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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 5°C.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 Elements
1. 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
θ
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 * speed
U = 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 <= xFLOOR(x)smallest 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
TK’s 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, Young’s modulus, Shear modulus, Fatigue
strength, resilience, toughness
00
0 , lll
ll>
−=ε
0AP
=σε=log(l/l0)
7/24/2009 30
Stress-strain Diagram forMetals
εσ
=E
modulussYoung'
tensile
tensile
EE
EE
>
=
ncompressio
ncompressio
Brittle
Ductileelasticyield
alproportionelastic
εε
εε
>
>
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
ππππ
ANS: Ferrite SS cannot be formed to the cylinder.
7/24/2009 34
Torsion Test
EGEG
lrG
5.0)1(2
0
≤→+
=
=
υ
θτ
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
υMaterialυMaterial
7/24/2009 35
Fatigue strengthTime varying loadsWohler’s 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.
∫=ε
εσ0
dU
7/24/2009 37
Resilience (energy per unit volume)
ES
U
E
dEdU
yR
R
el
elel
20
2
00
21
2
=
=
∫=∫=
ε
εε
ε
εεεσEx: 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
ε
εσε
+=
∫=
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 20°C, may be brittle at 4°C, yet creep rapidly at 100°C.
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 < 250°C)
• 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 σy )
Illustration of Mechanical properties
Too heavy (need lower ρ)
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
Rankσ, MPaRankρ 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
ng’s
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
1/2EρM =
22 M/ρE =
( ) ( ) ( )MLog2Log2ELog −ρ=
Contours of constantM are lines of slope 2
on an E-ρ chart
CE
=ρC
E 2/1 =ρ
CE 3/1 =
ρ
0.1
10
1
100
Metals
Polymers
Elastomers
Woods
Composites
Foams0.01
1000
1000.1 1 10Density (Mg/m3)
You
ng’s
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/ρ) Larger Better For Light & Stiff Tie-rod
Light & Strong σY/ ρ
Best material for a spring, regardless of its shape or the way it is loaded, are those with the greatest value of (σY)2 /E
Best thermal shock resistant material needs largest value of σY/Eα
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 INDICES•GROUPING 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/ρ
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
⎟⎟⎠
⎞⎜⎜⎝
⎛
σρ
=y
FLm
Chose materials with smallest⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
yσρ
m = massA = areaL = lengthρ = density
= yield strengthyσ
Function
Objective
Constraints
Free variables
Tie-rod: Rod subjected to tensile force.
Minimise mass m:m = A L ρ (1)
• Length L is specified• Must not fail under load F
• Material choice• Section area A; eliminate in (1) using (2):
(2)yAF σ≤/
Example 2: stiff, light beam
m = massA = areaL = lengthρ = densityb = edge lengthS = stiffnessI = second moment of areaE = Youngs Modulus
⎟⎠⎞
⎜⎝⎛ ρ
⎟⎟⎠
⎞⎜⎜⎝
⎛= 2/1
2/15
ECLS12m Chose materials with smallest ⎟
⎠⎞
⎜⎝⎛ ρ
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=
ρ=ρ= 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 = lengthρ = 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
⎟⎠⎞
⎜⎝⎛ ρ
⎟⎟⎠
⎞⎜⎜⎝
⎛= 3/1
23/12
EL
CwS12m
L
F
ρ=ρ= LtwLAm
Chose materials with smallest ⎟⎠⎞
⎜⎝⎛ ρ
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/ρ
Beam, minimum weight, stiffness E1/2 /ρ
Beam, minimum weight, strength σ2/3/ρ
Beam, minimum cost, stiffness E1/2/Cmρ
Beam, minimum cost, strength σ2/3/Cmρ
Column, minimum cost, buckling load E1/2/Cmρ
Spring, minimum weight for given energy storage σYS2/Eρ
Minimizing cost instead of weight is achieved by replacing density ρ by ρCm , where Cm=cost/mass
MATERIALS for SPRINGS
? OBJECTIVE: MAXIMIZE ENERGY STORAGE
? AXIAL SPRINGS, LEAF, HELICAL, SPIRAL, TORSION
? PRIMARY FUNCTION: STORING/RELEASING ENERGY
EWV
2σ∝
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 σ, 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 fσ=
7/24/2009 74
? ELASTIC ENERGY/COST ECM
m
f
ρσ 2
=
EM f
ρσ 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 f2σ=
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
2σ Eon basedSelection
2
ρσ
m
2
CEon basedSelection
ρσ
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
σ/ρ E/ρ
Carbon steel 3
Carbon steel 4
Carbon steel 5
σ > 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 σ/ρ and E/ρ
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
=
=
=
τ
σ
σ ,,
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
σσ
=
Plate with cross-sectional plane
Half of plate with stress distribution.
Stress Concentration
hdbP
)(avg −=σ
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 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=σ
( ) bbhole700
153050001.2 =⎟⎟
⎠
⎞⎜⎜⎝
⎛−
=σ
80=allowableσ 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 ππ
σ ×==
( )( )
10035012502.516.1 3max =×
==davg π
σσ 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 = ρ × (50-5)× h × 100 ; A = (50-5)× h
( ) ( ) hhhdbPKt
1500550
250007.2 =−
=−
=σ
σρ
σρ
6750M or,
1500 4500M or,
=
=
⎟⎠⎞
⎜⎝⎛=⎟
⎠⎞
⎜⎝⎛
σρ
1010 log6750
log M
d/b = 0.1; Kt=2.7;
7/24/2009 121
mmheh
8.0389.11500=⇒==σ
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 1brpp
∫ ∫=b
drrdpF0
2
0 ispatch contact on load applied Total θ
π
[ ]
( )
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
π
π
π
π
π
=
=
∫ −==−
∫ −=
∫⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=
5.1=tK r
7/24/2009 130
Cylindrical Contact
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=
2
max 1bxpp
R1
R2
L
X
Y
Z
Pressure variation along Y-axis is negligible,
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛−=
22
max 1ay
bxpp
( )
max
2
0
2max
0
2
max
2 or
cos2 sinlet
12 ispatch contact on load applied Total
pLbF
dbpFθ b x
dxbxpLF
b
π
θθ
π
=
==
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=
∫
∫
Cylindrical Contact…
Stress concentration factor = 4/π
max
max2
2
32
pLbF
pbF
contactlcylindrica
contactspherical
π
π
=
=How to determine b ???
7/24/2009 132
How to determine b
Assume pmax = σy and find value of b.
max
max
2
5.1
pLF
b
pF
b
contactlcylindrica
contactspherical
π
π
=
=
Criterion b << d1 needs to be checked.
7/24/2009 133
For axi-symmetric point load Timoshenko & Goodier suggested:
( )
( )
( )rF
E
yxEF
zG
F
EG
zy
z
πνδ
ν
νπ
δ
ρν
ρπδ
ν
ρ
21
10
)1(24
14
)1(2
x
2
1
221
3
2
222
−=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
+
−+
+
=
⎭⎬⎫
⎩⎨⎧ −
+=
+=
++=
Ref: S. Timoshenko and J.N.Goodier, Theory of elasticity, 2nd
Edition, 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
πνδ
θθνδ
θθνδ
θθθνδθ
νδ
ππ
νδ
θπ
νθδ
π
π
π
π
pE
b
pE
b
dpE
b
dbpE
r
drbrpE
drrr
brpE
drrdr
brpE
r
b
b
b
−=
⎥⎦⎤
⎢⎣⎡ +
−=
+−
=
−==
−−
=
−−=
−−=
∫
∫
∫
∫
∫ ∫
max2
32 pbF π=
( )22
1max
1
21
1πνδ p
Eb −
=max
2
32 pbF π=
FEb 1
21
11
83 νδ −
=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
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛−−=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
−−=
−−=
−=
δ
δ
δ
δ
δ
δ
FE
Rb1
21
13 175.0 ν−
=
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 Poisson’s ratio = 0.28 and E=207 GPa.
Ans: b=118 microns. Size=2*b
Example
FE
Rb1
21
13 175.0 ν−
=
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
νντ
ννσσ
σ
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 Poisson’s 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 poisson’s ratio of one body is reduced to 0.22. Prob 2: What will happen if Poisson’s ratio of
one body is increased to 0.32 and Young’s 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
νντ
ννσσ
σVariation of stresses with Z.
Four equations. Eight variables. We need four inputs.
Assume b = 100 μm, ν=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
ννπδ
πνδ( )22
1
1 sphere of Deflection
max1
21
1πνδ 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
ννπ
ννπ
δ
max2
32 pbF π=
7/24/2009 145
( ) ( )⎥⎦
⎤⎢⎣
⎡ −+
−
⎥⎦
⎤⎢⎣
⎡+
=2
22
1
21
21
2 11
21
21
5.1
4 or
EERR
bF
b ννππ
( ) ( )⎥⎦
⎤⎢⎣
⎡ −+
−
⎥⎦
⎤⎢⎣
⎡+
=2
22
1
21
21
3 11114
3 or EE
RR
Fb νν
Question: Two carbon steel balls (AISI 1030 tempered at 650°C), each 25 mm in diameter, are pressed together by a force F = 100N. Find the maximum value of compressive stress. Poisson’s ratio = 0.285, Young’s modulus = 208 GPa.
Answer: 1.85 GPa.
( ) ⎥⎥⎦
⎤
⎢⎢⎣
⎡
++−= 5.122
3
max 1zb
zpzσ
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 650°C ), Poisson’s ratio = 0.285 and Young’s modulus = 208 GPa. Other ball is made of synthetic rubber (Poisson’s ratio = 0.48 and Young’s 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 650°C), 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. Poisson’s ratio = 0.285, Young’s 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
==
−=
−==
⎟⎟⎠
⎞⎜⎜⎝
⎛ −+
−
⎟⎠⎞⎜
⎝⎛ +
=
=
⎟⎠⎞
⎜⎝⎛−=
τ
τ
νσ
σσ
νν
π
π
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 Poisson’s 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
νσ
σ
σ
Problem: A 200-mm diameter cast iron (ν=0.26, E = 80 GPa) wheel, 55 mm wide, rolls on a flat steel (ν=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
==
−=−=
−==
==
−=⎟⎟⎠
⎞⎜⎜⎝
⎛ −+
−
⎟⎠⎞⎜
⎝⎛ +
=
τ
νσ
σσπ
νν
π
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 (ν=0.35, E=90 GPa) material and cup is made of plastic (PEEK: ν=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 material’s 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
εσεσεσ
εσ
++=⇒
=
U
U
( )
( )
( )1233
3122
3211
1
1
1
σνσνσε
σνσνσε
σνσνσε
−−=
−−=
−−=
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
σσσσσσνσσσ
7/24/2009 158
Finding Distortion Energy
( )hd UUU
EU
+=
⎥⎦
⎤⎢⎣
⎡
++−++
=312321
23
22
21
221
σσσσσσνσσσ
( )
[ ]
3
2123
221
321
2
222
σσσσ
νσ
σσσσσσνσσσ
++=
−=
⎥⎦
⎤⎢⎣
⎡
++−++
=
h
hh
hhhhhh
hhhh
EU
EU
[ ]31232123
22
213
1 σσσσσσσσσν−−−++
+=
EUd
7/24/2009 159
von-Mises Theory
[ ][ ]312321
23
22
21
31232123
22
21
2
31
31
σσσσσσσσσ
σσσσσσσσσνν
−−−++=
−−−+++
=+
=
y
yd
SE
SE
U
2
31
yd SE
U ν+=
[ ]31232123
22
21
safety offactor consider weIf
σσσσσσσσσ −−−++≥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
maxσστ −
=
How to find principal stresses and estimate factor of safety.
7/24/2009 161
Principal Stresses
7/24/2009 162
Principal Stresses
( )( ) ( )( ) ( )( ) ( )( )( ) ( ) ( ) ( )φφτφφτφσφσσ
φφτφφτφφσφφσσ
φ
φ
cossincossinsincos
sincoscossinsinsincoscos0
22yxxyyx
yxxyyx AAAAAF
+++=
+++=
=∑
( )
( )φτφσσσσ
σ
φτφσφσσ
φ
φ
2sin2cos22
2sin2
2cos12
12cos
xyyxyx
xyyx
+⎟⎟⎠
⎞⎜⎜⎝
⎛ −+⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
+⎟⎠⎞
⎜⎝⎛ −
+⎟⎠⎞
⎜⎝⎛ +
=( )
( )
yx
xy
yxxy
yxxy
σστ
φ
φσσ
φτ
φσσ
φττφ
−=⇒
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−=
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−=
22tan
2sin2
2cos0
2sin2
2cos
Principal Stresses
( )φτφσσσσ
σφ 2sin2cos22 xy
yxyx +⎟⎟⎠
⎞⎜⎜⎝
⎛ −+⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
yx
xy
σστ
φ−
=2
2tan
( ) ( )
( )⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −±⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ +−±⎟⎟
⎠
⎞⎜⎜⎝
⎛ +=
22
21
22
22,
22
2
xyyxyx
xyyxyx
τσσσσ
σσ
τσσσσσφ
Similarly we can find σ3. In practice σ1 , σ2 , and σ3 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
π
μσ
σμ
7/24/2009 167
EX. NOMINAL SHAFT DIA. 4.5mmNUMBER OF SPECIMEN 34
4.58mm0.0097
d
d
σμ
4.59,4.34,4.5796,4.50, 4.582,4.5847……………4.5948
6
4.5294
0.0987
( )1
/22
−∑ ∑−
=N
Ndd iidσ
⎟⎟⎠
⎞⎜⎜⎝
⎛ −−
= d
did
d
edf σμ
πσ21
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 σ( ). Find the reliability of member.
Deterministic FOS = 40/30. 100% reliability.
ss σμ ,40=
σσ σμ ,30=
NOTE: Reliability is probability that machine element will perform intended function satisfactorily.
830
==
σ
σ
σμ
640
==
s
s
σμ10,10 == QQ σμ
1086
10304022 =+=
=−=
Q
Q
σ
μ
7/24/2009 169
xQyxQ
xyQyxQxCQ
CxQCQ
1===
±=+=
==
x
yx
yx
yx
x
x
CCC
μ
μμ
μμ
μμμ
μ
1
±+
ALGEBRAIC MEAN STD. DEVIATIONFUNCTIONS
2
22222
2222
22
0
xx
yyxxy
yxxy
yx
x
xC
μσ
μσμσμ
σμσμ
σσ
σσ
+
+
+
7/24/2009 170
Margin
( )f
f
P R
QP P
−=
<=
1yReliabilit
0failure ofy Probabilit
830
==
σ
σ
σμ
640
==
s
s
σμ10,10 == QQ σμ
σ−= SQ
( )2
21
21 ⎟
⎟⎠
⎞⎜⎜⎝
⎛ −−
= Q
Q
eQf σμ
πσQ
QQZ
σμ−
=
variablenormalLet
Q
Q
Z
Z
Q
Q
where
dZeR
QZ
σμ
π
σμ
−=
=
−=
∫∞+
−
0
21
Z
21
0
2
10
10
=
=
Q
Q
σ
μ
110
100
0at
0 −=−
=
=
Z
Q
∫=∞−
−0 221
21 z z
dZeFπ
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
== σ
dzeRz
243.22
211design ofy Reliabilit −−
∞−∫−=π
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
πσ
πμ
σμ
σσ ==
==
Q
Q
Z
Z
Q
Q
where
dZeRQ
Z
σμ
πσμ
−=
=−
= ∫+∞
−
0
21
Z
21;
0
2
2
22
2
7200040
880000540
⎟⎟⎠
⎞⎜⎜⎝
⎛+=
−=
d
d
Q
Q
πσ
πμ
2
2
22 880000540720004009.3
dd ππ−=⎟⎟
⎠
⎞⎜⎜⎝
⎛+ 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
ix12
2
:by expressed isfunction complex a ofdeviation Standard σφσμ
φ
( )( )22
21
3 344/ LLkdP +=σ
( ) ( ) ( ) ( )
[ ]
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
μσ
μσ
μμ
μμσ
σσσσσσσσσ
σ
σ
σ
σ
=
+++=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎠⎞
⎜⎝⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛+⎟⎟
⎠
⎞⎜⎜⎝
⎛=
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎟⎠
⎞⎜⎜⎝
⎛∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
+⎟⎠⎞
⎜⎝⎛
∂∂
= Statistically independent
( )( )3
22
21
3
34087000344/
d
LLdP k
μμ
μμμμμ
σ
σ
=
+=
( )
( )
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
σμ
μμ
μ
μ
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
NSut≤1σ
Maximum tensile stress Factor of
safety
Ultimate tensile strength. Often referred as tensile strength.
NSuc≤3σ
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 >>>≤ σσσσ ifNSut
3213 0 σσσσ >>>≤ ifNSuc
Coulomb Mohr Theory
3131 01 σσσσ
>>≤− ifNSS 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 τστστ
−==⇒ 31
i
and).( stress torsional tosubjected isBar :Given
( )NSS
or
NSSas
ut
i
ut
i
ucut
12
10 3131
≤−
−
≤−>>
ττ
σσσσ
NSut
i 32
≤τ
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. φ 20mm) dimensions.
Tolerance: “permissible variation in the dimensions of a component”.Tolerance: Unilateral or bilateral.
Inaccuracies of Manufacturing
methods
01.0;20 == dd σμ
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
81072063054046039033027022018014013
52046040035030025021018015012010012
32029025022019016013011090756011
210185160140120100847058484010
13011510087746252433630259
81726354463933272218148
52464035302521181512107
32292522191613119866
232018151311986545
1614121087654434
12108654432.52.523
875432.52.521.51.51.22
64.53.52.521.51.51.2110.81
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=220μm
Shaft 110e9...Maximum = 110mm – 0.072=109.928mm...Minimum = 110mm - (0.072 +0.087) = 109.841mmResulting limits 109.841/ 109.928Tolerance of shaft, tls=87μm
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=160μm
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 μm
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 ∝
θ
Interference Fit
δ=0.001d mm
δ=0.0005d mm
δ=0.00025d mm
δ=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
δrs
δrh
δrs
δrh
rf
rfrf
pf
pf
Base-line
7/24/2009 204
( ) ( )
( )
( )( ) 02
sin2 balance Force
strain Radial
strain ntialCircumfere
=⎟⎠⎞
⎜⎝⎛−−++=
−=
∂∂
=−
∂∂
+=
−==
−+=
dzdrddzrddzddrrd
Erdr
drr
Erdrdrdr
rrr
rrr
rr
r
rrr
θσθσθσσ
σνσδδδδε
σνσδθ
θθδε
θ
θ
θθ
( )( )
drdror
dzddzddzddr
drgrearrangin
dzdrddzrddzddrrd
rr
rr
rrr
σσσ
θσθσθσ
θσθσθσσ
θ
θ
θ
+=
=−+
=⎟⎠⎞
⎜⎝⎛−−++
0
02
sin2 ( )
( )Er
Errr
rr
θ
θ
σνσδ
σνσδ
−=
∂∂
−=
Edr
dr
r
Edr
dr
rr
rrr
rr
rr
⎟⎠⎞
⎜⎝⎛ −−
=∂
∂
⎟⎠⎞
⎜⎝⎛ −+
=
σνσνσδ
σνσσδ
⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜
⎝
⎛
−
++⎟
⎠⎞
⎜⎝⎛ −+=
∂∂
drd
drddr
dr
Er
drdr
Er rr
r
rr
rr
σνσ
σ
νσσσδ
2
1 2
2
03 2
2
=+dr
drdr
d rr σσ( ) 02 2
2
=+drrd
drd rr σσ
( ) 02 2
2
=+drrd
drd rr σσ ( ) 02 1 =++ C
drrd r
rσσ
( ) 01
2
=+ rCdrrd rσ
02
02
221
2
2
12
=++
=++
rCC
CrCr
r
r
σ
σ
Two conditions are required to express radial stress in terms of radius.
oor
iir
rratprratp
=−==−=
σσ
oo
ii
prCC
prCC
=+
=+
221
221
2
2
( ) ( )
( ) ( )22
222
22
222
stress ntialCircumfere
stress Radial
io
iooiooii
io
iooiooiir
rrpprrrrprpσ
rrpprrrrprpσ
−−−−
=
−−+−
=
θ
CASE I: Internally Pressurized (Hub)-
( )( )
( )( )22
22
22
22
1 stress Radial
1 stress ntialCircumfere
fo
offr
fo
off
rrrrrp
σ
rrrrrp
σ
−
−=
−
+=θ
( )
fr
fo
off
pσ
rrrrp
σ
−=
−
+=
max,
22
22
max,
θ
( )Er
rh
f
rh σνσδε θθ
−==strain ntialCircumfere
rf
f
rh
fo
off
rrrrr
Ep
hδ
νεθ =⎟⎟⎠
⎞⎜⎜⎝
⎛+
−
+= 22
22
max,
( )
( )⎟⎟⎠
⎞⎜⎜⎝
⎛
−−
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛
−+
−=
22
22
22
22
1 stress Radial
1 stress ntialCircumfere
if
iffr
if
iff
rrrrrpσ
rrrrrpσθ
CASE II: Externally Pressurized (shaft)-
rf
( )Er
rs
f
rs σνσδε θθ
−==strain ntialCircumfere
f
rs
if
fi
s
f
rrrrr
Ep
sδ
νεθ =⎟⎟⎠
⎞⎜⎜⎝
⎛−
−
+−= 22
22
max,
fr
ifff
pσ
rrrpσ
−=
⎟⎟⎠
⎞⎜⎜⎝
⎛
−−=
max,
222
max,2
θ
( ) ( ) ⎥⎥⎦
⎤
⎢⎢⎣
⎡−
−
+++
−
+=
−=
s
s
ifs
fi
h
h
foh
foff
rsrh
ErrErr
ErrErr
pr ννδ
δδδ
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 hub’s 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
δ
δ
ANS: pf =73 MPa
Through interference fit torque can be transmitted, which can be estimated with a simple friction analysis at the interface.
( )( )
LdpTTorque
LdpF
ApNF
f
fff
ff
2
2μπ
πμ
μμ
=
=
== μ = 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 π=
HrnW )*5.0( 2π= HnrhF )(=
θπ
μ cot2=
For θ = 45° μ = 0.6366For θ = 60° μ = 0.3676For θ = 80° μ = 0.1123
Slope of real surfaces are nearly always less than 10° (i.e. θ> 80°), therefore μ < 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 = 3σy & s = σy /1.7Expected value of μ =.2
HAW real= sAF real= Hs
=μ
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, τi.
maxmax AF iτ=
2max
22maxmax
)4( AA
WF
iy
i
τστμ−
≅=
)(2 22iy
i
τττμ
−≅
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
σσσ
σσσ
−=
+=
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
kNkS
becanSstrengthFatigue
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<N<106).
( ) ( )
( ) ( ) 26
126
1
62
312
31
3
10log)45.0log(10loglog
10
10log)75.0log(10loglog
10
kkSkkS
cycleatSstrengthFatigue
kkSkkS
cycleatSstrengthFatigue
utl
f
utl
f
+=⇒+=′
′
+=⇒+=′
′
K1=-0.07395 k2=3.13 (stress in MPa)
Slide 229
Slide 232
7/24/2009 235
Endurance limit modification factors
Endurance limit is measured under best circumstances, which cannot be guaranteed for design applications.
Component’s endurance limit must be modified or reduced from material’s 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
0.672550°C1.0250°C
0.768500°C1.02200°C
0.843450°C1.025150°C
0.900400°C1.02100°C
0.943350°C1.0150°C
0.975300°C1.0020°CKtempTemperatureKtempTemperature
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
⎩⎨⎧
≤<≤≤
=−
−
mmddmmddKsize 2545151.1
5179.224.1157.0
107.0
NOTE: A 7.5mm diameter beam specimen is used for testing fatigue strength. Larger the machine part, greater is the probability that a flaw exit somewhere in larger volume. Fatigue failure tendency ↑
Applicable only for cylindrical components.
Necessary to define “effective diameter” based on equivalent circular cross section for components having non-circular cross-section.
Effective diameter for non-rotating cross sections
Effective dimension is obtained by equating the volume of material stressed at and above 95% of maximum stress to the same volume in the rotating beam specimen.
Lengths will cancel out, so only areas are considered.
For a rotating round section, the 95% stress area is the area in a ring having outside diameter d and inside diameter of 0.95, so
( )[ ] 22295.0 0766.095.0
4dddA =−=
πσ
7/24/2009 243
50
±30 kN
±30 kN
Example: A hot rolled steel plate (Sut=400 MPa) at room temperature is subjected to completely reversed axial load of 30kN. Assume size factor and expected reliability as 0.85 and 95% respectively. Determine the thickness of plate for infinite life.
STEP 1: Estimate endurance limit of mat.
5
TorsionSSAxialSSbendingSS
ute
ute
ute
29.045.05.0
SteelFor
=′=′=′
MPaSS
SS
e
e
ute
18040045.0
45.0
=′×=′
=′
STEP 2: Estimate endurance limit of plate.
Find modification (i.e reliability, finish, temp., stress concentration and size) factors.
0.868 0.78
1.00.8970.868
509095
Reliability factor, kr
Probability of survival, %
Surface Finish Factor
-0.71857.7Hot rolledExponent bConstant aFinishing method
( )butfinish SaK MPain =
MPaSMPaS
e
e
9.12178.0868.0180
factorsfinish andy reliabilit including S Corrected 'e
=′××=′
Temperature Factor1.0020°C
KtempTemperature
MPaSMPaS
e
e
6.10385.019.121
factors size andre temperatufinish, ,y reliabilit including S Corrected '
e
=′××=′
50
±30 kN
±30 kN
5
Stress concentration factor2.5 1/2.5 =0.4
Thickness > 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
kNkS
becanSstrengthFatigue
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 σ1 for n1 cycles.Fully reversed stress σ2 for n2 cycles.Fully reversed stress σ3 for n3 cycles.……
ii
ii
stressat fail tocyclesN stressat cyclesn where
1
σσ
==
∑ =i
i
Nn
7/24/2009 251
Cumulative Fatigue DamagePalmgren-Miner cycle ratio summation rule.. Miner’s rule
∑ =NNi
i 1
(N) life fatigue total theof sproportion are ,..., if 21
α
αα
cyclesin life TotalN where11==∑ NN
nN i
i
∑ =⎟⎠⎞⎜
⎝⎛
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 100°C 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
kNkS
SstrengthFatigue
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
SSσσ
12
=⎟⎟⎠
⎞⎜⎜⎝
⎛+
ut
m
e
a
SSσσ
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
Cσm
σa
θr
SS
m
a
e
a
ut
m
==
=+
σσθ
σσ
tan
1
r
SSrSSr
am
eut
euta
σσ
σ
=
+=
1
1
=+
=+
y
a
y
m
e
a
ut
m
SS
SSσσ
σσ ( )
m
a
mya
eut
eyutm
SSS
SSS
σσθ
σσ
σ
=′
−=−
−=
tanSe
Syt
Syt SutO
A
B
Cσm
σa
θ’
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
==
=+
σσθ
σσ
tan
1
r
SSrSSr
am
eut
euta
σσ
σ
=
+=
[ ]
[ ]
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
θ
mm9.54dMPa 3.117
==aσ
Modified Goodman line
Area OABC represents region of safety.
MPaMPa 350&130 9.54,ddiameter For
ma <<≥
σσ
Design is safe
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
==
=+
σσθ
σσ
tan
1
r
SSrSSr
am
eut
euta
σσ
σ
=
+=
[ ]
[ ]
13535tan
kN 35070*5.0Fmean Force
kN 35070*5.0F range Load
m
a
=⇒=
=⇒+=
=⇒−=
r
F
FkN
m
a
θ
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
==
=+
σσθ
σσ
tan
1
r
SSrSSr
am
eut
euta
σσ
σ
=
+=
[ ]
[ ]
13535tan
kN 35070*5.0Fmean Force
kN 35070*5.0F range Load
m
a
=⇒=
=⇒+=
=⇒−=
r
F
FkN
m
a
θ
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
σ
σ
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/
+−Δ=−
Δ=−
Δ=−
Δ=
Δ=
+−
∫
∫
∫∫
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
σπβ
σπβ
σπβ
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 πσσβ minmax −=Δ
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
Δ=
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
+−Δ=−
+−
n
a
ANN
c
i
a
a
n
nicσπβ
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 ≤M≤135 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 β=1 and initial crack size is 0.1 mm. Estimate the residual life of bar in cycles.
MPayI
M 5.937/
=Δ
=Δσ
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 =⇒−=Δ πσσβ
( ) ( )12/1
12
+−Δ=−
+−
n
a
ANN
c
i
a
a
n
nicσπβ
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===Ω3
1
maxmax
pp
pkpk aa ==Ω
Hkp
pC
HCp
pk
a
a
⎟⎟⎠
⎞⎜⎜⎝
⎛=Ω
=Ω
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.
