predavanje iz zamora materijala
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Dynamic Loadingol id Mechan ics Ander s Ekber g
1 (8)
Dynamic loading
More difficult tomeasure, analyzeand estimatedynamic loading compared to static loading
Loads can vary in time and space
Load range(or more a ccurately stress range) is themost important parameter in fatigue analysis
This calls for a definition of a load cycle(or ra therstress cycle)
There can be large statistical scatter in the loading
High frequency contentof the loa ding is difficult tomea sure a nd/or ana lyze. This content mayhave an
effect on the fa tigue behaviourStandardized load spectraa re often used a s input to thefatigue models.
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Dynamic Loadingol id Mechan ics Ander s Ekber g
2 (8)
Varying amplitudes
Residual stresses
Overloads may introduce residual stressesdue toplastic deformations.
These stresses may supress the initiationof fa tiguecracks and/or lead to closureof existing cracks during(parts of) the loa d cycle
This is normallybeneficial
Add influence from different load cycles
This is far from obvious
In continuum approaches (initiation), this is done bydamage accumulation
In crack propagat ion a nalysis, it is done by the use of acrack propagation law.
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Dynamic Loadingol id Mechan ics Ander s Ekber g
3 (8)
Load cycles
A loa d cycle is a closedloop in loa d space
For harmonic loading, the load cycle
starts from a certain load magnitude
moves through a max-value and a min-value back tothe sta rt magnitude (or the other wa y a round)
The load cycle is then completely defined by theamplitudeand mid value
t
F
t
F
The problem in identifying a load cycle comes when weare not dea ling with harmonic loa dpaths
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Dynamic Loadingol id Mechan ics Ander s Ekber g
4 (8)
t
t
Rainflow counting
D epict the loading sequence as a function of time. Forconvenience
start with largest maximum or smallest minimum use straight lines between (local) minima and maxima
Sta rt from the top and let a drop sta rt from every maximumand minimum. A drop stops if:
it sta rts from max and passes a larger or equal max it st arts from min and passes a larger or equal min it reaches the run of another drop
Identify closed loops by joining drops
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Dynamic Loadingol id Mechan ics Ander s Ekber g
5 (8)
t
1
2
3
5
4
67
8
Rainflow counting example
1 passes an equally large maximum
2 passes a larger minimum
3 passes a larger maximum
4 reaches the run of drop 2
5 reaches the run of drop 1
6 fa lls out
7 fa lls out
8 reaches the run of drop6
1and 6, 2and 5, 3 and 4; 7and 8are running
the same distances in the opposite directions.
These couples are forming closed loopsin load space and can thus be identified asstress cycles with minimum and maximum
magnitudes and a mid va lue.
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Dynamic Loadingol id Mechan ics Ander s Ekber g
6 (8)
Rainflow counting notes
R ainflow counting is used to identify the most damagingstress cycles
When applying the ra inflow count method, the
chronological orderof applied loads diminishes. In otherwords, you no longer know at what instant of time the overloa ds and (their pertinent residual stresses) occur
More tha n one loa d component ra inflow counting
should be carried out on the stressThe ra inflow count method is only applicable foruniaxial loading. There are theories for how to extend themethod to multiaxial loading (however not generallyaccepted)
The rainflow count method is fairly easy to implementina computer code. Several such codes exist and are inpractical use
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Dynamic Loadingol id Mechan ics Ander s Ekber g
7 (8)
A Uniaxial Stress Cycle
Pulsating
compression
Alternatingtension/compression
time
Pulsatingtension
a
a
a
a
a
a
R =min
maxm =
1
2(max + min ) a =
1
2(max min )
Stress ratio Mid Stress Stress am litude
R = 1
m =0
a = max
R =0m =
max
2
a =max
2
R
m
a
=
=
=
min
min
2
2
(half the stress range)
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Dynamic Loadingol id Mechan ics Ander s Ekber g
8 (8)
10-20 -15 -10 -5 0 5-5
-4
-3
-2
-1
0
1
2
3
4
5R-ratio
R=min/
max
min
The R-ratio
R
R
RR
=
=
( )
=
=
min max
max min
max
max
a
a
a
1
2
2
1
21
The stress cycle isdefined by thestress amplitudeand the R-ratio
max max= + 10
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Uniaxial High Cycle Fatigueol id Mechan ics Ander s Ekber g
1 (8)
High cycle fatigue initiation
Two primary causes
From stress concentrations, such as pores, inclusions,initia l cracks etc.
There will be a local increase of the stress levels
D ue to a pile-up of dislocations, which will form slipbands, which will grow to form cracks
C racks form due to a local decrease in the fa tiguestrength
Which of this two mechanisms that will dominatedepends on the purity of the materia l, the nature of theloading, etc.
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Uniaxial High Cycle Fatigueol id Mechan ics Ander s Ekber g
2 (8)
For some ma terials, there is astress amplitude below which
no fa tigue fa ilures will occur
This is called the fatigue limit
The fa tigue limit can be
considered to be a materialparameter
Fatigue life at different stress levels
D esigning for infinite life is to assure tha t no stress levelsexceeds the fatigue limit
B ut how do we transla te the fa tigue limit to other types ofloading?
101
102 3
104
10 105 106
Stress cycles to failure
Stressamplitude
107
FL
Infinite fatigue life
Finite fatigue lifesteel
aluminum
a
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Uniaxial High Cycle Fatigueol id Mechan ics Ander s Ekber g
3 (8)
The Whler (S-N) curve
The Whler curve showsfatigue lifecorresponding to acertain stress amplitude
I t is a lso ca lled a S-N-curve
The diagram is primarily validfor uniaxial loading
The curve does not take intoaccount a ny effects of the midvalueof the stress during astress cycle
Consequently, the curve is only valid for loading with a certainR-ratio
There are also Whler curves for entire components (e.g.chains, wheel axles). Then, fatigue life is normally plotted
against applied load
101
102 3
104
10 105 106
Stress cycles to failure
Stressamplitude
107
FL
Infinite fatigue life
Finite fatigue lifesteel
aluminum
a
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Uniaxial High Cycle Fatigueol id Mechan ics Ander s Ekber g
4 (8)
10
FL
steel
aluminum
110
210
310
410
510
610
7
a
Given stress amplitude
Givespertinent
fatigue life
Givenservice life
Gives allowable stress amplitude
No fatigue damage is induced, the component
can sustain an infinite number of load cycles
The Whlerdiagram can beused to design forfinite (and infinite)
life
This can be doneeither for a givenservice loadingor a
given service life
This slope on the Whler curvecan be described by the eq uat ion
a
m N
f
= K
Using the Whler (S-N) curve
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Uniaxial High Cycle Fatigueol id Mechan ics Ander s Ekber g
5 (8)
How to Construct a Whler Diagram
10
FL
110
210
310
410
510
610
7
a
FRA
For alternating loading,FRA = UTS
For pulsating loadingFRA = UTS 2
Note that the Whler curve isonly valid for a certa in R-value( R = min max )
1000
0.9
UTS
FL 0.5 UTS
For low strength
wrought steel
For steel, thefa tigue limit
corresponds to106
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Uniaxial High Cycle Fatigueol id Mechan ics Ander s Ekber g
6 (8)
Y
FL
Y
UTS
FLP
Plasticity
a
m
The fa tigue limits for two cases
fully reversed tension/compression (orbending)
pulsatingtension (or bending) and theyield limit, a re needed to create thediagram
The diagram is validfor different R-ratios
The diagram is only validfor uniaxial loading
The Haigh diagram
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Uniaxial High Cycle Fatigueol id Mechan ics Ander s Ekber g
7 (8)
Reduction of the Haigh Diagram
Reduction is made on the amplitude axis
R eduction is normally ma de w ith respect to
Surface roughness(taking also the effect of corrosionintoaccount)
Size of the raw material
Loaded volume (no reduction in the case of a notch)
Fatigue notch factor Kf
The reduction factors are taken from diagrams (seeMaterial Fatigue,p.9-12)
The fatigue notch factor is determined from
Kf=1+q(Kt-1)whereKt is the stress concentration factor and qdepends on the notch
radius (q
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Uniaxial High Cycle Fatigueol id Mechan ics Ander s Ekber g
8 (8)
Create the Haigh diagram
a
R educe the H aigh diagrama
m
FL FLP
UTSY
m
FL
FLP
UTSY
Insert your service stress, P, inthe reduced Haigh diagrama
m
UTSY
P
Check if your in the sa fearea. Ca lcula te sa fety factorsa
mP
Using the Haigh Diagram
( , )K Kt m f a
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Damage AccumulationSol id Mechan ics Ander s Ekber g
1 (2)
Assume that, during the servicelife, we have 500 loadings of type 1(defined by mid-value and
magnitude), 1000 loa dings of type 2and 10000 loadings of type 3
The Palmgren Minerrule statesthat failureoccurs when
n
N
i
ii
I
=
=
11
where niis the number of applied
loa d cycles of type i, and Ni is the
pertinent fatigue life
Palmgren Miners Rule
1 2 3
11
12
13
14
15 6
7
a
1
2
3
ni
Ni
=500
103
i=1
I
+10
3
105
+10
4
= 0.51 < 1
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Damage AccumulationSol id Mechan ics Ander s Ekber g
2 (2)
Damage
D efine the damageinduced in the i:th loa d cycle as
DN
i
i
= 1
Then, Palmgren Miners rulesta tes that fa tigue fa ilureoccurs when
Di
i
I
=
=
1
1
Note that, in this case, the accumulation is made for allload cyclesand not all types of load cycles
The previous example can then be expressed as
Di
i
I
= + + = 3
2 ij ij c, ,
Crossland criterion
EQC a a a a a a C h,max e= ( ) + ( ) + ( ) + >1
2 1 2
2
2 3
2
3 1
2
, , , , , , c
EQS ad
ad
C h,max eC= + >3
2 ij ij c, ,
Dang Van criterion
EQDV1,a 3,a
DV h,max eDV2
=
+ >c
Val id onl y forpr opor t ional l oading
(in-phase an d f ixed
pr incipa l d ir ect ions)
14 (24)
Equivalent stress criteria components
Sh t
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
Shear stress measures
The shear stress initiates microscopic cracks(stage Icrack growth)
A static shear stress have no influence onfatigue damage the shear stress amplitudeis employed
Hydrostatic stress
Mean valueof normal stressestha t opensup cracks (Stage IIcrack growth)
h= = + +( )1
3
1
3 11 22 33ii
regardless of coordinate system(stress invariant)
15 (24)
The deviatoric stress tensor
The stress tensor can be split into deviatoric and volumetric part
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
The stress tensor can be split into deviatoricand volumetricpart
ij
xx xy xz
yz yy yz
zx zy zz
xx xy xz
yz yy yz
zx zy zz
ij ij kk
=
=
+
= + = +
h
h
h
h
dh
d
1 0 0
0 1 0
0 0 11
3 I
The volumetricpart conta ins the hydrostatic stressThe deviatoricpart reflects influence of shear stresses
Midvalue:
ij
xx xy xz
yx yy yz
zx zy zz
,m
d
m
d
dm
d
m
dm
d
m
d
m
d
mdm
d
m
dm
= =
(proportional loading)
Amplitude: ij ij ijt t,ad d
,md( )= ( ) (or a
d dmd
t t( )= ( ) )
16 (24)
Low Cycle Fatigue
Stresses close to (or a t) the yield limit
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
x
x
x
x
Stresses close to (or a t) the yieldlimitSmall stress increment large strainincrement. B est resolution if stra ins areemployed in fa tigue model
Induced fa tigue dama ge due to globa lplasticity
Loading above yield limit, (LCF) gives
With stress concentration factor
K
max
and strain concentra tionfactor
K
max
we get K K
K
Kt
K
K
K> K
YtK
17 (24)
Stress concentrations in LCF Neubers rule
At a stress concentra tion Neubers rule
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
At a stress concentra tion,Neuber s rulegives the ralation between stress andstrain as
max maxf2
max,a max,af2
,a
=
=
K
EK
E
2
2
This equation has two unknownStress and strain must also fulfil constitutiverelationship (for cyclic loading)
2 equat ions and 2 unknown
max
max
Constitutiverelation
Neuber
hyperbola
18 (24)
LCF Design Rules
According to Morrow the relationship between
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
According to Morrow, the relationship betweenstrain amplitude, a , and pertinent number of loadcycles to fa ilure, Nfcan be written as
af
f f f=
( ) + ( )E N Nb c
2 2
or, with a sta tic mean stress 0
af m
f f f=
( )( ) + ( )E N N
b c
2 2
According to Coffin Manson, the relationshipcan be simplified a s
a= 1.75UTS
E Nf0.12
+ 0.5D0.6
Nf0.6
Morrow
Mor r ow wit h mean
st r ess cor r ect ion
Coff in Man son
19 (24)
Fatigue crack growth
In experiments crack propaga tion ha s been measured a s af f h f
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
In experiments, crack propaga tion ha s been measured a s afunction of the stress intensity factor
I II III
logda
dN
log KKth KC
There exists a thresholdvalue ofK below which fatigue cracks
will not propagate
At the other extreme, Kmaxw ill
approa ch the fracture toughness
KC , and the material will failLinear relationshipbetween
logd da
N( )and Kin region II
d da Ndepends also on cracksize. This is not shown in theplot
20 (24)
Paris law
P aris law can be written asd
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
P aris law can be written asd
d
a
NC K
m= ( Cand ma re materia l parameters)
1. Find stress intensity factor fo r the current geometry
2. Find crack length corresponding to K Kmax= C3. (Check if LE FM is O K )4. Integrate P aris law
5. Solve for the number of stress cycles corresponding to fa ilure
P aris law does not a ccount formean stress effects(described by the R-ratio)
history effects(introduced by H)
and is only valid for
uniaxial loading
long cracks
LE FM-condit ions
21 (24)
Variable amplitude loading
A (tensile) overload will introduce (compressive)residual stresses
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
A (tensile) overload will introduce (compressive)residual stresses
These residual stresses will influence Kand thus the
rate of crack propagation
The Wheeler model is used to defines reduction of thecrack growth ra te due to overloa d
The reduction factorisdefined as
Rc
0d= +
a d
Reduced crack growth rate
is then calculated as
d
d
d
dRR
a
N
a
N
=
crack
d0
adc
22 (24)
Crack closure
Normally cracks only grow when they are open
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
Normally cracks only grow when they are open
The Elber accounts for crack closure,also for tensile loadsbydefining an effective stress intensity range
P aris law
K K KK K
= [ ]
max min
min minmax ,0
E lber correction for crack closure a tt K=Kop
K K Keff op maxModified P aris law
dd eff
aN C K
m=
Empirical relation
K R K
R R R R
op== + +
( )
( ) . . .
max , where
0 25 0 5 0 25 1 12
23 (24)
K
Crack closure
10
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
Crack closure
Kmax
Keff
Kop
Kmin
K
U sing E lber correction in Parislaw is non-conservative(predictsa longer fatigue life)compared tostandardParis law
The only difference whenusing Elber correction is
in a new, higher Kmin
1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1-10
-8
-6
-4
-2
0
2
4
6
8
R
Kop
andKmin
KopKmin
Smallest magnitude of
Kminin Paris law
KParisKElber
KParis
24 (24)
Crack arrest at different scales
A The loa d magnitude is
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Fatigue a survival kitSol id Mechan ics Ander s Ekber g
gbelow the fatigue limit
we will not initiateany(macroscopic cracks)
B The applied load gives astress intensitybelow thefatigue thresholdstressintensity ma croscopic
cracks will not continue togrow
A
B
Fatiguefailure
log a
KI,th= U a
No fatigue failure
No propagation
log
log e