Question 2.14: A close-coiled helical spring has stiffness 900 N/m in compr...

We know deflection (δ ) and stiffness (k ) of the spring are

$\delta=\frac{64 F R^3 n}{G d^4} \quad \text { and } \quad k=\frac{G d^4}{64 R^3 n}$

Putting k = 900 N/m or 0.9 N/mm and $G=0.4 \times 10^5  N / mm ^2$, we get

$0.9=\frac{0.4 \times 10^5 \times d^4}{64 R^3 n}$

or          $d^4=\left\lgroup \frac{0.9 \times 64}{0.4 \times 10^5} \right\rgroup R^3 n$          (1)

Torque ‘FR’ causes shear stress τ and is given by

$\tau=\frac{16 F R}{\pi d^3}$

Putting $\tau=\tau_{\max }$ and substituting values, we get

$\tau=\tau_{\max }=\frac{16 \times 45 \times R}{\pi d^3}$

or      $120=\frac{16 \times 45 \times R}{\pi d^3}$

or          R = 0.52d³          (2)

Now, solid length² of spring being 45 mm,

$n=\frac{45}{d}$            (3)

Combining Eqs. (1)–(3), we find

$\begin{aligned} d^4 & =\left\lgroup\frac{0.9 \times 64}{0.4 \times 10^5} \right\rgroup \times\left(0.52 d^3\right)^3 \times \frac{45}{d} \\ & =(109.75)^{1 / 4} \\ & =3.24  mm \end{aligned}$

The mean coil radius is

$R=0.52 d^3=0.52 \times(3.24)^3=17.68  mm$

The number of turns is

$N=\frac{45}{d}=\frac{45}{3.24}=13.88 \cong 14$

Therefore, the wire diameter is 3.24 mm, mean coil radius is 17.68 mm and number of turns is 14.

$^2$ Solid length of the spring is its length when it is fully compressed and the coils of the spring are touching each other.