Bearings: Sliding Contact
and Rolling contact
Subject: Machine Design
Code:MEC503
Prof. Nilotpal Banerjee
Introduction
What is a bearing:
Bearing is a mechanical element that permits relative motion between two members with
minimum friction.
Functions:
1. A bearing assists in free rotation of shafts/axles with less friction.
2. A bearing supports a shaft/axle in proper position.
3. A bearing also takes up the forces on the shaft/axle and may transmit it to the
foundation.
Types:
Bearings can be classified broadly under two categories based on the types of contact
they have between the rotating and the stationary member
a. Sliding contact
b. Rolling contact
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
2
Rolling Contact Bearings
Rolling contact bearings are also called anti-friction
bearing due to its low friction characteristics. These
bearings are used for taking up radial load, thrust load
and combination of thrust and radial load. These bearings
are extensively used due to its
(i) Relatively lower price,
(ii) Being almost maintenance free and
(iii) For its easy operation.
However, friction increases at high speeds for rolling
contact bearings and it may be noisy while running.
These bearings are of two types,
1.Ball bearing and 2. Roller bearing
A typical ball bearing is shown the figure on the right side
of this slide, with nomenclature.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
Fig. B1
3
Rolling Contact Bearings….
The bearing shown in the figure in the
previous slide is called Single row deep
groove ball bearing. It is used to carry
radial load but it can also take up
considerable amount of axial load. The
retainer keeps the steel balls in position
and the groove below the steel balls is the
inner ring and over it is the outer ring.
The outer ring, also called outer race, is
normally placed inside a bearing housing
which is fixed, while the inner race holds
the rotating shaft. Therefore, a seat of
diameter d and width B is provided on the
shaft to press fit the bearing. This
arrangement is shown in the schematic
diagram beside.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
Fig. B 2
4
Rolling Contact Bearings….
1. Single row Angular
Contact Ball Bearing. It is
mostly used for radial
loads and heavy axial
loads.
2. Double Row Angular
Contact Bearing, has two
rows of balls. Axial
displacement of the shaft
can be kept very small
even for axial loads of
varying magnitude.
3. Single thrust ball
bearing. It is mostly used
for unidirectional
axial load.
Fig. B 3.1
9 November 2022
4. A taper roller bearing which is generally
used for simultaneous heavy radial load and
heavy axial load. Roller bearings has more
contact area than a ball bearing, therefore,
they are generally used for heavier loads
than the ball bearings.
5. A spherical roller bearing, has self
aligning property. It is mainly used for heavy
axial loads. However, considerable amount
of loads in either direction can also be
applied.
6. A Cylindrical Roller Bearing Used for
heavy radial load and high speed use. Within
certain limit, relative axial displacement of
the shaft and the bearing housing is
permitted for this type of bearings.
Fig. B 3.2
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
5
Cross-sectional views….
Fig. B 4
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
6
Selection of Rolling Contact Bearings
In relation to selection, Rolling Contact Bearings have certain criteria on
which it is important:
1. Rating life:
Rating life is defined as the life of a group of apparently identical ball or
roller bearings, in terms of number of revolutions or hours, rotating at a
given speed, so that 90% of the bearings will complete or exceed before
any indication of failure.
Consider 100 apparently identical bearings. All the 100 bearings are put
onto a shaft rotating at a given speed while acted upon by same particular
load. After some time, one after another, failure of bearings will be
observed. When in this process, the tenth bearing fails, then the number
of revolutions or hours lapsed is recorded. These figures recorded give the
rating life of the bearings or simply L10 life (10 % failure). Similarly, L50
means, 50 % of the bearings are operational. It is known as median life.
Figure defines the life of rolling
contact bearings.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
Fig. B 5
7
Selection of Rolling Contact Bearings….
2. Bearing load:
If two groups of identical bearings are tested under loads P and P for respective lives of L and L , then,
=
where,
(1)
L : life in millions of revolution or life in hours and
a : constant which is 3 for ball bearings and 10/3 for roller bearings
3. Basic load rating:
It is that load which a group of apparently identical bearings can withstand for a rating life of one million
revolutions. Hence, if say, L1 is taken as one million then the corresponding load is,
C=P L
Where, C is the basic or dynamic load rating.
(2)
Therefore, for a given load and a given life the value of C represents the load carrying capacity of the
bearing for one million revolutions. This value of C, for the purpose of bearing selection, should be lower
than that given in the manufacturer’s catalogue. Normally the basic or the dynamic load rating as
prescribed in the manufacturer’s catalogue is a conservative value, therefore the chances of failure of
bearing is very less.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
8
Selection of Rolling Contact Bearings….
4. Equivalent radial load:
The load rating of a bearing is given for radial loads only. Therefore, if a bearing is subjected to both axial and radial
load, then an equivalent radial load is estimated as,
P =V × P or P =XVP +Y P
(3)
Where,
P : Equivalent radial load
P : Given radial load
P : Given axial load
V : Rotation factor (1.0, inner race rotating; 1.2, outer race rotating)
X : A radial factor
Y : An axial factor
The values of X and Y are found from the chart whose typical format and few
representative values are given the table on the right. The factor C is obtained
From the Bearing Catalogue.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
9
The selection procedure
• Depending on the shaft diameter and magnitude of radial and axial load a
suitable type of bearing is to be selected from the manufacturer’s catalogue,
either a ball bearing or a roller bearing. The equivalent radial load is to be
determined from equation (3).
• If it is a tapered bearing then manufacturer’s catalogue is to be consulted for the
equation given for equivalent radial load.
• The value of dynamic load rating C is calculated for the given bearing life and
equivalent radial load. From the known value of C, a suitable bearing of size that
conforms to the shaft is taken.
• However, some augmentation in the shaft size may be required after a proper
bearing is selected.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
10
The selection procedure
Step 1. Determine radial and axial forces and the Diameter of the bearing.
Step 2. Type of bearing.
Step 3. To determine the value of radial factor (X) and Thrust factor(Y) from the table.
and
, where C is static load carrying
These values depend on two ratios
capacity.
Step 4. Calculate equivalent dynamic load by the formula,
P =V × P or P =XVP +Y P
Step 5. Express the life in millions revolution L
Step 6. calculate dynamic load capacity by, C = P L
Step 7. Select a ball bearing from the table and check whether its dynamic capacity is
within limit, if not select the next series and go back to step 3 and continue.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
11
Example 1
Problem:
A simply supported shaft, diameter 50mm, on bearing supports carries a load of 10kN at its midpoint. The
axial load on the bearings is 3kN. The shaft speed is 1440 rpm. Select a bearing for 1000 hours of
operation.
Solution
The radial load P = 5 kN and axial load P = 3 kN. Therefore, a single row deep groove ball bearing may be
selected since radial load is predominant. This choice has wide scope, considering need, cost, future
changes etc.
×
×
Millions of revolution for the bearing, L =
= 86.4
For the selection of bearing, a manufacturer’s catalogue has been consulted. The equivalent radial load on
the bearing is given by, P =V × P or P =XVP +Y P
Here, V=1.0 (assuming inner race rotating)
From the catalogue, Co = 19.6 kN for 50mm inner diameter.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
12
Example 1….
From the catalogue, Co = 19.6 kN for 50mm inner diameter.
Therefore, value of “e” from the table and by linear interpolation = 0.327.
Here, = = 0.6 > e . Hence, X and Y values are taken from fourth column
of the sample table. Here, X= 0.56 and Y= 1.356
Therefore, P =XVP +Y P = 0.56x 1.0 x5.0 +1.356x 3.0 = 6.867 kN
Therefore basic load rating C = P × L =6.867 × 86.4 =30.36kN
Now, the table for single row deep groove ball bearing shows that
for a 50mm inner diameter, the value of C = 35.1 kN. Therefore, this bearing
may be selected safely for the given requirement without augmenting the
shaft size. A possible bearing could be SKF 6210.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
13
Static load carrying capacity
Static load is defined as the load acting on the bearing when the shaft is stationary.
• Static load produces permanent deformation in balls and races.
• The permissible value of static load depends on the allowable magnitude of permanent
deformation. In practice it is observed that a permanent deformation of 0.0001 of the ball or
roller diameter in the mostly stressed ball and race contact.
• Therefore, “The static load carrying capacity of a bearing is defined as the static load which
corresponds to a total permanent deformation of balls and races at the most heavily stressed
point of contact, equal to 0.0001 of the ball diameter”.
• Stribeck’s equation gives the static load capacity of the bearing.
• Basic assumptions for the above equation are:
1. The races are rigid and retain their circular shape under load.
2. The balls are equally spaced.
3.The rolling elements on the upper half do not support any load.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
14
Stribeck’s equation
The adjoining figure shows the forces on the inner race
through rolling elements, supporting the static load C .
Considering the equilibrium of forces in vertical direction,
C =P +2P cos β + 2P cos 2β +………
(4)
As it has been already assumed that the races are rigid only
balls are deformed. Let δ is the deformation at the mostly
stressed ball corresponding to force P and the centre of the
inner race moves point O to a point O’ through a distance δ
. Likewise say δ , δ , δ ….. Are the radial deflections in
other respective balls.
Therefore, δ =δ cos β or
0
0
= cos β ……..
Forces acting on Inner Race
(5)
By Hertz’s equation deflection at each ball is ,
1∝ 3
⁄
and
0
0
⁄
=
……………
(6)
Defection of Inner Race
Fig. B 6
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
15
Stribeck’s equation
⁄
= cos β and P = P cos β
From equations (5) and (6),
P = P cos 25
⁄
⁄
, similarly
, substituting these values in equation (4),
C =P +2 P cos β
=3 1 + 2 789 5
⁄
cos β + 2 P cos 2β
⁄
⁄
+ 2 789 25
⁄
cos 2β +…
………..
Or, C =P M ……
Where, M= 1 + 2 cos 5
(7)
⁄
+ 2 cos 25
Now if number of balls= z then 5 =
taken from
;
⁄
………..
(8)
the values of M for different values of z are
It is observed that (z/M) are practically constant. Stribeck suggested (z/M)= 5
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
16
Stribeck’s equation
Substituting this value in equation (7),
C =
zP
……….
(9)
Now, as force being proportional to the projected area of the ball,
P =kd
(10)
Where, d is the ball diameter and the factor k depends on the radii of curvature at
the point of contact and also on the moduli of elasticity of material, therefore,
From (9) and (10), C =
9 November 2022
?@
z
[ Stribeck’s Equation]
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
(11)
17
Dynamic load carrying capacity
Failure of bearing is mainly due to fatigue and therefore, life of a
bearing is based on the fatigue of the bearing and can be defined as “as
the number of revolutions ( or hours of service at some given speed),
which it endures before first evidence of fatigue crack in the balls or
races”.
The dynamic load carrying capacity is defined as “radial load ( thrust
load in case of thrust bearing ) which can be carried for a minimum life
of one million revolution”.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
18
Standard Bearing Life
While selecting a proper size of bearing
it is essential to specify the Bearing life
for that particular application. For that
normally past experience plays a major
role. As in the case of automobile, since
the speed is not always constant
bearing life is specified in number of
revolutions. In case of other industrial
application where the speed is more or
less constant the life is expressed in
numbers of hours. The two tables
given enables one to have rough
guideline.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
19
Load Factor
The forces acting on a bearing are normally
estimated considering the equilibrium of
forces acting. These forces are multiplied by
a “Load Factor” to take care of effects of
dynamic load. Load factors are used in case
of applications involving gear, chain and belt
drives. In chain and belt drives the dynamic
load comes due to vibrations. Further, for
gear drive, additional dynamic load is coming
due to inaccuracies in the tooth profile and
elastic deformation of gear teeth. Values
given in table are used as rough estimation
where exact analysis is not available.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
20
Example 2
Problem: A single row deep groove ball bearing is to be selected for a shaft
which is 75 mm in diameter and rotates at 125 rpm. It carries a radial load of
21kN and there is no thrust load. The expected life of the bearing is 10 000
hrs.
Solution:
Step 1. The radial load P = 21 kN and axial load P = 0 kN.
Step 2. type, single row deep groove ball bearing.
Step 3. Millions of revolution for the bearing, L =
Step 4. C = P L
9 November 2022
×
×
= 75
=21000× 75 =88 560.43 N
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
21
Example 2…
From the table for diameter 75 mm
the available bearings are,
Therefore, bearing number 6315 is
selected.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
22
Example 3,4 and 5 for practice
Problem 3: A single row deep groove ball bearing is subjected to a radial load of 8 kN
and a thrust load of 3 kN. The shaft rotates at 1200 rpm. The expected life is 20 000
hours. The minimum diameter is75 mm. select a suitable bearing.
Problem 4: A ball bearing with dynamic load capacity of 22.8 kN is subjected to a radial
load of 10 kN. Calculate,
(i) The expected life in million revolutions that 90% of the bearings will reach.
(ii) The corresponding life in hours.
(iii) The life that 50% of the bearings will complete before failure.
Problem 5: Select a bearing for a 40 mm diameter shaft which rotates 400 rpm due to
bevel gear mounted on the shaft the bearing will have to withstand a 5 kN radial load
and 3 kN thrust load. The life of bearing expected to be 1000 hrs. N at 400 RPM.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
23
Selection: convention
A rolling contact bearing is normally designated by three or four digits.
i) The last two digits multiplied by 5 indicate the bore diameter of the
bearing. Example, bearing 6315 indicates bearing of 75 mm (15x5=75)
bore diameter.
ii) The 3rd from the right indicates the series of the bearing.
a) Extra light series---1
b) Light series---2
c) Medium series--- 3( Example, 6315 is medium series)
d) Heavy series---- 4
iii) The fourth (and sometimes fifth) digit from right gives the type of
contact. Example, the digit 6 tells that it is a deep groove ball bearing.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
24
Roller Bearings
Types of Roller Bearing;
Apart from Ball bearings there are four more types of rolling elements:
• Cylindrical rollers
• Spherical rollers
• Tapered rollers
• Needle rollers
o Most rolling-element bearings feature in cages. The cages reduce friction, wear,
and bind the elements by preventing it from rubbing against each other.
o Roller bearing are usually used when shock and impact are present, or when
large bearing are needed.
o Tapered roller bearing can carry a large axial load. The magnitude depends on
the angularity of the rollers. The radial load will also produce a thrust component.
o Roller bearing in general can be applied only where the angular misalignment
caused by shaft deflection is very small. Spherical roller bearing has excellent load
capacity and carry a thrust component in either direction.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
25
Types of rollers
Fig. B 7
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
26
Cylindrical Rollers
When maximum load carrying capacity is required in a given space, the point
contact in ball bearing is replaced by the line contact of roller bearing. A cylindrical
roller bearing consists of relatively short rollers that are positioned and guided by
the cage.
Advantages:
• Due to line contact between rollers and races, the radial load carrying capacity of
the cylindrical roller bearing is very high.
• Cylindrical roller bearing is more rigid than ball bearing.
• The coefficient of friction is low and frictional loss is less in high-speed
applications.
Disadvantages :
• In general, cylindrical roller bearing cannot take thrust load.
• Cylindrical roller bearing is not self-aligning. It cannot tolerate misalignment. It
needs precise alignment between axes of the shaft and the bore of the housing.
• Cylindrical roller bearing generates more noise.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
27
Taper Rollers
• The taper roller bearing consists of rolling elements in the form of a frustum of
cone.
• They are arranged in such a way that the axes of individual rolling elements
intersect in a common apex point on the axis of the bearing.
• In kinematics’ analysis, this is the essential requirement for pure rolling motion
between conical surfaces.
• In taper roller bearing, the line of resultant reaction through the rolling elements
makes an angle with the axis of the bearing.
• Therefore, taper roller bearing can carry both radial and axial loads.
• A taper roller bearing subjected to pure radial load induces a thrust component
and vice versa.
• Taper roller bearings are always used in pairs to balance the thrust component.
• Taper roller bearing has separable construction. The outer ring is called ‘cup’ and
the inner ring is called ‘cone’. The cup is separable from the remainder assembly
of the bearing elements including the rollers, cage and the cone.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
28
Taper Rollers
Taper roller bearings offer the following advantages:
• Taper roller bearing can take heavy radial and thrust loads.
• Taper roller bearing has more rigidity and Taper roller bearing can be easily
assembled and disassembled due to separable parts.
Disadvantages:
• It is necessary to use two taper roller bearings on the shaft to balance the axial
force. It is necessary to adjust the axial position of the bearing with pre-load.
• This is essential to coincide the apex of the cone with the common apex of the
rolling elements.
• Taper roller bearing cannot tolerate misalignment between the axes of the shaft and
the housing bore.
• Taper roller bearings are costly. Taper roller bearings are used for cars and trucks,
propeller shafts and differentials, railroad axle- boxes and as large size bearings in
rolling mills.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
29
Lubrication of Bearing
Lubrication is technique of reducing friction between two surfaces by
the use of a substance called lubricant.
Three types of Lubricant:
• Liquid lubricant- Mineral/vegetable oils
• Semi-solid Lubricant- Grease
• Solid Lubricant- Graphite/Molybdenum Disulphide
Functions:
• To reduce friction
• To reduce /prevent wear
• To carry away the heat generated due to friction
• To prevent corrosion
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
30
Lubrication of Bearing
Basic modes of lubrication:
• Thick Film Lubrication: where two surfaces of bearing in relative motion are clearly
separated by a film of fluid. Only viscosity of the fluid is important and surface
finish of the surfaces has no effect.
It is further divided in two categories;
a) Hydrodynamic: load-supporting fluid film is created by shape and
relative motion between two surfaces.
b) Hydrostatic: load-supporting fluid film is created by an external pump,
supplying sufficient fluid under pressure. Externally pressurized Bearings.
• Thin Film Lubrication: Which is also called boundary lubrication is a condition
where the fluid film is relatively thin and having partial metal to metal contact.
• Zero Film: Here bearing operates without any Lubrication.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
31
Elastohydrodynamic Lubrication
Elastohydrodynamic Lubrication(EHL) – is a lubrication regime (a type
of hydrodynamic lubrication (HL)) in which significant elastic deformation of
the surfaces takes place and it considerably alters the shape and thickness of
the lubricant film in the contact. The term underlines the importance of the
elastic deformation of the bodies in contact under load in developing the
lubricant film. EHL, the same way as HL, is used to decrease friction and wear
in tribological contacts. It is achieved by the development of a thin lubricant
film between rubbing surfaces, which separates them and decreases friction.
Since the HL film is developed due to elastic deformation this mode of
lubrication is called Elastohydrodynamic Lubrication(EHL). This type of
lubrication occurs in gears, cams and also in rolling contact bearings.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
32
Viscosity: A recap
Viscosity is defined as that property of fluid by virtue of which it offers resistance
against any kind of shear force on it. An fluid film between two surfaces is shown
on the right hand side as an example. The lower plate is stationary while the
upper one is moving with a velocity “U” due to the force “P” towards right. The
intermediate layers moves with a velocity proportional to its distance from the
stationary plate such as,
B B
=
C C
B
C
= ,
(12)
the tangential force per unit area,
D
E
B
C
is the shear stress is proportional to rate of shear .
By Newton’s law of Viscosity ,
Shear Stress ∝ Rate of Shear,
therefore
9 November 2022
F
∝
G
H
Fig. B 8
(12A)
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
33
Viscosity: A recap
∴ P = μA
G
H
(13)
Where, μ is called coefficient of Dynamic Viscosity or Absolute Viscosity. When the velocity distribution
is non liner with respect to h the equation is modified as,
P = μA
@G
@H
(14)
The unit of Dynamic Viscosity or Absolute Viscosity is found as,
H
O.PP
=Ns/mm
.PP⁄Q
μ = FG=PP
=MPa-s, popularly, viscosity is often expressed as Poise(P) or Centi-Poise(cP).
1 Poise=1 dyne. cm/s
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
34
Fluid film Bearing
In fluid film bearing the entire load of the shaft is carried by a thin film of fluid present
between the rotating and non-rotating elements. The types of fluid film bearings are
listed below,
• Sliding contact type
• Journal bearing
• Thrust bearing
• Slider bearing
The figure shows a plot of Friction vs. Shaft speed
for three types of bearings. It is observed that,
Fig. B 9
• For the lower shaft speeds the journal bearing have more friction than roller and ball
bearing and ball bearing friction being the lowest.
• With the increase of shaft speed the friction in the ball and roller bearing
phenomenally increases but the journal bearing friction is relatively lower.
• Hence, it is advantageous to use ball bearing and roller bearing at low speed
applications. Journal bearings are mostly suited for high speeds and more loads.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
35
Journal Bearing
The black annulus represents the bush(Bearing) and grey circle
represents the shaft (Journal) placed within an oil film shown by
the shaded region. The journal, carries a load P on it.
Fig. B 10
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
36
Journal Bearing Operation
Journal bearings operate as per the principle stated below
• The journal being smaller in diameter than the bearing, it will always rotate
with an eccentricity.
• When the journal is at rest, due to bearing load P, the journal is in contact with
the bush at the lower most position and there is no oil film between the bearing
bottom and the journal.
• Now when the journal starts rotating, then at low speed condition, with the
load P acting, it has a tendency to shift to its sides.
• At this equilibrium position, the frictional force will balance the component of
bearing load. In order to achieve the equilibrium, the journal orients itself with
respect to the bearing .
• The angle θ, shown for low speed condition, is the angle of friction. Normally at
this condition either a metal to metal contact or an almost negligible oil film
thickness will be there.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
37
Journal Bearing Operation…..
• At the higher speed, the equilibrium position shifts and a continuous oil film
will be created as indicated in the third figure above. This continuous fluid film
has a converging zone, which is shown in the enlarged view.
• It has been established that due to presence of the converging zone or wedge,
the fluid film is capable of carrying huge load. If a wedge is taken in isolation,
the pressure profile generated due to wedge action will be as shown.
• Hence, to build-up a positive pressure in a continuous fluid film, to support a
• load, a converging zone is necessary. Moreover, simultaneous presence of the
converging and diverging zones ensures a fluid film continuity and flow of fluid.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
38
The History of hydrodynamic theory of lubrication
• Petroff (1883) carried out extensive experimental investigation and
showed the dependence of friction on viscosity of lubricant, load and
dimensions of the journal bearing.
• Tower (1883 ) also conducted experimental investigation on bearing
friction and bearing film pressure.
• After that Osborne Reynolds conducted experiments and published the
findings in the form of present day hydrodynamic theory of lubrication
and the corresponding mathematical equation is known as Reynolds’
equation.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
39
The Reynolds’ equation
Reynolds’ Equation in simplified form is given below,
0
0p
H
rs
q rp
0
0t
+
H
rs
q rt
=
G rH
rp
(15)
where,
U : surface speed of the wedge, in x-direction
p : pressure at any point(x,z) in the film
μ : Absolute viscosity of the lubricant
Fig. B 11
h : film thickness, measured in y-direction
The left hand side of the equation represents flow under the pressure gradient. The
right hand side represents a pressure generation mechanism.
In this equation the following assumptions are made,
• That the lubricant is incompressible and Newtonian.
• The wedge shape, is assumed to be a straight profile as shown.
• The bearing is very long in the Z direction and the variation of pressure is in the X and
Z direction.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
40
The Reynolds’ equation
Let us have a look at the right hand term in details.
r
rp
ρ
v wv
h +
r
rx
ρ
y wy
h +
ρ
rH
rz
+
squeeze film
h
r{
rz
(15.1)
compression
The First term of above equation can be written as,
|B rC
r}
+
Physical wedge
|C rB
r}
+
BC r|
r}
(15.2)
Fig. B 12
Stretch
There are two moving surfaces 1 and 2 as indicated in figure on RHS. For 1 the velocities are ~ , • and
€ along the three coordinate axes X, Y and Z respectively. For 2, similarly the velocities are ~ , • and
€ respectively.
Equation ((15.1) represents the full form of the right hand side of Reynolds’ equation (15). For the
purpose of explanation, partial derivative of only the first term of equation (15.1) is written in
equation (15.2). Here ~ − ~ have been replaced by U.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
41
The Reynolds’ equation
|B rC
The first term of equation (15.2)
, represents a physical wedge. The second
r}
|C rB
is known as the stretch. All the three terms contributes to pressure
term
r}
generation mechanism.
rH
The term ρ in equation (15.1) is called squeeze film; with respect to time and
rz
shows how the film thickness is changing.
r{
The term h in equation (15.1) shows the compressibility of fluid with time.
rz
The simplified form of Reynold’s equation(15), has only physical wedge term
|B rC
.
r}
There is no general analytical solution to equation (15); approximate solutions have
been obtained by using electrical analogies, mathematical summations and
numerical and graphical methods. One of the important solutions is due to
‚
ƒ
Sommerfeld and is expressed as, „ = …
‚
ƒ
†‡ˆ
‰
= φ‹
(15.3)
Where S= Smmerfeld’s Number and φ gives the functional relationship.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
42
Petroff’s Equation
Petroff’s Equation gives the estimate of co-efficient of friction in journal bearings. It is
based on certain simplifying assumptions:
(i) The shaft and the bearing are concentric.
(ii) The bearing is having light load.
Although the assumptions are highly impractical, Petroff’s equation defines a group of
dimensionless parameters which influences the frictional properties of the bearing.
r= Radius of the journal (mm)
l= length of the bearing (mm)
c= Radial clearance (mm)
Υ = Journal speed(revolution/s)
The velocity at the surface of the journal,
U = 2πr Œ•
(16)
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
43
Petroff’s Equation
G
From Newton’s law of viscosity, equation (14), P = μA H
P=Tangential frictional force
A= Area of friction surface, 2π•‘
U= Surface velocity, 2եΥ
h= Distance between the journal and the bearing= clearance, c
P=“ 2’•‘ 2’•Œ•
ƒ
=
” ‚ •†‡ˆ
ƒ
(17)
” ‚ •†‡
ˆ
Frictional Torque, –— =3• =
(18)
ƒ
If the radial load is W as shown, then the pressure intensity, p can be
given as,
p=
˜
™ šz @ F
9 November 2022
=
˜
›
∴ W = 2prl
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
(19)
44
Petroff’s Equation….
Denoting frictional co-efficient by “ f”, frictional force
is fW and frictional torque (–— )= fWr = f2p• l
From (18) and (20),
∴ f = 2π
š
” ‚ •†‡ˆ
=
ƒ
(20)
f2p• l,
qϥ
s
(21)
Fig. B 13
The Above equation is known as Petroff’s Equation
In the above equation it is observed that it includes two important dimensionless parameters
‚
†‡ˆ
and
which influence the value of co-efficient of friction and other frictional
ƒ
‰
properties such as frictional torque and power. The equation (21) is modified as,
‚
„
ƒ
Where ‹ =
9 November 2022
‚
ƒ
†‡ˆ
‰
= 2’
‚
ƒ
†‡ˆ
‰
= 2’ S,
is known as Sommerfeld’s Number
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
45
Design parameters of journal bearing
The first step for journal bearing design is to determine the bearing
pressure for given design parameters,
• Operating conditions (temperature, speed and load)
• Geometrical parameters ( length and diameter)
• Type of lubricant ( viscosity)
The design parameters, mentioned above, are to be selected for initial
design. The bearing pressure is known from the given load capacity and
preliminary choice of bearing dimensions. After the bearing pressure is
determined, a check for proper selection of design zone is required. The
selection of design zone is explained in next slide.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
46
Design parameters of journal bearing
The figure on the right hand side shows
a plot of variation of coefficient of
friction with bearing characteristic
†‡ˆ
number ( ). Bearing characteristic
‰
number is defined as =
†‡ˆ
‰
It is a non-dimensional number, where
μ is the viscosity, Œ• is the speed of the
journal bearing and p is the pressure
˜
which is given by p = @› , where d and l
are diameter and length of the journal
respectively.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
Fig. B 14
47
Design parameters of journal bearing
• The plot shows that beyond point B, with the increase in bearing
†‡
characteristic number ( ˆ ) friction increases and from point B to point A, with
†‡
reduction in( ˆ )the
‰
‰
friction again increases. So the point B is the limit i.e the
co-efficient of friction is minimum. The value of bearing characteristic number
corresponding to minimum co-efficient is called the bearing modulus and is
denoted by “K”. The zone between A to B is known as boundary lubrication or
imperfect lubrication zone.
• Imperfect lubrication means that metal to metal contact is possible. The portion
between B to D is known as the hydrodynamic lubrication zone .
• Calculated value of bearing characteristic number should lie somewhere in the zone
of C to D. This zone is characterized as design zone.
• For any operating point between C and D due to fluid friction certain amount of
temperature rise takes place. Due to which the viscosity of the lubricant decreases,
and bearing characteristic number also decreases.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
48
Design parameters of journal bearing
• Hence, the operating point will shift towards C, resulting in lowering of the friction
and the temperature. As a consequence, the viscosity will again increase and will pull
the bearing characteristic number towards the initial operating point. Thus a self
control phenomenon always exists. For this reason the design zone is considered
between C and D. The lower limit of design zone is roughly five times the value at B.
• On the contrary, if the bearing characteristic number decreases beyond B then
friction goes on increasing and temperature also increases and the operation
becomes unstable.
• Therefore, it is observed that, bearing characteristic number controls the design of
journal bearing and it is dependent of design parameters like, operating conditions
(temperature, speed and load), geometrical parameters ( length and diameter) and
viscosity of the lubricant.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
49
Viscous flow through Rectangular Slot
In figure flow through a rectangular slot is shown.
l= length of the slot in the direction of flow
b= depth and h= width of the block.
It is assumed that the dimension of ‘b’ is very large in
comparison to that of ‘h’.
Pressure difference between the two surfaces of central
slice of thickness of 2x is given by,
∆p = p − p
The downward force, P = 2xb × ∆p
(22)
The shear force on both surfaces of the slice is due to
viscosity of the fluid.
P = μA
@G
@H
=μ 2lb
@y
@p
(23)
where v is the velocity in Y direction.
Fig. B 15
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
50
Viscous flow through Rectangular Slot
By force balance in vertical direction,
@y
@p
2xb × ∆p = μ 2lb
or dv = −
¢s
q›
xdx
(24)
The negative sign indicates that v decreases as x increases.
Integrating,v = −
¢s p
q›
(25)
+C
From boundary conditions,
v = 0 when x = ±
From (25) and (26)
H
and C =
•=
¢‰ C
q•
¢s H
q› ¤
(26)
−¥
(27)
therefore the velocity distribution is parabolic and the max(at x=0) is given by,
vP
p
=
¢sH
¤q›
9 November 2022
(28)
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
51
Viscous flow through Rectangular Slot..
Since the velocity distribution is parabolic,
Average Velocity, v
y¦
= vP
p
=
¢sH
q›
The flow the fluid through the slot can be given as,
Q=v
y¦
9 November 2022
× Area =
¢sH
q›
× bh =
¢s H
q›
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
(29)
52
Methods of Journal Bearing Design
Mainly there are two methods for journal bearing design, they are,
(a) Method developed by M. D. Hersey
This method is based on dimensional analysis, applied to an infinitely long bearing. Analysis
incorporates a side-flow correction factor obtained from the experiment of S. A. and T. R. McKee.
McKee equation for coefficient of friction, for full bearing is given by,
Coefficient of friction, f = K
Where, p =
˜
™ šz @ F
=
@
š
˜
›
qϥ
s
+K
(30)
,
l = length of bearing,
d = diameter of journal = 2r ,
nQ = speed of the journal
μ = absolute viscosity of the lubricant,
c = difference bush and journal diameter(Clearance),
K = side-flow factor = 0.002 for (l/d) 0.75-2.8
K = Constant dependent on the system of units
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
53
Method developed by M. D. Hersey
The steps to be followed are,
Step 1: Basic design parameters are provided by the designer from the operating
conditions.
• Bearing load (W)
• Journal diameter (d)
• Journal speed (nQ )
Step 2: Depending upon type of application, selected design parameters are
obtained from a design handbook, these are,
• l/d ratio
• Bearing pressure(p)
• c/d ratio
• Proper lubricant and an operating temperature
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
54
Method developed by M. D. Hersey
Step 3: Value of
qϥ
s
should be within the design zone.
Equation (30) is used to compute f. Heat generation and heat
dissipation are computed to check for thermal equilibrium.
• Iteration with selected parameters is required if thermal equilibrium
is not established.
• Provision for external cooling is required if it is difficult to achieve
thermal equilibrium.
This method described here is relatively old. The second method is
more popular which is developed by (b) A. A. Raimondi and J. Boyd.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
55
Method by A. A. Raimondi and J. Boyd
This method is based on hydrodynamic theory. Reynolds’ Equation in
simplified form is given in equation (15) as,
0
0p
H
rs
q rp
0
+
0t
H
rs
q rt
=
G rH
rp
There is no general analytical solution to Reynold’s equation (15). One
of the important solutions is due to Sommerfeld and is expressed as,
š
f=φ
qϥ
s
š
Where S=
š
=φS
qϥ
s
(31)
is Smmerfeld’s Number (dimensionless) and
φ gives the functional relationship.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
56
Method by A. A. Raimondi and J. Boyd
The Sommerfeld number is helpful to the designers, because it includes
design parameters; bearing dimensions r and c , friction f , viscosity μ,
speed of rotation Υ and bearing pressure p. But it does not include
the bearing arc. Therefore the functional relationship can be obtained
for bearings with different arcs, say 360 , 60 etc.
Raimondi and Boyd (1958) gave a methodology for computer–aided
solution of Reynolds equation using an iterative technique. For l/d
ratios of 1, 1:2 and 1:4 and for bearing angles of 360 to 60 extensive
design data are available.
Charts have been prepared by Raimondi and Boyd for various design
parameters, in dimensionless form, and are plotted with respect to
Sommerfeld’s number.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
57
Charts
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
58
Method by A. A. Raimondi and J. Boyd
The design parameters which are given by
Raimondi and Boyd are as follows,
ℎª
=Minimum oil film thickness
h
= Film thickness at any other point
c
= radial clearance
ℎª ⁄7
= Minimum film thickness variable
r⁄c f
= Coefficient of friction variable
Q/ rcnQ l = Flow
= Flow ratio
Q Q ⁄Q
p⁄pP p = Maximum film pressure ratio
= Terminating position of film (deg)
θs«
θH«
= Minimum film thickness position
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
Fig. B 16
59
Method by A. A. Raimondi and J. Boyd
In the figures shown the center of the journal is at O and the center of the bearing at O′.
The distance between these centers is the eccentricity and is denoted by e. Eccentricity
ratio is denoted by ϵ and defined as,
ϵ=
(32)
š
• The bearing shown in the figure is known as a partial
bearing.
• If the radius of the bushing is the same as the
radius of the journal, it is known as a fitted bearing.
• If the bushing encloses the journal, as indicated by
the dashed lines, it becomes a full bearing.
• The angle β describes the angular length of a partial
bearing. For example, a 120◦ partial bearing has the
angle β equal to 120◦. For full bearing β equal to 360◦
Fig. B 17
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
60
The above design parameters are defined in the figure on the previous
two slides. The pressure profile shown is only for the positive part of
the bearing where converging zone is present. Negative part has not
been shown because it is not of use.
(33)
From the figures , c = R − r & R = e + r + h
from equation (32) and (33),
c = R − r = ¯ + ℎª = 7° + ℎª
Or, 7 1 − ° = ℎª ∴ ° = 1 −
where the quantity
9 November 2022
C±
ƒ
C±
ƒ
is known as minimum film thickness variable.
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
61
From equation (31) the Sommerfeld number is given by, S=
š
qϥ
s
.
Angle φ shown in Fig. B 16 is known as angle of eccentricity or angle of
attitude, it gives the position of minimum film thickness with respect to the
direction of load. The values of φ can be obtained from table (in degrees).
The coefficient Friction variable(CFV)= r⁄c f and The frictional torque as given
by equation (20), (–— )= fWr N-mm
The frictional power, kW ² = 2πnQ fWr 10³ kW
(34)
The flow variable, (FV)=Q/ rcnQ l , where, Q=flow of lubricant in mm ⁄s
A part of the lubricant escapes through side leakage which can be calculated as
flow rate =Q Q ⁄Q given in the table.
The maximum pressure pP p is calculated from the Maximum film pressure
ratio (p⁄pP p ).
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
62
Heat generation
The heat generated in the bearing is given by,
H¦ = fWrv = 2πnQ fWr 10³
kW or kJ/s, where v=rubbing velocity
š
CFV and W = 2plr
in the above equation,
Substituting,
f=
H¦ = 4π 10³ rcnQ lp CFV
Heat carried by the lubricant is given by, Hš = mCs ∆t
Where, m= mass of the oil pushing through the bearing in kg/s
µ‰ = specific heat capacity of oil kJ/kgK & ∆¶ = temperature rise (K),
· = ¸¹ 10³ kg/s, substituting Q = rcnQ l FV
· = ¸ rcnQ l FV 10³ kg/s.
Putting this value in the expression of heat carried, Hš = Cs ∆tρ rcnQ l FV 10³ .
Equating , H¦ = Hš ,
∆t=
ºs
{ »
¼
¼
Therefore, ∆t=
9 November 2022
. In most cases ρ = 0.86 and Cs = 1.76 kJ/kgK
¤. s
¼
¼
and Average Temperature, T
y
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
= T½ +
∆t
(35)
63
The Dimensionless performance
Parameters for a full journal bearing with
side flow is given in the table. The values
of the table is based on the assumption
that that lubricant is supplied at atmospheric
pressure.
1. Once l/d ratio is determined other
parameters can be estimated. When,
a) when(l/d) ratio is more than 1, the bearing
is called ‘long’ bearing.
a) when(l/d) ratio is less than 1, the bearing
is called ‘short’ bearing.
a) when(l/d) ratio is equal to 1, the bearing
is called ‘square’ bearing.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
64
2.Unit bearing pressure
Unit bearing pressure is the load per unit projected area of the bearing.
It depends on a number of factors, such as:
a) Bearing material
b) Operating temperature
c) Nature & frequency load
Depending on experiments the bearing
Pressure is tabulated as given.
3. Start-up load: The unit bearing pressure for
Starting condition should not be more than 2N/mm
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
65
4. Radial clearance: Ideally speaking, the radial clearance should be small.
However, radial clearance can be given by, c = 0.001 r.
5. Minimum oil film thickness: There is a limit of minimum oil film thickness,
below which there remains a possibility breaking of Hydrodynamic film and
metal to metal to contact, which is given by, h = 0.0002 r.
6. Maximum oil film temperature: Oil will oxidise when the operating
temperature increases beyond 120 C. The babbitt material softens at
125 C, for pressure of 7N/mm and at 190 C for 1.4N/mm . Practical
value is limited to 90 C.
Note: Bearings can be designed for (i) maximum load carrying capacity & (ii)
minimum friction. The optimum values can be obtained from the table
given.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
66
The Table for radial clearance
9 November 2022
The Table for minimum oil film thickness
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
67
Materials for bearing
The common materials used for bearings are listed below.
• Lead based babbits :
around 85 % Lead; rest are tin, antimony and copper. Pressure rating ≤ 14MPa
• Tin based babbits :
around 90% tin; rest are copper, antimony and lead. Pressure rating ≤ 14MPa
• Phosphor bronze :
major composition copper; rest is tin, lead, phosphorus. Pressure rating ≤ 14MPa
• Gun metal
: major composition copper; rest is tin and zinc. Pressure rating ≤ 10MPa
• Cast iron : pressure rating ≤ 3.5 MPa
Other materials commonly used are, silver, carbon-graphite, teflon etc.
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
68
Example
1.
A machine journal bearing has a journal diameter of 150 mm and length of 120mm. The
bearing diameter is 150.24 mm. It is operating with SAE 40 oil at ÀÁÂC. The shaft is carrying a
load of 8 kN and rotates at 960 rpm. Estimate the bearing coefficient of friction and power
loss using Petroff’s equation.
Solution: From the given condition, 2r = 0.15m; 2R =0.15024m;
l = 0.12 m; W=8kN; nQ = 960/60 = 16 ; c = (R-r) = 0.00012 m;
p = W/dl = 8000/ (150x 120) = 0.44 Mpa
Viscosity of SAE 40 at 65 C, μ = 30 mPa.s = 30x10³ Ns/m
qϥ
. Ã
×30x w
f = 2π
= 2π
= 0.0134
š
s
.
0.44x
(From equation 21)
frictional torque (T² )= fWr = 0.0134× 8000 × 0.075 = 8.067 Nm
Angular speed, ω = 2πnQ = 2π × 16 = 100.48 rad⁄s
Power loss=T² × ω = 8.067 × 100.48 = 811watt
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
69
Example
2. A journal of a stationary oil engine is 80 mm in diameter. and 40 mm long. The radial clearance
is 0.060mm. It supports a load of 9 kN when the shaft is rotating at 3600 rpm. The bearing is
lubricated with SAE 40 oil supplied at atmospheric pressure and average operating temperature is
about 65 C. Using tables analyze the bearing assuming that it is working under steady state
condition.
Solution: From the given data,
d = 80 mm; l =40 mm; c = 0.06 mm; W = 9kN; nQ = 3600rpm = 60 rps; SAE 40 oil at 65 C;
Step 1. p= W / ld = 9 x1000 /(40 x 80) = 2.813 Mpa
Step 2. From the table μ = 30 cP at 65 C for SAE 40 oil.
Step 3. Sommerfeld Number S=
qϥ
s
š
=
×
.
wÅ ×
.¤
= 0.284
Step 4. For S = 0.284 and l/d = ½, from the table(After interpolating) ho /c = 0.38 and ε = e /c = 0.62 .
therefore, ho = 0.38xc = 0.382x 0.06=0.023mm and
Step 5.
š
9 November 2022
e = 0.62 x c = 0.62 x 0.06 = 0.037 mm.
•
Æ
f = 7.5, for S = 0.284 for = ½ from Table. Therefore, f = 7.5
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
š
= 7.5
.
= 0.0113
70
Step 6. From table φ= 46 , for S=0.284 and l/d=1/2
Step 7. Similarly from the same table, the following can be obtained,
Q/ rcnQ l =4.9 and therefore,
Q = 4.9 x 0.04 x 0.00006 x 60 x 0.04= 2.82 × 10³ m ⁄s = 28.2 cm ⁄s
QQ ⁄Q = 0.75 and hence,
QQ =0.75 x 28.2 = 21.2 cm ⁄s
p⁄pP p=0.36
∴ pP p= p⁄0.36= 2.813 / 0.36 = 7.8 MPa
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
71
Example
3. The following data are related to a 360 deg hydrodynamic bearing:
Radial load= 3.2 kN
Journal speed= 1490 rpm
Journal Diameter= 50 mm
Bearing length= 50 mm
Radial clearance= 0.05mm
Viscosity of lubricant=25cP
Assuming that the total heat generated is carried by the lubricant calculate;
(i) Coefficient of friction
(ii) Power lost in friction
(iii) Minimum oil film thickness
(iv) Flow in lit/min
(v) Temperature rise
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
72
Given that, W=3.2kN, Υ =1490/60=25, d=50mm, l=50mm, c=0.05mm
and μ = 25cP
p = (W⁄ld)= 3.2 × 1000 ⁄ 50 × 50 = 1.28 N⁄mm
qϥ
s
Sommerfeld Number S=
=
š
.
‘ ⁄Ç = 50⁄50 = 1
From the Table,
r
h
Q
f = 3.22;
= 0.04;
= 4.33
c
c
rcnQ l
7
0.05
∴„=
= 3.22
= 0.00644
•
25
9 November 2022
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
×
wÅ ×
. ¤
= 0.121
73
Power lost in friction,
2’ × 25 × 0.00644 × 3.2 × 1000 × 25
ÉÊ — = 2’Œ• „Ê• ×
=
= 0.08
10
Minimum film thickness, ℎª = 0.4 × 7 = 4 × 0.05 = 0.02··
10³
Flow,
Q = 4.33rcnQ l = 4.33 × 25 × 0.05 × 25 = 6720.5 mm ⁄s = 6720.5 × 10³ × 60
= 0.403 litre⁄min
Temperature rise, ∆t =
=
¤. × . ¤× .
.
9 November 2022
¼
¤. s
¼
[ from equation (35)]
= 7.9 C
Professor Nilotpal Banerjee, Department of Mechanical
Engineering, NIT Durgapur
74