Question

A radial load of 6.5 kN is applied on a cylindrical roller bearing. The desired life of the bearing with 90% reliability is 18000 hours with the application factor 1.7. if the shaft rotates at 1600 rpm, then the required basic dynamic load rating for the bearing is given by:

1. 103.4 MN
2. 103.4 kN
3. 1.042 kN
4. 10.34 kN

Answer 1

Correct Answer - Option 2 : 103.4 kN

Concept:

Load life equation:

F^aL = Constant

where a = 3 For ball bearing and a = 10/3 For roller bearing.

For 90% reliability, L = 10⁶ cycles.

Load life in cycles is represented as 

L = 60 × L_H × N, where L = Life in cycles, L_H = Life in hour, N = RPM.

Calculation:

Given:

Application factor (k) = 1.7, N = 1600 rpm, L_H = 18000 hours, F_r = 6.5 kN

Now, we know that

F^aL = Constant

i.e. \({\left( {\frac{F}{k}} \right)^{\frac{{10}}{3}}} \times {L_{90}} = {\left( {{F_r}} \right)^{\frac{{10}}{3}}} \times Lr\;\)

\({\left( {\frac{{\bf{F}}}{{1.7}}} \right)^{\frac{{10}}{3}}} \times {10^6} = {\left( {6.5} \right)^{\frac{{10}}{3}}} \times 60 \times 18000 \times 1600\)

∴ F = 103.4 kN