Fatigue Failure

5/27/2018 Fatigue Failure

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Ken Youssefi MAE dept., SJSU 1

It has been recognized that a metal subjectedto a repetitive or fluctuating stress will fail at a

stress much lower than that required to cause

failure on a single application of load. Failures

occurring under conditions of dynamic loadingare called fatigue fai lures.

Fatigue Failu re

Fatigue failure is characterized by three stages

Crack Initiation

Crack Propagation

Final Fracture


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Jack hammer component,

shows no yielding before

fracture.

Crack initiation site

Fracture zone

Propagation zone, striation


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VW crank shaftfatigue failure due to cyclic bending and torsional stresses

Fracture areaCrack initiation site

Propagation

zone, striations


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928 Porsche timing pulley

Crack started at the fillet


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1.0-in. diameter steel pins from

agricultural equipment.

Material; AISI/SAE 4140 low

allow carbon steel

Fracture surface of a failed bolt. The

fracture surface exhibited beach marks,

which is characteristic of a fatigue failure.


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This long term fatigue crack in a high quality component took a

considerable time to nucleate from a machining mark between the spider

arms on this highly stressed surface. However once initiated propagation

was rapid and accelerating as shown in the increased spacing of the 'beach

marks' on the surface caused by the advancing fatigue crack.

bicycle crank spider arm


7/47Ken Youssefi MAE dept., SJSU 7

Gear tooth failure

Crank shaft



Hawaii, Aloha Flight 243, a Boeing 737, an upper part of the plane's cabin

area rips off in mid-flight. Metal fatigue was the cause of the failure.



Cup and Cone

DimplesDull Surface

Inclusion at the bottom of the dimple

Ductile

Fracture Surface Characteristics

Shiny

Grain Boundary cracking

Brittle Intergranular

Shiny

Cleavage fractures

Flat

Brittle Transgranular

Beachmarks

Striations (SEM)

Initiation sites

Propagation zone

Final fracture zone

Fatigue

Mode of fracture Typic al surface character ist ics



Fatigue FailureType of Fluctuat ing Stresses

a=

max

min

2

Alternating stress

Mean stress

m

=

max

min

2

+

min

= 0

a=

max/ 2

m=

a= max

max

= - min



Fatigue Failure, S-N Curve

Test specimen geometry for R.R. Moore

rotating beam machine. The surface is

polished in the axial direction. A constant

bending load is applied.

Motor

Load

Rotating beam machineapplies fully reverse bending stress

Typical testing apparatus, pure bending


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The standard machine operates at an

adjustable speed of 500 RPM to

10,000 RPM. At the nominal rate of

10,000 RPM, the R. R. Moore machine

completes 600,000 cycles per hour,

14,400,000 cycles per day.

Bending moment capacity

20 in-lb to 200 in-lb


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Fatigue Failure, S-N Curve

Finite life Infinite life

N 103

Se

= endurance limit of the specimenSe


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Relat ionsh ip Between Endurance L imit

and Ult imate Streng th

Steel

Se =

0.5Sut

100 ksi

700 MPa

Sut 200 ksi (1400 MPa)Sut>200 ksi

Sut>1400 MPa

Steel

0.4Sut

Se =

Sut


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Relat ionsh ip Between Endurance L imit and

Ult imate Strength

Aluminum alloys

Se =

0.4Sut

19 ksi

130 MPa

Sut


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Correction Factors for Specimens Endurance Limit

= endurance limit of the specimen (infinite life > 106)Se

For materials exhibiting a knee in the S-N curve at 106cycles

= endurance limit of the actual component (infinite life > 106)Se

N

S Se

10

6

103

= fatigue strength of the specimen (infinite life > 5x108)Sf

= fatigue strength of the actual component (infinite life > 5x108)Sf

For materials that do not exhibit a knee in the S-N curve, the infinite

life taken at 5x108cycles

N

S Sf

5x108103


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Se = Cload Csize Csurf Ctemp Crel(Se)

Load factor, Cload (page 326, Nortons 3rded.)

Pure bending Cload= 1

Pure axial Cload= 0.7

Combined loading Cload= 1

Pure torsion Cload= 1 if von Mises stress is used, use

0.577 if von Mises stress is NOT used.

Sf = Cload Csize Csurf Ctemp Crel(Sf)or


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Size factor, Csize (p. 327, Nortons 3rded.)

Larger parts fail at lower stresses than smaller parts. This is

mainly due to the higher probability of flaws being present in

larger components.

For rotating solid round cross section

d 0.3 in. (8 mm) Csize= 1

0.3 in. < d 10 in. Csize= .869(d)-0.097

8 mm < d 250 mm Csize= 1.189(d)-0.097

If the component is larger than 10 in., use Csize= .6


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For non rotating components, use the 95% area approach to calculate

the equivalent diameter. Then use this equivalent diameter in the

previous equations to calculate the size factor.

dequiv= (A95

0.0766)1/2

dd95= .95d

A95= (/4)[d2(.95d)2] = .0766 d2

dequiv

= .37d

Solid or hollow non-rotating parts

dequiv= .808 (bh)1/2

Rectangular parts


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Ibeams and Cchannels


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surface factor, Csurf (p. 328-9, Nortons 3rded.)

The rotating beam test specimen has a polished surface. Most

components do not have a polished surface. Scratches andimperfections on the surface act like a stress raisers and reduce

the fatigue life of a part. Use either the graph or the equation with

the table shown below.

Csurf = A (Sut)b


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Temperature factor, Ctemp (p.331, Nortons 3rded.)

High temperatures reduce the fatigue life of a component. For

accurate results, use an environmental chamber and obtain the

endurance limit experimentally at the desired temperature.

For operating temperature below 450 oC (840 oF) the

temperature factor should be taken as one.

Ctemp = 1 for T 450oC (840 oF)


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Reliability factor, Crel(p. 331, Nortons 3rded.)

The reliability correction factor accounts for the scatter and

uncertainty of material properties (endurance limit).


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Fatigue Stress Concentrat ion Facto r, Kf

Experimental data shows that the actual stress concentration factor is not as

high as indicated by the theoretical value, Kt. The stress concentration factor

seems to be sensitive to the notch radius and the ultimate strength of thematerial.

(p. 340, Nortons 3rded.)

Steel

Kf=1+(Kt1)qNotch sensitivity factor

Fatigue stress

concentration factor


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Fatigue Stress

Concentrat ion Facto r,

qfor Aluminum

(p. 341, Nortons 3rded.)


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Design p rocess Ful ly Reversed Loading fo r Inf in i te Life Determine the maximum alternating applied stress (

a)in terms of

the size and cross sectional profile

Select material Sy, Sut

Use the design equation to calculate the size

SeK

f

a=

n

Choose a safety factor n

Determine all modifying factors and calculate the endurance

limit of the component Se

Determine the fatigue stress concentration factor,Kf

Investigate different cross sections (profiles), optimize for size or weight

You may also assume a profile and size, calculate the alternating stress

and determine the safety factor. Iterate until you obtain the desired

safety factor


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Design fo r Fini te Life

Sn= a (N)b equation of the fatigue line

N

S

Se

106

103

A

B

N

S

Sf

5x108

103

A

B

Point ASn = .9Sut

N = 103Point A

Sn = .9Sut

N = 103

Point BSn = Sf

N = 5x108Point B

Sn = Se

N = 10

6


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Design fo r Fini te Life

Sn= a (N)b

log Sn= log a+ blog N

Apply boundary conditions for point A and B to

find the two constants a and b

log.9Sut= loga+ blog103

logSe= loga+ blog106

a=(.9Sut)

2

Se

b=

.9Sut

Se

1

3 log

SnKfa= n

Design equation

CalculateSnand replace Sein the design equation

Sn = Se (N

106) (

Se.9Sut

)log


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The Effect o f Mean Stress on Fat igue L ife

Mean stress exist if the

loading is of a repeating orfluctuating type.

Mean stress

Alternating

stress

m

a

Se

SySoderberg lineSut

Goodman line

Gerber curve

Mean stress is not zero


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The Effect of Mean Stress on Fat igue Life

Modi f ied Goodman Diagram

Mean stress

Alternating

stress

m

a

Sut

Goodman line

Sy

Sy

Se

Safe zoneC

Yield line

Th Eff t f M St F t i Li f


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- Syc



+m

a

Sut

Goodman line

Sy Yield line

Safe zone

- m

C

Sy

Se

Safe zone


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+m

a

Sut

Safe zone

- m

C

Sy

Safe zone

Se

- Syc

Finite life

Sn

1=

Sut

a m

+

Fatigue, m> 0Fatigue, m0

a=

Se

nf

a+

m=

Syc

ny

Yield

a+

m=

Sy

ny

Yield

nfSe

1=

Sut

a m

+Infinite life

A l i S C i f A l i


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Applying Stress Concentrat ion factor to A l ternat ing

and Mean Components of Stress

Determine the fatigue stress concentration factor, Kf, apply directly to

the alternating stress Kfa

If Kfmax< Sy then there is no yielding at the notch, use Kfm =Kf

and multiply the mean stress by Kfm Kfmm

If Kfmax> Sy then there is local yielding at the notch, material at the

notch is strain-hardened. The effect of stress concentration is reduced.

Calculate the stress concentration factor for the mean stress using

the following equation,

Kfm=

Sy Kfa

m

nfSe

1=

Sut

Kfa Kfmm+ Infinite life

Fatigue design equation


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Comb ined Loading

All four components of stress exist,

xa

alternating component of normal stress

xm mean component of normal stress

xya alternating component of shear stress

xym mean component of shear stress

Calculate the alternating and mean principal stresses,

1a, 2a=(xa /2) (xa/2)2

+(xya)2

1m, 2m=(xm /2) (xm/2)2+(xym)

2


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Comb ined Loading

Calculate the alternating and mean von Mises stresses,

a =(1a+2a- 1a2a)1/2

2 2

m =(1m+2m- 1m2m)1/2

2 2


nfSe

1

=Sut

a m

+ Infinite life


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Design Example

R1 R2

10,000lb.

6612

D= 1.5dd

r (fillet radius) = .1d

A rotating shaft is carrying 10,000 lb force

as shown. The shaft is made of steel with

Sut= 120 ksi and Sy= 90 ksi. The shaft

is rotating at 1150 rpm and has a

machine finish surface. Determine the

diameter, d, for 75 minutes life. Use

safety factor of 1.6 and 50% reliability.

Calculate the support forces, R1= 2500, R2= 7500lb.

A

The critical location is at the fillet, MA= 2500 x 12 = 30,000lb-in

a=Calculate the alternating stress,

Mc

I=

32M

d 3=

305577

d 3

m= 0

Determine the stress concentration factor

rd

= .1

D

d= 1.5

Kt= 1.7

Design E ample


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Design Example

Assume d= 1.0 in

Usingr = .1andSut= 120 ksi,

q(notch sensitivity) = .85Kf= 1+ (Kt1)q = 1 + .85(1.71) = 1.6

Calculate the endurance limit

Cload= 1 (pure bending)

Crel= 1 (50% rel.)

Ctemp= 1 (room temp)

Csurf = A (Sut)

b

= 2.7(120)

-.265

= .759

0.3 in. < d 10 in. Csize= .869(d)-0.097= .869(1)

-0.097= .869

Se = Cload Csize Csurf Ctemp Crel(Se) = (.759)(.869)(.5x120) = 39.57 ksi


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Design Example

Design life, N = 1150 x 75 = 86250cycles

Sn = Se (N

106)

(Se

.9Sut)log

Sn = 39.57(86250

106)

(

39.57

.9x120

)log

= 56.5 ksi

a=

305577

d 3= 305.577 ksi n=

SnKfa

=56.5

1.6x305.577= .116 < 1.6

So d= 1.0 in. is too small

Assume d= 2.5 in

All factors remain the same except the size factor and notch sensitivity.

Usingr = .25andSut= 120 ksi,

q(notch sensitivity) = .9Kf= 1+ (Kt1)q = 1 + .9(1.71) = 1.63

Csize= .869(d)-0.097= .869(2.5)

-0.097= .795 Se =36.2 ksi

D i E l


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Design Example

a=

305577

(2.5)3= 19.55 ksi

n=Sn

Kfa=

53.35

1.63x19.55= 1.67 1.6

d=2.5in.

Check yielding

n=Sy

Kfmax=

90

1.63x19.55= 2.8 > 1.6 okay

Se =36.2 ksi Sn = 36.20(86250

106) (

36.2.9x120

)log

= 53.35 ksi


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Design ExampleObservat ion s

n=Sn

Kfa=

56.5

1.6x305.577

= .116 < 1.6

So d= 1.0 in. is too small

Calculate an approximate diameter

n= SnKfa

= 56.51.6x305.577/d 3

= 1.6 d = 2.4in. So, your next guessshould be between

2.25 to 2.5

Mmax (under the load) = 7500 x 6 = 45,000lb-in

Check the location of maximum moment for possible failure

R1 R2= 7500

6612

D= 1.5dd

r (fillet radius) = .1d

A

MA(at the fillet) = 2500 x 12 = 30,000lb-in

But, applying the fatigue stress conc. Factor of 1.63,

KfM

A

= 1.63x30,000 = 48,900 > 45,000


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Example

A section of a component is shown.

The material is steel with Sut= 620 MPa

and a fully corrected endurance limit of

Se= 180 MPa. The applied axial loadvaries from 2,000 to 10,000 N. Use

modified Goodman diagram and find

the safety factor at the fillet A, groove B

and hole C. Which location is likely to

fail first? UseKfm= 1

Pa= (PmaxPmin) / 2= 4000 N Pm= (Pmax+ Pmin) / 2= 6000 N

Fil let

rd

= .16

D

d= 1.4

425

=

35

25=

Kt= 1.76


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Example

Kf= 1+ (Kt1)q = 1 + .85(1.761) = 1.65

Calculate the alternating and the

mean stresses,

a=

Pa

A

=4000

25x5= 52.8 MPaKf 1.65

m=

Pm

A=

6000

25x5= 48 MPa

nSe

1=

Sut

a m

+ Infinite life


n= 2.7n180

1=

620

52.8 48+

Usingr = 4andSut= 620 MPa,

q(notch sensitivity) = .85


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ExampleHole

d

w= .143

5

35= Kt= 2.6

Usingr = 2.5andSut= 620 MPa,


Kf= 1+ (Kt1)q = 1 + .82(2.61) = 2.3

Calculate the alternating and themean stresses,

a=

Pa

A=

4000

(35-5)5= 61.33 MPaKf 2.3

m=

Pm

A =

6000

30x5 = 40 MPa

n= 2.5n180

1=

620

61.33 40+

E l


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ExampleGroove

r

d= .103

D

d= 1.2

3

29=

35

29=

Kt

= 2.33

Usingr = 3andSut= 620 MPa,


Kf= 1+ (Kt1)q = 1 + .83(2.331) = 2.1

(35-6)5

Calculate the alternating and the

mean stresses,

a=

Pa

A=

4000= 58.0 MPaKf 2.1

m=

Pm

A=

6000

29x5= 41.4 MPa

n= 2.57n180

1=

620

58.0 41.4+

The part is likely to fail at the hole, has the lowest safety factor

E l


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Example

Fa=(FmaxFmin) / 2=7.5 lb. Fm=(Fmax+Fmin) / 2=22.5 lb.

The figure shows a formed round wire cantilever

spring subjected to a varying force F. The wire is

made of steel with Sut= 150ksi. The mounting

detail is such that the stress concentration couldbe neglected. A visual inspection of the spring

indicates that the surface finish corresponds

closely to a hot-rolled finish. For a reliability of

99%, what number of load applications is likely to

cause failure.

Ma= 7.5 x 16 = 120 in - lb Mm= 22.5 x 16 = 360 in - lb

a=

Mc

I=

32Ma

d3 =

32(120)

(.375)3 = 23178.6 psi

m=

Mc

I=

32Mm

d 3=

32(360)

(.375)3= 69536 psi

E l


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Example

Cload= 1 (pure bending)

Ctemp= 1 (room temp)

Calculate the endurance limit

Crel= .814 (99% reliability)

Csurf = A (Sut)b= 14.4(150)

-.718= .394

A95= .010462d

2

(non-rotating round section)

dequiv= A95/ .0766 = .37d = .37 x.375= .14dequiv= .14 < .3 Csize= 1.0

Se = Cload Csize Csurf Ctemp Crel(Se) = (.394)(.814)(.5x150) = 24.077 ksi

nSe

1=

Sut

a m

+ n24077

1=

150000

23178.6 69536+ n = .7< 1

Finite life

Sn1=

Sut

a m

+

Find Sn, strength for finite number of cycle

Sn1=

150000

23178.6 69536+ Sn= 43207 psi

E l


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Example

Sn = Se (N

106)

(Se

.9Sut)log

43207 = 24077 (N

106)

(24.077

.9x150)log

N = 96,000 cycles

Fatigue Failure

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