The design load on a ball bearing is 413 lbf and an application factor of 1.2 is appropriate.
The speed of the shaft is to be 300 rev/min , the life to be 30 kh with a reliability of 0.99. What is the $C_{10}$ catalog entry to be sought (or exceeded) when searching for a deep-groove bearing in a manufacturer’s catalog on the basis of $10^6$ revolutions for rating life? The Weibull parameters are $x_0 = 0.02 \ , \left(\theta − x_0 \right) = 4.439$ , and $b = 1.483$ .

Question

The design load on a ball bearing is 413 lbf and an application factor of 1.2 is appropriate.
The speed of the shaft is to be 300 rev/min , the life to be 30 kh with a reliability of 0.99. What is the [latex]C_{10}[/latex] catalog entry to be sought (or exceeded) when searching for a deep-groove bearing in a manufacturer’s catalog on the basis of [latex]10^6[/latex] revolutions for rating life? The Weibull parameters are [latex]x_0 = 0.02 \ , \left( heta − x_0 ight) = 4.439[/latex] , and [latex]b = 1.483[/latex].

Accepted Answer

[latex]x_D=\frac{L_D}{L_R}=\frac{60 \mathscr{L}_D n_D}{L_{10}}= \frac{60\left(30000 ight)300}{10^6}=540[/latex]

Thus, the design life is 540 times the [latex]L_{10}[/latex] life. For a ball bearing, [latex]a = 3[/latex]. Then, from Eq. (11–7),
[latex]C_{10}\doteq a_f\left[\frac{x_D}{x_0+\left( heta -x_0 ight)\left(1-R_D ight)^{{1}/{a}} } ight] \ \ \ R\geq 0.90[/latex]

[latex]C_{10}= \left(1.2 ight)\left(413 ight) \left[\frac{540}{0.02+4.439\left(1- 0.99 ight)^{{1}/{1.483}} } ight]^{{1}/{3}}=6696 [/latex] lbf.

Question 3.11.3: The design load on a ball bearing is 413 lbf and an applicat...

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