.243

Answer 1

The correct option is A 8.243
Stiffness of helical spring,

$k=\frac{G{d}^4}{64R^{3}n}$

d= spring wire diameter; R =Mean coil radius; n = Number of turns

$\therefore$$k\propto d^4$

$\frac{k^{\prime }}{k}={\left(\frac{d^{\prime }}{d}\right)}^4$

$k^{\prime }={\left(\frac{1.02d}{d}\right)}^{4}k$

$k^{\prime }=1.08243k$

% increase in stiffness
$=\frac{k^{\prime }-k}{k}\times 100\%$

$=\left(\frac{1.08243k-k}{k}\right)\times 100\text{\%}=8.243\%$