MODULE ONE PRPC19 Design of Machine Elements- 1 .pdf

By
Dr. Prakash Kumar
Assistant Professor
Department of Production Engineering
NIT Tiruchirappalli
PRPC19 Design of Machine Elements
Lecture Notes
On
Module 1
Design due to static and dynamic
load

Syllabus
•Introduction to the design process,
•factor influencing machine design,
•Mechanical properties of materials,
•direct stress, bending stress, torsional stress and variable
•stress in machine parts,
•theories of failure,
•stress concentration factor,
•factor of safety.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 2

What is Engineering Design?
•Adesignisaplanorspecificationfortheconstructionofanobjectorsystemorfor
theimplementationofanactivityorprocess,ortheresultofthatplanorspecification
intheformofaprototype,productorprocess.
•Theprocessofdevisingasystem,component,orprocesstomeetdesiredneeds.
•Itisadecision-makingprocess(ofteniterative),inwhichthebasicsciences,
mathematics,andengineeringsciencesareappliedtooptimallyconvertresourcesto
meetastatedobjective.
•Designisthecreationofthenewandbettermachines,machinecomponentsor
productsaswellasthedevelopmentormodificationoftheexistingdesignwhichis
moreeconomicalintheoverallcostofproductionandoperations.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 3

Some example of poor design in daily life
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 4

Disasters due to poor design
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 5
TheChernobylnucleardisasterof
1986inPripyat,Ukraine.Reportedly,
62peoplediedimmediatelyafterthe
explosion.Thelong-termdeathcount
isanestimated4,000–9,000people.
The Challenger space shuttle exploded
in 1986 after an O-ring seal in its right
solid rocket booster failed.
• Fuel tanks of Concorde airplanes:
The design layout of the fuel tanks
was the cause of the Concorde crash
in 2000, killing 113 people.
When the aircraft struck the debris on
the runway, the tire that subsequently
exploded caused a tank to rupture.

Reason for failures in most engineering designs
•Incorrect or overextendedassumptions
•Poorunderstanding of the problem to be solved
•Incorrectdesign specifications
•Faultymanufacturing and assembly
•Errorindesign calculations
•Incompleteexperimentation andinadequatedata collection
•Errorsindrawings
•Faultyreasoning from good assumptions
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 6

What is Machine Design?
•Machinedesignisdefinedastheuseofscientificprinciples,technicalinformationand
imaginationinthedescriptionofamachineoramechanicalsystemtoperform
specificfunctionswithmaximumeconomyandefficiency.Thisdefinitionofmachine
designcontainsthefollowingimportantfeatures:
•Adesignerusesprinciplesofbasicandengineeringsciencessuchasphysics,
mathematics,staticsanddynamics,thermodynamicsandheattransfer,vibrationsand
fluidmechanics.
•Thedesignerhastechnicalinformationofthebasicelementsofamachine.These
elementsincludefasteningdevices,chain,beltandgeardrives,bearings,oilsealsand
gaskets,springs,shafts,keys,couplings,andsoon.
•Thedesigneruseshisskillandimaginationtoproduceaconfiguration,whichisa
combinationofthesebasicelements.However,thiscombinationisuniqueand
differentindifferentsituations.
•Thefinaloutcomeofthedesignprocessconsistsofthedescriptionofthemachine.
Thedescriptionisintheformofdrawingsofassemblyandindividualcomponents.
•Adesigniscreatedtosatisfyarecognisedneedofcustomer.Theneedmaybeto
performaspecificfunctionwithmaximumeconomyandefficiency.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 7

Classification of Machine Design
•1) Adaptive Design: In this design, the designer shall adopt the existing design and
can be made minor alteration or modifications in the existing design. There is no need
of special knowledge or skills to the designer.
•Example: In the automobiles designer can not be change the design of engine but
designer can modify the engine by increasing the capacity of engine.
•2)Development Design: In this design, the designer shall adopt the existing design and
can be made modifications in the materials, manufacturing process, shapes and sizes
etc. The designer should have some sort of knowledge and trainings.In this case,
though the designer starts from the existing design, but the final product may differ
quite markedly from the original product.
•Example: In 1990 cars are having sharp edges at the corners but now a days these
sharp edges are replaced by the various curvatures and aerodynamic shapes.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 8
Machine Design
Adaptive Design
Development
Design
New Design

•3)NewDesign:Thistypeofdesignneedsalotofresearch,technicalabilityand
creativethinking.Thedesignisdependsonthepersonalskillsandknowledgeofthe
designerinthedesignprocesses.
•Example:Nowadaysduetolimitedresourcesoffossilfuelsandenvironmental
safetyautomobileindustriesaregoingtodevelopnewtechnologiesinautomobile
basedonpotentialenergies.
•Thedesigns,dependinguponthemethodsused,maybeclassifiedasfollows:
•(a)Rationaldesign.Dependsuponmathematicalformulaeofprincipleofmechanics.
•(b)Empiricaldesign.Dependsuponempiricalformulaebasedonthepracticeandpast
experience.
•(c)Industrialdesign.Dependsupontheproductionaspectstomanufactureany
machinecomponentintheindustry.
•(d)Optimumdesign.Itisthebestdesignforthegivenobjectivefunctionunderthe
specifiedconstraints.Itmaybeachievedbyminimizingtheundesirableeffects.
•(e)Systemdesign.Itisthedesignofanycomplexmechanicalsystemlikeamotorcar.
•(f)Elementdesign.Designofanyelementofthemechanicalsystemlikepiston,
crankshaft,connectingrod,etc.
•(g)Computeraideddesign.Thistypeofdesigndependsupontheuseofcomputer
systemstoassistinthecreation,modification,analysisandoptimizationofadesign.
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Standard Design Process
•Thecompletedesignprocessfrom
starttofinish,isoftenoutlinedasin
thefigure.
•Beginswithanidentificationof
needandadecisiontodosomething
aboutit.
•Aftermanyiterations,theprocess
endswiththepresentationofthe
plansforsatisfyingtheneed.
•Severaldesignphasesmaybe
repeatedthroughoutthelifeofthe
product.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 10

Phase of the Design Process
•Identification of need generally starts the design process. Recognition of the need
and phrasing the need often constitute a highly creative act, because the need may be
only a vague discontent, a feeling of uneasiness, or a sensing that something is not
right.
•The definition of problem is more specific and must include all the specifications
for the object that is to be designed.
•The synthesis of a scheme connecting possible system elements is sometimes
called the invention of the concept or concept design. This is the first and most
important step in the synthesis task.
•Analysesmust be performed to assess whether the system performance is
satisfactory.
•Synthesis, analysis and optimizationare intimately and iteratively related.
•Evaluationis the final proof of a successful design and usually involves the testing
of a prototype in the laboratory.
•Presentationis a selling job.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 11

Basic Requirements of Machine Elements
•Amachineconsistsofmachineelements.Eachpartofamachine,whichhasmotion
withrespecttosomeotherpart,iscalledamachineelement.Itisimportanttonote
thateachmachineelementmayconsistofseveralparts,whicharemanufactured
separately.Thebroadobjectiveofdesigningamachineelementistoensurethatit
preservesitsoperatingcapacityduringthestipulatedservicelifewithminimum
manufacturingandoperatingcosts.Inordertoachievethisobjective,themachine
elementshouldsatisfythefollowingbasicrequirements:
•(i)Strength:Amachinepartshouldnotfailundertheeffectoftheforcesthatacton
it.Itshouldhavesufficientstrengthtoavoidfailureeitherduetofractureordueto
generalyielding.
•(ii)Rigidity:Amachinecomponentshouldberigidthatis,itshouldnotdeflector
bendtoomuchduetoforcesormomentsthatactonit.
•(iii)WearResistance:Wearisthemainreasonforputtingthemachinepartoutof
order.Itreducesusefullifeofthecomponent.Wearalsoleadstothelossofaccuracy
ofmachinetools.
•(iv)MinimumDimensionsandWeight:Amachinepartshouldbesufficiently
strong,rigidandwearresistantandatthesametime,withminimumpossible
dimensionsandweight.Thiswillresultinminimummaterialcost.
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•(v) Manufacturability: Manufacturability is the ease of fabrication and assembly. The shape
and material of the machine part should be selected in such a way that it can be produced with
minimum labourcost.
•(vi) Safety: The shape and dimensions of the machine parts should ensure safety to the operator
of the machine. The designer should assume the worst possible conditions and apply ‘fail-safe’
or ‘redundancy’ principles in such cases.
•(vii) Conformance to Standards: A machine part should conform to the national or
international standard covering its profile, dimensions, grade and material.
•(viii) Reliability: Reliability is the probability that a machine part will perform its
intended functions under desired operating conditions over a specified period of time.
A machine part should be reliable, that is, it should perform its function satisfactorily
over its lifetime.
•(ix) Maintainability: A machine part should be maintainable. Maintainability is the ease
with which a machine part can be serviced or repaired.
•(x) Minimum: Life-cycle Cost: Life-cycle cost of the machine part is the total cost to be
paid by the purchaser for purchasing the part and operating and maintaining it over its
life span.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 13

DESIGN OF MACHINE ELEMENTS
•Designofmachineelementsisthe
mostimportantstepinthecomplete
procedureofmachinedesign.In
ordertoensurethebasic
requirementsofmachineelements,
calculationsarecarriedouttofind
outthedimensionsofthemachine
elements.Thebasicprocedureofthe
designofmachineelementsis
illustratedbelow
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 14

The Design Engineer’s Responsibilities
•In general, design engineering is required to satisfy the needs of customers
( management, clients, consumers, etc. ) and is expected to do so in a competent,
responsible, ethical, and professional manner.
•Careful attention to the following action steps will help you to organize your
solution processing technique.
✓Understand the problem.
✓Identify the known.
✓Identify the unknown and formulate the solution strategy.
✓State all assumption and decision.
✓Analyze the problem.
✓Evaluate your solution.
✓Present your solution
•The design engineer’s professional obligations include conducting activities in an
ethical manner.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 15

Standardization
•Standardization is defined as obligatory norms, to which various characteristics of a
product should conform.
•The characteristics include materials, dimensions and shape of the component, method
of testing and method of marking, packing and storing of the product.
•Standards Used in Machine Design:
•1) Standards for materials, chemical compositions, mechanical properties and
heat treatment. For example, Indian standard IS 210 for gray cast iron like FG 150,
FG 200, FG 220 etc. IS 1570 (part 4) specifies chemical composition like 55Cr3 has
0.5 –0.6 % carbon, 0.1-0.35 % Si, 0.6-0.8 Manganese and 0.6 –0.8% chromium.
•2) Standards for shapes and dimensions of commonly used machine elements.
For example, IS 2494 for cross section of V belts, IS 5129 dimensions of rotary shaft
oil seal units.
•3) Standards for fits, tolerances and surface finish of components.For example,IS
2709, IS 919, IS 10719.
•4) Standards for testing of products.For example, IS 807 is a code of practice for
design, manufacture, erection and testing of cranes and hoists.
•5) Standards for engineering drawing of components. For example, SP46 prepared
by BIS on ‘Engineering Drawing Practice for Schools and Colleges’ which covers all
standards related to engineering drawing.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 16

Standards and Types of Standards:
•A standardization is defined as a set of specifications for parts, materials or processes.
•Types of Standards:
•1) Company Standards:They are used in a particular company or a group of sister
concerns.
•2) National Standards: These are the IS (Bureau of Indian Standards), DIN
(German), AISI or SAE (USA) or BS (UK) standards.
•3) International Standards: These are prepared by the International Standards
Organization (ISO).
•Advantages of Standards:
•1) Interchangeability of machine component
•2) Better product quality, reliability and long service life
•3) Mass production at low cost and less time
•4) Ensures the safety
•5) Reduction in various sizes and grades
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 17

IS 210 IS 1570 part 4
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 18

Standards and Codes
•A standard is a set of specifications for parts, materials, or processes intended to
achieve uniformity, efficiency, and a specified quality.
•One of the important purposes of a standard is to place a limit on the number of items
in the specifications so as to provide a reasonable inventory of tooling, sizes, shapes,
and varieties.
•A code is a set of specifications for the analysis, design, manufacture, and
construction of something.
•The purpose of a code is to achieve a specified degree of safety, efficiency, and
performance or quality.
•All of the organizations and societies listed below have established specifications for
standards and safety or design codes.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 19
•Aluminum Association (AA)
•American Gear Manufacturers Association (AGM)
•American Institute of Steel Construction (AISC)
•American Iron and Steel Institute (AISI)
•American National Standards Institute (ANSI)
•ASM International
•American Society of Mechanical Engineers (ASME)
•American Society of Testing and Material (ASTM)
•American Welding Society (AWS)
•American Bearing Manufactures Association
(ABMA)
•British Standards Institute (BSI)
•Industrial Fasteners Institute (IFI)
•International Bureau of Weights and Measures
(BIPM)
•International Standards Organization (ISO)
•National Institute for Standards and Technology
(NIST)
•Society of Automotive Engineers (SAE)
•Institution of Mechanical Engineers (I. Mech. E.)

Modern Design Considerations
•Nowadays the design of machine elements is not limited to only strength, analysis,
reliability and durability of the machine element but various modern design
parameters are also considered in design process which are as given below,
•1) Design for safety
•2) Ecological design
•3) Design for society
•4) Ergonomics considerations
•5) Aesthetics considerations
•1)Design for safety: While designing the machine element designer should consider
the safety operations and use of the elements.
•2) Ecological Design: While designing any machine component designer should
consider the adverse effects of a design on the environment.
•3)Design for Society: Design is the solution for the problems occurred at the end user
i.e. society. The designer should develop the design by considering the satisfactory
level of the society while use or operate the designed machine element
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 20

Cont….
•4) Ergonomic considerations:
•Ergonomic is defined as the relationship between man and machine and application of
anatomical, physiological and psychological principles to solve the problems arising
from man-machine relationship.
•Ergonomics means the natural laws of work (‘ergo’ i.e. ‘work’ and ‘nomos’ i.e.
‘natural laws’).
•The man-machine joint system forming closed loop as shown in figure. The man-
machine interface at two points
•1) Display: Gives information to the operator
•2) Control: Operator adjust the machine
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 21
Topics of ergonomic studies are,
1) Anatomical factors
2) Layout of instrument dials and display panels
3) Design of hand lever and hand wheels
4) Energy expenditure in hand and foot
operations
5) Lighting, temperature and climate conditions

Cont…
•Ergonomic considerations in the design of displays:
•The displays are of three types, which giving quantitative measurements (eg.Pressure
gauge, Speedometer), giving the state of affair (eg.Coloredlamps) and indicating
predetermined settings (eg.Automobile gear shifter).
•1) The scale on dial indicator should be divided in suitable numerical progression.
•2) The number of subdivisions between numbered divisions should be minimum.
•3) The size of letter or height of letter or number should be greater than or equal to
reading distance/200.
•4) Vertical figures should be used for the stationary dials, while radially oriented
figures are suitable for rotating dials.
•5) The pointer should have a knife edge with mirror in the dial to minimize parallax
error.
•Factors of selection of controls:
•The type and size of control devices selected is depends on the factors as below,
•1) The required speed and feed of the operation
•2) The required operating force and accuracy of the control
•3) The required range and direction of movement
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 22

Cont….
•Ergonomic considerations in design of controls:
•The following considerations have made for the design of control devices,
•1) The control should be easily accessible and logically positioned.
•2) The control operation should be involved minimum motion and avoid outward
movements.
•3) The shape of control device which is in contact with the hands of an operator
should be with anatomy of human hand.
•4) The proper colorproduces beneficial psychological effect, hence the control for
attention should be painted with the red colorwith grey background.
•5) The sound attention should be used for the control operations.
•Working Environment and Safety:
•The working environment affects the man-machine relationship such as supervising,
controlling and maintaining the work and equipment. The environment affect the
efficiency and health of operator and machine. The factors of environment are,
•1)Lighting
•2)Noise
•3)Temperature
•4)Relative humidity and air motion
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 23

5) Aesthetic considerations:
•Aesthetic is defined as a set of principles of beauty. It deals with the appearance of the
product.
•Eachproducthastoperformspecificfunctionstothesatisfactionofcustomer.The
necessityoffunctionalrequirementsbringproductsandpeopletogether.
•However,whenthereareanumberofproductsinthemarkethavingthesame
qualitiesofefficiency,durabilityandcost,thecustomerisattractedtowardsthemost
appealingproduct.Thegrowingrealizationoftheneedofaestheticconsiderationsin
productdesignhasgivenrisetoaseparatedisciplineknownas‘industrialdesign’.
•Thejobofindustrialdesigneristocreatenewshapesandformsfortheproductwhich
areaestheticallyappealing.Forex.(1)Thechromiumplatingofautomobile
componentsimprovesthecorrosionresistancealongwiththeappearance.(2)the
aerodynamicshapeofthecarimprovestheperformanceaswellasgivesthepleasing
appearance
•Guidelines in aesthetic design: The appearance of the product should reflects,
•1)Performance: The aerodynamic shape of an automobile will decrease air resistance
and increases efficiency of engine.
•2)Function: The aerodynamic shape of an automobile indicate the speed.
•3)Quality:Thesurface finish of the shaft indicates the strength.
•4)Suitability to the Environment: Color codes are used in hazardous and safety areas
in industries.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 24

Selection of preferred size
•CharlesRenardfirstintroducedpreferrednumbersinthe19thcentury.
•Fordesigneritisimportanttospecifythesizeoftheproduct.The‘size’oftheproduct
isageneralterm,whichincludesdifferentparameterslikepowertransmittion,load
carryingcapacity,speed,dimension,lengthetc.e.g5kW,10kN,1000rpm.
•E.g:-Acompanymaybemanufacturingsevendifferentmodelsofelectricmotors
rangingfrom0.5to50kWtocatertotheneedofdifferentcustomers.Preferred
numbersareusedtospecifythe‘sizes’oftheproductinthesecases.
•The system is based on the use of geometric progression to develop a set of numbers.
•Five basic series R5, R10, R20, R40 and R80 which increases in steps of 58%, 26%,
12%, 6%, and 3 % respectively. Each series has its own series factor.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 25
•1
st
step:-seriesestablishbytaking1
st
numberand
multiplyingitbyaseriesfactortogetsecond
number.
•2
nd
step:-multiply2
nd
numberbyseriesfactortoget
3
rd
number.
•Processcontinueduntilseriesiscompleted.
•E.g:-Manufacturerintroducedifferentliftingtackle
modelrangefrom15to100kN-
>16,20,25,31.5,40,50,63,80,100kN.

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 26
Few examples
•R
10
, R
20
and R
40
: Thickness of sheet metals, wire diameter
•R
5
, R
10
, R
20
: Speed layout in a machine tool
•R
20
or R
40
: Machine tool feed
•R
5
: Capacities of hydraulic cylinder
•

Basic series and Derived series
•R5, R10, R20, R40 and R80 are called basic series.
•Any series that is formed on the basis of these five basic series is called derived series.
•Derivedseriesarederivedfrombasicseries.Therearetwomethodsofforming
derivedseries,namely,reducingthenumbersofaparticularbasicseriesorincreasing
thenumbers.
•Inthefirstmethod,aderivedseriesisobtainedbytakingeverysecond,third,fourthor
p
th
termofagivenbasicseries.
•(i)SeriesR10/3(1,…,1000)indicatesaderivedseriescomprisingofeverythird
termoftheR10seriesandhavingthelowerlimitas1andhigherlimitas1000.
•(ii)SeriesR20/4(…,8,…)indicatesaderivedseriescomprisingofeveryfourthterm
oftheR20series,unlimitedinbothsidesandhavingthenumber8insidetheseries.
•(iii)SeriesR20/3(200,…)indicatesaderivedseriescomprisingofeverythirdterm
oftheR20seriesandhavingthelowerlimitas200andwithoutanyhigherlimit.
•(iv)SeriesR20/3(…200)indicatesaderivedseriescomprisingofeverythirdtermof
theR20seriesandhavingthehigherlimitas200andwithoutanylowerlimit.
•Inthesecondmethod,thederivedseriesisobtainedbyincreasingthenumbersofa
particularbasicseries.Letusconsideranexampleofaderivedseriesofnumbers
rangingfrom1to1000basedontheR5series.
•Therefore,thecompletederivedseriesonthebasisofR5seriesisasfollows:
•1,1.6,2.5,4,6.3,10,16,25,40,63,100,160,250,400,630,1000
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 27

Numerical Problem 1
Q:-Find out the numbers of the R5 basic series from 1 to 10.
Answer. First number = 1
Second number = 1(1.5849)= 1.5849 = 1.6
Third number = 1(1.5849)(1.5849)=2.51= 2.5
Fourth number = 1(1.5849)
3
=3.98= 4
Fifth number = 1(1.5849)
4
= 6.3
Sixth number = 1(1.5849)
5
= 10
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 28

Numerical Problem 2
Q :-Find out number of R20/4(100,…1000) derived series.
Answer. Series factor for R20series is 1.122
Every forth term of R20 series is selected, the ratio factor (Ø) is given by,
Ø = (1.122)
4
= 1.5848
First number = 100
Second number = 100(1.5848)= 158.48 = 160
Third number = 100(1.5848)(1.5848)=251.16 = 250
Fourth number = 100(1.5849)
3
=398.04 = 400
Fifth number = 100(1.5849)
4
= 630.81 = 630
Sixth number = 100(1.5849)
5
= 999.71 = 1000
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 29

Numerical Problem 3
•Q:-Amanufacturerisinterestedinstartingabusinesswithfivedifferentmodelsof
tractorsrangingfrom7.5kWto75kWcapacities.Specifypowercapacitiesof
models.Thereisanexpansionplantofurtherincreasethenumberofmodelsfromfive
toninetofulfilltherequirementoffarmers.Specifythepowercapacitiesofthe
additionalmodels
•Power rating of five models will be as follows
•7.5 (Ø)
0
, 7.5 (Ø)
1
, 7.5 (Ø)
2
7.5 (Ø)
3
, 7.5 (Ø)
4
•Maximum power rating is 75 kW. Therefore, 7.5 (Ø)
4
= 75
•Hence, Ø = 1.7783
•First number = 7.5
•Second number = 7.5(1.7783)= 13.34 = 13
•Third number = 7.5(1.7783)
2
=23.72 = 24
•Fourth number = 7.5(1.7783)
3
=42.18 = 42
•Fifth number = 7.5(1.7783)
4
= 75
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 30

Cont..
•Answer. 2
nd
Part
•Power rating of nine models will be as follows
•7.5 (Ø)
0
, 7.5 (Ø)
1
, 7.5 (Ø)
2
7.5 (Ø)
3
…………………7.5 (Ø)
8
•Maximum power rating is 75 kW. Therefore, 7.5 (Ø)
8
= 75
•Hence, Ø = 1.3335
•First number = 7.5
•Second number = 7.5(1.3335)= 10.00 = 10
•Third number = 7.5(1.3335)
2
=13.34 = 13
•Fourth number = 7.5(1.3335)
3
=17.78 = 18
•Fifth number = 7.5(1.3335)
4
= 23.72 = 24
•Sixth number = 7.5(1.3335)
5
=31.62 = 32
•Seventh number = 7.5(1.3335)
6
=42.17 = 42
•Eighth number = 7.5(1.3335)
7
= 56.24 = 56
•Ninth number = 7.5(1.3335)
8
= 74.99 = 75 kW
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 31

Sequential Design VsConcurrent Engineering
•Sequentialengineeringisthetermusedtoexplainthe
methodofproductioninalinearsystem.Thevarioussteps
aredoneoneafteranother,withallattentionandresources
focusedonthatsingletask.
•Sequentialengineeringisasystembywhichagroup
withinanorganizationworkssequentiallytocreatenew
productsandservices.
•Thesequentialengineeringisalinearproductdesign
processduringwhichallstagesofmanufacturingoperate
inserial.
•Bothprocessandproductdesignruninserialandtake
placeinthedifferenttime.
•ProcessandProductarenotmatchedtoattainoptimal
matching.
•Decisionmakingdonebyonlygroupofexperts.
•Inconcurrentengineering,varioustasksarehandledatthe
sametime,andnotessentiallyinthestandardorder.This
meansthatinfofoundoutlaterinthecoursecanbeadded
toearlierparts,improvingthem,andalsosavingtime.
•Concurrentengineeringisamethodbywhichseveral
groupswithinanorganizationworksimultaneouslyto
createnewproductsandservices.
•Theconcurrentengineeringisanon-linearproductdesign
processduringwhichallstagesofmanufacturingoperateat
thesametime.
•Bothproductandprocessdesignruninparallelandtake
placeinthesametime.
•ProcessandProductarecoordinatedtoattainoptimal
matchingofrequirementsforeffectivequalityanddelivery.
•Decisionmakinginvolvesfullteaminvolvement.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 32

Engineering materials
•Theknowledgeofmaterialsandtheirpropertiesisofgreatsignificanceforadesign
engineer.Themachineelementsshouldbemadeofsuchamaterialwhichhas
propertiessuitablefortheconditionsofoperation.
•Classification of Engineering Materials
•The engineering materials are mainly classified as:
•1. Metals and their alloys, such as iron, steel, copper, aluminum, etc.
•2. Non-metals, such as glass, rubber, plastic, etc.
•The metals may be further classified as: (a) Ferrous metalsand (b) Non-ferrous
metals.
•The ferrous metals are those which have the iron as their main constituent, such as
cast iron, wrought iron and steel.
•The non-ferrousmetals are those which have a metal other than iron as their main
constituent, such as copper, aluminum, brass, tin, zinc, etc.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 33

Selection of Materials for Engineering Purposes
•The selection of a proper material, for engineering purposes, is one of the most
difficult problems for the designer. The best material is one which serves the desired
objective at the minimum cost. The following factors should be considered while
selecting the material:
•1. Availability of the materials,: Material should be readily available in market in
large enough quantities to meet the requirement
•2. Suitability of the materialsfor the working conditions in service, and
•3. The cost of the materials.
•The important properties, which determine the utility of the material, are physical,
chemical and mechanical properties.
•Manufacturing Considerations:
➢In some applications machinability of material is an important consideration in selection
➢Where the product is of complex shape, cast ability or ability of the molten metal to flow
into intricate passages is the criterion of material selection
➢In fabricated assemblies of plates & rods, weldabilitybecomes the governing factor
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 34

Mechanical Properties of Metals
•Themechanicalpropertiesofthemetalsarethosewhichareassociatedwiththe
abilityofthematerialtoresistmechanicalforcesandload.Thesemechanical
propertiesofthemetalincludestrength,stiffness,elasticity,plasticity,ductility,
brittleness,malleability,toughness,resilience,creepandhardness.
•1.Strength:Itistheabilityofamaterialtoresisttheexternallyappliedforces
withoutbreakingoryielding.Theinternalresistanceofferedbyaparttoanexternally
appliedforceiscalledstress.Dependinguponthetypeofstressesinducedbyexternal
loads,strengthisexpressedastensilestrength,compressivestrengthorshearstrength.
•2.Elasticity:Itisthepropertyofamaterialtoregainitsoriginalshapeafter
deformationwhentheexternalforcesareremoved.Thispropertyisdesirablefor
materialsusedintoolsandmachines.Allengineeringmetalsareelasticbutthedegree
ofelasticityvaries.Itmaybenotedthatsteelismoreelasticthanrubber.
•3.Plasticity:.Itispropertyofamaterialwhichretainsthedeformationproduced
underloadpermanently.Inthiscase,theexternalforcesdeformthemetaltosuchan
extentthatitcannotfullyrecoveritsoriginaldimensions.Thispropertyofthematerial
isnecessaryforforgings,instampingimagesoncoinsandinornamentalwork.
•Duringelasticdeformation,atomsofmetalaretemporarilydisplacedfromtheir
originalpositionsbutreturnbackwhentheloadisremoved.Duringplastic
deformation,atomsofmetalarepermanentlydisplacedfromtheiroriginalpositions
andtakeupnewpositions.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 35

Mechanical Properties of Metals
•4.Stiffness:Itistheabilityofamaterialtoresistdeformationundertheactionof
externalload.Foragivenstresswithinelasticlimit,thematerialthatdeformsleastis
thestiffest.Themodulusofelasticityisthemeasureofstiffness.Stiffnessisan
importantconsiderationinthedesignoftransmissionshafting.
•5.Resilience:Itistheabilityofthematerialtoabsorbenergywhendeformed
elasticallyandtoreleasethisenergywhenunloaded.Aresilientmaterialabsorbs
energywithinelasticrangewithoutanypermanentdeformationtoresistshockand
impactloads.Thispropertyisessentialforspringmaterials.
•6.Ductility:Itisdefinedastheabilityofamaterialtodeformtoagreaterextent
beforethesignofcrack,whenitissubjectedtotensileforce.Inotherwords,ductility
isthepropertyofamaterialenablingittobedrawnintowirewiththeapplicationofa
tensileforce.Mildsteel,copperandaluminiumareductilematerials.Ductilityisa
desirablepropertyinmachinecomponentswhicharesubjectedtounanticipated
overloadsorimpactloads.
•7.Malleability:Itisabilityofamaterialtodeformtoagreaterextentbeforethesign
ofcrack,whenitissubjectedtocompressiveforce.Malleablemetalscanberolled,
forgedorextrudedbecausetheseprocessesinvolveshapingundercompressiveforce.
Lowcarbonsteels,copperandaluminiumaremalleablemetals.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 36

•8.Toughness:Itisthepropertyofthematerialtoabsorbenergybeforefracturetakes
placeduetohighimpactloadslikehammerblows.Inotherwords,toughnessisthe
energyforfailurebyfracture.Thispropertyisdesirableinpartssubjectedtoshock
andimpactloads.Inpractice,toughnessismeasuredbytheIzodandCharpyimpact
testingmachines.
•9.Brittleness:Itispropertyofthematerialwhichshowsnegligibleplastic
deformationbeforefracturetakesplace.Brittlenessistheoppositetoductility.Cast
ironisanexampleofbrittlematerial.Brittlecomponentsfailbysuddenfracture.A
tensilestrainof5%atfractureinatensiontestisconsideredasthedividingline
betweenductileandbrittlematerials.
•10.Machinability:Itisthepropertyofamaterialwhichreferstoarelativecasewith
whichamaterialcanbecut.
•11.Creep:Whenapartissubjectedtoaconstantstressathightemperatureforalong
periodoftime,itwillundergoaslowandpermanentdeformationcalledcreep.This
propertyisconsideredindesigninginternalcombustionengines,boilersandturbines.
•12.Fatigue:Whenamaterialissubjectedtorepeatedstresses,itfailsatstresses
belowtheyieldpointstresses.Suchtypeoffailureofamaterialisknownas*fatigue.
Thispropertyisconsideredindesigningshafts,connectingrods,springs,gears,etc.
•13.Hardness:Propertyofthematerialthatenablesittoresistpermanent
deformation,penetration,indentationetc.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 37

Physical Properties of Metals
•The physical properties of the metals include luster, colour, size and shape, density,
coefficient of thermal expansion, thermal conductivity, and electrical resistivity.
•Density: Density is defined as the mass per unit volume of a material. Its usual units
are kg/m
3
. pounds-mass. The Greek letter rho (r) is the symbol for density. In some
applications, the term specific weight or weight density is used to indicate the weight
per unit volume of a material.
•Coefficient of Thermal Expansion: The coefficient of thermal expansion is a
measure of the change in length of a material subjected to a change in temperature.
For machines and structures containing parts of more than one material, the different
rates of Thermal Expansion can have a significant effect on the performance of the
assembly and on the stresses produced.
•ThermalConductivity:Thermalconductivityisthepropertyofamaterialthat
indicatesitsabilitytotransferheat.Wheremachineelementsoperateinhot
environmentsorwheresignificantinternalheatisgenerated,theabilityofthe
elementsorofthemachine’shousingtotransferheatawaycanaffectmachine
performance.
•Electrical Resistivity: For machine elements that conduct electricity while carrying
loads, the electrical resistivity of the material is as important as its strength.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 38

Plain Carbon Steel
•Dependinguponthepercentageofcarbon,plaincarbonsteelsareclassifiedas:
•i)Lowcarbonsteel–Lessthan0.3%carbon,popularlyknownasmildsteel,itssoft
&ductile,easilymachined&welded,howeverduetolowcarboncontent
unresponsivetoheattreatment
•ii)Mediumcarbonsteel–carboncontentintherangeof0.3%to0.5%,popularly
knownasmachinerysteel,easilyhardenedbyheattreatment,stronger&tougherthan
lowcarbonsteel,wellmachined,respondreadilytoheattreatment
•iii)Highcarbonsteel–morethan0.5%carbon,popularlyknownashardsteelsor
toolsteels,respondreadilytoheattreatment,whenheattreatedattainhighstrength
combinedwithhardness,lessductilethanlowcarbonsteels&mediumcarbonsteels,
difficulttoweld,excessivehardnessaccompaniedbyexcessivebrittleness
•Inapplicationslikeautomobilebodies&hoods,theabilityofthematerialtodeform
toagreaterextentor‘ductility’isthemostimportantconsiderationsoaplaincarbon
ispreferred.
•Inapplicationslikegears,machinetoolspindles&transmissionshaft,strength
toughness&responsetoheattreatmentareimportantconsiderations,medium&high
carbonsteelsarepreferred
•Springwiresaresubjectedtoseverestress&strengthisthemostimportant
considerationsohighcarbonsteelisselectedforhelical&leafsprings
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 39

Steels Designated on the Basis of Mechanical Properties
•These steels are carbon and low alloy steels where the main criterion in the selection
and inspection of steel is the tensile strength or yield stress.
•According to Indian standard IS: 1570 (Part–I)-1978 (Reaffirmed 1993), these steels
are designated by a symbol ‘Fe’ or ‘Fe E’ depending on whether the steel has been
specified on the basis of minimum tensile strength or yield strength, followed by the
figure indicating the minimum tensile strength or yield stress in N/mm
2
.
•By Ultimate Tensile Strength:
•For example ‘Fe 290’ means a Plain Carbon steel having minimum tensile strength of
290 Mpa.
•By Yield Strength:
•Fe E 220’ means a plain carbon steel having yield strength of 220 N/mm
2
.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 40
FeNumber
Value of Minimum Ultimate
Tensile Strength in N/mm
2
or
MPa
Designation of ‘Minimum
Ultimate Tensile Strength’
Fe ENumber
Value of Minimum Yield
Strength in N/mm
2
or MPa
Designation of ‘Minimum
Yield Strength’

Steels Designated on the Basis of Chemical Composition
•According to Indian standard, IS : 1570 (Part II/Sec I)-1979 (Reaffirmed 1991), the
carbon steels are designated in the following order :
•(a) Number indicating 100 times the average percentage of carbon content,
•(b) Letter ‘C’, and
•(c) Number indicating 10 times the average percentage of manganese content. The
figure after multiplying shall be rounded off to the nearest integer.
•For example 20C8means a plain carbon steel containing 0.15 to 0.25 per cent (0.2
per cent on average) carbon and 0.60 to 0.90 per cent (0.75 per cent rounded off to 0.8
per cent on an average) manganese.
•55C4:Plain Carbon Steel with average 0.55% (55/100) Carbon and average 0.4%
(4/10) Manganese.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 41
C Number
10 times average % of
‘Manganese’
Designation of ‘Plain
Carbon steel’
Number
100 times average %
of ‘Carbon’

Designation of Free Cutting Steel
•Example:
•25C12S14: Free Cutting Steel with average 0.25% (25/100) Carbon, average 1.2%
(12/10) Manganese and average 0.14% (14/100) Sulphur.
•20C12Pb15: Free Cutting Steel with average 0.20% Carbon, average 1.2%
Manganese and average 0.15% Lead.
•Cast Steel:
•Example:
•CS840: High tensile cast steel with an Ultimate tensile strength of 840 MPa
•CS1030:High tensile cast steel with an Ultimate tensile strength of 1030 MPa
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 42
CNumber
10 times average % of
‘Manganese’
Designation of ‘Plain
Carbon steel’
Number
100 times average %
of ‘Carbon’
S or PbNumber
100 times average %
of ‘S’ or ‘Pb’
Symbol of Element as
‘Sulphur (S)’ or ‘Lead (Pb)’
CsNumber
High Tensile Strength
in MPa
Designation of ‘High
Tensile Cast steel’

Alloy Steels
•Alloysteelisdefinedascarbonsteeltowhichoneormorealloyingelementsare
addedtoobtaincertainbeneficialeffects.Thecommonlyaddedelementsinclude
silicon,manganese,nickel,chromium,molybdenumandtungsten..
•Silicon:Siliconispresentinalmostallsteels.Itincreasesstrengthandhardness
withoutloweringtheductility.Siliconispurposelyaddedinspringsteeltoincreaseits
toughness.Siliconsteelscontainingfrom1to2%siliconand0.1to0.4%carbonand
otheralloyingelementsareusedforelectricalmachinery,valvesinI.C.engines,
springsandcorrosionresistingmaterials.
•Manganese:Whenmanganeseexceeds1percent,itisregardedasanalloying
element.Itincreaseshardnessandstrength.Italsoincreasesthedepthofhardening.
Themanganesealloysteelscontainingover1.5%manganesewithacarbonrangeof
0.40to0.55%areusedextensivelyingears,axles,shaftsandotherpartswherehigh
strengthcombinedwithfairductilityisrequired.Theprincipalusesofmanganese
steelisinmachinerypartssubjectedtoseverewear.
•Nickel:Nickelincreasesstrength,hardnessandtoughnesswithoutsacrificingductility.
Itincreaseshardenabilityofsteelandimpactresistanceatlowtemperature.Themain
effectofnickelistoincreasetoughnessbylimitinggraingrowthduringtheheat
treatmentprocess.Nickelsteelhasusedinthemanufactureofboilertubes,valvesfor
usewithsuperheatedsteam,valvesforI.C.enginesandsparkplugsforpetrolengines.
Anickelsteelalloycontaining36%ofnickelisknownasinvar.Ithasnearlyzero
coefficientofexpansion.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 43

Alloy Steels
•Chromium:Itisusedinsteelsasanalloyingelementtocombinehardness
withhighstrengthandhighelasticlimit.Italsoimpartscorrosion-resisting
propertiestosteel.Chromenickelsteelisextensivelyusedformotorcar
crankshafts,axlesandgearsrequiringgreatstrengthandhardness.
•Molybdenum:Molybdenumincreaseshardnessandwearresistance.Itresists
softeningofsteelduringtemperingandheating.Thesesteelspossessextra
tensilestrengthandareusedforair-planefuselageandautomobileparts.
•Vanadium:Vanadiumgivesawidehardeningrangetosteel,andthealloycan
behardenedfromahighertemperature.Thesesteelsarefrequentlyusedfor
partssuchassprings,shafts,gears,pinsandmanydropforgedparts.
•Tungsten:Itprohibitsgraingrowth,increasesthedepthofhardeningof
quenchedsteelandconfersthepropertyofremaininghardevenwhenheated
toredcolour.Tungstenandmolybdenumhavesimilareffects.Theprincipal
usesoftungstensteelsareforcuttingtools,dies,valves,tapsandpermanent
magnets.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 44

Designation of alloy steel
•Low Alloy Steels: (% of Element < 10%)
•Example:
•High Alloy Steel: (% of Element > 10%)
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 45
•25Cr4Mo2: Low Alloy Steel with 0.25%
(25/100) Carbon, 1%(4/4) Chromium and
0.2% (2/10) Molybdenum.
Example:
X20Cr18Ni2: High Alloy Steel, 0.2% (20/100), 18% Chromium and 2% Nickel.
LetterNumber
Average % of Element
Multiplied by Factor
Symbol of
Alloying Element
Number
100 times average %
of ‘Carbon’
LetterNumber
Average % of Element
Multiplied by Factor
Symbol of
Alloying Element
……
LetterNumber
Average % of Element
Symbol of
Alloying Element
Number
100 times average %
of ‘Carbon’
LetterNumber
Average % of Element
Symbol of
Alloying Element
……X
Designation of
High alloy steel

Cast Iron
•Cast iron is an alloy of iron & carbon, containing more than 2% of carbon
•Typical composition of ordinary cast iron is:
➢Carbon = 3-4%
➢Silicon = 1-3%
➢Manganese = 0.5-1%
➢Sulphur = up to 0.1%
➢Phosphorous = up to 0.1%
➢Iron = Remainder
•Advantages:
✓Available in large quantities,
✓higher compressive strength,
✓components can be given any complex shape without involving costly machining
operations,
✓excellent ability to damp vibrations,
✓more resistance to wear even under the conditions of boundary lubrication,
✓mechanical properties of parts do not change between room temperature and 350
degree centigrade
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 46

Designation of Cast Iron
•Grey Cast Iron
•Example:
•FG 300: Grey C.I. with Ultimate Tensile Strength 300 MPa
•FG 260: Grey C.I. with Ultimate Tensile Strength 260 Mpa.
•Nodular or Spheroidal Cast Iron:
•Example:
•SG300/10:Nodular C.I. with Ultimate Tensile Strength 300 Mpaand Minimum
Elongation 10%
•SG800/2:Nodular C.I. with Ultimate Tensile Strength 800MPa and Minimum
Elongation 2%.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 47
FGNumber
Ultimate Tensile Strength
in N/mm
2
or MPa
Designation of ‘Grey C.I.’
SG Number
Minimum
Elongation in %
Designation of
‘Nodular or
Spheroidal C.I.’
Number /
Ultimate Tensile
Strength in
N/mm
2
or MPa

Casting Materials
•GrayCastIron:themostwidelyusedbecauseithasaverylowcost,iseasilycastin
largequantities,andiseasytomachine.However,itisbrittleandweakintension.
Thegraphite,intheformofthinflakesdarkens,hencethestructurenamegraycast
iron.
•DuctileandNodularCastIron:thesameasmalleablecastiron,becausebothcontain
graphiteintheformofspheroids.Ductileironhasahighmodulusofelasticity
(172GPa)ascomparedwithgraycastiron,anditiselasticinthesensethataportion
ofthestress-straincurveisastraightline.
•CastSteels:whenpartsarecomplexandmustalsohavehighstrengthcasting
processesareusedforsteels.
•WhiteCastIron:whitecastironhasallintheformofcementiteandpearlite.White
castironisverybrittleandhardtomachinebutalsoveryresistanttowear.
•MalleableCastIron:Annealingofwhitecastironwithinacertaincompositionrange
formsmalleablecastiron.Thegraphiteinmalleablecastironhasanodularformand
isknownastempercarbon.Agoodgradeofmalleablecastironmayhaveatensile
strengthofover350MPa,withanelongationofasmuch18percent.
•AlloyCastIrons:Nickel,chromium,andmolybdenumarethemostcommonalloying
elementsusedincastiron.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 48

Aluminum
•Aluminum and its alloys have good strength-weight ratio, good resistance to
corrosion, and high thermal and electrical conductivity.
•Aluminum can be processed by sand casting, die casting, hot or cold working, or
extruding. Its alloys can be machined press-worked, soldered, brazed, or welded.
•The corrosion resistance of the aluminum alloys depends upon the formation of a thin
oxide coating.
•The most useful alloying elements for aluminum are copper, silicon, manganese,
magnesium, and zinc.
•Aluminiumalloys are designated by a particular numbering system.
•Castaluminiumalloysarespecifiedbya‘fourdigit’systemwhilewroughtalloysbya
‘five-digit’system.
•FirstdigitItidentifiesthemajoralloyingelement.
•SeconddigitItidentifiestheaveragepercentageofthemajoralloyingelement,halved
androundedoff.
•Third,fourthandfifthdigitsTheyidentifytheminoralloyingelementsinorderof
theirdecreasingpercentage.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 49
Aluminium–1Copper —2 Manganese —3 Silicon —4
Magnesium —5Magnesium silicide —6Zinc —7 Other elements —8

Nonferrous Metals
•Magnesium:thelightestofallcommercialmetals,itsgreatestuseisintheaircraft
andautomotiveindustries.Magnesiumalloysfindtheirgreatestuseinapplications
whenstrengthisnotanimportantconsideration.
•Titanium:Titaniumanditsalloysaresimilarinstrengthtomoderate-strengthsteel
butweighhalfasmuchassteel.Thematerialexhibitsverygoodresistanceto
corrosion,haslowthermalconductivity,isnonmagnetic,andhashigh-temperature
strength.
•Copper-baseAlloys:
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 50
✓Brass : alloyed with Zinc.
➢5%-15% : easy to cold work
➢20%-36% : cheaper alloy, better
machinability and slightly
greater strength; however, poor
corrosion resistance.
➢36%-40% : less ductile, and
cannot be cold-worked as
severely.
✓Bronze :
➢Silicon bronze
➢Phosphor bronze
➢Aluminum
➢Beryllium bronze

Plastics
•The plastics are synthetic materials which are mouldedinto shape under pressure with
or without the application of heat. These can also be cast, rolled, extruded, laminated
and machined. Following are the two types of plastics :
•(a) Thermosetting plastics, and (b) Thermoplastic.
•Thethermosettingplasticsarethosewhichareformedintoshapeunderheatand
pressureandresultsinapermanentlyhardproduct.Theheatfirstsoftensthematerial,
butasadditionalheatandpressureisapplied,itbecomeshardbyachemicalchange.
Example-phenolformaldehyde(Bakelite),phenol-furfural(Durite),urea
formaldehyde(Plaskon),etc.
•Thethermoplasticmaterialsdonotbecomehardwiththeapplicationofheatand
pressureandnochemicalchangeoccurs.Theyremainsoftatelevatedtemperatures
untiltheyarehardenedbycooling.Thesecanberemeltedrepeatedlybysuccessive
applicationofheat.Someofthecommonthermoplasticsarecellulosenitrate
(Celluloid),polythene,polyvinylacetate,polyvinylchloride(P.V.C.),etc.
•The plastics are extremely resistant to corrosion and have a high dimensional stability.
They are also used for making safety glasses, laminated gears, pulleys, self-lubricating
bearing, etc. due to their resilience and strength.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 51

Composite Materials
•Composite materials are formed from two or more dissimilar materials, each of which
contributes to the final properties.
•Most engineering composites consist of two materials : a reinforcement called a filler
and a matrix. The filler provides stiffness and strength; the matrix holds the material
together and serves to transfer load among the discontinuous reinforcements.
•Structures of composite materials are normally constructuredof multiple plies
(laminates) where each ply is oriented to achieve optimal structural stiffness and
strength performance.
•High strength-to-weight ratios, and high stiffness-to-weight ratios.
•The directionality of properties of composite materials increases the complexity of
structural analyses.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 52

Material Selection Process
•The selection of a material for a machine part or structural member is one of the most
important decisions the designer is called on to make.
•The selection process can be as physical, economical, and processing parameters.
•One of the basic technique is to list all important material properties associated with
the design, e.g., strength, stiffness, and cost.
•Next, for each property, list all available materials and rank them in order beginning
with the best material.
•Once the lists are formed, select a manageable amount of materials from the top of
each list.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 53

MANUFACTURING TOLERANCES
•Tolerance is defined as the permissible deviation from the basic dimension.
•The tolerance may be unilateral or bilateral. When all the tolerance is allowed on one
side of the nominal size then it is said to be unilateral system of tolerance.
•When the tolerance is allowed on both sides of the nominal size, then it is said to be
bilateral system of tolerance
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 54
•It is impracticable to get a machine part of an accurate basic size as
it increases the cost without increasing the utility of part. A large
tolerance decreases the cost
•The components are so manufactured that their dimensions lie
between two limits—maximum and minimum. The two limits are
sometimes called the upper and lower deviations.

TYPES OF FITS
•When two parts are to be assembled, the relationship resulting from the difference
between their sizes before assembly is called a fit. Depending upon the limits of the
shaft and the hole, fits are broadly classified into three groups.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 55
Types of Fits: (a) Clearance Fit (b)
Transition Fit (c) Interference Fit
•Clearancefitisafitwhichalwaysprovidesa
positiveclearancebetweentheholeandthe
shaftovertheentirerangeoftolerances.
•Interferencefitisafitwhichalwaysprovides
apositiveinterferenceoverthewholerange
oftolerances.
•Transitionfitisafitwhichmayprovide
eitheraclearanceorinterference,depending
upontheactualvaluesoftheindividual
tolerancesofthematingcomponents..
Thefollowingaretwobasesoflimitsystem:
1.Holebasissystem.Whentheholeiskeptasaconstant
memberanddifferentfitsareobtainedbyvaryingthe
shaftsizethenthelimitsystemissaidtobeonahole
basis.
2.Shaftbasissystem.Whentheshaftiskeptasaconstant
memberanddifferentfitsareobtainedbyvaryingthehole
size,thenthelimitsystemissaidtobeonashaftbasis.

Indian Standard System of Limits and Fits
•According to a system recommended by the Bureau of Indian Standards, tolerance is
specified by an alphabet, capital or small, followed by a number, e.g., H7 or g6.
•The description of tolerance consists of two parts—fundamental deviation and
magnitude of tolerance,
•The magnitude of tolerance is designated by a number, called the grade. The grade of
tolerance is defined as a group of tolerances, which are considered to have the same
level of accuracy for all basic sizes. There are eighteen grades of tolerances with
designations IT1, IT2, …, IT17, and IT18.
•The standard tolerances for grades IT 5 to IT 16 are determined in terms of standard
tolerance unit (i) in microns, where
•where D is the size or geometric mean diameter in mm. Diameter steps in I.S.I are 1-
3, 3-6, 6-10, 10-18, 18-30, 30-50, 50-80, 80-120, 120-180, 180-250, 250-315, 315-
400 and 400-500 mm
•IT6 –10i; IT7 –16i; IT8 –25i; IT9 –40i; IT10 –64i; IT11 –100i; IT12 –160i; IT13
–250i; IT14 –400i; IT15 –640i; IT16 –1000i.
•Fundamental deviation: It is the minimum difference between the size of a component
and its basic size. This is identical to the upper deviation for shafts and lower
deviation for holes.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 56

Numerical Problem 4
•Calculatethetolerances,fundamentaldeviationsandlimitsofsizesfortheshaft
designatedas40H8/f7.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 57

Numerical Problem 5
•journal of nominal or basic size of 75 mm runs in a bearing with close running fit (H
8/g7,). Find the limits of shaft and bearing. What is the maximum and minimum
clearance?
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 58

Stress
•When some external system of forces or loads act on a body, the internal forces (equal
and opposite) are set up at various sections of the body, which resist the external
forces. This internal force per unit area at any section of the body is known as unit
stress or simply a stress.
•It is denoted by a Greek letter sigma (σ). Mathematically,
•Stress, σ= P/A
•where P = Force or load acting on a body, and
•A = Cross-sectional area of the body.
•Tensile stress
•The stresses are called tensile when the fibers of the component tend to elongate due
to the external force. It is given by σ
t= P/A
•Compressive stress
•when the fibers tend to shorten due to the external force, the stresses are called
compressive stresses.. It is given by σ
c= P/A
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 59
A prismatic bar subjected to compressive loading.

Strain
•When a system of forces or loads act on a body, it undergoes some deformation. This
deformation per unit length is known as unit strain or simply a strain. It is denoted by
a Greek letter epsilon (ε). Mathematically,
•Strain, ε = δl/ l or δl= ε.l
•where δl= Change in length of the body, and
•l = Original length of the body.
•Young's Modulus or Modulus of Elasticity
•Hooke's law* states that when a material is loaded within elastic limit, the stress is
directly proportional to strain, i.e. ??????∝&#3627409152;or??????=??????.&#3627409152;
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 60
where E is a constant of proportionality
known as Young's modulus or modulus of
elasticity. It may be noted that Hooke's law
holds good for tension as well as
compression.
??????=
??????
&#3627409152;
=
??????×&#3627408473;
??????×&#3627409151;&#3627408473;

Linear and Lateral Strain
•Consider a circular bar of diameter d and length ‘l’, subjected to a tensile force P as
shown in Fig.
•A little consideration will show that due to tensile force, the length of the bar increases
by an amount δland the diameter decreases by an amount δd.
•It is thus obvious, that every direct stress is accompanied by a strain in its own
direction which is known as linear strain and an opposite kind of strain in every
direction, at right angles to it, is known as lateral strain.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 61
The ratio of lateral strain to linear strain is
a constant known as the Poisson's ratio,

SHEAR STRESS AND SHEAR STRAIN
•When the external force acting on a component tends to slide the adjacent planes with
respect to each other, the resulting stresses on these planes are called direct shear
stresses.
•τ=Tangential force/Resisting area = P/A
•where,
•τ= shear stress (N/mm
2
or MPa)
•A = cross-sectional area of the rivet (mm
2
)
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 62
•Two plates held together by means of a
rivet are shown in Figure. The average
shear stress in the rivet is given by
A plane rectangular element, cut from the component and
subjected to shear force, is shown in Figure. Shear stresses
cause a distortion in the original right angles. The shear
strain (γ) is defined as the change in the right angle of a
shear element. Within the elastic limit, the stress–strain
relationship is given by τ= Gγ
where, γ= shear strain (radians)
G is shear modulus or modulus of rigidity (in MPa).

Modulus of Rigidity and Bulk modulus
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 63
•Modulus of Rigidity :The ratio of shear stress to the corresponding shear strain
within the elastic limit is known as Modulus of Rigidity or shear Modulus. This is
denoted by C or G or N.
•Bulk Modulus: When a material is subjected to three direct stresses along the three
mutually perpendicular directions with equal intensity, then the ratio of the direct
stress to the volumetric strain is denoted as the bulk modulus, K.
•The ratio of the change in volume to the original volume is known as volumetric
strain. Mathematically, volumetric strain,&#3627409152;
&#3627408483;=
????????????
??????
•Where δV=Change in volume, and V = Original volume
•The bulk modulus (K) and Young's modulus (E) are related by the following relation,
•The Young's modulus (E) and modulus of rigidity (G) are related by the following
relation,

Numerical Problem 6
•Q. A cast iron link, as shown in Fig., is required to transmit a steady tensile load of 45
kN.Find the tensile stress induced in the link material at sections A-A and B-B.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 64

Numerical Problem 7
•Q. A hydraulic press exerts a total load of 3.5 MN. This load is carried by two steel
rods, supporting the upper head of the press. If the safe stress is 85 MPa and E = 210
kN/mm
2
, find : 1. diameter of the rods, and 2. extension in each rod in a length of 2.5
m.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 65

Numerical Problem 8
•Q.Thepistonrodofasteamengineis50mmindiameterand600mmlong.The
diameterofthepistonis400mmandthemaximumsteampressureis0.9N/mm
2
.
FindthecompressionofthepistonrodiftheYoung'smodulusforthematerialofthe
pistonrodis210kN/mm
2
.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 66

Stress strain Diagram
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 67
Point O to A
Point C to D
Point D to E
At point E
Normal or engineering stresscan be determined by dividing the
applied load by the specimen original cross sectional area.
True stress is calculated using the actual cross sectional area at
the instant the load is measured.

•(a) Limit of Proportionality (A):It is the limiting value of the stress up to which
stress is proportional to strain.
•(b) Elastic Limit: This is the limiting value of stress up to which if the material is
stressed and then released (unloaded) strain disappears completely and the original
length is regained. This point is slightly beyond the limit of proportionality.
•(c) Upper Yield Point (B):This is the stress at which, the load starts reducing and the
extension increases. This phenomenon is called yielding of material. At this stage
strain is about 0.125 per cent and stress is about 250 N/mm
2
.
•(d) Lower Yield Point (C): At this stage the stress remains same but strain increases
for some time.
•(e) Ultimate Stress (E):This is the maximum stress the material can resist. This stress
is about 370–400 N/mm
2
. At this stage cross-sectional area at a particular section
starts reducing very fast. This is called neck formation. After this stage load resisted
and hence the stress developed starts reducing.
•(f) Breaking Point (F): The stress at which finally the specimen fails is called
breaking point. At this strain is 20 to 25 per cent.
•The ratio of stress to strain in this linear region of stress-strain diagram is called
Young Modulus or the Modulus of Elasticity given
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 68

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 69
•Yield strength
•The yield strength of a material is closely associated with the
yield point.
•Yield strength is defined as the lowest stress at which extension
of the test piece increases without further increase in load.
•Ultimate strength
•Ultimate strength is the highest point on the stress strain curve.
At this highest point concentrated plastic deformation takes
place, resulting in the formation of neck or waist in the
specimen, resulting in decrease in the load.
•Rupture strength
•The rupture strength is the stress corresponding to the failure
point F of the stress strain curve. For structural steel it is some
what lower than the ultimate strength. This is because the
rupture strength is computed by dividing the rupture load by
the original cross sectional area.

Bearing Stress
•A localisedcompressive stress at the surface of contact between two members of a
machine part, that are relatively at rest is known as bearing stress or crushing stress.
•The bearing stress is taken into account in the design of riveted joints, cotter joints,
knuckle joints, etc.
•Let us consider a riveted joint subjected to a load P as shown in Fig
•σb(or σc)=P/d t n
•where d = Diameter of the rivet,
• t = Thickness of the plate,
• d.t = Projected area of the rivet, and
• n = Number of rivets per pitch length in bearing or crushing
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 70

STRESSES DUE TO BENDING MOMENT
•A straight beam subjected to a bending moment M
bis shown in Fig
•The bending stress at any fibreis given by,
• ??????
&#3627408463;=
&#3627408448;
&#3627408463;&#3627408486;
&#3627408444;
•where,
•σ
b= bending stress at a distance of y from the neutral axis (N/mm
2
or MPa)
•M
b= applied bending moment (N-mm)
•I = moment of inertia of the cross-section about the neutral axis (mm
4
)
•The bending stress is maximum in a fiber, which is farthest from the neutral axis.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 71

STRESSES DUE TO TORSIONAL MOMENT
•A transmission shaft, subjected to an external torque, is shown in Fig
•The internal stresses, which are induced to resist the action of twist, are called
torsional shear stresses.
•. The torsional shear stress is given by
• ??????=
&#3627408448;&#3627408481;&#3627408479;
&#3627408445;
•The angle of twist is given by
• ??????=
&#3627408448;&#3627408481;&#3627408473;
&#3627408445;??????
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 72
where,
τ= torsional shear stress at the fi bre(N/mm2
or MPa)
M
t= applied torque (N-mm)
r = radial distance of the fi brefrom the axis of
rotation (mm)
J = polar moment of inertia of the cross-section
about the axis of rotation (mm
4
)
where,
θ= angle of twist (radians)
l = length of the shaft (mm)

ECCENTRIC AXIAL LOADING
•A typical example of such an eccentric loading is shown in Fig.(a)
•According to the principle of statics, the eccentric force P can be replaced by a
parallel force P passing through the centroid axis along with a couple (P x e) as shown
in Figs(b)
•In Fig. (c), the couple causes bending stress of magnitude (P
ey/I). The resultant
stresses at the cross-section are obtained by the principle of superimposition of
stresses. They are given by,
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 73

Numerical Problem 9
•Amachinememberhavingdiameter50mmandlength250mmsupportedatoneend
asacantileversubjectedtodifferentforceasshowninfigure.Findthemaximum
stressatpointAandB.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 74

Numerical Problem 10
•Ajournal25mmindiametersupportedinslidingbearingshasamaximumend
reactionof2500N.Assuminganallowablebearingpressureof5N/mm
2
,findthe
lengthoftheslidingbearing.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 75

Bending Stress in Curved Beams
•A curved beam is defined as a beam in which the neutral axis in unloaded condition is
curved instead of straight.
•The application of curved beam principle is used in crane hooks, chain links and
frames of punches, presses, planers etc.
•In straight beams, the neutral axis of the section coincides with its centroidal axis and
the stress distribution in the beam is linear.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 76
•Butincaseofcurvedbeams,theneutral
axisofthecross-sectionisshifted
towardsthecentreofcurvatureofthe
beamcausinganon-linear(hyperbolic)
distributionofstress.
The eccentricity e between centroidal and neutral
axes is given by, e = R –R
N
The bending stress (σ
b)at a fibre, which is at a
distance of y from the neutral axis is given by,
The maximum stress occurs either at the inner fibre
or at the outer fibre.

Values of Rn and R for various commonly used cross-section
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 77

Numerical Problem 11
•The crane hook carries a load of 20 kNas shown in Fig. The section at X-X is
rectangular whose horizontal side is 100 mm. Find the stresses in the inner and outer
fibresat the given section.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 78

Numerical Problem 12
•The C-frame of a 100 kNcapacity press is shown in Fig. The material of the frame is
grey cast iron FG 200 and the factor of safety is 3. Determine the dimensions of the
frame.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 79

Principal Stress:
•The plane, on which, only normal stresses acts, but no shear stress, is called as
Principal Plane. The magnitude of normal stress acting on principal plane is called as
Principal Stress.
•The maximum value of stress acting on the principal plane is called as Maximum
Principal Stress. The minimum value of normal stress acting on principal plane is
called as Minimum Principal Stress.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 80
Theplanesofmaximum
shearstressareatright
anglestoeachotherandare
inclinedat45°toprincipal
plane.Themaximumshear
stressisgivenbyhalfofthe
algebraicdifferencebetween
theprincipalstresses.
Maximum and Minimum Principal Stress: Maximum Shear Stress:

Factor of Safety
•If some member of a structure be exposed to a stress somewhat less than the ultimate
stress of the material of which the member is composed, it may be supposed that the
member is quite safe to carry the load imposed upon it
•Failure could not occur if the actual stress never reached the ultimate stress.
•In actual practice, however, great care is always taken to ensure that the working
stress shall never be nearly so closet to the ultimatestress.
•In this way a certain margin of safety in working is ensured.
•The factor of safety is defined as
??????
&#3627408480;=
failurestress
allowablestress
or ??????
&#3627408480;=
failureload
allowableload
•The allowable stress is the stress value, which is used in design to determine the
dimensions of the component. It is considered as a stress, which the designer expects
will not be exceeded under normal operating conditions.
•For ductile materials, the allowable stress σis obtained by ??????=
??????
&#3627408486;&#3627408481;
&#3627408467;
&#3627408480;
•For brittle materials, the allowable stress σis obtained by ??????=
??????
&#3627408482;&#3627408481;
&#3627408467;
&#3627408480;
•where S
ytand S
utare the yield strength and the ultimate tensile strength of the material
respectively.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 81

Factors on which Factor of Safety depends:
•(i) Effect of Failure:The factor of safety is high in applications where failure of a
machine part may result in serious accidents.
•(ii) Type of Load: The factor of safety is low for static force while high for impact
load.
•(iii) Degree of Accuracy in Force Analysis: low factor of safety for precisely
determined force where as high FoSfor uncertain and unpredictable force.
•(iv) Material of Component: The factor of safety is small for homogeneous ductile
material and high for nonhomogeneous brittle material.
•(v) Reliability of Component: The factor of safety increases with increasing
reliability.
•(vi) Cost of Component: The factor of safety is low for cheap machine parts.
•(vii) Testing of Machine Element: A higher factor of safety is necessary, when it is
not possible to test the machine part.
•(viii) Service Conditions: When the machine element is likely to operate in corrosive
atmosphere or high temperature environment, a higher factor of safety is necessary.
•(ix) Quality of Manufacture: a higher factor of safety is required to compensate for
poor manufacturing quality.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 82

Selection of Factor of Safety
Before selecting a proper factor of safety, a design engineer should consider the
following points :
1.The reliability of the properties of the material and change of these properties
during service ;
2.The reliability of test results and accuracy of application of these results to actual
machine parts ;
3.The reliability of applied load ;
4.The certainty as to exact mode of failure ;
5.The extent of simplifying assumptions ;
6.The extent of localisedstresses ;
7.The extent of initial stresses set up during manufacture ;
8.The extent of loss of life if failure occurs ; and
9.The extent of loss of property if failure occurs.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 83

Guidelines for values of factor of safety
•(i) For cast iron components: a large factor of safety, usually 3 to 5, based on ultimate
tensile strength, is used in the design of cast iron components.
•(ii) For components made of ductile materials: The recommended factor of safety is
1.5 to 2, based on the yield strength of the material.
•(iii) For components made of ductile materials and those subjected to external
fluctuating forces, endurance limit is considered to be the criterion of failure. The
recommended factor of safety based on this endurance limit of component is usually
1.3 to 1.5.
•(iv) The design of certain components such as cams and followers, gears, rolling
contact bearings or rail and wheel is based on the calculation of contact stresses by the
Hertz’ theory. The recommended factor of safety for such components is 1.8 to 2.5
based on surface endurance limit.
•(v) Certain components, such as piston rods, power screws or studs, are designed on
the basis of buckling consideration. The recommended factor of safety is 3 to 6 based
on the critical buckling load of such components.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 84

Condition for higher factor of safety
•(i)Themagnitudeandnatureofexternalforcesactingonthemachinecomponent
cannotbepreciselyestimated.
•(ii)Itislikelythatthematerialofthemachinecomponenthasanon-homogeneous
structure.
•(iii)Themachinecomponentissubjectedtoimpactforceinservice.
•(iv)Thereispossibilityofresidualstressesinthemachinecomponent.
•(v)Themachinecomponentisworkinginacorrosiveatmosphere.
•(vi)Themachinepartissubjectedtohightemperaturesduringoperation.
•(vii)Thefailureofthemachinepartmayhazardthelivesofpeople(hoist,lifting
machineryandboilers)andsubstantiallosstoproperty.
•(viii)Itisnotpossibletotestthemachinecomponentunderactualconditionsof
serviceandthereisvariationinactualconditionsandstandardtestconditions.
•(ix)Higherreliabilityisdemandedinapplicationslikecomponentsofaircrafts.
•(x)Thereispossibilityofabnormalvariationinexternalloadonsomeoccasions.
•(xi)Thequalityofmanufactureofthemachinepartispoor.
•(xii)Theexactmodeoffailureofthecomponentisunpredictable.
•(xiii)Thereisstressconcentrationinamachinecomponent.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 85

Numerical Problem 13
•A rectangular bar 40mm x 20 mm is subjected to a steady tensile load of 200 kN.Find
the factor of safety, if yield strength of the material is 500 MPa.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 86

DESIGN OF SIMPLE MACHINE PARTS
•The following points should be remembered in design of simple machine parts
•(i) The dimensions of simple machine parts are determined on the basis of pure tensile
stress, pure compressive stress, direct shear stress, bending stress or torsional shear
stress.
•(ii) A higher factor of safety of up to 5 is taken to account the ignorance of stress
concentration, and reversal of stresses etc.
•(iii) The allowable stress is to be obtained from ultimate tensile strength and yield
strength.
•(iv) yield strength in shear is 50% of the yield strength in tension. Therefore,
Ssy= 0.5 Syt.
and the permissible shear stress (τ) is given by
??????=
??????
&#3627408480;&#3627408486;
&#3627408467;
&#3627408480;
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 87

Numerical Problem 14
•Twoplates,subjectedtoatensileforceof50kN,arefixedtogetherbymeansofthree
rivetsasshowninFig(a).Theplatesandrivetsaremadeofplaincarbonsteel10C4
withatensileyieldstrengthof250N/mm2.Theyieldstrengthinshearis50%ofthe
tensileyieldstrength,andthefactorofsafetyis2.5.Neglectingstressconcentration,
determine(i)thediameteroftherivets;and(ii)thethicknessoftheplates.
•Permissible shear stress for rivets ??????=
??????
&#3627408480;&#3627408486;
&#3627408467;
&#3627408480;
50
•Diameter of rivets No of rivet x c/s area x shear stress= Force applied 20.6 22
•Permissible tensile stress for plates ??????=
??????
&#3627408486;&#3627408481;
&#3627408467;
&#3627408480;
100
•Thickness of plates stress x area = force 3.73 4
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 88

Numerical Problem 15
•Figure shows a crank loaded by a force F = 1.3 kNthat causes twisting and bending of
a 20-mm-diameter shaft fixed to a support at the origin of the reference system. In
actuality, the support may be an inertia that we wish to rotate, but for the purposes of a
stress analysis we can consider this a statics problem.
•(a) Draw separate free-body diagrams of the shaft AB and the arm BC, and compute
the values of all forces, moments, and torques that act. Label the directions of the
coordinate axes on these diagrams.
•(b) Compute the maxima of the torsional stress and the bending stress in the arm BC
and indicate where these act.
•(c) Locate a stress element on the top surface of the shaft at A, and calculate all the
stress components that act upon this element.
•(d) Determine the maximum normal and shear stresses at A.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 89

COTTER JOINT
•A cotter joint is used to connect two co-axial rods, which are subjected to either axial
tensile force or axial compressive force.
•It is also used to connect a rod on one side with some machine part like a crosshead or
base plate on the other side.
•It is not used for connecting shafts that rotate and transmit torque.
•This is a temporary method of joining the rods.
•Cotter joints can be quickly assembled as well as disassembled
•Examples:
•Piston rod and crosshead of a steam engine
•Piston rod and pump rod
•Foundation bolt
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 90

TYPES OF COTTER JOINTS
•1.Socket and spigot cotter joint.
•2.Sleeve and cotter joints.
•3.Gib and cotter joints.
•In a socket and spigot cotter joints, one end of the rods is provided with a socket type
of end and the other end of the rod is inserted into a socket the end of the rod which
goes in to a socket is also called spigot .a rectangular hole is made in the socket and
spigot.
•Sleeve and cotter joint is a type of joint used to typically connect two similar coaxial
cylindrical rods it contains a sleeve and two wedge shaped tapered cotters appropriate
slots are cut in the sleeve and in the cylindrical rods the cotters are assembled in
•A gibis a piece of mild steel having the same thickness and taper as the cotter.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 91

Socket and spigot cotter joint.
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 92

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 93

Cotter joint –modes of failure
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 94
Spigot breaking in
Tension outside the Joint
Spigot breaking in
Tension across Slot
Socket Breaking in
Tension across Slot
Shearing away
of Socket
Shearing away of the
Collar in the spigot
Double shearing
of Socket end
Double shearing of Cotter pin

Design of Socket and Spigot Cotter Joint
•1. Failure of the rods(Spigot) in tension
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 95

2. Tensile failure of spigot
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 96
•

3. Tensile failure of socket
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 97

4. Shear failure of cotter
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 98

5. Shear Failure of spigot End
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 99

6. Shear Failure of socket End
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 100

7. Crushing Failure of spigot End
•8. Crushing Failure of Socket End
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 101
•8. Crushing Failure of Socket End

9. Failure of cotter in bending
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 102

•When all the parts of the joint are made of steel, the following proportions in terms of
diameter of the rod (d) are generally adopted :
•d1 = 1.75 d , d2 = 1.21 d , d3 = 1.5 d , d4 = 2.4 d , a = c = 0.75 d , b = 1.3 d, l = 4 d , t =
0.31 d ,
•t1 = 0.45 d , e = 1.2 d.
•Taper of cotter = 1 in 25, and draw of cotter = 2 to 3 mm.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 103

DESIGN PROCEDURE FOR COTTER JOINT
•The basic procedure to calculate the dimensions of the cotter joint consists of the
following steps:
•(i) Calculate the diameter of each rod by Eq.
•(ii) Calculate the thickness of the cotter by the empirical relationship t = 0.31 d.
•(iii) Calculate the diameter d2 of the spigot by Eq.
•(iv) Calculate the outside diameter d1 of the socket on the basis of tensile stress in the
socket,
•The diameter of the spigot collar d3 and the diameter of the socket collar d4 are
calculated by the following empirical relationships, d3 = 1.5 d and d4 = 2.4 d
•(vi) The dimensions a and c are calculated by the following empirical relationship,
a = c = 0.75 d
•(vii) Calculate the width b of the cotter
•(viii) Check the crushing and shear stresses in the spigot end
•(x) Calculate the thickness t1 of the spigot collar by the following empirical
relationship, t1 = 0.45 d The taper of the cotter is 1 in 32.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 104

Design a cotter joint
•Design and draw a cotter joint to support a load varying from 30 kNin compression to
30 kNin tension. The material used is carbon steel for which the following allowable
stresses may be used. The load is applied statically. Tensile stress = compressive stress
= 50 MPa; shear stress = 35 MPaand crushing stress = 90 MPa.
•Then we have to find
•d, d
2, t, d
1, b, d
4, c, a, d
3, t
1, l, e
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 105

Numerical Problem 16
•Design and draw a cotter joint to support a load varying from 50 kNin compression to
50 kNin tension. The material used is carbon steel 30C8. Design the joint and specify
its main dimensions.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 106

Knuckle Joint
•Knuckle Joint is used to connect two rods coincideor intersectand lie in one plane.
•The knuckle joint is used to transmit axial tensile force. The construction of this joint
permits limited angular movement between rods, about the axis of the pin.
•Typical applications of knuckle joints are as follows:
•(i) Joints between the tie bars in roof trusses.
•(ii) Joints between the links of a suspension bridge.
•(iii) Joints in valve mechanism of a reciprocating engine.
•(iv) Fulcrum for the levers.
•(v) Joints between the links of a bicycle chain.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 107

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 108

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 109

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 110

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 111

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 112

Tensile Failure of Rods
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 113

Shear failure of pin
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 114

Crushing failure of pin eye
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 115

Crushing failure of pin in fork
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 116

Bending Failure of pin
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 117

Tensile failure of Eye
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 118

Shear failure of eye
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 119

Tensile failure of fork
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 120

Shear failure of fork
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 121

DESIGN PROCEDURE FOR KNUCKLE JOINT
•The basic procedure to determine the dimensions of the knuckle joint consists of the
following steps:
•(i) Calculate the diameter of each rod by
•(ii) Calculate the enlarged diameter of each rod by empirical relationship D
1= 1.1 D
•(iii) Calculate the dimensions a and b by empirical relationship a = 0.75 D b = 1.25 D.
•(iv) Calculate the diameters of the pin by shear
•(v) Calculate the dimensions do and d1 by relationships using do = 2d d1 = 1.5d.
•(vi) Check the tensile, crushing and shear stresses in the eye
•(vii) Check the tensile, crushing and shear stresses in the fork by
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 122

Numerical Problem 17
•Itisrequiredtodesignaknucklejointtoconnecttwocircularrodssubjectedtoan
axialtensileforceof50kN.Therodsareco-axialandasmallamountofangular
movementbetweentheiraxesispermissible.Designthejointandspecifythe
dimensionsofitscomponents.Selectsuitablematerialsfortheparts.
•Takematerialplanecarbonsteel30C8(Syt=400N/mm
2
).
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 123

Failure of machine element
•Amaterialisconsideredfailedwhenapermanentornon-recoverabledeformationoccurs.
•Thereiseitherdirectseparationofparticlesasincaseofbrittlematerialsorslippingof
particlesasforductilematerialswhereplasticdeformationsalsotakeplace.
•Machinecomponentsandstructuralmembersaregenerallydesignedonthehypothesisthat
thematerialwillnotyieldduringtheexpectedloadingconditions.Forexample,ina
uniaxialloading,amachinecomponentwillbesafeaslongasthestressproducedbythe
loadislessthantheyieldstressofthematerial.
•Therearevariousmachinecomponentswhicharesubjectedtotheseveraltypesofloadsor
stressessimultaneously.Forexample,propellershaftofanautomobileissubjectedtothe
torsionalmomentaswellasbendingmomentwhiletransmittingthepower.
•Amachineelementcanbeconsideredtohavefailedwhenitcannolongerperformits
designfunction.
•Whydopartsfail?
•Partsfailwheninducedstressduetovariousloadingconditionexceedthestrengthof
material.
•Whatkindofstressescausethefailure?
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 124

Failure of machine element
•Failure generally occurs due to complete fracture (i.e. brittle failure ) or excessive
deformation, which may or may not result in rupture (ductile failure).
•How do one compare stresses induced to the material properties?
•Generally machine parts are subjected to combined loading and to find material
properties under real loading condition is practically not economical
•Thus, material properties are obtained from simple tension/torsion test
•Under uniaxial loading the stress strain curve can be used to represent the response
until failure.
•Under multiaxialstresses, failure theories are needed for representing the material
behaviour based on plasticity (or yielding) and fracture.
•Theories of failure provide a relationship between the strength of machine component
subjected to complex state of stress with the material properties obtained from simple
test (Tensile)
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 125

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 126
Strength of machine
component subjected
to complex state of
stress
Strength of standard
component subjected
to uniaxial state of
stress

•In general, ductile, isotropic materials are limited by their shear strengths.
•Brittle materials are limited by their tensile strengths.
•If cracks are present in a ductile material, it can suddenly fracture at nominal stress
levels well below its yield strength, even under static loads.
•Static loads are slowly applied and remain constant with time.
•Dynamic loads are suddenly applied (impact), or repeatedly varied with time
(fatigue), or both.
•In dynamic loading, the distinction between failure mechanisms of ductile and brittle
materials blurs.
•Ductile materials often fail like brittle materials in dynamic loading.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 127
Material Strength
S
y= Yield strength in tension, S
yt= S
yc
S
ys= Yield strength in shear
S
u= Ultimate strength in tension, S
ut
S
uc= Ultimate strength in compression
S
us= Ultimate strength in shear = .67S
u

•Events such as distortion, permanent set, cracking, and rupturing are among the ways
that a machine element fails.
•Unfortunately, there is no universal theory of failure for the general case of material
properties and stress state.
•Instead, over the years several hypotheses have been formulated and tested, leading to
today’s accepted practices.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 128
Failure along principal shear stress plane
Failure along principal normal stress plane

Ductile and Brittle Materials
•A ductile material deforms significantly before fracturing. Ductility is measured by %
elongation at the fracture point. Materials with 5% or more elongation are considered
ductile.
•Brittle material yields very little before fracturing, the yield strength is approximately
the same as the ultimate strength in tension. The ultimate strength in compression is
much larger than the ultimate strength in tension.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 129

Theories of Elastic Failure:
•1)Maximum Principal Stress Theory (Rankine’sTheory)
•2)Maximum Shear Stress Theory(Coulomb, Trescaand Guest’sTheory)
•3)Distortion Energy Theory (Huber Von MisesTheory)
•4)Maximum Strain Theory (Saint Venant’sTheory)
•5)Maximum Total Energy Theory (Haigh’sTheory)
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 130

Maximum principal or Normal stress theory
•Credited to the English scientist and educator W. J. M. Rankine(1802–1872)
•The theory states that the failure of a material subjected to bi-axial or tri-axial stresses
will occurs when the maximum principle stresses in the complex system reaches the
value of the maximum stress (yield or ultimate strength) at the elastic limit in simple
tension or the minimum principal stress (i.e., maximum compressive stress) attains the
elastic limit in simple compression.
•Thus, in this theory, maximum or minimum principal stress is the criterion of failure.
•If σ
1, σ
2and σ
3are the three principal stresses at a point on the component and
• σ
1> σ
2> σ
3
•Then failure occurs when ??????
1≥ ??????
??????&#3627408481;??????????????????
1≥ ??????
&#3627408482;&#3627408481;
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 131
Boundary for Maximum Principal
Stress Theory under Bi-axial Stresses
Suppose σ
1> σ
2As per this theory, we will consider only the
maximum of principal stresses (σ
1) and disregard the other
principal stress (σ
2)
Region of Safety
Experimental investigations suggest that the maximum principal stress
theory gives good predictions for brittle materials. However, it is not
recommended for ductile materials.

Numerical Problem 18
•Asolidcircularshaftofdiameterdissubjectedtoapuretorqueof200Nm.
Determinethediameteroftheshaftaccordingtothemaximumprincipalstresstheory,
takingaFOSof2,iftheyieldstrengthofthematerialis280N/mm
2
.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 132

Maximum shear stress theory
•This criterion of failure is accredited to CA Coulomb, H Trescaand JJ Guest.
•The theory states that the failure of a mechanical component subjected to bi-axial or
tri-axial stresses occurs when the maximum shear stress at any point in the component
becomes equal to the maximum shear stress in the standard specimen of the tension
test, when yielding starts.
•The maximum value of shear stress in terms of principal stresses σ
1and σ
2
•In the tension test, the specimen is subjected to uni-axial stress (σ
1) and (σ
2= 0).
•When the specimen starts yielding (σ
1= S
yt), then
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 133
Therefore, the maximum shear stress theory predicts that the
yield strength in shear is half of the yield strength in tension,
i.e., S
sy=0.5 S
yt

Cont..
•Suppose σ
1, σ
2and σ
3 are the three principal stresses at a point on the component, the
shear stresses on three different planes are given by,
•The largest of these stresses is equated to (τ
max) or (S
yt/2). Considering factor of
safety,
•Region of Safety For bi-axial stresses, σ
3= 0
•The above equations can be written as,
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 134
Boundary for Maximum Shear Stress
Theory under Bi-axial Stresses
The maximum shear stress theory of failure is widely
used by designers for predicting the failure of
components, which are made of ductile materials, like
transmission shaft.

Shear Diagonal
•Shear diagonal or line of pure shear is the locus of all points, corresponding to pure
shear stress.
•for pure shear stress,
•σ
1= –σ
2= τ
12
•The above equation can be written as,
•
σ
1
σ
2
=−1=−tan(45°)
•A line GH is constructed in such a way that it passes through the origin O and makes
an angle of –45°with the Y-axis. This line is called shear diagonal or line of pure
shear.
•Region of Safety graph point G is the point of intersection of lines FA and GH.
•σ
1–σ
2= S
ytand
σ
1
σ
2
=−1
•Solving two equations simultaneously,
•Since the point G is on the borderline, this is the limiting value for shear stress.
•??????
&#3627408480;&#3627408486;=
1
2
??????
&#3627408486;&#3627408481;
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 135

Numerical Problem 19
•At a section of a mild steel shaft, the maximum torque is 8437.5Nm and maximum
bending moment is 5062.5 Nm. The diameter of shaft is 90 mm and the stress at the
elastic limit in simple tension for the material of the shaft is 220 N/ mm
2
. Determine
whether the failure of the material will occur or not according to maximum shear
stress theory. If not, then find the factor of safety.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 136

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 137

Numerical Problem 20
•A load P = 44 kNis applied to a crankshaft at a distance of 200 mm from the bearing.
Material for the shaft is 30C4 steel with Sy= 276 MPa. Using factor of safety of 2 and
maximum shear stress theory find the diameter of the shaft.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 138

Maximum Principal Strain Theory (Saint -Venant)
•This theory is due to Saint Venant.
•According to this theory, the failure will occur in a material when the maximum
principal strain reaches the strain due to yield stress in simple tension or when the
minimum principal strain (i.e., maximum compressive strain) reaches the strain due to
yield stress in simple compression.
•Yield stress is the maximum stress at elastic limit.
•Consider a three dimensional stress system.
•Principal strain in the direction of principal stress σ
1is
•Principal strain in the direction of principal stress σ
3is
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 139
This theory is not used, in general, because it only gives reliable results in particular cases.

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 140

Numerical Problem 21
•Theprincipalstressesatapointinanelasticmaterialare200N/mm2(tensile),100
N/mm2(tensile)and50N/mm2(compressive).Ifthestressattheelasticlimitin
simpletensionis200N/mm
2
,determinewhetherthefailureormaterialwilloccur
accordingtomaximumprincipalstraintheory.Ifnot,thendeterminethefactorof
safety.Takepoisson’sratio0.3.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 141

Maximum / Total Strain Energy Theory (Haigh)
•The theory states that the failure or yielding of a material subjected to bi-axial or tri-
axial stresses occurs when the strain energy per unit volume stored in the machine
component exceeds the strain energy stored in a unit volume at the yield point stress
in a simple tensile test specimen.
•Strain energy in a body is equal to work done by the load(P) in straining the material
and it is equal to
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 142
This theory may be used for ductile materials.

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 143

Numerical Problem 22
•According to the theory maximum strain energy theory, determine the diameter of a
bolt which is subjected to an axial pull of 8 kNtogether with a transverse shear force
of 4 kN.Elastic limit in tension is 285 N /mm2, factor of safety 3 and Poisson’s ratio
0.3.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 144

Maximum shear strain energy (Distortion-energy) theory
•It is known as the Huber von Misesand Hencky’stheory.
•Thetheorystatesthatthefailureoryieldingofamaterialsubjectedtobi-axialortri-
axialstressesoccurswhenenergyusedinchangingtheshapeoftheunitvolumeofa
componentisequaltothedistortionenergyperunitvolumeattheyieldstressofa
specimensubjectedtosimpletensionorcompressiontest.
•Inanotherword,thefailureofamaterialoccurswhenthetotalshearstrainenergyper
unitvolumeinthestressedmaterialreachesavalueequaltotheshearstrainenergy
perunitvolumeattheelasticlimitinthesimpletensiontest.
•Totalstrainenergyconsistsoftwoparts,thatis,(a)volumetricstrainenergycausing
thechangeinvolumeand(b)shearingstrainenergyforchangingtheshape.
•Thetotalstrainenergy
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 145

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 146

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 147

•effective stress for the entire general state of stress given by σ1, σ2, and σ3is usually
called the von Mises stress,
•Considering the factor of safety,
•For bi-axial stresses (σ
3= 0),
•A component subjected to pure shear stresses and the corresponding Mohr’s circle
diagram is
•Therefore, according to the distortion-energy theory, the yield strength in shear is
0.577 times the yield strength in tension.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 148

Region of Safety
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 149
Experiments have shown that the distortion energy
theory is in better agreement for predicting
the failure of a ductile component than any other
theory of failure.

Numerical Problem 23
•A thin cylindrical shell of a diameter of 200 mm and a wall thickness t is subjected to
an internal pressure of 2.4 N/mm
2
. Neglecting the effect of p as principal stress,
determine the thickness of the shell if s
yp= 270 N/mm
2
. take fos2.5.
•Hints: principal stress are pd/2t and pd/4t.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 150

Numerical Problem 24
•The diameter of shaft in Fig. is 60 mm. The material
used is 50C4 steel with Syp= 400 MPa. Using Factor
of Safety = 2.5, find the load P applying
•(a) maximum shear stress theory
•(b) Maximum energy of distortion theory
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 151

COMPARISON
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 152

Forms of design equations to be used in practice
•Usually machine components are subjected to uniaxial normal stress σaccompanied
by shear stress τ. Under this loading condition the forms of design equation to be used
are:
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 153

Numerical Problem 25
•A bolt is acted upon by an axial pull of 16 kNalong with a transverse shear force of
10 kN.Determine the diameter of the bolt required according to different theories.
Elastic limit of the bolt material is 250 MPa and a factor of safety 2.5 is to be taken.
Poisson’s ratio is 0.3.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 154

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 155

Numerical Problem 26
•TheshaftofanoverhangcranksubjectedtoaforcePof1kNisshowninFig.The
shaftismadeofplaincarbonsteel45C8.Thefactorofsafetyis2.Determinethe
diameteroftheshaftusingthemaximumshearstresstheory.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 156

Coulomb-Mohr Theory for Ductile Materials
•Not all materials have compressive strengths equal to their corresponding tensile
values.
•Three Mohr circles, one for the uniaxial compression test, one for the test in pure
shear, and one for the uniaxial tension test, are used to define failure by the Mohr
hypothesis.
•Coulomb-Mohr theory, assumes that the boundary BCD is straight. With this
assumption only the tensile and compressive strengths are necessary.
•Consider the conventional ordering of the principal stresses such that σ
1≥ σ
2≥ σ
3. The
largest circle connects σ
1and σ
3, as shown in Fig
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 157

•Cross-multiplying and simplifying reduces this equation to
•For plane stress, when the two nonzero principal stresses are σ
A≥ σ
B, we have
•Case 1: σ
A≥ σ
B≥ 0. For this case, σ
1= σ
Aand σ
3= 0.
•Case 2: σ
A≥ 0 ≥ σ
B. Here, σ
1= σ
Aand σ
3= σ
B,
•Case 3: 0 ≥ σ
A≥ σ
B. For this case, σ
1= 0 and σ
3= σ
B,
•For pure shear τ, σ
1= −σ
3= τ . The torsional yield strength occurs when τ
max= S
sy.
Substituting σ
1= −σ
3= S
sy
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 158

Modifications of the Mohr Theory for Brittle Materials
•We will discuss two modifications of the Mohr theory for brittle materials: the Brittle-
Coulomb-Mohr (BCM) theory and the modified Mohr (MM) theory.
•Brittle-Coulomb-Mohr
•Modified Mohr
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 159

Stress Concentration
•Whenever a machine component changes the shape of its cross-section (or having
any discontinuity in a machine part), the simple stress distribution no longer holds
good and the stress distribution in neighborhood of the discontinuity is different.
•Therefore, the elementary stress equations no longer describe the state of stress in
the part at these locations.
•In neighborhood of the discontinuity, distributions of stress occur in which the peak
stress reaches much larger magnitudes than does the average stress over the section.
•This increase in peak stress near holes, grooves, notches, sharp corners, cracks, and
other changes in section is called stress concentration.
•Stress concentration is defined as the localization of high stresses due to the
irregularities present in the component and abrupt changes of the cross-section.
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 160
The section variation that causes the stress concentration
is referred to as a stress raiser.
Stress concentration

Causes of stress concentration
•(i)VariationinPropertiesofMaterials:
•(a)internalcracksandflawslikeblowholes;(b)cavitiesinwelds;
•(c)airholesinsteelcomponents;and (d)nonmetallicorforeigninclusions.
•Thesevariationsactasdiscontinuitiesinthecomponentandcausestress
concentration.
•(ii)LoadApplication:
•(a)Contactbetweenthemeshingteethofthedrivingandthedrivengear
•(b)Contactbetweenthecamandthefollower
•(c)Contactbetweentheballsandtheracesofballbearing
•(d)Contactbetweentherailandthewheel
•(e)Contactbetweenthecranehookandthechain
•theconcentratedloadisappliedoveraverysmallarearesultinginstress
concentration.
•(iii)AbruptChangesinSection:resultsinstressconcentrationatthesecross-
sections.
•(iv)DiscontinuitiesintheComponentCertainfeaturesofmachinecomponentssuch
asoilholesoroilgrooves,keywaysandsplines,andscrewthreadsresultin
discontinuitiesinthecross-sectionofthecomponent.Thereisstressconcentrationin
thevicinityofthesediscontinuities.
•(v)MachiningScratchesMachiningscratches,stampmarksorinspectionmarksare
surfaceirregularities,whichcausestressconcentration.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 161

Stress Concentration Factor
•In order to consider the effect of stress concentration and find out localized stresses, a
factor called stress concentration factor (k
t) is used.
•&#3627408472;
&#3627408481;=
&#3627408448;&#3627408462;&#3627408485;??????&#3627408474;&#3627408482;&#3627408474;&#3627408480;&#3627408481;&#3627408479;&#3627408466;&#3627408480;&#3627408480;&#3627408462;&#3627408481;&#3627408481;ℎ&#3627408466;&#3627408465;??????&#3627408480;&#3627408464;&#3627408476;&#3627408475;&#3627408481;??????&#3627408475;&#3627408482;??????&#3627408481;&#3627408486;
&#3627408449;&#3627408476;&#3627408474;??????&#3627408475;&#3627408462;&#3627408473;&#3627408480;&#3627408481;&#3627408479;&#3627408466;&#3627408480;&#3627408480;
•A theoretical, or geometric, stress-concentration factor Ktor Kts
• &#3627408472;
&#3627408481;=
??????
??????&#3627408462;&#3627408485;
??????
0
,&#3627408472;
&#3627408481;&#3627408480;=
??????
??????&#3627408462;&#3627408485;
??????
0
•SCF depends only on the GEOMETRY
and TYPE OF LOADING; not on material
Equilibrium requirement, the average stress be
the same on the section. Unusually high
stress near the stress raiser, but low stresses
in the remainder of the section.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 162
•where σ
0and τ
0are stresses determined by elementary
equations and σ
max. and τ
max. are localized stresses at the
discontinuities.

Stress Concentration Factors in tension
•When more than one load acts on a notched member (e.g. combined tension, torsion,
and bending) the nominal stress due to each load is multiplied by the SCF
corresponding to each load, and the resultant stresses are found by superposition
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 163
Stress Concentration Factor (Rectangular
Plate with Transverse Hole in Tension or
Compression)

Stress Concentration Factors in tension
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 164
Stress Concentration Factor (Flat Plate with
Shoulder Fillet in Tension or Compression)

Stress Concentration Factors in tension
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 165
Stress Concentration Factor (Round Shaft
with Shoulder Fillet in Tension)

Stress Concentration Factors in bending
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 166
Stress Concentration Factor (Round Shaft
with Shoulder Fillet in Bending)

Stress Concentration Factors in torsion
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 167
Stress Concentration Factor (Round Shaft
with Shoulder Fillet in Torsion)

Stress Concentration due to Elliptical Hole
•the theoretical stress concentration factor at the edge of hole is given by,
• &#3627408472;
&#3627408481;=1+2
&#3627408462;
&#3627408463;
•where,
•a = half width (or semi-major axis) of the ellipse perpendicular to the direction of the
load
•b = half width (or semi-minor axis) of the ellipse in the direction of the load
•As b approaches zero, the ellipse becomes sharper and sharper. A very sharp crack is
indicated and the stress at the edge of the crack becomes very large. As the width of
the elliptical hole in the direction of the load approaches zero, the stress concentration
factor becomes infinity.
•The ellipse becomes a circle when (a = b).
• K
t= 1 + 2 = 3
•Therefore, the theoretical stress concentration factor due to a small circular hole in a
flat plate, which is subjected to tensile force, is 3.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 168

How to Estimate SCF
•Difficult to determine SCF, theoretically, due to complex geometric shapes.
•SCFs are found using EXPERIMENTAL TECHNIQUES
•Photo-elasticity, grid methods, brittle-coating methods, electrical strain-gauge
methods
•Finite Element Methods -> use very fine mesh at the SC region and transition of
mesh from rest of the part to SC region must be gradual
•Similar cautions to be taken when using grid methods and strain gauges
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 169
•Photoelasticityisanexperimentalmethod
todeterminethestressdistributioninamaterial.The
methodismostlyusedincaseswheremathematical
methodsbecomequitecumbersome.Inthismethod,an
identicalmodeloftheplateismadeofepoxyresin.The
modelisplacedinacircularpolariscopeandloadedat
theedges.Itisobservedthatthereisasuddenrisein
themagnitudeofstressesinthevicinityofthehole.
•Themethodisanimportanttoolfordetermining
criticalstresspointsinamaterial,andisusedfor
determiningstressconcentrationinirregular
geometries.

Finite Element method
•The CAD model is subdivided into many small pieces of simple shapes called
elements.
•FEA program writes the equations governing the behavior of each element taking into
consideration its connectivity to other elements.
•These equations relate the unknowns, for example displacements in stress analysis, to
known material properties, restraints and loads.
•The program assembles the equations into a large set of simultaneous algebraic
equations (thousands or even millions).
•These equations are then solved by the program to obtain the stress distribution for the
entire model.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 170

How to mitigate SC/ Design to Minimize SC
•Force should be transmitted from point to point as smoothly as possible. The lines
connecting the force transmission path are sometimes called the force (or stress)
flow.
•Sharp transitions in the direction of the force flow should be removed by smoothing
contours and rounding notch roots.
•When stress raisers are necessitated by functional requirements, the raisers should be
placed in regions of low nominal stress if possible.
•When notches are necessary, removal of material near the notch can alleviate stress
concentration effects.
•A type of stress concentration called an interface notchis commonly produced when
parts are joined by welding.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 171

Methods of reducing stress concentration
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 172
3. If a notch is
unavoidable it is better to
provide a number of small
notches
rather than a long one.
This reduces the stress
concentration to a large
extent.
•1. Provide a fillet radius so
that the cross-section may
change gradually.
•2. Sometimes an elliptical
fillet is also used.
Additional Notches and Holes in
Tension Member

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 173
4. If a projection is
unavoidable from
design
considerations it is
preferable to
provide a narrow
notch than a wide
notch.
5. Stress relieving
groove are
sometimes
provided.
Fillet Radius,
Undercutting
and Notch for
Member in
Bending

•Reduction of Stress Concentration in Threaded Members
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 174
Reduction of Stress Concentration in
Threaded Components: (a) Original
Component (b) Undercutting (c)
Reduction in Shank Diameter
A type of stress concentration called
an interface notch is commonly
produced when parts are joined by
welding

THINK ??
•Which one has the lowest SC in each group?
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 175

Numerical Problem 27
•The circular shaft shown in Fig. is girdled by a U-shaped groove, with h = 10.5 mm
deep. The radius of the groove root r = 7 mm, and the bar diameter away from the
notch D = 70 mm. A bending moment of 1.0 kN-m and a twisting moment of 2.5 kN-
m act on the bar. The maximum shear stress at the root of the notch is to be calculated.
(Given: K
tB=1.78, K
tT=1.41)
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 176

Numerical Problem 28
•The flat bar shown in figure is 10 mm thick and is pulled by a force P producing a
total change in length of 0.2 mm. Determine the maximum stress developed in the bar.
Take E= 200 GPa.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 177

Numerical Problem 29
•Find the maximum stress developed in a stepped shaft subjected to a twisting moment
of 100 N-m as shown in figure. What would be the maximum stress developed if a
bending moment of 150 N-m is applied
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 178

Numerical Problem 30
•A non-rotating shaft supporting a load of 2.5 kNis shown in Fig. The shaft is made of
brittle material, with an ultimate tensile strength of 300 N/mm
2
. The factor of safety is
3. Determine the dimensions of the shaft.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 179

Types of Loading : what do mean by Fatigue ?
•Static-Load does not change in magnitude and direction and
normally increases gradually to a steady value
•Dynamic –Load may change in magnitude for example, traffic of
varying weight passing a bridge. Load may change in direction, for
example, load on piston rod of a double acting cylinder
•Cyclic, Impact, Random
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 180
•Fatigue-repeated, variable, alternating, fluctuating (stresses)
•Fatigue is a form of failure that occurs in structures subjected to
dynamic & fluctuating stress (e.g., bridge, aircraft & machine
components).

Introduction to Fatigue in Metals
•Static conditions : loads are applied gradually, to give sufficient time for the strain to
fully develop.
•Variable conditions : stresses vary with time or fluctuate between different levels, also
called repeated, alternating, or fluctuating stresses.
•When machine members are found to have failed under fluctuating stresses, the actual
maximum stresses were well below the ultimate strength of the material, even below
yielding strength.
•Since these failures are due to stresses repeating for a large number of times, they are
called fatigue failures.
•When machine parts fails statically, the usually develop a very large deflection, thus
visible warning can be observed in advance; a fatigue failure gives no warning!
•The fiber on the surface of a rotating shaft subjected to a bending load, undergoes
both tension and compression for each revolution of the shaft. Any fiber on the shaft
is therefore subjected to fluctuating stresses.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 181

Fatigue Failure in Metals
•There are mainly two characteristics of fatigue failures:
•(a) Progressive development of crack.
•(b) Sudden fracture without any warning since yielding is practically absent.
•Fatigue failures are influenced by
•(i) Nature and magnitude of the stress cycle. (ii) Endurance limit.
•(iii) Stress concentration. (iv) Surface characteristics.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 182
initiation
propagation
fracture
•A fatigue failure arises from three stages of
development:
-Stage I : initiation of microcracks due to cyclic plastic
deformation (these cracks are not usually visible to the
naked eyes).
-Stage II : propagation of microcracks to macrocracks
forming parallel plateau0like fracture surfaces separated by
longitudinal ridges (in the form of dark and light bands
referred to as beach marks).
-Stage III : fracture when the remaining material cannot
support the loads.

Cyclic stress
•There are three types of mathematical models for cyclic stresses—fluctuating or
alternating stresses, repeated stresses and reversed stresses.
•The fluctuating or alternating stress varies in a sinusoidal manner with respect to time.
It has some mean value as well as amplitude value.
•The repeated stress varies in a sinusoidal manner with respect to time, but the
variation is from zero to some maximum value.
•The reversed stress varies in a sinusoidal manner with respect to time, but it has zero
mean stress. In this case, half portion of the cycle consists of tensile stress and the
remaining half of compressive stress.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 183

Fatigue Life Methods in Fatigue Failure Analysis
•Let Nbe the number of cycles to fatigue for a specified level of loading
-For , generally classified as low-cycle fatigue
-For , generally classified as high-cycle fatigue
•Three major fatigue life methods used in design and analysis are
1.stress-life method: is based on stress only, least accurate especially for low-cycle
fatigue; however, it is the most traditional and easiest to implement for a wide
range of applications. It represent high cycle application adequately.
2.strain-life method: involves more detailed analysis of plastic deformation at
localized regions where the stresses and strains are considered for life estimates.
This method is especially good for low-cycle fatigue; however, idealizations in the
methods make it less practical when uncertainties are present.
3.linear-elastic fracture mechanics method: assumes a crack is already present
and detected. It is then employed to predict crack growth with respect to stress
intensity. Practical with computer codes in predicting in crack growth with respect
to stress intensity factor
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 184

Stress-Life Method : R. R. Moore-Endurance limit
•The fatigue or endurance limit of a material is defined as the maximum amplitude of
completely reversed stress that the standard specimen can sustain for an unlimited
number of cycles without fatigue failure.
•Since the fatigue test cannot be conducted for unlimited or infinite number of cycles,
10
6
cycles is considered as a sufficient number of cycles to define the endurance
limit.
•There is another term called fatigue life, which is frequently used with endurance
limit.
•The fatigue life is defined as the number of stress cycles that the standard specimen
can complete during the test before the appearance of the first fatigue crack.
•the endurance limit is determined by means of a rotating beam machine developed by
R RMoore.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 185
Rotating Beam Subjected to Bending Moment: (a) Beam, (b) Stress Cycle at Point A

S-N Curve
•In R. R. Moore machine tests, a constant bending load is applied, and the number of
revolutions of the beam required for failure is recorded.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 186
•Tests at various bending stress
levels are conducted by changing
the bending moment by addition or
deletion of weights.
•In each test, two readings are taken,
viz., stress amplitude (Sf) and
number of stress cycles (N).
•The S–N curve is the graphical
representation of stress amplitude (S
f)
versus the number of stress cycles (N)
before the fatigue failure on a log graph
paper.
•Ordinate of S-N curve is fatigue strength,
S
f, at a specific number of cycles

Characteristics of S-N Curves for Metals
•For ferrous materials like steels, the S–N curve becomes asymptotic at 10
6
cycles,
which indicates the stress amplitude corresponding to infinite number of stress cycles.
The magnitude of this stress amplitude at 10
6
cycles represents the endurance limit of
the material denoted as S
e.
•Non-ferrous metals and alloys do not have an endurance limit, since their S-N curve
never become horizontal.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 187
•For materials with no
endurance limit, the
fatigue strength is
normally reported at
•N= 1/2 is the simple
tension test

Endurance Limit for Steels
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 188
•Forsteels,theendurancelimitrelates
directlytotheminimumtensilestrength
asobservedinexperimental
measurements.
•Fromtheobservations,theenduranceof
steelscanbeestimatedas
•withtheprimemarkonS’
ethe
endurancelimitreferringtotherotating-
beamspecimen.

Fatigue Strength : Basics
•Low-cycle fatigue considers the range from N=1 to about 1000 cycles.
•In this region, the fatigue strength S
fis only slightly smaller than the tensile
strength S
ut.
•High-cycle fatigue domain extends from 10
3
to the endurance limit life (10
6
to 10
7
cycles).
•Experience has shown that high-cycle fatigue data are rectified by a logarithmic
transform to both stress and cycles-to-failure.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 189

Fatigue Strength : General
•For actual mechanical applications (valid only between 10^3 to 10^6 cycles), the
fatigue strength calculated above is extended to a more general form as
•For a completely reversed stress
•Low-cycle fatigue is often defined as failure that occurs in a range of 1 ≤ N ≤ 10
3
cycles.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 190
N: Cycle to failure1/
()
revb
a
N
s
=

Numerical Problems 31
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 191
•Acarbonsteelhavingultimatestrength630MPa.estimate
•(a)therotating-beamendurancelimitat10
6
cycles.
•(b)theendurancestrengthofapolishedrotating-beamspecimencorrespondingto10
4
cyclestofailure
•(c)theexpectedlifeofapolishedrotating-beamspecimenunderacompletely
reversedstressof385MPa.

Endurance Limit Modifying Factors
•The endurance limit of the rotating-beam specimen might differ from the actual application
due to the following differences from laboratory tests.
✓Material : composition, basis of failure, variability
✓Manufacturing : method, heat treatment, fretting corrosion, surface condition, stress concentration
✓Environment : corrosion, temperature, stress state, relaxation times.
✓Design : size, shape, life, stress state, stress concentration, speed, fretting, galling
•Modifyingfactorsofsurfacecondition,size,loading,temperature,andmiscellaneous
itemsareproposedbyMarintoquantifythesedifferences.
• ??????
&#3627408466;=&#3627408472;
&#3627408462;&#3627408472;
&#3627408463;&#3627408472;
&#3627408464;&#3627408472;
&#3627408465;&#3627408472;
&#3627408481;&#3627408466;&#3627408474;&#3627408477;&#3627408472;
&#3627408479;&#3627408472;
&#3627408474;??????
&#3627408466;
′
•where
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 192
✓&#3627408472;
&#3627408462;= surface condition modification factor
✓&#3627408472;
&#3627408463;= size modification factor
✓&#3627408472;
&#3627408464;= load modification factor
✓&#3627408472;
&#3627408465;= modifying factor to account for stress concentration
✓&#3627408472;
&#3627408481;&#3627408466;&#3627408474;&#3627408477;= temperature modification factor
✓&#3627408472;
&#3627408479;= reliability factor
✓&#3627408472;
&#3627408474;= miscellaneous-effects modification factor
✓??????
&#3627408466;
′
= rotary-beam test specimen endurance limit

Surface and Loading Factors
•Surface Factor : the surface modification factor depends on the quality of the finish
of the actual part surface and on the tensile strength of the part material. It can be
calculated as
•Loading Factor : the axial and torsional loadings results in different endurance limit
than that of a standard rotating-bending test. The load factor applies to other loading
conditions as
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 193

Size Factor
•Size Factor : the size factor has been evaluated using 133 set of data points in the
literature. For axial loading, . For bending and torsion can be expressed as
•Effective dimension is introduced for non-circular cross section by equating the
volume of the material stressed at and above 95 percent of the maximum stress to
the same volume in the rotating-beam specimen.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 194

Temperature Factor
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 195
•Attemperatureslowerthanroomtemperature,
brittlefractureofacomponentneedstobe
consideredfirst;atoperatingtemperatures
higherthanroomtemperature,yieldshouldbe
investigated.
•Ifonlytensile-strengthdataareavailable,
polynomialfittingtothedatacouldprovide
thetemperaturefactoratvarioustemperature
values.Forcarbonsteelsandalloysteel,we
have
•If the rotating-beam endurance limit is known
at room temperature, we have
&#3627408472;
&#3627408481;&#3627408466;&#3627408474;&#3627408477;
&#3627408472;
&#3627408481;&#3627408466;&#3627408474;&#3627408477;

Reliability Factor
•Most endurance strength data are reported as mean values.
•To account for the scatter of measurement data, the reliability modification factor is
written as
•Miscellaneous-Effects Factor k
m
•The miscellaneous factor intends to account for the reduction in endurance limit due
to all other effects, such as residual stresses, corrosion, different material treatments
(metallic coating, metal spraying, etc), directional characteristics of operations
(rolled or drawn parts), cyclic frequency and frettagecorrosion.
•One should also treat the miscellaneous-effect factor k
mas a reminder that these
must accounted for, because actual values of k
mare not always available.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 196
&#3627408472;
&#3627408479;

Stress Concentration and Notch Sensitivity
•It is observed that the actual reduction in the endurance limit of a material due to
stress concentration is less than the amount indicated by the theoretical stress
concentration factor Kt.
•The fatigue stress concentration factor from the existence of irregularities or
discontinuities in materials is defined a
•Notch sensitivity is defined as the susceptibility of a material to succumb to the
damaging effects of stress raising notches in fatigue loading.
• ??????=
&#3627408444;&#3627408475;&#3627408464;&#3627408479;&#3627408466;&#3627408462;&#3627408480;&#3627408466;&#3627408476;&#3627408467;&#3627408462;&#3627408464;&#3627408481;&#3627408482;&#3627408462;&#3627408473;&#3627408480;&#3627408481;&#3627408479;&#3627408466;&#3627408480;&#3627408480;&#3627408476;&#3627408483;&#3627408466;&#3627408479;&#3627408475;&#3627408476;&#3627408474;??????&#3627408475;&#3627408462;&#3627408473;&#3627408480;&#3627408481;&#3627408479;&#3627408466;&#3627408480;&#3627408480;
&#3627408444;&#3627408475;&#3627408464;&#3627408479;&#3627408466;&#3627408462;&#3627408480;&#3627408466;&#3627408476;&#3627408467;&#3627408462;&#3627408464;&#3627408481;&#3627408482;&#3627408462;&#3627408473;&#3627408480;&#3627408481;&#3627408479;&#3627408466;&#3627408480;&#3627408480;&#3627408476;&#3627408483;&#3627408466;&#3627408479;&#3627408475;&#3627408476;&#3627408474;??????&#3627408475;&#3627408462;&#3627408473;&#3627408480;&#3627408481;&#3627408479;&#3627408466;&#3627408480;&#3627408480;
•Since, σ
o= nominal stress as obtained by elementary equations
•Hence, actual stress = K
fσ
o, theoretical stress = K
tσ
o
•increase of actual stress over nominal stress = ( K
fσ
o–σ
o)
•increase of theoretical stress over nominal stress = (K
tσ
o–σ
o)
•Therefore, ??????=
K
fσ
o–σ
o
K
tσ
o–σ
o
=
K
f–1
K
t–1
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 197

•
•(i) When the material has no sensitivity to notches, q = 0 and K
f= 1
•(ii) When the material is fully sensitive to notches, q = 1 and K
f= K
t.
•In general, the magnitude of the notch sensitivity factor q varies from 0 to 1.
•In case of doubt, the designer should use (q = 1) or (K
t= K
f) and the design will be on
the safe side.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 198
Notch Sensitivity Charts (for Reversed
Bending and Reversed Axial Stresses)
Notch Sensitivity Charts (for Reversed
Torsional Shear Stresses)

Reversed stresses—design for finite and infinite life
•There are two types of problems in fatigue design—
•(i) components subjected to completely reversed stresses, and
•(ii) components subjected to fluctuating stresses.
•The design problems for completely reversed stresses are further divided into two
groups—
•(i) design for infinite life, and (ii) design for finite life.
•Case I: When the component is to be designed for infinite life, the endurance limit
becomes the criterion of failure. The amplitude stress induced in such components
should be lower than the endurance limit in order to withstand the infinite number of
cycles. Such components are designed with the help of the following equations:
•where (σ
a) and (τ
a) are stress amplitudes in the component and S
eand S
seare corrected
endurance limits in reversed bending and torsion respectively.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 199
mean stress is zero
always a mean stress

Case II:
•When the component is to be designed for finite life, the S–N curve as shown in Fig.
The curve is valid for steels.
•It consists of a straight line AB drawn from (0.9 S
ut) at 10
3
cycles to (S
e) at 10
6
cycles
on a log-log paper.
•AB, which is used as a criterion of failure for finite-life problems.
•Depending upon the life N of the component, draw a vertical line passing through
log
10(N) on the abscissa. This line intersects AB at point F.
•Draw a line FE parallel to the abscissa. The ordinate at the point E, i.e. log
10(S
f),
gives the fatigue strength corresponding to N cycles.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 200

Numerical Problem 32
•Aplatemadeofsteel20C8(Sut=440N/mm
2
)inhotrolledandnormalizedcondition
isshowninFig.Itissubjectedtoacompletelyreversedaxialloadof30kN.The
notchsensitivityfactorqcanbetakenas0.8andtheexpectedreliabilityis90%.The
sizefactoris0.85.Thefactorofsafetyis2.Determinetheplatethicknessforinfinite
life.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 201

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 202

Numerical Problem 33
•Arotatingbarmadeofsteel45C8(Sut=630N/mm
2
)issubjectedtoacompletely
reversedbendingstress.Thecorrectedendurancelimitofthebaris315N/mm
2
.
Calculatethefatiguestrengthofthebarforalifeof90,000cycles.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 203

Cumulative damage in fatigue (Miner’s equation)
•In certain applications, the mechanical component is subjected to different stress
levels for different parts of the work cycle.
•The life of such a component is determined by Miner’s equation.
•Suppose that a component is subjected to completely reversed stresses (σ
1) for (n
1)
cycles, (σ
2) for (n
2) cycles, and so on.
•Let N1 be the number of stress cycles before fatigue failure, if only the alternating
stress (σ
1) is acting.
•Miner’s equation can be expressed as
•Suppose that α
1, α
2,… are proportions of the total life that will be consumed by the
stress levels σ
1, σ
2,… etc.
•Let N be the total life of the component.
•Then, n
1= α
1N
• n
2= α
2N
•Substituting these values in Miner’s equation
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 204

Numerical Problem 34
•Arotating-beamspecimenwithanendurancelimitof350MPaandanultimate
strengthof700MPaiscycled20percentofthetimeat490MPa,50percentat385
MPa,and30percentat280MPa.Letf=0.9andestimatethenumberofcyclesto
failure.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 205

Numerical Problem 35
•Amachinepartwillbecycledat±350MPafor5(10
3
)cycles.Thentheloadingwill
bechangedto±260MPafor5(10
4
)cycles.Finally,theloadwillbechangedto±225
MPa.Howmanycyclesofoperationcanbeexpectedatthisstresslevel?Forthepart,
Sut=530MPa,f=0.9,andhasafullycorrectedendurancestrengthofSe=210MPa.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 206

Fatigue Failure Criteria for Fluctuating Stresses
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 207
Gerber
Goodman
Soderberg
ASME-elliptic
Langer
Is my machine part safe when subjected to
fluctuating stress state, i.e. a set of σaand σm
??
Diagram is good for design purpose, easy to
use.

Numerical Problems 36
•A40mm-diameterbarhasbeenmachinedfromanAISI1050cold-drawnbar.This
partistowithstandafluctuatingtensileloadvaryingfrom0to70kN.Becauseofthe
ends,andthefilletradius,afatiguestress-concentrationfactorKfis1.85for10
6
or
largerlife.FindSaandSmandthefactorofsafetyguardingagainstfatigueandfirst-
cycleyielding,using(a)theGerberfatiguelineand(b)theASME-ellipticfatigue
line.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 208
Assume Sut= 690 MPa, Sy= 580 MPa

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 209

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 210

Combination of Load Modes
•Fatigue problems can be categorized into
-completely reversing simple loads : can be handled using S-N curve; only one type
of loading is allowed with midrange stress being zero.
-fluctuating simple loads : can be resolved using a criterion to relate midrange and
alternating stresses; again, only one type of loading is allowed.
-combinations of loading modes : a combinations of different types of loading.
✓apply appropriate stress concentration factors to each type of stress
✓calculate equivalent von Misesstress
✓select a fatigue failure criterion
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 211

Combination of Bending, Torsion, and Axial Stresses
•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. PrakashKumar 212
✓apply appropriate stress concentration factors (Kfor Kfs) to each type of stress
✓calculate equivalent von Mises stresses for alternating and mean stress states
✓select a fatigue failure criterion
✓Conservative check for localized yielding using von Mises stresses
von Misesstress in 3D stress state
von Mises stress in plane stress state
σ'max=σ'a+σ'm

Numerical Problems 37
•Asolidroundbarwithdiameterof50mmhasagroovecuttoadiameterof45mm,
witharadiusof2.5mm.Thebarisnotrotating.Thebarisloadedwitharepeated
bendingloadthatcausesthebendingmomentatthegroovetofluctuatebetween0and
2825Nm.Thebarishot-rolledAISI1095,butthegroovehasbeenmachined.
Determinethefactorofsafetyforfatiguebasedoninfinitelifeusingthemodified
Goodmancriterion,andthefactorofsafetyforyielding.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 213
Assume Sut= 830 MPa, Sy= 460 MPa

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 214

Numerical Problems 38
•TheshaftshowninthefigureismachinedfromAISI1040CDsteel.Theshaftrotates
at1600rpmandissupportedinrollingbearingsatAandB.TheappliedforcesareF
1
=4.8kNandF
2=9.6kN.Determinetheminimumfatiguefactorofsafetybasedon
achievinginfinitelife.Ifinfinitelifeisnotpredicted,estimatethenumberofcyclesto
failure.Alsocheckforyielding.
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 215
Assume Sut= 590 MPa, Sy= 490 MPa

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 216

•
18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 217

18-09-2023 Lecture notes on Design due to static and dynamic load, by Dr. Prakash Kumar 218

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