Design of a helical compression spring

Data

$F = 800$ N, $\delta = 30$ mm, $C = 6$, $\tau = 450$ MPa, $G = 81.4$ GPa $= 81\,400$ N/mm².

Wire diameter

Wahl factor:

$$K = \frac{4C-1}{4C-4} + \frac{0.615}{C} = \frac{23}{20} + 0.1025 = 1.2525$$

From $\tau = K\dfrac{8FC}{\pi d^2}$:

$$d^2 = \frac{8KFC}{\pi\tau} = \frac{8\times1.2525\times800\times6}{\pi\times450} = 34.02 \;\Rightarrow\; d = 5.83\ \text{mm}$$

Select standard wire $d = 6$ mm. Check: $\tau = 1.2525\times\dfrac{8\times800\times6}{\pi\times 36} = 425\ \text{MPa} < 450$ MPa (safe).

Mean diameter

$$D = C d = 6\times6 = 36\ \text{mm}, \qquad D_o = D + d = 42\ \text{mm}$$

Active turns

From $\delta = \dfrac{8FC^3 n}{Gd}$:

$$n = \frac{G d\,\delta}{8FC^3} = \frac{81400\times6\times30}{8\times800\times216} = 10.6$$

Take $n = 11$ active turns. The actual stiffness is $k = \dfrac{Gd}{8C^3n} = \dfrac{81400\times6}{8\times216\times11} = 25.7$ N/mm, so the actual deflection is $800/25.7 = 31.1$ mm.

Total turns and lengths

Squared and ground ends add 2 inactive turns: $n_t = 11 + 2 = 13$.

• Solid length: $L_s = n_t d = 13\times6 = 78$ mm
• Clearance: $0.15\times31.1 \approx 4.7$ mm
• Free length: $L_f = L_s + \delta + \text{clearance} = 78 + 31.1 + 4.7 = 113.8$ mm

Pitch $p \approx \dfrac{L_f - 2d}{n} = \dfrac{113.8-12}{11} = 9.26$ mm.

Check for buckling: $L_f/D = 113.8/36 = 3.16$, which is below the usual limit of about 4, so the spring is stable without a guide. Squared and ground ends seated on parallel plates make buckling even less likely.

Answer: $d = 6$ mm, $D = 36$ mm, $n = 11$, total 13 turns, $L_s = 78$ mm, $L_f \approx 114$ mm.