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
Ken Youssefi MAE dept., SJSU 1
are called fatigue failures.
Fatigue failure is characterized by three stages
v Crack Initiationv Crack Propagationv Final Fracture
Jack hammer component, shows no yielding before fracture.
Crack initiation site
Ken Youssefi MAE dept., SJSU 2
Crack initiation site
Fracture zonePropagation zone, striation
VW crank shaft – fatigue failure due to cyclic bending and torsional stresses
Propagation zone, striations
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Fracture areaCrack initiation site
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928 Porsche timing pulley
Crack started at the fillet
Fracture surface of a failed bolt. The fracture surface exhibited beach marks, which is characteristic of a fatigue failure.
Ken Youssefi MAE dept., SJSU 5
1.0-in. diameter steel pins from agricultural equipment.Material; AISI/SAE 4140 low allow carbon steel
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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
Crank shaft
Ken Youssefi MAE dept., SJSU 7
Gear tooth failure
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.
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Cup and ConeDimplesDull SurfaceInclusion at the bottom of the dimple
Ductile
Fracture Surface Characteristics
ShinyGrain Boundary cracking
Brittle Intergranular
Mode of fracture Typical surface characteristics
Ken Youssefi MAE dept., SJSU 9
ShinyCleavage fracturesFlat
Brittle Transgranular
BeachmarksStriations (SEM)Initiation sitesPropagation zoneFinal fracture zone
Fatigue
Fatigue Failure – Type of Fluctuating Stressesσa = σmax
σmax = - σmin
Ken Youssefi MAE dept., SJSU 10
σa =σmax σmin
2
Alternating stress
Mean stress
σm=
σmax σmin
2+
σmin = 0
σa = σmax / 2σm =
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.
Ken Youssefi MAE dept., SJSU 11
Motor
Load
Rotating beam machine – applies fully reverse bending stress
Typical testing apparatus, pure bending
Fatigue Failure, S-N Curve
Finite life Infinite life
N < 103 N > 103
Ken Youssefi MAE dept., SJSU 12
S′e
= endurance limit of the specimenSe′
Relationship Between Endurance Limit and Ultimate Strength
Steel
Se =′0.5Sut
100 ksi700 MPa
Sut ≤ 200 ksi (1400 MPa)Sut > 200 ksiSut > 1400 MPa
Steel
Ken Youssefi MAE dept., SJSU 13
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 ksi130 MPa
Sut < 48 ksi (330 MPa)Sut ≥ 48 ksiSut ≥ 330 MPa
Aluminum
For N = 5x108 cycle
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Copper alloys
Se =′0.4Sut
14 ksi100 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
Ken Youssefi MAE dept., SJSU 15
= 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
Correction Factors for Specimen’s Endurance Limit
Se = Cload Csize Csurf Ctemp Crel (Se)′
• Load factor, Cload (page 326, Norton’s 3rd ed.)
Pure bending Cload = 1
Sf = Cload Csize Csurf Ctemp Crel (Sf)′ or
Ken Youssefi MAE dept., SJSU 16
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.
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.
For rotating solid round cross section
Ken Youssefi MAE dept., SJSU 17
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 = .95dA95 = (π/4)[d2 – (.95d)2] = .0766 d2
Ken Youssefi MAE dept., SJSU 18
dequiv = .37d
Solid or hollow non-rotating parts
dequiv = .808 (bh)1/2
Rectangular parts
Correction Factors for Specimen’s Endurance Limit
I beams and C channels
Ken Youssefi MAE dept., SJSU 19
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.
Ken Youssefi MAE dept., SJSU 20
Csurf = A (Sut)b
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.
Ken Youssefi MAE dept., SJSU 21
Ctemp = 1 for T ≤ 450 oC (840 oF)
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).
Ken Youssefi MAE dept., SJSU 22
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.)Kf = 1 + (Kt – 1)q
Notch sensitivity factorFatigue stress concentration factor
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Steel
Fatigue Stress Concentration Factor,
Kf for Aluminum
(p. 341, Norton’s 3rd ed.)
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Design process – Fully Reversed Loading for Infinite Life
• Determine the maximum alternating applied stress (σa ) 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
Ken Youssefi MAE dept., SJSU 25
• Use the design equation to calculate the size
SeKf σa = n
• 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 lineA A
Ken Youssefi MAE dept., SJSU 26
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 BSn = 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
Ken Youssefi MAE dept., SJSU 27
Se
SnKf σa = n Design equation
Calculate Sn and replace Se in the design equation
Sn = Se ( N106 )
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.
σa Mean stress is not zero
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Mean stress
Alternating stress
σm
Se
SySoderberg line Sut
Goodman line
Gerber curve
The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
σa
Sy
Se
Yield line
Ken Youssefi MAE dept., SJSU 29
Mean stress
Alternating stress
σmSut
Goodman line
Sy
Se
Safe zoneC
The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
σa
Sy Yield line
S
Ken Youssefi MAE dept., SJSU 30
- Syc +σmSut
Goodman line
Safe zone
- σm
C
Sy
Se
Safe zone
The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
σa
S
Finite lifeSn1=Sut
σa σm+
Fatigue, σm > 0Fatigue, σm ≤ 0
σa =Se
nf
Yield
nfSe
1=Sut
σa σm+ Infinite life
Ken Youssefi MAE dept., SJSU 31
+σmSut
Safe zone
- σm
C
Sy
Safe zone
Se
- Syc
σa + σm =Syc
ny
Yield
σa + σm =Sy
ny
Yield
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 σa
• If Kf σmax < Sy then there is no yielding at the notch, use Kfm = Kf
and multiply the mean stress by Kfm → Kfm σm
• If Kf σmax > Sy then there is local yielding at the notch, material at the notch is strain-hardened. The effect of stress concentration is reduced.
Ken Youssefi MAE dept., SJSU 32
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 σa
σm
nfSe
1=Sut
Kf σa Kfmσm+ Infinite life
Fatigue design equation
Combined LoadingAll four components of stress exist,
σxa alternating component of normal stress
σxm mean component of normal stress
τxya alternating component of shear stress
τxym mean component of shear stress
Ken Youssefi MAE dept., SJSU 33
τxym mean component of shear stress
Calculate the alternating and mean principal stresses,
σ1a, σ2a = (σxa /2) ± (σxa /2)2 + (τxya)2
σ1m, σ2m = (σxm /2) ± (σxm /2)2 + (τxym)2
Combined Loading
Calculate the alternating and mean von Mises stresses,
σa′ = (σ1a + σ2a - σ1aσ2a)1/22 2
σm′ = (σ1m + σ2m - σ1mσ2m)1/22 2
Ken Youssefi MAE dept., SJSU 34
Fatigue design equation
nfSe
1=Sut
σ′a σ′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 withSut = 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
Ken Youssefi MAE dept., SJSU 35
R1 = 2500, R2 = 7500 lb.
The critical location is at the fillet, MA = 2500 x 12 = 30,000 lb-in
σa =Calculate the alternating stress, McI =
32Mπd 3
= 305577d 3
σm = 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
C = 1 (pure bending)
Ken Youssefi MAE dept., SJSU 36
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 ( N106 )
⅓ (Se
.9Sut)log
Sn = 39.57 ( 86250106 )
⅓ (39.57
.9x120 )log= 56.5 ksi
σa = 305577d 3
= 305.577 ksi n =Sn
Kfσa=
56.51.6x305.577
= .116 < 1.6
So d = 1.0 in. is too small
Ken Youssefi MAE dept., SJSU 37
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
σa = 305577(2.5)3
= 19.55 ksi
n =Sn
Kfσa=
53.351.63x19.55
= 1.67 ≈ 1.6
Se = 36.2 ksi → Sn = 36.20 ( 86250106 )
⅓ (36.2
.9x120 )log= 53.35 ksi
Ken Youssefi MAE dept., SJSU 38
d = 2.5 in.
Check yielding
n = Sy
Kfσmax=
901.63x19.55
= 2.8 > 1.6 okay
Design Example – Observations
n =Sn
Kfσa=
56.51.6x305.577
= .116 < 1.6
So d = 1.0 in. is too small
Calculate an approximate diameter
n =Snσ =
56.5= 1.6 → d = 2.4 in. So, your next guess
R1 R2 = 7500
6˝6˝12˝D = 1.5dd
r (fillet radius) = .1d
A
Ken Youssefi MAE dept., SJSU 39
n = Kfσa=
1.6x305.577/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
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
P = (P – P ) / 2 = 4000 N P = (P + P ) / 2 = 6000 N
Ken Youssefi MAE dept., SJSU 40
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,
σa =Pa
A= 4000
25x5= 52.8 MPaKf 1.65
Using r = 4 and Sut = 620 MPa, q (notch sensitivity) = .85
Ken Youssefi MAE dept., SJSU 41
σm =Pm
A= 6000
25x5= 48 MPa
nSe
1=Sut
σa σm+ Infinite life
Fatigue design equation
n = 2.7n1801
=620
52.8 48+ →
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
Calculate the alternating and the
Ken Youssefi MAE dept., SJSU 42
Calculate the alternating and the mean stresses,
σa =Pa
A= 4000
(35-5)5= 61.33 MPaKf 2.3
σm =Pm
A= 6000
30x5= 40 MPa
n = 2.5n1801
=620
61.33 40+ →
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
Calculate the alternating and the
Ken Youssefi MAE dept., SJSU 43
(35-6)5
Calculate the alternating and the mean stresses,
σa =Pa
A= 4000 = 58.0 MPaKf 2.1
σm =Pm
A= 6000
29x5= 41.4 MPa
n = 2.57n1801
=620
58.0 41.4+ →
The part is likely to fail at the hole, has the lowest safety factor
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.
Ken Youssefi MAE dept., SJSU 44
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
σa = McI =
32Ma
πd 3=
32(120)π (.375)3 = 23178.6 psi
σm = 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
Ken Youssefi MAE dept., SJSU 45
nSe
1=Sut
σa σm+ → n24077
1=
15000023178.6 69536+ → n = .7 < 1
Finite life
Sn1=Sut
σa σm+
Find Sn, strength for finite number of cycle
Sn1=
15000023178.6 69536+→ → Sn = 43207 psi
Example
Sn = Se ( N106 )
⅓ (Se
.9Sut)log
43207 = 24077 ( N106 )
⅓ ( 24.077.9x150 )log
→
N = 96,000 cycles
Ken Youssefi MAE dept., SJSU 46