Values of X, Y and e for some types of bearings
Type of bearing | a° | Fa/C o | Fa/ (VFr)£ e | Fa /(VFr)> e | e | ||
X | Y | X | Y | ||||
Single Row radial Ball Bearing | 0.014 | 0.56 | 2.30 | 0.19 | |||
0.028 | 1.99 | 0.22 | |||||
0.056 | 1.71 | 0.26 | |||||
0.084 | 1.55 | 0.28 | |||||
0.11 | 1.45 | 1.30 | |||||
0.17 | 1.31 | 0.34 | |||||
0.28 | 1.15 | 0.38 | |||||
0.42 | 1.04 | 0.42 | |||||
0.56 | 1.00 | 0.44 | |||||
Single Row Angular Contact Ball Bearing | 0.014 | 0.45 | 1.81 | 0.30 | |||
0.029 | 1.62 | 0.34 | |||||
0.057 | 1.46 | 0.37 | |||||
0.086 | 1.34 | 0.41 | |||||
0.11 | 1.22 | 0.45 | |||||
0.17 | 1.13 | 0.48 | |||||
0.29 | 1.14 | 0.52 | |||||
0.43 | 1.01 | 0.54 | |||||
0.57 | 1.00 | 0.54 | |||||
– | 0.41 | 0.87 | 0.68 | ||||
– | 0.37 | 0.66 | 0.95 | ||||
Single row Tapered Roller Bearing | – | 0.4 | 0.4ctga | 1.5ctga |
11.2.3. Determine total axial forces Fa 1 and Fa 2. For that we should find the sum Fa + S 1 and compare with force S 2. There are 3 possible cases:
– if Fa + S 1 > S 2 Fa 1 = S 1 and Fa 2 = Fa + S 1;
– if Fa + S 1 = S 2 Fa 1 = S 1 and Fa 2 = S 2;
– if Fa + S 1< S 2 Fa 1 = S 2 – Fa and Fa 2 = S 2.
In our case Fa + S 1 = 346.7 +2247.5 = 2594.2 N < S 2 = 2961.1 N, therefore Fa 1 = 2961.1 – 346.7 = 2614.4 N and Fa 2 = 2961.1 N.
11.3. Determine factor V that takes into account what ring of the bearing is movable.
– for bearings with movable inner ring V = 1;
– for bearings with movable outer ring V = 1.2.
In general purpose speed reducers bearings with movable inner ring are used only.
11.4. Determine safety factor Ks that takes into account the load nature. It may be ranged from 1 to 2.5 depending upon the type of designing machine. For general purpose speed reducers the safety factor is assumed as 1.3.
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11.5. Determine temperature factor Kt according to table 11.2.