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Chapter 3

Q. 3.11.10

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

Step-by-Step

Verified Solution

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

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

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

 

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

 

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

 

for bearing B with F_D = F_{eB} = 2654 \ N , gives

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

Answer The reliability of the bearing pair is

R = R_A R_B = 0.994 791 \left(0.999 766\right) = 0.994 558

 

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.