7/30/2019 98-6
1/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
1 (20)
a
m
FL
FLP
UTSY
a FL=
m = 0
t ime
a FLP=
a
m FLP=
Plasticdeformations
FLP
a
mFLP
UTSY
FL
FLP
Loadedvolume
Surface roughness
Size of raw
material
Haigh diagram I
YY
Haigh diagramReduced Haigh diagram
7/30/2019 98-6
2/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
2 (20)
a
m
UTSY
P
a
mP
( , )K Kt m f a
Haigh diagram II
A
A
C
OB
SFaAA'
AP=
SFmOB'
OA=
SFam
OC'
OP=
m const=
a const=
K
K
f a
t m const
=
K q K
K K
f t
f t
= +
1 1( )
Service stress Safety factors
7/30/2019 98-6
3/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
3 (20)
Modern Fatigue Design
Background
Evolution in structural design due to increased computat ional power CA D /CA E - software
Need for new fatigue design methods that are va lid for a general type of loading easy to implement in a computer code
Several options, but no method with general validity
HCF: equivalent stress is defined and compared to afatigue limit (expressed in the equivalent stress)
LCF: ca lcula tion of damage connected to theconstitutive model of the materia l. Fat igue damageconnected to the plastic deformation
LEFM: effective stress intensity range
7/30/2019 98-6
4/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
4 (20)
Multiaxial high cycle fatigue initiation
Problem:
The H aigh diagra m is valid for
U niaxia l loading
O ne stress component
Solution:
Assume that, in the general case,
fatigue behaviour is influenced by
Applied shear stress amplitude
Hydrostatic stress
B ased on these assumptions, derive a fa tigue initiationcriterion that defines a limiting stress magnitude forwhich fatigue cracks will develop) for a general type ofloading.
Assumes undamaged material (continuum mechanics)
7/30/2019 98-6
5/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
5 (20)
Hydrostatic stress
The hydrostatic stress is the mean valueof normal stressesacting on the materialpoint (positive in tension)
A tensile (positive) hydrostatic stressopens up microscopic cracks (Stage IIcrack growth)
h = + +( )1
3x y z
The hydrostatic stress is a stress invariant
ij =
11 12 13
21 22 23
31 32 33
h = = + +( )1
3
1
311 22 33ii
regardless of coordinate system
7/30/2019 98-6
6/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
6 (20)
Shear stress measures
The shear stress initiates slip bands whichleads to microscopic cracks (stage I crackgrowth)
Since a static shear stress have noinfluence on the fatigue damage, theshear stress amplitude is employed
Two measures
Tresca shear stress
Tresca =1 32
von Mises stress
We need to define the amplitudes of these
vM = ( ) + ( ) + ( )1
21 2
22 3
23 1
2
7/30/2019 98-6
7/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
7 (20)
Equivalent Stress Measures
Uniaxial CaseOne stress component
Mid value and amplitude of this stresscomponent are taken to reflect thefatigue properties
The stresses during a load cycles are
defined by a service stress
Multiaxial CaseSix stress components (general case)
Hydrostatic stress and shear stressamplitude are taken to reflect thefatigue properties
The stresses during a load cycles are
defined by a closed curve
Y
FLP
m
Plastic
zone
UTSY
FL
Service
stress
a
FATIGUE
NO FATIGUE
FATIGUE
FATIGUE
NO
FATIGUE
Shear stressamplitude
NO
FATIGUE
Stressesduring oneload cycle
Plasticzone
Plasticzone
e3
c3
e3
e3
7/30/2019 98-6
8/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
8 (20)
Shear Stress Amplitude
General
I t ha s been found empirically tha t a superposed staticshear stressdoes not ha ve any influence on the fa tigue
initiation FL FLP= whereas FL FLP
In order to eliminate the influence of a superposed shear
stress, the shear stress amplitude is norma lly used inmultiaxia l H C F-criteria
This amplitude is the difference between the currentshear stressmagnitude and the mid value of the shear
stress for the current stress cycleFor the genera l case, this amplitude is rathercomplicated to compute (see Fat igue a Sur vey,Appendix I)
7/30/2019 98-6
9/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
9 (20)
Shear stress Uniaxial case
Mohrs stress circle for loa ding in a uniaxia l case
2 = 0
max
1 = max
x
2 = 0
max
1 = max
x
2 = 0
max
1 = max
x
time
Ma x normal and shear stress correspond to the samedirections throughout the loa d cycle
45 45 45
t i m e
maxt i m e
P
t i m e
maxmid
7/30/2019 98-6
10/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
10 (20)
The deviatoric stress tensor
The stress tensor
ij
xx xy xz
yz yy yz
zx zy zz
=
Split into volumetric and a deviatoricpart
ij
xx xy xz
yz yy yz
zx zy zz
xx xy xz
yz yy yz
zx zy zz
=
=
+
= +
h
h
h
h
hd
1 0 0
0 1 0
0 0 1
I
The volumetricpart conta ins the hydrostatic stress
The deviatoricpart reflects influence of shear stresses
7/30/2019 98-6
11/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
11 (20)
Mid value of the deviatoric stress tensor
In-phaseij ij ija c f t = + ( )
aij and cij are consta nts
f t( ) is a common t ime dependent functionFixed principal directionsE very component of corresponds to a fixed directionthroughout the loading
ij
xx xy xz
yx yy yz
zx zy zz
t
t t t
t t t
t t t
t
t
t
a
a
d
d d d
d d d
d d d
d
d
d
d
( ) =
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
=
( )
( )
( )
=
1
2
3
11
22
0 0
0 0
0 0
0 0
0dd
d
d
d
d
0
0 0
0 0
0 0
0 033
11
22
33a
c
c
c
f t
+
( )
7/30/2019 98-6
12/20
High Cycle Fatigue (HCF) part IISol id Mechanics An der s Ekber g
12 (20)
Movie 1 Click me!
http://www.am.chalmers.se/~anek/teaching/fatfract/Movie1.movhttp://www.am.chalmers.se/~anek/teaching/fatfract/Movie1.movhttp://www.am.chalmers.se/~anek/teaching/fatfract/Movie1.mov7/30/2019 98-6
13/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
13 (20)
Mid value of the deviatoric stress tensor III
Limitations In-phase loading
In in-phase loading, thestresscomponents have their
max- and min-magnitudes atthe same instant in time
In out-of-phase loa ding, max-and min magnitudes occur a t
different instants of time fordifferent stress components
time
stress
time
stress
The case of out-of-phase loading is much more difficult toana lyse, for insta nce due to difficulties in
D efining a stress cycle D efining a mid value of the shear stress
7/30/2019 98-6
14/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
14 (20)
Mid value of the deviatoric stress tensor IV
Limitations Fixed principal directions
Rotating principal directions
" "
ij,pd
d
d
d
=
1
1
3
0 0
0 0
0 0
corresponds to a rotating coordinate system
Instead w e have to look a t the full devia toric stress tensor a ndfind its mid va lue
ij
xx xy xz
yz yy yz
zx zy zz
,md
md
h
h
h m
= =
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
7/30/2019 98-6
15/20
7/30/2019 98-6
16/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
16 (20)
Amplitude of the deviatoric stress tensor
The mid value of the deviatoric stress tensor is found as
ij,md
md
dm
dm
dm
= =
1
1
3
0 0
0 0
0 0
(proportional loading)
( m denotes mid-value of component during stress cycle)
or as
ij
xx xy xz
yz yy yz
zx zy zz
,md
md
h
h
h m
= =
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
(general)
the amplitude of the deviatoric stress tensor is defined as
ij ij ijt t,ad d
,md( ) = ( ) (or a
d dmd
t t( ) = ( ) )
7/30/2019 98-6
17/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
17 (20)
Amplitude of the deviatoric stress tensor II
For in-phase loading with fixed principal directions (proportionalloading), we can express the amplitude of the Tresca a nd von Misesstress using the amplitude of the devia toric stress tensor
Tresca,a 1,a
d
3,a
d
2( ) ( ) ( )t t t=
where ( 1,ad 1d 1,md( ) ( )t t= etc)
vM,a a a a a a a( ) ( ) ( ) ( ) ( ) ( ) ( ), , , , , ,t t t t t t t = ( ) + ( ) + ( )1
21 2
22 3
23 1
2
(it can be shown tha t using a or ad
gives the same results)
The ma x values are given a s
Tresca,a1,a
d
3,a
d
2= where (
1,ad 1,max
d
1,min
d
2= )
vM,a a a a a a a= ( ) + ( ) + ( )1
21 2
22 3
23 1
2, , , , , ,
Amplitude of the deviatoric stress tensor II
For in-phase loading with fixed principal directions (proportionalloading), we can express the amplitude of the Tresca a nd von Misesstress using the amplitude of the devia toric stress tensor
Tresca,a 1,a
d
3,a
d
2( ) ( ) ( )t t t=
where ( 1,ad 1d 1,md( ) ( )t t= etc)
vM,a a a a a a a( ) ( ) ( ) ( ) ( ) ( ) ( ), , , , , ,t t t t t t t = ( ) + ( ) + ( )1
21 2
22 3
23 1
2
(it can be shown tha t using a or ad
gives the same results)
The ma x values are given a s
Tresca,a1,a
d
3,a
d
2= where (
1,ad 1,max
d
1,min
d
2= )
vM,a a a a a a a= ( ) + ( ) + ( )1
21 2
22 3
23 1
2, , , , , ,
7/30/2019 98-6
18/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
18 (20)
Equivalent stress criteria
Sines criterion
EQS ad
ad
ad
ad
ad
ad
S h,mid eS= ( ) + ( ) + ( ) + >1
21 2
2
2 3
2
3 1
2
, , , , , , c
Crossland criterion
EQC a a a a a a C h,max eC= ( ) + ( ) + ( ) + >1
2
1 2
2
2 3
2
3 1
2
, , , , , , c
Dang van criterion
EQDV
1,a 3,a
DV h,max eDV2=
+ >c
19 (20)
7/30/2019 98-6
19/20
High Cycle Fatigue (HCF) part IISol id Mechanics An der s Ekber g
19 (20)
Concluding remarks
Fatigue analysis
C alcula te the state of stress
Apply the equivalent stress criterion, fatigue if
eq e>
In the case of no f at igue, ca lculate safety coefficient a s
SF=
e
EQ
Pros Cons
Suitable for computer ana lysis C orrosion correction etc.
G eneral state of stress La ck of empirical knowledge
Id entify critical parts of component Separa tes betw een fa tigue /no fa tigue
Have a physical basis
7/30/2019 98-6
20/20
High Cycle Fatigue (HCF) part IISol id Mechan ics Ander s Ekber g
20 (20)
Lunch