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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

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© 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

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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


4.9. 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


4.11. 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


4.11.9. 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.


(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.


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.


(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


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.


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.


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.


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.


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.


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%


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).


(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.


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


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.


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.


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


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.


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


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


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


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).


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.


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.


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).


(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).


(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.


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.


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


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.


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


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


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’.


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.)


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.


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


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


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)


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


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


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


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.


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.


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


σ2 = Min. principal stress = σx + σy

2 √

σx − σy

2

2

+ τxy2

= 6.365 − 14.23 = − 7.865 N ⁄ mm2


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


σ2 = 45 − 5 = 40 N ⁄ mm2

Since σ1 and σ

2 are + ve, Consider σ

3 into account. (σ

3 = 0)


= σ

1 − σ

3

2 =

50 − 0

2 = 25 N ⁄ mm2


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.


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


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



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



= 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.)


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.


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



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


= 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


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


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


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.


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


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


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


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


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


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


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


= 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


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


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


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.


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.


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


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


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


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.


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


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


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.


(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


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


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


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


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


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


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]


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)


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:


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


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


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


(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]


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

*********


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

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