MECHANISM OF FATIGUE FAILUREMECHANISM OF FATIGUE FAILURE

�� Understanding fatigue mechanism will help improve fatigue resistUnderstanding fatigue mechanism will help improve fatigue resistanceance

�� There are three stages in fatigue failure: There are three stages in fatigue failure:

1.1. Crack initiation (Crack initiation (fatigue always starts at a crackfatigue always starts at a crack))

2.2. Crack propagationCrack propagation

3.3. Fast fracture, which can be ductile or brittle (Fast fracture, which can be ductile or brittle (failure with no warningfailure with no warning))

�� The initiation site is very small, extending only about two to fThe initiation site is very small, extending only about two to five grains around ive grains around

the origin.the origin. The location of the initiation is at a stress concentration and The location of the initiation is at a stress concentration and may be may be

difficult to distinguish from the succeeding stage of propagatiodifficult to distinguish from the succeeding stage of propagation, or crack n, or crack

growth.growth. The The crack initiation site is always parallel to the shear stress dircrack initiation site is always parallel to the shear stress directionection..

Stage I Crack GrowthStage I Crack Growth

�� Cyclic stress causes dislocation movementsCyclic stress causes dislocation movements

�� A slip band develops within a grain and then accumulates permaneA slip band develops within a grain and then accumulates permanent damagent damage

�� This damage results in This damage results in intrusionsintrusions and and extrusions, extrusions, and a and a crack is initiatedcrack is initiated

�� Slow propagation of crack along crystal planes with high resolveSlow propagation of crack along crystal planes with high resolved shear stressd shear stress

(the order of angstroms per cycle)(the order of angstroms per cycle)

Stage II Crack GrowthStage II Crack Growth

The crack that leads to fracture is usually The crack that leads to fracture is usually nucleated in a slip nucleated in a slip band on which the maximum shear stress is acting. band on which the maximum shear stress is acting.

This is because slip is a shear process so more plastic This is because slip is a shear process so more plastic deformation will occur in slip bands on which the maximum deformation will occur in slip bands on which the maximum shear stress is acting than in other slip bandsshear stress is acting than in other slip bands

In stage II, the crack propagates very fast normal to the applied stress(crack propagation rates, microns per cycle)

The crack will eventually reach a critical size and failure occurs

Mechanism of Fatigue Failure

�� Detailed examination of stage II Detailed examination of stage II fracture surfaces, using optical and fracture surfaces, using optical and electron microscopy, have provided electron microscopy, have provided evidence that a pattern of ripples or evidence that a pattern of ripples or striationsstriations are frequently present on are frequently present on these surfaces. these surfaces.

�� Each striation on the fracture surface Each striation on the fracture surface is produced by a single stress cycle. is produced by a single stress cycle. The presence of striations means that The presence of striations means that failure was produced by fatigue. failure was produced by fatigue.

�� The absence of striations however, The absence of striations however, does not meandoes not mean that failure was not that failure was not produced by fatigue. Ductility of a produced by fatigue. Ductility of a material plays a role in the presence material plays a role in the presence of striations. The lower the ductility of of striations. The lower the ductility of a material the less obvious are the a material the less obvious are the striations because there is insufficient striations because there is insufficient plastic deformation. plastic deformation.

No Load

Loaded

Max Load (tensile)

Load reduced

No Load

Loaded again

Slip

�� The The Striations are formed by a process called Striations are formed by a process called plastic bluntingplastic blunting: :

Advancing tip of the fatigue crack Advancing tip of the fatigue crack becomes blunt (not sharp) during the tensile becomes blunt (not sharp) during the tensile portion of the stress cycle followed by portion of the stress cycle followed by resharpening of tip during compression. This resharpening of tip during compression. This is illustrated in Figure below. is illustrated in Figure below.

(a) At the start the crack tip is sharp

(b) Tensile load applied: the small double

notch at the crack tip concentrates the slip along planes at ~ 45oC to the plane of the crack.

(c) As the crack widen to max its grows longer by plastic shearing and becomes

blunter.

(d) When load changed to compression, the slip direction in the end zone is reversed.

(e) The crack faces are crushed together and the new crack surface created in tension is

forced into the plane of crack

(f) The resharpened crack is ready to advance and blunted in the stress cycle.

Stage II Crack GrowthStage II Crack Growth

�� It is the It is the stage II crack growthstage II crack growth which is the controlling factor in fatigue failure of which is the controlling factor in fatigue failure of

most engineering materials. Stage II crack growth itself is contmost engineering materials. Stage II crack growth itself is controlled primarily by rolled primarily by

the the stress intensity at the crack tipstress intensity at the crack tip..

�� From a design viewpoint, the assessment of crack propagation ratFrom a design viewpoint, the assessment of crack propagation rates for stage II es for stage II

is of primary concern. The most effective approach to fatigue cris of primary concern. The most effective approach to fatigue crack growth has ack growth has

been through the use of been through the use of fracture mechanicsfracture mechanics. Here the crack growth is correlated . Here the crack growth is correlated

with the cyclic changes in the state of stress near the crack.with the cyclic changes in the state of stress near the crack.

�� The state of stress near the crack is described by the stress inThe state of stress near the crack is described by the stress intensity factor Ktensity factor KII. .

The rate of crack growth da/dN should correlate with The rate of crack growth da/dN should correlate with ∆∆K, the range over which K, the range over which

K varies during cyclic loading.K varies during cyclic loading.

Stage II Crack GrowthStage II Crack Growth

∆∆K = KK = Kmaxmax−− KKminmin

∆∆ The basic equation that relates da/dN and The basic equation that relates da/dN and ∆∆K is known as the K is known as the Paris lawParis law, and it , and it

has the formhas the form

∆∆ A and m are constants which depend on the materials. The value oA and m are constants which depend on the materials. The value of m is usually f m is usually

between 2between 2--4 (for steel, m=3, for Al, m=34 (for steel, m=3, for Al, m=3--4)4)

aK

aaK

r πβσ

πσβπσβ

=∆

−=∆ minmax

m

dN

da KA∆=

Paris region

Stable growth

Thre

sh

old

re

gio

n

Slo

w c

rack g

row

th

Fast

fractu

re

reg

ion

Rap

id-u

nsta

ble

cra

ck g

row

th

log(∆K)

Lo

g (

da/

dN

)

da

dN= A(∆K)m

A

m

∆Kth

Region I Region II Region III

The relationship between fatigue

crack growth rate and ∆K as shown in the graph

Divided into 3 regions:

• region I: No observable crack growth

• region II: linear relationship between crack growth rate and

stress intensity factor

• Region III: accelerated crack

growth.

fC

K

f

mmr

m

mr

m

dN

da

r

aKbecausea

aA

aAKA

aaaK

C πβσ

πσβ

πβσ

πβσπβσπβσ

βσπ max21

2/

minmax

)(

)()()(

)()(

max

==

=

=∆=

=−=∆

Paris region

Stable growth

Thre

sh

old

re

gio

n

Slo

w c

rack g

row

th

Fast

fractu

re

reg

ion

Rap

id-u

nsta

ble

cra

ck g

row

th

log(∆K)

Lo

g (

da/

dN

)

da

dN= A(∆K)m

A

m

∆Kth

�� The fatigue crack growth life The fatigue crack growth life can be obtained by integrating can be obtained by integrating this equationthis equation

�� Between the limits of initial and Between the limits of initial and final crack sizefinal crack size

m

dN

da KA∆=

Region I Region II Region III

1

( ) 2/

1)2/(1)2/(

1)2/(1)2/(1)2/(

2/

2/2/

2/

1)2/(

1)2/(1)2/(

)(

1

)()()(

0

2

mmr

m

mi

mf

mi

mfm

f

i

m

f

i

mmmr

m

f

i

mmr

m

f

Am

aa

f

m

aa

m

a

a

a

a

da

a

a

a

da

Af

a

a

aA

da

N

f

N

mIf

N

dNN

πσβ

πσβ

πσβ

+−

−

+−

−

+−

+−+−

+−+−+−

=

==

≠

=

==

∫

∫

∫∫The number of cycles to failure, NThe number of cycles to failure, Nff is:is:

• A mild steel plate is subjected to constant amplitude uniaxial fatigue loads to

produce stresses varying from σmax=180 MPa to σmin=-40 MPa. The static

properties of the steel are σo = 500, MPa, E=207 GPa, Kc=100 MPa.m1/2.

• If the plate contains an initial through thickness edge crack of 0.5mm, how

many fatigue cycles will be required to break the steel?

Assume an infinite wide plate with β =1.12. For ferritic-stainless steels a

general correlation gives

Thus, A= 6.9 x 10-12 MPa√m, m=3.0 and σr = 180-0 (since compressive

stresses are ignored and we shall neglect the small influence of mean

stress on the crack growth.

Example

3312 )()(109.6 mMPaKxdN

da∆= −

ai = 0.0005 m,

Nf = 261,000 cycles

Solution

mmmx

Ka c

f 78078.012.1180

1001122

max

==

=

=

πβσπ

( )

6

5.05.0

2/33312

1)2/3(1)2/3(

1)2/(

104.157

)0005.0()078.0(

)()180()12.1)(109.6)(1)2/3((

)0005.0()078.0(2/

1)2/(1)2/(

−

−−

−

+−+−

+−

−

−

−=

+−

−==

+−+−

x

xN mm

rm

mi

mf

Am

aa

fππσβ

• A fatigue test was conducted in which the mean stress was 70 MPa, and the stress amplitude was 210 MPa.

a) compute the max and min stress levels

b) Compute stress ratio

c) Compute the magnitude of the stress range.

Example 2

(a) Given the values of σm (70 MPa) and σa (210 MPa) we are asked to

compute σmax and σmin

σm=(σmax+σmin)/2=70 MPa

σmax + σmin = 140 MPa

σa=(σmax−σmin)/2=210 MPa

σmax – σmin = 420 MPa

Simultaneously solving these two expressions leads to

σmax=280 MPa

σmin=−140 MPa

Solution:

(b) Using Equation

the stress ratio R is determined as follows:

R=σmin/σmax=−140 Mpa/280 MPa=−0.50

(c) The magnitude of the stress range σr is determined using Equation

σr=σmax−σmin=280 MPa − (−140 MPa)=420 MPa

• The fatigue data for a brass alloy are given as follow:

a) Make an S-N plot (stress amplitude versus log cycles to failure) using these data.

b) Determine the fatigue strength at 4 x106 cycles

c) Determine the fatigue life for 120 MPa.

Example 3

Stress amplitude (MPa) Cycles to failure

170 3.7x104

148 1.0x 105

130 3.0x 105

114 1.0 x106

92 1.0x 107

80 1.0 x 108

74 1.0 x 109

Solution

(b) As indicated by one set of dashed lines on the plot, the fatigue strength at

4 x 106 cycles [log (4 x 106) = 6.6] is about 100 MPa.

(c) As noted by the other set of dashed lines, the fatigue life for 120 MPa is

about 6 x 105 cycles (i.e., the log of the lifetime is about 5.8).

• A cylindrical 1045 steel bar is subjected to repeated compression-tension stress cycling along its axis. If the load

amplitude is 66,700 N. Compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume safety factor of 2.0.

Example 4

This problem asks that we determine the minimum allowable bar diameter to

ensure that fatigue failure will not occur for a 1045 steel that is subjected to

cyclic loading for a load amplitude of 66,700 N. From the Figure, the fatigue

limit stress amplitude for this alloy is 310 MPa. Stress is defined as σ =F/A0.

For a cylindrical bar A0 = π(do/2)2

Substitution for A0 into the Equation to

We now solve for d0, taking stress as the fatigue limit divided by the factor of

safety. Thus

Solution

22

4

)2

( ooo d

F

d

F

A

F

ππσ ===

mm

mxmNx

N

N

Fdo

4.23

104.23

)2

/10310(

)700,66(44 3

26

=

=

=

= −

πσ

π

• A 6.4mm diameter cylindrical rod fabricated from a 2014 T6 aluminum alloy is subjected to reversed tension-compression load cycling along its axis. If the maximum tensile and compressive loads are +5340 N and -5340N respectively, determine its fatigue life. Assume the stress plotted is stress amplitude.

Example 5

• We are asked to determine the fatigue life for a cylindrical 2014-T6

aluminum rod given its diameter (6.4 mm) and the maximum tensileand compressive loads (+5340 N and –5340 N, respectively). The

first thing that is necessary is to calculate values of σmax and σmin

= 166 Mpa

σmin = -166 MPa

Stress amplitude , σa = (σmax-σmin)/2 = 166MPa

From figure, the number of cycles to failure is 1x107 cycles

Solution

2

maxmaxmax

)2/( oo d

F

A

F

πσ ==

Appearance of Fatigue failureAppearance of Fatigue failure

�� Fatigue results in a brittle appearance of fracture. On Fatigue results in a brittle appearance of fracture. On

a macroscopic scale, fatigue displays very little plastic a macroscopic scale, fatigue displays very little plastic

deformation.deformation.

�� Two different regions can be distinguished. Two different regions can be distinguished.

1.1. The first is a The first is a smooth regionsmooth region, through which the crack has , through which the crack has

grown under cyclic loading. grown under cyclic loading. ““Beach marksBeach marks”” are are

characteristics of fatigue failure and may indicate crack characteristics of fatigue failure and may indicate crack

originorigin

2.2. The second region is a The second region is a rough arearough area covered by the crack covered by the crack

during the final fracture, where the component has failed during the final fracture, where the component has failed

in a ductile manner when the cross section was no longer in a ductile manner when the cross section was no longer

able to carry the loadable to carry the load

Beach

marks

Rough Rough

regionregion

Smooth Smooth

regionregion

http://materials.open.ac.uk/mem/mem_ccf4.htm

Smooth regionSmooth region

Rough regionRough region

Bicycle crack failureBicycle crack failure

Rough regionRough region

Smooth regionSmooth region

Beach marks

The crack progressed slowly through the crank arm (dark area)

until the remaining fragment was incapable of supporting the

bending moment generated by

the force on the pedal and the crank arm fractured rapidly.

Smooth regionSmooth region

Rough regionRough region

Beach MarksBeach Marks

FATIGUE CONTROLFATIGUE CONTROL

�� Most fatigue failure can be traced to deficiencies in design ratMost fatigue failure can be traced to deficiencies in design rather than her than

inadequacies in materials or improper manufacture or maintenanceinadequacies in materials or improper manufacture or maintenance..

�� These fatigue failures are usually caused by These fatigue failures are usually caused by stress concentrationsstress concentrations (which may (which may

have been eliminated or minimised at the design stage)have been eliminated or minimised at the design stage)

�� Conditions which may reduce fatigue strength include:Conditions which may reduce fatigue strength include:

1.1. Stress concentrations due to improper designStress concentrations due to improper design

2.2. Stress concentrations due to improper manufacture (rough machineStress concentrations due to improper manufacture (rough machined surfaces)d surfaces)

3.3. Residual tensile stresses due to grinding or cold formingResidual tensile stresses due to grinding or cold forming

4.4. Plating Plating

5.5. Surface condition introduced by heat treatment (oxide penetratioSurface condition introduced by heat treatment (oxide penetration, decarburisation)n, decarburisation)

6.6. Size (most published fatigue date on materials are based on smalSize (most published fatigue date on materials are based on small laboratory l laboratory

specimens with do not adequately evaluate the fatigue strength ospecimens with do not adequately evaluate the fatigue strength of large parts)f large parts)

�� Fatigue starts at the surface of the material. This is because:Fatigue starts at the surface of the material. This is because:

�� Slip is easier at the surface than in the interior of the materiSlip is easier at the surface than in the interior of the materialal

�� The environment is in direct contact with the surfaceThe environment is in direct contact with the surface

�� Any change in the surface properties will result in a change in Any change in the surface properties will result in a change in fatigue strengthfatigue strength

�� Possible methods of designing against fatigue failure include:Possible methods of designing against fatigue failure include:

�� Avoid stress concentrations at the design stageAvoid stress concentrations at the design stage

�� Produce parts with polished surfaces (polishing eliminates machiProduce parts with polished surfaces (polishing eliminates machining marks)ning marks)

�� Surface treatment such as: surface hardening (nitriding, carburiSurface treatment such as: surface hardening (nitriding, carburising) produce sing) produce

layers in compression. layers in compression.

�� Any treatment that increases the hardness or yield strength of tAny treatment that increases the hardness or yield strength of the material will he material will

increase the stress level needed to produce slip, resulting in eincrease the stress level needed to produce slip, resulting in enhance fatigue nhance fatigue

strengthstrength

2924

• Occurs when a material experiences lengthy periods of cyclic or repeated stresses

• Failure at stress levels much lower than under static loading

• Fatigue is estimated to be responsible for approximately 90% of all metallic failures

• Failure occurs rapidly and without warning• There is no fixed ratio between materials Yield-and Fatigue

Strength

• Normally the ratio varies between 30-60%• Fatigue Strengths are usually average values

SUMMARY