CHAPTER 11

Rolling Contact Bearing

Introduction
Objective of bearing:

To provide relative positioning and rotational freedom while transmitting a load

between two parts. Example: a shaft and its housing.

Objective of lubrication:
To reduce:
1) The friction
2) The wear
3) The heat between two surfaces moving relative to each other.
It is done by: inserting a substance, called lubricant, between the moving surfaces.

There are two parts of bearings:

1) Rolling contact bearing Chapter 11, the study will cover:
types
dimension
load
life
It is used to transfer the main load through elements in rolling contact.
2) Journal bearing which is covered in Chapter 12

Rolling Contact bearing

Rolling contact bearings are also known as antifriction bearings.

The load, speed, and operating viscosity of the lubricant affect the friction
characteristics of a rolling bearing.
These bearings provide coefficients of friction between 0.001 and 0.002.
The designer must deal with such matters as fatigue, friction, heat, lubrication,
kinematics problems, material properties, machining tolerances, assembly, use
and cost.

129

Advantage of rolling-contact bearings:

1.
2.
3.
4.
5.
6.
7.

Low starting and good operating friction torque.

East of lubrication
Requiring less axial space.
Generally, taking both radial and axial loads.
Rapid replacement
Warning of impending failure by increasing noisiness.
Good low-temperature starting.

Disadvantages of rolling element bearings:

1.
2.
3.
4.
5.
6.
7.

Greater diametral space.

More severe alignment requirements.
Higher initial cost.
Noisier normal operation.
Finite life due to eventual failure by fatigue.
Ease of damage by foreign matter.
Poor damping ability.

11.1 Bearing Types

Most rolling bearings are categorized in one of the three groups:

1. Take pure radial loads

2. Pure thrust load (axial contact for supporting loads that combined axial and radial
loads are primarily)
3. Combination of the two kinds of loads (angular contact for carrying)

There are two types of rolling bearing:

1. Ball bearings
2. Roller bearings.

The essential parts of the bearing are illustrated

in figure 11.1 which includes:
1.
2.
3.
4.

outer ring
inner rings
bulls or rolling elements
Separator.

130

Note:
Some types of bearings are without separator; these types are low in price. And the
function of separator is to present the rubbing contact.

Bearing manufacturers provide engineering manuals and brochures containing lavish

descriptions of the various types available.

Figure 11-2 shows some of the various types of standardized by bearings.

For more information about these types read page 561-562

Figure 11-3 shows some types of roller

bearings. It shows
a)
b)
c)
d)
e)
f)

straight roller
spherical roller
tapered roller
needle
tapered
steep-angle tapered roller

131

11.2 Bearing Life

Contact stresses occur on the inner ring, the rolling element, and on the outer ring,
when the ball or roller of rolling-contact bearings rolls.
If:
1.
2.
3.
4.

Bearing is clean and properly lubricated

Mounted and sealed against the entrance of dust and dirt,
Maintained in this condition
Operated reasonable temperatures.

Then:
The only cause of failure is metal fatigue.

Common life measures for bearing are based on:

1. Number of revolutions of the inner ring (outer ring stationary) until the first
tangible evidence of fatigue.
2. Number of hours of use at a standard angular speed until the first tangible
evidence of fatigue.

The life of an individual bearing or any one group of identical bearings cannot be
accurately predicted
AFBMA established the following definitions associated with the life of bearing.
o Bearing life: the number of revolutions or hours at some uniform speed at which
the bearing operates until fatigue failure.
o Rating life L10: The number of revolutions ( or hours at a uniform speed) that
90% of a group of identical roller bearings will complete or exceed before the first
evidence of fatigue develops.
o Median life: refers to the life that 50% of the group of bearings would complete
or exceed. It is about 5 times of L10 life.
The definition of the rating life L10 is based on a 90% reliability(or 10% failure)
A typical curve of bearing life expectancy is shown in figure below.
Percentage of Bearings in Operation

100

50

L10 life

Median life

Life

132

11.3 Bearing Load Lives at Rated Reliability

Experiments show that two groups of identical bearings tested under different loads
F1 and F2 will have respectively lives L1 and L2 according to the relation:
L1 F2
=
L2 F1

(11-2)

Where
L = the life of millions of revolutions or the life of hours at a given constant speed
n, in rev/min.
a=

3
10/3

for ball bearings

for roller bearings (cylindrical and tapered roller)

If we represent the curve in log-log curve as shown in figure

The AFBMA has established a standard load rating for bearings in which speed is not
a consideration.
This rating is called the basic load rating C
The basic load rating is defined as: the constant radial load which a group of
apparently identical bearings can endure for a rating life of one million revolutions of
the inner ring (stationary load and stationary outer ring)
The life of a bearing subjected to any other load F to be:

C
L=
F

(11-1)

133

A bearing manufacturer may choose a rated cycle value of 106 revolutions

Timken Company choose a rate of 90(106) revolutions which is based on 3000 hours
at a speed of 500 rev/min.
L10 = (3000h )(60 min/ h )(500rev / min ) = 90(10 6 )rev.

Equation 11-2 can be written in more useful way as:

C = FL1 / a

(11-1)

To illustrate the use of the equation above, suppose that a certain Timken bearing has
a rated load of 2140 lb, find the basic load rating C?
C = FL1 / a
L = 90(10 6 )rev F = 2140lb a = 10 / 3
C = 2140(90 )

3 / 10

= 8254lb

We shall call this the catalog load rating and display it algebraically as C10, to denote
it as the 10% rating life for a particular bearing in the catalog.
From equation (11-1)
F1 L11/ a = F2 L12/ a
Put L1 as L10 and F1 as C10, we can write the above equation as:
C10 L110/ a = FL1 / a
Finally, we can write the equation in the general form as:
L n 60
C10 = FD D D
LR n R 60

1/ a

(11-3)

where
C10: catalog rating (lbf or kN)
LR: rating life (hours)
nR: rating speed (rev/min)
FD: desired radial load (lbf or kN)
LD: desired life (hours)
nD: desired speed (rev/min)

134

11.4 Bearing Survival: Reliability Vs Life (This section is not available in textbook, 7ed)

If the machine is assembled with a total of N bearings, each having the same
reliability R, then the reliability of the group must be:
RN=RN
Suppose we have a gear-reduction unit consisting of six bearings, all loaded so that
the L10 lives are equal.
For example, if the reliability of each bearing is 90%, the reliability of all the bearings
in the assembly is:
R6 = (0.90)6=0.531
This points up the need to select bearings having reliabilities greater than 90%
The reliability based on Weibull distribution:
For Ball bearing:
L / L10 0.02 1.483
R = exp

4.439

(1*)

If L/L10=0.5, the reliability is R=0.9637

For Tapered Bearing:

L / L10 1.5
R = exp

4.48

(2*)

11.5 The Reliability Goal

In this section, we wish to learn how to select bearings for any desired reliability.
Equations (1* and 2*) show to parameters: L which is the desired life and R which is
the desired reliability.
The problem is to find the value of the L10 life that will satisfy these two requirements
(L and R)
So we solve equations (1*, 2*) for L10, we have
For Ball Bearing
L10 =

L
1 / 1.483
0.02 + 4.439[ln (1 / R )]

(11-8a)

135

For Tapered Bearing

L10 =

L
1 / 1.5
4.48[ln (1 / R )]

(11-8b)

To determine the catalog load rating, we substitute equation (11-8) into equation (113) we have:
For Ball Bearing

(L D n D / LR n R )
C10 = FD
1 / 1.483

0.02 + 4.439[ln (1 / R )]

1/ 3

(11-9)

For Tapered Bearing

(L D n D / LR n R )
C10 = FD
1 / 1.5
4.48[ln (1 / R )]

3 / 10

(11-10)

Note:
The different between equation (11-3) and equations (11-9&11-10) is that:
Equation 11-3: if you need to find the load rating without the reliability factor (R)
Equations 11-9 & 11-10: if you need to find the load rating including the
reliability factor (R)

Example:
A certain application requires a bearing to last for 1800 h with a reliability of 99%. What
should be the rated life of the bearing selected for this application?
Solution:
Given:
Required:

L=1800 h, R=0.99
L10

For Ball Bearing

L10 =

L
= 8197h
1 / 1.483
0.02 + 4.439[ln (1 / R )]

For Tapered Bearing

L10 =

L
= 8627 h
1 / 1.5
4.48[ln (1 / R )]

136

Example
A roller bearing is to be selected to withstand a radial load of 4 kN and have an L10 life of
1200 h at a speed of 600 rev/min. What load rating would you look for in searching the
Timken Engineering Journal?
Solution
Given:
FD=4kN, LD=1200 h, nD=600 rev/min,
LR=3000 h, nR=500 rev/min,
Roller bearing a = 10/3
Required: load rating C10
L n 60
C10 = FD D D
LR n R 60

1/ a

(1200)(600 )(60)

= 4
(3000)(500)(60)

3 / 10

= 3.21kN

Example
What load rating would be used if the application in the pervious example is to have a
reliability of 99 percent?
Solution:
From equation 11-10:
(L D n D / LR n R )
C10 = FD
1 / 1.5
4.48[ln (1 / R )]

3 / 10

((1200)(600) /(3000)(500) )
= 4

1 / 1.5
4.48[ln(1 / 0.99)]

3 / 10

= 5.136kN

This value let us enter the catalog with C10=5.136kN

11.6 Selection of Ball and Straight Roller Bearings

Ball bearings are usually operated with some combination of radial and thrust load.

Since, catalog ratings are based only on radial load, it is convenient to define an
equivalent radial load Fe that will have the same effect on bearing life as do the
applied loads

137

An equivalent AFBMA radial load equation for ball bearings is the maximum of the
two values:
Fe = VFr
Fe = XVFr + YFa
Where
Fa : applied axial thrust load
Fr : applied radial load
Fe : equivalent radial load
V : a rotation factor
X : a radial factor
Y : a thrust factor
A rotation factor V is defined such that:
When
The inner ring rotates V = 1.0
The outer ring rotates V = 1.2

The equivalent radial equations can be plotted in Figure 11.6

The equivalent equation can be defined into two dimensionless groups as shown in
figure in which they are:

When

Fe
=1
VFr

when

Fa
e
VFr

F
Fe
= X +Y a
VFr
VFr

when

Fa
>e
VFr

The pervious equation can be written in one single equation as:

Fe = X iVFr + Yi Fa
when

Fa
e
VFr

i = 2 when

Fa
>e
VFr

i =1
where

(11-9)

Table 11-1, lists values of X2, Y1, X2, and Y2 as a function of , which in turn is a
function of Fa/C0, where C0 is the bearing static load catalog rating.
The X and Y factors in equation (11-9) depend upon the geometry of the bearing,
including the number of balls and the ball diameter.
138

The AFBMA has established standard boundary dimensions for bearings which
define the bearing bore, the outside diameter OD, the width, and the fillet sizes on the
shaft and housing shoulders.
It covers all ball and straight roller bearings in the metric sizes.
It is illustrated in figure 11-7
The bearings are identified by a 2-digit number called the dimension-series code.
First digit : the width series
Second digit: the diameter series

The dimension series code dose not reveal the dimensions directly, therefore it is
necessary to resort tabulations.
Table 11-2 is an example of the 02 series for Deep-Groove and Angular Ball
Bearings.
Table 11-3 gives the dimensions and basic load ratings for Cylindrical Roller Bearing
for the series 02 and 03.
Table 11-4 contains recommendations on bearing life for some classes of machinery
Table 11-5 gives the load-application factors which serve the same purpose as a
factor of safety. It can be used as increase in the equivalent load before selecting a
bearing.

Example
An SKF 6210 angular-contact ball bearing has an axial load Fa of 400 lbf and a radial
load Fr of 500 lbf applied with the outer ring stationary. The basic load rating C0 is 4450
lbf and the basic load rating C10 is 7900 lbf. Estimate the L10 life at a speed of 720
rev/min.
Solution
Outer ring stationary V=1, F0/C0=400/4450=0.09
From table 11-1, using interpolation e = 0.285
139

Fa / VFr = 400/ [(1)500]=0.8 > e From equation (11-9) i = 2

Fe=X2VFr+Y2Fa
X2=0.56, and Y2=1.527 by interpolation
Fe = (0.56)(1)(500)+1.527(400)=890.8 lbf
With LD=L10 and FD=Fe
Using equation 11-3 to find L10
L n 60
C10 = FD D D
LR n R 60
60 LR n R
LD = L10 =
60n D

1/ a

C10

Fe

10 6 7900
=

= 16150h
60(720) 890.8

11-9 Selection of Tapered Roller Bearings

Tapered roller bearings have a number of features that make them complicated
The tapered roller bearing assembly consist of four components as shown in figure
11.3:
o Cone (inner ring)
o Cup (outer ring)
o Tapered rollers
o Cage (spacer retainer)

140

A tapered roller bearing carries either radial and thrust (axial) loads, or any
combination of the two.
If there is no external thrust load, there is still a thrust reaction within the bearing due
to the taper shape of the bearing.
To avoid the separation of the races and the rollers, the thrust load must be resisted by
an equal and opposite force. This is can be done by use at least two tapered roller
bearings on a shaft
In this case, the two bearings can be mounted either:
o The cone fronts facing each other (It is called indirect mounting)
o The cone backs facing each other (It is called direct mounting)

The thrust component Fa produced by a pure radial load Fr is specified by the Timken
Company as:
Fa =

0.47 Fr
K

(3*)

Where K is the ratio of the radial rating of the bearing to the thrust rating
In design of bearing, the initial value of K is chosen to be
K = 1.50 for radial bearing
K = 0.75 for steep-angle bearings
After the analysis is done, the correction value for K is reselected from the
Timken Engineering journal as shown in figure 11-15 and the analysis is repeated.

141

Figure 11-9 shows a typical bearing mounting subjected to an external thrust load
Te.

G
FrA

G
Te

FrB

The radial reactions FrA and FrB are computed by taking moments about the effective
load centers G.
The distance for the value of a which is shown in figure 11.3 is obtained from the
catalog rating sheets.
We shall use subscripts A and B to designate each of the two bearings.
The equivalent radial load on the bearings
0.47 FrB

o Bearing A: FeA = 0.4 FrA + K A

+ Te
KB

0.47 FrA

Te
o Bearing B: FeB = 0.4 FrB + K B
KA

(4*)
(5*)

After we calculate the value of FeA and FeB, we check the following:
o IF FrA > FeA use FrA for design of baring A.
o IF FrA < FeA use FeA for design of baring A.
o IF FrB > FeB use FrA for design of baring B.
o IF FrB < FeB use FeA for design of baring B.

Example
The gear-reduction unit is arranged to rotate the cup while the cone is stationary. Bearing
A takes the thrust load of 250 lb and, in addition, has a radial load 875 lb. Bearing B is
subjected to a pure radial load of 625 lb. The speed is 150 rev/min. The desired L10 life is
90 kh. The desired shaft diameters are 1.125 in at A and 1 in at B. Select suitable tapered
roller bearings, using an application factor of unity.

142

Solution:
Given

At Bearing A: Te = 250 lb, FrA = 875 lb, dA =1.125 in.

At Bearing B: FrB = 625 lb, dB=1.125 in.
L10 = 90 kh, nD = 150 rev/min. R=.9

Assume an initial value for KA and KB as 1.5

0.47 FrB
FeA = 0.4 FrA + K A
+ Te

KB
0.47(625)

= 0.4(875) + 1.5
+ 250 = 1020lb
1.5

Since FeA > FrA use FeA for design of bearing A.

Form equation 11-3
L n 60
C10 = FD D D
LR n R 60

1/ a

(9000)(150)(60)

= 1020
(3000)(500)(60 )

3 / 10

= 1970lb

Now, go figure 11-15 and with (dA=1.125 in ) select form columns 4 and 5 the value
of C10 that is equal or near to the value that we calculate in equation above
Thus, we select 15590 cone, a 15520 cup, and KA=1.69
Calculate FeA and then C10 again with the corrected value of KA, we get: C10=2130 lb.
Again go to figure 11-15 and with (dA=1.125 in ) select form columns 4 and 5 the
value of C10 that is equal or near to the value that we calculate in equation above we
have the same values: 15590 cone, a 15520 cup, and KA=1.69.

For bearing B:
0.47 FrA

FeB = 0.4 FrB + K B

Te
KA

0.47(875)

= 0.4(625) + 1.5
250 = 240lb
1.69

Since FrB > FeB use FrB for design of bearing B

Form equation 11-3
L n 60
(9000)(150)(60)

C10 = FD D D = 625
= 1210lb
(3000 )(500 )(60)
LR n R 60
Now, go to figure 11-15 and with (dB = 1 in) there are five bearings from which to
choose. The one at the top of the list has the smallest rating, the smallest OD, and the
narrowest width.
Thus, we select 07100 cone, a 07196 lb cup, C10=1570 lb and KB=1.45
Calculate FeB and then C10 again with the corrected value of KB, we get the same
C10=1210 lb. So the selection for bearing B is: 07100 cone, a 07196 lb cup, C10=1570
lb and KB=1.45.
1/ a

3 / 10

143

Problem 11-13
A gear-reduction unit uses the countershaft depicted in the figure. Find the two bearing
reactions. The bearings are to be angular-contact ball bearings, having a desired life of
40 kh when used at 200 rev/min. Use 1.2 for the application factor and a reliability goal
for the bearing pair of 0.95. Select the bearings from Table 11-2.

Figure P11-13 Dimensions in inches.

Solution
The bearings are to be angular-contact ball
bearings, having a desired life of
LD = 40 kh
nD = 200 rev/min.
Application factor f = 1.2
Reliability goal R2 = 0.95.
Select the bearings from Table 11-2.
To find the reliability for each bearing:
R2 = (R)2 R=(R2)1/2==( 0.95)1/2 = 0.975
The torque at gear A: T = 240(12)(cos 20) = 2706 lbf in
2706
The force at gear B can be found by: F =
= 498 lbf
6 cos 25o
In xy-plane:
M O = -82.1(16) - 210(30) + 42RCy = 0
RCy = 181 lbf
ROy = 82 + 210 - 181 = 111 lbf

144

In xz-plane:

= 226(16) - 452(30) - 42 Rcz = 0

= -237 lbf
= 226 - 451 + 237 = 12 lbf
= (1112 + 122 )1/ 2 = 112 lbf Ans.
= (1812 + 237 2 )1/ 2 = 298 lbf Ans.
= 1.2(112) = 134.4 lbf
= 1.2(298) = 357.6 lbf
40 000(200)(60)
xD =
= 480
106
1/ 3

480
(C10 )O = 134.4
1/1.483
0.02 + 4.439[ln(1/ 0.975)]

= 1438 lbf or 6.398 kN

RCz
ROz
RO
RC
FeO
FeC

1/ 3

480
(C10 )C = 357.6
1/1.483
0.02 + 4.439[ln(1/ 0.975)]

= 3825 lbf or 17.02 kN

Bearing at O: Choose a deep-groove 02-12 mm with C10 = 6.89 kN and C0 = 3.10 kN.
Bearing at C: Choose a deep-groove 02-30 mm with C10 = 19.5 kN and C0 = 10.0 kN.
Note: It may be an advantage to use identical 02-30 mm bearings in a gear-reduction unit.
Problem 11-14
The worm shaft shown in part a of the figure transmits 1.35 hp at 600 rev/min. 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.

Figure P11-14: (a) Worm and worm gear; (b) force analysis of worm shaft, forces in pounds.

145

Solution
The worm shaft transmits 1.35 hp at 600 rev/min.
Bearing A is to be an angular-contact ball bearing mounted to take the 555-lbf thrust load.
Bearing B takes only radial load, so a straight roller bearing will be employed.
Application factor f = 1.3
nD =25 kh, R= 0.99.
Bearing at A (Ball)
Fr = (362 + 2122)1/2 = 215 lbf = 0.957 kN
Fa = 555 lbf = 2.47 kN
Trial #1:
Since we do not have a value for the specification of the bearing, we start by selecting
one of the bearing from the table.
Therefore, select a 02-85 mm angular-contact with C10 = 90.4 kN and C0 = 63.0 kN.
Fa / C0 = 2.47 / 63.0 = 0.0392
Form table 11-1: by interpolation e = 0.236
Fa / VFr = 2.47/ [(1)0.957] = 2.581 > e From equation (11-9) i = 2
Fe=X2VFr+Y2Fa
X2=0.56, and Y2=1.88 by interpolation
Fe = (0.56)(1)(0.957)+1.88(2.47)=5.18 kN
With the application factor FD=1.3(5.18)=6.73kN

( L D n D / LR n R )
C10 = FD
1 / 1.483
0.02 + 4.439[ln (1 / R )]

(25000)(600)(60)rev = 900
L
LD n D / LR n R =
=
L10
10 6 rev
1/ 3

( )

1/ 3

900
C10 = 6.73
1 / 1.483
0.02 + 4.439[ln (1 / 0.99)]

= 107.7 kN

We found that C10 > (C10)selected from the table

107.7 kN > 90.4kN, therefore we try another bearing with the specification:
Select a 02-95 mm angular-contact with C10 = 121 kN and C0 = 63.0 kN.
Trial #2:
Tentatively select a 02-95 mm angular-contact ball with C10 = 121 kN and C0 = 85 kN.
Fa / C0 = 2.47 / 85.0 = 0.02906
Form table 11-1: by interpolation e = 0.222
Fa / VFr = 2.47/ [(1)0.957]=2.581 > e From equation (11-9) i = 2
146

Fe=X2VFr+Y2Fa
X2=0.56, and Y2=1.98 by interpolation
Fe = (0.56)(1)(0.957) + 1.98(2.47) = 5.427 kN
With the application factor FD=1.3(5.427) = 7.05kN

( L D n D / LR n R )
C10 = FD
1 / 1.483

0.02 + 4.439[ln (1 / R )]
(25000)(600)(60)rev = 900
L
LD n D / LR n R =
=
L10
10 6 rev
1/ 3

( )

1/ 3

900
C10 = 7.05
1 / 1.483

0.02 + 4.439[ln (1 / 0.99)]

= 112.82kN

We found that (C10 )calculated < (C10)selected from the table

112.82 kN < 121kN, therefore the specification of the bearing A is:
A 02 series angular-contact with Bore 95 mm, C10 = 121 kN, and C0 = 63.0 kN.
Bearing at B (Roller):
Fr = (362 + 672)1/2 = 76.06 lbf = 0.3385 kN
FD =1.3(Fr) = 0.44 kN

( LD n D / L R n R )
C10 = FD
1 / 1.483
0.02 + 4.439[ln (1 / R )]

(25000)(600 )(60)rev = 900

L
LD n D / LR n R =
=
L10
10 6 rev
1/ 3

( )

900
C10 = 0.44
1 / 1.483
0.02 + 4.439[ln (1 / 0.99)]

3 / 10

= 5.34kN

From table 11-3, using a 02 series cylindrical roller bearing with Bore 25 mm, C10 = 16.8
kN, and C0 = 8.8 kN.

147