M2L01L02 Fatigue Basics
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Transcript of M2L01L02 Fatigue Basics
5192016
1
Aircraft Structural Integrity Module 2 Fatigue
Lecture 1
Todayrsquos LectureIntroduction to Fatigue Some Basics
Fatigue Crack in Boeing 7373-300 Emergency landing of Southwest flt 812 Apr 2013
5192016
2
F A T I G U E Basics
What is fatigue
Fatigue is hellipbull a process in which damage accumulates due to
repetitive loads which may be well below yieldstress or static strength
bull a fracture phenomenon occurring after a largenumber of load cycles where a single load of thesame magnitude will do no harm
bull a form of failure that occurs in structuressubjected to dynamic and fluctuating stressesUnder these circumstances it is possible forfailure to occur at a stress level considerablylower than the tensile or yield strength for astatic load
5192016
3
Stress ndash Life (S-N) Curve
Fatigue Cyclic (even if low) loads cause failures
Steel Ferrous Alloys
Many non-Ferrous Alloys
eg Al Alloys
5192016
4
Why should the cyclic loads lead to failure at lower loads
Origin of fatigue phenomenon
Fatigue Process Crack Initiation and Growth
Free Surface
STAGE I CRACK INITIATION Several micro-cracks originate and then coalesce into a macro-crack
STAGE II CRACK GROWTH
LOADING DIRECTION
Dislocations Slip Microcrack Macrocrack
5192016
5
Dislocations
Slip bands and micro-crack initiation
①
②
③
④
~ 01 μ
~ 01 μ
Persistent Slip Bands
5192016
6
Features of Fatigue Phenomenon
bull A gradually progressive process
bull Slow development of damage in the early stage followed by rapid growth towards the end
bull First stage crack initiation phase
ndashMay occupy most of the life in most cases (eg nicely machined parts small cyclic loads )
ndashUsually confined to a small area of high local stress
ndashAdjacent areas with only slightly lower stresses may not see any fatigue damage
ndashusually several independent micro-cracks grow and coalesce to form one dominant crack
hellip 1
Dislocations Slip Microcrack Macrocrack Crack Propagation Failure
Features of Fatigue Phenomenon
bull Second Stage Crack Propagation phasendashThis crack grows slowly under steady fatigue loading
but starts to accelerate as the net section decreases increasing the local stress at the crack-front
bull Third and Final Stage Failure ndashFailure occurs as an unstable fracture when the
remaining area is too small to support the load
hellip 2
bull Stages create distinctive features of the fracture surface - used to identify fatigue failure in failure investigations
5192016
7
Tell-tale marks of Fatigue - Striations
Features of Fatigue Phenomenon
bull Some materials exhibit endurance limits ie a stress below which the life is infinitendash Steels typical endurance limit - 40 -
60 of YS
ndash solutes (carbon nitrogen) present in the material pin the dislocations and prevent their motion at small strains
ndash Al alloys No endurance limits
ndash Related to the absence of dislocation-pinning solutes
hellip 3
bull At large Nf the lifetime is dominated by nucleation
ndash Therefore strengthening the surface (shot penning) is beneficial to delay crack nucleation and extend life
5192016
8
S
N
bull Typically load life and load life
bull Experimentally-determined ndash Rotating bending mc round specimens
ndash Electrohydraulic mc flat (or other) specimens
bull The practical limit of experimental testing has been 106 or 107 cycles due to impractical test times frequency limitations of machines and artefacts of higher frequencies
bull Adjustment for mean stress using Goodman Relation
Observations on S-N Curve
S
N
bull X-Axis No of Cycles On Log10 scale Log10 N
bull Y-Axis Stress ndash Either Stress Amplitude (Alternating stress) or Max
Stress
ndash Alternating Stress (ie Stress Amplitude) is the most dominant aspect of load that influences fatigue
ndash Either linear scale or Log10 scale
bull Curve is obtained for a given stress ratio R
bull R = (Min Stress Max Stress)
Observations on S-N Curve
5192016
9
S
N
bull S-N curve data is generally developed under Constant Amplitude ndash we will discuss later variable amplitude
bull The curve is dependent on many conditions ndash Loading ratio min load max load (R-ratio)
ndash Material
ndash stress concentrators
ndash surface finish
ndash frequency cyclic shape dwell etc hellip Not much effect unless in corrosive environment
Observations on S-N Curve
S
N
High cycle fatigue vs Low cycle fatigue
bull HCF Life usually more than 104 cycles ndash The distinction between high cycle fatigue and low
cycle fatigue is reflected in most design standards as high cycle fatigue design standards normally give S-N curves starting with N = 104 cycles
bull LCF Life usually less than 104 cycles ndash For low cycle fatigue design the stress range
concept is not immediately valid since typically the low cycle fatigue strength is governed by (large inelastic) strains
Arbitrary Could be 103
Arbitrary Could be 103
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
2
F A T I G U E Basics
What is fatigue
Fatigue is hellipbull a process in which damage accumulates due to
repetitive loads which may be well below yieldstress or static strength
bull a fracture phenomenon occurring after a largenumber of load cycles where a single load of thesame magnitude will do no harm
bull a form of failure that occurs in structuressubjected to dynamic and fluctuating stressesUnder these circumstances it is possible forfailure to occur at a stress level considerablylower than the tensile or yield strength for astatic load
5192016
3
Stress ndash Life (S-N) Curve
Fatigue Cyclic (even if low) loads cause failures
Steel Ferrous Alloys
Many non-Ferrous Alloys
eg Al Alloys
5192016
4
Why should the cyclic loads lead to failure at lower loads
Origin of fatigue phenomenon
Fatigue Process Crack Initiation and Growth
Free Surface
STAGE I CRACK INITIATION Several micro-cracks originate and then coalesce into a macro-crack
STAGE II CRACK GROWTH
LOADING DIRECTION
Dislocations Slip Microcrack Macrocrack
5192016
5
Dislocations
Slip bands and micro-crack initiation
①
②
③
④
~ 01 μ
~ 01 μ
Persistent Slip Bands
5192016
6
Features of Fatigue Phenomenon
bull A gradually progressive process
bull Slow development of damage in the early stage followed by rapid growth towards the end
bull First stage crack initiation phase
ndashMay occupy most of the life in most cases (eg nicely machined parts small cyclic loads )
ndashUsually confined to a small area of high local stress
ndashAdjacent areas with only slightly lower stresses may not see any fatigue damage
ndashusually several independent micro-cracks grow and coalesce to form one dominant crack
hellip 1
Dislocations Slip Microcrack Macrocrack Crack Propagation Failure
Features of Fatigue Phenomenon
bull Second Stage Crack Propagation phasendashThis crack grows slowly under steady fatigue loading
but starts to accelerate as the net section decreases increasing the local stress at the crack-front
bull Third and Final Stage Failure ndashFailure occurs as an unstable fracture when the
remaining area is too small to support the load
hellip 2
bull Stages create distinctive features of the fracture surface - used to identify fatigue failure in failure investigations
5192016
7
Tell-tale marks of Fatigue - Striations
Features of Fatigue Phenomenon
bull Some materials exhibit endurance limits ie a stress below which the life is infinitendash Steels typical endurance limit - 40 -
60 of YS
ndash solutes (carbon nitrogen) present in the material pin the dislocations and prevent their motion at small strains
ndash Al alloys No endurance limits
ndash Related to the absence of dislocation-pinning solutes
hellip 3
bull At large Nf the lifetime is dominated by nucleation
ndash Therefore strengthening the surface (shot penning) is beneficial to delay crack nucleation and extend life
5192016
8
S
N
bull Typically load life and load life
bull Experimentally-determined ndash Rotating bending mc round specimens
ndash Electrohydraulic mc flat (or other) specimens
bull The practical limit of experimental testing has been 106 or 107 cycles due to impractical test times frequency limitations of machines and artefacts of higher frequencies
bull Adjustment for mean stress using Goodman Relation
Observations on S-N Curve
S
N
bull X-Axis No of Cycles On Log10 scale Log10 N
bull Y-Axis Stress ndash Either Stress Amplitude (Alternating stress) or Max
Stress
ndash Alternating Stress (ie Stress Amplitude) is the most dominant aspect of load that influences fatigue
ndash Either linear scale or Log10 scale
bull Curve is obtained for a given stress ratio R
bull R = (Min Stress Max Stress)
Observations on S-N Curve
5192016
9
S
N
bull S-N curve data is generally developed under Constant Amplitude ndash we will discuss later variable amplitude
bull The curve is dependent on many conditions ndash Loading ratio min load max load (R-ratio)
ndash Material
ndash stress concentrators
ndash surface finish
ndash frequency cyclic shape dwell etc hellip Not much effect unless in corrosive environment
Observations on S-N Curve
S
N
High cycle fatigue vs Low cycle fatigue
bull HCF Life usually more than 104 cycles ndash The distinction between high cycle fatigue and low
cycle fatigue is reflected in most design standards as high cycle fatigue design standards normally give S-N curves starting with N = 104 cycles
bull LCF Life usually less than 104 cycles ndash For low cycle fatigue design the stress range
concept is not immediately valid since typically the low cycle fatigue strength is governed by (large inelastic) strains
Arbitrary Could be 103
Arbitrary Could be 103
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
3
Stress ndash Life (S-N) Curve
Fatigue Cyclic (even if low) loads cause failures
Steel Ferrous Alloys
Many non-Ferrous Alloys
eg Al Alloys
5192016
4
Why should the cyclic loads lead to failure at lower loads
Origin of fatigue phenomenon
Fatigue Process Crack Initiation and Growth
Free Surface
STAGE I CRACK INITIATION Several micro-cracks originate and then coalesce into a macro-crack
STAGE II CRACK GROWTH
LOADING DIRECTION
Dislocations Slip Microcrack Macrocrack
5192016
5
Dislocations
Slip bands and micro-crack initiation
①
②
③
④
~ 01 μ
~ 01 μ
Persistent Slip Bands
5192016
6
Features of Fatigue Phenomenon
bull A gradually progressive process
bull Slow development of damage in the early stage followed by rapid growth towards the end
bull First stage crack initiation phase
ndashMay occupy most of the life in most cases (eg nicely machined parts small cyclic loads )
ndashUsually confined to a small area of high local stress
ndashAdjacent areas with only slightly lower stresses may not see any fatigue damage
ndashusually several independent micro-cracks grow and coalesce to form one dominant crack
hellip 1
Dislocations Slip Microcrack Macrocrack Crack Propagation Failure
Features of Fatigue Phenomenon
bull Second Stage Crack Propagation phasendashThis crack grows slowly under steady fatigue loading
but starts to accelerate as the net section decreases increasing the local stress at the crack-front
bull Third and Final Stage Failure ndashFailure occurs as an unstable fracture when the
remaining area is too small to support the load
hellip 2
bull Stages create distinctive features of the fracture surface - used to identify fatigue failure in failure investigations
5192016
7
Tell-tale marks of Fatigue - Striations
Features of Fatigue Phenomenon
bull Some materials exhibit endurance limits ie a stress below which the life is infinitendash Steels typical endurance limit - 40 -
60 of YS
ndash solutes (carbon nitrogen) present in the material pin the dislocations and prevent their motion at small strains
ndash Al alloys No endurance limits
ndash Related to the absence of dislocation-pinning solutes
hellip 3
bull At large Nf the lifetime is dominated by nucleation
ndash Therefore strengthening the surface (shot penning) is beneficial to delay crack nucleation and extend life
5192016
8
S
N
bull Typically load life and load life
bull Experimentally-determined ndash Rotating bending mc round specimens
ndash Electrohydraulic mc flat (or other) specimens
bull The practical limit of experimental testing has been 106 or 107 cycles due to impractical test times frequency limitations of machines and artefacts of higher frequencies
bull Adjustment for mean stress using Goodman Relation
Observations on S-N Curve
S
N
bull X-Axis No of Cycles On Log10 scale Log10 N
bull Y-Axis Stress ndash Either Stress Amplitude (Alternating stress) or Max
Stress
ndash Alternating Stress (ie Stress Amplitude) is the most dominant aspect of load that influences fatigue
ndash Either linear scale or Log10 scale
bull Curve is obtained for a given stress ratio R
bull R = (Min Stress Max Stress)
Observations on S-N Curve
5192016
9
S
N
bull S-N curve data is generally developed under Constant Amplitude ndash we will discuss later variable amplitude
bull The curve is dependent on many conditions ndash Loading ratio min load max load (R-ratio)
ndash Material
ndash stress concentrators
ndash surface finish
ndash frequency cyclic shape dwell etc hellip Not much effect unless in corrosive environment
Observations on S-N Curve
S
N
High cycle fatigue vs Low cycle fatigue
bull HCF Life usually more than 104 cycles ndash The distinction between high cycle fatigue and low
cycle fatigue is reflected in most design standards as high cycle fatigue design standards normally give S-N curves starting with N = 104 cycles
bull LCF Life usually less than 104 cycles ndash For low cycle fatigue design the stress range
concept is not immediately valid since typically the low cycle fatigue strength is governed by (large inelastic) strains
Arbitrary Could be 103
Arbitrary Could be 103
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
4
Why should the cyclic loads lead to failure at lower loads
Origin of fatigue phenomenon
Fatigue Process Crack Initiation and Growth
Free Surface
STAGE I CRACK INITIATION Several micro-cracks originate and then coalesce into a macro-crack
STAGE II CRACK GROWTH
LOADING DIRECTION
Dislocations Slip Microcrack Macrocrack
5192016
5
Dislocations
Slip bands and micro-crack initiation
①
②
③
④
~ 01 μ
~ 01 μ
Persistent Slip Bands
5192016
6
Features of Fatigue Phenomenon
bull A gradually progressive process
bull Slow development of damage in the early stage followed by rapid growth towards the end
bull First stage crack initiation phase
ndashMay occupy most of the life in most cases (eg nicely machined parts small cyclic loads )
ndashUsually confined to a small area of high local stress
ndashAdjacent areas with only slightly lower stresses may not see any fatigue damage
ndashusually several independent micro-cracks grow and coalesce to form one dominant crack
hellip 1
Dislocations Slip Microcrack Macrocrack Crack Propagation Failure
Features of Fatigue Phenomenon
bull Second Stage Crack Propagation phasendashThis crack grows slowly under steady fatigue loading
but starts to accelerate as the net section decreases increasing the local stress at the crack-front
bull Third and Final Stage Failure ndashFailure occurs as an unstable fracture when the
remaining area is too small to support the load
hellip 2
bull Stages create distinctive features of the fracture surface - used to identify fatigue failure in failure investigations
5192016
7
Tell-tale marks of Fatigue - Striations
Features of Fatigue Phenomenon
bull Some materials exhibit endurance limits ie a stress below which the life is infinitendash Steels typical endurance limit - 40 -
60 of YS
ndash solutes (carbon nitrogen) present in the material pin the dislocations and prevent their motion at small strains
ndash Al alloys No endurance limits
ndash Related to the absence of dislocation-pinning solutes
hellip 3
bull At large Nf the lifetime is dominated by nucleation
ndash Therefore strengthening the surface (shot penning) is beneficial to delay crack nucleation and extend life
5192016
8
S
N
bull Typically load life and load life
bull Experimentally-determined ndash Rotating bending mc round specimens
ndash Electrohydraulic mc flat (or other) specimens
bull The practical limit of experimental testing has been 106 or 107 cycles due to impractical test times frequency limitations of machines and artefacts of higher frequencies
bull Adjustment for mean stress using Goodman Relation
Observations on S-N Curve
S
N
bull X-Axis No of Cycles On Log10 scale Log10 N
bull Y-Axis Stress ndash Either Stress Amplitude (Alternating stress) or Max
Stress
ndash Alternating Stress (ie Stress Amplitude) is the most dominant aspect of load that influences fatigue
ndash Either linear scale or Log10 scale
bull Curve is obtained for a given stress ratio R
bull R = (Min Stress Max Stress)
Observations on S-N Curve
5192016
9
S
N
bull S-N curve data is generally developed under Constant Amplitude ndash we will discuss later variable amplitude
bull The curve is dependent on many conditions ndash Loading ratio min load max load (R-ratio)
ndash Material
ndash stress concentrators
ndash surface finish
ndash frequency cyclic shape dwell etc hellip Not much effect unless in corrosive environment
Observations on S-N Curve
S
N
High cycle fatigue vs Low cycle fatigue
bull HCF Life usually more than 104 cycles ndash The distinction between high cycle fatigue and low
cycle fatigue is reflected in most design standards as high cycle fatigue design standards normally give S-N curves starting with N = 104 cycles
bull LCF Life usually less than 104 cycles ndash For low cycle fatigue design the stress range
concept is not immediately valid since typically the low cycle fatigue strength is governed by (large inelastic) strains
Arbitrary Could be 103
Arbitrary Could be 103
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
5
Dislocations
Slip bands and micro-crack initiation
①
②
③
④
~ 01 μ
~ 01 μ
Persistent Slip Bands
5192016
6
Features of Fatigue Phenomenon
bull A gradually progressive process
bull Slow development of damage in the early stage followed by rapid growth towards the end
bull First stage crack initiation phase
ndashMay occupy most of the life in most cases (eg nicely machined parts small cyclic loads )
ndashUsually confined to a small area of high local stress
ndashAdjacent areas with only slightly lower stresses may not see any fatigue damage
ndashusually several independent micro-cracks grow and coalesce to form one dominant crack
hellip 1
Dislocations Slip Microcrack Macrocrack Crack Propagation Failure
Features of Fatigue Phenomenon
bull Second Stage Crack Propagation phasendashThis crack grows slowly under steady fatigue loading
but starts to accelerate as the net section decreases increasing the local stress at the crack-front
bull Third and Final Stage Failure ndashFailure occurs as an unstable fracture when the
remaining area is too small to support the load
hellip 2
bull Stages create distinctive features of the fracture surface - used to identify fatigue failure in failure investigations
5192016
7
Tell-tale marks of Fatigue - Striations
Features of Fatigue Phenomenon
bull Some materials exhibit endurance limits ie a stress below which the life is infinitendash Steels typical endurance limit - 40 -
60 of YS
ndash solutes (carbon nitrogen) present in the material pin the dislocations and prevent their motion at small strains
ndash Al alloys No endurance limits
ndash Related to the absence of dislocation-pinning solutes
hellip 3
bull At large Nf the lifetime is dominated by nucleation
ndash Therefore strengthening the surface (shot penning) is beneficial to delay crack nucleation and extend life
5192016
8
S
N
bull Typically load life and load life
bull Experimentally-determined ndash Rotating bending mc round specimens
ndash Electrohydraulic mc flat (or other) specimens
bull The practical limit of experimental testing has been 106 or 107 cycles due to impractical test times frequency limitations of machines and artefacts of higher frequencies
bull Adjustment for mean stress using Goodman Relation
Observations on S-N Curve
S
N
bull X-Axis No of Cycles On Log10 scale Log10 N
bull Y-Axis Stress ndash Either Stress Amplitude (Alternating stress) or Max
Stress
ndash Alternating Stress (ie Stress Amplitude) is the most dominant aspect of load that influences fatigue
ndash Either linear scale or Log10 scale
bull Curve is obtained for a given stress ratio R
bull R = (Min Stress Max Stress)
Observations on S-N Curve
5192016
9
S
N
bull S-N curve data is generally developed under Constant Amplitude ndash we will discuss later variable amplitude
bull The curve is dependent on many conditions ndash Loading ratio min load max load (R-ratio)
ndash Material
ndash stress concentrators
ndash surface finish
ndash frequency cyclic shape dwell etc hellip Not much effect unless in corrosive environment
Observations on S-N Curve
S
N
High cycle fatigue vs Low cycle fatigue
bull HCF Life usually more than 104 cycles ndash The distinction between high cycle fatigue and low
cycle fatigue is reflected in most design standards as high cycle fatigue design standards normally give S-N curves starting with N = 104 cycles
bull LCF Life usually less than 104 cycles ndash For low cycle fatigue design the stress range
concept is not immediately valid since typically the low cycle fatigue strength is governed by (large inelastic) strains
Arbitrary Could be 103
Arbitrary Could be 103
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
6
Features of Fatigue Phenomenon
bull A gradually progressive process
bull Slow development of damage in the early stage followed by rapid growth towards the end
bull First stage crack initiation phase
ndashMay occupy most of the life in most cases (eg nicely machined parts small cyclic loads )
ndashUsually confined to a small area of high local stress
ndashAdjacent areas with only slightly lower stresses may not see any fatigue damage
ndashusually several independent micro-cracks grow and coalesce to form one dominant crack
hellip 1
Dislocations Slip Microcrack Macrocrack Crack Propagation Failure
Features of Fatigue Phenomenon
bull Second Stage Crack Propagation phasendashThis crack grows slowly under steady fatigue loading
but starts to accelerate as the net section decreases increasing the local stress at the crack-front
bull Third and Final Stage Failure ndashFailure occurs as an unstable fracture when the
remaining area is too small to support the load
hellip 2
bull Stages create distinctive features of the fracture surface - used to identify fatigue failure in failure investigations
5192016
7
Tell-tale marks of Fatigue - Striations
Features of Fatigue Phenomenon
bull Some materials exhibit endurance limits ie a stress below which the life is infinitendash Steels typical endurance limit - 40 -
60 of YS
ndash solutes (carbon nitrogen) present in the material pin the dislocations and prevent their motion at small strains
ndash Al alloys No endurance limits
ndash Related to the absence of dislocation-pinning solutes
hellip 3
bull At large Nf the lifetime is dominated by nucleation
ndash Therefore strengthening the surface (shot penning) is beneficial to delay crack nucleation and extend life
5192016
8
S
N
bull Typically load life and load life
bull Experimentally-determined ndash Rotating bending mc round specimens
ndash Electrohydraulic mc flat (or other) specimens
bull The practical limit of experimental testing has been 106 or 107 cycles due to impractical test times frequency limitations of machines and artefacts of higher frequencies
bull Adjustment for mean stress using Goodman Relation
Observations on S-N Curve
S
N
bull X-Axis No of Cycles On Log10 scale Log10 N
bull Y-Axis Stress ndash Either Stress Amplitude (Alternating stress) or Max
Stress
ndash Alternating Stress (ie Stress Amplitude) is the most dominant aspect of load that influences fatigue
ndash Either linear scale or Log10 scale
bull Curve is obtained for a given stress ratio R
bull R = (Min Stress Max Stress)
Observations on S-N Curve
5192016
9
S
N
bull S-N curve data is generally developed under Constant Amplitude ndash we will discuss later variable amplitude
bull The curve is dependent on many conditions ndash Loading ratio min load max load (R-ratio)
ndash Material
ndash stress concentrators
ndash surface finish
ndash frequency cyclic shape dwell etc hellip Not much effect unless in corrosive environment
Observations on S-N Curve
S
N
High cycle fatigue vs Low cycle fatigue
bull HCF Life usually more than 104 cycles ndash The distinction between high cycle fatigue and low
cycle fatigue is reflected in most design standards as high cycle fatigue design standards normally give S-N curves starting with N = 104 cycles
bull LCF Life usually less than 104 cycles ndash For low cycle fatigue design the stress range
concept is not immediately valid since typically the low cycle fatigue strength is governed by (large inelastic) strains
Arbitrary Could be 103
Arbitrary Could be 103
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
7
Tell-tale marks of Fatigue - Striations
Features of Fatigue Phenomenon
bull Some materials exhibit endurance limits ie a stress below which the life is infinitendash Steels typical endurance limit - 40 -
60 of YS
ndash solutes (carbon nitrogen) present in the material pin the dislocations and prevent their motion at small strains
ndash Al alloys No endurance limits
ndash Related to the absence of dislocation-pinning solutes
hellip 3
bull At large Nf the lifetime is dominated by nucleation
ndash Therefore strengthening the surface (shot penning) is beneficial to delay crack nucleation and extend life
5192016
8
S
N
bull Typically load life and load life
bull Experimentally-determined ndash Rotating bending mc round specimens
ndash Electrohydraulic mc flat (or other) specimens
bull The practical limit of experimental testing has been 106 or 107 cycles due to impractical test times frequency limitations of machines and artefacts of higher frequencies
bull Adjustment for mean stress using Goodman Relation
Observations on S-N Curve
S
N
bull X-Axis No of Cycles On Log10 scale Log10 N
bull Y-Axis Stress ndash Either Stress Amplitude (Alternating stress) or Max
Stress
ndash Alternating Stress (ie Stress Amplitude) is the most dominant aspect of load that influences fatigue
ndash Either linear scale or Log10 scale
bull Curve is obtained for a given stress ratio R
bull R = (Min Stress Max Stress)
Observations on S-N Curve
5192016
9
S
N
bull S-N curve data is generally developed under Constant Amplitude ndash we will discuss later variable amplitude
bull The curve is dependent on many conditions ndash Loading ratio min load max load (R-ratio)
ndash Material
ndash stress concentrators
ndash surface finish
ndash frequency cyclic shape dwell etc hellip Not much effect unless in corrosive environment
Observations on S-N Curve
S
N
High cycle fatigue vs Low cycle fatigue
bull HCF Life usually more than 104 cycles ndash The distinction between high cycle fatigue and low
cycle fatigue is reflected in most design standards as high cycle fatigue design standards normally give S-N curves starting with N = 104 cycles
bull LCF Life usually less than 104 cycles ndash For low cycle fatigue design the stress range
concept is not immediately valid since typically the low cycle fatigue strength is governed by (large inelastic) strains
Arbitrary Could be 103
Arbitrary Could be 103
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
8
S
N
bull Typically load life and load life
bull Experimentally-determined ndash Rotating bending mc round specimens
ndash Electrohydraulic mc flat (or other) specimens
bull The practical limit of experimental testing has been 106 or 107 cycles due to impractical test times frequency limitations of machines and artefacts of higher frequencies
bull Adjustment for mean stress using Goodman Relation
Observations on S-N Curve
S
N
bull X-Axis No of Cycles On Log10 scale Log10 N
bull Y-Axis Stress ndash Either Stress Amplitude (Alternating stress) or Max
Stress
ndash Alternating Stress (ie Stress Amplitude) is the most dominant aspect of load that influences fatigue
ndash Either linear scale or Log10 scale
bull Curve is obtained for a given stress ratio R
bull R = (Min Stress Max Stress)
Observations on S-N Curve
5192016
9
S
N
bull S-N curve data is generally developed under Constant Amplitude ndash we will discuss later variable amplitude
bull The curve is dependent on many conditions ndash Loading ratio min load max load (R-ratio)
ndash Material
ndash stress concentrators
ndash surface finish
ndash frequency cyclic shape dwell etc hellip Not much effect unless in corrosive environment
Observations on S-N Curve
S
N
High cycle fatigue vs Low cycle fatigue
bull HCF Life usually more than 104 cycles ndash The distinction between high cycle fatigue and low
cycle fatigue is reflected in most design standards as high cycle fatigue design standards normally give S-N curves starting with N = 104 cycles
bull LCF Life usually less than 104 cycles ndash For low cycle fatigue design the stress range
concept is not immediately valid since typically the low cycle fatigue strength is governed by (large inelastic) strains
Arbitrary Could be 103
Arbitrary Could be 103
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
9
S
N
bull S-N curve data is generally developed under Constant Amplitude ndash we will discuss later variable amplitude
bull The curve is dependent on many conditions ndash Loading ratio min load max load (R-ratio)
ndash Material
ndash stress concentrators
ndash surface finish
ndash frequency cyclic shape dwell etc hellip Not much effect unless in corrosive environment
Observations on S-N Curve
S
N
High cycle fatigue vs Low cycle fatigue
bull HCF Life usually more than 104 cycles ndash The distinction between high cycle fatigue and low
cycle fatigue is reflected in most design standards as high cycle fatigue design standards normally give S-N curves starting with N = 104 cycles
bull LCF Life usually less than 104 cycles ndash For low cycle fatigue design the stress range
concept is not immediately valid since typically the low cycle fatigue strength is governed by (large inelastic) strains
Arbitrary Could be 103
Arbitrary Could be 103
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
10
HCF and LCF
bull HCFndash Typical of rotating components machineriesndash 100 RPM = 6000 cycles per hr = 144x105 day or
taking 8 hrday operation5x107 yrndash Automobile crankshaft may accumulate 108 cycles in
100000 km runndash Reciprocating machineries springshellip etc etc
bull LCFndash Airplane Ground-Air-Ground Cyclesndash Shipsndash Vehicle Chassishellip
Various types of fatigue
bull Fretting fatiguendash Fatigue associated with or aided by repetitive rubbing
of surfaces causing wear and cracks also often accompanied by corrosion Such cracking provides sites for fatigue damage and decreases fatiguestrength of materials operating under cycling stress
bull Sonic fatigue (Acoustic fatigue) ndash Fatigue caused by acoustic vibrations acoustic
pressures engine noise can generate high stresses
bull Thermomechanical fatigue ndash Mechanical loads (cyclic) also accompanied by
thermal cycles
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
11
Endurance Limit Fatigue Limit Fatigue Strength
All these terms refer to ldquohow much stress amplitude can be applied in a cyclic loading without causing a failurerdquobull Endurance Limit A stress level below which a material can
be cycled indefinitely without failurendash Several Steels Ti-alloys Mo-alloys etc exhibit such endurance
limit
Many materials (eg Al-alloys Mg Cu Ni Alloys even some steels) do not have Endurance Limit For such materials we define bull Fatigue Strength as the max stress amplitude that can be
applied to get given life (say 106 cycles 5x108 cycles etc)bull Fatigue Limit as the limiting value of stress amplitude at
which the life (no of cycles to failure) becomes ldquovery largerdquo
[Theoretically there is no ldquoproofrdquo of endurance limithellipNo infinite lifehellip it may be just that we havenrsquot tested for large number of cycles So Fatigue Limit is a more realistic term]
22
Axial loading
S-N Curve Fatigue Tests Test Arrangements Rotating Bending loading
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
12
Fatigue Strengths for SteelsRotating bending 107 or 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Carbon Steels Alloy Steels
Fatigue Strengths for Al-alloysRotating bending 108 Cycles
Tensile Strength Su ksi
Alt
Fat
igu
e St
ren
gth
Sf
ksi
MP
a
MPa
Ref Ali Fatemi eFatigue Chap4
x Wrought Cast
Metal Fatigue in Engineering by Ralph I Stephens Ali Fatemi Robert R Stephens Henry O Fuchs John Wiley amp Sons Nov 2000
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
13
Some related ASTM test standards
Designation Title
E466 - 15Standard Practice for Conducting Force Controlled Constant Amplitude Axial Fatigue Tests of Metallic Materials
E468 - 11Standard Practice for Presentation of Constant Amplitude Fatigue Test Results for Metallic Materials
E606 E606M - 12
Standard Test Method for Strain-Controlled Fatigue Testing
E2714 - 13 Standard Test Method for Creep-Fatigue Testing
E2789 -10(2015)
Standard Guide for Fretting Fatigue Testing
E2948 - 14Standard Test Method for Conducting Rotating Bending Fatigue Tests of Solid Round Fine Wire
Cyclic Deformation and Fatigue Crack Formation
E739 -10(2015)
Standard Practice for Statistical Analysis of Linear or Linearized Stress-Life (S-N) and Strain-Life (ε-N) Fatigue Data
E1049 -85(2011)e1
Standard Practices for Cycle Counting in Fatigue Analysis
log 119878119886 = log 119886 + 119887 log119873there4 119878119886 = 119886119873119887
bull Often N is written as 2Nf
which is no of reversals
bull Constant a is generally representative of the ldquoFracture Strengthrdquo of material say 119878119891
prime (also called ldquoFatigue Strength Coeffrdquo)
there4 119878119886 = 119878119891prime 2119873119891
119887
119874119877 119878119886 = 119878119891prime119873119887
bull b is Basquin (Year 1910) Exponent
bull b -005 to -015
Observations on S-N Curve
Sa
Log N
Log N
log Sa
1
b
BasquinRelation
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
14
Nomenclature in Cyclic Loading
120590119898119886119909
120590119898119894119899
120590119898
120590119886 ∆120590
0
∆120590 ∶ 119878119905119903119890119904119904 119877119886119899119892119890 = 120590119898119886119909 minus 120590119898119894119899
120590119886 119860119897119905119890119903119899119886119905119894119899119892 119878119905119903119890119904119904 = ∆1205902 = (120590119898119886119909minus120590119898119894119899)2
120590119898 119872119890119886119899 119878119905119903119890119904119904 = (120590119898119886119909+120590119898119894119899)2
119877 119878119905119903119890119904119904 119877119886119905119894119900 =120590119898119894119899
120590119898119886119909
120590119898119894119899 119872119894119899119894119898119906119898 119878119905119903119890119904119904
120590119898119886119909119872119886119909119894119898119906119898 119878119905119903119890119904119904
Constant Amplitude Fatigue Loading
120590119898119886119909 = 100 120590119898119894119899= minus100120590119898 = 0 120590119886 = 100119877119886119899119892119890 = 200 119877 = minus1
120590119898119886119909 = 100 120590119898119894119899= 20120590119898 = 60 120590119886 = 40119877119886119899119892119890 = 80 119877 = 02
120590119898119886119909 = minus10 120590119898119894119899= minus120120590119898 = minus65 120590119886 = 55119877119886119899119892119890 = 110 119877 = 12
0
100
minus100
Tension - Compression
0
100
20
Tension - Tension
0minus10
minus120
Compression - Compression
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
15
Constant Amplitude Fatigue Loading
Comp ndash Comp R gt 1
0 ndash Comp R = infin
Tension ndash Comp R = -1
0 ndash TensionR = 0
Tension - Tension 0 lt R lt 1
Time
Stress
Effect of Mean Stress
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
16
Mean Stress Effect
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
bull The damaging portion of load cycle Tensile stresso this causes cracks to open and grow
bull Baseline Sa-N curve Zero mean stress (Sm=0) and cyclic stress amplitude (alternating stress) Sa
bull Let us superimpose an additional steady stress Sm on this
Sm lt 0
bull If Smlt0 (ie Compr) Max Tens Stress and N The curve shifts to right
Sm gt 0
bull If Smgt0 (ie tensile) Max Tens Stress and N The curve shifts to left
Goodman Relation
Sm = 0
Stre
ss A
mp
litu
de
Sa
Life Cycles N
Sa0
N
1
1
3
3
4
4Constant Life N
SmSu
SaSa0
1
1
2
2
Sm gt 0
Su
m = 1
m lt 1
m gt 1
119878119886
1198781198860+
119878119898
119878119906
119898
= 1
In general m ~ 06 ndash 2For most calculations we assume m=1
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
17
Mean Stress Effect Goodman amp Others
httpwwwfgguni-ljsi~pmozeesdepmasterwg12l0200htm
Goodman Gerber Soderberg LinesFailure envelopes
Gerber Parabola(m = 2)
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
18
Goodman Diagram Constant Life Lines
Constructing failure envelopes using Goodman Diagram
Constant Life Lines
Constant Load Lines
httpneilwimerweeblycombasic-fatigue-analysishtml
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
19
Effect of R-ratio
Al Alloy 2024 T3Axial Loading
S max
ksi
N
102 103 104 105 106 107 108
40
60
0
80
20
73 ksi = UTS
R= SminSmax
+060
+050
+040
+025
+010
0
-030
-10
Ref Fatiue of Aircraft Structures NAVAIR 01-1A-13 1966
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
20
Numerical Example 1
In a Tension-Compression (R=-1) fatigue test with an alloy following mean fatigue lives were found for various stress levels
The ultimate strength of the alloy was found to be 462 MPa(i) Plot the Sa-N curve (ii) Check whether this obeys Basquin relation Find
Basquin exponent(iii) Using Goodman relation derive the Sa- N curve for
R=0
Smax (MPa) 138 172 207 276 345 380
N 12589250 1273500 199526 15850 1778 1060
Solution
0
100
200
300
400
10E+02 10E+04 10E+06 10E+08
Sa vs N
0
05
1
15
2
25
3
0 2 4 6 8
log Sa vs log N
Since R=-1 Sa=Smax
Basquin Exponent -0107Sfrsquo = 102892 = 780 MPa
WE take m=1 For R=0 Sm=Sa so that the eqn is 119878119886
1198781198860+
119878119898
119878119906
119898
= 1
119878119886
1198781198860+
119878119886
119878119906= 1rArr 119904119886 =
11
1198781198860+
1
119878119906
Substituting values of Sa0 we get Sa for each value of N Then we can plot
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
21
Solutionhellip continued
N 12589250 1273500 199526 15850 1778 1060
Smax (MPa) 138 172 207 276 345 380
Sa0 (R=-1) 138 172 207 276 345 380
Sa (R=0) 106 125 143 173 198 209
0
50
100
150
200
250
300
350
400
10E+02 10E+03 10E+04 10E+05 10E+06 10E+07 10E+08
Sa vs N
R = -1
R = 0
Numerical Example 2
For an Aluminium Alloy the S-N behaviour in tension compression fatigue with zero mean stress is approximated by the relation
119878119886= 480119873minus012
(i) What would be the life of a structure made of this alloy if it is subjected to cyclic stress of 80 MPa
(ii) What would be life if there is an additional steady stress of +10 MPa
(iii) What would be life if the additional steady stress in -10 MPa
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
22
Solution to Numerical Example 2
(i) This is direct substitution in the given stress-life curve equationWe have Sa = 80 MPa Therefore
80 = 480119873minus012
This gives N = (48080)^(1012) = 3052065
(ii) With an additional steady stress of +10 MPaHere the mean stress is now Sm=10 MPa and Sa=80 MPaSince mean stress is not zero we need to use Goodman relation to arrive at the mean stress effect
With mean stress zero 1198781198860= 119878119891prime119873119887
Recall Goodman relation
In absence data we take Su = Sfrsquo Now we can just substitute for Sa(=80) and Sm (=10) and Su=Sfrsquo = 480 with b= -012 and get N
Ans N= 2560937
(iii) Similarly with Sm = -10 we can repeat calculation
Ans 3624244
119878119886
1198781198860+
119878119898
119878119906= 1rArr
119878119886
119878119891prime119873119887 +
119878119898
119878119906= 1
Stress ConcentrationHole in a large plate under uniaxial tension
bull Large plate Max stress is 3 times the nominal stress (applied stress far-field stress)
bull Stress concentration Factor Kt = SmaxSn = 3
Elliptic hole SCF = 1+ 2(ba) = 1+2(a)
smax
sns
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
23
Stress Concentration (contd)
Petersonrsquos Original
compilationndash updated and
enhanced
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
24
Fatigue under Stress Concentration
S
N
s
s Krsquo (KrsquoltKt)
s Kt
Notched
Un-notched
Effective Stress Concentration for fatigue is less than the Stress Concentration in Static
Many types of discontinuities which act as stress concentrators Holes cut-outs key-slots cracks notches hellip
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119904119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ
119870119891 = 1 + 119902 119870119905 minus 1
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
Fatigue notch sensitivity factor
Stress Concentration Factor
119870119905 =119872119886119909 119904119905119903119890119904119904 119908119894119905ℎ 119899119900119905119888ℎ
119878119905119903119890119904119904 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
Fatigue Concentration Factor
119870119891 =119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ119900119906119905 119899119900119905119888ℎ
119865119886119905119894119892119906119890 119878119905119903119890119899119892119905ℎ 119908119894119905ℎ 119899119900119905119888ℎ 1 lt 119870119891 lt 119870119905
Fatigue Notch Sensitivity Factor q119870119891 = 1 + 119902 119870119905 minus 1 and
119902 =119870119891 minus 1
119870119905 minus 1 0 lt 119902 lt 1
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
25
Estimates of 119902 ∶ All Empirical
bull Neuberrsquos Approx Formula (R=-1) 119954 =120783
120783+ Τ120646 119955(r = notch root radius 120588 = material characteristic length)bull For aluminium alloys
Ult Strength 150 300 600 MPa120588 20 06 04 mm
bull Petersonrsquos Approx Formula (R=-1) 119954 =120783
120783+119938
119955(r = notch root radius 119886 = another mat characteristic length)
bull For Steels 119886 = 002542070
119878119906
18
(Su in MPa a in mm)
bull For Al-alloys 119886 = 0635 119898119898 119886119901119901119903119900119909
NeuberParameter
0
005
01
015
02
025
03
0 500 1000 1500
Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500Ne
ub
er
Par
amet
er
(
mm
)
Su MPa
0
05
1
15
2
25
0 500 1000 1500
Ne
ub
er
Par
ame
ter
Su MPa
Al-Alloys
Steels
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
26
Goodman Diagram for notched fatigue
bull Rather simplisticbull Does not
account for variation of Kf
with mean stress or life cycles
S a
Sa
Accounting for means stress effectHaig Diagrams
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
27
WADC TECHNICAL REPORT 52-307 PART 1 FATIGUE PROPERTIES OF ALUMINUM ALLOYS AT VARIOUS DIRECT STRESS RATIOS ADS0007610 1955
A= SaSm
More about notched fatiguehellip
bull Mean stress effect is more pronounced in notched specimens than in smooth ones
bull Tensile mean stress accentuates the notch effect in fatigue hellip in fact Kf can exceed Kt in some cases Need to be catered for otherwise fatal in fatigue loading
bull Compressive mean stress can be beneficial It significantly reduces the effect of stress concentration and may in fact eliminate it
bull Residual stresses can contribute significantly to the level of mean stress This in turn may affect the notched behaviour
5192016
28
Questions
5192016
28
Questions