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A self-aligning ball bearing has a basic dynamic load rating (C10, for 106 revolutions) of 35 kN. If the equivalent radial load on the bearing is 45 kN, the expected life (in 106 revolutions) is


1. below 0.5
2. 0.5 to 0.8
3. 0.8 to 1.0
4. above 1.0

1 Answer

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Best answer
Correct Answer - Option 1 : below 0.5

Concept:

The approximate rating of service life of a ball or roller bearing is based on the fundamental equation.

\({{L}} = {\left( {\frac{{{C}}}{{{W}}}} \right)^{{k}}} \times {10^6}\ {{revolution}}\)

where L is rating lifeC is basic dynamic loadW is equivalent dynamic load

k = 3 for ball bearing

k = 10/3 for roller bearing

The relationship between the life in revolutions (L) and the life in working hours (LH)  is given by:

L = 60 N.LH  revolutions

where N is the speed in rpm

Calculation:

Given: C = 35 kN, W = 45 kN

\({\rm{L}} = {\left( {\frac{{\rm{C}}}{{\rm{W}}}} \right)^{\rm{k}}} \times {10^6}{\rm{\;rev}} = {\left( {\frac{{35}}{{45}}} \right)^3} \times {10^6} = 0.47 \times {10^6}\;rev\)

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