Correct Answer - Option 2 : 44 kN
Concept:
Load-life relationship:
The approximate rating of the service life of a ball or roller bearing is based on the given fundamental equation.
\({\rm{L_{10}}} = {\left( {\frac{{\rm{C}}}{{\rm{P}}}} \right)^{\rm{k}}} \)
L10 = Rated bearing life (in million revolutions)
C = Dynamic load capacity.
P = Load acting in bearing.
k = 3 for ball bearing and k = 10/3 for the roller bearing.
The relation between life in million revolutions and life in hours is given by:
\(L_{10}=\frac{L_{10h}\;\times \;N \;\times \;60}{10^6}\)
where, L10h = rated bearing life in hours and N = speed of rotation in rpm.
Calculation:
Given:
N = 1500 rpm, P = 6 kN, L10h = 8100 hours, roller bearing k = 10/3
We know that;
\({\rm{L_{10}}} = {\left( {\frac{{\rm{C}}}{{\rm{P}}}} \right)^{\rm{k}}} \times {10^6}{\rm\;{revolution}}\) and;
\(L_{10}=\frac{L_{10h}\;\times \;N \;\times \;60}{10^6}\)
Combining both equations:
\(\frac{L_{10h}\;\times \;N \;\times \;60}{10^6}= {\left( {\frac{{\rm{C}}}{{\rm{6}}}} \right)^{{\frac{10}{3}}}} \)
\(\frac{8100\;\times \;1500 \;\times \;60}{10^6}= {\left( {\frac{{\rm{C}}}{{\rm{6}}}} \right)^{{\frac{10}{3}}}} \)
C = 43.34 kN