A spur steel pinion (S0 = 200 MN/m2) is to drive a spur steel gear (S0 = 140 MN/m2). The diameter of the pinion is to be 100 mm and the center distance is 200 mm. The pinion is to transmit 5 kW at 900 rpm. The teeth are to be 20° full depth. Determine the necessary module and width of face to give greatest number of teeth. Design for strength only using Lewis equation.
S0 is the endurance strength corrected for average stress concentration?
The allowable stress S is given by $S=S_{0}\left ( \frac{3}{3+v} \right )$
where V is the pitch line velocity in m/sec.
The following table gives the form factor y for use in Lewis strength equation.
TABLE I—Form Factors y— for use in Lewis strength equation
|
Number of Teeth
|
Full-Depth Involute or Composite
|
20°
Full-Depth Involute
|
20o Sub Involute
|
|
12
|
0.067
|
0.078
|
0.099
|
|
13
|
0.071
|
0.083
|
0.103
|
|
14
|
0.075
|
0.088
|
0.108
|
|
15
|
0.078
|
0.092
|
0.111
|
|
16
|
0.081
|
0.094
|
0.115
|
|
17
|
0.084
|
0.096
|
0.117
|
|
18
|
0.086
|
0.098
|
0.120
|
|
19
|
0.088
|
0.100
|
0.123
|
|
20
|
0.090
|
0.102
|
0.125
|
|
21
|
0.092
|
0.104
|
0.127
|
|
23
|
0.094
|
0.106
|
0.130
|
|
25
|
0.097
|
0.108
|
0.133
|
|
27
|
0.099
|
0.111
|
0.136
|
|
30
|
0.101
|
0.114
|
0.139
|
|
34
|
0.104
|
0.118
|
0.142
|
|
38
|
0.106
|
0.122
|
0.145
|
|
43
|
0.108
|
0.126
|
0.147
|
|
50
|
0.110
|
0.130
|
0.151
|
|
60
|
0.113
|
0.134
|
0.154
|
|
75
|
0.115
|
0.138
|
0.158
|
|
100
|
0.117
|
0.142
|
0.161
|
|
150
|
0.119
|
0.146
|
0.165
|
|
300
|
0.122
|
0.150
|
0.170
|
|
Rack
|
0.124
|
0.154
|
0.175
|