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.
Question
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:
The value of bending stress on the pinion 35.38 M pa
Explanation:
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 mm
Velocity 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 pa
This is the value of bending stress on the pinion