Ball Bearings
nd
(2
Shift)
Problem 11
In a particular application, the radial load acting on a ball bearing is 6 kN and the
expected life for 90% bearings is 10000 hours at 1450 rpm. Calculate the
dynamic radial load rating of the bearing required.
(57278 N)
Problem 12
A single-row deep groove ball bearing is subjected to a radial force 10 kN
and an axial force of 4 kN. The shaft rotates at 1450 rpm and the expected
L10H life of the bearing is 15000 hours. Select a suitable ball bearing with a
minimum bore of 60 mm.
Problem 13
A deep groove ball bearing, ISI no. 60BC02, rotating at 1450 rpm is subjected to
a radial force of 2000 N and an axial force of 900 N. The radial and thrust factors
are 0.56 and 2.0, respectively. The load factor or service factor is 1.5. What will
be the expected life in hours for 90% reliability?
If four such bearings are used in an application, what will the combined
reliability?
(Calculate equivalent dynamic load Pr;
From catalogue read the value of C for 60BC02 bearing
𝑪 𝟑
𝑳𝟏𝟎 = ( ) = 𝟕𝟕𝟑. 𝟕 𝒎𝒊𝒍𝒍𝒊𝒐𝒏 𝒄𝒚𝒄𝒍𝒆𝒔
𝑷𝒓
L10H = 8893 hours
Reliability of 4 bearings combined = (0.90)4 = 0.656 or 65.6%)
Problem 14
A deep groove ball bearing has a dynamic load capacity of 35,000 N. It is
to operate on the following work cycle:
Radial load of 5000 N at 200 rpm for 25% time,
Radial load of 9000 N at 500 rpm for 20% time,
Radial load of 3000 N at 400 rpm for the remaining time,
Assuming that the loads are steady and the inner race rotates; find the
average expected life of the bearing in hours.
(L10H = 8398; L50H = 41990)
Problem 15
The radial load acting on a ball bearing is 2500 N for the first five revolutions and
reduces to 1500 N for the next ten revolutions. The load variation then repeats
itself. The expected life of the bearing is 20 million revolutions. Determine the
dynamic load carrying capacity of the bearing.
𝟏
𝑳𝒊 𝑷𝟑𝒊 𝟑
(𝑪 = (𝜮
) = 𝟓𝟑𝟎𝟑 𝑵)
𝟏𝟎𝟔
Problem 16
A ball bearing subjected to a radial load of 3000 N is expected to have a
satisfactory life of 10000 h at 720 rpm with a reliability of 95%. Calculate the
dynamic load carrying capacity of the bearing, so that it can be selected from a
manufacturer’s catalogue based on 90% reliability. If there are four such
bearings, each with a reliability of 95% in a system, what is the reliability of the
complete system?
𝑳𝟎𝟓 =
𝟏𝟎𝟎𝟎𝟎 × 𝟔𝟎 × 𝟕𝟐𝟎
= 𝟒𝟑𝟐
𝟏𝟎𝟔
(We need life with 90% Reliability because catalogues give C for 90% reliability)
𝑳𝟎𝟓
𝑳𝟏𝟎
𝟏 𝟏/𝒃
𝒍𝒏 𝑹
𝟗𝟓
=(
)
𝟏
𝒍𝒏 𝑹
𝟗𝟎
𝟒𝟑𝟐
𝑳𝟏𝟎
𝒍𝒏 𝟎.𝟗𝟓 𝟏/𝒃
= (𝒍𝒏 𝟎.𝟗𝟎)
L10 = 799 for b = 1.17 (Text book)
L10 = 698 for b = 1.5 (New Indian Standard)
𝟏
𝟏
𝑪 = 𝑷(𝑳𝟏𝟎 )𝟑 = 𝟑𝟎𝟎𝟎(𝟕𝟗𝟗)𝟑 = 𝟐𝟕𝟖𝟑𝟖 𝑵
𝟏
𝒐𝒓 𝟑𝟎𝟎𝟎(𝟔𝟗𝟖)𝟑 = 𝟐𝟔𝟔𝟏𝟐 𝑵 𝒇𝒐𝒓 𝒃 = 𝟏. 𝟓
Reliability of system having four bearings of 95% Reliability is 81%.
Problem 17
Select a single row deep groove ball bearing for a radial load of 4 kN and an axial
load of 5 kN, operating at a speed of 1500 rpm for an average life of 5 years at
10 hrs per day. Assume uniform and steady load.
Data is given to select a deep-groove ball bearings. The given data is to be
converted to:
(i)
(ii)
desired life to L10 number of rotation and
applied load to Pr, dynamic equivalent radial or axial load.
Then use
𝐶𝑟 = 𝑃𝑟 (𝐿10 )1/3 𝑓𝑜𝑟 𝑏𝑎𝑙𝑙 𝑏𝑒𝑎𝑟𝑖𝑛𝑔
or
𝑃𝑟 (𝐿10 )0.3 𝑓𝑜𝑟 𝑟𝑜𝑙𝑙𝑒𝑟 𝑏𝑒𝑎𝑟𝑖𝑛𝑔
(i)
To calculate L10 life:
Average Life is given and is
L50 = (5 years) (365 days / years) (10 hr / day) (60 min / hr) (1500 rpm)
= ……………………..million rev (Say N Million rev)
L10 life = 0.2 L50 million rev (you should get L10 life as 328.5)
(ii)
To calculate equivalent radial load, Pr
Pr = XVFr + YFa (Given Fr = 4 kN, Fa = 5 kN)
(Assumed inner race rotation as is the case in more than 95% cases V=1)
Pr = 4X + 5Y (Pr in kN)
First trial
X = 0.56 and Y = 1.55. On substitution
Pr = 9.99 kN = 9990 N
C OR Cr = Pr (L10)1/3 = 68930 N
Consult a catalogue or data book:
Bearing no.
C
C0
Bore d
outer Dia D etc
6409
76100
45500
45
120
6311
71500
41500
55
120
(From Bhandari)
𝑆𝑢𝑝𝑝𝑜𝑠𝑒 𝑤𝑒 𝑠𝑒𝑙𝑒𝑐𝑡 6311:
𝐹𝑎 5000
𝐹𝑎
5000
=
= 1.25;
=
= 0.12
𝐹𝑟 4000
𝐶𝑜 41500
From Table 15.4 of Bhandari
𝐹𝑜𝑟
𝐹𝑎
= 0.120, 𝑒 = 0.3033
𝐶𝑜
As such Fa / Fr > e; Y = 1.43 and X = 0.56 Close to assumed values
Pr = 0.56 (4000) + 1.43 (5000) = 9390
C = 9390 (328.5)1/3 = 64.777 N
Bearing no. 6409, 6410, 6311, 6215 are all OK with respective C of 76100,
87100, 71500, 66300. Final selection will depend shaft dia desired.
Problem 18
Select a suitable rolling element bearing for the following application:
Shaft dia. = 50 mm, Speed = 600 rpm, Required bearing life = 12000 hours with
reliability 95%, Radial load = 2 kN, Axial load = 0.5 kN, Operating temperature =
500C, Inner race revolves.
(Grease can be used upto an operating temperatures of about 1000C and speed
factor upto 200000(bore, mm x rpm)
Problem 19
A deep groove ball bearing has dynamic capacity of 20000 N and is to
operate on the following work cycle; Radial load of 6000 N at 200 rpm for
25% time, radial load of 9000 N at 500 rpm for 20% of the time and radial
load of 3500 N at 400 rpm for the remaining period. Assuming the loads
are steady and the inner race rotates, find the average expected life of the
bearing in hours.
Problem 20
Select a suitable roller bearing to carry a radial load of 10,000 N. The shaft
rotates at 1000 rpm, average life is 5000 hours. Inner race rotates. Take
mild shock.
Pr = (Shock factor) (radial load) = 15000
Taking mild shock factor = 1.5
L50 = (5000) (60) (1000) and L10 = 0.2 L50
C = Pr (L10)0.3
Select from a catalogue for the calculated C.