1

6-6 Linear-Elastic Fracture Mechanics Method •  Stage I – Initiation of micro-crack

due to cyclic plastic deformation •  Stage II – Progresses to macro-

crack that repeatedly opens and closes, creating bands called beach marks

•  Stage III – Crack has propagated far enough that remaining material is insufficient to carry the load, and fails by simple ultimate failure

6-6 Linear-Elastic Fracture Mechanics Method

Paris Law ( )

( )

( )

( )( ) constant=⋅⇒

=⇒

==

=

=−=

∫

∫∫

bf

a

a mfm

a

a m

N

0f

mI

minmaxI

Nσπaβ

daC1Nσ

πaσβ

daC1dNN

ΔKCdNda

πaσβπaσσβK

f

i

f

i

f

Δ

Δ

Δ

ΔΔ

Propagation of this class of cracks also follows the S-N diagram’s functional form.

•  Crack growth in Region II is approximated by the Paris equation

Quantifying Fatigue Failure:���Stress-Life Method for Zero Mean Stress ���

(Suitable for Elastic Stress and High Lifetimes—“S-N diagrams”)

Stress Life Testing: R. R. Moore Machine

[Failure of materials in mechanical design: analysis, prediction, prevention, by Jack A. Collins, Wiley, 1993, p. 189]

(material point on

the outside)

R. R. Moore specimen

R.R. Moore specimen (in inches)

Motivation: Train Axel Low Cycle Fatigue: 1≤N≥103

High Cycle Fatigue: N>103

2

R. R. Moore Exaggerated

[Modified from Failure of materials in mechanical design: analysis, prediction, prevention, by Jack A. Collins, Wiley, 1993, p. 189]

Deformation of R. R. Moore Specimen Consider Zero Mean Stress First

Stre

ss a

t A

t mid-range stress, σm  =  0

load reversals (2 reversals = 1 cycle)

stress amplitude, σa  

maximum stress, σmax  

minimum stress, σmin  

Follow a material point A on the outer fiber

6-4 Stress-Life Method

R.R. Moore machine fatigue test specimen (in inches).

S-N bands for aluminum alloy

Fig. 6-10. Fatigue diagram for 4130 (chromoly) steel,

zero mean stress

′

Experimental Correlations of Fatigue: S-N Diagrams

(to failure)

If stress amplitude σa  is low enough, the specimen lasts forever in cyclic loading

Log-log plots that correlate stress amplitude to number of cycles to failure are called S-N diagrams

The fatigue strength Sf (or endurance strength) is the stress amplitude σa corresponding to the number of cycles to failure

3

Fatigue Strength and Endurance Limit, Se (R. R. Moore Se′)

This lower limit of σa = Sf below which ∞ life is guaranteed is called the endurance limit Se

(to failure)

The endurance limit of an R. R. Moore specimen (i.e., carefully lab-controlled) is denoted by Se

'

R.R. Moore specimen (in inches)

′The fatigue strength Sf (or endurance strength) is the stress amplitude σa corresponding to the number of cycles to failure

6-7 “Sut Correlation” for the Endurance Limit

⎪⎪⎩

⎪⎪⎨

⎧

>

>

≤

=

MPa1400

kpsi200

Pa)kpsi(1400M200

MPa700

kpsi100

5.0

'

ut

ut

utut

e

S

S

SS

SFor Steels,

Martensite

Ductile

x

(See Eq. 6-8 and Table A-24)

6-8 Estimating the S-N Diagram High cycle (103 ≤ N ≤ 106):

aNbSaNS

f

bf

logloglog +=

=S-N Diagram

( ) ef SS ʹ′=610

( ) utf fSS =310

(use Sut correlation)

( )

⎟⎠

⎞⎜⎝

⎛−=

=

e

ut

e

ut

SfSb

SfSa

log31

2

Know two points (Sf)10e3, (Sf)10e6:

Solve for a and b: (Use f from Fig. 6-18 for (Sf)10e3)

(Valid for 103≤N≤106)

Sut

103 (lo-cycle boundary assumed)

106 (hi-cycle boundary assumed)

Sf

N

E.g. 1. (Example 6-2) Given a 1050 HR steel (Sut = 90 ksi (Table A-21)), estimate:

(a)  The (R. R. Moore) rotating-beam endurance limit at 106 cycle

(b)  The fatigue strength of a polished rotating-beam specimen corresponding to 104 cycles to failure

(c)  The expected life of a polished rotating-beam specimen under a completely reversed stress of 55kpsi.

kpsiS

S

e

ut

45)90(5.0

ksi90' ==

=

(c)

(a)

(b) ( ) ( )

( ) kpsi6.641013.133

13.133

785.0459086.0log

31log

31

13.133459086.0

)186Figure(86.0

0785.04

0785.0

22

==

=

−=⎥⎦⎤

⎢⎣⎡ ⋅

−=⎟⎠

⎞⎜⎝

⎛−=

=⋅

==

−=

−

−

f

f

e

ut

e

ut

S

NS

SfSb

SfSa

f

cycles)10(75.777500

13.13355

55

4

0785.01/1

failure

==

⎟⎠⎞

⎜⎝⎛=⎟

⎠

⎞⎜⎝

⎛=

==

−bf

af

aS

N

ksiσS

4

Exercise: Find safety Factor against fatigue failure, R. R. Moore specimen of E.g. 1, Mmax = 200 lbf-in

6-9 Endurance Limit Modifying Factors •  High-cycle Fatigue (103<N<107 cycles)

efedcbae SkkkkkkS ʹ′=

where Se′=endurance limit under idealized condition ka=surface factor kb=size factor kc=load modification factor kd=temperature factor ke=reliability factor kf=miscellaneous factor

Surface and Size Factor •  Surface factor:

•  Size factor

buta aSk =

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

≤<

≤≤

≤<

≤≤

=

−

−

−

−

mm2545151.1

mm5179.224.1

in10291.0

in211.0879.0

107.0

107.0

157.0

107.0

dd

dd

dd

dd

kb

For Torsion and Bending For Axial loading 1=bk

kb|R.R. Moore specimen = 1

Size factor: Non-R.R.Moore Loading/Geometry

•  Effective dimension, de –  Equate the volume of material stressed at and above 95%

of the maximum stress to the same volume in the rotating-beam sample

–  Bending without rotation: solid/hollow round

–  Rectangular section

dde 370.0=

( )[ ] 22295.0 0766.095.0

4 edddA =−=π

σ

hbde ⋅= 808.0

295.0 01046.0 dA =σ

hbA 05.095.0 =σ

5

Table 6-3. 95% Stress Areas

Loading and Temperature factors •  Loading Factor

•  Temperature factor

⎪⎩

⎪⎨

⎧

=

Torsion59.0Axial85.0Bending1

ck

( ) ( )( ) ( ) 41238

253

10595.010104.0

10115.010432.0975.0

FF

FFd

TTTTk

−−

−−

−+

−+=

RT

Td S

Sk =

If the endurance limit at room temperature is known

Data culled from 21 carbon and alloy steels for when temperature testing is not available (Standard Handbook of Machine Design, Shigley/Mischke, 2nd ed., McGraw Hill, 1996, p. 13.16)

Reliability factor

Reliability is defined as the probability of no failure. We say, “The actual value of the endurance limit is greater than ke(R)∙Se′ with a probability of R (left-most column).” (See also §6-17)

•  Manufacturing history –  Residual stresses

•  Corrosion •  Coatings •  Electrolytic Plating •  Metal Spraying •  Cyclic frequency •  Frettage Corrosion

Miscellaneous factors

6

Stress Concentration and Notch Sensitivity (Bending & Axial)

( ) qK

K

t

f

⋅−+=

⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

=

11

specimenfree-­‐notchforlimitEndurance

specimennotchedforlimitEndurance

11

−

−=

t

f

KK

q

Notch sensitivity factor

For cast iron, q=0.2

otK σσ =max

38253 1067.21051.11008.3246.0,1

1ututut SSSa

ra

q ⋅×−⋅×+⋅×−=+

= −−−

Eqn  6-­‐35a  (extends  Figure  6-­‐20)

r [in], Sut [kpsi]

Fatigue stress concentration, Kf, depends on geometry, material & loading

Static stress concentration, Kt, depends on geometry (see Table A-15):

Stress Concentration & Notch Sensitivity���(Torsion)

( ) sst

fs

qK

K

⋅−+=

⎭⎬⎫

⎩⎨⎧

⎭⎬⎫

⎩⎨⎧

=

11

specimenfree-­‐notchforlimitEndurance

specimennotchedforlimitEndurance

Fatigue stress concentration, Kfs, depends on geometry, material & loading

Static stress concentration, Kts, depends on geometry (see Table A-15):

otsK ττ =max

38253 1067.21035.11051.2190.0,1

1utututs SSSa

ra

q ⋅×−⋅×+⋅×−=+

= −−−

Eqn  6-­‐35b  (extends  Figure  6-­‐21)

r [in], Sut [kpsi]

11

−

−=

ts

fss K

Kq

Notch sensitivity factor

For cast iron, qs=0.2

E.g. 2. SN Diagram for Zero Mean Stress

( )MPa9.15MPa9.15

636620201.02

mN25mN25kN5.0kN5.0m 05.0

34

≤≤−

⋅====

⋅≤≤⋅−⇒≤≤−

⋅=⋅=

τ

Tπ

TcπTc

JTcτ

TFFdFT

MPa2.23MPa2.23or)9.15(46.1)9.15(46.1

toe

toe

≤≤−

≤≤−⇒

ττ

Static shear-stress concentration is Kts=1.6 at the toe of the 3-mm fillet: τtoe → Ktsτnom

Calculate the fatigue concentration factor with Kts=1.6 at the toe of the 3-mm fillet:

( )11 −+= tssfs KqK

( ) ( ) 46.116.1775.0111 =−+=−+= tssfs KqK

Shaft A: 1010 hot-rolled steel Kts= 1.6 at the 3-mm fillet F = −0.5 to 0.5 kN cycle Sut = 320 MPa Find safety factor against the finite-life fatigue failure of shaft A.

Sut = 320 MPa too low to register on Fig. 6-21. Use Eqn. (6-35b)

toe of fillet

( ) 21.232

21.2321.23=

−−=aτ

E.g. 2. Zero Mean Stress

Convert ideal to actual endurance limit:

( ) MPa1603205.05.0 ===ʹ′ ute SSApproximate ideal endurance limit:

36.32.239.77

stress servicestrength

====a

es

τSnSafety factor:

Stress amplitude:

(kc converts tensile endurance to shear endurance!)

( )

( )

MPa9.77)160)(59.0)(9.0(917.059.0

9.02024.1

917.03207.57

59.0

107.0

718.0

===

=

==

===

ʹ′=

=

−

−

ckees

c

b

buta

efedcbae

SSkk

aSk

SkkkkkkS

(95% stress area is the same as R. R. Moore specimen)

The diameter of shaft A proves to be overly conservative. Exercise: How much do we have to reduce the diameter to get a design factor of 1.75?