Haigh diagram I - Chalmersanek/teaching/fatfract/98-6.pdfSolid Mechanics High Cycle Fatigue (HCF)...

$: Haigh diagram I - Chalmersanek/teaching/fatfract/98-6.pdfSolid Mechanics High Cycle Fatigue (HCF) part II Anders Ekberg 3 (20) ÒModernÓ Fatigue Design Background Evolution in structural$
High Cycle Fatigue (HCF) part IISolid Mechanics Anders Ekberg

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σa

σm

σFLσFLP

σUTSσY

σ

σ σa FL=

σ m = 0

time

σ σa FLP=σa

σ σm FLP=

Plasticdeformations

σFLP

σa

σm

λκδσFLP

σUTSσY

λκδσFL

σFLP

Loaded volumeδ

Surface roughnessκ

Size of rawmaterial

λ

Haigh diagram I

σYσY

Haigh diagram Reduced Haigh diagram


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σa

σm

σUTSσY

P

σa

σm

P

( , )K Kt m f a⋅ ⋅σ σ

Haigh diagram II

A’

A

C’

O B’

SFaAA'AP

=

SFmOB'OA

=

SFamOC'OP

=

σ m const=

σa const=

KK

f a

t mconst

⋅⋅

=σσK q K

K Kf t

f t

= + −≤

1 1( )

Service stress Safety factors


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ÒModernÓ Fatigue DesignBackground

Evolution in structural design due to• increased computational power• CAD/CAE - software

Need for new fatigue design methods that are• valid 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 a fatigue limit (expressed in the equivalent stress)

• LCF: calculation of damage connected to the constitutive model of the material. Fatigue damage connected to the plastic deformation

• LEFM: effective stress intensity range


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Multiaxial high cycle fatigue initiation

Problem:

The Haigh diagram is valid for• Uniaxial loading• One stress component

Solution:

Assume that, in the general case,fatigue behaviour is influenced by

• Applied shear stress amplitude• Hydrostatic stress

Based on these assumptions, derive a fatigue initiation criterion that defines a limiting stress magnitude for which fatigue cracks will develop) for a general type of loading.

Assumes undamaged material (continuum mechanics)


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Hydrostatic stress

The hydrostatic stress is the mean value of normal stresses acting on the material point (positive in tension)

A tensile (positive) hydrostatic stress opens up microscopic cracks (Stage II crack growth)

σ σ σ σh = + +( )13 x y z

The hydrostatic stress is a stress invariant

σσ σ σσ σ σσ σ σ

ij =

11 12 13

21 22 23

31 32 33

σ σ σ σ σh = = + +( )13

13 11 22 33ii

regardless of coordinate system


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Shear stress measures

The shear stress initiates slip bands which leads to microscopic cracks (stage I crack growth) Since a static shear stress have no influence on the fatigue damage, the shear stress ”amplitude” is employed

Two measures

• Tresca shear stress

τ σ σTresca = −1 3

2• von Mises stress

We need to define the “amplitudes” of these

σ σ σ σ σ σ σvM = −( ) + −( ) + −( )12 1 2

22 3

23 1

2


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Equivalent Stress Measures

Uniaxial CaseOne stress component

Mid value and amplitude of this stress component are taken to reflect the fatigue properties

The stresses during a load cycles are defined by a service stress

Multiaxial CaseSix stress components (general case)

Hydrostatic stress and shear stress “amplitude” are taken to reflect the fatigue properties

The stresses during a load cycles are defined by a closed curve

σY

σFLP

σm

Plastic zone

σUTSσY

σFL

Service stress

σa

FATIGUE

NO FATIGUE

FATIGUE

FATIGUE

NO FATIGUE

Shear stressamplitude

NO FATIGUE

Stresses during one load cycle

Plastic zone

Plastic zone

σe3c3

σe3

−σe3


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Shear Stress ÒAmplitudeÓ

General

It has been found empirically that a superposed static shear stress does not have any influence on the fatigue initiation

τ τFL FLP= whereas σ σFL FLP≠

In order to eliminate the influence of a superposed shear stress, the shear stress ”amplitude” is normally used in multiaxial HCF-criteria

This ”amplitude” is the difference between the current shear stress magnitude and the mid value of the shear stress for the current stress cycle

For the general case, this “amplitude” is rather complicated to compute (see Fatigue – a Survey, Appendix I)


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Shear stress Ð Uniaxial case

Mohr’s stress circle for loading in a uniaxial case

τ

σσ2 = 0

τmax

σ1 = σmax

x

τ

σσ2 = 0

τmax

σ1 = σmax

x

τ

σσ2 = 0

τmax

σ1 = σmax

x

time

Max normal and shear stress correspond to the same directions throughout the load cycle

45° 45° 45°

timeσ max

time

P

time

τmax τmid


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The deviatoric stress tensor

The stress tensor

σσ τ ττ σ ττ τ σ

ij

xx xy xz

yz yy yz

zx zy zz

=

Split into volumetric and a deviatoric part


σσ σ τ τ

τ σ σ ττ τ σ σ

σ

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 s

The volumetric part contains the hydrostatic stress

The deviatoric part reflects influence of shear stresses


11 (20)

Mid value of the deviatoric stress tensor

In-phase σ ij ij ija c f t= + ⋅ ( )

• aij and cij are constants• f t( ) is a common time dependent function

Fixed principal directionsEvery component of corresponds to a fixed direction throughout the loading

⇒⇒⇒⇒


σσ

σij

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

0 dd

d

d

d

d0

0 0

0 0

0 0

0 033

11

22

33a

c

c

c

f t

+

⋅ ( )


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Movie 1 – Click me!

http://www.am.chalmers.se/~anek/teaching/fatfract/Movie1.mov


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Mid value of the deviatoric stress tensor III

Limitations Ð In-phase loading

In in-phase loading, the stresscomponents have their max- and min-magnitudes at the same instant in time

In out-of-phase loading, max- and min magnitudes occur at different instants of time for different stress components

time

stress

time

stress

The case of out-of-phase loading is much more difficult to analyse, for instance due to difficulties in

• Defining a stress cycle• Defining a mid value of the shear stress


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Mid value of the deviatoric stress tensor IV

Limitations Ð Fixed principal directions

Rotating principal directions ⇒

" "σσ

σσ

ij,pd

d

d

d=

1

1

3

0 0

0 0

0 0

corresponds to a rotating coordinate system

Instead we have to look at the full deviatoric stress tensor and find its mid value

σσ σ τ τ


ij

xx xy xz

yz yy yz

zx zy zz

,md

md

h

h

h m

= =−

−−

s

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )


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Mid value of the deviatoric stress tensor V

Finding the mid value in a general case – Click me!

http://www.am.chalmers.se/~anek/teaching/fatfract/Movie2.mov


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ÒAmplitudeÓ of the deviatoric stress tensor

The mid value of the deviatoric stress tensor is found as

σσ

σσ

ij,md

md

dm

dm

dm

= =

s1

1

3

0 0

0 0

0 0

(proportional loading)

(“m” denotes mid-value of component during stress cycle)

or as

σσ σ τ τ


ij

xx xy xz

yz yy yz

zx zy zz

,md

md

h

h

h m

= =−

−−

s

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

(general)

the “amplitude” of the deviatoric stress tensor is defined as

σ σ σij ij ijt t,ad d

,md( ) = ( ) − (or s s sa

d dmdt t( ) = ( ) − )


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ÒAmplitudeÓ of the deviatoric stress tensor II

For in-phase loading with fixed principal directions (proportional loading), we can express the “amplitude” of the Tresca and von Mises stress using the “amplitude” of the deviatoric stress tensor

τσ σ

Tresca,a1,ad

3,ad

2( )

( ) ( )t

t t=

− where (σ σ σ1,a

d1d

1,md( ) ( )t t= − etc)

σ σ σ σ σ σ σvM,a a a a a a a( ) ( ) ( ) ( ) ( ) ( ) ( ), , , , , ,t t t t t t t= −( ) + −( ) + −( )12 1 2

22 3

23 1

2

(it can be shown that using σa or σad gives the same results)

The max values are given as

τσ σ

Tresca,a1,ad

3,ad

2=

− where (σ

σ σ1,ad 1,max

d1,mind

2=

−)

σ σ σ σ σ σ σvM,a a a a a a a= −( ) + −( ) + −( )12 1 2

22 3

23 1

2, , , , , ,

ÒAmplitudeÓ of the deviatoric stress tensor II

For in-phase loading with fixed principal directions (proportional loading), we can express the “amplitude” of the Tresca and von Mises stress using the “amplitude” of the deviatoric stress tensor

τσ σ

Tresca,a1,ad

3,ad

2( )

( ) ( )t

t t=

− where (σ σ σ1,a

d1d

1,md( ) ( )t t= − etc)

σ σ σ σ σ σ σvM,a a a a a a a( ) ( ) ( ) ( ) ( ) ( ) ( ), , , , , ,t t t t t t t= −( ) + −( ) + −( )12 1 2

22 3

23 1

2

(it can be shown that using σa or σad gives the same results)

The max values are given as

τσ σ

Tresca,a1,ad

3,ad

2=

− where (σ

σ σ1,ad 1,max

d1,mind

2=

−)

σ σ σ σ σ σ σvM,a a a a a a a= −( ) + −( ) + −( )12 1 2

22 3

23 1

2, , , , , ,


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Equivalent stress criteria

Sines criterion

σ σ σ σ σ σ σ σ σEQS ad

ad

ad

ad

ad

ad

S h,mid eS= −( ) + −( ) + −( ) + >12 1 2

22 3

23 1

2, , , , , , c

Crossland criterion

σ σ σ σ σ σ σ σ σEQC a a a a a a C h,max eC= −( ) + −( ) + −( ) + >12 1 2

22 3

23 1

2, , , , , , c

Dang van criterion

σσ σ

σ σEQDV1,a 3,a

DV h,max eDV2=

−+ >c


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Concluding remarks

Fatigue analysis

Calculate the state of stress

Apply the equivalent stress criterion, fatigue if

σ σeq e>In the case of no fatigue, calculate safety coefficient as

SF = σσ

eEQ

Pros Cons

Suitable for computer analysis Corrosion correction etc.

General state of stress Lack of empirical knowledge

Identify critical parts of component Separates between fatigue / no fatigue

Have a physical basis


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Haigh diagram I - Chalmersanek/teaching/fatfract/98-6.pdfSolid Mechanics High Cycle Fatigue (HCF)...

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