Answer:
Radial load of 15 kN, remaining life is 2478519 rev
When load is 25.05, we have 485783 rev
Explanation:
For 02-30 mm angular contact bearing, then [tex]C_{10}=20.3 kN[/tex]
Since [tex] (F_1)^{a}L_1=K[/tex] then
[tex] (C_{10})^{a}L_{10}=K[/tex]
For a ball bearing, a=3 hence
[tex] (20.3)^{3}\times (10^{6})=K[/tex]
[tex]K=8.365\times 10^{9}[/tex]
Life at load 15 kN [tex]L_1=\frac {K}{F_1^{a}}=\frac {8.365\times 10^{9}}{15^{3}}=2478519 rev[/tex]
Life load at 25.05 kN [tex]L_2=\ frac {8.365\times 10^{9}}{25.05^{3}}= 532160.6 rev[/tex]
In such a scenario, the Palmgren-Miner cycle ratio summation rule is expressed as
[tex]\frac {I_1}{L_1}+\frac {I_2}{L_2}=1[/tex]
By substitution
[tex]\frac {216000}{2478519}+ \frac {I_2}{532160.6 }=1[/tex]
[tex]I_2=485783.4659 rev[/tex]
