Fatigue FailureIt has been recognized that a metal subjected to 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 loading are called fatigue failures.Fatigue failure is characterized by three stages

Crack Initiation Crack Propagation Final FractureMAE dept., SJSU

Ken Youssefi

1

Jack hammer component, shows no yielding before fracture.

Crack initiation site

Fracture zone Propagation zone, striationKen Youssefi MAE dept., SJSU

2

VW crank shaft fatigue failure due to cyclic bending and torsional stresses

Propagation zone, striations

Crack initiation site

Fracture area

Ken Youssefi

MAE dept., SJSU

3

928 Porsche timing pulley

Ken Youssefi

MAE dept., SJSU

Crack started at the fillet

4

Fracture surface of a failed bolt. The fracture surface exhibited beach marks, which is characteristic of a fatigue failure.

1.0-in. diameter steel pins from agricultural equipment. Material; AISI/SAE 4140 low allow carbon steel

Ken Youssefi

MAE dept., SJSU

5

bicycle crank spider arm

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.Ken Youssefi MAE dept., SJSU

6

Crank shaft

Gear tooth failure

Ken Youssefi

MAE dept., SJSU

7

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.

Ken Youssefi

MAE dept., SJSU

8

Fracture Surface Characteristics Mode of fracture Ductile Typical surface characteristics Cup and Cone Dimples Dull Surface Inclusion at the bottom of the dimple Shiny Grain Boundary cracking Shiny Cleavage fractures Flat Beachmarks Striations (SEM) Initiation sites Propagation zone Final fracture zoneMAE dept., SJSU

Brittle Intergranular

Brittle Transgranular

Fatigue

Ken Youssefi

9

Fatigue Failure Type of Fluctuating Stresses Wa = Wmax Wmax = - Wmin

Alternating stress

Wa = Wmin = 0

Wmax Wmin2

Mean stress

Wa = Wm = Wmax / 2 Wm=Ken Youssefi MAE dept., SJSU

Wmax + Wmin210

Fatigue Failure, S-N CurveTest specimen geometry for R.R. Moore rotating beam machine. The surface is polished in the axial direction. A constant bending load is applied.

Typical testing apparatus, pure bendingMotor

Load

Rotating beam machine applies fully reverse bending stressKen Youssefi MAE dept., SJSU

11

Fatigue Failure, S-N CurveN < 103 N > 103

Finite life

Infinite life

Se

e

= endurance limit of the specimen

Ken Youssefi

MAE dept., SJSU

12

Relationship Between Endurance Limit and Ultimate StrengthSteel Steel 0.5Sute=

Sut 200 ksi (1400 MPa) Sut > 200 ksi

100 ksi

700 MPa Sut > 1400 MPa

Cast iron 0.4Sut

Cast iron

Sut < 60 ksi (400 MPa) Sut 60 ksi

Se =

24 ksi

160 MPa Sut < 400 MPaKen Youssefi MAE dept., SJSU

13

Relationship Between Endurance Limit and Ultimate StrengthAluminumAluminum alloys

0.4Sut

Sut < 48 ksi (330 MPa) Sut 48 ksi

Se =

19 ksi

130 MPa Sut 330 MPa For N = 5x108 cycle

Copper alloysCopper alloys

0.4Sut

Sut < 40 ksi (280 MPa) Sut 40 ksi

Se =

14 ksi

100 MPa Sut 280 MPa For N = 5x108 cycleKen Youssefi MAE dept., SJSU

14

Correction Factors for Specimens Endurance LimitFor materials exhibiting a knee in the S-N curve at 106 cycles

Se = endurance limit of the specimen (infinite life > 106) Se = endurance limit of the actual component (infinite life > 106)S103

Se106

N

For materials that do not exhibit a knee in the S-N curve, the infinite life taken at 5x108 cycles

Sf = fatigue strength of the specimen (infinite life > 5x108) Sf = fatigue strength of the actual component (infinite life > 5x108)S103

Sf5x108

Ken Youssefi

MAE dept., SJSU

N

15

Correction Factors for Specimens Endurance Limit Se = Cload Csize Csurf Ctemp Crel (Se)or

Sf = Cload Csize Csurf Ctemp Crel (Sf)(page 326, Nortons 3rd ed.)

Load factor, Cload

Pure bending Pure axial Pure torsion Combined loading

Cload = 1 Cload = 0.7 Cload = 1 if von Mises stress is used, use 0.577 if von Mises stress is NOT used. Cload = 1

Ken Youssefi

MAE dept., SJSU

16

Correction Factors for Specimens Endurance Limit Size factor, Csize(p. 327, Nortons 3rd ed.)

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)0.3 in. < d 10 in. 8 mm < d 250 mm

Csize = 1 Csize = .869(d)-0.097 Csize = 1.189(d)-0.097

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

Ken Youssefi

MAE dept., SJSU

17

Correction Factors for Specimens Endurance LimitFor 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. A95 = (/4)[d2 (.95d)2] = .0766 d2 d95 = .95d d

dequiv = (

A95 0.0766

)1/2

Solid or hollow non-rotating parts

Rectangular parts

dequiv = .37dKen Youssefi MAE dept., SJSU

dequiv = .808 (bh)1/2

18

Correction Factors for Specimens Endurance LimitI beams and C channels

Ken Youssefi

MAE dept., SJSU

19

Correction Factors for Specimens Endurance Limit surface factor, Csurf(p. 328-9, Nortons 3rd ed.)

The rotating beam test specimen has a polished surface. Most components do not have a polished surface. Scratches and imperfections 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

Ken Youssefi

MAE dept., SJSU

20

Correction Factors for Specimens Endurance Limit Temperature factor, Ctemp(p.331, Nortons 3rd ed.)

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 450 oC (840 oF)

Ken Youssefi

MAE dept., SJSU

21

Correction Factors for Specimens Endurance Limit Reliability factor, Crel (p. 331, Nortons 3rd ed.)The reliability correction factor accounts for the scatter and uncertainty of material properties (endurance limit).

Ken Youssefi

MAE dept., SJSU

22

Fatigue Stress Concentration Factor, KfExperimental 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 the material. Notch sensitivity factor Fatigue stress Kf = 1 + (Kt 1)q concentration factor rd(p. 340, Nortons 3 ed.)

Steel

Ken Youssefi

MAE dept., SJSU

23

Fatigue Stress Concentration Factor, Kf for Aluminum(p. 341, Nortons 3rd ed.)

Ken Youssefi

MAE dept., SJSU

24

Design process Fully Reversed Loading for Infinite Life Determine the maximum alternating applied stress (Wa ) in terms of the size and cross sectional profile Select material Sy, Sut n

Choose a safety factor

Determine all modifying factors and calculate the endurance limit of the component Se Determine the fatigue stress concentration factor, Kf Use the design equation to calculate the size

Se Kf Wa = n 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 factorMAE dept., SJSU

Ken Youssefi

25

Design for Finite Life

Sn = a (N)b equation of the fatigue lineA S103

A B Se106

S Sf N103

B N

5x108

Point A

Sn = .9SutN = 103

Point A

Sn = .9SutN = 103

Point BKen Youssefi

Sn = SeN = 106MAE dept., SJSU

Point B

Sn = SfN = 5x10826

Design for Finite LifeSn = a (N)blog Sn = log a + b log N Apply boundary conditions for point A and B to find the two constants a and b

log .9Sut = log a + b log 103 log Se = log a + b log 106

a= b= log (

(.9Sut) Se1 3

2

log

.9Sut Se

Sn = Se ( 106 )Calculate Sn

N

Se ) .9Sut

and replace Se in the design equationSn Kf Wa = nDesign equation27

Ken Youssefi

MAE dept., SJSU

The Effect of Mean Stress on Fatigue LifeMean stress exist if the loading is of a repeating or fluctuating type.

Wa

Mean stress is not zero

Gerber curve Alternating stress Se Goodman line

Soderberg lineKen Youssefi MAE dept., SJSU

Sy Mean stress

Sut

Wm28

The Effect of Mean Stress on Fatigue Life Modified Goodman DiagramWaSy Yield line

Alternating stress

Se Goodman line Safe zoneC

Sy Mean stress

Sut

Wm

Ken Youssefi

MAE dept., SJSU

29

The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram

WaSy Yield line

Se Goodman line Safe zone - Wm - Syc Safe zoneC

Sy

Sut

+Wm

Ken Youssefi

MAE dept., SJSU

30

The Effect of Mean Stress on Fatigue Life Modified Goodman DiagramFatigue,

Wm 0

Fatigue,

Wm > 0 Sut= =1 1 Infinite life

WaSe

WaSe

+ +

Wm WmSut

nfFinite life Yield

Wa = n fYield Se Syc Safe zone - Wm - Syc

WaSn

Wa + Wm = n y

Safe zone

C

Wa + Wm = n ySy Sut +Wm

Sy

Ken Youssefi

MAE dept., SJSU

31

Applying Stress Concentration factor to Alternating and Mean Components of Stress Determine the fatigue stress concentration factor, Kf, apply directly to the alternating stress Kf Wa

If Kf Wmax < Sy then there is no yielding at the notch, use Kfm = Kf and multiply the mean stress by Kfm Kfm Wm

If Kf Wmax > Sy then there is local yielding at the notch, material at the Calculate the stress concentration factor for the mean stress using the following equation, Kfm =

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

Sy

Kf Wa

Wm1 Infinite life32

Fatigue design equation Kf Wa KfmWm = +

Se

Sut

nf

Ken Youssefi

MAE dept., SJSU

Combined LoadingAll four components of stress exist,

Wxa Wxm

alternating component of normal stressmean component of normal stress alternating component of shear stress mean component of shear stress

Xxya Xxym

Calculate the alternating and mean principal stresses,

W1a, W2a = (Wxa /2) W1m, W2m = (Wxm /2)Ken Youssefi

(Wxa /2)2 + (Xxya)2 (Wxm /2)2 + (Xxym)233

MAE dept., SJSU

Combined LoadingCalculate the alternating and mean von Mises stresses,

Wa Wm

= =

(W1a + W2a - W1aW2a)1/2 (W1m + W2m - W1mW2m)1/22 2

2

2

Fatigue design equation

WaSe

+

WmSut

=

1

nf

Infinite life

Ken Youssefi

MAE dept., SJSU

34

Design ExampleA rotating shaft is carrying 10,000 lb force as shown. The shaft is made of steel with 12 d A

10,000 lb. 6 6 D = 1.5d R2

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 r (fillet radius) = .1d

R1 = 2500, R2 = 7500 lb. MA = 2500 x 12 = 30,000 lb-in

The critical location is at the fillet, Calculate the alternating stress,

Wa =

Mc

I

=

32M

d

3

=

305577

d

3

Wm = 0

Determine the stress concentration factor d D = 1.5 dKen Youssefi

r = .1

Kt = 1.735

MAE dept., SJSU

Design ExampleAssume d = 1.0 in

Using r = .1 and Sut = 120 ksi, q (notch sensitivity) = .85Kf = 1 + (Kt 1)q = 1 + .85(1.7 1) = 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)0.3 in. < d 10 in.-.265

= .759

Csize = .869(d)-0.097 = .869(1)-0.097 = .869ksi36

Se = Cload Csize Csurf Ctemp Crel (Se) = (.759)(.869)(.5x120) = 39.57Ken Youssefi MAE dept., SJSU

Design ExampleDesign life, N = 1150 x 75 = 86250 cycles Se log ( .9S ) N 86250 ut Sn = 39.57 ( 6 Sn = Se ( 6 )10 10

log (

)56.5

39.57 ) .9x120

= 56.5 ksi

Wa =

305577

d

3

= 305.577 ksi

n=

Sn KfWa

=

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.

Using r = .25 and Sut = 120 ksi, q (notch sensitivity) = .9 Csize = .869(d)-0.097 = .869(2.5)Ken Youssefi

Kf = 1 + (Kt 1)q = 1 + .9(1.7 1) = 1.63

-0.097

= .795

Se = 36.2 ksi37

MAE dept., SJSU

Design ExampleSe = 36.2 ksi

Sn = 36.20 (= 19.55 ksi

86250106

)

36.2 log ( .9x120 )

= 53.35 ksi

Wa =

305577

(2.5)

3

n=

SnKfWa

=

53.35 1.63x19.55

= 1.67 1.6

d = 2.5 in.

Check yielding

n=

Sy 90 = 2.8 > 1.6 okay = KfWmax 1.63x19.55MAE dept., SJSU

Ken Youssefi

38

Design Example Observationsn= Sn KfWa= 56.5 1.6x305.577 = .116 < 1.6 d R1 r (fillet radius) = .1d A 12 6 6 D = 1.5d R2 = 7500

So d = 1.0 in. is too small Calculate an approximate diameter

n=

Sn KfWa

=

56.5 1.6x305.577/d3

= 1.6

d = 2.4 in.

So, your next guess should be between 2.25 to 2.5

Check the location of maximum moment for possible failure Mmax (under the load) = 7500 x 6 = 45,000 lb-in MA (at the fillet) = 2500 x 12 = 30,000 lb-in But, applying the fatigue stress conc. Factor of 1.63, Kf MA = 1.63x30,000 = 48,900 > 45,000Ken Youssefi MAE dept., SJSU

39

ExampleA 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 load varies 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? Use Kfm = 1 Pa = (Pmax Pmin) / 2 = 4000 N Pm = (Pmax + Pmin) / 2 = 6000 N

Fillet

r

d D 35 = = 1.4 d 25

=

4 = .16 25

Kt = 1.76

Ken Youssefi

MAE dept., SJSU

40

ExampleUsing r = 4 and Sut = 620 MPa, q (notch sensitivity) = .85Kf = 1 + (Kt 1)q = 1 + .85(1.76 1) = 1.65 Calculate the alternating and the mean stresses, Pa 4000 = 52.8 MPa = 1.65 Wa = Kf A 25x5 Pm 6000 = 48 MPa = Wm = A 25x5 Fatigue design equation

WaSe+

WmSut

=

1

n

Infinite life

52.8 48 = + 180 620

1

n

n = 2.7

Ken Youssefi

MAE dept., SJSU

41

Hole d 5 = .143 = w 35 Kt = 2.6

Example

Using r = 2.5 and Sut = 620 MPa, q (notch sensitivity) = .82Kf = 1 + (Kt 1)q = 1 + .82(2.6 1) = 2.3 Calculate the alternating and the mean stresses, Pa 4000 = 61.33 MPa = 2.3 Kf Wa = A (35-5)5 Pm 6000 = 40 MPa = Wm = A 30x5 61.33 40 = + 180 620 1

n

n = 2.542

Ken Youssefi

MAE dept., SJSU

ExampleGroove

r

d Kt = 2.33 D 35 = 1.2 = d 29 Using r = 3 and Sut = 620 MPa, q (notch sensitivity) = .83 Kf = 1 + (Kt 1)q = 1 + .83(2.33 1) = 2.1 Calculate the alternating and the mean stresses, Pa 4000 = 58.0 MPa = 2.1 Wa = Kf A (35-6)5

=

3 = .103 29

Wm =

Pm A

=

6000

29x51

= 41.4 MPa

58.0 41.4 = + 180 620Ken Youssefi

n

n = 2.57MAE dept., SJSU

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

43

ExampleThe figure shows a formed round wire cantilever spring subjected to a varying force F. The wire is made of steel with Sut = 150 ksi. The mounting detail is such that the stress concentration could be 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. Fa = (Fmax Fmin) / 2 = 7.5 lb. Fm = (Fmax + Fmin) / 2 = 22.5 lb. Ma = 7.5 x 16 = 120 in - lb Mm = 22.5 x 16 = 360 in - lb = 23178.6 psi = 69536 psi

Wa = Wm =

Mc

IMc

=

32Ma

d d

3

= =

32(120)

(.375)

3

I

=

32Mm3

32(360)

(.375)

3

Ken Youssefi

MAE dept., SJSU

44

Calculate the endurance limit

ExampleCsurf = A (Sut)b = 14.4(150)2

Cload = 1 (pure bending) Ctemp= 1 (room temp) Crel= .814 (99% reliability)

-.718

= .394

A95 = .010462 d (non-rotating round section)

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

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

WmSut

=

1

n

23178.6 69536 = + 24077 150000

1

n

n = .7 < 1Finite life

Find Sn, strength for finite number of cycle

WaSnKen Youssefi

Wm+

Sut

=1

23178.6

Sn

+

69536 =1 150000

Sn = 43207 psi

MAE dept., SJSU

45

ExampleSe log ( .9S ) ut

Sn = Se ( 106 )

N

43207 = 24077 (

N 106

)

log ( .9x150 )

24.077

N = 96,000 cycles

Ken Youssefi

MAE dept., SJSU

46