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

The value of bending stress on the pinion 35.38 M pa

Step-by-step 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( (N)/(60) )`

`V = 3.14 (0.034) ((1650))/(60)`

`V = 2.93 (m)/(s)`

Form factor for the pinion gear is

Y = 0.303

Now

`K_(v) = (6.1 +0.303)/(6.1) = 1.049`

Force on gear tooth

`F = (P)/(V)`

`F = (1200)/(2.93)`

F = 408.73 N

Now the bending stress is given by the formula

`\sigma = (K_(v) F)/(m b y)`

`\sigma = ((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