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.
Fatigue failure is characterized by three stages
Ken Youssefi

Crack Initiation

Crack Propagation

Final Fracture
MAE dept., SJSU
1
Jack hammer component,
shows no yielding before
fracture.
Crack initiation site
Fracture zone
Propagation zone, striation
Ken Youssefi
MAE dept., SJSU
2
VW crank shaft – fatigue failure due to cyclic bending and torsional stresses
Propagation
zone, striations
Crack initiation site
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Fracture area
MAE dept., SJSU
3
928 Porsche timing pulley
Ken Youssefi
MAE dept., SJSU
Crack started at the fillet
4
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
Ken Youssefi
MAE dept., SJSU
5
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.
Ken Youssefi
MAE dept., SJSU
6
Crank shaft
Gear tooth failure
Ken Youssefi
MAE dept., SJSU
7
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
8
Fracture Surface Characteristics
Mode of fracture
Typical surface characteristics
Ductile
Cup and Cone
Dimples
Dull Surface
Inclusion at the bottom of the dimple
Brittle Intergranular
Shiny
Grain Boundary cracking
Brittle Transgranular
Shiny
Cleavage fractures
Flat
Fatigue
Beachmarks
Striations (SEM)
Initiation sites
Propagation zone
Final fracture zone
Ken Youssefi
MAE dept., SJSU
9
Fatigue Failure – Type of Fluctuating Stresses
Alternating stress
a =
max min
2
Mean stress
m
=
Ken Youssefi
MAE dept., SJSU
max + min
2
10
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.
Typical testing apparatus, pure bending
Motor
Load
Rotating beam machine – applies fully reverse bending stress
Ken Youssefi
MAE dept., SJSU
11
Fatigue Failure, S-N Curve
N > 103
N < 103
Finite life
Infinite life
S′e
Se′ = endurance limit of the specimen
Ken Youssefi
MAE dept., SJSU
12
Relationship Between Endurance Limit
and Ultimate Strength
Steel
Steel
Se′ =
0.5Sut
Sut ≤ 200 ksi (1400 MPa)
100 ksi
Sut > 200 ksi
700 MPa Sut > 1400 MPa
Cast iron
Cast iron
Se′ =
0.4Sut
Sut < 60 ksi (400 MPa)
24 ksi
Sut ≥ 60 ksi
160 MPa Sut < 400 MPa
Ken Youssefi
MAE dept., SJSU
13
Relationship Between Endurance Limit and
Ultimate Strength
Aluminum
Aluminum alloys
Se′ =
0.4Sut
Sut < 48 ksi (330 MPa)
19 ksi
Sut ≥ 48 ksi
130 MPa Sut ≥ 330 MPa
For N = 5x108 cycle
Copper alloys
Copper alloys
Se′ =
0.4Sut
Sut < 40 ksi (280 MPa)
14 ksi
Sut ≥ 40 ksi
100 MPa Sut ≥ 280 MPa
For N = 5x108 cycle
Ken Youssefi
MAE dept., SJSU
14
Correction Factors for Specimen’s Endurance Limit
For materials exhibiting a knee in the S-N curve at 106 cycles
S ′ = endurance limit of the specimen (infinite life > 106)
e
Se = endurance limit of the actual component (infinite life > 106)
S
103
Se
106
N
For materials that do not exhibit a knee in the S-N curve, the infinite
life taken at 5x108 cycles
Sf′ = fatigue strength of the specimen (infinite life > 5x108)
Sf = fatigue strength of the actual component (infinite life > 5x108)
S
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103
Sf
5x108
MAE dept., SJSU
N
15
Correction Factors for Specimen’s Endurance Limit
Se = Cload Csize Csurf Ctemp Crel (S′e)
• Load factor, Cload
Ken Youssefi
Pure bending
Cload = 1
Pure axial
Cload = 0.7
Pure torsion
Cload = 1 if von Mises stress is used, use
0.577 if von Mises stress is NOT used.
Combined loading
Cload = 1
MAE dept., SJSU
16
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
17
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 = (
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A95
0.0766
)1/2
MAE dept., SJSU
d95
d
18
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
19
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
Ken Youssefi
for T ≤ 450 oC (840 oF)
MAE dept., SJSU
20
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
21
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.
Notch sensitivity
Kf = 1 + (Kt – 1)q
factor
Ken Youssefi
MAE dept., SJSU
22
Fatigue Stress
Concentration Factor,
Kf for Aluminum
Ken Youssefi
MAE dept., SJSU
23
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
•
Use the design equation to calculate the size
Se
Kf a =
n
•
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
24
Design for Finite Life
Sn = a (N)b equation of the fatigue line
A
A
S
S
B
B
Sf
Se
106
103
Point A
Point B
Sn = .9Sut
Point A
N
Sn = .9Sut
N = 10
N = 103
Sn = Se
Sn = Sf
3
Point B
6
N = 10
Ken Youssefi
5x108
103
N
MAE dept., SJSU
N = 5x108
25
Design for Finite Life
Sn = a (N)b
log 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 10
a=
3
log Se = log a + b log 106
b=
N
Sn = Se ( 106 )
Calculate Sn
2
Se
1
3
log
.9Sut
Se
Se
log ( .9S )
ut
and replace Se in the design equation
Kf a =
Ken Youssefi
⅓
(.9Sut)
Sn
n
Design equation
MAE dept., SJSU
26
The Effect of Mean Stress on Fatigue Life
Mean stress exist if the
loading is of a repeating or
fluctuating type.
a
Sy
Yield line
Gerber curve
Alternating
stress
Se
Goodman line
Sy
Soderberg line
Sut
m
Mean stress
Ken Youssefi
MAE dept., SJSU
27
The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
a
Sy
Alternating
stress
Yield line
Se
Goodman line
C
Safe zone
Sy
Sut
m
Mean stress
Ken Youssefi
MAE dept., SJSU
28
The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
a
Sy
Yield line
Se
Goodman line
Safe zone
- m
Ken Youssefi
Safe zone
C
Sy
- Syc
MAE dept., SJSU
Sut
+m
29
The Effect of Mean Stress on Fatigue Life
Modified Goodman Diagram
Fatigue,
m ≤ 0
Fatigue,
a
a
Se
a
Se
a = n
f
Sn
+
+
m > 0
m
Sut
m
Sut
=
= 1
Se
Yield
a + m = n
y
nf
Finite life
Sy
Safe zone
Ken Youssefi
Infinite life
Yield
Sy
- m
1
Safe zone
a + m = n
y
C
Sy
- Syc
MAE dept., SJSU
Sut
+m
30
Applying Stress Concentration factor to Alternating
and Mean Components of Stress
•
Determine the fatigue stress concentration factor, Kf, apply directly to
•
If Kf max < Sy then there is no yielding at the notch, use Kfm = Kf
the alternating stress → Kf a
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
Fatigue design equation
Kf a
Kfmm
=
+
Se
Ken Youssefi
Sut
Kf a
m
1
nf
MAE dept., SJSU
Infinite life
31
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,
Ken Youssefi
1a, 2a = (xa /2) ±
(xa /2)2 + (xya)2
1m, 2m = (xm /2) ±
(xm /2)2 + (xym)2
MAE dept., SJSU
32
Combined Loading
Calculate the alternating and mean von Mises stresses,
a′ = (1a2 + 2a2 - 1a2a)1/2
2
2
m′ = (1m
+ 2m - 1m2m)1/2
Fatigue design equation
′a
Se
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+
′m
Sut
=
MAE dept., SJSU
1
nf
Infinite life
33
Design Example
12˝
A rotating shaft is carrying 10,000 lb force
as shown. The shaft is made of steel with
D = 1.5d
d
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,
10,000 lb.
6˝
6˝
A
R1
r (fillet radius) = .1d
R2
R1 = 2500, R2 = 7500 lb.
The critical location is at the fillet,
Calculate the alternating stress,
MA = 2500 x 12 = 30,000 lb-in
a =
Mc
I
=
32M
πd
=
3
305577
d
3
m = 0
Determine the stress concentration factor
r
= .1
d
D
= 1.5
d
Ken Youssefi
Kt = 1.7
MAE dept., SJSU
34
Design Example
Assume 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)
0.3 in. < d ≤ 10 in.
-.265
= .759
Csize = .869(d)-0.097 = .869(1)-0.097 = .869
Se = Cload Csize Csurf Ctemp Crel (S′e) = (.759)(.869)(.5x120) = 39.57
Ken Youssefi
MAE dept., SJSU
ksi
35
Design Example
Design life, N = 1150 x 75 = 86250 cycles
Se
log ( .9S )
86250
N ⅓
ut
Sn = 39.57 ( 6
Sn = Se ( 6 )
10
a =
305577
d3
10
= 305.577 ksi
n=
Sn
Kfa
=
39.57
)
⅓ log ( .9x120 )
= 56.5 ksi
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
Ken Youssefi
MAE dept., SJSU
→
Se = 36.2 ksi
36
Design Example
Se = 36.2 ksi → Sn = 53.35 ksi
a =
305577
(2.5)
3
= 19.55 ksi
n=
Sn
Kfa
=
53.35
1.63x19.55
= 1.67 ≈ 1.6
d = 2.5 in.
Check yielding
n=
Ken Youssefi
Sy
90
= 2.8 > 1.6 okay
=
Kfmax 1.63x19.55
MAE dept., SJSU
37