DME - Unit 5.pptx

ME 8593-DESIGN OF
MACHINE ELEMENTS
DEPARTMENT OF MECHANICAL ENGINEERING

COURSE OBJECTIVES
• To familiarize the various steps involved in the
Design Process
• To understand the principles involved in
evaluating the shape and dimensions of a
component to satisfy functional and strength
requirements.
• To learn to use standard practices and standard
data.
• To learn to use catalogues and standard machine
components.
(Use of P S G Design Data Book is permitted)

TEXT BOOKS:
• 1. Bhandari V, “Design of Machine Elements”, 3rd Edition, Tata McGraw-Hill Book Co, 2010.
• 2. Joseph Shigley, Charles Mischke, Richard Budynas and Keith Nisbett “Mechanical
• Engineering Design”, 8th Edition, Tata McGraw-Hill, 2008.
REFERENCES:
• 1. Sundararajamoorthy T. V. Shanmugam .N, “Machine Design”, Anuradha Publications,
Chennai, 2003.
• 2. Robert C. Juvinall and Kurt M. Marshek, “Fundamentals of Machine Design”, 4th Edition,
Wiley, 2005
• 3. Alfred Hall, Halowenko, A and Laughlin, H., “Machine Design”, Tata McGraw-Hill
BookCo.(Schaum’s Outline), 2010
• 4. Bernard Hamrock, Steven Schmid,Bo Jacobson, “Fundamentals of Machine Elements”,2nd
Edition, Tata McGraw-Hill Book Co., 2006.
• 5. Orthwein W, “Machine Component Design”, Jaico Publishing Co, 2003.
• 6. Ansel Ugural, “Mechanical Design – An Integral Approach", 1st Edition, Tata McGraw-Hill
Book Co, 2003.
• 7. Merhyle F. Spotts, Terry E. Shoup and Lee E. Hornberger, “Design of Machine Elements” 8th
Edition, Printice Hall, 2003.

Units
• UNIT I : STEADY STRESSES AND VARIABLE
STRESSES IN MACHINE MEMBERS
• UNIT II : SHAFTS AND COUPLINGS
• UNIT III : TEMPORARY AND PERMANENT
JOINTS
– Weld Joints, Riveted Joints, Knuckle joints
• UNIT IV : ENERGY STORING ELEMENTS AND
ENGINE COMPONENTS
– Spring, Connecting rod, Flywheel
• UNIT V : BEARINGS

Unit 1
• Introduction to the design process

Thought bites
Is an Engine, a Machine?

Is an Engine, a Machine?
• All engines can be called machines, but not
all machines can be called engines.
• Engine is basically a prime mover which
generates power using some fuel i.e. diesel,
petrol etc. A machine needs power to do work
which must be created by hand, engine or
electric motor. Engine could be a component
of machine.

What is Machine elements?
What are the parts in an engine?

Machine elements
(Day today Life examples)

Machine Design
• Machine design is defined as the use of
scientific principles, technical information
& imagination in the description of a
machine or a mechanical system to
perform specific functions with maximum
economy & efficiency.
• Machine Design is defined as the creation
of new design (Machines) or improving the
exist one.

•Mathematics
•Engineering Mechanics
•Strength of Materials
e
e
g
• Math matics
• Engin ering Mechanics
• Stren th of Materials
• Workshop Processes
• Engineering Drawing
What is the basic knowledge required for Machine Design?
•Mathematics
•Engineering Mechanics
•Strength of Materials
•Workshop Processes
•Engineering Drawing
• Mechanics of Machines
• Mechanics of Materials
• Fluid Mechanics & Thermodynamics
17

4 C’s in Design Process
• Creativity
• Complexity
• Choice
• Compromise

Classifications of Machine Design
1. Adaptive design (Old design)
2. Development design (Modification in old design)
3. New design (Creating a new design)
a. Rational Design (Mathematical formulae)
b. Empirical design (Empirical formulae – Practice & Past
Experience)
c. Industrial design (Production aspect)
d. Optimum design (Best design)
e. System design
f. Element design
g. Computer Aided design

Basic Requirement of Machine Element
(DESIGN CONSIDERATIONS IN MACHINE DESIGN)
• Strength
• Type of Load and stresses
• Rigidity
• Maintenance
• Flexibility
• Size and shape
• Stiffness
• Reliability
• Kinematics of machine
• Safety of operation
• Weight
• Manufacturing considerations
• Selection of Materials
• Corrosion of Materials
• Friction and wear
• Frictional resistance and lubrication
• Life
• Assembly considerations
• Conformance to standards
• Vibrations
• Thermal considerations
• Workshop facilities
• Ergonomics
• Aesthetics
• Cost
• Noise
• Environmental factors

General procedure in Machine Design
Detailed drawing
Need or aim
Synthesis
Analysis of the FORCES
Material selection
Design of elements
Recognize and specify the problem
Select the mechanism that would give the desired
motion and form the basic model with a sketch etc
Determine the stresses and thereby the sizes of
components s.t. failure or deformation does not
occur
Modify sizes to ease construction & reduce overall cost
Modification
Production

Procedure of Design of Machine Elements

Material Selection
• The best material is one which will serve the
desired purpose at minimum costs
• Factors Considered while selecting the Material
– Availability
– Cost
– Mechanical properties:
– Manufacturing considerations – Shaping, Machining,
Joinimg, surface finishing, FoS, Assembly cost

Factor of safety
• Is used to provide a design margin over the
theoretical design capacity to allow for
uncertainty in the design process.
– In the calculations,
– Material strengths,
– Manufacturing process
• FoS = Strength of the component (Max load)
Load on the component (Actual load)

Unit 5 - Bearings
Syllabus:
• Sliding contact and rolling contact bearings -
Hydrodynamic journal bearings, Sommerfeld
Number, Raimondi and Boyd graphs, -
Selection of Rolling Contact bearings.

What are Bearings?
• A bearing is a device to permit constrained
relative motion b/w two parts typically
rotation or linear movement.
• Bearing may be classified broadly according to
the motion they allow and according to their
principle of operation.
Bear – கரடி, தாங்கு.

In a ball bearing, the load is transmitted from
the outer race to the ball, and from the ball to
the inner race. Since the ball is a sphere, it only
contacts the inner and outer race at a very
small point, which helps it spin very smoothly.
But it also means that there is not very much
contact area holding that load, so if the bearing
is overloaded, the balls can deform or squish,
ruining the bearing.
Roller bearings like the one illustrated above are
used in applications like conveyer belt rollers,
where they must hold heavy radial loads. In these
bearings, the roller is a cylinder, so the contact
between the inner and outer race is not a point but
a line. This spreads the load out over a larger area,
allowing the bearing to handle much greater loads
than a ball bearing. However, this type of bearing is
not designed to handle much thrust loading.

Bearing alloys
A bearing is a device to allow
constrained relative motion between
two parts, typically rotation or linear
movement.
Bearings may be classified broadly
according to the motions they allow and
according to
their principle of operation as well as by
the directions of applied loads they can
handle.

A bearing is a device to allow constrained relative motion
between two parts, typically rotation or linear movement.
Bearings may be classified broadly according to the motions they
allow and according to their principle of operation as well as by
the directions of applied loads they can handle.
Outer surface
Inner surface
ball

https://www.youtube.com/watch?v=_cN1E8K
Uvmk

Bearing materials
1. White metals
2. Cu-base alloys
3. Al- base alloys
4. Plastic materials
5. Ceramics
Lead base
Tin base (Babbit Metals – after Issac babbit)
Teflons
Nylons
Sb 10%, Sn 82%, Cu 4%, Pb 4% - automotive industries
Sb 13%, Sn 12%, Cu 0.75%, As 0.25% Pb- 74%
Plain tin Bronze, Phosphor bronze, Leaded bronze,
Sintered bronze
Good Load bearing capacity – Aero engines, Automobiles,
Domestic equipment
Sn 7%, Cu 1.3%, Ni 1.3%, Balance Al - Automobiles
Where Oil lubrication is established
Alumina – Large speed precision

Function of bearing
• The main function of rotating shaft (Journal) is
to transmit power from one end of the line to
the other.
• Supports the load.
• It needs a good support to ensure stability
and frictionless rotation. The support for the
shaft is know as “Bearing”.
• https://gfycat.com/yearlyshoddyarmyworm

Types of Bearings
• Based on direction of Load:
– Radial bearing
– Thrust/Axial bearings
– Combined bearing
• Based on Nature of contact
– Sliding contact
– Rolling contact

Types of bearing
• Sliding contact bearing or Plain bearing or Journal bearing
or Sleeve bearing
– Hydrodynamic bearing (Thick film bearing)
– Thin film bearing (Boundary lubricated bearings)
– Hydrostatic bearing (Externally pressurized lubricated bearing)
• Rolling contact bearing or Anti-friction bearing or simply
ball bearing:-
– (1)Deep groove ball bearing
– (2)Cylinder roller bearing
– (3)Angular contact bearing
– (4)Taper roller bearing
– (5)Self aligning bearing

Unit 5
Bearings
Hydrodynamic
/Journal
Bearings
Roller/Ball
Bearings
Design of
Journal
Bearing
(Hd, Hg,
m)
Finding out other
parameters using
Somerfield’s
number and
Raimondi and
Boyd’s curve
(β S, h0 )
Selection of
Ball Bearing
(Bearing SKF
no.) –
Indirect
method
Life related
problems on
Ball/Roller
Bearing –
Direct method

Types of Sliding Contact Bearing
(Based on sliding action)
When the angle of
contact of the bearing
with the journal is
360° as shown in (a),
then the bearing is
called a full journal
bearing.
When the angle of
contact of the bearing
with the journal is 120°,
as shown in Fig (b), then
the bearing is said to be
partial journal bearing.
the diameters of the
journal and bearing
are equal, then the
bearing is called a
fitted bearing, as
shown in Fig. (c).

1. Sliding contact bearing
• In these bearing load is transferred though a
thin film of lubricant coils (oils).

1.1 Hydrodynamics bearing
• A journal bearing , in its simplest
form is a cylinder bushing made of
a suitable material and containing
property machine inside and
outside diameters. The journal is
usually the part of a shaft or pins
that rotates inside the bearing.
• Its handle high load and velocity
because metal to metal contact is
minimal due to the oil films.
• They are require large supply of
lubrication oil.

1.2 Hydrostatic bearing
• Hydrostatic bearings are externally
pressurized fluid bearings, where the fluid is
usually oil, water or air, and the pressurization
is done by a pump.

2. Rolling contact bearing
• A load is transfer though rolling elements such
as balls straight and tapered cylinders and
spherical rollers.
• The designer must deal with such matter as
fatigue, friction, heat , lubrication etc.

Babbit metal
Tin base babbits : Tin 90% ; Copper 4.5% ; Antimony 5% ; Lead 0.5%.
Lead base babbits : Lead 84% ; Tin 6% ; Anitmony 9.5% ; Copper 0.5%.
Bronzes.
The gun metal (Copper 88% ; Tin 10% ; Zinc 2%) is used for high grade bearings
subjected to high pressures (not more than 10 N/mm2 of projected area) and high
speeds.
The phosphor bronze (Copper 80% ; Tin 10% ; Lead 9% ; Phosphorus 1%) is used for
bearings subjected to very high pressures (not more than 14 N/mm2 of projected area)
and speeds.
Cast iron. The cast iron bearings are usually used with steel journals. Such type of
bearings are fairly successful where lubrication is adequate and the pressure is limited
to 3.5 N/mm2 and speed
to 40 metres per minute.
Silver. The silver and silver lead bearings are mostly used in aircraft engines where the
fatigue strength is the most important consideration.
Non-metallic bearings. The various non-metallic bearings are made of carbon-graphite,
rubber, wood and plastics. The carbon-graphite bearings are self lubricating,
dimensionally stable over a wide range of operating conditions,
Materials used for Sliding Contact Bearings

Assumptions in Hydrodynamic
bearings
• Obeys newton law of viscous flow
– Relationship between the shear stress and shear rate
of a fluid subjected to a mechanical stress. The ratio
of shear stress to shear rate is a constant, for a given
temperature and pressure.
• Pressure is constant through out the film
thickness
• Lubricant is incompressible
• Viscosity is constant
• Flow is one dimensional – Side leakage is
neglected

Wedge film formation in
Hydrodynamic bearing
• In fully hydrodynamic (or "full-film") lubrication,
the moving surface of the journal is completely
separated from the bearing surface by a very
thin film of lubricant (as little as 0.0001" with
isotropic-superfinished {ISF} surfaces). The
applied load causes the centerline of the
journal to be displaced from the centerline of
the bearing. This eccentricity creates a circular
"wedge" in the clearance space.
• The lubricant, by virtue of its viscosity, clings to
the surface of the rotating journal, and is drawn
into the wedge, creating a very high pressure
(sometimes in excess of 6,000 psi), which acts to
separate the journal from the bearing to support
the applied load.

Terms used in Hydrodynamic Journal
Bearing

PSG databook Pg no: 7.30
(Hydrodynamic bearing starts)

Terms used in Hydrodynamic Journal
Bearing
C = Db - D
D
Db - D

Coefficient of Friction for Journal Bearings
By McKee
By Petroff’s equation or Petroff's law
The Petroff’s equation and McKee's equations are employed for lightly loaded bearings.

Critical Pressure of the Journal Bearing
Sommerfeld Number
The Sommerfeld number is also a dimensionless parameter used extensively in
the design of journal bearings , Mathematically

Heat Generated in a Journal Bearing
Heat Dissipated in a Journal Bearing
Heat generated < Heat dissipated

Design Procedure of Journal Bearing
1. Diameter of journal/Shaft (Use P= 2NT/60 & T =  D^3 /16)
2. Select L/D ratio from PSG databook 7.31
3. Find Clearance from C/D ratio - PSG databook 7.32
4. Bearing Pressure P=W/LD (Check at PSG databook 7.31 – If not satisfied
change L/D ratio)
5. Absolute viscosity, Z (PSG databook 7.31)
6. Lubricant oil used and it’s viscosity (PSG databook 7.41)
7. Bearing characteristics number (Zn/P)
8. Film thickness (if needed)
9. Coefficient of Friction (From Mckee’s Equation - PSG databook 7.34)
10. Materials used
11. Additional cooling system required or not
12. Somerfield number (if needed)
13. Heat generated (PSG databook 7.34) and Heat dissipation (Hd = q A Δt)
14. Mass flow rate (Qt = m Cp Δt)

Unit Conversion
Solving DME problems be like:

Problems
• Design a journal bearing for a centrifugal
pump running at 1440 rpm. Dia of journal is
10cm and the load on each bearing is 2000kg.
The factor (Zn/p) may be taken as 2800 for
pump bearings. Assume Atm temp as 30 ͦC.
Operating temperature as 75 ͦC. Energy
dissipation coefficient as 1250 W/m2/ ͦC. C/R =
0.001, L/D = 1.5.

Design a journal bearing for a centrifugal pump running at 1440 ypm. Dia of journal is
10cm and the load on each bearing is 2000kg. The factor (Zn/p) may be taken as 2800
for pump bearings. Assume Atm temp as 30 ͦC. Operating temperature as 75 ͦC. Energy
dissipation coefficient as 1250 W/m2/ ͦC. C/R = 0.001, L/D = 1.5.
• Given:
– N (Pump Speed) = 1440 rpm
– D (Dia of journal)= 10 cm
– W (Load) = 2000 kg
– Zn/P (Factor) = 2800
– Atm temp, ta= 30 ͦC
– Operating temperature, top= 75 ͦC
– Energy dissipation coefficient,
C (q) = 1250 W/m2/ ͦC
– C/R (Clearance to Radius ratio)= 0.001
– L/D (Length to dia of journal ratio)= 1.5
• To Find:
Design a
Journal
bearing

• Solution:
C/R = 0.001
C = 0.001 * R
C= 0.001 * (D/2)
C= 0.001 * (10/2)
C= 0.005 cm = 0.05 mm
C = 0.05 mm
Diametrical Clearance = 0.05 mm
Given: N (Pump Speed) = 1440 rpm D (Dia of journal)= 10 cm, W (Load) = 2000 kg,
Atm temp, ta= 30 ͦC, Operating temperature, top= 75 ͦC, Energy dissipation coefficient C
(q) = 1250 W/m2/ ͦC, C/R (Clearance to Radius ratio)= 0.001, L/D (Length to dia of
journal ratio)= 1.5
Note : Convert
cm to mm (x10)

• Solution:
Bearing Pressure, P = W/LD
= 2000kg / L*100
L/D = 1.5  L = 1.5 * 100 = 150 mm
Now, Bearing Pressure, P = 2000 kg /150 * 100 mm2
journal ratio)= 1.5
Note : Convert
mm to cm (/10)

Kg is the unit of mass.
Kgf is a unit of force (Obsolete)

• Solution:
Bearing Pressure, P = W/LP
= 2000kg / L*100
L/D = 1.5  L = 1.5 * 100 = 150 mm
P = 2000 kg / 15 * 10 cm2
Bearing Pressure, P = 13.33 kgf/cm2
(OR)
P = 2000 * 10 N / 150 * 100 mm2
P = 1.33 N/mm2
journal ratio)= 1.5

• Solution:
Bearing Pressure, P = W/LP
= 2000kg / L*100
L/D = 1.5  L = 1.5 * 100 = 150 mm
P = 2000 kg / 15 * 10 cm2
Bearing Pressure, P = 13.33 kgf/cm2
From PSG DATABOOK Pg no 7.31, For Centrifugal pump the
bearing pressure allowable is 7 – 14 Kgf/cm2, For the
centrifugal pump given the P = 13.33 kgf/cm2. Hence the
design is safe.
journal ratio)= 1.5

• Solution:
Zn/P = 2800 (Given)
Absolute viscosity of the oil, Z = 2800 * 13.33 / 1440
Z = 25.9 ≈ 26 CP (Centi Poise)
[Poise - A unit of dynamic viscosity]
From PSG Databook Pg No 7.41
journal ratio)= 1.5

• Solution:
Zn/P = 2800 (Given)
Z = 2800 * 13.33 / 1440
Absolute viscosity, Z = 25.9 ≈ 26 CP (CentiPoise)
[Poise - A unit of dynamic viscosity]
From PSG Databook Pg No 7.41, For 26CP and
Operating temperature top= 75 ͦC, the oil of viscosity
grade SAE 40 is selected for lubrication.
journal ratio)= 1.5

• Solution:
The coefficient of Friction, μ
The equation is taken from PSG databook Pg no
7.34 – McKEES equation
μ = 33.25/10^10 (26 CP* 1440 rpm / 13.33 kgf/cm2) *
(10cm/0.05mm) + k
(Since μ has no unit, every units must be
converted in to same unit to cancel all units)
journal ratio)= 1.5

• Solution:
μ = 33.25/10^10 (26 CP* 1440 rpm / 13.33 kgf/cm2) *
(10cm/0.05mm) + k
(Since μ has no unit, every units must be
converted in to same unit to cancel all units)
μ = 33.25/10^10 (26/100P* 1440 rpm / 1.333N/mm2)
* (10cm/0.05/10cm) + k
journal ratio)= 1.5
1 Poise = 100 CP , 1 CP = 0.01 Poise

• Solution:
μ = 33.25/10^10 (26/100 P* 1440 rpm / 13.33
kgf/cm2) * (10cm/0.05/10cm) + k
From PSG databook pg no 7.34, the graph
between L/D and k, for L/D= 1.5 vs curve, the k
obtained is 0.0025
journal ratio)= 1.5

• Solution:
μ = 33.25/10^10 (26/100 P* 1440 rpm / 1.333) *
(10cm/0.05/10cm) + k
μ = 33.25/10^10 (26/100 P* 1440 rpm / 1.333) *
(10cm/0.05/10cm) + 0.0025
The coefficient of Friction, μ = 0.0026
journal ratio)= 1.5
Note: 1 poise = 100centipoise

• Solution:
Heat generated, Hg = μ . W . v
[From PSG Databook Pg. No 7.34]
Hg = μ . W . v Kgf m/min
= 0.0026 * (2000) * (π D N)
= 8.8* π * 10cm * 1440
= 8.8* π * 10/100 m * 1440
Heat generated, Hg = 1,267.20 Kgf m/min
journal ratio)= 1.5

• Solution:
Heat generated, Hg = μ . W . v
[From PSG Databook Pg. No 7.34]
Hg = μ . W . v Kgf m/min (Or Watt)
= 0.0026 * (2000*10) * (π D N/60)
= 52* π * 10cm * 1440/60
= 52* π * 10/100 m * 1440/60
Heat generated, Hg = 392.07 W
journal ratio)= 1.5
Always use Watt calculation only
for such Heat generation and
dissipation problem

Solution:
Heat dissipation, Hd = q A Δt Watt
Δt = ½ (top - ta) = ½ (75-30) = 22.5
A = L x D = 15 * 10
Hd = 1250 * 15/100 * 10/100 * 22.5
Heat dissipation, Hd = 421.875 W
Heat to be removed = Hg – Hd
= 663.50 – 392.07
Heat to be removed = 271.43 W
Inference : To remove this heat, an artificial cooling system
arrangement is needed.
journal ratio)= 1.5

Solution(If mass flow rate is separately asked):
Heat added or Heat that has to be removed, Qt = Hg - Hd
Qt = 241.625
Qt = m Cp Δt
m = Qt / Cp Δt
m = 271.43 / 2000* 22.5
= 0.00603 kg/s
Mass flow rate, m =0.3618 Kg/min
journal ratio)= 1.5
Cp = 1840 to 2100
J/Kg/C

Results
1. Diametrical Clearance = 0.05 mm
2. Bearing Pressure, P = 13.33 kgf/cm2
3. Absolute viscosity, Z = 25.9 ≈ 26 CP (CentiPoise)
4. Oil of viscosity grade SAE 40 is selected for lubrication.
5. The coefficient of Friction, μ = 0.0026
6. Heat generated, Hg = 392.07 W
7. Heat dissipation, Hd = 421.875 W
8. Since Hd < Hg ,To remove this excess heat, an artificial
cooling system arrangement is needed.
9. Mass flow rate, m =0.3618 Kg/min

Assignment
• Design a journal bearing for centrifugal pump.
Dia of journal = 75mm
• Load on journal = 11500 N
• Speed of journal = 1140rpm
• Operating temp = 65 C
If length or L/D ratio is not given we need
to assume from PSG databook 7.31
Assume clearance from PSG databook 7.32

Given Data
• Dia of journal, D = 75mm
• Load on journal, W= 11500 N
• Speed of journal, N = 1140rpm
• top = 65 C
To Find:
• Design the Journal bearing

Dia of journal, D = 75mm
Load on journal, W= 11500N
Speed of journal, N = 1140rpm, ta = 65 C
Assume L/D = 1.5
[From PSG databook Pg No 7.31, for centrifugal
pump L/D ratio is 1 – 2]
L = 1.5 * 75 = 112.5 mm
L = 11.25 cm
D = 7.5 cm

• P = W/ L D
= 11500 / 112.5* 75
P = 1.363 N/mm2
OR
P = 1150 kg / 11.25*7.5 cm2
P = 13.63 kg/cm2
Design is safe from PSG databook pg no 7.31
allowable bearing pressure 7- 14kg/cm2

• Zn/P = 2844.5 [PSG databook 7.31]
• Z = 2844.5 * 13.63 / 1140 = 34 CP
• From PSG databook 7.41, Opt temp = 65 C and
Absolute viscosity = 34 CP, SAE 40 grade is
selected.
Load on journal, W= 11500W

• Coefficient of Friction,
– μ = 33.25/10^10 (ZN/P) (D/C) + k
– For diametrical clearance, PSG databook Pg no :
7.32, for shaft dia 75mm and electric motor,
assume C = 75 (in microns).
– D/C – convert to same unit and cancel
– μ = 0.00945

• Heat generated, Hg = μ . W . v
Hg = 0.00945 * 11500 *ПDN
v = П (75/1000) 1140/60
Hg = 486.51 W
• Heat dissipated, Hd = (Δt + 18)2 L D / K
Δt = 65 -30 / 2 [Assume ambient temp]
Hd = 137.20 W (change L & D in ‘m’)

For K
• For finding out K in Hd
• From PSG databook Page no 7.35, 775 for light
construction
• Convert 775 as 0.755 (to make the answer in
Watt)
• Hd = 137.20 W

Assignment
• A journal bearing of 100mm dia and 151mm long
supports a radial load of 6kN. The shaft rotates at
560rpm. The diametrical clearance is 0.15mm. The
room temp is 25C and operating temp is 70C. The
bearing is well ventilated and no artificial cooling is
required. Suggest suitable oil to meet requirement.
• Hint : Find Hd and correlate with Hg beacause no
cooling required. Then substitute Hg value in this
formula Hg = μ . W . v to find μ. Then, find Z by using
McKees equation and find grade of Oil

Step for this problem
• Find Hd
• Hg = Hd
• Hg = μ . W . V  Find μ
• μ = 33.25/10^10 (ZN/P) (D/C) + k
• In graph, Z vs opt temp, find SAE oil grade

Problem
• Design a journal bearing to support a load of
7000 N at 700 rpm using a hardened steel
journal and bronze backed babbit bearing.
Room temp = 30 C, oil temp = 85 C.

Design a journal bearing to support a load of 7000 N at 700 rpm
using a hardened steel journal and bronze backed babbit
bearing. Room temp = 30 C, oil temp = 85 C.
• Given
– W = 7000 N
– N = 700 rpm
– ta = 30 C
– top = 85 C
– Bearing Material : bronze backed babbit bearing
– Journal Material : hardened steel journal

W = 7000 N, N = 700 rpm, ta = 30 C, top = 85 C
Bearing Material : bronze backed babbit bearing
Journal Material : hardened steel journal
• From PSG databook pg No 7.30, for heavy
babbit material, Motor is one of the
application. Hence it is assumed to be a
bearing of motor.
• Then from PSG databook pg no:7.31, for
motor, the L/D ratio is 1 – 2.
Assume L/D = 1
• Diameter of Journal not given, hence Assume
Dia of journal, D = 100 mm

W = 7000 N, N = 700 rpm, ta = 30 C, top = 85 C
• L/D = 1 & D = 100 mm
• Hence, L = 100 mm
• Bearing Pressure, P = W/LD
P = 7000 / 100*100
P = 0.7 N/mm2
• (OR) P = 700/10*10 = 7 Kgf/cm2
• From PSG databook Pg no: 7.31, allowable
bearing pressure, The design is safe.

• From PSG databook Pg no: 7.31, for motor the
Z = 25 CP.
• Hence Assume Z = 25CP
• Given operating temp = 85 C
• From PSG databook Graph Pg no: 7.41, SAE 40
grade oil is selected
W = 7000 N, N = 700 rpm, ta = 30 C, top = 85 C

• From PSG databook Pg no: 7.32, for motor the
C range is 50 to 100 for D = 90 mm
• Hence Assume C = 100 μm
• From PSG databook Pg no: 7.34, find μ
• Then find Hg and Hd (Assignment)
W = 7000 N, N = 700 rpm, ta = 30 C, top = 85 C

Problem
• Following data is given for a 360 ͦ
hydrodynamic bearing. Journal dia = 100mm,
Radial clearance = 0.12mm, Radial load=50kN,
Bearing length = 100mm, Journal speed =
1440rpm, Viscosity of lubricant is 16 CP,.
Calculate i) Min film thickness, ii) Coefficient
of friction, iii)Power lost in friction
Following data is given for a 360º hydrodynamic bearing : Radial load = 3.2 kN, Journal speed = 1490 rpm,
L/D ratio = 1, Unit bearing pressure = 1.3 Mpa, Radial clearance = 0.05 mm, Viscosity of the lubricant = 25
CP
Assuming that the total heat generated in the bearing is carried by the total oil flow in the bearing,
calculate (i) Journal diameter and bearing length, (ii) coefficient of friction, (iii) power lost in friction and
(iv) minimum oil film thickness. (APR/MAY 2019) (NOV/DEC 2020 AND April/May 2021)

Following data is given for a 360 ͦ hydrodynamic bearing. Journal dia =
100mm, Radial clearance = 0.12mm, Radial load=50kN, Bearing length =
100mm, Journal speed = 1440rpm, Viscosity of lubricant is 16 CP,. Calculate i)
Min film thickness, ii) Coefficient of friction, iii)Power lost in friction
Given:
• 360 ͦ hydrodynamic bearing , β = 360°
• Journal dia, D = 100mm
• Radial clearance, C/2= 0.12mm
• Radial load, W = 50,000N
• Bearing length, L = 100mm
• Journal speed, N = 1440 rpm
• Viscosity of lubricant, Z = 16 CP
To Find:
1. Min film thickness, h0
2. Coefficient of friction, μ
3. Power lost in friction, Hg

360 ͦ hydrodynamic bearing , β = 360°, Journal dia, D =
100mm, Radial clearance, C/2= 0.12mm, Radial load, W =
50,000N, Bearing length, L = 100mm, Journal speed, N = 1440
rpm, Viscosity of lubricant, Z = 16 CP
• Bearing Pressure, P = W/LD
= 50,000 / 100*100
P = 5 N/mm2
[Note : Since no application is given in question,
unable to find the design is safe or not. Also, the
question asked is not to design the bearing.
Hence we don’t need to check the safety of
bearing]

• From PSG databook pg. no 7.34,
• Somerfield number, S = Z n’/P (D/C)2
S = 16*10^-3 * 1440/60 rps / (5*10^5)
(100/0.24)2
S = 0.133

• From PSG databook pg. no 7.40, β = 360° & S
= 0.133, for L/D = 1
• Min film thickness 2h0 / C = ____
• h0 = _______ mm
• From PSG databook pg. no 7.40, β = 360° & S
= 0.133, for L/D = 1
• μD/C = 1  μ = 0.008
• Hg = μ W v (Assignment)

Assignment
• Following data is given for a 360º hydrodynamic bearing :
• Radial load = 3.2 kN
• Journal speed = 1490 rpm
• L/D ratio = 1
• Unit bearing pressure = 1.3 MPa
• Radial clearance = 0.05 mm
• Viscosity of the lubricant = 25 CP
• Assuming that the total heat generated in the bearing is carried by
the total oil flow in the bearing, calculate (i) Journal diameter and
bearing length, (ii) coefficient of friction, (iii) power lost in friction
and (iv) minimum oil film thickness. (APR/MAY 2019) (NOV/DEC
2020 AND April/May 2021)

QB 9
• The following data is given for a full hydrodynamic bearing used for
electric motor:
• Radial load = 1200 N
• Journal speed = 1440 rpm
• Journal diameter = 50 mm
• Static load on the bearing = 350 N
• The values of surface roughness of the journal and the bearing are
2 and 1 micron respectively. The minimum oil film thickness should
be five times the sum of surface roughness of the journal and the
bearings. Determine (i) length of the bearing; (ii) radial clearance;
(iii) minimum oil film thickness;(iv) viscosity of lubricant; and (v)
flow of lubricant. Select a suitable oil for this application assuming
the operating temperature as 65°C. (APR/MAY 2018)

Procedure for solving
Journal bearing Problem
Note : The continuity of the steps may be modified as
per the given data in the problem

Rolling contact bearing - Evolution

Parts of Single row deep groove Ball
bearing

Design of Ball Bearing
Pressure/ Equivalent
load
Capacity of the Bearing
Selection of the
Bearing more than the
Capacity
Life of the Bearing
Selection of the
Bearing (Assuming the
dia or some other way)
Pressure/ Equivalent
load
Capacity of the Bearing Safe or Unsafe
Life of the Bearing
Indirect
Approach
Direct Approach

Problem 1
• Select a single row angular contact ball
bearing to support a shaft of 50mm dia,
carrying an equivalent load of 8.2kN. Shaft
rotates at 1000 rpm and life of bearing should
be above 4000 hrs.

Select a single row angular contact ball bearing to support a shaft of 50mm
dia, carrying an equivalent load of 8.2kN. Shaft rotates at 1000 rpm and life of
bearing should be above 4000 hrs.
• Given:
– d = 50 mm
– P = 8.2 kN = 8200 N
– N = 1000rpm
– LH = 4000hrs
– Single row angular contact ball bearing
• To Find:
– Select suitable bearing (Bearing number)

d = 50 mm; P = 8.2 kN = 8200 N; N = 1000rpm; LH
= 4000hrs; Single row angular contact ball bearing
• Solution:
– From PSG databook pg no: 4.6, for the
corresponding speed N = 1000rpm & Life, LH =
4000hrs

d = 50 mm; P = 8200 N; N = 1000rpm; LH = 4000hrs;
Single row angular contact ball bearing
• From PSG databook pg no: 4.6, for the
corresponding speed N = 1000rpm & Life, LH =
1000hrs
• C/P = 6.20
• C = 6.20 x 8200 = 50,840 N
• From PSG databook pg no: 4.19, in angular
contact ball bearing, for d = 50mm &
C= 50,840 N = 5,084 kgf ≈ 5300kgf

• Result:
• Bearing number is SKF 7310 B is
selected
• (Note: You can check the same at 4.18 also to
get 7214 B, but d doesn’t match)

Problem 2
• A ball bearing of 95 kN dynamic capacity
carries a shaft rotates at 1500rpm. Calculate
the max equivalent load that can be carried to
a life of 5yrs. Assuming the bearing operates
for 10 hrs per day & 6 days/per week as
working days.

A ball bearing of 95 kN dynamic capacity carries a shaft rotates at 1500rpm.
Calculate the max equivalent load that can be carried to a life of 5yrs.
Assuming the bearing operates for 10 hrs per day & 6days/per week as
working days.
• Given:
– C = 95,000 N
– N = 1500rpm
– L= 10hrs/day x 6 days/per week x 52 weeks/yr x 5 yrs
= 10 x 6 x 52 x 5
L= 15,600 hrs
• To Find:
– P

C = 95,000 N; N = 1500rpm; L= 15,600 hrs
• From PSG databook pg no: 4.2,
• Dynamic capacity,
• C = (L/L10)^1/K x P
• C is given in question, L must be in terms of
revolutions
• Converting Speed in rph = 1,500 rpm x 60 =
90,000 rph
Note: Relationship b/w life in million rev and life in working hours ,
Lh = 60 N LH / 106 mr (Not available in data book)

C = 95,000 N; N = 1500rpm; L= 15,600 hrs
• L = 90,000 rph x 15,600 hrs
• L = 1404 x 10^6 rev
• L= 1404 mr
• K= 3 for ball bearings (From PSG databook pg no: 4.2)
• L10 = 1 mr (From PSG databook pg no: 4.2)
• C = (L/L10)^1/K x P
• 95,000 = (1404/1) ^ 1/3 x P
• P = 8,484 N

Result
• The max equivalent load = 8,484 N

Problem 3
• A deep groove ball bearing SKF 6308 carries a
shaft of centrifugal pump rotating at 800 rpm.
Bearing is subjected to radial load of 5KN,
thrust load of 2kN. Calculate the equivalent
radial load on the bearing and the life
expected for 90% survival in million revolution.

A deep groove ball bearing SKF 6308 carries a shaft of centrifugal pump
rotating at 800 rpm. Bearing is subjected to radial load of 5KN, thrust load of
2kN. Calculate the equivalent radial load on the bearing and the life expected
for 90% survival in million revolution.
Given:
• Deep groove ball bearing SKF 6308
• N = 800 rpm
• FR = 5,000 N
• FA = 2,000 N
To Find:
• P & L

Deep groove ball bearing SKF 6308; N = 800
rpm; FR = 5,000 N; FA = 2,000 N
Solution:
• To find Equivalent load, from PSG databook pg
no :4.2
• P = (X FR + Y FA ) S [Since C is not given, we chose this formula]
• We need to find X, Y and S
• For S, from PSG databook pg no :4.2
• S = 1.1

Deep groove ball bearing SKF 6308;
P = (X FR + Y FA ) S
• For Axial and Radial load factors (X,Y)
• From PSG databook pg no :4.14
• C0 = 2200 kgf = 22,000N
• C = 3200 kgf = 32,000 N

FR = 5,000 N; FA = 2,000 N
• For FA / C0 = 2000/ 22000 = 0.09

FR = 5,000 N; FA = 2,000 N
• For FA / C0 = 2000/ 22000 = 0.09
• FA / C0 = 0.09 is near by 0.07 in PSG databook
4.4, hence values aligned with 0.07 can be
considered for calculation.
• Now, For FA / FR = 2000/5000 = 0.4 > 0.22 (“e”)

FR = 5,000 N; FA = 2,000 N; S = 1.1
• Corresponding to FA / C0 = 0.07 & FA / FR > e,
corresponding in PSG databook 4.4
• X = 0.56
• Y = 1.6
• P = (X FR + Y FA ) S
• =(0.56 x 5000 + 1.6 x 2000) 1.1
• P = 6,600 N

C = 3200 kgf = 32,000 N; P = 6600 N
Life of bearing
• 32,000 = (L/1)^(1/3) x 6600
• L = 114 million revolutions (mr)

Results
• Equivalent load on the bearing is 6,600 N
• Life of bearing is 114 million revolutions

Problem 4
• The rolling contact ball bearing are to be selected
to support the overhang counter shaft. The shaft
speed is 720 rpm. The bearing are to have 99%
reliability corresponding to a life of 24,000 hours
the bearing is subjected to an equivalent load of
1kN. Consider life adjustment factors for
operating condition and materials has 0.9 and
0.85 respectively.
• Find the basic dynamic loading rating of the
bearing from manufacturers catalogue specified
at 90% reliability.
Note: Relationship b/w life in million rev and life in working hours ,
Lh = 60 N LH / 106 mr (Not available in data book)

Problem 5
• A deep groove ball bearing of SKF series 62 is
chosen for the shaft of dia 40mm, rotating at
800rpm. Bearing is expected to carry a radial
load 1kN & axial load 300N. Calculate the life
of bearing having 95% reliability.

A deep groove ball bearing of SKF series 62 is chosen for the shaft of dia
40mm, rotating at 800rpm. Bearing is expected to carry a radial load 1kN &
axial load 300N. Calculate the life of bearing having 95% reliability.
• Given:
– d = 40mm
– N = 800 rpm
– FR = 1000 N
– FA = 300 N
– Ball bearing of SKF series 62
• To Find:
– Life of bearing at 95% reliability, L’

d = 40mm; N = 800 rpm; FR = 1000 N; FA = 300 N;
Ball bearing of SKF series 62
• From PSG databook pg no: 4.2
• Life for 95% reliability:
L5
L′10
=
ln
1
𝑝5
ln
1
𝑝10
1/𝑏
• L10 – Need to find (90% survival)

d = 40mm; N = 800 rpm; FR = 1000 N; FA =
300 N; Ball bearing of SKF series 62
• From PSG databook pg no: 4.2
• From PSG databook pg no: 4.13, for Series 62 & d= 40mm

d = 40mm; N = 800 rpm; FR = 1000 N; FA =
• From PSG databook pg no: 4.13, for Series 62
& d= 40mm
• C0 = 1600 kgf = 16,000 N
• C = 2280 kgf = 22,800 N
• From PSG Databook pg no: 4.2
– To find X, Y & S
– From PSG Databook pg no: 4.2,
– S = 1.1

d = 40mm; N = 800 rpm; FR = 1000 N; FA =
• FA/C0 = 300/16000 = 0.0018 ≈ 0.0025
• For FA/C0 = 0.0025, e value is 0.22
• Now, FA/FR = 300/1000 = 0.3
• The value we got is 0.3 > e (e = 0.22)
• Hence, X = 0.56
Y = 2

X = 0.56, Y = 2, S = 1.1, FR = 1000 N; FA = 300 N
• P = (X FR + Y FA ) S
• P = (0.56 x 1000 + 2 x 300) x 1.1
• P = 1,276 N
• Substitute the value of P & C in rating life of
bearing formula

P = 1,276 N, C = 22800N
To Find: Life of bearing at 95% reliability, L
• 22800 = (L/10^6)^1/3 x 1276 [K=3 for ball bearing]
• L =5,705 mr (90% survival)
• L′10 = L = 5,705 mr = 5,705 x 10^6
•
L5
L′10
=
ln
1
𝑝5
ln
1
𝑝10
1/𝑏
• b = 1.34 (From PSG databook pg no: 4.2, for deep
groove ball bearing)

N = 800 rpm
• L5 = 3333.76 mr
• Convert in to Hours
– L5 = 3333.76 x 10^6 / 60(min/hr) x 800 rpm
– L5 = 69,453 hours
• Result:
• The life of bearing having 95% reliability =
69,453 hours or 3,334 mr

Design Procedure for Ball/Roller bearings
1. Assumption & noting down of Bearing dimensions –
PSG 4.12 to 4.15 (If d is not provided, need to assume either from 60
or 62 series – Start from Series 62 & d = 40mm)
2. Calculation of Equivalent load, P – PSG 4.2 (X,Y, S - FR & FA will
be given)
3. Find out the Dynamic Capacity using the C/P ratio for N & L
given - PSG 4.6
4. Check whether the assumed bearing capacity and calculated
are equated such that Cassumed > Ccalculated (if so design is
safe or redesign it again)
5. If redesign – compare the calculated C value with next series
and assume the bearing the number.
6. If Life asked – find out life – PSG 4.2

Design Procedure for Ball/Roller bearings
(Life of Bearing)
1. Assumption & noting down of Bearing dimensions esp
Capacity – PSG 4.12 to 4.15 (If d is not provided, need to assume
either from 60 or 62 series – Start from Series 62 & d = 40mm)
2. Calculation of Equivalent load, P – PSG 4.2 (X,Y, S - FR & FA will
be given)
3. After finding out the C & P, find the required Life (mr) using
- PSG 4.2
4. Substitute the required Life in below equation
Lx
L′10
=
ln
1
𝑝100−𝑥
ln
1
𝑝10
1/𝑏
- PSG 4.2

Problems on Ball bearing
• Select a single row deep groove ball bearing
for a radial load of 4000N and an axial load of
5000N; operating at a speed of 1600rpm for
an average life of 5 years at 10hrs/day.
Assume uniform and stead load.

Select a single row deep groove ball bearing for a radial load of 4000N and an
axial load of 5000N; operating at a speed of 1600rpm for an average life of 5
years at 10hrs/day. Assume uniform and stead load.
• Given:
– FR = 4000 N
– FA = 5000 N
– N = 1600 rpm
– LH = 5 yrs; at 10hrs/day
– LH = 5yrs x 10hrs x 300 days (Assuming 300 working days)
– LH = 15,000 hrs
• To Find:
– Select a single row deep groove ball bearing

FR = 4000 N, FA = 5000 N, N = 1600 rpm, LH = 15,000 hrs
STEP 1
• Solution:
1. Assumption of Bearing dimensions:
– From PSG Databook pg no: 4.12 to 4.15 is for
Deep grove ball bearing
– i.e., from SKF (name of the company which
manufactures bearings) series 60 to 64
– Since dia is not given in our question, we gonna
assume the dimensions from any one of the
series.

STEP 1
• Solution:
– Start Assuming from Series 60 or 62 (Mere guess,
if design not satisfied, it has to be changed) from
PSG databook page no : 4.13
– In that, let us Assume Series 62, d = 40mm
For medium duty Ball bearings : 30 – 120 mm dia can
be chosen

STEP 1
• Solution:
– Assume Series 62 (Mere guess, if design not
satisfied, it has to be changed)
– In that, let us Assume d = 40mm
For medium duty Ball bearings : 30 – 120 mm dia can
be chosen

STEP 1
• Solution:
– Copy down all the data given in PSG databook for
Dia 40mm and Series 62
Dia 40 mm
ISI NO 40BC02
Bearing basic design no SKF 6208
Outer dia, D 80 mm
Bearing Width, B 18 mm
Ball radius 2 mm
Static Capacity, C0 1600 kgf = 16,000 N
Dynamic Capacity, C 2280 kgf = 22,800 N

STEP 2
Solution:
• Calculation of Equivalent Load
– From PSG Databook pg no: 4.2
– To find X, Y & S
– From PSG Databook pg no: 4.2,
Assume it is for rotary M/c with no impact.
– Hence, Assume S = 1.5 (Choosing max and checking)

• First find : FA/C0 = 5,000 / 16,000 = 0.3125
• Then, Find : FA /FR = 5000/4000 = 1.25
• To find e < or e >,
– The e value for deep grove ball bearing lies
between 0.22 to 0.44 from PSG DB pg 4.4
– But FA /FR is 1.25 which is always greater than 0.44
(Max value of “e”)

• But FA /FR is 1.25 which is always greater than
0.44 (Max value of “e”)
• Hence choose FA /FR > e column
• X = 0.56
• For Y, Choose FA/C0 = 0.3125 which is between
0.25 to 0.5,
• Hence approximately assume 1.15

STEP 2
Solution:
• Calculation of Equivalent Load
P = (0.56 x 4000 + 1.15 x 5000) 1.5
P = 11,985 N

STEP 3
Solution:
• Checking of Bearing Capacity (Which we
choose. Else re-assume the dia of bearing
again)
• From PSG databook pg no : 4.6
• Life of Bearing, LH = 15,000 hrs and N =
1600rpm

P = 11,985 N, N = 1600 rpm, LH = 15,000 hrs
STEP 3
• From PSG databook pg no : 4.6
• Life of Bearing, LH = 15,000 hrs and N =
1600rpm
• The value of C/P = 11.50
• Now, C = 11.50 * 11,985
= 1,37,827.50 N = 13,783 kgf
• But, Dynamic capacity for the series 62 & dia
40mm which we assumed is 2,800kgf

• Cassumed < Ccalculated
• Cassumed = 22,800 N (2280kgf) for Series 62 & dia 40mm
• Ccalculated = 1,37,827.50 N = 13,783 kgf
• Hence our assumption of series 62 must be changed
to next series 63 as the design become unsafe.

For the C=13,783
• C h e c k i n g S e r i e s 6 3
• The C value occurring around the calculated C
is for d = 105 mm
• C = 14,300 kgf = 1,43,000 N
• C0 = 14,300 kgf = 1,43,000 N
• Now repeating the procedure we did
previously

FA = 5000 N & C0 = 1,43,000 N
• For, FA / C0 = 5,000/1,43,000 = 0.035
• X = 0.56 & Y = 1.9 (approximately)
• Now, P = (X Fr + Y FA ) S
= (0.56 * 4000 + 1.9 *5000) 1.1
P = 12,914 N
• Next, C/P ratio
• We already know C/P = 11.50

P = 12,914 N
• C = 11.50 * 12914
• C = 1,48,511
• Again Cassumed < Ccalculated
• Our design is unsafe
• Hence repeat the steps once again
for next dia (don’t need to change
the series here as the value is much
nearer)

C = 1,48,511
• C h e c k i n g S e r i e s 6 3
• The C value occurring around the calculated C
is for d = 120 mm
• C = 16,300 kgf = 1,63,000 N
• C0 = 17,300 kgf = 1,73,000 N
• Now repeating the procedure we did
previously

FA = 5000 N & C0 = 1,73,000 N
• For, FA / C0 = 5,000/1,73,000 = 0.028
• X = 0.56 & Y = 2
• Now, P = (X Fr + Y FA ) S
= (0.56 * 4000 + 2 *5000) 1.1
P = 13,464 N
• Next, C/P ratio
• We already know C/P = 11.50

P = 13,464 N
• C = 11.50 * 13464 = 1,54,836 N
• C = 15,483 kgf
• Now Cassumed > Ccalculated
• Hence our design is safe
Result:
• T h e b e a r i n g n u m b e r i s S K F 6 3 2 4
i s s e l e c t e d .

• A shaft transmitting 50 kW at 125 rpm from the gear G1
to the gear G2 and mounted on two single-row deep
groove ball bearings B1 and B2 is shown in Fig. The gear
tooth forces are Pt1 = 15915 N, Pr1 = 5793 N, Pt2 = 9549
N, Pr2 = 3476 N The diameter of the shaft at bearings B1
and B2 is 75 mm. The load factor is 1.4 and the
expected life for 90% of the bearings is 10000 hrs.
Select suitable ball bearings. (APR/MAY 2018)

A shaft transmitting 50 kW at 125 rpm from the gear G1 to the gear G2 and mounted on two single-row deep
groove ball bearings B1 and B2 is shown in Fig. The gear tooth forces are Pt1 = 15915 N, Pr1 = 5793 N, Pt2 = 9549
N, Pr2 = 3476 N The diameter of the shaft at bearings B1 and B2 is 75 mm. The load factor is 1.4 and the
expected life for 90% of the bearings is 10000 hrs. Select suitable ball bearings.
Given:
• Power = 50 kW
• N = 125 rpm
• d = 75 mm
• L10h = 10,000 hrs
• Load Factor = 1.4
To Find:
• Select a suitable ball bearing

Methodology to solve these
kind of problems
• Step 1 : To find Radial and Axial forces
• Step 2: To find dynamic load capacity
(C = (L/L10)^1/K x P)
• Step 3: Selection of SKF bearing number based
on Capacity and Diameter

• To find radial and axial forces, resolve the given
gear and bearing arrangement (Basics of SOM
from previous semester)
• Considering vertical plane,
taking moments about
Bearing B1
• Pr1(125) + Pt2(775) – Rv2(625) = 0
• Rv2 = 13,000 N
• Considering vertical forces,
• Pt2 + Pr1 = Rv2+ Rv1
• Rv1 = 2,350 N

• To find radial and axial forces, resolve the given
gear and bearing arrangement (Basics of SOM
from previous semester)
• Considering horizontal plane,
taking moments about
Bearing B1
• Pt1(125) + Pr2(775) – RH2(625) = 0
• RH2 = 7,500 N
• Considering horizontal forces,
• Pt1 + Pr2 = RH2+ RH1
• RH1 = 11,900 N

• Radial (Resultant) Forces at the two bearings
are:
• Fr1 = [(Rv1)2+(RH1)2
• = 12,150 N
• Fr2 = [(Rv2)2+(RH2)2
• = 15,000 N

Given: kW = 50 kW, N = 125 rpm, d = 75
mm, L10h = 10,000 hrs, Load Factor = 1.4
• Since there is no Axial thrust provided for the
shaft,
• Fa1 = Fa2 = 0
• Now,
Fr1 = 12,150 N
Fr1 = 15,000 N
Fa1 = Fa2 = 0

kW = 50 kW, N = 125 rpm, d = 75 mm, L10h = 10,000
hrs, Load Factor = 1.4
Fr1 = 12,150 N, Fr1 = 15,000 N
• Converting L from hrs to mr
• L10 = 10,000 x 125 x 60 =75 mr
• From PSG DB Pg no : 4.2
• C1 = (L/L10)^1/K x P x Load factor
= (75)^ 1/3 * 12,150 x 1.4 = 17,600 N
• C2 = (L/L10)^1/K x P x Load factor
= (75)^ 1/3 * 15,000 x 1.4 = 88,600 N

Results
• Step III : Selection of Bearing:
• From PSG db pg 4.12 to 4.14
• For d = 75 mm,
Bearing 1 (C = 1,760 kgf) Bearing 2 (C = 8,860 kgf)
SKF no 6015 (C = 3100) SKF no 6315 (C = 9,000)
SKF no 6215 (C = 5200) SKF no 6415 (C = 12,000)

Problems on Cyclic Loads and Speed
• A deep groove ball bearing has dynamic capacity of
20,200 N and is to operate on the following work
cycle.
• Radial load of 5800 N at 200 rpm for 25% of the time
• Radial load of 8900 N at 500 rpm for 20% of the time
• Radial load of 3500 N at 400 rpm for remaining time
• Assuming the loads are steady and the inner race
rotates, find the expected average life of the bearing
in hours. (NOV/DEC 2019)

Problems on Cyclic Loads and Speed
• Equivalent Load 𝑃𝑒 =
3 𝑃1
3.𝑁1+𝑃2
3.𝑁2+𝑃3
3.𝑁3+⋯
𝑁1+𝑁2+𝑁3…..
• N = % of time x rpm

Lubrication of Ball and Roller bearings
• Purpose
– To reduce friction and wear between the sliding parts
– To prevent rusting
– To protect bearing surface from water, dirt
(grease),etc.,
– To dissipate the heat
• Pure mineral oil or light grease
– Na or K based greases
• Too much of Oil or grease may lead to temp rise
of bearing due to churning (agitation).
[Range – below 90 or above 150 C]

DME - Unit 5.pptx

In this document

More Related Content

What's hot

Similar to DME - Unit 5.pptx

More from Charunnath S V

Recently uploaded

DME - Unit 5.pptx