The same 02-30 angular-contact ball bearing as in Prob. 11–38 is to be subjected to a two-step loading cycle of 4 min with a loading of 18 kN, and one of 6 min with a loading of 30 kN. This cycle is to be repeated until failure. Estimate the total life in revolutions, hours, and loading cycles.

Question

Accepted Answer

Total life in revolutions

Let:
l = total turns
[latex]f_{1}[/latex] = fraction of turns at [latex]F_{1}[/latex]
[latex]f_{2}[/latex] = fraction of turns at [latex]F_{2}[/latex]

From the solution of Prob. 11-38, [latex]L_{1}=1.434\left(10^{6} ight) ext { rev and } L_{2}=0.310\left(10^{6} ight) ext { rev }[/latex].

Palmgren-Miner rule:

[latex]\frac{l_{1}}{L_{1}}+\frac{l_{2}}{L_{2}}=\frac{f_{1} l}{L_{1}}+\frac{f_{2} l}{L_{2}}=1[/latex]

from which

[latex]\begin{aligned}l &=\frac{1}{f_{1} / L_{1}+f_{2} / L_{2}} \l &=\frac{1}{\left\{0.40 /\left[1.434\left(10^{6} ight) ight] ight\}+\left\{0.60 /\left[0.310\left(10^{6} ight) ight] ight\}} \&=451585 rev .\end{aligned}[/latex]

Total life in loading cycles

4 min at 2000 rev/min = 8000 rev/cycle

6 min at 2000 rev/min = 12 000 rev/cycle

Total rev/cycle = 8000 + 12 000 = 20 000

[latex]\frac{451585 rev }{20000 rev / cycle }=22.58 ext { cycles }[/latex]

Total life in hours

[latex]\left(10 \frac{ min }{ cycle } ight)\left(\frac{22.58 cycles }{60 min / h } ight)=3.76 h[/latex]

Question 11.39: The same 02-30 angular-contact ball bearing as in Prob. 11–3...

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