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Fatigue Failure

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Title: 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 pin 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 (1988), Aloha Flight 243, a Boeing 737, an
upper part of the plane's cabin area rips off in
mid-flight.
Metal fatigue compounded by corrosion
9
Fracture Surface Characteristics
Mode of fracture
Typical surface characteristics
Cup and ConeDimplesDull SurfaceInclusion at
the bottom of the dimple
Ductile
10
Brittle Fracture
The cracks usually travel so fast that you can't
tell when the material is about to break. In
other words, there is very little plastic
deformation before failure occurs. In most cases,
this is the worst type of fracture because you
can't repair visible damage in a part or
structure before it breaks.
Cleavage fractures
Grain Boundary cracking
11
Brittle fracture surfaces

slightly bumpy crack surface
Chevron Fracture Surface
Brittle fracture in a mild steel (shinny surface)
12
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13
Low temperatures can severely embrittle steels.
The Liberty ships, produced in great numbers
during the WWII were the first all-welded ships.
A significant number of ships failed by
catastrophic fracture. Fatigue cracks nucleated
at the corners of square hatches and propagated
rapidly by brittle fracture.
14
Striations (beach markesInitiation
sitesPropagation zoneFinal fracture zone
Fatigue
15
Fatigue Failure Type of Fluctuating Stresses
?a Alternating Stress
?m Mean Stress
16
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.
17
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
18
Fatigue Failure, S-N Curve (Strength vs. of
cycles to failure)
N lt 103
N gt 103
19
Relationship Between Endurance Limit and Ultimate
Strength
Steel
20
Relationship Between Endurance Limit and Ultimate
Strength
Aluminum
Aluminum alloys
For N 5x108 cycle
21
Correction Factors for Specimens Endurance Limit
For materials exhibiting a knee in the S-N curve
at 106 cycles
22
Correction Factors for Specimens Endurance Limit
Material exhibits a knee in S-N curve, infinite
life at 106 cycle
Se Cload Csize Csurf Ctemp Crel (Se)
  • Load factor, Cload (page 330, Nortons 4th ed.
    or page 362 in 5th ed.)

23
Correction Factors for Specimens Endurance Limit
  • Size factor, Csize (p. 331, Nortons 4th ed. or
    page 363 in 5th 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
24
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
25
Correction Factors for Specimens Endurance Limit
I beams and C channels
26
Correction Factors for Specimens Endurance Limit
  • surface factor, Csurf (p. 332/3, Nortons 4th
    ed. or page 363/4 in 5th 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.
27
Correction Factors for Specimens Endurance Limit
  • Temperature factor, Ctemp (p.335, Nortons 4th
    ed. or page 367 in 5th 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 840 oF (450 oC)
Ctemp 1 0.0032(T 840)
for 840 oF lt T 1020 oF
28
Correction Factors for Specimens Endurance Limit
  • Reliability factor, Crel (p. 335, Nortons 4th
    ed. or page 367 in 5th ed.)

The reliability correction factor accounts for
the scatter and uncertainty of material
properties (endurance limit).
29
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 (depends on the
geometry only). The stress concentration factor
seems to be sensitive to the notch radius and the
ultimate strength of the material.
Theoretical stress concentration factor, Kt
30
Fatigue Stress Concentration Factor, q for
Aluminum
(p. 345, Nortons 4th ed. or p. 376 in 5th ed.)
31
Design process Fully Reversed Loading for
Infinite Life
Mean stress 0
  • Determine the maximum alternating applied stress
    (?a ) in terms of the size for the selected 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 (infinite life
    106) or, Sf (infinite life 5x108)
  • Determine the fatigue stress concentration
    factor, Kf , q and Kt
  • 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

32
Fatigue Failure, S-N Curve (Strength vs. of
cycles to failure)
N lt 103
N gt 103
33
Design for Finite Life
34
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
35
Nortons book (Machine Design) notations
36
Nortons book (Machine Design) notations
37
For material exhibiting an endurance-limit knee
38
Review of fatigue strength notations
'
Se endurance limit of a specimen, material
showing a knee
Se corrected endurance limit, material showing
a knee
39
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
40
The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
Sy
Se
Goodman line
Sut
Sy
Mean stress
41
The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
?a
Sy
Yield line
Se
C
Safe zone
?m
Sy
- ?m
42
The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
?a
Se
C
Safe zone
Safe zone
?m
Sut
Sy
- ?m
- Syc
43
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.

44
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
45
Combined Loading
Calculate the alternating and mean von Mises
stresses,
46
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
47
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)
48
Design Example
Design life, N 1150 x 75 86250 cycles
Csize .869(d)-0.097 .869(2.5)-0.097 .795
49
Design Example
Se 36.2 ksi
?
50
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
51
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
52
Example
Using r 4 and Sut 620 MPa, q (notch
sensitivity) .85
Kf 1 (Kt 1)q 1 .85(1.76 1) 1.65
53
Example
Hole
Kf 1 (Kt 1)q 1 .82(2.6 1) 2.3
54
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
55
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.
56
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
57
Example
N 96,000 cycles
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