Chapter 5 Fcp(10jun2014)

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Chapter 51

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Overview

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Fatigue definition A material subjectedto a repetitive or

fluctuating stress willfail at a stress muchlower than thatrequired to causefailure on a singleapplication of load.Failures occurringunder conditions of dynamic loading arecalled fatigue failures.

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Fatigue Mechanisms - two steps;Crack Initiation and

Crack Propagation

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Characterisation of FatigueThere are three commonlyrecognized forms of fatigue

!igh cycle fatigue "!CF#$%ow cycle fatigue "%CF#$Thermal mechanicalfatigue "T&F#

Fatigue strength isdetermined by runningmultiple specimen tests at anumber of differentstresses.The objective is to identifythe highest stress that will

produce a fatigue lifebeyond ten million cycles.

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This stress is also known as the material's endurance limit or fatigue limit.

Gas turbines are designed so that the stresses in engine components do not

exceed this value including an additional safety factor.

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'hat are the important parameters to characterize agiven cyclic loading history(

&ean stress

)tress range

)tress amplitude

)tress *atio

+ , *ange 5

minmax σ σ σ −=∆

minmax K K K −=∆

( )minmax σ σ σ −= 21

a

( )minmax σ σ σ += 21

m

max

min

max

min

K K R ==

σ

σ

Fatigue Crac- rowth , %/F& approach

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Frequency$ f in units !z. For rotating machinery at 0111 rpm$ f 2 31 !z. 4n general only influences fatigue crac- growth if

there are environmental effects present$ such as humidity orelevated temperatures.'aveform$ is the stress history a sine wave$ square wave orsome other waveform( As with frequency$ generally onlyinfluences fatigue crack growth if there are environmental effect.

Cyclic vs. )tatic %oading.+ey difference between static and cyclic loading

Static – Until applied K reaches Kc (30 MPa √ m for example)

the crack will not grow.Cyclic , + applied can be well below +c " 3 MPa √ m fore5ample#. Over time the crac- grows.The design may be safe considering static loads$ but

any cyclic loads must also be considered.


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4n 6786 the ideas of %/F& were applied to fatigue crac-growth by 9aris$ omez and Anderson .For given cyclic loading$ define ∆+ as + ma5 , + min which

can be found ∆σ and the geometry of the crac- body.)ay that the crac- grows an amount ∆a during : cycles.9aris$ omez and Anderson said that the rate of crac-

growth depends on∆

+ in the following way

Thus plot of log "da;d:# vs. log " ∆+# should be straight

line with a slope m .The actual relationship between crac- growth rate and∆+ is depicted on the following slide. There are threedifferent regime of fatigue crac- growth A$ < and C.


( )mK C dN da

N a ∆=→

∆∆

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F AT4 =/ C*AC+ *O'T!REGIME A B C

Terminology Slow-growth rate(near-threshold)

Mid-growth rate(Paris regime)

High-growth rate

Mi ros o!i mode Stage I" single

shear

Stage II"

(striations)

Additional stati

modes#ra t$re s$r%a e%eat$res

#a eted orserrated

Planar withri!!les

Additional lea&age ormi ro&oid oales ene

Mi rostr$ t$ral

e%%e ts

'arge Small 'arge

'oad ratio e%%e ts 'arge Small 'arge

En&ironmentale%%e ts

'arge 'arge

Stress state e%%e ts - 'arge 'arge

ear-ti! !lasti ity * r c < d g r c > d g r c >> d g9

* large influence on crack growth for certain combinations of environment, load ratio and frequency.

† r c and d g refer to the cyclic plastic zone size and the grain size, respectively.

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F AT4 =/ C*AC+ *O'T! >Regime AConcept of the threshold stress intensity ∆+ th

'hen ∆+ is ? ∆K th, , where ∆K th is the threshold stressintensity factor, the rate of crack growth is so slow that the crackis often assumed to be dormant or growing at an undetectablerate.

An operational definition for ∆K th often used is that ifthe rate of crac- growth is 61 >@mm;cycle or less the

conditions are assumed to be at or below ∆K th. An important point is that these e5tremely slow crac-growth rates represent an average crac- advance of lessthan one atomic spacing percycle. !ow is this possible('hat actually occurs is that there are many cycles with

no crac- advance$ then the crac- advances by 6 atomicspacing in a single cycle$ which is followed again by

many cycles with no crac- advance.

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F AT4 =/ C*AC+ *O'T! >Regime B'hen we are in regime < "9aris regime# the followingcalculation can be carried out to determine the number ofcycles to failure.From 9aris %aw

∆+ can be e5pressed in terms of ∆σ

'here B depends on the specific specimen geometry.

Thus the 9aris %aw becomes

Assume that B is a constant. )olve for d a and integrateboth sides

( )mK C dN da ∆=

aY K π σ ∆=∆

( )m

aY C dN da

π σ ∆=

( ) ∫∫ ∆= f f

o

N mmma

a m dN CY ada

0

2

2

/

/ π σ

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F AT4 =/ C*AC+ *O'T! >Regime BCON’T

For m D

For m 2 D

The constant C and m are material parameters that mustbe determined e5perimentally. Typically m is in the rangeD , E for metals and E , 611 for ceramics and polymers.For cases where B depends on crac- length$ theseintegration generally will be performed numerically.Important: :ote that in the 9aris regime the rate ofcrac- growth is wea-ly sensitive to the load ratio *. The-ey parameter governing crac- growth is ∆+.

( ) ( ) ( )( )

( )( )

−∆−

=−− 22222

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2

2 / / / m

f

m

o

mmm f

aaCY m

N π σ

( ) o f

f

a

a

CY N ln

π σ

22

1

∆

=

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From those e5pressions$ we need to determine the initialcrac- length a o and the final crac- length a f "some timescalled the critical crac- length#.!ow do we determine the initial crac- length a o (Crac- can be detected using a variety of techniques$ rangingfrom simple visual inspection to more sophisticated

techniques based on ultrasonics or 5>rays. 4f no crac-s aredetectable during inspection$ we must assume that a crac- just at the resolution of our detection system e5ists.!ow do we determine the final crac- length a f ( 'e -now

that eventually the crac- can grow to a length at which thematerial fails immediately$ i.e.

orC K K →max

C f K aY →π σ max


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Thus we may solve a f as follows

A very important idea that comes from this analysisis the following even if a component has adetectable crac-$ it need not to be removed fromservice =sing this framewor- the remaining lifecan be assessed. The component can remain inservice provided it is inspected periodically. This isthe crack-tolerant or damage tolerant designapproach.

22

2

1

maxσ π Y K a C f =


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An advancing fatigue crac- leavescharacteristic mar-ings calledstriations in its wa-e. These canprovide evidence that a given failure

was caused by fatigue.The striations on the fracture surfaceare produced as the crac- advancesover one cycle$ i.e. each striation

correspond to d a .)triations are close togetherindicating low stress$ many cycles.'idely spaced striations mean highstress few cycles .

Fatigue failure is brittle in nature$even in normally ductile materials

there is very little plastic deformationassociated with the failure.

F AT4 =/ ) T*4AT4O:)

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/GA&9%/ 3.6 A wide thic- plate$ + 6c 2 E1 -si>in 1.3 $ contains anedge crac- 1.6 inch long. The plate is subjected toalternating stresses between 1 and D1 -si. Hata onsimilar material under similar environmental

conditions e5hibited 9aris law with coefficients m 2E and C 2 I5 61 >61 for + in -si>in 1.3 . !ow manycycles will the plate support before failure(

Solution4t is first necessary to determine the length of thelongest crac- the plate can support withoutfailure. A suitable stress intensity factor is

+ 6c 2 6.6D σma5 √"πa c#E1 2 6.6D "D1# √"πa c#a c 2 6. 0 D inches

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SolutionFor this case$ 9aris %aw has the form

da;d: 2 C" ∆+# m 2 C J6.6D ∆σ #√"πa c#Km

)ubstituting this materialLs 9aris law parameters$we obtain

da;d: 2 I 5 61 >61 "6.6D#E"D1#EπDa D

*earranging the forgoing equations yieldsda;a D 2 I 5 61 >61 "6.6D#E"D1#EπDd:

Thus

4ntegrate from a i 2 1.6 inch to a c2 6.1D inch$: ? 36@3 cycles ans

∫ −−=02.1

1.0

2

24410)(

)20()12.1(1071

daa x

N π

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/GA&9%/ 3.D

A large aluminium alloy plate contains a

central crac- of length 61 mm. The plate issubjected to a constant amplitude tensile cyclicloading from 8 &9a to 81 &9a. The 9aris lawe5ponent is 0$ and ∆+ at da;d: 2 61 >I m;cyclesis D@ &9a √m. Assuming that B is 6.1D$calculate how many loading cycles must beapplied for the crac- to grow to D1 mm(

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First calculate C in 9arisL lawC 2 61 >I;D@02 E.33 5 61 >66$for crac- in m;cycleObtain stress range$ ∆σ 2 3E &9a)ubstitute into the given e5pression

−

∆−

=−− 2 / )2(2 / )2(2 /

)(

1

)(

1

)()2(

2m

f m

ommm f

aaCY m N

π σ

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Solution

−=− 2121233311 010

1

0050

1

5402110554

2 / / /

).().()(.. π

x

N f

Thus : f 2 673 8I3 cycles

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Chapter 5 Fcp(10jun2014)