Question

A steel spur pinion has a module of 2 mm, 17 teeth cut on the 20° full-depth system, and a face width of 20 mm. At a speed of 1650 rev/min, this pinion is expected to carry a steady load of 1.2 kW. Determine the bending stress.

Answer 1

Answer:

The value of bending stress on the pinion 35.38 M pa

Explanation:

Given data

m = 2 mm

Pressure angle $\phi$ = 20°

No. of teeth T = 17

Face width (b) = 20 mm

Speed N = 1650 rpm

Power = 1200 W

Diameter of the pinion gear

D = m T

D = 2 × 17

D = 34 mm

Velocity of the pinion gear

$V =\pi D( \frac{N}{60} )$

$V = 3.14 (0.034) \frac{(1650)}{60}$

$V = 2.93 \frac{m}{s}$

Form factor for the pinion gear is

Y = 0.303

Now

$K_{v} = \frac{6.1 +0.303}{6.1} = 1.049$

Force on gear tooth

$F = \frac{P}{V}$

$F = \frac{1200}{2.93}$

F = 408.73 N

Now the bending stress is given by the formula

$\sigma = \frac{K_{v} F}{m b y}$

$\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}$

$\sigma$ = 35.38 M pa

This is the value of bending stress on the pinion