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Fatigue failure is characterized by three stages. Crack Initiation. Crack Propagation. Final Fracture. Fatigue Failure.

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fatigue failure

Fatigue failure is characterized by three stages

  • Crack Initiation
  • Crack Propagation
  • Final Fracture
Fatigue Failure

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.

MAE dept., SJSU

slide2

Crack initiation site

Fracture zone

Propagation zone, striation

Jack hammer component, shows no yielding before fracture.

MAE dept., SJSU

slide3

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

Propagation zone, striations

Crack initiation site

Fracture area

MAE dept., SJSU

slide4

928 Porsche timing pulley

Crack started at the fillet

MAE dept., SJSU

slide5

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

MAE dept., SJSU

slide6

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.

MAE dept., SJSU

slide7

Crank shaft

Gear tooth failure

MAE dept., SJSU

slide8

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.

MAE dept., SJSU

slide9

Fracture Surface Characteristics

Mode of fracture

Typical surface characteristics

Ductile

Cup and ConeDimplesDull SurfaceInclusion at the bottom of the dimple

Brittle Intergranular

ShinyGrain Boundary cracking

Brittle Transgranular

ShinyCleavage fracturesFlat

Fatigue

BeachmarksStriations (SEM)Initiation sitesPropagation zoneFinal fracture zone

MAE dept., SJSU

fatigue failure type of fluctuating stresses

max

a =

- min

max =

Alternating stress

min

max

a =

2

min = 0

Mean stress

m =

a =

max / 2

min

max

+

m=

2

Fatigue Failure – Type of Fluctuating Stresses

MAE dept., SJSU

fatigue failure s n curve

Typical testing apparatus, pure bending

Motor

Load

Rotating beam machine – applies fully reverse bending stress

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.

MAE dept., SJSU

slide12

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

MAE dept., SJSU

fatigue failure s n curve1

Se

Infinite life

Finite life

S′e

= endurance limit of the specimen

Fatigue Failure, S-N Curve

N > 103

N < 103

MAE dept., SJSU

relationship between endurance limit and ultimate strength

Steel

Se =

Se =

Sut ≤ 200 ksi (1400 MPa)

0.5Sut

Sut> 200 ksi

100 ksi

Sut> 1400 MPa

700 MPa

Cast iron

Cast iron

0.4Sut

Sut< 60 ksi (400 MPa)

Sut ≥60 ksi

24 ksi

Sut< 400 MPa

160 MPa

Relationship Between Endurance Limit and Ultimate Strength

Steel

MAE dept., SJSU

relationship between endurance limit and ultimate strength1

Se =

Se =

Sut < 40 ksi (280 MPa)

Sut < 48 ksi (330 MPa)

0.4Sut

0.4Sut

Sut ≥ 40 ksi

Sut ≥ 48 ksi

14 ksi

19 ksi

Sut ≥ 330 MPa

Sut ≥ 280 MPa

130 MPa

100 MPa

Copper alloys

Copper alloys

For N = 5x108 cycle

Relationship Between Endurance Limit and Ultimate Strength

Aluminum

Aluminum alloys

For N = 5x108 cycle

MAE dept., SJSU

correction factors for specimen s endurance limit

Se

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

Se

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

Sf

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

S

Se

Sf

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

N

106

103

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

S

Sf

N

5x108

103

Correction Factors for Specimen’s Endurance Limit

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

MAE dept., SJSU

correction factors for specimen s endurance limit1

Se = Cload Csize Csurf Ctemp Crel (Se)

Sf = Cload Csize Csurf Ctemp Crel (Sf)

Pure bending

Cload = 1

Pure axial

Cload = 0.7

Pure torsion

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

Combined loading

Cload = 1

Correction Factors for Specimen’s Endurance Limit

or

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

MAE dept., SJSU

correction factors for specimen s endurance limit2

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

Correction Factors for Specimen’s Endurance Limit
  • 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.

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

MAE dept., SJSU

correction factors for specimen s endurance limit3

A95

d

d95 = .95d

dequiv = (

)1/2

0.0766

Rectangular parts

Solid or hollow non-rotating parts

dequiv = .37d

dequiv = .808 (bh)1/2

Correction Factors for Specimen’s Endurance Limit

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.

A95 = (π/4)[d2 – (.95d)2] = .0766 d2

MAE dept., SJSU

correction factors for specimen s endurance limit4
Correction Factors for Specimen’s Endurance Limit

I beams and C channels

MAE dept., SJSU

correction factors for specimen s endurance limit5

Csurf = A (Sut)b

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.

MAE dept., SJSU

correction factors for specimen s endurance limit6
Correction Factors for Specimen’s Endurance Limit
  • 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)

MAE dept., SJSU

correction factors for specimen s endurance limit7
Correction Factors for Specimen’s Endurance Limit
  • Reliability factor, Crel(p. 331, Norton’s 3rd ed.)

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

MAE dept., SJSU

fatigue stress concentration factor k f

Notch sensitivity factor

Fatigue stress concentration factor

Kf=1+ (Kt–1)q

(p. 340, Norton’s 3rd ed.)

Steel

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.

MAE dept., SJSU

fatigue stress concentration factor q for aluminum
Fatigue Stress Concentration Factor, q for Aluminum

(p. 341, Norton’s 3rd ed.)

MAE dept., SJSU

design process fully reversed loading for infinite life

Use the design equation to calculate the size

Se

Kf a =

n

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
  • 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

MAE dept., SJSU

design for finite life

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

A

A

S

S

B

B

Sf

Se

Sn = .9Sut

N

N

106

5x108

103

103

Point A

N = 103

Sn = .9Sut

Sn = Se

Sn = Sf

Point A

Point B

Point B

N = 103

N = 106

N = 5x108

Design for Finite Life

MAE dept., SJSU

design for finite life1

log .9Sut = loga + blog103

logSe = loga + blog106

(.9Sut)2

Se

.9Sut

(

)

a

1

=

b

log

.9Sut

=

Se

log

Se

3

N

(

)

Se

Sn

=

106

Sn

Kf a =

n

CalculateSnand replace Sein the design equation

Design equation

Design for Finite Life

Sn = 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”

MAE dept., SJSU

the effect of mean stress on fatigue life

a

Gerber curve

Se

Alternating stress

Goodman line

m

Sut

Sy

Soderberg line

Mean stress

The Effect of Mean Stress on Fatigue Life

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

Mean stress is not zero

MAE dept., SJSU

the effect of mean stress on fatigue life modified goodman diagram

Yield line

a

C

Safe zone

Alternating stress

m

The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram

Sy

Se

Goodman line

Sut

Sy

Mean stress

MAE dept., SJSU

the effect of mean stress on fatigue life modified goodman diagram1

Safe zone

Goodman line

- Syc

Sut

The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram

a

Sy

Yield line

Se

C

Safe zone

+m

Sy

- m

MAE dept., SJSU

the effect of mean stress on fatigue life modified goodman diagram2

m > 0

m≤0

Fatigue,

Fatigue,

Infinite life

Se

a

m

1

a=

Finite life

=

+

nf

nf

a

m

Se

Sut

1

=

+

Yield

Sy

Syc

Sn

Sut

a+ m=

a+ m=

Yield

ny

ny

The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram

a

Se

C

Safe zone

Safe zone

+m

Sy

Sut

- m

- Syc

MAE dept., SJSU

applying stress concentration factor to alternating and mean components of stress

Calculate the stress concentration factor for the mean stress using the following equation,

Kfa

Sy

Kfm=

m

Fatigue design equation

Kf a

Kfmm

1

Infinite life

=

+

nf

Se

Sut

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.

MAE dept., SJSU

combined loading

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

All four components of stress exist,

xaalternating component of normal stress

xmmean component of normal stress

xyaalternating component of shear stress

xymmean component of shear stress

MAE dept., SJSU

combined loading1

a′ = (1a+ 2a - 1a2a)1/2

m′ = (1m+ 2m - 1m2m)1/2

2

2

2

2

′a

′m

1

=

+

nf

Se

Sut

Fatigue design equation

Infinite life

Combined Loading

Calculate the alternating and mean von Mises stresses,

MAE dept., SJSU

design example

Calculate the support forces,

R1= 2500, R2= 7500 lb.

The critical location is at the fillet,

MA= 2500 x 12 = 30,000 lb-in

32M

305577

Mc

a=

Calculate the alternating stress,

=

=

I

πd 3

d 3

Determine the stress concentration factor

D

r

= 1.5

= .1

Kt = 1.7

d

d

10,000 lb.

Design Example

12˝

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.

d

D = 1.5d

A

R1

R2

r (fillet radius) = .1d

m= 0

MAE dept., SJSU

design example1

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

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

Design Example

Assume d = 1.0 in

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)

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

MAE dept., SJSU

design example2

39.57

(

)

log

.9x120

86250

(

)

Sn

39.57

=

= 56.5 ksi

106

Sn

56.5

305577

n =

a=

= .116 < 1.6

=

= 305.577 ksi

Kfa

1.6x305.577

d 3

So d = 1.0 in. is too small

Se

(

)

.9Sut

log

Assume d = 2.5 in

N

(

)

Se

Sn

=

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

106

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

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

Se =36.2 ksi

Design Example

Design life, N = 1150 x 75 = 86250 cycles

Csize = .869(d)-0.097 = .869(2.5)-0.097 = .795

MAE dept., SJSU

design example3

36.2

(

)

log

.9x120

86250

(

)

Sn

36.20

=

= 53.35 ksi

305577

a=

106

= 19.55 ksi

(2.5)3

Sn

53.35

n =

= 1.67 ≈ 1.6

=

Kfa

1.63x19.55

d=2.5 in.

Check yielding

Sy

90

n=

2.8 > 1.6 okay

=

=

Kfmax

1.63x19.55

Design Example

Se =36.2 ksi

MAE dept., SJSU

design example observations

12˝

d

D = 1.5d

A

R1

R2 = 7500

Sn

r (fillet radius) = .1d

56.5

n =

Calculate an approximate diameter

= .116 < 1.6

=

Kfa

1.6x305.577

Sn

56.5

n =

→ d = 2.4 in.

=

= 1.6

So d = 1.0 in. is too small

Kfa

1.6x305.577/d 3

Check the location of maximum moment for possible failure

Design Example – Observations

So, your next guess should be between 2.25 to 2.5

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 > 45,000

MAE dept., SJSU

example

Fillet

r

4

= .16

=

25

d

Kt = 1.76

D

35

=

= 1.4

d

25

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 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

Pm = (Pmax + Pmin) / 2 = 6000 N

Pa = (Pmax – Pmin) / 2 = 4000 N

MAE dept., SJSU

example1

Calculate the alternating and the mean stresses,

Pa

4000

a=

1.65

= 52.8 MPa

Kf

=

A

25x5

Pm

6000

m=

= 48 MPa

=

A

25x5

Fatigue design equation

a

m

1

Infinite life

=

+

1

n

Sut

52.8

48

Se

n = 2.7

=

+

n

180

620

Example

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

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

MAE dept., SJSU

example2

d

5

→ Kt= 2.6

=

= .143

w

35

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

Calculate the alternating and the mean stresses,

Pm

6000

m=

= 40 MPa

=

A

30x5

Pa

4000

a=

2.3

= 61.33 MPa

Kf

=

A

(35-5)5

1

61.33

40

n = 2.5

=

+

n

180

620

Example

Hole

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

MAE dept., SJSU

example3

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

Calculate the alternating and the mean stresses,

Pm

6000

m=

= 41.4 MPa

=

A

29x5

Pa

4000

a=

2.1

= 58.0 MPa

Kf

=

A

(35-6)5

1

58.0

41.4

n = 2.57

=

+

n

180

620

Example

Groove

r

3

= .103

=

29

d

→ Kt= 2.33

D

35

=

= 1.2

d

29

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

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

MAE dept., SJSU

example4

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

32Ma

32(120)

Mc

a=

=

= 23178.6 psi

=

I

πd 3

π(.375)3

32Mm

32(360)

Mc

m=

= 69536 psi

=

=

I

πd 3

π(.375)3

Example

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.

MAE dept., SJSU

example5

A95= .010462d 2 (non-rotating round section)

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

1

23178.6

69536

n = .7< 1

a

m

=

+

1

n

24077

150000

=

+

Finite life

n

Sut

Se

Find Sn, strength for finite number of cycle

a

m

23178.6

69536

Sn = 43207 psi

1

=

1

+

=

+

Sut

Sn

Sn

150000

Example

Calculate the endurance limit

Csurf = A (Sut)b = 14.4(150)-.718 = .394

Cload = 1 (pure bending)

Ctemp= 1 (room temp)

Crel= .814 (99% reliability)

dequiv= .14 < .3 → Csize = 1.0

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

MAE dept., SJSU

example6

24.077

(

)

log

.9x150

N

(

)

24077

=

43207

106

Se

(

)

.9Sut

log

N

(

)

Se

Sn

=

106

Example

N = 96,000 cycles

MAE dept., SJSU