Download ppt - Fatigue Failure FOEMEF 17-06-2011

It 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

Fatigue failure is characterized by three stages

Crack Initiation Crack Propagation Final Fracture

Prepared By Anchit Kaneria Machine Design

Jack hammer component, shows no yielding before fracture.

Crack initiation site

Fracture zonePropagation zone, striation


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

Fracture areaCrack initiation site

Propagation zone, striations


928 Porsche timing pulley

Crack started at the filletPrepared By Anchit Kaneria Machine Design

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.


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


Gear tooth failure

Crank shaft


Machine Design

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.

Prepared By Anchit Kaneria

Cup and ConeDimplesDull SurfaceInclusion at the bottom of the dimple

Ductile

Fracture Surface Characteristics

ShinyGrain Boundary cracking

Brittle Intergranular

ShinyCleavage fracturesFlat

Brittle Transgranular

BeachmarksStriations (SEM)Initiation sitesPropagation zoneFinal fracture zone

Fatigue

Mode of fracture Typical surface characteristics


Fatigue Failure – Type of Fluctuating Stresses

a =max min

2

Alternating stress

Mean stress

m =max min

2+

min = 0

a = max / 2m =

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 machine – applies fully reverse bending stress

Typical testing apparatus, pure bending













Finite life Infinite life

N < 103 N > 103

S′e

= endurance limit of the specimenSe′


Relationship Between Endurance Limit and Ultimate Strength

Steel

Se =′0.5Sut

100 ksi

700 MPa

Sut ≤ 200 ksi (1400 MPa)Sut > 200 ksiSut > 1400 MPa

Steel

0.4Sut

Se =′

Sut < 60 ksi (400 MPa)

Sut ≥ 60 ksi24 ksi

160 MPa Sut < 400 MPa

Cast iron Cast iron


Relationship Between Endurance Limit and Ultimate Strength

Aluminum alloys

Se =′0.4Sut

19 ksi

130 MPa

Sut < 48 ksi (330 MPa)Sut ≥ 48 ksiSut ≥ 330 MPa

Aluminum

For N = 5x108 cycle

Copper alloys

Se =′0.4Sut

14 ksi

100 MPa

Sut < 40 ksi (280 MPa)Sut ≥ 40 ksiSut ≥ 280 MPa

Copper alloys

For N = 5x108 cycle


Correction Factors for Specimen’s Endurance Limit

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

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

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

N

S Se

106103

= 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 5x108 cycles

N

S Sf

5x108103



Se = Cload Csize Csurf Ctemp Crel (Se)′

• Load factor, Cload (page 326, Norton’s 3rd ed.)

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



• Size factor, Csize (p. 327, Norton’s 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) 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


Correction Factors for Specimen’s 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.

dequiv = (A95

0.0766)1/2

dd95 = .95d

A95 = (π/4)[d2 – (.95d)2] = .0766 d2

dequiv = .37d

Solid or hollow non-rotating parts

dequiv = .808 (bh)1/2

Rectangular parts



I beams and C channels


Correction Factors for Specimen’s Endurance Limit• surface factor, Csurf (p. 328-9, Norton’s 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



• Temperature factor, Ctemp (p.331, Norton’s 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)



• Reliability factor, Crel (p. 331, Norton’s 3rd ed.)

The reliability correction factor accounts for the scatter and uncertainty of material properties (endurance limit).



Fatigue Stress Concentration Factor, 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 the material.

(p. 340, Norton’s 3rd ed.)

Steel

Kf = 1 + (Kt – 1)qNotch sensitivity factor

Fatigue stress concentration factor

Machine Design

Fatigue Stress Concentration Factor,

Kf for Aluminum

(p. 341, Norton’s 3rd ed.)

Prepared By Anchit Kaneria

Design process – Fully Reversed Loading for Infinite 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

SeKf 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


Design for Finite Life

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

N

S

Se

106103

A

B

N

SSf

5x108103

A

B

Point ASn = .9Sut

N = 103Point A

Sn = .9Sut

N = 103

Point BSn = Sf

N = 5x108Point B

Sn = Se

N = 106


Design for Finite LifeSn = a (N)b log 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 =(.9Sut)

2

Se

b =.9Sut

Se

13

log

SnKf a = n Design equation

Calculate Sn and replace Se in the design equation

Sn = Se ( N

106 )⅓ (

Se

.9Sut)log


The Effect of Mean Stress on Fatigue Life

Mean stress exist if the loading is of a repeating or fluctuating type.

Mean stress

Alternating stress

m

a

Se

SySoderberg line Sut

Goodman line

Gerber curve

Mean stress is not zero


The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram

Mean stress

Alternating stress

m

a

Sut

Goodman line

Sy

Sy

Se

Safe zoneC

Yield line


- Syc


+m

a

Sut

Goodman line

Sy Yield line

Safe zone

- m

C

Sy

Se

Safe zone



+m

a

Sut

Safe zone

- m

C

Sy

Safe zone

Se

- Syc

Finite lifeSn1=

Sut

a m+

Fatigue, m > 0Fatigue, m ≤ 0

a =Se

nf

a + m =Syc

ny

Yield

a + m =Sy

ny

Yield

nfSe

1=

Sut

a m+ Infinite life


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 a

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

and multiply the mean stress by Kfm → Kfm m

• If Kf max > 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 Kf a

m

nfSe

1=Sut

Kf a Kfmm+ Infinite life

Fatigue design equation


Combined LoadingAll 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


Combined 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


Design Example

R1 R2

10,000 lb.6˝6˝12˝

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 = 7500 lb.

A

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

a = Calculate the alternating stress, McI =

32Mπd 3 =

305577

d 3m = 0

Determine the stress concentration factor

rd = .1

Dd

= 1.5Kt = 1.7


Design ExampleAssume d = 1.0 in

Using r = .1 and Sut = 120 ksi, q (notch sensitivity) = .85

Kf = 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)-.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′


Design ExampleDesign life, N = 1150 x 75 = 86250 cycles

Sn = Se ( N

106 )⅓ (

Se

.9Sut)log

Sn = 39.57 ( 86250

106 )⅓ (

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.

Using r = .25 and Sut = 120 ksi, q (notch sensitivity) = .9

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

Csize = .869(d)-0.097 = .869(2.5)-0.097 = .795 Se = 36.2 ksi→


Design Example

a = 305577

(2.5)3= 19.55 ksi

n = Sn

Kfa=

53.351.63x19.55

= 1.67 ≈ 1.6

d = 2.5 in.

Check yielding

n = Sy

Kfmax=

901.63x19.55

= 2.8 > 1.6 okay

Se = 36.2 ksi → Sn = 36.20 ( 86250106 )

⅓ (36.2

.9x120 )log= 53.35 ksi


Design Example – Observations

n = Sn

Kfa=

56.51.6x305.577

= .116 < 1.6

So d = 1.0 in. is too small

Calculate an approximate diameter

n = Sn

Kfa=

56.51.6x305.577/d 3 = 1.6 → d = 2.4 in. So, your next guess

should be between 2.25 to 2.5

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

Check the location of maximum moment for possible failure

R1 R2 = 7500

6˝6˝12˝D = 1.5dd

r (fillet radius) = .1d

A

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 > 45,000


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

rd

= .16

Dd

= 1.4

425

=

3525

=Kt = 1.76


Example

Kf = 1 + (Kt – 1)q = 1 + .85(1.76 – 1) = 1.65

Calculate the alternating and the mean stresses,

a = Pa

A=

400025x5

= 52.8 MPaKf 1.65

m = Pm

A=

600025x5

= 48 MPa

nSe

1=Sut

a m+ Infinite life


n = 2.7n180

1=

62052.8 48+ →

Using r = 4 and Sut = 620 MPa, q (notch sensitivity) = .85


ExampleHole

dw

= .143535

= → Kt = 2.6

Using r = 2.5 and Sut = 620 MPa, q (notch sensitivity) = .82

Kf = 1 + (Kt – 1)q = 1 + .82(2.6 – 1) = 2.3


a = Pa

A=

4000(35-5)5

= 61.33 MPaKf 2.3

m = Pm

A=

600030x5

= 40 MPa

n = 2.5n180

1=

62061.33 40+ →


Machine Design

ExampleGroove

rd

= .103

Dd

= 1.2

329

=

3529

=→ Kt = 2.33

Using r = 3 and Sut = 620 MPa, q (notch sensitivity) = .83

Kf = 1 + (Kt – 1)q = 1 + .83(2.33 – 1) = 2.1

(35-6)5


a = Pa

A=

4000 = 58.0 MPaKf 2.1

m = Pm

A=

600029x5

= 41.4 MPa

n = 2.57n180

1=

62058.0 41.4+ →

The part is likely to fail at the hole, has the lowest safety factorPrepared By Anchit Kaneria

Example

Fa = (Fmax – Fmin) / 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 = 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.

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

a = McI =

32Ma

πd 3 =32(120)

π (.375)3 = 23178.6 psi

m = McI =

32Mm

πd 3 =32(360)

π (.375)3= 69536 psi


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 = .010462 d 2 (non-rotating round section)

dequiv = √ A95 / .0766 = .37d = .37 x.375 = .14

dequiv = .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=

15000023178.6 69536

+ → n = .7 < 1Finite life

Sn1=Sut

a m+

Find Sn, strength for finite number of cycle

Sn1=

15000023178.6 69536

+→ → Sn = 43207 psi


Example

Sn = Se ( N

106 )⅓ (

Se

.9Sut)log

43207 = 24077 ( N

106 )⅓ ( 24.077

.9x150 )log→

N = 96,000 cycles
