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)

The correct answer is 8.24.

To find the percentage change in the spring stiffness when the wire diameter is increased, we can analyze the relationship between the stiffness and the wire diameter of a compressive helical spring.

The stiffness of a helical spring, denoted by K, is given by the formula:
$K=\frac{G{d}^4}{64R^{3}n}$
where:
• G is the modulus of rigidity of the wire material,
• d is the wire diameter of the spring,
• R is the mean coil radius, and
• n is the number of active turns.

From this formula, keeping all other parameters constant (material modulus of rigidity, mean coil radius, and number of turns), the stiffness K is directly proportional to the fourth power of the wire diameter d:
$K\propto d^4$

Let the initial wire diameter be d and the initial stiffness be K.
If the wire diameter is increased by 2%, the new wire diameter d' is:
$d^{\text{'}}=d+0.02d=1.02d$

Let the new stiffness be K'. The ratio of the new stiffness to the initial stiffness is:
$\frac{K^{\text{'}}}{K}={\left(\frac{d^{\text{'}}}{d}\right)}^4$

Substituting d' into the equation:
$\frac{K^{\text{'}}}{K}={\left(\frac{1.02d}{d}\right)}^4={\left(1.02\right)}^4$

Calculating the value:
${\left(1.02\right)}^4\approx 1.082432$
So, K' ≈ 1.082432 K.

The percentage change in spring stiffness is given by:
$\text{Percentage Change}=\frac{K^{\text{'}}-K}{K}\times 100\%$
$\text{Percentage Change}=(1.082432-1)\times 100\%=8.2432\%$

Rounding to two decimal places, the percentage increase in stiffness is 8.24%.