DME 1 module-1

DESIGN OF MACHINE ELEMENTS - I
[AS PER CHOICE BASED CREDIT SYSTEM (CBCS) SCHEME]
SEMESTER – V
Dr. Mohammed Imran
B. E. IN MECHANICAL ENGINEERING
Course Code 18ME52 CIE Marks 40
Teaching Hours / Week (L:T:P) 3:2:0 SEE Marks 60
Credits 04 Exam Hours 03

Text Books:
 Design of Machine
Elements, V.B. Bhandari,
Tata McGraw Hill
Publishing Company Ltd.,
New Delhi, 2nd Edition
2007.
 Mechanical Engineering
Design, Joseph E Shigley
and Charles R. Mischke.
McGraw Hill
International edition, 6th
Edition, 2009.
Dr. Mohammed Imran
2
By Dr. Mohammed Imran

Design Data Handbook:
 Design Data Hand
Book, K. Lingaiah,
McGraw Hill, 2nd Ed.
 Data Hand Book, K.
Mahadevan and
Balaveera Reddy, CBS
Publication
 Design Data Hand
Book, S C Pilli and H.
G. Patil, I. K.
International Publisher,
2010.
Dr. Mohammed Imran
3

Design Data Hand Book, K. Lingaiah, McGraw Hill, 2nd Ed.
4

Design Data Hand Book, K. Lingaiah, McGraw Hill, 2nd Ed.
5

Reference Books:
 Machine Design, Robert L. Norton, Pearson Education Asia,
2001.
 Engineering Design, George E. Dieter, Linda C Schmidt,
McGraw Hill Education, Indian
Edition, 2013.
 Design of Machined Elements, S C Pilli and H. G. Patil, I. K.
International Publisher, 2017.
 Machine Design, Hall, Holowenko, Laughlin (Schaum’s
Outline series) adapted by S.K
Somani, tata McGraw Hill Publishing company Ltd., New
Delhi, Special Indian Edition,
2008
 Design of Machine Elewments -1, Dr.M.H.Annaiah, Dr. C.N.
Chandrappa, Dr.J.Suresh
Kumar, New Age International Publishers.
Dr. Mohammed Imran
6

Module-1
 Introduction: Design Process: Definition of design, phases of design.
Review of engineering materials and their properties. Manufacturing
processes. Use of codes and standards. Selection of preferred sizes.
Review of axial, bending, shear and torsion loading on machine
components, combined loading, two- and three dimensional stresses,
principal stresses, stress tensors, Mohr's circles.
 Design for static strength: Factor of safety and service factor.
Failure mode: definition and types, Failure of brittle and ductile
materials; even and uneven materials; Theories of failure: maximum
normal stress theory, maximum shear stress theory, distortion energy
theory, strain energy theory, Columba –Mohr theory and modified
Mohr’s theory. Stress concentration, stress concentration factor and
methods of reducing stress concentration.
10 Hours
Fundamentals of Mechanical Engineering Design
Dr. Mohammed Imran
7

MODULE -1
PART-A
INTRODUCTION: DESIGN PROCESS
Dr. Mohammed Imran

Mechanical engineering:
 The branch of engineering dealing with the design, constructions and use of machine.
The discipline that applies engineering, physics, engineering mathematics and material
science principles to design, analyze, manufacturing and maintain mechanical
systems. It is one of the oldest and broadest of the engineering disciplines.
 The mechanical engineering field requires an understanding of core areas including
mechanics, dynamics, thermodynamics, materials science, structural analysis and
electricity. In addition to these core principles, the mechanical engineers use tools such
as computer aided design (CAD), computer aided manufacturing (CAM) and product
life cycle management to design and analyze manufacturing plants, industrial
equipment and machinery.
 The industrial equipments and machineries such as heating and cooling systems,
transport systems, aircraft, watercraft, robotics, medical devices, weapons, etc.
 It is the branch of engineering that involves the Machine and their Design (machine
design), production, and operation of machinery.
9

Definitions
 Machine: It is an mechanical device that change the direction or magnitude of
a force by relative motions of mechanical components of device. Common
devices are I.C engine, lathe, turbines etc.
 Design: It is nothing but a series of activities to gather all the information
necessary to realize the designer's idea as a real products shown in figure.1.
 Machine Design: Machine design is the art of developing new ideas for the
construction of machine and expressing those ideas in the form of plans and
drawing. It may be entirely new in concept or an improvement of an existing
machine for their better performance and economy.
10

Classification of design:
 Adoptive design: In this the designer’s work is concerned
with adaptation of existing designs. The designer only
makes minor alternation or modification in the existing
designs of the product.
 Development design: This type of design needs scientific
training and design ability in order to modify the existing
designs into a new idea by adopting a new material or
different method of manufacture.
 New Design: This type of design needs lot of research,
technical ability and creative thinking. Only those designers
who have personal qualities of a sufficiently high order can
take up the work of a new design.
 Rational design, Empirical design, Industrial Design, etc
11

Classification of design:
 New Design: This type of design needs lot of research, technical ability
and creative thinking. Only those designers who have personal qualities of
a sufficiently high order can take up the work of a new design.
 Rational design: This type of design depends upon mathematical formulae of
principle of mechanics.
 Empirical design: This type of design depends upon empirical formulae
based on the practice and past experience
 Industrial Design: This type of design depends upon the production aspects to
manufacture any machine component in the industry.
 Optimum Design: It is the best design for the given objective function under
the specified constraints. It may be achieved by minimizing the undesirable
effects.
 System Design: It is the design of any complex mechanical system like a motor
car.
 Element Design: It is the design of any element of the mechanical system like
piston, crankshaft, connecting rod, etc.
 Computer aided Design : This type of design depends upon the use of
computer systems to assist in the creation, modification, analysis and optimization
of a design.
Cont……
12

Phases of design process / basic
procedure of machine design:
Depending on the
nature of the design
task, several design
phases may be
repeated throughout
the life of the product,
from inception to
termination.
Identification of needs
Definition of problem or
Analysis of forces
Synthesis
Analysis and Optimization
Evaluation
Presentation
Iteration
Fig.2. Phases of design process
13

Phases of design process
 Identification of needs: First of all, make a complete statement of the problem, indicating
the need, aim or purpose for which the machine is to be designed.
 Definition of problem: The definition of problem is more specific and must include all the
specifications for the object that is to be designed. The specifications are the input and
output quantities, the characteristics and dimensions of the space the object must occupy,
and all the limitations on these quantities.
 Synthesis: It is a process to select the possible mechanism or group of mechanisms which
will give the desired motion.
 Analysis and Optimization: The optimization is a called altering, modification and
redesigning the size of the existing machines in an innovative ideal design of machine
which may have best performance, efficiency and low cost. The analysis of selected
machine mechanism carried out by the analysis of forces, material selection and design
of elements (size and stresses). These analysis parameters are discussed below.
 Evaluation : Evaluation is a significant phase of the total design process. Evaluation is the
final proof of a successful design and usually involves the testing of a prototype in the
laboratory. Here we wish to discover if the design really satisfies the needs.
 Presentation : The engineer, when presenting a new solution to administrative,
management, or supervisory persons, is attempting to sell or to prove to them that their
solution is a better one. Unless this can be done successfully, the time and effort spent on
obtaining the solution have been largely wasted.
Cont……
14

Design considerations or requirements
 Sometimes the strength required of an element in a
system is an important factor in the determination of
the geometry and the dimensions of the element. In
such a situation we say that strength is an important
design consideration. When we use the expression
design consideration, we are referring to some
characteristic that influences the design of the
element or the entire system.
15

Design considerations or requirements
 Many of the important ones are as follows (not necessarily in order
of importance):
 Some of these characteristics have to do directly with the
dimensions, the material, the processing, and the joining of the
elements of the system. Several characteristics may be
interrelated, which affects the configuration of the total system.
1 Good functionality 14 less noise
2 High Strength/stress 15 Styling
3 Distortion/deflection/stiffness 16 Shape
4 Wear 17 Size
5 Corrosion 18 Control
6 Safety 19 Thermal properties
7 Reliability 20 Surface
8 Manufacturability 21 Lubrication
9 Utility 22 Marketability
10 Minimum life cycle cost 23 Maintenance
11 Friction 24 Volume
12 Weight 25 Liability
13 Life 26 Remanufacturing/resource recovery
Cont……
16

Engineering Materials and their
Mechanical properties
 The knowledge of materials and their properties is of
great significance for a design engineer.
Engineering Materials:
17

Properties of Engineering Materials
Physical Properties of Metals
The physical properties of the metals include luster, colour, size and shape,
density, electric and thermal conductivity, and melting point. The following table
shows the important physical properties of some pure metals.
Mechanical Properties of Metals
The mechanical properties of the metals are those which are associated with the ability of the
material to resist mechanical forces and load. These mechanical properties of the metal include
strength, stiffness, elasticity, plasticity, ductility, brittleness, malleability, toughness, resilience, creep
and hardness. We shall now discuss these properties as follows:
18

 Strength. It is the ability of a material to resist the externally applied forces without breaking or yielding. The
internal resistance offered by a part to an externally applied force is called stress.
 Stiffness. It is the ability of a material to resist deformation under stress. The modulus of elasticity is the measure of
stiffness.
 Elasticity. It is the property of a material to regain its original shape after deformation when the external forces are
removed shown in fig.4.(a) point E. This property is desirable for materials used in tools and machines. It may be noted
that steel is more elastic than rubber
 Plasticity. It is property of a material which retains the deformation produced under load permanently shown in
fig.4.(a) point U. This property of the material is necessary for forgings, in stamping images on coins and in ornamental
work.
 Ductility. It is the property of a material enabling it to be drawn into wire with the application of a tensile
force. A ductile material must be both strong and plastic. The ductility is usually measured by the terms, percentage
elongation and percentage reduction in area. The ductile material commonly used in engineering practice (in
order of diminishing ductility) are mild steel, copper, aluminium, nickel, zinc, tin and lead.
 Brittleness. It is the property of a material opposite to ductility. It is the property of breaking of a material with little
permanent distortion. Brittle materials when subjected to tensile loads snap off without giving any sensible elongation.
Cast iron is a brittle material.
 Malleability. It is a special case of ductility which permits materials to be rolled or hammered into thin sheets. A
malleable material should be plastic but it is not essential to be so strong. The malleable materials commonly used in
engineering practice (in order of diminishing malleability) are lead, soft steel, wrought iron, copper and aluminium.
 Toughness. It is the property of a material to resist fracture due to high impact loads like hammer blows. The toughness
of the material decreases when it is heated. It is measured by the amount of energy that a unit volume of the material
has absorbed after being stressed upto the point of fracture. This property is desirable in parts subjected to shock and
impact loads shown in fig.4.(b).
19

 Machinability. It is the property of a material which refers to a relative case with which a material can be cut. The
machinability of a material can be measured in a number of ways such as comparing the tool life for cutting different
materials or thrust required to remove the material at some given rate or the energy required to remove a unit volume
of the material. It may be noted that brass can be easily machined than steel.
 Resilience. It is the property of a material to absorb energy and to resist shock and impact loads. It is measured by the
amount of energy absorbed per unit volume within elastic limit. This property is essential for spring materials. shown in
figure.4.(a)
 Creep. When a part is subjected to a constant stress at high temperature for a long period of time, it will undergo a
slow and permanent deformation called creep. This property is considered in designing internal combustion engines,
boilers and turbines.
 Fatigue. When a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type
of failure of a material is known as *fatigue. The failure is caused by means of a progressive crack formation which
are usually fine and of microscopic size. This property is considered in designing shafts, connecting rods, springs, gears,
etc.
 Hardness. It is a very important property of the metals and has a wide variety of meanings. It embraces many
different properties such as resistance to wear, scratching, deformation and machinability etc. It also means the ability of
a metal to cut another metal. The hardness is usually expressed in numbers which are dependent on the method of
making the test. The hardness of a metal may be determined by the following tests:
 Brinell hardness test,
 Rockwell hardness test,
 Vickers hardness (also called Diamond Pyramid) test, and
 Shore scleroscope.
Cont……
20

Manufacturing processes.
21

Standards and Standardization
A standard is a set of specifications for parts, materials, or processes intended (anticipated) to
achieve uniformity, efficiency, and a specified quality as defined by a certain body or an
organization. One of the important purposes of a standard is to limit the multitude of variations
that can arise from the arbitrary creation of a part, material, or process (Characteristics are
include the dimensions, shapes, tolerances, surface finish, manufacturing methods etc.).
Types of Standards Used In Machine Design:
 Based on the defining bodies or organization, the standards used in the machine design can be divided into
following three categories:
 Company Standards: These standards are defined or set by a company or a group of companies for their use.
 National Standards: These standards are defined or set by a national apex body and are normally followed
throughout the country. Like BIS, AWS.
 International Standards: These standards are defined or set by international apex body and are normally followed
throughout the world. Like ISO, IBWM.
Advantages:
 Reducing duplication of effort or overlap and combining resources
 Bridging of technology gaps and transferring technology
 Reducing conflict in regulations
 Facilitating commerce
 Stabilizing existing markets and allowing development of new markets
 Protecting from litigation
22

Standards and Codes
A standard is a set of specifications for parts, materials, or processes intended
(anticipated) to achieve uniformity, efficiency, and a specified quality as defined
by a certain body or an organization.
A code is a set of specifications for the analysis, design, manufacture, and
construction of something. The purpose of a code is to achieve a specified degree
of safety, efficiency, and performance or quality.
All of the organizations and societies listed below have established specifications
for standards and safety or design codes. The name of the organization provides
a clue to the nature of the standard or code. The organizations of interest to
mechanical engineers are:
 Aluminum Association (AA)
 American Society of Mechanical Engineers (ASME)
 American Society of Testing and Materials (ASTM)
 American standards for materials ASM
 Society of Automotive Engineers (SAE)
 International Standards Organization (ISO)
23

Selection of preferred sizes
 Size of the machine element with preferred size.
 Round off to whole numbers, we get the series
 The ISO has defined four basic series of preferred
numbers
24

Selection of preferred sizes
 To users of a decimally-oriented system of units, such as SI, Renard's series
is much more useful than these other geometric series, because it begins on
10 and ends on 100. The ISO adopted Renard's series as the basis of the
preferred numbers for use in setting metric sizes. The designations of the
series they have defined begin with “R” as a tribute to Renard, and the
series are called “renard series.”
 The ISO has defined four basic series of preferred numbers:
R5: 10, 16, 25, 40, 63, 100.
R10: 10, 12.5, 16, 20, 25, 31.5, 40, 50, 63, 80, 100.
R20: 10, 11.2 12.5, 14, 16, 18, 20, 22.4, 25, 28, 31.5, 35.5, 40, 45, 50, 56,
63, 71, 80, 90, 100.
R40: 10, 10.6, 11.2, 11.8, 12.5, 13.2, 14, 15, 16, 17, 18, 19, 20, 21.2, 22.4,
23.6, 25, 26.5, 28, 30, 31.5, 33.5, 35.5, 37.5, 40, 42.5, 45, 47.5, 50, 53, 56,
60, 63, 67, 71, 75, 80, 85, 90, 95, 100.
25

Material selection.
Selection of a proper material for the machine component is one of the most important steps in the
process of machine design. The best material is one which will serve the desired purpose at minimum cost
and better performance. The factors which should be considered while selecting the material for a
machine component are as follows:
Availability: The material should be readily available in the market, in large enough quantities
to meet the requirement. Eg: Cast iron, Aluminium alloys and copper etc.
Cost: For every application, there is a limiting cost beyond which the designer cannot go. In
cost analysis, there are two factors, namely,
 Low cost of material and
 Low cost of processing the material into finished goods.
It is likely that the cost of material might be low, but the processing may involve costly
manufacturing operations.
Mechanical properties: Mechanical properties are the most important technical factor
governing the selection of material. They include strength under static and fluctuating loads,
elasticity, plasticity, stiffness, resilience, toughness, ductility, malleability and hardness.
Depending upon the service conditions and the functional requirement, different mechanical
properties are considered and a suitable material is selected.
Manufacturing considerations : The manufacturing processes, such as casting, forging, extrusion,
welding and machining govern the selection of material. The major factors influence for the
manufacturing process for selection of material should have good castability, machinability
and weldability.
26

Stresses and strains
Stress: When some external system of forces or loads acts on a body, the internal forces (equal and opposite)
are set up at various sections of the body, which resist the external forces. This internal force per unit area at
any section of the body is known as unit stress or simply a stress. It is denoted by a Greek letter sigma
(σ). Mathematically,
Where, P = Force or load acting on a body, and
A = Cross-sectional area of the body.
In S.I. units, the stress is usually expressed in Pascal (Pa) such that 1 Pa = 1 N/m2. In actual practice, we use
bigger units of stress i.e. megapascal (MPa) and gigapascal (GPa), such that
Strain: When the system of force is act upon the body, it undergoes some deformations This deformation per
unit length is known as unit strain or simply a strain. It is denoted by a Greek letter epsilon (ε). Mathematically,
Where δl = Change in length of the body,
l= Original length of the body.
27

Stress–Strain diagrams: (Relationship)
Stress–Strain diagrams:
The behaviour of material and its usefulness for engineering applications can be
obtained by making a tension test and plotting a curve showing the variation of
stress with respect to strain. A tension test is one of the simplest and basic tests and
determines values of number of parameters concerned with mechanical properties
of materials such as
 Proportional limit
 Elastic limit
 Modulus of elasticity
 Yield strength
 Ultimate tension strength
 Modulus of resilience
 Modulus of toughness
 Percentage elongation
 Percentage reduction in area
28

Stress–Strain diagrams: (Relationship)
Cont……
29

MODULE -1
PART-B
INTRODUCTION: STATIC LOAD AND
STRESSES TYPE-1
Dr. Mohammed Imran

Static loads and stresses
A static load is defined as a force, which is gradually applied to a
mechanical component and which does not change its magnitude or direction
with respect to time. A mechanical component may fail, that is, may be unable
to perform its function satisfactorily, as a result of any one of the following
three modes of failure
i. Failure by elastic deflection;
ii. Failure by general yielding; and
iii. Failure by fracture.
Load: The external force is acting on machine part called load
The different types of load are
1. Dead or study load: When the load does not change in magnitude or
direction.
2. Variable Load: When the load is change continuously.
3. Suddenly applied or shock load: When load is applied suddenly and
removed.
4. Impact Load: When the load is applied with an initial velocity.
31

Normal or simple stresses:
32
2.1 DDHB pp:35
2.1 DDHB pp:35
2.2 DDHB pp:35

Bending stresses:
33
2.52 DDHB pp:37

Bending stresses:
Cont……
34

Shear stresses
35
2.5 DDHB pp:35

Torsional stresses
36
2.51(a) DDHB pp:37
2.50 DDHB pp:37

Eccentric axial stresses
37
2.56 DDHB pp:38

Combined normal and shear stress
The stress analysis of an element subjected to the
simultaneously action of a normal stress σ and a shear
stress  shows that the combined action of these
stresses produces an internal normal stress σmax and a
tangential stress max which is given by
38
2.53 DDHB pp:38
2.54 DDHB pp:38

Problems on combine loads
1. A beam of uniform rectangular C/S of fixed at one end and carries a
load 1000N at a distance of 300mm from the fixed end. The
maximum bending stress in the beam is 80 N/mm2. Find the width
and depth of beam. If depth is twice of width.
2. A circular rod of diameter 50mm is subjected to loads as shown in
fig. Q (2). Determine the nature and magnitude of stresses at the
critical points.
Fig. Q(2)
39

3. A 50mm diameter steel rod supports a 9.0kN load and in addition is
subjected to a torsional moment of 100N-m as shown in fig. Q(3).
Determine the maximum tensile and the maximum shear stress.
Fig. Q(3)
40

4. Determine extreme fiber stress at section x-x of the machine member
loaded as shown in fig.Q(4). Also show the distribution of stresses at
this section.
Fig. Q(4)
All dimensions in mm.
41

42

43

44

MODULE -1
PART-B
INTRODUCTION: STATIC LOAD AND
STRESSES TYPE-2
Dr. Mohammed Imran

Two- dimensional stresses (Biaxial stresses)
A two-dimensional state of stress in which only
two normal stresses are present is called biaxial
stress.
The more general stress condition in which the
stresses on an element acts in both x and y
directions and all stress components in z- direction
vanish is known as biaxial stress system. It is
different from one dimensional or uniaxial stress
condition considered in the previous section. Thus
it is assumed that all the components in the z-
direction (xz = yz = σz =0) shown in Fig.1. are
zero. The stress element shown in Fig.2 is
obtained; it is the most general condition that can
exist. Biaxial stresses arise in the analysis of
beams, pressure vessels, shafts and many other
structural members.
Figure:1.
Figure:2.
46

Two- dimensional stresses (Biaxial stresses)
Figure:2.
Cont….
A class of common
engineering problems
involving stresses in a thin
plate or free surface of a
structural element, such as
the surfaces of thin walled
pressure vessels under
external or internal pressure,
the free surface of shafts in
torsion or beams under
transverse loads.
47

General Biaxial stresses analysis
Figure:2.
Cont….
In general if a plane
element is removed from a
body it will be subjected to
the normal stresses σx and
σy together with the
shearing stress xy as shown
in fig 3.
It is desirable to investigate
the state of stress on any
inclined plane  defined by
angle θ, positive ccw as
shown in fig 4.
Figure:3.
48
Fig. 2.4 DDHB pp:46

General Biaxial stresses analysis Cont….
Summation of forces in Fig. along the direction of σn ΣFn = 0.
Summing the forces along  (t-direction) Σft = 0.
Resultant stress on plane
Oblique angle
Figure:4.
1. Normal stress and Shear stress
49
2.30 DDHB pp:36
2.31 DDHB pp:36

2. Principal Stresses (Magnitude ) and location of angles / planes
Cont….
In the design and stress analysis, the maximum stresses are desired in order to ensure safety of
the load carrying member. The above equations can be used if we know at what angle θ it
occurred. The angle θ can be found by differentiating the function and setting the result equal to
zero.
Principal plane Shear stress or tangential stress on this plane is zero
There are two principal planes right angle each other
Principal stresses
Where
50
2.34 DDHB pp:37
2.32 DDHB pp:36 2.33 DDHB pp:37

3. Maximum shear Stresses and location of angles / plane
Cont….
There are certain values of angles θ that leads to maximum value of shear stress for a given set
of stresses σx, σy, xy, the orientation of the planes on which maximum shear stress can be found
using the same technique i.e. differentiating above equation w.r.t. θ and setting it equal to zero.
Where
The plane of maximum shear will be 1 +45 and 1 +135 w.r.t Plane x
51
2.35 DDHB pp:37
2.29 DDHB pp:36

Three dimensional stresses
 Three dimensional stresses are
refers three normal stresses (x
y and z ) and three shear
stresses (xy, xz, and yz) act on
the element called 3D stresses
 Triaxial stress refers to a
condition where only normal
stresses act on an element and
all shear stresses (xy, xz, and
yz) are zero.
 Under equilibrium xy = yx; xz
= zx; and yz = zy;
52

 The r orientation of the
stress element, all the
shear stress components
are zero. At this r
orientation, the normal to
the faces are called
principal directions and
the normal stresses acting
on this faces are called
principal stresses. Six
faces are there, 3
direction and 3 stresses
i.e, 1, 2 , & 3 .
53

 The three roots to cubic equation for
six stress components such as
The Maximum shear stress is equal to half of the greatest difference between any
two of the principal stresses
54

 In three dimensions, in each orthogonal direction X,
Y & Z, there could be one normal and two shear
stresses. Thus the most generalized state stress at a
point in 3D is as shown below. It is also conveniently
described by a stress tensor as follows:
The six stress components to
establish the state at a point
can be represented as stress
matrix called stress tenor
Stress tensors
55

Principal Stresses (Magnitude ) and location of angles / planes
In the design and stress analysis, the maximum stresses are desired in order to ensure safety of
the load carrying member. The above equations can be used if we know at what angle θ it
occurred. The angle θ can be found by differentiating the function and setting the result equal to
zero.
Principal plane Shear stress or tangential stress on this plane is zero
There are two principal planes right angle each other
Principal stresses
Principal stresses
Where
56

Mohr's circles for 3D
57
•Mohr’s circle is a graphical
interpretation of the
transformation equations
for plane stress.
•The process involves the
construction of a circle such
a manner that the
coordinates of each point
on the circle represents the
normal and shearing
stresses on one plane
through the stressed

Graphical stress analysis
or 2D Mohr's circles
58
Mohr’s Circle for Plane Stress Sign Conventions
Figure 2.5 DDHB pp:46

59
Graphical stress analysis
Mohr’s Circle for Plane Stress
Drawing Procedure for Mohr’s Circle
1. Choose a set of x-y coordinate axes.
2. Find the stresses σx, σy and xy = yx and list them with proper sign.
3. Draw a set of σ-coordinate axes with σ and 
positive to the right and upward, respectively.
4. Plot the point (σx, -xy) and label it point V
(vertical plane).
5. Plot the point (σy, yx) and label it point H (horizontal plane).
6. Draw a line between V and H. This establishes the center and the radius
R of Mohr’s circle.
7. Draw the circle.
8. An extension of the radius between C and V can be identified as the x-
axis or reference line for the angle measurements (I.e., θ =0).

60
2θ2
2θ1
σx, xy
σy, -xy
σ2
σ1
x-axis
v, v1 plane
x
y-axis
H,
H
1
plane
y







 
2
y
x


σx







 
2
y
x


σy
2θ’
1
max
min
2θ’2
Mohr's circles for 2D Figure 2.5 DDHB pp:46

Factor of safety and service factor.
Factor of safety,
While designing a component, it is necessary to provide sufficient reserve
strength in case of an accident. This is achieved by taking a suitable factor of
safety (F.O.S).
The factor of safety is defined as the ratio of the maximum or failure stress to
the working or allowable stress.
The allowable stress is the stress value, which is used in design to determine the
dimensions of the component. It is considered as a stress, which the designer
expects will not be exceeded under normal operating conditions.
For ductile materials, the allowable stress  is obtained by the following
relationship:
For brittle materials, the relationship is,
61

Factor of safety and service factor.
Factors affecting the selection of factor of safety:
The various factors that affect the selection of factor of
safety are
1. Uncertainity of loading
2. Uncertainity of material properties
3. Type of loading and stress
4. Effect of machining or mode of manufacture
5. Effect of environment in which the component I expected to operate
6. The effect of size in material strength properties.
7. Effect of wear upon the functions and life of a machine member.
8. Human life at risk.
9. Severity of damage due to the failure
10. Uncertainity of working conditions
11. Specific requirements for life and reliability
62

MODULE -1
PART-C
THEORIES OF FAILURE
Dr. Mohammed Imran

Failure mode and types
Failure: Failure of a machine component while the
machine is operating under its design
constraints (design envelope).
Example: The mechanical failure was found to be a
broken crank shaft.
64

65
 Commonly observed modes of failure are
- Yielding - Creep
- Excessive deformation - Environmental degradation of stiffness and strength
- Buckling - Vibration and Noise
- Fatigue - Wear, etc.,
- Fracture
 Designing components / structures to avoid these failure
modes is not a new idea.
 Design against FRACTURE (Failure due to Crack Propagation)
is a relatively new approach. So also Fatigue Analysis based
on Fracture Mechanics concepts.
 The use of Fracture Mechanics has undoubtedly prevented a
substantial number of component / structural failures in
service.
Failure mode and types

Failure of ductile and brittle materials
 Ductile materials usually fail
by yielding and hence the
limiting strength is the yield
strength of material as
determined from simple
tension test which is assumed
the same in compression also.
 For brittle materials limiting
strength of material is
ultimate tensile strength in
tension or compression.
66

Failure of ductile and brittle materials
67

Even and uneven materials
 Even materials have equal responses to tension and compression,
while uneven materials are stronger in one than the other.
 Most ductile materials are even materials.
 Brittle materials may be even or uneven.
 These are typically stronger in compression and shear than in tension.
 If they are even, their point of failure can be found easily by comparing the
normal stress they experience to their tensile strengths.
 Some wrought materials, such as fully hardened tool steel, can be brittle.
These materials tend to have compressive strength equal to their tensile
strengths . They are called EVEN materials
Many cast materials, such as gray cast iron, are brittle but have compressive
strengths much greater than their tensile strengths. These are called
UNEVEN materials
68

Theories of failure
 In designing machine components to resist failure we usually assume ourselves that the internal stress does
not exceed the strength of the material.
 If the material is ductile then yield stress is used as the criterion of failure in design. .
 If the material is brittle such as cast iron, ultimate strength is used as the criterion of failure in
design.
 In designing components of brittle materials it is also necessary to remember that the ultimate compressive
strength is much greater than the ultimate tensile strength. Whereas the ductile materials the strength is
same in tension and compression.
 To predict the failure of materials different designers have purposed different criterion for design which
are known as theories of failure. The following theories are very commonly used in design practice
 Maximum (Principal) Stress Theory (Rankine’s Theory)
 Maximum Shear Stress Theory (Tresca’s or Guest’s Theory)
 Maximum Distortion Energy Theory (Octahedral or Hencky and Van-mises Theory)
 Maximum Strain energy theory
 Columba –Mohr theory
 Modified Mohr’s theory
69

Theories of failure
70
Ductile materials (yield criteria)
1. Maximum Shear Stress Theory (Tresca’s or Guest’s Theory)
2. Maximum Distortion Energy Theory (Octahedral or Hencky and Van-mises Theory)
3. Maximum Strain energy theory
Brittle materials (fracture criteria)
1. Maximum (Principal) Stress Theory (Rankine’s Theory)
2. Columba –Mohr theory
3. Modified Mohr’s theory

Maximum normal (Principal)stress theory
 It is assumed that the failure or yield occurs at a point in a member when the max.
principal or normal stress in the biaxial stress system reaches the limiting strength of the
material in a simple tension test.
 In this case max. principal stress is calculated in a biaxial stress case and is equated to
limiting strength of the material.
Maximum principal stress
Minimum principal stress
71
2
2
1
2
2
xy
y
x
y
x









 










 




2
2
2
2
2
xy
y
x
y
x









 










 




For ductile materials
1 should not exceed in tension, FOS=Factor of safety
For Brittle materials
1 should not exceed in tension,
This theory is basically applicable for brittle materials which are relatively stronger in shear and not applicable to ductile
materials which are relatively weak in shear

72
Suppose we arrange the 3 principal um normal stress theory states that “Failure occurs whenever the largest
principal stress equals the strength”.
stresses for any stress sate in the form of σ1 > σ2 > σ3 then if yielding is criterion of failure, this theory
predicts that failure occurs whenever
σ1 = σyt. or σ3 = -σyc -------- For ductile materials
σ1 = σut. or σ3 = -σuc -------- For brittle materials
Where σyt and σyc are yield tensile and compressive strengths and σut and σuc are ultimate tensile and
compressive strengths respectively.
Where σyt and σyc are yield tensile and compressive strengths and σut and σuc are ultimate tensile and
compressive strengths respectively.
For a pure tension case this theory predicts that a component will fail when σy = , but experiment shows that
the component loaded in tension will permanently deform when the maximum shear stress is about 60% of
the yield strength i.e
 = 0.6 σy
FACTOR OF SAFETY
As per this theory the factors of safety is given by
Otherwise
Cont……

Cont……
73
+σ2
-σ1
-σ2
+σ1
σyc
σyt
σyc
σyt
o
• Failure occurs when one of the three principal
stresses reaches a permissible strength (TS). σ1 > σ
2 Failure is predicted to occur when σ1=σt and
σ2<-σc
• Where σt and σc are the tensile and compressive
strength
• For a biaxial state of stresses for even material
For a biaxial state of stresses for uneven material
Boundary for maximum – normal – stress theory
under bi – axial stresses
+σ2
-σ1
-σ2
+σ1
σyc
σyt
σyc
σyt
o

Maximum shear stress theory
74
•The failure or yielding is assumed to take place at a point in a
member where the max shear stress in a biaxial stress system
reaches a value equal to shear strength of the material obtained
from simple tension test.
•In a biaxial stress case max shear stress developed is given by
where
This theory is mostly used for ductile materials.
2
2
max
2
xy
y
x



 







 
 or
For 3D stress state max is the largest among the three values of

75
For 3D stress state max is the largest among the three values of
For 2D stress state max is the largest among the three values of (3 = 0)
For opposite sign of principal stresses 1, and 2 state max is equal
2
2
max
2
xy
y
x



 







 

Where y =0; Then max is equal
In the case of principal stresses 1,
and 2 are opposite sign.
Cont……

76
Cont……

Cont……

Distortion energy theory / Hencky Mises theory
• It is assumed that failure or yielding occurs at a point the member
where the distortion strain energy (also called shear strain energy) per
unit volume in a biaxial stress system reaches the limiting distortion
energy (distortion energy at yield point) per unit volume as determined
from a simple tension test.
• The maximum distortion energy is the difference between the total
strain energy and the strain energy due to uniform stress.
78

Distortion energy theory
 The criteria of failure for the distortion – energy theory is expressed as
 Considering the factor of safety
 For bi – axial stresses (σ3=0),
79
or
Cont……

 A component subjected to pure shear stresses and the corresponding Mohr’s
circle diagram is
80




Y
X
o σ1
-σ2


σ
Mohr’s circle for pure shear stresses
Element subjected to pure shear stresses
Cont……

81
From the figure, σ1 = -σ2 = and σ3=0
Substituting the values in the equation
We get
Replacing by y, we get
In the biaxial stress case, principal stress 1, 2 are calculated based on x ,y &
xy which in turn are used to determine whether the left hand side is more than
right hand side, which indicates failure of the component.
Cont……

82
Cont……

• Failure is assumed to take place at a point in a member where strain energy per unit
volume in a biaxial stress system reaches the limiting strain energy that is strain
energy at yield point per unit volume as determined from a simple tension test.
• Strain energy per unit volume in a biaxial system is
• The limiting strain energy per unit volume for yielding as determined from simple
tension test is
 Equating the above two equations then we get
83
Maximum Principal Strain / Saint Venant’s theory

 In a biaxial case 1, 2 are calculated based as x, y & xy
 It will be checked whether the Left Hand Side of Equation is less than Right
Hand Side of Equation or not. This theory is used for ductile materials.
• It is assumed that the failure or yielding occurs at a point in a member
where the maximum principal (normal) strain in a biaxial stress exceeds
limiting value of strain (strain at yield port) as obtained from simple
tension test.
• In a biaxial stress case
• One can calculate 1 & 2 given x , y & xy and check whether the material
fails or not, this theory is not used in general as reliable results could not be
detained in variety of materials.
84
Maximum Principal Strain / Saint Venant’s theory
Cont……

 This theory states a material will fails when the maximum
strain energy in a multi axial stress state exceeds the strain
energy obtained in a simple tensile test.
 This theory is very rarely used
85
Maximum Strain energy theory (Heigh’s Thoery):

 This theory suggests that principal Mohr’s circles be drawn
representing every available test condition and that the envelope of
these circles be taken as the envelope of any or all principal Mohr
circles representing stress state on the verge of failure.
 For biaxial stresses if 1 and 2 are alike then Coulomb Mohr
theory is used and hence the design equation is,
86
For torsion
 This theory is suitable for brittle materials
Columba –Mohr theory

 The Mohr’s theory predicts failure for any other stress state in which the largest
of the three Mohr’s circles corresponding to 1, 2 and 3 is tangent to the line
AE in figure.
 This theory can be used to predict either the beginning of yielding or the
beginning of fracture
87
Columba –Mohr theory
 The basis of Mohr’s theory of failure is shown in figure. Three circles are
representing the strength c uniaxial compression test, second circle is 
uniaxial tension test and third is at center taken from a test of pure shear at
failure.

Modified Mohr’s theory.
This theory is a modification of the Coulomb-Mohr theory and is the preferred theory for
brittle materials

Limitations of Theory of Failure
90
 The Maximum normal stress theory is in the poorest agreement
with test results, but is most often use for brittle materials and
for condition of fatigue loading. The Maximum normal stress
theory is also useful when a difference exist between the values
of tensile and compressive strengths.
 For ductile materials having the same tensile and compressive
strength either the Maximum shear stress theory or distortion
energy theory is used. The distortion energy theory gives the
most accurate results.
 For quick solution Maximum shear stress theory can be used.
 Maximum normal stress theory fails when the principal stresses
are of opposite sign.

Design data hand book chapter required for
Theories of failure problems
91
 Chapter- 1, 2 and 5
 Stress concentration in Machine Members
 Page number from 1 to 47 and from pp: 71 to 81

Problem on theories of failure
92
 The load on a bolt consists of an axial pull of 10kN together with a transverse
shear force of 5kN. Find the diameter of bolt required according to
1. Maximum principal stress theory
2. Maximum shear stress theory
3. Maximum principal strain theory
4. Maximum strain energy theory
5. Maximum distortion energy theory
Permissible tensile stress at elastic limit =100MPa and Poisson’s ratio =0.3
Given Data
Bolt is called round shaft has diameter ‘d’ = 
Axial pull = Tensile (P) = 10kN =10 103 N = 104 N
Shear force (Ps) = 5kN = 5 103 N
Permissible tensile stress at elastic limit = yt =100MPa
Poisson’s ratio =  =0.3

Solution
93
 Cross – sectional area of the bolt,
 Axial stress,
 And transverse shear stress,
2
2
7854
.
0
4
d
d
A 


2
2
2
1
/
73
.
12
7854
.
0
10
mm
kN
d
d
A
P




2
2
/
365
.
6
7854
.
0
5
mm
kN
d
A
Ps





MODULE -1
PART-D
STRESS CONCENTRATION
Dr. Mohammed Imran

Stress concentration
95
Figure 4.5 and 4.2 DDHB pp:59

Determination of stress concentration factor
Theoretical stress concentration factor is determined by
using two methods such as
 Mathematical analysis based on theory of elasticity
 Experimental methods
 Photo-elasticity
 Soap film
 Plaster model
 Electrical strain gauge method
 Brittle coating method
 Finite element Method
 Electrical flow analogy
96

Method of reducing stress concentration
97

Method of reducing stress concentration of
Mitigation of stress concentration
Cont……
98

Method of reducing stress concentration of
Mitigation of stress concentration
Cont……
99

Design data hand book chapter required for
Stress concentration problems
100
 Chapter-1 and 4
 Stress concentration in Machine Members
 Page number from 52 to 70

Problems on stress concentration
1. Determine the maximum stress in the following cases taking stress
concentration into account.
a) A Rectangular plate of 50mm x 80mm with a hole of 10mm diameter in the center
is loaded in axial tension of 10kN. Thickness of the plate is 10mm.
b) A circular shaft of 45mm diameter stepped down to 30mm diameter having fillet
radius of 6mm subjected to a twisting moment of 15Nm.
(VTU, July/Aug. 2003, June/July 2013)
2. A round shaft of 50mm diameter is subjected to a bending moment of
100Nm. If A transverse circular hole of 10mm diameter is drilled on the shaft,
determine the maximum stress induced on the shaft.
3. A rectangular plate of 50mm wide with a circular hole of diameter 10mm in
the centre is subjected to a bending moment of 10 Nm. If the thickness of the
plate is 10mm, determine the maximum stress induced in the plate.
101

4. A rectangular plate 15mm thick made of a brittle
material ia shown in fig Q(11), Calculate stresses at
each of three holes.
Fig Q(11)
102

5. A flat bar shown in fig Q(12), is subjected to axial load
of 100kN. Assuming that the stress in the bar is limited
to 200 N/m2, Determine the thickness of bar.
Fig Q(12)
103

5. A stepped shaft shows in fig Q(13), is subjected to
a transverse load. The shaft is made of steel with
ultimate tensile strength of 400 Mpa. Determine the
diameter d of the shaft based on the factor of
safety of 2.
Fig Q(13)
104

6. A flat plate subjected to a tensile force of 5kN is shown in fig
Q(14). The plate material is grey cast iron good. Determine the
thickness of the plate. Factor of safety is 2.5.
Fig Q(14)
105

8. Determine the safe load that can be carried by a bar of
rectangular C/S in fig. Q(15) limiting the maximum stress to
130 MPa taking stress concentration into account.
Fig Q(15)
106

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