Tooth force components on a bevel pinion

Force components in a bevel gear

The resultant tooth force acts at the mean pitch radius $r_m$ of the face width, normal to the tooth surface. With pitch cone angle $\gamma$ and pressure angle $\phi$:

$$W_t = \frac{T}{r_m}, \qquad W_r = W_t\tan\phi\cos\gamma, \qquad W_a = W_t\tan\phi\sin\gamma$$

The radial force $W_t\tan\phi$ acts normal to the pitch cone; its components are along the radial direction ($\cos\gamma$) and along the axis ($\sin\gamma$). On the pinion the axial force pushes it away from the apex.

Geometry

Pitch cone angle of the pinion:

$$\tan\gamma = \frac{z_1}{z_2} = \frac{18}{54} \Rightarrow \gamma = 18.43^\circ \quad (\sin\gamma = 0.3162,\ \cos\gamma = 0.9487)$$

Large-end pitch diameter: $d = mz_1 = 5 \times 18 = 90$ mm.

Mean pitch diameter:

$$d_m = d - b\sin\gamma = 90 - 40 \times 0.3162 = 77.35\ \text{mm}$$

Torque

$$\omega = \frac{2\pi \times 900}{60} = 94.248\ \text{rad/s}, \qquad T = \frac{P}{\omega} = \frac{7500}{94.248} = 79.58\ \text{N m}$$

Components

$$W_t = \frac{T}{d_m/2} = \frac{79.58}{(77.35/2)\times10^{-3}} = 2058\ \text{N}$$ $$W_r = W_t\tan20^\circ\cos\gamma = 2058 \times 0.3640 \times 0.9487 = 710\ \text{N}$$ $$W_a = W_t\tan20^\circ\sin\gamma = 2058 \times 0.3640 \times 0.3162 = 237\ \text{N}$$

The resultant normal force is $W_n = W_t/\cos20^\circ = 2190$ N. On the gear (shaft at 90 degrees) the pinion's radial force becomes its axial force and vice versa: the gear's axial force $= 710$ N and radial force $= 237$ N.

Answer: $W_t = 2058$ N, $W_r = 710$ N, $W_a = 237$ N on the pinion.