78 - 6 - Rolling Contact Bearing

Rolling Contact Bearing EXAMPLE-1 The radial load acting on a ball bearing is 2500 N or the irst i!e re!ol"tion s and red"ces to 1500 N to ne#t ten re!ol"tions$ The load !ariation then re%eat itsel the e#%ected lie o  the bearing is 20 &illion re!ol"tions deter&ines the d'na&ic load carr'ing ca%acit' o the bearing$

SOLUTION:

 L10

Given data:

=20 million revolution

Step -1: Equivalent load for omplete !or" #le 3

 Pe

3

 N 1 P1 + N 2 P2  N 1+ ¿ N  = 2

3

√ ¿ 2500

¿ ¿

1500 =

¿ ¿ ¿3 5¿ ¿ 3 √ ¿

=1$%&'( N Step -2: d#nami load arr#in) apait# *= Pe ( L10)

1 3

  =1$%&'(+20, =%&0&'& N

( 20 )

1 3

..... +/n,

EXAMPLE-2 A ball bearing s"b(ected to a radial load o )000 N is e#%ected to ha!e satisactor' lie o  1000h at *20 r%& +ith a reliabilit' o ,5 calc"late the d'na&ic load carr'ing ca%acit' o  the bearing$ so that it can be selected ro& a &an"act"res catalog"e based on ,0 reliabilit'$ . there are o"r s"ch bearing each +ith a reliabilit' o ,5 in a s'ste& +hat is the reliabilit' o the co&%lete s'ste& SOLLUTION =&000N

 L10

=1000

STE3 1 4earin) life for $%5 relia6ilit#  R95

60  NL95 h =

10

=&2

6

STE3 2 4earin) life of $05 relia6ilit#

+

 L95  L10

 L10

[ ] ( )

1 log  R 95

,=

=

log (

 L95 0.5405

1 )  R 90

=

1 1.17

=0'%0%

432 0.5404 =7$$'28

STE3 2 9#nami load ¿ 1/ 3

*=3 + L10

=27(0'$

STE3  #tem relia6ilit#

 Rs

=+ R

4

4

,=   0.95

=0'(1% O (1'%5'''''''''''''''''''' /NS''

EXAMPLE-) A one &in"te +or/ c'cle consists o the ollo+ing %arts

"rationsecond3  Pr  radialload3

 Pa

a#ial load3

6%eed

Part-1 40 s 00 N

Part-2 20 s )00 N

200 N

120 N

1440 r%&

*20 r%&

6tatic load and d'na&ic load carr'ing ca%acities are ,5 N and 14)0 N res%ecti!el'7 or the bearing$ The lie o the bearing in ho"rs$ Consider that the inner race o the ball bearing is rotating$

SOLUTION: 3/T 1 Stati load *o= 8$%N /;ial load 3a=200N adial load 3r=800N

 Pa C o  P a  Pr

200 = 695 =0.288

200 = 600 =0'&&

0.44− 0.37 e = 0'&7< 0'0&( 0.280

  =0'&(  P a

 Pr =0'&&>0'&(>e

Terefore ?=1@ A=0 Equivalent load 3 1=800 N

PART 2

3a=120N 3r=&00N

 Pa  Pr =120B&00=0'  Pa Co =120B8$%=0'172  Pa or  Co  Pa  Pr

@ te value of e=0'&1