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:

The value of bending stress on the pinion 35.38 M pa

Answer:The value of bending stress on the pinion 35.38 M paExplanation:Given data

m = 2 mm

Pressure angle [tex]\phi[/tex] = 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 mmVelocity of the pinion gear

[tex]V =\pi D( \frac{N}{60} )[/tex]

[tex]V = 3.14 (0.034) \frac{(1650)}{60}[/tex]

[tex]V = 2.93 \frac{m}{s}[/tex]

Form factor for the pinion gear is

Y = 0.303

Now

[tex]K_{v} = \frac{6.1 +0.303}{6.1} = 1.049[/tex]

Force on gear tooth

[tex]F = \frac{P}{V}[/tex]

[tex]F = \frac{1200}{2.93}[/tex]

F = 408.73 N

Now the bending stress is given by the formula

[tex]\sigma = \frac{K_{v} F}{m b y}[/tex]

[tex]\sigma = \frac{(1.049)(408.73)}{(0.002)(0.02)(0.303)}[/tex]

[tex]\sigma[/tex]

= 35.38 M paThis is the value of bending stress on the pinion