Fatigue Analysis of Structures (Aerospace Application)

1

Aero Structure-Fatigue Analysis

By

Dr. Mahdi Damghani

2016-2017

2

Suggested Readings

Reference 1 Reference 2

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

Reference 3

3

Topics

• Familiarisation with fatigue of metals

• Life estimates

• Endurance limit

• Fluctuating stresses

• Stress concentration and Notch sensitivity

4

Your Questions

• Feel free to ask your questions in the following link:

https://padlet.com/mahdi_damghani/Fatigue_Analysis

5

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.

6

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

7

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

8

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

9

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

10

Introduction

Growth of surface micro-cracks of aircraft towbar as the result of fatigue

loading leading to failure

11

Introduction

12

Introduction

Fatigue crack at stress raiser (discontinuity, i.e.

hole) due to stress concentration

13

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

14

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)

15

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

16

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

17

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

18

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

19

Watch

20

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

21

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).

22

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

23

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

24

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

25

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

26

How SN curve is established

w1w1w1

31

minmax,25.0

max24

ranw

IMy ryrI

27

S-N bands for Aluminium alloys for completely reversed cycling

28

Fatigue strength in log-log S-N curve (Sf)

e

e

SSSSb 1000

731000 log

31

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

29

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

30

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)

31

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

32

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

33

Endurance limit (Se)

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

34

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

35

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

36

Stress concentration

NotchHole Hole

37

Stress concentration in aircraft

38

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

39

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)

40

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

41

Stress concentration value

42

Stress concentration value

43

Stress concentration for a shaft under bending load and axial load

44

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

45

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

46

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

47

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

48

Notch sensitivity

49

Notch sensitivity

50

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

51

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

52

Industrial examples of fatigue critical locations (fillets in flap track)

Fillet A

53

Industrial examples of fatigue critical locations (buttstraps)

54


55


Cover

Panel

Buttstrap

56

Industrial examples of fatigue critical locations (strut brackets)

57

Industrial examples of fatigue critical locations (baffle panels)

58

Industrial examples of fatigue critical locations (baffle panels)

59

Industrial examples of fatigue critical locations (joints)

60

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 .

61

Solution

093.032/3/18.132/38/

drdD

65.1tK

62

Solution

84.0q

55.1165.184.01

11

f

tf

K

KqK

63

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.

64

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

65

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?

66

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

67

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

68

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

69

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

70

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

71

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

72


?11_ tBf KqK

?_ BtK

55.1165.184.01

11_

tBf KqK

We solved this in previous example,

see below

73


• Let’s calculate fatigue stress concentration at point C, too.

086.035/3/08.135/38/

drdD

6.1_ CtK

74


84.0q

5.116.184.01

11

_

_

Cf

tCf

K

KqK

75


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



).(5.50212502.4 mNM B

76


MPaC

MKS BBfB 1.33510

217.35.69555.1 6

_ MPaC

MKS CCfC 1.17910

209.45.50250.1 6

_



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


bfa NaS

ba

f aSN

1

)(680001437

1.335 1308.01

cyclesN f

78

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

79

Tutorial 2 (real industrial problem)

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

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

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

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)

83

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)

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

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)

86

Note

Safe

Unsafe

Infinite life

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

88

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

89

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

90

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?

91

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

92

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

93

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

94

Cumulative damage

95

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;

96

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.

97

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

98

Example 4

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

99

Solution

iD

0.1sRepetition 56.239.00.1sRepetition

100

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)

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

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

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).

104

Example 6

105

Solution

)(6.5210255.1

5010001096100

10106

3

22

3

max MPaPPPS

0min S

1y

m

ar

a

SS

SS

11

y

mara S

SSS

106

Solution

bfar aNSS

107

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

108

Tutorial 3

109


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

110


• From S-N curve we remember;

bfar aNSS 2.0160011.145 N cyclesN 310972.162

111


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

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;

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

113


• 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

Fatigue Analysis of Structures (Aerospace Application)

Engineering

Transcript of Fatigue Analysis of Structures (Aerospace Application)