In Ex. 11–8 bearings A and B (cone 15100 and cup 15245) have C10 = 12 100 N.What is the reliability of the pair of bearings A and B?

Question

In Ex. 11–8 bearings [latex]A \ and \ B[/latex] (cone 15100 and cup 15245) have [latex]C_{10} = 12 100[/latex] N.
What is the reliability of the pair of bearings [latex]A \ and \ B[/latex]?

Accepted Answer

The desired life [latex]x_D[/latex] was [latex]{5000\left(800 ight)60}/{\left[90\left(10^6 ight) ight]}= 2.67[/latex] rating lives. Using Eq. (11–21)

for bearing [latex]A[/latex], where from Ex. 11–8, [latex]F_D = F_{eA} = 4938 \ N \ , \ and \ a_f = 1[/latex] , gives

[latex]R_A\doteq 1-\left\{\frac{2.67}{4.48\left[{12100}/{\left(1 imes 4938 ight) } ight]^{{10}/{3}} } ight\} ^{{3}/{2}}=0.994791[/latex]

which is less than 0.995, as expected. Using Eq. (11–21)

[latex]R\doteq 1-\left\{\frac{x_D}{ heta \left[{C_{10}}/{\left(a_f F_D ight) } ight]^a } ight\} ^b=1-\left\{\frac{x_D}{4.48\left[{C_{10}}/{\left(a_f F_D ight) } ight]^{{10}/{3}} } ight\}^{{3}/{2}}[/latex]

for bearing [latex]B [/latex] with [latex]F_D = F_{eB} = 2654 \ N[/latex] , gives

[latex]R_B\doteq 1-\left\{\frac{2.67}{4.48\left[{12100}/{\left(1 imes 2654 ight) } ight]^{{10}/{3}} } ight\} ^{{3}/{2}}=0.999766[/latex]

Answer The reliability of the bearing pair is

[latex]R = R_A R_B = 0.994 791 \left(0.999 766 ight) = 0.994 558[/latex]

which is greater than the overall reliability goal of 0.99. When two bearings are made identical for simplicity, or reducing the number of spares, or other stipulation, and the loading is not the same, both can be made smaller and still meet a reliability goal. If the loading is disparate, then the more heavily loaded bearing can be chosen for a reliability goal just slightly larger than the overall goal.

Question 3.11.10: In Ex. 11–8 bearings A and B (cone 15100 and cup 15245) have...

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