Chapter 6
Fatigue Failure Resulting from
Variable Loading
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Introduction
F
Non rotating shaft
Under static failure, the stress on the member is constant
Problem: static failure, a very large deformation will occur on the structure or machine
members.
.
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F
F
Rotating shaft
- F to F
F
-F
Non rotating shaft
Alternating or fluctuating stresses on member will cause the member subjected to fatigue
failure.
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R. R Moore Rotating Beam: Figure 6-9 pp 274
87 mm
7.6 mm
250 mm
Testing specimen
Subject to pure bending (no traverse shear)
Very well machined and polished
No circumferential scratches
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Fatigue Strength and Cycle Graph (S-N Graph): Figure 6-10 pp274
103
Low Cycle
High Cycle
Cycle
Se
Fatigue Strength Sf
Finite Life
Infinite Life
Life
A
Endurance
limit Se
NA
100
101
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102
103
104
105
Cycle, N
106
107
108
Terminology
N : Cycle : one rotation of the specimen = 1 cycle of alternating stress
Sf : fatigue strength is the limit of strength where failure occurs when the alternating
stress above the fatigue strength. However, when cycle is larger than 106, the fatigue
strength is constant to a value Se. This value is called Endurance Limit
Cycle: Cycle is related to the cycle of the specimen: boundary is 103
Life: The life is related to the stress of the specimen: boundary is Se.
Significance to design
 Designers have a choice to set the design life whether finite life or infinite life.
 Designers can predict the failure to occur (finite life): i.e. when the stress of the
member is sA the predicted life will be NA cycles.
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Endurance limit
Steel
 0.5Sut
S 
700MPa
'
e
 0.5Sut
Se'  
100kpsi
Sut  1400 MPa
Sut  1400 MPa
Eq 6.8 pp 282
Sut  200 kpsi
Sut  200kpsi
Note: Sut = 3.41 HB Mpa
Eq 2.17 pp 37
Notes:
Various class of cast iron, polished and machined: Table A-24 pp 1046
Aluminum alloy: does not have an endurance limit, fatigue strength from Table A-24 is set
to 5(108) cycles
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Relationship between fatigue strength Sf and cycle N
S f  aN b
 
N  a
 a 
where
1
b
Eq 6.13 and 6.14 pp 285
a: slope of the curve (log vs log)a1
b: Sf @ 103
f S ut
1
b   log(
)
3
Se
(f S ut ) 2
a
Se
Note: f is fatigue strength factor @103
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Eq 6.14 and 6.15 pp 285
How to determine fatigue strength factor f @103
1.
From Figure 6.18 pp 285
2.
Calculation
f 
 'F
Sut
where
(2  10 3 ) B
log( 'F / Se )
log( 2  Ne )
'F  Sut  345 Mpa
B
Eq 6-10 pp 284
Eq 6.12 pp 284
Eq 6.11 pp 284
 'F  Sut  50 kpsi
*Please differentiate between b (6.15) and B (6.12)
General equation for fatigue strength
Sf   ' F (2N )B
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Eq 6-9 pp 284
ENDURANCE LIMIT (Se)
Se  ka kb kc kd keS'e
where
S’e : endurance limit based on R.R. Moore experiment
ka: surface factor
kb: size factor
kc: loading factor
kd: temperature factor
ke:miscellaneous-effects factor
Discuss:
Why do we need these variables?
What do you expect the value of these variables and why?
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Example
Calculate the endurance limit Se, if Material AISI 1010 HR, surface is machined until D =
30 mm, d = 20mm and r (fillet radius) = 2 mm is achieved.
d
D
r
Surface Factor ka
k a  aSutb
Eq 6.19 pp 287
a and b: Table 6-2 pp 280
Discuss: should the factor a and b be hot rolled or machined.
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Size Factor kb
1.24 d 0.107
kb  
0.157
 1.51d
0.879 d 0.107
kb  
0.157
 0.91d
2.79  d  51mm
51  d  254mm
Eq 6.20 pp 288
0.11  d  2in
2  d  10in
Note: only applicable to round rotating shaft due to torsion or bending
Other conditions
i.
Axial load
kb = 1 (for axial load)
* Please refer to kc
ii.
Round non-rotating shaft
95% stress area is applied
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A 0.95   0.0766d2
A0.95  0.0105D 2
0.0766d e  0.0105D 2
2
d e  0.37D
iii.
For non circular non rotating beam
iv.
Common non-rotating structural
d e  0.808(hb)0.5
*Refer to Table 6-3 pp 290 for common non rotating structural
Loading factor kc
Bending
Axial
kc = 1
kc = 0.85
Temperature Factor
Refer to Table 6-4 pp 291
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Eq 6.26 pp 290
Reliability Factor (ke)
Refer to Table 6-5 pp 293
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Stress concentration and notch sensitivity
See Fig 6.3, 6,4 : fatigue failure originates from stress concentration
Table A-13: Charts for theoretical Stress Concentration Factor (Kt)
For the example above the suitable chart is Figure A-13-9
3.0
Figure A-13-9
2.6
Roundshaft with shoulder fillet
Mc
inbending, o 
,where
I
d
d4
c  ,andI 
Kt
2
64
2.2
D/d=1.5
1.8
1.54
1.4
geometrical property to be used
in calculating the stress.
1.0
0
0.05
0.10
0.15
r/d
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0.20
0.25
0.30
Fatigue stress concentration factor (Kf)
K f  1  q(K t  1)
Eq 6.32 pp 295
Kts: shear stress concentration factor
Kfs: fatigue shear stress concentration factor
K f s  1  qshear (K ts  1)
Eq 6.32 pp 295
q, qshear : notch sensitivity from Figure 6-20 and 6-21 respectively.
 If the value of notch sensitivity can not be retrieved from the respective figures, the
most conservative assumption is that Kf = Kt or Kfs = Kts
Or using calculation
K f  1
KT  1
1
a
r
And Refer to 6-15 pp 335
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Eq 6-33 pp 296
Therefore the maximum stress will be
max  K f 0
or
Studio: example 6-17
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max  K f 0
Eq 6.30 pp 285
Combination of loading modes
Combination purely alternating bending, purely alternating axial and alternating
torsional stresses
Alternating axial stress
( axial)  K f ( axial)  0( axial)
Alternating bending stress
 (bend )  K f (bend )  0(bend )
Alternating torsional stress
 ( torsion)  K f s( torsion)  0( torsion)
Total alternating stress (DET)
2
( axial) 


  3( ( torsion) )2
 t   (bend ) 
0.85 

*Please note that kc for axial and bending are 0.85 and 1 respectively
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Application of S-N curve to fatigue design.
Infinite life
 all 
Se
 T
n
Finite life
S f  aNb
 
N a 
 a 
1
b
where
Discuss Example 6-8 and 6-9
Question 6-11
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a 
T
n
Contoh
Y
X
A
Z
Rajah 1
Rajah 1 menunjukkan sebahagian dari pemegang untuk satu mesin. Pemegang merupakan rod yang dikimpal di A pada mesin
tersebut. Rod ini diperbuat dari AISI 1020 HR dan berdiameter 15mm.
Jika momen pada A menghasilkan tegasan lenturan ulang-alik a =30 MPa. Daya paksi di A menghasilkan tegasan paksi ulang-alik
(avea 20MPa.
Berdasar kepada data di atas jawab soalan berikut
a)
b)
Kirakan ketahanan lesu (endurance limit) Se bahan tersebut
Kirakan faktor keselamatan jika tumpuan tegasan yang disebabkan oleh kimpalan ialah Kf untuk lenturan = 1.7 dan Kf untuk
daya paksi = 1.2.
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Stress
Fluctuating Stresses
max : max imumstress
a : amplitude stress
m : midrange stres
min : minimumstress
Time
Midrange Stress
Alternating Stress
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 max   min
2
   min
 a  max
2
m 
Similarly for the fluctuating forces
Midrange Force
Alternating Force
Fmax  Fmin
2
F  Fmin
Fa  max
2
Fm 
These forces will generate respective midrange and alternating stresses.
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How to determine the midrange or alternating stress?
Moment on rotating shaft.
Stress
2
2
3
A
I
M
4
1
2
1
3
4
1
3
Time
4
When the shaft rotates in ccw direction, element A will rotate from position 1, 2, 3 and 4
for one full cycle. At the same time, the stresses of A will fluctuate from 0, smax, 0 and
smin. Therefore, moment on rotating shaft will generate sa = Mc/I and sm = 0.
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3
Stress
Torsion on rotating shaft
2
A
I A
4
ta = 0 and tm = Tr/J.
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Time
Determine the alternating and midrange stresses for the
following cases?
Rotating shaft
i.
Axial load (P)
ii.
Combination of T and M
Non rotating shaft
Moment (M)
Torsion (T)
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between -10Nm and 10Nm
20 Nm and 50 Nm
-30 Nm and 10Nm
between 10Nm and 10Nm
20 Nm and 50 Nm
-30 Nm and 10Nm

min
Stress Ratio R  
max
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
a
Amplitude Ratio A  
m
Fatigue Failure Criteria: graphical and empirical method
a
Load Line
Langer Line
Sy
A : (  a , m )
B : (a1, m1 )
C : ( a2 , m2 )
Load line
D
C
Se
D : ( a3 , m3 )
Gerber Line
Modified Goodman Line
B
A
Sy
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Sut
m
Graphical Method
Goodman line
n
Sa1 Sm1

a
m
Gerber Failure Theory
n
S a 2 Sm 2

a
m
Langer Line
n
S a 3 S m3

a
m
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Empirical Method
Goodman Line
Sm S a

1
Sut S e
Basic line equation
m
For a known (sa, sm)
Sut

a
Se

Gerber Failure Locus (Parabolic Equation)
 Sm

Basic parabolic equation  S
 ut
2

S
  a  1
Se

2
For a known (sa, sm)
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 n m 
n a

 
1
Se
 S ut 
1
n
Langer Line (failure due to yielding)
Basic line equation
Sm Sa

1
Sy Sy
m  a 1


Sy Sy n
For a known (sa, sm)
Sy
 a  m
n
Other failure theories, same principles can be applied.
*In the following discussion, these three theories will be used to demonstrate and
similar principle can be applied to other theories.
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Langer line vs. other theories (i.e. Modified Goodman and Gerber Theory)
a
a
Sy
Sy
Se
Se
Sy
Sut
m
Sy
Sut
m
If the stresses on the member is plotted on the shaded area, the load line of the member
will intersect with Langer line first. Therefore, failure due to yielding will occur first. This is
type of failure is called Langer-first-cycle failure. Empirically, it can be determined when
nlanger < nf(other).
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Procedure in solving the fatigue problem
1.
Determine sa, sm
2.
Combine the alternating stresses using DET
3.
Combine the mean stresses using DET
4.
Using the theories to solve the problem
Question 6-39 (pp 352)
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ta and tm.
Higher Education
a
a
Sy
Sy
Se
Se
Sy
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Sut
m
Sy
Sut
m
103
Low Cycle
Cycle
High Cycle
Se
Fatigue Strength S f
Finite Life
Infinite Life
Life
A
Endurance limiteS
100
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101
102
103
104
NA
105
Cycle, N
106
107
108
Notes
Purely alternating bending stresses (  0 )
Se
0
n
S e  k a k d k c k d k e S 'e
 all 
 k a k d kc k d
1
S 'e
Kf
Combination purely alternating bending and purely alternating axial stress
Alternating axial stress
 o (a )
Alternating bending stress
 o (b )
Total alternating stress
 o   o ( a )   o (b )
 all 
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Se
0
n
What is the value of kc and Kf as both factors have different values
Bending
Axial :
:
kb (eq) kc = 1 and Kf from bending
kb, k c
 1.23S ut0.0778 and Kf from axial
Soln : let kb,kc and Kf is common in Se which is equal to 1
Axial stress (assuming n =1)
 all 
Se
  0a
n
k a kb ( a ) kc ( a ) k d
1
K f (a)
S 'e   o ( a )
kb  1
k a 11k d 1S 'e 
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K f (a)
kc ( a )
 o(a)
Bending stress
 all 
Se
  ob
n
k a kb ( b ) k c ( b ) k d
k a 11k d 1S 'e 
1
K f (b )
S 'e   o ( b )
K f (b )
k c ( b ) kb ( b )
 o (b )
Final equation with factor of safety n
K f (b )
k a k d S 'e K f ( a )


 o(a) 
 o (b )
n
kc ( a )
k c ( b ) kb ( b )
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CONTOH
Y
X
A
Z
Rajah 1
Soalan
Rajah 1 menunjukkan sebahagian dari pemegang untuk satu mesin. Pemegang merupakan rod yang dikimpal di A pada mesin
tersebut. Rod ini diperbuat dari AISI 1020 HR dan berdiameter 15mm.
Jika momen pada A ialah (30 10%)Nm dan daya paksi ialah (8 10%) kN. Untuk memastikan getaran diambil kira, 10%
ditambah.
Berdasar kepada data di atas jawab soalan berikut
c) Kirakan ketahanan lesu (endurance limit) Se bahan tersebut
d) Tentukan a, m,a dan m jika faktor tumpuan tegasan yang disebabkan oleh kimpalan ialah Kf untuk lenturan = 1.7 dan Kf
untuk daya paksi = 1.2.
Kirakan faktor keselamatan terhadap kegagalan alahan dan kegagalan lesu. Yang mana akan gagal dahulu : alahan atau lesu?
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