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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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

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