asme fatigue for engineers part iii
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Fatigue fo r Eng ineers
Prepared b y
A. F. Grandt, J r.
Professor of Aeronautics and Astronautic sPurdue University
W. Lafayette, IN 47907
June 1999
2
Object ive
Overview nature/consequences of the
fatigue failure mechanism
Determine number of cycles required to
develop a fatigue crack
propagate a fatigue crack
Discuss implications of fatigue on
design and maintenance operations
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3
Struct ural Failure Modes
Excessive Deformation
Elastic
Plastic Buckling
Fracture
Creep
Corrosion
Fatigue
Forc
e
Displacement
Yield
Permanent
displacement
displacement
Force
4
Fatigue Failure Mechanism
Caused by repeated (cyclic) loading
Involves crack formation, growth, and final
fracture
Fatigue life depends on initial quality, load, . . .
St
ress
Time
Crack Nucleation
Fracture
Crack Growth
Elapsed Cycles N
CrackLength(a)
a
Crack
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Paper Cl ip Exp er iment
Bend wire repeatedly until fracture
Note:
life (number of applied load cycles)depends on:
applied stress amplitude
component quality (notches, scratches, etc.)
heat emitted >> plastic deformation
6
Character is t ics o f Fat igue
Brittle fracture surface appearance Cracks often form at free surface
Macro/micro beach marks/ striations
0.3 in
Beach marks
20 m
Striations
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Fatigue is problem for m any
types of s t ruc tures
8
Exerc ise
Describe fatigue failures from yourpersonal experience
What was cause of fatigue failure?
What was nature of cyclic load?
Was initial quality an issue?
How was failure detected?
How was problem solved?
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Exerc ise
Estimate the fatigue lifetime needed for:
Automobile axle
Railroad rail
Commercial aircraft components landing gear
lower wing skin
Highway drawbridge mechanism
Space shuttle solid propellant rocket motorcases
10
Exerc ise
Give an example of a High Cycle
Fatigue (HCF) application.
What is the required lifetime?
What are consequences of failure?
Given an example of a Low Cycle
Fatigue (LCF) application.
What is the required lifetime?
What are consequences of failure?
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Fatigue Crack Form ation
12
Crack Formation
Fracture
Crack Growth
Elapsed Cycles N
CrackLength(a)
Fatigue Crack Form ation
Objective Characterize resistance to fatigue crack formation
Predict number of cycles to initiate small*fatigue crack
in component
*crack size ~ 0.03 inch
= committee crack
Approach Stress-life concepts
(S-N curves)
Strain-life concepts
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Stress-l i fe (S-N) Appro ach
Concept: Stress range controls fatigue life
S
S
Log cycles N
S/2
Note:
Life increases as load amplitude decreases Considerable scatter in data Run-outs suggest infinite life possible
Life N usually total cycles to failure
S
time
S
14
Model Stress-life (S-N) Curv e
Se = endurance limitfor steels
Se ~ 0.5 ultimate stress Sult
Se ~ 100 ksi if Sult 200 ksi
Log reversals 2N
LogS
/2
Se
S/2 = f (2N)b
f = fatigue strength coefficient b = fatigue strength exponent
typically -0.12 < b < -0.05
Note: Measure life in terms of reversals 2N
(1 cycle = 2 reversals)
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S-N Curv e: Mean Stress
Mean stress effects lifestress ratio R = S
min/ S
max
Smean
= 0.5(Smin
+ Smax
)
Sa = 0.5(Smax - Smin) = S/2Mean stress models
Sa/Se + Sm/Sult = 1
S/2 = (f - Smean)(2N)b
Mean StressStressAmplitude
N = 106
N = 103
Haigh constant life diagram
S
timeSmin
Smax
S = 2Sa
16
S-N Curve: Other Factors
S-N curves are very sensitive to surface finish, coatings, notches
prior loading, residual stresses
specimen size effects, etc.
Many empirical knock-down factors
S-N approach best suited for HCF (HighCycle Fatigue) applications
limited by local plastic deformation
strain-life approach better for LCF (Low
Cycle Fatigue)
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Strain -life (- N) Appro achConcept: Strain range controls lifeExperiment
Control
Measure Reversals (2Nf)to failure (1 cycle
= 2 reversals)
Stable stress range needed to maintain
Note: stable usually occursby mid-life (2Nf/2)
time
time
18
Cyclic Stress-Strain Curve
Relate stable cyclic stress and strain ranges
t ime
time
Hystersis l oop
/2
/2
/2 = /2E + (/2K )1/n
Cyclic stress-strain curve
E = elastic modulus
K = cyclic strength coefficient
n = strain hardening exponent
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Plast ic Strain-Life Cu rve
Relate plastic strain amplitude p/2with reversals to failure 2Nf
Compute p/2 = /2 - /2E = total - elastic strain amplitudes
Log
p/2
Log 2Nf
p/2 = f (2Nf)c
f = fatigue ductility coefficient
c = fatigue ductility exponent
typically -0.7 < c < -0.5
20
Total Strain-Life Curv e
Plot total strain amplitudes versus life 2Nf
total
/2 = /2 = 0.5 elastic
+0.5 plastic
=/2E + 0.5
plasti c
/2 = {(f - Smean )/E}(2N)b + f (2Nf)c
p/2 = f (2Nf)c
/2E = {( f - Smean)/E} (2Nf)b
Log 2N f
Logstrainamplitude
2Nt = transition life
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Total Strain-Li fe
Note:
Plastic strain dominates for LCF
Elastic strain dominates for HCF
Transition life 2Nt separates LCF/HCF
p =f (2Nf)c
/2 = {(f - Smean )/E} (2N)b + f (2Nf)c
Log 2N f
Logstrainamplitude
/2E = {( f - Smean)/E}(2Nf)b
2N t = transition life
LCF
HCF
22
Variable Am pl i tude Loading
Load amplitude varies in many applications
Use of constant amplitude S - N or- Ndata requires damage model
Miners rule*
(Ni/Nf) = 1
Ni= number of applied cycles of stress amplitude S
ai
Nf= fatigue life for Sai cycling only
*Use with caution!
S
time
N i
2Sai
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Example Problem
Assume:
f = 220 ksi, b = - 0.1 stress history shown (1 block of loading)
Find: number of blocks to failure
+ 80 ksiS
time
- 80 ksi
- 100 ksi
+ 100 ksi
2N = 100
2N = 1000
2N = 1000S
S
24
Solut ion
(Ni/Nf) = 1 2Nf = {(S/2) / (f - Smean)}1/b
(Ni/Nf) = 1
When:
1/0.0089= 112.5
Answer
112 blocks
S/2(ksi)
Smean(ksi)
2Nf 2Ni Ni/Nf
80 0 24,735 100 0.0040
50 +50 206,437 1000 0.0048
50 -50 21 E6
1000 4.74 E-6
0.0089
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Notch Fatigue Notches can reduce life
Define Fatigue Notch Factor
Kf
Kf = Smooth/notch fatigue
strength at 106 cycles
= Ss /Sn1 < Kf < Kt
(K t = elastic stress
concentration factor)
Kf= 1 no notch effectK
f= K
t full notch effect
Smooth
Notch
S/2
Log cycles N
S s /2
Sn /2
106
25
Load Sequence Effects
Hi-lo strain sequenceresults in tensile meanstress when last large peak in compressionas shown here
decreases life! If last peak had been
tension, would result incompressivemeanstress
increaselife
Load sequence important!
t
t
Mean stress
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Neuber s Rule
Kf= fatigue notch concentration factor
(s,e) = nominal stress/strain ranges(away from notch)
(,) = notch stress/strain rangesNeubers rule relates notch and
nominal stress/strain behavior
Solve with:
Kf
2se =
/2 = /2E + (/2K )1/n
/2 = {(f - Smean)}(2Nf)b + f (2Nf)c
(,)
(s,e)
28
Summ ary Ini t iat ion Methods
Total strain-life approach combines: original S-N curve (best suited for HCF) and
plastic strain-life method developed for LCFproblems
S-N and strain-life often viewed as crack
initiation approaches
actually deal with life to form small crack
crack size implicit in specimen/test procedure
typically assume committee crack ~ 0.03 in.
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In i t iat ion Summary Cont
Notches increase local stress/strain andoften are source for crack formation
complex problem leads to local plasticity
characterize by fatigue notch concentrationfactor Kf,, Neubers rule
Load interaction effects result in local
mean stress
can increase/decrease life
invalidate Miners rule
30
Fatigue Crack Grow th
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Crack Growth App roach
Assumes entire life
fatigue crack growth
ignores initiation assumes component
cracked before cycling begins
Used with damage tolerant design
protects from pre-existent (or service) damage
based on linear elastic fracture mechanics
Elapsed Cycles N
Crack Growth
CrackLength(a)
Fracture
Initial crack
32
Damage Tolerance
The ability of a structure to resist priordamage for a specified period of time
Initial damage
material
manufacturing
service induced
size based on
inspection capability,
experience, . . .time
Cracksize
Desired Life
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Fatigue Crack Grow th
Objective Characterize material resistance to fatigue crack growth
Predict catastrophic fracture and subcritical crack
growthApproach
Assume crack growth
controlled by stress
intensity factor K
fracture
growth rate da/dN
Elapsed Cycles N
Crack Growth
CrackLength(a)
Fracture
Initial crack
34
Stress Intensi ty Factor KI
KI is key linear elastic fracture mechanicsparameter that relates:
applied stress: crack length: a
component geometry: (a)((a) is dimensionless) a
Crack
= 1.12
aK=I
Note units: stress-length1/2
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Stress Intensi ty Factors
2a
W
K a Seca
W=
12
=Remote Stress
20 95
a
W .
W
a
h
a
W
0 6.
a
W
h
W
10.
K a
a
W
a
W
=
=
+
112 0 231 10.55. .a
W
a
W
a
W
+
21 73 30 39
2 3 4
. .
For and
Many K Isolutionsavailable
36
Crack t ip Stress Fields
( )
+==
==
=
+=
=
yxz
z
yzxz
Ixy
Iy
Ix
r
Kr
Kr
K
strainplane
0stressplane
02
3cos
2cos
2sin
2
2
3sin
2sin1
2cos
2
2
3sin
2sin1
2cos
2
Theory of elasticity gives elastic stresses near crack tip in
terms of stress intensity factor K I
All crack configurations have same singular stress field at tip(are similar results for other modes of loading, i.e., modes II and III)
Crack
x
y
r
xy
y
x
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KcFracture Cri terion
Fracture occurs when
K > constant = Kc K
c= material property
= fracture toughness Criterion relates:
crack size: a
stress: geometry: (a) material: Kc
Plasticity limits smallcrack applications
2a
ult
FractureStress
Crack Size a
( )K a ac =
38
Fracture Toug hness Kc
Typical Kc values (thick plate)
Note Kcdepends on:
specimen thickness -- Kc decreases as
thickness increases until reaching minimum -
KIc = plane strain toughness
crack direction (material anisotropy)
M aterial
(thickplate)
2024-T351
Aluminum
7075-T651
Aluminum
Ti-6Al-4V
Titanium
300M steel
(235ksiyield)
18Nickel
(200ksiyield)
Kc
(ksi-in1/2)
31 26 112 47 100
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Fracture Examp le
Member A fractures whencrack length a = 2.0 inch
and remote stress = 5 ksi
What stress will fracturemember B (assume samematerial)?
2.0 in
4.0 in
5 ksi
5 ksi
A
5 in
8 in
= ?
= ?
B
40
Fracture Examp le Solut ionEdge crack
K = (a)1/2(a) = Kc at fracture
a/w = 2/4 = 5 a = 2 = 2.83K
c= 35.5 ksi-in1/2= constant
Center Crack
K = ( a)1/2(a) (a) = [Sec ( a/W)]1/2
a = 2.5 W = 8 = 1.34K = Kc at fracture = 35.5
2.0 in
4.0in
5 ksi
5 ksi
5 in
8 in
= ?
= ?
a
W
a
W
=
+ 1 12 0 231 10. 5 5. .
a
W
a
W
a
W
+
21 73 30 39
2 3 4
. .
f = 9.5 ksi
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Fat igue Crack Grow th
Goal: show cyclic stress intensity factorKcontrols crack growth rate da/dN
P = constant
time
P
2a
P
Crack Face Load
2a
Remote Load
= constant
time
Same material
Different loadings
42
Measure Crack Growth
2a
Remote Load
2a
P
Crack Face Load
da
dN
CrackLength(a)
Number of Cycles (N)
=K PBa
K = a
CrackLength(a)
Number of Cycles (N)
da
dN
a*
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Corr elate Rate da/dN vsK
CrackLength(a)
Number of Cycles (N)
da
dN
2a
2a
CrackLength(a)
Number of Cycles (N)
da
dN
a*
KthK
c
Log K
Logda/
dN
K a=
K PB a
=
44
da/dN VsK
KthKc
Log K
Logda/dN
Note: K correlates fatiguecrack growth rate da/dN
K accounts for crackgeometry
No crack growth for
da/dN
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Sample Crack Grow th Data
da/dN - K data for7075-T6 aluminum
Note effect of stress
ratio R = min/maxstress (da/dN as R) Reference: Military
Handbook-5
Other handbook dataare available
46
Model d a/dN -K Curve
Fit test data with numericalmodels such as:
KthKc
Log K
Logda/dN
da
dNF K= ( )
da
dNC Km=
da
dN
C K
R K K
m
c
=
( )1
Here C, m, Kc are
empirical constantsR = min/max stress
(are many other models)
Paris
Forman
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Compute Fatigue Li fe Nf
ao, af = initial, final crack sizes
F(K) = function of:
cyclic stress: , R, . . . crack geometry: (a) crack length: a
material
Nda
F Kf
a
a
o
f
= ( )da
dNF K= ( )
time
2a
48
Example L ife Calculat ion
a
Crack
= constant
time
Given: edge crack in wide plate
Kc= 63 ksi-in1/2initial crack ai = 0.5 inchcyclic stress = 10 ksi, R = 0
( = max = 10 ksi)
da/dN = 10-9K4
Find: a) cyclic life Nf b) life if initial crack size
decreased to ai= 0.1 inch
Note: at fractureK = Kc = 63 = 1.12max (a)1/2
final crack af = 10 inch
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Solut ion
[ ] = =
da
C K
da
C am ma
a
a
a
o
f
o
f
112. Nf
( ) ( )[ ]N
C ma af m f
m
o
m=
1
1 12 1 5
1 5 1 5
. .
. .
K a= 112.da
dNC K
m=
a) Nf= 12,234 cycles (ai = 0.5)
b) Nf= 63,747 cycles (ai = 0.1)
Note: big influence of initial crack length!
50
Fat igue Crack Retardat ion
Time
AppliedStress(
)
Overload
Without Overload
With Overload
RetardationCrackLength(a)
Elapsed Cycle (N)
Note load interaction effect Tensile overload can retard crack growth (increase life) Life increase due to crack tip plasticity
Depends on magnitude/sequence of overload, material, Are empirical retardation models
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Crack Growth Summary
Fracture mechanics approach assumesentire fatigue life is crack growth
Stress intensity factor K controls fractureand growth rate da/dN
K = [a]1/2(a) Fracture: K = Kc Fatigue: da/dN = F(K) Integrate da/dN for life
Are load interaction and other effects (see
references)
51
Cycle-by-Cycle Calculat io n
Compute cycle-by-cycle growth in crack length a
acurrent = aprior+ da/dNcurrent
da/dNcurrent = F(Kcurrent) * Retardation term
Sum for all cycles in spectrum
Powerful technique for computer programming
n
n+1AppliedStress()
Time (t)
Variable amplitudeloading prevents
simple life integration
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Fatigue Desig n/Repair
Concepts
54
Des ign Phi losophies
Fatigue Design Criteria
Infinite Life
Safe-Life
Damage Tolerant
Fail-safe
Slow crack growth
Retirement-for-cause
a
Crack
St
ress
Time
Crack Formation
Fracture
Crack Growth
Elapsed Cycles N
Pre-CrackCrackLength(a)
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Inf in ite Life Criter ion
Design Goal: prevent fatigue damage from ever
developing (i.e. infinite life)
Usually based on endurance limit
Could also employ threshold K concepts
Leads to small design stresses/heavy members
Limited to simple components/loading
Often impractical/not achievable in practice
Weight critical structure
Complex loads
56
Safe-Life Criter io n
Design goal: component is to remain crack free forfinite service life
Assumes initial crack-free structure
Establish mean life by test/analysis
Safety factors account for scatter
predicted mean
Desired life = mean/S.F.
Design Life
Failure
Occurrence
1 32 4
Problems:
large safety factor no protection from
initial damage
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Fail-Safe CriterionDesign goal: contain single component failure
without losing entire structure
Assumes crack is present
Provide alternate load paths, redundant structure, crack
stoppers, etc. Requires detection of 1st failure
Time
Cracksize
1st member
2nd memberCrack arrest
58
Slow Crack Growth Cri ter ion
Design goal: prevent initial crack from growing tofracture during life of structure
Pre-existent crack size specified by inspectionlimits, experience
Crack growth life
> service life x S.F.
Based on fatigue
crack growth
resistance
Emphasizes nondestructive inspection
Cracksize
Desired Life
time
Fracture
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Retirement-for-Cause
Failure size
Crack
Length
Time
inspect/repair
Design goal: Use periodic inspection/repair
to achieve desired fatigue lives
Limited by repeated maintenance economics
60
Life Extension Concepts
Shot peenHole coldwork
Interference fastenersOverstress, etc.
Introduce BeneficialResidual Stresses
MetalComposite
Mechanical FastenBond
Doublers
HCF damping materials
Reduce Stressvia Reinforcement
Weight limitsFlight restrictions
etc.
Reduce OperatingLoads
No Cracks Found
(assume small cracks)
MetalComposi te Mechanical Fasten
Bond
Patches
Replace componentStop drill cracks
Welding
Repair CrackedStructure
Cracks Found
ComponentInspection
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Summary
Fatigue is complex problem that
involves many disciplines
Fatigue affects design and operation of
many types of structures Fatigue may be treated by several
methods/philosophies
Assume component cracked
Assume component uncracked
Probabilistic methods
62
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