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Chapter 11: Rolling Element Bearings
ME/MF F241:
Machine Design and Drawing
Dr. Regalla Srinivasa Prakash
Scope
• Bearing types
• Bearing life
• Bearing load(F)-life(L) trade off at constant reliability
• Bearing survival: the reliability(R)-life(L) trade-off
• Load(F)-life(L) –reliability(R) trade-off
• Combined radial (Fr) and thrust (Fa) loading
• Variable loading
• Selection of ball and roller bearings
• Selection of tapered roller bearings
• Adequacy assessment for selected rolling contact bearings
• Lubrication in rolling element bearings
• Mounting and enclosure in rolling element bearings
ROLLING ELEMENTS BEARINGS – THE TWO TYPES:
• Rolling element bearings are two types:
– Ball bearings (balls are the rolling elements)
– Roller bearings (cylinders are the rolling elements)
Ball Bearing Nomenclature:
Different types of ball bearings:
Different types of roller bearings
a) Straight roller
b) Tapered roller, thrust
c) Spherical roller, thrust
d) Needle
e) Tapered roller (both radial
and thrust)
f) Steep-angle tapered roller
Bearing load (F) - Life (L) trade-off at constant (rated, 90%) reliability (R):
1/ aFL =constant
a=3 for ball bearings
a=10/3 for roller bearings
Experimentally
obtained data
plotted, for 90%
reliability
Rating Life
Rating life
Rating life for different manufacturers:
• SKF: 106 revolutions
• Timken: 90(10)6 revolutions
•The rating life is a term sanctioned by the Anti-friction
Bearing Manufacturers Association (AFBMA)
• “the rating life of a group of nominally identical ball or roller
bearings is defined as the number of revolutions (or hours at
a constant speed) that 90% of the group of bearings will
achieve or exceed before the failure criterion develops.”
•For Timken company, the criterion is a wear area of 6.5
mm2.
Contd.
a
DDD
a
RR nLFnLC/1/1
10 6060
aa LFLF1
22
1
11
Associating the load F1 with C10, the catalogue rating that you need to look at, and
the life measure in revolutions L1 with the L10, which is the manufacturer specific
quantity, we can write,
a
DD
aLFLC
/1/1
1010
Here FD and LD refer to the design quantities for the bearing to be selected.
If we want to specify in the life hours, then we can write, rpm (nR & nD) values:
What the different terms in the above equation mean?
Rating
Life=L10
Desired
Life=LD
Desired Load=FD
Rating Load=C10
Contd.
1/
10
60( )
60
aD DD
R R
L nC F
L n
Inverting the equation,
Catalogue load rating=
a
DDD
a
RR nLFnLC/1/1
10 6060
Catalog rating, kN
Rating life in hours
Rating speed in RPM Desired radial load, kN
Desired life in hours
Desired speed in RPM
Example:
The SKF rates its rolling contact bearings as 106 revolutions whereas Timken rates as 90*(106) revolutions. Select a ball bearing for a motorcycle for a life of 5000 hours to work at a speed of 1800 RPM under a radial load of 3000 N with a reliability of 90% from the SKF catalogue.
Solution:
kNN
nL
nLFC
a
RR
DDD 43.2476.24429
10
60180050003000
60
603/1
6
/1
10
From the table 11-2, for the above load rating, the nearest ball bearing is 35 mm
bore, 72 mm OD, 17 mm width, 1 mm fillet radius, 41 mm shaft diameter and 65
mm housing shoulder diameter (it has C10 of 25.5 kN).
0
0
1 1 exp[ ( ) ]bx xF R
x
0
0
exp[ ( ) ]bx xR
x
R=reliability
x=life measure dimensionless variate,
L/L10
x0=guaranteed, or minimum value of
the variate
Using the Weibull distribution, along
any constant load line (horizontal
line in the graph on the right):
=characteristic parameter corresponding to the 63.2121 percentile value of the
variate; b= shape parameter that controls the skewness
Bearing load (F) - Life (L) - reliability (R) three-way relationship
(What if more or less than 90% reliability is desired?):
Failure (not force)
probability =
Contd.
1/
10 1/
0 0
( ) , 0.90( )(1 )
aDD b
D
xC F R
x x R
aDD
aBB xFxF
11
a
B
aD
DB
x
xFF
1
1
Along a constant load line (AB),
ngsubstituti
Rxxx
xSolving
x
xxR
b
D
B
B
b
BD
/1
00
0
0
1ln
exp
a
b
D
DD
aB
aD
DBRxx
xF
x
xFCF
/1
/1
00
1
1
10/1ln
The natural logarithmic function can be series-expanded and simplified to yield
Revisit to the previous example:
• The SKF rates its rolling contact bearings as 106 revolutions whereas Timken rates as 90*(106) revolutions. Select a ball bearing for a motorcycle for a life of 5000 hours to work at a speed of 1800 RPM under a radial load of 3000 N, now with a reliability of 95% from the SKF catalogue. The pure radial load is not steady and hence use an application factor (AF) of 1.5. Use Weibull distribution and Weibull parameters, guaranteed or minimum value of the dimensionless variate x as x0=0.02, characteristic parameter minus the minimum guaranteed value as (-x0)=4.439 and the shape parameter as b=1.483.
Solution: desired value of the dimensionless variate
xD=L/L10=(60*LD*nD)/(60*LR*nR) = (60*5000*1800)/(106)=540
This means that the design life is to be 540 times the L10 life. Hence the
necessary C10 is
kNNC 24.4343236
95.01439.402.0
540)3000)(5.1(
31
483.1110
From the table 11-2, for the above load rating, the nearest ball bearing is 55 mm
bore, 100 mm OD, 21 mm width, 1.5 mm fillet radius, 63 mm shaft diameter and
605 mm housing shoulder diameter. The C10 itself is 43.6 kN.
Exercise Problem:
An angular-contact, inner ring rotating, 02-series ball bearing is required for an
application in which the life requirement is 50000 h at 480 rev/min. The design
radial load is 2745. The application factor is 1.4. The reliability goal is 0.90. Find the
multiple of rating life xD required and the catalog rating C10 with which to enter
Table 11–2. Choose a bearing and write down all of its specifications. Also
estimate, the actually chosen ball bearing, the actual existing reliability in service.
Two different applications having and not having a thrust load:
No thrust load
Thrust load present
Accounting for thrust force:
1e
r
F
VF when
e
r
Fe
VF
e a
r r
F FX Y
VF VF when e
r
Fe
VF
e i r i aF X VF Y F
Purpose is to find the equivalent radial
load Fe, that would do the same damage
as that done by the existing radial and
thrust loads together. V is the rotation
factor. V=1 for inner ring rotation, V=1.2
for outer ring rotation.
e
Generalizing for both zones,
For horizontal line zone, i=1 and for
inclined line zone, i=2.
Table 11-1 gives the values of Xi and
Yi.
THE ITERATIVE SOLUTION METHOD WITH BOTH Fr AND Fa
• Calculate xD.
• Ignore Fa, and for FD=Fr, find the C10 as well as C0 from catalogue for given reliability. C0 is the bearing’s static load catalog rating. Assign (C10)old = C10.
• Find Fa/C0.
• Find “e” from Table 11-1 using interpolation for this Fa/C0.
• For this Fa/C0, is Fa/(VFr) greater than “e”?, if Yes note down the X2 and Y2 values. Interpolation may be needed. If No, ignore Fa, solution ends.
• Estimate the equivalent load Fe. Apply V only to Fr. Calculate the desired load FD=af(Fe). The af 1 is the load application factor accounting for unsteady nature of loading.
• Calculate the new (C10)new value. Compare with the earlier (C10)old value. If (C10)new < (C10)old, then (C10)old is the final desired rating of the bearing to be selected. The solution ends. If (C10)new > (C10)old we need to go for another iteration.
• Assign (C10)old = (C10)new . Find the new C0 for the (C10)new .
• Continue the iterations until the (C10)new is less than preceding (C10)old.
RELIABILITY GOAL OF THE MECHANICAL SYSTEM
• The combined reliability goal is normally specified, say, Rt. It is the for the shaft mounted on those bearings.
• Then each of the two bearings, if both of them are same type, must possess a reliability of:
t
t
t
tBA
BAt
RRThus
R
Rge
RRor
RRRRRRRIf
RRR
,
9746.095.0
,95.0.,.
, 2
Example problem:
The worm shaft shown in part a of
the figure transmits 1.35 hp at 600
rpm. A static force analysis gave the
results shown in part b of the figure.
Bearing A is to be an angular contact
ball bearing mounted to take the
555-lbf thrust load. The bearing at B
is to take only the radial load, so a
straight roller bearing will be
employed. Use an application factor
of 1.3, a desired life of 25 kh, and a
reliability goal, combined of 0.99.
Specify each bearing.
The coding method for standard bearings:
02 bearings means…
Shoulder dimensions:
What ds and dH in the
catalogues mean…