fatigue failure foemef 17-06-2011
TRANSCRIPT
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
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Jack hammer component, shows no yielding before fracture.
Crack initiation site
Fracture zonePropagation zone, striation
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VW crank shaft – fatigue failure due to cyclic bending and torsional stresses
Fracture areaCrack initiation site
Propagation zone, striations
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928 Porsche timing pulley
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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
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Gear tooth failure
Crank shaft
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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.
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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
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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
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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
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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.
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Typical testing apparatus, pure bending
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.
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Typical testing apparatus, pure bending
Rotating beam machine – applies fully reverse bending stress
Fatigue Failure, S-N Curve
Finite life Infinite life
N < 103 N > 103
S′e
= endurance limit of the specimenSe′
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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
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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
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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
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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
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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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
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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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
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Correction Factors for Specimen’s Endurance Limit
I beams and C channels
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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
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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)
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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).
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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.)
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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
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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
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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
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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
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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
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- Syc
The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
+m
a
Sut
Goodman line
Sy Yield line
Safe zone
- m
C
Sy
Se
Safe zone
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The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
+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
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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
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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
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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
Fatigue design equation
nfSe
1=
Sut
′a ′m+ Infinite life
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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
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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′
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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→
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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
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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
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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
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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
Fatigue design equation
n = 2.7n180
1=
62052.8 48+ →
Using r = 4 and Sut = 620 MPa, q (notch sensitivity) = .85
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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 mean stresses,
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+ →
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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
Calculate the alternating and the mean stresses,
a = Pa
A=
4000 = 58.0 MPaKf 2.1
m = Pm
A=
600029x5
= 41.4 MPa
n = 2.57n180
1=
62058.0 41.4+ →
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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
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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 = .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
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Example
Sn = Se ( N
106 )⅓ (
Se
.9Sut)log
43207 = 24077 ( N
106 )⅓ ( 24.077
.9x150 )log→
N = 96,000 cycles
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