machine elements design

281
MEL 311: Machine Element Design Harish Hirani Department of Mechanical Engineering Design Innovation & Manufacturing MEP 202 Mechanical Engineering Drawing MEP 201 Mechanics of Solids AML 140 Kinematics & Dynamics of Machines MEL 211 Pre-requisites

Upload: santiagogaborov

Post on 18-Nov-2015

281 views

Category:

Documents


18 download

DESCRIPTION

machines elements

TRANSCRIPT

  • MEL 311: Machine Element Design

    Harish HiraniDepartment of Mechanical Engineering

    Design Innovation & ManufacturingMEP 202

    Mechanical Engineering Drawing MEP 201

    Mechanics of SolidsAML 140

    Kinematics & Dynamics of MachinesMEL 211

    Pre-requisites

  • 7/24/2009 2

  • Identification of need

    Problem formulation

    Mechanism/Synthesis

    Analysis

    Verification/Validation

    Presentation

    Design Innovation & Manufacturing

    Mechanical Engineering Drawing

    Mechanics of Solids

    Kinematics & Dynamics of Machines

  • Purchase a safe lathe machineLow risk of injury to operatorLow risk of operator mistakeLow risk of damage to workpiece/toolAutomatic cut-out on overload

    Problem: Design a reliable and simple test rig to test shaft connections subjected to impulse loads.

    SafetySimple

    Minimum no. of componentsSimple design of components

    Low complexityDesign for standards

    Reliable operationGood reproducibility

    Low wearLow susceptibility to external noise

    Tolerance for overloadingEasy handling

    Quick exchange of test connectionsGood visibility of measuring system

    Problem formulation

  • 7/24/2009 7

    Can we increase speed of Jute Flyer ??

    Flyer Spinning Machine

    Design Innovation & Manufacturing

    Current speed 4000 rpm

    Target speed 6000 rpm

    Bobbin

  • 7/24/2009 8

    Can we increase speed of Jute Flyer ??

    Flyer Spinning Machine

    Design Innovation & Manufacturing

    Increase rotational speed

    Constraints: Stress < ??

    Deflection < ???

    Mechanics of Solids

  • 7/24/2009 9

    Increase operating speed wharve assembly

    Bearing life must be at least 3 years The wharves must be lighter than the current wharvesTemperature rise must be within 5C.Cost of new wharveassembly 1.5 times cost of existing assembly

  • Identification of need

    Problem formulation

    Mechanism/Synthesis

    Analysis

    Verification

    Presentation

    Design is an iterative process

    Analysis requires mathematical model of system/component.

  • Machine Element Design: SystemElements

    Power transmission System Gears, Bearings, Shaft, Seals.

  • 7/24/2009 12

    Machine Elements1. Design of shafts2. Design of couplings3. Design of belt and chains4. Design of springs5. Design of Clutches & Brakes6. Design of Screws7. Design of bolted joints8. Gear (spur, helical, bevel and worm) design9. Bearing Selection of Rolling contact bearing10.Design of journal bearings

    Text books:

    1. Mechanical Engineering Design. Shigley and Mischke..

    2. Machine Design: An Integrated Approach.. R. L. Norton

    25-30 HoursMinor II

    Major

  • 7/24/2009 13

    Basics required to design Machine Elements1. Solid Mechanics 2. Factors of safety3. Standards and Design Equations4. Selection of Materials and Processes5. Standard numbering system (i.e BIS

    designations of materials).6. Applications of failures theories7. Introduction to design for fatigue8. Surface strength9. Introduction to CAD. Computer Assistance

    12-15 HoursMinor IMajor

  • 7/24/2009 14

    Computer aided..Design of gears.Design of hydrodynamic bearings.

    ecc.75

    .00002clearance

    length.01

    visco.005

    93.6load

    speed1000

    radius.02

    104.72omega

    2.094U

    OutputNameInput

  • 7/24/2009 15

    .5 .55 .6 .65 .7 .75 .8 .85 .9 .950

    250

    500

    750

    1000

    1250

    1500

    1750

    2000

    2250

    2500

    2750Load capacity versus eccentricity ratio

    Eccentricity ratio

    Load

    , N

  • 7/24/2009 16

    1 2 3 4 5 6 7 8 9 100

    100

    200

    300

    400

    500

    600

    700Effect of clearance on load

    0.001 R * Factor

    Load 2

    1

    rCLoad

  • 7/24/2009 17

    clearance = 0.001 * radius

    load = U * visco * (length ^ 3) /(clearance ^2) * pi()/4 * ecc/((1-ecc^2)^2) *sqrt((16/(pi()^2)-1)*(ecc^2) + 1)

    omega = 2 *pi()/60 * speedU = omega * radius

    ecc.75.00002clearance

    length.01visco.005

    93.6loadspeed1000radius.02

    104.72omega2.094UOutputNameInput

    PARAMETRIC STUDY: Hydrodynamic BearingIterative study to desirable results

  • 7/24/2009 18

    What is TK Solver?Package for solving numerical equations:

    linear or nonlinear, single or multiple equations - up to 32,000.

    No need to enter the equations in any special order-- TK Solver is based on a declarative (as opposed to procedural) programming language..No need to isolate the unknowns on one side of the equations

    2^2^2^ cba =+ Input (a,b) or (b,c) or (c,a)

    Output c or a or b

  • 7/24/2009 19

    Enter EquationsThis sheet shows the relationship between variables in the models. This is where model is controlled from.

    Variable sheet shows the input or output value, with units if relevant, and the status of each variable

  • 7/24/2009 20

    largest integer = xCEILING(x)nearest integer to xROUND(x)

    -1 if x < 0, 0 if x=0, 1 if x > 0SIGNUM(X) or SGN(X)remainder of x1/x2MODULUS (x1,x2) or MOD(x1,x2)integer part of xINTEGER(x) or INT(x)

    ROOT(X,N) nth root of x; SQRT(x) , ABS(x), COSH(), ACOSH(), SINH(), ASINH(), TANH(), ATANH()ATAN2(y,x), ATAN2D(y,x) {4-Quadrant arc tangent of y/x }EXP(), LN() {base e}, LOG() {base 10}COSD(), ACOSD(),SIND(),ASIND(),TAND(),ATAND()COS(), ACOS(), SIN(), ASIN(), TAN(), ATAN()

    BUILT IN FUNCTIONS

    TKs built-in functions are NOT case-sensitive; SIN(x)=sin(x)=Sin(x)

    User-defined function names ARE case-sensitive.

  • List Function Sheet

    Comment:Domain List:Mapping:Range List:

    returns the weight density of a matmatlTabledensity

    Element Domain Range123

    'alum'steel'copper

    2.768057.750548.580955

    Expresses functional relationship between the corresponding elements of two lists

  • 7/24/2009 22

    Material Selection using TKSolver

    Machine Design: An Integrated Approach..

    by Robert L. Norton

  • 7/24/2009 23

    Evaluation Scheme

    Minor I 15%Minor II 15%Major 30%Laboratory 25%Tutorial 15%

  • Introduction to machine elements design..

    Machine: Structure + Mechanisms

    Combination of rigid bodies which do not have any relative motion among themselves Automobile chassis Machine tool bed Machine columns

    Slider crank mechanism Cam and follower mech. Gear train

    Shafts, couplings, springs, bearings, belt and gear drives, fasteners, and joints are basic elements of machines..

    1 2

    3 4

  • 7/24/2009 25

    Scientific procedure to design machine elements

    Ultimate goal is to size and shape the element so that elements perform expected function without failure.

    1. Predict mode & conditions of failure.2. Force/Moment/Torque analysis.3. Stress and deflection analysis.4. Selection of appropriate material.

    Thorough understanding of material prop essential

    Iterations

    1

    2

    3

    QUANTIFICATION

    Wear, Vibration, misalignment, environment

  • Journal bearing test rig

    Acrylic bearing

  • Brass bearing

    0500

    100015002000

    0 30 60 90 120 150 180

    Angle (Degree)

    Flui

    d pr

    essu

    re

    (kPa

    )

    Acrylic bearing

    0

    500

    1000

    1500

    0 30 60 90 120 150 180

    Angle (Degree)

    Flui

    d pr

    essu

    re

    (kPa

    )

    Max pressure = 1800 kPa

    Max pressure = 1300 kPa

    Estimating stress

    Selecting material

  • Evaluation of Materials in Vacuum Cleaners

    $ 954800Moulded ABS, polypropylene

    Cylindrical shape, 1985

    $1506300Mild SteelMotor driven, 1950

    $ 3801050Wood, canvas, leather ,Mild steel

    Hand powered, 1905

    Cost*Weight (kg)

    Power (W)

    Dominant material

    Cleaner & year

    Wooden SteelPolymeric

    Costs have been adjusted in 1998 values, allowing for inflation [Ref. M. Ashby]

  • 7/24/2009 29

    Material PropertiesGenerally determined through destructive testing of samples under controlled loading conditions.Tensile test

    Apply load & measure deflectionPlotting of stress & strain

    Strength, Youngs modulus, Shear modulus, Fatigue

    strength, resilience, toughness

    00

    0 , lll

    ll>

    =

    0AP

    ==log(l/l0)

  • 7/24/2009 30

    Stress-strain Diagram forMetals

    =E

    modulussYoung'

    tensile

    tensile

    EEEE

    >

    =

    ncompressio

    ncompressio

    Brittle

    Ductileelasticyield

    alproportionelastic

    >

    >

  • Ultimate strength: Largest stress that a material can sustain before fracture

    True stress Engineering stress

    Ductility: Material elongation > 5%.A significant plastic region on the stress-strain curveNecking down or reduction in area.Even materials.

    Brittleness: Absence of noticeable deformation before fracture.

    NOTE: Same material can be either ductile or brittle depending the way it is manufactured (casting), worked, and heat treated (quenched, tempered). Temperature plays important role.

  • 1810373025206030

    415345395520615552552483

    265220295350380345207275

    165172207207207200193200

    Nodular cast ironMalleable cast ironLow carbon steelMedium carbon steelHigh carbon steelFerrite SSAustenite SSMartensitic SS

    Ductility (% EL)Su (MPa)Sy (MPa)E (GPa)Material

    Remark: Choice of material cannot be made independently of the choice of process by which material is to be formed or treated. Cost of desired material will change with the process involved in it.

  • 206030

    552552483

    345207275

    200193200

    Ferrite SSAustenite SSMartensitic SS

    Ductility (% EL)Su (MPa)Sy (MPa)E (GPa)Material

    Ex: A flat SS plate is rolled into a cylinder with inner radius of 100mm and a wall thickness of 60 mm. Determine which of the three SS cannot be formed cold to the cylinder?

    ( ) ( )( ) ( )

    ( ) %1.23100%

    1005160228.8163010025.02

    0

    0

    0

    =

    =

    ====+=+=

    lll

    EL

    rlmmtrl

    fr

    ofr

    i

    ANS: Ferrite SS cannot be formed to the cylinder.

  • 7/24/2009 34

    Torsion Test

    EGEG

    lrG

    5.0)1(2

    0

    +

    =

    =

    Stress strain relation for pure torsion is defined by

    Radius of specimen

    Angular twist in radians

    0.280.330.34

    SteelMagnesiumTitanium

    0.340.350.28

    AluminumCopperIron

    MaterialMaterial

  • 7/24/2009 35

    Fatigue strengthTime varying loadsWohlers strength-life (S-N) diagram

    Tensile & torsion tests apply loads slowly and only once to specimen. Static

    NOTE: Strength at 106 cycles tend to be about 50-60% of static strength

  • 7/24/2009 36

    Impact resistanceIf the load is suddenly applied, the energy absorption capacity (strain energy)

    Resilience: Strain energy present in the material at the elastic limit.Toughness: Strain energy present in the material at the fracture point.

    =

    0

    dU

  • 7/24/2009 37

    Resilience (energy per unit volume)

    ES

    U

    E

    dEdU

    yR

    R

    el

    elel

    20

    2

    00

    21

    2

    =

    =

    ==

    Ex: In mining operation the iron ore is dumped into a funnel for further transport by train. Choose either steel (E=207 GPa, Sy=380 MPa) or rubber (E=4 GPa, Sy=30 MPa) for the design of funnel.

    0.3488, 0.1125

  • 7/24/2009 38

    Toughness (energy per unit volume)

    [ ] futy

    T

    SSU

    dUf

    +=

    =

    21

    toughnessofion approximatan on,intergrati actualfor availableseldom is curvestrain and stressfor expression analytical Since

    T

    0

  • Product must meet all government regulations & societal concerns.

    Substituting a new material needs appropriate design change

    Induction Motor casing

    Grey cast iron. Increasing cost & decreasing availability

    Safety regulations imposed by government.

  • 7/24/2009 40

    Material Selection: Expectations

    Economic & weightless materialsHigh strengthLow temperature sensitivity High wear & corrosion resistanceEnvironmental friendlyControllable friction, stiffness, damping

  • 7/24/2009 41

    There are more than 100, 000 materials???

    How many materials can be accommodate ???

  • Classes of Engineering Materials

    PE, PP, PCPA (Nylon)

    Polymers,elastomers

    Butyl rubberNeoprene

    Polymer foamsMetal foams

    FoamsCeramic foams

    Glass foams

    Woods

    Naturalmaterials

    Natural fibres:Hemp, Flax,

    Cotton

    GFRPCFRP

    CompositesKFRP

    Plywood

    AluminaSi-Carbide

    Ceramics,glasses

    Soda-glassPyrex

    SteelsCast ironsAl-alloys

    MetalsCu-alloysNi-alloysTi-alloys

    Members of class have common features:

    Similar chemical composition Similar properties Similar processing units

  • Relatively High Moduli (E, G, K) & Mechanical STRONG &

    STIFF.

    High ductility allows them to be formed by deformation

    process; accommodate stress concentration by deforming

    and redistributing load more evenly.

    Preferable in cyclic/ Fatigue Load Conditions

    Least resistance to corrosion

    Good Conductors of Electricity & Heat

    METALS

  • Glasses typically have no clear crystal structure

    High moduli

    Hard and wear resistant

    Low thermal conductivity

    Insulate against Passage of Electricity

    Typically 15 times stronger in compression than in tension

    Resist corrosion (low chemical reactivity)

    Brittle and low tolerance for stress concentrations (like holes or cracks) or for high contact stresses (at clamping points).

    CERAMICS, GLASSES

  • CERAMICS

    Strength depends strongly on mode of loading.

    In tension, Fracture strength

    In compression Crashing strength

    Crashing S.= 10-15 Fracture S.

  • 7/24/2009 46

    Electrical Insulating

    Little stronger (~20%) in compression than in tension

    EASY TO SHAPE: complicated parts performing several functions can be mould in a single operation. Generally no finishing is required. Corrosion resistance & low friction coefficient.

    Polymers are roughly 5 Times Less Dense than Metal, which make Strength/Weight Ratio (specific strength) equal to Metals Moduli (~2% of metals).

    POLYMERS, ELASTOMERS

    Large elastic deflections allow snap-fit, making assembly fast & cheap.

  • Thermoplastic POLYMERS

    Strength is identified as the stress at which strain is approximately 1%.

    At Glass transition temperature, upon cooling a polymer transforms from a super-cooled liquid to a solid

    Temperature sensitive properties ( to be used < 200 C)

    Polymer which is tough & flexible at 20C, may be brittle at 4C, yet creep rapidly at 100C.

  • COMPOSITES: Designed for Combination of Best Characteristics (light, strong, stiff, etc.) of Each Component Material

    Graphite- Reinforced Epoxy Acquires Strength from Graphite Fibers while Epoxy Protects Graphite from Oxidation & provides Toughness

    High Price- Relatively Difficult to Form & Join

    Upper temperature limit decided by polymer matrix (generally < 250C)

    Little (~30%) weaker in compression than tension because fiber buckle

  • 7/24/2009 49

    Not tough enough (need bigger Kic)

    Not stiff enough (need bigger E)

    Not strong enough (need bigger y )

    Illustration of Mechanical properties

    Too heavy (need lower )

    Stiff, Strong, Tough, Light

  • Relationships: property bar-charts

    Covalent bond is stiff (S= 20 200 N/m) Metallic & Ionic (15-100 N/m)

    Polymers having Van der Waals bonds (0.5 to 2 N/m). r0~ 3*10-10m)

    Metals Polymers Ceramics Composites

    PEEK

    PP

    PTFE

    WC

    Alumina

    Glass

    CFRP

    GFRP

    Fibreboard

    Y ou n

    g s

    mod

    ulus

    , GPa

    Steel

    Copper

    Lead

    Zinc

    Aluminum

    orSE /= ATOMIC SIZE

    Remark: Property can be displayed as a rank list or bar chart.

    Each bar represents the range of E that material exhibits in its various forms.

  • 7/24/2009 51

    Rank list

    853064400Ti-6-4

    423082880Al-Sic composite

    17592700Al 539

    959037850Steel 4140

    525047150Nodular cast iron

    Rank, MPaRank kg/m3Material

    1-5 ; 1-10

    1-100

  • 7/24/2009 52

    Material Selection

    Best material needs to have maximum overall score (rank)

    OS = weight factor 1 * Rank of Material property 1+ weight factor 2 * Rank of Material property 2+ weight factor 3 * Rank of Material property 3+Weight factor 1+weight factor 2+ = 1.0

  • 7/24/2009 53

    Material Selection: Deciding weighting factorsMaterial Selection: Deciding weighting factors

    1

    1

    1

    1

    4

    4

    1

    2

    3

    5

    Total

    15 Total

    0.26611105

    0.066100004

    0.13310003

    0.210102

    0.33311111

    normalizedDummy5321Attribute

    Fatigue strength, Corrosion resistance, Wettability, Conformability, Embeddability, Compatibility, Hardness, Cost, etc.

  • 7/24/2009 54

    Ex: Components of ring spinning textile machine go through unlubricated sliding at low load but high relative speed (20,000 rpm).

    (1) Increase hardness, (2) Reduce surface roughness, (3) Minimize cost, (4) Improve adhesion to substrate, and (5) Minimize dimensional change on surface treatment/coating

    1

    1

    1

    1

    1

    Dummy

    0.2-0110Dimension

    0.3331-111Adhesion

    0.06700-00Cost

    0.133001-0Roughness

    0.2671011-Hardness

    Weighting factor

    DimensionAdhesionCostRoughnessHardnessDesign

    property

  • Four methods to fulfill the required functions:(1) Plasma sprayed Al2O3 (polished), (2) Carburizing, (3) Nitriding, (4) Boronizing

    7.8779678Boronizing7.298794Nitriding6.8788974Carburizing5.2735529P S Al2O3

    0.20.3330.0670.1330.267Weighting factor

    Weighted total

    DimensionAdhesionCostRoughnessHardnessSurface improvement method

  • 7.8779 320 MPa67 1 microns8 72 HRCBoronizing7.298 300 MPa79 0.5 microns4 50 HRCNitriding6.8788 300 MPa97 1 microns4 52 HRCCarburizing5.2735 100 MPa52 3 microns9 78 HRCP S Al2O3

    0.20.3330.0670.1330.267Weighting factor

    Weighted total

    DimensionAdhesionCostRoughnessHardnessSurface improvement method

    Subjective ranking and weighting impairs the material selection process.

  • Material property- charts: Modulus - Density

    0.1

    10

    1

    100

    Metals

    Polymers

    Elastomers

    Ceramics

    Woods

    Composites

    Foams

    0.01

    1000

    1000.1 1 10Density (Mg/m3)

    You

    ngs

    mod

    ulus

    E, (

    GP

    a)

    Modulus E is plotted against density on logarithmic scale.

    Data for one class are enclosed in a property envelop.

    Some of Ceramics have lower densities than metals because they contain light O, N, C atoms..

  • Optimised selection using chartsIndex

    1/2EM =

    22 M/E =

    ( ) ( ) ( )MLog2Log2ELog =

    Contours of constantM are lines of slope 2

    on an E- chart

    CE

    =C

    E 2/1=

    C

    E 3/1=

    0.1

    10

    1

    100

    Metals

    Polymers

    Elastomers

    Woods

    Composites

    Foams0.01

    1000

    1000.1 1 10Density (Mg/m3)

    You

    ngs

    mod

    ulus

    E, (

    GP

    a)

    Ceramics

    12 3

  • 7/24/2009 59

  • 7/24/2009 60

  • Best material for a light stiff rod, under tension is one that have greatest value of specific stiffness

    (E/) Larger Better For Light & Stiff Tie-rod

    Light & Strong Y/

    Best material for a spring, regardless of its shape or the way it is loaded, are those with the greatest value of (Y)2 /E

    Best thermal shock resistant material needs largest value of Y/E

    PERFORMANCE INDEX

    Combination of material properties which optimize some aspects of performance, is called MATERIAL INDEX

  • Design requirements

    What does the component do ?

    What essential conditions must be met ?

    What is to be maximised or minimised ?

    Which design variables are free ?

    Function

    Objectives

    Constraints

    Free variables

    PERFORMANCE INDICESGROUPING OF MAT. PROPERTIES REPRESENT SOME

    ASPECTS OF PERFORMANCE

    To support load, transmit power,

    store energy

    Cost, energy storage

  • 7/24/2009 63

    FUNCTION

    TIE

    BEAM

    SHAFT

    COLUMN

    Contain pressureTransmit heat

    OBJECTIVEMIN. COST

    MIN. WEIGHT

    MAX. ENERGYSTORAGE

    MIN. IMPACT

    SAFETY

    CONSTRAINTSSTIFFNESSSPECIFIED

    STRENGTHSPECIFIED

    FAILURELIMIT

    GEOMETRY

    INDEX

    M=E0.5/

    WHAT DOES COMPONENT DO?

    WHAT IS TO BE MAX./MIN.?

    WHAT NEGOTIABLEBUT DESIRABLE.?

  • Example 1: strong, light tie-rod

    Strong tie of length L and minimum mass

    L

    FF

    Area A

    Tie-rod is common mechanical component.

    Tie-rod must carry tensile force, F, without failure.

    L is usually fixed by design.

    While strong, need lightweight.

    Hollow or solid

  • =y

    FLm

    Chose materials with smallest

    y

    m = massA = areaL = length = density

    = yield strengthy

    Function

    Objective

    Constraints

    Free variables

    Tie-rod: Rod subjected to tensile force.

    Minimise mass m:m = A L (1)

    Length L is specified Must not fail under load F

    Material choice Section area A; eliminate in (1) using (2):

    (2)yAF /

  • Example 2: stiff, light beam

    m = massA = areaL = length = densityb = edge lengthS = stiffnessI = second moment of areaE = Youngs Modulus

    = 2/1

    2/15

    ECLS12m Chose materials with smallest

    2/1E

    b

    b

    L

    FBeam (solid square section).

    Stiffness of the beam S:

    I is the second moment of area:

    Material choice. Edge length b. Combining the equations gives:

    3LIECS =

    12bI

    4=

    == LbLAm 2Minimise mass, m, where:

    Function

    Objective

    Constraint

    Free variables

  • 7/24/2009 67

    Outcome of screening step is to shortlist of candidates which satisfy the quantifiable information

  • 7/24/2009 68

  • Example 3: stiff, light panel

    m = massw = widthL = length = densityt = thicknessS = stiffnessI = second moment of areaE = Youngs Modulus

    Panel with given width w and length L

    Stiffness of the panel S:

    I is the second moment of area:

    3LIECS =

    12twI3

    =

    tw

    = 3/1

    23/12

    EL

    CwS12m

    L

    F

    == LtwLAm

    Chose materials with smallest

    3/1E

    Material choice. Panel thickness t. Combining the equations gives:

    Minimise mass, m, where

    Function

    Objective

    Constraint

    Free variables

  • Function, Objective, and Constraint Index

    Tie, minimum weight, stiffness E/

    Beam, minimum weight, stiffness E1/2 /

    Beam, minimum weight, strength 2/3/

    Beam, minimum cost, stiffness E1/2/Cm

    Beam, minimum cost, strength 2/3/Cm

    Column, minimum cost, buckling load E1/2/Cm

    Spring, minimum weight for given energy storage YS2/E

    Minimizing cost instead of weight is achieved by replacing density by Cm , where Cm=cost/mass

  • MATERIALS for SPRINGS

    ? OBJECTIVE: MAXIMIZE ENERGY STORAGE

    ? AXIAL SPRINGS, LEAF, HELICAL, SPIRAL, TORSION

    ? PRIMARY FUNCTION: STORING/RELEASING ENERGY

    EWV

    2

    Yield strength for metals and polymers, compressive crushing strength for ceramics, tear strength of elastomers and tensile strength for composites.

  • 7/24/2009 72

  • If there is limit on , rubber ????

    Better than spring steel20-50Rubber

    Cheap & easily shaped1.5-2.5Nylon

    --10-12GFRP

    Comparable in performance with steel, expensive

    15-20CFRP

    Expensive, corrosion resistant

    15-20Ti alloys

    Traditional choice: easily formed and heat treated.

    15-25Spring steel

    Brittle in tension; good only in compression

    10-100Ceramics

    CommentMATERIAL ( )32 .... mMJEM f=

  • 7/24/2009 74

  • ? ELASTIC ENERGY/COST ECM mf

    2

    =

    EM f

    2

    =

    Check minimum required strength.

    Better than spring steel20-50, 20-50Rubber

    Cheap & easily shaped1.5-2.5, 1.5-2Nylon

    --10-12, 3-5GFRP

    Comparable in performance with steel, expensive

    15-20, 4-8CFRP

    Expensive, corrosion resistant15-20, 2-3Ti alloys

    Traditional choice: easily formed and heat treated.

    15-25, 2-3Spring steel

    Brittle in tension; good only in compression

    10-100, 5-40Ceramics

    CommentMATERIAL EM f2=

  • 7/24/2009 76

    GPa

    20-50Rubber

    15-25Spring steel

    10-100Ceramics

  • 7/24/2009 77

  • 7/24/2009 78

  • 7/24/2009 79

    Eon basedSelection

    2 Eon basedSelection

    2

    m

    2

    CEon basedSelection

  • 7/24/2009 80

    Using Minimum criterion on E (> 6.89 GPa)

  • 7/24/2009 81

  • 7/24/2009 82

    Chromium steel

  • 7/24/2009 83

  • ?

    AISI: American Iron and Steel Institute 1019 (?)

  • 7/24/2009 85

    Hardness

    Surface property. Resistance to indentation. Resistance to wear.401 HB, 425 HV and 43 HRC.Sut 3.45 HB 0.2 HB MPa (used for low- or medium carbon steel)Large or thick part Case hardening.Coating..

    Question: Steel member has 250 HB hardness. Estimate ultimate strength.

  • 7/24/2009 86

    AISI: American Iron and Steel Institute 1019 (?)

    Sut 3.45 HB 0.2 HB

    346.75 308.75

    383.25 341.25

    405.15 360.75

    422.4 377.00

  • 7/24/2009 87

    Steel Numbering Systems

    AISI numbers define alloying elements and carbon contents of steel.

    Question: What is composition of AISI 4340.

  • Carbon steel 2

    / E/

  • Carbon steel 3

  • Carbon steel 4

  • Carbon steel 5

    > 1 GPa

    YS > 50% of UTS

  • Low carbon percentage. But high %

    Relatively low E & G

  • Stainless Steels

    Harden to 58-60 HRC for cutting devices, punches and dies

    440CS44004

    Harden to 50-52 HRC for tools that do not require high wear resistance (e.g. injection-molding cavities, nozzles, holding blocks, etc)

    420S42000

    Hardened to 30 HRC and use for jigs, fixtures and base plates

    416S41600

    For rust resistance on decorative an nonfunctional parts

    430S43000

    UsesType

  • Stainless steel 2

    Relatively low / and E/

  • Molybdenum steel

    Nickel chromium Molybdenum steel

  • Strength > 2 GPa

  • 7/24/2009 97

    Free Body Diagrams

    =

    =

    0

    0 Fmequilibriu static of Equations

    M

    Segmenting a complicated

    problem into manageable

    P = 1000 N

    0.25 0.75

    1000 N4000 N3000 N

  • Question: Draw a free body diagram of each component of brake shown in following figure.

  • STRESS

    (a) Normal, tensile (b) normal, compressive; (c) shear; (d) bending; (e) torsion; (f) combined

    JyTI

    yMAP

    b

    sct

    =

    =

    =

    ,,

    Elementary equations. No discontinuity in cross-section. Holes, shoulders, keyways, etc.

    Critical section

  • a. Before assembly

    b. After assembly

    Finite element model to calculate stresses

    High concentration of elements are required to estimate stress level.

  • 7/24/2009 102

    Axial Load on Plate with Hole

    avg

    maxtK

    factorion concentrat Stress

    =

    Plate with cross-sectional plane

    Half of plate with stress distribution.

    Stress Concentration

    hdbP

    )(avg =

  • Geometric discontinuities are called stress raiser. Stress concentration is a highly localized effect.

  • Stress concentration factor for rectangular plate with central hole.

    EX: A 50mm wide and 5mm high rectangular plate has a 5mm diameter central hole. Allowable stress is 300 MPa. Find the max. tensile force that can be applied.

    Ans: d/b = 0.1; Kt=2.7

    A = (50-5)5

    P = 25 kN

  • Stress concentration factor under axial load for rectangular plate with fillet

    EX: Assume H=45mm, h=25mm, and fillet radius r=5mm. Find stress concentration factor.

    Ans: ~1.8

  • Stress concentration factor under axial load for rectangular plate with groove

  • Stress concentration factor under axial load for round bar with fillet

    Gap between lines decrease with increase in r/d ratio.

  • Stress concentration factor for round bar with groove

  • 7/24/2009 109

    Ex: Assuming 80 MPa as allowable strength of plate material, determine the plate thickness

    Maximum stress near fillet

    Maximum stress near hole

    Allowable

    Kt=1.8 Kt=2.1

    bbfillet300

    3050008.1 =

    =

    ( ) bbhole700

    153050001.2 =

    =

    80=allowable b=8.75 mm

  • Stress concentration factor under bending for rectangular plate with fillet

    EX: Assume H=45mm, h=25mm, and fillet radius r=5mm. Find stress concentration factor.

    Ans: ~1.5

  • Stress concentration factor under bending for rectangular plate with central hole

    Concentration factor for thick plate with central hole is higher compared to thin plate with same size hole.

  • Stress concentration factor under bending for rectangular plate with groove

    Decrease in Kt for r/h > 0.25 is negligible.

  • Stress concentration factor under bending for round bar with fillet

  • Stress concentration factor under bending for round bar with groove

  • 7/24/2009 115

    Ex: Assuming 100MPa as allowable stress, determine the shaft dia, d.

    Due to symmetry, reaction force at each bearing = 1250 N.Stress concentration will occur at the fillet.Kt=1.6

    ( )( )33

    35012503232dd

    Mavg

    ==

    ( )( )

    10035012502.516.1 3max =

    ==davg

    Diameter d=41.5 mm

  • Stress concentration factor under torsion for round bar with fillet

    Stress concentration under torsion loading is relatively low.

  • Stress concentration factor under torsion for round bar with groove

  • 7/24/2009 118

    Notch SensitivityRefer slide 43, Metals can accommodate stress concentration by deforming & redistributing load more evenly.

    Some materials are not fully sensitive to the presence of geometrical irregularities (notch) and hence for those materials a reduced value of Kt can be used. Notch sensitivity

    parameter q = 0 means stress concentration (Kf ) factor = 1; and q=1 means Kf = Kt.

    11

    =

    t

    f

    KK

    q

  • Material selection for a plate having central hole and is subjected to Tensile force

    EX: A 50mm wide (b) and h mm high rectangular plate has a 5mm diameter central hole. Length of plate is equivalent to 100mm. Select a lightest but strong material which bear tensile force P = 25 kN. Ans: Mass = (50-5) h 100 ; A = (50-5) h

    ( ) ( ) hhhdbPKt

    1500550

    250007.2 =

    =

    =

    6750M or,

    1500 4500M or,

    =

    =

    =

    1010 log6750log M

    d/b = 0.1; Kt=2.7;

  • 7/24/2009 121

    mmheh

    8.0389.11500 ===

    Commonly available. Economic.

    Stress concentration ???

    Mass reduction ????

  • L1

    P

    L2

    Question: Draw a free body diagram of each component of assembly shown in following figure.

  • 7/24/2009 123

    Contact StressesTwo rolling surfaces under compressive load experience contact stresses.

    Ball and roller bearingsCams with roller followerSpur or helical gear tooth contact

    Gear

    Pinion

  • 7/24/2009 124

    Contact Stresses

    Compressive load elastic deformation of surfaces over a region surrounding the initial point of contact.

    Stresses are highly dependent on geometry of the surfaces in contact as well as loading and materialproperties.

    Stress concentration near contact region is very high. Stress concentration factor ????

  • R1

    R2

    R1

    R2

    Roller against cylindrical line of zero width.Theoretical contact patch is point of zero dimension.

    2

    1

    dbdb

  • 7/24/2009 126

    Contact stresses

    Zero areas Infinite stress. Material will elastically deform and contact geometry will change.Deformation b will be small compared to dimensions of two bodies.

    High stress concentration

  • Contact stresses ..Two special geometry cases are of practical interest and are also simpler to analyze are: sphere-on-sphere & cylinder-on-cylinder.

    By varying radii of curvature of one mating surface, sphere-plane, sphere-in-cup, cylinder-on-plane, and cylinder-in-trough can be modeled.

    Radii of curvature of one element infinite to obtain a plane.Negative radii of curvature define a concave cup or concave trough surface.

  • 7/24/2009 128

    R1

    R2

    R1

    R2

    Finite positive value of R1 & R2

    Infinite values of R1, but finite positive value of R2.

    Positive value of R1, but negative value of R2.

  • Spherical contact

    =

    2

    max 1 brpp

    =b

    drrdpF0

    2

    0 ispatch contact on load applied Total

    [ ]

    ( )

    max2

    3max

    0max222

    0

    22max

    0

    2

    max

    32 or

    32 or

    2 assumingon

    2 or

    12 ispatch contact on load applied Total

    pbF

    bbpF

    dtttbpFtrb

    drrrbbpF

    drrbrpF

    b

    b

    b

    =

    =

    ==

    =

    =

    5.1=tK r

  • 7/24/2009 130

    Cylindrical Contact

    =

    2

    max 1 bxpp

    R1

    R2

    L

    X

    Y

    Z

    Pressure variation along Y-axis is negligible,

    =

    22

    max 1 ay

    bxpp

  • ( )

    max

    2

    0

    2max

    0

    2

    max

    2 or

    cos2 sinlet

    12 ispatch contact on load applied Total

    pLbF

    dbpF b x

    dxbxpLF

    b

    =

    ==

    =

    Cylindrical Contact

    Stress concentration factor = 4/

    max

    max2

    2

    32

    pLbF

    pbF

    contactlcylindrica

    contactspherical

    =

    =How to determine b ???

  • 7/24/2009 132

    How to determine b

    Assume pmax = y and find value of b.

    max

    max

    2

    5.1

    pLF

    b

    pF

    b

    contactlcylindrica

    contactspherical

    =

    =

    Criterion b

  • 7/24/2009 133

    For axi-symmetric point load Timoshenko & Goodier suggested:

    ( )

    ( )

    ( )rF

    E

    yxEF

    zG

    F

    EG

    zy

    z

    21

    10

    )1(24

    14

    )1(2

    x

    2

    1

    221

    3

    2

    222

    =

    +

    +

    +

    =

    +=

    +=

    ++=

    Ref: S. Timoshenko and J.N.Goodier, Theory of elasticity, 2ndEdition, McGraw Hill.

    X

    Y

    Z

  • ( ) ( )

    ( )

    ( ) ( )

    ( ) ( )

    ( ) ( )

    ( )

    ( )22

    1 or

    22sin

    21 or

    12cos21 or

    coscos1 sinb assumingon

    /11or

    /12

    21or

    /1

    21,

    1 sphere of Deflection

    max1

    21

    1

    2

    0max

    1

    21

    1

    2

    0max

    1

    21

    1

    2

    0max

    1

    21

    1

    2

    0max

    1

    21

    1

    0

    2max

    1

    21

    1

    0

    2

    0

    2max

    1

    21

    1

    pE

    b

    pE

    b

    dpE

    b

    dbpE

    r

    drbrpE

    drrr

    brpE

    drrdr

    brpE

    r

    b

    b

    b

    =

    +

    =

    +

    =

    ==

    =

    =

    =

    max2

    32 pbF =

  • ( )22

    1max

    1

    21

    1 p

    Eb

    =max

    2

    32 pbF =

    FEb 1

    21

    11

    83 =

    O

    AB

    C

    21

    2

    11

    2

    111

    2

    111

    22111

    2211

    1

    2 or,

    termsnegligible2111 or,

    11 or,

    or,

    or,

    RbR

    RbR

    RbR

    bRR

    ACOAR

    OCOB

    =

    +

    =

    =

    =

    =

    =

    FE

    Rb1

    21

    13 175.0 =

  • 7/24/2009 136

    A ball thrust bearing with 7 balls is loaded with 700N across its races through the balls. Diameter of spherical balls is 10mm. Assume load is equally shared by all balls. Determine the size of contact patch on the race. Assume Poissons ratio = 0.28 and E=207 GPa.

    Ans: b=118 microns. Size=2*b

    Example

    FE

    Rb1

    21

    13 175.0 =

  • 7/24/2009 137

    Static stress distribution in spherical contact

    ( )

    ( ) ( )

    ( ) ( )

    +

    +++=

    +

    ++++==

    ++=

    3

    2222max

    3

    2222max

    5.122

    3

    max

    5.11215.05.0

    12215.0

    1

    zb

    z

    zb

    zp

    zb

    z

    zb

    zp

    zb

    zp

    yx

    z

    Example: A ball thrust bearing with 7 balls is loaded with 700N across its races through the balls. Diameter of spherical balls is 10mm. Assume load is equally shared by all balls, Determine the stresses developed in balls. Assume Poissons ratio = 0.28 and E=207 GPa.

    Ans: pmax=3.34 GPa. Maximum stress at z=0, 3.34 GPa

    Prob 1: What will happen if poissons ratio of one body is reduced to 0.22. Prob 2: What will happen if Poissons ratio of

    one body is increased to 0.32 and Youngs modulus is reduced to 180 GPa.

    NOTE: All the stresses diminish to < 10% of pmax within z = 5*b.

  • ( )

    ( ) ( )

    ( ) ( )

    +

    +++=

    +

    ++++==

    ++=

    3

    2222max

    3

    2222max

    5.122

    3

    max

    5.11215.05.0

    12215.0

    1

    zbz

    zbzp

    zbz

    zbzp

    zbzp

    yx

    z

    Variation of stresses with Z.

    Four equations. Eight variables. We need four inputs.

    Assume b = 100 m, =0.28, pmax = 2 GPa and z = 0.

  • 7/24/2009 140

    Parametric variation

  • 7/24/2009 141

  • 7/24/2009 142

  • 7/24/2009 143

    Graphs help to find whether function is monotonic or uni-modal.

  • 7/24/2009 144

    ( )

    ( ) ( )

    +

    =

    =

    2

    22

    1

    21

    max

    max2

    22

    2

    114

    deflection Total

    41 similarly

    EEbp

    pE

    b

    ( )22

    1

    1 sphere of Deflection

    max1

    21

    1 p

    Eb

    =

    Two spherical contacting surface

    ( ) ( )

    ( ) ( )

    +

    +

    =

    +

    =+

    +=

    2

    22

    1

    21

    21

    max

    2

    22

    1

    21

    max2

    2

    1

    2

    2

    2

    1

    2

    11

    21

    214

    or

    11422

    or

    22 radii, geometric of in terms presented becan deflection Total

    EERR

    pb

    EEbp

    Rb

    Rb

    Rb

    Rb

    max2

    32 pbF =

  • 7/24/2009 145

    ( ) ( )

    +

    +

    =2

    22

    1

    21

    21

    2 11

    21

    21

    5.1

    4 or

    EERR

    bF

    b

    ( ) ( )

    +

    +

    =2

    22

    1

    21

    21

    3 11114

    3 or EE

    RR

    Fb

    Question: Two carbon steel balls (AISI 1030 tempered at 650C), each 25 mm in diameter, are pressed together by a force F = 100N. Find the maximum value of compressive stress. Poissons ratio = 0.285, Youngs modulus = 208 GPa.

    Answer: 1.85 GPa.

    ( )

    ++= 5.122

    3

    max 1zb

    zpz

  • Question: Two balls, each 25 mm in diameter, are pressed together by a force F = 100N. Find the maximum value of compressive stress. For one material (AISI 1030 tempered at 650C ), Poissons ratio = 0.285 and Youngs modulus = 208 GPa. Other ball is made of synthetic rubber (Poissons ratio = 0.48 and Youngs modulus = 2.0 MPa)

    Maximum stress is < 1.5 MPa, but b ~ 45% of ball radius.

  • Question: One carbon steel balls (AISI 1030 tempered at 650C), having diameter = 25, is pressed against a AISI 1030 steel flat surface by a force of F = 100N. Find the maximum value of compressive stress. Poissons ratio = 0.285, Youngs modulus = 208 GPa.

    Conclusion: Increase radius of one of surface, reduces the value of maximum compressive stress.

  • 7/24/2009 148

    Cylindrical Contact

    bzp

    p

    p

    EERRL

    Fb

    pLbF

    bxpp

    y

    zx

    786.0304.0

    2

    1111

    4

    2

    1

    max@

    maxmax

    maxmax

    maxmaxmax

    2

    22

    1

    21

    21

    max

    2

    max

    ==

    =

    ==

    +

    +

    =

    =

    =

    Example: An overhead crane wheel runs slowly on a steel rail. Find the size of the contact patch, and stresses? What is the depth of max shear stress?

    Given: Diameter of wheel and length are 150 mm and 20mm respectively. Assume radial load is 10000N. Assume Poissons ratio = 0.28 and E=207 GPa.

  • 7/24/2009 149

    Stress distribution in Cylindrical Contact

    +=

    +

    +=

    +

    =

    bzbzp

    bz

    bzb

    zp

    bzp

    y

    x

    z

    2/12

    2/1

    21

    /1

    22max

    22

    22

    max

    22max

    Problem: A 200-mm diameter cast iron (=0.26, E = 80 GPa) wheel, 55 mm wide, rolls on a flat steel (=0.29, E = 210 GPa) surface carrying a load of 10.0 kN. Find the maximum value of all stresses. Evaluate all three compressive stresses (in x-, y- and z- directions) at z = 0.2 mm below the wheel rim surface.

  • 7/24/2009 150

    Answer

    MPap

    MPap

    p

    MPaLb

    Fp

    meEE

    RRL

    Fb

    y

    zx

    76.57304.0

    992

    1902

    409.61111

    4

    maxmax

    maxmax

    maxmaxmax

    max

    2

    22

    1

    21

    21

    ==

    ==

    ==

    ==

    =

    +

    +

    =

  • 7/24/2009 151

    Problem

    The figure shows a hip prosthesis containing a femur (ball shaped having diameter 50 mm) and cup (having diameter 54 mm). The femur is coated with 500 microns thick titanium (=0.35, E=90 GPa) material and cup is made of plastic (PEEK: =0.378, E=3.7 GPa) . Assume normal load transferred from femur to cup is 300 N. Find the maximum values of stresses.

  • 7/24/2009 152

    Failure of Machine ElementThere are only two ways in which an element fails:

    ObsolescenceLoss of function

    Element losses its utility due to:Change in important dimension due to wear.

    Change in dimension due to yielding (distortion)

    Breakage (fracture).

    Jamming (friction)Brittle material, fatigue

    Ageing, wrong choice of materials

  • Yielding (distortion)

    Wear

    FractureJamming

  • 7/24/2009 154

    Failure Theories

    Often failure mechanisms are complicated involving effect of tension, compression, shear, bending and torsion.

  • 7/24/2009 155

    Failure Theories for yielding & fracture

    First step towards successful design is obviating every possible failure.Failures are often associated with multi-axial stress states. On the basis of comparative study between theoretical and experimental work, few theories to predict failure have emerged. Each theory has its own strengths and shortcomings and is best suited for a particular class of material and kind of loading (static/dynamic).

  • 7/24/2009 156

    Failure of Ductile Materials under Static Loading

    Distortion energy (von Mises) theory and the maximum shear stress theory agree closely with experimental data.Distortion energy theory is based on the concept of relative sliding of materials atoms within their lattice structure, caused by shear stress and accompanied by shape distortion of the element.

  • 7/24/2009 157

    Von-Mises (Distortion energy) Theory

    ( )33221121

    21 umeenergy/volStrain

    ++=

    =

    U

    U

    ( )

    ( )

    ( )1233

    3122

    3211

    1

    1

    1

    =

    =

    =

    E

    E

    E

    To avoid complexity, the principal Stresses and principal strainThat act on planes of zero

    Shear stress have been considered.

    ( )hd UUU

    EU

    +=

    ++++

    =312321

    23

    22

    21

    221

  • 7/24/2009 158

    Finding Distortion Energy

    ( )hd UUU

    EU

    +=

    ++++

    =312321

    23

    22

    21

    221

    ( )

    [ ]

    3

    2123

    221

    321

    2

    222

    ++=

    =

    ++++

    =

    h

    hh

    hhhhhh

    hhhh

    EU

    EU

    [ ]31232123222131 +++=

    EUd

  • 7/24/2009 159

    von-Mises Theory

    [ ][ ]312321232221

    31232123

    22

    21

    2

    31

    31

    ++=

    +++

    =+

    =

    y

    yd

    SE

    SE

    U

    2

    31

    yd SEU +=

    [ ]312321232221safety offactor consider weIf

    ++NSy

  • 7/24/2009 160

    Maximum Shear Stress Theory (Tresca Theory)

    Evaluate maximum shear stress

    Compare with shear strength of material (Sys)If we consider factor of safety (N) then compare with (Sys/N)

    231

    max =

    How to find principal stresses and estimate factor of safety.

  • 7/24/2009 161

    Principal Stresses

  • 7/24/2009 162

    Principal Stresses

    ( )( ) ( )( ) ( )( ) ( )( )( ) ( ) ( ) ( )

    cossincossinsincos

    sincoscossinsinsincoscos0

    22yxxyyx

    yxxyyx AAAAAF

    +++=

    +++=

    =

    ( )

    ( )

    2sin2cos22

    2sin2

    2cos12

    12cos

    xyyxyx

    xyyx

    +

    +

    +=

    +

    +

    +=

    ( )

    ( )

    yx

    xy

    yxxy

    yxxy

    =

    =

    =

    22tan

    2sin2

    2cos0

    2sin2

    2cos

  • Principal Stresses

    ( ) 2sin2cos22 xyyxyx +

    +

    +=

    yx

    xy

    =2

    2tan

    ( ) ( )

    ( )

    +

    +=

    +

    +=

    22

    21

    22

    22,

    22

    2

    xyyxyx

    xyyxyx

    Similarly we can find 3. In practice 1 , 2 , and 3 are arranged in descending order of magnitude.

  • 7/24/2009 164

    Factor of Safety

    FOS is a ratio of two quantities that have same units:

    Strength/stress Critical load/applied loadLoad to fail part/expected service loadMaximum cycles/applied cyclesMaximum safe speed/operating speed.

    NOTE: FOS is deterministic. Often data are statistical and there is a need to use Probabilistic approach.

  • 7/24/2009 165

    Variation in Material Strength (MPa)

    29.17812.5725 - 9001060

    25.00725.0650 - 8001050

    20.83627.5565 - 6901040

    19.17522.5495 - 6101030

    34.17967.5865 - 10701095

    18.33920865 - 9751080

    St. DeviationMeanRangeMaterial (AISI, rolled)

  • Probability density functionEx: Measured ultimate

    tensile strength data of nine specimen are: 433 MPa, 444, 454, 457, 470, 476, 481, 493, and 510 MPa. Find the values of mean, std. dev., and coefficient of variation. Assuming normal distribution find the probability density function.

    ( )

    ( ) 1

    234.241

    05194.0 C variationof Coeff.

    34.2467.468

    2

    34.2467.468

    21

    s

    sS

    =

    =

    ===

    ==

    +

    dSSf

    eSf

    MPaMPa

    S

    s

    s

  • 7/24/2009 167

    EX. NOMINAL SHAFT DIA. 4.5mmNUMBER OF SPECIMEN 34

    4.58mm0.0097

    d

    d

    4.59,4.34,4.5796,4.50, 4.582,4.58474.5948

    6

    4.5294

    0.0987

    ( )1

    /22

    =

    NNdd ii

    d

    = ddid

    d

    edf

    21

    21)(

    Conclusion: Variation in stress level occurs due to variation in geometric dimensions.

  • 7/24/2009 168

    Ex: Consider a structural member( ) subjected to a static load that develops a stress ( ). Find the reliability of member.

    Deterministic FOS = 40/30. 100% reliability.

    ss ,40=

    ,30=

    NOTE: Reliability is probability that machine element will perform intended function satisfactorily.

    830

    ==

    640

    ==

    s

    s

    10,10 == QQ

    1086

    10304022 =+=

    ==

    Q

    Q

  • 7/24/2009 169

    xQyxQ

    xyQyxQxCQ

    CxQCQ

    1===

    =+=

    ==

    x

    yx

    yx

    yx

    x

    x

    CCC

    1

    +

    ALGEBRAIC MEAN STD. DEVIATIONFUNCTIONS

    2

    22222

    2222

    22

    0

    xx

    yyxxy

    yxxy

    yx

    x

    xC

    +

    +

    +

  • 7/24/2009 170

    Margin

    ( )f

    f

    P RQP P

    =

  • Q

    Q

    Z

    Z

    Q

    Q

    where

    dZeR

    QZ

    =

    =

    =

    +

    0

    21

    Z

    21

    0

    2

    10

    10

    =

    =

    Q

    Q

    110

    100

    0at

    0 =

    =

    =

    Z

    Q

    =

    0 221

    21 z z dZeF

    0

  • 7/24/2009 172

    Z-Table provides probability of failure

    In the present case Probability of failure is 0.1587 & reliability is .8413.

  • 7/24/2009 173

  • 7/24/2009 174

    Comparison

    FOS equivalent to 1.33 is insufficient for the present design, therefore there is a need to increase this factor.Selecting stronger material (mean value of strength = 50 units!!!!)

  • 7/24/2009 175

    ( ) ( )MPaMPaS y 15,184 & 32,270:arebar tensilea of Stress andStrength :Ex

    ==

    dzeRz

    243.22

    211design ofy Reliabilit =

    R = 1-0.0075 ???? Ref: Probabilistic Mechanical Design, Edward B. Haugen, 1980.

    Prob: A steel bar is subjected to compressive load. Statistics of load are (6500, 420) N. Statistics of area are (0.64, 0.06) m2. Estimate the statistics of stress.Ans: (10156, 1156.4) Pa.

  • 7/24/2009 176

    Ex: A round 1018 steel rod having yield strength (540, 40) MPais subjected to tensile load (220, 18) kN. Determine the diameter of rod reliability of 0.999 (z = -3.09).

    MPad

    MPad

    MPaMPa s

    22

    s

    4/18000;

    4/220000

    40;540Given

    ==

    ==

    Q

    Q

    Z

    Z

    Q

    Q

    where

    dZeRQ

    Z

    =

    =

    = +

    0

    21

    Z

    21;

    0

    2

    2

    22

    2

    7200040

    880000540

    +=

    =

    d

    d

    Q

    Q

    2

    2

    22 880000540720004009.3

    dd =

    + d = 26 mm

  • Example: Stress developed in a machine element is given by:

    Given P = (1500, 50) N, Strength = (129, 3) MPa, L1=(150, 3) mm, L2=(100, 2) mm. Assume std. dev. of d is 1.5% mean value of d. k = 0.003811.

    Determine distribution of d if the maximum probability of machine-element-failure is 0.001

    = =ni xi

    ix1

    22

    :by expressed isfunction complex a ofdeviation Standard

    ( )( )22213 344/ LLkdP +=

    ( ) ( ) ( ) ( )

    [ ]

    3

    2/13

    2/1

    22

    32

    2

    3

    22

    42

    2

    3

    2/1

    22

    2

    22

    1

    22

    22

    1136200

    290472614204183012291.11

    002.085216003.0170430015.04136355022724

    21

    d

    d

    dd

    dd

    LLdP

    e

    de

    LLdP

    =

    +++=

    +

    +

    +

    =

    +

    +

    +

    = Statistically independent

  • ( )( )3

    22

    21

    3

    34087000344/

    d

    LLdP k

    =

    +=

    ( )

    ( )

    m 001.0 m 06686.0

    11031417482.11363000

    113620063

    340870006129009.3

    2

    3

    2

    32

    21

    2

    32

    3

    ==

    =

    +

    +

    ==

    d

    d

    dd

    d

    d

    e

    eZ

    Calculating FOS = Strength/stress FOS =129/114=1.13

  • 7/24/2009 179

    Question: Estimate all the stress at point A of L shape rod (diameter = 6 mm), which is made of steel (yield strength = 300 MPa). Assume plate is rigidly mounted (deflection of plate is negligible). Estimate the safety of plate.

    Plate

    L shape rod

  • 7/24/2009 180

    Question: Determine the diameter of L shape rod, which is made of steel (yield strength = 300 10 MPa). Assume plate is rigidly mounted (deflection of plate is negligible), standard deviation of load components is 5% of mean values, standard deviation in dimensions is 0.1% of mean values, and expected reliability of rod is 99%.

    Plate

    L shape rod

  • 7/24/2009 181

    Failure Theories for Brittle material under Static loading

    Brittle material fracture rather than yield.Fracture in tension is due to normal tensile stress.

    Shear strength of brittle material can be greater than their tensile strength, falling between their compressive and tensile values.

    Conclusion: Different failure modes are due to the difference in relative shear and tensile strengths between the ductile and brittle materials.

  • 7/24/2009 182

    Maximum Normal Stress Theory

    NSut1

    Maximum tensile stress Factor of

    safety

    Ultimate tensile strength. Often referred as tensile strength.

    NSuc3

  • 7/24/2009 183

    Compressive & Tensile Strength

    19201.68Tool steel900.58High Si Cast iron4767.93Silicon Nitride1653.2Silicon400.397Boron Nitride5155.158Boron Carbide2182.183Aluminium Nitride1281.667Alumina

    Tensile (MPa)

    Compressive (GPa)

    Material

  • Tensile

    Tensile

    Compressive

    Compressive

    0

    Coulomb Mohr theory

  • 7/24/2009 185

    03211 >>> ifNSut

    3213 0 >>> ifNSuc

    Coulomb Mohr Theory

    3131 01 >> if

    NSS ucut

  • 7/24/2009 186

    Ex: A round cantilever bar made of brittle material experience torsion applied to the free end. Assume that the compressive strength is twice the tensile strength. Express failure stress in terms of strength.

    ii

    == 31i

    and).( stress torsional tosubjected isBar :Given

    ( )NSS

    or

    NSSas

    ut

    i

    ut

    i

    ucut

    12

    10 3131

    >>

    NSut

    i 32

  • 7/24/2009 187

    Tolerances

    03.002.004.0

    04.000.0

    00.004.0

    20202020 +

    +

    Machine elements are manufactured / fabricated with some tolerance on their basic (normal size, i.e. 20mm) dimensions.

    Tolerance: permissible variation in the dimensions of a component.Tolerance: Unilateral or bilateral.

    Inaccuracies of Manufacturing

    methods

    01.0;20 == dd

  • 7/24/2009 188

    FitsCareful decision on tolerance is important for assembling two components.

    Relationship resulting from the difference between sizes of components before assembly is called a Fit.Clearance fit: positive gap between hole and shaft. Relative movement is possible. Interference fit: Negative gap. Relative movement is restricted.Transition fit: border case. Either a clearance or interference fit, depending upon actual values of dimensions of mating components.

  • :Calculate assembled. are )(20pin -crank a and )(20 bearingA :Prob 061.0

    0.040000.0

    0.013+

    Maximum and minimum diameters of the crank-pin and bearing.

    Maximum and minimum clearance between crank-pin and bearing.

    Known as 20H6-e7

    939.19 96.19 00.20 013.20

  • 7/24/2009 190

    :Calculate ).(20 housing ain inserted is )(20 A valve :Prob 000.0

    0.021035.0

    0.048 +++

    Maximum and minimum diameters of the valve seat and housing-hole.

    Maximum and minimum interference between the seat and its housing.

    Known as 20H7-s6

    048.20035.2000.20 021.20

  • 7/24/2009 191

    B.I.S. (Bureau of Indian Standards) System of Tolerances

    As per B.I.S. tolerance is specified by two parts (i.e. H6, e7). :

    Fundamental deviation: Location of tolerance zone w. r. t. Zero line.Represented by an alphabet (capital or small). Capital letters describe tolerances on hole, while small letters describe tolerance on shaft.Magnitude: by a number, often called grade. There are eighteen grades of tolerance with designations IT1, IT2,, IT 18. IT is acronym of International Tolerance.

  • Letter Symbols for Tolerances

    H6-e7

    H7-s6

    a

    c

    e

    g

    j

  • 7/24/2009 193

  • 7/24/2009 194

  • 25813050314200-22523612250314180-20021010843273160-18019010043273140-1601709243273120-1401447937233100-120124713723380-100102593220265-8087533220250-6570432617240-5060432617230-4048352215224-3041352215218-2433281812114-1833281812110-142823151016-10231912813-618146400-3

    uspnkBasic series

  • 7/24/2009 19613001150100087074062052043036030025014

    81072063054046039033027022018014013520460400350300250210180150120100123202902502201901601301109075601121018516014012010084705848401013011510087746252433630259817263544639332722181485246403530252118151210732292522191613119866232018151311986545161412108765443412108654432.52.523875432.52.521.51.51.2264.53.52.521.51.51.2110.81

    IT Grade

    315250180120805030181063inc.

    2501801208050301810631over

    Nominal Sizes (mm)

  • Hot rolling, Flame cutting

    Sand Casting

    Forging

    Die Casting

    Drilling

    Cold Rolling, Drawing

    Extruding

    Planning, Shaping

    Milling

    Sawing

    Boring, Turning

    Reaming

    Broaching

    Plan grinding

    Diamond turning

    Cylindrical grinding

    Super finishing

    Honing

    Lapping

    1615141312111098765432IT Grade

  • 7/24/2009 198

    Hole 110H11 Minimum = 110mm + 0mm = 110.000mm ...Maximum = 110mm + (0+0.220) = 110.220mm Resulting limits 110.000/110.220Tolerance of hub, tlh=220m

    Shaft 110e9...Maximum = 110mm 0.072=109.928mm...Minimum = 110mm - (0.072 +0.087) = 109.841mmResulting limits 109.841/ 109.928Tolerance of shaft, tls=87m

    Examples

  • 7/24/2009 199

    Hole 34H11 Minimum = 34mm + 0mm = 34.000mm ...Maximum = 34mm + (0+0.160) = 34.160mm Resulting limits 34.000/34.160Tolerance of hub, tlh=160m

    Shaft 34c11...Maximum = 34mm 0.120=33.880mm...Minimum = 34mm - (0.120 +0.160) = 33.720mmResulting limits 33.880/ 33.720Tolerance of shaft, tls=160 m

    Examples 34H11/c11

  • 7/24/2009 200

    Examples:Clearance Fit: In hydrodynamic bearings a critical design parameter is radial clearance between shaft and bearing. Typical value is 0.1% of shaft radius. Tolerances cause additional or smaller clearance. Too small a clearance could cause failure; too large a clearance would reduce load capacity.Interference Fit: Rolling-element bearings are generally designed to be installed on a shaft with an interference fit. Slightly higher interference would require significant force to press bearing on shaft, thus imposing significant stresses on both the shaft and the bearing.

  • 7/24/2009 2011 2 3 4 5 6 7 8 9 10

    0

    100

    200

    300

    400

    500

    600

    700Effect of clearance on load

    0.001 R * Factor

    Load

    2

    1

    rCLoad

  • Interference Fit

    =0.001d mm

    =0.0005d mm

    =0.00025d mm

    =0.00 mm

    Semi-permanent jointHeavy

    Considerable pressure is required to assemble /disassemble joints.

    Medium

    Suitable for low speed and light duty joints

    Light

    Require light pressure. Suitable for stationary parts

    Wringing

    For 20mm shaft dia, interference = 20 microns

    Utilized to minimize the

    need for keyways.

  • 7/24/2009 203

    Press FitPressure pf is caused by interference between shaft & hub. Pressure increases radius of hole and decreases radius of shaft.

    rf

    rs

    rh

    rs

    rh

    rf

    rfrfpf

    pf

    Base-line

  • 7/24/2009 204

    ( ) ( )

    ( )

    ( )( ) 02

    sin2 balance Force

    strain Radial

    strain ntialCircumfere

    =

    ++=

    =

    =

    +=

    ==

    +=

    dzdrddzrddzddrrd

    Erdr

    drr

    Erdrdrdr

    rrr

    rrr

    rr

    r

    rrr

  • ( )( )

    drdror

    dzddzddzddr

    drgrearrangin

    dzdrddzrddzddrrd

    rr

    rr

    rrr

    +=

    =+

    =

    ++

    0

    02

    sin2 ( )

    ( )Er

    Errr

    rr

    =

    =

    Edr

    dr

    r

    Edr

    dr

    rr

    rrr

    rr

    rr

    =

    +

    =

    ++

    +=

    drd

    drddr

    dr

    Er

    drdr

    Er rr

    r

    rr

    rr

    2

    1 22

    03 22

    =+dr

    drdr

    d rr ( ) 02 2

    2

    =+drrd

    drd rr

  • ( ) 02 22

    =+drrd

    drd rr ( ) 02 1 =++ Cdr

    rd rr

    ( ) 012

    =+ rCdrrd r

    02

    02

    221

    2

    2

    12

    =++

    =++

    rCC

    CrCr

    r

    r

    Two conditions are required to express radial stress in terms of radius.

    oor

    iir

    rratprratp

    ====

    oo

    ii

    prCC

    prCC

    =+

    =+

    221

    221

    2

    2

  • ( ) ( )

    ( ) ( )22

    222

    22

    222

    stress ntialCircumfere

    stress Radial

    io

    iooiooii

    io

    iooiooiir

    rrpprrrrprp

    rrpprrrrprp

    =

    +

    =

    CASE I: Internally Pressurized (Hub)-

    ( )( )

    ( )( )22

    22

    22

    22

    1 stress Radial

    1 stress ntialCircumfere

    fo

    offr

    fo

    off

    rrrrrp

    rrrrrp

    =

    += ( )

    fr

    fo

    off

    p

    rrrrp

    =

    +=

    max,

    22

    22

    max,

    ( )Er

    rh

    f

    rh

    ==strain ntialCircumfere

    rf

    f

    rh

    fo

    off

    rrrrr

    Ep

    h

    =

    +

    += 22

    22

    max,

  • ( )

    ( )

    =

    +

    =

    22

    22

    22

    22

    1 stress Radial

    1 stress ntialCircumfere

    if

    iffr

    if

    iff

    rrrrrp

    rrrrrp

    CASE II: Externally Pressurized (shaft)-

    rf

    ( )Er

    rs

    f

    rs

    ==strain ntialCircumfere

    f

    rs

    if

    fi

    s

    f

    rrrrr

    Ep s =

    += 22

    22

    max,

    fr

    ifff

    p

    rrrp

    =

    =

    max,

    222

    max,2

  • ( ) ( )

    +++

    +=

    =

    s

    s

    ifs

    fi

    h

    h

    foh

    foff

    rsrh

    ErrErr

    ErrErr

    pr

    22

    22

    22

    22

    r

    r

    or

    ceinterferen Total

    Ex: A wheel hub is press fitted on a 105 mm diameter solid shaft. The hub and shaft material is AISI 1080 steel (E = 207 GPa). The hubs outer diameter is 160mm. The radial interference between shaft and hub is 65 microns. Determine the pressure exercised on the interface of shaft and wheel hub.

    ( ) ( )

    ( )

    =

    ++

    +=

    22

    2

    r

    22

    22

    22

    22

    r

    2 :solid isshaft If

    :materials same of made areshaft and hub If

    fo

    off

    if

    fi

    fo

    foff

    rrr

    Epr

    rrrr

    rrrr

    Epr

    ANS: pf =73 MPa

  • Through interference fit torque can be transmitted, which can be estimated with a simple friction analysis at the interface.

    ( )( )

    LdpTTorque

    LdpFApNF

    f

    fff

    ff

    2

    2

    =

    =

    == = coefficient of friction

    Abrasion Adhesion

  • 7/24/2009 211

    C.A.Coulomb 17811)Clearly distinguished between static & kinetic friction

    2)Contact at discrete points.

    3)Friction due to interlocking of rough surfaces

    4)No adhesion5)f func(v)

  • 7/24/2009 212

    PLOUGHING Effect

    Assume n conical asperities of hard metal in contact with flat soft metal, vertically project area of contact:

    ( )2*5.0 rnA =HrnW )*5.0( 2= HnrhF )(=

    cot2=

  • For = 45 = 0.6366For = 60 = 0.3676For = 80 = 0.1123

    Slope of real surfaces are nearly always less than 10 (i.e. > 80), therefore < 0.1.

  • ADHESION Theory

    Two surfaces are pressed together under load W.

    They deformed until area of contact (A) is sufficient to support load W. A = W/H.

    To move the surface sideway, must overcome shear strength of junctions with force F F = A s

  • 7/24/2009 215

    For most of materials H = 3y & s = y /1.7Expected value of =.2

    HAW real= sAF real= Hs

    =

    On steel (0.13%C)Silver 0.5Copper 0.8Indium 2.0Lead 1.2

    Metals on it self Gold 2Silver 1Copper 1Chromium 0.4Lead 1.5

    Shear stress of softer of contacting materials

  • Junction Growth

    Constant F A ????

  • Limiting Junction GrowthPresence of weak interfacial films. Assume shear stress, i.

    maxmax AF i=

    2max

    22maxmax

    )4( AA

    WF

    iy

    i

    =

    )(2 22 iyi

    Average shear strength

  • 7/24/2009 218

  • 7/24/2009 219

  • 7/24/2009 220

  • 7/24/2009 221

    Fatigue FailureFatigue failure looks brittle even in ductile metals. Parts often fail at stresses well below the ultimate strength of mat.

    High factor of safety.

    Rankine published Causes of unexpected breakage of railway axles in 1843, postulating that materials experience brittleness under fluctuating stresses.

    Aloha Airlines flight 243, a Boeing 737-200, lost about 1/3 of its cabin top while in flight at 8.5 km. This failure, which happened in 1988, was caused by corrosion assisted fatigue.

  • Machine elements subject to time varying repeated loading

    2

    2minmax

    minmax

    =

    +=

    a

    m

    Ex: A particular fiber on surface of shaft subjected to bending loads undergoes both tension & compression for each revolution of shaft. If shaft is part of electric motor rotating at 1440 rpm, the fiber is stressed in tension & compression 1440 times each minute.

    Stresses repeat a large number of times, hence failure is named as Fatigue failure.

  • 7/24/2009 223

    Fatigue Failure

    Fatigue is a concern whenever cyclic/fluctuating loading is present.

    Loading may be axial (tensile or compressive), flexural (bending) or torsional.

    Appearance similar to brittle fracture

    Damage accumulating phenomenon (progressive fracture).

  • 7/24/2009 224

    Beach marks highlight advances of a fatigue crack (s)

    Crack initiation

    Crack growth

    Fracture

  • Crack initiation, propagation, and fracture.

    Crack growth

    FastFracture

  • 7/24/2009 226

    Crack initiation CG FF

    CI CG FF

    Normal Element

    Faulty (stress raisers, material defects) Element

    CI: Crack initiation

    CG: Crack growth

    FF: Final fracture

    Relative time

  • 7/24/2009 227

    Normal element

    Life 32,000 Hours

    Removed before final fracture

    Faulty element

    Life 100 hours

    Unexpected final fracture

  • 7/24/2009 228

    Low nominal stress results in a high ratio of fatigue zone to FF zone

    High nominal stress is indicated by low ratio of fatigue zone to FF

  • 7/24/2009 229

    Fatigue RegimesLow cycle fatigue ( 103 cycles)

    Latches on automobile glove compartmentStuds on truck wheels

    Since static design often uses Yield strength (< Sut) in defining allowable stresses, therefore static approaches are acceptable for designing low cycle component.

    High cycle fatigue (> 103 cycles)Car door hinges Aircraft body panels

    axialSSbendingSS utlutl 75.0;9.0 ==

  • 7/24/2009 230

    Fatigue StrengthMeasured by testing idealized (R. R. Moore) standard specimen on rotating beam machine.

    Highly polished surface.If specimen breaks into two equal halves, test is indicative of mat. Fatigue strength. Otherwise, it is indicative that material or surface flaw has skewed results.Test specimen is subjected to completely reversed bending stress cycling at 66% Sut and cycles to fatigue are counted.

    Procedure is repeated on other identical specimens subjected to progressively decreasing stress amplitude.

    Dimensions in inches

  • 7/24/2009 231

    S-N (Wohler) diagram

    Plot of fatigue strength (S) vslogarithm of number of cycles (N)

    Indicate whether material has endurance limit (possibility of infinite life) or not.

    Strength - Cycles German engineer

  • 7/24/2009 232

    Endurance Limit ( )eS

    TorsionSSAxialSSbendingSS

    ute

    ute

    ute

    29.045.05.0

    SteelFor

    ===

    ute

    ute

    ute

    ute

    ute

    SScyclesalloys

    SScyclesalloys

    SScyclesalloysNickel

    SScyclesalloysCopper

    SScyclesalloysMagnesium

    45.0)10*5(Aluminum

    55.0)10(Titanium

    42.0)10(

    38.0)10(

    35.0)10(

    8

    7

    8

    8

    8

    =

    =

    =

    =

    =NOTE: It is always good engineering practice to conduct a testing program on materials to be employed in design.

  • 21 loglog

    by expressed

    kNkSbecanSstrengthFatigue

    f

    f

    +=

    Number of cycles to failure, N

    Fatigu

    e st

    reng

    th

  • 7/24/2009 234

    Example: The ultimate tensile strength of an axially loaded steel member is 1080 MPa. Find out fatigue strength as a function of number of cycles (103

  • 7/24/2009 235

    Endurance limit modification factors

    Endurance limit is measured under best circumstances, which cannot be guaranteed for design applications.

    Components endurance limit must be modified or reduced from materials best-case endurance limit.

    Stress concentration factor, surface finish factor, size factor, reliability factor, temperature factor, etc.

    Design factors

  • 7/24/2009 236

    Reliability FactorReliability factor obtained from Table can be considered only as a guide (academic) because actual distribution varies from one material to other. For practical applications, originally determined data are required.

    1.00.8970.8680.8140.7530.7020.6590.620

    50909599

    99.999.99

    99.99999.9999

    Reliability factor, kr

    Probability of survival, %

  • 7/24/2009 237

    Surface Finish Factor

  • 7/24/2009 238

    Surface Finish Factor

    -0.995272Forged-0.71857.7Hot rolled

    -0.2654.51Machined or cold drawn-0.0851.58Ground

    Exponent bConstant aFinishing method

    ( )butfinish SaK MPain =

    Ex: A steel has Sut = 520 MPa. Estimate Kfinish for a machined surface.

    ANS: 0.86

  • 7/24/2009 239

    Temperature Factor

    0.672550C1.0250C

    0.768500C1.02200C0.843450C1.025150C

    0.900400C1.02100C0.943350C1.0150C

    0.975300C1.0020CKtempTemperatureKtempTemperature

    NOTE: Initially increase in temperature causes the redistribution of stress-strain profiles at notches or stress concentration features, hence increases the fatigue strength.

  • 7/24/2009 240

    Stress Concentration FactorSCF is slightly lesser than SCF under static loading.

    Many mat. Relieve stress near a crack tip through plastic flow.

    To avoid complexity in the present course assume, SCF under fatigue loading = SCF under static loading.

  • 7/24/2009 241

    Size factor, Ksize

    18.1 mm

    MPaSMPaS

    e

    e

    5.414.06.103

    factorsion concentrat stress and size re, temperatufinish, ,y reliabilit including S Corrected 'e

    ==

  • Example: A rod of steel (Sut=600 MPa) at room temperature is subjected to reversed axial load of 100 kN. The rod is machined on lathe and expected reliability is 95%. There is no stress concentration. Determine the diameter of rod for an infinite life.

    STEP 1: Estimate endurance limit of mat. 0.45*600 = 270 MPa.

    STEP 2: Estimate endurance limit of plate.

    Find modification (i.ereliability, finish, temp., stress concentration and size) factors.

    0.868, 0.77, 1, 1, 1.24 d-0.107

    ANS: Diameter > 30 mm

  • Example: A rotating bar made of steel (Sut=600 MPa) is subjected to a completely reversed bending stress. The corrected endurance limit of component is 300 MPa. Calculate the fatigue strength of bar for a life of 80,000 rotations.

    ( ) ( )( ) ( )

    ( )MPaS

    NSkk

    kk

    kNkSbecanSstrengthFatigue

    f

    f

    f

    f

    372

    9877.2log0851.0log10log600*9.0log

    10log300log

    loglog

    by expressed

    23

    1

    26

    1

    21

    =

    +=+=

    +=

    +=

    NOTE: We can state that at stress value = 372 MPa, life of bar is 80,000 rotations.

  • Question: Ultimate tensile strength of a bolt, subjected to axial tensile loading, is 1080 MPa. A 20% decrease in its stress would increase its life by 50000 cycles.

    Determine the bolt-life.

    ( ) ( ) ( ) ( )[ ]

    +=

    +=

    +=

    50000log)8.0log(

    50000log*

    *8.0log

    50000loglog**8.0loglog

    1

    1

    1

    NN

    k

    NNk

    SS

    NNkSS

    f

    f

    ff

  • 7/24/2009 250

    Cumulative Fatigue DamageSuppose a machine part is subjected to:

    Fully reversed stress 1 for n1 cycles.Fully reversed stress 2 for n2 cycles.Fully reversed stress 3 for n3 cycles.

    ii

    ii

    stressat fail tocyclesN stressat cyclesn where

    1

    ==

    =i

    i

    Nn

  • 7/24/2009 251

    Cumulative Fatigue DamagePalmgren-Miner cycle ratio summation rule.. Miners rule

    = NNii 1

    (N) life fatigue total theof sproportion are ,..., if 21

    cyclesin life TotalN where11 == NNn

    N ii

    =

    NNN

    n

    i

    i1

  • Example: A component is made of steel having ultimate strength of 600 MPa and endurance limit of 300 MPa. Component is subjected to completely reversed bending stresses of:

    350 MPa for 75% of time;

    400 MPa for 15% of time;

    500 MPa for 10% of time;

    Determine the life of the component.

    ( ) ( )( ) ( )

    ( ) 9877.2log0851.0log10log600*9.0log

    10log300log

    23

    1

    26

    1

    +=+=

    +=

    NSkk

    kk

    f

    247134010163333

    3

    2

    1

    ===

    NNN

    ANS: 20214 cycles N1

    247110.

    3401015.

    16333375. =++

  • Question: A component is made of AISI 1008 cold drawn steel. Assume there is no stress concentration, size factor = 0.87, and expected reliability is 99%. The component at temperature of 100C is subjected to completely reversed bending stress of:

    140 MPa for 60% life 180 MPa for 25% life 200 MPa for 15% life

    Determine the life of component. ANS: Sut=340MPa. Determine Ktemp=1.02Kfinish=0.9624 and Kr=0.814.

    Corrected endurance strength for 103 cycles = 212.7 MPaCorrected endurance strength for 106 cycles = 118.2 MPa

  • 583.2&0851.010 and 10for strengths calculated Using

    loglog

    express to233 no. slideRefer

    21

    63

    21

    ==

    +=

    kk

    kNkSSstrengthFatigue

    f

    f

    Using fatigue strength equation:

    N1 cycles to fail component at stress 140 MPa = 136200

    N2 cycles to fail component at stress 180 MPa = 7104

    N3 cycles to fail component at stress 200 MPa = 2059

    Using Palmgren Miner rule (refer slide 246)

    Life of component, N = 8893 cycles

  • 7/24/2009 255

    Fatigue strength depends on

    Type of loadingSize of component Surface finishStress concentrationTemperatureRequired reliability

    NOTE: Factor of safety depends on the mean and alternating applied stresses and fatigue and yield/ultimate strengths

  • 7/24/2009 256

    Axial loading

    Difficult to apply axial loads without some eccentricity bending & axial.Whole critical region is subject to the same maximum stress level.

    Therefore, it would be expected that the fatigue strength for axial loading would be less than rotating bending.

  • 7/24/2009 257

    Fluctuating Stresses

    Fatigue failure criteria for fluctuating stresses ???

  • 7/24/2009 258

    Fatigue failure criteria for fluctuating stressesWhen alternating stress =0, load is purely static. Criterion of failure will be Syt or Sut.When mean stress=0, stress is completely reversing. Criterion of failure will be endurance limit. When component is subjected to mean as well alternating stress, different criterions are available to construct borderline dividing safe zone and failure zone.

    Remark: Gerber parabola fits failure points of test data. Soderberg line is conservative.

    1=+yt

    m

    e

    a

    SS

    12

    =

    +

    ut

    m

    e

    a

    SS

  • 7/24/2009 259

    Goodman line Failure criterion

    Widely used, because

    It is completely inside failure points of test data, therefore it is safe.

    Equation of straight line is simple compared to equation of parabola.

    Se

    Syt

    Syt SutO

    A

    B

    Cm

    a

    r

    SS

    m

    a

    e

    a

    ut

    m

    ==

    =+

    tan

    1

    r

    SSrSSr

    am

    eut

    euta

    =

    +=

  • 1

    1

    =+

    =+

    y

    a

    y

    m

    e

    a

    ut

    m

    SS

    SS

    ( )

    m

    a

    mya

    eut

    eyutm

    SSS

    SSS

    =

    =

    =

    tanSe

    Syt

    Syt SutO

    A

    B

    Cm

    a

    Example: A cantilever beam is made of steel having Sut=600 MPa, Syt =350 MPa and Se =130 MPa. The moment acting on beam varies from 5 N.m to 15 N.m. Determine the diameter of the beam.

    r

    SS

    m

    a

    e

    a

    ut

    m

    ==

    =+

    tan

    1

    r

    SSrSSr

    am

    eut

    euta

    =

    +=

    [ ]

    [ ]

    25

    10tan

    N.m 5)5(15*5.0Mmean Moment

    N.m 10)5(15*5.0M rangeMoment

    m

    a

    ==

    =+=

    ==

    r

    M

    M

    m

    a

    mm9.54dMPa 3.117

    ==a

    Modified Goodman line

    Area OABC represents region of safety.

  • MPaMPa 350&130 9.54,ddiameter For

    ma

  • 7/24/2009 262

    Ex: A cylindrical bar is subjected to 0 to 70 kNtensile load. Assume UTS = 690 MPa, YS = 580 MPa, and EL = 234 MPa. Assume stress concentration factor as 1.85. Find diameter of bar.

    r

    SS

    m

    a

    e

    a

    ut

    m

    ==

    =+

    tan

    1

    r

    SSrSSr

    am

    eut

    euta

    =

    +=

    [ ]

    [ ]

    13535tan

    kN 35070*5.0Fmean Force

    kN 35070*5.0F range Load

    m

    a

    ==

    =+=

    ==

    r

    F

    FkN

    m

    a

  • 7/24/2009 263

    Ex: A cylindrical bar (dia = 40 mm) is subjected to 0 to 70 kN tensile load. Assume UTS = 690 MPa, YS = 580 MPa, and EL = 234 MPa. Assume stress concentration factor as 1.85. Find FOS.

    r

    SS

    m

    a

    e

    a

    ut

    m

    ==

    =+

    tan

    1

    r

    SSrSSr

    am

    eut

    euta

    =

    +=

    [ ]

    [ ]

    13535tan

    kN 35070*5.0Fmean Force

    kN 35070*5.0F range Load

    m

    a

    ==

    =+=

    ==

    r

    F

    FkN

    m

    a

  • 7/24/2009 264

    Linear Elastic Fracture Mechanics(LEFM) Method

    Assumption: Cracks exist in parts even before service begins.Focus: Predict crack growth and remove parts from service before crack reaches its critical length.

    Griffith 1921

    Energy release rate is energy required rate

  • 7/24/2009 265

    Modes of Crack Displacement

    Figure Three modes of crack displacement. (a) Mode I, opening; (b) mode II, sliding; (c) mode III, tearing.

    Mode I is the most common & important mode. Stress intensity factor depends

    on geometry, crack size, type of loading & stress level.

  • 7/24/2009 266

    Design for Finite/Infinite LifeFatigue / Wear

    Attempt to keep local stresses --crack initiation stage never comes.Pre-existing voids or inclusions.Tensile stress opens crack (growth), while compressive closes (sharpen) it.

  • 7/24/2009 267

    Linear Elastic Fracture Mechanics Method..

  • 7/24/2009 268

    Linear Elastic Fracture Mechanics Method..

    2b

    d

    2a

    A B

  • 7/24/2009 269

    Linear Elastic Fracture Mechanics Method..

  • Life PredictionParis equation (for region II)

    Linear Elastic Fracture Mechanics Method..

    ( )

    ( )

    ( )

    ( )

    ( ) ( )12/1

    1

    1

    constants mat. aren &A

    12

    2/

    2/

    +=

    =

    =

    =

    =

    +

    n

    a

    ANN

    ada

    ANN

    a

    daA

    NN

    KAdadN

    KAdNda

    c

    i

    c

    i

    c

    i

    c

    i

    c

    i

    a

    a

    n

    nic

    a

    annic

    a

    annic

    a

    an

    N

    N

    n

  • 7/24/2009 272

    Austenitic cast iron, flakes 21 MPa.m^0.5Austenitic cast iron, nodular 22 MPa.m^0.5High silicon cast iron 9 MPa.m^0.5Carbon steel, AISI 1080 49 MPa.m^0.5Low Alloy steel, AISI 3140 77 MPa.m^0.5Cast Austenitic SS 132 MPa.m^0.5Tin based babbit 15 MPa.m^0.5Alumina 3.3 MPa.m^0.5Silicon carbide 2.3 MPa.m^0.5

    Fracture toughness

    ( ) cc aK minmax =

  • 7/24/2009 273

    Ex: Aluminum alloy square plate (width= 25mm), having internal crack of size 0.125 mm at center, is subjected to repeatedly tensile stress of 130 MPa. Crack growth rate is 2.54 microns/cycle at stress intensity range = 22 MPa(m)0.5. Crack growth rate at stress intensity range = 3.3 MPa(m)0.5 is 25.4 nm/cycle. How many cycles are required to increase the crack size to 7.5mm?

    mm 0.125 2a mm 25 2h mm 25 2b Given

    ===

    ( ) constants mat. aren &A nKAdNda

    =

    2.54e-6/2.54e-8 = (22/3.3)^nOr n = log10(100)/log10(22/3.3)n=2.4275.

  • 7/24/2009 274

    ( ) ( )12/1

    12

    +=

    +

    n

    a

    ANN

    c

    i

    a

    a

    n

    nic

    ANS: 24500 cycles.

  • 7/24/2009 275

    Question: A rectangular cross-section bar (width 6mm, depth = 12 mm) is subject to a repeated moment 0 M135 N.m. Ultimate tensile strength, yield strength, fracture toughness, constant A and c are equal to 1.28 GPa, 1.17 GPa, 81 MPa.m^0.5, 114e-15, and 3.0 respectively. Assume =1 and initial crack size is 0.1 mm. Estimate the residual life of bar in cycles.

    MPayI

    M 5.937/

    =

    =

    The maximum tensile stress is below the yield strength, therefore bar will not fail under static moment. We need to find the size of critical crack size using value of stress range and fracture toughness.

    ( ) maaK ccc 0024.0minmax ==

    ( ) ( )12/1

    12

    +=

    +

    n

    a

    ANN

    c

    i

    a

    a

    n

    nic

  • 7/24/2009 276

    Reference: Professor E. Rabinowicz, M.I.T

    Death of machine inevitable. Design considering yielding & fracture

  • 7/24/2009 277

    Adhesive (frictional) wear

    Mechanical interaction at real area of contact

  • Laws of Adhesive WearWear Volume proportional to sliding distance (L)

    True for wide range of conditions

    Wear Volume proportional to the load (N)

    Dramatic increase beyond critical load

    Wear Volume inversely proportional to hardness of softer material

    HNLkV

    31=

    Transition from mild wear to severe depends on relative speed, atmosphere, and temperature.

  • 7/24/2009 279

    Approach followed by M. F. Ashby

    pkHNk

    LV

    a=== 31

    maxmax

    pp

    pkpk aa ==

    Hkp

    pC

    HCp

    pk

    a

    a

    =

    =

    max

    max

  • 7/24/2009 280

  • 7/24/2009 281

    Ex: Ship bearings are traditionally made of bronze. The wear resistance of bronze is good, and allowable maximum pressure is high. But due to its chemical activity with sea water galvanic corrosion occurs and wear occurs. Material chart shows that filled PTFE is better than Bronze material.