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DESIGN OF MACHINE ELEMENTS – I
Design of Machine Members – I
(For B.E./B.Tech. Mechanical Engineering Students)
Dr. E.V.V. RAMANAMURTHY, M.Tech., Ph.D.,
Professor – Mech.
Raghu Institute of Tech.
Visakhapatnam.
Dr. S. RAMACHANDRAN, M.E., Ph.D.,
Professor – Mech.
Sathyabama Institute of Science & Technology
Chennai – 119
AIRWALK PUBLICATIONS
(Near All India Radio)
80, Karneeshwarar Koil Street,
Mylapore, Chennai – 600 004.
Ph.: 2466 1909, 94440 81904
Email: [email protected], [email protected]
www.airwalkbooks.com, www.srbooks.org
© First Edition: 8th
July – 2018
This book or part thereof should not be reproduced in any form without the
written permission of the publisher.
Price: Rs. 250/-
ISBN: 978-93-88084-08-6
Typesetting by: Akshayaa DTP, 48E, Sri Gangaiya Avenue, 2nd
Cross Street, Ramapuram Chennai – 89, Mobile: 9551908934.
Printed at:
ME 401 – DESIGN OF MACHINE ELEMENTS – IKTU Syllabus − VII Sem. Mech.
MODULE I
Introduction to Design − Definition, steps in design process, preferred numbers,
standards and codes in design − Materials and their properties − Elastic and
plastic behaviour of metals, ductile and brittle behaviour, shear, bending and
torsional stresses, combined stresses, stress concentration factor.
MODULE II
Theories of Failure − Guest’s Theory, Rankine’s Theory, St. Venant’s Theory,
Haigh’s Theory, and Von Mises and Hencky Theory. − Shock and Impact
loads, fatigue loading, endurance limit stress, factors affecting endurance limit,
factor of safety.
MODULE III
Threaded Joints − Terminology, thread standards, types of threads, stresses in
screw threads − Bolted joints − effect of initial tension, eccentric loading,
design of bolts for static and fatigue loading, gasketed joints, power screws.
MODULE IV
Design of riveted joints − Material for rivets, modes of failure, efficiency of
joint, design of boiler and tank joints, structural joints − Cotter and Knuckle
joints − Gib and Cotter Joint, analysis of knuckle joint. − Design of welded
joints- welding symbols, stresses in fillet and butt welds, Butt joint in tension,
fillet weld in tension, fillet joint under torsion, fillet weld under bending,
eccentrically loaded welds.
MODULE V
Springs − classification, spring materials, stresses and deflection of helical
springs, axial loading, curvature effect, resilience, static and fatigue loading,
surging, critical frequency, concentric springs, end construction. − Leaf springs
− Flat springs, semi elliptical laminated leaf springs, design of leaf springs,
nipping.
MODULE VI
Shafting − material, design considerations, causes of failure in shafts, design
based on strength, rigidity and critical speed, design for static and fatigue
loads, repeated loading, reversed bending. Design of Coupling − selection,
classification, rigid and flexible coupling, design of keys and pins.
*********
DESIGN OF MACHINE MEMBERS – IJAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY
KAKINADA – ANDHRA PRADESHSYLLABUS
UNIT – I
INTRODUCTION: General considerations in the design of Engineering
Materials and their properties − selection − Manufacturing consideration in
design, tolerances and fits − BIS codes of steels.
STRESSES IN MACHINE MEMBERS: Simple stresses − combined stresses
− torsional and bending stresses − impact stresses − stress strain relation −
various theories of failure − factor of safety − design for strength and rigidity
− preferred numbers. The concept of stiffness in tension, bending, torsion and
combined situations − static strength design based on fracture toughness.
UNIT – II
STRENGTH OF MACHINE ELEMENTS: Stress concentration − theoretical
stress concentration factor − fatigue stress concentration factor notch sensitivity
− design for fluctuating stresses − endurance limit − estimation of endurance
strength − Goodman’s line − Soderberg’s line − modified Goodman’s line.
UNIT – III
Riveted and welded joints − design of joints with initial stresses − eccentric
loading. Bolted joints − design of bolts with pre-stresses − design of joints
under eccentric loading − locking devices − both of uniform strength, different
seals.
UNIT – IV
KEYS, COTTERS AND KNUCKLE JOINTS: Design of keys-stresses in
keys-cotter joints-spigot and socket, sleeve and cotter, jib and cotter joints-
knuckle joints.
SHAFTS: Design of solid and hollow shafts for strength and rigidity − design
of shafts for combined bending and axial loads − shaft sizes − BIS code. Use
of internal and external circlips, gaskets and seals (stationary & rotary).
UNIT – V
SHAFT COUPLING: Rigid couplings − muff, split muff and flange couplings,
flexible couplings − flange coupling (modified).
UNIT – VI
MECHANICAL SPRINGS: Stresses and deflections of helical springs −
extension -compression springs − springs for fatigue loading, energy storage
capacity − helical torsion springs − co-axial springs, leaf springs.
ACKNOWLEDGEMENT
It has been a long time dream to write this book for the young
budding Design of Machine Elements – I, which came true. Keeping
in mind, the importance of fundamentals of the subject, and as well as
an approach towards the examination point of view, this book has been
written and compiled with easily understandable format. Simple
drawings have been drawn and many important problems have been
solved which are frequently asked in various University Examinations.
Crisp and precise explanations are rendered for the problems and as
well as the theory associated with it. This book is an eye opener for
the beginner with no prior knowledge on this subject. It is quite relevant
to state here that our parents’ blessings gave us the required courage
to write this book.
We express our sincere gratitude to the honourable Chairman
Thiru Dr. B. BABU MANOHARAN, M.A., M.B.A., Ph.D., Chairman,
St. Joseph’s Group of Institutions, who is the GOD FATHER for us,
who has given strong support and encouragement to write many number
of books and thanks to beloved Director Mr. B. SHASHI SEKAR, M.Sc.,
and Managing Director, Mrs. B. JESSIE PRIYA, M.Com., St. Joseph’s
Group of Institutions, for their constant encouragement and support to
bring out this book a success one.
Any errors, omissions and suggestions for the improvement of this
book, brought to our notice will be thankfully acknowledged and
incorporated in the next edition.
AUTHORS
Contents ii
CONTENTS
MODULE – I
INTRODUCTION TO DESIGN1.1. – 1.88
1.1. Introduction to design 1.1
1.1.1. Steps in design process 1.2
1.2. Classification of design 1.3
1.3. Factors influencing machine design 1.4
1.4. Classification of engineering materials 1.5
1.4.1. Mechanical properties of materials 1.6
1.4.2. Selection of materials 1.8
1.5. Ferrous metals 1.9
1.5.1. Cast iron 1.9
1.5.2. Steel 1.10
1.5.3. Cast steel 1.11
1.5.4. Alloy steels 1.11
1.5.5. System of designation for steels 1.12
1.6. Non-ferrous metals 1.14
1.7. Preferred numbers 1.15
1.8. Standards and codes 1.17
1.9. Simple stresses 1.19
1.9.1. Types of the load 1.19
1.10 Direct, bending, torsional, and shear stresses 1.20
1.11. Elastic and plastic behaviour of metals 1.24
1.11.1. Stress-strain diagram for ductile materials 1.24
1.11.2. Plastic behaviour of metals 1.26
1.11.3. Ductile fracture 1.26
1.11.4. Brittle fracture and stress – strain diagram for
brittle materials
1.29
1.12. Factor of safety 1.30
1.13. Impact stress 1.35
1.14. Principal stresses 1.37
1.15. Combined stresses due to eccentric loading 1.68
1.16. Stress concentration 1.72
1.16.1. Definition of stress concentration 1.74
1.16.2. Nominal stress (σo) cross section 1.74
1.16.3. Stress concentration factors 1.75
1.16.4. Reduction of stress concentration effects 1.76
MODULE – II
THEORIES OF FAILURE2.1. – 2.110
2.1. Design principles 2.1
2.1.1. Common modes of failure 2.1
2.1.2. Factor of safety 2.2
2.2. Theories of failure 2.3
2.3. Fatigue failure 2.39
2.4. Shock and impact loads 2.40
Contents ii
2.5. Design for fatigue (variable) loading 2.41
2.6. Endurance limit stress 2.43
2.6.1. Factors affecting endurance limit 2.45
2.7. Soderberg and Goodman diagrams 2.47
2.8. Goodman method for combination of stresses 2.49
2.9. Soderberg method for combination of stresses: (Ductile
materials)
2.51
2.10. Modified goodman’s diagram [For brittle materials] 2.52
2.11. Design equations for fatigue (variable) loading 2.54
2.11.1. Combined loading (bending and torsion) 2.56
2.11.2 Design for finite life 2.56
2.11.3. Cumulative fatigue damage 2.57
MODULE – III
DESIGN OF FASTENERS3.1. – 3.82
3.1. Threaded joints 3.1
3.1.1. Advantages of threaded joints 3.1
3.1.2. Disadvantages of threaded joints 3.2
3.1.3. Terminology (Nomenclature) of screw threads 3.2
3.1.4. Common types of screw fastening 3.4
3.1.5. Bolt of uniform strength 3.6
3.1.6. Thread standards (Designation of screw threads) 3.8
iii Contents
3.1.7. Design of bolts for cylinder cover 3.8
3.1.8. Stresses in screw threads and effect of initial tension 3.13
3.2. Bolted joints 3.17
3.2.1. Effect of initial tension 3.17
3.2.1.1. Direct tensile stress 3.18
3.2.1.2. Torsional shear stress 3.18
3.2.1.3. Shear stress in the threads 3.19
3.2.1.4. Compression stress 3.19
3.2.2. Design of bolted joints under eccentric loading 3.19
3.2.2.1. Eccentric load acting parallel to the axis
of bolt
3.19
3.2.2.2. Eccentric load acting perpendicular to the
axis of bolt
3.23
3.2.2.3. Eccentric load acting in the plane
containing the bolts
3.26
3.3. Gaskets 3.31
3.3.1. Gasketed joint 3.32
3.4. Design of power screws 3.39
3.4.1. Introduction 3.39
3.4.2. Application of power screw 3.39
3.4.3. Types of power screw threads 3.40
3.5. Friction in power screw (screw jack) 3.41
Contents iv
3.5.1. Action of the power screw in screw jack 3.41
3.5.2. Efficiency of a power screw in screw jack 3.44
3.6. Friction in V-threads (ACME thread) 3.49
3.7. Stresses in power screws 3.54
3.8. Design of screw jack 3.74
MODULE – IV
DESIGN OF RIVETED JOINTS – COTTER
AND KNUCKLE JOINTS – WELDED JOINTS
4.1. – 4.154
4.1. Riveted joints 4.1
4.1.1. Material for Rivets 4.2
4.1.2. Types of riveted joints 4.4
4.1.3. Terms used in riveted joints 4.6
4.1.4. Caulking and fullering 4.6
4.2. Modes of failures in riveted joint 4.7
4.3. Design of structural joints 4.10
4.4. Procedure for design of rivetted joint using PSG data book 4.62
4.4.1. Standard diameter of rivet and rivet hole diameter 4.63
4.4.2. Design stresses 4.63
4.4.3. Strength of a Riveted Joint 4.64
4.4.4. Efficiency of a Riveted Joint 4.64
4.5. Design of structural joints 4.64
v Contents
4.6. Design of boiler joints and tank joints 4.67
4.6.1. Design of longitudinal butt joint for a boiler 4.68
4.6.2. Design of circumferential joint 4.71
4.7. Knuckle joint 4.79
4.7.1. Proportions 4.80
4.7.2. Design procedure 4.80
4.8. Cotter joints 4.84
4.8.1. Design procedure 4.86
4.9. Sleeve and Cotter joint 4.106
4.9.1. Design of sleeve and cotter joint 4.106
4.10. Design of Gib and Cotter joint 4.113
4.10.1. Design procedure 4.114
4.11. Welded joints 4.118
4.11.1. Classification of welding 4.119
4.11.2. Types of welded joints 4.119
4.11.3. Stresses in Fillet Weld (Strength of Fillet Weld) 4.119
4.11.4. Fillet weld in tension (parallel fillet weld) 4.120
4.11.5. Method of indication of welding symbol 4.122
4.11.6. Combination of transverse and parallel fillet welds
in Tension
4.122
4.11.7. Fillet joint under torsion and bending – eccentric
load on welded joints
4.134
Contents vi
4.11.8. Design procedure 4.136
4.11.9. Stresses in butt weld 4.152
4.11.10. Butt joint in tension 4.153
MODULE – V
DESIGN OF SPRINGS5.1. – 5.120
5.1. Springs 5.1
5.2. Spring materials 5.1
5.3. Classification – important types of springs 5.3
5.4. Helical springs 5.5
5.4.1. Terminology used for helical springs 5.5
5.4.2. Close-coiled helical spring 5.6
5.4.3. Open-coiled helical spring 5.6
5.5. Stresses and deflection of helical compression springs
subjected to axial loading
5.6
5.5.1. Curvature effect 5.9
5.5.2. To find angular deflection θ 5.10
5.5.3. End connections for helical springs 5.10
5.5.4. Eccentric loading of springs 5.11
5.5.5. Buckling of compression springs 5.11
5.5.6. Helical tension springs (extension springs) 5.12
5.5.7. Types of ends for extension springs 5.12
vii Contents
5.5.8. Load on the extension spring 5.13
5.5.9. Resilience 5.13
5.5.10. Static loading 5.13
5.5.11. Critical frequency 5.14
5.6. Advantages of compression springs over extension springs 5.16
5.7. Surging (spring surge) and critical frequency 5.16
5.8. Energy stored in helical springs of circular wire 5.17
5.9. Fatigue loading of helical springs 5.65
5.10. Springs in series 5.69
5.11. Springs in parallel 5.70
5.12. Helical torsion spring 5.71
5.13. Flat spiral spring 5.78
5.14. Disc springs (or) belleville springs 5.81
5.15. Concentric springs 5.87
5.15.1. Design procedure of concentric springs 5.88
5.15.2. Coaxial springs 5.89
5.15.3. Advantages of concentric springs 5.96
5.16. Conical and volute springs 5.96
5.17. Leaf spring 5.97
5.17.1. Constructional details 5.98
5.17.2. Nipping of leaf springs 5.103
Contents viii
5.17.3. Initial gap (C) 5.104
5.17.4. Initial pre-load (Pi) 5.105
5.17.5. Factors not considered when deriving the design
formulae
5.107
5.17.6. Design formulae 5.107
5.17.7. Leaf spring material 5.108
5.17.8. Permissible stresses 5.108
MODULE – VI
DESIGN OF SHAFTS, KEYS AND COUPLINGS6.1. – 6.162
6.1. Introduction to design of shafts 6.1
6.2. Types of shaft (according to use) 6.2
6.3. Standard shaft diameters 6.2
6.4. Stress in shafts 6.3
6.5. Shafting materials 6.3
6.6. Causes of failure in shafts 6.3
6.7. Design of shafts – design considerations 6.4
6.8. Design of shaft based on strength 6.4
6.9. Design of shaft based on torsional rigidity 6.9
6.9.1. Stiffness (k) 6.9
6.9.2. Comparison of stiffness 6.10
6.9.3. Percentage saving of material 6.10
6.9.4. Calculation of twisting moment (Mt) 6.11
6.9.5. In case of gear drives 6.11
ix Contents
6.10. Design based on critical speed 6.12
6.11. Torque diagrams 6.14
6.12. Introduction to design of keys 6.85
6.13. Forces acting on a key 6.89
6.13.1. Assumptions made in the design of keys 6.90
6.13.2. Failures modes of keys 6.90
6.13.3. Design procedure of key 6.93
6.13.4. Effects of keyways 6.93
6.14. Introduction to design of couplings 6.97
6.15. Design of rigid coupling 6.98
6.15.1. Box (or) Sleeve (or) Muff coupling 6.98
6.15.2. Clamp (or) Compression (or) Split sleeve coupling 6.108
6.15.3. Flange coupling 6.111
6.16. Design of flexible couplings 6.116
6.16.1. Bushed pin flexible coupling 6.117
6.16.2. Oldham Coupling 6.160
6.16.3. Universal or Hooke’s coupling 6.160
Short Questions and Answers S.Q.A.1 – S.Q.A.40
Index I.1 – I.3
Contents x
MODULE – I
INTRODUCTION TO DESIGN
* Introduction to Design − Definition, steps in design process,
preferred numbers, standards and codes in design −Materials and their properties − Elastic and plastic
behaviour of metals, ductile and brittle behaviour, shear,
bending and torsional stresses, combined stresses, stress
concentration factor.
1.1. INTRODUCTION TO DESIGN
Definition
Design is an innovative and highly iterative process of formulating a
plan for the realisation of a specified need or to solve a specific problem
resulting in creation of a product that is functional, safe, reliable, competitive,
usable, manufacturable and marketable. A designer’s personal resources,
communicative ability and problem solving skills are interlinked with the
knowledge of technology and engineering tools to produce the product.
Machine design is creation of new machines and improving the existing
machines which are more economical in overall cost of production and
operation.
1.1.1. Steps in design process
Designing of a machine component or solving a design problem involves
the various steps as shown in the Fig. 1.1.
Fig. 1.1: Steps in Design Process
There is no general rigid rule but design can be made in several methods
and procedure is as follows.
(i) Identification of need/design problem
The first step in design process involves identifying the need or defining
of a design problem for which a machine needs to be designed. Definition of
a problem is more specific and must include all the specifications for the
object that is to be designed.
(ii) Synthesis
Synthesis is the scheme of connecting possible elements or mechanisms
or group of mechanism which gives the desired motion. Synthesis is sometimes
called the invention of the concept or concept design.
1.2 Design of Machine Elements – I - www.airwalkbooks.com
(iii) Analysis and optimisation
It is the process of calculating the various forces acting on each machine
element and the energy transmitted by each member. Each element has to
survive the analysis and elements which have the higher margins are optimised
to determine the best performance and this process is an iterative with
subsequent process.
(iv) Material selection
Based on the different material physical and functional properties suitable
material is selected for each machine element being designed.
(v) Design of size and stresses of elements
Sizes of each machine element member is defined based on the analysis
of various forces acting on each member causing stresses which are within
limits of the permissible stresses of material used. It is ensured that no element
deflects or deforms within the permissible limit.
(vi) Evaluation and modification
Evaluation is the final proof of successful design and usually involves
testing of a prototype in the laboratory. Any deficiency noticed, the process
gets iterated with the synthesis or analysis.
(vii) Detailed design and drawings
Once the design has satisfactorily completed the evaluation and
modification process, the detailed design and drawings of each machine
component, assemblies are made with complete specification of manufacturing
to reduce the overall cost.
(viii) Production and product
The component as per the drawing is manufactured, assembled and the
product is launched.
1.2. CLASSIFICATION OF DESIGN
Machine design is classified as follows
(a) Adaptive design
Adaptive design is one in which designer’s work is concerned with the
adaptation of the existing design requiring no special skills and knowledge.
Examples are bicycles and IC engines where development has practically
ceased except for certain minor modification and alterations.
Introduction to Design 1.3
(b) Developed design
In developed design, a high standard of scientific training is essential
when the proven existing designs are to be modified to their method of
manufacture, material, appearance etc. In this case, designer starts with an
existing design and the final product outcome is remarkably different from the
original product.
(c) New design
New design requires a lot of research, technical capability and creativity.
A few designers bring out new machines by making use of basic scientific
principles. Designers with high personal capabilities of higher order can only
take up new design.
1.3. FACTORS INFLUENCING MACHINE DESIGN
Machine design is greatly influenced by the factors arising from the
customer’s requirement and factors concerned with the manufacture. Machine
component design should meet the following important requirements.
1. Functional
2. Operational
3. Maintenance
4. Material used
5. Manufacturing methods used
(i) Factors related to customer requirement
(a) Material cost, production cost, operating cost.
(b) Mechanical and chemical environmental influence.
(c) Ease of maintenance.
(d) Economy of energy consumption.
(e) Handling, shipping and transportation.
(f) Size, weight and form.
(g) Appearance and Aesthetics.
1.4 Design of Machine Elements – I - www.airwalkbooks.com
(h) Quantity and delivery schedules.
(i) Spare part availability.
(ii) Factors related to manufacturing
(a) Loading and stress limits.
(b) Working principle and design.
(c) Strength, wear resistance and corrosion resistance.
(d) Material selection, its condition and availability.
(e) Manufacturing method and assembly method.
(f) Limits, fits and tolerance.
(g) Type and quality of surface finish.
(h) Protective coating requirements.
(i) Type of standards required.
(j) Jigs, fixtures and tools required.
(k) Gauges and inspection method.
(l) Design for manufacture.
(m) Type of scrap generated and utilisation.
(n) Interchangeability.
1.4. CLASSIFICATION OF ENGINEERING MATERIALS
Engineering materials are broadly classified as
(i) Metal and its alloys.
Eg: Iron, steel, copper, aluminium
(a) Ferrous metals containing iron as major constituent.
(b) Non-ferrous metal containing other than iron like Cu, Al, etc.
(ii) Non metals like plastic, fibre, rubber, glass, composite, etc.
Introduction to Design 1.5
MATERIALS AND THEIR PROPERTIES
1.4.1. Mechanical Properties of Materials
(a) Strength is the ability of the material to resist the externally applied loads
without failure (fracture or yielding). Measure of strength is ultimate strength
for brittle materials and yield point stress for ductile materials.
(b) Elasticity is a property of the material to regain the original shape after
deformation on removal of load.
(c) Plasticity enables the material to permanently retain the deformation
produced by the externally applied loads.
(d) Stiffness (Rigidity) enables the material to resist deformation under loads.
(e) Ductility enables the material to be drawn into wire when tensile force is
applied. Steel, aluminium and copper are ductile materials. Ductile material
has large plastic deformation before rupture while brittle material has a small
plastic deformation (Fig. 1.2 (a) & (b)).
Ductility helps the material to absorb large overloads. Operations like
bending, drawing, heading, etc., require ductility in the materials.
(f) Brittleness means lack of ductility. Cast iron is a brittle material.
(g) Malleability is ability of material to be drawn into thin sheets under
compressive force. Eg: Gold, Aluminium
(h) Resilience of a material is its capacity to absorb energy with in the elastic
range. Resilience enables material to resist shock and impact and hence it is
desired in springs. The shaded area (Fig. 1.3 (a)) represents modulus of
resilience, i.e., strain energy stored per unit volume when the stress is at the
proportional limit.
1.6 Design of Machine Elements – I - www.airwalkbooks.com
(i) Toughness enables the material to absorb energy in the plastic range
(Fig. 1.3 (b)), it enables the material to be twisted or bent under a sudden
load before rupture. Shaded area in Fig. 1.3 (b) represents the modulus of
toughness.
(j) Hardness enables the material to resist indentation, wear or plastic
deformation.
(k) Creep: At elevated temperatures, materials yield and undergo permanent
deformation at a stress lower than the yield point stress. In addition to the
loss of strength, there is a continuous gradual elongation of the members at
high temperature over a long period of time, known as creep. Steam and gas
turbine castings, turbine blades, rocket engines, missile nose cones and nuclear
reactor components are subjected to creep.
(l) Strain hardening: When drawing ductile materials like mild steel, copper,
brass and aluminium through dies or when rolling them between rollers, plastic
deformation takes place and this increases the yield point stress and ultimate
strength. This is known as strain hardening.
(m) Damping capacity is the ability of a material to damp vibrations by
absorbing the kinetic energy of vibration. Cast iron has greater damping
capacity than steel. Hence C.I is used in machine tools to decrease vibrations.
(n) Hardenability is the ability of steel to through harden. Hardenability can
be improved by using alloying elements like boron, vanadium, manganese,
chromium and molybdenum.
(o) Machinability is the ease with which the metal can be removed in
machining operations like turning, drilling, etc. When selecting materials for
mass production of components, machinability is a deciding factor. Good
machinability results in less tool wear, good surface finish and less power
Introduction to Design 1.7
consumption. Good machinability is obtained by adding sulphur and lead in
steel. However, there is a reduction in tensile strength.
1.4.2. Selection of Materials
Selection of a proper material for the machine component is one of the
most important steps in the process of machine design. The selection process
may involve trial and error method.
The following factors should be considered while selecting the material.
1. Availability: The material should be readily available in large enough
quantities to meet the requirement.
2. Cost: When the limiting cost of the component exceeds, the designer has
to consider other alternative materials. In cost analysis, there are two factors
(a) Cost of Material.
(b) Cost of Processing the material into finished goods.
3. Mechanical Properties: The important mechanical properties from the
consideration of design are strength, rigidity, ductility, hardness, toughness.
Depending upon the service conditions and the functional requirements,
different mechanical properties are considered and a suitable material is
selected.
H For example, the material for the connecting rod of an internal
combustion engine should be capable to withstand the fluctuating
stresses induced due to combustion of fuel. In this case, the endurance
strength becomes the criterion of design.
H The piston rings should have hard surface to resist the wear. In this
case, surface hardness is the design criterion.
H In case of bearing materials, low coefficient of friction is desirable.
H In case of clutch or brake lining, high coefficient of friction is
required.
4. Manufacturing Considerations: Some times, an expensive material is more
economical than a low priced material, which is difficult to machine.
When the product is of a complex shape, casting properties are important.
1.8 Design of Machine Elements – I - www.airwalkbooks.com
The manufacturing processes, such as casting, rolling, forging, extrusion,
welding and machining govern the selection of the material.
1.5. FERROUS METALS
1.5.1. Cast Iron
Cast iron is an alloy of iron, carbon and silicon with carbon content
around 3%. The type of cast irons are grey iron, white iron, chilled cast iron,
malleable iron, spheroidal or modular graphite iron, alloy cast iron.
Advantages
1. It is available in large quantities and is produced on a mass scale.
The tooling required for the casting process is relatively simple and
inexpensive. This reduces the cost of Cast iron products.
2. Cast iron components can be given any complex shape without
involving costly machining operations.
3. Cast iron has a higher compressive strength compared to steel.
4. Cast iron has an excellent ability to damp vibrations, which makes
it an ideal choice for machine tool guides and frames.
5. Cast iron has more resistance to wear even under the conditions of
boundary lubrication.
6. Mechanical properties of cast iron parts do not change between room
temperature and 350°C.
Disadvantages
1. It has a poor tensile strength compared to steel.
2. Cast iron does not offer any plastic deformation before failure, and
exhibit no yield point. The failure of cast-iron parts is sudden and
total.
3. Cast iron is brittle and has poor impact resistance.
4. The machinability of cast iron parts is poor compared to parts made
of steel.
Introduction to Design 1.9
Applications
1. Machine tool-beds, Frames and Guideways, Hydraulic cylinders,
Pulleys, Gears, Anvils etc.
2. I.C. engine-cylinder block, cylinder head, Flywheel, Brake drums etc.
1.5.2. Steel
Steel consists of iron, carbon and manganese. Carbon content is less than
1.7%.
1. Effects of various elements in steel
Carbon: Increase in carbon content from 0% to 0.83% increases the ultimate
strength. If the carbon content is going to be more than 0.83%, the increase
in carbon content reduces the strength. Hardness increases with carbon content
but ductility and weldability decrease as carbon content increases.
Manganese: As manganese content increases, ultimate strength and hardness
increase and weldability decreases.
Sulphur: Sulphur lowers toughness and makes the steel soft (adds to free
cutting).
Silicon: Silicon is added to steel as a deoxidiser to minimize the last traces
of oxygen.
2. Classifications
Low carbon steels/mild steels Carbon content 0.05 to 0.25%.
Medium carbon steels Carbon content 0.3 to 0.83%.
High carbon steels Carbon content 0.9 to 1.3%.
3. Applications
Carbon 0.1 to 0.2% Tubing, forgings, pressed steel parts, rivets, screws and
for case hardened parts.
Carbon 0.2 to 0.3% General purpose grade. Forged and machined parts,
structural members, boiler plates.
Carbon 0.3 to 0.55% Forged and machined parts, automotive bolts, shafts.
Heat treated to a hardness of 200 − 450 BHN.
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Carbon 0.55 to 0.75% Rails, hammers.
Carbon 0.65 to 0.85% Coil and flat springs.
Carbon 0.6 to 0.95% Tools, punches, dies, saws. Heat treated to a hardness
of 375 − 500 BHN.
1.5.3. Cast steel
Cast steel has higher strength, higher endurance limit, much higher
ductility and greater toughness than cast iron. Cast steel weighs less than cast
iron for the same strength and stiffness. Steel castings are used for heavy
machinery bases, machine frames, gears, wheels, etc.
Designation
Example: CS 130 Unalloyed steel castings with minimum tensile strength
1300 N ⁄ mm2.
1.5.4. Alloy steels
In alloy steels alloying elements are added to impart special effects like
higher tensile strength, increased toughness and hardness, greater resistance to
corrosion etc.
Most common alloying elements and their effects
1. Chromium improves hardenability, corrosion resistance and increases wear
resistance and hardness. If the chromium content is more than 11%, the steel
is called stainless steel. Stainless steels offer high resistance to corrosion.
2. Nickel increases strength (ultimate strength) without decreasing ductility.
Nickel steels have good impact properties.
Nickel and chromium are mostly used together to obtain the toughness
and ductility provided by nickel and hardness and wear resistance provided
by chromium.
3. Molybdenum improves hardenability and creep strength.
4. Vanadium improves hardenability, imparts toughness, retains strength and
hardness at elevated temperature, improves shock and fatigue resistance
(increases resilience) and retards softening during tempering.
Introduction to Design 1.11
5. Tungsten retains hardness even at elevated temperature, improves wear
resistance and imparts toughness and hardness.
1.5.5. System of designation for steels
1. Carbon steel
Example: C 30
Prefix C stands for carbon, 30 for the average percentage of 0.30%.
Steels specified by tensile properties
Example: St 40 − Steel having a minimum tensile strength of 400 N/mm2
2. Alloy steel
Prefix C is not used.
Chemical symbols of significant elements are arranged in the descending
order of their average percentages.
Underlining by a bar indicates percentage is in the decimal value.
Example: 15 Ni 13 Cr 1 Mo 12
Carbon 0.15% average
Nickel 1.3% average
Chromium 1% average
Molybdenum 0.12% average
3. Carbon tool steel
Example: T 80 − Letter T for tool steels. Carbon 0.8% average.
Designation of Steels
A large number of varieties of steel are used for machine components.
Steels are designated by a group of letters or numbers indicating any of the
following three properties:
1. Tensile strength
2. Carbon content
3. Composition of alloying element.
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Steels which are standardised on the basis of their tensile strength without
detailed chemical composition.
Ex: Fe 360 indicates a steel with a minimum tensile strength of 360 N/mm2.
(or) Fe E250 indicates a steel with a minimum yield strength of 250 N/mm2.
4. Designation of Plain carbon steels
(a) A number indicating 100 times the average percentage of carbon.
(b) A letter C.
(c) A number indicating 10 times the average % of Mn.
Example: 55C4.
Indicates plain carbon steel with 0.55% of Carbon and 0.4% of
Manganese.
Ex: A steel with 0.35 − 0.45% C and 0.7 to 0.9% Mn is designated as 40C8.
In case of cast alloy steels, chemical symbol of significant alloying
elements are arranged in descending order of percentage content. The average
percentage of each alloying element is indicated by the number following its
chemical symbol. When the alloying element is less than one percent, it is
written upto two decimal places underlined by a bar.
Ex Carbon 0.12 to 0.81% 15 Cr 65 Silicon and Manganese are not
important alloying elements and
they are deletedSilicon 0.10 to 0.35%
Manganese 0.4 to 0.6%
Chromium 0.5 to 0.8%
Ex Carbon 0.15 to 0.25% Average Carbon content − 0.2%
or 20 hundredth of a percent.Silicon 0.1 to 0.5%
Manganese 0.3 to 0.5% 20 Cr 18 Ni 2
Nickel 1.5% to 2.5%
Chromium 16 to 20%
Ex 40 Cr 14 Carbon 0.4%
Chromium 0.14%
Introduction to Design 1.13
1.6. NON-FERROUS METALS
These are used to meet the following requirements:
(a) Resistance to corrosion
(b) Ease of casting and cold working
1. Copper base alloys: Copper is alloyed with zinc to produce brass. If alloyed
with tin, aluminium, manganese, silicon or phosphorous, it is called bronze.
Brass is used in application where moderate strength and ductility,
resistance to corrosion and resistance to wear are required. Bronze is superior
to brass in the above qualities but it is more costly.
Materials for bearing linings
Tin babbit 87.75% tin, 4% copper, 8% antimony and 0.25% bismuth
Lead babbit 80% lead, 20% antimony
2. Aluminium and its alloys
Pure aluminium is highly ductile and has good forming properties, but
it has poor casting and machining properties.
When alloyed with copper, ultimate strength and endurance strength
increased and there is an improvement in machinability and casting
characteristics. Aluminium-Copper alloys are used in crank cases, transmission
housing, etc.
Aluminium-Silicon alloys have better mechanical properties and corrosion
resistance than Aluminium-Copper alloy, but they have poorer machinability.
These alloys are used in marine castings, water jacket housings and castings
where machining is minimum.
Duralumin
Duralumin is an Al-Cu-Mg-Mn alloy and it has good corrosion resistance
and strength.
3. Non-metallic materials
Non-metallic materials used are
(a) Plastics (light weight housings, panels, flexible hoses).
(b) Fibre Reinforced Plastics (car bodies, boat hulls).
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(c) Rubber (insulators, belts, piping, tyres).
(d) Leather (belts).
(e) Asbestos (friction lining for clutches and brakes).
1.7. PREFERRED NUMBERS
1.7.1. Preferred Numbers
When a machine is to be made in several sizes with different powers
or capacities, it is necessary to decide what capacities will cover a certain
range efficiently with minimum number of sizes. It has been shown by
experience that a certain range can be covered efficiently when it follows a
geometrical progression with a constant ratio. The preferred numbers are the
conventionally rounded off values derived from geometric series including the
integral powers of 10 and having as common ratio of the following factors:
Ratio’s Series
5√10 = 1.58 (R5)
10 √10 = 1.26 (R10)
20 √10 = 1.12 (R20)
40 √10 = 1.06 (R40)
These four series are called basic series. The other series called derived
series may be obtained by simply multiplying or dividing the basic sizes by
10, 100, etc. The preferred numbers in the above four series shown in
Table 1.1 as per standard IS: 1076 (Part I).
Notes
1. The standard sizes (in mm) for wrought metal products are shown
in Table 1.2 according to IS: 1136 − 1990. The standard G.P. series
used correspond to R10, R20 and R40.
2. The hoisting capacities (in tonnes) of cranes are in R10 series, while
the hydraulic cylinder diameter are in R40 series and hydraulic
cylinder capacities are in R5 series.
Introduction to Design 1.15
3. The basic thickness of sheet metals and diameter of wires are based
on R10, R20 and R40 series. Wire diameter of helical springs are
in R20 series.
4. Standard spindle speeds for machine tools are given in Table 1.3.
5. Also preferred basic and design sizes are given in PSG design data
book Pg. No. 3.12.
Table 1.1: Preferred numbers of the basic series. IS: 1076−1990
Basic
seriesPreferred numbers
R5 1.00, 1.60, 2.50, 4.00, 6.30, 10.00
R10 1.00, 1.25, 1.60, 2.00, 2.50, 3.15, 4.00, 5.00, 6.30, 8.00, 10.00
R20 1.00, 1.12, 1.25, 1.40, 1.60, 1.80, 2.00, 2.24, 2.50, 2.80, 3.15,
3.55, 4.00, 4.50, 5.00, 5.60, 6.30, 7.10, 8.00, 9.00, 10.00
R40 1.00, 1.06, 1.12, 1.18, 1.25, 1.32, 1.40, 1.50, 1.60, 1.70, 1.80,
1.90, 2.00, 2.12, 2.24, 2.36, 2.50, 2.65, 2.80, 3.00, 3.15, 3.35,
3.55, 3.75, 4.00, 4.25, 4.50, 4.75, 5.00, 5.30, 5.60, 6.00, 6.30,
6.70, 7.10, 7.50, 8.00, 8.50, 9.00, 9.50, 10.00
Table 1.2: Preferred sizes for wrought metal products. IS: 1136-1990
Size range Preferred sizes (mm)
0.01 − 0.10 mm 0.02, 0.025, 0.030, 0.04, 0.05, 0.06, 0.08 and 0.10
0.10 − 1 mm 0.10, 0.11, 0.12, 0.14, 0.16, 0.18, 0.20, 0.22, 0.25,
0.28, 0.30, 0.32, 0.35, 0.36, 0.40, 0.45, 0.50, 0.55,
0.60, 0.63, 0.70, 0.80, 0.90 and 1
1-10 mm 1, 1.1, 1.2, 1.4, 1.5, 1.6, 1.8, 2.22, 2.5, 2.8, 3, 3.2, 3.5,
3.6, 4, 4.5, 5, 5.5, 5.6, 6, 6.3, 7.8, 9 and 10
10 − 100 mm 10 to 25 (in steps of 1 mm), 28, 30, 32, 34, 35, 36,
38, 40, 42, 44, 45, 46, 48, 50, 52, 53, 55, 56, 58, 60,
62, 63, 65, 67, 68, 70, 71, 72, 75, 78, 80, 82, 85, 88,
90, 92, 95, 98 and 100
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Size range Preferred sizes (mm)
100−1000 mm 100 to 200 (in steps of 5 mm), 200 to 310 (in steps of
10 mm), 315, 320, 330, 340, 350, 355, 360, 370, 375,
380 to 500 (in steps of 10 mm), 520, 530, 550, 560,
580, 600, 630, 650, 670, 700, 710 and 750 − 1000 (in
steps of 50 mm)
1000−10000 mm 1000, 1100, 1200, 1250, 1400, 1500, 1600, 1800, 2000,
2200, 2500, 2800, 3000, 3200, 3500, 3600, 4000, 4500,
5000, 5600, 6000, 6300, 7000, 7100, 8000, 9000 and
10000
Table 1.3: Standard spindle speeds for machine tools
Basic series Preferred numbers
R20 φ = 1.12 100, 112, 125, 140, 160, 180, 200, 224, 250, 280, 315,
355, 400,450, 500, 560, 630, 710, 800, 900, 1000
R20/2 φ = 1.25 112, 140, 180, 224, 280, 355, 450, 560, 710, 900
R20/3 φ = 1.4 11.2, 16, 22.4, 31.5, 45, 63, 90, 125, 180, 250, 355,
500, 710, 1000, 1400, 2000, 2800, 4000, 5600, 8000
R20/4 φ = 1.6 112, 140, 180, 224, 280, 355, 450, 560, 710, 900
R20/6 φ = 2 11.2, 22.4, 45, 90, 180, 355, 710, 1400, 2800, 5600
1.8. STANDARDS AND CODES
A standard is a set of specifications for parts, materials or processes
intended to achieve uniformity, efficiency and specified quality. The main
purpose of standardization is to establish mandatory (or) obligatory norms for
the design and production of machines so as to reduce variation in their types
and grades, and to achieve quality characteristics in Raw materials, semi
finished and finished products.
The benefits of standardization are given below.
H Better product quality, reliability and longer service life.
H Mass production of components at low cost.
Introduction to Design 1.17
H Easy availability of parts for replacement and maintenance.
H Less time and effort required to manufacture.
H Reduction in variation in size and grades of an article.
H The Bureau of Indian Standards (BIS) has standardized a number of
items for the benefits of designers and users.
H Some Examples for standards are given below, most commonly
referred standards by piping engineers are:
IS 210 : Grey iron castings.
IS 226 : Structural steel (Superseded by IS 2062).
IS 554 : Dimensions of pipe thread.
IS 778 : Specification for Copper Alloy gate, globe and check
valves.
IS 1363 : Hexagonal Bolts, Screws and Nuts − Grade C.
IS 1364 : Hexagonal Bolts, Screws and Nuts − Grade A and B.
IS 1538 : Cast Iron fittings.
IS 1979 : High Test line pipe
IS 2002 : Steel plates.
IS 3114 : Code of practice for laying pipes.
IS 13095 : Butterfly Valves.
IS 13257 : Ring type joints gasket and grooves your flanges.
CODE
H A code is a set of specifications for analysis, design, manufacture and
construction of something.
H The purpose of a code is to achieve a specific degree of safety,
efficiency and performance or quality.
H Examples for standard and code in design are given here.
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IS 3935 : 1966 Code practice for composites construction.
IS 3201 : 1988 Criteria for design and construction of pre cast trusses
and purling.
IS 6332 : 1984 Code of practice for construction of floor and roofs
using pre-cast doubly − curved shell units.
IS 14215 : 1994 Code of practice for design and construction of floor
and roof with RC channel units.
IS 11447 : 1985 Code of practice for construction with large panel
prefabricates.
1.9. SIMPLE STRESSES
LOAD: It is an external force acting on machine member.
1.9.1. Types of the Load
1. Static load (or) Steady load
2. Variable load (or) Dynamic load (Load varying with time)
3. Suddenly applied load (or) Shock load (When load is suddenly
applied or removed)
4. Impact load (When load is applied with some initial velocity)
STRESS
The internal resistance force per unit area at any section of the body is
known stress.
Mathematically,
Stress = load
unit area in N ⁄ mm2
Introduction to Design 1.19
1.10. DIRECT, BENDING, TORSIONAL, AND SHEAR STRESSES
1. Direct stress stresses due to Axsial loading (or) Direct loading
(i) Tensile Stress (σt)
Tensile Stress = σt = P
A
[K.M. Data Book Pg. No. 2] Eqn. No. 1.1(a)
P − Tensile load in N
A − Cross-sectional area in mm2
[K.M. Means K. Mahadevan, K. Balaveera Reddy − Design Data Hand Book]
(ii) Compressive stress (σc)
Compressive stress (σc) = −
P
A
P − Compressive load in N
A − Cross-sectional area in mm2
If
l Length of specimen in m
E Modulus of elasticity in N/m2
G Modulus of rigidity in N/m2
γ Poisson’s ratio
e Elongation in m
We have
(iii) Elongation (e) = P.l
A .E
[K.M. Data Book Pg. No. 3, Eqn. No. 1.2(b)]
(iv) Poisson’s ratio (γ) = E
2G
[Refer PSG Design Data Book Page No 7.1]
* In general, for tensile load, consider + ve sign.
for compressive load, consider − ve sign.
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1.(a) Contact stress, Bearing stress (or) Crushing stress
It is a localised compressive stress at the surface of contact between two
members.
Bearing stress = Pb =
P
l × d
where P − Radial load acting on the
journal.; l − length of the contact
between journal & bearing; d − Dia. of
the journal; (l × d) − represents the
projected area. (It is not cross-sectional
area).
Bearing stress (or) Crushing stress
= Radial load
Projected area
= P
l × d
2. Bending Stress (Stresses due to Bending load)
(i.e., Due to transverse load)
According to theory of simple bending.
σb
y =
M
I =
E
R
[PSG Databook Pg. No. 7.1]
[K.M. Databook Pg. No. 3 Eqn. No. 1.3(a)]
σb = M ⋅ y
I
where σb − Bending stress; y − Distance between Neutral axis and Extreme
outer fibre; M − Bending moment; I − Movement of Inertia, R − Radius of
curvature.
In Fig 1.7(a) at (A), (σb) Bendingstress is compressive σ
b =
M ⋅ yl
at (B), (σb) Bending stress is tensile σ
b = +
M ⋅ yI
Introduction to Design 1.21
In Fig. 1.7(b) at (A), Bending stress is tensile (σb) = + M ⋅ y
I
at (B) Bending stress is compressive (σb) = − M ⋅ y
I
* In general, Tensile bending stress, consider + ve sign.
Compressive bending stress, consider − ve sign.
H For Circular Cross-Section
σb = bending stress = ± M ⋅ y
Iy =
d
s
= ± M ⋅ d ⁄ 2π ⁄ 64 d4
= ± 32 M
π d3 I =
π64
d4
σb = ± 32 M
π d3
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3. Torsional stress (Stresses due to torque)
Consider a shaft of radius (r) and diameter (d) subjected to a twisting
movement or torque (T) as shown in Fig. 1.8.
If
T Torque or twisting moment in Nm
r Radius of shaft in m
J Polar moment of inertia in m4
l Length of shaft in m
G Modulus of rigidity in N/m2
N Speed of shaft in RPM
τ Shear stress in (kgf/cm2 in MKS and in N/m2 in SI)
θ Angle of twist in radians
We have Torsional equation
T
J =
Gθl
= τr
[Refer PSG DB Pg. No. 7.1]
[K.M. Databook Pg. No. 3
Eqn. No. 1.3(b)]
∴ Shear stress τ = Tr
J
Angle of twist θ = Tl
GJ
[Refer PSG DB Pg. No. 7.1]
[K.M. Databook Pg. No. 3
Eqn. No. 1.3(c)]
Polar moment of Inertia (J)
For solid shaft J = π d4
32
Hollow shaft J = π (d
04 − d
i4)
32
d0 outer diameter
di inner diameter
Introduction to Design 1.23
4. Shear Stresses
(i) Direct Shear Stress
Ex: Shearing of Rivet.
Let d − Dia. of the rivet.
P − shear load.
Direct Shear Stress = τd =
P
π4
d2
(ii) Torsional Shear τs
When a body is subjected to twisting moment or torque (T)
According to torsion equation T
J =
τs
r
T
π32
d4
= τs
d
2
where T − Twisting moment.; J − Polar moment of inertia.;
r − Radius of the shaft = d
2 ,
JS =
π32
d4 (Solid shaft); JH
= π32
(d04 − d
i4) (Hollow shaft)
τs = Torsional shear stress = 16 T
π d3
1.11. ELASTIC AND PLASTIC BEHAVIOUR OF METALS
1.11.1. Stress-Strain diagram for ductile materials
In designing various parts of a machine, it is necessary to know the
mechanical properties of the material. These properties are commonly
determined by conducting a standard tensile test, on UNIVERSAL TESTING
MACHINE (UTM).
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This test consists of gradually loading a standard specimen of a material
and recording the corresponding values of load and elongation until the
specimen fractures. The load is applied gradually and measured by a testing
machine. The stress is determined by dividing the load value by the original
cross-sectional area of the standard specimen.
The elongation is measured by determining the distance between the two
reference points on the specimen which are moved apart by the application
of gradual load. The original length between two reference points is known
as Gauge length.
The strain is determined by dividing the elongation value by the gauge
length.
The values of the stress and strain are used to draw the stress-strain
diagram of the material used.
The stress-strain diagram for a mild steel under tensile test is shown in
the Fig. 1.10.
Proportional limit: From point O to A is a straight line, which represents
that the stress is proportional to strain. The Hook’s law holds good upto point
A, and it is known as proportional limit.
Elastic limit: It may be noted that even if the load is increased beyond point
A upto the B, the material will regain its shape and size, when the load is
removed. The material has elastic properties upto the point B. This point is
A – Proportional limit
B – Elastic limit
C – Upper yield point
D – Lower yield point
E – Ultimate (or) Maximum stress
F – Breaking point.
Introduction to Design 1.25
known as elastic limit. It is defined as the stress developed in the material
without any permanent deformation.
1.11.2. Plastic Behaviour of Metals
Yield point: If the material is stressed beyond point B, the plastic stage will
reach, i.e., on removal of the load, the material will not be able to recover
its original size and shape. Beyond point B, the strain increases at a faster
rate with any increase in the stress until the point C is reached. At this point,
the material yields before the load and there is an appreciable strain without
any increase in stress. The stress corresponds to yield point is known as yield
point stress.
C − Upper yield stress,
D − Lower yield stress,
E − Ultimate stress,
F − Breaking stress.
1.11.3. Ductile fracture
A material is said to be ductile if it can withstand large plastic
deformation. A ductile fracture can be defined as a fracture which is the result
of intense localised plastic deformation of the metal at the tip of the crack.
At elevated temperatures, all fractures tend to become ductile because slip can
occur more easily.
The ductile fracture is defined as the fracture which takes place by a
slow propagation of crack with extensive plastic deformation. The crack is
stable resisting further extension unless applied stress is increased.
The ductile fracture takes place in metals which do not harden much
and is the end result of extensive plastic deformation of a specimen in a tensile
test.
The deformation of the specimen with the corresponding force (i.e. load)
is noted and a graph is drawn as shown in Fig. 1.11.
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From the graph, we see that from O to A, the ratio of stress to strain
is constant. After the point A, the ratio is not constant and is changing. The
point B is the yield point after which the strain increases more quickly than
the stress. The ultimate tensile stress is at point E and if the stress is increased
more than this, fracture will occur.
1.11.3.1. Mechanism of Ductile Fracture
When the tensile stress across the specimen is increased beyond the
elastic limit, there is an uniform reduction in its cross-sectional area.
Steps in Ductile Fracture
(a) Necking.
(b) Formation of microvoids.
(c) Coalescence of microvoids to form a crack.
(d) Crack propagation by shear deformation.
(e) Fracture.
(a) Necking
When the tensile stress is increased beyond ultimate tensile strength
value, a neck is formed some where near the middle of the specimen. This
process of formation of Neck is called as Necking as shown in Fig. 1.12 (a).
Introduction to Design 1.27
(b) Formation of microvoids
The continuation of the loading causes the plastic deformation that
produces many fine cavities in the specimen called as microvoids as shown
in Fig. 1.12 (b).
Here, the small cracks are created due to the combination of dislocations,
which were formed during the manufacture of the material.
(c) Formation of crack
As the stress increases, the cracks join together to form larger cavities.
It has been observed that during the formation of small cracks, the neck
propagates due to the combination of dislocation as shown in Fig. 1.12 (c).
(d) Crack propagation by shear deformation
These cavities keep growing outwards due to the increase of plastic
deformation and a central crack is formed.
It has been observed that the continuation of plastic deformation produces
bigger cracks in the specimen as shown in Fig. 1.12 (d).
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(e) Fracture
On further increase of stress, the crack propagates on the surface of the
specimen which results in a ‘Cup and Cone’ type of fracture as shown in
Fig. 1.12 (e).
The ‘cup’ region of the fracture has a very fibrous appearance. The
appearance is as if the individual elements are split into longitudinal fibers
and drawn to a point before fracture. The outer ‘cone’ is a region of highly
localised shear. Extensive localised deformation occurs by sliding of grains
one over the others.
Ductile fracture is a less serious problem and compared to brittle fracture
it is slow and occurs with the expenditure of large amount of energy.
1.11.4. Brittle fracture and Stress – Strain diagram for brittle materials
H Brittle fracture may be defined as a fracture which takes place by the
rapid propagation of crack with a negligible deformation.
H It has been observed that in amorphous materials, the fracture is
completely brittle.
H But in crystalline materials, it occurs after small deformation.
H The brittle fracture generally occurs in body centred cubic and
hexagonal close packed single crystal at very low temperatures.
Introduction to Design 1.29
H In crystalline materials, the fracture takes place normal to the specific
crystallographic planes called cleavage planes.
H Brittle fracture also occurs along the grain boundaries in
polycrystalline materials. The stress strain curve shows the occurrence
of brittle fracture.
H A completely brittle fracture is shown in the Fig. 1.13(a) where very
little plastic deformation occurs while Fig. 1.13(b) shows the brittle
fracture with some deformation where cracks can be seen on the
fractured surface.
1.12. FACTOR OF SAFETY
While designing a component, it is necessary to ensure sufficient reserve
strength in case of an accident. It is ensured by taking a suitable factor of
safety. (FOS)
Factor of safety = Failure stress
Allowable stress =
Failure load
Allowable load
The allowable stress is the stress value which is used in design to
determine the dimensions of the component.
For Ductile materials,
FOS = Yield stress
Allowable stress (or) Permissible stress
For Brittle materials,
FOS = Ultimate stress
Allowable stress
There are number of factors which are difficult to evaluate accurately
in design analysis.
Some factors are as follows:
1. Uncertainty in the magnitude of external force acting on the component.
2. Variations in the properties of materials like yield strength or ultimate
strengths.
3. Variables in the dimensions of the component due to imperfect
workmanship.
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In addition to these factors, the no. of assumptions made in analysis, in
order to simplify the calculations, may not be exactly valid in working
conditions. The factor of safety ensures against these uncertainties and
unknown conditions.
Problem 1.1: A steam engine cylinder of diameter 200 mm, the maximum
pressure across the piston is 50 kN/m2. Design the diameter of piston rod if
the maximum tensile or compressive stress on piston rod is limited to
42 N/mm2.
Given:
σmax
= 42 N ⁄ mm2 = 42 × 10+ 6 N ⁄ m2 ; dc = 200 mm = 0.2 m,
Pmax
= 50 kN ⁄ m2 = 50 × 103 N ⁄ m2
Solution:
(i) Load on piston rod
Load (P) = Pmax
× Area of cylinder
= Pmax
× π d
c2
4 =
50 × 103 × π × 0.22
4 = 1.57081 × 103 N
Load (P) = 1571 N
(ii) Find diameter of piston rod
σmax ≥ Load (P)
Area of piston rod
42 × 106 ≥ 1571
π dp2
4
dp2 ≥
1571 × 4
42 × 106 × π = 4.762 × 10− 5
dp ≥ 6.9 × 10− 3 m
(or) dp ≥ 6.9 mm
From nearest standard diameter R20 series we get dp = 7.10 mm
Introduction to Design 1.31
Problem 1.2: A link as shown in figure is subjected to a steady tensile force
of 50 kN. Find the tensile stress induced in link.
Solution:
Given:
Load P = 50 kN = 50 × 103 N
Tensile stress at section B – B
Area at section B − B (AB) = l × b = 10 × 50 = 500 mm2
Tensile stress σtB
= Load (P)Area (A
B) =
50 × 103
500 = 100 N ⁄ mm2 = 50 MPa
Tensile stress induced at section A – A
Area at section A − A (AA) = 30 (75 − 50) = 750 mm2
Tensile stress σtA
= Load (P)Area (A
A) =
50 × 103
750 = 66.67 N ⁄ mm2 = 66.67 MPa.
Problem 1.3: Two rectangular plates are fastened by two bolts of 25 mm
diameter and nut. There is a washer whose ID = 27 mm and OD = 55 mm
placed between the plates and there is an another washer placed between the
nut and upper plate of dimensions ID = 27 mm and OD = 49 mm. The base
plate carries a load of 100 kN. Calculate the stress on both washers before
nut is tightened. When nut is tightened so as to produce a tension of 10 kN
an each bolt, what are the stresses in each washer.
1.32 Design of Machine Elements – I - www.airwalkbooks.com
Solution:
Given:
Bolt dia (d) = 25 mm, washer 1: d01
= 55 mm, di1
= 27 mm;
washer 2 d02
= 49 mm, di2
= 27 mm; load (P) = 100 kN = 100 × 103 N
(i) Stresses without nut tightening
Area of washer 1 A1 =
π4
d012 − d
i12 =
π4
552 − 272 = 1803.27 mm2
Area of washer 2 A2 =
π4
do2
− di22 =
π4
492 − 272 = 1313.18 mm2
(ii) Load acting and stress before tightening of nut
Load (P) is acting on two bolts, so load on each washer
P1 = 100 × 103
2 = 50 × 103 N
Upper washer (2) load = 0
Lower washer (1) load = P1 = 50 × 103 N
Stress on washer between plates (σ1) =
P1
A1
= 50 × 103
1803.27 = 27.73 N ⁄ mm2
(iii) Nuts are tightened
Load on upper washer (2) = P2 = 10 kN = 10 × 103 N
Stress on upper washer (σ2) =
P2
A2
= 10 × 103
1313.18 = 7.615 N ⁄ mm2
Load on lower washer (1) = P3 = P
1 + 10 kN
P3 = 100 + 10 = 110 kN
Stress on washer between plates (σ3) =
P3
A1
= 110 × 103
1803.27 = 61 N ⁄ mm2
Introduction to Design 1.33
Problem 1.4: A simply supported beam of rectangular cross section having depth
three times width is subjected to a point load of 20 kN at 300 mm from the left
support. The span of beam is 700 mm. Determine the dimensions of section if
the allowable strength of material is 200 MPa. (FAQ)
Solution:
Given:
Simply supported beam,
L = 700 mm,
a = 300 mm, b = 700 − 300 = 400 mm
P = 20 kN = 20 × 103 N, [ σ ] = 200 N ⁄ mm2 , d = 3b
From PSG DB pg no 6.5 for SSB with a, b and point load we have
maximum bending moment
Mmax = Pab
L =
20 × 103 × 300 × 400
700
Mmax = 3.428 × 106 N−mm
Bending stress equation Mb
I =
σb
y
[PSG D.B Pg. No. 7.1]
[K.M. D.B. Pg. No. 3]
I = bd3
12 =
b ⋅ (3b)3
12 =
27b4
12 mm4
y = d
2 =
3b
2
∴ 3.428 × 106
27 b4
12
= σ
36 ⁄ 2 (or) σ
b =
3.48 × 106 × 3b × 12
27 b4 × 2 ≤ 200
b3 ≥ 3.428 × 106 × 3 × 12
27 × 2 × 200 ≥ 11426.66
1.34 Design of Machine Elements – I - www.airwalkbooks.com
b ≥ 22.52 mm say 23 mm
∴ width b = 23 mm
depth d = 23 × 3 = 69 −~ 70 mm
1.13. IMPACT STRESS
When machine members are subjected to the load with impact, then the
stress produced in the member due to the falling load is known as ‘impact
stress’.
Consider a load ‘W’ falling on a body from a height ‘h’.
The load is suddenly applied
and the type of loading is called
impact loading.
Let A = area of cross-section
of the member.
The stress on the member
for gradually applied load is
σ = W
A .
But due to the application of
sudden load, the stress induced in
the member will be greater than
the value W
A.
Let δ = Deflection due to the
impact.
P = Equivalent load which
will produce the deflection ‘δ’.
Introduction to Design 1.35
Energy gained by the system in the form of strain energy = Area of
triangle OAB.
= 1
2 P δ .... (1)
Potential energy lost by weight
= W (h + δ) .... (2)
But, the energy gained by the system in the form of strain energy is
equal to the potential energy lost by the weight.
Equate equations (1) and (2)
1
2 P δ = W (δ + h) .... (3)
Let σ = stress induced in the member due to the application of impact
load.
where E = young’s modulus; σ = P
A
(or) P = (σ ⋅ A)
Deflection δ =
P
A
l
E ⇒
δ = σ
l
E
Replace P = σ ⋅ A and δ = σ ⋅ l
E in the equation (3)
1
2 ⋅ (σ ⋅ A)
σ ⋅ l
E = W
σ ⋅ l
E + h
expand and simplify the equation.
Al
2E σ2 −
Wl
E σ − Wh = 0 (is a Quadratic equation.)
1.36 Design of Machine Elements – I - www.airwalkbooks.com
Solving the equation,
σ = W
A
1 ± √1 +
2hAE
Wl
(or) σ ⋅ A = W
1 ± √1 +
2hAE
Wl
P = W
1 ± √1 +
2hAE
Wl
1.14. PRINCIPAL STRESSES
Machine components are subjected to several external loads of different
nature.
Therefore, it is necessary to find the equivalent single stress by using
principal stresses.
Introduction to Design 1.37
At any point in a strained material, there are three mutually perpendicular
planes on which only direct stresses are acting, and there are no shear stresses.
These planes are principal planes and the direct stresses are called Principal
stresses or Normal stresses. Out of three Principal stresses (σ1 , σ
2 , and σ
3)
one is maximum, one is minimum and the other one is intermediate.
Two-dimensional
Max. principal stress (K.M. D.B. Pg. No. 5, Eqn. No. 1.8(c))
= σ1 = σx + σy
2 + √
σx − σy
2
2
+ τxy2 =
σx + σy
2 +
1
2 √(σx − σy)2 + 4 τxy
2
Min. principal stress
= σ2 = σx + σy
2 − √
σx − σy
2
2
+ τxy2 =
σx + σy
2 −
1
2 √(σx − σy)2 + 4 τxy
2
Max. shear stress (K.M. D.B. Pg. No. 5, Eqn. No. 1.8(d))
= τmax
= σ
1 − σ
2
2[Refer PSG D.B Pg. No. 7.2]
tan 2 θ = 2 τ
xy
σx − σ
y
where θ = angle between σ1 and x − axis.
Note: If σ1 and σ2 both are + ve, consider σ3 into account.
Let σz be stress in z-direction. If σ
z = 0, then σ
3 = 0
Then τmax
= σ
1 − σ
3
2 =
σ1 − 0
2
τmax
= σ
1
2
1.38 Design of Machine Elements – I - www.airwalkbooks.com
Problem 1.5: Calculate normal stresses at (A) and (B)
Also calculate max. shear stresses at (A) and (B).
(FAQ)
Solution:
Given:
d = 50 mm, T = 1 kNm = 1 × 103 Nm, F = 3 kN = 3000 N
P = 15 kN = 15 × 103 N, l = 250 mm
(i) Consider axial load
Axial load induces direct stress.
σt = Direct stress =
P
A =
15 × 103
1963.49 (tensile) = 7.639 N ⁄ mm2
P = Axial load = (Tensile) = 15 kN = + 15 × 103kN
Area (A) =
π4
d2 = π4
(50)2 = 1963 (A) = π4
d2 = π4
(50)2 = 1963.49 mm2
(ii) Consider transverse (or) Bending load
Bending load = F = 3 kN = 3 × 103 N
Bending moment = F × l = 3 × 103 × 250 = 750 × 103 N−mm
Introduction to Design 1.39
Bending load induces bending stress,
σb = ± M ⋅ y
Iy =
d
2 =
50
2 = 25 mm
[PSG D.B. Pg. No. 7.1] I = π d4
64
I = π64
(50)4 = 306.79 × 103 mm4
at (A)
(σb)tensile = + 750 × 103 × 25
306.79 × 103
= + 61.115 N ⁄ mm2 (tensile)
at (B) (σb)comp. = − 61.115 N ⁄ mm2 (comp)
Total stress = σx = σ
t + σ
b
at (A) σx = σt + (σb)t
= + 7.639 + 61.115 = + 68.75 N ⁄ mm2
at (B) σx = σt + (σb)c
= + 7.693 + ( − 61.115) = − 53.422 N ⁄ mm2
(iii) Consider torsion
Twisting moment T = 1 kN − m = 1000 × 1000 N−mm
T
J =
τr [PSG D.B Pg. No. 7.1] r =
50
2 = 25 mm
τxy = τ = T ⋅ r
J J =
π32
d4 [K.M. D.B. Pg. No. 14]
= 106 × 25
613.59 × 103 J =
π32
× 504
τxy = 40.74 N ⁄ mm2 J = 6134.59 × 103 mm4
At (A) At (B)
1.40 Design of Machine Elements – I - www.airwalkbooks.com
σx = + 68.75 N ⁄ mm2 σx = − 53.476 N ⁄ mm2
σy = 0 σy = 0
τxy = + 40.74 N ⁄ mm2 τxy = + 40.74 N ⁄ mm2
At (A)
σ1 = Max. normal stress = σx + σy
2 + √
σx − σy
2
2
+ τxy2
[PSG D.B. Pg. No. 7.2]
[K.M. D.B. Pg. No. 5]
= 68.75
2 + √
68.75
2
2
+ (40.74)2
= 34.375 + 53.30
= + 87.675 N ⁄ mm2
σ2 = Min. normal stress = σx + σy
2 − √
σx − σy
2
2
+ τxy2
= 34.375 − 53.30
= − 18.925 N ⁄ mm2
Max. Shear stress = (τmax)at A = σ1 − σ2
2
= 87.675 − ( − 18.925)
2
τ
max at (A)
= 53.3 N ⁄ mm2
Introduction to Design 1.41
At (B)
σ1 = Max. normal stress = σx + σy
2 + √
σx − σy
2
2
+ τxy2
= − 53.476
2 + √
− 53.476
2
2
+ (40.74)2 = − 26.738 + 48.730
= + 21.992 N ⁄ mm2 (tensile)
σ2 = Min. normal stress = − 26.738 − 48.73
= − 75.468 N ⁄ mm2 (comp.)
Max. Shear stress = (τmax
)at (B) =
σ1 − σ
2
2
= 21.992 − ( − 75.468)
2
τ
max at (B)
= 48.73 N ⁄ mm2
Problem 1.6: For the stress state given, find the principal normal and shear
stresses and determine the angle from the x-axis to σ1. Draw the stress element
and label all details.
Solution:
σx = 16 MPa = 16 N ⁄ mm2 where θ1 = Angle from σ1 to x-axis
σy = 9 MPa = 9 N ⁄ mm2 θ2 = Angle from σ2 to x−axis
τxy = 5 MPa = 5 N ⁄ mm2
tan 2 θ1 = 2 τxy
σx − σy
[K.M. D.B. Pg. No. 5, Eqn. No. 1.8(g)]
= 2 × 5
16 − 9 =
10
7 = 1.428
1.42 Design of Machine Elements – I - www.airwalkbooks.com
2 θ1 = tan− 1 1.428
θ1 =
55°2
= 27.5°
θ2 = 90 + 27.5 = 117.5°
Max. Principal stress (or) Max. normal stress
= σ1 = σx + σy
2 + √
σx − σy
2
2
+ τxy2
[Refer PSG D.B Pg. No. 7.2]
[K.M. D.B. Pg. No. 5]
= 16 + 9
2 + √
16 − 9
2
2
+ 52 = 12.5 + 6.1032 = 18.60 N ⁄ mm2
Min. principal stress (or) Min. normal stress
= σ2 = σx + σy
2 = √
σx − σy
2
2
+ τxy2
= 12.5 − 6.1032 = 6.3968 N ⁄ mm2
Introduction to Design 1.43
Since σ1 and σ
2 are + ve, ∴ Consider σ
3 into account.
σ3 = 0
Max. shear stress = τmax
= σ
1 − σ
3
2 =
18.6 − 0
2
τmax
= 9.3 N ⁄ mm2
Problem 1.7: Determine the required thickness of the steel bracket at section
A-A. When loaded as shown in Fig. in order to limit the tensile stress to
60 MN/m2. (FAQ)
Solution:
At section A-A imagine two forces
F1 and F
2 equal to F
i.e., F1 = F
2 = 4500 N.
The force F and F1 constitute
couple (F × e). The effect of couple
produces bending, which induces
bending stress.
σb = M ⋅ y
I
[K.M. D.B Pg. No. 2]
M = F × e
= 4500 × 50 N−mm
= 4500 × 50 × 25
1
25 ⋅ t ⋅ 503
y = 50
2 = 25 mm
= 540
l N ⁄ mm2
I = 1
12 × t × 503
The force F2 produces a direct
tensile load which induces direct tensile
stress.
1.44 Design of Machine Elements – I - www.airwalkbooks.com
σt = P
A =
F2
A =
4500
(50 × t) = 90
t [A = (50 × t)]
Total stress
= σtotal = σb + σt = 540
t +
90
t =
630
t
i.e., σx = 630
t
σy = 0; τxy = 0; θ1 = 60 N ⁄ mm2 (given) (since it is tensile)
σ1 = σx + σy
2 + √
σx − σy
2
2
+ τxy2
[Refer PSG D.B. Pg. No. 7.2]
[K.M. D.B. Pg. No. 3]
= 630
2t + √
630
2t
2
= 60
630
2t +
630
2t = 60
t = 630
60 = 10.5 mm.
Thickness of steel bracket = t = 10.5 mm
Problem 1.8: Determine the maximum shear stress in the member loaded
shown in Fig. (FAQ)
Solution:
Assume point (C) and image (F1 and F
2) two equal and opposite force
i.e., F1 = F
2 = F = 500 N.
The force F and F1 produces couple, the effect of couple produces
twisting, which induces torsional shear stress.
Introduction to Design 1.45
τs =
T ⋅ rJ
τs = 5000 (500) × 50
π32
× 1004 T = 5000 × 500 N−mm
= τxy r = 50 mm
τs = τxy = 12.73 N ⁄ mm2 J =
π32
d4 = π32
× 1004
The force F2 produces bending, which induces bending stress
σb = M ⋅ y
I
M = F2 × 250
= 5000 × 250 N−mm
= 5000 × 250 × 50
π64
(100)4
= 12.73 N ⁄ mm2
y = d
2 =
100
2 = 50 mm
I = π64
d4 = π64
1004 mm4
σx = σb = 12.73 N ⁄ mm2 ; σy = 0
τxy
= 12.73 N ⁄ mm2[Refer PSG DB Pg. No. 7.2]
[K.M. D.B. Pg. No. 5]
σ1 = Max. principal stress
= σ
x + σ
y
2 + √
σ
x − σ
y
2
2
+ τxy2
= 12.73
2 + √
12.73
2
2
+ (12.73)2
= 6.365 + 14.23 = + 20.595 N ⁄ mm2
1.46 Design of Machine Elements – I - www.airwalkbooks.com
σ2 = Min. principal stress = σx + σy
2 √
σx − σy
2
2
+ τxy2
= 6.365 − 14.23 = − 7.865 N ⁄ mm2
Introduction to Design 1.47
Max. Shear stress = τmax = σ1 − σ2
2
= 20.595 − ( − 7.865)
2 = 14.23 N ⁄ mm2
Problem 1.9: Stresses in a wheel hub are found to be 40 N/mm2 and
50 N/mm2 tension at a point as shown in the Fig. Calculate the max. shear
stress at the point.
Solution:
Given data:
σx = 40 N ⁄ mm2; σy = 50 N ⁄ mm2;
σz = 0 since σ3 = 0; τxy = 0
[Refer K.M. D.B. Pg. No. 5]
Max. principal stress
σ1 = σx + σy
2 + √
σx − σy
2
2
+ τxy2
= 40 + 50
2 + √
40 − 50
2
2
= 45 + 5 = 50 N ⁄ mm2
Min. principal stress
σ2 = 45 − 5 = 40 N ⁄ mm2
Since σ1 and σ
2 are + ve, Consider σ
3 into account. (σ
3 = 0)
Max. shear stress = τmax
= σ
1 − σ
3
2 =
50 − 0
2 = 25 N ⁄ mm2
1.48 Design of Machine Elements – I - www.airwalkbooks.com
Problem 1.10: A 50 mm diameter rod is subjected to a 10 kN force and a
torsional moment of 100 N-m as shown in the Fig. Determine the maximum
tensile and maximum shear stress at point (A).
Solution:
Assume point ‘C’ and imagine two
equal and opposite forces
F1 and F2 (i.e., F1 = f2 = f = 10,000 (N))
The force F and F1 constitute a
couple. The couple produces bending, which
induces bending stress.
∴ At (A) the bending stress is tensile.
σb = M ⋅ yI
M = 10,000 × 25 N−mm
σb = 10,000 × 25 × 25
π64
× 504
= 20.37 N ⁄ mm2
y = 50
2 = 25 mm
I = π64
504
[K.M. D.B. Pg. No. 13]
The remaining force F2 produces a direct
tensile stress.
σt = F2
A =
10,000
π4
502
= 5.0929 N ⁄ mm2
total stress = σx = σb + σt = 20.37 + 5.09
= 25.46 N ⁄ mm2
The torsional moment induces
torsional shear stress.
Introduction to Design 1.49
τxy = T ⋅ r
J [Refer K.M. D.B. Pg. No. 2]
= 100 × 1000 (25)
613592.3
= 4.07 N ⁄ mm2
J = 100 N−m
= 100 × 1000 N−mm
[K.M. D.B. Pg. No. 14]
J = π32
d4
=
π32
× 504 = 6135923 mm4
r = 50
2 = 25 mm
σ1 = σx + σy
2 + √
σx − σy
2
2
+ τxy2 (σy = 0)
= 25.46
2 + √
25.46
2
2
+ (4.07)2
= 12.73 + 13.36 = 26.09 N ⁄ mm2
σ2 = 12.73 − 13.36
= − 0.63 N ⁄ mm2
Max. shear stress = τmax = σ1 − σ2
2
= 26.09 − ( − 0.63)
2
τmax = 13.36 N ⁄ mm2
1.50 Design of Machine Elements – I - www.airwalkbooks.com
Problem 1.11: Determine the maximum principal stress, min. principal stress
and max. shear stress at the centre of the crank shaft bearing for the load
as shown in the Fig.
Solution:
The force 10 kN is acting perpendicular to the crank pin and this force
induces bending stress and torsional shear stress at the axis of the crank shaft.
P = 10 kN = 10 × 103 N
Bending moment M = 10 × 103 × 100 N−mm
Twisting moment T = 10 × 103 × 120 × 104 N−mm
Bending stress = σb = M ⋅ y
I I =
π64
d4 [K.M. D.B. Pg. No. 13]
σx = σb = 106 × 30
636.17 × 103
= 47.15 N ⁄ mm2
= π64
× 604 = 636.17 × 103 mm4
y = 60
2 = 30 mm
Shear stress = τxy = T ⋅ r
J
= 120 × 104 × 30
1.27 × 106r =
d
2 =
60
2 = 30 mm
= 28.34 N ⁄ mm2 J =
π32
d4 = π32
× 604
[K.M. D.B. Pg. No. 14]
σy = 0 = 1.27 × 106 mm4
[Refer K.M. D.B. Pg. No. 5]
Introduction to Design 1.51
Max. principal stress = σ1 =
σx + σ
y
2 + √
σ
x − σ
y
2
2
+ τxy2
= 47.15
2 + √
47.15
2
2
+ (28.34)2
= 23.575 + 36.863 = 60.4 N ⁄ mm2
Minimum principal stress = σ2 = σx + σy
2 = √
σx − σy
2
2
+ τxy2
= σ2 = 23.575 − 36.863
= − 13.25 N ⁄ mm2
1.52 Design of Machine Elements – I - www.airwalkbooks.com
Max. shear stress = τmax
= σ
1 − σ
2
2 =
60.4 − ( − 13.25)2
τmax = 36.825 N ⁄ mm2 .
Problem 1.12: Determine the maximum normal and max. shear stresses at
section A-A as shown in the Fig.
Solution:
The load F is resolved into vertical and horizontal components.
FV and FH.
F = 10 kN = 10 × 103 N
FV = Vertical load = F ⋅ sin θ = 10,000 × sin 30° = 5,000 N
FH = Horizontal load = F ⋅ cos θ = 10,000 × cos 30° = 8660.25 N
Both vertical and horizontal loads produces bending moments
(i.e., vertical bending moment and horizontal bending moment.)
Introduction to Design 1.53
Therefore, it is necessary to find resultant bending moment.
Vertical bending moment
= BMV = FV (30 + 25 + 65) = 5000 (120) = 600 × 103 N−mm
Horizontal bending moment
= BMH = FH (30 + 25 + 65) = 8660.25 × 120 = 1.039 × 106 N−mm
Resulting bending moment = BMR = √(MBV)2 + (BMH)2
(BM)R = √(600 × 103)2 + (1.039 × 106)2 = 1.199 × 106 N−mm
This resultant bending moment induces bending stress σb
Bending stress = σb = (BMR) y
I y =
80
2 = 40 mm
σb = σx = 1.199 × 106 × 40
2.01 × 106 I =
π64
× 804 [K.M. D.B. Pg. No. 13]
σb = σx = 23.85 N ⁄ mm2 = 2.01 × 106 mm4
And also, the horizontal load induces torsional shear stress.
[Refer K.M. D.B. Pg. No. 2]
Torsional shear stress T = Twisting moment due to horizontal load.
= τxy = T ⋅ r
J = FH × 150
τxy = 1.29 × 106 × 40
4.02 × 106 = 8660.25 × 150 = 1.29 × 106 N−mm
τxy = 12.83 N ⁄ mm2 (σy = 0) r = 80
2 = 40 mm
J = π32
d4 = π32
× 804 = 4.02 × 106 mm4
[Refer K.M. D.B. Pg. No. 5]
1.54 Design of Machine Elements – I - www.airwalkbooks.com
Max. principal stress = σ1 = σx + σy
2+ √
σx + σy
2
2
+ τxy2
= 23.85
2 + √
23.85
2
2
+ (12.83)2 = 11.925 + 17.517
σ1 = 29.442 N ⁄ mm2
Min. principal stress
= σ2 = σx + σy
2 − √
σx − σy
2
2
+ τxy2 = 11.925 − 17.517
σ2 = − 5.592 N ⁄ mm2
Max. shear stress at A-A = τmax
A − A =
σ1 − σ2
2 =
29.442 − ( − 5.592)2
τmax
A − A = 17.517 N ⁄ mm2
Problem 1.13: A wall bracket is loaded as shown in the Fig. The cross-section
of the bracket is rectangular having b = 3t. Determine the dimensions of the
cross-section of the bracket if the permissible stress is limited to 28 N/mm2.
(FAQ)
Solution:
Mark the angle with respect to x-axis. i.e., 30° with x-axis.
Resolve the force F = 5 kN into horizontal and vertical components
Horizontal component of F = FH = F ⋅ cos 30° = 5 × 103 × cos 30° = 4330.12 N
Vertical component of F = FV = F ⋅ sin 30° = 50 × 103 × sin 30° = 2500 N
Introduction to Design 1.55
Consider a point ‘C’.
At ‘C’ imagine two forces FH1 & FH
2 equal and opposite forces with
magnitude of FH; (i.e., FH = FH1 = FH
2 = 4330.12 N)
1. The force FV induces bending stress.
σb1 =
M ⋅ yI
M = FV × 120
= 2500 × 120 = 300 × 103 N−mm
= 300 × 103 × 3t × 12
27 t4 × 2 y =
b
2 =
3t
2 But b = 3t
σb1 =
200 × 103
t3 N ⁄ mm2 I =
1
12 t b3 =
1
12 t (3t)3
[K.M. D.B. Pg. No. 12]
2. The force FH and FH2 constitute a couple the effect of couple produces
bending. The couple induces bending stress.
σb2 =
M ⋅ yI
M = (F ⋅ cos 30°) × 60
= (5 × 103 cos 30°) × 60 = 259.8 × 103 N−mm
=
259.8 × 103 × 3t
2
1
12 × 27 × t4
σb2 =
173.2 × 103
t3 N ⁄ mm2
y = b
2 =
3t
2 But b = 3t
I = 1
12 (t) b3 =
1
12 t (3t)3
1.56 Design of Machine Elements – I - www.airwalkbooks.com
3. The force FH1 (tensile) induces direct tensile stress.
σt = FH
1
A =
5 × 103 cos 30°b × t
= 4330.12
(3t) t = 1443.37
t2
Introduction to Design 1.57
At – A
Total stress = σx = σ
b1
+ σb
2
+ σt
= 200 × 103
t3 +
173.2 × 103
t3 +
1443.37
t2
= 373.2 × 103
t3 +
1443.37
t2
σy = 0; τxy = 0;
σx = σ1 = 28 N ⁄ mm2 (Given)
∴ 373.2 × 103
t3 +
1443.37
t2 = 28.
Calculate ‘t’ by trail and error method
Let t = 10 t = 15
L.H.S. 373.2 + 14.43 = 387.63 L.H.S. 110.57 + 6.41 = 116.98
L.H.S. > R.H.S. L.H.S. > R.H.S.
t = 20 t = 25
L.H.S. 46.65 + 3.60 = 50.25 L.H.S. 23.88 + 2.309 = 26.18
L.H.S. > R.H.S. L.H.S. ≈ R.H.S.
Therefore, take thickness t = 25 mm
and b = 3 × t = 3 × 25 = 75 mm.
Problem 1.14: Determine the maximum shear stress and the principal stress
for the member shown in Fig. (a).
Solution:
Introduce equal and opposite ver tica l forces Fv1
and
Fv2
(Fv1
= Fv2
= Fv) perpendicular to the axis of the circular bar and equal and
opposite horizontal forces FH1
and FH2
(FH1
= FH2
= FH
), along the axis. This
does not disturb the equilibrium.
1.58 Design of Machine Elements – I - www.airwalkbooks.com
Effects of various loads (Fig. (b))
Fv and F
v2 constitute a couple F
v l
1 and twist the circular bar. F
v1 bends
the circular bar in the vertical plane. FH
and FH2
constitute a couple FH
l1
and bends the circular bar in the horizontal plane. FH1
gives tensile load to
the circular bar. Twisting moment,
T = Fv l1 = 2000 × 500 = 10 × 105 N−mm
Introduction to Design 1.59
Shear stress,
τ = 16 T
π d3 =
16 × 10 × 105
π × 503 = 40.7 N ⁄ mm2
Bending moment (vertical) = BMv = F
v1 l
2
= 2000 × 125 = 25 × 104 N−mm
Bending moment (horizontal) = BMH
= FH
l1
= 1000 × 500 = 50 × 104 N−mm
Bending moment (resultant) = √BMV2 + BM
H2
BMR = √(25 × 104)2 + (50 × 104)2 = 55.9 × 104 N−mm
Bending stress σb = 32 BMR
π d3 =
32 × 55.9 × 104
π × 503 = 45.6 N ⁄ mm2
Tensile stress due to FH1
= F
H1
(π d2 ⁄ 4)
σt =
1000
(π × 502 ⁄ 4) = 0.509 N ⁄ mm2
σT = σb + σt = 45.6 + 0.509 = 46.1 N ⁄ mm2
Maximum shear stress = τmax = √ σT
2
2
+ τ2
[Refer K.M. D.B.
Pg. No. 3
Eqn. No. 1.5(b)]
= √ 46.1
2
2
+ (40.7)2 = 46.8 N ⁄ mm2
Maximum normal stress = σ1 = σT
2 + √
σT
2
2
+ τ2
= 46.1
2 + √
46.1
2
2
+ (40.7)2 = 69.8 N ⁄ mm2
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Problem 1.15: An unknown weight falls through 10 mm on to a collar rigidly
attached to the lower end of a vertical bar 3 m long and 600 mm2 cross
section. The maximum instantaneous extension is 2 mm. What is the
corresponding stress and the value of the weight. Take E = 200 kN ⁄ mm2.
(FAQ)
Refer Fig.
1. Equate the energy of impact to the strain energy of the bar and determine W.
W (h + δ) = 1
2 P δ .... (1)
δ = instantaneous extension = 2 mm
P = equivalent static load = δ AE
L
E = 2 × 105 N ⁄ mm2
P = 2 × 600 × 2 × 105
3000 = 80,000 N
Now, using (1) solve for W.
W (10 + 2) = 1
2 × 80,000 × 2
W = 6666.7 N
2. Instantaneous stress σinst = P
A =
80,000
600 = 133.3 N ⁄ mm2
Problem 1.16: An unknown weight falls through 10 mm on to a collar rigidly
attached to the lower end of a vertical bar 3 m long and 600 mm2 cross-section.
The maximum instantaneous extension is 2 mm. What is the corresponding stress
and the value of the weight; Take E = 200 kN ⁄ mm2. (FAQ)
Given data:
h = 10 mm; l = 3 m = 3000 mm; A = 600 mm2 ; δ = 2 mm (δl);
E = 200 kN ⁄ mm2 = 200 × 103 = 2 × 105 N ⁄ mm2
Introduction to Design 1.61
Solution:
Young’s modulus = E = σe
= stress
strain(e = ε)
Strain = δl =
δ l
l =
2
3000 = 6.66 × 10− 4
Instantaneous stress σ = E ⋅ e = 2 × 105 × 6.66 × 10− 4 = 133.33 N ⁄ mm2
But σ = P
A .
Instantaneous static load P = σ ⋅ A = 133.33 × 600 = 80 × 103 N
1
2 P δ = W (h + δ)
1
2 × 80 × 103 × 2 = W (10 + 2)
W = 80 × 103
12 = 6666.66 N
W = 6.666 kN.
Problem 1.17: An I-section beam of depth 250 mm is supported at two points
4 m apart. It is loaded by a weight 4 kN falling through a height h and
striking the beam at mid span. Moment of inertia of the section is
8 × 107 mm4. Modulus of Elasticity is 210 kN ⁄ mm2. Determine the permissible
value of h if the stress is limited to 120 N ⁄ mm2. (FAQ)
Solution:
depth = d = 250, mm ; y = d
2 =
250
2 = 125 mm; l = 4 m = 4,000 mm
W = 4 kN = 4 × 103 N ; I = 8 × 107 mm4 ; E = 210 kN ⁄ mm2 = 210 × 103 N ⁄ mm2
Instantaneous stress = σ = 120 N ⁄ mm2
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Instantaneous stress = σ = M ⋅ y
I σ =
M
Z where Z =
I
y
[From PSG D.B.
Pg. No. 7.1]
But Max. BM = M = P ⋅ l
4 (for a simply supported beam load acting at centre)
σ =
P ⋅ l4
× y
I
[From PSG D.B Pg. No. 6.5]
Instantaneous static load P = σ ⋅ I ⋅ 4
l ⋅ y = 120 × 8 × 107 × 4
4000 × 125 = 76.8 × 103 N
Instantaneous deflection, δ for a simply supported beam with a
concentrated centre load.
δ = P ⋅ l348 EI
δ = 76.8 × 103 × 40003
48 × 210 × 103 × 8 × 107
[From PSG D.B. P.No. 6.5]
[K.M. D.B. Pg. No. 15]
[–ve sign indicates the deflection
in downward direction]
Instantaneous deflection δ = 6.095 mm
We know that
1
2 P δ = W (h + δ)
1
2 × 76.8 × 103 × 6.095 = 4 × 103 (h + 6.095)
h = 58.514 − 6.095 mm
h = 52.419 mm
Introduction to Design 1.63
Problem 1.18: A weight of 6000 N falls through a distance ‘h’ at the middle
of a beam of span 4.5 m. The end connections of the beam may be considered
as simply supported. Determine the value of h, such that the maximum induced
stress in the beam does not exceed 160 N ⁄ mm2. The modulus of section of
the beam = 2 × 10− 4 m3, second moment of area = 1 × 10− 5 m4 (FAQ)
Given data:
W = 6000 N; l = 4.5 m = 4500 mm; Z = 2 × 10− 4 m3 = 2 × 10− 4 × (1000)3
= 200 × 103 mm3; I = 1 × 10− 5 m4 = 1 × 10− 5 × (1000)4 = 107 mm4.
Let M = Max. BM for a simply supported beam, concentrated load acting
at the centre.
Max. BM = M = P ⋅ l
4 =
P × 4500
4[From PSG D.B Pg. No. 6.5]
160 =
P × 4500
4
200 × 103
bending stress σ = M
z
P = 28444.44 N
Instantaneous deflection, δ = Pl3
48 EI =
28.44 × 103 × 45003
48 × 210 × 103 × 107
= 25.71 mm E is not given Assume
E = 120 × 102 N ⁄ mm2
[From PSG D.B Pg. No. 6.5]
[K.M. D.B. Pg. No. 15]
1
2 P δ = W (h + δ)
1
2 × 28.44 × 103 × 25.71 = 6000 (h + 25.71)
h = 60.93 − 25.71 = 35.22 mm
h = 35.22 mm
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Problem 1.19: A cantilever shaft of 60 mm diameter and 0.3 m length in
subjected to following load (a) point load of 5 kN at free end (b) Axial load
of 20 kN and (c) Torque of 1 kN-m as shown in Fig. Determine the stresses
at A and B.
Solution:
1. Axial stress due to 20 kN, σa = F
A =
20,000 × 4
π × 602 = 7.07 N ⁄ mm2
Bending stress due to the tip load 5 kN,
σbt = σbc = 32 × M
π d3 =
32 × 5000 × 300
π × 603 = 70.7 N ⁄ mm2
Shear stress due to torque 1 kN-m,
τ = 16T
π d3 =
16 × 1,000,000
π × 603 = 23.6 N ⁄ mm2
2. At A state of stress is depicted in Fig. (c)
Total normal stress σT = σ
bt + σ
a = 70.7 + 7.07 = 77.8 N ⁄ mm2
τ = 23.6 N ⁄ mm2
σ1, 2 = σx + σy
2 ± √
σx − σy
2
2
+ τxy2
[Refer K.M. D.B. Pg. No. 5]
= σT
2 ± √
σT
2
2
+ τ2 = 77.8
2 ± √
77.8
2
2
+ 23.62 = 38.9 ± 45.5
σ1 = 84.4 N ⁄ mm2 , σ2 = − 6.6 N ⁄ mm2
τmax
= σ
1 − σ
2
2 =
84.4 − ( − 6.6)2
= 45.5 N ⁄ mm2
3. At B state of stress is depicted in Fig. (d)
σT = − σ
bc + σ
a = − 70.7 + 7.07 = − 63.6 N ⁄ mm2
τ = 23.6 N ⁄ mm2
Introduction to Design 1.65
σ1.2 = σT
2 = ± √
σT
2
2
+ τ2 = − 63.6
2 ± √
− 63.6
2
2
+ 23.62
σ1 = 7.8 N ⁄ mm2 ; σ2 = − 71.4 N ⁄ mm2
τmax
= σ
1 − σ
2
2 =
7.8 − ( − 71.4)2
= 39.6 N ⁄ mm2
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Problem 1.20: Calculate the strain energy stored per unit volume,
i.e., resilience is increased for the same maximum stress by turning down the
shank of the bolt to the core diameter of the thread. Take
E = 2.1 × 105 N ⁄ mm2.
Solution:
1. Calculation of stresses in the shank and in the threaded portion
At = Area of the core section =
π dc2
4 =
π × 16.62
4 = 216.4 mm2
Stress in the threaded portion = Load
Area
σ1 = 20000
216.4 = 92.42 N ⁄ mm2
As = Area of the shank section
= π × d2
4 =
π × 202
4 = 314.2 mm2
σs = Stress in the shank section = 20000
314.2 = 63.65 N ⁄ mm2
2. Strain energy calculation
U = σ2
2E × volume [Refer K.M. DB Pg. No. 29, Eqn. No. 2.27(b)]
U = σ
t2 A
t l
t
2E +
σs2 A
s l
s
2E =
92.422 × 216.4 × 40
2 × 2.1 × 105 +
63.652 × 314.2 × 80
2 × 2.1 × 105
= 176.05 + 242.46 = 418.5 N−mm
3. Strain energy after the shank is turned to the core diameter
σ = σt = 92.42 N ⁄ mm2
U = σ
t2
2E × A
t (l
t + l
s) =
92.422 × 216.4 × (40 + 80)2 × 2.1 × 105
= 528.11 N−mm
Introduction to Design 1.67
1.15. COMBINED STRESSES DUE TO ECCENTRIC LOADING
Frames of punching machines, hydraulic riveter, clamps etc. are subjected
to combined stresses due to eccentric loading.
In Fig. 1.17(b), the loading is eccentric. In Fig. 1.17(c), equal and
opposite forces (F1 = F
2 = F) are introduced along the axis of the member.
This results in a direct force F and a couple C = Fe, acting on the member
as shown in Fig. 1.17(d). The stress distribution are shown in Fig. 1.18.
1.68 Design of Machine Elements – I - www.airwalkbooks.com
Stresses on section m − n
At m, tensile stress
σm = F e Ct
I −
F
ACt = Distance between axis and
outer most layer in leftside
At n, compressive stress
σn = F e Cc
I +
F
A
Cc = Distance between axis and
outermost layer in right side
Problem 1.21: Determine the required thickness of the steel bracket at section
X – X when loaded as shown in Fig. The permissible tensile stress is
100 MN ⁄ m2.
Introduction to Design 1.69
Solution:
This is a problem in eccentric loading.
Along the axis, two equal and opposite forces F are introduced (Fig.(b)).
The upward F at A and downward F at B constitute a couple C = F e.
Therefore, at B, we have the upward F and a couple C. This equivalent loading
is depicted in Fig. (c).
Induced bending stress due to the couple
σb = F e
Z
F e = 5000 × 50 = 250,000 N−mm
Z = 1
6 tb2 =
1
6 × t × 502 = 416.7 t
= 250,000
416.7 t ≈
600
t N ⁄ mm2
Direct tensile stress due to F
σt = F
bt =
5000
50 × t =
100
t N ⁄ mm2
The stress distribution is shown in Fig. (d)
Maximum stress is on side 1
σT1 = σbt + σt (both are tensile)
= 600
t +
100
t =
700
t
700
t ≤ σ
t ( = 100 N ⁄ mm2)
t ≤ 7 mm, take t = 8 mm
Note: Stress on side 2
σ12
= − σbc
(compressive stress) + σt =
− 600
8 +
100
8 = − 62.5 N ⁄ mm2
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Problem 1.22: A wall bracket shown in Fig. is subjected to a pull of 5 kN
at 60° to the vertical. The cross section of the bracket is rectangular having
b = 3t. Determine the dimensions of the cross section of the bracket if the
permissible stress is limited to 28 N ⁄ mm2. (FAQ)
1. Refer to Fig. (b)
F is resolved into FH
and FV
FV bends the bracket and bending stresses are induced across AB.
Introduce equal and opposite forces FH1
and FH2
( FH1
= FH2
= FH
) along
the axis. FH
and FH2
constitute a couple inducing bending stresses. FH1
induces
uniform tensile stress across the section.
2. Direct tensile stress
σ1 = FH
bt =
4330
3t2 =
1443.3
t2
FH = 5000 sin 60° = 4330 N
FV = 5000 cos 60° = 2500 N
b = 3t
3. Bending stress due to FV (at (A & B)
σ2 = M
1
6 t b2
= 300,000
1
6 t (et)2
= 200,000
t3 M = FV × 120 = 250 × 120
= 300,000 N−mm
Introduction to Design 1.71
4. Bending stress due to couple FH × 60
Couple = C = 4330 × 60 = 259,800 N−mm
σ2 =
C
1
6 t b2
= 259,800
1
6 t (3t)2
= 173,200
t3
5. Total normal stress at A – Tensile stress
σ = σ1 + σ
2 + σ
3 =
1443.2
t2 +
200000
t3 +
173200
t3
σ = 373200
t3 +
1443.3
t2
6. Now σ ≤ [ σ ]
i.e. 373200
t3 +
1443.3
t2 ≤ [ σ ] = 28 N ⁄ mm2
t is evaluated by trial and error method
trail 1: If t = 20 mm, we get σ = 50.25 N ⁄ mm2 > 28 N ⁄ mm2
trail 2: If t = 25 mm, we get σ = 26.18 N
mm2 < [ σ ]
= 28
N
mm2
Hence, it is safer to take
t = 25 mm and
b = 3 × 25 = 75 mm
1.16. STRESS CONCENTRATION
Elementary equations for stresses,
Tensile stress = σt =
P
A
Bending stress = σb =
M ⋅ yI
Shear stress = τ = T ⋅ r
J
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The above equations are based on assumptions, that there are no
dis-continuities in the cross-section of the component.
A plate, with a small circular hole subjected to tensile stress is shown
in Fig. 1.19.
It is observed from the nature of stress distribution at the section passing
through the hole, that there is a sudden rise in the magnitude of stresses in
the neighbourhood of the hole.
The localised stresses in the neighbourhood of the hole are far greater
than the stresses obtained by the elementary equations.
Introduction to Design 1.73
1.16.1. Definition of Stress Concentration
It is defined as the localisation of high stresses due to irregularities (or)
abrupt changes of the cross-section.
Theoretical stress
concentration
factor
= K
t =
Highest value of actual stress
near discontinuity
Nominal stress obtained by elementary
equations for min. cross−section
= σ
max
σo
(or) τ
Max
τo
σmax = max. stress; τmax = Max. shear stress
σo = Nominal stress; τo = Nominal shear stress
H σo and τ0 are the stresses calculated by the elementary equations for
minimum cross-section.
H The values of Kt depends upon the geometry of the component. (From
PSG Databook P.No. 7.9 to 7.16). (From K.M. Data Book Pg. No.
36 to 45)
1.16.2. Nominal Stress (σo) Cross Section
σo = P
(W − d) t
σo = P
d × t
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1.16.3. Stress Concentration Factors
The stress concentration factors are determined by two methods.
1. Mathematical analysis based on the theory of elasticity.
2. Experimental methods like photo-elasticity.
There are limitations for the techniques of the theory of elasticity. For
more complex shapes, the stress concentration factors are determined by
photos-elasticity.
The chart represents stress concentration factors for different geometric
shapes and conditions of loading. (PSG databook – 7.9 to 7.16) (K.M. D.B.
Pg. No. 36 to 45)
Example: Pg. No. 7.10 in PSG Databook
Rectangular plate with a transverse hole loaded in tension or
compression.
τo = Mt ⋅ r
J
σo = P
π4
d2
σo = Mb ⋅ y
I
y = d
2
I = π64
d4
Introduction to Design 1.75
Kt = σmax
σo
.... (1) σo = Nominal stress; where t = Plate thickness
σo = P
(W − a) t .... (2) (PSG Pg. No. 7.10)
Calculate a / W ratio; From chart, corresponding to the value of a / W
ratio, take the value of Kt (stress concentration factor). Substitute the value of
Kt and σo in equation (1) and find the maximum stress.
H The effect of stress concentration depends upon the material of the
component.
H Under static loads, ductile materials are not affected by stress
concentration.
H Therefore, stress concentration factors are not used for ductile
materials under static loading.
H The effect is more severe in case of brittle materials, due to their
inability to plastic deformation.
H Stress concentration factors are used for components made up of
brittle materials subjected to static loads.
H When the load is fluctuating, the endurance strength of the ductile
material is greatly reduced due to stress concentration.
1.16.4. Reduction of Stress Concentration Effects
In many cases, it is not possible to avoid the discontinuities or abrupt
changes of the cross-section. However, it is possible for the designer to reduce
the severity of stress concentration by selecting the correct geometric shape.
The original component with a V-notch is shown in Fig. 1.21. It is obtained
that a single notch results in a high degree of stress concentration.
The severity of stress concentration is reduced by using the principle of
minimisation of the material.
There are three methods for achieving this effect:
(i) Use of multiple notches.
1.76 Design of Machine Elements – I - www.airwalkbooks.com
(ii) Removal of undesired material.
(iii) Drilling additional holes.
Problem 1.23: Determine the maximum stress produced in a rectangular plate
50 mm wide, 8 mm thick with a central hole of 10 mm diameter. It is loaded
in an axial tension of 1 kN. (FAQ)
Solution:
Given data:
Width = W = 50 mm
Thickness = h = 8 mm
Dia. of centre hole = a = 10 mm
Axial tension = P = 1 kN = 1 × 103 N
Stress concentration = Kt = σmax
σo
To find Kt
Kt = Stress concentration (Refer Databook Page No. 7.10)
(K.M. D.B. Pg. No. 36)
σmax = Maximum stress = N ⁄ mm2
Introduction to Design 1.77
σ0 = Nominal stress
a
W =
10
50 = 0.2
= P
(W − a) h Corresponding to the value of a
W = 0.2, take the
value of Kt which is equal to 2.5.
= 1 × 103
(50 − 10) 8 = 3.125 N ⁄ mm2
∴ Kt =
σmax
σ0
,
2.5 × 3.125 = σmax
,
The maximum stress = σmax = 7.8125 N ⁄ mm2
Problem 1.24: A flat plate is subjected to a tensile force of 5 kN as shown
in the Fig. Take FOS = 2.5; ultimate stress = 200 N ⁄ mm2. Calculate the plate
thickness. (FAQ)
Solution:
Given:
Flat plate, tensile force (P) = 5 kN = 5 × 103 N,
FOS = 2.5; σu = 200 N ⁄ mm2
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The stresses are critical at two sections,
1. Section at the fillet A-A
2. Section at the hole B-B
1. SECTION – A-A (Fillet section)
D = 45 ; d = 30
Fillet radius r = 5
D
d =
45
30 = 1.5
r
d =
5
30 = 0.167
Kt = σmax
σo
[From PSG databook, Pg. No. 7.9]
[K.M. Databook Pg. No. 38]
σmax = Kt ⋅ σo Kt = 1.8
= 1.8 × 5 × 103
30 × t
σo = Nominal stress
= 300
t N ⁄ mm2 .... (1) σo =
P
d × t =
5 × 103
30 × t
Introduction to Design 1.79
2. SECTION – B-B: (Hole section) (From PSG DB Pg. No.7.10)
σo = P
(W − a) t W = 30 mm
= 5 × 103
(30 − 15) t a = 15 mm
t = thickness
a
W =
15
30 = 0.5
at a
W = 0.5 take Kt value, Kt = 2.16
σmax = Kt ⋅ σo
= 2.16 × 5000
15 × t
σmax
= 720
t N ⁄ mm2 .... (2)
From (1) and (2) it is seen that maximum stress is induced at the hole
section. Equating permissible stress to equation (2)
i.e., 720
t =
200
FOS =
σu
FOS
700
t =
200
2.5
or t = 700 × 2.5
200 = 8.75 mm
∴ t = 8.75 mm Take t = 9 mm
Problem 1.25: A stepped shaft has maximum dia = 45 mm, minimum
dia. = 30 mm, fillet radius = 6 mm. If the shaft is subjected to an axial load
of 10 kN, find the maximum stress induced.
Solution:
Given Stepped Shaft
D = 45 mm ; d = 30 mm
r = 6 mm ;
P = 10 kN = 10 × 103 N
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The maximum stress occurs
at fillet cross-section at A-A
D
d =
45
30 = 1.5
r
d =
6
30 = 0.2 [K.M. D.B. Pg. No. 42]
[PSG D.B. Pg. No. 7.11]
Kt from Graph,
Kt = 1.45
Nominal stress = σo = P
A =
10 × 103
π4
(30)2
σo = 14.147 N ⁄ mm2
Kt =
σmax
σo
Kt ⋅ σ
o = σ
max
σmax
= Max. stress = 1.45 × 14.147 = 20.513 N ⁄ mm2
σmax
= 20.513 N ⁄ mm2
Problem 1.26: A stepped shaft has maximum dia = 50 mm, minimum dia. =25 mm, fillet radius 5 mm. If the shaft is subjected to a twisting moment of
1500 N-m, find the maximum stress induced.
Solution:
Given:
Stepped shaft: D = 50 mm
d = 25 mm
Fillet radius r = 5 mm
T = Mt = 1500 N−mm
Introduction to Design 1.81
Calculate Max. shear
stress τo =
Mt ⋅ rJ
(Refer PSG DB Pg. No. 7.1),
(K.M. D.B. Pg. No. 2)
J = π32
d4
D
d =
50
25 = 2
r
d =
5
25 =
1
5 = 0.2
Kt = 1.35
Kt = τmax
τo
(From PSG D.B Pg. No. 7.14), (K.M. D.B. Pg. No. 43)
where r = minimum shaft diameter
τo = Nominal shear stress = Mt ⋅ r
J
d
2 =
25
2 = 12.5 mm
= 1500 × 12.5
π32
× 254
= 0.488 N ⁄ mm2
∴ Kt = τmax
τo
τmax = 1.35 × 0.4889 = 0.66 N ⁄ mm2
1.82 Design of Machine Elements – I - www.airwalkbooks.com
Problem 1.27: A Non-rotating shaft supporting a load of 2.5 kN is shown in
the Fig. The material of the shaft is brittle, with an ultimate tensile strength
of 300 N ⁄ mm2. Calculate the dia. of the shaft. Take FOS = 3.
Solution:
The stresses are critical at
1. fillet
2. at the centre of the span
1. At fillet
r
d =
0.1 d
d = 0.1
[PSG Databook Pg. No. 7.11]
[K.M. Databook Pg. No. 42]
D
d =
1.1 d
d = 1.1
Kt = 1.47 from Graph
σo = M ⋅ y
I =
1250 × 350 × d
2
π64
d4
= 4.4563 × 106
d3
(From PSG D.B. Pg. No. 7.1)
(K.M. D.B Pg. No. 2)
σmax = kt ⋅ σo = 1.47 × 4.4563 × 106
d3 =
6.55 × 106
d3 N ⁄ mm2
.... (1)
Introduction to Design 1.83
2. At centre
The beam is subjected to max. bending moment at centre.
σb = M ⋅ y
I =
625 × 103 × d
2
π64
d4
Simply supported beam, concentrated
load acting at its centre, Max. B.M.
σb = 6.366 × 106
d3
N ⁄ mm2 .... (2) M = PL
4 =
2.5 × 103 × 1000
4
= 625 × 103 N ⁄ mm2
[From PSG D.B Pg. No. 6.5], [K.M. D.B. Pg. No. 15]
From equations (1) and (2)
The stress is max. at the fillet section.
∴ Equating it with permissible stress,
Permissible stress = σu
FOS =
300
3 = 100 N ⁄ mm2
∴ 100 = 6.5508 × 106
d3
dia. of shaft d = 40.31 mm
Take d = 42 mm
Problem 1.28: Calculate the dimension of a stepped cylinder with D
d = 1.5
and the ratio of fillet radius to the dia. ‘d’ as 0.25, when the cylinder is
subjected to a BM of ± 1500 N−mm; Material for the cylinder is C 15 steel.
(FAQ)
Solution:
1.84 Design of Machine Elements – I - www.airwalkbooks.com
Given: Stepped cylinder
Kt = σmax
σo
D
d = 1.5
σmax = Kt (σo) r
d = 0.25
120 = 1.25 × M ⋅ y
I
(From PSG D.B Pg. No. 7.14), Kt = 1.25
(K.M. Data Book Pg. No. 43)
= 1.25 ×
1500 × d
2
π64
d4
(σb)nominal = M ⋅ y
I
d3 = 1.25 × 15 W × 32
π × 120For C15 materials
d = 5.419 mm Take σy = (PSG Pg. No. 1.9)
D = 1.5 d = 1.5 × 5.419 = 240 N ⁄ mm2 ; Assume FOS = 2
= 8.128 mm. σmax = σy
FOS =
240
2 = 120 N ⁄ mm2
Problem 1.29: What maximum axial force can be applied on a plate of width
50 mm and thickness 10 mm with a central hole of 10 mm diameter without
exceeding the yield point stress of 62.5 MPa across its width. (FAQ)
Solution:
Given data
Let P = Maximum axial force
Width = w = 50 mm
Introduction to Design 1.85
Thickness = h = 10 mm
Central hole diameter = a = 10 mm
Stress = 62.5 MPa
= 62.5 N ⁄ mm2
(Refer PSG Databook Page No. 7.10). To find Kt
Stress concentration = Kt =
σmax
σo
Kt = stress concentration
σmax = maximum stress
= 62.5 MPa = 62.5 N ⁄ mm2
σ0 = nominal stress in N ⁄ mm2
= P
(w − a) h = P
(50 − 10) 10
= P
400 N ⁄ mm2
Kt = σmax
σ0
a
w =
10
50 = 0.2
2.5 = 62.5
P
400
Corresponding to the value of a
w = 0.2, take
the value of Kt. It’s value is 2.5.
Max. axial P = 10000 N
Force = P = 10 kN
1.86 Design of Machine Elements – I - www.airwalkbooks.com
Problem 1.30: Taking stress concentration into account, find the maximum
stress induced when a tensile load of 20 kN is applied to
(i) a rectangular plate of 80 mm wide and 12 mm thick with a transverse
hole of 16 mm diameter.
(ii) a stepped shaft of diameter 60 mm and 30 mm with a fillet radius of
6 mm. (FAQ)
Solution: Given data
Tensile load = P = 20 kN = 20 × 103 N
(a) A rectangular plate of 80 mm wide and 12 mm thick with a
transverse hole of diameter 16 mm.
Kt = σmax
σnom
.... (1) based on net section.
w = width = 80 mm
h = thickness = 12 mm
a = hole diameter = 16 mm
Finite width-plate with a transverse
hole. Refer Graph P. No. 7.10 PSG
a
w =
16
80 = 0.2
σmax = maximum stress
Kt = stress concentration
From Graph, But , σnom =
P
(w − a) h
for a
w = 0.2, the value of Kt = 2.5
Substitute the values in above eq. (1)
Kt = σmax
P
(w − a) h
2.5 = σmax
20 × 103
(80 − 16) 12
Maximum stress
= σmax = 65.104 N ⁄ mm2
Introduction to Design 1.87
(b) a stepped shaft of diameter 60 mm and 30 mm with a fillet radius of
6 mm.
[Refer PSG Databook, Page No. 7.11]
[K.M. Databook Pg. No. 43]
D = 60 mm
d = 30 mm
r = 6 mm
Shaft with a shoulder fillet in tension.D
d =
60
3 = 2
Refer Graphr
d =
6
30 =
1
5 = 0.2
From Graph,
for r
d = 0.2 and
D
d = 2
the value of Kt = 1.5
σnom = nominal stress = P
A
A = π4
d2
(Consider smaller diameter)
Kt = σmax
σnom
π4
(30)2 = 706.858 mm2
1.5 = σmax
28.294σnom =
20 × 103
706.858 = 28.294 N ⁄ mm2
Maximum stress = σmax = 42.4413 N ⁄ mm2
*********
1.88 Design of Machine Elements – I - www.airwalkbooks.com
INDEX
A
Alloy steels, 1.11
Application of power screw, 3.39
Applications of the springs, 5.2
B
Bolt of uniform strength, 3.6
Bolted joints, 3.17
Buckling of compression springs, 5.11
Bushed pin flexible coupling, 6.117
Butt joint in tension, 4.153
C
Cast iron, 1.9
Cast steel, 1.11
Caulking and fullering, 4.6
Causes of failure in shafts, 6.3
Classification of couplings, 6.97
Classification of design, 1.3
Classification of welding, 4.119
Close-coiled helical spring, 5.6
Coaxial springs, 5.89
Common types of screw fastening, 3.4
Common modes of failure, 2.1
Compression stress, 3.19
Concentric springs, 5.87
Conical and volute springs, 5.96
Cotter joints, 4.84
Critical frequency, 5.14
Cumulative fatigue damage, 2.57
Curvature effect, 5.9
D
Design of shaft based on strength, 6.4
Design of gib and cotter joint, 4.113
Design procedure of key, 6.93
Design of screw jack, 3.74
Design of structural joints, 4.10
Design of circumferential joint, 4.71
Design of structural joints, 4.64
Design stresses, 4.63
Design of sleeve and cotter joint, 4.106
Design of flexible couplings, 6.116
Design of bolts for cylinder cover, 3.8
Design of rigid coupling, 6.98
Design of power screws, 3.39
Design for finite life, 2.56
Design based on critical speed, 6.12
Direct tensile stress, 3.18
Disc springs (or) belleville springs, 5.81
Ductile fracture, 1.26
E
Eccentric loading of springs, 5.11
Effect of initial tension, 3.17
Effects of keyways, 6.93
Index I.1
Efficiency of a Riveted Joint, 4.64
End connections for helical springs, 5.10
Endurance limit stress, 2.43
F
Factor of safety, 1.30, 2.2
Factors influencing machine design, 1.4
Factors affecting endurance limit, 2.45
Failures modes of keys, 6.90
Fatigue loading of helical springs, 5.65
Fatigue failure, 2.39
Features of shaft coupling, 6.97
Ferrous metals, 1.9
Flange coupling, 6.111
Flat spiral spring, 5.78
Forces acting on a key, 6.89
Friction in V-threads (ACME thread), 3.49
Friction in power screw, 3.41
G
Gasketed joint, 3.32
Gaskets, 3.31
H
Helical springs, 5.5
Helical torsion spring, 5.71
I
Impact stress, 1.35
Initial gap, 5.104
Initial pre-load, 5.105
K
Knuckle joint, 4.79
L
Leaf spring material, 5.108
Leaf spring, 5.97
M
Material for Rivets, 4.2
Mechanical properties of materials, 1.6
Mechanism of ductile fracture, 1.27
Modes of failures in riveted joint, 4.7
N
Nipping of leaf springs, 5.103
Non-ferrous metals, 1.14
O
Oldham Coupling, 6.160
Open-coiled helical spring, 5.6
P
Permissible stresses, 5.108
Plastic behaviour of metals, 1.26
Preferred numbers, 1.15
Principal stresses, 1.37
R
Resilience, 5.13
Riveted joints, 4.1
I.2 Design of Machine Elements – I - www.airwalkbooks.com
S
Selection of materials, 1.8
Shafting materials, 6.3
Shear stress in the threads, 3.19
Shock and impact loads, 2.40
Simple stresses, 1.19
Sleeve and cotter joint, 4.106
Soderberg and Goodman diagrams, 2.47
Spring materials, 5.1
Springs in series, 5.69
Springs in parallel, 5.70
Springs, 5.1
Standard shaft diameters, 6.2
Standards and codes, 1.17
Static loading, 5.13
Steel, 1.10
Stiffness (k), 6.9
Strength of a Riveted Joint, 4.64
Stress in shafts, 6.3
Stress concentration, 1.72
Stresses in power screws, 3.54
Stresses in butt weld, 4.152
System of designation for steels, 1.12
T
Theories of failure, 2.3
Threaded joints, 3.1
Torque diagrams, 6.14
Torsional shear stress, 3.18
Types of ends for extension springs, 5.12
Types of welded joints, 4.119
Types of riveted joints, 4.4
Types of power screw threads, 3.40
Types of shaft (according to use), 6.2
U
Universal or Hooke’s coupling, 6.160
Welded joints, 4.118
*********
Index I.3