PSI - Issue 40

Ksenia Syzrantseva et al. / Procedia Structural Integrity 40 (2022) 418–425 Ksenia Syzrantseva at al. / Structural Integrity Procedia 00 (2022) 000 – 000

421

4

25600

( )   P C dC

,

1

0.9003243



0

that confirms the reliability of the obtained results.

Fig. 2. Histogram of the sample dynamic load capacity C and its density function P(C) in accordance with the Weibull distribution.

At the second stage of the solution of the bearing reliability evaluation problem the computer simulation is carried out for samples of the random value F – the radial load on the bearing in accordance with the required loading modes. The bearing safety factor on the dynamic load capacity is a sampling with the elements calculated by the formula similar to (1), but its components are not average values of samples for the dynamic load capacity { C } and load { F }, but specifically the elements of their sampling:

k 1,  .

i

(4)

, 1/ p

i F L n C   i

i

At the third stage we reconstruct the unknown density function of the random value of the bearing safety factor by the nonparametric statistics method implemented as the program in MathCad based on the Parzen-Rosenblatt evaluation described in detail by Syzrantseva et al. (2020). The probability of the bearing failure implies, by definition, the probability that the safety factor by the dynamic load capacity n will be less than unity, which corresponds to the value of the following integral:

Q n P n dn   1 0 ( ) ( ) .

(5)

Parameters of safety factor samples by the dynamic load capacity for the bearing 2207 are shown in the Table. The failure probabilities calculated by integrating by formula (5) are presented in the seventh column of the Table. The last column shows the values of the design load correction factor, which is taken into account to obtain the probability of failure equal to the probability of bearing failure under the average normal operating mode. To illustrate the capabilities of the developed method, we will use the following torque distribution function:

   j 2 1 2 j 

  

  

  j

6

 

    , i

    i  ( ) 1 5 j

(6)

2 sin 2 1

 





