ASME Fatigue for Engineers Part III

7/28/2019 ASME Fatigue for Engineers Part III

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1

Fatigue fo r Eng ineers

Prepared b y

A. F. Grandt, J r.

Professor of Aeronautics and Astronautic sPurdue University

W. Lafayette, IN 47907

June 1999

2

Object ive

Overview nature/consequences of the

fatigue failure mechanism

Determine number of cycles required to

develop a fatigue crack

propagate a fatigue crack

Discuss implications of fatigue on

design and maintenance operations


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3

Struct ural Failure Modes

Excessive Deformation

Elastic

Plastic Buckling

Fracture

Creep

Corrosion

Fatigue

Forc

e

Displacement

Yield

Permanent

displacement

displacement

Force

4

Fatigue Failure Mechanism

Caused by repeated (cyclic) loading

Involves crack formation, growth, and final

fracture

Fatigue life depends on initial quality, load, . . .

St

ress

Time

Crack Nucleation

Fracture

Crack Growth

Elapsed Cycles N

CrackLength(a)

a

Crack


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5

Paper Cl ip Exp er iment

Bend wire repeatedly until fracture

Note:

life (number of applied load cycles)depends on:

applied stress amplitude

component quality (notches, scratches, etc.)

heat emitted >> plastic deformation

6

Character is t ics o f Fat igue

Brittle fracture surface appearance Cracks often form at free surface

Macro/micro beach marks/ striations

0.3 in

Beach marks

20 m

Striations


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Fatigue is problem for m any

types of s t ruc tures

8

Exerc ise

Describe fatigue failures from yourpersonal experience

What was cause of fatigue failure?

What was nature of cyclic load?

Was initial quality an issue?

How was failure detected?

How was problem solved?


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Exerc ise

Estimate the fatigue lifetime needed for:

Automobile axle

Railroad rail

Commercial aircraft components landing gear

lower wing skin

Highway drawbridge mechanism

Space shuttle solid propellant rocket motorcases

10

Exerc ise

Give an example of a High Cycle

Fatigue (HCF) application.

What is the required lifetime?

What are consequences of failure?

Given an example of a Low Cycle

Fatigue (LCF) application.

What is the required lifetime?

What are consequences of failure?


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Fatigue Crack Form ation

12

Crack Formation

Fracture

Crack Growth

Elapsed Cycles N

CrackLength(a)

Fatigue Crack Form ation

Objective Characterize resistance to fatigue crack formation

Predict number of cycles to initiate small*fatigue crack

in component

*crack size ~ 0.03 inch

= committee crack

Approach Stress-life concepts

(S-N curves)

Strain-life concepts


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13

Stress-l i fe (S-N) Appro ach

Concept: Stress range controls fatigue life

S

S

Log cycles N

S/2

Note:

Life increases as load amplitude decreases Considerable scatter in data Run-outs suggest infinite life possible

Life N usually total cycles to failure

S

time

S

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Model Stress-life (S-N) Curv e

Se = endurance limitfor steels

Se ~ 0.5 ultimate stress Sult

Se ~ 100 ksi if Sult 200 ksi

Log reversals 2N

LogS

/2

Se

S/2 = f (2N)b

f = fatigue strength coefficient b = fatigue strength exponent

typically -0.12 < b < -0.05

Note: Measure life in terms of reversals 2N

(1 cycle = 2 reversals)


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S-N Curv e: Mean Stress

Mean stress effects lifestress ratio R = S

min/ S

max

Smean

= 0.5(Smin

+ Smax

)

Sa = 0.5(Smax - Smin) = S/2Mean stress models

Sa/Se + Sm/Sult = 1

S/2 = (f - Smean)(2N)b

Mean StressStressAmplitude

N = 106

N = 103

Haigh constant life diagram

S

timeSmin

Smax

S = 2Sa

16

S-N Curve: Other Factors

S-N curves are very sensitive to surface finish, coatings, notches

prior loading, residual stresses

specimen size effects, etc.

Many empirical knock-down factors

S-N approach best suited for HCF (HighCycle Fatigue) applications

limited by local plastic deformation

strain-life approach better for LCF (Low

Cycle Fatigue)


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Strain -life (- N) Appro achConcept: Strain range controls lifeExperiment

Control

Measure Reversals (2Nf)to failure (1 cycle

= 2 reversals)

Stable stress range needed to maintain

Note: stable usually occursby mid-life (2Nf/2)

time

time

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Cyclic Stress-Strain Curve

Relate stable cyclic stress and strain ranges

t ime

time

Hystersis l oop

/2

/2

/2 = /2E + (/2K )1/n

Cyclic stress-strain curve

E = elastic modulus

K = cyclic strength coefficient

n = strain hardening exponent


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Plast ic Strain-Life Cu rve

Relate plastic strain amplitude p/2with reversals to failure 2Nf

Compute p/2 = /2 - /2E = total - elastic strain amplitudes

Log

p/2

Log 2Nf

p/2 = f (2Nf)c

f = fatigue ductility coefficient

c = fatigue ductility exponent

typically -0.7 < c < -0.5

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Total Strain-Life Curv e

Plot total strain amplitudes versus life 2Nf

total

/2 = /2 = 0.5 elastic

+0.5 plastic

=/2E + 0.5

plasti c

/2 = {(f - Smean )/E}(2N)b + f (2Nf)c

p/2 = f (2Nf)c

/2E = {( f - Smean)/E} (2Nf)b

Log 2N f

Logstrainamplitude

2Nt = transition life


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Total Strain-Li fe

Note:

Plastic strain dominates for LCF

Elastic strain dominates for HCF

Transition life 2Nt separates LCF/HCF

p =f (2Nf)c

/2 = {(f - Smean )/E} (2N)b + f (2Nf)c

Log 2N f

Logstrainamplitude

/2E = {( f - Smean)/E}(2Nf)b

2N t = transition life

LCF

HCF

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Variable Am pl i tude Loading

Load amplitude varies in many applications

Use of constant amplitude S - N or- Ndata requires damage model

Miners rule*

(Ni/Nf) = 1

Ni= number of applied cycles of stress amplitude S

ai

Nf= fatigue life for Sai cycling only

*Use with caution!

S

time

N i

2Sai


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Example Problem

Assume:

f = 220 ksi, b = - 0.1 stress history shown (1 block of loading)

Find: number of blocks to failure

+ 80 ksiS

time

- 80 ksi

- 100 ksi

+ 100 ksi

2N = 100

2N = 1000

2N = 1000S

S

24

Solut ion

(Ni/Nf) = 1 2Nf = {(S/2) / (f - Smean)}1/b

(Ni/Nf) = 1

When:

1/0.0089= 112.5

Answer

112 blocks

S/2(ksi)

Smean(ksi)

2Nf 2Ni Ni/Nf

80 0 24,735 100 0.0040

50 +50 206,437 1000 0.0048

50 -50 21 E6

1000 4.74 E-6

0.0089


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Notch Fatigue Notches can reduce life

Define Fatigue Notch Factor

Kf

Kf = Smooth/notch fatigue

strength at 106 cycles

= Ss /Sn1 < Kf < Kt

(K t = elastic stress

concentration factor)

Kf= 1 no notch effectK

f= K

t full notch effect

Smooth

Notch

S/2

Log cycles N

S s /2

Sn /2

106

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Load Sequence Effects

Hi-lo strain sequenceresults in tensile meanstress when last large peak in compressionas shown here

decreases life! If last peak had been

tension, would result incompressivemeanstress

increaselife

Load sequence important!

t

t

Mean stress


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Neuber s Rule

Kf= fatigue notch concentration factor

(s,e) = nominal stress/strain ranges(away from notch)

(,) = notch stress/strain rangesNeubers rule relates notch and

nominal stress/strain behavior

Solve with:

Kf

2se =

/2 = /2E + (/2K )1/n

/2 = {(f - Smean)}(2Nf)b + f (2Nf)c

(,)

(s,e)

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Summ ary Ini t iat ion Methods

Total strain-life approach combines: original S-N curve (best suited for HCF) and

plastic strain-life method developed for LCFproblems

S-N and strain-life often viewed as crack

initiation approaches

actually deal with life to form small crack

crack size implicit in specimen/test procedure

typically assume committee crack ~ 0.03 in.


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In i t iat ion Summary Cont

Notches increase local stress/strain andoften are source for crack formation

complex problem leads to local plasticity

characterize by fatigue notch concentrationfactor Kf,, Neubers rule

Load interaction effects result in local

mean stress

can increase/decrease life

invalidate Miners rule

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Fatigue Crack Grow th


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Crack Growth App roach

Assumes entire life

fatigue crack growth

ignores initiation assumes component

cracked before cycling begins

Used with damage tolerant design

protects from pre-existent (or service) damage

based on linear elastic fracture mechanics

Elapsed Cycles N

Crack Growth

CrackLength(a)

Fracture

Initial crack

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Damage Tolerance

The ability of a structure to resist priordamage for a specified period of time

Initial damage

material

manufacturing

service induced

size based on

inspection capability,

experience, . . .time

Cracksize

Desired Life


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Fatigue Crack Grow th

Objective Characterize material resistance to fatigue crack growth

Predict catastrophic fracture and subcritical crack

growthApproach

Assume crack growth

controlled by stress

intensity factor K

fracture

growth rate da/dN

Elapsed Cycles N

Crack Growth

CrackLength(a)

Fracture

Initial crack

34

Stress Intensi ty Factor KI

KI is key linear elastic fracture mechanicsparameter that relates:

applied stress: crack length: a

component geometry: (a)((a) is dimensionless) a

Crack

= 1.12

aK=I

Note units: stress-length1/2


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Stress Intensi ty Factors

2a

W

K a Seca

W=

12

=Remote Stress

20 95

a

W .

W

a

h

a

W

0 6.

a

W

h

W

10.

K a

a

W

a

W

=

=

+

112 0 231 10.55. .a

W

a

W

a

W

+

21 73 30 39

2 3 4

. .

For and

Many K Isolutionsavailable

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Crack t ip Stress Fields

( )

+==

==

=

+=

=

yxz

z

yzxz

Ixy

Iy

Ix

r

Kr

Kr

K

strainplane

0stressplane

02

3cos

2cos

2sin

2

2

3sin

2sin1

2cos

2

2

3sin

2sin1

2cos

2

Theory of elasticity gives elastic stresses near crack tip in

terms of stress intensity factor K I

All crack configurations have same singular stress field at tip(are similar results for other modes of loading, i.e., modes II and III)

Crack

x

y

r

xy

y

x


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KcFracture Cri terion

Fracture occurs when

K > constant = Kc K

c= material property

= fracture toughness Criterion relates:

crack size: a

stress: geometry: (a) material: Kc

Plasticity limits smallcrack applications

2a

ult

FractureStress

Crack Size a

( )K a ac =

38

Fracture Toug hness Kc

Typical Kc values (thick plate)

Note Kcdepends on:

specimen thickness -- Kc decreases as

thickness increases until reaching minimum -

KIc = plane strain toughness

crack direction (material anisotropy)

M aterial

(thickplate)

2024-T351

Aluminum

7075-T651

Aluminum

Ti-6Al-4V

Titanium

300M steel

(235ksiyield)

18Nickel

(200ksiyield)

Kc

(ksi-in1/2)

31 26 112 47 100


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Fracture Examp le

Member A fractures whencrack length a = 2.0 inch

and remote stress = 5 ksi

What stress will fracturemember B (assume samematerial)?

2.0 in

4.0 in

5 ksi

5 ksi

A

5 in

8 in

= ?

= ?

B

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Fracture Examp le Solut ionEdge crack

K = (a)1/2(a) = Kc at fracture

a/w = 2/4 = 5 a = 2 = 2.83K

c= 35.5 ksi-in1/2= constant

Center Crack

K = ( a)1/2(a) (a) = [Sec ( a/W)]1/2

a = 2.5 W = 8 = 1.34K = Kc at fracture = 35.5

2.0 in

4.0in

5 ksi

5 ksi

5 in

8 in

= ?

= ?

a

W

a

W

=

+ 1 12 0 231 10. 5 5. .

a

W

a

W

a

W

+

21 73 30 39

2 3 4

. .

f = 9.5 ksi


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Fat igue Crack Grow th

Goal: show cyclic stress intensity factorKcontrols crack growth rate da/dN

P = constant

time

P

2a

P

Crack Face Load

2a

Remote Load

= constant

time

Same material

Different loadings

42

Measure Crack Growth

2a

Remote Load

2a

P

Crack Face Load

da

dN

CrackLength(a)

Number of Cycles (N)

=K PBa

K = a

CrackLength(a)


da

dN

a*


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Corr elate Rate da/dN vsK

CrackLength(a)


da

dN

2a

2a

CrackLength(a)


da

dN

a*

KthK

c

Log K

Logda/

dN

K a=

K PB a

=

44

da/dN VsK

KthKc

Log K

Logda/dN

Note: K correlates fatiguecrack growth rate da/dN

K accounts for crackgeometry

No crack growth for

da/dN


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Sample Crack Grow th Data

da/dN - K data for7075-T6 aluminum

Note effect of stress

ratio R = min/maxstress (da/dN as R) Reference: Military

Handbook-5

Other handbook dataare available

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Model d a/dN -K Curve

Fit test data with numericalmodels such as:

KthKc

Log K

Logda/dN

da

dNF K= ( )

da

dNC Km=

da

dN

C K

R K K

m

c

=

( )1

Here C, m, Kc are

empirical constantsR = min/max stress

(are many other models)

Paris

Forman


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Compute Fatigue Li fe Nf

ao, af = initial, final crack sizes

F(K) = function of:

cyclic stress: , R, . . . crack geometry: (a) crack length: a

material

Nda

F Kf

a

a

o

f

= ( )da

dNF K= ( )

time

2a

48

Example L ife Calculat ion

a

Crack

= constant

time

Given: edge crack in wide plate

Kc= 63 ksi-in1/2initial crack ai = 0.5 inchcyclic stress = 10 ksi, R = 0

( = max = 10 ksi)

da/dN = 10-9K4

Find: a) cyclic life Nf b) life if initial crack size

decreased to ai= 0.1 inch

Note: at fractureK = Kc = 63 = 1.12max (a)1/2

final crack af = 10 inch


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49

Solut ion

[ ] = =

da

C K

da

C am ma

a

a

a

o

f

o

f

112. Nf

( ) ( )[ ]N

C ma af m f

m

o

m=

1

1 12 1 5

1 5 1 5

. .

. .

K a= 112.da

dNC K

m=

a) Nf= 12,234 cycles (ai = 0.5)

b) Nf= 63,747 cycles (ai = 0.1)

Note: big influence of initial crack length!

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Fat igue Crack Retardat ion

Time

AppliedStress(

)

Overload

Without Overload

With Overload

RetardationCrackLength(a)

Elapsed Cycle (N)

Note load interaction effect Tensile overload can retard crack growth (increase life) Life increase due to crack tip plasticity

Depends on magnitude/sequence of overload, material, Are empirical retardation models


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Crack Growth Summary

Fracture mechanics approach assumesentire fatigue life is crack growth

Stress intensity factor K controls fractureand growth rate da/dN

K = [a]1/2(a) Fracture: K = Kc Fatigue: da/dN = F(K) Integrate da/dN for life

Are load interaction and other effects (see

references)

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Cycle-by-Cycle Calculat io n

Compute cycle-by-cycle growth in crack length a

acurrent = aprior+ da/dNcurrent

da/dNcurrent = F(Kcurrent) * Retardation term

Sum for all cycles in spectrum

Powerful technique for computer programming

n

n+1AppliedStress()

Time (t)

Variable amplitudeloading prevents

simple life integration


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53

Fatigue Desig n/Repair

Concepts

54

Des ign Phi losophies

Fatigue Design Criteria

Infinite Life

Safe-Life

Damage Tolerant

Fail-safe

Slow crack growth

Retirement-for-cause

a

Crack

St

ress

Time

Crack Formation

Fracture

Crack Growth

Elapsed Cycles N

Pre-CrackCrackLength(a)


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Inf in ite Life Criter ion

Design Goal: prevent fatigue damage from ever

developing (i.e. infinite life)

Usually based on endurance limit

Could also employ threshold K concepts

Leads to small design stresses/heavy members

Limited to simple components/loading

Often impractical/not achievable in practice

Weight critical structure

Complex loads

56

Safe-Life Criter io n

Design goal: component is to remain crack free forfinite service life

Assumes initial crack-free structure

Establish mean life by test/analysis

Safety factors account for scatter

predicted mean

Desired life = mean/S.F.

Design Life

Failure

Occurrence

1 32 4

Problems:

large safety factor no protection from

initial damage


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Fail-Safe CriterionDesign goal: contain single component failure

without losing entire structure

Assumes crack is present

Provide alternate load paths, redundant structure, crack

stoppers, etc. Requires detection of 1st failure

Time

Cracksize

1st member

2nd memberCrack arrest

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Slow Crack Growth Cri ter ion

Design goal: prevent initial crack from growing tofracture during life of structure

Pre-existent crack size specified by inspectionlimits, experience

Crack growth life

> service life x S.F.

Based on fatigue

crack growth

resistance

Emphasizes nondestructive inspection

Cracksize

Desired Life

time

Fracture


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Retirement-for-Cause

Failure size

Crack

Length

Time

inspect/repair

Design goal: Use periodic inspection/repair

to achieve desired fatigue lives

Limited by repeated maintenance economics

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Life Extension Concepts

Shot peenHole coldwork

Interference fastenersOverstress, etc.

Introduce BeneficialResidual Stresses

MetalComposite

Mechanical FastenBond

Doublers

HCF damping materials

Reduce Stressvia Reinforcement

Weight limitsFlight restrictions

etc.

Reduce OperatingLoads

No Cracks Found

(assume small cracks)

MetalComposi te Mechanical Fasten

Bond

Patches

Replace componentStop drill cracks

Welding

Repair CrackedStructure

Cracks Found

ComponentInspection


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Summary

Fatigue is complex problem that

involves many disciplines

Fatigue affects design and operation of

many types of structures Fatigue may be treated by several

methods/philosophies

Assume component cracked

Assume component uncracked

Probabilistic methods

62


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ASME Fatigue for Engineers Part III

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