Tooth forces on a spur gear pinion

Force on a spur gear tooth

The tooth force acts along the line of action (common normal), at the pressure angle $\phi$ to the common tangent of the pitch circles. The normal force $W_n$ is resolved into:

• tangential component $W_t$ (transmits the torque),
• radial component $W_r = W_t\tan\phi$ (separates the gears, loads the bearings).

        W_r
         ^   W_n
         |  /
         | /  phi
   ------+------> W_t   (tangent at pitch point)

$$W_t = \frac{T}{d/2}, \qquad W_r = W_t\tan\phi, \qquad W_n = \frac{W_t}{\cos\phi}$$

Calculation

Angular speed: $\omega = \dfrac{2\pi \times 720}{60} = 75.398$ rad/s.

Torque:

$$T = \frac{P}{\omega} = \frac{10000}{75.398} = 132.63\ \text{N m}$$

Pitch circle diameter: $d = mz = 5 \times 20 = 100$ mm.

Pitch line velocity:

$$v = \omega\frac{d}{2} = 75.398 \times 0.050 = 3.770\ \text{m/s}$$

Tangential force:

$$W_t = \frac{T}{d/2} = \frac{132.63}{0.050} = 2653\ \text{N} \quad \left(= \frac{P}{v} = \frac{10000}{3.770}\right)$$

Radial force:

$$W_r = W_t\tan20^\circ = 2653 \times 0.3640 = 965\ \text{N}$$

Normal force:

$$W_n = \frac{W_t}{\cos20^\circ} = \frac{2653}{0.9397} = 2823\ \text{N}$$

The driven gear experiences equal and opposite forces, and each shaft bearing must support the vector sum of $W_t$ and $W_r$ (that is, $W_n$) in addition to the overhung weight.

Answer: $T = 132.63$ N m, $v = 3.770$ m/s, $W_t = 2653$ N, $W_r = 965$ N, $W_n = 2823$ N.