Self-Lube Hand book - NSK Europe - #10

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10 Cor Pr Pr nx60 Examples of bearing calculations Example 1 What nominal life can be obtained from NP55 with a steady radial load Fr = 3900N at speed of 1500 rev/min? The dynamic load rating Cr of the unit from page 25 is 43500N. Since the bearing is not subject to axial load the equivalent load Pr = Fr according to the formula on page 7. Therefore applying the service factor of 1.2 for a steady load. Pr = Fr • 1.2 = 3900 • 1.2 = 4680N. From page 7, L10 life in hours = ( Cr )3 • 106 = ( 43500 )3 • 106 = 8923 hours Alternatively, using the loading ratio tables on page 9 an approximate life can be obtained by locating the nearest Cr value in the appropriate rev/min column. Therefore Cr = 43500 = 9.29 Under the 1500 rev/min column the nearest Cr value is 9.65 which gives an approximate life of 10000 hours. Example 2 With a radial load Fr = 2940N and an axial load Fa = 1470N at 300 rev/min with moderate shock present, what nominal L10 life can be obtained from unit reference SF40? The dynamic radial load rating Cr of the unit from page 39 is 32500N and the static load rating Cor is 19900N. Since the bearing is subject to radial and axial loads we have to establish the equivalent load Pr according to page 7. First, we establish the value of f0 Fa f0 Fa = 14.0 • 1470 = 1.03 Using this value, from table 18.2 we establish a value for Y = 1.55. From page 7 we then calculate the value of Pr Pr = 2940N or Pr = 0.56 (2940) + 1.55 (1470) = 3925N Using the greater value of Pr and applying an application factor of 1.7 (page 7) for moderate shock loads: Pr = 3925 • 1.7 = 6673N From page 7: L10 life hours = ( Cr )3 • 106 = ( 32500 )3 • 106 = 6418 hours Alternatively, using the loading ratio tables on page 9, an approximate life can be obtained by locating the nearest Cr /Pr value in the appropriate rev/min column. Therefore, Cr /Pr = 32500/6673 = 4.87. Under the 300 rev/min column on page 9, calculated value of 4.87 is approximately mid-way between table values of 4.48 and 5.13. By interpolation, this gives an approximate life of 6250 hours. Housing strength To check the housing strength for example 2 when the axial load Fa = 1470N and applying an application factor of 1.7 then: Axial load = 1470 • 1.7 = 2499N From page 19 we see that the maximum axial loads for the above unit are: 0.45 Cor in one direction, and 0.25 Cor in the opposite direction. Calculating these two maximum axial loads that may be applied to housing: 0.45 • 19900 = 8955N 0.25 • 19900 = 4975N From the above it can be seen that the housing will support the axial load of 2499N in either direction. Therefore, the unit above is satisfactory for the loading conditions stated. Note It is advisable to shoulder the shaft for high axial loads. 4680 1500x60 4680 Pr Cor 19900 Pr 6673 60n 60x300 Pr