Fatigue Failure
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Transcript of Fatigue Failure
Ken Youssefi MAE dept., SJSU 1
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
Ken Youssefi MAE dept., SJSU 2
Jack hammer component, shows no yielding before fracture.
Crack initiation site
Fracture zonePropagation zone, striation
Ken Youssefi MAE dept., SJSU 3
VW crank shaft – fatigue failure due to cyclic bending and torsional stresses
Fracture areaCrack initiation site
Propagation zone, striations
Ken Youssefi MAE dept., SJSU 4
928 Porsche timing pulley
Crack started at the fillet
Ken Youssefi MAE dept., SJSU 5
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.
Ken Youssefi MAE dept., SJSU 6
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
Ken Youssefi MAE dept., SJSU 7
Gear tooth failure
Crank shaft
Ken Youssefi MAE dept., SJSU 8
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 9
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
Ken Youssefi MAE dept., SJSU 10
Fatigue Failure – Type of Fluctuating Stresses
a =max min
2
Alternating stress
Mean stress
m =max min
2+
Ken Youssefi MAE dept., SJSU 11
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
Ken Youssefi MAE dept., SJSU 12
Fatigue Failure, S-N Curve
Finite life Infinite life
N < 103 N > 103
S′e
= endurance limit of the specimenSe′
Ken Youssefi MAE dept., SJSU 13
Relationship Between Endurance Limit and Ultimate Strength
Steel
Se =′
0.5Sut
100 ksi
700 MPa
Sut ≤ 200 ksi (1400 MPa)
Sut > 200 ksi
Sut > 1400 MPa
Steel
0.4Sut
Se =′
Sut < 60 ksi (400 MPa)
Sut ≥ 60 ksi24 ksi
160 MPa Sut < 400 MPa
Cast iron Cast iron
Ken Youssefi MAE dept., SJSU 14
Relationship Between Endurance Limit and Ultimate Strength
Aluminum alloys
Se =′
0.4Sut
19 ksi
130 MPa
Sut < 48 ksi (330 MPa)
Sut ≥ 48 ksi
Sut ≥ 330 MPa
Aluminum
For N = 5x108 cycle
Copper alloys
Se =′
0.4Sut
14 ksi
100 MPa
Sut < 40 ksi (280 MPa)
Sut ≥ 40 ksi
Sut ≥ 280 MPa
Copper alloys
For N = 5x108 cycle
Ken Youssefi MAE dept., SJSU 15
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
Ken Youssefi MAE dept., SJSU 16
Correction Factors for Specimen’s Endurance Limit
Se = Cload Csize Csurf Ctemp Crel (Se)′
• Load factor, Cload
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.
Ken Youssefi MAE dept., SJSU 17
Correction Factors for Specimen’s Endurance Limit
• Size factor, Csize
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 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
Ken Youssefi MAE dept., SJSU 18
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.
dequiv = (A95
0.0766)1/2 dd95
Ken Youssefi MAE dept., SJSU 19
Correction Factors for Specimen’s Endurance Limit
• surface factor, Csurf
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 Specimen’s Endurance Limit
• Temperature factor, Ctemp
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 Specimen’s Endurance Limit
• Reliability factor, Crel
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.
Kf = 1 + (Kt – 1)qNotch sensitivity factor
Ken Youssefi MAE dept., SJSU 23
Fatigue Stress Concentration Factor,
Kf for Aluminum
Ken Youssefi MAE dept., SJSU 24
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
Ken Youssefi MAE dept., SJSU 25
Design for Finite Life
Sn = a (N)b equation of the fatigue line
N
S
Se
106103
A
B
N
S
Sf
5x108103
A
B
Point ASn = .9Sut
N = 103Point A
Sn = .9Sut
N = 103
Point BSn = Sf
N = 5x108Point B
Sn = Se
N = 106
Ken Youssefi MAE dept., SJSU 26
Design for Finite Life
Sn = a (N)blog Sn = log a + b log N
Apply 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
1
3log
SnKf a = n Design equation
Calculate Sn and replace Se in the design equation
Sn = Se (N
106)⅓ (
Se
.9Sut
)log
Ken Youssefi MAE dept., SJSU 27
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 lineSut
Goodman line
Gerber curve
SyYield line
Ken Youssefi MAE dept., SJSU 28
The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
Mean stress
Alternating stress
m
a
Sut
Goodman line
SyYield line
Sy
Se
Safe zoneC
Ken Youssefi MAE dept., SJSU 29
- 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
Ken Youssefi MAE dept., SJSU 30
The Effect of Mean Stress on Fatigue Life Modified Goodman Diagram
+m
a
Sut
Safe zone
- m
C
Sy
Safe zone
Se
- Syc
Finite lifeSn
1=Sut
a m+
Fatigue, m > 0Fatigue, m ≤ 0
a =Se
nf
a + m =Sy
ny
Yield
a + m =Sy
ny
Yield
nfSe
1=
Sut
a m+ Infinite life
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 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
Ken Youssefi MAE dept., SJSU 32
Combined Loading
All 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
Ken Youssefi MAE dept., SJSU 33
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
Ken Youssefi MAE dept., SJSU 34
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
=32M
πd 3=
305577
d 3m = 0
Determine the stress concentration factor
rd
= .1
Dd
= 1.5
Kt = 1.7
Ken Youssefi MAE dept., SJSU 35
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′
Ken Youssefi MAE dept., SJSU 36
Design ExampleDesign life, N = 1150 x 75 = 86250 cycles
Sn = Se (N
106)⅓ (
Se
.9Sut
)logSn = 39.57 (
86250
106)⅓ (
39.57.9x120
)log
= 56.5 ksi
a = 305577
d 3= 305.577 ksi n =
Sn
Kfa
=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→
Ken Youssefi MAE dept., SJSU 37
Design Example
a = 305577
(2.5)3= 19.55 ksi
n = Sn
Kfa
=53.35
1.63x19.55= 1.67 ≈ 1.6
d = 2.5 in.
Check yielding
n = Sy
Kfmax
= 90
1.63x19.55= 2.8 > 1.6 okay
Se = 36.2 ksi Sn = 53.35 ksi→