Fatigue initiation Low Cycle Fatigue FATIGUE - Chalmersanek/teaching/fatfract/98-11.pdf · Low Cycle Fatigue Stresses close to (or at) the yield limit Small stress increment large

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  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    FATIGUE

    A Crash CourseCrack propagation

    Retardation effects

    Fatigue initiation

    Limited High Cycle Fatigue life

    Low Cycle Fatigue

    Stress Concentrations in LCF

    Stress cycle identification

    Damage accumulationMultiaxial fatigue initiation

    Crack closure

    Crack growth arrestment

    Crack growth in mixed mode loading

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Damage mechanisms

    ab

    cd

    el

    pl

    YFL

    pl

    FL

    Y

    el

    Plastic shakedown limit

    Elastic shakedown limit

    Yield limit

    Fatigue limit

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Fatigue design methods

    Parameters

    Magnitude of loading

    Complexity of loading

    Damage of material

    Environment

    Possibility of experimentalvalidations

    Experience

    ...Low StressMagnitudes

    No Cracks Large Cracks

    Multia

    xial

    Stres

    s

    High StressMagnitudes

    Uniax

    ial

    Stre

    ssA D

    B CE

    F G

    H

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Philosophies in Design

    Safe life means that the component is designed to last a previously determined life

    Equivalent stress criterion (no cracks initiated) Fracture mechanics analysis (crack does not

    propagate enough to cause failure in operational life)

    Crcaks are not allowed to grow to failure.Also, (partial) failure of a part of the component through crack development must not endanger safety

    Design steps A flaw size, that will lead to fracture, is estimated. A tolerable flaw size is defined An initial flaw size is defined The time to grow the crack from initial to

    tolerable size is calculated Based on this, inspection schemes are defined

    Fail Safe

    Safe Life

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Fatigue

    How?

    Micro structural changes which cause nucleation of permanent damage. Persistent slip bands (PSB)

    Creation of microscopic cracks

    Growth and coalescence of microscopic flaws to form "dominant" cracks

    Stable propagation of dominant macro crack(s)

    Structural instability or complete failure

    Where?

    Initiation where s > 1

    Normally weak spots

    - inclusions- small cracks

    stress concentrations- inclusions- corrosion pits

    Fatigue will then grow (propagate) in the direction of max sInitiation propagation

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    Fatigue crack initiation and propagation

    Small cracks Shear driven Interact with microstructure Mostly analyzed by continuum

    mechanics approaches

    Large cracks Tension driven Fairly insensitive to

    microstructure Mostly analyzed by fracture

    mechanics models

    Stage IITension driven crack

    (propagation)

    Stage IShear driven crack

    (initiation)

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Influencing factors

    Stress concentrations

    Give rise to increased stress levels

    Volumetric effects

    Increased volume subjected to high load magnitude gives larger probability of a weak, highly stressed spot

    Large raw material gives reduced fatigue resistance due to manufacturing quality

    Environmental effects

    Corrosion Corrosion pits which act as stress concentrators. Cracks will always form due to corrosion -> no fatigue limit

    Heat will often decrease the fatigue strength and the fatigue limit may diminish at higher temperatures.

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    a

    m

    FLFLP

    UTSY

    a FL=

    m = 0

    time

    a FLP=a

    m FLP=

    Plasticdeformations

    FLP

    a

    mFLP

    UTSY

    FL

    FLP

    Loaded volume

    Surface roughness

    Size of rawmaterial

    Haigh diagram I

    YY

    Haigh diagram Reduced Haigh diagram

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    a

    mUTSY

    P

    a

    mP

    ( , )K Kt m f a

    Haigh diagram II

    A

    A

    C

    O B

    SFaAA'AP

    =

    SFmOB'OA

    =

    SFamOC'OP

    =

    m const=

    a const=

    KK

    f a

    t mconst

    =K q K

    K Kf t

    f t

    = +

    1 1( )

    Service stress Safety factors

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    N

    10

    FL

    steel

    aluminum

    1 102 103 104 105 106 107

    a

    Given stress amplitude

    Gives pertinent

    fatigue life

    Given service life

    Gives allowable stress amplitude

    No fatigue damage is induced, the component can sustain an infinite number of load cycles

    The Whler diagram can be used to design for finite (and infinite) life

    This can be done either for a given service loading or a given service life

    This slope on the Whler curve can be described by the equation

    am Nf = K

    Whler (S-N) curve

    1k

    or be approximated as a straight line

    Valid only for a certain R-ratio

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    t

    1

    2

    35

    4

    678

    Stress cycle identification rainflow counting

    1 passes an equally large maximum2 passes a larger minimum3 passes a larger maximum4 reaches the run of drop 25 reaches the run of drop 16 falls out7 falls out8 reaches the run of drop 61 and 6, 2 and 5, 3 and 4; 7 and 8 form closed loops (i.e. stress cycles)

    Depict the loading sequence as a function of time. start with largest max or smallest min use straight lines between min and max

    Let drops start from every max and min nd stop if: it starts from max and passes a larger or equal max it starts 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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    Assume that, during the service life, we have 500 loadings of type 1 (defined by mid-value and magnitude), 1000 loadings of type 2 and 10000 loadings of type 3

    The Palmgren Miner rule states that failure occurs when

    DnNjj

    Ji

    ii

    I= =

    = =

    1 1

    1

    where Dj is the damage of a load cycle, ni is the number of applied load cycles of type i, and Ni is the pertinent fatigue life

    Damage accumulation Palmgren - Miners Rule

    1 2 3

    11

    12

    13

    14

    15

    16 N71

    a

    DnNjj

    Ji

    ii

    I= = + +

    = 12 1 22

    2 32

    3 12

    , , , , , , c

    EQS ad

    ad

    S h,mid eS= + >32 ij ij

    c, ,

    Crossland criterion

    EQC a a a a a a C h,max eC= ( ) + ( ) + ( ) + >12 1 22

    2 32

    3 12

    , , , , , , c

    EQS ad

    ad

    C h,max eC= + >32 ij ij

    c, ,

    Dang Van criterion

    EQDV1,a 3,a

    DV h,max eDV2=

    + >c

    Valid only for proportional loading

    (in-phase and fixed principal directions)

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    Equivalent stress criteria components

    Shear stress measures

    The shear stress initiates microscopic cracks (stage I crack growth) A static shear stress have no influence onfatigue damage the shear stress amplitude is employed

    Hydrostatic stressMean value of normal stresses that opensup cracks (Stage II crack growth)

    h = = + +( )13

    13 11 22 33ii

    regardless of coordinate system(stress invariant)

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    The deviatoric stress tensor

    The stress tensor can be split into deviatoric and volumetric part

    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 0 113

    s I

    The volumetric part contains the hydrostatic stressThe deviatoric part reflects influence of shear stresses

    Midvalue:

    ij

    xx xy xz

    yx yy yz

    zx zy zz

    ,md

    md

    dm

    dm

    dm

    dm

    dm

    dm

    dm

    dm

    dm

    = =

    s (proportional loading)

    Amplitude: ij ij ijt t,ad d

    ,md( ) = ( ) (or s s sa

    d dmdt t( ) = ( ) )

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    x

    x

    x

    x

    Low Cycle Fatigue

    Stresses close to (or at) the yield limitSmall stress increment large strain increment. Best resolution if strains are employed in fatigue model

    Induced fatigue damage due to global plasticity

    Loading above yield limit, (LCF) gives

    With stress concentration factor

    K

    max

    and strain concentrationfactor

    K

    max

    we get K K

    K

    Kt

    K

    K

    K > K

    YtK

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Stress concentrations in LCF Neubers rule

    At a stress concentration,Neubers rulegives the ralation between stress and strain as

    max maxf2

    max,a max,af2

    ,a

    =

    =

    KEK

    E

    2

    2

    This equation has two unknown

    Stress and strain must also fulfil constitutiverelationship (for cyclic loading)

    2 equations and 2 unknown

    max

    max

    Constitutive relation

    Neuber hyperbola

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    LCF Design Rules

    According to Morrow, the relationship between strain amplitude, a , and pertinent number of load cycles to failure, Nf can be written as

    a f f f f= ( ) + ( )

    EN Nb c2 2

    or, with a static mean stress 0

    a f m f f f= ( ) ( ) + ( )

    EN Nb c2 2

    According to Coffin Manson, the relationship can be simplified as

    a = 1.75UTS

    ENf

    0.12 + 0.5D0.6Nf 0.6

    Morrow

    Morrow with mean stress correction

    Coffin Manson

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Fatigue crack growth

    In experiments, crack propagation has been measured as a function of the stress intensity factor

    I II III

    logdadN

    log KKth KC

    There exists a threshold value of K below which fatigue cracks will not propagate

    At the other extreme, Kmax will approach the fracture toughness KC, and the material will fail

    Linear relationship between log d d

    aN( ) and K in region II

    d da N depends also on crack size. This is not shown in the plot

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Paris law

    Paris law can be written asdd

    aN

    C K m= (C and m are material parameters)

    1. Find stress intensity factor for the current geometry

    2. Find crack length corresponding to K Kmax = C3. (Check if LEFM is OK)4. Integrate Paris law5. Solve for the number of stress cycles corresponding to failure

    Paris law does not account for mean stress effects (described by the R-ratio) history effects (introduced by H)

    and is only valid for uniaxial loading long cracks LEFM-conditions

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Variable amplitude loading

    A (tensile) overload will introduce (compressive) residual stresses

    These residual stresses will influence K and thus the rate of crack propagation

    The Wheeler model is used to defines reduction of the crack growth rate due to overload

    The reduction factor is defined as

    R c0d

    = +

    a d

    Reduced crack growth rate is then calculated as

    dd

    ddR

    RaN

    aN

    =

    crack

    d0

    adc

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Crack closure

    Normally cracks only grow when they are open

    The Elber accounts for crack closure, also for tensile loads by defining an effective stress intensity range

    Paris lawK K K

    K K

    = [ ]max min

    min minmax ,0

    Elber correction for crack closure att K=KopK K Keff op max

    Modified Paris lawd

    d effa

    N C Km=

    Empirical relation

    K R K

    R R R Rop =

    = + +

    ( )

    ( ) . . . max , where

    0 25 0 5 0 25 1 12

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    Crack closure

    Kmax

    Keff

    KopKmin

    K

    Crack closure

    Using Elber correction in Paris law is non-conservative (predicts a longer fatigue life)compared to standardParis law

    The only difference when using 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

    10

    RK

    op a

    nd K

    min

    KopKmin

    Smallest magnitude of Kmin in Paris law

    KParisKElber

    KParis

  • Fatigue a survival kitSolid Mechanics Anders Ekberg

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    Crack arrest at different scales

    A The load magnitude is below the fatigue limit we will not initiate any (macroscopic cracks)

    B The applied load gives a stress intensity below the fatigue threshold stress intensity macroscopic cracks will not continue to grow

    A

    B

    Fatiguefailure

    loga

    KI,th = U a

    No fatigue failure

    No propagation

    log

    log e