Helical spring under fluctuating load

Loads

$$F_m = \frac{450+150}{2} = 300\ \text{N}, \qquad F_a = \frac{450-150}{2} = 150\ \text{N}$$

Spring index $C = 6$:

$$K_w = \frac{23}{20}+\frac{0.615}{6} = 1.2525, \qquad K_s = 1 + \frac{0.5}{6} = 1.0833$$

Stresses (with $D = Cd$)

$$\tau_a = K_w\frac{8F_aC}{\pi d^2}, \qquad \tau_m = K_s\frac{8F_mC}{\pi d^2}$$

Soderberg line

$$\frac{1}{n} = \frac{\tau_a}{S_{se}} + \frac{\tau_m}{S_{sy}}$$

Substitute and solve for $d^2$:

$$d^2 = \frac{8C\,n}{\pi}\left(\frac{K_wF_a}{S_{se}} + \frac{K_sF_m}{S_{sy}}\right)$$ $$d^2 = \frac{8\times6\times1.4}{\pi}\left(\frac{1.2525\times150}{300} + \frac{1.0833\times300}{680}\right) = 21.39\,(0.6263 + 0.4779) = 23.62$$ $$d = 4.86\ \text{mm} \;\Rightarrow\; \text{use } d = 5\ \text{mm}$$

Check with $d = 5$ mm

$D = 30$ mm.

$$\tau_a = 1.2525\times\frac{8\times150\times30}{\pi\times125} = 114.8\ \text{MPa}, \quad \tau_m = 1.0833\times\frac{8\times300\times30}{\pi\times125} = 198.6\ \text{MPa}$$ $$\frac{1}{n} = \frac{114.8}{300} + \frac{198.6}{680} = 0.3827 + 0.2921 = 0.6748 \;\Rightarrow\; n = 1.48 \ (>1.4)$$

Maximum stress $= 1.2525\times\dfrac{8\times450\times30}{\pi\times125} = 344$ MPa, well below the yield 680 MPa.

Number of active turns

Stiffness required: $k = \dfrac{450-150}{12} = 25$ N/mm.

$$n = \frac{G d}{8C^3 k} = \frac{79300\times5}{8\times216\times25} = 9.18 \;\Rightarrow\; 9\ \text{turns}$$

With $n = 9$: $k = 25.5$ N/mm and the deflection between loads is $300/25.5 = 11.8$ mm.

Answer: $d = 5$ mm ($D = 30$ mm), safety factor 1.48, 9 active turns.