Fatigue Failure

1
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.
2
Jack hammer component, shows no yielding before
fracture.
3
VW crank shaft fatigue failure due to cyclic
bending and torsional stresses
Propagation zone, striations
Crack initiation site
Fracture area
4
928 Porsche timing pulley
Crack started at the fillet
5
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
6
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.
7
Gear tooth failure
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.
9
Fracture Surface Characteristics
Mode of fracture
Typical surface characteristics
Cup and ConeDimplesDull SurfaceInclusion at
the bottom of the dimple
Ductile
ShinyGrain Boundary cracking
Brittle Intergranular
ShinyCleavage fracturesFlat
Brittle Transgranular
BeachmarksStriations (SEM)Initiation
sitesPropagation zoneFinal fracture zone
Fatigue
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Fatigue Failure Type of Fluctuating Stresses
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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.
12
Fatigue Failure, S-N Curve
N lt 103
N gt 103
13
Relationship Between Endurance Limit and Ultimate
Strength
Steel
14
Relationship Between Endurance Limit and Ultimate
Strength
Aluminum
Aluminum alloys
For N 5x108 cycle
15
Correction Factors for Specimens Endurance Limit
For materials exhibiting a knee in the S-N curve
at 106 cycles
16
Correction Factors for Specimens Endurance Limit
or

• Load factor, Cload (page 326, Nortons 3rd ed.)

17
Correction Factors for Specimens Endurance Limit

• Size factor, Csize (p. 327, Nortons 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
18
Correction Factors for Specimens 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 (p/4)d2 (.95d)2 .0766 d2
19
Correction Factors for Specimens Endurance Limit
I beams and C channels
20
Correction Factors for Specimens Endurance Limit

• surface factor, Csurf (p. 328-9, Nortons 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.
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Correction Factors for Specimens Endurance Limit

• Temperature factor, Ctemp (p.331, Nortons 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)
22
Correction Factors for Specimens Endurance Limit

• Reliability factor, Crel (p. 331, Nortons 3rd
  ed.)

The reliability correction factor accounts for
the scatter and uncertainty of material
properties (endurance limit).
23
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.
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Fatigue Stress Concentration Factor, Kf for
Aluminum
(p. 341, Nortons 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

• 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

26
Design for Finite Life
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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
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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 is not zero
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The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
Sy
Se
Goodman line
Sut
Sy
Mean stress
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The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
?a
Sy
Yield line
Se
C
Safe zone
?m
Sy
- ?m
31
The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
?a
Se
C
Safe zone
Safe zone
?m
Sut
Sy
- ?m
- Syc
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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 lt Sy then there is no yielding at the
  notch, use Kfm Kf and multiply the mean stress
  by Kfm ? Kfm ?m

• If Kf ?max gt Sy then there is local yielding at
  the notch, material at the notch is
  strain-hardened. The effect of stress
  concentration is reduced.

33
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
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Combined Loading
Calculate the alternating and mean von Mises
stresses,
35
Design Example
10,000 lb.
6?
6?
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 1.5d
d
A
R1
R2
r (fillet radius) .1d
?m 0
36
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
37
Design Example
Design life, N 1150 x 75 86250 cycles
Csize .869(d)-0.097 .869(2.5)-0.097 .795
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Design Example
Se 36.2 ksi
?
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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 gt 45,000
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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
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Example
Using r 4 and Sut 620 MPa, q (notch
sensitivity) .85
Kf 1 (Kt 1)q 1 .85(1.76 1) 1.65
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Example
Hole
Kf 1 (Kt 1)q 1 .82(2.6 1) 2.3
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Example
Groove
r
3
.103

d
29
? 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
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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.
45
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 lt .3 ? Csize 1.0
Se Cload Csize Csurf Ctemp Crel (Se)
(.394)(.814)(.5x150) 24.077 ksi
46
Example
N 96,000 cycles