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Ken Youssefi MAE dept., SJSU 1
It has been recognized that a metal subjectedto 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 loadingare called fatigue fai lures.
Fatigue Failu re
Fatigue failure is characterized by three stages
Crack Initiation
Crack Propagation
Final Fracture
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Ken Youssefi MAE dept., SJSU 2
Jack hammer component,
shows no yielding before
fracture.
Crack initiation site
Fracture zone
Propagation zone, striation
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Ken Youssefi MAE dept., SJSU 3
VW crank shaftfatigue failure due to cyclic bending and torsional stresses
Fracture areaCrack initiation site
Propagation
zone, striations
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Ken Youssefi MAE dept., SJSU 4
928 Porsche timing pulley
Crack started at the fillet
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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.
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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
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7/47Ken Youssefi MAE dept., SJSU 7
Gear tooth failure
Crank shaft
5/27/2018 Fatigue Failure
8/47Ken 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.
5/27/2018 Fatigue Failure
9/47Ken Youssefi MAE dept., SJSU 9
Cup and Cone
DimplesDull Surface
Inclusion at the bottom of the dimple
Ductile
Fracture Surface Characteristics
Shiny
Grain Boundary cracking
Brittle Intergranular
Shiny
Cleavage fractures
Flat
Brittle Transgranular
Beachmarks
Striations (SEM)
Initiation sites
Propagation zone
Final fracture zone
Fatigue
Mode of fracture Typic al surface character ist ics
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10/47Ken Youssefi MAE dept., SJSU 10
Fatigue FailureType of Fluctuat ing Stresses
a=
max
min
2
Alternating stress
Mean stress
m
=
max
min
2
+
min
= 0
a=
max/ 2
m=
a= max
max
= - min
5/27/2018 Fatigue Failure
11/47Ken 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 machineapplies fully reverse bending stress
Typical testing apparatus, pure bending
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Ken Youssefi MAE dept., SJSU 12
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
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Ken Youssefi MAE dept., SJSU 13
Fatigue Failure, S-N Curve
Finite life Infinite life
N 103
Se
= endurance limit of the specimenSe
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Ken Youssefi MAE dept., SJSU 14
Relat ionsh ip Between Endurance L imit
and Ult imate Streng th
Steel
Se =
0.5Sut
100 ksi
700 MPa
Sut 200 ksi (1400 MPa)Sut>200 ksi
Sut>1400 MPa
Steel
0.4Sut
Se =
Sut
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Ken Youssefi MAE dept., SJSU 15
Relat ionsh ip Between Endurance L imit and
Ult imate Strength
Aluminum alloys
Se =
0.4Sut
19 ksi
130 MPa
Sut
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Ken Youssefi MAE dept., SJSU 16
Correction Factors for Specimens Endurance Limit
= endurance limit of the specimen (infinite life > 106)Se
For materials exhibiting a knee in the S-N curve at 106cycles
= endurance limit of the actual component (infinite life > 106)Se
N
S Se
10
6
103
= 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 5x108cycles
N
S Sf
5x108103
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Ken Youssefi MAE dept., SJSU 17
Correction Factors for Specimens Endurance Limit
Se = Cload Csize Csurf Ctemp Crel(Se)
Load factor, Cload (page 326, Nortons 3rded.)
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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Ken Youssefi MAE dept., SJSU 18
Correction Factors for Specimens Endurance Limit
Size factor, Csize (p. 327, Nortons 3rded.)
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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Ken Youssefi MAE dept., SJSU 19
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.
dequiv= (A95
0.0766)1/2
dd95= .95d
A95= (/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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Ken Youssefi MAE dept., SJSU 20
Correction Factors for Specimens Endurance Limit
Ibeams and Cchannels
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Ken Youssefi MAE dept., SJSU 21
Correction Factors for Specimens Endurance Limit
surface factor, Csurf (p. 328-9, Nortons 3rded.)
The rotating beam test specimen has a polished surface. Most
components do not have a polished surface. Scratches andimperfections 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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Ken Youssefi MAE dept., SJSU 22
Correction Factors for Specimens Endurance Limit
Temperature factor, Ctemp (p.331, Nortons 3rded.)
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 450oC (840 oF)
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Ken Youssefi MAE dept., SJSU 23
Correction Factors for Specimens Endurance Limit
Reliability factor, Crel(p. 331, Nortons 3rded.)
The reliability correction factor accounts for the scatter and
uncertainty of material properties (endurance limit).
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Ken Youssefi MAE dept., SJSU 24
Fatigue Stress Concentrat ion Facto r, 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 thematerial.
(p. 340, Nortons 3rded.)
Steel
Kf=1+(Kt1)qNotch sensitivity factor
Fatigue stress
concentration factor
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Ken Youssefi MAE dept., SJSU 25
Fatigue Stress
Concentrat ion Facto r,
qfor Aluminum
(p. 341, Nortons 3rded.)
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Ken Youssefi MAE dept., SJSU 26
Design p rocess Ful ly Reversed Loading fo r Inf in i te 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
SeK
f
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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Ken Youssefi MAE dept., SJSU 27
Design fo r Fini te Life
Sn= a (N)b equation of the fatigue line
N
S
Se
106
103
A
B
N
S
Sf
5x108
103
A
B
Point ASn = .9Sut
N = 103Point A
Sn = .9Sut
N = 103
Point BSn = Sf
N = 5x108Point B
Sn = Se
N = 10
6
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Ken Youssefi MAE dept., SJSU 28
Design fo r Fini te Life
Sn= a (N)b
log Sn= log a+ blog N
Apply boundary conditions for point A and B to
find the two constants a and b
log.9Sut= loga+ blog103
logSe= loga+ blog106
a=(.9Sut)
2
Se
b=
.9Sut
Se
1
3 log
SnKfa= n
Design equation
CalculateSnand replace Sein the design equation
Sn = Se (N
106) (
Se.9Sut
)log
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Ken Youssefi MAE dept., SJSU 29
The Effect o f Mean Stress on Fat igue L ife
Mean stress exist if the
loading is of a repeating orfluctuating type.
Mean stress
Alternating
stress
m
a
Se
SySoderberg lineSut
Goodman line
Gerber curve
Mean stress is not zero
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Ken Youssefi MAE dept., SJSU 30
The Effect of Mean Stress on Fat igue Life
Modi f ied Goodman Diagram
Mean stress
Alternating
stress
m
a
Sut
Goodman line
Sy
Sy
Se
Safe zoneC
Yield line
Th Eff t f M St F t i Li f
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Ken Youssefi MAE dept., SJSU 31
- Syc
The Effect of Mean Stress on Fat igue Life
Modi f ied Goodman Diagram
+m
a
Sut
Goodman line
Sy Yield line
Safe zone
- m
C
Sy
Se
Safe zone
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Ken Youssefi MAE dept., SJSU 32
The Effect of Mean Stress on Fat igue Life
Modi f ied Goodman Diagram
+m
a
Sut
Safe zone
- m
C
Sy
Safe zone
Se
- Syc
Finite life
Sn
1=
Sut
a m
+
Fatigue, m> 0Fatigue, m0
a=
Se
nf
a+
m=
Syc
ny
Yield
a+
m=
Sy
ny
Yield
nfSe
1=
Sut
a m
+Infinite life
A l i S C i f A l i
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Ken Youssefi MAE dept., SJSU 33
Applying Stress Concentrat ion factor to A l ternat ing
and Mean Components of Stress
Determine the fatigue stress concentration factor, Kf, apply directly to
the alternating stress Kfa
If Kfmax< Sy then there is no yielding at the notch, use Kfm =Kf
and multiply the mean stress by Kfm Kfmm
If Kfmax> 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 Kfa
m
nfSe
1=
Sut
Kfa Kfmm+ Infinite life
Fatigue design equation
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Ken Youssefi MAE dept., SJSU 34
Comb ined 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
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Ken Youssefi MAE dept., SJSU 35
Comb ined 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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Ken Youssefi MAE dept., SJSU 36
Design Example
R1 R2
10,000lb.
6612
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= 7500lb.
A
The critical location is at the fillet, MA= 2500 x 12 = 30,000lb-in
a=Calculate the alternating stress,
Mc
I=
32M
d 3=
305577
d 3
m= 0
Determine the stress concentration factor
rd
= .1
D
d= 1.5
Kt= 1.7
Design E ample
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Ken Youssefi MAE dept., SJSU 37
Design Example
Assume d= 1.0 in
Usingr = .1andSut= 120 ksi,
q(notch sensitivity) = .85Kf= 1+ (Kt1)q = 1 + .85(1.71) = 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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Ken Youssefi MAE dept., SJSU 38
Design Example
Design life, N = 1150 x 75 = 86250cycles
Sn = Se (N
106)
(Se
.9Sut)log
Sn = 39.57(86250
106)
(
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.
Usingr = .25andSut= 120 ksi,
q(notch sensitivity) = .9Kf= 1+ (Kt1)q = 1 + .9(1.71) = 1.63
Csize= .869(d)-0.097= .869(2.5)
-0.097= .795 Se =36.2 ksi
D i E l
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Ken Youssefi MAE dept., SJSU 39
Design Example
a=
305577
(2.5)3= 19.55 ksi
n=Sn
Kfa=
53.35
1.63x19.55= 1.67 1.6
d=2.5in.
Check yielding
n=Sy
Kfmax=
90
1.63x19.55= 2.8 > 1.6 okay
Se =36.2 ksi Sn = 36.20(86250
106) (
36.2.9x120
)log
= 53.35 ksi
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Ken Youssefi MAE dept., SJSU 40
Design ExampleObservat ion s
n=Sn
Kfa=
56.5
1.6x305.577
= .116 < 1.6
So d= 1.0 in. is too small
Calculate an approximate diameter
n= SnKfa
= 56.51.6x305.577/d 3
= 1.6 d = 2.4in. So, your next guessshould be between
2.25 to 2.5
Mmax (under the load) = 7500 x 6 = 45,000lb-in
Check the location of maximum moment for possible failure
R1 R2= 7500
6612
D= 1.5dd
r (fillet radius) = .1d
A
MA(at the fillet) = 2500 x 12 = 30,000lb-in
But, applying the fatigue stress conc. Factor of 1.63,
KfM
A
= 1.63x30,000 = 48,900 > 45,000
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Ken Youssefi MAE dept., SJSU 41
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 loadvaries 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? UseKfm= 1
Pa= (PmaxPmin) / 2= 4000 N Pm= (Pmax+ Pmin) / 2= 6000 N
Fil let
rd
= .16
D
d= 1.4
425
=
35
25=
Kt= 1.76
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Ken Youssefi MAE dept., SJSU 42
Example
Kf= 1+ (Kt1)q = 1 + .85(1.761) = 1.65
Calculate the alternating and the
mean stresses,
a=
Pa
A
=4000
25x5= 52.8 MPaKf 1.65
m=
Pm
A=
6000
25x5= 48 MPa
nSe
1=
Sut
a m
+ Infinite life
Fatigue design equation
n= 2.7n180
1=
620
52.8 48+
Usingr = 4andSut= 620 MPa,
q(notch sensitivity) = .85
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Ken Youssefi MAE dept., SJSU 43
ExampleHole
d
w= .143
5
35= Kt= 2.6
Usingr = 2.5andSut= 620 MPa,
q(notch sensitivity) = .82
Kf= 1+ (Kt1)q = 1 + .82(2.61) = 2.3
Calculate the alternating and themean stresses,
a=
Pa
A=
4000
(35-5)5= 61.33 MPaKf 2.3
m=
Pm
A =
6000
30x5 = 40 MPa
n= 2.5n180
1=
620
61.33 40+
E l
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Ken Youssefi MAE dept., SJSU 44
ExampleGroove
r
d= .103
D
d= 1.2
3
29=
35
29=
Kt
= 2.33
Usingr = 3andSut= 620 MPa,
q(notch sensitivity) = .83
Kf= 1+ (Kt1)q = 1 + .83(2.331) = 2.1
(35-6)5
Calculate the alternating and the
mean stresses,
a=
Pa
A=
4000= 58.0 MPaKf 2.1
m=
Pm
A=
6000
29x5= 41.4 MPa
n= 2.57n180
1=
620
58.0 41.4+
The part is likely to fail at the hole, has the lowest safety factor
E l
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Ken Youssefi MAE dept., SJSU 45
Example
Fa=(FmaxFmin) / 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= 150ksi. The mounting
detail is such that the stress concentration couldbe 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=
Mc
I=
32Ma
d3 =
32(120)
(.375)3 = 23178.6 psi
m=
Mc
I=
32Mm
d 3=
32(360)
(.375)3= 69536 psi
E l
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Ken Youssefi MAE dept., SJSU 46
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= .010462d
2
(non-rotating round section)
dequiv= A95/ .0766 = .37d = .37 x.375= .14dequiv= .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=
150000
23178.6 69536+ n = .7< 1
Finite life
Sn1=
Sut
a m
+
Find Sn, strength for finite number of cycle
Sn1=
150000
23178.6 69536+ Sn= 43207 psi
E l
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Example
Sn = Se (N
106)
(Se
.9Sut)log
43207 = 24077 (N
106)
(24.077
.9x150)log
N = 96,000 cycles
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