fatigue analysis of structures (aerospace application)
TRANSCRIPT
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Aero Structure-Fatigue Analysis
By
Dr. Mahdi Damghani
2016-2017
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Suggested Readings
Reference 1 Reference 2
Chapter 10 of Ref [1], Chapter 6 of Ref [2], Chapter 4 of Ref [3]
Reference 3
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Topics
• Familiarisation with fatigue of metals
• Life estimates
• Endurance limit
• Fluctuating stresses
• Stress concentration and Notch sensitivity
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Your Questions
• Feel free to ask your questions in the following link:
https://padlet.com/mahdi_damghani/Fatigue_Analysis
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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.
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Introduction
Taken from:http://science.howstuffworks.com/transport/flight/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
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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
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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
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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
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Introduction
Growth of surface micro-cracks of aircraft towbar as the result of fatigue
loading leading to failure
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Introduction
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Introduction
Fatigue crack at stress raiser (discontinuity, i.e.
hole) due to stress concentration
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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
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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• Beyond the scope of this lecture
• Generally speaking we have;• Low cycle fatigue (1<=N<=103)
• High cycle fatigue (N>103)
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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
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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
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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
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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
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Watch
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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
• The specimen is very carefully machined and polished, with a final polishing in an axial direction to avoid circumferential scratches and minimize surface roughness
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R. R. Moore machine
• 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).
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Electro-hydraulic axial fatigue machine (for all R values)
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Output from fatigue test (see Ref [3])
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
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Fluctuating stresses
Stress ratio Amplitude ratio
2
22minmax
minmax
SSS
SSSS
m
ra
minmax SSSr
RR
SSA
m
a
11
max
min
SSR
component amplitudestress maximum
stress minimum
aSSS
max
min
stress of rangestress midrange
r
m
SS
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S-N/Wohler diagram (see Ref [2])
Results of completelyreversed 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
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How SN curve is established
w1w1w1
31
minmax,25.0
max24
ranw
IMy ryrI
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S-N bands for Aluminium alloys for completely reversed cycling
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Fatigue strength in log-log S-N curve (Sf)
e
e
SSSSb 1000
731000 log
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10log10logloglog
Endurance limit
Fatigue life, i.e. life required to nucleate and grow a small crack to visible crack length
bfa NaS
eS
a and b are constants for cycles 103-107
1000S
Cycles to the failure
a is fatigue strength when Nf is 1 cycle only
710
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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
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Note
• 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)
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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
See how steel (graph A) has pronounced endurance limit whereas aluminium (graph B) does not have a very clear endurance limit
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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
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Endurance limit (Se)
• In real life endurance limit can be different due to many factors (see Ref [2] for more details);
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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
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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
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Stress concentration
NotchHole Hole
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Stress concentration in aircraft
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Stress Concentration around a circular hole subjected to remote loading in an infinite composite panel
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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)
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How to get stress concentration value?
• Experimental procedures• Photoelasticity
• grid methods
• brittle-coating methods
• Electrical strain-gauge methods
• Finite element analysis• Not exact
• Analytical approaches based on conformal mapping and complex algebra
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Stress concentration value
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Stress concentration value
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Stress concentration for a shaft under bending load and axial load
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Stress concentration for a notched shaft under bending load and axial load
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Stress concentration for a prismatic bar under bending load and axial load
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Stress concentration for a plate with a central hole under bending load and axial load
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Fatigue Stress Concentration Factor
Fatigue stress concentration factor
Notch sensitivity
10 q
In analysis or design work, find Kt first, from the geometry of the part. Then specify the material, find q, andsolve for Kf from the equation
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Notch sensitivity
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Notch sensitivity
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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
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Industrial examples of fatigue critical locations (lugs in flap track)
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Industrial examples of fatigue critical locations (fillets in flap track)
Fillet A
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Industrial examples of fatigue critical locations (buttstraps)
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Industrial examples of fatigue critical locations (buttstraps)
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Industrial examples of fatigue critical locations (buttstraps)
Cover
Panel
Buttstrap
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Industrial examples of fatigue critical locations (strut brackets)
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Industrial examples of fatigue critical locations (baffle panels)
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Industrial examples of fatigue critical locations (baffle panels)
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Industrial examples of fatigue critical locations (joints)
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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 .
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Solution
093.032/3/18.132/38/
drdD
65.1tK
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Solution
84.0q
55.1165.184.01
11
f
tf
K
KqK
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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.
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Solution We know:
bfa NaS
2
22minmax
minmax
SSS
SSSS
m
ra
MPaA
PS 802504
80000maxmax
MPa
APS 10
250410000min
min
MPaSa 352
1080
5.22.025050 tKw
d
35.215.29.01 fK
2.0160035 fN cyclesN f 24.199645576
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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?
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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.
20 30
25 12.5
33 11.5
11 29.5
60 20
90 25
)(ksiS )(ksiSm
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Solution
• Let’s calculate for one entry only;
max
minminmax , S
SRSSS
20 40
0 25
-5 28
24 35
-10 50
-20 70
)(min ksiS )(max ksiS
RSRSSS 1maxmaxmax RSS 1max
2
, minmaxminmax
SSSSSS m
minmax
maxmin
2 SSSSSS
m
minmin2 SSSSm
min22 SSSm min5.0 SSSm
ksiS 20205.030min
ksiSSS 402020minmax
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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.
52.0max 1log09.983.20log RSN
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Solution
• Let’s calculate for one entry only;
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 ksiS )(max ksiS R fN
4899855410
52.0
4020140log09.983.20
N
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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
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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
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Solution for Tutorial 1
?11_ tBf KqK
?_ BtK
55.1165.184.01
11_
tBf KqK
We solved this in previous example,
see below
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Solution for Tutorial 1
• Let’s calculate fatigue stress concentration at point C, too.
086.035/3/08.135/38/
drdD
6.1_ CtK
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Solution for Tutorial 1
84.0q
5.116.184.01
11
_
_
Cf
tCf
K
KqK
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Solution for Tutorial 1
kNR 78.28.6550225
1 kNR 02.478.28.62
).(5.69525078.2 mNM B
)(10217.3
32/3232/
33
33
max
mm
dyICB
MPaC
MKS BBfB 1.33510
217.35.69555.1 6
_ MPaC
MKS CCfC 1.17910
209.45.50250.1 6
_
Based on these stresses where does
fatigue likely to happen?
)(10209.4
32/3532/
33
33
max
mm
dyICC
Based on these stresses where does
fatigue likely to happen?
).(5.50212502.4 mNM B
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Solution for Tutorial 1
MPaC
MKS BBfB 1.33510
217.35.69555.1 6
_ MPaC
MKS CCfC 1.17910
209.45.50250.1 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
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77
Solution for Tutorial 1
bfa NaS
ba
f aSN
1
)(680001437
1.335 1308.01
cyclesN f
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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
CSBAN eqf loglog
max
minmax ,1SSRR
KDFSS D
eq
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79
Tutorial 2 (real industrial problem)
Note: Stresses are in MPa and obtained for flap load of 1000N (unit load)
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80
Solution for Tutorial 2
• 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
01458.8841
10004.16
0
max
min
R
MPaS
S
ksiSMPaS eqD
eq 3.3289.6
7.2227.22201651.0
0.145
CSBAfeqf
eqNCSBAN log10loglog
cyclesN f 83.36555910 103.32log81.37.10
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81
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
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82
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)
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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)
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84
Example of constant life diagrams
• This can be obtained from S-N diagram
Increase in life
2
22minmax
minmax
SSS
SSSS
m
ra
This is SN curve for various mean stress
values
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85
Fatigue failure criteria for fluctuating stresses (Haigh plot)
Mid-range strength
Yield strength
Ultimate tensile strength
Effective alternating stress at failure for a life time of Nf cycles(modified fatigue strength)
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Note
Safe
Unsafe
Infinite life
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87
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
1y
m
ar
a
SS
SS
12
ut
m
ar
a
SS
SS 1
ut
m
ar
a
SS
SS
122
y
m
ar
a
SS
SS
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General observationsMost 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
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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
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Complex loading and cycle counting
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?
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Methods of cycle counting
• Beyond the scope of this lecture
• 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
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Output form cycle counting
• A typical result of cycle counting would look like;After cycle counting it becomes fairly straightforward to calculate stress range and mean stress values and proceed as normal
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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
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Cumulative damage
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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;
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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.
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Solution
• We use miner’s cumulative damage theory as below;
i
i
NnD
4
4
3
3
2
2
1
1
Nn
Nn
Nn
NnD
139.01012
101024
101010
10510
7
7
7
6
6
5
4
4
D
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Example 4
• Calculate the total number of repetitions required for the loading of example 4 to reach the fatigue failure point.
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Solution
iD
0.1sRepetition 56.239.00.1sRepetition
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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)
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101
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;
cyclesN f 83.36555910 103.32log81.37.10
i
i
NnD 61073.2
83.3655591 D
i
i
NnD 31013.1
83.365559414 D
i
i
NnD 31066.5
83.365559414
300015000
D
CSBAN eqf loglog
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102
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;
i
i
NnD 21053.4
83.365559313
3000150008
D
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103
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).
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Example 6
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Solution
)(6.5210255.1
5010001096100
10106
3
22
3
max MPaPPPS
0min S
1y
m
ar
a
SS
SS
11
y
mara S
SSS
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Solution
bfar aNSS
![Page 107: Fatigue Analysis of Structures (Aerospace Application)](https://reader038.vdocuments.mx/reader038/viewer/2022102715/587866fb1a28ab18098b77d3/html5/thumbnails/107.jpg)
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Tutorial 3 • 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
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Tutorial 3
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Solution for Tutorial 3
MPaI
yMA
F 9.23840
6420101010040
4101004
3
2
3maxmax
max
MPaI
yMA
F 75.4740
642010102040
410204
3
2
3minmin
min
MPaa 6.952
75.479.2382
minmax
MPam 3.1432
75.479.2382
minmax
1y
m
ar
a
SS
1420
3.1436.95
arSMPaSar 11.145
Cycle ID Fmin (kN)
Fmax (kN)
Tmin (kNm)
Tmax (kNm)
Cycles repetiti
on1 20 100 0 0 5000
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Solution for Tutorial 3
• From S-N curve we remember;
bfar aNSS 2.0160011.145 N cyclesN 310972.162
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Solution for Tutorial 3
MPaI
yMA
F 45.11940
642010105040
410504
3
2
3maxmax
max
MPaI
yMA
F 89.2340
642010101040
410104
3
2
3minmin
min
MPaDT
D
DT
JTr 15.159
401021616
32
23
6
34max
0min
MPaa 78.472
89.2345.1192
minmax
MPam 67.712
89.2345.1192
minmax
MPaa 6.792
015.159
MPam 6.792
015.159
Cycle ID Fmin (kN)
Fmax (kN)
Tmin (kNm)
Tmax (kNm)
Cycles repetiti
on2 10 50 0 2 1000
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Solution for Tutorial 3
• Since we have both shear and direct stress we convert then into equivalent von-Mises stress
• Plug these into Soderberg criterion
• Finally we get;
MPaxyxeqa 9.1456.79378.473 2222
MPaxyxeqm 4.1556.79367.713 2222
1y
eqm
ar
eqa
SS 1
4204.1559.145
arSMPaSar 6.231
bfar aNSS 2.016006.231 N cyclesN 31074.15
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Solution for Tutorial 3
• Damage is calculated as;
• For part to fail we must have;
• This means that the component can go through;
i
i
NnD 094.0
1074.151000
1097.1625000
33
D
1DN nrepeatitio 614.101
DN nrepeatitio
2 Cyclefor 1000614.101 Cyclefor 5000614.10
cyclescycles