Roller chain drive: length, speed and factor of safety

Data: $p = 19.05$ mm, $z_1 = 19$, $z_2 = 48$, $n_1 = 1000$ rpm, $P = 15$ kW, $C \approx 40p$, breaking load $= 31.1$ kN.

Driven speed

$$n_2 = n_1\,\frac{z_1}{z_2} = 1000 \times \frac{19}{48} = 395.8\ \text{rpm}$$

Pitch diameters

$$d_1 = \frac{p}{\sin(180^\circ/z_1)} = \frac{19.05}{\sin 9.474^\circ} = 115.7\ \text{mm}, \qquad d_2 = \frac{19.05}{\sin 3.75^\circ} = 291.3\ \text{mm}$$

Chain length in pitches

$$L_p = \frac{2C}{p} + \frac{z_1 + z_2}{2} + \left(\frac{z_2 - z_1}{2\pi}\right)^2 \frac{p}{C}$$ $$L_p = 80 + 33.5 + \left(\frac{29}{2\pi}\right)^2\frac{1}{40} = 80 + 33.5 + 0.53 = 114.03$$

Take 114 pitches (even number, so that no offset link is needed).

Exact centre distance

$$\frac{C}{p} = \frac{1}{4}\left[\left(L_p - \frac{z_1 + z_2}{2}\right) + \sqrt{\left(L_p - \frac{z_1 + z_2}{2}\right)^2 - 8\left(\frac{z_2 - z_1}{2\pi}\right)^2}\right]$$ $$\frac{C}{p} = \frac{1}{4}\left[80.5 + \sqrt{80.5^2 - 8(21.30)}\right] = \frac{1}{4}[80.5 + 79.43] = 39.98$$ $$C = 39.98 \times 19.05 = 761.7\ \text{mm}$$

Chain velocity and pull

$$v = \frac{z_1 p n_1}{60\times 1000} = \frac{19 \times 19.05 \times 1000}{60\,000} = 6.03\ \text{m/s}$$ $$F = \frac{P}{v} = \frac{15\,000}{6.03} = 2487\ \text{N}$$

Factor of safety

$$FS = \frac{31\,100}{2487} = 12.5$$

This is above the usual minimum (about 8 to 11 for ordinary speeds), so the chain is adequate. Chain drives give a constant average velocity ratio with no slip, but the velocity varies slightly with the chordal action, and they need lubrication.

Answer: n2 = 395.8 rpm; d1 = 115.7 mm, d2 = 291.3 mm; 114 pitches; C = 762 mm; v = 6.03 m/s; F = 2.49 kN; FS = 12.5.