07 Fatigue Failure I

1

Chapter 6 Fatigue Failure

All machine and structural designs are problems in fatigue because the forces of Nature are always at work and each object

must respond in some fashion.

Fatigue

• Fatigue failure is the fracture of a structural member due to repeated cycles of loading or fluctuating loading.

• Fatigues is the single largest cause of failure in metals, estimated to be the cause of 90% of all metallic failures.

• Fatigue failure are catastrophic and insidious, occurring suddenly and often without warning. Static loading provides

ffi i t ti f d fl tisufficient time for deflection.

• The fatigue failure occurs at relatively low stress levels to a component or structure subjected to fluctuating or cyclic stresses.

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Fatigue

• Fatigue is a complex phenomenon, and no universal theories to describe the behavior of materials subjected to cyclic loadings exist; instead, there are a large number of theories toloadings exist; instead, there are a large number of theories to describe the behavior of particular materials.

• Most of the engineering design experience in fatigue is based on an experimental understanding of the behavior of carbon steels. Much effort has been directed toward extending these semi-empirical rules to other ferrous and nonferrous metals, as well as ceramics polymers and composite materialswell as ceramics, polymers, and composite materials.

• For the most part, fatigue involves the accumulation of damage within a material. Damage usually consists of cracks that can grow by a small distance with each stress cycle.

Fatigue

• Experiments have found that fatigue cracks generally begin at a surface and propagate through the bulk. Therefore, mucha surface and propagate through the bulk. Therefore, much attention is paid the quality of surfaces in fatigue-susceptible machine elements.

• Fatigue cracks begin at several sites simultaneously and propagate when one flaw becomes dominant and grows more rapidly than others.

• Fatigue testing is imperative to confirm safe mechanical design.

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Stages of Fatigue Life

Schematics of Fatigue Life

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Examples of Fatigue Failure

Pure tension with no stress concentration

Summary of Fatigue Failure

Thus far we’ve studied STATIC FAILURE STATIC FAILURE of machine elements.The second major class of component failure is due to DYNAMIC DYNAMIC LOADINGLOADINGLOADINGLOADING

Repeated stressesAlternating stressesFluctuating stresses

The ultimate strength of a material (Su) is the maximum stress a material can sustain before failure assuming the load is applied only once and held.

i h i f i l f il d li l diFatigue strength Resistance of a material to failure under cyclic loading. A material can also FAILFAIL by being loaded repeatedly to a stress level that is LESSLESS than (Su)

Fatigue failure

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Approach to Analysis Fatigue-Life

Fatigue-Life Methods

Fatigue Strength and the Endurance Limit

Endurance Limit Modifying Factors

Stress Concentration and Notch Sensitivity

Fluctuating Stresses

Combinations of Loading Modes

Varying, Fluctuating Stresses; Cumulative Fatigue Damage

Fatigue-Life Methods

Three major fatigue life methods used in design and analysis for safe lifedesign and analysis for safe life estimation:

1. Stress life method (S-N Curves)

2 Strain life method (ε N Curve)2. Strain life method (ε-N Curve)

3. Linear elastic fracture mechanics method

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Fatigue Regimes

Regimes

C l i h C l

Finite Life Infinite Life

Low-Cycle Fatigue

High-Cycle Fatigue

31 10 cyclesN≤ ≤ 310 cyclesN >

Stress-Life Methods

based on stress levels onlyIt is the least accurate approach, especially for low-cycle applications. Most traditional method:• It is the easiest to implement for a wide range of

design applicationsdesign applications

• It has ample supporting data

• It represents high-cycle applications adequately

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Strain-Life Methods

Based on strain amplitude

Involves more detailed analysis of the plastic deformation at localized regions where the stresses and strains are considered for life estimates.

G d f f i li iGood for low-cycle fatigue applications.

Some uncertainties exist in the results.

Linear Elastic Fracture Mechanism

Assumes a crack is already present andAssumes a crack is already present and detected.

Predicts crack growth with respect to stress intensity.

Most practical when applied to largeMost practical when applied to large structures in conjunction with computer codes and a periodic inspection program.

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Stress-Life Methods

To determine the strength of materials under the action of fatigue loads, specimens are subjected to repeated or varying forces of specified magnitudes while the cycles or stress reversals are counted to destruction. The most widely used fatigue-testing device is the R. y g gR. Moore high-speed rotating-beam machine. The specimen is very carefully machined and polished, with a final polishing in an axial Direction to avoid circumferential scratches.

Figure: Test-specimen geometry for the R. R. Moore rotating-beam machine. The bending moment is uniform over thecurved at the highest-stressedcurved at the highest stressed portion, a valid test ofmaterial, whereas a fracture elsewhere (not at the highest-stress level) is grounds for suspicion of material flaw.

Fatigue Test Machine

R. R. Moore rotating-beam fatigue testing machine

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Fatigue Test (Ferrous Metals and Alloys)

With an endurance limit!

Fatigue Test (Aluminum Alloys)

Without an endurance limit!

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Fatigue Strength of Polymer

Figure 7.7 Fatigue strengths as a function of number of loading cycles. (c) Selected properties of assorted polymer classes.

S-N Diagram Under Cyclic Stress

Cyclic stress is a function of time, but the variation is such that the stress sequence repeats itself

Nc=1/2Nc=1

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Strain-Life: Hysteresis under Cyclic Load

Diagram of Reversals to FailureFatigue Ductility coefficient (N=1)

Fatigue Strength g gcoefficient (N=1)

Slope of plastic strain line

Slope of elastic strain line

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Strain Life Theory

Strain (Crack)Strain (Crack)

Manson-Coffin Relationship

ElasticElastic PlasticPlastic


'Fatigue ductility Total strain

( ) ( )cfbf NN

E''' 22

2ε

σε+=

∆

Number of cycle

exponent

Fatigue ductility coefficient

Stress at fracture (one cycle) Fatigue strength

exponent

Cyclic Properties of Metals

b c

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Limitation of Strain Life Theory Strain-life theory gives insight into important properties in

fatigue strength determination: as long as there is a cyclic plastic strain, no matter how small, eventually there will be f il

ElasticElastic PlasticPlastic


'

failure.

( ) ( )αεσε ''' 22

2NN

E faf +=

∆

• Total Strain at failure is the difficult to determine• Strain concentration factors are nowhere

Linear Elastic Fracture Mechanics

R i A l k th It i t l ff t d b t i l i t tRegime A: slow crack growth. It is strongly affected by material microstructure, environment effects, and stress ratio Rs.

Regime B (Paris Regime): related to micro-structure, mechanical load variables, and environment.Regime C: high growth rate. micro-structural effects and loadings, cleavage

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Crack Growth

( ) aaKI πσβπσσβ ∆=−=∆ minmax

Initial crack length

Crack Growth

( )∆= mIK

dNda

( )∫∫∆

== f

i

f a

a mf

N

a

daC

NdNπσβ

10

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Fatigue Strength 1725 rpm 106 cycle -1.5day108 cycle -40day

The fatigue is affected by• Stress concentration• Residual stress• Surface roughness• Environment (temperature and corrosion)

Fatigue Test: Best-Case-Scenario

Endurance Limit

>>

≤=

MPaSMPakpsiSkpsi

MPakpsiSSS

ut

ut

utut

e

1400700200100

)1400(2005.0'

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S-N Equation under Given Failure Stress

( ) ( )bFNf NS 2'' σ=

Basic S-N Eq.

'Fσ

MPakpsiS

or

utF

mF

345/50

500Hfor '

B0'

+=

<=

σ

εσσ

True Failure stress

( )( )

eF

NSb

2log/log ''σ

−=

For exponent b

( )eN2log

( ) ( )

( )but

F

utb

Ff

Sf

fSS

3'

3'10

'

102

1023

⋅=

=⋅=

σ

σ

For N=103

S-N Equation under Given Fatigue Strength Fraction f

9.070for 20070for plot See=<≤≤fkpsiS

kpsiS

ut

ut

Fraction f:

S-N Eq.

( ) fSfS 12

For a and b

bf aNS =

( )

−==

e

ut

e

ut

SfSb

SfSa log

31;

For Nb

rev

aN

/1

=σ

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Fatigue Strength Example

A steel rotating beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at a completely reversed stress amplitude of 70 kpsi.

>

≤= kpsiSkpsi

)MPa(kpsiSS.S

utut' 200100

140020050

>>

MPaSMPakpsiSkpsiS

ut

ute

1400700200100

Fatigue Example

A shaft in bending is made of AISI steel. It has a tensile strength of 95 ksi and yield strength of 74

k i E ti t ( ) d li it (b) f tiksi. Estimate (a) endurance limit, (b) fatigue strength for 103, 104 , 105 and 106 cycles of life

≤ )MPa(kpsiSS. utut 140020050

>>=

MPaSMPakpsiSkpsiS

ut

ut'e

1400700200100

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High-Cycle Fatigue Example

The pressure vessel lids of nuclear power plants are bolted down to seal the high pressure g pexerted by the pressured water. The ultimate strength is 157 kpsi. Find

(1) how low the stress has to be for a life of 10,000 cycles;

(2) A 5% decrease in this stress would give how many cycles of life.

High-Cycle Fatigue Example

The maximum compressive stress in the jack is 190 MPa when the car is jacked up so high that both wheels on one side of the car are in the air and the load on the jack is 8000 N How many times can the jack beand the load on the jack is 8000 N. How many times can the jack be used for a small truck that weighs 6 tons and loads the jack to 17,000 N before it fails from fatigue? The jack material is AISI 1080 steel.

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Fatigue Strength under Low Cycle

9.070for 20070for plot See=<≤≤fkpsiS

kpsiS

ut

ut

Fraction f:

S-N Eq.

1

For a and b

bf aNS =

( )fbSa ut log31; ==

For S-N

==

ut

ff

utf SS

fNNSS log

log3,3

log

Endurance Limit Modifying Factors

Fatigue experiments assume that the best circumstances exist for promoting long fatigue lives. However, this

situation cannot be quarantined for design applications.situation cannot be quarantined for design applications. Component’s endurance limit must be modified.

'efedcbae SkkkkkkS =

'eS = endurance limit from experimental apparatus

ka = surface finish factorkb = size factorb

ke = reliability factorkd = temperature modification factor

kf = miscellaneous factor

Besides Stress Concentration Effect….

kc = load modification factor

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Surface Finish Factor'efedcbae SkkkkkkS =

buta aSk =

Example: A steel has a minimum ultimate strength of 520 MPa and a machined surface. Estimate ka.


Size Factor kb

For bending and torsion:

<<<<<<<<

=

−

−

−

−

mm2d51dmm5d 2.79d.in 10d in 2d.in 2d in 0.11d.

k

.

.

.

.

b

545111241

9108690

1570

1070

1570

1070

For bending and torsion:

<< mm2d51d. 54511

For axial loading:

1=bk

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Equivalent Diameter for Size Factor kb

Equivalent diameter de: equating the volume of material stressed at any above 95 percent of the maximum stress to the same volume in the rotating-beam specimen.

For rotation round cross-section, For rectangle

( )( )2

22950

07660

9504d.

d.dA .

=

−=π

σ

For non-rotation round cross-section,

d.dd.A

e

.

3700010460 2

950

==σ

cross-section,

hb..

hb.d

hb.A

e

.

808007660050

050950

==

=σ

Equivalent Diameter for Size Factor kb

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'SkkkkkkS

Loading Factor kc

efedcbae SkkkkkkS =

= Axial.

Bendingkc 850

1

Torsion.

c

590


Temperature Factor kd

( ) ( ) ( ) ( )FT

T.T.T.T..k

F

FFFFd

o100070

1059501010401011501043209750 41238253

≤≤

−+−+= −−−−

kd can be applied to St or Se.

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Reliability Factor ke

ae z.k 0801−=

Miscellaneous Effects


Figure: The use of shot peening to improve fatigue properties. (a) Fatigue strength at two million cycles for high strength steel as a function of

ultimate strength; (b) typical S-N curves for nonferrous metals.

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Fatigue Stress Concentration Bending/Axial Load

factor ionconcentrat stressfatigue is K specimenfree notch for limit endurance

specimennotched for limit enduranceK

f

f =

Notch sensitivity

11

−−

=t

f

KK

q

( )11 −+= tf KqK

Fatigue stress concentration factor

0σσ fmax K=

( )tf q

Concentrated stress

( ) ( ) ( ) kpsiSS.S.S..a:constant Neuberr/a

q:equation Neuber

utututut −−+−=

+=

−−− 38253 10672105111008324601

1

Fatigue Stress Concentration Torsion Load

factor ionconcentrat stressfatigue is K specimenfree notch for limit endurance

specimennotched for limit enduranceK

f

f =

Notch sensitivity

11

−−

=ts

fsshear K

Kq

( )11 −+= tsshearfs KqK

Fatigue stress concentration factor

0ττ fsmax K=

( )tsshearfs q

Concentrated stress

( ) ( ) ( ) kpsiSS.S.S..a:constant Neuberr/a

q:equation Neuber

utututut −−+−=

+=

−−− 38253 10672103511051219001

1

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Fatigue Stress Concentration Example

The rotating shaft is machined and subjected to F=6 kN. Find the minimum factor of safety for fatigue based on infinitethe minimum factor of safety for fatigue based on infinite

life. If the life is not infinite, estimate the number of cycles. Check for yielding as well.

Fatigue Stress Concentration Example

The driveshaft for a Formula One racing car has a diameter of 30mm and a half-circular notch with a 1-mm radius. The shaft was dimensioned for equal shear and bending stresses. The shaft material has an ultimate tensile strength of 965 MPa. Assume the equivalent stress is proportional to 22 3τσσ +=e

Determine the fatigue stress concentration factors for bending and torsion of the driveshaft Also determine if increasedand torsion of the driveshaft. Also, determine if increased acceleration or increased curve handling will give the higher risk of driveshaft failure.

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Modified Endurance Limit Example

The bar is machine-made of low-carbon steel (AISI 1020). Find the fatigue at 104 cycle for

the notched and un-notched bars.

Tensile loaded bar. (a) Un-notched; (b) notched.

Characterizing Fluctuating Stresses

2minmax σσσ +

=mMean stress

Sminmax σσσ −=r

Stress range

22minmax σσσσ −

== ra

Stress amplitude

minσRStressCommon cyclic patterns

max

min

σ=sRStress

ratio

s

s

m

aa R

RA+−

==11

σσAmplitude

ratio

1. Completely reversed2. Nonzero mean 3. Released tension4. Released compression

),1,0( ∞=−== asm ARσ)0( ≠mσ

)1,0,0( min === as ARσ

)1,,0( max −=∞== as ARσ

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Cyclic Stress Example

A tuning fork is hit with a pencil and starts to vibrate with a frequency of 440 Hz The maximum bending stress in thefrequency of 440 Hz. The maximum bending stress in the tuning fork is 2 MPa at the end positions.

• Calculate the mean stress, the range of stress, the stress amplitude, the stress ratio, and the amplitude ratio.

• Calculate how much stress the tuning fork can sustain without being plastically deformed if it is made of AISIwithout being plastically deformed if it is made of AISI 1080 steel.

Fatigue Failure Criteria for Fluctuating Stress

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Fatigue Failure Criteria for Fluctuating Stress

Failure Criteria Under Fluctuating Stress

Soderberg Line nSS y

m

e

a 1=+

σσ

Gerber Line

ASME-Elliptic

Goodman Line

nSn

Sn

y

m

e

a 122

=

+

σσ

nSn

Sn

ut

m

e

a 12

=

+

σσ

nSS ut

m

e

a 1=+

σσ

ye

Langer static yieldnSy

ma =+σσ

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Safety Factor under Fluctuating Stress

Goodman Failure Example

The bar is made of cold-drawn1040 steel. The cyclic non-zero axial load varies from -100 kN to 290 kN. Using G d f il th t d t i th f t f tGoodman failure theory to determine the safety factor

60mm 40mm

R=10mm

Thickness =40mm

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Goodman Line Example IA straight, circular rotating beam with a 30-mm diameter and 1-m length has an axial load of 30,000N applied at the end and a stationary radial load of 400N. The material is AISI 1040 steel, k =0 75 k =k =k =k =k =1 Find the safety factor for infinite lifeka=0.75, kb=kc=kd=ke=kf=1. Find the safety factor for infinite life by using the Goodman line.

Goodman Line

SSma 1=+

σσnSS ute

Goodman Diagram Example IIThe cantilever shown in sketch j carries a downward load F that varies from 300 to 700 lbs. (a) Compute the resulting safety factor for static and fatigue failure if the bar is made from AISI 1040 steel. (b) What fillet radius is needed for a fatigue failure safety factor of 3.0 (use the constant notch sensitivity)?

Notes: This solution assumes that the shoulder is machined, but it may be reasonable to use a ground surface if the application is critical.

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Fatigue Failure of Brittle Material

The fatigue for a brittle material differs markedly from that of a ductile material because

• Yielding is not involvedg• Suc>Sut• No enough work down on brittle failure

−=

+−

=

111

utma

utm

utm

e

a

S/nSnS/SS/SS

σσ

( )

+++−

+=

=+

=

2

4112

1

eut

euteuta

ma

utme

SrSSrSSrSS

/rS/nSS

σσ

Influence of Multi-Axial Stress Status

Simple Multi-axial Stress Complex Multi-axial Stress

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Combinations of Loading Modes

• Completely Reversing Simple Loads

• Fluctuating Simple Loads• Fluctuating Simple Loads

• Combinations of Loading Modes

( ) ( ) ( ) ( ) ( ) ( )[ ]21

22

3850

/

torsionatorsionfsaxiala

axialfbendingabendingf'a K

.KK

+

+= τ

σσσ

( ) ( ) ( ) ( ) ( ) ( )[ ]21

22

3850

/

torsionmtorsionfsaxialm

axialfbendingmbendingf'm K

.KK

+

+= τσ

σσ

Cumulative Damage

Instead of a single fully reverse stress history block composed of n cycles, support a machine part, at a critical location, is subjected to • a fully reversed stress σ1 for n1, σ2 for n2, ….or • a ‘wiggly’ time line of stress exhibiting many

and different peaks and valleys

Linear Damage Rule (Miner’s Rule): Failure is predicted if

1'3

'3

'2

'2

'1

'1 ≥+++ L

Nn

Nn

Nn

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Cumulative Damage Example

For the un-notched bar with the machine-made of low-carbon steel (AISI 1020), the fatigue stress is 25 ksi for 20% of the time 30 ksi for 30% and 35 ksi for 40%20% of the time, 30 ksi for 30%, and 35 ksi for 40%, and 40 ksi for 10%. Find the number of cycles until cumulative failure.

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07 Fatigue Failure I