Page 330 - Analysis and Design of Machine Elements

Analysis and Design of Machine Elements
308
where life-adjustment factors for reliability of 90%, 95%, 96%, 97%, 98% and 99% are
1
selected as 1, 0.62, 0.53, 0.44, 0.33 and 0.21, respectively [6, 7].
Thenthe ratedlifeatadifferentreliability is
6
10 1 ( f C )
t
L = ≥ [L ] (11.18)
n h
60n f P
p
The basic dynamic load rating with a rated life of L corresponding to a different reli-
n
ability is
) 1
(
f P 60nL n
p
C = ≤ [C] (11.19)
6
f t 10 1
11.3.2 Life Prediction under Variable Loads
The analysis discussed so far has assumed that bearings operate at a constant speed
under a constant load throughout service life. More often than not, in many applications
bearings are subjected to a spectrum of different loads at different speeds during each
duty cycle. In such cases, the Palmgren linear cumulative damage rule, or Miner’s rule
for short introduced in Chapter 2, may be utilized for bearing life prediction.
According to Miner’s rule, if a bearing is subjected to a load spectrum, each load con-
tributes to the eventual failure of the bearing. Assume equivalent dynamic loads P ,
1
P ,…P are applied to a bearing. The rotational speed corresponding to each load is n ,
2 k 1
n ,…n . The percentage of operating time under each load is a ,a …a .
1
2
2
k
k
Assume the bearing reaches life limit after operation under the load of P ,P ,…P for
1
k
2
a total of H hours. The actual number of cycle under the load P is
i
′
L = a n H (11.20)
i i i
And the number of cycles when failure happens under the load of P is L . Therefore,
i i
according to Miner’s rule, when the bearing fails under the variable loads, we have
k
∑ L ′
i
= 1 (11.21)
L
i=1 i
The total number of cycles under all the load is
L = a n H + a n H +…+ a n H =(a n + a n +…+ a n )H = n H
k k
2 2
k k
1 1
m
2 2
1 1
m
(11.22)
Therefore, the average rotational speed n is
m
n = a n + a n +…+ a n (11.23)
m 1 1 2 2 k k
Assuming the load spectrum is substituted by an average equivalent dynamic load P ,
m
under the load of which the bearing fails after operating L number of cycles, that is,
m

P L = P L i (11.24)
m m
i
Therefore,
( )
P m
L = L (11.25)
i m
P
i