lec10lec11fatigueanalysis-161211121313.pdf

Aero Structure-Fatigue Analysis
By
Dr. Mahdi Damghani
2016-2017
1

Suggested Readings
Reference 1 Reference 2
2
Chapter 10 of Ref [1], Chapter 6 of Ref [2], Chapter 4 of Ref [3]
Reference 3

Topics
• Familiarisation with fatigue of metals
• Life estimates
• Endurance limit
• Fluctuating stresses
• Stress concentration and Notch sensitivity
3

Your Questions
• Feel free to ask your questions in the following link:
https://padlet.com/mahdi_damghani/Fatigue_Analysis
4

Introduction
• In most testing of those properties of materials that relate to the
stress-strain diagram, the load is applied gradually, to give
sufficient time for the strain to fully develop.
• Furthermore, the specimen is tested to destruction, and so the
stresses are applied only once
• Testing of this kind is applicable, to what are known as static
conditions
• The condition frequently arises, however, in which the stresses
vary with time or they fluctuate between different levels.
• For example, a particular fiber on the surface of a rotating shaft
subjected to the action of bending loads undergoes both tension
and compression for each revolution of the shaft.
5

Introduction
6
Taken from:
http://science.howstuffworks.com/transport/flig
ht/modern/air-traffic-control1.htm
Push back from
the gate, taxi to
the runway
Speeding down
the runway
Climbing to
cruising altitude Descending and
Manoeuvring to
the destination
Aligning the
plane with
runway
Landing, taxis to the
gate and parking at
terminal

Introduction
• Some cyclic loads in aircrafts;
• Push back (on landing gear)
• Turning (high stresses on landing gear)
• Taxy
• Take off
• Cabin pressurisation
• Landing impact
• Undercarriage loading
• Global/local turbulence
• Gust and manoeuvre
7

Introduction
• Often, machine members are found to have failed
under the action of repeated or fluctuating stresses
• The most careful analysis reveals that the actual
maximum stresses were well below the ultimate
strength of the material, and quite frequently even
below the yield strength
• The most distinguishing characteristic of these
failures is that the stresses have been repeated a
very large number of times. Hence the failure is called
a fatigue failure
8

Introduction
• Fatigue failure
• Gives no warning (no large deflection etc) and it is very
sudden=brittle fracture
• Dangerous (because it is sudden)
• Complicated phenomenon (partially understood)
• Life of a component must be obtained based on empirical
methods
9

Introduction
10
Growth of surface micro-
cracks of aircraft towbar
as the result of fatigue
loading leading to failure

Introduction
12
Fatigue crack at stress
raiser (discontinuity, i.e.
hole) due to stress
concentration

Introduction
• Micro-cracks formation on the surface of part due to
cyclic plastic deformations
• During cyclic loading, these cracked surfaces open
and close, rubbing together
• The beach mark appearance depends on the
changes in the level or frequency of loading and the
corrosive nature of the environment
• Cracks keep growing up to a critical length and
suddenly part fails
13

Fatigue analysis approaches
• Stress-life method
• Will be covered in the lecture
• Strain-life method
• Beyond the scope of this lecture
• Linear elastic fracture mechanics method
14
• Generally speaking we have;
• Low cycle fatigue (1<=N<=103)
• High cycle fatigue (N>103)

Stress-life method
• The least accurate method
• Specially for low cycle fatigue
• Good approximation for high cycle fatigue
• Assumes little plastic deformation due to cyclic loading
• The most widely used since mid-1800s
• A lot of data and understanding is available
• Good predictions for high cycle fatigue
• Easy to perform
15

Strain-life method
• Detailed analysis of the plastic deformation at
localized regions where the stresses and strains are
considered for life estimates
• Good for low-cycle fatigue
• Several idealizations must be compounded, and so
some uncertainties will exist in the results
16

Fracture mechanics method
• The fracture mechanics method assumes a crack is
already present and detected
• It is then employed to predict crack growth with
respect to stress intensity
• It is most practical when applied to large structures in
conjunction with computer codes and a periodic
inspection program
• Currently Airbus uses this method in combination with
stress-life method
17

Stress-life method
• To determine the strength of materials under the
action of fatigue loads, specimens are subjected to
repeated or varying forces of specified magnitudes
while the cycles or stress reversals are counted to
destruction
• Therefore graphs of subsequent slides (S-N
diagrams) can be produced
18

Fatigue machine (for R=-1 only)
• R. R. Moore high-speed
rotating-beam machine
• Specimen is subjected to
pure bending (no
transverse shear) by
means of weights
20
• The specimen is very carefully machined and
polished, with a final polishing in an axial direction to
avoid circumferential scratches and minimize surface
roughness

R. R. Moore machine
21
• When rotated one half
revolution, the stresses in
the fibers originally below
the neutral axis are
reversed from tension to
compression and vice
versa
• Upon completing the
revolution, the stresses
are again reversed so that
during one revolution the
test specimen passes
through a complete cycle
of flexural stress (tension
and compression).

Electro-hydraulic axial fatigue
machine (for all R values)
22

Output from fatigue test (see Ref [3])
23
The number of cycles to failure is called
fatigue life Nf
Each cycle is equal
to two reversals
Stress in the component some
books use symbol σ instead
time
Stress

Fluctuating stresses
24
Stress ratio
Amplitude
ratio
2
2
2
min
max
min
max
S
S
S
S
S
S
S
m
r
a





min
max S
S
Sr 

R
R
S
S
A
m
a




1
1
max
min
S
S
R 
component
amplitude
stress
maximum
stress
minimum



a
S
S
S
max
min
stress
of
range
stress
midrange


r
m
S
S

S-N/Wohler diagram (see Ref [2])
25
Results of completely
reversed axial fatigue tests.
Material: UNS G41300 steel,
normalized; Sut = 116 kpsi;
maximum Sut = 125 kpsi.
Often plotted in log-log or semi-log graphs. If
plotted in log-log, y axis is in terms of stress
amplitude or stress range and x axis in terms
of number of reversals or cycles to failure

How SN curve is established
26
w1
w1
w1
3
1
min
max
,
25
.
0
max
2
4
r
a
nw
I
My r
y
r
I



 






 

 


S-N bands for Aluminium alloys for
completely reversed cycling
27

Fatigue strength in log-log S-N curve
(Sf)
28













e
e
S
S
S
S
b 1000
7
3
1000
log
3
1
10
log
10
log
log
log
Endurance limit
Fatigue life, i.e. life
required to nucleate
and grow a small
crack to visible crack
length
 b
f
a N
a
S 
e
S
a and b are
constants for
cycles 103-107
1000
S
Cycles to
the failure
a is fatigue
strength when Nf
is 1 cycle only
7
10

Note
• S-N curve in the
previous slide was for
one load ratio only
(R=Smin/Smax) but in
practice we have so
many load ratios (R) so
we need more graphs
29

Note
30
• Real S-N curve for
various load ratios
for un-notched
plate of Aluminium
7050-T7451
tested in long
transverse
direction
Source:
Metallic Materials Properties Development and Standardization (MMPDS)

Endurance limit of steel vs aluminium (Se)
• Endurance limit
• The maximum
stress which can
be applied to a
material for an
infinite number of
stress cycles
without resulting in
failure of the
material
31
See how steel (graph A)
has pronounced endurance
limit whereas aluminium
(graph B) does not have a
very clear endurance limit

Endurance limit (Se)
• For example for steel material the endurance limit can
be obtained by the following relationship;
• Note that this is achieved from experiments and is for
fatigue test specimen not real life loading
32

Endurance limit (Se)
• In real life endurance limit can be different due to
many factors (see Ref [2] for more details);
33

Fatigue prone locations
• The following locations in the structure are prone to
fatigue failure (formation of crack) due to stress
concentration and therefore illustrated in subsequent
slides;
• Holes
• Notches
• Lugs
• Pins
• Fillets
• Joints
34

Stress Concentration and Notch
Sensitivity
• Any discontinuity in a machine part alters the stress
distribution in the immediate vicinity of the
discontinuity
• Elementary stress equations no longer describe the
state of stress in the part at these locations
• Such discontinuities are called stress raisers,
• Regions in which they occur are called areas of stress
concentration
35

Stress concentration
36
Notch
Hole
Hole

Stress concentration in aircraft
37

Stress Concentration around a circular hole subjected to
remote loading in an infinite composite panel
38

Stress concentration factor
• A theoretical, or geometric, stress-concentration
factor Kt or Kts is used to relate the actual maximum
stress at the discontinuity to the nominal stress
(remote stress)
• Stress concentration factor depends on;
• The geometry of structure
• Type of loading (uni-axial, bi-axial, bending moment etc)
39

How to get stress concentration value?
• Experimental procedures
• Photoelasticity
• grid methods
• brittle-coating methods
• Electrical strain-gauge methods
40
• Finite element analysis
• Not exact
• Analytical approaches based on conformal mapping
and complex algebra

Stress concentration for a shaft under
bending load and axial load
43

Stress concentration for a notched shaft
under bending load and axial load
44

Stress concentration for a prismatic bar
under bending load and axial load
45

Stress concentration for a plate with a central
hole under bending load and axial load
46

Fatigue Stress Concentration Factor
47
Fatigue stress
concentration factor
Notch sensitivity
1
0 
 q
In analysis or design work, find Kt
first, from the geometry of the part.
Then specify the material, find q, and
solve for Kf from the equation

Note
• In using these charts it is well to know that the actual
test results from which the curves were derived
exhibit a large amount of scatter
• It is always safe to use Kf = Kt if there is any doubt
about the true value of q
• Often q is not far from unity for large notch radii
50

Industrial examples of fatigue critical
locations (lugs in flap track)
51

locations (fillets in flap track)
52
Fillet A

locations (buttstraps)
53

54

55
Cover
Panel
Buttstrap

locations (strut brackets)
56

locations (baffle panels)
57

locations (baffle panels)
58

locations (joints)
59

Example 1
• A steel shaft in bending has an ultimate strength of
690MPa and a shoulder with a fillet radius of 3mm
connecting a 32mm diameter with a 38mm diameter.
Estimate Kf .
60

Solution
61
093
.
0
32
/
3
/
18
.
1
32
/
38
/




d
r
d
D
65
.
1

t
K

Solution
62
84
.
0

q
 
  55
.
1
1
65
.
1
84
.
0
1
1
1







f
t
f
K
K
q
K

Example 2
• A plate with thickness h=4 mm is subjected to an
alternating load P. The plate is subjected to a load cycle
Pmin=10 kN and Pmax=80 kN. Considering the notch
sensitivity factor of q = 0.9 calculate the life of panel.
Assume a=1600 MPa and b=-0.2 and material ultimate
strength is 600 MPa.
63

Solution
64
We know:
 b
f
a N
a
S 
2
2
2
min
max
min
max
S
S
S
S
S
S
S
m
r
a









 MPa
A
P
S 80
250
4
80000
max
max 



 MPa
A
P
S 10
250
4
10000
min
min
MPa
Sa 35
2
10
80



5
.
2
2
.
0
250
50 


 t
K
w
d
  35
.
2
1
5
.
2
9
.
0
1 



f
K
  

 2
.
0
1600
35 f
N cycles
N f 24
.
199645576


Solution
• Solve the same question assuming that the hole
exists on a plate with infinite length and width
• What is the difference between this situation and
previous situation?
65

Example 3
• Calculate maximum and minimum stresses and the
load ratios for the following set of cycles, expressed
as a stress range and mean stress.
66
20 30
25 12.5
33 11.5
11 29.5
60 20
90 25
)
(ksi
S
 )
(ksi
Sm

Solution
• Let’s calculate for one entry only;
67





max
min
min
max ,
S
S
R
S
S
S
20 40
0 25
-5 28
24 35
-10 50
-20 70
)
(
min ksi
S )
(
max ksi
S
 




 R
S
RS
S
S 1
max
max
max  
R
S
S



1
max






2
, min
max
min
max
S
S
S
S
S
S m









min
max
max
min
2 S
S
S
S
S
S
m




 min
min
2 S
S
S
Sm



 min
2
2 S
S
Sm min
5
.
0 S
S
Sm 


ksi
S 20
20
5
.
0
30
min 



ksi
S
S
S 40
20
20
min
max 






Example 4
• Use the following equation to calculate the number of
cycles to failure for the maximum stress and load
ratios calculated in Example 3. Note that the equation
is for Aluminium 2024-T3 material for Kt=1.
68
 
 
52
.
0
max 1
log
09
.
9
83
.
20
log R
S
N 



Solution
• Let’s calculate for one entry only;
69
20 40 0.5 48998554
0 25 0.0 132655518
-5 28 -0.18 21779524
24 35 0.69 1480700983
-10 50 -0.2 102823
-20 70 -0.29 3485
)
(
min ksi
S )
(
max ksi
S R f
N
48998554
10
52
.
0
40
20
1
40
log
09
.
9
83
.
20


















N

Tutorial 1
• A rotating shaft simply supported in ball bearings at A
and D and loaded by a nonrotating force F of 6.8 kN.
Estimate the life of the part assuming;
• All shoulder fillets have radius of 3mm
• Sut = 690MPa and Sy = 580MPa (ultimate and yield strength)
• Se = 236MPa (endurance limit)
• q=0.84
• a=1437MPa
• b=-0.1308
70

Solution for Tutorial 1
• The first task is to see where the fatigue failure could
potentially happen
• We know fatigue usually happens at discontinuity,
notches, fillets etc where stress is high
• In this problem potential places are points B and C
• So we investigate these locations
71

72
  ?
1
1
_ 


 t
B
f K
q
K
?
_ 
B
t
K
 
  55
.
1
1
65
.
1
84
.
0
1
1
1
_






 t
B
f K
q
K
We solved this in
previous example,
see below

• Let’s calculate fatigue stress concentration at point C,
too.
73
086
.
0
35
/
3
/
08
.
1
35
/
38
/




d
r
d
D
6
.
1
_ 
C
t
K

74
84
.
0

q
 
  5
.
1
1
6
.
1
84
.
0
1
1
1
_
_







C
f
t
C
f
K
K
q
K

75
kN
R 78
.
2
8
.
6
550
225
1 
 kN
R 02
.
4
78
.
2
8
.
6
2 


)
.
(
5
.
695
250
78
.
2 m
N
MB 


)
(
10
217
.
3
32
/
32
32
/
3
3
3
3
max
mm
d
y
I
CB




 

MPa
C
M
K
S B
B
f
B 1
.
335
10
217
.
3
5
.
695
55
.
1 6
_ 


 
MPa
C
M
K
S C
C
f
C 1
.
179
10
209
.
4
5
.
502
50
.
1 6
_ 


 
Based on these
stresses where does
fatigue likely to
happen?
)
(
10
209
.
4
32
/
35
32
/
3
3
3
3
max
mm
d
y
I
CC




 

Based on these
stresses where does
fatigue likely to
happen?
)
.
(
5
.
502
125
02
.
4 m
N
MB 



76
MPa
C
M
K
S B
B
f
B 1
.
335
10
217
.
3
5
.
695
55
.
1 6
_ 


 
MPa
C
M
K
S C
C
f
C 1
.
179
10
209
.
4
5
.
502
50
.
1 6
_ 


 
Based on these
stresses where does
fatigue likely to
happen?
These stresses are both
greater than endurance
limit of 236MPa and
yield stress of 580MPa.
What does this mean?
We have finite life and
material does not
yield

77
  

b
f
a N
a
S








b
a
f
a
S
N
1
)
(
68000
1437
1
.
335 1308
.
0
1
cycles
N f 









Tutorial 2 (real industrial problem)
• Flap track of an aircraft undergoes 414 occurrences of
cyclic loading that brings about stresses at fillet locations.
Stresses are obtained from detailed finite element analysis
for unit load cases as shown in the next slide. Fatigue life
of component follows the relationship
in which A=10.7, B=3.81, C=10 and
It can be further assumed that minimum stresses are zero
and KDF=0.651 (knock down factor due to surface
treatment). Note that above equations are in ksi units.
Calculate the life of component assuming that the fatigue
load on flap is 8841.8 N
78
 
C
S
B
A
N eq
f 

 log
log
 
max
min
max
,
1
S
S
R
R
KDF
S
S
D
eq 



Tutorial 2 (real industrial problem)
79
Note: Stresses are in MPa
and obtained for flap load of
1000N (unit load)

80
• Note that max stress need to be adjusted based on
surface treatment
• In this example we did not need to use Kf as we are
directly extracting max stresses from detailed FEA
0
145
8
.
8841
1000
4
.
16
0
max
min











R
MPa
S
S
  ksi
S
MPa
S eq
D
eq 3
.
32
89
.
6
7
.
222
7
.
222
0
1
651
.
0
0
.
145






   







 C
S
B
A
f
eq
f
eq
N
C
S
B
A
N
log
10
log
log
 
cycles
N f 83
.
365559
10 10
3
.
32
log
81
.
3
7
.
10

 


Important Notes [3]
• Fatigue damage of components correlates strongly with;
• Applied stress amplitude/range
• Applied mean stress (mostly in high cycle fatigue region)
• In high cycle fatigue, direct (normal) mean stress is
responsible for opening and closing micro-cracks
• Normal tensile mean stresses are detrimental and
compressive mean stress are beneficial for fatigue
strength (why?)
• Shear mean stress has little effect on crack propagation
• There is no/little effect of mean stress in low cycle fatigue
due to large amount of plastic deformation
81

Important Notes [3]
• So far, what we have been doing was based on not taking
into account the effects of tensile normal mean stresses on
the high cycle fatigue strength of components
• We have been assuming Sm=0
• Therefore, below scientists proposed empirical equations
to take such effect into account;
• Gerber (1874)
• Goodman (1899)
• Haigh (1917)
• Soderberg (1930)
82

Influence of tensile normal mean stress on
fatigue strength
• To compensate and understand the influence of tensile normal
mean stress on high cycle fatigue strength, several empirical
plots can be established (constant life plot as below and also see
next slide)
83

Example of constant life diagrams
84
• This can be obtained from S-N diagram
Increase in life
2
2
2
min
max
min
max
S
S
S
S
S
S
S
m
r
a





This is SN curve for
various mean stress
values

Fatigue failure criteria for fluctuating
stresses (Haigh plot)
85
Mid-range
strength
Yield
strength
Ultimate tensile
strength
Effective alternating stress at
failure for a life time of Nf cycles
(modified fatigue strength)

Note
86
Safe
Unsafe
Infinite life

Note
• Extension of previously mentioned fatigue criteria will
allow the use of Sar instead of Se
• Sar is a fully reversed stress amplitude corresponding
to a specific life in the high-cycle fatigue region
87
1


y
m
ar
a
S
S
S
S
1
2










ut
m
ar
a
S
S
S
S 1


ut
m
ar
a
S
S
S
S
1
2
2


















y
m
ar
a
S
S
S
S

General observations
88
Most actual test data tend to
fall between the Goodman
and Gerber curves
For most fatigue situations R<1 ( i.e. small
mean stress in relation to alternating stress),
there is little difference in the theories
In the range where the theories show large
differences (i.e. R values approaching 1)
there is little experimental data
The Soderberg line is very conservative and
seldom used

Complex loading and cycle counting
• Instead of a single fully
reversed stress history block
composed of n cycles,
suppose a machine part, at a
critical location, is subjected to
either of;
• A fully reversed stress S1 for n1
cycles, S2 for n2 cycles
• A “wiggly” time line of stress
exhibiting many and different
peaks and valleys
89

Complex loading and cycle counting
90
Source: N.E. Dowling, Mechanical behaviour of
materials, 3rd edition, (Pearson / Prentice hall)
• What stresses are significant?
• What counts as a cycle?
• What is the measure of damage incurred?

Methods of cycle counting
• Interested readers are recommended to refer to
chapter 3 of Ref. [3];
• Level crossing cycle counting
• Peak-valley cycle counting
• Range counting
• Three-point cycle counting method
• Four-point cycle counting method
• Rain-flow counting technique
91

Output form cycle counting
• A typical result of cycle counting would look like;
92
After cycle counting it
becomes fairly straightforward
to calculate stress range and
mean stress values and
proceed as normal

Cumulative damage
• Palmgren-Miner cycle-ratio summation rule (1924), also
called Miner’s rule
• where ni is the number of cycles at stress level Si and
Ni is the number of cycles to failure at stress level Si
• The parameter D has been determined by experiment
• Usually 0.7<D<2.2 with an average value near unity
93

Note on cumulative damage
• Damage parameter (D), that is defined in previous
slide, is the ratio of instantaneous to critical crack
length, i.e. D = a / af
• There are other damage models in the literature that
are not linear such as those proposed by
Subramanyan (1976) and Hashin (1980) as below;
95

Example 3
• A structural member is to be subjected to a series of
cyclic loads which produce different levels of
alternating stress as shown in table below. Determine
whether or not a fatigue failure is probable.
96

Solution
• We use miner’s cumulative damage theory as below;
97

 
i
i
N
n
D 




4
4
3
3
2
2
1
1
N
n
N
n
N
n
N
n
D
1
39
.
0
10
12
10
10
24
10
10
10
10
5
10
7
7
7
6
6
5
4
4









D

Example 4
• Calculate the total number of repetitions required for
the loading of example 4 to reach the fatigue failure
point.
98

Solution
99


 i
D
0
.
1
s
Repetition 56
.
2
39
.
0
0
.
1


s
Repetition

Example 5
• For the question of Tutorial 2 calculate damage for
the following cases;
• Damage calculation for single occurrence at each cycle
• Damage calculation for all occurrences at each cycle
• Damage calculation for 15000 hours (3000 hours per block)
100

Solution
• We know from solution that;
• Damage is calculated for single occurrence;
• Damage is calculated for all occurrences;
• Damage is calculated for 15000 hrs flight;
101
 
cycles
N f 83
.
365559
10 10
3
.
32
log
81
.
3
7
.
10

 


 
i
i
N
n
D 6
10
73
.
2
83
.
365559
1 



D

 
i
i
N
n
D 3
10
13
.
1
83
.
365559
414 



D

 
i
i
N
n
D 3
10
66
.
5
83
.
365559
414
3000
15000 









D
 


 C
S
B
A
N eq
f log
log

Solution
• Due to scatterings of fatigue data and also unknowns,
it is common place to use a factor of safety known as
scatter factor
• This factor can sometimes reach values of 8
• If we assume scatter factor of 8 then we have for the
damage;
102

 
i
i
N
n
D 2
10
53
.
4
83
.
365559
313
3000
15000
8 










D

Example 6
• The squared hollow box girder belongs to an aircraft and it is subjected
to a tensile force of 10P and a transverse shear force P at the beam’s
tip. The following cycles were recorded for the load P on the box girder
for one year of aircraft’s service:
The squared box girder has length L = 1000mm and the hollow squared
cross-section has an edge length of 100mm and uniform thickness of 2
mm.
• Calculate the second moment of area of the cross-section about the
horizontal axis passing through the centroid
• Assuming yield stress of material is 500MPa and using the Soderberg
fatigue criteria, determine the number of repetitions (years) to failure
due to fatigue.
(Note: consider the material parameters for the S−N curve as being A =
1800MPa and B = −0.2 and use the Miner’s rule for the most critical point
in the box girder).
103

Solution
105
)
(
6
.
52
10
255
.
1
50
1000
10
96
100
10
10
6
3
2
2
3
max MPa
P
P
P
S 








 0
min 
S


 1
y
m
ar
a
S
S
S
S










 1
1
y
m
ar
a
S
S
S
S


Solution
106


 b
f
ar aN
S
S 


Tutorial 3
107
• A solid shaft of circular cross-section with a diameter of 40mm is subjected to an
eccentric axial load F at a distance of 10mm from the center of the cross-section
and it is also subjected to a torque T. The loads are applied repeatedly and, for
each repetition, the following load pulsations and number of cycles applies:
Using the Soderberg fatigue criteria, calculate;
A) Damage to the component
B) Number of repetitions before the rod fails by fatigue
(Note: consider the material yield stress 420MPa and the following material
parameters for the S-N curve: a = 1600MPa and b = -0.2)
Cycle ID Fmin (kN) Fmax (kN)
Tmin
(kNm)
Tmax (kNm)
Cycles
repetition
1 20 100 0 0 5000
2 10 50 0 2 1000

109
MPa
I
y
M
A
F
9
.
238
40
64
20
10
10
100
40
4
10
100
4
3
2
3
max
max
max 















MPa
I
y
M
A
F
75
.
47
40
64
20
10
10
20
40
4
10
20
4
3
2
3
min
min
min 















MPa
a 6
.
95
2
75
.
47
9
.
238
2
min
max








MPa
m 3
.
143
2
75
.
47
9
.
238
2
min
max










 1
y
m
ar
a
S
S




 1
420
3
.
143
6
.
95
ar
S
MPa
Sar 11
.
145

Cycle
ID
Fmin (kN) Fmax (kN)
Tmin
(kNm)
Tmax
(kNm)
Cycles
repetition
1 20 100 0 0 5000

• From S-N curve we remember;
110


 b
f
ar aN
S
S 
  2
.
0
1600
11
.
145 N cycles
N 3
10
972
.
162 


111
MPa
I
y
M
A
F
45
.
119
40
64
20
10
10
50
40
4
10
50
4
3
2
3
max
max
max 















MPa
I
y
M
A
F
89
.
23
40
64
20
10
10
10
40
4
10
10
4
3
2
3
min
min
min 















MPa
D
T
D
D
T
J
Tr
15
.
159
40
10
2
16
16
32
2
3
6
3
4
max 










 0
min 

MPa
a 78
.
47
2
89
.
23
45
.
119
2
min
max








MPa
m 67
.
71
2
89
.
23
45
.
119
2
min
max








MPa
a 6
.
79
2
0
15
.
159




MPa
m 6
.
79
2
0
15
.
159




Cycle
ID
Fmin (kN) Fmax (kN)
Tmin
(kNm)
Tmax
(kNm)
Cycles
repetition
2 10 50 0 2 1000

112
• Since we have both shear and direct stress we convert
then into equivalent von-Mises stress
• Plug these into Soderberg criterion
• Finally we get;
MPa
xy
x
eq
a 9
.
145
6
.
79
3
78
.
47
3 2
2
2
2





 


MPa
xy
x
eq
m 4
.
155
6
.
79
3
67
.
71
3 2
2
2
2





 




 1
y
eq
m
ar
eq
a
S
S




 1
420
4
.
155
9
.
145
ar
S
MPa
Sar 6
.
231



 b
f
ar aN
S
S 
  2
.
0
1600
6
.
231 N cycles
N 3
10
74
.
15 


113
• Damage is calculated as;
• For part to fail we must have;
• This means that the component can go through;

 
i
i
N
n
D 094
.
0
10
74
.
15
1000
10
97
.
162
5000
3
3





D
1

 D
N n
repeatitio 614
.
10
1


D
N n
repeatitio
2
Cycle
for
1000
614
.
10
1
Cycle
for
5000
614
.
10
cycles
cycles



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