Page 342 - Analysis and Design of Machine Elements

Analysis and Design of Machine Elements
320
Steps Computation Results Units
For bearing 1 X = 1.0, Y = 0
1 1
For bearing 2 X = 0.4, Y = 1.4
2 2
P = X R + Y A = 1.0 × 3118
1 1
1
1
1
+0 × 2062 = 3118 N
P = X R + Y A = 0.4 × 1853
2
2
2
2 2
+1.4 × 2114 = 3701 N
5. Estimate the life of For moderate shock, select f = 1.5. L = 90 291 h
p h
bearing
Since P > P ,check thelifeof
2 1
bearing by P
2
( )
10 6 f C
t
L = =
h 60n f P
p 1
10
10 6 ( 1 × 73200 ) 3
× =
60 × 1000 1.5 × 3701
90291h > 25000 h
The bearings are satisfactory.

Example Problem 11.3
A single-row, deep-groove ball bearing is required to carry a stable radial load of 10 kN
at rotational speed of 1000 rpm for 1500 hours with a reliability of 99%, determine
1. the rated bearing life at the reliability of 90%;
2. the required basic dynamic load rating of the bearing.
Solution

Steps Computation Results Units
1. The rated life at the Find the reliability life-adjustment factor as L = 7142 h
10
reliability of 90% = 0.21. Calculatethe corresponding ratedlifeby
1
Eq. (11.17)
L 1500
L = n = = 7142 h
10 0.21
1
2. The basic dynamic From Eq. (11.19), compute the basic dynamic load C = 75.4 kN
load rating of the capacity
bearing Assume f = 1.0, f = 1.0
t p
1
( )
f P 60nL
p
C = n =
f t 10 1
6
1
( )
60 × 1000 × 1500 3
10 × = 75.4 kN
6
10 × 0.21