The Practical Use of Fracture Mechanics

The Practical Use

of Fracture Mechanics

by
DAVID BROEK
FractuREsearch Inc., Galena, OR, USA

Kluwer Academic Publishers

Dordrecht / Boston / London
Library of Congress Cataloging in Publication Data

Brock, David.
The pract,ical use of fr'actUJ'e mechanics.

Bibliography: p.
1. Fract,ure mechanics. 1. TiUe.
TA409.B773 1988 620.1'126 88-9336

ISBN-13: 978-0-7923-0223-0 e-ISBN-13: 978-94-009-2558-8

DOI: 10,1007/978-94-009-2558-8

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first published in hardbound edition only in 1988

second printing with minor corrections in 1989
third printing 1991
reprinted 1994
reprinted 1996
reprinted 1997
hardback edition ISBN 90-247-3707-9
paperback edition ISBN 90-247-0223-0
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 Preface

This book is about the use of fracture mechanics for the solution of practical
problems; academic rigor is not at issue and dealt with only in as far as it
improves insight and understanding; it often concerns secondary errors in
engineering. Knowledge of (ignorance of) such basic input as loads and stresses
in practical cases may cause errors far overshadowing those introduced by
shortcomings of fracture mechanics and necessary approximations; this is
amply demonstrated in the text.
I have presented more than three dozen 40-hour courses on fracture
mechanics and damage tolerance analysis, so that I have probably more
experience in teaching the subject than anyone else. I learned more than the
students, and became cognizant of difficulties and of the real concerns in
applications. In particular I found, how a subject should be explained to appeal
to the practicing engineer to demonstrate that his practical problem can indeed
be solved with engineering methods. This experience is reflected in the presenta-
tions in this book. Sufficient background is provided for an understanding of the
issues, but pragamatism prevails. Mathematics cannot be avoided, but they are
presented in a way that appeals to insight and intuition, in lieu of formal
derivations which would show but the mathematical skill of the writer. A
practicing engineer does care little about how a crack tip stress field is derived;
he accepts that it can be done, as long as he can understand that the result must
be of the form it is. His real concern is what it means for the solutions to
practical problems.
Mathematical background is of use to future scientists, but few engineering
students taking fracture mechanics courses will become researchers in fracture
mechanics. My advice is that indeed very few should. Fracture mechanics has
matured to a useful engineering tool as has e.g. buckling analysis. Certainly, it
is not perfect, but no engineering analysis is. Not much buckling research is
practiced today; the present number of researchers in fracture mechanics is far
out of proportion to the remaining engineering problems.
Despite the acclaimed solid education of engineers, it is my experience in
teaching fracture mechanics to literally hundreds of practicing engineers, that
most have only a vague idea of such subjects as plastic deformation and design;
to many Mohr's circle is an enigma; at most one in a class knows the stress
concentration factor of a circular hole; fewer even remember yield criteria
and their significance. For this reason Chapter 2 of the text discusses the
effect of notches and local yielding and provides a simplified look at yield

v
vi
criteria. The treatment is necessarily compromising rigidity, but it serves the
purpose of providing the insight without which fracture mechanics cannot be
understood.
My research work covered fundamental fracture and fatigue mechanisms,
experimental evaluation of criteria for fatigue, fracture, and combined mode
loading, the development of engineering procedures for arrest analysis in
stiffened panels, collapse conditions, and damage tolerance analysis in general.
My engineering background however, has always prevailed and forced me to
consider the practicality of procedures. This book reflects a lifetime of
experience in research and practical applications. No subject is discussed on the
basis of hearsay. Instead the basis is "hands-on" experience with virtually every
issue from the fundamental to the practical.
I am aware of my shortcomings, prejudices and opinionations, but believe to
be entitled to these on the basis of my engineering experience. This text reflects
them, and I do not apologize. Too many "refinements" in engineering solutions
pertain to secondary errors; they increase the complexity, but do not improve
the solution. One does not improve the strength of a chain by improving the
strong links. The weak links in the fracture mechanics analysis are the
unknowns, not the procedures. This book is for engineering students and for
engineers, who must solve urgent problems yesterday. Engineering solutions are
always approximative, no matter what the subject is. Such is the nature of
engineering. Necessary assumptions are far more influential than those due to
limitations of fracture mechanics.
The text is intended for the education of 'engineers'. At the same time it serves
as a reference. For this reason there is some duplication and extensive cross-
references are provided. This may be objectionable to the reader going through
the text from A to Z, but it will be of help to those who read sections here and
there. It is not perfect as no human effort ever is, and I shall welcome construc-
tive criticism with regard to the engineering applications. My haste in accom-
plishing things (enforced by the unfortunate situation that I have to make a
living, while writing a book is an extraneous effort which is not very profitable)
may be reflected in the text. Again, I am not apologizing, just explaining.
I am grateful to my wife, Betty, for putting up with my preoccupations and
moods while writing this text, and for submitting all writing to a word-process-
or. I am also thankful to my son Titus, who spent numerous hours in producing
solutions to exercises and in drawing figures.
I dedicate this book to the memory of my father, Harm Broek. Many
sons see their father as the ultimate example. So do I. His unfailing support has
always been a driver of my ambitions.

Galena, Ohio, February 1988

 Notice

Extensive computer software for fracture mechanics analysis was developed by

the author of this book. This software is capable of
- performing residual strength analysis in accordance with Chapters 3, 4 and lO
both for LEFM and EPFM.
- performing fatigue crack growth analysis for constant amplitude, random
loading and semi random loading in accordance with Chapters 5 and lO, with
or without retardation. There are options for various retardation models, rate
equations and tabular rate data (Chapter 7).
- automatically generating semi-random stress histories on the basis of
exceedance diagrams (Chapter 6) and performing clipping and truncation
upon command.
- determining inspection intervals and cumulative probability of detection in
accordance with the procedure discussed in Chapter 11, using the calculated
crack growth curves, and accounting for specificity and accessibility.
- providing professional plots.
An extensive library of materials data is included, as well as an extensive
library of geometry factors. Besides a pre-processor can generate geometry
factors, using most of the procedures discussed in Chapter 8.
The above software is available for personal computers. Because of the large
size of the software, it is split up in seven modules, each of which fIts in a
personal computer. The modules communicate through disket files that are
generated automatically. The software can be obtained from FractuREsearch
Inc, 9049 Cupstone Drive, Galena, OH 43021, USA.
A much simplified version of the same software (also by the author) is
available from the American Society of Metals (ASM), Metals Park, OH 44073,
USA. This simplified version has no data library, no preprocessor for geometry
factor, cannot do retardation, and does not generate semi-random stress
histories.

Vll
 Contents

Preface v

Notice VB

Chapter 1. INTRODUCTION
1.1. Fracture control 1
1.2. The two objectives of damage tolerance analysis 3
1.3. Crack growth and fracture 8
1.4. Damage tolerance and fracture mechanics 15
1.5. The need for analysis: purpose of this book 17
1.6. Exercises 20

Chapter 2. EFFECTS OF CRACKS AND NOTCHES: COLLAPSE 22

2.1. Scope 22
2.2. An interrupted load path 22
2.3. Stress concentration factor 25
2.4. State of stress at a stress concentration 28
2.5. Yielding at a notch 31
2.6. Plastic collapse at a notch 35
2.7. Fracture at notches: brittle behavior 41
2.8. Measurement of collapse strength 44
2.9. Exercises 46

Chapter 3. LINEAR ELASTIC FRACTURE MECHANICS 48

3.1. Scope 48
3.2. Stress at a crack tip 48
3.3. General form of the stress intensity factor 52
3.4. Toughness 55
3.5. Plastic zone and stresses in plane stress and plane strain 57
3.6. Thickness dependence of toughness 61
3.7. Measurement of toughness 67
3.8. Competition with plastic collapse 70
3.9. The energy criterion 73
3.10. The energy release rate 75
3.11. The meaning of the energy criterion 79
3.12. The rise in fracture resistance: redefinition of toughness 79
3.13. Exercises 86

ix
x

Chapter 4. ELASTIC-PLASTIC FRACTURE MECHANICS 88

4.1. Scope 88
4.2. The energy criterion for plastic fracture 88
4.3. The fracture criterion 90
4.4. The rising fracture energy 93
4.5. The residual strength diagram in EPFM: collapse 97
4.6. The measurement of the toughness in EPFM 98
4.7. The parameters of the stress-strain curve 102
4.8. The h-functions 106
4.9. Accuracy 109
4.10. Historical development of J 112
4.11. Limitations of EPFM 116
4.12. CTOD measurements 118
4.13. Exercises 121

Chapter 5. CRACK GROWTH ANALYSIS CONCEPTS 123

5.1. Scope 123
5.2. The concept underlying fatigue crack growth 123
5.3. Measurement of the rate function 126
5.4. Rate equations 130
5.5. Constant amplitude crack growth in a structure 133
5.6. Load interaction: Retardation 136
5.7. Retardation models 145
5.8. Crack growth analysis for variable amplitude loading 149
5.9. Parameters affecting fatigue crack growth rates 157
5.10. Stress corrosion cracking 163
5.11. Exercises 165

Chapter 6. LOAD SPECTRA AND STRESS HISTORIES 168

6.1. Scope 168
6.2. Types of stress histories 169
6.3. Obtaining load spectra 175
6.4. Exceedance diagram 176
6.5. Stress history generation 180
6.6. Clipping 192
6.7. Truncation 195
6.8. Manipulation of stress history 198
6.9. Environmental effects 204
6.10. Standard spectra 205
6.11. Exercises 205
 xi

Chapter 7. DATA INTERPRETATION AND USE 208

7.1. Scope 208
7.2. Plane strain fracture toughness 209
7.3. Plane stress and transitional toughness, R-curve 212
7.4. Toughness in terms of J and JR 214
7.5. Estimates of toughness 215
7.6. General remarks on fatigue rate data 218
7.7. Fitting the da/dN data 222
7.8. Dealing with scatter in rate data 232'
7.9. Accounting for the environmental effect 236
7.10. Obtaining retardation parameters 238
7:11. Exercises 241

Chapter 8. GEOMETRY FACTORS 243

8.1. Scope 243
8.2. The reference stress 244
8.3. Compounding 247
8.4. Superposition 249
8.5. A simple method for asymmetric loading cases 255
8.6. Some easy guesses 258
8.7. Simple solutions for holes and stress concentrations 260
8.8. Simple solutions for irregular stress distributions 267
8.9. Finite element analysis 271
8.10. Simple solutions for crack arresters and multiple elements 274
8.11. Geometry factors for elastic-plastic fracture mechanics 278
8.12. Exercises 279

Chapter 9. SPECIAL SUBJECTS 282

9.1. Scope 282
9.2. Behavior of surface flaws and corner cracks 282
9.3. Break through: leak-before-break 290
9.4. Fracture arrest 293
9.5. Multiple elements, multiple cracks, changing geometry 300
9.6. Stop holes, cold worked holes and interference fasteners 311
9.7. Residual stresses in general 316
9.8. Other loading modes: mixed mode loading 319
9.9. Composites 327
9.10. Exercises 329

Chapter 10. ANALYSIS PROCEDURES 332

10.1. Scope 332
10.2. Ingredients and critical locations 332
Xll

10.3. Critical locations and flaw assumptions 334

10.4. LEFM versus EPFM 339
10.5. Residual strength analysis 345
10.6. Use of R-curve and JR-curve 353
10.7. Crack growth analysis 355
10.8. Exercises 361

Chapter II. FRACTURE CONTROL 362

ILl. Scope 362
11.2. Fracture control options 362
11.3. The probability of missing the crack 369
11.4. The physics and statistics of crack detection 373
11.5. Determining the inspection interval 377
11.6. Fracture control plans 379
11.7. Repairs 384
11.8. Statistical aspects 385
11.9. The cost of fracture and fracture control 387
11.1 O. Exercises 389

Chapter 12. DAMAGE TOLERANCE SUBSTANTIATION 391

12.1. Scope 391
12.2. Objectives 391
12.3. Analysis and damage tolerance substantiation 393
12.4. Options to improve damage tolerance 395
12.5. Aircraft damage tolerance requirements 397
12.6. Other requirements 404
12.7. Flaw assumptions 408
12.8. Sources of error and safety factors 410
12.9. Misconceptions 417
12.10. Outlook 420
12.11. Exercises 422

Chapter 13. AFTER THE FACT: FRACTURE MECHANICS AND

FAILURE ANALYSIS 424
13.1. Scope 424
13.2. The cause of service fractures 425
13.3. Fractography 428
13.4. Features of use in fracture mechanics analysis 430
13.5. Use of fracture mechanics 436
13.6. Possible actions based on failure analysis 440
13.7. Exercises 440
 xiii

Chapter 14. APPLICATIONS 443

14.1. Scope 443
14.2. Storage tank (fictitious example) 443
14.3. Fracture arrest in ships 447
14.4. Piping in chemical plant (fictitious example) 462
14.5. Fatigue cracks in railroad rails 465
14.6. Underwater pipeline 476
14.7. Closure 483

Chapter 15. SOLUTIONS TO EXERCISES 485

SUBJECT INDEX 515

 CHAPTER I

Introduction

1.1. Fracture control

Fracture control of structures is the concerted effort by designers, metallurgists,

production and maintenance engineers, and inspectors to ensure safe operations
without catastrophic fracture failures. Of the various structural failure modes
~buckling, fracture, excessive plastic deformation) fracture is only one. Very
seldom does a fracture occur due to an unforeseen overload on the undamaged
structure. Usually, it is caused by a structural flaw or a crack: due to repeated
or sustained "normal" service loads a crack may develop (starting from a flaw
or stress concentration) and grow slowly in size, due to the service loading.
Cracks and defects impair the strength. Thus, during the continuing develop-
ment of the crack, the structural strength decreases until it becomes so low that
the service loads cannot be carried any more, and fracture ensues. Fracture
control is intended to prevent fracture due to defects and cracks at the
(maximum) loads experienced during operational service.
Iffracture is to be prevented, the strength should not drop below a certain safe
value. This means that cracks must be prevented from growing to a size at which
the strength would drop below the acceptable limit. In order to determine which
size of crack is admissible, one must be able to calculate how the structural
strength is affected by cracks (as a function of their size); and in order to
determine the safe operational life, one must be able to calculate the time in
which a crack grows to the permissible size. For this, one must first identify the
locations where cracks could develop. Analysis then must provide information
on crack growth times and on structural strength as a function of crack size.
This type of analysis is called damage tolerance analysis.
Damage tolerance is the property of a structure to sustain defects or cracks
safely, until such time that action is (or can be) taken to eliminate the cracks.
Elimination can be ·affected by repair or by replacing the cracked structure or
2

component. In the design stage one still has the options to select a more crack
resistant material or improve the structural design, to ensure that cracks will not
become dangerous during the projected economic serivce life. Alternatively,
periodic inspections may be scheduled, so that cracks can be repaired or
components replaced when cracks are detected. Either the time to retirement
(replacement), or the inspection interval and type of insepction, must follow
from the crack growth time calculated in the damage tolerance analysis.
Inspections can be performed by means of any of a number of non-destructive
inspection techniques, provided the structure is inspectable and accessible; but
destructive techniques such as proof-testing are essentially also inspections. If a
burst occurs during hydrostatic testing of e.g. a pipe line, then there was
apparently a crack of sufficient size to cause the burst. Although this may be
troublesome, the proof test is intended to eliminate defects under controlled
circumstances (e.g. with water pressure) to prevent catastrophic failure during
operation when the line is filled with oil or gas. After the burst and repair, the
line can continue service. If no burst occurs during the proof test, then
apparently any cracks were smaller than the critical size in the proof test. A
certain period of safe operation is then possible before such cracks would grow
to the permissible size.
Fracture control is a combination of measures such as described above
(including analysis), to prevent fracture due to cracks during operation. It may
include all or some of these measures, namely damage tolerance analysis,
material selection, design improvement, possibly structural testing, and main-
tenance/inspection/replacement schedules. The extent of the fracture control
measures depends upon the criticality of the component, upon the economic
consequences of the structure being out of service, and last but not least, the
consequential damage caused by a potential fracture failure (including loss of
lives). Fracture control of e.g. a hammer may be as simple as selecting a material
with sufficient fracture resistance. Fracture control of an airplane, includes
damage tolerance analysis, tests, and subsequent inspection and repair/replace-
ment plans.
Damage tolerance analysis and its results form the basis for fracture control
plans. Inspections of whichever nature, repairs and replacements, must be
scheduled rationally using the information from the damage tolerance analysis.
This book deals with practical damage tolerance analysis. Fracture control
measures are discussed in general, but the execution, use, and interpretation of
the damage tolerance analysis for scheduling fracture control measures are
discussed in detail.
The mathematical tool employed in damage tolerance analysis is called
fracture mechanics; it provides the concepts and equations used to determine
how cracks grow and how cracks affect the strength of a structure. During the
last 25 years fracture mechanics has evolved into a practical engineering tool. It
is not perfect, but no engineering analysis is. The equation for bending stress
(0' = Mh/I) is rather in error when used to calculate structural strength, because
 3

it ignores plastic deformation. Nevertheless it has been used successfully for

many years in design. Similarly, fracture mechanics can be used successfully.
Acclaimed inaccuracies are due to inaccurate inputs much more than due to
inadequacy of the concepts, which will become abundantly clear in the course
of this book. Naturally, the results of damage tolerance analysis must be used
judiciously, but this can be said of any other engineering analysis as well.
Although further improvements of fracture mechanics concepts may well be
desirable from a fundamental point of view, it is unlikely that damage tolerance
analysis can be much improved, as its accuracy is determined mostly by the
accuracy of material data and predicted loads and stresses.
Fracture mechanics can give useful answers to questions that hitherto could
not be answered at all. The answers may not be perfect, but a reasonable anwser
is better than none. Unfortunately, it is rather easy to obtain a wrong answer:
pitfalls are numerous. This book is intended to explain the engineering usage of
fracture mechanics, and to point out the pitfalls in detail. Some of the pitfalls
are so treacherous that one sees things more often done wrong than right. These
and other things have led to some myths about fracture mechanics and its uses,
which are hard to eliminate. They will be addressed in this book.
Although the basis of fracture mechanics concepts will be discussed, this book
focuses on how the analysis should be performed, on how to solve practical
problems and on how to avoid errors. Attainable accuracy, and the factors
affecting accuracy, are discussed in detail. A brief recapitulation is given of those
concepts of fracture mechanics that are actually used in practical damage
tolerance analysis. For more in-depth treatment of those subjects, the reader is
referred to other text-books on the matter [1,2, 3]. Yet, the present text provides
sufficient background for a proper understanding of the practical methods
discussed.
As this book deals with the practical use of established fracture mechanics
concepts, references have been kept to a minimum. They can be found in more
extensive texts. It is not necessary to reference established concepts; e.g.
references to Hooke's law are superfluous, and so are references to accepted and
established fracture mechanics concepts. Hence, references are provided only in
those cases where relevance or extent do not warrant complete treatment in this
text, so that use of the original publication might be desirable.

1.2. The two objectives of damage tolerance analysis

Establishment of a fracture control plan requires knowledge of the structural

strength as it is affected by cracks, and of the time involved for cracks to grow
to a dangerous size. Thus, damage tolerance analysis has two objectives, namely
to determine
1. the effect of cracks on strength (margin against fracture)

2. the crack growth as a function of time.

4

These two objectives are discussed below.

Figure 1.1 shows diagrammatically the effect of crack size on strength. In
fracture mechanics crack size is generally denoted as a. In Figure 1.1 the strength
is expressed in terms of the load, P, the structure can carry before fracture
occurs (fracture load). Supposing for the moment that a new structure has no
significant defects (a = 0), then the strength of the new structure is Pu , the
(ultimate) design strength (load). It should be emphasised that the strength of
the new, crack-free structure is finite. Fracture will, and must occur when it is
subjected to a load Pu , otherwise the structure was over-designed.
In every design a safety factor (ignorance factor) is used. This factor may be
applied in different ways, but the result is always the same. In some areas of
technology the safety factor is applied to load. For example if the maximum
anticipated service load is P" the structure is actually designed to sustain}Ps =
Pu , where} is the safety factor. The designer sizes the structure in such a manner
that the stress is equal to or slightly less than the ultimate tensile strength when
the load is Pu (checks against plastic deformation are usually necessary as well).
Alternatively, the safety factor is applied to the allowable stress: if the actual
material strength (ultimate tensile strength) is Flu, the structure is sized in such
a way that the stress at the highest service load, P" is less than or equal to Flu /}

P.es
(LOAD)
P"

CRACK SIZE (a)

Figure 1.1. Residual strength in the presence of cracks; strength of new structure (a = 0) is P u =
jP,.
 5

where} is again the safety factor. Hence, since load and stress are usually
proportional, the structure is actually capable of carrying iPs = Pu. Plasticity
may well prohibit proportionality, but since plasticity is generally limited to
small areas at notches and stress concentrations, the above is approximately
correct. But, even if it is not correct in actual numbers, it is true in spirit: the
structure is designed to carry a load higher than the highest anticipated service
load by a factor }, and the structural strength is Pu = iPs. The value of} is
between 3 (many civil engineering structures) and 1.5 (airplanes).
It is emphasized that P s is the'highest service load. If the service load varies,
the load may well be much less than P s during most of the time. For example
the loads on cranes, bridges, off-shore structures, ships and airplanes are usually
much less than Ps • Only in exceptional circumstances (e.g. storms) does the load
reach Ps • At other times the load may be only a fraction of P,., so that the margin
against fracture is much larger than}, except in extreme situations. The loads on
some structures, e.g. pipelines, pressure vessels, rotating machinery are reaching
more nearly always the same level (Ps ), as shown in Figure 1.2.
The new structure has a strength Pu with safety factor j. Its strength is finite,
so that the probability of fracture is not entirely zero. If the load should reach
Pu (e.g. in a storm) the structure fails. The probability of this occurring is
non-zero, but experience has shown that it is acceptably low. If cracks are
present the strength is less than Pu • This remaining strength under the presence
of cracks is generally referred to as the 'residual strength', Pres; the diagram in
Figure 1.1 is called the residual strength diagram. With a residual strength
Pres < Pu the safety factory has decreased: } = PreslPs which is less than
} = PulPs. In concert, the probability of fracture failure has become higher.
Fracture is the catastrophic break-up of the structure into two or more pieces.
With a crack of size a, the residual strength is Pres. Should a load P = Pres occur
then fracture takes place. The fracture process may be slow and stable initially,
the crack extending (by fracture), but the structure still hanging together.
Eventually, the fracture becomes unstable and the structure breaks into two or
more pieces. The whole process of stable-unstable fracture may take place in a
fraction of a second. If the load P = Pres does not occur, service loading
continuing at loads at or below Pre" the crack will continue to grow, not by
fracture but by cracking mechanisms such as fatigue, stress-corrosion or creep.
Due to continual growth the crack becomes longer, the residual strength less,
the safety factor lower, and the probability offracture higher. If nothing is done
and the structure remains in service, the residual strength evenutally will become
equal to Ps (or even equal to the average service load Pa in Figure 1.2). Then the
safety factor is reduced to I and fracture occurs already at P,., i.e. at the (highest)
service load, or even at Pa • This is what must be prevented: the crack should not
be allowed from becoming so large that fracture occurs at the service loads.
6

p
P,. - - - - - - - - - - - - - ----

lime
.(a)

_ _ MILD WEATHER .. , .. STORM -I" MILD WEATHER_

START STOP/START STOP/START STOP/START time

(b)

Figure 1.2. Schematic examples of load histories. (a) Typical loading of offshore structures, ships,
airplanes; (b) typical loading of rotating machinery.

Hence, the structure or component must be replaced before the crack becomes
dangerous, or the crack must be detected and repaired before such time.
The above implies that the limit should be set somewhat above Ps • For
example, one may require that the residual strength never be less than Pp = gP"
where g is the remaining safety factor, and Pp the minimum permissible residual
strength. The design engineer or user does not decide what should be the initial
safety factor j. This factor is prescribed by rules and regulations issued by
engineering societies (e.g. ASME) or Government authorities. Similarly, these
rules or requirements should prescribe g. This has not been done for all types
of structures yet, while e.g. the ASME rules approach the problem somewhat
differently (rules and regulations are discussed in Chapter 12). However, some
rule or goal must be estabished, some decision made, to set the minimum
permissible residual strength, so that the maximum permissible crack size, ap '
can be determined from the residual strength diagram.
Provided the shape of the residual strength diagram is known, and Pp
 7

prescribed, the maximum permissible crack size follows from the diagram. In
order for damage tolerance analysis to determine the largest allowable crack, the
first objective must be the calculation of the residual strength diagram of Figure
1.1. If ap can be calculated directly from Pp it may not be necessary to calculate
the entire residual strength diagram, but only the point (ap , Pp ). However, this
is seldom possible and rarely time saving. In general, the calculation of the entire
diagram is far preferable. The maximum permissible crack size follows from the
calculated residual strength diagram and from the prescribed minimum permiss-
ible residual strength, Pp • The residual strength diagram will be different for
different components of a structure and for different crack locations; permissible
crack sizes will be different as well.
The permissible crack size is sometimes called the critical crack size. However,
the objective of fracture control is to prevent 'critical' cracks. A critical crack is
one that would cause fracture in service. Cracks are not allowed to grow that
long. Instead, they are permitted to grow only to the permissible size ap • They
would be critical only in the event that a load as high as Pp would occur.
Knowing that the crack may not exceed ap is of little help, unless it is known
when the crack might reach ap • The second objective of the damage tolerance
analysis is then the calculation of the crack growth curve, shown diagrammatic-
ally in Figure 1.3. Under the action of normal service loading the cracks grow
by fatigue, stress corrosion or creep, at an ever faster rate leading to the convex
curve shown in Figure 1.3.
Starting at some crack size ao the crack grows in size during time. The
permissible crack ap following from Figure 1.1 can be plotted on the curve in
Figure 1.3. Provided one can calculate the curve in Figure 1.3 one obtains the

a H

~ -------------------

time
H

Figure 1.3. Crack growth curve (schematically).

8

time H of safe operation (until ap is reached). If ao is for example an (assumed

or real) initial defect, then the component or structure must be replaced after a
time H. Alternatively, ao may be the limit of crack detection by inspection. This
crack ao will grow to ap within a time H. Since crack growth is not allowed
beyond ap ' the crack must be detected and repaired or otherwise eliminated
before the time H has expired. Therefore, the time between inspections must be
less than H. At an inspection at time II' the crack will be missed, because ao is
the detection limit. If the next inspection were to take place H hours later, the
crack would have reached ap already, which is not permitted; i.e. the inspection
interval must be less than H; it is often taken as H/2. In any case, the time of
safe operation by whatever means of fracture control follows from H. In turn,
H emerges from the damage tolerance analysis, provided both the residual
strength diagram (ap ) and the crack growth curve can be calculated to obtain H.

1.3. Crack growth and fracture

The residual strength and crack growth diagrams are essentially different, not
only in shape but also in significance. Fracture is the final event, often taking
place very rapidly, and resulting in a breaking-in-two. Crack growth on the
other hand occurs slowly during normal service loading. Also the mechanisms
of crack growth and fracture are different.
Crack growth takes place by one of five mechanisms:
a. Fatigue due to cyclic loading.
b. Stress corrosion due to sustained loading.
c. Creep.
d. Hydrogen induced cracking.
e. Liquid metal induced cracking.
Of these, the first two, and combinations thereof are the most prevalent, while
the last is hardly of interest for load-bearing structures. Crack growth is
sometimes referred to as 'sub-critical crack growth', a pleonasm. Fracture is
critical, and fracture is not the same as crack growth.
A crack may cause a fracture. There are only two mechanisms by which
fracture can occur, namely:
a. Cleavage.
b. Rupture.
A third 'mechanism', namely intergranular fracture, requires operation of some
form of either cleavage or rupture.
A mechanism for fatigue crack growth is shown in Figure 1.4. Other
mechanisms are possible but not essentially different [4, 5, 6]. Even at very low
loads there is still plastic deformation at the crack tip because of the high stress
concentration. Plastic deformation is slip (due to shear stresses; see Chapter 2) of
 9
I

,--=====:::::::=-~l~
A:_
________ ________
~a
~ ~~1

&a

~
~
z
~---------~:;~

~::::::=::::======'1
is
«
o....I

I I
-l6a!-

~
3
:------
~ ~-h--===_N ~~
I I
9 : :
5 I :
~ ! I I
~z ':.. I 16a I
1+-1
~ I I I
9 I I I

~ C:'='==-=-=== = ==-!--[)j
~ I I I
9 I I I
z I I !
::J G : I I I
~
I I I 6a I
z I I I- ~
is I I I I
«

~
9 I :: :

~
z
is
«
C::=======+H
-
) I I I I
9
z
I I I I
::J ! I I I
I I I I
i ,

Figure 1.4. One of various possible mechanisms of fatigue crack growth.

10
atomic planes, depicted in Figure 1.4, stage B. Continued slip on complement-
ary planes results in a blunted crack tip (Figure 1.4, stages B-D). The very first
slip step in stage 2 has already caused a very small crack extension !J.a. Upon
unloading (or if necessary compressive loading) the crack tip again becomes
sharp. Mechanistically, the whole process of slip could be reversed so that the
end result after unloading would again be as stage A. However, because of
oxidation of the freshly exposed material along the slip steps, and the general
disorder due to the slip, the process is irreversible in practice; the crack extension
!J.a remains. In the next load cycle the process is repeated: the crack grows again
by !J.a.
Growth per cycle, !J.a, is extremely small as can be judged immediately from
the mechanism in Figure 1.4. Typically, the growth is on the order of lO- s_IO- 4
inches (10- 7 -10- 3 mm); however if the load is cycled for 104 _10 8 cycles, the
crack will have grown by an inch. The repeated blunting and sharpening gives
rise to marks on the crack surface (often erroneously referred to as fracture
surface), which can be made visible at high magnification in an electron
microscope, as shown in Figure 1.5. The marks, called fatigue striations,
represent the successive positions of the crack front, i.e. the blunting/sharpening
steps. If the crack grows by 10- 5 inch, then the striation spacing is 10 5 inch (i.e.
0.05 inch at a magnification of 5000 x). Not all materials exhibit striations as
regular as in Figure 1.5. [4, 5, 6] (Chapter 13).
Crack growth by stress corrosion is a slow process as well. The crack extends
due to corrosive action (often along the grain boundaries) facilitated by the high
stretch and consequent atomic disarray at the crack tip. A common mechanism
of creep cracking is the diffusion of vacancies (open atomic places), a con-
glomerate of vacancies forming a hole, which subsequently joins up with the
crack tip.
A crack by itself is only a partial failure, but it can induce a total failure by
fracture. Fracture occurs by either of two mechanisms, cleavage or ductile
rupture. Cleavage is the splitting apart of atomic planes. From grain to grain the
preferred splitting plane is differently oriented causing a faceted fracture.
(Figure 1.6). The facets by themselves are flat and, therefore, good reflectors of
incident light. This causes the cleavage fracture to sparkle when fresh, but the
glitter may soon fade due to oxidation.
The alternative fracture mechanism of ductile rupture is shown in Figure 1.7.
All structural materials contain particles and inclusions. These particles are
generally complex compounds of the alloying elements. Some alloying elements
are used to improve castability and machineability; others are specifically
included to improve the alloy's strength. First the large particles let loose or
break, forming widely spaced holes close to the crack tip. In the final phase,
holes are formed at myriads of smaller particles; these holes or voids join up to
complete the fracture. Because of its irregularity the fracture surface diffuses the
 II

Figure 1.5. Fatigue striations on crack surface of aluminium alloy. Spacing of striations coincides
with cyclic loading (insert). Magnification 12000 x.

light and looks dull grey. The holes of the large particles and those of the small
particles (Figure 1.7) are visible at high magnification in the electron
microscope, as shown in figure 1.8. As the fracture surface shows the halves of
the holes, the fracture is referred to as dimple rupture or just rupture. Elimina-
tion of the large particles (by selective alloying and heat treatment) can improve
the fracture resistance of an alloy [4]. On the other hand, alloying is necessary
to provide strength in the first place, so that not all particles can be avoided.
That they playa role in the final fracture process becomes rather of secondary
importance; without the particles, the strength would be less.
Both cleavage fracture and rupture are fast processes. A cleavage fracture
may run as fast as I mile/sec (1600 m/s), a dimple rupture as fast as 1500 ft/sec
(500 m/s), although it may be slower. Fracturing is sometimes stable. The crack
12

GRAINS

Figure 1.6. Cleavage fracture starting at (blunted) crack tip. Bottom: flat facets glitter due to
reflection of incident light.

(fracture) then extends by one of the fracturing processes (cleavage or rupture)

instead of by one of the cracking mechanisms. Usually stable fracture is
immediately followed by the final unstable event.
The infamous brittle fracture of steel below the transition temperature occurs
by cleavage. Cleavage is often referred to as brittle fracture, while dimple
rupture is described as ductile fracture. This may be adequate for metallurgists,
but in general one should be very careful using these terms. By far the majority
of service fractures occur by dimple rupture, but most of these exhibit very little
overall plastic deformation, so that they are brittle from an engineering point of
view (see Chapter 2). Reference to cleavage and rupture avoids confusion. A
brittle fracture is one with little (overall) plastic deformation, whether cleavage
or rupture.
Similarly, fractographers sometimes refer to rupture as the 'overload fracture'
to distinguish it from the fatigue or stress corrosion crack surface (Chapter 13).
This can also cause confusion. Distinction between 'fracture surface' and 'crack
surface' is more appropriate. Practically all service failures are due to cracks or
 13

......
.••. ®
~-,~) . . . 0 .. j
~
'NCREASING
STRESS
BLUNTING

.. . .
I

e . .'. .
.. e · ·
.... e
'. .
LARGE
PARTICLES
CRACKING

CONCENTRATED
SLIP BE1WEEN
CRACKED LARGE
PARTICLES

..
FINAL
VOID FORMATION
AND COALESCENCE
• AT SMALL PARTICLES
[FRACTURE]

I
CRACK SURFA<tE I FRACTURE SURFACE

-j-t--
+I ~co£o~t VIEW FROM DIRECTION A

~~~. ".'
IL __ ON CRACK AND FRACTURE
SURFACE
~c;. 0 0

o
C5Q~ 0 I' Soo
STRIATIONS

Figure 1.7. Four stages of ductile rupture. Bottom: fracture surface shows holes formed by large and
small particles (dimples).

defects, a true overload fracture being very rare indeed (Chapter 13). It is
equally wrong (though not as confusing) to speak of fatigue fracture or stress
corrosion fracture. The fracture may be a consequence of a fatigue crack or a
stress corrosion crack, but it occurs by cleavage or rupture.
There is a third fracture type, namely intergranular fracture. The distinction
here is on the basis of fracture path (along the grain boundaries). Nevertheless,
intergranular fracture is not an altogether different type. Whether intergranular
or transgranular, the mechanism of separation either resembles cleavage or
rupture, be it that cleavage at a grain boundary is not as well defined as
transgranular cleavage.
14

Figure 1.B. Fracture surface of ductile rupture showing large holes with cracked large particles and
interconnecting dimples; Aluminium aHoy; Top: Scanning electron microscope; 1400 x . Bottom:
Transmission microscope (replica): 5000 x .
 15

1.4. Damage tolerance and fracture mechanics

Damage tolerance analysis must provide a capability for the calculation of both
the residual strength diagram (fracture due to cracks) and of the crack growth
curve. Fracture mechanics methods have been developed to analyze fracture to
obtain the fracture stress (residual strength), and to analyze fatigue crack
growth and stress corrosion crack growth. The treatment of creep crack growth
by fracture mechanics is still largely in the research phase. As this book is
dealing with practical and applicable methods and not with potential methods,
creep crack growth will not be discussed.
Although it is possible to deal with stress corrosion cracking in principle, the
crack growth times are usually so short (from 1 to lOOO hrs) that crack growth
is rather uninteresting from a practical point in view. Fracture control in the
case of stress corrosion is often aimed at crack growth prevention. Fracture
mechanics methods to accomplish this will be discussed.
In the treatment of crack growth emphasis will be on fatigue cracking. In
many structures fatigue cannot be prevented altogether; cracks then will occur.
These must be dealt with by means of fracture control, i.e. they must be
eliminated before they can cause a fracture. The time to the initiation of a
fatigue crack will not be considered for two reasons; first because fracture
mechanics deals with existing cracks, and second because it is not necessary to
know exactly when cracks appear to exercise control of cracks (Chapter II).
Besides, in cases offracture control by means of replacement and retirement, an
initial crack (hypothetical or real) is usually assumed present in the new
structure (Chapter 12). In some cases rules and regulations actually specify that
an initial crack be assumed.
Fracture mechanics (as all engineering mechanics) uses stresses, rather than
loads. Thus, the residual strength diagram of Figure 1.1 is normally based upon
O"re" the stress the structure can sustain (instead ofload), before fracture occurs.
The residual strength diagram based upon stress is shown in Figure 1.9. For this
purpose the engineering stress is used, as in the case of the uncracked structure,
the residual strength being given as O"res (Figure 1.9). Note that O"res is a strength
(like the tensile strength or the yield strength) and not a stress. Fracture occurs
when the stress equals O"res: fracture if 0" = O"res). Residual strength should not
be confused with residual stress. (Residual stress is a stress residing in a structure
while there are no loads applied).
Stress can be used as a basis for the analysis if there is a relationship between
the applied stress and the processes taking place at the crack tip. As the crack
tip events are governed by the local stresses at the crack tip, it is required that
the local crack tip stress be described as a function of the applied stress. Such
relationships can be derived, provided the problem is defined clearly. For this
purpose one must distinguish the 3 modes of loading shown in Figure 1.10.
16

ap - - __ - -

Figure 1.9. Residual strength diagram on the basis of nominal engineering stress.

MODE MODE " MODE "'

Figure 1.10. The modes of loading. Mode I opening mode. Mode II shear mode. Mode III tearing
mode.

These modes of loading (not modes of cracking), are usually referred to simply
by Roman numerals I, II and III. Other descriptions used are opening mode or
tension mode for mode I, (in-plane) shear mode for mode II, and (out-of-plane)
shear mode or tearing mode for mode III.
It turns out that the crack-tip stress equations are very similar for each of the
modes. As a matter of fact the format of the equations is exactly the same.
Consequently, the fracture and crack growth analysis procedures for each of the
modes individually, turn out to be identical, be it that different numbers apply.
If one knows how to deal with mode I, one knows essentially how to analyse
modes II and mode III individually. This is one reason why this book deals
primarily with mode I. However, there are other more compelling reasons.
In practice, by far the majority of cracks result from mode I loading. The other
 17

two modes do not occur individually, but they may occur in combination with
mode I, i.e. I-II, I-III or I-II-III. However, if the loading of these modes is in
phase, cracks will rapidly choose a direction of growth in which they are
subjected to mode I only (Chapter 9). Thus, the majority of apparent combined
mode cases are reduced to mode I by nature itself. There are few cases left then
which cannot be treated as pure mode I. These occur when for example in a
mode I-II combination (where the crack would normally select a mode I path),
there is a direction in the material where the resistance to cracking and fracture
is substantially lower. A case in point is a circumferential weld in a torque tube
(Chapter 9). Normally, a crack in a torque tube will develop and grow under
45 degrees (tension only) and thus be in mode I. But if a circumferential weld
confines the crack to the circumferential direction (shear plane) mode II must
be considered. Another case occurs when mode I and e.g. mode II loading are
out of phase (bending and torsion cycles with different frequencies; see Chapter
9).
Although individual loading modes are easy to deal with, a combined mode
loading case is rather more difficult. Several combined mode (or mixed mode)
analysis concepts have been proposed. A practical approach will be presented
in Chapter 9 for the few instances of 'true' mixed mode loading described above.

1.5. The need for analysis: purpose of this book

During the first half of the industrial era structural failures were numerous;
railroad accidents, elevator accidents, boiler explosions, etc., were a common
occurrence; accidents due to structural fracture were reported weekly if not
daily. The first powered flight was postponed because of a broken propellor
shaft. Orville Wright returned from Kitty Hawk to Dayton to machine a new
shaft, which caused a 2-week delay of the first flight. Due to improved materials,
and refinements of design procedures, and not in the least, due to the enforce-
ment of design safety factors and quality control measures, the number of
strucutral failures has abated, but not been reduced to zero.
More than two dozen major bridges have collapsed during this century. Since
World-War II in excess of 200 civil airplanes had fatal accidents due to fatigue
cracks; the number of cracks discovered in commercially operated jets is
estimated at well over 35 000. Many cracks have been reported in nuclear power
structures. Although less frequent, catastrophic structural failures have far more
serious consequences now than in the past. At the same time society has become
less tolerant and abundantly more litigious. Thus, no manufacturer, nor
operator of larger structures, can afford to ignore the issue of fracture control.
Rational fracture control being based on damage tolerance analysis, the use of
fracture mechanics and damage tolerance analysis is well-nigh imperative.
Rules and requirements have been issued for Military airplanes as well as for
18

commercial airplanes. Damage tolerance analysis of an airplane subject to these

rules, typically requires 20000-60000 man hours. (An equal or larger effort is
in the associated testing for material's data and analysis substantiation). The
ASME boiler and pressure vessel code requires damage tolerance analysis, at
least for nuclear power structures. Before long, requirements are likely to be
implemented for railroads, ships, bridges, pipelines and other structures, but
many manufacturers are already using some form of damage tolerance analysis.
Fracture mechanics concepts and damage tolerance analysis may not be ideal
(they never will be), but they provide answers where none were available before.
Judicious use of these can certainly reduce the risk of fracture. (No risk in life
is ever reduced to zero). The purpose of this text is to present the procedures of
practical damage tolerance analysis as they are (or can be) applied to almost any
structure, including ships, pipelines, aircraft, pressure vessels, cranes, bridges
and rotating machinery. Each of these structures presents its own specific
problems, but there is enough common ground for a universal procedure, which
can be modified, simplified or extended to serve a certain purpose. The various
components of the universal procedure will be discussed. Specific use and
interpretation for various types of structures will be reviewed; examples of
applications given.
A theoretical text this is not. The emphasis is on how damage tolerance
analysis is and should be performed, on how the necessary information is
obtained and used, on how the results of the analysis are employed to exercise
fracture control. The reliability of the analysis and the major sources of error
are discussed. Engineering approaches and approximations are given ample
space: in view of the general inaccuracies caused by material data and projected
loads and stresses, approximative solutions are quite adequate as the errors so
introduced are secondary. In many instances, approximative solutions are the
only feasible ones from an economic point of view. It may well be reasonable
to spend large sums on the analysis of one crack in a nuclear pressure vessel, the
cost of a power plant being extreme and the consequential cost of a failure
beyond imagination. However, the analysis of a crack in a hand-tool must be
cheap and the best engineering methods are indicated. For an aircraft structure
with hundreds of potential crack locations in very complex details, so many
cases must be considered that approximations and engineering judgement must
be used. The same holds for ships, offshore structures, bridges, etc. It will be
shown in this text why and where approximations are useful and suitable. The
order of magnitude of the error in the final solution will be the guideline.
The 'how' cannot be understood without some knowledge of the 'why'. It
would be a disservice to the reader to refer her or him to other texts for the
'why', in particular where these texts are often intended for theoreticians rather
than practitioners. Thus the present text provides the concepts and theoretical
background in an abridged and simplified form, adequate for the understanding
 19

of applications and use. The background is contained in Chapters 2-5 which

present the concepts of fracture mechanics for the analysis of fracture (elastic as
well as elastic-plastic) and for the analysis of crack growth. These chapters,
although providing sufficient basis for what follows, can serve as an introduc-
tion to more comprehensive texts [e.g. 1, 2, 3] for readers desiring more theoreti-
cal information.
Chapter 6 is devoted to the analysis and interpretation of load and stress
histories (the sequence of cyclic stresses causing fatigue crack growth). The
stress history is of eminent importance in damage tolerance analysis, and yet,
this subject is rather ignored in theoretical texts.
Mathematics have been kept to the barest minimum. Lengthy derivations do
not appear if they are irrelevant to the understanding of the 'why'. Instead,
dimensional arguments are used to show why an equation must be of the form
in which it appears. Nevertheless, as in any analysis, a certain amount of
mathematics is unavoidable and necessary.
Following the background are chapters dealing with obtaining the ingredients
for the analysis, and the performance of the analysis. Chapters 7, 8 and 9
respectively provide information on the interpretation and use of the materials
data needed for input, simple methods to obtain geometry factors for complex
structures and cracks, and the effects of special conditions such as mixed mode
loading, cold work of holes, interference fits, residual stresses, etc. Approxima-
tions and the assessment of their errors are discussed in extenso. The analysis
procedure is illustrated on the basis of examples of applications to different
types of structures, cracks and load spectra (Chapters 10 and 14). Chapters II,
12 and 13 show how damage tolerance analysis is used for the implementation
of fracture control measures, while addressing sources of error, pitfalls,
common misconceptions, rules and regulations, and the establishment of
fracture control plans.
The reader may find that certain subjects and problems are touched upon at
several places in this book. Cross references are then made to the pertinent
chapter where the subject is discussed in detail. This was done on purpose, at
the risk of offending those readers who go through the entire text in sequence.
It is anticipated that many readers will use this as a reference text; cross
references to related subjects and problems are included for their convenience.
For those readers and students desiring to sharpen their skill, exercise problems
are provided at the end of each chapter; solutions are given in Chapter 15.
A final remark about the use of units is appropriate. Although metric units
become more and more accepted, English units and kg-force are still used at
many places. Since the unit system used is of no consequence whatsoever for the
essence of the engineering procedures, various unit systems are being used in this
text. No attempt is made to provide other units between parentheses, as this
requires a lot of space and subtracts more than it adds. The only units of interest
20

are those oflength, stress, force, and stress intensity. Should the reader want to
do so, she or he can easily make the conversions.
Material properties such as tensile strength, yield strength and collapse
strength are consistently referred to as strength and denoted as F tu , F tv and
Feol respectively. Although they are expressed as stress, denoted as (J, they are
critical values (material properties).

1.6. Exercises

1. Design a tension member with circular cross section to carry a load of

1000000lbs (1000 kips) with a safety factor of 2 against yield, and a safety
factor of 3 against static fracture. Consider two materials: Material A with
F ty = 50 ksi and F tu = 60 ksi; Material B with F ty = 50 ksi and
F tu = 80 ksi. What is the ultimate design load in each case.
2. Design the same member as in problem 1 with a safety factor on load of 2.5
for static fracture, while no yielding is allowed at 1.3 times the maximum
service load.
3. Assume that for the members designed in problems I and 2 a damage
tolerance rule applies requiring that the residual strength may never drop
below 1.2 times the maximum service stress. For all cases calculate (Jp and Pp •
How large is the remaining safety factor in each case?
4. Define the difference between cracking and fracture. By which mechanisms
do cracks develop and grow?
5. Assume Material A and the design of problem 1, and assume the damage
tolerance requirement of problem 3 is in effect. Further assume that the
maximum permissible crack size following from the damage tolerance re-
quirement is 2 inch. Sketch the residual strength diagram in terms of stress
and load, and identify all known points (nomenclature and numbers) in the
diagram.
6. Using the result of problem 5, and assuming that cracking occurs by fatigue,
estimate graphically the fatigue crack size that would cause a fracture if a
load equal to 1.5 times the maximum anticipated service load would occur
(is this likely to happen?). Also estimate the crack size which would cause a
fracture at the maximum anticipated service stress (under which circum-
stances might such a fracture occur?).

References
(1) D. Broek, Elementary engineering fracture mechanics, 4th Edition, Nijhoff (1985).
 21

[2] M.F. Kanninen and C.H. Popelar, Advancedfracture mechanics, Oxford University Press (1985).
[3] J.F. Knott, Fundamentals offracture mechanics, Butterworths (1973).
[4] D. Broek, Some contributions of electron fractography to the theory of fracture, Int. Met.
Reviews. Review 185, 9 (1974), pp. 135-181.
[5] C.Q. Bowles and D. Broek, On the formation of fatigue striations, Eng. Fract. Mech., 8 (1972),
pp.75-85.
[6] V. Kerlins, Modes of fracture, Metals Handbook, 12 (1987), pp. 12-71.
 CHAPTER 2

Effects of cracks and notches: collapse

2.1. Scope

In Linear Elastic Fracture Mechanics (LEFM; Chapter 3) as well as in Elastic-

Plastic Fracture Mechanics (EPFM; Chapter 4) the analysis of fracture is based
on a parameter representing the crack tip stress field; while the quantity used in
EPFM is actually the strain energy release rate, this can be shown to be
equivalent to a stress field parameter. In neither LEFM nor EPFM the possibil-
ity of so-called plastic collapse is implicitly evaluated. Fracture mechanics
analysis may provide a fracture stress (residual strength) higher than the stress
for plastic collapse; since the maximum load carrying capacity is reached at the
time of collapse, the fracture stress calculated with fracture mechanics may be
in error (too high). Similarly, in such a situation, the calculated critical crack size
would be too large. Fracture parameters measured in a test where fracture
occurs due to collapse would be too low.
No fracture analysis is complete, without the evaluation of collapse
conditions. Collapse and fracture are competing conditions and the one satisfied
first will prevail. Before the fracture mechanics analysis procedures are discussed
in Chapters 3, 4 and 5, the conditions for collapse will be reviewed in this
chapter.

2.2. An interrupted load path

Discontinuities, fillets and notches, and cracks in particular, give rise to a stress
concentration, i.e. a local region where the stresses are higher than the nominal
or average stress. The following is a brief and limited discussion of stress
concentrations due to geometrical discontinuities in the structure. Stress con-
centrations playa decisive role in virtually all structural cracking problems.
Consider two parallel bars of the same size and of the same material, as in
Figure 2.1. Each carries half of the total load; the strain in the bars is equal,
causing an elongation AL. If the left bar is cut in two, the right bar will have to

22
 23
p

I J
2(J
(J (J

I I
p p

Figure 2.1. Effect of cutting one of two parallel bars.

carryall the load: the stress, the strain and the elongation will be twice as high
as before. Consequently, the gap between the two halves of the left bar will be
2A.L. The left bar carries no stress at all (Figure 2.l.c).
Next consider the case where the left and right bar are attached (e.g. welded)
as depicted in Figure 2.2. If the bars are intact, the situation is identical to the
previous. However, if the left bar is cut a different situation develops. If the top
half of the right bar is strained, the top half of the left bar must necessarily
undergo approximately the same strain. Both bars being of the same material
(same modulus of elasticity, E), equal strain in the bars dictates equal stress
(0' = eE). Thus, each bar still carries the same stress and therefore each bar
carries half the load. However, since the left bar is cut, the right bar alone must
carry all load across the cut. Below the cut the two bars are again attached and
must strain equally. Consequently the bottom halves of the bar again share the
load equally.
The attachment sets the condition for approximately equal strain and equal
stress in both bars, almost all the way to the cut. Close to the slit, the load of
the left bar must be transferred to the right bar which, over a short distance, will
then carry the total load. The nominal stress in the right bar in the section of
the slit is then PIA. However, the load from the left must be transferred to the
right and back over such a small distance that the additional load cannot be
distributed evenly. Instead, most of the extra load will be carried by a small
portion of the right half cross section, so that higher stresses occur close to the
cut (Figure 2.2c): there is a stress concentration at the cut. It does not make a
24
p
p

I I 1T 1Ll.l,
(T
Ll.l

!
.1 .10-
Ll.l<Ll.l,<2
Iii 0-
l
I
I
I
I I
p

Figure 2.2. Effect of cutting one of two welded bars or half of one (wider) integral bar.

difference whether the bars are welded together or whether they are one piece,
cut until midway. Transfer of the load from the cut half to the right half takes
place by shear. This can be appreciated by imagining how a material element of
the left deforms in the area of the slit as depicted in figure 2.2.
As long as the bar is intact, the strain is uniform and the total elongation is
11L. If the bar IS half-way cut the stress and strain will still be uniform in the
extreme upper and lower portions; they will be the same as in the bar without
the cut. Only in the area of the slit, where the total load must be carried by half
of the bar, the stresses and strains are higher. Hence, the total elongation will
be somewhat larger than ilL, but much less than 21lL (compare Figures 2.1 and
2.2).
It is helpful to consider load-path (load-flow) lines: imaginary lines indicating
e.g. how one unit of load is transferred from one loading point to the other
(Figure 2.3). For uniform load the flow lines are straight and equally spaced,
indicating that the load is evenly distributed (uniform stress). If the load path
is interrupted by a cut, the flow lines must go around this slit within a short
distance as shown in Figure 2.3. Were the load lines rubber hoses and the cut
a wedge, a similar pattern would develop.
At the tip of the cut the flow lines are closely spaced, indicating that more load
is flowing through a smaller area, which means higher stress. Load flow lines are
also useful for obtaining a rough indication of the direction of stress. In
by-passing the cut the flow lines are bent, i.e. the load changes direction. The
direction of the load flow line is an indication of the direction of the local tensile
 25

----- ...

(a) (b)

(c)

Figure 2.3. 'Load-flow' lines.

stress, as indicated in Figure 2.3.c. It appears that the direction of load around
the notch is not the same as in the uniformly loaded part: the local stress has a
vertical as well as a horizontal component, and it must be concluded that in the
vicini ty of the cut the stress field is biaxial (a y and ax) while the applied load is
uniaxial. In the absence of the cut the stress field is uniform, the state of stress
uniaxial throughout. Due to the slit a biaxial stress field develops locally. Not
only does the slit cause a stress concentration, it also gives rise to a transverse
stress.

2.3. Stress concentration factor

In the case of a notch (instead of a sharp cut) the situation is very similar. Every
discontinuity forms an interruption of the load path, will therefore deviate the
load-flow lines and, hence, cause a stress concentration (Figure 2.4). If the notch
is blunt, e.g. a round hold, its dimension in the direction of load is larger,
causing an earlier deviation of the load-flow lines. Redistribution of this load
can then take place over a larger distance. As a consequence, the area of stress
concentration is more extended than in the case of a cut, and the highest stresses
are less, i.e. the stress concentration is lower.
As a general rule, blunt notches produce lower local stresses, sharp notches
cause higher local stresses. The highest local stress a/is a number of times higher
than the nominal stress anorn' The ratio between local stress and nominal stress
is called the theoretical stress concentration factor, denoted as k t (elastic
stresses). The local stress is
26

17?
110

tt
Figure 2.4. 'Load-flow' lines around notches and consequent stress concentrations.

(2.1)
The nominal stress is not always defined in the same manner. Sometimes one
uses the (uniform) stress in the full section away from the notch, sometimes the
average stress in the section through the notch. As the local stress is the same
in either case, the use of a different definition for the nominal stress leads to
different values of the stress concentration factor. This presents no problem as
long as the proper combinations are used to obtain the local stress.
For an elliptical notch (Figure 2.5), the stress concentration factor is:
b
k/ = 1 + 2-
a
(2.2)

where a and b are defined as in Figure 2.5. The radius of curvature, e, of the
ellipse at the end of the transverse axis is e = d Ib, so that Equation (2.2) can
also be written as:

kt = 1 + 2 A = 1 + 0( A· (2.3)

In the case of a circle, b = a, (and b = e = R; R being the radius of the hole)

so that the stress concentration factor of a circular hole is equal to k/ = 3.
Stress concentration factors for many notch shapes have been determined.
They can be found in handbooks, the best known being the one by Peterson [1].
Equation (2.3) is a more or less general form for the stress concentration factor
if b is interpreted as a relevant geometrical dimension (0( depends upon notch
geometry). The equation shows the large effect of the notch-root radius e; the
 27

f
p

2b

(a) (b)
p

yl
I
I
I
I

(c)

Figure 2.5. Elliptical notches. (a) High stress concentration; (b) Low stress concentration; (c) Stress
free notch faces.

sharper the notch (smaller (2), the larger is k,. For an ellipse with bla = 3 (12 =
b19) the stress concentration factor is k, = 7, while for an ellipse with alb = 3,
the stress concentration is only k, = 1.67.
28

2.4. State of stress at a stress concentration

As was shown in Section 2.2, a stress concentration also causes a change in the
state of stress; even if the stress is uniaxial throughout the remainder of the
body, the state of stress in the area of the notch will be at least biaxial (Figure
2.3). At the free surface where no external loads are acting the state of stress will
be plane stress (there are no stresses on a free surface). Since the free surface
carries no shear either, it is a principal plane with a principal stress equal to zero.
A state of stress in which one of the principal stresses is zero is a state of plane
stress.
The face of the notch (Figure 2.5c) is a free surface: it carries no stress (plane
stress). The root of the notch (if there is a radius) is also a free surface. This
means that at the free surface of the notch root (Jx must be zero, because there
is no stress on (perpendicular to) that free surface. Slightly inwards from the
notch root, however, (J x will be non-zero. The faces of the plate are still stress
free, so that (Jz = O. The state of stress is biaxial (plane stress). At some small
distance from the notch root, the state of stress may be triaxial as explained
below.
Because the surface (face of plate) is a principal plane, and because the three
principal planes are mutually perpendicular, it follows that the stresses (Jx and
(Jy at the notch root in the plane of the notch are the principal stresses (JI and

(J2. Due to the stress concentration the local values of (JI and (J2 are very high

and so are the strains 81 and 8 2 • According to Hooke's law, this will lead to a
strain in Z-direction, given by:

8z = (2.4)

assuming that the stress (Jz = o. This negative strain, 8 n indicates a thinning of
the plate.
The stresses are high only in the vicinity of the notch root. Further away (Jx
vanishes and (J v is much lower. Hence further away 8z would become very small.
The faces of the notch are stress free, so that 8 z = 0 along the faces. It appears
that there will only be a small amount of material at the notch root for which
8 z should be very large, while around it 8 z is either zero (notched faces), or small
(further inwards). Assuming that the material wanting to undergo large 8 z is
approximately a cylinder (roll), the situation is as shown in Figure 2.6.
If the roll is very long (large thickness) and thus relatively thin, such a large
8 z cannot take place. The surrounding material, being attached to this roll, will
not permit the contraction to occur, apart from a little bit at the face of the plate.
Imagine that the roll of material is a steel bar cast in concrete, where the concrete
represents the surrounding material. If the steel is cooled, it wants to contract.
However, contraction will be prevented by the concrete, which would lead to
 29

Figure 2.6. Contraction at notch.

(thermal) tension stresses in the bar. The same happens to the roll of material
in Figure 2.6: contraction is prevented by the surrounding material and a tensile
stress develops in the roll.
Apparently, constraint of the contraction causes a tension {Jz in the z-
direction. Should the contraction be completely constrained then the strain 6z is
zero. Writing the complete equation for 6: provides:

(2.5)

and then:
(2.6)
Equation (2.6) shows that a high tension develops in thickness direction when
no contraction is permitted. This stress is exerted on the roll by surrounding
30

material acting to constrain the contraction. At the free surface this stress
cannot exist, but it builds up rapidly, going inward (Figure 2.7). Due to the
absence of a {!z at the surface, an 8 z occurs there, so that a small dimple develops
at the surface.
It follows then that at the surface, with stresses only in X and Y-direction,
there is a biaxial state of stress (plane stress). Further inward there is a triaxial
state of stress. Should there be complete constraint then this triaxial state of
stress is plane strain, because 8 z = 0 for complete constraint. (A state of stress
where one of the principal strains is zero is called plane strain).
Now consider a thin plate with a notch as in figure 2.7b. In this case the roll

t h ic kness
B
.. -- B-

t
cr

-
cr

+
thick plate thin plate
thin cylinder no contraction free contraction
plane strain plane stress

Figure 2.7. Contraint of long, thin cylinder (large thickness) and free contraction of short, thick
cylinder (small thickness).
 31

---
-, ,
~
---
-.. .....
, I
\
\
I

Figure 2.8. Length of contracting cylinder bears no relation to thickness in case of surface notch.

of material wanting to undergo contraction is short and relatively thick. con-

traction can occur freely and will be in accordance with Equation (2.4). The
stress in thickness direction will be zero (O'z = 0; plane stress).
In cases between those of Figure 2.7a and b, there will be some, but not
complete constraint. These transitional cases have a triaxial state of stress, but
not one of plane strain.
Apparently the state of stress at the notch root depends upon the length of
the roll. As the length of the roll is equal to the thickness, one can argue that
the state of stress depends upon thickness. However, this dependence upon
thickness is a coincidence. If the notch is of the type of Figure 2.8, the length
of the roll has no relation to the thickness; in such a case the thickness is
irrelevant for the state of stress. Although this may seem trivial, its importance
is emphasized; in the use of fracture mechanics serious errors are possible if it
is assumed that the state of stress is always dictated by thickness.

2.S. Yielding at a notch

The stress required for plastic deformation depends strongly upon the state of
stress. In plane stress yielding occurs when the highest principal stress is ap-
32

proximately equal to the yield strength, but much higher stresses are required
in the case of a triaxial state of stress.
For readers not familiar (any more) with yield criteria, the following brief
summary may be of help. Yielding is plastic deformation which takes place by
slip; it is therefore caused by shear stresses. Plastic deformation will not take
place unless the shear stress is sufficient to cause slip. In a tensile bar (uniaxial
tension) the state of stress is as shown in Figure 2.9a. Yielding occurs when (J =
FlY where FlY is the uniaxial yield strength as measured in a tensile test. Since
plastic deformation takes place in the case of Figure 2.9a, the shear stress

3
(a) (1,
(1, (1,

fl
(b) 1
v,

Triaxial equivalent
(hydrostatic state
of stress)

~~~ g,~~~ 0~
(~ ~ ~ 0; v, (1,

v,
v, (1, Triaxial

(1"~2~Y2
v"v\h

(d)~
(1,

Figure 2.9. Postponement of plastic deformation due to combined stresses. (a) Uniaxial (longitudi-
nal); shear followed by slip (plastic deformation); (b) Uniaxial (transverse); shear followed by slip;
(c) Biaxial; no shear, no slip, no plastic deformation; (d) Biaxial (0"2 = O"tf2; plastic deformation,
but not until stress is twice as high as in case a.
 33

required for slip must have been exceeded. The maximum shear stress acting on
a plane at 45 degrees is" = a/2. Apparently then the shear stress required for
slip (yield) equals rty = Fty /2.
Figure 2.9b shows a similar case of uniaxial loading; again yielding occurs
when the shear stress equals Fty /2. Next consider Figure 2.9c assuming for the
moment that the material is two-dimensional (existing in the plane of the figure
only). Since there is equal tension in both directions the two shear stresses of
Figures 2.9a and b cancel, so that there is no shear stress in the case of Figure
2.9c. Without shear stress there can be no slip and no plastic deformation. In
order for slip to occur one of the stresses must be larger than the other by F ty ,
in which case the same shear stress will be present as in Figure 2.9a. For
example, if the horizontal ~tress is Fty , the vertical stress must be 2Fty for yielding
to occur, so that the highest stress must be twice the yield strength before plastic
deformation begins. If the horizontal stress were 3Fty , the vertical stress would
have to be 4Ftv before the shear of Figure 2.9a would be restored. In both of
these examples the difference between the two principal stresses is F tv •
Naturally, a real material is three dimensional as in Figure 2.9, and out of
plane shear stresses are possible. But it is easy to see now that there still will be
no shear when all three principal stresses are equal (Mohr's circle becoming a
point), and hence, there will be no plastic deformation. Apparently, yielding
requires that the difference between the largest and smallest principal stress is
equal to Fty , because only in that case will there be a shear stress r = Ftv /2 as
required for yield. Thus, yielding (slip) will take place when (r Fty /2):
or (2.7)
depending upon which is lower a2 or a3. Were for example a 2 = a 3 = 0.8 ai,
then it would follow from Equation (2.7) that the stress required for yielding is:
or
i.e. the stress would have to be 5 times the yield strength for plastic deformation
to occur. If a l = a2 = a3' the shear stress is zero, and yielding will never occur,
regardless how high the stress. The above is the Tresca yield criterion. Instead,
the Von Mises yield criterion is more generally used, but its results differ very
little from those of Tresca. As it makes no difference for the essence of the
discussion, the Tresca criterion is used here, because it is the easiest to
understand.
The above has important repercussions for the yielding at a notch, since the
state of stress may be triaxial. For example, consider a case of plane strain,
where the third stress a z is given by Equation (2.6). Note again that for a
material element in the section through the notch, the principal stresses ai' a2 ,
a3 are an aX' a z respectively. Consider a case in which ax = a v as would occur
at a very sharp notch. Given that v ~ 0.33, the lowest principal stress a 3 = a z
34

would be a z = 0.33 (ax + a y) = 0.66 ay. Plastic deformation would then

require a stress ay = 3Fly according to Equation (2.7).
Let Figure 2.10 represent the notch and the distribution of ay, on the section
through the notch. At the notch root the local stress is ay- At low loads the local
stress is less than 3Fly , and therefore the strains remain elastic. Further increase
of the load will raise all stresses (Figure 2. lOb) until aY = 3Fly , upon which
plastic deformation commences.
Next consider a notch in a thin plate with plane stress as in Figure 2.11. In
this case plastic deformation occurs when ay = FlY' because (ay - aJ = (ay -
0) = ay. During further increase of the load the stress distribution develops in
a similar manner as in the previous case. Eventually the entire remaining section
will yield, unless fracture occurs earlier.
It is important that the notch tip stresses are much higher in plane strain than
in plane stress. In the latter case they are limited to FlY, in the former case to 3Fly '
Thus a plane strain condition is more severe and can more easily lead to fracture
and cracks as will be shown in detail in later chapters.
If the notch is blunt, the stress aA(2) cannot exist at the notch root (Figure
2.5). Although there could be a a z (a3), the state of stress would still be plane
stress, because of a2 = O. Hence, at the root of the notch the stress a y would
be limited to the yield strength Fly' Further inwards a x will exist; in a thick
plate with constraint, the state of stress will then be triaxial and possibly plane
strain. Further inward therefore the stress might reach as high as 3Fly , as shown
in Figure 2.12, before yielding occurs.

INCREASING
APPLIED STRESS

Figure 2.10. Yielding at sharp notch in plane strain.

 35

INCREASING
APPLIED STRESS

-ELASTIC

Figure 2.11. Yielding at notch in plane stress.

Assuming the material does not exhibit strain hardening (horizontal stress-
strain curve beyond yield), the stress cannot increase further after yielding
occurs. Away from the notch there is again a uniaxial state of stress (ux = u z =
0) so that the stress will again be limited to F,y, Hence, by the time the entire
section is yielding (fracture may occur before this can happen) the stress distri-
bution in the section is as shown in Figure 2.12. Whether there is plane stress
or plane strain at the notch, the final stress distribution will be about the same
(compare Figures 2.11 and 2.12 and note that triaxial stress area is very small),
apart from a small area where the stress reaches 3F,y in the plane strain case.
Note that these stress distributions may not be reached if fracture occurs before
the entire ligament yields.

2.6. Plastic collapse at a notch

Not only may the high local stresses cause cracks by fatigue, stress corrosion (or
creep) which may eventually lead to a fracture, they may cause fracture to
proceed immediately from the notch, in particular when the notch is sharp (e.g.
crack). Alternatively, failure can occur by plastic collapse which is always
followed by fracture.' If the stress distributions of figure 2.11 or 2.12 can be
reached before fracture then plastic collapse could occur. Obviously, with the
given stress-strain curve without work hardening, the cross section with the
notch cannot carry any more load once the entire cross section is yielding,
because the yielding will continue unihibited until fracture results. This is called
plastic collapse. Thus, in plane stress where the stress in the entire cross section
is equal to yield strength at the time of collapse, (Figure 2.11) the maximum load
36

EXTEND OF HEAVY LINE

SHOWS EXTENT OF
YIELDING

~ .''1 Triaxial "''1 Plane

I Strain but not Plane Strain Stress

Plane Stress
at surface
of notch (crx=O)

Figure 2.12. Progress of yield at blunt notch.

carrying capability is:

Collapse: Pmax B(W - a)Fly (2.8)

if a is the notch depth, W the total width and B the thickness. This failure load
is called the collapse load or limit load. The nominal stress in the full width part
IS (J = PIBW. Hence, the part in Figure 2.11 fails when the nominal stress is:

W-a
Collapse: (Jfe = Pmaxi W = W FlY' (2.9)

This is the equation of a straight line (as a function of the notch depth a. If a =
W, failure will occur already when the nominal stress is (Jre = O. This failure
stress is plotted as a function of notch depth in Figure 2.13. If fracture indeed
 Notch depth a

Figure 2.13. Net section collapse.

occurs as a consequence of collapse, the strength, atn would be the residual

strength as defined in Chapter I.
If the material work hardens, the notched cross section can carry a higher
load. In general however, it cannot reach a situation where the entire cross
section carries a stress equal to the tensile strength. This can be understood if
it is realized that the strains in the cross section are not uniform. Yielding starts
at the notch root (Figure 2.11) and proceeds through the ligament. Hence, the
strains at the notch root are always much higher than elsewhere. This is depicted
in Figure 2.14. Clearly, even though after yielding the stress is almost uniform,
there is still a strain concentration at the notch.
As long as the stress is elastic there is a stress concentration given by a, =
k,a nom ' In the elastic case, a = eE. therefore the strain concentration will be:

(2.10)

where the strain concentration, k" is equal to the theoretical stress concentratin,
k, = k,. When the whole ligament is yielding, the stress concentration has
disappeared (k a = 1 in Figure 2.14b), but there is still a strain concentration.
Neuber [2], writes for the stress concentration factor ka, and for the strain
concentration factor k" and postulates that:
(2.11 )

In this equation k, stands for the theoretical stress concentration factor, as

discussed. In the elastic case ka = k" so that both are then equal to k,. The
38

fA - - - - - -- - -- - - - - - - - - - - - - - 1J.J4.

A B

(c)
(b)

Figure 2.14 .. Stress and strain distribution at notch (plane stress). (a) Strain; (b) Stress with
horizontal stress-strain curve; (c) stress with rising stress-strain curve.

stress concentration factor, kG, is reduced when the material yields, but the
strain concentration increases. Taking the case of Figure 2.14b where ku has
become kG = I, the strain concentration factor has become k, = k; / kG = k~ /
I = k;. In the case of e.g. k, = 3, the stress concentration decreases from
kG = k, = 3 in the elastic case to kG = I in the fully plastic case; the strain
concentration on the other hand increases from k, = k, = 3 in the elastic case
to k, = k; = 9 in the plastic case.
Now consider a work hardening material in which the stress in the ligament
could be raised somewhat above F,y- Some stress concentration will remain
(Figure 2.14c) but the strain concentration will become very high. As the whole
ligament is plastic, even the strain at the plate edge is above yield, but at the
notch the (plastic) strain is many times higher. Eventually the high notch root
strain will cause fracture long before the stress further away has reached values
much above F,y. Once fracturing has begun a much sharper notch has formed,
so that the situation becomes worse and fracture continues.
Given the strain distribution and the stress-strain curve, the stress distribu-
tion can be sketched as shown in Figure 12.14. When tearing or plastic collapse
commences at the notch tip, the stresses in most of the ligament are still close
 39

to F,y, because the strain gradient is very steep. Thus, even in a strain hardening
material and plane stress, the (average) stress in the cross section cannot reach
Ftu , where Ftu is the tensile strength (see stress-strain curve in Figure 12.14).
Apparently, collapse will occur at an average ligament stress somewhat higher
than F,y but less than Ftu . The average ligament stress at which collapse occurs
is called the collapse strength Fcol . Note that for a non-workhardening material
Fcol = F,y, and that at best Fcol = F,u. It is not very well possible to determine
the values of F,y and F,u other than by a (tensile) test; similarly, it is not very well
possible to determine Fcol other than by a test on a notched sample. However,
its value depends upon the severity of the notch (ku and k,). For circular holes
with k, = 3, the collapse strength is often close to the tensile strength
(Fcol ~ F,u), but for a sharp crack, the value is very close to the yield strength
(Fcol ~ F,y).
It follows that for a work hardening material Equation (2.9) changes into:
W- a
ufe = W Fcol (2.12)

which is also a straight line as a function of notch depth (Figure 2.13).

If there is plane strain, or in general non-plane-stress, the stress distribution
after yield is not uniform (Figure 2.12). Since the stress peak is local, the average
stress in the section cannot become much higher than in the case of plane stress.
In most cases - depending upon notch acuity - Equation (2.12) still applies.
It should be noted that the above discussion was strictly for the case of
uniform applied loading. If there is bending (or other stress gradients in the
applied stresses), the conditions for collapse are slightly more complicated. The
fully plastic stress distribution in the ligament for the case of bending is shown
in Figure 2.15c. The maximum bending moment occurs when the net setion
stress is equal to F col. Taking the moment around point A one obtains:

Mmax = 2FcolB (w2- a)(w 4- a) I

= 4 2
FcoIB(W - a) . (2.13)

Using the collapse strength of e.g. 50 ksi and W = 2in, B = 0.5 in (Figure 2.15)
and a = 0.5 in, the collapse moment is:
Mmax = t x 50 x 0.5 (2 - 0.5)2 = 14.1 inkips = 14100 inlbs.

One can establish the entire collapse curve by solving Equation (2.13) for
various crack sizes, as shown in Figure 2.15d.
Analysis of combined bending and tension requires determination of the
point of stress reversal, D, in Figure 2.15e. Since the stresses in the cross section
have to be in equilibrium with the external load, one can establish 2 equilibrium
40

tp
50
a
a (ksi) 40
37.5
Feol
30
..."..,

I-----f 20
1.5

10
6

(a) (b)
~P
2 3 4 5 6
2a (in)

-----....
M
------..
M
---......
M
M (103 in-Ibs)

25
W
20

15

10

(e) ~ ~ ~ 1 2
(d) 2a (in)

/'
/' /V B
/j
./"
la
Feol
W
d
0

(e)
I (W-a- d)

Figure 2./5. Collapse analysis. (a) Center crack; (b) Collapse stress; (c) Stress distribution for
increasing M. (d) Collapse strength; (e) Combined bending and tension.
 41

equations, one for P and one for M. For a given P one finds D and then M by
solving the equilibrium equations. This is shown in Table 2.1.
Generally speaking, in engineering all solutions must be obtained in terms of
the nominal applied stress, as was done going from Equation (2.8) to (2.9). For
the bending case the same can be done, because the nominal bending stress at
Mmax is (J = 6 Mmaxl BW2. Substitution in Equation (2.13) leads to:

(Jle =
3(W W- a)2 F
2 eol •
(2.14)

Note that in contrast to Equations (2.9) and (2.12), the Equation (2.14) is not
a straight line, but a curve (Figure 2.15d). For complete derivation see Table 2.1.

2.7. Fracture at notches: brittle behavior

The collapse load or limit load is the highest load that can ever be reached, i.e.
the absolute maximum load carrying capability is defined by the collapse load.
Fracture follows automatically when the collapse conditions are reached.
However, fracture may occur already before the collapse conditions are
attained. (This is the concern of fracture mechanics analysis as discussed in
Chapters 3 and 4). In cases where fracture occurs due to collapse, the residual
strength (Chapter 1) is determined by Equation (2.9) or (2.14): (Jres = (Jel. The
residual strength can never be higher. If fracture occurs before collapse, the
residual strength is lower than the nominal stress at collapse ((Jres < (Je/).
Strains and, up to point, stresses are higher at the notch than anywhere else.
In the case of very high k, this may lead to local fracture long before the
remainder of the section reaches collapse. As a matter of fact, with very sharp
notches (cracks) the condition for fracture at the notch root may already be met
while virtually the entire notched section is still elastic. Once fracturing starts,
the notch has beome longer and sharper, so that the fracture proceeds
throughout the ligament. Cracks due to fatigue or stress corrosion constitute
very sharp notches; in some cases a cracked part will be capable of reaching the
collapse load, but in many cases fracture occurs at much lower loads (stresses).
This problem will be discussed in detail in Chapters 3 and 4. Nevertheless,
collapse, if it occurs before the fracture condition is reached, will cause failure.
According to Equation (2.9) collapse occurs when the stress in the full section
has not yet reached the yield strength. Thus all plastic deformation is confined to
the section through the notch. If the collapse strength is higher than the yield
strength Equation (2.12) applies. In that case the stress in the full section could
reach or exceed the yield if the collapse stress is high and the notch size small.
In general however, the stress at failure in the full section will still be below yield.
Thus, in the vast majority of fractures occuring at notches and cracks, plastic
deformation takes place in the notched section only, in particular if fracture
occurs before collapse conditions can be attained (Chapters 3 and 4).
42

Table 2.1. Collapse conditions for cases with bending

Pure bending: unit thickness:

6M
(a)
W2

From Moment equilibrium around A:

W- a
Collapse: M 2 --
2- x .1
4
(W - a) Fco I

M = 0.25 (W - a)2 Fcol . (b)

Substitute (a) in (b) for (M = a b W 2 /6):

(c)

Note: quadratic curve; also arc 1.5 Fcol for a = 0, collapse curve starts at a = 1.5 Fcol (not at Fcol
as in tension).

o
(j
w
w-o P=(JW

"Fcol
Combined bending and tension (bending due to excentricity; unit thickness, total load is P = aW
in center; hence eccentricity with respect to cracked section.
Point B (i.e. Xs) is unknown:
Horizontal equilibrium:
aW = (W - a - xs)Fcol - xsFcol (d)
Moment equilibrium point B:

(e)

Solve (d) and (e) to obtain XB = 0.5 W - 0.25 J4W2 - 8a(W - a), and more important:
afe = (a/W + 0.5 J4 - 8a(W - a)/W2) ~ol.
 43

The limitation of yielding to the notched section has considerable effect on the
total elongation at the time of failure. In a normal tensile test on an unnotched
sample the strain at the time of necking is often on the order of 10% or more.
Thus the bar has become 10% longer (Figure 2.16). During necking still higher
strains occur. In a notched sample plastic deformation occurs only in the
notched section. Even if this small section would deform 10% before fracture,
the total elongation would be much smaller than of the unnotched bar (Figure
2.16). Consequently, elongation after fracture is hardly noticeable. If little
overall plastic deformation occurs, the fracture is called 'brittle' from an engi-
neering point of view. The text in this book strictly adheres to the engineering
use of the word. In that sense virtually all service fractures are brittle: fractures
occur almost always at notches or cracks. Thus plastic deformation is always
restricted to the notched section and the fractures are brittle in the engineering
sense. The fracture mechanism may be either by dimple rupture (often called
ductile) or by cleavage (often called brittle; see chapter 1), but the great majority
of fractures occurs by rupture; from an engineering point of view they are
brittle because of little overall deformation in the cracked section (Figures 2.16).

p p
P

t
AFTER AFTER AFTER
FAILURE FAILURE FAILURE

I-::-l ~
i Wo i .
Wo •
I I
La Lr La
_{ 6
La >- ..... 6{-

® _i6 @
JJ
~
p p La

--------------p

Figure 2.16. Apparent 'brittleness' due to notches and cracks as a consequence of localized plastic
deformation in area of high stress.
44
2.S. Measurement of collapse strength
The easiest way to measure the collapse strength, Fcol , is by means of a center
cracked panel. In view of the fact that constraint will still play a role, the
specimen should have a thickness close to the thickness used in the structure of
interest. As this may lead to unwieldy specimens, a compact tension specimen
(Chapter 3) can be used where circumstances so dictate, provided it is analyzed
for combined bending and tension (Table 2.1).
It should be noted here, that in the case of center cracks, the definition of
crack size is 2a instead of a. This is a convention, by which all cracks with one
tip are defined as a, all cracks with two tips as 2a (see also Chapter 3). Note that
with this definition Equations (2.9) and (2.12) become for the center cracked
panel:

(2.15)

The specimen may be fatigue cracked, but blunting before collapse usually is so
extensive that a sharp saw cut and a fatigue crack are undistinguishable at the
onset of fracture. The load at fracture (maximum load) is recorded. From this
the net section stress can be calculated directly as:
PfraClure I (2.16)
w- 2aB
Note that the measured Fcol should be greater than or equal to the yield strength,
otherwise fracture occurred before collapse was reached and the measured net
section stress is not the collapse stress. In that event the test should be analyzed
using fracture mechanics as discussed in the following chapters.
Instead of calculating the net section stresses directly, one may calculate the
remote stress, (5= PIA, at the time of fracture, which is the residual strength. By
plotting the latter in a diagram such as in Figure 2.17, and by drawing a straight
line through the data point and the point (W, 0), the collapse strength Fcol is
found at the intercept of this straight line with the vertical axis, as can be seen
from Equation (2.15). The advantage of this procedure is that results obtained
from specimens with different crack sizes can be plotted as well, thus providing
a check of the applicability of the collapse criterion. If all data fall on the same
straight line as per Equation (2.15) then collapse occurred in all tests; if they do
not, the tests below the line represent cases where fracture occurred before
collapse and these tests should be analyzed using fracture mechanics concepts
(Chapters 3 and 4).
Figure 2.17c presents test data [3] for 304 stainless steel showing a collapse
strength of 67 ksi. (The yield strength of this material is 30 ksi and the ultimate
strength is 90 ksi). Indeed, the collapse strength is considerably higher than
 45

Fracture
Stress
a

•
(a) W 28

(b)
W 28

a
(ksi)

304 Stainless Steel

12 in. Wide

10

(c) 2 4 6 8 10 12 2a(jn)

Figure 2.17. Measurement of Fccl • (a) Plot of test data; (b) Determination of Feol ; (c) Actual test data
(Ref. 3).
46

yield. For all cracks smaller than 5-in the residual strength is more than 30 ksi.
This means that the entire panel was above yield (O'r/ is the nominal remote
stress) and not just the net section. Data points at various crack sizes all fall on
or close to the straight line, thus confirming that collapse occurred in all tests.
Other metal alloys usually have a collapse strength much closer to, or equal
to the yield strength; in this respect the example of 304 stainless steel is not
representative.

2.9. Exercises

1. Calculate the theoretical stress concentration factor of an elliptical notch

with a semi major axis of2 inch perpendicular to the applied load, and a semi
minor axis of 0.4 inch. What is the stress concentration factor; what is the
strain concentration factor. What are these factors after the notched section
has fully yielded in plane stress, assuming no strain hardening?
2. For the case of Exercise 1 calculate the nominal stress in the full section at
the time of collapse if the yield strength is 50 ksi. Do the same for a material
with a collapse strength of 75 ksi. Calculate the fracture load in the two cases
if the notch is a center notch and W = 12 inch, B = 0.5 inch.
3. Cut a strip of paper 4 inches wide and 10 inches long: At three inches from
one end cut a circular hole of 1.6 inch diameter in the center. At three inches
from the other end cut a slit of 1.6 in long and two holes at both ends of the
slit of 0.2 inch diameter. After cutting the holes cut the notch into an elliptical
shape (semi minor axis 0.2 in). Calculate the stress concentration factors of
the two notches and predict where the strip will fail. Roll the ends around 2
pencils and pull the strip to failure to prove your prediction.
4. If a crack appears in service one sometimes drills a so called stop hole at the
crack tip as a temporary repair. Suppose a crack has started at the edge of
a strip. Its size is a. A stop hole is drilled with diameter d, its center exactly
on the crack tip. The crack tip root radius is almost zero. Assume the crack
with the stop hole to be an ellipse (why is this permissible). Calculate the
theoretical stress concentration factor before and after stop drilling for the
general case. If the crack is 1 inch long calculate which size stop hole is
needed to give a theoretical stress concentration factor of 5; and 7?
5. What is the stress at yield at a notch in plane strain if Fry = 30 ksi? What is
it at the free surface of the notch root? Sketch the stress distribution.

6. Using the solution of Exercise 5 calculate the nominal stress in the full section
at which yielding begins at the free surface of the notch. Sketch the stress
distribution in these two cases. Assume kr = 4.
 47

7. Consider a cylindrical bar with a circumferential groove of half circular

shape. In order to reduce the stress concentration one sometimes machines
relief grooves: similar grooves but of smaller depth at short distance on both
sides of the main groove. Demonstrate by means of load-flow lines that this
indeed gives a lower stress concentration factor.
8. Using load-flow lines demonstrate the high stress concentration factor (as
high as 10) at the fillet radius of the head of a tension bolt.
9. Calculate and construct the diagram for fracture due to collapse for a center
cracked panel as well as for an edge cracked panel. W = 600 mm; Fcol =
350 MPa. At which stress will collapse occur if the panels contain a crack of
150mm? For the edge cracked panel assume that the side edges are con-
strained (kept straight, so that no bending occurs).

References
[I] R.E. Peterson, Stress concentration design/actors, John Wiley (1953).
[2] H. Neuber, Theory of stress concentration for shear strained prismatical bodies with arbitrary
non-linear stress-strain law of J. App. Mech. 28 (1961) pp. 544-550
[3] M.F. Kanninen, et aI., Towards an elastic-plastic fracture mechanics capability for reactor
piping, Nuclear Eng. and Design (48) (1978) pp. 117-134
 CHAPTER 3

Linear elastic fracture mechanics

3.1. Scope

In this and the following chapter, the fracture mechanics concepts for the
analysis of fracture will be discussed. With these concepts it will be possible to
obtain the residual strength diagram and the maximum permissible crack size
(Chapter I). Following the arguments used in Chapter I, the discussion will be
limited to mode I loading (mixed mode loading is considered in Chapter 9).
Materials with relatively low fracture resistance fail below their collapse
strength and can be analysed on the basis of elastic concepts through the use of
Linear Elastic Fracture Mechanics (LEFM). Such materials are, among others,
practically all high strength materials used in the aerospace industry, high-
strength-low-alloy steels, cold worked stainless steels, etc. For the fracture
analysis of many tlther materials, the use of Elastic-Plastic Fracture Mechanics
(EPFM) as discussed in Chapter 4, is often indicated. Understanding of EPFM
concepts requires familiarity with LEFM concepts.

3.2. Stress at a crack tip

Consider (Figure 3.1) a body of arbitrary shape with a crack of arbitrary size,
SUbjected to arbitrary tension, bending, or both, as long as the loading is mode
I (Chapter I). The material will be considered elastic, following Hooke's law.
For such a case the theory of elasticity can be used to calculate the stress field.
The stresses, (Ix and (I, can be obtained as well as the shear stress r n ' Details of
the derivation can be found in more extensive texts [1,2, 3]. In the following only
the solution will be given, and it will be shown that the result is in accordance
with what would be expected.
As was explained already in Chapter 2, the crack tip stress field is at least
biaxial (load-flow lines) and it may be triaxial if contraction in thickness
direction is constrained. Hence, there will be stresses in at least X and Y
direction, (Ix and (Iy. From the stress field solution it appears that the stresses on

48
 49

a material element as shown Figure 3.1 can be described by (in the absence of
constraint):

. 0 . 30)
= -K- cos -0 ( 1 "2 SIn 2"
J2iU'
(1x - SIn
2
. 0 . 30)
K
--cos -0 ( 1 "2 SIn 2"
J2iU' 2
(1y - SIn
(3.1)
(1z 0
K
- - cos -
oSIn. -0 cos-
30
'xy
J2iU' 2 2 2
Indeed, both (1x and (1y exist (as should be the case as discussed in Chapter 2).
For the case that 0 = 0 (plane through the cracked section), the shear stress, 'xy,
is zero, as should be expected for a plane of symmetry. Although it does not
make any difference for the following discussion, it is convenient to confine the
considerations to the plane through the crack with 0 = 0; in that case the
functions of 0 will be either 0 or 1, so that they essentially disappear: (note also
that x = r for 0 = 0):
K
(1x =
.j2nx
(3.2)
K
(1y =
.j2nx
It appears that, at least along the plane Y = 0 for which Equations (3.2) hold,
the transverse stress, (1x, is equal in magnitude to the longitudinal stress, (1.v' The
stresses depend upon the distance x from the crack tip; note that at greater
distances (larger x) the stresses are lower. The stresses also depend upon a
parameter K, which is as yet undefined.
The expressions for the stresses appear to be remarkably simple. Even more
remarkable is that a stress field solution could be obtained at all, since the body,
crack, and the loading are arbitrary. Because of this arbitrariness it is not
surprising that the equations contain an unknown parameter K. As Equations
(3.2) are for an arbitrary case, they must describe the stresses at each and every
crack in each and every elastic body in the universe. Apparently, the equations
are universal and can be used for all crack problems. Therefore, also the
parameter K will appear in all crack problems, reason to give it a name: K is
called the stress intensity factor, not to be confused with the stress concentration
factor, k" discussed in Chapter 2.
Similar solutions can be obtained for other modes of loading (Chapter 9), be
50

CTyy [OR CTy]

Crack t x

Figure 3.1. Arbitrary body with arbitrary crack and arbitrary mode I loading.

it that the goniometric functions are different for different modes. A stress
intensity factor appears in all solutions, but the definition of K is slightly
different for different modes. For this reason, the stress intensity factors are
often labeled in accordance with the mode of loading (Chapter I), namely K),
K II , and Kill' In the above equations K should then be labeled as K). However,
when there can be no confusion about which mode is meant, the subscript is
often omitted; this is what will be done in the following. Should more than one
loading mode be operable, the labels must be carried to avoid confusion, as will
be done in Chapter 9.
Since Equations (3.2) are for all crack problems, there is no objection against
selecting a familiar geometry. Let this be a very large (infinite) panel, subjected
to uniform unaxial loading with a nominal stress (T, and a central crack, as in
Figure 3.2a. The size of the crack will be called 2a. This is a convention: In
fracture mechanics all cracks with 2 tips are called 2a, all cracks with one tip a.
There is no reason why it could not be done differently, but this convention is
necessary to compare notes. The convention should be adhered to if one wants
to use data generated elsewhere, otherwise erroneous results will be obtained.
Equations (3.2) apply to the problem of Figure 3.2a, and it is now possible
to examine the significance of the stress intensity factor, K. It should be noted
then that the stresses everywhere in an elastic body are proportional to the
applied load. If the load is increased by a factor 2, all stresses everywhere will
 51

,
(a) (b)
Figure 3.2. Center crack with uniform loading. (a) Infinite plate; (b) finite plate.

increase by a factor 2. Thus, it can be concluded that the crack tip stress must
be proportional to the applied stress, (I, i.e. (see Equation (3.2))
(I

(Iy -;- .j2nx' (3.3)

It stands to reason that the crack tip stresses will also depend upon crack size.
The stresses will certainly he higher when a is larger; hence, the crack size, a,
must appear in the numerator in Equation (3.3). There is only one way in which
this can happen. Both sides of the equation must have the dimension of stress,
but in the denominator at the right there appears a square root of a length, the
distance x. In order to cancel the square root of length in the denominator, the
crack size must appear in the numerator as square root of a:
. (IJ{I
(Iy -;- .j2nx' (3.4)

Equation (3.4) still contains a proportionality sign instead of an equal sign,

because dimensional analysis does not show whether there is a dimensionless
number involved. Calling this dimensionless number C, one finally arrives at an
equation:
C(IJ{I
(3.5)
.j2nx .
Clearly, it is simple to find the format of Equation (3.5), but a formal solution
[1, 2, 3] would be necessary to obtain the actual value of C. It turns out that
C = Jii, for the case depicted in Figure 3.2a. Hence:
52

(3.6)

Comparison of Equations (3.6) and (3.2), leads to the conclusion that for the
configuration of Figure 3.2a:
K = afo. (3.7)

3.3. General form of the stress intensity factor

The manner used in the previous section to demonstrate the significance of K is

not limited to the case shown in Figure 3.2a. For example consider (Figure
3.2b): the plate of finite width W. From the same arguments as used above, it
follows that:
CaJa (3.8)
ay = .j2nx·

In this case it must be expected that the size of the plate also will affect the crack
tip stresses. It must be anticipated that the stresses will increase when W
becomes smaller. The only manner in which the effect of W can appear is in the
factor C. Hence, C must be a function of (depend upon) the width W. However,
C must be dimensionless, and hence, it can depend upon W only through
dependence upon a dimensionless parameter such as W/a or a/W. For the
configuration of Figure 3.2b the expression for C appears to be [1]:

(3.9)

so that:

RaJa
(3.10)
.j2nx
and

K = In sec ~ aJa. (3.11 )

If W is very large, or a very small, the value of .jsec (na/W) = 1. This is in

accordance with anticipation, because for small a/ W the configuration of Figure
3.2b is identical to that of Figure 3.2a. Indeed, in that case Equation (3.11)
reduces to Equation (3.7).
Now it becomes obvious that for ANY configuration the crack tip stress will
 53

always be:

K
(3.12)
,J2nx
and the stress intensity factor is always:

K = c(I) (1y'a (3.13)

where L is a (unified) length dimension describing the geometry of the cracked

part.
In the practical use of these equations all C's are divided by J1i, and Fa
is
substituted for y'a to compensate. The function C(a/L)/J1i is then renamed {J,
the geometry factor (often erroneously called correction factor):

(3.14)

Note that Equations (3.14) are identical to the previous equations. The
Equations (3.14) present the stress and the stress intensity factor in comparison
to those for the infinite panel: {J = 1 for the infinite panel, but for the finite
width panel:

{J = Jsec n;. (3.15)

It must be emphasized that Equations (3.14) represent the crack tip stresses and
stress intensity for all crack problems, the equations having been derived from
the general solution for an arbitrary crack in an arbitrary body with arbitrary
mode I loading. For any crack in any practical problem only the function {J, or
its functional value, need be derived. For many configurations the function, {J,
has been calculated already; the results can be found in handbooks [4, 5, 6].
Examples of {J functions for a few common crack cases are shown in Figure 3.3.
Note that in the Equations (3.1) through (3.14) the stress, (1, is the nominal stress
in the uncracked section. The fact that the stresses are higher in the cracked
section when e.g. W becomes smaller, is wholly accounted for by {J. Although
handbooks are a source for {J expressions for many generic configurations,
obtaining {J for a practical problem may be rather involved. Fortunately there
are many simple ways in which it can be obtained with good accuracy. Such
simple procedures to derive {J for practical cases are discussed in Chapter 8. At
54

-
28

,6=1 +0.256
8
Vi <J!~2 {8~3
-1.15,W} +12.200 \WJ

8 L L/W= 2 ONLY:

,6 =1.12- 0.23 ! +10.56(~)2 -21.74(8)3

W
(8\4
+ 30.42 Vi)

(SEE FIGURE 8.12)

W

8 8 8 (8'l
,6=1.12+0.43 W -4.79 W) +15.46 (8~3
Vi)

Figure 3.3. Examples of p-functions.

 55

this point it is sufficient to note that {3 can always be obtained and that
Equations (3.14) are universal.

3.4. Toughness
Several difficulties arise with the direct use of Equations (3.14), and these may
have serious consequences, as discussed in subsequent sections. Nevertheless,
although somewhat prematurely, the practical use of the equations for the
analysis of fracture problems, will be considered first, be it that some of the
conclusions will have to be modified later.
Fracture will occur when the stresses at the crack tip become too high for the
material to bear. As the stress intensity factor determines the entire crack tip
stress field, the above statement is equivalent to: fracture will occur when K
becomes too high for the material. How high the stress intensity can be depends
upon the material; it must be determined from a test.
For example let a test be performed on a 30 inches (762 mm) wide plate of a
certain steel X, with a central crack 2a = 4 inches (101.6mm). Let the plate
have a thickness of 0.2 inch (5.08 mm). The plate is pulled to fracture in a tensile
test machine. Suppose that fracture occurs at a load of 180 kips or 180000 lbs
(800000 N). The nominal stress in the uncracked section at the time of fracture
was then 180/(0.2 x 30) = 30ksi (800000/(5.08 x 762) = 207N/mm2
= 207 MPa).
This information permits calculation of the value of the stress intensity at
fracture. Note from equation (3.15) that {3 ~ 1 for the small a/Wused, and that
a = 2 in (0.0508 m). Hence:

K = I x 30 x J1CX2 = 75 ksi Jlrl.

or
K 1 x 207 x ..)n x 0.0508 83 MPa Jill.
or
K = 1 x 207 x ..)n x 51 = 2620N/mm 3/2 •
Apparently, when K reached the value of 75 ksi Jill (83 MPa Jffi), the stresses
at the crack tip were too high for the material and fracture ensued. This value
of K is called the 'toughness' of the material. Hence, the toughness of a material
is the highest stress intensity that can be supported by a cracked component
made of that material.
The unit of toughness is ksi Jill or MPa Jffi; it follows directly from the
dimension of K, which is stress x Jcrack length. Conversion of units from
English to metric provides: I ksi Jill = 6.89 MPa ..)0.02540 m = 1.09 MPa Jffi,
or 1 MPa Jffi = 0.92 ksi Jill.
56
Once the toughness of a material has been measured in a test, its value can
be recorded in a data sheet or data handbook as in the case of e.g. Fir' The latter
is the value of the stress at which yielding occurs. Toughness is the value of the
stress intensity at which fracture occurs in the case of cracks.
Now the universality of Equations (3.14) becomes important: the equations
hold for ALL cracks. At a crack in a structure made of steel X, the stresses at
the crack tip will be equal to those in the test specimen at the time of fracture
when K is equal to the tougness. Since the material of the test specimen could
not sustain this stress field, it follows that the structure of the same
material also will fracture when K is equal to the toughness, because then the
crack tip stress field in the structure will be identical to that in the specimen; if
the specimen failed as a consequence of this stress field, the structure will fail
when this same stress field occurs.
Consider a simple structure in uniform tension with an edge crack of a = 3
inch (76.2 mm), a width of 40 inches (1016 mm), and a thickness of 0.2 inch
(5.1 mm), made of the same material as the above specimen. For this configura-
tion and small a/W, it is found from Figure 3.3 that f3 ~ 1.12. The fracture
condition is:
Fracture if: K = Toughness (3.16)
with K = f3(J Fa, it follows that fracture occurs when:
Fracture if: f3(J Fa = Toughness. (3.17)
The nominal stress at which fracture takes place will be denoted as (Jf,' It follows
from Equation (3.17) that: .
Toughness
(3.18)
f3Fa
Substitution of f3 ~ 1.12, a 3 inch (0.0762 m) and toughness 75ksi Fn
(83 MPa Jill), gives:
Toughness 75
22 ksi
f3Fa 1.12~
or
Toughness 83
= 151 MPa.
f3Fa 1.12~n x 0.076

The fracture stress, (Jf" is the residual strength (remaining strength under the
presence of a crack; Chapter I). Using the equation for f3 in Figure 3.3, one can
calculate the residual strength from Equation (3.18) for a number of crack sizes.
A plot of the results provides the residual strength diagram shown in Figure 3.4.
 57

W=40in
50 1016mm.

40

30

• ../" Point calculated

•
in example
*
~
Other points
20
•
calculated

•
10
•
•
•
2 4

Figure 3.4. Fracture stress for example in text.

(The fracture stress, (Jfr' would indeed be the residual strength defined in
Chapter I unless failure by collapse as discussed in Chapter 2 prevails. As this
remains to be established, (Jf' is not yet called (Jres).

3.5. Plastic zone and stresses in plane stress and plane strain
Equations (3.2) and (3.14) predict that far from the crack tip where x is large,
the stresses (Jx and (Jy become zero. That the transverse stresses, (J x' becomes zero
is not surprizing but (Jy becoming zero is incorrect because, e.g. in Figure 3.2 the
stress (Jv far away from the crack is equal to the applied stress (J. The reason is
that Equations (3.2) do not provide the complete solution of the stress field. The
complete solution is a series of terms:

(3.19)

In particular the second term will ensure that (Jy = (J for large x.
Approaching the crack tip, x becomes very small, so that all terms except for
the first become very small. Actually, for x = 0 all terms become zero, except
for the first term, which then becomes infinite. Hence, for small x, close to the
crack tip, all terms can be neglected with respect to the first term, and for x = 0
58

Equations (3.1) are the exact solution. Fracture begins at the crack tip and not
somewhere else, so that the use of only the first term is justified.
For x = 0 the stresses become infinite. This is correct from the point of view
of the theory of elasticity, because Hooke's law (0' = BE) puts no limitations on
stress nor strain. The crack (tip) is a sharp discontinuity, which causes very high
stress (according to Eq. (2.3), a notch with a tip radius of zero causes an infinite
stress concentration). Note however, that the infinite stress occurs at only one
single point (i.e. in an infinitely small area) and not over a certain distance. Yet,
in a real material plastic deformation will occur so that stresses cannot increase
much further after yielding begins.
As explained in Chapter 2, the stress field at notches and cracks and the state
of stress depend upon thickness. Contraction in thickness direction of the highly
stressed roll at the crack tip can take place freely when the plate is thin, leading
to plane stress. In thick plates however, contraction is constrained which leads
to plane strain (see Chapter 2, Figures 2.6 and 2.7). According to Equations
(3.2) the transverse stress, O'x, equals the longitudinal stress, 0'" at the crack tip.
Plane strain (Chapter 2) then leads to: .

(3.20)

which provides (with v ~ 0.33):

O'z = v(O'x + O'Y) ~ 0.660'y. (3.21)
According to the Tresca yield criterion (Chapter 2) yielding begins when the
difference between maximum and minimum principal stress is equal to the yield
strength, FlY' This leads to the following condition for plastic deformation in
plane strain:
(3.22)
Hence, the stress O'y must rise to three times the yield strength before plastic
deformation will commence. In plane stress on the other hand, where O'z = 0,
yielding will being when O'y = FlY, (Chapter 2).
Assuming that the stress does not increase much after yielding (again, see
Chapter 2), the stress distributions will be as shown in Figure 3.5. The curved
part of the stress distribution is still described by Equations (3.2), and it is the
same in plane stress and plane strain, because further away from the crack tip
constraint is not a problem (Chapter 2). Apparently, within some distance rp '
from the crack tip the yield conditions are met, so that there is always a small
area at the crack tip where plastic deformation takes place (Note that O'y is
infinite according to Equation 3.14). This area is called the plastic zone. It is
possible to estimate the distance from the crack tip over which plastic deforma-
 59

cry

----
CRACK~~~~~~~ ________________________________

PLASTIC X, f
ZONE
(a)

cry \

CRACK~~~----------------------------------

(b)
PLASTIC fp
ZONE
- X, f

Figure 3.5. Stress Distribution (a.,) at crack tip. (a) Plane stress; (b) Plane strain.

tion occurs, i.e. the size of the plastic zone. For plane stress this follows from
the Tresca yield criterion (Chapter 2) and the condition (Iy = FlY' and for plane
strain from the condition (Iy = 3Fly ' Using Equation (3.2) for (Iy one obtains:
60

K K2
for plane stress: J2iU"v F,y or rp

)
2nrp 2nF,~.
(3.23)
K2
for plane strain: ~ = 3F,y or rp
2nrp l8nF,~

In reality, the plastic zones are about twice as large as in Equations (3.23). This
is due to the fact that actually a much higher stress should have been carried by
the plastic zone. This extra stress must be bypassed around the plastic zone thus
increasing the stress there and causing more material to yield [1]. By limiting the
crack tip stresses because of plastic deformation, the dashed part of the stress
distribution in Figure 3.5 was essentially eliminated. However, the total stress
distribution, including the dashed part, was calculated from the theory of
elasticity on the basis of equilibrium. Cutting off the dashed part is
violating equilibrium. This must be mended by restoring equilibrium. The
dashed areas in Figure 3.5 represent a load which cannot be carried by the
material in the plastic zone because the load carrying capacity of that material
is limited by yielding. However, the cut load must still be carried through
(equilibrium) and it must bypass the crack, or it must bypass rp ' as the plastic
zone cannot carry it. This implies that yielding actually extends beyond rp' (Note
that the high stresses were due in the first place due to a load path interruption
by the crack, so that the load had to bypass the crack as shown in Chapter 2.
Now it appears that the load represented by the shaded areas in Figure 3.5 must
bypass further away, because once it is loaded to yield, the area rp is no longer
a viable load path either). The consequence is that the plastic zones are actually
larger than those represented by Equations (3.23). Restoring equilibrium will
lead to plastic zones exactly twice as large [1], as those in Equations (3.23). For
other reasons [1], the plastic zones are usually taken as:
K2
r
p
= - -
emF,;
(3.24)

with (X = 2 for plane stress and (X = 6 for plane strain.

It appears from Figure 3.5 and Equations (3.23) and (3.24) that the plastic
zone in plane strain is much smaller than in plane stress. Also, the crack tip
stresses are higher in plane strain than in plane stress. The longitudinal stress,
<T" differs by a factor of 3, while <T z = 0.66<T v in plane strain and <T z = 0 in plane
stress (see also Chapter 2). .
In the example of the hypothetical test given in the previous section, fracture
took place at a value of K of 75 ksi .Jill. Supposing that the material had a yield
strength of F,y = 100 ksi, the size of the plastic zone at the time of fracture
would have been according to Equation (3.24):
 61

plane stress: rp 0.09 inch

75 2
plane strain: rp = 6n x 1002 = 0.03 inch.

These are very small plastic zones indeed. At this point it cannot be decided
whether there was plane stress or plane strain in the example in Section 3.4. This
question is discussed in the following section.

3.6. Thickness dependence of toughness

The true size of the plastic zone is not important for the following discussion,
but the format of Equation (3.24) is relevant; it shows that the size of the plastic
zone depends upon K only. Thus for all cracks in any configuration, and at the
same K, the stress distributions at the crack tip will still be completely dictated
by K and by K only. It follows that the fracture criterion then still holds
regardless of the plasticity; if fracture occurs in a test specimen at a certain value
of K (the toughness) then it will do so in any other configuration at the same K,
because in that other configuration the stress distribution as well as the size of
the plastic zone will still be the same as in the test specimen, provided there is
equal constraint.
One problem has arisen however. The stress distributions are not the same in
plane stress and plane strain; as shown in Figure 3.5. Also a: differs in the two
cases. As a consequence, the above fracture criterion cannot be used indiscrimi-
nately. If fracture in plane strain occurs at a certain value of K, then it will not
in plane stress at that same value of K, because then the stress distributions will
be different (different a z and different ay at the crack tip due to different plasticity
as shown in Figure 3.5). It follows then that the toughness will depend upon the
state of stress. As the latter depends upon thickness, the toughness must be
expected to depend upon thickness.
It should be noted here that Equation (3.24) was derived using only the first
term of the stress distribution in Equation (3.19). Should the plastic zone be so
large that the other terms come into play then its size will depend not only upon
K, but also upon C, D etc. In such a situation equal K is not enough to guarantee
equal stress distribution and fracture is not likely to occur at always the same
value of K. Apparently, for the fracture criterion to be useful the plastic zone
must be very small. Since rp depends upon K, this implies that the stress intensity
at fracture, i.e. the toughness, must be low. As rp also depends upon the yield
strength, the fracture criterion will be better applicable to materials of high yield
strength and low toughness. Although the criterion does not work when the
yield strength is very low and the toughness very high, satisfactory results can
be obtained for many such cases, provided one realizes that plastic collapse
62

(Chapter 2) may prevail. This will be discussed in detail in a later section. In the
case of extremely high toughness and/or low yield strength, EPFM may have to
be used (Chapter 4).
Consider two plates, a thin one with plane stress, and a thick one with plane
strain. Both have cracks and are loaded to have equal stress intensity factors.
In the case that the prevailing stress intensity is equal to the toughness of the
thick plate fracture ensues in the thick plate. The stress distributions in the thin
plate will be much more benign (see Figure 3.5 and note the difference in o-z) at
this same K, so that it must be anticipated that the stress intensity in the thin
plate can be further increased; i.e. the expectation is that the toughness is higher
in plane stress than in plane strain (Figure 3.6a).
In plates of intermediate thickness the state of stress gradually changes from
plane stress, via more and more triaxility, to plane strain in thick plates.
Therefore, the toughness will not suddenly, at one particular thickness, drop
from its high plane stress value to its low(er) plane strain value, but the change
will be gradual; i.e. the toughness decreases with thickness as depicted in Figure
3.6b. Beyond a certain thickness there will be plane strain. As the state of stress
cannot become worse than plane strain (complete constraint of contraction), the
toughness will not decrease further.
For plane strain cases the toughness is called the plane strain fracture
toughness. It is generally denoted as Kin i.e. the critical value of the mode I
stress intensity, K" at which fracture occurs. (Compare: FlY is the critical value
of the stress 0- at which yield occurs). Similarly one speaks of the plane stress
fracture toughness, and of the transitional fracture toughness for intermediate
situations. Plane stress and transitional toughness are generally denoted as K,
or K I ,. Unfortunately, this introduces a confusing inconsistency. In plane stress
or in transitional cases, the loading is still mode I, the toughness is still the
critical value of the mode I stress intensity. Thus, the general denotation of
toughness should be Kin regardless of the state of stress. One could speak of
THE toughness Kin and indicate whether the value is for plane stress, plane
strain, or a transitional case. Be that as it may, the above denotations have
become the 'accepted' ones. Once the user gets used to these denotations, they
are, in a way, helpful. If the toughness is given as KIn it is immediately clear that
this toughness is for plane strain; while a toughness K c or K lc is obviously for
a thickness in which full constraint does not occur (plane stress or transitional).
The toughness as a function of thickness can be measured in tests of the type
discussed in Section 3.4 on plates of different thickness. It can then be used to
calculate the residual strength of cracked structures:
Fracture if: K = Kc or K - K }
i.e. Fracture if: po-Fa = K, ~or ;IC) (3.25a)
 63
K
AT
FRACTURE

t
High Stress

~,:~,:'d;'"
PLANE STRESS ~
~...
_ __ Fracture

DEVELOPING \

/S~~~M~~
Low Stress /
Equal Plast ic
Zones

." '4
B B

(a) ~Thickness

Toughness

K IC -- - - - - - - - - - - - -::-,-"..-.------

llane
Stress
/_ Transitional I Plane

(b) Thickness

Figure 3.6. Dependence of toughness upon thickness. (a) Effect of thickness on plastic zone and
state of stress; (b) Effect of thickness on toughness.
64
from which the fracture strength follows as
Kc (or K/c)
(3.25b)
pFa
For the configuration at hand p can be obtained from a handbook or
otherwise (Chapter 8), and the residual strength calculated provided one uses
the toughness for the thickness at hand, in the same manner as discussed in
Section 3.4.
It is of interest to know at which thickness full constraint and plane strain
occur. As discussed in Chapter 2, plane strain develops when the roll of material
at high stress at the crack tip is long and thin (Figure 3.7). Hence, LID, where
L is the roll's length and D its thickness, must be large, e.g. larger than a certain
value Q. When the crack is all the way through the thickness, the length of the
roll is equal to the thickness, B, i.e. L = B. Taking D equal to the size of the
plastic zone, D = rp ' from Equation (3.24) one obtains as the condition for
plane strain:
L B
(3.26)
D (Klc!r:J.Ft~, > Q.
In this equation r:J. and Q are numbers. Rewriting Qlr:J. q, the condition can be
given as:

B Klc
> q'F 2 •
(3.27)
ty

The value of q cannot be derived mathematically. It is obtained from toughness

tests on plates of different thicknesses as described, and by finding the thickness
at which the curve levels off. Such experiments have shown [7] that this happens
for q = 2.5, but the results are not very conclusive as can be seen in Figure 3.8.
The scatter is considerable and the most that can be said from Figure 3.8 is that
the leveling off occurs somewhere between q = 2 and q = 4. Nevertheless the
number 2.5 is often considered sacrosanct. The reader is advised that it is but
a rough indication (note: instead of q the symbol IX is used in Figure 3.8).
It should be emphasized that the thickness is only of relevance for through-
the-thickness cracks. Constraint is determined by the length of the roll of highly
stressed material. In the case of a part through crack such as in Figure 3.9,
contraction is always fully constrained and the thickness has no relevance to the
problem (note that the thickness came into the problem because the length L of
the roll happened to be equal to the thickness in the case of a through-the-thick-
ness crack, and in that case only; this was discussed already in Chapter 2).
Figure 3.7. Thickness and State of Stress. (a) Attempt to contraction at crack tip; (b) Effect o f -
thickness upon contraction.
 65

cylinder of material that

wants to contract

(a)

thickness
B --- - B

t CT t
CT

- -
~ ! ~
--- 0 - crack
plane
t
L
CT
~CT
+
Low stress thick plate, Low stress thin plate,
thin cylinder no contraction free contraction
plane stress
(b) plane strain

Figure 3.7.
66
measured

i
K rc (ksiv'Trir
(apparent)

120

"'- -+- ~tests

.....0
100 - - - - ------- -
~
0 ~ 0-- 0 1
0 K 1c
0 't---..
"-
80
I 23 tests"
1

60
2.5 I
2 3 4 5 6
a=_B_
- (~)2
°Y'
Figure 3.S. Effect of thickness on measured K" of a Marageing Steel [7].

THROUGH
- CRACK
L=B

~SURFACE
FLAW

\
CORNER
CRACK

..-l_
--

Figure 3.9. Constraint condition for part-through cracks.

Therefore part-through cracks (surface flaws and corner cracks) are in plane
strain, at least in the center. For such cracks, one must use the plane strain
fracture toughness, KIn regardless of the thickness.
 67

3.7. Measurement of toughness

In principle the toughness can be measured on any kind of cracked specimen.

Fracture occurs when Equation (3.25) is satisfied. With knowledge of f3 for the
specimen at hand, and the measured fracture stress (load) the stress intensity at
fracture can be calculated; the result is the toughness. In the reverse operation,
Equation (3.25b) is used to calculate the residual strength; the toughness then
being known , the crack size given, and f3 obtained for the structure at hand, (ffr
follows from Equation (3 .25b).
Although any kind of specimen can be used a few simple test specimens have
been standardized by ASTM [7, 8] for the measurement of plane strain fracture
toughness. The most universally used is the so-called compact tension (CT)
specimen depicted in Figure 3.10. It contains a notch in the form of a chevron

\
I

I
: Tp 1.25W I
I- -I I

Figure 3.10. Standard K,c specimen; left: configuration; right: notch.

68

and is loaded through two pin's in a tensile machine. Cyclic loading is applied
to introduce a fatigue crack. When the crack is at the desired length the cycling
is stopped, and the load is raised until fracture occurs. The stress intensity at
fracture can then be calculated; i.e. the toughness obtained.
The stress intensity factor for the compact specimen is usually written as:

K = B:l!2 [29.6(~r2 - l85.5(~J2 + 655.7(~r2

- 10 17 W (a)7/2 + 639 (aW )9/2J . (3.28)

This expression seems to be rather different from Equation (3.14), but it is

essentially the same. It is written in this manner because the nominal stress is
hard to define. However, there is no objection against defining a (somewhat
hypothetical) nominal stress as a = PI BW. Then Equation (3.28) indeed reverts
to:

K [16.7 - 104.6(~) + 370(~J - 574(~J

+ 361 (~JJaFa = paFa· (3.29)

During the test the so-called crack mouth opening displacement is measured by
inserting a gage in the notch (Figure 3.11). Since there is hardly any strain in the
material between the loading pins, the crack mouth opening displacement is
essential equal to the displacement of the loading pins. Load and displacement
are plotted in a load-displacement diagram (Figure 3.11). The maximum load
is substituted in Equation (3.28) to find the toughness.
The test standard prescribes that the load-displacement line must be straight
or nearly so. When extensive plastic deformation occurs the diagram will
become non-linear. However, yielding is often confined to such a small plastic
zone that non-linearity is hardly perceivable. Should the diagram be significant-
ly non-linear then the plastic zone is large: as discussed in the previous sections
the fracture criterion based on K is then no longer valid and "the" toughness
cannot be defined. Hence, the standard sets a limitation on non-linearity for the
measurement of 'valid' toughness numbers. Should the entire ligament yield
then collapse may occur as discussed in Chapter 2.
The standard test is specifically for the measurement of the plane strain
fracture toughness KJc. This means that there must be plane strain. Therefore
the thickness of the specimen must be sufficient to satisfy Equation (3.27) with
q = 2.5. The toughness is not known in advance (otherwise measurement
would not be necessary), so that one cannot determine how thick the specimen
should be. It is possible that the specimen was made too thin, in which case the
 69

Clip Gage
~

I I
'irl
Strain
Gage

Leaf
Sprmg
(a)

Figure 3.11. Standard Kh test. (a) Clip gage mounted in specimen; (b) Straight P - ,5 record; (c)
Curvature due to plasticity; (d) Curvature due to plasticity and growth.

measured toughness would not be the plane strain fracture toughness. For this
reason the toughness obtained from Equation (3.28) is initially called the
candidate value, KQ , of the plane strain fracture toughness, K/c. One must check
whether B > 2.5K6/Ft~,. If this is the case then indeed KIc = KQ; if not then a
new test must be done on a thicker specimen if one insists on knowing the plane
strain toughness. It should be noted however, that Kc = KQ (Figure 3.6), is a
perfectly useful toughness number for the actual thickness of the specimen. (See
also Chapter 7).
As was pointed out already, the number q = 2.5 is rather dubious, and
therefore rigorous application of Equation (3.27) somewhat ludicrous. The
number 2.5 was obtained from an interpretation ofa limited number of test data
(Figure 3.8). It was then agreed upon by a committee that the number was
reasonably representative, and as a consequence it appears in the test standard
[8]. However, a committee decision, nor an ASTM standard, is a sufficient
guarantee that the number is indisputable. If there is general concurrence that
the number is useful then there are no objections against its use, but it still
remains an arbitrary number. Also the uncommon significance attached
to the standard test is largely exaggerated. Toughness can be measured on any
cracked specimen, provided the plastic zone is small. If it could not, the result
could not be used to calculate residual strength of any other structural crack
(based upon the generality of Equations (3.2», which is the purpose of the
measurement. In that case the standard test would have no use in the first place.
(See also Chapter 7).
70

Although the compact specimen is cheap and simple, it is totally unsuitable

for the measurement of plane stress and transitional toughness. For such
toughness measurements a center cracked panel is the most advisable. In order
to obtain useful numbers, the plastic zone at fracture must be kept small. This
often requires very large panels, as will appear in the following section. Due to
the fact that stable fracture often preceeds fracture instability, a revised
definition of toughness is needed as well (Section 3.12).

3.8. Competition with plastic collapse

In Chapter 2 the conditions for collapse were discussed. It turned out that
collapse occurs in a center cracked panel when:
W - 2a
(lIe = W Feol ' (3.30)

In this equation Feol is the collapse strength which may be as low as the yield
strength and almost as high as the tensile strength. Recall from Chapter 2 that
Equation (3.30) is a straight line as a function of crack size a.
Also discussed in Chapter 2 is the fact that Equation (3.30) represents the
absolute highest load carrying capability: at collapse plastic deformation
becomes unbounded and fracture follows, regardless of the toughness. Hence,
(fracture) failure by collapse may occur before K reaches the toughness.
Naturally, K lc is the result of a semi-elastic concept while at collapse the entire
cracked section is yielding. Therefore, the two failure conditions cannot be
directly compared; but they can be in an indirect way.
If the toughness is high, Equations (3.25b) will predict a very high fracture
stress (residual strength). This may be as high or higher than the stress for failure
by collapse. Then failure by collapse will prevail, because, of two competing
failure modes, the one which first becomes possible will prevail. This may
happen when the toughness is high, but under other conditions as well.
Should the fracture stress (ll" be higher than the stress causing failure by
collapse, then collapse will prevail. This is the case when the result of Equation
(3.25b) is more than the result of Equation (3.30), i.e. when (for a center crack):

w- 2a
(3.31 )
or W Feol <

The lowest of the two is the actual residual strength discussed in Chapter 1.
From Equation (3.31) it appears that there are 3 situations in which a collapse
failure could prevail, namely when
 71

- the toughness is very high or;

- the crack is very small (a ~ 0);
- the width W is very small.
These situations are illustrated in Figure 3.12. For the material with
K/c = 160 ksi Jill a panel, of 12 inch width always fails by collapse, but a 60
inch wide panel of the same material fails by K/c. Apparently, for a particular
structural configuration failure may occur by collapse (unconditionally leading
to fracture) in certain instances and by fracture in other instances. Collapse
always prevails if the cracks are small, regardless of how low the toughness. At
any toughness Equation (3.25b) puts the fracture stress at infinite when the
crack size approaches zero. Thus the residual strength curve will always rise

70
are.
(ksiJ
\
\
F ly ·50
\
\
,
"-

30

20

10

(a) 4 8 10 11 12
2a (in)

70
ares
(k.il
Fly = 60

20

10

10 15 20 25 30 35 ~o 45 50 55 60
(b) 2slin)

Figure 3.12. Effect of panel size on Residual Strength. (a) Residual strength of l2-in wide panels;
(b) Residual strength of 60-in wide panels.
72

asymptoticalIy to infinity for smalI a; this means that there will always be a point
at a certain small a, where colI apse failures will prevail, regardless how low the
toughness.

will become infinite. Clearly, when a = °

For a crack size a --> 0, the fracture equations predict that the fracture stress
the fracture stress is equal to the
ultimate tensile strength FlU, or less. Thus, the most left-hand part of the curve
represented by Equation (3.25b) will always be in error, whether the toughness
is high or low. For a = 0, the strength is F,u (or less), while for large a the

behavior between a = °
Equation (3.25b) applies (which is the curve in Figure 3.12). Obviously, the
with (J ::::: F/j, and the curve for large a, cannot be as
A or B in Figure 3.12a. Hence, if one assumes the 'eye-balled' curve C (tangent
to the curve), the assumption cannot be far from the truth; it certainly will be
adequate for engineering analysis.
It is important that of two plates of different sizes - but of the same material
- one may fail by collapse, the other by fracture. This is depicted also in Figure
3.12 as well as in Figure 3.13. From this it can be concluded that a specimen for
the measurement of toughness must be of sufficient size. If in the case of Figure
3.13 a plate of the smaller size W] would be used in a test, the failure would occur
by collapse. Hence, the value of the stress intensity at the time of failure would
stilI be lower than the toughness, because K has not yet reached the toughness.
This problem also exists for plane strain toughness testing with the CT
specimen. The specimen may fail by collapse if the toughness is very high or if
the specimen is too small (specimen size is rather ignored in the test standard).
If one insisted t.o calculate KQ from such a test by using Equation (3.28), the KQ

Residual
Strength
[Lowest TANGENT APPROXIMATION
of O'fr'
at and Gfe }

W3 W, W, 2a

Figure 3.13. Effect of plate size on failure mode.

 73

would be LESS than the actual K lc • Fortunately, the test would be invalid
because the standard requires a 'straight' load displacement diagram. If the
specimen were to fail by collapse, the load displacement diagram would not be
straight because of large scale plastic deformation in the ligament.
Apparently, Equation (3.25b) is not sufficient to determine the residual
strength diagram, nor is Equation (3.25a) enough to determine the toughness. It
is always necessary that the stress for collapse be determined as well, because the
lower of the two prevails, and the residual strength, are" is the lower of afe and
afro For short cracks, the tangent approximation, aft' may prevail ifit is lower
than afe and afe. Determination of the collapse curve is simple in the case of
uniform tension loading. If the loading is non-uniform such as for the compact
tension specimen, the collapse condition is somewhat more difficult to evaluate
(see Table 2.1). The calculation of residual strength diagrams is discussed
further in Chapter 10.

3.9. The energy criterion

This and the following sections present another look at the fracture criterion.
This is of interest for understanding LEFM, but it is essential for the
understanding of EPFM as discussed in Chapter 4. Finally, it will appear that
the definition of toughness may have to be revised (Section 3.12).
Conservation of energy demands that the work, F, done by the load on a
body, is not lost. It is conserved as strain energy, U, so that:
F - U = o. (3.32)
The work done by the load is F = JP d<5 (where P is the load and <5 the load
displacement), which is the area under the load-displacement curve. As long as
the material is elastic, the load-displacement diagram is a straight line, so that
the work done by the load is tP<5 (see Figure 3.14). It follows from Equation
(3.32) that U = tP<5 as well. But U can be determined in a different manner.
Consider a small material element of unit size, subjected to a uniaxial tension
a, as shown in Figure 3.14b. The stress does work to deform the material over
J
de the total work done is a de, which for a linear elastic material is the area of
the triangle shown in Figure 3.l4b, namely tae. By substituting e = alE from
ta
Hooke's law, the work becomes 2/E. As this work must be equal to the strain
ta
energy, the value of 2/E also represents the strain energy in the material
element.
The stress on each volume element is not always the same. Therefore the total
strain energy in the entire body or structure is obtained by taking the integral
over the volume of the body:

(3.33)
74
p

(a) p

(J

(J

(J

(b)
Figure 3.14. Strain energy. (a) Load and load-displacement diagram; (b) Stress on material element.

in which a may depend upon x, y and z. In the case of a simple tensile bar the
stress is the same on each and every volume element. In that case the total strain
energy is simply equal to ta2 /E times the volume. If the cross section of the bar
is A and its length L, then the strain energy is:
a2
U = 2ELA. (3.34)

With knowledge of F ( = tPt5) one can substitute both F and U in Equation

(3.32) to obtain:

o. (3.35)
 75

By noting that, in this case, J = eL = (JLjE, and P = (JLA, it can be readily

seen that Equation (3.35) is true. As already stated, the criterion provides
U = tPc5.
Equation (3.32) also holds when the body has a crack, 2a. In the case of
limited plasticity the load-displacement diagram is still a straight line as in
Figure 3.15. If a crack of somewhat larger size, a + da, exists then it will take
less load to cause the same displacement: the stiffness is less. This, the load
displacement line is lower as shown in Figure 3.15d.
Let the load increase to PI, the displacement to c5 I, as in Figure 3.15d. In
the event that fracture takes place at this load, the crack extends over a small
increment da from a to a + da. During this process there must be energy
conservation, but instead of two energy terms F and U there is now a third
energy term W, where W is the work expended in fracturing material over da.
The energy conservation equation now covers only the changes of the energy.
During fracture over da the load may do some work dF, the strain energy may
change somewhat, dU, and some energy, d W, will be required for fracture.
Hence, the energy conservation criterion reads:
d
da (F - U - W) = 0 (3.36)

which can be written as:

d dW
da (F - U) = da· (3.37)

The equality must hold when fracture occurs. Conversely when the equality
cannot hold, fracture does not yet take place and Equation (3.32) remains valid.
Apparently then, Equation (3.37) is a fracture criterion. Fracture will occur
when enough energy can be delivered to provide for the fracture energy d Wjda.
Energy delivery must come from a surplus of d(F - U)jda. If this surplus is
sufficient to cover the energy required for fracture, d Wjda, then fracture will
occur; ifnot Equation (3.32) continues to govern.
The fracture criterion of Equation (3.37) can be used if the energy delivery
d(F - U)jda can be quantified. This will be done in the next section.

3.10. The energy release rate

Consider again the situation of Figure 3.15d with a load, PI, resulting in a
displacement J I . First examine the situation in which the displacement does not
change when the fracture occurs over da. In that case the load decreases to P 2 ,
because it is easier to maintain the displacement J I , with a longer crack.
However, the load does not move and therefore it does no work: dF = O. The
strain energy does change from tP I c5 1 (area OAD) to tP2 J I (area OBD) after the
76

(b) (c)
p

(a)

(d) (e)

Figure 3.15. Energy release during constant displacement and constant load. (a) Plate with crack
a under load P; displacement b; (b) Load displacement record; (c) Pb record for small and larger
crack; (d) Fracturing at constant displacement; (e) Fracturing at constant load.

process. The strain energy decreases, so the change is negative. Hence, it turns
out that the deliverable energy is - dUjda, and Equation (3.37) becomes:
dU dW dU
- -
da
= -
da
where da < o. (3.38)

Next examine the case in which the load remains constant during the fracture
over da. In that case the displacement increases from 15, to 152 as shown in Figure
 77

3.15e. Now the load does work, namely an amount P( D2 - D1); note that the
load remains constant during the process. The strain energy also changes,
namely from tP1D 1to tP1D2. This time the strain energy increases, so its change
is positive. Equation (3.37) becomes:

P 1(D 2 - D1) - tP 1(D 2 - D1) = dd~).

or (3.39)

tP 1(D 2 - D1) = ~~
In one case the deliverable energy is - dU/da, in the other casetP(D 2 - DI).
Note that the latter amount is exactly equal to dU/da. It appears then that the
deliverable energy is always the same, namely dU/da, regardless of whether
fracture occurs under constant load or constant displacement. In either case the
deliverable energy is equal to the change in strain energy dU/da. For constant
displacement the deliverable energy is coming directly from a release of strain
energy. In the other case the strain energy increases, but the load does twice that
much in work, so that the surplus is still equal to the change in strain energy,
be it with opposite sign
Clearly then, the deliverable energy is always equal to the change in strain
energy regardless of its sign. Instead of d(F - U)/da one may then use the
absolute change of the strain energy dU/da in Equation 3.37. Then the fracture
criterion is:
dU dW
da . (3.40)
da
The left hand side is called the strain energy release rate, the right hand side the
fracture energy or fracture resistance. Fracture will occur if, due to the extended
slit by da, sufficient strain energy is released to cover the energy to create a
fracture over da.
The fracture criterion is somewhat simpler now than before, but it remains
necessary to find an expression for dU/da. As the strain energy is affected by the
crack one can write:

U = Ubody with no crack + Udue to crack· (3.41)

Consider a very large plate of length L, width Wand thickness B with a small
center crack of 2a. If the loading is uniform tension then the strain energy of
each element of the uncracked body is the same and equal to t(,z / E, so that the
strain energy of the uncracked body becomes:
(J2
Ubody with no crack = 2E LBW. (3.42)
78

The stress due to the crack will depend upon the crack tip stresses. These crack
tip stresses are proportional to the applied stress u. Thus the strain energy due
to the crack must be proportional to u 2j E. The term will also be proportional
to the thickness B (twice as thick a plate has twice the energy). Hence,
u2
Udue to crack -7- E B. (3.43)

The contributions by Equations (3.42) and (3.43) must have the same
dimension. They differ by a length squared (LW). Since the contribution due to
the crack must depend upon crack size, a, it becomes inescapable that it depends
upon a2, otherwise the dimensions would not be correct. Therefore:
u2
Udue to crack = C E Ba 2 • (3.44)

There can be a dimensionless coefficient C in the equation, the value of which

cannot be obtained from simple dimensional analysis. Formal calculation of the
strain energy will show that C = 7t. Using this and combining Equations (3.42)
and (3.44) gives the total strain energy U. As everything is evaluated for unit
thickness, one must divide by B to obtain:

(3.45)

Now dUjda follows from differentiation as:

dU
(3.46)
da
This is for a crack with two tips. Since all consideration are for one crack tip,
one must use
dU
(3.47)
da E
per crack tip and per unit thickness.
This permits writing the fracture criterion of Equation (3.40) as:
7tu 2 a dW
Fracture if: --y da . (3.48)

Usually, the fracture energy per unit crack extension dWjda is denoted by R (for
fracture Resistance), while the energy release rate dUjda is denoted as G. Then
Equation (3.48) in shorthand is given as:
7tu.2 a
-- = R or G = R (3.49)
E
which is the same as Equation (3.48) except for the short hand notation.
 79

3.11. The meaning of the energy criterion

Equation (3.49) shows that fracture occurs when (Jta 2 a) reaches a certain value,
namely ER. Now notice that Jta 2 a is exactly the square of the stress intensity
factor: K. Hence, the equation pronounces that fracture occurs when K2 reaches
a certain value, which is the same as K reaching a certain value:
Fracture if: K = JER. (3.50)

It follows then that JER

must be the toughness Kn and apparently the fracture
resistance is R = K: /E.
It may be concluded that, fortunately, the fracture criterion derived from the
energy conservation law appears to be identical to the one derived before on the
basis of crack tip stress. Had this not been the case, one of the two would have
to be declared wrong. Apply this criterion to the same test and example as in
Section 3.4. From the test one can obtain the material's fracture energy R.
Assuming a steel for which E = 30000 ksi, and knowing that f3 ~ I, a = 2
inch, and the fracture stress 30 ksi, Equation (3.49) provides the fracture energy:
302 X 2 2
O 19 ksi in
Jt X
R = 30000 . ksi

= O. 19 in. kip
2 0.19kip/in = 190Ibs/in.
10

Note that indeed R = K: /E, because with K = 75 ksi Jill (Section 3.4) one
obtains R = 75 2/30000 = 0.19 kips/in. The fracture energy is an energy (in/lbs
or Nm) per unit crack extension (in or m) and per unit thickness (in or m), so
that its unit is in Ibs/in or Nm2/m = N/m.
One can apply this result to calculate the residual strength for the same case
as in Section 3.4. The fracture stress follows from Equation (3.49) as:

atr = Jf3::a . (3.51)

Substituting the value of R = 0.19 kips/in obtained above, f3 = 1.12, and

a = 3 in, one obtains afr = 22 ksi, which is the same result as obtained in
Section 3.4. The two fracture criteria are indeed equivalent.

3.12. The rise in fracture resistance: redefinition of toughness

It would seem reasonable to assume that the fracture resistance R is constant

and independent of how far fracture has progressed. This is assuming that it
takes equally much energy to fracture the first da as the adjacent, next da. It
would cause R to be a horizontal line as a function of l1a, as shown in Figure
3.16a.
The deliverable energy (dU/da), usually denoted as G is equal to K 2/E. For
80

G,R G,R

G, G,

R
R
G, /
/

,/,/ / G1
,/ ~1//
/ /
/ / /

(/

~__ ~~t~ ____ ___

(a) (b)

Figure 3.16. R-Curve. (a) Horizontal R-curve; (b) Rising R-curve.

the case that f3 = I, G would be G = 1U,2 a/ E, which is a straight line as a

function of a, with a slope depending upon (J. Assume a crack of size a present,
and take a second origin in the diagram by taking a to the left. One can then
draw the dashed line for G, the actual value of G for crack size a being given by
G1 on the ordinate.
Clearly G1 is less than R, and fracture will not take place at the stress (JI .
Raising the stres's to (J2 will increase the value of G to G2 • Since G2 = R, fracture
will indeed occur. The 'crack' extends during fracture, so that G increases from
G2 to G3 , while R remains the same. Hence, G remains larger than Rand
fracturing will continue uncontrollably.
In reality R is not a horizontal line. The fracture energy appears to increase
somewhat during fracture [I]. If the increase is small, such as in Figure 3.16b,
the fracture will still proceed as above. This is generally the case when the
toughness is low or in plane strain. When the toughness is high, the R-curve rises
more steeply, as shown in Figure 3.17.
At the stress (Ji' the energy release Gi , is equal to R for the first time.
Fracturing will commence but not continue; it is stable. If the fracture proceeds
over l1a, the value of G rises from Gi to G1 • But R rises steeper, and although
Gi was equal to R, the new G 1 is less than (the new increased) R. Apparently, the
fracture cannot proceed (it is stable).
Further increase of the stress to (J2 will cause fracture to continue. At G2 the
deliverable energy is again equal to R. Finally, at the stress (Jlr' the fracture will
be unstable, because G3 increasing to G4 remains larger than R. Fracture will
then proceed uncontrollably (instability).
 81
G,R G,R

I
Aa
a2 a2
a2
(a)
(b)
Figure 3.17. Rising R-Curve. (a) Stable fracture from G, to GJ ; (b) Stable fracture at cracks of
different length.

During the stable phase of the fracture the crack extends from a! to a2' This
should not be interpreted to mean that a crack of size a 2 has a strength (ffr! .
Although this seems trivial, serious errors in fracture calculations often occur
due to this misinterpretation as shown later in this section. Extension of a! to
a2 occurs by fracture and not by the process by which a! was formed as e.g. by
fatigue or stress corrosion. A fatigue crack of length a 2 cannot sustain the same
high stress. Fracturing will commence at a lower stress than (ffr!' namely (ffr2'
(Figure 3.17b). Fracture instability occurs at (ffr2' which is lower than (ffr!' Note
that the smaller slope of the G-lines indicates lower stress.
The interpretation is that a fatigue crack of size a! causes the strength to be
(ffr!' That the fracture proceeds to a 2 before it becomes unstable is rather
immaterial for the end result (two half structures). Similarly, a crack of size a2
has a lower strength (ffr2' Given the presence of a certain fatigue crack say a!,
in e.g. an airplane wing, the strength of the wing is (ffr!' That the fracture
proceeds first to a2 before the wing comes off is rather immaterial for the
passenger (and for the engineer). It is the strength that counts. Naturally, if the
stress would rise above (fi (Figure 3.17) but not reach (ffr then the wing would
stay together and the crack would be somewhat longer than a!, due to some
stable fracture. However, this is trivial: if the fracture stress is not reached the
structure always stays together, whether there is stable fracture or not.
Clearly, fracture instability occurs when the G line is tangent to the R-curve,
i.e. when the two have equal slopes (Figure 3.17). As the slope of the line is
described by the first derivative, the condition for instability is:
82

G(a;) = R(a;) }
and also (3.52)
(dG/da)u; = (dR/da)u,
If only the first equation is satisfied fracture will occur, but in order for fracture
to become unstable (uncontrollable), the second equation must be satisfied as
well.
This has important consequences for the definition of toughness. In previous
sections toughness was defined as the stress intensity at which fracture occurs.
It now appears that (unstable) fracture occurs at different values of G for
different crack sizes. Figure 3.17 shows that with crack of size a l fracture occurs
at GA , while with a crack size a2, fracture occurs at GB • Since K2 = JEG it
follows that the values of K are different at the time of fracture in the two cases.
In previous sections it was assumed that fracture always takes place at the same
value of K, namely the toughness K,. Apparently, this is not true if there is stable
fracture first. This might cause serious complications for the analysis of fracture,
but it will be shown below that an engineering approach can solve the problem.
It should first be mentioned that Figures 3.16 and 3.17 are for the case that
/3 = I, namely the infinite panel. For real structural cracks /3 is not equal to one,
so that the G-lines are curved (note that G = K2jE = /3 2u 2najE (The curvature
being larger for large cracks, the values of G, and hence of K, at fracture tend
to be more or less equal to different crack sizes as illustrated in Figure 3.18a.
From this one might conclude that the assumption of a constant K (namely K,)
at fracture was not so objectionable from an engineering point of view.
Next consider the residual strength diagram as shown in Figure 3.18b. In the
presence of a fatigue crack of say one inch, fracture starts at a stress indicated
by A, and fracture instability occurs at a stress indicated by B (compare previous
figures). This means that if a crack of 1 inch is present, the strength of the
structure is defined by B. It does not mean that the residual strength with a crack
size of 3 inches is defined by B. The figure shows that such a crack would cause
fracture instability at D, and its residual would therefore be defined by D.
Although the one inch crack shows stable fracture until 3 inches, a fatigue crack
of three inches would cause fracture at D, and not at B.
The two curves in Figure 3.18b can be defined by a 'critical' stress intensity.
The top curve is determined by the critical G (or K) for fracture instability, the
bottom curve by the value of G (or K) for the onset of fracture:

Onset of fracture: /3u;~ = K; } (3.53)

Instability: /3u, JiUi: = K, .

From this the stresses at onset of fracture, U;, and the stresses at fracture
instability, Un would follow from:
 83

_ a l1a _
(a)

a
ksi

K i =27ksiv'in

Kc=65 ksiv'in

30

20

10

234 5 6 7 8 9 10 11 12
(b) 2a (in)

Figure 3.18. Approximate Constancy of G; and K,. (a) Curved as opposed to straight G-lines; (b)
Residual strength curves for plane stress.

I
Kj
(Jj
p.[iUi;
(3.54)
Ke
(Je
PFa:
84
The curves represented by Equations (3.54) are shown as the top and bottom
curve in Figure 3.19a. Clearly, the value of Kc in the above equation is not a
constant (see previous figures), but the value of Kc depends upon crack size. On
the other hand, as shown in Figure 3.18a, the value of Kc will not vary a great
deal in actual structures.
Neither the top curve, nor the bottom curve in Figure 3.19a provides the
actual residual strength. On the basis of Figure 3.18, it was shown that the actual
strength of a structure with a crack of say one inch is B. Hence, to obtain the
actual residual strength of this crack one should plot the strength B at the
original crack size. This has been done in Figure 3.19.

Fictitious Test
CJ I \ Data:
Initiation 0
Fracture •

CT,

CT,

(a) 2a

Test Data
CJ 48 in. Wide Panels
(ksi)

~
30 o c-~
-~
o 0
20
--
10

2 4 6 6 10 20 30
(b) 2a (j~)

Figure 3.19. Use of Keff in Fracture Analysis. (a) Residual strength and initial crack size; (b)
Predicted curve and test data on basis of K eff •
 85

The residual strength in the presence of cracks is determined neither by the

top curve, nor the bottom curve in Figure 3.19a, but by the middle curve. The
latter gives the actual residual strength for the case that a crack of size a is
present. One could assign a critical Keff (effective toughness value) to this curve,
and describe the curve by the following equations:
Fracture if:
K Keff
f3ac~ Keff (3.55)
and
Keff
afr =
f3~
In these equations, indeed a i is the length of the crack as caused by fatigue
or stress corrosion, the fracture stress, afr, is indeed the strength of the structure
with this crack present. However, the Equations (3.55) combine a crack and
stress not occurring simultaneously. As such, the equations are physically in
error. However, from an engineering point of view, they are perfectly
acceptable, as long as Keff is more or less a constant (material property). As was
shown in Figure 3.18a this is indeed the case.
The upper curve in Figure 3.19a could be defined by:
(3.56)
This would be a good definition if Kc were approximately constant, which is the
case as shown in Figure 3.18a. However, if this were used in a fracture analysis,
an erroneous result would be obtained:
Kc
afr = f3J1W'
(3.57)

The fracture stress calculated by Equation (3.57) would be for e.g. a fatigue
crack size of three inches in Figure 3.18b, and therefore would be B, while the
REAL fracture stress is at D. Clearly, the use of Kc as defined would lead to
wrong answers. The use of Equation (3.55) would lead to the correct answer if
an engineering approximation is accepted.
It must be concluded that if the toughness is defined as K.rr, the engineering
answer will be correct, while erroneous answers are obtained with Kc. Neverthe-
less, in many instances reviewed by the author, Kc was used instead of Kerr. As
it is somewhat cumbersome to use K.rr for toughness, the discussion of residual
strength analysis in Chapter 10 will make use of the denotation Kn but it is
strongly emphasized that Keff is meant; i.e. the numbers used for the toughness
are Kerr, and NOT Kc.
86
It seems almost superfluous to mention that the above discussion is not
restricted to plane stress or transitional cases. All arguments hold for plane
strain, although this is often not realized. As the R-curve in plane strain usually
rises only moderately, the problem is not acute; it nevertheless exists. Hence, all
of the above applies to plain strain as well as to plane stress.
Clearly, the above engineering approximations need not be made if fracture
analysis is based on the R-curve. The procedure would then be to solve the
problem of Figures 3.15 through 3.18 either graphically or mathematically. This
can be done rather easily (Chapter 10). The question is whether the result will
be any more reliable. From an academic point of view, it will be, but from a
practical point of view it is not. Such a solution requires knowledge of the
R-curve. Certainly, the R-curve can be measured, but it requires the measure-
ment of the progress of fracture. Such measurements by their nature are so
inaccurate that the R-curve can be determined only in an approximate way. Its
subsequent use in analysis does not provide more accurate answers than the
engineering solution discussed above. Academic rigor does not necessarily lead
to better answers; it only complicates problems. This will become abundantly
clear in the next chapter.

3.13. Exercises
1. Calculate the fracture toughness of a material for which a plate test with a
central crack gives the following information: W = 20in, B = 0.75 inch,
2a = 2 inch, failure load P = 300 kip. The yield strength is FlY = 70 ksi. Is
this plane strain? Check for collapse. How large is the plastic zone at the time
of fracture?
2. Using the result of problem I, calculate the residual strength of a plate with
an edge crack W = 5 inch, a = 2 inch. Do the same for W = 6 inch and
a = 0.5 inch (See Figure 3.3 for /3). Check for collapse.

3. Given is a toughness of K = 70 ksi Jill

and a collapse strength equal to the
yield strength with FlY = 75 ksi. Determine the residual strength of a center
cracked plate of 20 inch width with a crack of 2a = 3 inch, and of a center
cracked plate of two inch width with a crack of 2a = I inch; check for
collapse.
4. In a toughness test on a center cracked plate one obtains the following
results: W = 6 inch, B = 0.2 inch, 2a = 2 inch, P m• x = 41 kips;
FlY = 50 ksi. Calculate the toughness. How large is the plastic zone at
fracture? Can this problem be solved with Equation (3.25), and why not? Is
the calculated toughness indeed the true toughness; why not? What caused
the fracture?
 87

5. In a plane strain fracture toughness test on a compact tension specimen the

failure load is 5 kip. W = 2 in, a = 1 in, B = 1 inch, FlY = 80 ksi. Calculate
the plane strain fracture toughness, using both Equations (3.28) and (3.29).
Is this a valid number? What is the size of the plastic zone at fracture?
6. In a plane strain toughness test one obtains a value of KQ = 50 ksi Is Jlil.
this a valid number if the material's yield strength is 100 ksi and the specimen
thickness is 0.5 inch? What is the maximum toughness this specimen can
measure? If the toughness is not valid, then estimate the plane strain fracture
toughness with an accuracy of 5%. How thick should one take the specimen
in order to measure the plane strain toughness under any conditions?
7. Cut a strip of paper four inches wide. With a knife or scissors cut a center
crack (slit) of length 2a = 1.5 inches in the top half. In the bottom half cut
two edge cracks with a = 0.60 inch. Predict where the strip will fail in
tension. Confirm your answer by rolling the ends around pencils and by
pulling.
8. (a) Calculate the complete residual strength diagram for a center cracked
plate of a material with FlY = 70 ksi, Kc = 90 ksi Jill
and W = 24 inch.
(b) Do the same for a panel of W = 6 inch.
(c) Do the same for a panel with a single edge crack assuming f3 of Figure 8.3
is applicable; W = 10 inch. Note: Think of tangent and collapse. For each
case determine the permissible crack size for a stress of 58 ksi, a stress of
50 ksi, and a stress of 10 ksi, by reading from your diagrams.

References
[I] D. Broek, Elementary engineering fracture mechanics, 4th Edition, Nijhoff (1985).
[2] M.F. Kanninen and C.H. Pope\ar, Advancedfracture mechanics, Oxford University Press (1985).
[3] J.F. Knott, Fundamentals of fracture mechanics, Butterworths (1973).
[4] H. Tada, P.C. Paris and G.R. Irwin, The stress analysis of cracks handbook, Del Research (1973).
[5] G.C. Sih, Handbook of stress intensity factors, Lehigh University Press (1973) ..
[6] D.P. Rooke and DJ. Cartwright, Compendium of stress intensity factors, Her Majesty's
Stationery Office (I976).
[7] W.F. Brown and J.E. Srawley, Plane strain crack toughness testing of high strength metallic
materials, ASTM STP 410 (\966).
[8] Anon., The standard K/c test, ASTM Standard E-399.
 CHAPTER 4

Elastic-plastic fracture mechanics

4.1. Scope

If fracture is accompanied by considerable plastic deformation, a concept

known as Elastic-Plastic Fracture Mechanics (EPFM) is used. The literature on
this subject is very confusing. The fracture parameter used is often referred to
as the 'J-integral'. However, J is simply the strain energy release rate, and on this
basis the equation for J can be readily obtained, without the lengthy derivations,
from a generalization of the energy criterion discussed in the previous chapter.
In the following the reader is assumed to be familiar with the strain energy
release criterion for LEFM. Instead of following the historical development of
EPFM, it will be shown that the strain energy release criterion can be used
directly for elastic-plastic fracture by the simple expedient of an equation for the
non-linear stress-strain curve. This leads directly to an expression for J. Its
application and use will be discussed. In order to avoid confusion, the argu-
mentation will be in small steps at a time. This leads to some repetition but will
facilitate understanding. It also leads to a few more equations, but these can be
followed easily because no steps are omitted. In a later section the integral form
of J will be considered. The use of the crack opening displacement (COD)
criterion will be discussed as well.

4.2. The energy criterion for plastic fracture

Whether there is plasticity or not, the energy conservation criterion must hold.
It was demonstrated in Chapter 3 that this leads to the following fracture
criterion:
dU dW
da = da or G = R. (4.1)

For elastic behavior it could be demonstrated that

88
 89

dW
da . (4.2)

With the use of Hook's law (a = eE), the energy release rate, dUlda, the
above equation can be written in terms of stress and strain:
f3 2 naea = R or G = R. (4.3)
The strain energy release is usually denoted in shorthand by G, and the fracture
energy by R. Although the expression for G was derived formally in the previous
chapter, it is important to notice that Equation (4.3) must be true on the basis
of dimensional arguments. The factor f3 is dimensionless. Since strain energy is
f a de = Cae (Chapter 3), the equation for G, the strain energy release rate,
must contain the factor ae. Obviously, G depends also upon the crack size a.
As G is an energy per unit thickness and per unit extension, its dimension is
in-lbs/in2 = lbs/in. The dimension of a is Ibs/in2 (e is dimensionless). Therefore,
the crack size must appear in linear form as a (not a2 or a3 or otherwise). In that
case the dimension of G becomes indeed Ibs/in2 x in = lbs/in. Thus is can be
understood without analysis that Equation (4.3) must be of the format shown.
Equation (4.3) can be changed into (4.2) because the stress-strain equation
(e = a/E) is known. Expressing G in terms of stress only, permits the use of the
criterion in engineering analysis where stresses are of interest:
nf3 2 a2 a
-- = R (4.4)
E
so that the fracture stress can be calculated from:

(4.5)

Provided f3 is known for the geometry at hand, Equation (4.5) can be used for
engineering fracture analysis to calculate the fracture stress of a cracked
structural component.
In the case of plastic deformation the geometry factor f3 may change, but there
will still be a dimensionless geometry factor; let this factor be denoted as H.
Whether the stress-strain curve of the material is linear or not, the strain energy
release rate is stiIl given by Equation (4.3) (same dimensional arguments):
Haea = R. (4.6)
Note again that the equation must contain a factor ae, the dimension of which
is lbs/in, so that the crack size must appear in linear form (a) for the dimension
to become lbs/in (strain energy per unit thickness and per unit crack extension).
Indeed, Equations (4.3) and (4.6) are identical regardless of the form of the
90
strain-stress curve, be it that another dimensionless geometry factor may
appear.
For a linear-elastic material dU/da is generally called G, but for a non-linear
material the shorthand denotation J is used, which is a little confusing and for
which there is no good reason. Since the denotation J is used throughout the
literature it will be used here as well. Similarly, while the symbol R is used for
the fracture energy of a linear-elastic material, the fracture energy for a non-
linear material is denoted in shorthand as JR' The fracture criterion is still
dU/da = d W/da, which is written in shorthand as G = R for elastic materials,
but in accordance with the above new symbols, it becomes for a non-linear
material:
J = JR (4.7)
or alternatively:
H(Jea = JR' (4.8)
The previous Equations (4.1) through (4.6) have not changed; only the symbols
have.
It was possible to use Equation (4.3) for fracture analysis because [; = (JI E, so
that Equation (4.3) becomes (4.5), thus permitting calculation of the fracture
stress. By the same token, Equation (4.8) can be used for fracture analysis, if G
can be expressed in (J, i.e. if an equation is available for the non-linear stress-
strain curve.
Any form of stress-strain equation will be useful, provided it properly
describes the material's stress-strain curve. It will have to be a curve-fitting
equation (empirical). There is no objection against this; also Hooke's law is
empirical as it is no more than a linear curve fit, where E follows from the slope
of the line.
The most convenient curve fit for a non-linear stress-strain curve (Figure 4.1)
is a power function. The resulting stress-strain equation is known as the Ram-
berg-Osgood equation:
(J (In
G = E+ F or G = Gel + Gpl • (4.9)

For most materials, but not all, Equation (4.9) provides a good fit of the
stress-strain curve. The equation can be used in various forms, but Equation
(4.9) is the most convenient. Other forms of the equation will be discussed in
Section 4.7.

4.3. The fracture criterion

The Ramberg-Osgood stress-strain equation can be used to evaluate the

fracture criterion of Equation (4.8). First note that the first part of Equation
 91

&
Figure 4.1. Stress-Strain diagram.

(4.9) represents the linear (Hooke's) part of the stress-strain curve. For this part
it is already known that dUjda = n{32u2ajE. Therefore, only the effect of the
plastic part needs consideration:

(4.10)

Note that this equation simplifies to Hooke's law for the case that n = 1
(F = E).
Substitution of Equation (4.10) in Equation (4.8) provides:
Hun+1a
F = JR' (4.11)

For n = 1 (F = E) the Equation (4.11) indeed reduces to Equation (4.4), as it

should; this implies that for n = 1: H = n{32.
Now dUjda for the elastic part, G, and dUjda for the non-linear part, J, are
both known. The fracture condition is (combining Eqs; 4.1-1 and 4.4):
dW
(4.12)
da
or
G+J=JR·
It has become general practice to denote the fracture energy d Wj da as JR' This
is again confusing because actually it is R + J R , which is now shorthanded as
JR'
Because G is equivalent to J for n = 1, as pointed out on the basis of
92

Equation (4.11), it is more consistent to call it Jel , and to drop the shorthand
denotation G altogether:
Fracture if: J el + Jpl = J R (4.13)
or

(4.14)

This equation indeed provides a useful fracture criterion for engineering ap-
plications, because it is expressed in stress. Fracture will occur when the stress
is sufficiently high to satisfy Equation (4.14). The fracture energy J R in the
equation represents the material's fracture resistance; it is therefore essentially
the material's toughness expressed in a somewhat different form. It can be
measured in a test, if the geometry factors p and H are both known for a certain
specimen configuration. The modulus E, as well as nand F can be derived from
a stress-strain curve measured in a regular tensile test. By performing a fracture
test on a cracked specimen, the fracture stress can be measured. Values for all
parameters in the left hand side of Equation (4.14) are then available, so that
the 'toughness' J R can be calculated.
For example, assume that for a certain material E = 30000ksi, n = 7 and
F = 2 X 10 12 . (Section 4.7 explains how nand F are obtained from a tensile
test). Also assume that for a certain specimen with a crack of a = 2 inch and
the values of p and Hare 1.1 and 6 respectively. The specimen is tested and it
is observed that fracture starts at a stress (1 = 30 ksi. Substitution of all data in
Equation (4.14) provides:
7t x 1.1 2
X 302 X 2 6 X 307+1 X 2
30000 + 2 X 10 12 JR

so that
JR = 0.23 + 3.94 = 4.17in-ksi = 41701bs/in.
The fracture energy now being known for the material at hand, Equation (4.14)
can be used to calculate the fracture stress of a cracked structure by solving for
(1. Naturally, both geometry functions p and H must be known for the structure

at hand. Unfortunately, Equation (4.14) cannot be solved directly for (1; an

iteration procedure must be employed, but this is merely a tedium. As the
solution usually is obtained by a computer, there is no serious difficulty. In the
event that Jel is small compared to Jpl , the first term in Equation (4.14) can be
neglected; it reduces to:

(4.15)
F
 93

which can be solved directly for the fracture stress:

= (FJR)I/(n+l)
(ffr Ha . (4.16)

Once more, note that for n = 1 (F = E) this equation reduces to Equation

(4.5), with apparently H = n/3 2 •
Assume that for a certain structure with a three inch crack the geometry
factors are p = 1.2 and H = 7. The structure is made of the material on which
the above hypothetical test was done, so that E, F and n are as given above. The
material's fracture resistance was found to be 4.17 kips/in = 4170 Ibs/in. Indeed
the elastic part of J was small, so that Equation (4.16) can be used to calculate
the fracture stress of the structure as:

(ffr
_ (2 x 107 x x3 4.17)1 /
-
12 (1+7) _
-
•
28.2 ksl.

Clearly then EPFM is no more difficult to use than LEFM; the only complica-
tion is the necessary iterative solution when Equation (4.14) must be used. (One
other complication arises as discussed in Section 4.4.) Of course the 'toughness'
J R is to be measured in a test, but also the LEFM toughness must be measured.
The geometry factors p and H must be obtained for the structure at hand. The
elastic geometry factor, p, has been calculated for many geometries and
compiled in handbooks [1, 2, 3], or can be derived using one of the simple
procedures discussed in Chapter 8. Similarly, the plastic geometry factors, H,
have been calculated for a number of geometries. These are also available in
handbooks [4, 5]. One complication arises here because H depends upon n also:
H = B(aIL, n). Obtaining H is more cumbersome than obtaining p, but once
calculated H can be presented in handbooks for later use.

4.4. The rising fracture energy

EPFM deals with high toughness materials. For such materials the fracture
energy, J R , tends to increase during the fracture process. This was discussed
already on the basis of R in Chapter 3; its consequence is that fracture may be
slow and stable initially, until at some point an instability occurs causing
fracture to become fast and uncontrollable. The procedure to distinguish
between stable and unstable fracture in EPFM is the same as in LEFM, and also
in this respect EPFM presents no new problems. The LEFM procedure for
R-curves as reviewed in Chapter 3 will be assumed known for the following
discussion.
Figure 4.2 shows the rising JR-curve. Fracture commences when J = JR' For
a certain existing crack, the curve for J can be drawn for a number of values of
the stress, on the basis of the left part of Equation (4.14). Several such curves
94

Figure 4.2. J-curves for different stresses and Jrcurve.

are shown in Figure 4.2. At the stress (Ja the value of J(a) is given by point A,
which is lower than B. Hence J < J R , and fracture is not possible. Increase of
the stress to (Ji raises J(a) to point B. Fracture now takes place because J = JR'
But this fracture is stable; it cannot proceed if the stress remains equal to (Ji'
because then J would increase to point C, while J R would increase to point D,
so that J would be less than JR' For fracture to continue the stress must be
increased to (Jb' which brings Jto point D. During the increase of the stress from
(Ji to (Jb the fracture proceeds in a stable manner from a to a + l1ab' Further

increase of the stress causes fracture to proceed . The fracture is controIIable

(stable); it can be stopped at any time simply by keeping the stress constant.
FinaIIy, when the stress reaches (Jf" the fracture become unstable (point E),
because from there on J will remain larger than JR' Fast, complete fracture wiII
cause the structure to break in two. If at any time between Band E the structure
would be unloaded, it would stay intact, although the damage would be larger
by l1a.
The stable fracture is often referred to as stable crack growth. This is a
misnomer, because the crack extends by fracture and not by one of the crack
growth mechanisms discussed in Chapter I. Stable fracture is fracture in
progress, although it is slow and can be stopped if the stress is not further
increased. After instability fracture is uncontrollable. Also the term 'sub-critical
crack growth' is sometimes used for the stable fracture. The description may be
clear in the context, but in its general use it refers to cracking by fatigue and
stress corrosion. The crack becomes critical when fracture begins, whether
stable or not. The use of more consistent language in fracture mechanics would
certainly avoid confusion. One may argue about the preciseness of the terms
 95

used here, but at least different phenomena are identified consistently by

different names.
It is apparent from Figure 4.2 that the condition for instability is:

(4.17)

which signifies tangency between the J(a) curve and the JR-curve. The instability
point can be found graphically by plotting the JR-curve and the J(a) curves for
a number of stress values as in Figure 4.2. Algebraic solutions are discussed in
Chapter 10. (See also Eqs. 3.52)
The procedure is valid when there is a condition of load control; i.e. if the
stress does not drop once the fracture becomes unstable. If there is displacement
control, the stress may drop. For example, when the loading is due to thermal
stress, the end displacement is fixed. During fracturing the stiffness decreases
which may cause a very rapid decrease of the stress, because less stress is needed
to maintain the fixed displacement. Since J depends upon (J, it may not be
possible to maintain J > J R , so that the instability is postponed. In that case the
drop in stress must be calculated from the decreased structural stiffness, and the
calculation be continued until Equation (4.17) is satisfIed (Chapter 10).
Both in load control as well as in displacement control, the stress will increase
to a certain maximum. Instability in load control coincides with the maximum
stress. In displacement control the fracture may remain stable after the
maximum is reached (Figure 4.3); instability mayor may not ensue later. Up till
the maximum stress there is no difference between load control and displace-
ment control. Displacement control is governed by the stiffness of the system
(loading + structure); instability is therefore system dependent and not a
'material property' (Figure 4.3).
Most structures are under load control. For example, the wave load on a ship,
the gust load on an airplane or the load on a crane does not abate when the
structure is fracturing. Hence, a load control analysis is usually applicable. In
the event of displacement control, the only thing of interest often still is the
maximum stress that can be sustained. In that case no account for displacement
control is needed either, because up till the maximum stress there is no distin-
ction anyway.
In most cases the difference between (Ji and the stress at instability is very
small, and solving Equation (4.14) once for (Ji (point B in Figure 4.2) may be
adequate (conservative). Examples of analysis are provided in Chapter 10.
Finally it should be mentioned that Equation (4.17) has been modified by
96

QO
DISPLACEMENT FIXED
LOAD CONTROL

(J
(J

a a
Figure 4.3. Load control (left) versus displacement control (right).

multiplying both sides by E/F~r leading to:

E dJ _' E dJR or •
III
h h an d
sort T (4.18)
F12, da FI~ dQ

Instability ' O"fr

Beginning of
Fracture, (Jj

a
Figure 4.4. Residual strength diagram in EPFM.
 97

The quantities at both sides are then renamed. The left side in shorthand is
called the tearing modulus, T, the right side T mat for lack of a descriptive word.
Hence, the instability equation is written as T = Tmato but nothing has changed.
To the uninitiated only the fog intensified.

4.5. The residual strength diagram in EPFM; collapse

By solving Equation (4.14) for a number of crack sizes one obtains the residual
strength diagram, as shown in Figure 4.5. Solution of Equation (4.17), either
graphically or otherwise (Chapter 10), provides the stresses for instability as
well. Hence, there will be a line for the onset of fracture, and one for instability
(compare Figures 3.l9 and 3.20). For convenience only the lower of these
(Figure 4.5) is taken into account in the following discussion, because in general
there is little difference between U i and ufr.
Now consider again Equation (4.14). For the case of a -+ 0, the calculated
residual strength will be infinite. This same problem was encountered in LEFM;
it has not been solved by EPFM. An approximation is still necessary as in
LEFM. For small crack sizes the calculated fracture stresses will be very high;
for very small cracks it will be infinite, i.e. higher than F tu , which is clearly
wrong. Apparently, as in LEFM, collapse as a competing fracture condition
cannot be ignored. Collapse determines the highest load carrying capability of
98
the structure, regardless of any fracture criterion, whether LEFM or EPFM.
Thus collapse must be analysed as a separate condition, as in the case of LEFM.
This problem was discussed extensively in Chapters 2 and 3. The condition
satisfied first (at the lowest stress) determines the strength of the structure.
For an ideally plastic material (horizontal stress-strain curve beyond F,y), the
exponent in the Ramberg-Osgood equation (Equation 4.9) is n = 00. Thus the
left hand side of Equation (4.14) becomes indeterminate; J is undefined and
attains any value necessary (collapse) to satisfy the equation. This always will
occur at plastic instability. When plastic deformation becomes uncontrollable J
attains the necessary value for collapse. Attempts have been made to incor-
porate the collapse condition in the EPFM criterion, but this cannot be done by
just considering J. It would require a separate evaluation of plastic instability
criteria. Although this might be possible in principle, it is hardly feasible for
engineering analysis. Concocted modifications of J are doomed to failure as are
plastic zone corrections to K [6]. The practical solution is to treat collapse in the
manner discussed in Chapters 2 and 3, and to assess both the collapse condition
and Equation (4.14). The one leading to the lowest failure stress determines the
residual strength. Then the problem is the same as in LEFM, discussed exten-
sively in the previous chapter.

4.6. The measurement of the toughness in EPFM

Any kind of specimen can be used to measure the Kc or K Jc ; If this were not the
case, applications to structures would not be possible either. There is no
objection against a standard specimen for which f3 is known very accurately, but
any other specimen with known f3 is appropriate. In the same vein, J R can be
measured with any specimen for which f3 and H are known. An example was
given already in Section 4.3. In order to obtain the JR-curve the stress should be
measured during the stable fracture as well as the progress of fracture
a = a + A.a. Equation (4.14) then provides JR. There is definitely a specimen
size requirement, because J becomes indeterminate when collapse occurs, as
discussed.
It is unfortunate that a test procedure to measure the JR-curve was standar-
dized [7] prematurely, at a time Equation (4.14) had not evolved. Thus the test
seems to bear no relation to what was discussed in the previous sections.
Recall from Figure 3.12 that the energy release rate is equal to dU/da, the
small area between two load-displacement curves for a and a + da, regardless
of whether the load or the displacement is constant over da. If the displacement
is constant there is a decrease of strain energy (negative), so that the available
energy is - dU/da (see Chapter 3). In the elastic case the area between the curves
is easily determined as G = - dU/da = - tPb. This is still the case when the
 99

load-displacement curve is non-linear as in Figure 4.6; the area between the

curves is dU/da (negative), so that J = - dU/da is:

J = - Jo (oP)
oa b
dt5. (4.19)

If an equation for the load-displacement curve is available, J can be obtained

from Equation (4.19). The standard for J measurements is based upon an
estimated equation for the load-displacement curve.
Consider a bend specimen in which the remaining ligament is fully plastic as
in Figure 4.7. The elastic deformations can then be neglected. The specimen will
deform as two rigid bars connected by a plastic hinge. The bend angle if! depends
upon the bending moment, which is equal to PL/2 (Figure 4.6). If the stress-
strain curve is assumed horizontal beyond F,y the resistance of the plastic hinge
is equal to F,y" The equation for I/J then must be as shown in Figure 4.7.
Obviously, if! is proportional to P and inversely proportional to F,y, the
resistance of the hinge. If the specimen is twice as thick, if! will be half as large
for the same load, so that if! must be inversely proportional to B. Hence, for the
equation to be dimensionless the ligament must appear as b2 in the denominator.
A dimensionless constant C will complete the equation. By noting that if! = 15/
L, the equation for the dependence of 15 on P is obtained as in Figure 4.7.
Apparently, this is the sought equation for the load-displacement curve.
Differentiation of the load displacement equation with respect to b provides
oP/ob, and it follows from substitution of P that oP/ob = 2P/b (Figure 4.7).
When the crack extends over da the ligament will decrease by db, which means
that da = - db, and therefore oP/oa = - 2P/b. Substitution of this in
Equation (4.19), and noting that J must be determined for a unit thickness
(which requires division by B), the equation for J becomes:

J = b~ f P dt5 or J = !~. (4.20)

p G'-J)

Figure 4.6. Strain energy release at constant displacement (infinitesimal da). left: linear; right:
non-linear. Compare with Figure 3.15.
100

(a) Bend specimen with plastic hinge

PL b
'" = C Fty Bb2 L

FtyBb 2 b
P =
CL 2
8P FtyBb P
2 Ce 2-
8b b
8P -2~
8a b
2A
(b) J = - 2- f8P db = ~ f Pdb
B 8a Bb Bb

'~"U A Bb

b
(c) J for bend specimen

Figure 4.7. Standard J le and J R test. (a) Specimen; (b) Analysis; (c) Result.

The integral part of the equation is simply the area, A, under the load displace-
ment diagram (Figure 4.7c), which can be measured - by using a planimeter or
otherwise - from the test record of P - b.
Surprisingly, the small area between the curves in Figure 4.6b is found by
taking twice the total area, A, under the curve and by dividing by band B.
Naturally, this is true only under the assumption made, namely that the entire
ligament is yielding and that the stress-strain curve is horizontal beyond FlY
(note that n does not appear in the equation). These are in fact the collapse
conditions under which J becomes indeterminate as argued above. The latter
means that Equation (4.20) is unsuitable for the independent calculation of J,
but since J = J R during fracture, the equation can be used to determine JR'
Figure 4.7 is for a ligament in bending; it applies to a compact tension specimen
 101

as well, be it that a small correction must be made [7] as there is some tension
superposed on the bending.
At the time of fracture J = JR' Hence the fracture resistance J R is derived
from the load-displacement diagram measured in a test on a compact specimen
by applying Equation (4.20). The fracture resistance increases (Section 4.4).
In a test fracture is first stable. At various stages during the test the amount of
stable fracture, /).a, is measured and marked on the load-displacement diagram
(Figure 4.8). By the use of Equation (4.20), J R can be obtained as a function of
/).a, yielding the sought JR-curve as in Figure 4.8b (compare with Figures 3.18,
3.19 and 4.6).
The standard recommends use of a so-called 'blunting' line, which forces the
JR-curve to go zero for a = O. Obviously, with a = 0 in Equation (4.14), the
stress for fracture would become indeterminate. Thus the blunting line cannot be
used in a fracture analysis. It is an artificiality; blunting is due to plastic
deformation (Figure 1.4), not fracture (Figure 1.7). It has no place in the
JR-curve, because it ignores the physics of the problem.
Constraint is essentially ignored in the standard. In LEFM the toughness K
or R depends upon constraint (thickness), and the standard for Klc tests puts
great emphasis on the constraint condition through the thickness requirement.

(a) P-b

BLUNTING
LINE

(b)

Figure 4.8. Measurement of JR' (a) Load displacement diagram with indications of fracture
extension; (b) J R curve.
102

Clearly, this problem exists equally in EPFM. The fracture resistance, J R ,

depends upon whether contraction in thickness direction can take place freely
or whether it is constrained by the surrounding material (Chapters 2 and 3).
The standard for K lc tests requires the thickness to be larger than B > 2.5
(KlcIF,yl Since in the elastic case J = K2/ E, this translates into:
B > 2.5 EJRIF,~, (4.21)
There is no reason why this requirement should be less severe in elastic plastic
fracture. Instead, the following condition has been proposed [8]:
B > 25 JIFty" (4.22a)
When this is applied in LEFM, it yields with E = 30000 ksi, and e.g.
FlY ~ 50ksi and Equation (4.21):

B > 0.04 (K/c)2.

F,y
(4.22b)

Obviously, this is an inadequate condition (Figure 3.8). The condition of

Equation (4.22a) puts C ~ 1500 rather than C ~ 25, in order for the require-
ment to be as stringent as the one for LEFM. There is no reason why it should
be any less for J R tests. Naturally, regardless of constraint, the test will provide
J R for the thickness used, like an LEFM test would provide Kc (Chapter 3).
It seems strange that ASTM embarked on standardizing a test while the
subject was still in the research phase (a standard may be useful for engineering
applications; it has no place in research), and at a time Equation (4.14) was not
available. Knowledge of JR is of no use but for calculation of the fracture stress
of a structure. The latter became possible only due to Equation (4.14). Last but
not least, Equation (4.14) has made the equation in the standard obsolete. A
thorough revision of the standard for engineering rather than research seems
opportune.

4.7. The parameters of the stress-strain curve

This section could be very short if the stress-strain equation used in EPFM had
been given the simple form of Equation (4.9). Unfortunately, new confusion has
been introduced as will appear in Section 4.8. It is for this reason that a longer
discussion of the stress-strain equation is necessary here.
Hooke's law is a simple mathematical description of the experimental fact
that elastic stress and strain are proportional; it is an empirical law. In the same
vein one can use mathematical equations that fit the remainder of the stress-
strain curve. The most useful equation is the Ramberg-Osgood equation, which
covers the plastic strain as:
 103

(4.23)

Its exponent n is called the strain hardening exponent. Further, F is a propor-

tionality constant like E is a proportionality constant. For lack of a better word
F might be called the 'plastic modulus'. For n = I and F = E the equation
covers elastic behavior as well.
Combination of the elastic and plastic strain provides the total strain:

(4.24)

The modulus E is measured as the slope of the linear part of the stress-strain
curve. Values for nand F can be obtained rather easily as well. A tensile test
provides the stress-strain curve in terms of strain. At any stress the plastic strain
is obtained from the measured total strain as Gpl = Gtot - alE (Figure 4.1); for
an example see the solution to Exercise 1).
Taking the logarithms in Equation (4.23) yields:
log Gpl = n log a - log F. (4.25)
Hence, in a plot of log(stress) versus log(plastic strain), the data should fall on
a straight line as shown in Figure 4.9a. The slope of the line is n, the intercept
with the abscissa is -log(F) as shown. If the data do not fall on a straight line,
the material does not obey a Ramberg-Osgood equation, which is sometimes
the case. There is nothing that can be done about this.
With nand F, the total stress-strain curve can be calculated with Equation
(4.24). If the data (Figure 4.9) are reasonably on a straight line Equation (4.24)
will be a good representation of the measured stress-strain curve as is demon-
strated in Figure 4.9b: the curve through the data points was obtained using n
and F from Figure 4.9a.
The Ramberg-Osgood equation was intended to describe the true stress-true
strain curve. However, up to the point of maximum load, the equation can be
used ju~t as well for the engineering stress-strain curve (Figure 4.9), be it that n
and F are different even though the material is the same. The latter is demon-
strated in Figure 4.10 which is for the same data as Figure 4.9. The only
restriction is that the equation cannot be used beyond maximum load in the case
of the engineering stress-strain curve. It can be readily demonstrated [9] that n
is equal to the inverse of the true plastic strain at maximum load. Although this
would be an easier way to find n, its accuracy is poor. The value ofn so obtained
from Figure 4.10 would be n = 8.49 which is different from the n derived from
the slope of the line in Figure 4.9. As F still would have to be determined as in
Figure 4.10, the plot is necessary anyway. Besides, this simpler method would
not apply to the engineering stress-strain curve.
104

1.8

1.6

~ 1. 4

~ 1. 2
'3
1.0

LOG (EPSILON PLASTIC) • 6.27 LOG (SIGMA) - LOG(2. 096E+13)

0.8

n ' 6.27, F· 2.096 E+13 (KSJ ·6.27)

0.6

o. 4

0.2

-4.5 -4.0 -3.5 -3.0 -2.5 -2.0 -1.5 -1.0 -0.5

(a) LOG (EPS J LON PLAST I CI

90
600

80
:::: 52
Ul SOD ~
~ 70
en
en
w 60
g: 400
en
50
300
40

30 200

20
100
10

(b) .025 .050 .075 .100 .125 . ISO .175 .200 .225
STRAIN

Figure 4.9. Ramberg-Osgood equation for engineering stress-strain curve. (a) Log (plastic strain)
versus log (stress); (b) Engineering stress-strain curve equational fit.

The 'plastic modulus', F, tends to have rather unwieldy values, so that it may
be more convenient to replace F by (Sf)":

with (Sf)" =F (4.26)

 105

1.2

1.0

LoGCEPSILON PLASTIC> = 5.77 LoGCSIGMAJ - LoGC3.o96E+12)

0.8
n = 5.77 , F = 3.096 E+12 CKSI·S.77)
0.6

0.4

0.2

-4.5 -4.0 -3. 5 -3.0 ·2.5 -2.0 -I. 5 -I. 0 -0.5

(a) LOC CEPSI LON PLAS TI [)

90
600

80
3'
U1 500 ~
25
U1
U1
w
e: 400
'" 50
300
40 E • 3 a 000, F = 3. 096 E+12, n = 5. 77

30 200

20
100
ID

. 025 . 050 . 075 . 100 . 125 . 150 . 175 . 200 . 225

(b) STRAIN

Figure 4.10. Fit of true stress-strain curve. (a) Long (true plastic strain) versus log (true stress); (b)
True stress - true strain equational fit.

It appears from Equation (4.26) that for (J = Sf' the strain reaches 1 (100%
strain). Sometimes Sf is called the 'flow stress', a somewhat unfortunate name,
because it suggests that Sf has physical significance as a stress. As a strain of
100% is hardly ever possible, the stress will never reach the value Sf.
This section could end here if the developers of EPFM geometry factors had
used the Ramberg-Osgood equation in the form discussed. Instead they
106

employed a more complex form, the implications of which must be understood

by the users of the geometry factors.
It should be realized that the Ramberg-Osgood equation has only three
parameters, namely E, F and n or E, Sf' n. No further parameters are needed,
or even permitted. Nevertheless, a fourth parameter was introduced as follows:

(4.27)

There is no objection against replacing Fby (ao)" as was done before in Equation
(4.26). However, introduction of the fourth parameter is permitted only if a and
aoare dependent, because by definition: a3/a = F. As a matter of fact, one may
now take any value for ao , as long as a is adjusted accordingly to a = a3/F.
For no reason at all, also the modulus was eliminated by DEFINING a strain
eo, such that eo = ao/E. Thus, Equation (4.27) was written as:

(4.28)

The above equation is permissible as long as one strictly adheres to the

dependence of parameters, namely:
a3
eo = ao/E and a = -. (4.29)
eoF
The literature on EPFM often refers to ao as the flow stress, which is a very
disturbing and confusing misnomer, because ao can be given any arbitrary value
as long as Equations (4.29) are adhered to. It is generally taken equal to the yield
strength, and the suggestion is raised that the latter is significant for the
equation. Nothing could be further from the truth when any arbitrary value is
appropriate.
This can be readily demonstrated by the example in Table 4.1. From Figure
4.9 it appears that E= 30 000 ksi, n=6.27 and F=2.l El3 for the given
material. Taking arbitrary values of 50, and 100 ksi for ao, and using the
mandatory Equations (4.29) the stress-strain equations become as shown in
Table 4.1. Any of these equations leads to the same strain for the same stress,
as shown. Indeed, ao can be chosen at will. It may be taken equal to the yield
strength if so desired, as long as it is realized that this is an arbitrary choice.
It can be argued that the yield strength depends upon the shape of the curve
(E, F, n), and vice versa, the yield strength depends upon F and n. This is true,
and as such taking ao equal to the yield strength is certainly defendable.

4.8. The h-functions

The geometry factors H(a/L, n) were developed [4,5] for a number of structural
geometries and n-values, using the plastic stress-strain equation of Equation
 107
Table 4.1. Effect of different definitions of Ramberg-Osgood equations.

E = 30000ksi; F = 2.1 E 13 ksi 6 .27 ; n = 6.27

([ ([6.27
As Equation (4.9) (A)
e = 30000 + 2.1 E 13

With SJ-27 = F one obtains sf = 133 ksi

As Equation (4.26). (B)

With arbitrary ([0 = 50 ksi; eo = 50/30000 = 0.00167

a = ([Z·27/ eo F = 506.27 /0.00167 x 2.1 E 13 = 1.28,

e ([ ( ([ )6.27 (C)
0.00167 = 50 + 1.28 50 As Equation (4.28)

With arbitrary ([0 = 100 ksi; eo = 100/30000 = 0.0033

a = 1006.27 /0.0033 x 2.1 E 13 = 50

e ([ ( ([ )6.27
0.0033 = 100 + 50 100 As Equation (4.28). (D)

Results

G E E E E

ksi Equation (A) Equation (B) Equation (C) Equation (D)

50 0.00381 0.00383 0.00381 0.00379
70 0.01998 0.02021 0.01996 0.01994
80 0.04342 0.04395 0.04339 0.04336

(4.28). This requires adherence to Equations (4.29) as discussed in the previous

section.
The equation for Jpl was written as:

Jpl = a([oeoch l (~J+I. (4.30)

In this equation P is the load, Po is the load at collapse supposing ([0 were the
collapse strength. Instead of the crack size, a, the unbroken ligament, c, is used.
Finally, hi is the geometry factor.
108

Clearly, the load P is related to the stress, Po to 0"0 and the ligament c to the
crack size a:
P

(4.31)
c =
where g, k, and f are just functions of the geometry. With this knowledge, and
with Equation (4.29) the complicated Equation (4.31) readily turns into:

Jpl
0"0 0"0 eo fa hi (g)n+1
= eoF k (:o)n+1
v
(4.32)

which immediately reduces to:

(4.33)

with H = fh(gjk)n+l. Equation (4.33) is the basic form of the equation already
known as Equation (4.8) from simple arguments. Equation (4.30) is just a
complicated version of the same. Obviously, cx, 0"0' and eo can be divided out;
they are superfluous. Indeed, the solution to Exercise 6 shows that the same
results are obtained with Equation (4.32) regardless of the choice of 0"0
(arbitrary). Equation (4.30) suggests that J depends upon a collapse load Po, but
obviously 0"0 is divided out as well. This should be expected, because a collapse
load cannot enter into J since the stress-strain equation used has no limit. The
elastic energy release, G, could be expressed in the same manner by using
P = gO", Po = kO"o and 0"0 = eoE
k2 (P)2
G = n/320"0eOt a Po . (4.34)

Bringing in the collapse load or the collapse strength does not make G
dependent upon same; it merely amounts to multiplying numerator and denomi-
nator by the same number, which does not change the basic equation. Collapse
does not enter LEFM equations, nor does it enter EPFM equations in their
present form. An artificial introduction does not change this fact. Collapse is a
competing condition which must be assessed separately, at least for the time
being.
Two other objections can be raised against Equation (4.30). Instead of just
one geometry parameter H, it must use four geometry parameters: h, g, k, and
f. Every time a calculation is performed double work is necessary: parameters
must be derived and are subsequently divided out. Naturally, one could, once
and for all, calculate H from H = fh(gjk)"+ I and from then on use Equation
 109

(4.33), but a computer does not object to unnecessary work and for it the form
of Equation (4.30) need not be changed.
The second objection is that J is expressed in the load P, while in engineering
one works with stress. Therefore Equation (4.33) is more useful; all other
fracture mechanics equations are expressed in stress for this very reason. For
complicated structures the conversion from load to stress is done in the design
stage not at the time of fracture analysis.

4.9. Accuracy
Researchers have expressed great concern about the large variability of J and
(consequently) JR. The reason for the large variability is obvious. As J depends
upon stress to the n-th power, according to Equation (4.11), a slight difference of
5% in stress with e.g. n = 9, leads to a difference of (1.05)10 = 1.63 or a
difference of 63% in JR. (Note that this occurs also in a standard tests: the
load-displacement diagram becoming almost horizontal, the area under the
curve, which determines J R , changes dramatically with a slight change in ().
This may seem bothersome but it is of little practical importance. The value
of J R is of no interest as long as the predicted fracture stress is reasonably
accurate. This the case, because in a fracture analysis the situation is reversed:
a difference of 63% in J with n = 9 will lead to only a difference of 5% in the
predicted fracture stress: (1.63)1/10 = 1.05. (See Eq. (4.16»
For a difference in J by a factor of 2, and for n = 7, the predicted fracture
stresses would be different by 2 1/8 = 1.09; hence the error (difference) would be
9% only. This is clearly demonstrated in Figure 4.11, showing the results of two
calculations with exactly the same input and n = 7. Two JR-curves were used
differing by a factor of approximately two throughout. The predicted fracture
stresses differ only by a small amount (9%). In general the stresses in a structure
will not be known with better accuracy, so that any of the predictions in Figure
4.11 would be satisfactory from an engineering point of view. Even the predicted
amounts of stable crack growth at maximum load do not differ appreciably, as
shown in Figure 4.12.
Most alloys satisfy the Ramberg-Osgood equation fairly well. However, a
material of great interest to some industries, namely annealed 304-SS, exhibits
a stress-strain curve that cannot be fitted with the equation. Yet it must be fitted
to such an equation, otherwise an EPFM fracture analysis cannot be performed.
In that case there is a choice as to whether the equation should fit the lower or
the upper part of the stress-strain curve.
There is no categoric answer to this problem. Any conclusions reached are
necessarily material specific. Although it may be argued that most of the crack
tip material is subject to relatively small plastic strains, so that a fit of the lower
part of the stress-strain curve is the most important, it cannot be denied that the
110

4.5 CASE 2 CASE 1

-.-
.-.-/' -'- _.- -'--
4.0
/----
3.5 /.
,/
3.0
crI //
..... 2.5
I
2.0 ,/
/
1.5

1.0

0.5

2 3 456 7 8 9
DELTA 0 (101M)

450 CASE 2 CASE 1

400

350
c:--__
I
VI
1II
W
cr
300 I
I
~
1II 250

200

150

100
Ii
50
I

10 20 30 40
I
50 60 70 80 90
CRACK LENGTH (101M)
Figure 4.11. Fracture stress prediction for different R-Curves for center cracked panel [15]. Above:
R-curves; Below: predicted fracture stress and fracture progress (Courtesy EMAS).
 111

450 OJ/O ... OJR/O ...

400

350

300

250

-- --
200

150

100 .r"
-'-
50

2 4 6 8 10 12 14 16 18
OELTA Q (MM)

450 OJ/OA OJR/OA

400
o
"0
;:;;:
...., 350
"0

o
a 300
.....
....,
"0 250

200

150

100

50

2 4 6 8 10 12 14 16 18
DELTA Q (MM)

Figure 4.12. Instability prediction according to Equations (4.17) for the two cases in Figure 4.11
[I5]; Above: Case I; Case 2 of Figure 4.11 (Courtesy EMAS).
112

crack tip material is subject to strains in accordance with the upper part of the
stress-strain curve. Common sense indicates that the decision must be made on
a material-by-material basis; there is no categoric answer. Any conclusions
arrived at certainly should not be generalized, and if they are, they are still
restricted to materials not obeying an exponential stress-strain curve; for other
materials there is no choice. When there is a choice, the criterion for the choice
is whether the fracture stress is predicted correctly; the value of J is irrelevant.
The problem would not exist if the measurement of J R would use Equation
(4.15) instead of (4.20), because the reverse operation of Equation (4.16) would
then automatically lead to the correct answer (use of same F and n as in
Equation (4.15).

4.10. Historical development of J

Eshelby [10] defined a number of contour integrals which are path independent
by virtue of the energy conservation theorem. The two-dimensional form of one
of these integrals can be written as:

J. V dy - T ~ ds = 0
j ox
where (4.35)
e

V = Iade
~ 0

V being strain energy per unit volume.

The integral is taken along a closed contour, S, followed counter clockwise
(Figure 4.13a) in a stressed solid, T is the tension perpendicular to S, u is the
displacement, and ds is an element of S.
Although the equation is somewhat elusive, it can be seen that the first term
represents strain energy, while in the second term T is the 'force', and dujdx a
strain, so that (dujdx) ds is a displacement. As 'force' times displacement equals
J
the work, F, done by the force, the equation essentially states that V - F = 0,
which is energy conservation criterion of Equation (3.32).
Applying this integral to a cracked body [II] one can construct a closed
contour ABCDEF around the crack tip, as shown in Figure 4.13b. The integral
of Equation (4.35) along this contour must equal zero; it consists of the sum of
four parts:

I+ I + I+ I
rl CD r2 FA
= o. (4.36)

Since T = 0 and dy = 0 along CD and FA the contribution of these parts is

zero, so that for the remaining parts:
 113
y

, a\
,'

c'
,J
x

Figure 4.13. Contour integrals. (a) Elastic body; (b) Body with crack; (c) Path independent contour.

(4.37)

Therefore, the contribution of ABC must be equal (but opposite in sign) to the
contribution of DEF. Note that one is clockwise, the other counter-clockwise.
This means that the integral, if taken in the same direction along r I and r 2 will
have the same value: Sll
= Sl2 in (Figure 4.13c). As r l and r 2 were arbitrary
paths, the integral over r is apparently path-independent (one may take any
path, beginning and ending at opposing crack faces, and the integral will always
have the same value). The value of the integral was called J:

f v dy - T
au
-a ds = J. (4.38)
I X

It should be noted that there is no proof as yet that this J is the same as the one
used in previous sections; thus, for the time being it should be considered as the
114

definition of a new quantity, defined only by the value of the above integral. The
path-independence is not of apparent relevance to fracture.
If the integral is path independent any convenient contour may be taken to
determine its value. The simplest contour is a circle, as in Figure 4.14, with
radius r, its center at the crack tip. For this case y = r sin e, so that
dy = r cos e de, and ds = r de. Then Equation (4.38) becomes:

J ~ UUn &) cos 8 - T::}'d8 (4.39)

No matter what the relationship between u and e, the integral Jude at any point
always evaluates to a l ue, where al is dimensionless. Since T is a stress, it can
always be expressed as Clu, and du/dx being a strain can always be expressed
as C2 e. Then T (du/dx) evaluates as C l C 2 ue = a2ue. Both a l and a2 may depend
upon e, but regardless of how complicated this dependence, the integral of
Equation (4.39) will be:

f {uea (e) cos e -

1t

J = l uea2(e)} r de. (4.40)

-1t

No matter what the functions of e are, the integral reduces to:

f f(e) de
1t

J = uer = uer F(e) 1

-1t
= uerQ. (4.41 )
-1t

Figure 4.14. Simple circular path for contour integral.

 115

The determined integral between - nand n will evaluate to a dimensionless

number. The solution therefore will be as shown in Equation (4.41), while (1 and
B may be defined at any convenient point. Taking another point as a reference
will merely change III and 1l2' and therefore Q, but Q is only a number anyway.
If one defines (1 and B at the location y = 0, x = r (Figure 4.14) then:
J = (1,B,rQ. (4.42)
By using the Ramberg-Osgood equation for the stress-strain curve, namely
B = (1n/F, Equation (4.42) yields:
J = (1~+lrQ/F. (4.42)
Then finally:

(1, = (~)I/(n+ I). (4.43)

If n = I (F = E) this reduces to:

JE.i
(1, = .jQ Jr' (4.44)

Compare this equation with Equation (3.2) as replicated below:

K
(1, = J2iCr' (4.45)

Clearly, Equation (4.44) is the same as the very original Equation (3.2). It shows
that for n = I we have Q = J2iC, and JE.i = K. This means for n = I that
J = K2/E = dU/da. Thus the path-independent contour integral is but the
strain energy release rate. It is now apparent why this integral was denoted by
J, a symbol already used for the energy release rate.
The above being the case for n = I, it will be true in general, because nowhere
above was a restriction made with regard to the shape of the stress-strain curve.
Thus, the only significance of the 'path-independent' contour integral is to show
that it represents the strain energy release rate, J = dU/da, so that the J-integral
is indeed the same as J defined previously. Indeed, that is of secondary
importance only: the energy release rate can be defined in a much simpler way
as was shown earlier in this chapter.
Nevertheless, the integral has its use. As it is known now that it equals dU/da,
it can be used to calculate G, J or K. It can be applied to the results of a finite
element analysis (most codes have post-processors to do this). If the analysis is
for the elastic case one can obtain K from K = JE.i. When the analysis uses the
Ramberg-Osgood it can provide the geometry factor H from:
116

FJ (4.46)
H = (In+la'

This is essentially the way in which H (and h; Section 4.9) were obtained.

4.11. Limitations of EPFM

Although the concepts discussed in this chapter are generally referred to as

elastic-plastic, they are in fact elastic. The non-linear stress-strain curve used
must apply for loading as well as for unloading. This is illustrated in Figure 4.15.
Truly elastic-plastic behavior is shown for comparison. In the latter case linear
unloading occurs, and there will be a remaining plastic strain when the stress is
reduced to zero. The stress-strain relation used in EPFM, upon unloading, must
produce zero strain at zero stress. The curve may be non-linear, but not
elastic-plastic. The return to zero strain means that the material is merely
non-linear-ELASTIC.
This does not put any restrictions on the use of J as long as there is no
unloading. Without any unloading it would never be known that the curve for
unloading is different from the loading curve. However, if there is unloading
anywhere, the assumption of non-linear elasticity will cause errors. Consider for
example Figures 3.15 and 4.16a. It was shown in Chapter 3 on the basis of
Figure 3.15 that for linear elasticity the energy release during an infinitesimal
extension of a by da is always equal to the change in strain energy regardless of
whether there is constant load or constant displacement. In one case the load
does work which is twice the increase in strain energy, so that the remaining part
of the work by the load (the energy available for fracture) is exactly equal to the
change in strain energy. In the case of constant displacement the load does no
work, but the strain energy decreases, so that the released energy is again equal
to the change in strain energy.
The above is still true if the material is non-linear-elastic (Figure 4.16a), but
not if the material is elastic-plastic. During an extension da of a under constant

LEFM EPFM

(J (J

Non- Linear
Elastic

!:

Figure 4.15. Basis of EPFM.

 117

p p
RELEASE

(a) (b)
Figure 4.16. Strain energy release upon fracture by da under constant displacement. (a) Non-linear
elastic; (b) Elastic plastic.

load the energy available for fracture is equal to the work done by the load
minus the increase in (plastic) strain energy. However, during linear elastic
unloading the plastic strain energy remains in the material. Only the small elastic
part of the strain energy is indeed released: the available energy is much less
during constant displacement than during constant load (Figure 4.16b). Hence,
strictly speaking, the resulting equations are valid only in load control, so that
fracture analysis is meaningful only up to the point of maximim load: this is not
a severe restriction, because in most engineering analysis one will be interested
only in the maximum stress a cracked structure can sustain.
Yet, even the latter is somewhat questionable. The material at the crack tip
is highly stressed. When stable fracture is occurring (up to maximum load), it
is this highly stressed crack tip material that is unloading after the fracture has
passed through. Hence, the errors due to the unloading assumptions are felt
most strongly where it counts most: at the crack tip. The errors will be small
initially, but increase with increasing l1a. This sheds doubt on the analysis of
stable growth and instability. As a matter of fact the analysis will be only
meaningful when the 1 curve rises very steeply, so that there is very little stable
fracture before maximum load (Figure 4.2). It has been suggested [12] that stable
fracture should be limited to just a few percent of the remaining ligament,
otherwise the errors become considerable.
Using the path-independence of the l-integral it was shown that the crack tip
stress field in EPFM is described by Equation (4.44). For an ideally plastic
material (n -+ 00) the equation leads to a finite crack tip stress, but for all other
values of n the stress is still infinite at r = 0. It was pointed out already that the
use of Equation (4.14) leads to an infinite fracture stress for a -+ 0, and
therefore will be increasingly in error for smaller cracks; approximations for
118

small cracks will be necessary just as in the case of LEFM. For this and other
reasons, as discussed, collapse still must be evaluated separately as a competing
condition.
Apparently, EPFM has cured none of the ills ofLEFM; it is a mere extension
of LEFM for n #- I. A host of modifications to J have been proposed [13]; most
of these belong in the category of 'patch work' as much as do plastic zone
corrections to K [6]; they provide few new insights, complicate the procedure
and lead to only marginal improvements.
Nevertheless, EPFM is still very useful and has a definite place in conjunction
with LEFM. It has extended the use of fracture mechanics to non-linear
materials at least up to a point. Judicious use will provide meaningful engineer-
ing answers, provided collapse analysis is done as well, just as in the case of
LEFM, and provided appropriate approximations are made for short cracks
(Section 3.8). It may be noted again that the accuracy of fracture analysis need
only be as good as that of general accuracy of engineering procedures (Chapters
10, 12, 14). With the emergence of geometry factors [4, 5], and the possibility for
simple estimates of these (Chapter 8), EPFM has become a useful engineering
tool.

4.12. CTOD measurements

An alternative approach to EPFM has been based upon the Crack Opening
Displacement (COD). Although referred to as COD, the method actually
employs the Crack-Tip Opening Displacement (CTOD).
Consider a crack tip in a stressed body as in Figure 4.17. Let forces be applied
over a distance da behind the crack tip in such a manner that the crack just
closes over a distance da. The crack is now shorter by da and, therefore, the
required closing forces must be equal to the stresses normally present when da
is uncracked. For the time being, assume that these stresses are approximately
equal to the yield strength.
During their closing action the forces travel over the distance v. Therefore,
they do work to the amount of dF = 2 x O.5Fty v da (plate of unit thickness).
v

da
I..
Figure 4.17. Closing forces to close the crack tip over da.
 119

Since v is related to CTOD, the work will be dF = a CTOD FlY da. Upon
release of these forces the same amount of energy is released, and the crack will
'grow' again by da. This energy release, dFjda = a CTOD FlY, is what has been
called the strain energy release rate G or J. Therefore:
G = a FlY CTOD or J = a FlY CTOD. (4.47)
The first of these expression would be applicable for LEFM, the second for
EPFM.
Naturally, the stresses over the future da are not uniformly equal to FlY as
assumed, but if they are not, only the dimensionless factor a will be affected. It
turns out [6] that a is approximately equal to unity, but various interpretations
would put it between nj4 and 4jn in LEFM. However in the case of J the value
of a depends upon n [4]. In any case, the above equations lead to:

CTOD ~ ~
Fly
= K2
EFIy
(LEFM) )
(4.48)
CTOD ~ ~ (EPFM)
FlY
Fracture occurs at a critical value of G (or K) or a critical value of J. Then,
according to Equations (4.48) fracture takes place at a critical value of CTOD,
defined as CTOD c ' Consequently, CTOD c should be a material property
characterizing fracture resistance; as such it is a descriptor of 'toughness', and
the measurement of CTOD in a test would provide the material's propensity to
fracture.
A test for CTOD measurements was standardized first in Great Britain in
British Standard BS-5762. Essentially, the test is performed on a small three-
point bend specimen, (Figure 4.18). As in other toughness tests a record is made
ofload versus crack mouth opening displacement. The critical crack tip opening
displacement CTOD n usually referred to as COD, can be obtained as follows.
The ligament is assumed fully plastic so that all specimen deflection' can be
considered to be due to rotation around a plastic hinge, the specimen limbs
rotating by rigid-body motion. If it is assumed that the center of the plastic hinge
coincides with the center of the ligament, then (TOD can be obtained from the
crack mouth opening in the manner shown in Figure 4.18. It is not necessary to
make the assumption that the center of the hinge coincides with the center of the
ligament. One possibility is to determine its location experimentally. Other
options are open [14, 6].
Knowledge of the critical CTOD per se, is of no use for damage tolerance
analysis. The number will have significance for engineering only ifit can be used
to predict fracture in a structure. For this to be possible CTOD must be
expressable in terms of the stress acting in a structure, so that the (fracture) stress
120

(a)

Center of Hinge

(f)
b
2"
I ,
,, crOD = COQ.~ _ _1_)
"\ 2ba + 1

a Crack Face

CODm
~I

(b)
Figure 4.18. Crack opening displacement test. (a) COD specimen; (b) Measured CODm and inferred
CTOD.

p p
[kips] [kips]
S, S,
10 10 j j

-- i
7.56 ...
F G H
o E
C B
5.67
5

(a) 5 (b) .05

Figure 4.19. Data for exercise 1. (a) Load-displacement measured in test; (b) Enlarged view of small
b regime.

can be calculated as that stress at which the CTOD of the structural crack
reaches the critical value. So far only empirical relations between stress (strain)
and CTOD have been developed. There is no objection against their use, as long
as they are reasonably general. However, with the aid of Equations (4.47)
 121

the measured CTOD can be converted into J R or G and then the result can be
used in accordance with Eqs. (4.16) or (3.25). Certainly, Equation (4.47) is an
approximation, but IX can be obtained as a function of n [4]. Besides, an approxi-
mate general equation is more likely to be useful than a specific empirical one.
The development of the simple EPFM procedures discussed in this chapter
have made the semi-empirical CTOD approach somewhat obsolete for damage
tolerance analysis. However, the CTOD test is a useful extension of EPFM, as
it can provide J R or K/c through Equation (4.47) from tests on small specimens.

4.13. Exercises

I. For a test bar of a certain steel one measures a load displacement curve as
in Figure 4.19. The original diameter of the cylindrical bar is 0.4 inch; the
original length is 4 inch. The final thickness in the neck is 0.28 inch.
(a) Determine the engineering stress-strain curve, FlY and Flu.
(b) Determine E, F and n.
(c) Determine ex, (Jo and eo for (Jo = 100 ksi.
(d) Do the same for (Jo = 50 ksi.
(e) Check whether (a) and (b) lead to the same strain for at least two different
stress values, e.g. 50 and 55 ksi.
2. Determine the true stress-true strain curve for the problem in Exercise 1 up
to the point of maximum load. Repeat questions (b), (c), (d) of Exercise 1.
3. The bar of Exercise I is unloaded at P = 7.3 kips and tested as a new bar in
a new test. Determine the new load-displacement curve. For this cold-worked
material, assuming that FlY coincides with the load at unloading, calculate FlY
and FlU.
4. A test on a center cracked panel with 2a = 4 inch, W = 32 inch shows a
fracture stress of 50 ksi. The collapse strength is 55 ksi. Given that W = 20
inch (500mm), B = 0.4 inch and assuming E, F and n as in Exercise I,
calculate J, at fracture. Or did failure occur by collapse? (Neglect Jel )·
H = 8.72.
5. Using the results of Exercises I and 4, calculate the failure stress of a panel
24 inch wide, with a center crack of 2a = 6 inch. H = 13.5; assume that the
J el can be neglected. Does failure occur by fracture or by collapse?
6. Given that Equation (4.30) for a center cracked panel is: J = CX(Joeo(1 -
(22/W»ah l (P/Po)" + I with P = (JWB and Po = (Jo(W - 2a)B, and that hI for
a/ W = 0.125 is hI = 4.13. Calculate the fracture stress with the information
obtained in Exercise Ic (0"0 = 100 ksi) and in Exercise Id «(Jo = 50 ksi) and
show that the results are the same, independent of the arbitrarily selected (Jo.
Assume J = 2 kips/in and neglect J el •
122
7. In a J R test on a CT specimen the load rises almost linearly to 5 kip upon
which fracture begins, and upon which the load-displacement diagram
becomes essentially horizontal. The initial crack size is one inch; the displace-
ment is 0.15 inch when fracturing begins. The crack (fracture) size reaches
1.05 inch when (j = 0.20 inch and 1.15 inch when (j = 0.27 inch. Calculate
the JR-curve. Thickness is 0.5 inch, W = 2 inch.

References
[I] D.P. Rooke and OJ. Cartwright, Compendium of Stress intensity factors, H.M. Stationery
Office, London (1976).
[2) G.C. Sih, Handbook of stress intensity factors, Inst. Fract. Sol. Mech, Lehigh Un (1973).
[3) H. Tada et aI., The stress analysis of cracks handbook, Del Res. Corp (1973).
[4) V. Kumar et aI., An engineering approach for elastic-plastic fracture analysis, Electric Power
Res. Inst., Rep NP-1931 (1981).
[5] V. Kumar et aI., Advances in elastic-plastic fracture analysis, Electric Power Res. Inst., Rep NP
3607 (1984).
[6) D. Broek, Elementary engineering fracture mechanics, 4th Edition, Nijhoff (1986).
[7) Anon., Standard method for the determination of J, a measure of fracture toughness, ASTM
Standard E-813.
[8] C.F. Ghih and M.D. German, Requirements for one-parameter characterization of crack tip
fields by the HRR singularity, GE Tech Rep (1978).
[9] A.S. Tetelman and A.J. McEvily, Fracture of structural materials, John Wiley (1967).
[10) J.D. Eshelby, Calculation of energy release rate prospects offracture mechanics, Sih et al. (eds),
Noordhoff (1974) pp. 69-84.
[II) J.R. Rice, A Path independent integral and the approximate analysis of strain concentration
by notches and cracks, J Appl. Mach (1968) pp. 379-386h.
[12] J.W. Hutchinson and P.e. Paris, Stability of J-controlled crack growth, ASTM STP 668 (1979)
pp.37-64.
[13) M.F. Kanninen and C.H. Popelar, Advancedfracture mechanics, Oxford Un. Press (1985).
[14) e.e. Veerman and T. Muller, The location of the apparent rotational axis in notched bend
testing, Eng. Fract. Mech. 4 (1972) pp. 25-32.
[15] D. Broek, J. astray and back to normalcy, ECF6-Fracture control of structures, Vol. II EMAS
(1986) pp. 745-760.
 CHAPTER 5

Crack growth analysis concepts

5.1. Scope
In this chapter the concepts and procedures for crack growth analysis are
discussed. Fatigue, being technically the most important crack growth
mechanism covers most of the chapter (environmentally assisted growth or
combined stress corrosion and fatigue is integrated into this discussion). Stress
corrosion cracking by itself is covered in but one section; this is not because it
is not considered important, but because stress corrosion cracking is practically
covered by prevention and not by control, while analysis procedures are essenti-
ally similar to those either for residual strength (Chapter 3) or fatigue. Fatigue
crack growth on the other hand can hardly be prevented in many structures; it
must be controlled.
The discussions cover the concepts of crack growth analysis, retardation and
special effects, as well as the analysis procedure. Examples of analysis are given.
However, fatigue crack growth analysis is a complicated subject, and the
discussions in Chapters 6 and 7 should be read as well, before attempts to
analysize crack growth are made.

5.2. The concept underlying fatigue crack growth

Cyclic stresses resulting from constant or variable amplitude loading can be

described by two of a number of alternative parameters, as shown in Figure 5.1.
Constant amplitude cyclic stresses are defined by three parameters, namely a
mean stress, am' a stress amplitude, a., and a frequency w or v. The frequency
is not needed to describe the magnitude of the stresses. Only two parameters are
sufficient to describe the stresses in a constant amplitude loading cycle. It is
possible to use other parameters; for example, the miniumum stress, amin, and
the maximum stress, ama" describe the stresses completely, and so does the stress
range, da = a max - amin, in combination with any of the others, except a a.
Almost any combination of two of the above parameters can completely define

123
124

time

time
(b)
(a)

(c)

Figure 5.1. Parameters for fatigue crack growth. (a) Blunting and resharpening; (b) ~a and ~K; (c)
Stress ratio.

the cycle. Note that in the above the Greek letter ~ is used to indicate a RANGE
of the stress. This is not in accordance with the normal use of ~ in mathematics,
where ~ indicates a small change. In this case ~ stands for the total range of
stress in a cycle: ~(J = (Jmax - (Jmin' which need not be small at all. It would be
better to use e.g. (J, for the stress range, but since the denotation ~(J has become
common practice, it will be used here as well.
Another parameter is often convenient. This is the so-called stress ratio, R,
defined as R = (Jmin/(Jmax' One of the above parameters can be replaced by R to
define the cycling. For instance, any of the following combinations fully defines
the stresses: ~(J and R, (Jmin and R, (Jmax and R, (Ja and R, (Jm and R. The case of
R = 0 defines a situation in which the stress always rises from, and returns to
O. When R = - 1, the stress cycles around zero as a mean, which is called fully
 125

reversed loading. Note that R = (Jmin/(Jmax = «(Jmax - 11(J)/(Jmax = 1 - 11(J/

(JmaX'so that 11(J = (l - R) (Jmax and conversely: (Jmax = 11(J/(l - R).
Crack growth life is expressed as the number of cycles to grow a fatigue crack
over a certain distance. The number of cycles is denoted by N.
The crack growth mechanism, as discussed in Chapter 1, shows that a fatigue
crack grows by a minute amount in every load cycle; the mechanism is shown
schematically again in Figure 5.1 a. Growth is the geometrical consequence of
slip and crack tip blunting. Resharpening of the crack tip upon unloading, sets
the stage for growth in the next cycle. It can be concluded from this mechanism
that the crack growth per cycle, l1a, will be larger if the maximum stress in the
cycle is higher (more opening) and if the minimum stress is lower (more re-shar-
pening), so that:
l1aper cycle i for (Jmaxl i and/or (Jminll· (5.1 )
The subscript I indicates the local stresses, at the crack tip, i stands for larger,
1for lower. Note that in this case the Greek letter 11 is used in its 'normal' sense,
meaning that it indicates a small change: growth from a to a + l1a. In the
previous paragraphs the stress range, 11(J, was defined as (Jmax-(Jmin' The stress
range will be larger when (Jmin is less, so that the above equation can also be
written as:
(5.2)
The local stresses at the crack tip can be described in terms of the stress intensity
factor K, discussed in Chapter 3, where K = {3(J Fa, if (J is the nominal applied
stress. In a cycle, the applied stress varies from (Jmin to (Jmax over a range 11(J.
Therefore, the local stresses vary in accordance with:

K min = {3 (J min Fa }
Kmax = {3 (J max Fa . (5.3)

11K = {311(J Fa
Again, the Greek letter 11 stands for range and not for a small increment; the
denotation Kr would be better and less confusing, but 11K is used here III
conformance to general practice (see also Figure 5.1 b).
With the use of Eqs (5.3) the Equation (5.2) for crack growth becomes:
l1apcrcyclci for Kmaxi and/or I1Ki. (5.4)
Further, the stress ratio is defined as R = (Jmin/(Jmax' It appears from Eqs (5.3)
that at any given crack size a, the stress ratio is also equal to Kmin/ KmaX' since
{3(Jmin = Fa/{3(JmaxFa = (Jmin/(Jmax = R, so that:

Kmax - 11K 11K

R = or
1 - R'
(5.5)
126

According to Equation (5.4) there is more crack growth when Kmax is higher. It
follows from Equation (5.5) that this is the case when 11K is larger and/or R is
higher, so that Equation (5.4) can be written as:
l1aper cycle i for I1Ki and/or R i· (5.6)
In this equation l1a is the amount of crack growth in one cycle, which would be
expressed in inch/cycle or mm/cycle. If growth were measured over e.g.
I1N = 10000 cycles, the average growth per cycle would be 11 a/ I1N, which is the
rate of crack propagation. In the limit where N --> 1, this rate can be expressed
as the differential da/dN. Equation (5.6) indicates that the rate is a rising
function of 11K and R, so that the proper mathematical form of Equation (5.6)
IS:

da
dN = f(I1K, R). (5.7)

As shown, Equation (5.7) derives directly from the model of crack growth
discussed in Chapter I and shown in Figure 5.1 a.

5.3. Measurement of the rate function

According to Equation (5.7) the rate of crack growth will larger for higher 11K
and higher R. The actual functional form of Equation (5.7) can be derived
from the crack growth model in Chapter 1. However, this model - although
qualitatively in order - is a two-dimensional simplification of a three-dimension-
al process that is extremely complicated due to the presence of grains with
different orientations, grain boundaries, particles, etc. As a consequence, a
rigorous mathematical description of the model is not well possible, and a
reliable functional form of Equation (5.7) cannot be obtained from theoretical
analysis. This leaves only one possibility to obtain the function: interrogation of
the material in a test. Although this might seem objectionable to theoreticians,
it should be noted that ALL material data are obtained from tests, such as FlU
and Fry, and even the modulus of elasticity E.
Crack growth data are obtained by subjecting a laboratory specimen to cyclic
loading. The specimen may be of any kind as long as f3 is known, so that the
stress intensity factors can be evaluated. Most commonly used are center
cracked panels and compact tension specimens. The following examples are for
a center cracked panel. As long as cracks are small with respect to panel size (e.g.
a/ W < 0.4) the geometry factor, f3 is approximately equal to one, so that
K = aJ1W.
A panel as in Figure 5.2 is provided with as small but sharp central notch, so
that cracks at both sides will start almost immediately. The specimen is
subjected to a cyclic stress of constant amplitude in a fatigue machine. First
 127

~a;R=O

w
time

(a)

da
a dN
1~a~ ______________ _ ~2 . - - - - - - - - - - -.'
• I
a2 r------ ----------,1
,I

I'
"

l~:'_______ _
a, T- - - - - - - -
,,, ,,I
,
"
"
,I
"
------"
.. ,,
,I
, ;
", '

(b) (c)

Figure 5.2. Obtaining the rate function (a) Specimen and loading; (b) Measured data; (c) Rate data.

consider a stress cycle with Urnin = 0, so that R = 0 throughout the test. Also,
U rna• = l1u in that case.
Crack growth is monitored throughout the test by measuring the length of
the crack at intervals of e.g. 10 000 cycles. The results are plotted to optain the
crack growth curve as in Figure 5.2b. This is all the information that can be
extracted directly from the test. It must be interpreted for the determination of
the form of Equation (5.7).
Consider a small crack increment, l1a 1, on the curve. (Figure 5.2b) According
to the measured curve, it took I1N1 cycles for the crack to grow over l1a 1. Thus,
the rate of growth is (11 a/11N)1. For example, if the crack increment is 0.1 inch
and this growth took I1N = 10000 cycles, then the rate is (l1a/I1N) =
0.1/10000 = 1 x 1O- 5 in/cycle.
The objective is to obtain the growth rate dependence upon 11K, which
requires determination of the stress intensity range. The average crack size at
l1a1 is a1. The stress range is l1u, so that I1K1 = P1 l1u J1Ui;. Apparently, a value
of 11K = I1K1, produced crack growth at a rate of (l1a/I1N)1. This result is
 128
plotted as a data point in a diagram with da/dN( = Aa/AN) and AK along the
axes, as shown in Fig. 5.2c.
The above procedure is repeated for a number of points along the crack
growth curve. At a larger crack size a2, an amount of growth Aa2takes only AN2
cycles. Because the curve is steeper, the rate is higher: as a2 > ai' also
AK2 > AKI • Hence, a larger AK indeed produces a higher rate of growth. A
plot of the data points as in Figure 5.2c confirms this.
Because differentiation is a very inaccurate procedure, large 'scatter' may
occur in da/dN. This problem is discussed in Chapter 7. It is the reason why in
practice da/dN is obtained as a running average of 5-7 points along the crack
growth curve.
Figure 5.2c provides the growth rate for any given AK. In Chapter 3 it was
shown that the crack tip stress distribution is unique and depends only upon the
stress intensity factor. If at two different cracks in the same material have equal
stress intensity then the two crack tip stress fields are identical; there is
similitude. Hence, if the stress intensities are equal the response of the cracks
must be the same. This means that the crack growth rate will always be the same,
if AK is the same. Thus, Figure 5.2c is the material's rate response in all cases.
It can be used to analyze crack growth in a structure built of this material.
The validity of this similitude argument can be checked by performing a
second test on a similar (or different) panel, but with a different Au. The crack
growth is measured (Figure 5.3), the results analysed in the same manner as
before, and the data of both tests plotted in the same rate diagram (Figure 5.3).
The rate data of.-the second test will fall on the same line as the data of the first
test. This confirms that the same rate was obtained at the same AK in both tests.
For example, take a point on crack growth curve 1 at a crack size of a l = 0.2 in.

da -3,--_ _ _ _ _ _ _ _ _ _ _---,
(jn~h) 1 . o . . - - - - - - - - - - - - - - - - , iiN 10
w= 6 in. (in/c)
AI-Cu Alloy
.8

00

.6

;0-'1.....--'---'---6'----'10--!:20::--:3~u------'100
50,000 100,000 150,000
N
AK kshJin
Figure 5.3. Data for two tests. Tests at Au = 17.6 and 11.4 ksi; R = 0 on center cracked panels;
left: measured data; right: reduced data.
 129
With a stress range of 1l(1 = 17.6 ksi, the value of the stress intensity range is
ilK = 17.6-J0.2 x 7t = 14ksi,fifl. In the second test the stress range was
1l(1 = 11.4 ksi. This would produce a stress intensity of 14 ksi.Jffi, if
a = 2.38 x 0.2 = 0.48 in (assuming f3 = 1) in the second test, i.e.
ilK = 11.4-J0.48 x n = 14.0ksi.Jffi, still assuming f3 = 1. This means that at
a crack size 0.2-in in test 1 and 0.48-in in test 2, the stress intensities were the
same, so that the rates should be the same. Figure 5.3a shows that the slopes
(rate) of the curves at these two crack sizes are indeed equal (naturally, this
followed immediately from the fact that the two tests led to the same da/dN
- ilK diagram). Similitude in behaviour is hereby established. The results can
be used to analyze crack growth in a structure.
The tests discussed so far were all at the same stress ratio, namely R = o.
According to Equation (5.7) the rates also depend upon R. This dependence can
be assessed by performing tests at different values of R. The results are plotted
versus ilK, to obtain data such as in Figure 5.4. Indeed, higher R produces
higher growth rates as should be anticipated on the basis of Equation (5.7). The
data in Figure 5.4 show that the effect of R is smaller than that of ilK, but that
is simply the way it comes out.
Data are always plotted on logarithmic scales of 10g(llK) and 10g(da/dN),

10 -.
LOWEST R-RAT 10 ~ o. 0
Gl INCREASING IN EgUAL INTERVAL
--'
~ UP TO 0.5
u 10 -s
"'"~
'"'"
z 10 -6
~
0
-0

10 -7

10 -8

10 -9

lOll f--------+------------j

10-12 1L - - - ' - - - ' - - - - ' - - . . . . l I O - -....l0--•

2 ....lO-6....l0---'IOO

OELT A K ( MPA RT M )

Figure 5.4. Crack growth rates for Ti-6AL-4V; Mill Anneal; Lab. Air; 70F (21C). Effect of R-Ratio.
130

because the rates vary over several orders of magnitude. A log scale for 11K is
not strictly necessary, but it has become standard practice to use a logarithmic
scale for 11K as well.
Crack growth properties of a number of structural alloys are compared in
Figure 5.5. Environment, loading frequency, and temperature may have a
significant effect on growth rates. Examples of some of these effects are shown
in Figure 5.6. For a discussion of crack growth at negative R see Chapter 7.

5.4. Rate equations

The form of Equation (5.7) follows from the test data; it cannot be obtained
from a theoretical model. Naturally, a functional form can be established by
fitting a curve through the test data. The resulting equations are sometimes
useful as they eliminate the necessity of using a graph.
From Figures 5.3 through 5.6 it appears that the rate data for one particular
R-ratio fall more or less on a straight line in a logarithmic plot. The equation
for a straight line is y = mx + h. In the present case y = log (da/dN) and
x = log (11K), so that:

log (:~) = mp log (11K) + log (Cp ). (5.8)

10 ~K (kslvin )
10' rTl---.---.-,.......,4-r........--r--;-"
a. A 537
b. WELD METAL
C. (3 Ti ALLOY ~~ (In/c)
d. AI. ALLOY
10- 4

_1
10

Figure 5.5. Typical rate properties of different alloys.

 131

AUSTENITIC STAINLESS
STEEL
a.RT
da b.600°C
~~ (In/c)
(I'm/c)d"N Al ALLOY
c, DRY AIR
d.HUMID AIR

-1
10

Figure 5.6. Typical effects of environment and temperature.

Taking the anti-log provides:

da = C (AK)mp (5.9)
dN P •

This equation is generally known as the Paris equation.

The parameters Cpand mp can be determined easily. For example, using the
two points A and B in Figure 5.7, yields:
da
dN
(in/cycle)

10- 4

IA
I
I
I
I
10_8L....4-...i.6-'-8~10--2~O-........
40--80
........- - - '

t.K (ksivfn)
Figure 5.7. Paris equation.
132

point A: log (1.6 x 10- 7 ) mp log 6.3 + log C p

(5.10)
point B: log (4 x 10- 5 ) mp log 40 + log Cpo
Taking the logarithms provides:
- 6.8 = 0.8mp + log C p }

-4.4 = 1.6m p + log Cp • (5.11)

2.4 = 0.8mp
This provides mp = 3; substitution of which in one of the equations leads to
log Cp = -9.2 or Cp = 6.3 X 10- 10 • The rate equation becomes:

: ; = 6.3 x 10- 10 dK 3 (5.12)

for this particular material. For most materials the value of mp is between 3 and
5. The value of Cp is more strongly material dependent; it has also largely
different values in different unit systems (see Chapter 7).
The Paris equation covers only one R-value. The lines for different R are often
parallel, i.e. have equal slope as e.g. in Figure 5.4. Thus all these lines would
have the same m-value, but different C; the latter depending upon R as C(R).
Hence, the following equation could cover all R-values:

:; = C(R)dK"R. (5.13)

For many materials, the dependence of Con R can be described in a simple

manner, as e.g.
da
(5.14)
dN
Where Cw is the value of C when R = O. Equation (5.14) can be used as is. Often
it is further modified by substituting K;,~x = (dK/(1 - R))"w, so that:

da C A vI'lR-n .. TflIw C A vI'lw vii ..

(5.15)
dN = w LlA Amax = w LlA Amax,

where mw = mR - nw, which is known as the Walker equation. Note that

Equation (5.15) essentially reverts back to the original Equation (5.4); of course
Equations (5.14) and (5.15) are equivalent, and both are in use.
One may argue that fracture occurs when the maximum stress intensity in a
cycle equals the toughness, i.e. if Kmax = Kc or K Ic • Since Kmax = dK/(l - R),
this would happen when dK = (l - R) Kc. At fracture the growth rate would
tend to infinity. A functional value can be made to go to infinity through
division by zero:
 133

da A.K"F
(5.16)
dN - CF (l - R)Kc - A.K"
At fracture, where A.K = (l - R) Kn the above equation indeed provides an
infinite growth rate. This equation is known as the Forman equation. It shows
the growth rate to depend upon R and should therefore apply for all R-values:
the equation 'pretends' to 'know' how dajdN depends upon R. A strong
objection against the equation is that in many cases fracture is not governed by
the toughness, because of collapse (Chapters 2 and 3).
In addition to Equations (5.9) through (5.16) many different curve fitting
equations can be developed to describe the test data. As a matter of fact, there
are probably as many equations as there are researchers in the field. Several
others are in common use. But none of these, nor the above equations, have any
physical significance; they are merely curve fitting equations. If they do fit the
data properly, there is no objection against their use. But no equation can fit all
data, so that religious adherence to one equation is not advisable. One should
use the equation providing the best fit in a particular case. An equation may be
used if convenient, but direct graphical use of the rate diagram is just as reliable.
Most crack growth analysis is done by computer, which can be supplied the rate
diagram in tabular form. It makes little difference to a computer whether it
interpolates in a table or uses an equation. It should be noted that the
parameters for the various equations are different, even if they cover the same
data set. For this reason the coefficients Cp , Cw , and CF and exponents m p , m w ,
and m F are used to indicate that they are specifically for a certain equation
(Paris, Forman, Walker). Use of the parameters of one equation for another-
even for the same material - may lead to dramatic errors. Conversion of
parameters to other unit systems requires great care (Chapter 7).

5.5. Constant amplitude crack growth in a structure

Most structures experience some form of variable amplitude loading in which

case the crack growth analysis is considerably more complicated than for
constant amplitude, as will appear later in this chapter. However, in the few
cases of constant amplitude loading the analysis can be readily performed with
or without the use of a computer. A crack in a structure will grow at the rates
indicated by the rate diagram because of the similitude discussed. Analysis of
structural crack growth can be carried out if the geometry factor is known
(Chapter 8) for the structural configuration at hand. The crack growth (curve)
in the structure follows from an integration of the rates:
da da
dN = f(A.K, R) or dN = f(A.K, R) (5.17)
134

Integration provides:

N = ("p da (5.18)
Jao J(AK, R)'
Generally, the integration is done numerically; it can seldom be performed in
closed form, because of the complexity of the functions J and P in AK, and of
the stress history. The function J might be as simple as the Paris equation:

N = ~ s.a p da (5.19)
Cp ao {p(a/w)AuFar p '
The P for a structural crack is usually a lengthy polynomial in a/W or known
only in tabular form, so that numerical integration is indicated even ifJis simple
and Au is constant (independent of a). Integration is performed most easily
through the use of a computer, but in the case of constant amplitude loading a
hand computation is very well possible. The principle for a simple numerical
integration in the case of constant amplitude loading is shown below.
If the loading is of constant amplitude, the integration can be done in small
steps with little error; the step size might be taken as e.g. a crack increment of
one percent of the current crack size. Assume for example (Figure 5.8) a case of
constant amplitude loading at e.g. Au = 20 ksi; further assume that a Paris
equation applies with da/dN=6.17E-10 !t..K3. Let f3 for the structural crack
be given as P = 1.12 + (a/W)2 (a hypothetical case) and let W = 4 inches
(Figure 5.8). The first two steps of a calculation starting at a crack size of 0.75
inch are shown below:
Initially:
a = 0.75 in; N = 0 cycles;
Aa = 0.01 x 0.75 = 0.0075 in (one percent increase);
AK = [1.12 + (0.75/4l] x 20 x .j0.75n = 35.5ksiFn;
da/dN = 6.17 x 10- 10 x 35.5 3 = 2.75 X 10- 5 in/cycle;
AN = = 0.0075/(2.75 x 10- 5) =
Aa/(da/dN) 273 cycles;
N = N + !t..N = 0 + 273 = 273 cycles;
a = a + Aa = 0.75 + 0.0075 = 0.7575 in;
Aa = 0.01 x 0.7575 = 0.007575 in (one percent);
AK = [1.12 + (0.7575/42 ] x 20 x .j0.7575n = 35.7ksiFn;
da/dN = 6.17 x 10- 10 x 35Y = 2.81 x 1O- 5 in/cycle;
AN = 0.007575/(2.81 x 10- 5 ) = 269 cycles;
 135

1.5

.5 a
w

(a) (b)

da 104~-----------------------------,
dN
(in/cycle) -5 ~=6.17X10-10AK3
10 dN
R=O

10-8L -____ ____________ ________

10 100 AK (ksi Yin)

(c)
Figure 5.B. Crack growth analysis; Example for constant amplitude. (a) Structure; (b) Hypothetical
p-curve; (c) Material's rate data.

N = 273 + 269 = 542 cycles;

a = 0.7575 + 0.0076 = 0.76510.
This process is continued until the crack size a is reached. It can be programmed
for execution by a computer:

pj = f(adw);
AKj = pjA(J~;
da
dN = f(AK j , R)

Aaj = ~aj (e.g. ~ = 0.01);

136

AM = Aa)(da/dN)i

~ = M + ANi;

if aj < a p then return to beginning.

With the steps on the order of one percent of the current crack size (or fixed step
sizes if so desired) good accuracy can be obtained. If large steps are taken an
integration rule such as Simpson or Runge-Kutta should be used. These inte-
gration rules were divised for numerical integration prior to the computer era,
when large steps had to be taken for hand calculations. In true integration where
in the limit da ~ 0 no such rule is required. The computer can indeed let the
step size go to zero, so that the integration rules are not necessary.
In an emergency constant amplitude analysis as above can be done by hand
in rather large steps as shown in Table 5.1. Even without the use of an integra-
tion rule, reasonable accuracy is obtained when the steps are not made too large,
especially at the beginning (Table 5.1). The results of Table 5.1 are compared in
Figure 5.9 with a computer analysis of the same problem, showing that the hand
calculation gives a reasonable approximation. The procedure illustrated in
Table 5.1 lends itself very well for execution with a spread sheet program, or a
dedicated program such as shown above can be used. Integration is a forgiving
procedure in contrast to differentiation. The latter problem is discussed in
Chapter 7.
Most structu,"es being SUbjected to variable amplitude loading (Chapter 6),
the above procedure for constant amplitude integration is seldom applicable.
Crack growth analysis for variable amplitude loading will be discussed later in
this chapter. Such analysis is much complicated by the problem of load inter-
action and retardation as discussed in the following section.

5.6. Load interaction: retardation

When one single high stress is interspersed in a constant amplitude history, the
crack growth immediately after the 'overload' is much slower than before the
overload. Figure 5.10 shows how three single overloads increase the crack
growth life by almost a factor of five (compare A and B). After a period of very
slow growth immediately following the overload, gradually the original growth
rates are resumed. This phenomenon is known as retardation. A negative load
following the overload reduces retardation but does not eliminate it (compare
Band C in Figure 5.10). Crack growth analysis for variable amplitude loading,
is not very well possible without an account of retardation effects. Before such
an account can be made, retardation must be explained.
Consider (Figure 5.1I) a crack subjected to constant amplitude loading at
 Table 5.1. Hand calculation. Case = Center cracked panel; W = 6 inch; da/dN = 4 x 10- 9 tJ.K 35 ; tJ.u = 12 ksi; R = o.
a tJ.a Average a in f3 = Jsec(nam/W) tJ.K = f3tJ.u Jna a" da/dN = 4 x 10 9 tJ.K 35 tJ.N = tJ.a/da/dN N=N+tJ.N
(in) (in) step am (ksiJIii) (in/cycle) (cycles) cycles
(in)
0.1 0
0.4 0.30 1.006 11.72 2.20 x 10- 5 18182
0.5 18182
0.5 0.75 1.040 19.16 1.23 x 10- 4 4065
22247
0.5 1.25 1.123 26.70 3.93 x 10- 4 1272
1.5 23519
0.5 1.75 1.282 36.07 1.13 x 10- 3 442
2 23961
0.8 2.40 1.799 59.28 6.42 x 10- 3 125
2.8 24085

Same calculation with finer steps during early growth

0.1 0
0.1 0.15 1.002 8.25 6.45 x 10- 6 15504
0.2 15504
0.2 0.30 1.006 11.72 2.20 x 10- 5 9091
0.4 24595
0.3 0.55 1.021 16.11 6.71 x 10- 5 4471
0.7 29066
0.3 0.85 1.053 20.65 1.60 x 10- 4 1875
30941
0.5 1.25 1.123 26.70 3.93 x 10- 4 1272
1.5 32213
0.5 1.75 1.282 36.07 1.13 x 10 3 442
2 32655
0.8 2.40 1.799 59.28 6.42 x 10- 3 125
...-
2.8 32780 w
-..l
138

4.5 COMPUTER
U1
w HAND <BOTTOM PART TABLE 5. 1)
:r:
Ll 4.0 HAND (TOP PART OF T"SLE 5. 1)
:z:
:::;
w 3.5
N
;J;
3.0
'"
I,
Ll
"'"
a::
Ll 2.5 I

I
J
2.0 J
J
1.5
I
I
I
/
1.0
/
/
/
0.5 .., ./
........-:

5 10 15 20 25 30 35 40 45
LIFE (1000 CYCLES)
Figure 5.9. Comparison of computer analysis and the hand analysis of Table 5.1.

A. JV\fV\MJ\. Sm=8.2, Sa=3.3

B.
f\
.MI\J\J\M Overload (0), Sma. =19.2

c. J
f\ f\ f\
V V V
nU
1\ f\ n fOverload, Sma.=19.2
v v ""'\..Cytle (0), Smln=-2.9

40r-----~~--------------------~~
a B
(mm)

20

10
B
6

4~~ __ ____
~ ______ ____-L____--J
~ ~

o 100 200 300 400 500

KILOCYCLES

Figure 5.10. Retardation after overload (Courtesy Aircraft Engineering).

R = 0; during the first cycle the load varies from A to C through B. Before the
loading starts, imagine a little (dashed) circle (Figure 5.11a) at the crack tip,
indicating the material that will undergo plastic deformation in the future plastic
zone (Chapter 3). The plate is then loaded to B. Imagine that one could now
 139

~,.-, FUTURE
• PLASTIC
~ ~ ZONE
(a) -- B

?i) PLASTIC
ZONE

(b)
time
PLASTIC ZONE

~
HOLE LEFT BY
PLASTIC ZONE REMOVED AND
REMOVAL SET ASIDE

(c) "

~ HOLE; SAME SIZE

AS CIRCLE
IN a.
• PLASTIC ZONE

(d)

+
(e) RESIDUAL
STRESS

Figure 5.11. Residual stress at crack tip (a) Load at A; (b) Load at B; (e) Load at B; plastic zone
removed; (d) Load at C; plastic zone still out; (e) Load at C; plastic zone back in.

remove the plastic zone and put it aside (Figure 5.llc). After unloading to C the
situation of Figure 5.11d is reached: all material is elastic - the plastic material
having been cut out - so that all strains and displacements are zero after
unloading. Hence, after unloading the hole at the crack tip in Figure 5.11d is
equal in size to the dashed circle in Figure 5.11a. The plastic zone has become
permanently deformed; it is larger than before the loading started. Thus it will
not fit in the hole of Figure 5.11 d. In order to make it fit it must be squeezed
back to its original size for which a stress at least equal to the yield strength is
needed (it must be deformed plastically again in compression to be squeezed
back to its original size; this requires stresses at least equal to the yield).
The plastic zone in fatigue loading is very small. Most of the fatigue crack
growth takes place at low values of ilK as can be appreciated from the data in
previous figures. At high ilK the rates are so high that little crack growth life
is left; most of the life is covered at low ilK. If for example, ilK = 10 ksi.Jln in
a material with a yield strength F,y = 50 ksi, the plastic zone size (Chapter 3)
would be rp = 102/(6n502) = 0.0021 inch. As the remainder of the plate is
elastic and returns to zero strain after unloading, this small plastic zone indeed
will be squeezed back to its original size and made to fit in its surroundings. In
order to squeeze the permanently deformed material back to its original size
140

reverse yielding is necessary: the compressive stress must be at least equal to the
yield strength. Whether the plastic zone is taken out as hypothesized above, or
whether it remains in place as it normally will, the end result will be the same.
It follows that after unloading, there is a compressive stress at least equal to
the yield strength at the crack tip. This is a residual stress (no external load)
which must be internally equilibrated by tension further away. The residual
stress distribution is as shown in Figure 5.11e. During subsequent cycling this
residual stress system will be present upon each unloading and it will have to be
added to (act together with) the applied stress. The crack growth response by
the material automatically accounts for this residual stress system; the data in
e.g. Figures 5.4-5.6 already reflect its effect, because the material 'knows' about
these residual stresses and shows growth rates in accordance with their presence.
If an overload occurs, a much larger plastic zone is formed (Figure 5.12).
After the overload a more extensive residual stress system is present than before
the overload. This more extensive system, acting against the applied stress,

>:+-+~--'-----.::+~VV\N
(a) l:/

(b)

Figure 5.12. Residual stresses before and after overload. (a) Before; (b) After; (c) Situation after
overload.
 141
causes subsequent growth to be slower (retarded). Once the crack has grown
through the overload residual stress field ( ~ overload plastic zone), the original
residual stress field is restored (Figure 5.12) and the 'normal' growth is resumed.
(In reality the compressive stresses do not extend throughout the plastic zone,
because the compressive yield zone is smaller. However, the principle is still
maintained).
Another way of looking at this problem is illustrative. It was shown in
Chapter 1 that crack growth occurs by plastic deformation (slip). As a conse-
quence, the crack tip plastic strain range is the best measure of crack growth.
The stress and strain being singular at the crack tip, it is somewhat hard to
envisage how they vary. However, the singularity occurs only in an infinitesim-
ally small material element, while it will effectively disappear due to crack tip
blunting. Therefore, crack growth phenomena can be discussed on the basis of
finite values of stress and strain, their values being bounded by the (cyclic)
stress-strain curve.
First, consider a crack with no previous history (no previous plastic zones)
with a very small plastic zone, cyclically loaded at R = 0 (Figure 5.13). At
maximum load, the stress and strain are defined by point A. Assuming an
extremely small plastic zone in a large plastic plate, the elastic deformations of
the plate will completely dominate the problem. Hence, all deformations will
come back to almost zero upon load release: the elastic plate will squeeze the
permanently deformed material in the (extremely small) plastic zone at zero
load. This implies that the crack tip strain will be reduced to almost zero (point
B in Figure 5.13).

fT

Figure 5.13. Stress-strain loop at crack tip. [7]. Copyright ASTM reprinted with permission.
142

Upon reloading, the stress and strain must conform to point A, because that
is the condition dictated by the surrounding elastic material. Obviously, the
crack tip is subjected to a plastic strain range /).ep with R, ~ 0 and Ra close to
- 1. The material experiences a stress-strain cycle BCDAEB.
The crack tip strain at R = 0 will return to almost zero, but not exactly so,
because there must be equilibrium of residual stresses, as illustrated in Figure
5.14. The compressive stresses in the plastic zone have to be equilibrated by
tension stresses around the plastic zone, which give cause to a small remaining
positive strain. Note however that the extent of residual tension stresses is
actually less than shown in Figure 5.12. This is due to the fact that upon return
from point A in Figure 5.13 yielding begins earlier, because AE = BD in Figure
5: 13 due to the Bauschinger effect.
The return of the crack tip strain to almost zero is a result of the action of the
large elastic plate, or rather of the remaining ligament. Note that the stiffness
of the plastic zone is very small compared to the stiffness of the ligament.
Now consider a real fatigue crack with a previous history (Figure 5.15); in its
wake along the crack edges is a strip of material representing the accumulation
of all previous crack tip plastic zones through which crack has progressed. This
material is no longer loaded, but at one time it underwent plastic deformation.
Closure stresses arise because the permanent elongation of the crack lips will

CT yy \~ Elastic
Elastic-plastic

It, r

It, r

Figure 5.14. Crack tip stresses at R = 0 loading. top: at max load. Bottom: at zero load. Compare
with Figure 5.13. [13]. Courtesy EMAS.
 143

U'yy \_.....-Elastic

Elastic-plastic

I, r

l,r

Figure 5.15. Crack tip stresses at R = 0 loading while accounting for previous plastic deformation
in wake of the crack, causing closure of crack tip at zero load. [13]. Courtesy EMAS.

close the crack before the load is zero. As such, they are similar to the com-
pressive stresses built up in the plastic zone. As a matter of fact, both stress
systems result from the same action of the surrounding elastic field. Therefore,
Figure 5.14 can be redrawn by including the crack lips as shown in Figure 5.15.
The closure stresses at the crack tip never exceed the yield. Thus, it requires
only an elastic strain to accommodate the remaining strain at the crack tip (in
case of complete closure), which means that the remaining crack tip strain is
elastic and the return point in Figure 5.13 is still B. Upon reapplication of the
load, the crack will remain closed until the closure stresses and the compressive
stresses are relaxed. But, the crack tip material is already straining. ·Its stress-
strain condition moves from B to C in Figure 5.13 while the crack is still closed.
Figure 5.16 shows the consequences of an overload. Depending upon the
minimum stress in the cycle and upon the relative stiffness of previously plastic
material with respect to the elastic material, the remaining strain after the
overload will be larger or smaller (point F in Figure 5.l6a). Therefore,
depending upon return to Fl or F 2 , the straining during the subsequent cycles
will be as in Figure 5.16b. In any case, the cyclic strain range (i.e. the opening
of the (J-e loop) is considerably reduced and the crack growth rate will decrease
accordingly (retardation).
According to some retardation models (see Section 5.7) the growth rate
reduction is proportional to the ratio between the overload plastic zone and the
144

tN!\ F

CT

,."
-----7
I
/" I
/ I
/ I
,IJ I I
I
'A~::~
1\ 1-1 ~
'\' 'I, I I I
I
I ~ n I I
I
I I
c

Figure 5.16. Consequences of overload for crack tip plastic strain loop. Top: during overload;
Bottom: after overload. [7]; Copyright ASTM. Reprinted with permission.

current plastic zone. Although the retardation models may be somewhat

artificial, they do contain the relevant parameters. With increasing crack size
(and plastic zone size) it becomes increasingly difficult for the elastic material to
restore the zero strain field after unloading (the stiffness ratio between ligament
and plastic zone decreases). Hence, at small crack sizes the return point may be
F, (Figure 5.16), but at large crack sizes, it shifts to F 2 , so that retardation
becomes more pronounced (see also Figure 5.10). This may explain why various
investigators find different retardation effects in the same material: retardation
is crack size dependent and panel size (ligament) dependent.
The above discussion explains why a compressive stress following the
overload will reduce the retardation, but cannot completely annihilate it
 145

(compare C and B in Figure 5.10); during compression the closed crack is no

stress raiser and therefore all compressive stresses and strains are elastic (elastic
strains are negligible as compared to plastic strains).

5.7. Retardation models

The complexities of the retardation phenomenon so far have precluded the
development of an all-encompassing mathematical-physical treatment of the
problem in a retardation model. More than half a dozen models have been
proposed [1-7], none of which covers all aspects of the problem.
A number of models are based on crack closure. As was explained in the
previous section, crack tip closure occurs even in constant amplitude loading.
Some closure models consider only that part of the cycle effective over which the
crack is open. However, even during that part of the cycle the crack tip material
is straining (B-C in Figure 5.13) and this straining is just as much part of the
crack-tip strain loop as is the part associated with the open crack. Hence,
considering only that part of the cycle during which the crack is open, is
physically incorrect. Certainly, closure changes after an overload and this affects
the general residual stress field at the crack tip, but this residual stress field is
changed more due to the larger plastic zone. In view of this, pure closure models
do not cover all aspects of the problem. There is no doubt that overloads affect
closure; hence there will be a correlation between closure and retardation. But
retardation and changes in closure are both consequences of the overload.
Closure changes are not the cause of retardation. Both are symptoms, and one
symptom cannot explain another; both are caused by the 'disease'. A proper
treatment of the model must consider the effect of overloads upon the total
residual stress field in the wake of the crack and ahead of the crack, and in
particular upon the plastic strain range.
Other retardation models attempt to account for the residual stress field
directly by superposing it on the stress field due to the external load. The
residual stresses themselves cause a certain stress intensity which can be added
to the stress intensity due to the applied loads. Such models have several
draw-backs and short-comings. In the first place, although the residual stress
field can be assessed qualitatively in an easy manner, the quantitative evaluation
(and thus evaluation of the resulting K) is difficult. A second problem is, that
a residual stress field already exists even in constant amplitude loading. Also this
field should be accounted for if the model is to be based on residual stresses.
(Normally, it will be automatically accounted for in the data base).
Despite the mathematical complexity of some, all models are two-dimension-
al over-simplifications of a complicated three-dimensional problem, and full of
assumptions. For example, all models must consider plastic deformation, but
even plastic deformation is not easily treated analytically, further the yield
146

strength, F;y, is arbitrarily defined (e.g. 0.2% plastic strain), while it is an

essential number in the calculation of the plastic zone. Plastic zone size
equations by themselves are subject to doubt, especially in the case of a changing
state of stress. Some models use empirical equations for changes of the closure
stress.
All of this leads to the conclusion that any model must contain one or several
unknown parameters which must be obtained from tests (i.e. adjusted empiric-
ally). In one model such a parameter was later included to make empirical
adjustment possible in the first place. There is no practical objection against the
use of empirical parameters (E, v and F;y follow from experiments and so does
after all da/dN), provided the use of these empirical parameters leads to useful
crack growth predictions for structural applications.
If all models are simplifications, none can be preferred over another.
Moreover, if they all contain adjustable parameters, they can all be made to
work if the parameters are adjusted appropriately. All claims that one model is
better than another are improper. Each model can be made to work if empiric-
ally adjusted; ifit does not work, it was not adjusted properly. Generality of the
adjustment may be a problem. In that respect some models may be somewhat
better than others. Clearly, the adjustment parameters will be material
dependent. But should they also depend strongly upon the stress history, as they
do (see Chapters 6, 7) then they cannot be used generally. Attempts to make
general use of these parameters then lead to false claims with regard to a model's
inadequacy.
There are so many retardation models that a discussion of all would be
beyond the scope of this book. Review of just a few might suggest that these
would be better than others. The interested reader is therefore referred to the
relevant literature. The general reader probably has no use for such a review at
all, because crack growth analysis with retardation requires the use of a
computer anyway and the general user will employ existing software. Such
software should have the option for the use of several retardation models. If just
one model is available generality may be hampered, because one model may be
somewhat more appropriate for certain applications and vice versa. Which
models are included in the software is rather secondary, because all can be made
to work if properly calibrated.
More important than which retardation model is used, may be the following:
(a) As demonstrated in the previous section retardation will depend upon the
relative stiffness of plastic zone and elastic material. This may become important
for larger cracks in small components. No known retardation model considers
this problem but some account could be made in the computer code.
(b) All models must consider crack tip plasticity and therefore will use
arbitrary numbers. As was explained in Chapters 2 and 3, the state of stress has
a great influence on plasticity. The plastic zone is rp = K!ax/rx.nF;; where rx. is
commonly considered to be 2 for plane stress and 6 for plane strain. Further,
 147

the condition for plane strain is B > 2.5K~ax/Ft~, in which 2.5 is a somewhat
arbitrary number (Chapter 3). 'Normal' stress cycles at low Kmax may give plane
strain, but an overload may cause plane stress due to its larger Kmax. This also
causes a more extensive residual stress field and more retardation. Thus, in
accounting for retardation the computer code should assess the state of stress
in each cycle. If it does not, even the 'so-called' sophisticated retardation model
may give large errors. For example, the larger retardation at longer cracks in
Figure 5.10 may be caused by a change from plane strain to plane stress during
the overloads. But even if the state of stress is evaluated, the value of a will be
arbitrary, the factor of 2.5 is arbitrary and F,y is more or less arbitrary. Some
models use different plastic zone formulations than above, but these still contain
arbitrary numbers.
In the following crack growth analysis for variable amplitude loading will be
illustrated on the basis of the Wheeler model. This model is used here not
because it is believed to be better than any other, but because it is very simple,
so that it can be used easily for illustrations. It is worth mentioning however that
if all models are simplifications anyway, the simplest certainly is the most
appealing. If all models must be calibrated for general use, even the simplest
model can be made to work by calibration.
Wheeler introduces a retardation parameter ¢ R' It is based on the ratio of the
current plastic zone size and the size of the plastic enclave formed by an
overload (Figure 5.17a). An overload occurring at a crack, size ao will cause a
crack tip plastic zone of size
K~ p2(JJao
rPO = --2 = --2-' (5.20)
anF,y aF,y
where (Jo is the overload stress. When the crack has propagated further to a
length ai the current plastic zone size will be
p2(Jimax ai
(5.21)
aF,~
where (Jimax is the maximum stress in the i-th cycle. This plastic zone is still
embedded in the plastic enclave of the overload: the latter proceeds over a
distance g in front of the current crack (Figure 5.17b). Wheeler assumes that the
retardation factor ¢R will be a power function of rpc/g. Since
g = ao + rPO - ai' the assumption amounts to:
da
dN
with (5.22)
148

Figure 5.17. The model of Wheeler. (a) Situation immediately after overload; (b) After some crack
growth; (c) Situation after second overload.

If rpc = {!, the crack has grown through the overload plastic zone, and the
retardation factor becomes <p R = I by definition. The exponent in Equation
(5.22) has to be determined empirically. This is the adjustable calibration factor.
°
Note that if y = there will be no retardation at all under any circumstances.

°
Hence, the minimum value ofy is zero. Typically for variable amplitude loading
< y < 2, depending upon the material but also upon the spectrum.
For the case of a single overload in a constant amplitude test the retardation
factor gradually decreases to unity while the crack progresses through the plastic
enclave (Figure 15.17b). If a second high load occurs, producing a plastic zone
extending beyond the border of the existing plastic enclave, the boundary of this
new plastic zone will have to be used in the equation (Figure 5.17c), and the
instantaneous crack length will then become the new ao.
Calibration of the above model (and all other models) proceeds as follows. A
test is performed under variable amplitude loading. The test result is then
're-predicted' several times using the proper dajdN - 11K data and the proper
/3, but with different values of the adjustable parameters; (in the case of Wheeler
y-values taken are e.g. 0,0.5, 1, 1.5 etc.). The parameter value(s) that produce(s)
the best coverage of the test data, is (are) the values to be used in analysis. An
 149

example of such a calibration [8] is shown in Figure 5.18. Clearly, in this case
}' = 1.4 is the parameter value to be used.
Unfortunately, the parameter calibration is not general. It depends upon the
load-history and spectrum (Chapters 6 and 7). A different spectrum with a
different mixture of high and low loads requires a different calibration factor.
E.g. the non-linear man-induced exceedance diagram (Chapter 6) requires
different calibration parameter(s) than a nature-induced log-linear exceedance
diagram; the calibration parameters are suitable for one type of spectrum, but
they cannot be generalized for all spectra. Failure to perform this re-calibration
and subsequent general use of calibration factors, is the main cause of claims
that one retardation model is better than another. If proper calibration is
performed for each spectrum type, any model can be as good as any other.
Calibration for a certain spectrum type and material generally gives good results
[8] for all variations of the same type of spectrum (Chapter 6), as shown in
Figure 5.19. This figure shows results of about 70 predictions for random
loading with the spectrum and calibration as used in Figure 5.18. More informa-
tion on model calibration can be found in Chapter 7.

5.8. Crack growth analysis for variable amplitude loading

Most structures are subjected to variable amplitude loading, i.e. of the type
shown in Figure 1.2 (For a detailed discussion of load histories see Chapter 6).
In such cases the crack growth rate dajdN varies from cycle to cycle, depending
upon AK and R of the cycle involved, and upon retardation (any cycle can be

28
lin)
O. TEST DATA

2.5

OL-------l0~O-O------2~OO-O------3~OO-O------4~O~OO------5-0~OO-----F-LlG~HTS

Figure 5.18. Calibration of Wheeler model for flight-by-flight aircraft simulation loading [8).
150
25~----------------------,

20

15

10

OL-L-LL~UL~~~~~~~~
0.4 0.5 0.6 0.7 O.S 0.9 1.0

NpREDICTED
N TEST

Figure 5.19. Accuracy of 70 predictions as compared to tests [8], Courtesy Engineering Fracture
Mechanics.
an overload cycle with respect to the subsequent cycle). Therefore, the computer
program must evaluate and add crack growth on a cycle-by-cycle basis. For
example if crack growth takes 200000 cycles, the computer must perform the
'operation' 200000 times. This may take considerable computer time. Typically,
a mainframe computer can perform at about 50000 cycles a minute, so that the
above computation would take about four minutes. A personal computer might
take as much as two to three hours for the same job (1987).
A logic diagram for the integration is shown in Figure 5.20. This is again
based on the Wheeler model because of the latter's simplicity, but it is not
essentially different for other retardation models. Naturally, in order to perform
the calculation for a structural crack, the computer must be provided with
applicable da/dN data for the material at hand (Chapter 7), and last but not
least, the stress history for the structure (Chapter 6).
A more detailed representation of the algorithm involved is shown in Table
 151

°i+I-Oj
Ni+I-N j
"'i+I-O'j

Figure 5.20. Logic diagram for crack-growth computation.

5.2. A hand-calculation for a few successive cycles is shown in Table 5.3. These
tables show the basic algorithm which is quite simple; a useful computer code
with many options, especially for the complicated book keeping for stress
histories, is rather involved and will contain approximately 3000 statements
(lines ).
The accuracy of the computation depends somewhat on the retardation
model, but useful results can be obtained with any well-calibrated model. Most
influential to the accuracy are the input of the stress sequence (Chapter 6) and
material data (Chapter 7). A general discussion of accuracy and errors is
presented in Chapter 12. The accuracy problem involved in the simple algorithm
of Table 5.2 is in the addition of a very small da to a relatively large a. For
example, 8-bit personal computers evaluate a number into 8 digits. If a = I
inch and da in a given cycle is 0.000001 inch, the new crack size will become
152

Table 5.2. Crack growth analysis in variable amplitude

Subroutine or preprocessor; fJ-Library or methods of Chapter 8

Subroutine; generation of stress history (Chapter 6), random-

ization, counting if necessary (Chapter 6).

Kmax = I1K;/(1 - R;)

rp; = K~",/(1.nF,~, +--- Subroutine state of stress for (1..

if Xe new > X eo1d then

(! = Xe - a j

Subroutine various retardation models. Only Wheeler shown

l/!R; = (!/rp;
da
-
dN = {(11K R)
' , Subroutine or data library

da da
dN = l/!Rj dN (retarded)

I1N = 1
l1a = I x da/dN

N=N+I
if ai < ap then return to line 2
a, N
a (years, voyages, flights) Subroutine conversion
Output Subroutines
Plots Subroutines
End

a = 1.0000001. However, if da = 0.000,0001, the computer evaluates

1 + 0.000,00001 = 1, i.e. the crack has not grown due to a computer rounding
error. This means that a and da must be evaluated in double precision which in
an 8-bit personal computer provides 16 digits. Hence a = landda = IE - 15
Tuble 5.3. Example of retarded crack growth by hand calculation: on the basis of Table 5.2.
Note a: fi assumed equal to 1 throughout; b: (X = 2 (plane strain assumed); c: dujdN = 2£-9 flK2 Km " assumed: y assumed as 1.5; F,y = 50 ksi;
d: Stress history assumed as in columns 3 and 4; e: Table must be worked horizontally.

2 3 4 5 6 7 8 9 10 II 12 \3 14 15 16
Cycle a fl(T R flK KmJx r ,Ji Xe new X vo1d Xe Q 1 (dajdN) Retarded N fla a = a + flu
= 1 dajdN new a for
next line
0.1 10 0 5.61 5.61 0.0020 0.10020 - 0.10020 0.0020 3.53£-7 1 3.53E-7 0.100000353
2 0.1 ' 10 0 5.61 5.61 0.0020 0.10020 0.10020 0.10020 0.0020 3.53£-7 2 3.53 £-7 0.100000706
3 0.1 + 15 0.1 8.42 9.35 0.0056 0.10056 0.10020 0.10056 0.0056 1.32£-6 3 1.32 £-6 0.100002026
4 0.1 + 10 0 5.61 5.61 0.0020 0.10020 0.10056 0.10056 0.0056 0.21 7.41 £-8 4 7.41 £-8 0.100002100
5 0.1+ 12 0 6.73 6.73 0.0029 0.10029 0.10056 0.10056 0.0056 0.37 2.25E-7 5 2.25£-7 0.10002325
6 0.1 + 10 0 5.61 5.61 0.0020 0.10020 0.10056 0.10056 0.0056 0.21 7.41 £-8 6 7.41 £-8 0.100002399
7 0.1 + 19 0 10.66 10.66 0.0072 0.10072 0.10056 0.10072 0.0072 2.42.£-6 7 2.42£-6 0.100004822
8 0.1 + 15 0.1 8.42 9.35 0.0056 0.10056 0.10072 0.10072 0.0072 0.69 9.11 £-7 8 9.11 £-7 0.100005733
9 0.1 j 10 0 5.61 5.61 0.0020 0.10020 0.10072 0.10072 0.0072 0.15 5.17 £-8 9 5.17 £-8 0.100005785
10 0.1 + 12 0 6.73 6.73 0.0029 0.10029 0.10072 0.10072 0.0072 0.26 1.59 £-7 10 1.59 £-7 0.100005944

Vl
W
154
will be evaluated correctly, but if da = IE - 16 and a = 1, the computer will
still not recognize growth. There is nothing that can be done about this rounding
problem. Usually it is not serious, but it may become a problem in evaluating
retardation. Computer programs working in single precision may be one cause
of claims regarding the accuracy of retardation models.
It should be noted, that the above does not change when a is evaluated in
meters. For example if a = 0.01 m and da = l-EI6m, the addition will be
performed properly in double precision because leading zeros do not count.
Mainframe computers already carry 16 decimals in single precision; they carry
32 in double precision. Even in that case rounding errors may occur, but they
are even less significant.
Although the algorithm in Table 5.2 is simple, computer codes are generally
rather complicated [e.g. 9], because there must be
(a) preprocessors for p, or a p library.
(b) options for various rate equations, da/dN table and/or library of data;
(c) options for various retardation models;
(d) accounting for state of stress;
(e) options for random loading;
(f) accounting procedures for stress history;
(g) options for cycle counting.
If all of the above are included, the main code of 3000 statements as
mentioned above can easily triple or quadruple in size. Of the above (e), (f) and
(g) may be the most important and most involved; they are discussed separately
in Chapter 6. Further discussions of the subject are found in Chapter 7 (data and
calibration) and Chapter 12 (errors and accuracy).
It is well-known that fatigue predictions, in general, have a low accuracy. In
the case of crack propagation, a linear integration (without interaction effects
or retardation) will generally yield results which are on the safe side. As shown
in Figure 5.10 negative loads reduce the retardation caused by positive loads,
but the net effect is usually a deceleration of crack growth, so that retardation
models must be used.
Figure 5.21 shows results of crack growth in rail steel [10] under simulated
train-by-train (Chapter 6) loading. Retardation hardly plays a role in rail steels,
therefore predictions were made by means of linear integration. The figure
shows that they are within a factor 2 of the experimental data.
Better accuracy can be obtained in general, provided the retardation model
is adjusted. Predicted crack growth for a titanium alloy sUbjected to aircraft
service loading [8] is shown in Figure 5.22 together with experimental data.
Generally, part of the discrepancy between computation and test may be caused
by scatter in crack growth properties. Most retardation models can be empiric-
ally adjusted. In this respect, the Wheeler model is attractive, because it contains
only one adjustable constant.
 155
42.5~---------------------------,

•
40.0 •
•
37.5 •
•
E
E 35.0 •
en
~ •
'" e 32.5 •
<.)
•
30.0 •
•

25.0L-_-L_ _...L.._ ___'I..-_....L._ _.l-_---':~-....L---I...--....L.---J

o 2 4 6 20
Million Gross Tons
Figure 5.21. Predictions and test data for service simulation loading in rail steel [10].

2.5

v •• LL=55 ksi ...

Test doto Ti-6AI-4V, Spectrum 8

0 LL =60 ksi 60 ksi

• 65 ksi 55 ksi
LL=65 ksi
•
0
2.0 " ... LL=70 ksi •
- - Wheeler, "j'=1.6

~
. v
.~ 1.5
C
N

.g
V

(f)

'"eu 1.0 v
v
<.)

vv
v
0.5

oL-_ _ _ _ _L-________L-__________I..-________ ___'~ ___________'

o 1000· 2000 3000 4000 5000

Flights

Figure 5.22. Predicted crack growth and test data for aircraft spectrum [8]. Courtesy Engineering
Fracture Mech.

Apparently, a crack growth prediction can be substantially more accurate

than a fatigue life prediction. Admittedly, a few experiments are necessary with
a spectrum of certain shape to empirically adjust the retardation model
parameters. From then on, predictions can be made for the same general
spectrum shape and variations thereof, for structural parts subjected to lower
156

and higher stresses and for cracks of different types (different fJ). Crack growth
properties of most materials show considerable scatter. The rail steels that are
the subject of Figure 5.21 showed variations of almost a factor often in constant
amplitude crack growth (see Chapter 14). Therefore, discrepancies between
predicted and experimental crack growth are not a shortcoming of the predictive
method per se, but are due to anomalies in material behaviour.
Fortunately, most materials are well-behaved in comparison, by showing less
scatter in crack growth. Nevertheless, there is enough scatter that predictions
will always have some uncertainty. This would still be the case if better retarda-
tion models were developed.
However, the prediction procedure in general contains many more uncertain-
ties, which may be just as detrimental to the final results as are the shortcomings
of the retardation model. These are:
(a) Uncertainty in the local stress level.
(b) Uncertainty in the stress intensity calculation.
(c) Insufficient knowledge of the load spectrum.
(d) Possible environmental effects.
Consider first the uncertainty in stress level and stress intensity. In the case of
a complex structure consisting of many elements, an error of five percent in the
stress analysis would be quite normal. The subsequent determination of the
stress intensity can easily add another five percent, especially in the case of
corner cracks or surface flaws. Thus, the final inaccuracy of the stress intensity
may be in the order of 10%. If the crack growth rate is roughly proportional to
the fourth power of 11K, the error in the crack growth prediction will be on the
order of (1.1)4 = 1.45 (45%); see also Chapter 12.
Despite extensive load measurements, the prediction of the load spectrum is
still an uncertain projection in the future. Slight misjudgements of the spectrum
can have a large effect on crack growth.
Even if possible environmental effects are disregarded, the errors in crack
growth prediction due to uncertainties in stress analysis and loads analysis can
be just as large or larger than the errors due to the crack growth integration.
Development of better crack growth integration techniques will not improve
this situation. Therefore, the shortcomings of the retardation models can hardly
be used as an argument against crack growth predictions.
Taking into account all errors that can enter throughout the analysis, it is
obvious that a safety factor should be used. This safety factor should not be
taken on loads or stresses or dajdN data. Doing this would make some
predictions more conservative than others. The complexity of crack growth
behaviour does not permit an easy assessment of the degree of conservatism
attained through the application of such safety factors. A safety factor should
rather be applied to the final result, i.e. to the crack growth curve, by dividing
 157

the number of cycles to any given crack size by a constant factor. The problem
of accuracy and sources of error is discussed further in Chapter 12.

5.9. Parameters affecting fatigue crack growth rates

When predictions of crack propagation have to be made, data should be

available relevant to the conditions prevailing in service. Such data may be hard
to find (for pragmatic solutions see Chapter 7). Fatigue crack propagation is
affected by an endless number of parameters, and the circumstances during the
test will seldom be the same as in service. The influence of the environment is
the most conspicuous.
The effect of environment on crack growth rates has been the subject of many
investigations on a variety of materials; the rate of fatigue crack propagation in
wet air may be an order of magnitude higher than in vacuum, the effect being
attributed to water vapour. The influence of salt water (seawater) is of particular
interest to marine structures. An example of its effect will be shown in Chapter
7. It is generally accounted for in the crack growth analysis by submitting the
computer program with the actual data in tabular form. It is then assumed that
during each cycle in a variable amplitude sequence the rate will immediately
adjust to the one found in the constant amplitude test data at the same 11K. This
is assuming that chemical/load equilibrium will be immediately attained. More
elaborate accounting can be implemented however. No single model can explain
the influence of the environment on the rate of propagation of fatigue cracks.
Different explanations apply to different materials. The effect is certainly a result
of corrosive action and as such it is time-dependent. Therefore, the environmental
effect is dependent upon the cycling frequency.
Among the many factors that affect crack propagation, the following should
be taken into consideration for crack growth predictions:
(a) thickness;
(b) type of product;
(c) heat treatment;
(d) cold deformation;
(e) temperature;
(f) manufacturer;
(g) batch-to-batch variation;
(h) environment and frequency.
For the factors lower in this list it is less likely that they can be properly
accounted for. No attempt will be made to illustrate the effects of all these factors
with data, because some have greatly different effects on different materials.
Many of these effects cannot be accounted for properly in the analysis of
structural cracks, primarily because the data are simply not available. A
pragmatic approach to solve the problem is discussed in Chapter 7. At this point
158
It IS sufficient to note that the necessary use of estimated data may be of
considerable influence on the accuracy of the analysis. With this in mind, the
acclaimed inaccuracies of e.g. retardation models may well become secondary
(see also Chapter 12).
In sheets there is a systematic, effect of thickness on crack propagation,
especially before the fracture mode transition. Fatigue cracks in sheets start
perpendicular to the sheet surface. When the crack grows the size of the plastic
zone increases and plane stress develops. This causes the fatigue crack to change
to single or double shear, as depicted in Figure 5.23. Plane stress develops when
the size of the plastic zone is in the order of the sheet thickness (Chapter 3). In
thi"ker sheets the transition will require a large plastic zone and occur at a
greater length of crack. The data suggest that crack growth is slower in plane
stress than in plane strain at the same stress intensity.
Although the effect of thickness on crack growth has been recognized for over
20 years, little effort has been expended in developing a useful model for
everyday damage tolerance analyses. Figure 5.24 emphasizes the necessity to
include this effect in the crack growth analysis. A tentative semi-empirical model
has been proposed [11], but the best way to account for thickness effects is
probably to submit the proper data to the computer code. A factor of two error
(due to thickness) in da/dN data may overshadow any effects of 'inaccurate'
retardation models.
Many investigators hold that there is a threshold for fatigue cracking: below
a certain 11K the rate da/dN is supposed to be essentially zero. At least one
conference, resulting in a two-volume book [12] was devoted to this subject. The
threshold would be reflected in a vertical da/dN - 11K curve at low 11K, as
shown in Figure 5.25. The threshold is usually determined by gradually
decreasing the stress in a test until crack growth comes to a halt. In view of
possible retardation this procedure is subject to some doubt. Besides, the
threshold is definitely crack size dependent; it is not unique. However, if one
accepts the presence of a threshold, the practical question is "what is the effect
on predicted crack growth". Figure 5.26 shows what it may amount to: the two
curves are indistinguishable. Generally speaking, the effect is hardly worthwhile
considering, but of course, in each case one would have to use judgement. If for
example the initial 11K is below threshold no growth occurs at all. In this respect
Figure 5.26 is somewhat deceiving; one could select a case where the effect is
larger. However in random loading many cycles will be above threshold and the
effect on life (in years or hours) is small, especially when there is retardation.
Most computer codes provide an option to use a threshold. Referring back to
Section 5.8 it should be noted that the computer uses a threshold automatically
because it rounds any da smaller than a certain value, depending upon the
current crack size.
There is often much concern regarding the anomalous behaviour of very small
 159

final
failure

tensile mode

(b)

Figure 5.23. The transition of a fatigue crack in sheet. (a) Transition of fatigue cracks to double
shear (top) and single shear (center and bottom) in AI-alloy specimens; (b) Single shear (A) and
double shear (B).

cracks [13]. As shown in Figure 5.27 small cracks tend to show growth rates
much higher than would be expected on the basis of the acting 11K. Various
explanations have been put forward but the 'cure' proposed is mainly artificial
use of an apparent crack size a + ) .where A is a fixed quantity determined
empirically.
Most of the short crack data stem from strain control fatigue tests at R = - I
160

i
30.----------------------
crock size (mm)

sheet
thickness 0.6 mm
20

10

sm =8 kg/mm2

sa = 6.5 kg/ mm 2

10 20 30 40
___ N (103 cycleS)
Figure 5.24. Effect of sheet thickness on crack growth.

E
dN

(1-R)K 1C !;K

Figure 5.25. Threshold in rates (schematic).

on small notched coupons (usually with central holes). Consider a short crack
at such a hole as in Figure 5.28a and compare it with a long crack at the same
/).K (Figure 5.28b). By the nature of the test, the plastic zone at the hole is much
larger than the crack tip plastic zone. Since completely reversed plastic strain is
enforced in most of these fatigue tests, the crack tip will also be subjected to
completely reversed strain (R, = -1). A larger crack at R = -1 will close
during compressive loading, so that its strain range will still be as if R, = O. The
small crack at the hole is experiencing a strain range twice as large as a regular
crack at Ra = O. Hence, the small crack should show a rate of growth as if its
/).Kwere approximately twice as large as the calculated value. This is indeed the
 161

4.5
U;
W
I
u 4.0
z
SOLID LINE: PARIS: COEFnCIENT 2£-9: EXPONENT 3.5

w 3.5 CASH-DOT SAME WITH THRESHOLO OF 5 KSI RT IN

N
U1
CENTER CR .... CK: W - 6 INCH; DELTA-SIGM.-\ - 12 KSJ; R 0
3.0 m

""
u
-<
rr
u 2.5

2.0

1.5

1.0

0.5

10 20 30 40 50 60 70 80 90
LIFE (1000 CYCLES)
Figure 5.26. Effect of using threshold in computation; threshold is 5 ksi; Curves indistinguishable.

case as can be judged from Figure 5.27. When the small crack grows, it will
gradually move away from completely reversed straining so that the growth rate
decreases. Once the crack tip is outside the plastic zone of the hole, the small
crack behaves as a normal crack.
The small crack behaviour depends upon the loading and upon the ratio
between ligament and hole diameter (notch depth). A hole, unlike a crack, does
not close in tension and it is equally as much a stress raiser in compression as
in tension, so that even under load control reverse plastic strains can occur.
Therefore, similar, but smaller, effects should be anticipated under general
loading conditions. However, if the hole is filled with a fastener, it will essentially
'close' and little or no small crack effects should be anticipated. In technical
problems many cracks start from holes, but these are often "filled" fastener holes,
so that the 'small crack problem' may be of little relevance. In most other cases
these so-called 'small cracks' are well below the detection limit in practical
inspections. They are then irrelevant because crack growth below the detectable
crack size is hardly of interest. Naturally, one can always give examples of cases
where the problem might appear, but generally speaking, it is an interesting
research subject, but its technical relevance is small.
As a final note in this section consider the use of J (Chapter 4) as opposed to
K for representing da/dN data. By far the larger part of fatigue crack growth
lives is spent at low AK. Let e.g. AK = IOksiy'iil and R = O. Taking a (low)
yield strength of F;sy = 40 ksi, the plastic size would be r = 10/(6 x
n x 402 ) = 0.003 in. If this is not a small enough plastic zone, none will ever be.
162

Kmolil ' ksi../i"-

10

Growth Rate versus Kmaa in

SAE 1015 St~1

Stress and Edge Strain

Control: R •• R".=-I

Rc=-I -K, R.. =-I

0 2_5
0 4.4 <>
-I A 6_2 V
10

Note: Solid symbols denote cracks

~ 01 length less than 250l'm
~

~ -2
10
Z
'I!
..,
0

!F
~
: 10-3
~
'"~
u

-<
10

-5
10

10 '00
K""",. MPa,/iii

Figure 5.27. Crack growth rates of small cracks [13] (Courtesy EMAS).

Naturally, at large crack sizes the plastic zone will be larger, but does this justify
the use of EPFM. Consider Figure 5.29. The crack growth life might be either
N], N2 or N 3 , but practically this makes little difference on the total. The
difference is only caused at the high AK. During most of the crack growth life,
the stress intensity is well below Klc (fortunately). Thus, the use of K for the
analysis is well justified. Finally, the champions of using J for representing
da/dN data have always used approximations of J rather than actual J values
obtained properly (Chapter 4). By using 'appropriate' approximations, one can
always make data look better. However, should the data at high AK indeed be
better represented by J, the differences in analysis results (Figure 5.29) would
still not justify the use of a more complicated parameter; a parameter moreover
that has more drawbacks than advantages (see Chapter 4). In short, the use of
 163

Notch zone
~crock lip zone ---e
Crock lip
Plastic zones Plaslic zone

E E

Figure 5.28. Stress-strain loop for small crack at notch. (left) as compared to regular (large crack;
right); both R = 0 loading. [7, 13]. Copyright ASTM. Reprinted with permission.

a
I J
II
I I
/,
1/
/1.
/-/.
.;<"

Figure 5.29. Effect on life of using different parameter in high rates regime.

J for fatigue crack growth analysis is another interesting research subject, but
not of practical interest.

5.10. Stress corrosion cracking

Crack growth can occur by other mechanisms than fatigue (Chapter 1). The
most important one is stress corrosion cracking. Given a specific material-
environment interaction the stress corrosion cracking rate is governed by the
stress intensity factor. Specimens with the same initial crack but loaded at
164

different levels (different initial K-values) show different times to failure as

shown diagrammatically in Figure 5.30. The specimen initially loaded to K lc fails
immediately. Specimens subjected to K values below a certain threshold level
never fail. This threshold level is denoted as K lsce ' The subscript see standing for
stress corrosion cracking.
During the stress corrosion cracking test the load is kept constant. Since the
crack extends, the stress intensity gradually increases. As a result the crack
growth rate per unit of time, da/dt, increases according to:
da
dt = f(K). (5.23)

This is shown in Figure 5.31. When the crack has grown to such a size that K
becomes equal to Klr> final failure occurs, as indicated in Figure 5.30b.
Obviously, if failure does not occur (infinite time to failure) the crack did not
grow at all, because if it did, K would increase causing more growth, etc. Thus,
the K giving an 'infinite' life is indeed a threshold.
The stress corrosion cracking threshold, Kiser> and the rate of crack growth
depend upon the material and the environment. From Figure 5.30 it follows that
a component with a certain size of crack loaded such that K = Klr> fails
immediately. Components loaded to K values at or above K lscc will show crack
growth to failure. No failure (no crack growth) occurs when K < K lsee '
Fracture mechanics can deal with stress corrosion crack growth in the same
manner as with fatigue crack growth. Obtaining the crack growth curve as a
function of time amounts to integration of da/dt in much the same manner as
integration of da/dN (see previous sections). However, the total times to failure
in stress corrosion are usually from 1 to 1000 hours or they are infinite (Figure
5.30). If the crack growth life is less than 1000 hours there is hardly time to

K -------::=-'"=-"!:"-=---- ~--~--~-~----~-~-~--+--
K
Iscc Isec

(a) _ time to failure (b) - time to failure

Figure 5.30. Stress corrosion test data. (a) Stress corrosion time to failure, upon loading to initial
K-level; (b) Cracking during test.
 165

~
dt
(in/hr) 1

0.1

d.
dt

0.01 L--------:--"7"'7
time (b) K (kSivln)
(a)
Figure 5.31. Stress Corrosion Crack growth. (a) Crack growth as a function of time; (b) Growth
rate as function of K.

inspect or deal with cracks otherwise. Thus crack growth analysis is seldom
worthwhile.
Stress corrosion cracking should be prevented rather than 'controlled'. Crack
growth is prevented by keeping K < K Iscc ' Mathematically, this problem is
identical to keeping K < Klc or Kc (i.e. the fracture problem as discussed in
Chapter 3). Thus prevention of stress corrosion cracking can be dealt with in
damage tolerance analysis in the same manner as fracture.
The discussions in Chapters 3 and 10 apply if K/c is replaced by KIscc ' By the
same token any computer software that can solve fracture problems (residual
strength) can solve stress-corrosion prevention problems.
For example if K Iscc = 15ksiy'1n, would an edge crack of a = 0.5 inch in a
structure subjected to u = 18 ksi present any danger? Assuming f3 = 1.15
(Chapter 3) the stress intensity is
K = f3u Fa = 1.15 x 8 x ..)0.5 x n = 11.5 ksiy'1n. This stress intensity is
below K Iscc and therefore crack growth by stress corrosion should not occur.

Exercises

1. A fatigue crack grows in constant amplitude from a = 1 to a = 1.05 inch

in 7000 cycles. What is the rate of growth?
2. The crack growth properties of a certain material can be described by
da/dN = 1 x 10- 8 AK2K~~x' If f3 = 1, what is the rate of growth of a crack
oflength a = 0.5 inch if U max = 12 ksi and R = 0.2. How many cycles does
it take for this crack to grow to a = 0.51 inch?
3. If da/dN = 1 x 10- 8 m/cycle at AK = 8 MPa..jffi and da/
dN = 4 x 10- 6 m/cycle at AK = 30 MPa..jffi, determine the parameters of
the Paris equation. Convert the data to in/cycle and ksiy'1n and then
determine the parameters again.
166

4. The following data are obtained from crack growth tests at constant
amplitude; center crack with W = 20 inches.

a (inch) N (cycles)
/J.(J = 16ksi; R = 0 /J.(J = IOksi; R = 0.5
0.1 o o
0.105 1100 2000
1.5 k
1.55 i + 100 k + 170

Establish the rate diagram using two data points for each R-value; assume
straight lines between data points.
5. For the data in Exercise 4 determine a Paris equation (calculate Cp and m p )
for each of these R-values.
6. Using the results of Exercise 4 determine the Walker equation.

7. Assuming Kc = 80 ksiy'ffi, determine the parameters of the Forman

equation for the data in Exercise 4. Hint: first replot the data as
{[(I - R)Kc - ilK]da/dN} versus ilK.
8. Using each of the equations of Exercises 5, 6 and 7 calculate the growth rates
for 10 values of ilK and 2R-values. Plot the calculated results in three
separate rate diagrams and compare the results with the original test data.
9. Using the equation obtained in Exercise 6 calculate the crack growth curve
of a center crack, starting at a = 0.5 up to a = 3 inch; W = 100 inches.
Use 5 integration steps (0.5-0.7, 0.7-1.0, 1.0-1.5, 1.5-2.0, 2.0-3).
a max = 15 ksi; R = 0.3; constant amplitude. Plot the crack growth curve.
Hint: use a similar procedure as in Table 5.1.
10. Repeat Exercise 9 using the Forman equation of Exercise 7. Compare with
results of Exercise 9 by plotting in the same graph.
11. For a structural crack of a = 0.1 inch we find p = 3. At a = 0.1, a max = 11 ksi
and R = 0.2 what is the crack growth rate predicted by the equations of
problems 6 and 7?
12. Assuming that y for Wheeler retardation is y = 1.2, calculate the rates for
the case of Exercise II with the Walker equation of Exercise 6 if the given
cycle was preceded by an overload cycle with a max = 17 ksi at R = O. Also
calculate the rates after the crack has grown by 0.003 inch after the same
overload. What would these rates be if y = O? Assume plane strain plastic
zone as K!ax/6nF,;; F,y = 70 ksi.
 167

13. If K lsec = 15 ksi.J\n, what is the highest permissible stress for the configura-
tion of Exercise 11 for stress corrosion cracking to be prevented?
14. A material with K le = 30 ksiFn is subjected to stress corrosion tests in salt
water. Three specimens with a single edge crack are loaded in uniform
tension at loads of 18, 24 and 30 kips respectively. All cracks are 0.5 inch
long; W = 3 inch and B = 0.5 inch; F,y = 70 ksi. The times to failure of the
three specimens are respectively: 5000, 100 and 3 hours. Estimate K lsee and
calculate the amount of crack growth that occurred in each of the tests.
Ignore collapse. Assume f3 = 1.12 throughout.

References
[l] B.J. Habibie, Eine Berechnungsmethode zum Voraussagen des Fortschritten von Rissen, Messer-
schmidt-Bolkow-Blohm Rep. UH-03-71 (1971).
[2] J. Willenborg et aI., A crack growth retardation model using an effective stress concept, AFFDL-
TM-71-1-FBR (1971).
[3] P.D. Bell and A. Wolfman, Mathematical modeling of crack growth interaction effects. ASTM
STP 595 (1976), pp. 157-171.
[4] J.C. Newman, A crack closure model for predicting fatigue crack growth under aircraft
spectrum loading, ASTM STP 748 (1981) pp. 53-84.
[5] A.U. de Koning, A simple crack closure model for prediction of fatigue crack growth, ASTM
STP743 (1981) pp. 63-85.
[6] H. Fuhring and T. Seeger, Structural memory of cracked components under irregular loading.
ASTM STP667 (1979) pp. 144-167.
[7] D. Broek, A similitude criterion for fatigue crack growth modeling, ASTM STP 868 (1985),
pp. 347-360.
[8] D. Broek and S.H. Smith, Fatigue crack growth prediction under aircraft spectrum loading,
Eng. Fract. Mech.ll (1979) pp. 123-142.
[9] D. Broek, Fracture Mechanics Software, FracturREsearch (1987).
[l0] D. Broek and R.C. Rice, Prediction of fatigue crack growth in railroad rails, SAMPE Nat.
Symp. Vol. 9 (1977) pp. 392-408 11. A. Nathan et aI., The effect of thickness on crack growth
rate, paper presented at 14th symposium on aeronautical fatigue (ICAF) (1987).
[l2] Various Authors, Fatigue thresholds, Eng. Mat. Adv. Servo (EMAD), (1982), 2 volumes.
[13] D. Broek and B.N. Leis, Similitude and anomalies in crack growth rates, Mat. Exp. and Design
in Fatigue, Westbury (1981) pp. 129-146.
 CHAPTER 6

Load spectra and stress histories

6.1. Scope

In this chapter the word spectrum will be used to mean any statistical re-
presentation of loads or stresses. A stress history (or load history) is the
particular sequence of stresses (or loads) experienced by a structure or
component in service or as used in crack growth analysis.
For the application of damage tolerance concepts, a reliable prediction must
be made of the number of load cycles that will propagate a crack from a certain
starting size to the permissible size. Inspection intervals or any other fracture
control measures envisaged will be based on this prediction (chapters 11, 12).
The prediction of fatigue crack propagation rates and propagation time of a
structural crac~ requires the input of relevant crack propagation data, geometry
factors and stress history. Although often great efforts are expended in
obtaining geometry factors (to great accuracy) and material data, and much
significance is attached to the applicability of retardation models, the stress
history frequently is the stepchild in the analysis. Yet, the effects of stress history
are so significant, that slight misinterpretations may lead to errors in crack
growth life far overshadowing those in data and geometry factors.
This chapter presents a general discussion on the generation of stress histories
for use in crack growth analysis. Most load spectra can be represented by
exceedance diagrams; therefore the latter will be used as the basis for the stress
history development for structures subjected to random or semi-random
loading, but the essentials of the discussions apply just the same to load spectra
of different form and/or different formats. The representation by an exceedance
diagram has many advantages for random and semi-random loading, not in the
least because of its simplicity. Examples of exceedance diagrams will be given
for many types of structures.
In most cases the stress history used in the crack growth analysis is a
simplified version of the stress history anticipated to be experienced in service.
The simplification is necessary because (I) the actual stress cycles cannot be

168
 169

cr

(a) - time

~ll~~\YWV:\
START

(b~
STOP/START STOP/START

-
STOP/START

time

Figure 6.1. Loading of Rotating machinery. (a) Typical; (b) Idealization.

known in advance, especially in the case of random loading, and because (2) the
stress history must be derived as an interpretation of past experience on other,
similar structures, obtained either from load/stress measurements or from
calculations.
All major considerations in the development of a representative stress history,
including selection of clipping and truncation levels, will be reviewed and
examples will be given to illustrate the effects on the results of the crack growth
analysis. The use of a representative stress history is extremely important
-probably more so than any other ingredient of the analysis- in particular when
there is retardation.

6.2. Types of stress histories

Few structures or components are subjected to constant amplitude loading. An

example of these few is perhaps a buried pipeline subjected to occasional de- and
re-pressurization cycles; but even in that case loadings due to hydrostatic tests,
possible seismic events, and perhaps frost heave, thaw settlement and land
slides, give cause to a stress history of variable amplitude. For those structures
subjected to loading of constant amplitude, there is hardly an interpretation
problem, provided the magnitude of the loads and consequent stresses are
known with sufficient accuracy.
170
(J

1 BLOCK

time

Figure 6.2. Proposed stress history for nuclear power generation components (2).

A class of structures with relatively simple load histories is rotating

machinery, among which one can reasonably categorize all turbomachinery,
propellers, helicopter blades and so on. The major load cycles consist of the
start-up-shut-down cycle (mostly centrifugal force), upon which small cycles are
superposed as shown in Figure 6.1. The latter cycles may be due to hydro-
dynamic/aerodynamic bending loads on the blades of a turbine, or propeller,
and/or vibratory loads. The superposed small cycles are usually not of constant
amplitude (because of changing power input or output). Neither is the start-stop
cycle always of the same magnitude because it depends upon the r.p.m. In some
cases however, one may be justified in using a constant amplitude start-stop
cycles and superposed constant amplitude small cycles as shown in Figure 6.1 b.
The input to the computer program then could be a simple table of stresses e.g.:
o - 11m ax I time;
I1 j - I1max n times.
This sequence can be repeated throughout the crack growth analysis, provided
the computer program offers the facility to take tabular stress input and
interpret its meaning.
Should the above simplification not be justified, then a longer table, with
more values of the superposed stress cycle (and number of occurrences) can be
used. The sequence as given in the table could be applied as is and repeated, or
if the computer code offers this possibility, the sequence could be randomized.
Piping components of nuclear power structures are subjected to a stress
history of very similar nature. A more or less standardized (the meaning of the
word here being: agreed upon by a group of people) form of the history [I, 2]
is shown in Figure 6.2, which has been used for tests and analysis development.
Although the above histories are simple, there still may be interpretation
problems. The problems of counting, clipping, trunction, number of levels, etc.,
 171

(a)

(b) (c)

Figure 6.3. Loads on ship. (a) Still water; (b) Sagging; (c) Hogging.

discussed in this chapter still apply. Also the discussion on 'errors' in Chapter
12 are of relevance.
Many structures experience loads (stresses) of variable amplitude in 'random'
sequence. Actually, the load sequence most often is semi-random, and
sometimes interspersed with deterministic loads (stresses). Typical examples are
airplanes, ships, offshore structures, railroad rails etc. The load sequence on a
ship is not the same during every voyage, because storms and high seas are
encountered only occasionally. Similarly, the loads on an offshore structure
depend upon winds and waves, and are different from month to month. Such

= WIND

Figure 6.4. Loads on offshore platform.

172

Ll2

L
u

~v
w
(b)

L !Q V~ SC L
, dC L
!J.L loV"S-Lh
2- dO(
U/V

!J.L !nVUS dC L
2~ dO(
L +!J.L
n --W-; S = Area

Figure 6.5. Loads on aircraft wing. (a) Gust loads on commercial aircraft; (b) Maneuver loads on
fighter aircraft.

sequences are not truly random; high loads are clustered in certain periods
(storms). The word semi-random is used here to describe such sequences.
The loads on e.g. a crane derive from the anticipated weights of various
magnitudes lifted. Loads on a ship arise from hogging and sagging across the
waves. In still water the buoyancy forces on a ship are more or less evenly
distributed (apart from bow and stern). The resulting bending moment then
depends upon the distribution of masses such as engine, fuel and cargo (see
Figure 6.3). In waves the buoyancy forces are unevenly distributed, causing
bending of the ship, and leading to tension in the deck during hogging and
compression during sagging, and to tension in the bottom during sagging.
Loads on off-shore structures arise from winds and waves (Figure 6.4), on
automobile structures from uneven roads, on railroad rails from different wheel
loads and different car weights, as well as from impacts due to wheel flats and
gaps (bolted rails and points).
Aircraft wings and tails experience loads due to maneuvers and gusts
(turbulence), as shown in Figure 6.5. Commercial airliners are subject to some
maneuver loading, but gust loading is the most important. Fighters on the other
hand do experience gusts, but maneuver loading is the most important. In
upward wing bending the lower surface of the wing is in tension.
The variable bending due to gust and maneuvers is superposed on the steady
bending of stationary horizontal flight (I g), where the wing just carries the
weight of the aircraft.
 173

(a)

(b)

(c) I
Figure 6.6. Landing loads on aircraft. (a) Ground loads; (b) Moment-line; (c) Stress in lower wing
skin during flight and landing.

Following each flight is a ground-air-ground cycle (g.a.g.), sometimes

combined with taxiing loads (Figure 6.6). Of course the g.a.g. cycle depends
upon the airplane configuration, size, landing gear location and is different at
different locations. For example, in a wing (Figure 6.6) the ground loads may
cause compressive stresses at A, and tension at B depending upon weight
distribution and location of the landing gear. The g.a.g. cycle is a deterministic
event in an otherwise semi-random stress history. It is known from experiments
that the g.a.g. cycle causes faster crack growth, in particular when the stresses
are compressive. This is largely due to the annihilation of residual compressive
stresses introduced during tension overloads, and consequently, the elimination
of retardation (Chapter 5). Some retardation models attempt to account for the
effect of compressive stress and thus for g.a.g. Because of its sometimes large
influence, the g.a.g. has to be accounted for properly.
As it is very important, it is emphasized again, that the loading on most of
these structures is not truly random. High loads are clustered (in periods of bad
weather); this must be accounted for if retardation is significant, because
complete randomization would spread out all high loads throughout the history
so that they would cause much more retardation than when occurring in
clusters. Moreover the history may be interspersed with deterministic loads,
such as g.a.g. cycles in aircraft, unloading and refourageing of ships after each
voyage, etc. If these deterministic loads are significant, they must be included in
the history at the proper intervals (not at random).
174
Level count
6 ------------------- Level Counts Exceedances

2
4
5 12
I 4 16
o 20
-I
-2

Exceedances not relevant for fatigue analysis.

Mean crossing peak count
6 ----.--------------
Level Counts Exceedances

4
4
-4 -------------------
-5 - - - - - - - - - - - - - - - - - - - -I
-2
Ignores small cycles not crossing mean
Peak count
6 --------------.---- Level Counts Exceedances

4
4
-1
-2
-3 .---.--------------
~ -------------------
-5 -----------.------- -I
-2

Range count (upward)

6 -------------------
Range Counts Exceedances

6J
50
40
-1 30
-2
-3 -------------------
20 o
~ ------------------- 10
-5 - - - - - - - - - - - - - - - - - - - o
Range-pair count
Range Count Exceedances
6 -----.-------------
6J
50 o
40 o
30
20
10 4
o
Solid symbols: up range
Open symbols: down range
O/same magnitude

Figure 6.7.. Examples of some of the counting procedures.

 175

6.3. Obtaining load spectra

If the loading is man-induced the load spectrum often can be calculated on the
basis of (anticipated) usage. This would be the case for cranes, fighter aircraft
(maneuvers) and rotating machinery. If, on the other hand, loading is induced
by nature (waves, winds, gusts, rough roads), the load spectrum is usually based
upon measurements, from which the loads must be inferred. Subsequently, the
stress history must be obtained from the load spectrum.
Load spectra can be measured in service by indirect means only, for example
from continuous strain gage records over long periods of time. The measure-
ment is indirect because the loads must be inferred from the measured strains
or from e.g. measured wave heights or wind force. As such records are very
extensive, a representation in some concise form is desirable: they must be
interpreted so that an envelope can be established. The envelope is called the
spectrum. Power spectrum density analysis can be used or one can establish
exceedance diagrams from counts of the records.
Many counting methods have been developed (peak count, mean-crossing
peak count, range pair count, range-pair-mean count or rainflow count, etc.), a
few of which are shown in Figure 6.7. At present, it is generally agreed that the
latter method gives the best representation for fatigue, but in many cases the
differences among the better counting methods are small [3]. It should be
mentioned here that it is rather immaterial whether different methods produce
somewhat different spectra; the only criterion is whether such a spectrum when
used in a crack growth analysis produces the correct crack growth result. This
brings about the major problem of the counting procedures, namely what is
important for crack growth: the maxima and minima or the ranges and means.
On the basis of counts of past measurements, the envelope of the future load
history can be estimated, but the actual load experience of a new structure
cannot be known until the service life is expired. All counting methods have
shortcomings and the tendency to neglect certain small load reversals. Their
usefulness depends upon their purpose. When the data are to be used for design,
the usefulness may be judged by how well the counting method has described
the actual loads, in particular the highest loads. For fatigue crack growth
calculations the usefulness depends on how well the methods describe those
loads which are the most relevant to the crack growth process. Although the
question regarding the use of maxima and minima instead of ranges is
important, the most important difficulty in the use of spectra is that after the
counting all information about the sequence is lost. A new sequence must be
generated for use in damage tolerance analysis.
The interpretation (counting) of measured load experience is a very interest-
ing problem by itself, but a detailed discussion is beyond the scope of this book.
The reader is referred to the excellent review of the problem by Schijve [3]. It
should be noted here that the counting procedure presently known as the
176
STRESS
PEAK
lOR RANGEl
1000 hrs
(j

5,
I

: 51
I
I I
- - - - -i- -
,
I
:
--4. -
,
- - -

TIME

103 EXGEEOANCES

(a) (b)

Figure 6.8. Significant counts for fatigue analysis. (a) Exceedances of maxima (or ranges: not levels);
(b) Levels and maxima.

'rainfiow' method is virtually identical to the much older range-pair-mean count

method [3]. For the purpose of this book, it is sufficient to start with the
end-result of the measurements, namely the load spectrum. Nevertheless, the
problem of counting will arise again later in this chapter

6.4. Exceedance diagrams

Although power spectra are useful, in particular when there are environmental
and frequency effects, exceedance diagrams are more convenient because they
lend themselves easily for the (re-) generation of stress histories. A typical
exceedance diagram is shown in Figure 6.8. The diagram shows how many times
a certain stress (or load) level is exceeded. This by itself is a 'loaded' statement.
More properly expressed, the diagram gives the number of times certain maxima
and minima (or ranges) are exceeded. For example in Figure 6.8b the level R is
exceeded twice when the load varies from Q through R to SI and S2. These
exceedances are NOT and should not be represented in the diagram, because
they would be of no use for crack growth analysis; only Q and S are of interest.
Rather, e.g. point SI in Figure 6.8a indicates that the maximum stress in A
CYCLE exceeded level SI a certain number (1000) times during the time (1000
hours) for which the diagram was established.
Thus, the exceedances must be interpreted as exceedances of peaks or
RANGES, depending upon the counting method, and not as level-crossings. A
level-crossing diagram (first case in Figure 6.7) might look very different and
would be rather useless for fatigue crack growth analysis without further
interpretation. On the other hand, level crossings may provide some informa-
 177
STRES
~ax

A
B

a;;teady ~O-~--'*'--t--r-:::"'r--:-LOG EXCEEDANCES

(a) o;"in
(b)

(d)
(e)

Figure 6.9. Typical exceedance diagrams. (a) typical for weather induced; ships, aircraft; earth-
quakes; (b) Typical weather induced; offshore structures; (c) Typical for mechanically induced;
railroad rails; (d) Typical for mechanically induced; fighter aircraft.

tion about the time spent at or above certain load levels, which maybe of interest
if there are significant environmental (time-dependent) effects. Note that ex-
ceedances are generally presented on a logarithmic-scale, the loads (stresses) on
a linear scale.
Exceedance diagrams can be categorized roughly in two classes, namely
(almost) semi-log linear and non-linear. These types are shown in Figure 6.9.
The almost semi-log-linear diagram is typical for nature induced (e.g. weather)
loading: very similar diagrams apply to ships, offshore structures and
commercial airplanes. The diagram may be asymmetric, or symmetric around
a non-zero 'steady' load or stress. The non-linear diagram is typical for mechan-
ically induced loads. This type of diagram applies to fighter airplanes, cranes
and e.g. railroad rails. It is almost always asymmetric. For lack of a better word,
the level at which top and bottom part of the diagram meet, will be called the
'steady stress' (load). The exceedance diagram may be expressed in loads,
g-levels, wave heights, gust velocity or any measure of load. The diagram gives
the load experience for a certain time interval, number of flights or voyages,
amount of traffic (rails or bridges), number of years or hours; it is a statistical
average for that time interval. It is usually assumed that for each such time
interval the same load envelope applies, be it that the load sequence may be
different in each interval.
178
24r------------------------------------------,

DIFFERENT SHIPS

50 years experience for Great-Lake ships

16

...·iIi
::f 12
~
in

0
0 10
Log Exceedonces

Figure 6.10. Stress exceedances for five ships [6].

Typical exceedance diagrams for ships are shown in Figure 6.10. These are
induced by nature and are indeed nearly semi-log-linear (note that they are
symmetric, but only the top half is shown). Spectra for off-shore structures are
very similar, but often asymmetric (Figure 6.11). An aircraft gust spectrum
(nature induced) is shown in Figure 6.12 (only top half is shown; diagram is
symmetric); indeed it is almost semi-log-linear, while the man-induced fighter

WAVE
HEIGHT
1ft. I
15

10

0.1 10 100
EXCEEDANCES 1%1

Figure 6.11. Offshore spectrum.

 179

.5

o 2 3 4 5 6
LOG IEXCEEDANCESI

Figure 6.12. Aircraft gust spectrum [8].

maneuver spectrum in Figure 6.13 is non-linear. Similarly, the man-induced

spectrum for railroad rails in Figure 6.14 is non-linear (and asymmetric). It must
be emphasized that the spectra in Figures 6.10-6.14 are examples only; they may
differ from structure to structure and should not be used as the general spectrum
for the above types of structures; they show the general trend.
1.1
(0
1.0 •
0.9
•
100 flight hours
0.8

0.7
o 0.6
i
1;;
0.5

"0 0.4
5
13 0.3
~
en 0.2
-;:.::_::.::ll:=~=~=:J
---I.l!.----- ____________________
(/)~ 0.1

o
-0.1

-0.2Ii=1i:=I:::=::I:::::::LLJ-t.._..L_J......LLL_...L_..L...Ll.l~...L-...L....L~
10 100

Exceedances

Figure 6.13. Fighter aircraft manoeuver spectrum [9]. Courtesy Eng. Fract. Mech.
180

o
a..
::;;
b
<J

Exceedcnces

Figure 6.14. Spectrum for railroad rails [10]. Courtesy SAMPE.

No load experience is available for a new structure. Hence, the fatigue crack
growth analysis must be based on past experience and be a projection in the
future. In some cases the loads (or stresses) in a structure are continously
monitered, so that the analysis can be updated from time to time depending
upon the actual service experience. (The United States Air Force 'tracks' many
of its aircraft for this specific purpose). The projection ofload experience in the
future introduces an uncertainty in the crack growth prediction. Nothing can be
done about this problem other than performing multiple computer runs, using
various alternative load spectra, to bound the fatigue performance for best,
worse and average scenarios.

6.5. Stress history generation

As fatigue crack growth analyses is based upon stresses, the spectrum, whatever
its basis, must be converted into stresses. If the stresses are proportional to the
quantity given in the exceedance diagram (load, wave height, gust velocity and
so on), the diagram can be converted easily into a stress exceedance diagram
through a multiplication factor. This is often the case or nearly so, because the
nominal stresses during fatigue are generally elastic. In some cases more com-
plicated transfer functions will be involved.
The exceedance diagram gives the number of times that a certain stress level
is exceeded during a given time interval, the level being a maximum, minimum
or range, not a level crossing (see previous section). In the example in Figure
6.15, the exceedances are for one year of operation. This means that the stress
 181
Stress
leve.L

5 EXCEEDANCES
10
-6 ~ ______________________ ~~~

-5 ~ ________________-.~£-I

- J I-------~~~

-1

Figure 6.15. Determination of stress levels from exceedance diagram.

Level Exceedances Occurences
1 4 4
2 20 16
3 100 80
4 800 700
5 3000 2200
6 20000 17000

level 4 will be exceeded 250 times per year, and level 3 will be exceeded 80 times
per year. As a result, there will be 250-80 = 170 events in which the maximum
stress reaches a level somewhere between 3 and 4.
One can identify a stress level i that is exceeded 99 times, and a levelj that is
exceeded 98 times. Thus, there would be one stress excursion to j. Similarly in
the case of Figure 6.15, one could identify 60000 different stress levels, each of
which would occur just once per year. In general, one could define as many
stress levels as there are exceedances in the diagram. Obviously, it is impractical
to consider so many different stress levels. Not only would it be impractical, it
would also ignore the fact that the spectrum is a statistical representation of past
experience and that the analysis is a prediction of the future. Accounting for so
many stress levels would be presuming that the stresses are known to occur in
the future exactly as they did in the past, which they will not.
182
LIFE IhrsJ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _- - ,
I thousands I 7 r-

6.5

5.5

4.5

3.5

2.5 L...._ _ _ _......_ _ _ _ _....._ _ _ _ _ _ _ _ _____

5 8 12 16
NUMBER OF LEVELS

Figure 6.16. Effect of levels in exceedance diagram approximation; computed number of hours for
crack growth as a function of number of levels. One level is constant amplitude.
Instead, the exceedance diagram is approximated by a limited number of
discrete levels; six such levels are shown in Figure 6.15. The stress levels do not
have to be evenly spaced, but they usually are selected that way. Alternatively,
one may select a number of exceedances and start constructing the levels from
the abscissa. This has the advantage that for a change of the load-stress
conversion factor always the same number of cycles would occur; only the stress
values would change. Experience shows that 10 to 12 levels (each for positive
and negative excursions) are usually sufficient to give the desired accuracy; the
use of more than 12 levels hardly changes the results. This can be appreciated
from the results of actual fatigue crack growth analyses for one particular
exceedance diagram, shown in Figure 6.16.
For clarity only 6 levels (6 positive and 6 negative) are shown in the example
in Figure 6.17. At each level a line is drawn that intersects the exceedance curve.
Usually the steps are completed by placing the vertical AB in such a manner that
the shaded areas above and below the exceedance curve are equal, as shown in
Figure 6.17. If the exceedance diagram is semi-log-linear, this vertical AB will
be at the logarithmic average of C and D, i.e midway between C and D on the
log scale of exceedances. Other procedures have been suggested, but these do
not lead to essentially different results, as can be inferred from Figure 6.16: if
more and more levels are selected, the approximation has less and less signifi-
cance. Besides, great sophistication (complication) is not warranted, because the
exceedance diagram of the past will not be repeated exactly in the future
anyway. Figures 6.15 and 6.17 shows how the exceedances and from these the
number occurrences of each level are obtained.
 183
a
6 a
5 A 61----=---.,=
C 0
B 4~----------~~

Steady

Exceedances
2
3

10 10' 103 10' 10 5 10' 10 7

Exceedances

(a) (b)

Figure 6.17. Example oflevel approximation (only 6 shown for clarity).

Level Exceedances Occurences
(a) Symmetric and log linear
6-6 30 30
5-5 300 270
4-4 3000 2700
3-3 30000 27000
2-2 300000 270000
I-I 3000000 2700000
(b) Asymmetric
0-6 60 60
0-5 8000 7940
0-4 100000 92000
0-3 900000 800000
0-2 2000000 1100000
0-1 5000000 3000000

If the exceedance diagram is for ranges of stress, the approximation is now

essentially complete but a problem remains as shown below. If the diagram is
for peaks (maxima and minima), positive and negative excursions still have to
be combined. One might be tempted to select positive and negative excursions
in random combinations. This could introduce a new and serious problem, as
shown on the basis of Figure 6.18.
Four cyclic sequences A, B, C and D are shown in Figure 6.18. Let it be
assumed that the material follows a Paris equation for crack propagation with
a power of 4 (Chapter 5), merely to facilitate the discussion. Sequence A has a
small reversal which certainly will not contribute to the crack growth. Hence,
only the large excursion of 8<5 would have to be accounted for. The growth
during this cycle is then C(8<5)4 = 4096 C<5 4 • The little reversal in sequence B
cannot be neglected. If this sequence were interpreted as two stress ranges of 5<5
each, the total growth would be 2 x C(5<5)4 = 1250 C<54 • Similarly, for
sequence C the result would be 2 x C(M)4 = 2592 C<54 • With this interpreta-
184

USING RANGES 1 & 2 :

!. AI.' C(~K1)4. C(86)4.
~ Alb' C(~K2)4 + C(~K3)4 • 2C(56)4 • 1250 C64
£. Ale • • 2C(66)4 • 2592 C64
• 2C(756)4 • 6328 C64

USING RAINFLOW COUNT:

!. AI.· C(~K1)4. C(86)4 •
II. Alb' C(~K1)4 + C(~K4)4. C(86)4 + C(26)4 - 4112 C64
£. ~Ic; • - C(B6)4 + C(46)4 • 4352 C64
l!. Ald- - C(86)4 + C(76)4 • 6497 C64

Figure 6.18. Two interpretations of ranges for crack growth analysis.

tion sequence B :"ould cause only about 1/3 of the growth of sequence A, while
even sequence C would cause less growth than A. Clearly, Band C are more
severe than A, and this should be reflected in the growth, otherwise erroneous
answers will be obtained. The solution to this problem is in the interpretation,
as shown below.
In all sequences the material is subjected to the large excursions of 815, causing
a growth of C(86)4 = 4096 C6 4 in ALL FOUR cases. In addition there is a small
reversal with a range of 26 in sequence B causing a growth of C(2bt = 16 C6 4 ,
and bringing the total to 4112 Cb 4 . Similarly, the small range of 415 in sequence
C causes an additional growth of C(4b) = 256 Cb 4 to bring the total at 4352 Cb 4 .
With this correct interpretation sequences Band C indeed produce more growth
than A. The conclusion is, that in an arbitrary sequence of peaks and valleys,
the identification of significant stress ranges requires a new interpretation. A
counting procedure such as the rainflow count will accomplish this (Figure 6.18
bottom); effectively the example above is the consequence of a rainflow count.
Should an arbitrary sequence of peaks and valleys be used in the analysis, then
the computer code must have a facility to count the stress history again. Besides,
a new problem of sequencing arises: should the large range in sequences Band
C be applied first and then the small range, or vice versa?
 185
Positive and negative excursions following from the exceedance diagram must
be combined and sequenced. If this is done randomly, rainflow counting of the
history will again be necessary as shown in Figure 6.18 to determine what
constitutes a range. Although this is a legitimate approach, a simpler procedure
is often employed. Because the spectrum was developed from a counted history
in the first place, why would it be necessary to disarrange it again, and then
count it again, if the result was counted already a priori. Besides, the effective
result of the second counting will be that the largest positive peak will be
combined with the lowest valley. Foreseeing this, one may combine positive and
negative excursions of equal frequency of occurrence immediately. The ranges
are then easily established as shown in Figure 6.17 top, and these can now be
applied randomly because they are already pre-counted and interpreted. This
leads to the largest possible load cycles, which is conservative; the computer
code does not need a counting routine. Sea and air are continous. A down-gust
must soon be followed by an up-gust of approximately equal magnitude
otherwise air would disappear in space. Similarly, when the waves are high, a
ship is likely to experience hogging and sagging of the same magnitude in close
succession. Hence, a priori combination of positive and negative excursions of
the same frequency of occurrence is realistic; it avoids the necessity of renewed
counting and sequencing.
The content of the stress history is now known, but the sequence must still be
determined. If retardation is not an issue, sequencing of stresses is rather
irrelevant. Fatigue crack growth analysis without load interaction is virtually
insensitive to load sequence, provided the stress history is more or less random.
If the material exhibits little or no retardation, the problem of sequencing as
discussed below does not apply. The same holds, iffor reasons of conservatism,
retardation is not considered. In the latter case it should be recognized that the
ANALYSIS is indeed insensitive to the sequencing, but if load interaction
actually occurs, the analysis without retardation does not give answers that
have any bearing upon reality.
On the other hand, if load interaction is (or must be) considered, stress
sequencing becomes of eminent importance. In many computer analyses the
loads are simply applied in random order. However, because load interaction is
indeed important, a random sequence is certainly not providing correct answers
when actual service loading is semi-random. Not all voyages of a ship are of the
same severity; a ship experiences storms only occasionally, and so does an
offshore structure. A jet liner experiences many smooth flights and occasionally
a rough flight. This means that the loading is not truly random, but clusters of
high loads occur (Figure 6.19). Were these high loads randomly distributed
throughout, they would cause large retardation continually. Because of the
clustering, the retardation will be much less. The fatigue crack growth analysis
must account for these effects: a mixture of PERIODS or flights or voyages of
different severity must be applied. This is defined here as semi-random loading.
 186

Stress
Moderate Mild I Mild I Severe I Mild Semi-severe I Mild I
!..-ccP'~rI",Od~_ _---->l--,-p~.r-"io=-=-d -.ar--Pe:r-'iO=-::d_ _---...j~.-period .. __ -+ Perio~__ Period !II .. Period ..

~Time
o.

Mod£'rate
-->1 I
• ..--_- -
Storm I Mild .1. High Winds .1. Mild .1
Weather

Stress

I---s_m_o_ot_h_~mO~~b"I.I_le-"n_ce_~.oI<lc~smooth. .
~! Thunderstorm .. I. Smooth
·1·
Flight

Landing ~Time

Figure 6.19. Semi-random loading in different periods (voyages, months, flights). (a) General case;
(b) ship or offshore structure; (c) Commercial aircraft.

Of course there will be fewer severe periods than mild periods, as shown in
the example in Table 6.1 and in Figure 6.19. In the computer analysis periods
of different severity are to be applied in random sequence, the cycles in each
period to be applied randomly, but different from the random sequence in
previous applications of the same period.
A semi-random sequence can be developed in many ways. A simple algorithm
is shown below. The mild and severe periods are constructed by recognizing that
 187

Table 6.1. Periods of different severity

(a) Exceedance diagram
Stress conversion: 1 unit = 1 ksi.
Max. stress Min. stress Steady stress Exceedances
15.000 -5.000 5.000
13.000 -3.000 5.000 5
11.000 -1.000 5.000 30
9.000 1.000 5.000 166
7.000 3.000 5.000 910
5.000 5.000 5.000 5000
Exceedances for 120 blocks.

(b) Stresses for 120 periods

I unit of stress is 1 ksi
Exceedances Occurr. Min. str. Max. str. Periods type/number
Total 120 periods
Cycles Cycles
1/96 2/19 3/4 4/1
1 1 -5.000 15.000 0 0 0 1
3 2 -4.000 14.000 0 0 0 2
8 5 -3.000 13.000 0 0 1 1
19 11 -2.000 12.000 0 0 2 3
46 27 -1.000 11.000 0 1 1 4
108 62 0.000 10.000 0 2 4 8
253 145 1.000 9.000 2 2 3
594 341 2.000 8.000 2 6 7 7
1393 799 3.000 7.000 6 9 10 12
3266 1873 4.000 6.000 15 18 18 19

the exceedance diagrams for the individual periods are of the same shape [11],
but with a different severity factor (slope), their total making up the diagram of
total exceedances. The example will be based upon an exceedance diagram for
100 periods (be it voyages, hours, months or flights) as shown in Figure 6.20,
and again only six levels are used, but the same procedure can be followed with
exceedance diagrams for fewer or more periods, and/or more levels.
The different periods are constructed as illustrated in Figure 6.20 and
Table 6.2. The total number of exceedances is 100000, so that the average
number of exceedances per period 1 000000/100 = 1000. The highest level
occurs 3 times (Table 6.2, column 8). Naturally, it will occur only in the severest
period denoted as period A. Letting this level occur once in period A: the
exceedance diagram for period A is established as shown in figure 6.20b. Note
that the total exceedances are 1000 (per period) and that level 6 must occur once
(step at level 6 must be at 1), so that the exceedance diagram of period A can
indeed be constructed as in Figure 6.20b; then all levels can be drawn as
188

10'

(b) (e)

Figure 6.20. Stress history generation (see also Table 6.2). (a) Total exeeedances; (b) Exeeedanees
for period A: only top half shown; (e) Exceedances for other periods: only top half shown.

shown. Period A can occur only three times, because then the cycles of level 6
will be exhausted. The exceedances for period A are read from the exceedance
diagram of A (Figure 6.20b), and from these the occurrences (number of cycles)
are determined as in Table 6.2 columns 4 and 5. There being three periods A,
the total cycles for all A are as in column 6. These cycles are subtracted from
the total (column 3) so that the remainder for all other 97 periods is as shown
in column 7. The next most severe period is B. Its highest will be level 5 which
will occur once. This information permits construction of the exceedance
diagram for B in the same way as for A as shown in Figure 6.20c. The
exceedances and occurrences are determined as in columns 8 and 9 in Table 6.2.
As there were only 12 cycles of level 5 left after subtraction of three periods A
(column 7), there can be 12 periods B. These 12 periods will use the numbers of
cycles shown in column 10, which must be subtracted from those in column 7
 Table 6.2. Generation of stress history with different periods based in Figure 6.20.

2 3 4 5 6 7 8 9 10 11
3 x 5 3-6 12 x 9 6-10

Level Exceed- Occurrences Type A Occurr Occur in Remainder Type B Occur Occur in Remainder
ances exceed in A 3 types A exceed in B 12 types B
Fig.6.20a Fig.6.20b Fig. 6.20c

6 3 3 1 3
5 21 18 3 2 6 12 12
7 132 111 12 9 27 84 5 4 48 36
3 830 698 48 36 108 590 20 15 180 410
2 5750 4920 158 110 330 4590 100 80 960 3630
43650 37900 575 417 1251 36649 480 380 4560 32089

Total number of periods 3 12 + 3 15

If considered necessary

12 13 14 15 16 17 18 19 20 21 22
36 x 13 11-14 15/49 16 x 49 15-17 18/12 9 + 19

Type C Occur Occur in Remainder Occur Occur in Remains Distributed New Exceed According
exceed in C 36 types C for III 49 type D in type B orD to
Fig.6.20c 49 type D type 12 type B diagram
D Fig.6.20c

1 36 4
8 7 252 158 3 147 II 16 3
52 44 1584 2046 42 2058 -12 -I 79 45 20 00
'Cl
400 348 12528 19561 399 19551 10 381 444 300

15 + 36 = 51 51 + 49 = 100
Same for negative levels if applicable. Periods: 3A + 12B + 36C + 49D 100 totaL
190
to leave the remaining cycles in column 11. Period C is constructed in the same
manner. There can be 36 periods C and then the cycles of level 4 are exhausted.
One could go on in this manner, but since there now are only 49 periods left,
it is better to divide the remaining cycles in column 15 by 49 in order to
distribute them evenly over 49 types D. This is done in columns 15-17. There
are some cycles unaccounted for, and also a few too many cycles were used as
shown in column 18. These are cycles oflower magnitude contributing little to
crack growth, and since the diagram is only a statistical average - and the
procedure an approximate one - this little discrepancy could be left as is.
However, if one would want to be precise, they could be accounted for by a little
ch~nge in the content of period C, as shown in columns 18-20.
By means of the manipulation in column 16 the exceedance diagram of type
D is determined as shown in column 21. Were the exceedance diagram for D
determined in accordance with the above principle, it would be as shown in
Figure 6.20c and the exceedances would be as in column 22. The latter are
certainly somewhat different from those in column 21, but since they concern
only the lower stresses, the effect on crack growth analysis will be minor (note
again that the procedure is based upon statistical averages).
The above is an example only. If more than 6 levels are used, more (and
different) types of periods can be generated. However, there is no need to go to
extremes as long as a semi-random history is obtained, recognizing that periods
of different severety do occur and that the higher loads are clustered in those
periods. No matter how refined the procedure, the actual load sequence in
practice will be different anyway. No mater how many levels are used and how
many different types of periods are generated, there comes a point where the
remaining cycles must be lumped into the remaining periods. Thus the mildest
period will never be entirely representative; but since this concerns only the
lowest stresses, no great harm is done.
Clearly, the total number of cycles in the stress history as generated above is
less than the total number of exceedances of the steady level. This is
entirely acceptable, because the very smallest cycles certainly will have no effect
on crack growth, the procedure being for crack growth with retardation. The
larger the number of levels used, the more (small) cycles there will be. But as
shown in Figure 6.16 the result of the analysis will be the same (see also section
6.7 on truncation).
In accordance with the nature of the loading, there are only three severe
periods A and 12 periods B in the total of 100 in Table 6.2. The majority consists
of mild periods D (49) and C (36). Regardless of the number oflevels chosen and
the number of periods concerned, the above procedure will reflect this reality.
Naturally, other procedures can be devised, but the above is certainly a rational
one and easy to implement; besides it is based upon reality [11].
In the crack growth analysis the various periods should be applied in random
order and the cycles within each period should be applied randomly. Thus the
 191

lovel level exceedances occurrences

1
2 I 2
3 2 7
3 15

Q.
10 lOa E xceedon ces

level level exceedonces occurrences

1 .... ..... I a a
2 2 7 7
3

JJ;"
3 15 8

cycle

10 100 Exceedances
b.

Figure 6.21. Clipping.

second occurrence of e.g. period C would apply the stresses in a different
sequence than the first occurence, but the total cycle content of period C would
be the same. If the 'basket' with lOO periods is empty, it is 'refilled', and the
process starts anew; yet because of the randomization the periods appear in
different order.

Aa 1
~ax
.9

,I
________ 1 ____ _
I
I I
\

.1
\
'-
\
,
4 10'
LOG (EXCEEDANCES) LIFE (FLIGHTS)

Figure 6.22. Effect of clipping on life; test data [8].

192
The above procedure can be implemented by hand calculations, and by
submitting a resulting table of stresses and occurrences for use in a computer
analysis. Aternatively, it can be programmed and included in the computer
code. If this is done, the input is very simple: only a few data describing the
exceedance diagram need be submitted; the computer does the rest.
If there are deterministic loads these must be interspersed in the above
sequence. For example, in case of aircraft there will be taxing cycles and
ground-air-ground cycles after each period (flight). They should not be applied
randomly among the other cycles.

6.6. Clipping

The spectrum is a statistical average of previous load experiences. A levelj may

appear to occur e.g. 10000 times. It would cause no surprise if it would actually
occur 9900 or 10 100 times in a future case. There will be a certain level in the
stress history that occurs only once. Since this is an extreme statistical number,
it is very well possible that the level is never reached in service. For example, it
would be quite natural iflevel 6 in Figure 6.20 and Table 6.2 were never reached.
Two high loads might still occur, but only go as high as level 5. In that case
there would be twenty-one cycles to level 5 instead of eighteen to level 5 plus
three to level 6. The total number of cycles would not change, but the level 6
cycles would be 'clipped' to level 5. Also the exceedance diagram would be
clipped as shown in Figure 6.21.
Since the highest stresses cause most of the retardation, clipping can cause
dramatic effects on crack growth [8]. In the older literature, clipping is often
called 'truncation', but presently the word truncation is reserved for the
procedure discussed in the following section. Note that clipping merely reduces
the magnitude of the highest loads to the clipping level; no loads are omitted,
as shown in Figure 6.21 b, insert.
The first tests on the effect of clipping were performed by Schijve [8] using
aircraft flight simulation loading. Figure 6.22 presents a summary of his test
results. Various clipping levels were used by reducing the size of the largest
cycles-which are small in number-to the size of the next highest level (no cycles
omitted). In tests with lower clipping levels all stresses above the clipping level
were reduced in magnitude to the clipping levels. The exceedance diagram in
Figure 6.22 shows how many cycles would be affected (up to about 800 at the
very lowest clipping level, out of a total of over 300000). According to Figure
6.22 clipping of the two highest levels (affecting only 80 of the 200000 cycles),
already reduced the crack propagation lives by almost a factor of about 2. It
appears that crack propagation may be faster than expected if the structure
encounters less severe service loading than was foreseen. The effect of clipping
on crack growth can be demonstrated also in crack growth analysis as shown
 193

Linear AnalysIs
Symbol Spectrurn (Flights)

0 •
l>
Willenborg
Wheeler, 2.3
Fighter 270

b •
a
Willenborg
Wheeler> 2.3
Trainer 460

c •
a
Willenborg
Wheeler 1 2.3
8-1 class bomber 140

d
. Willenborg
C-transport 1270
" Wheeler, 2.3

1.2.--------

II

1.0
~
iii
'E 0.'
::::;
'0
c 0,8
,2
g Gust spectrum test
It data by Schljve
0.7
"
~

"
~
~ 0,6

'"c ~

:~ 0,5 0
U

0,4
-b-'

0,3
100 1000 10,000 100,000
Flights for Crock Growth to 2 in.

Figure 6.23. Effect of clipping [9], Courtesy Eng. Fract. Mech.

in Figure 6.23. These are computational results [9] for four different spectra.
Clearly, clipping has an effect only if there is retardation; in an analysis without
retardation the differences would not be noticeable.
As discussed above, it should be expected that clipping occurs in service. The
exceedance diagram is a statistical average, and loads that are occurring only a
few times might reach to a slightly lower level only. Should this occur then crack
growth would be faster; if the analysis did not account for clipping the crack
growth in service might be much faster than predicted. It is sometimes argued
that clipping is unrealistic and that all those load levels should be included that
may be anticipated to occur in service. The latter part of the argument is crucial;
if they indeed occur, they should be included. However, whether they will occur
is questionable. The spectrum is only a conjecture or, at best, an interpretation
of measured data. Slight variations of the spectrum may be unimportant in the
lower part, but they are very significant in the upper part.
194
level (lxceedances occurrences
level 8 8
50 42
250 200
2000 1750
10000 8000
80000 70000

log (lxceedances

Q.
: level exceedances occurrences
1 8 8
I 2 50 42
level I 3 250 200
I 4 2000 1750
I 5 10000 8000
I
6 a
I
I
I
I
I 5
I
b. I
I
Ilevel exceedances occurrences
I 1 8 8
level
I 2 50 42
I 43 250 200
2000 1750
I 5 20000 18000
I 6 a
I
4~----------~~
I
5r---------------~~~= I
I

c.

Figure 6.24. Truncation. (a) Complete spectrum; (b) Improper truncation (omission oflower level);
(c) Proper truncation at level 5.

Clipping of the spectrum should be a factor for serious consideration in crack

growth analysis when retardation is taken into account. It is easy to produce
optimistic crack growth curves if high enough stresses are included (Figures 6.22
and 6.23), but the objective of the analysis is to obtain realistic information.
Engineering judgement is the only guideline for the selection of an appropriate
clipping level. A reasonable level might be the one that is exceeded ten times.
Although this level is often selected, it is too arbitrary. Depending upon the
clipping level, and upon the retardation properties of the material and the
spectrum shape, the life may vary by a factor of two or three as shown in Figures
6.22 and 6.23) so that a categorical selection of ten exceedances would still be
giving disputable results for some spectra and some materials.
 195

The best solution to the problem is to perform multiple computer runs. A

well-designed computer code will have options for automatic clipping. Once the
preliminary work is done, multiple runs do not require any further labor other
than by the computer. Multiple runs using different clipping levels will establish
the sensitivity to clipping in any particular case. If the effect is small, no special
problems arise, but if it is large the upper and lower bound of the crack growth
curve can be established, as well as an average.

6.7. Truncation
It takes as much computer time (and testing time) to deal with one small cycle
as with one large cycle. Thus the small cycles at the lower end of the exceedance
diagram consume most of the time (cost) while their effect may be very small.
Therefore, it would be advantageous if their number could be reduced. This is
called truncation.
In essence the spectrum approximation by discrete levels already causes
truncation of lower stressess. Figure 6.24a shows truncation from 100000
exceedances to 80000 due to the selection of the lower level. Further truncation
could be achieved by raising the lower level to level 5. This is sometimes
understood to mean elimination of all levels 6 as shown in Figure 6.24b. In that
case 70000 cycles would be simply thrown out without any account of their
effect. This is an improper procedure.
True truncation involves reconstruction of the lower step as shown in Figure
6.24c. This is in accordance with the stepwise approximation of the idagram
which is known to be legitimate. In this way the 70000 cycles of level 6 are
replaced by an additional 10 000 cycles of level 5, and as such they are still
accounted for in a manner consistent with the entire procedure. Yet there is a
savings of 60000 cycles. The total cycles are reduced from 80000 to 20 000 and
the computation time is reduced accordingly by 75%. A proper truncation
procedure (not elimination) must be included in software.
Truncation requires judgement and it is recommended that the effect of
truncation on analysis results be evaluated by making different computer runs
to determine whether truncation is permissible (giving the same results as the full
history). The larger the number of stress levels the less the effect of truncation
on crack growth. Hence, truncation is better justifiable if the user selects e.g.
12-16 levels instead of the smallest possible number of levels. Figure 6.25 and
Table 6.3 show that proper truncation is indeed permissible (conservative),
improper truncation leading to unconservative results. (Note labels in Fig. 6.25
are reversed)
If truncation is understood to mean the elimination of small cycles. Then
concerns about the effect of truncation are legitimate, and a sensitivity analysis
should be performed. On the other hand, if done properly, true truncation is
196

Table 6.3. Stress levels in case of truncation (compare Figure 6.24. Results potted in Figure 6.25).
Unit of stress is ksi
Case I; no truncation; 10 levels
Exceedances Occurr. Min. str. Max. str.
cycles cycles
I I -5.000 25.000
4 3 -3.500 23.500
10 6 -2.000 22.000
25 15 -0.500 20.500
63 38 1.000 19.000
158 95 2.500 17.500
398 240 4.000 16.000
999 601 5.500 14.500
2511 1512 7.000 13.000
6309 3798 8.500 11.500

Case 2; proper truncation; 9 levels

Exceedances Occurr. Min. str. Max. str.
cycles cycles
I I -5.000 25.000
4 3 -3.500 23.500
10 6 -2.000 22.000
25 15 -0.500 20.500
63 38 1.000 19.000
158 95 2.500 17.500
398 240 4.000 16.000
999 601 5.500 14.500
3980 2981 7.000 13.000

Case 3; improper truncation; 9 levels

Occurr. Min. str. Max. str.
cycles /Ioad /Ioad
1 -5.000 25.000
3 -3.500 23.500
6 -2.000 22.000
15 -0.500 20.500
38 1.000 19.000
95 2.500 17.500
240 4.000 16.000
601 5.500 14.500
1512 7.000 13.000
All cycles below 13 ksi omitted (compare case I).
 197

I I
Ui
w
:r
u
Z
3. Ei

:3. ~~
- - NO TRUNCATION,

- - - - PROPER "UNW10N,
10 LEVEeS,

2ND em
1ST CASE "BLE 6. 3

TAB,E 6.3
I
I
I I
II
:::;
'"-'
N
U;
,~ f - . - ._. ,"PROPER ,"UNCATION, 3RO CASE TABLE 6.3 I I
C5:
2 ..: I I
/ I
u
-<
Ct:
u 2.e
I I
1. tJ / /
/ /
1. 2
/ /
O. '1
/// ///

O. 4
...--'-- - -- ---- --
~.--.:::
2.5 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5
Ll FE (1800 FLl GHTS)

Figure 6.25. Effect of proper and improper truncation (unconservative) on piedicted crack growth.
Warning: labels for proper and improper truncation are inversed in above figure.
perfectly legitimate as can be seen from Figure 6.25. A well designed computer
code includes provisions for proper truncation, the truncation being performed
automatically upon user specification of the truncation level. The issue is then
of little importance. If the computer code understands truncation to be the
elimination of the lower levels multiple runs will always be necessary to prove
that this 'improper' truncation is justifiable. Moreover, multiple runs must then
be made in every case, because the effect of the 'improper' truncation depends
upon spectrum shape, material and geometry.

stress
A is for 100 years
n is for 10 years (120 months)
45 C is for 1 year (52 weeks )
CLIP 40 ,,-:---...~"'I....

10 8 10 9
exceedances

-10 -1

r
~
-13.34""- _ -
-16.67 ~

Figure 6.26. Manipulation of Spectrum.

198
Table 6.4. Stress history for 100 years

Max. stress Min. stress Steady stress Exceedances

50.00 -16.67 10.00 I
42.00 -11.34 10.00 40
34.00 -6.00 10.00 1535
26.00 -0.67 10.00 63096
18.00 4.67 10.00 2511876
10.00 10.00 10.00 100000000
I unit of stress is 1.00 ksi.

History for 100 years 1 unit of stress = l.00 ksi

Max. str. Min. str. Occurr Years Years Years Years Years
79 16 3 1 I

40.0 -16.7 I 0 0 0 0
40.0 -14.4 9 0 0 1 2 4
40.0 -12.2 36 0 I 3 4 7
40.0 -10.0 169 I 4 4 5 8
36.7 -7.8 785 7 10 13 12 20
33.3 -5.6 3642 35 40 46 38 60
30.0 -3.3 16903 164 181 204 170 268
26.7 -I.l 78455 764 830 939 780 1222
23.3 I.l 364156 3550 3840 4341 3604 5639
20.0 3.3 1690264 16480 17818 20134 16712 26141
16.7 5.6 7845510 76493 82705 93460 77572 121331
13.3 7.8 36415624 355052 383875 433780 360038 563138

6.S. Manipulation of stress history

The stress history generated in the manner discussed in Section 6.5 may become
a little awkward if the number of exceedances is very small or the number of
flights, time, voyages etc., covered by the diagram is very large. The best results
are obtained when the total exceedances are on the order of 2 000 to 100000, and
the number of periods (voyages, months, etc.) on the order 50-1000. Therefore
it may be advantageous to reduce exceedance diagrams for smaller or larger
numbers to the above ranges.
For the following discussion all exceedance diagrams are presumed to start at
one exceedance. If this requires extrapolation to undesireable higher stresses, the
spectrum should be clipped at the desired level. Examples of how the exceedance
diagram can be manipulated to obtain a realistic stress history follow below.
Let the exceedance diagram A in Figure 6.26 be for 100 years of usage (1200
months or 5200 weeks). The diagram starts at 100 exceedances and ends at
100000000 exceedances. Extrapolation to one exceedance brings the highest
stress level to 50, which cannot be justified. The problem can be rectified by
 199

Table 6.5. Stress history 1200 months

Exceedance table for 1200 months
Max. stress Min. stress Steady stress Exceedances
50.00 -16.67 10.00
42.00 -11.34 10.00 40
34.00 -6.00 10.00 1585
26.00 -0.67 10.00 63096
18.00 4.67 10.00 2511876
10.00 10.00 10.00 100000000
1 unit of stress is 1.00 ksi

History for 1200 months 1 unit of stress = 1.00ksi

Max. str. Min. str. Occurr Months Months Months Months Months
959 192 39 9 1
40.0 -16.7 1 0 0 0 0
40.0 -14.4 9 0 0 0 0 9
40.0 -12.2 36 0 0 0 2 18
40.0 -10.0 169 0 0 3 3 24
36.7 -- 7.8 785 0 3 4 4 16
33.3 --5.6 3642 2 6 11 11 43
30.0 -- 3.3 16903 \3 17 22 24 97
26.7 --1.\ 78455 63 72 82 79 305
23.3 1.\ 364156 295 328 356 342 1313
20.0 3.3 1690264 1373 1509 1634 1567 5999
16.7 5.6 7845510 6374 7003 7571 7253 27722
13.3 7.8 36415624 29587 32500 35132 33660 128603

clipping the spectrum at 40. Using a stress history with different periods of years
(i.e. years with different cycle content) leads to the results in Table 6.4. The
period 'year' may be too long, and it may be more realistic to construct 1200
different months, or even 5200 different weeks. This would result in stress
histories as shown in Tables 6.5 and 6.6. Obviously, the latter two stress histories
are not realistic, because too many cycles are concentrated in one period (month
or week). A much better result can be obtained by using a spectrum for 10 years
(i.e. 120 months), which is spectrum B in Figure 6.26. Note that B and A are still
the same spectrum; a factor of ten in exceedances is made up for by a ten times
smaller time interval. As can be seen from Table 6.7 a more realistic stress
history is obtained. If weeks are used as a period, a spectrum for one year (52
weeks) can be used, which is spectrum C in Figure 6.26. The resulting stress
history is shown in Table 6.8. Note that the highest stress is now 40 without
clipping.
It is advisable to reduce the spectrum (or stress table) to one that covers
50-1000 periods in the above manner, in order to arrive at a realistic stress
history. All of the above may seem a little artificial, but it should be realized that
200
Table 6.6. Stress history for 5200 weeks
Exceedance table for 5200 weeks
Max. stress Min. stress Steady stress Exceedances
50.00 -16.67 10.00
42.00 -11.34 10.00 40
34.00 -6.00 10.00 1585
26.00 -0.67 10.00 63096
18.00 4.67 10.00 2511876
10.00 10.00 10.00 100000000
I unit of stress is 1.00 ksi
History for 5200 weeks I unit of stress = 1.00 ksi.
Max. str. Min. str. Occurr Weeks Weeks Weeks Weeks Weeks
4159 832 167 41 I
40.0 -16.7 0 0 0 0 1
40.0 -14.4 9 0 0 0 0 9
40.0 -12.2 36 0 0 0 0 36
40.0 -10.0 169 0 0 0 3 45
36.7 -7.8 785 0 0 3 5 78
33.3 -5.6 3642 0 3 5 5 105
30.0 -3.3 16903 3 4 5 4 98
26.7 -1.1 78455 14 18 24 23 302
23.3 l.l 364156 68 76 84 75 1009
20.0 3.3 1690264 316 352 386 347 4466
16.7 5.6 7845510 1471 1617 1774 1597 20542
13.3 7.8 36415624 6827 7511 8231 7398 95084

the purpose of crack growth analysis is to make a projection in the future, so

as to exercise fracture control (Chapter 11). As the future is not known in detail,
any reasonable projection that accounts for the salient features of the load
history is as good or better than any other. Furthermore, such seemingly trivial
things as clipping may have a greater influence on the predictions than all other
painstaking efforts for accuracy and realism. Clipping levels are often selected
on the basis of a mere assumption, yet unduly 'sophisticated' (read time
consuming and complex) procedures are often used to generate e.g. flight-by-
flight stress histories. All of the latter's 'sophistication' can be overridden by one
'simple' assumption with regard to clipping. More often than not, the 'sophis-
ticated' stress history is than applied randomly with total disregard for the fact
that high loads are clustered as discussed extensively in this chapter.
It should be clear from the foregoing discussions that there are only a few
issues that count in the generation of a stress history. These are:
(a) Different periods of severity (flights, voyages, etc.) must be used and
applied semi-randomly if in practice the loading is semi-random. Random
 201

Table 6.7. Stress history for 120 months

Exceedance table for 120 months
Max. stress Min. stress Steady stress Exceedances
45.00 -13.34 10.00 1
38.00 -8.67 10.00 25
31.00 -4.00 10.00 631
24.00 0.66 10.00 15849
17.00 5.33 10.00 398106
10.00 10.00 10.00 10000000
1 unit of stress is 1.00 ksi

History for 120 months 1 unit of stress = 1.00 ksi

Max. str. Min. str. Occurr Months Months Months Months Months
95 19 4 1 1
40.0 -13.3 0 0 0 0
40.0 -11.4 7 0 0 0 2 4
39.2 -9.5 21 0 0 3 3 6
36.2 -7.5 81 0 3 3 4 8
33.3 -5.6 312 2 4 7 6 11
30.4 -3.6 1194 9 13 14 13 22
27.5 -1.7 4574 37 41 43 42 66
24.6 0.3 17524 142 157 163 155 243
21.7 2.2 67138 545 599 621 583 1914
18.7 4.2 257216 2089 2291 2376 2233 3495
15.8 6.1 985446 8006 8770 9086 8541 13361
12.9 8.1 3775436 30675 33591 34800 32713 51169

application of stresses derived by complicated means will negate all complicated

efforts to determine stress histories.
(b) Deterministic loads must be applied at the point where they occur. For
example g.a.g. cycles must occur between flights; random application may defy
all other sophisticated procedures.
(c) A reasonable number of stress levels (10-12 positive and negative) must
be selected. More levels will complicate the procedure without improving the
usefulness of the results.
(d) Largest positive and negative excursions must be combined, and so on.
Random combinations will require subsequent counting, the result of which can
be foreseen, while the stress history was based on an already counted history in
the first place.
(e) The total number of periods and cycles must be in accordance with the
exceedance diagram.
The above criteria account for what may be called the signature of the
loading. Small changes in these, including clipping, will usually have more effect
202
Table 6.8. Stress history for 52 weeks
Exceedance table for 52 weeks

Max. stress Min. stress Steady stress Exceedances

40.00 -10.00 10.00
34.00 -6.00 10.00 16
28.00 -2.00 10.00 251
22.00 2.00 10.00 3981
16.00 6.00 10.00 63095
10.00 10.00 10.00 1000000
1 unit of stress is 1.00 ksi

History for 52 weeks 1 unit of stress = 1.00 ksi

Max. str. Min. str. Occurr Weeks Weeks Weeks Weeks
41 8 2 1
40.0 -10.00 0 0 0 1
37.5 -8.3 5 0 0 2
35.0 -6.7 12 0 1 2
32.5 -5.0 38 0 3 3 8
30.0 -3.3 122 2 3 3 9
27.5 -1.7 385 7 8 8 17
25.0 0 1216 22 27 25 47
22.5 1.7 3845 71 80 76 142
20.0 3.3 12159 225 254 234 434
17.5 5.0 38451 714 795 732 1353
15.0 6.7 121593 2258 2514 2314 4275
12.5 8.3 384511 7140 7953 7318 13 511

on crack growth than any complex means of establishing stress levels.

Hypothesize for a moment that it would be necessary to use a complicated
procedure to establish e.g. the loads or stresses in every segment of a flight
(ascend, cruise, descend etc.) or of a voyage of a ship (river, port, locks, cruise,
etc.). Any such procedure would be full of assumptions, and it needs no
explanation that no airplane or ship would ever encounter the assumed circum-
stances. If it were necessary to go through this 'sophisticated' procedure, it
would only be because the predicted crack growth would be different if other
assumptions were made. But in that case the prediced crack growth would have
no relevance to the service behavior in the first place. Actual service loading
would be different from the assumptions made, and if the results were that
critical to the assumptions the predictions would be useless, regardless of all
sophisticated procedures used. The simple fact that one would consider the
result usefull nevertheless, implies that one makes the assumption that it does
not really matter whether the loads are as derived from this sophisticated
procedure. If so, why would the 'sophisticated' procedure be necessary. Sur-
 203

prizingly, if such complicated procedures are used, the stresses often are applied
in random sequence, while they should be semi-random, and clipping is ignored.
It is very easy to perform many computer runs with different stress levels,
simple and sophisticated stress histories, etc. If the results do come out about
the same (which they do), the above point is proven. One might ask whether
tests would prove the same. Not enough test data are available to conclude that
they do. But, after all, crack growth prediction are made by analysis, and hence,
only the analysis results count. If the computer cannot distinguish the difference,
the results apply.
Next hypothesize that there would be a difference, depending upon the
sophistication of the stress history generation. Then of course, also the behavior
in service would be different. In other words: no matter how sophisticated the
stress history generation, it would have no bearing upon the service behavior,
and the effort a waste.
The above discussion shows that procedures to derive stress histories in a
complicated way are defying themselves. If they were necessary, they would be
useless by implication. If they are not necessary, they are useless from the start.
It must be concluded that only those effects count that could make a signifi-
cant difference in the calculated crack growth. Those are the effects listed above
concerning the signature of the loading and clipping. They have to be accounted
for as was demonstrated by test and analysis results. All others are secondary;
they complicate procedures without adding to their usefulness.
A final problem in the generation of the stress history is in the definition of
stress. Should one use nominal stress, local stress, or hot spot (welded
structures) stress? Should secondary stresses, residual stresses and dead stresses
be accounted for, and if yes, how? To a certain degree, this can be dealt with
through the baseline data used in the analysis. However, there is more here than
meets the eye, the main problem being that local plasticity during the highest
load cycles tends to make even the most sophisticated load-interaction models
go awry. This problem can be solved only by pragmatic engineering judgement.

6.9. Environmental effects

If there is load- or load-environment interaction it is especially important that
clustering of high loads in storms is recognized by comparing analyses for fully
random and clustered loading (mild weather or storms). Computer runs with
similar clustering, but with a totally different sequence, should provide essenti-
ally the same result.
For marine structures especially, the effect of environment should receive
ample consideration. Quite obviously, the main environments to consider are
'salt air', sea water, and the 'splash zone'. The question arises: if going from one
cycle to the next (e.g. high AK -low AK or vice versa) in successive cycles, does
204
the new rate immediately fall in place with the (constant amplitude) baseline
data. (In other words: is there no environment interaction).
Models for environment interaction have been proposed. As the environ-
mental effect is time dependent, these models bring in the element of time, or the
frequency of the loading. They are certainly of scientific interst, but the practical
problem is one of load (retardation) and environment interaction. As the latter
problem has not been addressed at all, the complication of accounting for
environment equilibrium is hardly worthwhile for practical crack growth
analysis. Thus, the models mentioned above remain in the realm of research and
'theory' (hypothesis) and are not too relevant to practical applications as
discussed in this book at this time. A pragmatic approach calls for the
assumption that chemical equilibrium is indeed established immediately. In that
case, the problem is solved by using crack growth rate data (da/dN-flK) for the
relevant environment. Retardation effects are accounted for as discussed. Re-
tardation parameters must be determined empirically, and therefore, these will
automatically account for any chemical interaction effects, included in the test
data used for calibration.

6.10. Standard spectra

So-called 'standard' spectra have been developed in Europe for general use for
a variety of structures. The word standard is to be interpreted to mean a general
norm based upon a very great number of measurements; it is not meant to be
a design standard or specification. These standard spectra were intended
primarily for use in tests, so that results of various experimental investigations
might be better comparible. There are standard spectra [12-19] for airplanes
(Falstaff and Twist), for helicopters (Helix and Felix), for offshore structures
(Wash), etc.
Stress histories for these spectra were obtained by using algorithms very
similar to those discussed in the previous sections using semi random loading
with periods of different severity. In essence, not the exceedance diagrams but
the stress histories derived in this manner are considered to be the standard. A
limitation of the standard spectra is that they essentially always perform in the
same way; this is useful for testing and data comparisons, but may be too severe
a limitation for practical use.
Using the standard stress histories in a crack growth analysis, would necess-
itate input of all stress cycles in the sequence generated, and thus require special
input facilities. Fortunately, the algorithm as discussed here (which is somewhat
simpler than the one used for the standard histories), can be rather easily
incorporated in a computer code for crack growth analysis. In that case only a
few data points describing the exceedance diagram would be needed as input,
upon which the analysis code would generate a load history, perform clipping
 205

and truncation as prescribed, and subsequently the crack growth calculation. In

this manner, the standard spectra would be more useful and more easy to use.
Their exceedance diagrams could be applied for crack growth analysis if a
specific exceedance diagram were lacking. Most stresses are proportional to
loads, so that the stress levels in different parts of the structure can be obtained
through a multiplication factor. The stress levels are given as relative numbers;
a conversion factor is sufficient to determine all stress levels. Hence, judiciously
used, the standard exceedance diagrams may provide spectrum information for
damage tolerance analysis where none is available. Suitable crack growth
analysis software only needs input of the exceedance diagram.

6.11. Exercises
I. Approximate the exceedance diagram of Figure 6.27 by 8 equally spaced
positive and 8 negative stress levels for equal exceedances as the positive
levels. Determine the number of occurrences of each level. Assuming that the
100% level represents a stress of 15 ksi, determine the stress ranges for each
level by combining positive and negative excursions of equal frequency of
occurrence.

Relative
Stress 1.2

300 DAYS

.8

.6

.4

.2

0
10 10' 106 Exceedances

_.2 4.5X10 5

_.4

_.6

Figure 6.27. Exercises.

206
2. Repeat exercise I by constructing 6 levels starting with the exceedances: 2, 10,
100, 1000, 10 000, 100000. If the stresses of exercise I and 2 were used in a
crack growth an81ysis, would there be a difference in results?
3. Using the results of Exercise I, generate a stress history with five 'periods' of
different severity.
4. Truncate the spectrum of Exercise I properly at the lowest level but one.
Generate a new stress history with five different 'periods'. Next use the
improper truncation procedure discussed in the text and compare the results
of the two procedures (Note that for the latter part the results of Exercise I
can be used directly; only for the first part a new history must be generated.)
5. How would clipping to the highest level but one change the stress histories of
exercise 4?
6: Repeat Exercises I, 3,4,5 using the spectrum of Figure 6.27 and by selecting
7 levels.
7. Change the stress history developed in Exercise 3 for the case that the
maximum level represents a stress of 21.5 ksi.
8. Why is clipping not important if there is no retardation?

References
[I] M.E. Mayfield et aI., Cold leg integrety evaluation, USNRC Report NUREGjCR-1319,
February 1980.
[2] International group on Crack Growth in Nuclear, Structures (lCCGR).
[3] J. Schijve, The analysis of random load-time histories with relation to fatigue tests an life
calculations, Fatigue of Aircraft Structures, p. 115, Pergamon (1963).
[4] J.B. de Jonge, The monitoring offatigue loads, ICAS congress, Rome (1970), Paper 70-3\.
[5] G.M. van Dijk, Statistical load data processing, ICAF symposium Miami (1971).
[6] D. Broek et aI., Fatigure strength of great-lakes ships, Battelle Rept to Am. Bur. Shipping
(1979).
[7] Private communication.
[8] J. Schijve, Cummulative damage problems in aircraft structures and materials, The aeron. J.
74 (1970) pp. 517-532.
[9] D. Broek and S.H. Smith, Fatigue crack growth prediction under aircraft spectrum loading,
Eng. Fract. Mech. 11 (1979) pp. 122-142.
[10] D. Broek and R.C. Rice, Prediction of fatigue crack growth in railroad rails, SAMPE Nat.
Symposia, 9 (1977) pp. 392-408.
[II] N.!. Bullen, The chance of a rough flight, Royal Aircraft Est. TR 65039 (1965).
[12] G.M. van Dijk and J.B. de Jonge, Introduction to afighter aircraft loading standardfor fatigue
evaluation, FALSTAFF, NLR-Report MP 75017 (1975).
[13] J.B. de Jonge, D. Schuetz, H. Nowack, and J. Schijve, A standard load sequence for flight
simulation testing, NLR- Report TR 73029, or LBF-Report FB - 106, or RAE-Report TR
73183 (1973).
[14] J.J. Gerharz, Standardized environmental fatigue sequence for the evaluation of composite
components in conbat aircraft (ENSTAFF), Lab. fiir Betriebsfestigkeit, Fraunhofer Inst. fiir
Betriebsfestigkeit FB-179 (1987).
 207

components in conbat aircraft (ENSTAFF), Lab. fiir Betriebsfestigkeit, Fraunhofer Inst. fiir
Betriebsfestigkeit FB-179 (1987).
[IS] G.E. Breithopf, Basic approach in the development of TURBISTAN, a loading standard for
fighter aircraft engine disks, ASTM conference Cincinati OH (1987).
[16] M. Huck and W. Schutz, A standard load sequence of Gaussian type recommended for general
application infatigue testing, Lab. fiir Betriebsfestigisellschaft IABG rept TF-570 (1976).
[17] J. Darts and W. Schutz, Helicopter fatigue life assessment, AGARD-CP-297, (1981), pp. 16.1
- 16.38.
[18] P.R. Edwards and J. Darts, Standardizedfatigue loading sequence for helicopter rotors (HELIX
and FELIX), Royal Aircraft Establishment RAE TR 84084, Part I and 2, Augutst 1984.
[19] W. Schutz Standardized stress-time histories-An overview. ASTM conference, Cincinnati OH
(1987).
 CHAPTER 7

Data interpretation and use

7.1. Scope

Material data are an essential input to all fracture and crack growth analysis;
without applicable data analysis is not possible. Misinterpretation and misuse
of data are major contributors to the acclaimed shortcomings of fracture
mechanics, because the data interpretation problem is not a trivial one,
especially where it concerns fatigue crack growth. The phrase: 'data are data,
and cannot be argued with', is commonly misapplied. The statement may be
true for the raw data, i.e. a load-COD curve, or a measured crack growth curve,
but cannot be applied to the derived data, Ke , K(C' J R , and da/dN - AK. The
latter are obtained from the raw data through an interpretation process full of
assumptions such as the data reduction procedures stipulated in the relevant
ASTM specifications [1-5]. Although these reduction procedures are probably
the best available, they are not indisputable.
This chapter is not intended to argue the shortcomings of data reduction
procedures. Rather, it is concerned with how these data are subsequently
interpreted and used in the analysis. Questions arise regarding constraint,
scatter, equation fitting, data errors and inaccuracies, retardation parameters,
mixed or changing environments, etc. These are the type of problems addressed
here, because they may affect the accuracy of the damage tolerance analysis,
more than the shortcomings of fracture mechanics.
Problems in the use of toughness data, both in terms of K and J, will be briefly
addressed first. As the toughness affects the calculated permissible crack size, ap '
inaccuracies change ap only. From the point of view of fracture control this is
often not very important as will be discussed in Chapter 11. A small change in
a" does not affect the life much, unless at' is already small - especially when
in rare cases it is below the detection limit, or where it affects arrest and leak-
before break (Chapter 9). Errors in rate data on the other hand, affect the life
directly. A factor of two difference in rates (which is not uncommon) changes
the life by a factor of two, which is significant. For this reason, the main

208
 209
emphasis in this chapter is on the interpretation and use (estimation sometimes)
of fatigue rate data.
In many cases the analysis must make use of data provided in handbooks [6]
or the general literature [7], or of data in magnetic files in software libraries.
These are almost always reduced data, as opposed to raw data: they have been
manipulated. Often some interpretation has been done as well, such as
averaging or curve fitting. The user of such data should not overlook this
'triviality', and if possible, check whether reduction and interpretation was
appropriate for the application envisaged. In practically all cases a decision
must be made on how to deal with scatter, in particular for rate data. Data for
the exact condition or alloy at hand (regarding temperature, environment,
material direction, R-ratios, etc.) often are not available. They then must be
estimated on the basis of available data for similar alloys or circumstances.
These kinds of problems are addressed in this chapter as well. Recipes cannot
be given; only guidelines can be provided. In the end the user must exercise
engineering judgement, possibly honed by the following discussions.
No attempt will be made to provide any material data here as this is not a
materials handbook. Considering that the DT-handbook [6] consists of
thousands of pages, any attempt to present data here would be inadequate and
selective. Real data will be used for illustrative purposes; in some cases hypo-
thetical data will be used to better demonstrate a specific point.

7.2. Plane strain fracture toughness

The plane strain fracture toughness, Kin is commonly obtained from a standard
test [I] on a compact tension (CT) specimen. Standardization can be defended
on many grounds (also the tensile test and the hardness test are standardized).
Does this mean that data obtained from non-standard tests are inadmissable?
If this question had to be answered affirmatively, the result of the test would be
of no use in the first place. It is assumed that the test result, Kin can be used for
the prediction of fracture in a structure, on the basis of the similitude argument
that fracture takes place at the same value of K as in the laboratory test. Hence,
if a specimen other than the standard could not be used, the implied assumption
would be that the other configuration does not fracture at K/c; ergo, it would be
impossible to predict fracture in a structure on the basis of the K/c from a
standard test, which would render the standard test useless as well.
The concept of fracture mechanics is that fracture in plane strain occurs when
K = K lc regardless of the configuration. Therefore, K lc is obtainable from a test
on any configuration, provided one knows the expression for K, or rather for
{3. Similarly, fracture can be predicted for any configuration for which the
expression for {3 is known. The main justification for the standard test is that {3
for the CT specimen has been calculated to great accuracy, for straight-front
210

through-the-thickness cracks. The standard declares a test invalid if the fatigue

crack front is not straight. This is because the fJ-expression provided is not valid
for such a crack, but fracture still took place at K = K le , and the test result
could still be used if fJ were known for the actual crack shape.
The test also is declared invalid when the thickness is less than 2.5 (KdFtyf
As was shown already in Chapter 3, the factor of 2.5 was determined by
committee agreement on the basis of test data, but it appears from Figure 3.8
that a factor of 2 or 4 would be just as defendable. Yet, the number of 2.5
appears in the standard, which does not make the number indisputable and
certainly not sanctimonious. The 'candidate toughness' obtained for a test
where the factor is 2.4 is not by definition worse than the one obtained for a test
where the factor is 2.7. As constraint depends upon the yield strength and K, the
factor in reality will depend upon Fry and K lc and upon configuration (i.e. it is
material dependent and configuration dependent).
It should be pointed out also, that the toughness obtained when the factor is
e.g. 2 may not be declared a valid plane strain toughness by the standard,
but it is the toughness for the thickness at hand. For the given thickness the
number is at least as reliable as Klc obtained in a valid test. Thus, if one is not
interested in K lc per se, but in the toughness for the chosen thickness, the result
is useful. If one insists on knowing the plane strain toughness the test must be
'valid', but one may want to exert caution with regard to the factor 2.5.
As pointed out the required factor for plane strain is likely to be material
dependent. Almost certainly, it is also geometry dependent. This can be appreci-
ated from the data [6] for large center cracked panels shown in Figure 7.1. The

TOUGHNESS 90
G
KSlffl
80 •
70
~
0

",,-.
-- a--.... ----.......
-
60 '" p-----....
50 ~~ ~ I:ls,
~ ~
~
0
40
~s
~ ::::::::-... "'I---- I---
~O
..::::::......
20

10

.1 .2 .3 .4 .5 .6 .7 .8 .9 1.0 1.1 12 1.3

THICKNESS (IN)

Figure 7.1. Toughness data [6] for 7075 Aluminum alloy (large center cracked panels).
 211

same reference [6] quotes K lc ~ 30 ksiJill for the same material, obtained from
a standard test. As the alloy's yield strength is FlY ~ 80 ksi, the standard would
predict plane strain at a thickness of 2.5 (30/80)2 = 0.35 inch. Yet, the center
cracked panels of 0.35 inch thickness have a toughness much higher than
30 ksiJill. Similar discrepancies are found for other materials. Constraint in CT
specimens is higher than in some other configurations. This is convenient
because it permits the use of small CT specimens and tends to lead to conser-
vative toughness values. But clearly, if the factor 2.5 is applied to the center
cracked panel, a toughness of only 30 ksiJill would be counted on for a plate
of 0.35 inch thickness, while the actual toughness is as high as 50-60 ksiJill. For
a given crack size this would be underestimating the residual strength by almost
50%, certainly not a negligible error.
Clearly, plane strain in center cracked panels has not been reached yet at a
thickness equal to 2.5 (KdF,yt At this thickness the toughness is higher
(transitional). Plane strain occurs at a thickness of 4-5 times (KIc/Flyt Besides,
there is another reason why the center-cracked panel behaves differently, namely
the rate of change of K(a) as explained below.
Every material, except a truly brittle one, exhibits a rising R-curve (Chapter
3), as shown in Figure 7.2. This is usually not a problem in a standard test on
a CT specimen, because G(a) and K(a) rise very sharply with crack size in this
specimen (G = K2/E; Chapter 3), so that the beginning of fracture usually
coincides with the instability. The standard prohibits excessive non-linearity of
the load displacement diagram in order to exclude cases in which instability is
not immediate [8]. In a center cracked panel G(a) or K(a) increase only
moderately with a. Hence, fracture is first stable, and instability occurs at higher
values of K, especially for larger cracks (Figure 7.2). Usually, the question of
interest is how much load a structure can carry and at which stress it breaks.
This is determined by the instability, and thus the K at instability is the relevant
number, called Kc or K.ff (see also chapter 3).
Which cases warrant the use of a valid KIc? It is obvious from the above
arguments that there is no categoric answer to this question. Engineering
G,R G,R

R __ l ___________ .s G
G
R

DIFFERENCE
IN Gc (Kelt)
G

,,
,;
6a 6a
..<a} (b)
Figure 7.2. Different behavior of compact tension specimen and center cracked panel. (a) CT
specimen; (b) Center crack.
212

judgement must be exercised unless conservatism is desirable. Cases where the

use of Klc is always indicated concern surface flaws and corner cracks. It is
emphasized that in these cases the thickness plays no role (Chapter 3). For the
analysis of part-through cracks refer to Chapter 9.
Although possibly superfluous, it is pointed out again that even in plane
strain, and regardless how low the toughness, the predicted residual strength for
small cracks is too high, the predicted permissible (or critical) crack size for high
stresses too large. For a -+ 0, the predicted stress tends to infinity. Using EPFM
in this regime (for an otherwise LEFM material) would not solve this problem
either. An approximate solution must be used. This is amply discussed in
Chapters 3, 4 and 10.
Finally, in using toughness data from handbooks, the relevant conditions
(temperature and direction) should apply. Like most other material properties,
toughness depends upon temperature. Engineers are used to the material
properties being temperature dependent, but the (often) strong dependence
upon direction of fracture and crack growth properties is sometimes not
recognized. The going nomenclature for crack direction and material direction
is shown in Figure 7.3. The first letter, L, Tor S indicates the direction of the
loading, the second letter (again L, T or S) the direction of the crack. Toughness
values for the ST-direction may be 30-60% lower than for the LT direction. Not
accounting for such effects may lead to errors. As the problem is most often
associated with the growth of part-through cracks, the reader is referred to the
more detailed discussion of the matter in Section 7.6.
Many times, toughness data for the temperature and direction of interest are
not available. Section 7.5 provides some guidelines for estimating the toughness
in such cases.

7.3. Plane stress and transitional toughness, R-curve

Data on plane-stress and transitional toughness should always be assessed with
extreme caution. There are many ways in which raw data can be erroneously
interpreted and derived data misquoted. There are equally many ways to misuse
them.
Many Kc data were derived from panels of insufficient size, which renders
them useless. As was shown in Chapter 3, when the toughness is high - which
it usually is in plan_e stress - small panels will fail by net section collapse, i.e. at
K < Kc. Thus the Kc derived from such a test is too low. It is sometimes quoted
as an apparent toughness, K app , but this practice is suggesting more than it
should. Fracture due to collapse should be treated differently as discussed in
Chapters 3, 4 and 10, and not on the basis of Kc nor Klc ' A test on a panel of
insufficient size yields no useful result other than Fcol , and the knowledge that
Kc is higher than the 'apparent Kapp'. A good data handbook [6] quotes the panel
 213

L = Long iludinal
T = Transverse
: "" S =Short transverse
L.__ _-
Spec Axis Crack
1 T L
2 T S
T 3 L T
~ --- 4 L S
5 S T
6 S L

Figure 7.3. Nomenclature for loading and Crack Directions.

size as well as the raw data so that the user can readily check whether collapse
or net section collapse (yield) occurred, and then make a judgement. If no panel
size is quoted the numbers are suspect.
For example if the toughness provided is 80 ksiFn for a 6-inch wide center
cracked panel of a material with FlY = 60 ksi, the number is useless. A 2-inch
crack (a = 1 inch) in such a panel would provide (ffr = 80jl.07.J1W = 42 ksi,
but failure by net section yield is at (ffc = (6 - 2)/6) x 60 = 40 ksi.
Obviously, the test panel failed at net section yield. The value of this to the user
is only that it is certain that Kc ~ 80 ksiFn and that all center cracked panels
smaller than 6-inch will fail by collapse.
Another problem arises because of the definition of toughness. The value
quoted may be the actual Kc or the effective toughness, Kcff ' which is sometimes
called K,. As explained in Section 3.12 neither of these is actually a material
property; both depend upon crack size and configuration, because they are
derived for fracture instability. In principle, the actual K, is not a very useful
number: if the amount of stable fracture is unknown, the 'crack' size itself is
unknown; the residual strength cannot be calculated. Conversely the calculated
'critical' crack size is always too large (Chapter 3). If some information on the
amount of stable fracture is available Keff can be estimated from Kc. Without
such information, a safe assumption is that Keff = 0.9-0.8 Kc. The use of Keff is
not without problems either, but it is the most rational solution short of using
the R-curve. The criteria for panel size apply.
Using the R-curve and an instability analysis may seem to solve the difficulties
with K cff , but in practice it does not. First of all, R-curve data are scarce. More
important, accurate measurement of R-curves are difficult. As a consequence,
the inaccuracy of the data usually negates the advantages of the more refined
procedure. Furthermore, collapse considerations apply to R-curves as well.
Like the plane strain toughness, K, and Keff depend upon temperature. Direc-
tional dependence is usually small (TL versus LT), because plane stress and
transitional toughness always apply to through-the-thickness cracks and these
214
do seldom, if ever, occur in ST direction. The problem of estimating the
toughness where no data are available, is discussed in Section 7.5.

7.4. Toughness in terms of I and I

Toughness data for use with EPFM are scarce. Available data pertain mostly
to materials used in nuclear power structures such as 304 SS, A-533B etc., many
of which can be found in a series of reports issued by the U.S. Nuclear
Regulatory Commission (NRC - NUREGs). Data for other materials are
scattered throughout the literature. No systematic data compilation is as yet
available.
Practically all data have been obtained from CT specimens and in almost all
cases through the J-estimation procedure prescribed in the test standard [2]. The
latter is based on the assumption of full net section yield, which by itself is
subject to some doubt (Chapter 4). Although many estimation schemes for J
have been devised, the only viable general procedure for the application of
EPFM in engineering fracture analysis is through the use of J = H(Jn+l a /F, or
equivalent, as discussed in Chapter 4 (excluding finite elements for routine
fracture assessments). Since H is known for CT specimens, it would be better to
obtain Jle and JR data through the use of the above equation rather than from
a doubtful approximation, especially if the data are to be used in this manner.
Fortunately EPFM equations are so forgiving that the discrepancies seldom are
of great relevance in the application (see Chapter 4, Figure 4.11). This has the
additional advantage that the user need not seriously worry if the data appear
to show large scatter; the scatter is logical and rather immaterial in the fracture
analysis (Chapter 4).
Apart from scarcity there remain two significant problems with the toughness
data. The first and most important is the constraint. All data are obtained from
through the thickness cracks in relatively thin specimens. Often constraint is not
made an issue, but in most tests there will not be plane strain. This does not
matter if the data are to be used for through-the-thickness cracks in the same
thickness. But most cracks are surface flaws or corner cracks where plane strain
prevails. In such cases the toughness lie and IR will be less than for the test
specimens. One can only speculate on the magnitude of the difference, but it
would be reasonable to expect similar differences as in LEFM. Constraint
criteria for plane strain have been proposed [9], but (in comparison with the
LEFM plane strain conditions) these seem to be too optimistic (favoring plane
strain in small thickness, as discussed in Chapter 4).
Another problem is that the J R data are obtained from small specimens and
hence, usually are only for small values of Aa (generally for Aa up to 0.2 inches
at best). In large structures the amount of stable fracture will be considerably
more than 0.2 inches (Figure 7.2 applies just as well to the JR-curve). Analysis
 215

then can be performed only if the J R curve is extrapolated. In view of the fact
that the analysis is very forgiving, such extrapolations can be made rather easily
without sacrificing much accuracy. Nevertheless, this is an unfortunate
situation which should be corrected through a thorough revision of the
standard test procedure with regard to specimen size requirements.

7.5. Estimates of toughness

As shown in the foregoing sections, estimates of toughness are often necessary;

even if data are available they may not be for exactly the circumstances
prevailing in the structure. Considerations upon which such estimates might be
based will be given, but the actual estimate remains the responsibility of the user.
If no data at all are available, the problem of estimating a reasonable toughness
value, JR or R-curve becomes more difficult; the success of the estimate depends
upon the availability of comparable data. The procedures discussed below lead
to ROUGH estimates; they are not recommended for general use. When there
is a lack of data, tests are always preferable.
Consider a structure built of alloy Y, and suppose that a handbook provides
data for alloy X for LT and TL direction, and for alloy Y for LT, while the value
for the TL direction is sought. If it can be ascertained that alloy Y is very similar
to X and that yield strength, tensile strength, grain size and production process
are similar, the estimate might be based on:

Toughness (YTL ) = Toughness (XTd x Toughness (YLT ) /Toughness (XLT )

following common rules for estimates based on reference values.

If no data at all are available for alloy Y, the estimate is more precarious.
Usually, if the alloys are similar
Toughness (Y) = F ,yx x Toughness (X) / F ,yy

should provide a safe (conservative) estimate, but only if F ,yy > F ,yx , This
should not be used if F',yy < F',yx. In that case it would be safer to use the same
toughness as for X. For dissimilar alloys such estimates should not be attempted
at all.
Estimating the effect of temperature is more dangerous because temperature
affects the yield strength. One may try:
Toughness (Xn) F,y YT1
Toughness (Yn ) = T h (X) x F YT x Toughness (Yn )
oug ness n Iy 2

but it is advisable to base the estimate on more extensive comparisons of

available data.
Data handbooks (e.g. [6]) provide an abundance of toughness data for high
216

strength alloys. But even for those materials estimates are often necessary;
because of the enormous number of parameters not all cases are covered. The
situation is worse for the most widely used materials: common structural steels.
On the other hand for those materials one often knows some Charpy data. The
lower shelf Charpy-value is essentially a fracture energy and can thus be
expected to correlate with K!c. The Charpy test measures the total fracture
energy of the specimen, which is essentially the integral of R(a) over the
ligament. If the R-curve would be horizontal the value of this integral divided
by the ligament would indeed be Rand K = JER.
For the low toughness at the
lower shelf the R-curve will be nearly horizontal, so that a correlation between
K 1c and Charpy energy is indeed likely. However, there are some essential
differences between a Charpy test and a toughness test. The most important of
these are the difference in loading rate (affecting the yield strength) and the
difference in notch acuity (affecting the state of stress at the notch root, and as
such the stress at yield).
From empirical comparisons [10, 11, 12], it appears that
K/( 12 jCv for K in ksi Fn and Cv in ftIbs;
K/( 11.4 jCv for K in MPa rm and Cv in Joules.
A conservative lower bound [11] is claimed to be:
K 22.5 (Cv) 0.17 for ksi Fn and ftlbs;
K }1.6 (Cv) 0.17 for MPa rm and Joules.
Toughness values so obtained would be for the same high loading rates (impact)
as prevalent in the Charpy test.
The toughness for slower loading rates may be obtained from the same
equations, while accounting for a transition temperature shift given by:
AT 215 - 1.5 Fly for OF and ksi and 36 < FI , < 140 ksi;
AT 0 if FlY> 140 ksi;
AT 119 - 0.12 FlY for °C and MPa and 250 < FlY < 965 MPa;
AT 0 if FI\ > 965 MPa.
Note that the slower loading will cause the transition temperature to be lower
(lower F;J, so that the estimated toughness values will be useful - if the loading
rate is low - even at these lower temperatures T - AT. Other empirical correla-
tions have been derived [12].
If for example the Charpy value is 20 ftlbs, the yield strength F;, = 60 ksi at
a temperature of 65F, the toughness for high loading rates would be estimated
as Klc = 12,j20 = 54 ksiFn. This would be a safe value to use, because the
 217

toughness would be higher for slower loading. One could probably use this
toughness value for temperature as 65 - (215 - 1.5 x 60) = - 60F (see
above equations for AT).
Empirical correlations to estimate the toughness at the upper shelf from
Charpy data have been developed as well [10]. The generality of these is less
certain, because the upper shelf Charpy energy is not directly related to
toughness. If the R-curve rises steeply - which it does at the upper shelf - the
integral of R(a) over the ligament is not universally relatable to R. Besides, much
of the Charpy energy on the upper shelf is energy used for general specimen
deformation rather than for fracture: in the extreme case Charpy specimens do
not fracture at all, but are simply folded double, so that only deformation
energy is measured. Nevertheless, in some cases the empirical correlations may
be the only way to arrive at an estimate.
Toughness estimates can also be based upon the results of COD tests. Cor-
relations between CTOD and toughness were discussed in Chapter 4.
Estimating K", or the effect of thickness is usually somewhat easier. Assuming
that true plane stress occurs when the thickness is equal to the plastic zone, plane
stress will develop during loading when B = (Klc /F;y)2/2n. Note that Klc is used
in this equation, because at very low stress (low K) there will be plane strain, but
plane stress must have developed when K = KIn otherwise K cannot be
increased to the full plane stress toughness Kc. Further assume that plane strain
is reached at a thickness of B = 2.5 (KIc!F;yf These two points can be identified
in the diagram of toughness versus thickness as shown in Figure 7.4. A straight
line approximation between the full plane stress and full plane strain value is
usually a conservative one (compare Figure 7.1). The straight line can then be
used as the basis for the estimate, provided the plane strain toughness and the

Figure 7.4. Conservative estimate of transitional toughness on the basis of two data points
(compare with Figure 7.1.).
218

toughness for at least one other thickness are known. If only Klc is known, one
may estimate that K.tr for full plane stress is between 2 and 2.5 times Klc (Kc is
another 10-20% higher), and then follow the above procedure. (For examples
see solutions to Exercises.)
Actual and reliable data are preferable above estimates. In the lack of
handbook data a test on a specimen of sufficient size is always preferable.
However, most engineers, unlike researchers, do not have easy access to a
laboratory; even if they do, economic conditions and/or pressures for immediate
answers often preclude obtaining data for each and every alloy, condition, heat
treatment, temperature, material direction and thickness. Thus, estimates are
often necessary. Used judiciously, the above procedures should provide conser-
vative answers.

7.6. General remarks on fatigue rate data

Usually da/dN data obtained from handbooks or the literature need further
interpretation and so, naturally, do test data. A typical data set as might appear
in a handbook [6] is shown in Figure 7.5. The set shown is rather complete as
it covers several R-ratios. Not always are such complete sets available; in general
R = 0 is covered (or R = 0.05 or 0.1), but often this is the only R-value for
which data are available.
There may be a question about the effect of negative R. Some schools hold
that negative stresses hardly affect crack growth, while others maintain that data
clearly demonstrate a substantial effect. Both schools are essentially right, the
dichotomy being a matter of interpretation.
Consider Figure 7.6. A small negative stress may still have some effect until
the crack is fully closed. However, after complete closure there is no longer a
stress concentration: the compressive stress can be carried through and does not
have to by-pass the crack. In tension the crack forms a load-path interruption
which must be bypassed and which causes the high crack tip stresses, crack
growth and fracture in the first place. In compression the crack faces carry
through the load, as a loose pile of bricks can carry compressive loads. Thus
indeed the compressive part of the cycle has no effect (low elastic crack tip
stresses, as opposed to a very high stress concentration and yielding in tension),
and e.g. crack growth curves obtained for R = 0 and R = - 1 with the
loading as in Figure 7.7a, would be almost identical as shown in Figure 7.7b. But
the da/dN data can be represented in two ways.
By accounting for the above argument the negative part of the cycle could be
ignored and the stress ratio be defined as R = 0, with J1(J = J1(J] and
J1K] = /3J1(J1Fa. The da/dN data will then be as shown in Figure 7.7c (school
1), and there will no apparent effect of negative R. On the other hand, in a
'formal' interpretation, R = - 1, the stress range J1(J = J1(J" and J1K1 = /3J1(J1
 219

~ Stress Frequency, No. of No. of

I- Ratio, R ~ Specimens Data Points
I- 0 0.080 600 I 37
10., I--
0 0300 6.00
1=
l- t; 0.500 GOO
I
I
61
31
I-
.. 10' 2

~
.~

o
.
a:
4
.~ IcY
o
~
"-
g 10· • ~
U
~
.!1'

'I
o
u... Icr6

~J

10- 8
I 10 10C
Stress Intensity Factor Range, ilK, ksi-in~/2

7049·T7352 AI, 3.00 IN. FORGING, CT SPECIMENS, L-T DIRECTION

Environment 70 F, Low Humidity Air

Figure 7.5. Example of rate data in MCIC Handbook [6); Courtesy MCIC.

J1W would be employed. Since the rates are obviously the same ones as in the
previous case (same data as in Figure 7.7b) the same rate data are plotted at
twice as large a 11K, namely I1Kt = 211Kj , as in Figure 7.7c (school 2). Now
there is a considerable effect of negative R; nevertheless the crack growth curves
are identical; both rate diagrams in Figure 7.7c are based on the same data of
Figure 7.7b.
Both interpretations are tenable as long as each is interpreted in the same
manner in a crack growth analysis. In a loading case such as in Figure 7.7d
school 1 must take R = 0,110'3 and I1K3 = [3110'3 and use data set 1 to find J1W,
da/dN, but school 2 must take the larger 110'4 and I1K4 = [3110'4 with J1W
negative R, to find da/dN from data set 2. (Note that both will find the same
da/dN as they should). Consistent interpretation and usage will prevent errors.
220

stress distribu\ion stress distribution

in cracked sect on in cracked section

-t--: -

(a) (b)
Figure 7.6. Crack Tip Stresses in Tension and Compression. (a) Tension; (b) Compression.

a
R =0 R =-1

(a)

(b)
school 1
da/dN set 1 da/cfN

~a.!!!e

--
I
I
I
da/dN l _I_- _ _ ~a~N3

I tI da/dNl

tI
t
11K IIK3 11K IIK t 11K, IIK t
R=O R=-1 IIKIII R=Q R=-l

(e) (d)

Figure 7.7. Effect of Negative R. (a) Loading, (b) Test data (hypothetical); (c) Rate plots for two
schools; (d) Use of data in analysis.
 221

One should be aware however, of the 'school' built into computer software.
Some programs use the school 1 interpretation, others use school 2, some
provide the option for either one.
Fatigue crack growth properties depend upon direction T-L, L- T etc. (Figure
7.3). In particular in the case of surface flaws one must be cautious in the
selection of the appropriate data set, especially in forgings where due to the
different grain flow the identification of L, T and S direction is not always trivial.
Figure 7.8 shows a part produced by two different methods. Machining to
provide for the seat of the bolt head will expose grain endings in different ways.
The crack will select the weakest path along the long exposed grain boundaries
and tend to grow in the ST (or SL) direction, so that ST-data should be used
for the analysis. The analyst must be cognizant of the production procedure and
use the appropriate data. A crack assumed in the wrong direction (i.e. reversing
the directions in the forged and machined-out-of-plate parts), results in
erroneous predictions, because ST toughness values and rates are usually much
lower than those for LT or TL (cross grain growth).
It is crucial that the proper crack direction and data set are selected for
fracture and crack growth predictions. The analyst must be aware of how the
part is made in order to identify crack location and direction of growth.
Arbitrarily choosing the crack in the same direction in all cases (and using the
wrong data) will lead to predictions with no bearing upon practice. 'The
computer says so', is no excuse; neither the computer code nor fracture
mechanics can be blamed if wrong assumptions are made.
When the environment to be considered is different from air, the rates may

LINE:
TOP
-
i31
RAWPROD:TS

(b)~ (C) 11111

IMPROVEMENT,

(a)

CRACKING *
ALONG "EXPOSED"
GRAIN
* SURFACE
FOLLOWS
GRAIN FLOW
BOUNDARIES

·a
NO GRAIN BOUNDARIES
CUT OR EXPOSED
BOTTOM (CROSS - GRAIN CRACKI
LINE,
FINISHED PRODUCT WITH MACHINED BOLT SEAT.

Figure 7.8. Effect of Production Procedure on Crack direction. (a) Rolled plate; (b) Forging; (c)
Oversize Forging.
222
be substantially different and also the effect of R-ratio. A typical data set for a
pipeline steel in seawater [13] is shown in Figure 7.9. It needs no emphasis that
the data used, must be for the proper environment. The damage tolerance
requirements may prescribe a data set for environmental effects (ASME). But
in general, an estimate must be made for the 'average' environmental effect as
will be discussed later.

7.7. Fitting the da/dN data

Data sets such as shown in Figure 7.5 still must be interpreted before they can
be used in analysis. Clearly, the 'scattered' data points cannot be used directly.

AK, ksi • in.1/2

20 40 6080 100
10-3

10-4 a>
U

--z·'"
>-
.~

10-5
--'"
"C

"C

10-6

10-5 o 10 Hz
I>. 1 Hz
0 0.1 Hz
v 0.01 Hz 10-7

Air
10-6

~~-L~~____-L____~__L-~~10-8
4 6 8 10 20 40 60 80 100

AK, MPa' m 1/2

Figure 7.9. Effect of Sea Water on Crack Growth in Pipe Steel [13] (Courtesy ASME).
 223

The problem of how to deal with scatter per se will be discussed in Section 7.8.
This section concerns the first step of the interpretation.
In some cases a growth rate equation may be desirable; in particular if the
data appear to fall on straight lines, the Paris or Walker equations may be
convenient. Most computer programs have options for the use of a number of
equations (Walker, Forman and some threshold equation are the most
common), but accept tabular data as well. The latter eliminates the need for
force-fitted (sometimes poorly-fitted) equations. Whatever the shape of the rate
curves, they can be represented in tabular form. This is especially convenient if
the data can be read from a permanent magnetic data file that can be called by
the program. However, the program does not interpret the data, so that one
cannot submit the data points as they appear in the da/dN - ilK data plot.
First lines must be drawn through the data for different R and points of these
lines must appear in the table, be it that this line may have any form without
being fixed by an equation.
Whether equations or tabular data are used, interpretations must be made.
All equations derive from 'curve' fitting and have no physical basis. None of
them is fundamentally better than any other; none is more universally useful
than any other. The most appropriate equation is the one giving the best fit for
the case at hand, which in turn depends upon the material, environment etc.
Also for the use of tabular data, the best fitting line must be drawn through the
data.
Commonly, equations are obtained by using e.g. a least squares fit of the da/
dN - ilK data. This is all that can be done if the original raw test data (a versus
N) are not available; the original crack growth data usually are not reported in
the handbooks or the literature. The best fit through the da/dN - ilK data
must then be used, although this 'best' fit may not give the best predictions as
shown below.
A measured crack growth curve is shown in Figure 7. lOa, the da/dN data in
Figure 7.10b. In this particular case a straight line is appropriate (Paris
equation). A least-squares fit of these data provides Cp = 6.496E-ll, and
mp = 3.43. When these parameters are used in a crack growth analysis routine
to re-predict the original crack growth curve, the result is as shown in Figure
7.l0a, which is certainly not the best fit to the actual crack growth curve. In this
case the curvature of the predicted curve seems appropriate (which means that
the value of mp is correct), but there is a more-or-less proportional error, which
can be corrected by adjusting Cpo The predicted life is too long as compared to
the test life by a factor of 1 937483/1871080 = 1.035. Multiplication of Cp by
this factor to obtain a new Cp = 1.035 x 6.496E-ll = 6.72E-ll, will result in
a better prediction as shown in Figure 7.lOa.
Apparently the regression fit of the da/dN data is not necessarily the best fit
for analysis. The reason for this is that in all curve fitting procedures every data
224

Iu
:z

""
N
(f]

i.2rI

I
I
0.2 O. 4 O. 6 O. 8 1. 0 1. 2 1. 4 1. 6 1. 8
LI FE (i E6 CYCLES)
(a)

d
CD
~ -4.3 f
Q
:z r
-4. 6 .
~
'2i
-Ui ./
-5. 2:-
I"" ,;/
CJ ft'/
2S I *
-5. 5 ~ //

-5.8
i
1 /4/'
-6. j r . ~<fr'+ "'t- test
I ~
-5.41.- .. ~ - - linear regression
I *./
-6. 7~!
I
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9
(b) LOG COELTA - K) (KSJ R! IN)

Figure 7.10. Finding the best representation of rate data. Material: A-533 Bat 550 F; R = 0.1. (a)
Test Data and Predicted curves (see text); (b) Regression Fit of Rate data.

point gets equal weight. But the points for high da/dN affect only a small portion
of the life, while those for low da/dN are of much more influence on the life (they
are the relevant data during most of the life); the latter should weigh more in the
curve fitting.
 225

In the above examples, the curvature of the predicted curves was appropriate,
and only Cp needed adjustment. If the curvature is found to be incorrect, mp
must be adjusted first by trial and error, as shown in Figure 7.11. This usually
causes wild gyrations in the predicted curve, but these can be ignored. The
objective is to arrive at a line with the proper curvature, which will have the
proper mp. Once this mp-value is found (TRY 2), the curve can be adjusted by
adjusting Cp in the manner discussed above (Figure 7.11).
Clearly, fitting an equation is not trivial; it requires judgement, and re-predic-
tion of the original data using a predictive computer code. If the original crack
growth curve is not available, there is no other option than to fit the da/dN data
as well as possible. This can be done by regression, but this is not necessarily the
best; regression analysis may be a refinement and not necessarily an improve-
ment over a hand-drawn best fit line. With the latter the parameters for some
equations can be derived very easily by hand.
The Paris equation for example is a straight line on a log scale, with the equation
y = mpx + b, where y = 10g(da/dN), x = 10g(L1K), and b = log Cpo By
taking two points (as far apart as possible) on the hand-drawn fit, the values of
da/dN and L1K can be substituted in the equation. This provides 2 equations
with two unknowns (Cp and m p ), which can be solved to obtain the values of the
parameters. An example was given in Chapter 5.
If data are available for more than one R-ratio (Figure 7.12), the Walker
equation can be derived in an equally simple manner. All data sets are fitted with

2.8
U1
UJ
15
z
2.6

UJ 2.4
N
U1
"" 2.2
u
<
5 2.0

1.8

1.6

1.4

1.2

o. 5 I. 0 I. 5 2. 0 2. 5 3. 0 3. 5 4. 0 4. 5
LIFE (lE6 CYCLES)

Figure 7.11. Adjustment of data for best predictions. Try 1 provides best m; adjustment of C as in
Figure 7.10 (see text) Material: A-533 Bat 550 F; R = 0.7.
226
parallel straight lines (same m). For each line m and CR are determined as for
the Paris equation, CR being the value of C for each particular R-ratio and C w
is the value of CR for R = O. The Walker equation reads:
da
dN (7.1)

or with Km•x = tJ.K/(1 - R):

da = C tJ.Kw (tJ.Ktrnw (7.2)

dN w (1 _ R)"w

so that:
da
dN = Cw tJ.K'w K.;.x· (7.3)

By plotting the calculated CR values versus the corresponding values of (l - R)

on a double-logarithmic scale and drawing a straight line, one obtains Cwand
nw in the same manner as above. Table 7.1 and Figure 7.13 show the derivation
for the data set in Figure 7.12. With these parameters the Walker equation can

2
10- LOWEST R-RATIIl" 0.0
w INCREASING BY 0.1
--'
u UP TO 0.5
~
u
"-
I
U
Z

10-10 L-_...J...._ _"--...J...._....L..._---J_ _....L...---J_--'

1 10 20 40 60 1DC
DELTA K ( KSI RT iNCH)

Figure 7.12. Rate data used for examples of equation fits (see text) Ti-alloy (see Tables 7.1 and 7.2).
 227
Table 7.1. Derivation of Walker equation for data of Figure 7.12

Paris equation for R = 0, using !J.K = 2 ksi Jin and 20 ksi Jin.
da
Log dN = m log !J.K + log C

Log 4.7 E-9 = m log 2 + log C } -8.32 0.301 m + log C

Log 5.6 E-6 = m log 20 + log C -5.25 1.301 m + log C
3.07 m
Log C = - 8.32 - 0.301 x 3.07 = - 9.24 C = 5.7 £-10 (R = 0).
Same for other R-ratio's or CR can be read directly from Figure at !J.K I (Iog!J.K 0).

Results

R CR I-R log (I - R log C R Predicted Ratio

C R from C R pred/
solution below C R figure
0 5.7 £-10 0 -9.24 6.30 £-10 1.10
0.1 7.6 £-10 0.9 -0.046 -9.12 7.56 £-10 0.99
0.2 8.9 £-10 0.8 -0.097 -9.05 9.27 £-10 1.04
0.3 1.1 £-9 0.7 -0.155 -8.96 1.17 £-10 1.06
0.4 1.3 £-9 0.6 -0.222 -8.88 1.52 £-10 1.17
0.5 1.6 £-9 0.5 -0.301 -8.79 2.08 E9 1.30
Average 1.11

log (1 - R) versus log C R plotted in Figure 7.13. From Figure 7.13 and Equation (7.1):

log C R = log C w - n. log (I - R); -8.78 log C •. + 0.3 n•.

-9.20 log C•. +0
0.52 0.3 nw
n. = 0.52/0.3 = 1.73; log C w = - 9.20; Cw 6.3 £-10

6.3 £-20 6.3 £-10

CR = (I _ R)1.73 For R = 0.5: C R = ~ 2.08 £-9

for other predicted values see last column in table.

3.07 - 1.73 = 1.34

da 6.3 £-10 3.07

Equation 7.1: dN (1 - R)1.73 !J.K

da
Equation 7.3: dN 6.3 £-10 !J.K 134 ~~~.
228
-8,---------------------------__--,
log C R

-9

-10L-____ ~ ____ ~ ____ ~ ______L __ _ ~

-.3 -.2 -.1 0 10g11-RI

Figure 7.13. Fit of Walker equation for data of Figure 7.12 (see Table 7.1).

then be used in the form Equation (7.1), or in the alternative form of Equation
(7.3) with mw = m R - n w.SomecomputercodesuseEquation(7.1),othersuse
Equation (7.3), so that the required input may be different (either C w , m R , nw
or C w , m w, nw).
Also the parameters of the Forman equation can be obtained in a simple
manner. The Forman equation is:

(7.4)

which can be written as:

{(1 - R) Kc - t1K} :~ (7.5)

Hence, the Forman equation ASSUMES that in a plot of

log [(1 - R) Kc - t1K)da/dN} versus log (t1K) all data for all R ratios con-
solidate to one line and that this line is straight. Figure 7.14 and Table 7.2
demonstrate this. From this line CF and mF can be determined in the same
manner as for the Paris equation. Should the data (for more than one R-ratio)
not fall on a single line as e.g. in Figure 7.15 then the Forman equation is not
a good fit, at least not with the value of Kc that was used. Note that Kc is the
only adjustable parameter in the equation, the effect of R-ratio is implied. Since
fracture of the crack growth specimens may have occurred by collapse, K/c or
Kc would not be appropriate to fit the data and the value used becomes an
adjustment parameter indeed, although the generality of the equation becomes
doubtful as it would apply to the specimens only. Yet, it can then be tried to
consolidate the data using a different Kc value.
If the data do not provide a single straight line as is the case in Figure 7.15
 229
10- 3

DATA R
0 0
z
'"0 0.1

~ •.
0.2
10- 4 0.3
0.'
•
~

'"
<I
I
0.5

"
a:'"
I 10- 5

10- 6

10- 7

-8
10
10 20 100

6K
Figure 7.14. Derivation of Forman equation for data set of Figure 7.12 (see Table 7.2).

then the Forman equation is not a good fit at all. Similarly, if log (CR ) versus
log (1 - R) data do not fall on a straight line, the Walker equation is not a good
fit. In that case a more complicated curve fit could be used, but curve fitting is
appealing only if the equation is a simple one. Ifnot, it is much more convenient
to use tabular data inputs instead of an equation. A computer works just as
easily with a data table (11K, da/dN for various R) as with an equation.
This section would not be complete without a warning with regard to
conversion of equation parameters to other equations or to other unit systems.
It appears that Cpo m p. C F • m F • are NOT transferable, nor are Cwo mw. nw. A
little exercise deriving all parameters for all equations from the same data set
(regardless of fit) will easily demonstrate the difference. For example, Tables 7.1
and 7.2 show the Walker and Forman equations for the same data sets. Note
that C F f= C w , and mw f= m F •
Unit conversions can be made, but they are treacherous. Fortunately the n
and m values do not change for different unit systems, because exponents are
dimensionless by nature. But C does change, because C is not dimensionless.
For example, in a Paris equation with 11K in ksi.Jln and da/dN in in/c, the
units for Cp follow from (note that 'cycle' is dimensionless):

in/cycle

(7.6)
230
Table 7.2. Derivation of Forman equation from data set of Figure 7.12

Assumption: Kc = 100 ksi ~ (other values may be assumed to obtain better fit).
R da/dN from Figure 7.12 K, = 100 ksi ~
(in/c) da
{(I - R) K, - ~K} dN

~K = 2ksi~ ~K 20ksi~ ~K = 2ksi~ ~K = 20ksi~

2 3 4 5
0 4.55 £-9 4.81 £-6 4.46 £-7 3.85 £-4
0.1 5.39 £-9 6.03 £-6 4.74 £-7 4.22 £-4
0.2 6.75 £-9 7.14 £-6 5.27 £-7 4.28 £-4
0.3 8.94 £-9 8.64 £-6 6.08 £-7 4.32 £-4
0.4 9.45 £-9 1.00 £-5 5.48 £-7 4.00 £-4
0.5 1.12 £-8 1.25 £-5 5.38 £-7 3.75 £-4
Average 5.24 £-7 4.07 £-4
Data of columns 4 and 5 are plotted in Figure 7.17. On a log-log-plot, these cannot be properly
separated and must be plotted next to each other instead of at proper ~K.
Equation of straight line in Figure 7.14:

2 points:
log (5.24 £-7) mF log 2 + log C F } -6.28 0.301 m F + log CF
log (4.07 £-4) m F log 20 + log CF -3.39 1.301 m F + log CF

2.89 mF

CF = 7.08 £-8

. da ~K2.89
EquatIon (7.4): - = 7.08 £-8 ------::---,--
dN (1 - R) x 100 - ~K

This strange unit comes about by the curve fit, and because Cp therefore has no
physical significance. Conversion of the units to psiyin and in/c would require
a conversion factor of:

l in __ in -
( - I )mp In (7.7)
(ksi yintp (1000 psi yin)mp - 1000 (psi yint p
i.e. the conversion depends upon m p , and Cp should be multiplied by (lOOO)-mp
for use of the same equation with psi yin. If Cp = lE-9 and mp = 3.2 for ksi
yin, its value becomes lE-9/(l000) 3.2 = 2.5E-19 for psiyin.
Conversion to ilK in MPa Jffl da/dN in m/c would require a conversion factor
of:
 231

-2.5

-3.0

_0. S
X
C
E -".0
'"
u -4. 5

'Z"
"0 -5.0
"-
c
:3 * *
~
-5.5
'-"
Cl
-' -6. 0

-6.5

O. 2 O. 4 O. 6 O. 8 I. 0 I. 2 I. 4 I. 6 I. 8
LOG mELT A - K) (KS J RT J N)

Figure 7.15. Data for A-533 B steel for various R-ratios plotted to fit Forman equation. Data do
not consolidate to one line and do not fit equation.

I in = 0.0254 m = 0.0254 m
(ksi Jffitp (6.86 MPa ..j0.0254 mtP (1.09t (MPa Jffitp '
p

(7.8)
In the above example the conversion to MPa Jiii and m/cycle, Cp would become
IE-9 x 0.0254/(1.09)32 = 1.93E-I1.
Conversions of equations other than the Paris equation are even more tricky.
They can be made, but usually it is easier to convert the da/dN and 11K data fIrSt
and then derive the new equation parameters from the converted data set. It is
advisable to make it a common practice to specify the unit system used when
quoting equation parameters, a practice not often followed in toe literature.
Equation parameters without specification of the unit system are useless; the
effects of the unit system are such that indiscriminate use of the parameters will
be disastrous.
In the use of computer software some caution is needed when working with
MPa Jiii for 11K. In that case the crack size, a, must be given in meters (instead
of the more customary mm). Since I MPa = IN/mm2, it is more rational to use
the unit MPa ..jmm, as some codes do. This may require unit conversion for the
coefficients in the rate equations or even for the 11K values in tabular input.
Although these seem trivial matters, little slips can lead to such large errors in
predictions that the problem is worth mentioning.
It may be obvious from this section that indeed the fitting of rate data is not
232
trivial. As was shown in Table 7.1 and Figure 7.13, the Walker equation is a
good fit to the data set in Figure 7.12. Yet, the resulting Cw-values are off by a
factor of l.ll on the average (Table 7.1). Thus, all rates will be II % higher, and
therefore predicted lives will be II % too low (conservative). Naturally, this is
largely due to the difficulty of reading data from a log-log-plot. However, it is
not really the reading of data that counts; after all rate data vary as strongly as
they do and direct use of data would not improve the situation. The reader is
challenged to derive more accurate equation parameters (Exercise II). It should
be noted that these inaccuracies are not due to shortcomings of fracture
mechanics; they are caused by material behavior. The latter is the 'real world';
it is 'the way it is', theory or not. No better 'theory' would improve the situation
(see also Chapter 12 on accuracy).
As may be clear from the examples, the Forman equation often will be a poor
fit (Figure 7.15). This is because the equation implicity accounts for the effect
of R, which is usually more complicated as can be seen from Table 7.1.
Any equation fitting leads to errors, even if it may seem (on a log-scale) that
the fit is very good. On the other hand, the use of tabular data may not be much
better, because the reading of data from a log-plot is full of error also. It should
be realized however, that this is not due to the log-plot, but due to the material
behaving in the manner shown. No theory or equation can accurately account
for material behavior if it depends as strongly as shown upon 11K. Any small
error in 11K will cause large errors in dajdN.

7.S. Dealing with scatter in rate data

Despite regular practice a regression fit is seldom the appropriate fit of rate data,
as was explained in the previous section. Similarily, use of statistical procedures
to account for the scatter are seldom in accord with physical reality. Most
statistical procedures ignore the physics and mechanics of the problem.
Commonly e.g. the 90% confidence curves are determined by using the
individual dajdN data points as the statistical population sample. However,
applying correct mathematics does not lend credence to the physical result.
First consider the nature of the scatter. The three main sources of scatter are:
(a) Consistent difference between heats A and B of the same alloy.
(b) Local differences due to inhomogeneities and 'weak spots'.
(c) Errors in measurement.
Essentially, only the first source of scatter is relevant; the other two have little
bearing on the problem. The three sources of scatter will be considered indepen-
dently as if the others were absent. Consistent differences between various heats
or batches of materials, and to a certain degree differences from location to
location in one batch or plate, will be reflected in a more-or-Iess consistent
difference in crack growth curves. As a consequence, the rate also will be
 233

consistently different as shown in Figure 7.16. This is true material scatter and
it must be accounted for in an analysis because it is not known what the exact
properties are of the batch used in the structure.
Next consider the scatter due to inhomogeneities and 'weak spots'. These may
occur anywhere in the material, but only locally. At some locations the crack will
accelerate, but soon afterward it resumes normal behavior. Another crack (in a
different specimen) will encounter such 'weak' spots at other locations and will
speed up locally (at different stages in life than the first one), and then resume
normal growth; similarly, it slows down sometimes locally. On the whole, the
two crack growth curves will be essentially identical as shown in Figure 7.17a.
Also the rate data will be identical, except that each test provides a few outlying
data points due to local higher or lower rates as indicated in Figure 7.17b. The
outlying data points occur at different locations in the two tests. If instead of
two, many tests are performed on specimens from the same plate, more and
more outlying data points appear and the scatter seems to become well estab-
lished (Figure 7.l7c).
Taking the upper (or 90%) and lower bound of this scatter (Figure 7.17c), and
reconstructing crack growth curves on this basis, results in Figure 7.17d.
Clearly, the upper and lower bound data lead to unrealistic results, as all
measured crack growth curves are essentially identical. This is caused by the
implicit assumption that it is possible that in some cases all crack growth could
be through a continuous string of weak spots (note that this scatter is caused by
local weak spots). This is an unt~nable assumption. Weak spots are local; in
each case only a few will be encountered; the entire bulk will not be one great
weak spot. Hence, the average curve is the relevant one; the 'scatter' is apparent
only and not real.
The third type of scatter is due to measurement errors, shown in Figure 7.18.

da..tlN
BATCH 2

ilK

Figure 7.16. Typical scatter due to batch-to-batch and heat-to-heat variations.

234

In general the crack size will not be consistently over-measured or consistently

under-measured; both over- and under-measurements occur. Even if the error
would always be to one side, it would result only in a shift of the crack growth
curve; the rates would not be much wrong:

(7.9)

Actual crack growth measurements, sometimes even show the crack to become
smaller. This indicates that the measurement interval is too small: it is erroneous
to believe that more measurements are always better. If the crack appears to
become smaller the data point is useless (negative da/dN). Even if the crack
'appears' not to grow the data point is useless: zero cannot be plotted on a
logarithmic scale. The above shows that wild gyrations can occur if measure-
ments are made too close and scatter can actually get worse because too many
data are taken, each of them having an error.
The main reason for the problem is the differentiation (obtaining - da), an
inherently inaccurate procedure. It tends to exaggerate measurement inac-
curacies. For example consider a case with a measurement accuracy of
0.005 inch, which is about as good an accuracy as can be attained by any

a Tests 1& 2 da/dN

Aberation B
in test 2

_ Aberation A
in test 1

11K
(a) (b)

'!
90~

da/dN r 90~
j - :/ /90~
/.
/. :
; .. "/'
.',
'-1

'"
'"
"I. ••• '';
' .../
/. ' / '
/.' ··i
'j

11K
(d)

Figure 7.17. Scatter due to local inhomogenieties. (a) Measured data for two tests; (b) Rate data for
two tests; (c) Multiple tests; (d) Predictions.
 235
Actual
crack growth
a curve
da/dN
• A~tual average

o Measurements -Data

N 11K
(a) (b)

90% Actual average 90% Actual

y/ / test data
da/dN ;--90%
/
I .I
j - :/

.',
90%
/ • • • of /
/
;" : /
i .' ."/'
"I. ••• ''/
' . . ·1
I ' '/'
I.' .. i
'j

ilK N

(e) (d)

Figure 7.18. Scatter due to measurement errors. (a) Measured data for one test; (b) Rate data for
one test; (c) Multiple tests; (d) Predictions.

measurement of crack size. If a = 0.5 inch nominally, its value is between 0.495
and 0.505 inch; the possible error is only one percent. Let the next measurement
be 0.52inch, indicating a crack size between 0.515 and 0.525, again with an
accuracy of one percent. The value of A.a is then between 0.525-0.495 = 0.030
and 0.515-0.505 = 0.010, with an expected value of 0.52-0.50 = 0.020. Hence,
the error in rate will be as large as 50% (0.020 ± 0.010). This problem is not
unique for crack growth; it always occurs where (numerical) differentiation is
performed.
To counteract this problem, the ASTM procedure recommends to take a
moving average for the rate determination. Yet the differentiation will still
exaggarate measurement inaccuracies, which will appear as 'scatter' in the rate
diagram. If a sufficient number of such data is accumulated the scatterband
becomes impressive, even if the. other sources of scatter were completely absent.
The resulting problem is the same as was discussed on the basis of Figure 7.17.
Upper and lower bound would give crack growth curves bearing no relation
to the tests (Figure 7.18d), since all tests essentially showed the same crack
growth curves. Using the errors of all measurements and assuming they all
worked in one direction, ignores the mechanics and physics of what causes
236

this 'apparent' scatter. Determining a 90% band with a statistical treatment

that does not account for this reality is inappropriate.
The question now arises what to do about the scatter in practical damage
tolerance analysis. Most of the scatter observed in data plots can be ignored,
because it is apparent scatter only. The line representing the average of the data
is the only realistic one (Figures 7.17 and 7.18). However, it is prudent to
account for the batch-to-batch, heat-to-heat, and manufacturer-to-manufac-
turer variation (Figure 7.16), because it is not known a priori which batch of
material will be used in the structure. As a rule of thumb, these effects can cause
a difference of about a factor of 2 in crack growth rates between worst and best.
Thus the worst would be about a factor of 1.41 higher than the average, the best
a factor of 1.41 lower than the average (1.41 x 1.41 = 2). On the logarithmic
scale of the rate plot this will only be a slight shift. If a Walker, Paris or Forman
equation is used, the correction can be effected by mUltiplication of C{ by a
factor of 1.4, after the coefficient has been determined (see previous section)
from the average of the data. If a threshold equation is used, an assumption
must be made on how the threshold is affected. The threshold being a disputable
property, no general rule can be given. The damage tolerance analyst will have
to use judgement. In the case of tabular data, a shift of about 1.4 could be used
as well, and if a threshold is used the same remarks apply as above.
With regard to the threshold in tabular data it is worthwhile noting that most
computer codes cannot handle da/dN = 0, because log (0) is undefined.
Therefore the rate at the threshold should be given as a small but non-zero
number, e.g. as 10- 2°, otherwise the computer may declare an input error and
the run may be terminated prematurely.
In conclusion, it is worthwhile mentioning that integration is as forgiving as
differentiation is inaccurate. Integration, being the inverse of differentiation,
tends to 'cover-up' (i.e. compensate for) the errors of differentiation. Examples
of this were given in Chapter 5, comparing hand analysis with computer
analysis.

7.9. Accounting for the environmental effect

Should the environmental effect result in data of a form such as [13] in Figure
7.9, then none of the common equations is applicable. It might well be possible
to use a more complex equation if the computer code is equipped to deal with
these, but most general computer codes can deal only with the common
equations or with tabular data. Thus a tabular representation of such data
would be indicated.
In the case of variable amplitude loading 11K varies from one cycle to the next.
In one cycle the 11K may call for rates at the 'plateau' level of Figure 7.9, in the
next cycle 11K may be low and call for rates close to the threshold. The baseline
 237

data were obtained at continually increasing !:iK. As the environmental effect is

time dependent and measured at 'chemical equilibrium' at the crack tip, it is
questionable whether in variable amplitude loading the effect is the same; the
equilibrium condition for the given !:iK cannot be reached immediately in one
cycle. Although procedures have been proposed to deal with the problem of
time-dependence, they do not appear in general computer codes, nor do most
of them consider the problem of variable amplitude loading. Thus, the user may
take a risk when indiscriminately using the data for variable amplitude without
the benefit of some variable amplitude test data. The latter can be used as a
check. If the analysis can reasonably reproduce the test result, one may have
some confidence that it can deal with the structural problems to be analysed.
In many cases the environmental effect is less complicated and common
equations can still be used if so desired. The questions about equilibrium in
variable amplitude loading and of time dependence, as mentioned above,
remain however.
Should the environment change during crack growth, a new and major
problem arises. This occurs in many structures exposed to weather. In winter the
environment is cold, dry air, in summer it is warm, wet air. It should be noted
here that there may be 100% relative humidity in winter, but this cold air (even
if saturated) does contain much less moisture than the summer's warm air; for
example air with 100% relative humidity at room temperature contains
approximately 30000ppm of water, but air with 100% relative humidity at
- 55°C contains only about 200 ppm of water (relative humidity is the fraction
of the possible moisture content).
The problem occurs for transport vehicles, bridges, airplanes, and many other
structures. Thus, while the following example is for an airplane, it applies in the
same manner to other structures. A wing full of fuel may warm up considerably
when the airplane is serviced, standing on the tarmac. During ascend when
many of the gust and maneuver loadings are encountered, the material is warm
(heat content of fuel), the air is warm and the moisture content high; at this
temperature saturated air contains 30000 ppm of water. When flying in the
stratosphere the structure and fuel cools to - 55°C. Cyclic loading still occurs,
but at this temperature even saturated air contains only 200 ppm of water. If the
flight is over the ocean, the air will contain a certain amount of salt. Upon
descend, the structure is cold (due to cold fuel) but the air warm with much more
water. On top of this, the air may contain sulphuric acid (pollution). One side
of the crack (if it is a through crack in the skin) is exposed to air, the other side
to jet fuel. Clearly, such an environment and the changes thereof, defies any
theory and any modeling. Not even tests can conceivably provide a solution to
this problem. Only pragmatic engineering is of use.
Pragmatism can be exercised in many ways, depending upon goals, outlook,
and desirable conservatism. Assumptions must be made and they can have far
238

reaching consequences. No categoric recipes can be given, only an example.

Consider a case where the data set can be represented by a Paris or Walker
equation, all with the same exponent(s). Consequently all environmental effects
would be only in the coefficient C. Estimating the relative times spent in each
environment, one could take a weighted average of C as shown in Table 7.3. The
final equation then will be as shown (note: the numbers in Table 7.3 are
fictitious).
Naturally, these assumptions are disputable, but so are all alternative assump-
tions. Clearly, the time spent in the different environments can be estimated
only, so that more 'refined' assumptions are as good as the above estimates.
Will different assumptions and refinements do any more than complicate the
proc~dure while- not giving better results? The reader be the judge.
The above procedure is an example only, but it shows roughly how the
problem is handled in some cases. Every analyst encountering similar situations
must make such approximations and judgements, but the problem cannot be
ignored. Obviously, a problem such as this is not likely to be solved by theoreti-
cal considerations. Fracture mechanics cannot be blamed, since this is an
engineering problem. With the potential inaccuracies following from this ines-
capable engineering approach, it is hardly realistic to require extreme accuracy
in stress and geometry factors.

7.10. Obtaining retardation parameters

All retardation models contain 'unknown' parameters; if they do not, too many
assumptions were made, e.g. about the accuracy of the plastic zone size and
constraint equations, the yield strength (arbitrarily defined), the strain
Table 7.3. Pragmatic estimate of da/dN for changing environment for the case of equal mp for all
curves
mp = 3 (Paris)
Cp = I £-8 (salt air)
Cp = 3 £-9 (air at RT)
Cp = 5 £-10 (cold air)
All data fictitious.
Experience Environment Weighted Cp
(estimated)
% time

10 Salt air 1£-8 0.1 x 1£-8 = 1.0 £-9

30 Air RT 3 £-9 0.3 x 3 £-9 = 0.9 £-9
60 Cold air 5 £-10 0.6 x 5 £-10 = 0.3 £-9
Total Weighted average: 2.2 £-9

Average Cp 2.2 £-9; mp 3.

 239

hardening etc. Models without any unknown parameters can hardly be expected
to have general applicability other than by happenstance. Any model that does
contain such parameters can be made to work because the parameters can be
obtained (adjusted) empirically. (Chapter 5). '
From an engineering point of view there is no objection against empirical
adjustment, provided the result is reasonably general. Objections against
empericism are somewhat exaggerated.
ALL material properties used in engineering are obtained from tests. Crack
growth data (dajdN) must be obtained from tests as well. Then it is hardly
objectionable to use empirical retardation parameters. The only objection can
be that retardation models are primitive and may not reflect the actual physics
of retardation, so that the empirical parameter may not be applicable for general
usage.
For the latter reason a model with one and only one (adjustable) parameter
is the most attractive. Without any such parameters, the model can hardly be
expected to be general; with too many parameters, it will always be possible to
fit a specific case, but the generality becomes more dubious.
Regardless of the number of parameters, an experimental determination of
their values is necessary. This is called calibration. For any such calibration at
least one test is needed for the type of variable amplitude loading relevant to the
structure. It must contain the signature of that loading as discussed in Chapter
6. The test can be done on a simple specimen (P is not important as it is not the
p-analysis that is at issue, but the retardation model). CT specimens should not
be employed for this purpose unless the stress history contains no compressive
stresses at all. In the case of compressive loading, the stress distribution (load
path) in a CT specimen bears no resemblance to that around structural cracks
in compression. (The specimen can be used in tension because of the uniqueness
of the near crack tip stress field.) Although one test is necessary, results of
several tests are preferable for a reliable calibration.
Performing the calibration is a simple matter, especially if there is only one
parameter in the model. Predictive crack growth analysis is performed, using the
proper loading spectrum with the appropriate dajdN data, and assumed
parameters for the retardation model. The analysis is repeated several times with
different parameter value(s), and the results compared with the test data. The
parameter value giving the best reproduction of the test data is the sought value.
An example for an aircraft spectrum loading case using the Wheeler retar-
dation model was shown in Figure 5.18. Note that the case with a Wheeler
exponent, Y = 0, represents the case without retardation, called the linear case.
The results in Figure 5,18 show that y = 1.4 gives the best representation of the
test.
The generality of this calibration must now be brought to trial. In practice it
must be assumed to be rather general, but in a research project generality can be
240

investigated by performing analysis for many different stress histories and by

comparing the results with data obtained from tests using similar stress
histories. Examples of such a generality check were presented in Figures 5.19
and 5.22. Data from tests with considerably different spectra (causing largely
different crack growth) were all covered very well by the analysis. It should be
noted that an analytical result within 20% of the test life is very good, in
comparison with errors caused by the data - interpretation (see previous
sections) and those arising from inaccuracies in stress history prediction and
clipping (Chapter 6).
Unfortunately, true generality cannot be claimed. First of all, retardation
depends upon the yield strength, and therefore the parameters are material
specific. They must be determined for each alloy separately (as must the da/dN
data). Second, and more important they are spectrum dependent. They can be
used for variations of the same type of spectrum as shown in Figures 5.22, but
they are not applicable to a spectrum of altogether different shape (Chapter 6):
retardation parameters determined for an off-shore spectrum cannot be used for
a pipe-line (even if the material were the same). Retardation depends upon the
mixture of high and low loads. A different spectrum shape (Chapter 6) will give
a different mixture of high and low loads. Its retardation parameters must be
determined separately, because after all, retardation models are simplifications.
Hence, retardation parameters are spectrum specific. Nature induced log-linear
spectra need different parameters than man-induced spectra (Chapter 6), and so
do spectra for e.g. rotating machinery. Carrying over retardation parameters
from one type of spectrum to another is not permitted; it is one of the
main reasons why some retardation models are acclaimed to be better than
others. Each model should be calibrated for the spectrum relevant to the
application. If it is, any model will perform satisfactorily, if the provisions
stipulated in Chapter 6 are implemented.
Naturally, one can ignore retardation altogether, which is almost always
conservative. Doing this is also making an assumption, namely that there is no
retardation. It certainly will not provide accurate analysis results (though
conservative). Ignoring retardation on the basis of the argument that it cannot
be properly accounted for, is naively believing that the implied crude
assumption of no retardation is better than an engineering approximation.
Calibration of the retardation model has several additional advantages. For
example, the problem of data interpretation is partially resolved. In the cali-
bration analysis one already uses the interpreted data, so that any misinterpre-
tations will be automatically compensated for in the calibration parameter(s).
Also the problem of rainflow counting (Chapter 6) is largely resolved. If the
stress history is not 'counted' in the calibration analysis, potential errors due to
counting are automatically compensated for in the parameters derived from the
calibration analysis. Hence, in subsequent predictions the problem can be
 241

ignored as well. Should one elect to use counting in the calibration analysis the
parameters may be found to have different values if there happens to be an effect
of counting at all (Chapter 6). Then of course, one must use counting in
subsequent analysis as well.
A problem often not recognized is that the calibration is also specific to the
computer code used. Some computer codes account for changes in state of
stress, others do not. Some use different equations for the plastic zone size than
others. Consequently, the calibration parameters obtained with different
computer codes will be different as well; they depend upon the equations used.
Every retardation model must be calibrated with the same computer code
as used for analysis. Transfer of calibration parameters from one code to
another is not permitted, unless the codes use exactly the same equations for
plastic zone, state of stress, etc., and use the same definitions of F,y- This may
seem cumbersome and unscientific, but the reason is that the above equations
and definitions are arbitrary. If one specific computer code is used however, for
calibration as well as for predictions, this is only a one-time problem. The
transfer of calibration factors may be another reason for acclaimed inaccuracy
of some models.
Despite the arguments in this section, objections may still be raised against
retardation handling and calibration. Unfortunately then there is no other
alternative than using linear analysis, which is not very accurate either. In the
extreme one might forego all analysis; this would provide no information at all.
Waiting for the perfect retardation model is no way out. Even the perfect model
must work with interpreted estimates of future stress histories. In that
respect, they are hardly different from weather predictions; the reader be the
judge of the latter's accuracy. Again, the resulting inaccuracies are not due to
fracture mechanics, but due to assumptions made for input.

7.11. Exercises

1. Estimate Kc for a material with Fty = 80 ksi and B = 0.3 inch, if the quoted
plane strain toughness is K/c = 40 ksi fo, and the plane stress toughness
90ksi fo.
2. A center cracked panel test shows a Kc = 75 ksi fo. The panel was 16 inch
wide, and the fatigue crack size was eight inches. Estimate KelT' if there was
a stable fracture causing Aa = 0.3 inch, and F,y = 80 ksi.
3. Supposing F ty = 35 ksi for the material in Exercise 3, estimate Kc.
4. If the Charpy energy is 16 ft/lbs at room temperature and the yield strength
Ftv = 70 ksi, estimate the toughness. For which temperature can this
toughness still be used if the loading rate is low?
242

5. The LT toughness for material A is 50 ksi Jill.

Estimate the toughness of
material B, assuming both materials have the same composition and if
material A has a yield strength of 60 ksi and B a yield strength of 75 ksi.
Estimate the toughness for B if the yield strength values are reversed.
6. The transitional toughness for a thickness of 0.5 inch of a material with
F,y = 70 ksi has been determined as Kc = 60 ksi Jill,
the plane strain
toughness as Klc = 40 ksi Jill.
Estimate the toughness for a through-crack
in a plate of 0.3 inch thickness.
7. Determine the parameters for the Walker equation for the data in Figure
7.12 in your own way by reading the data yourself. Then calculate the data
lines with your equation and draw your own conclusions.
8. Determine the parameters for the Forman equation for the data in Figure'
7.12 in your own way by reading the data and draw your own conclusions
for recalculating the data with your own equation.
9. For the cases in Exercises 7 convert the parameters to units of psi Jill.
10. Rederive the parameters in Exercises 9 by first converting the scales.
Compare with the results of exercise 9.

II. Using the data of Table 7.3 obtain the best constant for an equation
governing five percent usage in salt air, 25% in room temperature air,
remainder in cold air.

References
[I] Standard test method for plane-strain fracture toughness of metallic materials ASTM standard
E-399.
[2] A standard method for the determination of J, a measure of fracture toughness, ASTM
standard E-813.
[3] Methods for crack opening displacement (COD) testing, British Standards Institution BS.5762.
[4] Standard recommended practice for R-curve determination ASTM standard C-561.
[5] Tentative test method for constant-load-amplitude fatigue crack growth rates above 10- 8
m/cycle ASTM standard E-647.
[6] Anon., Damage tolerant design handbook. Mat & Ceramics Info Center, (Columbus) MCIC
HB-OJ; yearly updates.
[7] J.E. Campbell et al. (ed.), Application offracture mechanicsfor selection of metallic structural
materials, Am. Soc. Metals (1982).
[8] D. Broek, Elementary engineering fracture mechanics, 4th Ed, Nijhoff (1986).
[9] M.F. Kannenin and C.H. Popelar, Advancedfracture mechanics, Oxford Un. Press (1985).
[10] S.T. Rolfe and 1.M. Barson, Fracture and fatigue control in structures, Prentice-Hall (1977).
[11] R. Roberts and C. Newton, Interpretive report in small scale test correlation with K" data,
Welding Res. Council. Bulletin 265 (1981).
[12] B. Marandet and G. Sanz, Evaluation of the toughness of thick medium strength steels by using
LEFM and correlations, ASTM STP 631 (1977) pp. 72-84.
[13] O. Vosikovsky, Fatigue crack growth in X-65 line pipe steel at low frequencies in aquous
environments, ASME trans 997 (1975) pp. 298-305.
 CHAPTER 8

Geometry factors

8.1. Scope

For the solution to any fra~ture or crack growth problem the analyst must know
the geometry factors for either K, J, or both, for the structural crack of interest.
Geometry factors for many generic configurations already have been obtained
and compiled in handbooks [1, 2, 3]. This can be done a priori for generic
loading and geometries, but actual structural details are often unique so that
ready made handbook solutions cannot be expected to be available.
In such a case a formal solution can be obtained in principle, but due to the
complexity of structural details formal analysis may be too costly or prohibitive
for any number of other reasons. Structural cracks usually are of the part-
through type so that 3-D analysis would be indicated. If the structure is expen-
sive, the cost of fracture high and only one type of crack is of interest, a finite
element analysis may be a solution. If on the other hand one must consider
literally hundreds of potential crack locations, it is not possible to obtain
geometry factors in this manner other than for a few of the most critical cases,
Finite element analysis of models with cracks for hundreds of crack locations
is prohibitive.
There is a great need for simple (be it approximative) methods to obtain
geometry factors. Fortunately, there are many practical and easy procedures
which can provide geometry factors with good accuracy; they are almost
always adequate for engineering applications in view of the general accuracy of
damage tolerance analysis (Chapter 12).
Methods to obtain geometry factors are the following:
(a) Direct use of handbook solutions.
(b) Indirect use of handbook solutions through superposition and compound-
ing.
(c) Methods based on insight and engineering judgement, combined with (b)
and (a).

243
244

(d) Use of Green's functions or weight functions, if necessary in combination

with finite element stress analysis of the uncracked structure.
(e) Detailed finite element analysis of models with cracks.
Although all these will be discussed in this chapter, emphasis will be on simple
methods. Where possible, the success and accuracy of these will be illustrated
through application to cases for which solutions are known.
Methods (a), (b), (c), and (d) are already included in some of the general
purpose fracture mechanics software [4], but they can be done easily by hand in
a short time. Use of all methods requires access to one of the handbooks [1, 2,
3].
In many cases one must define a reference stress for the stress intensity factor,
and this reference stress must be used consistently throughout the analysis.
Large errors may occur when this is not done correctly. For this reason the
problem of the reference stress will be discussed first in Section 8.2.

8.2. The reference stress

The stress intensity factor is defined as K = [J(5JiUi, in which (5 is the nominal

stress away from the crack. The geometry factor fJ accounts for the fact that
average stresses in the cracked section are higher, as well as for all free
boundaries affecting the crack tip stress as expressed by K. Thus fJ = fJ(a/ W,
a/D, a/S, ... ), where W, D, and S are relevant structural dimensions (Chapter
3), or in general fJ = fJ(a/L), L being a generalized length. Determining
geometry factors means derivation of the function fJ(a/L) for the specific loading
and geometry details relevant to the crack to be analyzed. Simple methods to
derive these functions are the subject of this chapter. The way in which the
function is to be used in crack growth and fracture analysis depends upon the
definition of stress as well, as shown below.
In the case of uniform applied stress there is no problem in the definition of
(5 in the stress intensity factor. But if the stress distribution is non-uniform it may

not be immediately obvious which stress should be used in the expression for K.
An example of this was given already in Chapter 3 where the stress intensity for
the compact tension specimen was discussed. Commonly the latter is expressed
in load instead of stress, because 'the stress' in a compact tension specimen
cannot be defined easily. Use of the load presents no problems in the evaluation
of a toughness test result, but in structural analysis stresses are used, not loads.
Especially in crack growth analysis for complex structures, the use of loads
would be awkward; more specifically, computer crack growth analysis codes are
based upon stress. It was shown in Chapter 3 that the problem can be mended
for the compact tension specimen, by simply defining a reference stress as
(5 = P/WB. Although this reference stress has no physical significance, the

stress intensity is evaluated correctly, provided fJ is changed in accordance with

 245

Equation (3.29). All analysis can then be based upon this reference stress.
Before turning to the case of non-uniform stress distributions, it may be
worthwhile to consider another example. For a central crack of size 2a in a plate
of width Wunder uniform tension it has been shown that 13 = ..}sec(na/W), and
K = f3(JFa = ..}sec(na/ W) (J Fa.
Should one insist on using a reference stress
different from (J in this expression, one can legitimately do so. For example, one
could use the average stress in the cracked section, (Jnel' which is given by
(J = (JneIW/(W - 2a). Then K would become:
..}sec na/W
K = 1 _ 2a/W (Jnel Fa = 13 (Jnel Fa

with (S.I)

13 = .Jsec na/W
1 - 2a/W
Obviously, the values of K obtained would be identical to those based upon (J.
Consistent use of Equation (S.l) in a residual strength analysis would provide
as output the fracture stress in terms of (Jnel (from which (J could be obtained).
Similarly, cyclic stress input in a crack growth analysis should then be in terms
of (Jnel> and the input for 13 should be in accordance with Equation (S.l). In this
case a reference stress other than (J offers no advantage, but the example
illustrates the principle. Any reference stress can be used provided 13 is adjusted
accordingly, so that the product f3(J remains unaffected.
The simplest case of a non-uniform stress distribution is a bending moment
(Figure S.1). The 13 for this case can be found in handbooks, but different
handbooks may provide different f3's. This difference is due to the use of a
different reference stress, as shown in the figure. In one case the reference is the
maximum bending stress in the outer fiber, (Jrna., while in the other case it is the
bending stress at x = a given as (Jm = (1 - 2a/W)(Jrna,. Defining 13 as 13m and
f3xa respectively for the two cases, a conversion can be made as follows:

K = 13m (Jrna, Fa = f3xa (Jxa Fa }

13m . (S.2)
13 13 (Jrna,
xa = rn· (Jm = (I - 2a/W)

Clearly f3w goes to infinity for a = W/2, because (Jm goes to zero, while the stress
intensity is finite. Numerical evaluation of a few cases will readily demonstrate
that the two lead to the same value of K. In this case, the maximum bending
stress is obviously the easiest to use. In employing handbook solutions one
should ascertain which reference stress is used, and make conversions if
desirable. It is good practice to define the reference stress when quoting 13.
246
10
(j
9

7
(j (jm
xa=i=li
w
6

0~---0~j----~~~2----~~3----~~4-----~L5----~·6----a~~

Figure 8.1. Geometry factors for different reference stresses in bending.

The need for a clearly defined reference stress is most obvious in the case of
a generally non-uniform stress distribution, such as in Figure 8.2. The highest
stress is usually the best choice. But all stresses are proportional to load (elastic),
so that the stress at any point is proportional to the highest stress in the
distribution. Hence, any other local stress can be used as the reference, by
multiplying p by the proportionality factor. With the nomenclature of Figure

K =(j, cr, oJ;;

K = (j2cr2 oJ;;
cr,
0;= a

Figure 8.2. Reference stress in case of non-uniform stress distribution.

 247

8.2:

(8.3)

It is emphasized once more that the reference stress must be used consistently.
The p-input to the crack growth analysis must be for the proper reference stress,
and the input of stress ranges, exceedance diagrams, stress histories and/or stress
occurrence tables must be in terms of the reference stress. The output of a
residual strength analysis will be in terms of the reference stress and may require
further interpretation. For example, let P2 in Equation (8.3) for a crack size of
e.g. two inches be 1.1, and U 2/U I = 2, so that PI = 2.2. Supposing
K/c = 60 ksifo, the residual strength would be evaluated for the two cases as
U I = 60/(2.2..jn.2) = 10.9 ksi, and U2 = 60/(1.1..jn.2) = 21.8 ksi. The former
solution predicts fracture to occur when the local stress U I has the value 10.9 ksi.
This is the case when the maximum stress in the distribution is
2 x 10.9 = 21.8 ksi. Although this may seem trivial in this example, misinter-
pretations are easily made when input and results of a lengthy analysis are
reconsidered at some later time or reviewed by others. Providing the definition
of the reference stress in input and output tables is a good practice.

8.3. Compounding

In the general expression for the stress intensity factor K = puJ/W, the
geometry factor P accounts for the effect of all boundaries: P = f(a/W,
a/D . .. ), where W, D, etc. are relevant dimensions of the structure. In many
cases the individual effects of these boundaries can be found in handbooks;
Their composite effect is obtained by compounding, which is multiplication of
all individual effects.
Possibly, the most prominent example of compounding is demonstrated by
the classical solution (other solutions have since been obtained [5-8]) for the
elliptical surface flaw (Figure 8.3). The various boundary effects are due to: back
free surface (BFS), front free surface (FFS), width (W), and crack front
curvature (CFC), i.e.

K = PBFS PFFS Pw PCFC U Fa = pu Fa } . (8.4)

P = PBFS PFFS Pw PCFC
If W is large, Pw = 1, and PBFS = 1.12; then the classical solution provides:
(sin2 cp + a2/ c2 cos 2 cp y/4
K = 1.12 PFFS ,,/2 u J1Ui. (8.5)
fo {I - (1 - a2 jc 2 ) sin2 cp} dcp
248

cross section

}
.. W

K= (3 uv:;;.
(a)

Caution: The abscissa

a
2C
is (j)l, not rP i3FFS
.4 1.8

.3 1.6

.2 1.4

.1 1.2

1.5 .1 .2' .3 .4 .5
J!..
(b) (c) 2c

Figure 8.3. Stress~Intensity of Surface Flaw under uniform tension. (a) Surface flaw; (b) qi versus
a/2c; (c) f3FFs versus a/2c.

The effect of the back free surface is simple, and almost always taken as
/3BFS = 1.12 regardless of crack type (Section 8.5). The effect of the front free
surface depends upon crack shape (Figure 8.3c). Finally, the effect of curvature
/3CFS = l/¢ is a complicated function of the parametric angle, cp, and the flaw
aspect ratio ale. The elliptical integral in the denominator can be and has been
evaluated once and for all for various ale as is shown in Figure 8.3b. Its value
is ¢, but the common representation is as in Figure 8.3b, providing Q = ¢2
instead of ¢. For a crack with a certain aspect ratio ale (or aI2e), the value of
Q can be read from this diagram and then ¢ = JQ is obtained. Alternatively,
the curve can be fitted with an equation so that ¢ can be calculated in a simple
manner.
The function of cp in the numerator must be evaluated for a certain point at
the crack front. Note that Equation (8.5) implies that the stress intensity varies
along the crack front. For the deepest point A (cp = n12; Figure 8.3), the value
of f( cp) equals 1 because cos (nI2) = 0, so that:
 249

P: ~ L12Pm~} (8.6)
K = f3A (J Fa
in which f3FFS and Q follow from the graphs in Figure 8.3.
F or point C at the surface (<p = 0; sin <p = 0; cos <p = 1) the numerator in
Equation (8.5) becomes equal to JlifC, and therefore:
1
1.12 f3FFS JQ JlifC
(8.7)
KC f3c (J Fa .
Note that f3c = f3AJlifC, so that with a < c, the stress intensity at the deepest
point of the flaw is higher than at the surface (f3~ > /3J.
There is no objection in expressing K' in terms of c instead of a; in that case:

[(! = 1.12 f3FFS ~ ~alc ,,1ajC fo

1
(J = f3c
a
(J 0} . (8.8)
f3c = 1.12f3FFSJQa1c = f3cJlifC = f3u-;;

Equations (8.7) and (8.8) are equivalent, provided f3 is properly adjusted as in

Equation (8.8). In crack growth analysis it is often more convenient to express
both KA and K in terms of a by using Equations (8.6) and (8.7). The different
C

stress intensities at a and c have significant consequences for the behavior of

surface flaws, as discussed in Chapter 9.
In obtaining f3 for a structural crack in a complex geometry, the effect of the
individual boundaries can often be found in handbooks or determined
otherwise (see following sections). By compounding these effects the 'total' /3 is
obtained. It should be noted that rigorous compounding adheres to slightly
different rules [9], but the procedure shown here is generally used and accepted.
For example if a free edge increases the crack tip stress by /3 F, then it will still
do so if there is a hole, so that /311 may be multiplied by Ih to get the total /3.

8.4. Superposition

While compounding is multiplication of geometry factors, superposItion is

addition of stress intensity factors due to various mode !loadings. For example
(Figure 8.4) in a combination of bending and tension the total crack tip stress
from Equation (3.1) is:
250

t
-
--

0.1
i}' 0.5 a/w 0.1

Figure 8.4. combination of tension and bending.

O"ref =<Yb

fJ·
0.5 ajW

(S.9)

Because the solution of the crack tip stress field is universal, the functions hi e)
in both terms of Equation (S.9) are identical for each i - j combination, so that
the equation can be written as:

(iij
= K ben
~27rr
+ K ten r (()
Jij =
K tot I' ()
jii;Jij ( .
(S.lO)

The total stress intensity is:

(S.ll)
Apparently the total K follows from the addition of stress intensity factors,
owing to the fact that both terms in Equation (S.9) for any of the individual
stresses contain the same function of e. It should be emphasized that super-
position of stress intensities of different modes of loading is NOT possible. In
such a case different 8-functions apply to the different modes and the step from
Equations (S.9) to (S.lO) cannot be made. For a discussion of combined mode
loading see Chapter 9.
As long as all loading is mode I superposition is permissible. For example, if
there is tension, in and out of plane bending, plus pressure inside the crack, the
total stress intensity becomes:
 251

K tot = Kbeno.u.p. + Kbeni.p. + K ten + Kpr· (8.12)

After the superposition is completed, Pmust be obtained by selecting a suitable
reference stress. Generally, all damage tolerance analysis is based on
K = P(fJiW" and Equation (8.12) is not very suitable for analysis.
Assume a combined bending and tension case as in Figure 8.4; the geometry
factors for the two cases are as shown. The stress intensity is:
(8.13)
In order to obtain Pfor the combination a reference stress must be selected. This
can be any of (ften, (fben' (ftot = (fben + (ften, or any other suitable stress (Section
8.2). Selection of (ften leads to:

(8.14)
(fben
P= Pben -
(ften
+ Pten
while the use of (ftot provides:

K = (p ben -(fben + Pten -(ften) (f,ot "na

(ftot (ftot
c:

(8.15)
P = Pben (fben + Pten (ften
(ftot (ftot

Equations (8.14) and (8.15) are equivalent and both can be used as long as one
is aware that all solutions are obtained in terms of the reference stress. For
example, if one wants to find the residual strength for a crack size of
a = 2.4 inch (in an 8-inch wide panel with Kc = 70 ksiFn) the solution is as
shown in Table 8.1. The same result is obtained in either case, provided the
numbers are interpreted correctly. The solution in terms of (ftot is least likely to
lead to interpretation errors, but careful application of the rules will provide
correct answers in all cases.
Equation (8.13) can be used directly, provided one of the stresses is defined
in value. The fracture condition K = Kc (or K/c ) leads to:Kben + K ten = K"
which represents a straight line in K ten - K ben - space, as depicted in Figure
8.5. It shows the combinations of tension and bending that lead to fracture. In
the above example of a 2.4-inch crack and Kc = 70 ksiFn, one obtains the
solutions as in Figure 8.5b. If the tension stress in e.g. 12.62 ksi, an additional
bending of 8.45 ksi would cause fracture (compare with results in Table 8.1).
Although Figure 8.5 is illustrative, a solution by means of Equations (8.14)
 N
Ul
N

Table 8.1. Residual strength analysis with different reference stresses. (Case of Figure 8.4). Compare results of columns 7 and 9
Specifics: W = 8 inch; K, = 70 ksi $n
ube.lute• = 0.67; Ute. lUbe. = 1.5
Utot = 2.5 Ute.
Ute.IUtot = 0.6
ube.lutot = 0.4
Example: a = 2.4 inch; Uf'te. = 12.62 ksi
Kte• = 1.27 x 12.62 x ~ = 44 ksi $no
K be• = 70 - 44 = 26 ksi $n
U be• = 26/(1.12 ~) = 8.45 ksi; U tot = 12.62 + 8.45 = 21 ksi
2 3 4 5 6 7 8 9
a alw Pte. Pbe. Ute. reference 70 U tot reference 70
(in) Figure 8.4 Figure 8.4 P Uf'te. = pJ;(; Uf'tot = 0.6 Uf'te. P UFtot = PFa
From Equation (8.14) From Equation (8.15)
(ksi) (ksi) (ksi)
0.8 0.1 1.15 1.045 1.850 23.87 39.78 1.108 39.85
1.6 0.2 1.20 1.055 1.907 16.37 27.28 1.142 27.34
2.4 0.3 1.27 1.120 2.020 12.62 21.03 1.210 21.07
3.2 0.4 1.35 1.255 2.191 10.08 16.80 1.312 16.83
4.0 0.5 1.45 1.500 2.455 8.04 13.40 1.470 13.43
 253

Kben

70

50

26--+-_.--_1-,.
10

10 50 70 Kten
(a) (b)
44

Figure B.5. Combination of mode I loadings (right diagram shows specific example of Table 8.\).

or 8.15 is usually more convenient. Most software is based on this solution,

the user deciding the reference stress. In the case of (fatigue) crack growth
analysis Equation (8.13) cannot be used and some form of Equations (8.14) or
(8.15) must be applied. All stresses, whether constant amplitude or variable
amplitude must be expressed in terms of the reference stress, usually taken as
O"tot. In most loading cases, the bending and tension will be in phase (e.g.
eccentric load), so that the ratio O"ben/O"ten is always the same (and so is O"ben/O"tot
or O"ten/O"tot), regardless of the actual stress values. Hence f3 can be evaluated a
priori with either Equations (8.14) or (8.15). As long as the table of stresses or
the exceedance diagram and the f3-values submitted to the crack growth analysis
computer program are compatible for the proper reference stress, correct
answers will be obtained.
If there is more than one boundary to be considered, compounding should be
performed first, and then the superposition executed. E.g. for a crack at a hole
in a plate of width W:

f3ben f3hole,ben f3w,ben

PteD Phole,ten Pw,ten

(8.16)
f3 f3 hole. ben f3 w.ben -0"ben + f3 hole. ten f3 w,ten-
0"ten
O"tot O"tot

The analysis then proceeds as above.

The principle of superposition has an important and very interesting con-
sequence as is demonstrated on the basis of Figure 8.6. Consider a plate under
uniform tension with no crack (case A). A similar plate with a crack of2a (case
B) can be 'fooled' into believing that there is no crack by applying stresses to the
faces of the crack. Originally the material at the crack location was carrying a
254

Ta Ta ja II
..
II
Ta
fBa = -- + tEa ..
tEa
UNCRACKED
II
A B c 0
/1
E

t L L
II
Figure 8.6. Superposition for uncracked plate.
uniform stress (case A). Hence, applying this uniform (J to the crack faces as in
case B will result in the identical situation as case A. Since case B is the
superposition of 2 mode I loading cases, the stress intensity of B (and of A) is
equal to the sum of the stress intensities of cases C and D:
KA = x.c + KD. (8.17)

The stress intensity of case A is zero (K A = 0), the case of no crack, so that:
~ = - x.c. (8.18)
Case D cannot exist by itself as the crack faces would interfere, but if C is applied
first then D can be superposed.
If we reverse the applied stresses in case D, the sign of the cracks tip stresses
(and hence of K) also will change (case E), and hence:
(8.19)
Combination of Equation (8.18) and (8.19) yields:
KE = x.c. (8.20)
Equation (8.20) provides a new stress intensity factor for a crack loaded by
internal pressure. This is useful for surface flaws along the internal wall of
pressure vessels, where the pressurized medium enters the crack (Figure 8.7).
However Equation (8.20) has more important implications. Note first that the
picture of Figure 8.6 can be redrawn for any geometry and stress distribution,
as long as the proper stresses are applied to the crack faces (Figure 8.8). The
following rule emerges:
The stress intensity for any loading case is equal to the stress intensity
obtained by applying to the faces of the crack the stresses that used to be there
when there was no crack (K = K E ). C

This rule can be used to obtain stress intensities and geometry factors in many
 255

(a) (b)

Figure 8.7. Cracks in pressurized container. (a) Internal: pressure inside crack; (b) External.

II
'.

II
m II fB
"

= -- 't

"

II
" E

II
Figure B.B. 'Uncracked stress distribution' rule.

cases by simple means as will be demonstrated at various places in the following

sections. In the following the rule will be referred to as the uncracked stress
distribution rule; it means that K C can be obtained from the solution to KE.

8.5. A simple method for asymmetric loading cases

Pictorial superposition as used in Figures 8.6 and 8.8 is often very helpful in
obtaining geometry factors, especially in the case of asymmetric loading. The
most prominent asymmetric loading case is the lug (Figure 8.9). This case can
be built up [10] by superposition of two symmetric loading cases (B and C).
Then of course the downward load P and the upward stress system (J are
256

I I! (]

(] T T [ (] i l l

=
o +

A ~B~rT~ c D
(b) 'L L (] (]'"
Figure 8.9. Superposition to Derive K for Asymmetric case. (a) Basic configuration; (b) Inverse
superposition.

superfluous and must be subtracted (case D). Superposition yields:

(8.21 )
It is obvious that KD = KA (inversing the picture does not change the stresses).
Then it follows that the stress intensity of the lug crack is:

2 KA = KB + XC or KA = KB; XC . (8.22)

The geometry factors for both KB and K C can be found in handbooks, so that
indeed K can be derived. Should the hole be negligably small, and W large, then
Equation (8.22) leads to:

(8.23)

where P is the load per UNIT thickness, Equilibrium requires that P = (1 W, so

that:
 257

KA = (Fa + u Fa)/2 = (2: + ~) u Fa = PU Fa)

W I . (8.24)
P = - + -2
2rca

For larger holes with diameter D and small W, compounding of P must be

effected first:

KA = (PDP Pwp Fa + PDu Pwu UFa)/2 = PUJ1W}. (8.25)

1 W 1
P = 2" PDP PwP rca + 2" PDu Pwu
As the various compounded geometry factors may be based on different
reference stresses, the equation should take proper account of, and be modified
for, the selected reference stress in the way shown in previous sections.
Geometry factors for almost all asymmetric loading cases can be obtained in
this manner. The procedure is to build-up the case from symmetric cases and to
subtract superfluous loadings until the original case is obtained with opposite
sign. (This is the same procedure as sometimes used in partial integration, where
eventually the opposite of the original integral is obtained and the solution
follows from 2 I = sum of partials).
A solution for the general case of load bypass is shown in Figure 8.10. In the
case of rows offasteners (rivets, bolts, spotwelds) each fastener transfers part of
the total load. The stress intensity is (see also Figure 8.9):

[(12 + ~ (KP + [('1-U2 ) (8.26)

2
, 0"2 ,

P=lq-qIW

1 W
+ 1-
A B
, c

0",
t 0"2 q-02
t

Figure B.10. General case of asymmetric loading (combine with Figure 8.9).
258
which can be manipulated in the manner shown in Equations (8.22) through
(8.25) to obtain p. Proper accounting for the reference stress is essential.

8.6. Some easy guesses

Geometry factors for complicated configurations often can be obtained in a
simple manner, provided one develops some insights. Admittedly, the accuracy
may be somewhat limited, but errors are frequently less than a few percent.
Since the accuracy of damage tolerance analysis is decided primarily by loads,
stresses, material data, and assumptions (Chapters 6, 12), a 5-10% accuracy in
p is usually adequate. Several examples of these simple procedures are
provided in the following sections, but first some trivial cases are discussed here,
upon which the reader can explore other possibilities.
Consider a uniformly loaded plate as in Figure 8.lla. No stresses are acting
along the center line. Cutting the plate in two (Figure 8.llb) makes no
difference: two half plates can carry the load just as well as one full plate, a
fail-safe or multiple-load-path feature often used. Next consider the cracked
plate of Figure 8.llc. Provided a/W is small, the geometry factor is approxi-
mately p = I. Cutting this cracked plate in two is not permissible because there
are stresses acting across the center line. If the cut is made these stresses are
released (Figure 8.lld, e). It is easy to see that this will open the crack a little

i T T t

(a)
• • -L
(b)

1 1 T T T _1

!!~
--
-.1!- -~
- :--

~-
f
W W=W/2 W=W/2 W=W/2 W=W/2

(c)
• • (d)
• (e)
-*- ~

Figure 8.11. Case a = case b; Case c = case d ,,; case e. Cutting in two of uniformly loaded plates.
 259

more: K is higher. Since K = fJ(JJiW, and (J is the same for the two halves (W
is large), this implies that fJ > 1. The cut stresses existed very close to the crack
only (Chapter 2), hence the difference from fJ = 1 cannot be large. It is certainly
less than a factor of 2. On the other hand one would expect the difference to be
more than one percent (1.01 < fJ < 2). Intuitively, one would guess the
difference to be about 10 percent: fJ = 1.1. In actuality the geometry factor for
the edge crack is fJ = 1.12 for small ajW (Figure 3.3). Hence, if one would guess
fJ = 1.1, the estimate would have an accuracy of two percent. Better accuracy
is certainly not required. Note that a guess of 1.05 would have been within seven
percent and a guess of 1.15 within three percent.
For large ajW the differences might be larger. However, let us explore the
premise holds for larger ajW. Then fJ for the single edge crack would be
evaluated as 1.12-../secnaj2 W, while the proper solution is the polynomial shown
in Figure 3.3 (note that W = Wj2). It is easy to evaluate these two alternatives
for various aj W; results are shown in Figure 8.12. The agreement, for the case
of LjW = 00, is within a few percent up till ajW >::: 0.5. But cracks larger than
half the structural size are of no technical interest. Hence, the above 'guesses'
are perfectly acceptable from an engineering point of view. (See also Chapter 12
on accuracy.)
Section 8.3 showed the classical solution for a surface flaw. The geometry
factor for the back free surface appeared to be fJ = 1.12. Clearly, this geometry

t
L/W=1
4 L/W=2
Isee equation
in fig. 3.3)
L
-a
i++

w
•
3

2 L/W= (XI or restra(ned

bending

IL/W=oo)

0.1 o.s 8/W

Figure 8.12. Geometry factors for edge cracks (estimated for L/W = 00; see text).
260
factor has the same origin as the one for the edge crack discussed above. The
classical solution is for an embedded elliptical crack (Figure 8.13). Cutting
the plate in two to obtain a surface flaw, cuts some stresses acting across the
center plane. Again, if one estimated these to be 10%, on would only introduce
a two percent error, if any. It is not difficult to figure that the actual factor of
1.12 (Equation 8.5) was simply taken from the edge crack solution above, and
as such was an estimate as well. This estimate is not necessarily better than the
10% estimate. Continuing this procedure, the same arguments can be used for
a corner crack (Figure 8.13). Again some interface stresses are cut, so that a
geometry factor for the side free surface must be introduced. From the above
one would estimate /3SFS/3BFS = 1.12 = 1.21. Indeed, most handbooks provide
the free surface geometry factor for corner cracks as 1.21. The above 'guess' is
'error free' as compared with the classical solution.

8.7. Simple solutions for holes and stress concentrations

Consider the symmetric case of two very small cracks at a hole in a wide plate
(no effect of width). The stress concentration at a hole is k t = 3 (Chapter 2), so
that the local stress is 30" (Figure 8.14a). Hence, the case is equivalent to the one
shown in Figure 8.14b (note that the radius of the hole is very large with respect
to the crack size if a --+ 0). The stress intensity is K = /3O"Fa, where 0" is the
nominal remote stress. Using the uncracked stress distribution rule discussed in
Section 8.4, it follows immediately that /3 = 3. One may apply the free surface
correction (Section 8.6) and obtain /3 = 3 x 1.12 = 3.36.
Next consider [10] the same situation with long cracks (a/ W still small), as
depicted in Figure 8.15a. It will not make any difference whether there is a hole
or not. If the hole were filled with material (Figure 8.15b) the latter would not
carry any load because the crack face is a free surface; the filler material would
simply 'go along for the ride' undergoing rigid body motion only. Hence the case
of Figure 8.l5b is identical to that of Figure 8.15a: the crack behaves as if the

-r-.~--f-

(a) (b) (c)

Figure 8.13. Solution for embedded crack applied to part-through cracks. (a) Embedded crack; (b)
Surface crack; (c) Comer crack.
 261

UNIFORM STRESS CJ'

CJ'

-
~a_

3CJ'

r ---T - - - - -- -I
(a) (b)

Figure 8.14. Case of two very small cracks at hole. (a) hole; (b) Equivalent case for a Rj O.
--

.Q.
,- '
, I
\
I
\
, ,I
I

f-
a a
-, I'
a
·1·
' -0 ;
.1- a
-I
2aeff
t· -I
(a) (b)
Figure 8.15. Large cracks at hole. (a) Physical crack; (b Effective crack.

hole is part of the crack, ~nd as far as the plate is concerned there is a crack of
an effective length 2aeff = 2a + D. If W is large, the stress intensity is:
K = f3wu ..}rtaeff in which 13 = 1 for large W. (8.27)
Expressing K on the basis of the true crack size a, yields (with f3w 1):

K u J1W:rr = u ..}rt(a + D/2

..}1 + D/2a u Fa = f3u Fa (8.28)

13 ..}1 + D/2a
262
5n------------------------, 5r------------------------,
fj

4
-0 D
I, ...
a

0.1 0.5 aiD 1 0.1 0.5 aiD 1

(a) (b)

Figure 8.16. P for symmetric cracks at hole. (a) Simple solution; (b) comparison with Bowie's
solution.

This Pcan be calculated for various a/D values and plotted as in Figure 8.16a
(curve). Also the point P = 3.36 for a = 0 is plotted. The curve is accurate for
large a, but not for small a. The point 3.36 is accurate for small a. The curve
between is certainly not like curves A or B in Figure 8.16a. Hence, a line C faired
between (0, 3.36) and the curve (accurate for large a) must be very close to the
truth, i.e. within a few percent. The result, shown in Figure 8.16b, can be
compared with a solution by Bowie [II]. Clearly, the two are very close. The
Bowie solution is considered the best available, but it was obtained by numerical
methods, and hence, it will have some error. Then it is hard to say whether the
Bowie solution or the above simple solution is the better one. But even if one
were to accept the Bowie solution as the absolute standard, the simple solution
is within a few percent; the former was obtained with great effort and cost, while
the latter can be derived literally 'on the back of an envelope', as shown above.
One might argue that the simple solution is hardly worthwhile if a SEMI-rig-
orous solution is available anyway. However, the above example demonstrates
that accurate results can be obtained by simple means. More important, the
simple procedure can now be used with confidence to generate solutions to more
complicated problems for which no semi-rigorous answers have as yet been
obtained, as shown below.
For a single crack at a hole (Figure 8.17) the above procedure leads to
P = 3.36 for small a, and for large a (large W):
 263
-- ..,
0-
~
~
I
\
I
,..
I

.
I

-
I
, I

-'
I-
0
-,-- a
-I
2a eff
(a) I. -I

5 1
~ 1
I
I
4
,
1

(b)
0.1 0.5
aID

Figure 8.17. One crack at a hole. (a) Single long crack; (b) fJ for single crack.

K = u .Jnaeff = u .In(a + D)/2 = .Jl/2 + D/2a u .jiia }

= pu.jiia .(8.29)
P = .Jl/2 + D/2a
which results in Figure 8.17b. For large a this provides p < I, which can be
understood if it is recognized that the physical crack has only one tip and is
defined as a (not 2a).
Next, consider biaxial loading, as shown in Figure 8.18. Taking one step back,
in the uniaxial case, (Figure 8.14) there will be a compressive stress (- u; u
being the nominal applied stress) at the poles of the hole. This can easily be
demonstrated by pulling a sheet of paper with a circular hole: the paper buckles
above and below the hole as a consequence of these compressive stresses. Thus,
in a case of biaxial stresses (Figure 8.18a) with u, = u/ = u, there is a tension
3u due to u, at the equator, and a compression - u due to u/ = u, resulting in
a total stress of 3u - u = 2u. This leads to p = 1.12 x 2 = 2.24 for small
264

-01
* 30

- - '----'---t ---- I - '-----'--I

0,=0"£ I 0,=-0"£
I

(b)

(e) (d)

Figure S.lS. Stresses at hole due to biaxial loading. (a) Tension-tension; (b) Tension-compression;
(c) Load path deviation at short crack (stress component perpendicular to crack); (d) No load path
deviation at tip of long crack.

cracks (a = 0). For long cracks Figure 8.15 and Equation (8.28) apply, because
the transverse stress is not deviated at the crack tip: one can cut the plate
in two horizontally if only (Jr is present (only when the crack is small will the load
be deviated around the hole and have an effect on K causing f3 = 2.24 as shown
in Figure 8.l8c, d). Hence for long cracks Equation (8.28) does apply, and the
geometry factor is as shown in Figure 8.19.
This procedure can be extended to any kind of biaxial loading. If (J, = O.S(J"
as in a pressure vessel with end caps, it follows that the equator stress is
3(J - O.S(J = 2.5(J. Hence, for small cracks f3 = 1.12 x 2.5 = 2.80. For long
cracks again Equation (8.28) applies, and the resulting f3 is as shown in Figure
8.19. Similarly, geometry factors for other biaxiality ratios can be obtained as
shown in Figure 8.19. The same procedure can be followed for single cracks
using Equation (8.29).
Should the width W be small, compounding will be necessary, as may be
demonstrated by the following example for a symmetric crack (both sides). For
small a the geometry factor will still be f3 = 3.36 (a - t 0; (Jr = 0). For large
cracks:
K = .Jsec naeff/W (J .Jnaeff' . (8.30)
Substitution of aeff= DI2 + a yields:
 265
sn-----------------------,

(ftl(fi
1
O.S
~______1
o

0.1 o.S aiD

Figure 8.19. Geometry factors for cracks at hole under biaxial loading.

K 13(1 Fa
(8.31)
13 ,Jsec n(D/2 + a)/W ,JI + D/2a·
A curve for the above 13 can be drawn for various aiD, and the results for small
cracks are obtained by fairing between 13 = 3.36 for a = 0 and the curve in the
same manner as described above.
The above method as applied to uniaxial loading with large W was shown to
be in excellent agreement with the Bowie solution. Thus it may be expected that
the extension of the procedure to more complicated cases will provide geometry
factors of good accuracy. The method can be used for cracks at any stress
concentration. In the case of Figure 8.20 the stress concentration is
k, = 1 + 2JT{e; Chapter 2). If for example I = 2 and (! = 0.5 then
k, = 1 + 2.J2/0.5 = 5. Using the uncracked stress distribution rule (Section
8.4) as above, one obtains 13 = 5 for small cracks. For sharp notches it becomes
questionable whether one should apply the factor 1.12 as in the case of holes.
However, it should be noted that this would introduce at maximum a 12%
error. Using judgement and taking the factor as 1.06 would reduce the error
to 6 percent. Clearly, the sharper the notch, the smaller the correction
should be. A rule of thumb could be to take the factor as f3FS = I + 0.12/
(k, - 2) for k, > 3, and 13 = 1.12 for k, < 3. The equation provides 13 = 1.12
for k, = 3 (circular hole), but some engineering judgement is required here.
For large cracks, the notch can be considered part of the effective crack
(aetf = I + a; Figure 8.20) as in the case of the holes. Hence:
266

a
a

(a) (b)
Figure 8.20. Crack at stress raiser. (a) Short crack; (b) Long crack.

K PwU ..)rcaeff = Pw (aeff/ W ) u .In(a + l) = PU Fa}

P Pw (aeff/W).Jl + I/a . (8.32)

P .Jl + I/a .Jsec n(1 + a/W for a central notch

P ..)1 + I/a {1.12 - 0.23 I -:v a + 10.56 (I;' ay

_ 21.74 C;, ay + 30.42 (I ;, ay

for an edge notch.

By plotting P = PFsk/ for a = 0 and the curve of Equation (8.32), and by

fairing for small cracks as in the case of holes a very good approximation of P
will be obtained.
A case in point is a crack developing from a hole and causing a cracked
ligament as illustrated in Figure 8.21, indeed a very common case. The con-
figuration is evaluated as discussed above.IOnce the ligament is broken, a crack
will emanate from the other side of the hole. In essence this will be equiva-
lent to a crack at a stress raiser for which 1= d + DI2and (} = DI2 (Figure 8.21).
Hence, k/ = 1 + 2..)(d + D/2)/(D/2), which provides P for small a, while
Equations (8.32) provide Pfor larger a. Using the procedures as described will
 267

(a) (b)

Figure 8.21. Cracking at hole with edge distance d from free surface. (a) First crack; (b) Broken
ligament plus crack.

lead to geometry factors with good accuracy (compounding and superposition

for various loading cases are to be performed as described).

8.8. Simple solutions for irregular stress distributions

Even if the stress distribution is non-uniform, several simple procedures to
obtain geometry factors can be used. the first, and easiest approach would be
to approximate the stress distribution by a superposition of uniform tension and
a (any) number of bending moments. An example is shown in Figure 8.22. Once
this is accomplished, all methods of compounding and superposition as
discussed in previous sections can be applied to arrive at the appropriate
geometry factors, provided the selected reference stress is adhered to and
accounted for conscientiously.
Alternatively, use can be made of the uncracked stress distribution rule,
discussed in Section 8.4. The rule states that the stress intensity factor (and thus

! I
~
~ + +
-8-
~
W W Weff W
• • • • • • •
(a) (b) (c) (d)

Figure 8.22. Approximation of stress distribution; superposition of b, c, d, should provide good

results up to a/W 0.5, which is usually a large enough crack size for practical problems.
268
P) can be obtained by applying to the faces of the crack stresses that used to be
there when there was no crack. Hence, if the stress distribution is known (as it
was before cracking), these stresses can be applied to the crack faces to obtain
K and p, be it that some work is involved. Naturally, only that part of the stress
distribution is used that covers the crack.
Suppose the stress distribution in the uncracked section is as shown in Figure
8.22a. By applying this stress distribution to the crack faces a stress intensity
factor can be derived using so-called Green's functions. The latter are shown for
central and edge cracks in Figure 8.23. They provide the stress intensity for a
point load per unit thickness (stress) at one particular point on the crack face.
The total stress intensity can be obtained by (numerical) integration of the
Green's functions, using the stresses in the uncracked section as a series of point
forces. A certain caution is necessary, because the stresses close to the crack tip
are the most influential.
This can be demonstrated by using an example for uniform stress distribution
for which p is known. An approximation could be made, using the Green's
functions of Figure 8.23 and the resulting rough analysis (performed by hand)
would be as in Table 8.2. A more refined approximation would lead to the
analysis of Table 8.3. Clearly, the results depend strongly on the approximations
close to the crack tip. Solutions of sufficient accuracy can be obtained only if the
approximations at the crack tip are adequate, where the adjective 'adequate' is
not well defined; engineering judgement is necessary. Computer software [4]
using Green's functions will account for this problem.
It should be noted that the above procedure provides the stress intensity
factor K. What is needed is the geometry factor p. As K = P(JJiW, the geometry

12-
I

B~A
~
I
2a

K=2P- .J1 +!CX/8! 2 [ -0·4!X/8! 2 +1.3 I

.Ji[;

X/O e
<0·6 1
0.6-0.7 1.01
0.7-0.8 ,.03
0.8-0.9 1.07
>0.9 1.11

Figure 8.23. Most useful Green's functions.

 Table 8.2. Hand analysis using Green's functions with crude steps

Right side (Figure 8.23a)'1 Left side (Figure 8.23a)a)

X 8X P, = 0"",8X b) K) P x = (javc~XC) K2d) K) K2

Xa\erage P, P, +
Fa ~ a-x Fa J: ~:
0.0
0.1 0.2 0.2 0.113 l.l06 0.125 0.2 0.113 0.905 0.102 0.227
0.2
0.3 0.2 0.2 0.113 1.363 0.154 0.2 0.113 0.734 0.083 0.237
0.4
0.5 0.2 0.2 0.113 1.732 0.196 0.2 0.113 0.577 0.065 0.261
0.6
0.7 0.2 0.2 0.113 2.380 0.269 0.2 0.1l3 0.420 0.047 0.316
0.8
0.9 0.2 0.2 0.113 4.359 0.493 0.2 0.113 0.229 0.026 0.519
1.0
\36+

f3 = ~ = 1.56
0"
Fa I~ -_ O.SS,. known f3 I; error 12%, but see Table 8.3.

Notes: (I.) Better result is obtained in Table 8.3 with finer steps. (2.) Assumptions: central crack 2a = 2 inch; uniform stress, 0" = I. (3.) With above
assumptions integration can be done in closed form, but not if stress is non uniform; this is an example for a case with known f3 so that results can
be checked.

a There are two sides of the crack ( - x and x), both must be accounted for.
b In case of non-uniform stress O"ave is average stress over 8X; in this case 0" I overall.
'0"." at - X; in this case 0" = I overall.
d K at right crack tip due to stresses on left side of crack (Figure 8.23).

tv
0\
\0
Table 8.3. More refined land integration of case of Table 8.2. N
-...J
Same notes apply as in Table 8.2. 0

Right side (Figure 8.23a) Left side (Figure 8.23a)

X Xaverage ~X P, = aave~X KJ P, = aave~X K2 K = K J + K2
P, P,

Fa ~ a - x Fa ~ a + x

0.0
0.1 0.2 0.2 0.113 1.106 0.125 0.2 0.113 0.905 0.102 0.227
0.2
0.3 0.2 0.2 0.113 1.363 0.154 0.2 0.113 0.734 0.083 0.237
0.4
0.5 0.2 0.2 0.113 1.732 0.196 0.2 0.113 0.577 0.065 0.261
0.6
0.65 0.1 0.1 0.056 2.171 0.122 0.1 0.056 0.461 0.026 0.148
0.7
0.75 0.1 0.1 0.056 2.646 0.148 0.1 0.056 0.378 0.021 0.169
0.8
0.85 0.1 0.1 0.056 3.512 0.197 0.1 0.056 0.284 0.016 0.213
0.9
0.925 0.05 0.05 0.028 5.066 0.142 0.05 0.028 0.197 0.006 0.148
0.95
0.96 0.02 0.02 0.011 7.000 0.077 0.02 0.011 0.143 0.002 0.079
0.97
0.98 0.02 0.02 0.011 9.950 0.109 0.02 0.011 0.101 0.001 0.110
0.99
0.9945 0.009 0.009 0.0051 19.043 0.097 0.009 0.0051 0.053 0.097
0.999
0.9995 0.0001 0.0001 0.0006 63.238 0.038 0.0001 0.0006 0.016 0.038
1.000 Ktotal = 1.727+

Klal 1.727
fi = Fa = ~ = 0.976; known fi = I; error 3.5%.
a na I n1
 271

factor is obtained as fJ = K/afo" which means that a has to be defined. As

discussed any reference stress, aref, can be used as long as the damage tolerance
analysis is based consistently on this same reference stress. The problem of
reference stress has been addressed at several places in the preceding sections
and specifically in Section 8.2. In the case of a complicated stress distribution,
the best option is to select the highest stress in the cracked section as the
reference stress. Note: this is not meant to be the highest stress at the crack face,
because the latter may vary with crack size. The geometry factor always must
be derived for a number of crack sizes. Hence, the highest stress in the section,
and not the highest stress on the crack face, is the best reference. Well-designed
software will account for this problem as well.
An alternative procedure to obtain geometry factors in the case of com-
plicated stress distributions is to make use of so-called weight functions.
Although this is a sound method, it requires a great deal more knowledge of the
effect of cracks on strains and displacements, because it is based upon a displace-
ment reference for a known case [12, 13, 14]. It is not particularly suited for hand
calculations. Some software [15] includes the use of weight functions and as such
it is in the realm of engineering applications. Their use requires more expertise
and should not be attempted lightly, unless through a reputed software package.

8.9. Finite element analysis

Finite element analysis can be used in two ways to determine geometry factors,
namely indirectly and directly. For the solution of common damage tolerance
problems the indirect use of finite element analysis is the most worthwhile. In
that case the finite element solution is obtained for the uncracked structure only,
and the stress distribution in the section of the future crack is calculated.
Subsequently, the uncracked stress distribution rule (Section 8.4) is used to
calculate the stress intensity factor with one of the approximate methods
described in previous sections, or through the use of Green's functio~s or weight
functions (Section 8.8), especially if software is available to perform either of
these tasks.
It is worthwhile pointing out that stress distributions obtained with finite
elements are of limited accuracy. Claims that finite element analysis is the most
rigorous stress analysis available are exaggarated and naive; a close look at
many finite element solutions bears this out immediately. The only 'rigorous'
solution is obtained from the differential equations in which the element size
(dx dy dz) literally approaches zero. Unfortunately, these differential equations
usually cannot be solved. The mere fact that the elements in finite element
analysis are of finite size, indicates that the solution is an approximation.
Provided the elements are sufficiently small, especially in areas of large stress
272

gradients, very good solutions can be obtained. However, in many practical

solutions for complicated structures, model size (degrees offreedom) limitations
and coarse modeling in areas of large stress gradients often are cause of limited
accuracy. In areas of stress concentrations and of load transfer to other
members, accuracies are seldom better than 10 percent for the calculated local
stresses. Errors larger than 10 percent are not uncommon.
In the direct use of finite element analysis for the derivations of stress intensity
and geometry factors, solutions must be obtained for models with cracks. As
there is an extremely large stress gradient at the crack tip, the element sizes
around the crack tip must be very small, unless use is made of higher order
elements which can model a stress singularity.
The finite element model can provide stresses, strains, displacements, and
strain energy only. From these the stress intensity and geometry factor can be
obtained by a variety of methods. For example, the universal crack tip stress
field solution provides the crack tip stress uy for e = 0 (Chapter 3) as:
K
uy = ,,)2nx (e = 0). (8.33)

The finite element solution provides uy at various locations (x). By substituting

the calculated uy, and the distance r for which it applies, in Equation (8.33), the
stress intensity is calculated as:
K = U YFEM ,,)2 n X FEM • (8.34)
As Equation (8.33) is valid only at very small x (Chapter 3), the stress intensity
obtained from Equation (8.34) is in error unless x is extremely small. On the
other hand, the calculated value of uy contains a larger error the smaller x
(unless singular elements are used). In order to circumvent this problem
Equation (8.34) can be solved a number of times using the calculated stresses
uy(r) at distances Xl' x 2 , x 3 , etc. For each combination (uy ; x) an apparent value
of K is obtained from Equation (8.34). None of these is correct. One may
plot the apparent values as a function of the distance x for which they were
calculated. A line drawn through the data can be extrapolated to x = 0 - at
which point Equation (8.33) is rigorous - , as shown in Figure 8.24. The
extrapolated value at x = 0 is the sought value of K. For an example see
solution to Exercise 14.
Finally, the geometry factor must be obtained. Selecting a reference stress Uref
(this can be the stress as applied to the model or a stress calculated by the model
at some convenient location) the geometry factor follows from:

p = K (8.35)
Uref J;=;; ,
where a is the crack size in the model. Naturally, the following damage tolerance
 273

Kapp

C\ K

(1Y2 -~'
x, X2 x, X2 X, X.
X
a (b)
(a)
Figure 8.24. Obtaining K from finite element model with crack. (a) Stresses from finite element
model; (b) Stress intensity.

analysis should be based upon the same reference stress (Section 8.2). The above
provides f3 for one crack size only, but f3 must be obtained for a range of crack
sizes. Hence, the procedure must be repeated for a number of crack sizes
(multiple finite element solutions). It should be noted that in each case the
same reference stress must be used - in terms of both location and magnitude
- to obtain f3 from Equation (8.35); any inconsistency may lead to serious errors.
There are a multitude of other methods to obtain f3 from finite element models
with cracks. In analogy with the above approach, use can be made of the
displacements. For example, the displacement u at a distance r from the crack
tip follows from the general crack tip filed solution as:
u = CK v'r fee), (8.36)
which can be used in the same manner as Equation (8.33) to obtain K from the
calculated displacements.
Alternatively, f3 may be derived from the strain energy U. The total strain
energy U in the model is calculated for crack size a. Subsequently the crack in
the model is extended by one element size (Lia) to a + Lia, and again the total
strain energy calculated. The change in strain energy dU/da is approximated as:
dU LiU
(8.37)
da Lia

from which K is obtained as (Chapter 3):

K = .JE dU/da (8.38)
and finally f3 in the manner described above.
The above procedure allows larger elements and is acclaimed to have better
accuracy than those discussed before. However, some caution is advisable
because Equation (8.37) is a differentiation, an inherently inaccurate process. It
274

involves subtraction of two large numbers of equal magnitude. Ifboth and Va+!'.a
are accurate within one percent, the accuracy of L1 V is certainly no better than
10%.
If there is a single load on the model the strain energy can be obtained as
Va = O.5Pc5 a, where c5a is the load-point displacement for crack size a. Similarly,
for crack size a + L1a the strain energy is Va +Aa = O.5Pc5 a+Aa' Calculation of K
and p then proceeds as above from Equations (8.37) and (8.38). For an example
see solution to Exercise 15.
Since J = G = dVjda (Chapter 4) the strain energy release rate can be
obtained by evaluating the J-integral (Chapter 4) along a convenient path in the
finite element model. Subsequently, K and pagain are obtained as shown before.
Most finite element codes have post-processors which provide Kby several or
all of the above methods. It cannot be said a priori which method will provide
the best results; this will depend upon the configuration, and especially upon the
modelling. As in the case of the indirect use of finite element models, the
accuracy depends mostly upon modelling of areas with stress concentrations
and load-transfer, and upon assumptions made for the boundary conditions.
Damage tolerance analysis always requires knowledge of p (not K) for a
range of crack sizes. Hence, several finite element solutions must be obtained for
a number of crack sizes, which requires a number of different models; indeed a
costly proposition for complex structural configurations. In the literature the
accuracy of such analysis is often demonstrated on the basis of simple configur-
ations such as center cracked panels. In the first place those solutions are always
known anyway, but more important, structures seldom resemble center cracked
panels. Finite element models of cracked real structures are rather more difficult
and expensive; besides they require many more assumptions with regard to
boundary conditions, load transfer, etc., and small elements in areas of stress
concentrations, so that their final accuracy is very limited despite the effort. The
simple procedures discussed in this chapter, possibly employing finite element
analysis of the uncracked structure, presently are the most viable methods for
general engineering applications. In view of the inaccuracies introduced by other
assumptions (Chapter 12) they are also the most sensible solutions.
Several other numerical analysis procedures to obtain K are availabe, but they
are not within the realm of general engineering applications and as such they are
beyond the scope of this book; for a discussion of these and further references,
the reader is referred to more basic standard texts.

8.10. Simple solutions for crack arresters and multiple elements

The stress intensity factor is affected by the presence of second elements

(stringers, doublers, flanges) or crack arresters to which load can be transferred.
If the second element is intact, load transfer from the cracked part to this
element will cause a decrease of K and therefore of p.
 275

Consider two wide plates, one without and one with stringers (or doublers),
as in Figure 8.25, both subjected to uniform stress (1. For the plate without
stringers the geometry factor will be f3 = 1 for crack sizes up to alb = 1
(W ~ b), as shown in Figure 8.25d. In the plate with stringers, the effect of the
latter will be negligible when the crack is small; f3 = I for alb ~ O. However,
if the crack extends from stringer-to-stringer, the situation is quite different. In
an un stiffened plate all load in the section of the crack must bypass the crack
inside the plate, which gives rise to f3 = 1 in the first place. If the stringers are
present a second load path is available: part of the load can now by bypassed
outside the plate via the fasteners into the stringers and then back into the plate,
again via the fasteners (Figure 8.25c). If part of the load bypasses outside the
plate, then the stresses at the crack tip (inside the plate) will be lower. This means
that K is lower, which is reflected in a lower f3 then in the unstiffened plate. Since
for the latter f3 = I, the f3 for alb = I in the stiffened plate will be less than I,
as shown in Figure 8.25d. For intermediate crack sizes f3 will gradually decrease
from 1 at a = 0 to the lower value at alb = 1.
Thus the stress intensity is lower than in the absence of stringers, which can

CJ f'-_ _ _ _---'_ _ _ _--'

2a
.. 2a

2b

(a) 'w'"
(b) (c)
~ L

0.5

alb alb
(d) (e)
Figure 8.25. Effect of stringers on stress intensity. (a) Unstiffened panel: load bypass in plate; (b)
Stiffened panel; (c) Side view; alternative load bypass through stringer; (d) Geometry factor; (e)
Stress concentration in stringer.
276

have a dramatic effect on crack growth and residual strength, as will be

discussed in Chapter 9; fracture arrest becomes a distinct possibility. The above
example is for mechanically fastened stringers, but similar results are obtained
if the second element (doubler or arrester) is welded or an integral part of the
structure. For example a transverse web crack in an I-beam will show a similar
decrease in f3 when the crack approaches the flanges. The decrease can be
significant: f3 = 0.3-0.5, depending upon the stiffness of flanges or stringers, the
stringer spacing, and the fastener spacing.
The decreases in K (or f3) is beneficial for the crack tip stresses (plate), but the
side effect is that the stresses in the stringer will be higher and that the fasteners
will be subjected to very high shear loads. If the crack is small, the stresses in
plate and stringer will be roughly equal; since stringer and plate are attached
they must undergo equal strains and equal strains require equal stresses if the
moduli of plate and stringer are equal. If the local stress in the stringer at the
location of the crack is denoted as (JI' then (JI = (J as long as cracks are small,
but (JI > (J when alh ---> 1.
A stress concentration factor L will be defined as L = (Jd(J. From the above
arguments it follows that L = I for alb = 0, and that L ~ I for alb ---> I, as
shown in Figure 8.25e. Depending upon stringer stiffness and spacing and
fastener spacing, L can attain values of 2 to 3 (i.e. the local stress in the stringer
will be 2 to 3 times higher than the applied stress). In the same vein the shear
on the fasteners (or welds) will increase from essentially zero at alh = 0 to a
significant shear load (stress) when a/ h reaches 1. Because the load bypass occurs
very close to the crack (Chapter 2) only the fasteners close to the cracked section
carry the shear. The so called stiffening ratio, 11, which reflects the ratio of
cross-sectional area of stringers and skin, and the fastener spacing determine the
magnitude of f3 and L.
For any configuration, f3, L, and fastener shear can be readily calculated by
means of numerical analysis of closed form solutions [16, 17, 18], or by finite
element analysis. The former are preferable because they permit parametric
analysis. Due to the high stringer stress and fastener loads plastic deformation
may occur, which may affect load transfer and therefore alter {3 and L. The
closed form analysis procedures can properly account for these effects, but then
only specific solutions are possible. For cases without plastic effects generic and
parametric solutions have been obtained and the results are readily available in
handbooks [1, 19], so that they can be used for general damage tolerance
analysis (Chapters 9 and 14).
Unfortunately, the handbook solutions [I, 19] only provide f3, and neither L
nor fastener shear, while all three are necessary for residual strength analysis.
However, good estimates of L can be obtained from f3 as follows. Define the
geometry factor for the un stiffened plate as f3u, and for the stiffened panel as f3s.
 277

The reason that f3s < f3u is the load transfer to the stringer. If there is no stringer,
the load carried by the plate in front of the crack tip is:

P = f (Jy B dr. (8.39)

For (J = 0 the stress field solution provides (Chapter 3):

f3(JFa
J2iU . (8.40)

Carrying the integration over a distance equal to r = a from the crack tip is
sufficient to obtain the bypassed load. Hence Equation (8.39) becomes:
r~fa Fa
B f3(J
dr = Bf3(J JGfi f r- I /2
a

dr (8.41)
P
r~O J2iU o
J2 f3(J aBo
Clearly, the additional load carried by the stringer is P, (stiffened) minus Pu
(unstiffened) and the additional stringer stress is (Ps - Pu)IA, is the stringer
cross sectional area. The total stress in the stringer is (J, = (J + (P, - PJIA"
so that the stress concentration L = IJ,I(J becomes, with Equation (8.41):
L = I + aB J2 (f3u - f3')IA,. (8.42)
The handbook provides f3u and f3,.; the value of L can then be obtained from
Equation (8.42). By taking f3u and f3s from the handbook for a number of crack
sizes between alb = 0 and alb = 1, the stringer stress concentration can be
calculated as a function of crack size. An example is shown in the solution to
Exercise 15.
Obtaining fastener shear loads is somewhat more precarious, but a sensible
estimate can be made as follows. All load is transferred by the fasteners closest
to the crack plane: assume that three fasteners above the crack transfer the load
into the stringer, three fasteners below the crack transfer the load back into the
plate. Together the three fasteners transfer the total load which is ((J, - (J)A,.
Typically, the fastener closest to the crack transfer most of the load (e.g. 60%),
the other two transfer e.g. 30 and 10% respectively. Hence, the highest fastener
shear load would be:
Pfastcncr = 0.6 (L - 1) (JA" (8.43)
where L follows from Equation (8.42). Clearly, the fasteners must be made
strong enough to carry this shear otherwise the whole scheme will not work
(Chapters 9). It turns out that the actual fastener load on the average is only in
the order of 60-70% of the value in Equation (8.43) due to fastener hole
ovalization (plasticity).
278
8.11. Geometry factors for elastic-plastic fracture mechanics

The definition of J is (Chapter 4):

J = J el + Jpl = p2 n ff2 alE + Haf.pla. (8.44)
The geometry factor Pis obtained by means of any of the procedures discussed
in the foregoing sections. If a stress-strain equation is available, the plastic part
of J can be expressed in a only. The commonly used equation (Chapter 4) is the
Ramberg-Osgood equation which provides for the plastic strain: f. = an/F.
Then the plastic part of J becomes:

Jpl = H(y: n) a n+ 1 a/F. (8.45)

By performing non-linear finite element analysis on the cracked structure (direct

method in Section 8.9), using the proper nand F values, Jpl can be calculated
from the integral formulation. With Jpl thus known, the stress applied to the
model known, and a known, the geometry factor can be extracted as

H(~ ) = F JplFEM (8.46)

L' n an + 1 a
The procedure must be repeated for various values of a in order to obtain
H(a/L) for a certain material with given nand F. If H has to be determined for
various materials with different n, the procedure must be repeated for all these
different n-values, a costly proposition.
Geometry factors, H, have been determined in this manner for a number of
configurations and n-values. These have been compiled in handbooks [20, 21],
which provide hI instead of H, but H can be derived from hI if so desired as
discussed in Chapter 4.
Simple methods to obtain H have not yet been devised, but the following
procedure might be used. If n = 1 then F = E and Equation (8.45) reduces to:
J = H a2 a/E. (8.47)
This is obviously the linear elastic case for which (Chapters 3, 4)
J = n p2 a 2 a/E. (8.48)
Clearly, in this case H = n p2. Since n + = 2, the square in a2comes from
n + 1. Assuming that the square of P also comes from n + 1, it follows that
either H = n pn+1 or H = (Jiip)" + I. At present there is no proof whether either
of the two expressions is correct. As a matter of fact reasonable agreement is
obtained only for small n and small a, so that the approximation may not be
very useful. It appears that
 279

(S.49)

may be used if a is small and n is low, but the solution is up to the user. As
forgiving as EPFM analysis is, the results are usually within acceptable engi-
neering accuracy. Hence, H can be obtained from fJ through Equation (S.49),
where fJ is derived by any of the procedures discussed in the previous sections.
The fact that Equation (S.49) does not exactly cover the computed H values does
not make the equation invalid, because there is equal reason to suspect the finite
element analysis, which will contain increasing errors for larger aj L and larger
n.

8.12. Exercises

1. Using the maximum bending stress as the reference stress determine fJ for
a crack with ajW = 0.5 from Figure S.la. Calculate the residual strength
given Kc = 70 ksiJffi, W = 10 inch, and F;y = 60 ksi.

2. Calculate and plot the fJ-curve for a plate of with an edge crack subjected
to combined tension and in-plane bending, using the maximum total stress
as a reference; note take six values of aj W at increments of 0.1, starting at
0.1. A remote load P is applied 2 inches from the center of the plate to the
side of the crack; W = 10 inch.

3. Repeat Exercise 2 using the uniform tension stress as a reference.

4. Given that W = 10 inches, a = 2 inches, F;y = 100 ksi and

K,c = 50 ksi.Jffi, calculate the residual strength for all cases in Exercises 2
and 3. (Ignore collapse).

5. Calculate and plot the fJ-curve for a single crack at a hole of diameter D
in a plate of width W subjected to uniform tension. Take 10 values of ajD
increasing with increments of 0.2.

6. Use the result of Exercise 5 to calculate the rate of fatigue crack growth
of a crack of 0.3 inch if D = 1 inch, W = 6 inches, Aa = 10 ksi at
R = 0.2, dajdN = 2E-9 AK2.3 K~!x.

7. Modify Exercise 5 for biaxial loading with aLja T = 3 and for aLj
aT = - 0.5 (aT negative). Then repeat exercise 6 for both cases, but with
a crack size of a = 0.1 inch.
280

8. Determine the fl-curve for symmetric cracks at a fastener hole where the
fastener takes out 20% of the load for five values of aiD at increments of
0.4 (Note from Exercise 5 that for aiD> 0.4 the hole may already be
considered part of the crack. Assume that W = 8 inch, D = 1 inch, B =
0.5 inch and that the stress distribution at the ends is uniform.

9. Calculate the residual strength diagram for the case of Exercise 8 up to

aiD = 2; K, = 50 ksiyln; F;y = 80 ksi.

10. Determine the fl-curve for cracks emanating from a semi-elliptical edge
notch with a depth d and radius r in a plate of width W subjected to
uniform loading. Use 5 aid values at increments of 0.1; dlr = 5.

II. A crack emanated from a hole with diameter D at a distance of e from the
edge of the plate. The entire ligament has cracked. Determine the fl-curve
for the crack emanating at the other side of the hole.

12. A shouldered part with a stress concentration factor of k t = 1.5 at the fillet
radius develops a through crack. What is fl for small cracks.

13. An elastic finite element analysis with elements of 0.1 inch at the crack tip
and a crack of 2 inches produces the following results for the longitudinal
stress in the plane of the crack: 48.9 ksi in element 1, 30.0 ksi in element 2,
25.2 ksi itt element 3. The applied stress is non-uniform. The highest stress
applied in the model is 5 ksi. Calculate fl, assuming the given stresses act
in the center of the elements, and taking highest stress as a reference.

14. A finite element analysis of a model with a crack is subjected to a point

load P. The crack size is 1 inch and the crack tip elements are 0.1 inch. The
calculated displacement of the loading point is 0.02 inch. In another run
the crack is extended over one element, and the calculated displacement is
0.021 inch. Calculate K if the applied load is 10 000 Ibs. Assume unit
thickness and E = 10000 ksi.

15. Assuming J1 = 0.4 and a fastener spacing sib = 0.1, calculate the L curve
and fastener load curve for 5 values of alb in increments of 0.1; let Figure
8.25c apply and assume that 60% of the load is transferred by the first
fastener. B = 0.2 inch, b = 8 inch; J1 is defined as A,lbB, where A, is
stringer cross section.
 281

References
[I) D.P. Rooke and D.J. Cartwright, Compendium of stress intensity factors, H.M. Stationery
Office, London (1976).
(2) O.C. Sih, Handbook of stress intensity factors, Lehigh University (1973).
(3) H. Tada et aI., The stress analysis of cracks handbook, Del Research (1973, 1986).
(4) D. Broek, GEOFAC. a pre-processor for geometry factor calculation, Fracturesearch software
(1987).
(5) J.e. Newman and I.S. Raju, Stress intensity factors equations for cracks in three-dimensional
finite bodies, ASTM STP 791 (1983) pp. 1-238-1-265.
(6) I.S. Raju and J .C. Newman, Stress intensity factors for circumferential cracks in pipes and rods
under tension and bending loads, ASTM STP 905 (1986) pp. 789-805.
(7) J.C. Newman and I.S. Raju, Analysis ofswface cracks infinite plates under tension and bending
loads, NASA TP-1578 (1979).
(8) 0.0. Trantina et aI., Three dimensional finite element analysis of small surface cracks, Eng.
Fract. Mech. 18 (1983) pp. 925-938.
(9) D. Broek, Fracture mechanics software, Fracturesearch (1987).
(10) D. Broek et aI., Applicability offracture toughness data to surface flaws and corner cracks at hole.
Nat. Airspace Lab (Amsterdam) NLR-TR 71033 (1971).
[Il) O.L. Bowie, Analysis of an infinite plate containing radial cracks originating at the boundary
of an internal circular hole, J. Math. and Phys. 25 (1956), pp. 60-71.
(12) D.P. Rooke et aI., Simple methods of determining stress intensity factors, AOARDograph 257
(1980) Chapter 10.
(13) M.F. Buckner, A Novel principle for the computation of stress intensity factors, Z. Angell'.
Math. Mech. 50 (1970) pp. 529-546.
(14) P.e. Paris et al., The weight function method for determining stress intensity factors, ASTM
STP 601 (1976) pp. 471-489.
(15) Anon. Crack growth analysis software, Failure Analysis Associates.
(16) H. Vlieger, Residual strength of cracked stiffened panels, Eng. Fract. Mech. 5 (1973) pp.
447-478.
(17) T. Swift, Development of the fail safe design features of the DC - 10, ASTM STP 486 (1974)
pp. 164-214.
(18) T. Swift, Design of redundant structures, AGARD LSP 97 (1978), Chapter 9.
(19) C.e. Poe, The effect of riveted and uniformly spaced stringers on the stress intensity factor of a
cracked sheet; AFFDL-TR-79-144 (1970) pp. 207-216.
(20) V. Kumar et aI., An engineering approach for elastic-plastic fracture analysis, Electric Power
Res. Inst. NP-1931 (1981).
(21) V. Kumar et aI., Advanced in elastic-plastic/racture analysis, Electric Power Res. Inst. NP-3607
(1984).
 CHAPTER 9

Special subjects

9.1 Scope
This chapter covers a number of special subjects. Although the procedures
discussed in Chapters 3 and 4 for residual strength analysis, and those for crack
growth analysis discussed in Chapters 5 and 7 remain unaffected in principle,
slight complications arise e.g. in the analysis of surface flaws, corner cracks and
multiple cracks, or in the case that residual stresses are present intentionally or
inadvertently. In other cases, such as leak-break analysis or in a situation where
load transfer to other members may set up conditions for fracture arrest, the
interpretation of the analysis results may be somewhat different than usual.
Engineering procedures to deal with such problems are discussed in this chapter.
Examples are presented
The final sections provide a brief review of engineering solutions to mixed-
mode loading cases and a short discussion on damage tolerance of composites.

9.2. Behavior of surface flaws and corner cracks

The classical solutions for stress intensity and geometry factors of elliplical
surface flaws and corner cracks was discussed in Chapter 8, Section 3. More
recently solutions for part-through cracks of all kinds (surface flaws and
cotner cracks in tension and bending, surface flaws in circular bars, corner
cracks at holes, etc) were obtained by Newman et al. [1, 2, 3, 4]. The latter
solutions are generally acclaimed to be of better accuracy, and may be
preferable above the classical solution.
For the following discussion, it does not matter which solution is used. The
only issue of importance is that the geometry factor (and therefore the stress
intensity) varies along the crack front. In contrast to the case of a through-the-
thickness crack with an essentially straight front where K and {3 are the same
everywhere along the crack front and where one can speak of THE stress
intensity, in the case of part-through cracks the stress intensity and {3-depend

282
 283

Figure 9.1. Non-elliptical surface flaws and corner cracks.

upon location . To illustrate the behavior of part-through cracks, a surface flaw

under uniform tension will be used as an example and use will be made of the
classical solution. In principle these issues are the same for all other part-
through cracks regardless of the loading and the geometry factor solution used.
Surface flaws and corner cracks are not necessarily elliptical. In many
practical cases (Figure 9.1) the shape of a surface flaw is irregular; especially
when there are multiple crack initiation points, the linking up of the various
small cracks often leads to a non-elliptical crack (Figure 9.1). Provided
geometry factors for these irregular cracks are available, the damage tolerance
analysis can proceed in a similar manner as for elliptical flaws. However,
ready-made geometry factors are available only for elliptical cracks, reason why
in practical analysis all part-through cracks are assumed to be elliptical. If the
flaw has indeed an irregular front, this assumption may cause considerable error
in the analysis.
The geometry factors for a surface flaw in tension were discussed in Section
8.3. For the following illustration we will consider a case where alB is small so
that the front free surface factor, PFFS ::::: 1. The stress intensities for the crack
extremities (Section 8.3) are then

KA = ~aFa
JQ }
1.12
K c = ___ Ar:::::.
a "\Ina
JQc
(9.1)
284
In this equation, Q = q/, is a function of crack aspect ratio a/c. Its value can
be obtained from Figure 8.3. Some literature provides a set of curves for Q
where the lines are labled for various ratios of applied stress/yield strength
(a / FlY)' The curves stem from the time that attempts were made to compensate
for plasticity by means of a so-called 'plastic zone correction' to the crack size
[5]. This practice has been long abandoned as it is cumbersome and inadequate.
Nevertheless some analysts and computer codes still use these curves. However,
if the practice is not followed for other cracks, there is no reason to use it for
surface flaws. Moreover, if a plastic zone correction is made for the surface flaw,
it should also be made in the evaluation of Klc and of 11K in the rate diagram,
which it is not. Thus, only the line for a/Fty = 0 (elastic solution) is applicable,
which is the one shown in Figure 8.3. (Modern solutions [1-4] do not use a
plastic zone correction either).
For residual strength analysis the fracture criterion (Chapter 3) is:
Fracture if K = Klc (9.2)

Indiscriminate use of this criterion would lead to:

1.12 ~
JQ av na = K/c. (9.3)

The assumption is implied that fracture indeed occurs when the highest stress
intensity anywhere equals the toughness. As the highest stress intensity occurs
at the deepest point.. it is KA that would be used in Equation (9.3). If the flaw
is circular (JG7C = I) the stress intensity is the same everywhere (KA = K C ),
and the use of Equation (9.3) is certainly justified. But, in the case of a long
elongated flaw (JG7C small), the stress intensity at C, K C ,is still be considerably
less than the toughness when Equation (9.3) is satisfied. One could then argue
that fracture probably will be postponed, as is indeed borne out by some test
data [5, 6, 7]. Thus the use of Equation (9.3) may be somewhat conservative. As
it cannot be assessed theoretically how much fracture would be postponed, it is
safe engineering practice to use the conservative Equation (9.3)
It is emphasized that the toughness used must be (the plane strain fracture
toughness), Kif' At through-the-thickness cracks constraint is dictated by
thickness because the length of the roll of highly stressed material wanting to
undergo (Poisson) contraction is equal to the thickness (Chapters 2 and 3). In
the case of part-through cracks, the length of this role, as determined by the
length of the crack front, bears no relation to thickness. The contraction of the
roll is fully constrained by surrounding elastic material (Figure 2.8), so that
plane strain prevails at least in the interior at the deepest point of the flaw.
Hence, the use of the plane strain fracture toughness is indicated for ALL
part-through cracks, regardless of thickness.
Fatigue crack growth (and stress corrosion cracking) is also dictated by the
 ~ c

Figure 9.2. Development of surface flaws by fatigue . Lower right: originally circular flaw becomes elliptical due to bending (stress gradient).
N
00
V1
286
stress intensity. As KA is larger than K C , more growth will occur in a certain
cycle at A than at C (Figure 9.2). Hence, the flaw shape, a/c, will change and
therefore Q will change accordingly. In the next cycle KA is still less than K C and
again a will grow more than c. As a consequence, the flaw aspect ratio decreases
and the shape begins to approach a circle. Once the flaw has become circular,
the stress intensity will be the same everwhere, KA being equal to K C because
JiiC = 1. Then also, the growth at a will be the same as at c and the flaw will
remain circular. This development from elliptical to circular can be commonly
observed in cases of uniform stress (Figure 9.2), provided the thickness is large
enough for a circular shape to be reached. If the latter is not the case, there will
not be enough material space and the crack may reach the front free surface
before it has attained the circular shape. In the case of out-of-plane bending, the
stre~s gradient through the thickness may cause KA to decrease; in such a case
a circular shape may not be reached either (Figure 9.2). Similarly, for corner
cracks at holes, where KA is affected by k, more than is K C , the end situation may
not be a circular shape. (Figure 9.1 lower right)
For example consider a flaw under uniform stress with an aspect ratio of
a/c = 0.25. In that case K C = JiiC KA = 0.5 KA by Equation (9.1), i.e. the
stress intensity at C is only half that at A. Assuming that e.g. a Paris equation
(Chapter 5) applies with an exponent of 4, (da/dN = CAK4 ), the growth at C
will be only (0.5)4 = 0.063 times the growth at A: the growth in depth will be
16 times faster than the growth in length. This is demonstrated by the results of
an actual crack growth analysis displayed in Figure 9.3a.
This presents a complication for the crack growth analysis. In order to obtain
the growth of a, the flaw shape must be known because Q must be evaluated
first. But the new flaw shape cannot be known unless the growth of c is known
and vice versa. This problem can be solved by modifying the crack growth
analysis to account for the growth of a and c simultaneously. In every cycle KA
and K C are calculated for the current flaw shape. Then growth da at a and dc
at c are assessed; the new flaw shape a + da/(c + dc) permits evaluation of KA
and K C for the next cycle. The analysis then automatically provides the changing
flaw shape as in Figure 9.3. A good computer code for crack growth analysis will
include this feature. It should be noted that growth of a and c cannot be
computed independently as the flaw shape must be known in each stage of the
analysis; simultaneous analysis is a pre-requisite. Also, the analysis may have to
make use of different rate data for a andc if the crack growth properties in depth
direction (LS) are markedly different from those in width direction (L T).
The above problem can be avoided if the flaw is assumed to be circular to
begin with. Note however (Figures 9.3a and b) that this causes dramatic
difference in projected crack growth life. This is another demonstration of the
fact that assumptions have much more effect on the results of an analysis than
small errors in geometry factors for example. When assuming a circular flaw it
 287

1.8
C;
LW
I ELLIPTICAL
1.6 CIRCULAR
'-'
z a/c=1 a/c=o.25

1.4 1
LW
N
I
U;
1.2 I
'"
u
/
'"u"" 1.0
/
a,c/
0.8
//
0.6 '/''/

0.4 .::::-::::: --
0.2

10 20 30 40 50 60 70 80 90
(a)
LIFE (1000 CYCLES)

1.8
C;
LW
I
u 1.6 CIRCULAR ELLIPTICAL CIRCULAR
z a/c=1 a/c=o.25 8/C=1
I I
LW
N
1.4
I I
if)
1.2 I i
'"
u
I I
'"u"" 1.0 a,c/ a,ci
0.8

0.6 /
I
/
/. I
/
/
/ c;.... a
0.4
///

0.2
----------------------.---.~-
(b) 25 50 75 100 125 150 175 200 225
LIFE (!ODD CYCLES)

Figure 9.3. Effect of flaw shape assumptions. (a) Elliptical versus circular flaw assumption; (b)
Alternative assumptions.
is hardly worthwhile worrying about the accuracy of /3. The cicular flaw
assumption may lead to a conservative answer, it does not provide a realistic
answer. Besides, if there is a stress gradient through the thickness then even for
a circular flaw KA # KC, and growth of a and c still would have to be assessed
simultaneously. Whether the assumption of a circular flaw is conservative in
that case will depend upon structure and loading.
 Table 9.1. Hand calculation for change of shape of surface flaw. tv
00
00
2 3 4 5 6 7 8 9 10 11
a e al2e alB <P fJFFS fJA = 1.12 fJ FFS I<P !'J.a !'J.e = (Fc)m!'J.a New a Newe
Figure 8.3 Figure 8.3 (a + !'J.a) (e + !'J.e)
0.1 0.300 0.17 0.10 1.08 1.07 1.11 0.05 0.010 0.15 0.310
0.15 0.310 0.24 0.15 1.23 1.07 0.97 0.05 0.017 0.20 0.327
0.20 0.327 0.31 0.20 1.32 1.05 0.89 0.05 0.024 0.25 0.351
0.25 0.351 0.36 0.25 1.42 1.06 0.84 0.05 0.030 0.30 0.381
0.30 0.381 0.39 0.30 1.48 1.05 0.80 0.10 0.070 0.40 0.451
0.40 0.451 0.44 0.40 1.55 1.06 0.77 0.10 0.083 0.50 0.534
0.50 0.534 0.47 0.50 1.58 1.05 0.72 0.10 0.091 0.60 0.625
Almost circular

18 '58 ·58 18

Notes:
I. Find power m in rate equation (in this case m = 3).
2. In this example thickness B = I inch; start is ale = 0.1/0.3 = 0.333.
3. Work table horizontally; e in first line is starter value.
4. Take small increments !'J.a.
5. Calculate new a and c and use in next line.
6. Submit columns I and 7 as fJ-table to computer program for crack growth.
 289

u 50e
Cl

/
/
::s- 4:]0
,;
is J7lJ

301l

5 10 15 20 30 3S

Figure 9.4. Comparison of computer calculation operating on a and c simultaneously, and result
for a, using f3 of Table 9.1 in computer analysis.

In many analyses and in some damage tolerance requirements (e.g. military

airplanes) part-through flaws indeed are assumed to be circular. The reason for
the assumption may be either the desire for conservatism, or simply the lack of
a computer capability to assess growth of a and c simultaneously. If the lack of
a computer capability is the problem one is not necessarily forced to assume a
circular flaw, because a quick engineering judgement of flaw shape development
can be made as follows.
One approximates the rate data by an equation, preferably a Paris equa-
tion. Then a hand-calculation is performed as in Table 9.1. The depth a is
increased by small increments l'ia and the associated growth of c is assessed,
going through the table horizontally. In the next line l'ia is assessed for the
new flaw shape, etc. The end result is a table of f3(a) which accounts for the
changing flaw shape. This table of f3(a) is used as input in the crack growth
analysis, which then provides the 'proper' growth of a. The associated c values
follow from the table. Of course, the proper Paris exponent should be used (the
value of 4 in the table is only an example). The growth curve for a was
recalculated in this manner and the result is compared with a similtaneous
computer analysis of a and c as shown in Figure 9.4. Clearly there is reasonable
agreement. A few more steps as in Table 9.1 will further improve the result. The
procedure still works when there is retardation because retardation is mostly
determined by the ratio of stress intensities in successive cycles, and of course
these ratios are dictated by the stressess only, and not by [3. Similarly, if there
290
are stress gradients through the thickness, the procedure of Table 9.1 can be
used by including the effect in f3A (columns 5-7 in the table),
Circular flaw assumptions may lead to results far different from crack growth
in service. On the other hand if that assumption is not made, another one may
be necessary. Unless the initial flaw shape is known from tests or service cracks,
one must make an assumption for the initial value of ale. It needs no further
explanation that the analysis results will depend strongly upon the assumption
made, and there is a risk that it leads to unconservative results. In this respect
an assumption of ale = I (circular), may well be preferable. Unrealistic results
are not due to shortcomings of the procedure. With the proper input the
analysis result is reliable. It is the assumption that determines the magi tude of
the error.

9.3. Break-through; leak-before-break.

Whether leak-before-break is at issue or not, a clear distinction must be made
between break through of a part-through crack due to crack growth and break
through due to fracture. Crack growth of a surface flaw by fatigue or stress
corrosion may be terminated by fracture before break-through, upon which the
flaw extends by the fracture processes such as rupture or cleavage. The latter
happens when K reaches KIc due to the gradually increasing stress intensity or
because one higher load occurs in a sequence of smaller ones. If fracture does
not occur, the flaw will extend gradually by crack growth until the front reaches
the front free surface (Figure 9.5). When this situation is reached, the stress
intensity at D and E will be very high (protruding wedge of material) and crack
growth will occur rapidly from D to F and from E to G. The latter takes so few
cycles that it is safe to assume that F and G are reached immediately upon
break-through. The crack has then become a through crack with a length 2a
(equal to the original 2e at break-through). Continued growth is dictated by the
stress intensity of the through-crack (different 13). It is possible that the new
stress intensity of the through-crack is higher than the toughness. In that case
fracture ensues immediately and subsequent flaw extension occurs rapidly, the
results being 2 half structures.
If the flaw is in the wall of a pressure vessel, its break-through causes a leak,
but when fracture immediately follows, this leak actually constitutes a break. If
on the other hand the stress intensity of the through crack is still less than the
toughness, further crack extension will occur by crack growth, the structure
remains intact for the time being, and a simple leak will occur. This is a case of
leak-before-break, which is desirable because a leak usually can be readily
retected. Provided leaks would not cause fires or other dangers, inspections for
cracks would not be necessary; one could simply wait for a leak and then repair.
Indeed leak-before-break is a desirable damage tolerance property.
 291

I"
THROUGH CRACK
2a = 28
I..
THROUGH CRACK
2a=2C

-
2C 2C
I ..

Figure 9.5. Break-through of fatigue crack. Left: circular; right: elliptical.

The smallest through-the-thickness crack that can be formed is of a length

equal to twice the thickness (2a = 2B); this will happen when the crack was
already circular (a = c = B) at the time of break-through (Figure 9.5 left). If
the initial flaw is very slender, there may not be enough room for the crack to
become circular. In that case the through crack that develops will have a length
equal to the major axis of the ellipse at break-through, 2a = 2c. Should the
stress intensity of the through crack still be below the toughness, the through
crack will continue to grow by one of the cracking mechanisms. As the crack
now has another configuration, the appropriate f3 must be used. Various
possible configuration changes of different part-through cracks are shown in
Figure 9.6. Most computer codes for crack growth analysis have a facility that
effects this geometry change automatically.

2a a

B N B N B N

.....a N=NOTCH

Figure 9.6. configuration changes due to break-through.

-- a
292

Should the stress intensity reach the toughness before the crack has grown
through the thickness then a (fast) fracture will ensue. The flaw development is
essentially the same as in the case of crack growth, except that the break through
now occurs suddenly from a crack smaller than break-through size (Figure 9.7).
Upon break-through a leak occurs in the same manner as before and a con-
figuration change takes place. Again the question is whether the stress intensity
for the new configuration is above or below the toughness. If K > Kc (or KIc )
fracture will continue (break), but if K < Ke (or KJc) the through crack will be
sub-critical, fracture will stop and only a leak is formed: leak-before-break, and
further growth would be by crack growth instead of fracture.
Fracture of the part-through crack is governed by plane strain. However,
the constraint for the through crack is governed by thickness, and there is a
distinct possibility that plane stress or a transitional condition prevails for the
through crack. In that case the toughness is higher (Figure 3.6) as well, (leak
only). Clearly then, the chances for leak-before-break are better when the
wall is thinner. The longer the through crack, the smaller the chances that
K < Kn because K depends upon the through-crack size a. Thus, a circular flaw
has a better chance of leading to a leak than an elongated crack, the former
causing a through crack of size 2a = 2B, the latter a through crack of larger
size.
Suppose that break through occurs by fracture of a small crack. Also suppose
that K of the ensuing through crack is less than the toughness, Kc or Kin
whichever is applicable. Fracture being a fast event, the question arises whether
it is indeed arrested or whether it will continue even though K is less than the
toughness. Although dynamic fracture analysis [5,8] in principle can answer this
question, the analysis is rather involved and beyond the realm of day-to-day
engineering fracture mechanics. Besides, the solution depends upon the
assumptions. Pragmatically, a simple assumption can be made. If K is equal to
the toughness or slightly below, the dynamic effects will probably prevail and a
break occur. Accounting for a reasonable dynamic effect of 15%, a leak-before-
break will occur when K is less than 85% of the toughness Kc or Kle , whichever
is applicable. Instead of using 0.85 Kc (or 0.85 K 1c ) sometimes the so-called
arrest toughness, K a , is used (ASME requirements). Even with the use of the
arrest toughness [5] the question of state of stress of the through crack remains,
as the through crack could be in plane stress. In view of the fact that measure-
ment of the arrest toughness is difficult and data are scant, the pragmatic
approach of using a 15% dynamic effect may be the only possible avenue. An
example of analysis is given in the Solutions to Exercise 1-4.
It should be noted here that leak-before-break assessment requires establish-
ment of complete residual strength diagrams (Chapter 3) for the part-through
crack as well as the through crack. Various ale ratios must be considered.
Considering only one crack size or stress can easily lead to erroneous
 293

Figure 9.7. Break-through by fracture. Left: circular; right: elliptical.

conclusions, especially in this case, where geometry factors depend upon crack
size as well as shape.

9.4. Fracture arrest

In all cases of load transfer to second elements, fracture arrest is possible; arrest
occuring when the fracture reaches the second element. The resulting large
damage is more easily detectable which is desirable for fracture control. If a
structure has crack arrest capabilities, superficial inspections for large damage
might be adequate. In the literature both the terms 'crack arrest' and 'fracture
arrest' are used. As the condition relates to fracture and not to crack growth
only the term 'fracture arrest' is used here.
Fracture arrest principles will be illustrated first on the basis of stiffened
panels. Subsequently, the idea will be generalized to multiple element structures
and 'crack' arresters. Arrest analysis procedures were developed mainly by the
aircraft industry [5, 9-11]. At present, most large commercial aircraft are
designed with fracture arrest capability. As a matter of fact, damage tolerance
requirements for commercial aircraft (Chapter 12) tend to make this almost
inescapable. Thus, it is justified to start off with an illustration of aircraft
practice. An application to ship hulls is shown in Chapter 14.
It was shown in Section 8.10 (Figure 8.25) that load transfer to stiffeners
causes a decrease of /3, as shown in Figure 9.8. In turn, this load transfer causes
a stress concentration L in the stringer. As in all fracture analysis a complete
residual strength diagram must be determined to avoid misinterpretation. The
residual strength diagram is calculated (Chapter 3,) on the basis of the criterion
Fracture if K = Kc or /3aJ1W = Kc. (9.4)
Note that the use of the plane stress or transitional toughness is appropriate
here, because the plate will usually be of too small a thickness for plane strain,
while all cracks considered are through the thickness. For large W, the geometry
factor will be essentially equal to unity for unstiffened panels (/3 = I). Using
Equations (9.4) the fracture stress for the un stiffened panel is calculated for a
number of crack sizes and plotted as the dashed line in Figure 9.8c. Appropriate
294
/3 L
2

'~
0·6 ~

0.2 0.5

0.2 0.6 1 ajb 0.2 0.6 1 alb

(a) (b)

RESIDUAL
STRENG

I ~
2b
I
1
1 cr
\
\
\

B
WITH STRINGERS

-
f
i 0: Kc
I I t"/3"fira
, tI ...!IiI -WITHOUT
I I
STRINGERS
I I

,I
0.6
alb

(c)
Figure 9.8. Residual strength of stiffened panel. (a) Geometry factor; (b) Stringer stress concentra-
tion; (c) Residual strength diagram.

corrections should be made for small crack sizes where (J is close to Fry, as
discussed in Chapter 3.
Subsequently, the residual strength diagram for the stiffened panel is esta-
blished, also using Equation (9.4). For small cracks (13 ~ 1) the residual
strength curve is the same as for the unstiffened panel (Figure 9.8c), but for the
case of alb = 1, the 13 for the un stiffened panel is much less than 1 (Chapter 8).
If for example 13 = 0.4 then the residual strength of the stiffened panel will be
1/0.4 = 2.5 times as high as that of the stiffened panel. This defines point Bin
Figure 9.8c to be 2.5 times as high as point A. For intermediate crack sizes the
residual strength curve for the stiffened panel will be as shown (see 13 in Figure
9.8a).
Apparently the residual strength diagram of the stiffened panel exhibits a
relative minimum and a relative maximum at alb = 1. Paradoxically, the
 295

residual strength under the presence of a long crack of a = b is higher than for
certain smaller crack sizes; the strength increases beyond the relative minimum
when the crack is longer.
Consider a case where a (fatigue) crack of size al has developed (Figure 9.8c).
Should a high stress (load) occur of magnitude 0'1 a complete fracture will ensue.
Next consider a case where the crack grows by fatigue until size a2' At a stress
of magnitude 0'2 a fracture will occur, but it will only run to a3 and be ar-
rested at a3' Note that the curve identifies all points where K = K,. The
fracture will proceed to a 3 , but then again (K < K c ); although the crack size
increases, the decrease in p is such that pJ1W, decreases. In order for fracture to
continue from a 3 the stress must be increased to iI. The damage will then increase
by fracturing to ii, but this fracturing will be controlled tearing. At the point (ii,
a) fracture will again be fast and unstable.
For ALL cracks larger than as' fracture will be arrested, and the residual
strength will still be ii, as illustrated in Figure 9.8c. Naturally, for all cracks
smaller than as an immediate total fracture would result, but this would require
a very high stress. If the minimum permissible residual strength, up (Chapters I,
11, 12) is up < iI the structure can always sustain a damage of a as shown by
the horizontial line in Figure 9.8c. Of course, a fracture can occur at stresses
lower than iI (a > as), but such a fracture would be partial only: it would be
arrested at or close to the stringer upon which the residual strength still would
be iI.
Indeed, if the structure is designed such that iI > up, large damage can be
sustained under all circumstances where 0' < up, so that fracture control is
facilitated by inspection for large damage. For reasons of stiffness airframes are
built of skin-stiffened structure anyway; it then is a small step to provide the
above fail-safety by apropriate selection of stringer size and spacing. (As
discussed in Chapter 8, the reduction of P by load transfer depends upon
stringer stiffness and spacing, i.e. upon the stiffening ratio Ji..)
Up till this point the stiffeners and the fasteners have been ignored. When the
fracture approaches the stringer the stresses in the stringer will increase signifi-
cantly because of load transfer from plate to stringer, the very reason for which
the plate experiences lower crack tip stresses and lower p. This was demon-
strated in Figure 8.25. The stringer stress concentration, L, is depicted in Figure
9.8b. Load transfer to the stringer has to be accomodated by the fasteners
causing high fastener shear. As the stringer is uncracked it will fracture when its
local stress is equal to its tensile strength O'{ = LO' = Ftu . Thus stringer fracture
occurs when the remote stress is O'fs = Ftu/L. Using L from Figure 9.8b, this
criterion provides the stringer fracture line as in Figure 9.9a.
A fatigue crack a l in Figure 9.9a will cause a fracture at a stress 0'1' and arrest
at a2' The stress can then be increased to iI (fracture proceding to a) upon which
the stringer will break. Then load transfer to the stringer is no longer possible
296

STRINGER
FRACTURE

PLATE
FRACTURE

I '

, '
I I , I

(a)

~
2b

as
(b)
Figure 9,9, Effect of stringer on residual strength diagram. (a) Stringer critical; (b) Plate critical
(stringer material with higher F ru than in Figure a).

and total fracture ensues, This means that (j is reduced to the level of the
intersection between stringer fracture line and plate fracture line, (Compare
Figures 9.8c and 9,9a). This is an undesirable situation, It can be improved by
selecting a stringer material with a higher F,,.., The diagram of Figure 9.9a will
then change into that of Figure 9.9b, Again, proper design will affect the crack
arrest capabilities to the degree required; other ways to change a stringer critical
case into a skin critical case will be shown later in this section (Figure 9.17)
It is important to note here that the stringer fracture is detemined by F,u
because the stringer is uncracked. Should the stringer be cracked already (which
is unlikely), its strength would be much less and be determined again by the
toughness, so that it would fail at lower stress, In such a case arrest may not be
possible,
 297

Increased {3 due to
1 - - 1 _ ... ~rOken stringer 0

alb
(a) (b)

~~r
.
0 2

~
2a

1--. b
.
b

Broken st,inger 0

•
1 2
~ 1m
(e)

Figure 9.10. Crack strting at fastener; with broken stringer. (a) Geometry factor; (b) Stress con-
centration in stringers I and 2; (c) Configuration.

Fastener failer due to high shear must be considered as well. The fastener
forces can be calculated accurately or estimated as discussed in Chapter 8. In
an uncracked structure the fasteners carry hardly any shear, so that they are
often light, their primary function being to hold plate any stringers together.
However, in the case of a crack the fasteners must transfer load to the stringer
so that they are subjected to high shear. Thus, in order to make fracture arrest
possible, heavier fasteners may be necessary.
It may seem that the case discussed above is not a relevant one, because cracks
will not occur between stringers. If cracks develop they will do so at a fastener
hole at a stringer (Figure 9.10). The stringer crossing the crack will be highly
stressed, and develop a crack as well. It will not take long before this stringer
breaks. The plate then will have to carry the load of the broken central stringer,
so that K is increased; f3 > 1. However, from then on the problem is as before:
there is a crack between the two adjacent stringers. By simple redefining the
stringer spacing as b, instead of 2b as before, all previous arguments will hold
(Figure 9. lOb), with slight modifications to f3 and L; these parameters can be
obtained from handbooks as discussed in Chapter 8. Typically, the stringer or
frame spacing in a commercial jet is from 8-12 inches. These structures are
designed to sustain a two-bay crack (16-24 inches) with the central frame or
stringer broken.
If stringers have higher stiffness, or are more closely spaced, they will transfer
more load and be more effective in reducing {3. The smaller the fastener spacing,
298
_ (FATIGUE) CRACK

,
~ FRACTURE

~~INTACT

i ?H
ft
(a) (b)

~V¥(;'fl'_(

(e)
2f
(d)
t" ¥ .# ,
Figure 9.11. Behavior of stiffened plates with 'zero' fastener spacing. (a) Integral; fracture severs
stringer; (b) Welded; fracture severs stringer; (c) Adhesively bonded (physical separation between
plate and stringer; (d) fillet weld (physical separation).

the more effective load transfer to the stringers, This can be understood intuit-
ively, because the closer the fasteners at either side of the crack tip the more the
stringer will prevent the crack from opening (lower K and lower [3). Hence, small
fastener spacing improves arrest capabilities, as long as the fasteners are of
sufficient size to carry the shear.
The smallest 'fastener' spacing is obtained if the stringer is integral or con-
tinuously welded (Figure 9.11 a, b). Unfortunately, this does permit fracture to
proceed into the stringer, severing the latter in the process. There being no
physical separation between plate and stringer the fracture will include the
stringer, upon which total fracture ensues. An alternative way to obtain 'zero'

- - - - - - -----i

---------

Figure 9.12. Possible arresters for pipeline.

 299

2b

I--_ _"'-b_ _~~

I
~ b----t

Figure 9.13. Various kinds of second elements.

fastener spacing is adhesive bonding or fillet welding (Figure 9.l1c, d). Because
of the physical separation the running fracture cannot penetrate the stringer.
The bonds or welds must be capable of transferring the load to the stringer by
shear.
Although in the above discussions the word stringer was used, one need only
replace the work 'stringer' by 'arrester' or 'second element', and the arguments
will be equally applicable. For example, a crack arrester for a pipeline could be
designed as in Figure 9.12. and the above discussions would apply. Similarly the
effects of second elements such as shown in Figure 9.13, whether integral or not

~~
alb

alb

cr 1

Integrated t
::=C6K" ~
2b
AK =(3Acr..{ia
0.5
L cr

LIFE (HOURS. CYCLES)

Figure 9.14. Slow down of fatigue crack growth due to arrester.

300
can be assessed in the manner described. Analysis procedures are already in
place and Pfor many configurations can be obtained from handbooks, while L
can be obtained from P (Chapter 8).
In crack growth (as opposed to fracture), the p-reductions discussed are
effective as well. If K remains less than Kc the crack will continue to grow by
fatigue or stress corrosion. Crack growth is also dictated by K; it will be slower
if K decreases due to decreasing p. If fracture does not occur crack growth will
be as shown in Figure 9.14.
Note that the decrease in p will slow down crack growth even when the
stringers are integral or welded (Figure 9.11), although these configurations may
not be effective for fracture arrest. Thus the effect of second elements on fatigue
crack growth will be significant also in configurations such as in Figure 9.13,
provided no fracture occurs. In this respect integral or welded stringers will be
more effective than mechanically fastened stringers; only in the case of fracture
will their effectiveness be less to non-existent.
The possibility that crack growth may be intermitted by fracture must be
counted on. If the loading is of variable amplitude a high load in the sequence
may cause fracture and arrest (Figure 9.15). Subsequent growth again occurs by
fatigue, be it that all of a sudden the crack is much longer due to the fracture.
An intermediate fracture will reduce the 'life' considerably. The long cracks
between arresters can be sustained (residual strength) so that inspections might
focus on the detection of obvious damage, but the latter can be sustained only
for a short time, because crack growth continues. Hence, the inspections
(though more easy) must be repeated frequently.
Fracture is a fast process. The dynamics of the problem might suggest that
fracture would continue even if K < Kc. However, the only way the fracture
could be driven past the stringer (or arrester) would be by the kinetic energy of
the fracturing structure. This kinetic energy is relatively small and the question
really is whether this energy can be absorbed by the arrester. If the arrester is
uncracked, which is normally the case, it easily can absorb this energy because
the plastic deformation energy of an uncracked part is very large. This is
demonstrated by a numerical example in Chapter 14. It is also borne out by
many experiments on stiffened panels [9-11] of which some results are shown in
Figure 9.16. Design for specific arrest requirements is possible, because-as
discussed in Chapter 8-the effect of the various parameters in the problem can
be readily assessed. Essentially all possibilities for design and design improve-
ment are shown in Figure 9.17. A detailed numerical example of arrester design
appears in Chapter 14.

9.5. Multiple elements, multiple cracks, changing geometry

Analysis of structures with multiple elements proceeds as discussed in the
previous section. The second element may be mechanically fastened, welded or
 301
integral (flanges of L, lor U sections, as in Figures 9.11 and 9.13). For many
of these cases 13 can be found in a handbook [12], although superposition and
compounding are usually necessary (Chapter 8). Note: the case of an lor L
section may not appear specifically in the handbook; instead there will be a case
of a crack in a thin member approaching an area of increased thickness; the
flange width should then be interpreted as the increased thickness; alternatively
the flange can be treated as a stringer; if load transfer does occur its effect on
13 should be included in the crack growth analysis.
Figure 9.18 illustrates cases where the presence of one crack influences the
other, i.e. extension of crack 1 increases/decreases 13 of the other crack.
Therefore, these cracks must be analysed simultaneously. The procedure for
simultaneous growth is the same as for the simultaneous growth of a and c of
a surface flaw as discussed on Section 9.2. If the computer code does not have
this capability the following shortcut can be made in a manner similar as in
Table 9.1. Both cracks are subject to the same stress history (relatively, as there
only will be a proportional difference in stresses). As a consequence the growth
of crack 2 during a certain small interval or cycle can be prorated to the growth
of crack I. At a certain stage with a, and a2 , the geometry factors are 13, and 132,
both of which now depend upon a, as well as upon a2' A simple rate equation
(e.g. Paris) is used to determine a priori how da, of a, changes due to the growth
of a2, and a table is made just as in the case of a surface flaw (Table 9.1). The
end result is a table of 13 for a, in which the growth of a2 is accounted for. This

LONGEST POSSIBLE LIFE~

I NTERMEDIATE FRACTURE ........

SHORTEST POSSIBLE LIFE

i-- - --- -- (~-~-~--

., ________ J~:c~w-// I
I
r

:J!:
0- j

l
2b

j. 0- j.

LIFE (HOUHti, CYCLES)

Figure 9.15a. Life as affected by intermediate fracture and arrest. Intermediate fracture and arrest.
 w
o
tv

__ II ----=-::: ,. __ undamaged material

strenght

____+-__ Design ultimate

(1,5' Ll)
=-=--=--=-~::::-===~=::::~"i{ '=¥th)&.
----T->----+-----c LL
I I I I;
-----~------~-----J-4--
I I : :I I ®:
I I I' I
@ ) I It

10' "0~10J 104 10" 10· 107

10 17 6 51 i. 3 II 2 1.10' 2.10,' 3"10'4 ii 4~'1041 6"0' 1'0. 2 1('-' 100
/ I I J I I
I .,. II I Exceedings
ICrock S,zeio, Inches II I I I Flight Hours I I
,,-i I / 1/ I :
I I II
I ! I, / / 1/ I I
1x 10' / I II I I
I I,
I II / / / I I
/
I II / / / I I
I II ,," / // / I
I II I 2~.!.t- . . . // ;1/1 I;'
: II I '" / I I I
I II I
I II I //,,':'/ / I
/' /,./ / /
I II I / /./
II I 3)1104 ..-_- . . . . . . . . . . . . / ; ; / /
II
a:t.Jt _____ _ ------::. ............ - .,// /
.:::---- ,/ /' //
F,IC------
/ ......... ///

4)110 4 - - -
D"
-~~
~~

5'10'---
D'

6.10

Figure 9.15b. Life as affected by fracture and arrest. Actual case in aircraft structure.
 303

40
,', ~>' JEI- I
cr 35
WATERIAL.7015-T6 I
• TOTAL FAILURE OF PANEL
'30

25

20

15
SPEC'MeI ~ 5 1 , , ,
o 80 120 160 200

Figure 9.16. Test data for stiffened panels [10].

table is submitted to the crack growth program, which now calculates the
growth curve for a j • The associated a2-sizes follow from the pre-established
table, as does c in Table 9.1
A changing geometry is also a common problem. For example, (consider the
case of Figure 9.19 ) with a crack starting in the thin flange (plane stress; high
Kc). Although for such cases f3 sometimes can be found in a handbook, an

Increasing
stiffening rallo c.
o.

Increasing fastener
d. e. stiffness I

2. 2.

Figure 9.17. Options for improvement of arrest capability.

304

~INTACT

mCAACKED

~Z=Z=2z~~ilII!lem!!!m.m::::27:=Z:=2=Z=2=2C
(a)

4&,>5»> ?? ? Z Z Z Z

(b)

~ _a_,_
:Z:Z-Z ZZ 22/1 MiN ZZ 2 2%

(c)

a a, (6 <:<: «i s s " S S "' s S S s S "

(d)

Figure 9.18. Cases oLmultiple cracks. (a) Increased f3 due to load transfer of broken 2nd element;
(b) Decreased f3 due to intact 2nd element: (c) Interacting cracks in same element: (d) Interacting
cracks in different elements.

engineering procedure is shown below; similar approaches for other configura-

tions are possible. For an inspector the crack size would be a in both cases, but
it would be too conservative to use the same definition of a in the residual
strength analysis, as this would be implicitly assuming that the load path
represented by area A is cut and that this load has to bypass the crack. But area
A does not carry load because it does not exist and therefore only the cut load
of area B (the flange) has to bypass and crack. Thus it would be more reasonable
to assume the configuration of Figure 9.19c where only the real cut load path
is added to a fictitious crack size, aeif' rather than to the real crack size. When
executed carefully, the procedure presents no difficulty as shown below.
As long as the crack is in the flange (high K, ), the problem is essentially the
one of Figure 9.19b for which the residual strength diagram can be calculated
(Figure 9.19d) in accordance with the standard procedure. The right part of the
curve for large a would be incorrect but that is irrelevant because it is not used
anyway. Subsequently, the residual strength diagram is calculated for the body
(Figure 9.19c) with an effective crack size, {[elf, the flanges accounted for in {[ell'
 305
Wb .
~\-;~I-~
\)
.-..
,1 l 1tl

... T
': a
Wf

(a) I

(b) ---
Illllllllll
I a

(e)

! ;jody - KC b or Kid
I
i

;~
o
,<
r--!-
i " lOf'J

a I i
(d)

Figure 9.19. Changing geometry (approximation). (a) Configuration; (b) Flange; (e) Body; (d)
Resid ual strength.

The residual strength diagram is plotted on the same scale as that of the flange
(Figure 9.19d) but in such a manner that Geff = 0 coincides with the location of
the fictitious edge of the body. Note that Kc for the body is lower (larger
thickness) which must be accounted for in the analysis; the body may be so thick
that plane strain prevails (K1c )'
The two residual strength curves are now combined; it is of course the lowest
residual strength curve that is of interest. The discontinuity is artificial and the
change-over can be considered as the average of the two curves. In this
particular case, a more rigorous analysis can be made, because a handbook
solution for a thickness change is available [12].
306
This example demonstrates that the complete residual strength diagram is
required to determine the permissible crack size. Also note that the abscissa of
the residual strength diagram is indeed on the scale of the physical crack size,
the only one of interest for inspection. Naturally, the problem can be solved by
using handbook solutions, but the above example demonstrates that where such
solutions are not available good engineering approximations can be made.
A changing geometry may also be encountered when multiple elements are
cracking simultaneously, as in the case of Figure 9.20. It may not be realistic to
consider only the case that the entire reinforcement is cracked. The engineering
approach illustrated below can provide a reasonable and slightly conservative
result. First consider the two members more or less as independent parts and
then the combination.
When the cracks are small, there hardly will be any load transfer between the
members. Thus a residual strength curve can be calculated for each. However,
for longer cracks there will be load transfer and therefore the net section yield
(collapse) condition is meaningless for the reinforcement and can be ignored.
The independent residual strength curves are shown in Figure 9.20b. Finally,
the residual strength curve is obtained for the case that A is completely severed;

8][_~._1.__..
(a) -a I

A independent but no net sect jon

yield or right hand tangent

B independent

(b)
Physical crack size a

Figure 9.20. Changing geometry with two members (approximation). (a) configuration; (b)
Residual strength.
 307
b

,
.------------;::jj,--
L -_ _ _

t
~Ij-

t
,-. L_'
t t J 1.
:1 : ..
I .. It ..
1 :, 1--
III"
L_ Failed Plan\'l:; /Most load
1 "'-:.
transfer
~~ . ~' .: here
1'- /t--- -
e
I..
II
I,'r'
III' .
I
..
__
..L
s
1 T
: : I..
, , ,
I
C I :J"
• • • I I ~

° c
Continuous panel

r) !::
+IV
c::::.=========~
••_ _..::;2a:.:=:..::b_~·r
.Is
1
T" _K_ cos ~ sin!!. cos ;lll 1
,f21rr 2 2 2
1
0"'[ K"o,hrtv2 1
2

Figure 9.21. Estimate of fastener forces in case of failed plank (top: planks; bottom: full plate
equivalent).

/3 for this can be found in handbooks, the resulting curve is shown in Figure
9.20b as well.
When cracks are small member A will be intact and curve c is not
applicable. Thus the curve starts out as the lowest of curves a and b (in this case
a and b coincide). Where curves band c intersect, member A will be failed and
curve c applies. On the basis of these two criteria, the solid line can be drawn
as a good approximation of the residual strength curve. The result is certainly
approximative, but the top of b and the tail of c are accurate. Not too many
different curves can be drawn between the top part of b and the tail of c.
Therefore the interpolation must be reasonable.
In the example of Figure 9.21 the centre plank may experience fracture
instability. The structure may qualify as fail safe if the remaining planks can still
308
carry (Jp. In the case of three planks this would cause an average stress of
1.5 x (Jp in the two remaining planks which will be less than the actual
strength of these (intact) panels, if they were initially designed with a safety
factor of 1.5 or more. The question is, however, whether the load of the failed
panel can be transferred to the intact planks. Load transfer has to occur through
the fasteners. Displacement compatibility requires (see discussion Figure 2.2)
that the load in the center plank remains there until close to the break and that
almost all its load will be transferred by shear by a few fasteners close to the
break. If these fasteners would fail under this shear the longitudinal splice would
zip open and the structure would still fail. Thus fastener strength is the crucial
point, and two rows of fasteners are usually required.
A good way to find the fastener forces is through finite element analysis, but
this will require assumptions with regard to the fastener stiffness; the fasteners
will yield and must therefore be represented as non-linear springs. If they are
simply assumed elastic the result of a finite element analysis will be no better
than of the following engineering approach to obtain a first estimate.
Compare the structure and the continuous panel in Figure 9.21 b. There will
be shear stresses along line A-C. This is essentially the Slime in the case of three
planks, but the shear in the vertical plane would have to be transmitted by the
fasteners. The shear stresses in the continuous panel are known from Equations
(3.1) as shown in Figure 9.2Ib. For e = nl2 they are: r = 0.35KIJ2iIT. As a
first approximation it may be assumed that the total shear force over the
fastener distance, rBs, should be transmitted by the adjacent fastener or, if there
are two rows of fasteners, shared equally by two fasteners. Then the forces on
the fasteners nearest the fracture can be estimated as P = 0.35 BfJ-.jbI2s, since
a = band fJ is the applied stress. This is probably a conservative estimate,
because fastener and hole deformation will tend to spread the shear somewhat
to more remote fasteners. The estimate can be improved by integrating
over s instead of taking r = s12.
In cases of rapidly decreasing K fracture arrest conditions may be set up. The
most prominent examples are stringer-stiffened structures as discussed in
Section 9.4. In general, it applies to all built-up structures where load transfer
to uncracked members or flanges is possible. Other cases of decreasing K may
be encountered if cracks are in fields with negative stress gradients, or in cases
of displacement control, when stresses decrease owing to reduced stiffness due
to the presence of the crack (as e.g. in the case of thermal stresses). Finally,
fracture arrest conditions may occur due to changing constraint (plane strain to
plane stress), as discussed in Section 9.3, or if a crack in a thick part starts
penetrating a thinner part. In such cases arrest is not due to decreasing K but
due to increasing toughness, Kc.
As in the discussions of stiffened panels, the term fracture arrest is used. Crack
arrest, (arrest of crack growth) will seldom occur: a decreasing K will simply
 309

4.5

4.0

l )
3.5

3.0

.- 2. S
~
2.0

1.5

i.O

0.5

0.5 1.0 LS 2. 0 2. 5 3. 0 3.5 4.0 4. 5

(a) CRACK LENGTH ( INCHES)

4S

I
40
~
z _RES.SIR.
35
~

3D •• EL~STl[
<
~ '~B
-cr
25

_ HNG€)IT
20
a, 0
IS

10

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8

(b) CRACK LENGTH ( INCHES)

1.8
~
w
I
u 1.6
z
::;
w 1.4
N
u;
~
1.2
u
<
'"u 1.0

0.8

0.6

0.4

0.2

l.O 20 30 40 50 60 70 80 90
(C)
LIFE (1000 CYCLES)

Figure 9.22. Case of decreasing p. Lug with bad bypass. (a) Geometer factor; (b) Residual strength;
arrest condition between a, and ii; (c) Crack growth; lower line, but upper line iffracture burst from
a, to ii (smaller jump possible).
310

slow down crack growth by fatigue (Figure 9.14). In the case of stress corriosion
cracking crack arrest may occur if K drops below KIscc '
Cases of decreasing K are quite common, but fracture arrest is not always of
significance. For the case of a 1ug the geometry factor was derived in Equations
(8.25). This geometry factor decreases rapidly because of the term Wjna. When
the crack is long, the effect of the finite Width, W, will act to increase p again.
Hence, for a particular case as in Figure 9.22, p may be as shown. The residual
strength diagram follows from (1 = KclPFa with appropriate corrections for
small a (Chapter 3). Thus the residual strength diagram is as shown in Figure
9.22b.
Clearly, there is a dip in the residual strength diagram, so that fracture arrest
will occur for all cracks between as and ii, the residual strength over this range
of crack sizes being essentially equal to jj as in the case of stiffened panels
discussed in Section 8.4. This arrest is not of much consequence for fracture and
residual strength other than that the residual strength does not drop below jj
until the crack exceeds ii. However, in the interpretation of the fatigue crack
growth curve this effect should not be ignored.
The fatigue crack growth curve is shown in Figure 9.22c. Indeed a slow down
between as and ii is discernible because of the decreasing K. It is quite possible,
however, in the case of variable amplitude loading that a high load occurs
between as and ii. Should this load have sufficient magnitude to cause fracture,
the damage will progress by fracture from e.g. as to ii (Figure 9.22b). Further
growth at lower stresses now occurs by fatigue as shown in Figure 9.22c, but the
fife between as and a is lost. Such 'fracture bursts' are sometimes observed in
service failures of structures with decreasing K. Thus, it may be prudent in the
analysis of such a case to ignore the crack growth life between as and ii (Figure
9.22c) for a case of variable amplitude loading. Whether or not such prudence
is taken must remain an engineering decision. Fracture mechanics can analyse
the problem, but cannot predict which loads will occur in the future.

9.6. Stop holes, cold worked holes and interference fit fasteners.

If a service crack is detected an immediate repair is not always opportune. In

such a case drilling so called stop holes may be a good temporary repair. A stop
hole eliminates the sharp crack tip, but leaves a notch. It will take some time
before the crack reinitiates, as shown schematically in Figure 9.23a. This reiniti-
ation time is the 'gain' effected by the stop hole.
The sharper the notch the higher is the stress concentration, and the sooner
the crack reinitiates. The stophole configuration can be considered an elliptical
noth (Figure 9.24) for which;
 311

k, = 1 + if, (9.5)

Where r = Dj2 is the radius, of the stop hole (Chapter 2), I = a for the
configuration of Figure 9.23b. Hence, larger diameter stop holes are more
effective. As a matter of fact, the configuration of Figure 9.23c would be the
most effective, as can be seen from Equation (9.5), but it is not a desirable one
for other reasons. Also, stop holes will have more effect for smaller cracks as
borne out by Equation (9.5).
Once the crack reinitiates the effective length of the new crack will be
an = a + Dj2 for the configuration of Figure 9.23d. See also Chapter 8 on
cracks emanating from notches. This longer crack will grow faster than the
original crack. From this, as well as from Equation (9.5) and Figures 9.23 a
through d, it can be concluded that the best position for the stop hole would be
the one of Figure 9.23b. However, in that situation one would never be sure that
the crack tip is eliminated; if it is not, the 'stop hole' becomes just a hole and
will have no effect at all. Hence, the best compromise is a stop hole centered at
the crack tip (Figure 9.23d).
The beneficial effect of a stop hole can be improved considerably by cold
work. Plastic deformation of the rim of the hole will stretch the material
permanently. However, since this material must fit in the surrounding
elastic material, it must be compressed to its original size. Thus the cold work
(plastic deformation) will create a residual compressive stress around the rim of
the hole in much the same manner as in the case of an overload (Chapter 5). Its
workings are discussed later in this section.
There are many ways to effect the cold work (plasticity). Methods as crude
as hard blows with a blunt instrument are effective and may be useful in tests,
but somewhat more sophisticated procedures are indicated for practical appli-
cations. The procedure most uniformly used is hole expansion by pulling a
tapered pin through the hole. Figure 9.24 shows how this expansion (cold work)
improves the beneficial effect of stop holes [13]. The figure shows cases of very
large hole expansions; normally one would use 3-5% in order to avoid introduc-
ing new cracks.
The above arguments are not limited to stop holes. If all structural holes
are cold worked to begin with, the residual compressive stresses around the rim
would postpone crack initiation as well as reduce crack growth rates once the
crack has started. It has become common practice in aircraft production to cold
work significant holes. Obviously,the procedure is not limited to aircraft
structures, but can be used just as effectively in any other structure, because it
is based on a principle, not a specific material or structure.
Devices for hole expansion are commercially available, but simple tools can
312

..
WITHOUT WITH
STOPHOLE 20
•
SlOfIHOLE
~
e-+---""""G"'"
(b) .....

(e)

N
(d)G---0
(a)

Figure 9.23. Temporary repair with stopholes. (a) Principle, (b), (c), (d) Various configurations.

16~--------------~--~--~--~--~~M---~--~
repair codes I, ! 71
E unrep I I
--+-1'
0 I ,

E 14 ~ 1~ :::
j
ex p ~.n si onJ- - - 6 - - - l.----¢t--+--.Q-+:
j ----1

J:
i

~~ , 2 <> 20 /0 0
..
5 mm stop holes
I
I )
' T,j

1
steel sheetj:;
10~--~--~--~--~--~----~~~--~--+--~-~H
,;L.
u
o
'II 1
1~
y~~-4~~~=*~~--ri'-
L---'LT I_V
L
8~~'----1 ---I~~l~j~!---+---+--~--~~~~
I
U
I I : I 1

6 1--__:~-__1I--+___<l:xZ1I>_-~,---11---__1'- ----+----+--- - + - - - j

I ~
4~-~--_+-~~~4_---L--__1~--~--~-__1~--+_-~
i : I
~~~ i
2~~

o 50 100 150 200 250 300 350 400 450 500

• endurance in kilocycles
Figure 9.24. Effect of stophole expansion on effectivenes of stopholes [13].
 313
(j) HIS",T PIN FROM TOP
(2) INSERT H'.~YES Dc BuSHING

l3)~PP _ Y WASHE, AND NUT

(4) INSERT FORKfcD RESTRAINER

RES TRA I NER ~R I GHT ONE FORKEDl

"""
"
SUBJECi

(a) (b)

Figure 9.25. Simple devices for expanding holes (more sophisticated devices commercially
available). (al Two side access; (bl One side access from top.

be made without great expenditure. If the hole is accessible from both sides, the
device can indeed be very simple as shown in Figure 9.25a. A tapered pin for e.g.
three or five percent hole expansion is inserted from the bottom. A common
wrench is used to pull the pin through. As the pin will cause some damage
(scratches) to the hole subsequent slight reaming is advisable, but the detriment-
al effect of such damage usualIy wilI be outweighed by the beneficial effect of the
residual stresses.
If the hole is accessible from one side only, a device with a split bushing must
be used. This is illustrated in Figure 9.25b. The tapered pin in this case has a
maximum diameter equal to or less than the hole diameter. It is inserted from
the top. Subsequently, the two halves of the split bushing are dropped into the
hole from above. The remaining hole is now too smalI for the pin. If the pin is
pulled through, the hole wilI be expanded because the bushing is split and can
expand. The bushing is removed after the pin is through. Commonly, the
bushing is not re-usable. Again, more refined tools are commercialIy available.
The principle underlying the effect of cold work will be discussed later,
because it is more illustrative to consider interference fits first. Oversize
fasteners, such as e.g. taper locks (trade name) improve life to crack initiation
and decrease the rate of growth of small cracks. In such a case the hole is also
stretched and it is often argued that the effect is the same as of cold work, which
is not the case. In cold work, the oversized pin is removed, allowing the material
to relax and causing the build-up of residual compressive stresses. In the case of
an interference fit the oversized fastener is left in, causing residual tensile
stresses.
314

t CYCLE OF
O"f WITH
INTEFERENCE

r~
" -'t
: I I I
CYCLE OF CYCLE OF
, I I I
O"f, I 11 ,
O"f WITHOUT
INTEFERENCE
I l'
I '/

(Ja(APPLIED STRESS) O"aOlPPLIED STRESS)

(a) (b)

Figure 9.26. Effect of interference fit fastener on cycle oflocal stress. (a) Open hole; (b) Interference
fit fastener.

In order to simplify the explanation, consider a rectangular hole, as in Figure

9.26a. If the applied stress varies between 0 and O"al then the local stress cycles
between 0 and 0"/1 (stress concentration k, = O"IiO"a). Now put an oversize spring
L in the hole I (Figure 9.26b insert). As the spring is too large (interference),
there will be a pre-tension, O"prt at the edge of the hole as shown in Figure 9.26b.
Assuming elastic behavior it can be established how the local stress will vary
during cycling of the applied stress. If the plate is stretched, the hole will stretch.
Eventually the size of the hole will be I + AI = L. Then the spring will be loose
and it can be taken out. From then on the behavior will be as that of an open
hole. Hence, eventually point A will be reached while the stress starts in O"prt.
Whether the line O"prt-A is straight or curved does not matter for the explanation.
If the applied stress varies from O-O"al, then the local cycles from O"prt-O"I2, instead
of from 0-0"/1 • Hence, the stress range is reduced, R is increased. Because !l0" has
more effect than R, the life is increased and crack growth is slower.
Cold work does exactly the opposite, as shown in Figure 9.27. Consider a rim
of material around the hole. This ring is expanded and is then too large so that
it is squeezed by the surrounding elastic material after the pin is through. The
edge of the hole is then under a compressive stress. Because the rim was yielded,
forcing it back to its original size requires yielding in opposite direction. Hence,
 315

Ofl

e, " - A-
C1

CYCLE
II II
WITHOUT III I - Of,
COLD WORK

0
I 1/ I
L L J
1
EFFECTIVE
CYCLE

TRUE
CYCLE

I I
II
II
L_J

Figure 9.27. Effect of cold-worked hole on local stress cycle.

CYCLE FOR •
INTERFERENCE 1'&-= - -
CYCLE FOR COMBINED
I INTERFERENCE AND
COlDWORK

CYCLE FOR COMBINED

INTERFERENCE AND
COLDWORK

I
II I II
I I
II I
I
I 1/ \ I II
1_ _L_J
I II I
L_L_I

APPLIED CYCLE

(a) (b)

Figure 9.28. Combination of cold work and interference. (a) Net positive effect; (b) Cancellation of
effects.
316

the residual compressive stress will be equal to at least F II, (Chapter 5). If the
applied stress is high, the compressed ring will eventually be relieved (point B
in Figure 9.27) and from there on the line for an open hole is applicable. If the
applied stress varies from 0 to (J a1 the local stress varies from - FlY to (J/3' Hence
R is reduced, but !l(J is much larger than before (it was O-(JII)' However, most
of this cycle ( - F,y to 0) is in compression. Only the tension part is effective, so
that the local stress is from 0-(J/3 which is less than from O-(JII . Hence, the life
is longer; crack growth is slower.
It is sometimes argued that the combination of cold work and interference
gives an even larger improvement. However, a simple assessment of the
situation shows that in general this cannot be so, as shown in Figure 9.28. The
local stress range is effectively larger. Indeed if (Jprt = F,y the two effects cancel
completely (Figure 9.28b). Hence, the combination is worse than either inter-
ference or cold work alone. Claims that combinations are better are fortuitous,
for which there can be two reasons (Figure 9.28):
(a). The range from zero in cold work is somewhat larger than with the
combination (R = 0), but the difference may be small in certain cases. If the
interference reduces fretting (which it will) the combination may give longer life.
(b). Figures 9.26-9.28 are not entirely true to nature because they are for
elastic behavior but plastic deformation occurs locally.
Yet, it is fortuitous if the combination is better in a certain test. Whether it
is or not depends upon spectrum (plastic deformation at high loads), the
material and the hardness of the interference pin or bushing (fretting). It will be
easy to show cases where the opposite is true. The foregoing explanations,
showing that the effects are due to opposite causes bears this out. Claims that
cold work and interference are basically the same, clearly are unfounded.

9.7. Residual stresses in general

The previous section shows how residual stresses affect crack growth. Residual
stresses are static. Thus, in general their only effect is a change in R, not in !l(J,
unless they are at a stress concentration as in the case of holes discussed in the
Section 9.6. In the case of uniform residual stresses, the situation is in
accordance with one of the cases shown in Figure 9.29.
If the residual stress is negative R will be lower. Since only the positive part
of a cycle is effective for crack growth (Chapter 7) also the effective !l(J (with
R = 0) may be less as is shown in figure 9.29. It should be noted here that
residual stresses may gradually disappear as a consequence of the cycle stresses,
a phenomenon called 'shake down'. If crack growth is by stress corrosion only
the sustained stress is of importance. The sustained stress is equal to the residual
stress plus the applied stress. Hence, tensile residual stresses will always promote
 317
RESIDUAL STRESS APPLIED ACTIVE
CYCLIC STRESS CYCLIC STRESS

-M- ----

b
oL------

M
01------
c
M

d
01----- .
, 1'1
\'

-_/_~-~- - -- --

Figure 9.29. Combination of residual stress and applied cyclic stress.

stress corrosion crack growth and compressive residual stresses will supress it.
This implies that all measures taken to introduce compressive residual stresses
(shot peening, hole expansion, etc.) will be beneficial. Interference fit fasteners
(tension) are detrimental for stress corrosion. It is stressed once more that in the
case of stress concentrations residual stresses may affect t1cr as well as R (Figures
9.26 and 9.27).
For residual strength (fracture) the applied stress is always additive to the
residual stress (superposition). Tensile residual stress is adverse, compressive
residual stress beneficial. These are qualitative statements. It is quite another
matter to quantify the residual stress effect. This requires determing f3 and K for
the residual stresses. In principle, this can be done using Green's functions or
weight functions (chapter 8), but principle is not practice. The stress intensity (or
318

f3) can be determined easily enough when the residual stresses are known, but
usually, they can be estimated only.
The above arguments apply to welded structures as well. Residual stresses are
always present especially in the heat affected zone (HAZ). If the user knows the
residual stresses at a weld, the procedures discussed in this book are applicable.
The work by Rybicki et al. [14, 15] may be helpful in determining residual
stresses at welds. Thus fracture mechanics is applicable to welded structures
provided the input is correct. Admittedly, there are some complications, but
there are complications for other structures as well as was clearly shown in
foregoing sections. A crack (or fracture) propagating in a weld or a HAZ must
be analyzed using the properties (KIf' da/dN) of the weld metal or of the HAZ.
Again, this is a matter of input, not a shortcoming of fracture mechanics. It may
be difficult to obtain the properties of welds and HAZ, but it is not impossible;
it may be costly and objectionable, it does not constitute a shortcoming of
fracture mechanics. Naturally, some questions will always remain, but many of
these can be answered by engineering procedures (see examples in Chapter 14).
The problem of crack and fracture in welds is no more complicated than those
In many other structures. Each have their specific difficulties. A proper solu-

Table 9.2. Crack tip stresses for Modes II and III

Mode II

(J,
- KI/ () (
~ sin"2 2
0
+ cos "2 cos "2
3())

KI/ . () 0 3()
- - SIn - cos - cos -
~ 2 2 2

In- =
KI/ () (
~ cos"2 1 -
. 0 .
SIn"2 SIn
3())
"2

(J_ = v((J, + (Jr)

KI/ = flllrFa
Mode III

Kill . ()
- --SIn-
~ 2

KI/I ()
--cos-
~ 2

ax = (I)' = (J;: = '[X)' 0

KI/I' = fllllTFa
 319
tion may require some expenditures, but servIce failures are usually more
expensive.

9.8. Other loading modes; mixed mode loading

The crack tip stress fields for modes IIand III can be derived in the same way
as for mode I. The solutions even have the same format as for mode I, as can
be seen in Table 9.2; they contain a stress intensity factor K. Throughout this
book the denotation K has been used for stress intensity, and since all
discussions concerned the same loading mode, there could be no confusion.
However, when considering different modes a distinction must be made between
the stress intensity factors of different modes, K I , KII and Kill.
Dimensional analysis in the manner used in Chapter 3, readily shows what the
format of KII and Kill must be. All crack tip stresses must be proportional to the
applied stress (elastic loading), hence, K+!. Certainly, the crack tip
stresses must be larger if the crack size a is larger, and since there is a root of
a length (r) in the denominator, the only possibility is that K -:- rJa, otherwise
the expression for stress would not have the proper dimensions (note that all
functions of () are dimensionless).
Dimensional analysis does not provide any constants that might appear.
These would have to be derived from a formal analysis of the problem. Again,
it turns out that for infinite panels the proportionality factor is J1i, so that
KII = rFa. The dimensional argument remains the same when the body is of
finite size, so that as in the case of mode I (see Chapter 3) the general expression
for K becomes KII = PllrFa, and similarly Kill = PIII!Fa. Again, P must be
dimensionless and still depend upon geometry; such dependence can be only in
the form p(ajL), where L is a generalized length parameter. All geometrical
effects must be reflected in P, so that the structural complexity will be accounted
for in the geometry factors. They are obtained by means of the same procedures
as discussed for mode I (Chapter 8). Although handbooks contain P/l and 13m
for some geometries, the mode II and III geometry factor are not available in
such abundance as PI
Fracture will occur when the crack tip stresses become too high, which leads
to the fracture conditions: Fracture if KII = K IIe , or if Kill = Kille. The
toughness K lle or Kille in principle could be measured in a test on any specimen,
provided PII or Pili is known for the configuration. Analysis of the residual
strength for any other structure would then proceed on the basis of the above
fracture criterion in the same manner as for mode I. Hence, for the analysis of
the SEPARATE modes no new analysis techniques would be needed.
In practice modes II and III do not occur separately, but always in combina-
tion with mode I, e.g. I-II, or I-III or I-II-III. The treatment of combined
modes is somewhat more complicated. Before considering these combinations
320
K 1ben

(a) (b)

Figure 9.30. Fracture locus for combined loading. (a) Combined modes I; (b) Combined modes I
and II.

of different modes, a brief review of combined mode I loadings (e.g. tension plus
bending), is in place; it was discussed in detail in Chapter 8.
In combined mode I loadings, addition of crack tip stresses leads to:

(J = K lten fee) + K lben fee)

.J2iii .J2iii
K tot fee) (9.6)
.J2iii
so that

This is the superposition principle (Chapter 8): stress intensity factors of the
SAME mode are additive, because the crack tip stress field solution is universal.
This means that the functions of e in the terms of Equation (9.6) are identical,
so that the terms can be combined by makingf(e) explicit. Fracture occurs when
Ktot = Kc (or KIc)' In combined bending and tension for example, this leads to
K ten + K ben = Kn which is represented by a straight line in Figure 9.30a. Each
combination (Kten , K ben ) falling on this straight line represents a fracture case.
Now compare this with a combination of different modes, e.g. I-II. Addition
of the crack tip stresses leads to:

(J = ~ fee) + KI/ gee) (9.7)

.J2iii .J2iii
Because the functions f and g of e are different, the equation cannot be man-
ipulated like equation (9.6) to provide an explicit stress intensity. Hence, in
 321

60

50
-----K 2c

---....
2021, -T3
PlanE'strlE'ss
40

30

...
KI]C OlD 5050
20
;~' "". o
10

(a) , ",I.,
10 20 30 40 50 60 70 80

500

400

•
300
•••
200

100 •

100 200 300 400 600

(b) ~ (ps.v;n)

2 .. r-----~----------------------~

2.2

I
2.0 I
1.8
I
I /
I /
1.6 I
...
o w= 12-in.
I 6 w= 6-in

K,
ksiKn. 1.2

2.2 2.4 2.6 2.8 3.0 3.2

(e) K1 • ksi..fin.

Figure 9.31. Mixed mode fracture test data. (a) aluminium alloys; (b) Plexiglass [16] (Courtesy
ASME); (e) Paper. Courtesy Mead Paper Co.
322

combine mode loading, stress intensities cannot be superposed. Another route

must be followed to solve the fracture problem. ..
The energy conservation criterion, being a general law of physics, should
apply to the problem. It was shown in Chapter 3 that the energy release rate is
equal to the derivative of the strain energy, d U/da. Fracture will occur when the
available (release) energy is sufficient to deliver the fracture energy, d W/da.
Hence, the fracture criterion is:

(9.8)

Since energy is a scalar, superposition is freely permitted, so that the equation

becomes:

( dU\ (dU) dW (9.9)

daJ J + da /I = Ta·
As shown in Chapter 3, the value of (dU/da)J is KJ /E. Obviously, (dU/da)/1 must
have the same dimension (units), which means that it must be proportional to
KJJ/G, where G is the shear modulus. But since G = EI2(\ - v), it follows that
(dU/da) = CKUE. The proportionality factor C can be obtained only from a
formal analysis of the problem; it turns out that C = 1. Thus Equation (9.9) can
be written as (plane stress).
KJ KJJ dW
-
E
+-
E da·
(9.10)

This would be the fracture criterion for combined modes I and /I. For it to be
valid in general, it must hold for the case that K/I = o. Then it follows readily
(see also Chapter 3) that dW/da = K~c/E. The equation must also hold when
K J = 0, in which case dW/da = Kic/E. This would lead to the conclusion that
K2c = K ,c . (Since apart from a factor (I-v) the above applies to plane stress as
well as plane strain, the toughness is denoted here as K 1c and K 2C' where the
Arabic 1 and 2 signify mode I and mode /I respectively, independent of the state
of stress.)
Equation (9.10) then reduces to:
(9.11 )
Equation (9.11) represents a circle with radius K 1c as shown in Figure 9.30b.
Every combination of K/I and K J falling on this circle represents a fracture case.
(Compare with Figure 9.30a.)
Application of two modes of loading in a test generally requires two loading
axes. Since most testing machines have only one loading axis, most tests to
vaJidate the criterion of Equation (9.11) have been performed on plates with
oblique cracks (Insert Figure 9.31 b), where both modes I and /I occur due to one
 323

loading axis. The stress intensities are KI = PI (TN Fa, and Kl/ = Pl/7:Fa,
where both PI and Pll are aproximately equal to 1 when the plate is large. In a
fracture test one measures the value of (T at fracture. From this (TN and 7: at
fracture can be determined as in Figure 9.31 b, and subsequently the values of
KI and Kl/ at fracture calculated. The data point can be plotted in KI-Klrspace.
For the fracture criterion of Equation (9.11) to be true, data points of tests with
different KdKl/ (different angles) must fall on a circle with radius Klc (or K2c)'
Three such data sets are shown in Figures 9.31. A test in which the crack
angle = 90 degrees is not a viable possibility. Thus the test data for large crack
angle (large Kl/) are dubious. But even if such data points are ignored, the
hypothesis does not seem to be substantiated: the data do not fall exactly on a
circle.
The question is now where the hypothesis could be wrong. Equation (9.8) is
certainly true, and therefore Equation (9.9) must be correct. Hence, the error
must have been introduced between Equations (9.9) and (9.10). It was shown in
Chapter 3 that in mode I, indeed dUjda = KJ/E. In the derivation of this
equation an implicit assumption was made however, namely that fracture
occurs in a self-similar manner: da in the same direction as a. This
seemed trivial at the time: it is everbody's experience that the fracture will
proceed in the manner indicated (Figure 9.32a). Pulling a piece of paper with a
tear will confirm this. In the same vein, self-similar fracture was assumed when
writing (dUjda)/l = KJdE. However, it is not obvious that the fracture proceeds
self-similarly in a combined mode I-II case. One feels intuitively that it does not;

1 1

! 1 (c) (d)

(a) (b)

Figure 9.32. Fracture direction. (a) Anticipated fracture path in mode I; (b) Anticipated fracture
path for mode I-II; (c) vertical da; (d) Direction of da.
324

I
I
I
I CRACK ;f'TENSION I

Figure 9.33. Crack perpendicular to principal stress (tension) (a) Shear web. (b) Pure torsion.
 325
it will proceed as shown in Figure 9.32b. This can be readily demonstrated by
pulling a sheet of notepad paper with an oblique (e.g. 45 degrees) slit.
The energy release rate (dU/da)tot depends upon the angle at which fracture
proceeds. It is obvious that a vertical da (figure 9.32c) will cause much less
energy release than a horizontal da. Indeed a horizontal da gives the highest
energy release in mode /, reasons why fracture proceeds as in Figure 9.32a. In
combined mode loading fracture will proceed in that direction for which the
total energy release is the largest, because Equation (9.8) is first satisfied for this
direction. Hence, in developing Equation (9.9), one should write:
KJ KJ[ dW
1Y.[(IjJ) E + 1Y./l(IjJ) E = da' (9.12)

where IY.[ and IY.I/ are both functions of IjJ (Figure 9.32) which can be formally
evaluated by calculating dU/da as a function of the angle IjJ of da. It can then
be determined for which IjJ the left hand side of Equation (9.12) becomes a
maximum. This provides not only the angle IjJ at which the fracture proceeds,
but also the values of K[ and KII required for fracture. The question is whether
this is still necessary.
There is no doubt that a fracture in mode / will proceed in a self-similar
manner. Neither is there any doubt that the fracture as in Figure 9.32b will
proceed as shown (simple tests on paper bear this out). In both cases this is the
direction perpendicular to the maximum principal stress. This is immediately
clear for pure mode /, but it can be demonstrated for the combination as well
[5, 16].
This being the case, the question must be asked: how could the oblique crack
(mode IIII) in the tests represented in Figure 9.31 develop in the first place. The
answer is simple: the cracks did not develop; they were cut intentionally for the
purpose of the tests. Had the crack developed by fatigue, they would have grown
horizontally and not under an angle. As a matter of fact this is a common
observation. Cracks in shear webs and pure-torsion bars develop under 45 0 as
shown in Figures 9.33. Even if such cracks initially are forced in another
direction, they will soon turn and find a path in which they experience mode /
only. Cracks tend to turn away from the shear loading so that the mode /
loading is the only relevant one. Hence, cracks such as in Figure 9.31 would
never develop.
The above behavior of service cracks is also observed in tests. A fatigue crack
in a combined shear-tension field [17] followed a curved path. Finite element
analysis of a model in which the crack was represented in the way it actually
developed in the test showed that Kl/ dropped immediately to zero. Its initial
value was high because it was forced by the notch, but the crack turned readily
away into a direction of mode / only. Thus all tests data in Figures 9.31 and
behavior of forced slant cracks or forced combined mode tests in general, are
326
of academic interest but of no practical value if natural cracks develop in mode
I, i.e. perpendicular to the maximum principal stress.
Essentially then the entire problem of combined mode loading seems of little
practical significance. This categorical statement must be blunted somewhat,
however. There are cases where the crack may indeed be forced into a combined
mode situation. A good example would be a torque tube with a circumferential
weld. Because the weld HAZ is often the weakest path, the crack will stay in the
HAZ where it experiences both mode I and II. Other examples of this forced
behavior can be thought of. In such cases combined mode crack growth and
fracture criteria may have to be used. One may then turn to formal analysis [5]
which leads to considerable complication. Instead, engineering pragmatism
would dictate simpler and more cost-effective ways. The simplest approach
would be to use the circle of Equation (9.11). In most engineering cases K/l will
be small with respect to K/ and for small ratios K/l/K/ the circle is a very good
approximation as can be readily seem from Figures 9.31. It could be improved
somewhat by recognizing that data suggest that K2c ~ 0.8 K 1c and that the data
fall more closely on an ellipse:

(KK/)2 + (~)2
1c 0.8 K 1c
_
-
1
•
(9.13)

The use of the criterion is simple. For example, if K 1c = 50 ksi JIn, it follows
that K 2c = 40 ksi JIn. When considering a case where Ih = P/I ~ 1 (other cases
can be treated in the same manner after determination of p/ and P/l), it can be
calculated how much shear can be applied if e.g. (j = 16 ksi and a = 2 inches.
It follows that Kl = 16~ = 40 ksiJIn. Fracture occurs in accordance
with Equation (9.13):

G~Y + (~;Y = 1

so that Kl/ = 24ksiJIn and r = 24/~ = 9.6ksi. Thus if the tension is

16ksi, an additional shear of 9.6ksi will cause fracture. Similarly, if the shear
is e.g. 16 ksi, and a = 1 inch, the value of Kl/ is 16 J1C = 28.4 ksi JIn, and
fracture occurs when K/ = 50v'1 - (28.4/40)2 = 35.2 ksi JIn, i.e. at (j = 35.2/
J1C = 19.9ksi.
The treatment of fatigue crack growth could follow similar lines, the 'effective'
11K/ being determined by:
(9.14)
and by using the mode I rate data (da/dN) with 11K = I1K/ eff •
In the above it was assumed that K/ and Kl/ are in phase, as would occur in
the shear web of a beam or a torque tube with bending and torsion. If they are
not in phase, the problem is different. For example consider the span of a bridge.
 327
Tension, Kl> is at a maximum when a truck passes over the crack location. The
shear, KlI , on the other hand changes sign when the truck passes the crack
location. Thus K[ and KII would be out of phase. Similar conditions occur when
torsion and bending vary independently. An example would be an aircraft wing.
Normally torsion and bending are in phase, but a new situation develops with
'flaps down'. Again torison and bending are in phase, but their ratio is different
than with 'flaps up'.
Although the above cases are realistic engineering problems, none of the
mixed mode hypotheses have even begun to address this situation; up till now,
virtually all research has addressed either academic problems of forced mode
IjII or IjIII, or concentrated on in phase loading for crack growth. Thus the
problems discussed in the previous paragraph must be addressed by engineering
methods. The only sensible procedure would seem to be to treat the problem as
mode I, with the crack growing perpendicular to the highest principle stress for
the worst mode I case, and to assume the stress range is the largest difference
in principal stresses.
In conclusion, it can be stated that the combined mode problem is largely
academic. In most cases of combined mode loading the crack develops into a
direction of only mode I. In practice, probably 90% of the engineering problems
are mode I to begin with. Of the 10% combined mode loading cases, probably
80% of the cracks will turn immediately upon initiation and experience mode
I only. This brings the total of engineering problems to 98 % mode l. Of the
remaining few per cent, most will be in phase, so that the engineering analysis
of Equations (9.19), (9.13) and (9.14) would apply. In those few cases where out
of phase loading is significant no hypothesis is available at all. The pragmatic
approach discussed above is the only way out.

9.9. Composites

Fracture mechanics methods to deal with crack growth and fracture in

composites are practically non-existent. Summaries of many ongoirig research
programs [18] reveal that the 'development' is still in the stage of explaining and
interpreting experimental observations. Research programs involving 'finite
element analysis' were not very successful either. Crack growth and fracture
analysis require a crack growth and fracture criterion based on the physics of
the problems. Finite element analysis, by itself does not provide this; it provides
stresses and strains but these are useless without a fracture criterion.
There is a risk that tests performed to investigate fracture criteria use
specimens of inadequate size, incorporating similar problems of interpretation
as in the case of small metal specimens. (Chapter 3 and 10). As composites are
expensive, the quest for small specimens is understandable.
It is not surprising that the damage tolerance assessment of composite
aircraft structures [18] is entirely based on full-scale tests. The component is
328

subjected to a simulated lifetime of cyclic stresses and environmental changes.

At the end of this 'life' it must still be able to sustain the design load. Sub-
sequently, damage is inflicted by a weight of prescribed size dropped from a
prescribed height. The testing is continued, the damage increasing in size during
load and environmental cycling for another lifetime. At the end of the next 'life'
the structure must still have the minimum permissible residual strength, (Jp.
Tn the case of composites various failure mechanisms can be operative,
namely: matrix cracking, fiber fracture, fiber decohesion and finally laminate
decohesion. Compare this with a stringer stiffened structure as discussed exten-
sively in Section 9.4. There, various failure mechanisms are possible as well,
namely: skin cracking (matrix cracking), stringer failure (fiber fracture) and
fastener failure or adhesive debonding (fiber dechohesion). Thus, the composite
problem is similar in many ways to the stringer-plate problem for which
solutions are readily available. Composite research might make use of these
established procedures. Composite fibers are essentially finely distributed
stringers. Admittedly, multiple laminates with different fiber orientations will
complicate the problem. But building upon the skin-stringer solution would
seem more promising than finite element analysis.
Paper and cardboard are composites, be it that they have short, randomly
oriented fibers. Clusters of fIbers, and sometimes individual fIbers can be seen in
a sheet of bond held against the light. No fracture tests are cheaper than tests

2.B

2.6
• W=18 in.

2.4
o W=12 in.
... W=6in.

2.2
W
2.0

I.B 20
:21 1.6
b"
,,' 1.4
go
~
Vi
1.2

1.0
Predicted curves "e- g T;
.jrlJ sec

O.B

0.6

0.4

0.2

0
0 2 4 6 10 12 14 16 18
Crock Size. 20e• inches

Figure 9.34. Residual strength prediction and data for random fiber composite (paper board).
Courtesy Mead Paper Co.
 329
on paper and cardboard to check analysis procedures. Figure 9.34 shows test
data for center-cracked card-board panels. A 'toughness' Kc = 1.34~sec 2n/12
Jfii = 3.6 ksi Fn follows from the average fracture stress for the three data
points indicated by arrows. Using common procedures of residual strength
analysis as discussed in Chapters 3 and 10, the residual strength curves for other
panel sizes were predicted. Figure 9.34 compares predictions (curves) with tests
(data points). Clearly, the results are in accord with predictions. There are
discrepancies, but these are well-explained by common material scatter, and no
larger than in metals. In the case of card board, one must account for aniso-
trophy and inhomogeneity of the material and the problem seems unmanage-
able from the point of view of analysis. However, if one can make predictions
(curves in Figure 9.34) as well as shown using a very simple concept, the
engineering problem is solved. Similarly the problem of crack growth and
fracture in composites can be solved. However, solutions as e.g. the one
discussed above must probably come from engineering-oriented (development)
research. Academic fracture mechanics research may help to improve under-
standing, but its track record for solving engineering problems is not encourag-
mg.

9.10. Exercises

1. Using the values for a, determine the residual strength diagrams for lon-
gitudinal suface flaws (internal as well as external) with ale = 1 and al
e = 0.3 in a pressurized cylinder of 10 inch diameter with a wall thickness
of 0.5 inch; assume that the hoop stress is pR/B (thin wall solution.
K/c = 35 ksi Fn; F,y = 70 ksi. Assume f3 of figure 8.3 is applicable. Note:
use a for residual strength analysis, and e = al(ale).
2. Given that the pressure in the cylinder of Exercise 1 cycles between 0 and
3000 psi, will the cylinder leak or break? Assume f3 = 1 for the through
cracks.
3. Assuming that F,y = 50 ksi instead of 70 ksi in Exercises 1 and 2, and
Kc = 80 Ksi Fn, will the cylinder leak or break?
4. If da/dN = C I1K 3 , determine the changing shape of the crack starting at
ale = 0.3 and ale = 1 starting at a = 0.1 inch. Assume same conditions as
in previous exercises.
5. Assume Figure 8.25 applies. Given is: B = 0.2 inch, b = 8 inches, W large,
J1. = 0.4, F,y = 60ksi, F,u = 75ksi, Kc = 70ksiFn. Calculate the residual
strength diagram (including stringer failure line). Is this case stringer critical
or skin critical? What would you change to make this a skin critical case? If
330

the allowable shear stress of the fasteners is 100 ksi what size offasteners do
you need? What is the bearing stress?
6. Suppose that in the case of Exercise 5 the normal fatigue stresses have a
range of 15 ksi at R = 0.2. Occasional higher stresses occur. One of these
occurs when the crack size is a = 2 inches, causing a fracture. Assume that
da/dN = 3£-9I1K21K!~x.
(a) At which stress did the fracture occur?
(b) At which crack size does arrest occur?
(c) What was the rate of fatigue crack growth before arrest?
(d) What is the rate of fatigue crack growth after arrest?
(e) Why could you have obtained the result of d without any calculation,
and what is the general consequence of this observation?
7. A stop-hole repair is made of an edge crack of 0.75 inch length for which
f3 = 1.12. Calculate f3 for the crack emanating from the stop hole. Diameter
stop hole is 0.25 inch, its center at the crack tip.

8. Given that Kc = 50 ksi -JIn,

what was the residual strength in the case of
the crack in Exercise 7 before stop hole drilling, and what is the residual
strength immediately after reinitiation of the crack from the hole? ~\' = 100
ksi.
9. A hole with a diameter of 0.25 inch and an edge distance of 0.625 inch
develops a crack that grows through the entire ligament. Calculate f3 for the
crack emanating on the other side.
10. Supposing a residual tensile stress of 20 ksi exists at a potential crack
location. The applied stress cycles between 0 and 10 ksi. Calculate the rate
of growth with the Forman equation if CF = 2 x 10 £-7, m F = 2 and
Kc = 80 ksi.Jill; Assume f3 = 1; a = 1 inch.
11. Would the crack of Exercise 10 grow by stress corrosion if
= 20 ksi .Jill?
KIser

12. To a pre-existing crack a mode II shear stress of T = 20 ksi, as well as a

tension are applied. The toughness of the material is 80 ksi .J1n. Assuming
f3 = 1 calculate allowable tension if the crack is of 2 inch length.

References
[II J.e. Newman and I.S. Raju. Stress intensity factors equations for cracks in three-dimensional
finite bodies, ASTM ATP 791 (1983) pp. I-238-I-265,
[21 I.S. Raju and J.e. Newman, Stress intensity factors for circumferential cracks in pipes and rods
under tension and bending loads ASTM STP 905 (1986) pp. 189-805.
 331
[3] J .C. Newman and I.S. Raju. Analysis of surface cracks in finite plates under tension and bending
loads, NASA TP-1578 (1978)
[4] G.G. Trantina et al., Three dimensional finite element analysis of small surface cracks, Eng.
Fract. Mech. 18 (1983).
[5] D. Broek, Elementary engineering fracture mechanics, Nijhoff (1986) 4th edition.
[6] D. Broek et al., Applicability of toughness data to surface flaws and corner cracks at holes. Nat.
Aerospace Lab. NLR TR 71033 (1971)
[7] L.R. Hall and R.W. Finger. Fracture and fatigue crack growth of partially embedded flaws,
AFFDL TR 70-144 (1970) pp.235-2626.
[8] M.F. Kanninen and C.H. popelar. Advanced fracture mechanics, Oxford University Press
(1985).
[9] T. Swift. Development of fail safe design features of the DC-IO, ASTM STP 486 (1974)
pp.I64-214.
[10] H. Vlieger and D. Broek. Residual strength of cracked stiffened panels. AGARDograph 176
(1974) Chapter V.
[II] H. Vlieger. Residual strength of cracked stiffened panels. Eng. Fract. Mech. 5 (1973) pp. 447-
478.
[12] D.P. Rooke and DJ. Cartwright, Compendium of stress intensity factors HM Stationery Office,
London (1976)
[13] H.P Van Leeuwen et al., The repair offatigue cracks in low alloy steel sheet, National Aerospace
Institute, Amsterdam, Rep NLR TR-70029 (1970)
[14] E.F. Rybicki and R.B. Stonesifer, Computation of residual stresses due to multi-pass welds in
piping systems, J. of Press. Vessel Techn. 101 (1979) pp. 149-154.
[15] E.F. Rybicki et al., A finite element model for residual stresses and deflections in girth-butt
welded pipes, J. of Press Vessel Techn 100 (1978) pp.256-262.
[16] F. Erdogan and G.S. Sih. On the crack extension in plates under plane loading and transverse
shear. J. Basic Eng. 85 (1963) pp.519-527.
[17] D. Broek and R.C. Rice. Prediction offatigue crack growth in rails. Battelle rept to US DOT
TSC (1977)
[18] Various Authors, Tough composite materials, Noyes (1985).
[19] J.R. Soderquist Certification of civil composit aircraft structure SAE paper 811061 (1981).
 CHAPTER 10

Analysis procedures

10.1. Scope
This chapter is essentially a summary of all foregoing chapters and serves as an
introduction to those following on fracture control and damage tolerance
requirements. As such it is repetitive in many ways and contains many cross-
references to other chapters. Nevertheless, despite its repitition, the reader may
find it useful as a general tie-in of the general procedures, and as a reference to
other parts of this book where the issues are discussed in more detail.

10.2. Ingredients and critical locations

The block diagram in Figure 10.1 shows the general outline of the damage
tolerance analysis. After selection of the critical location to be analysed, the
ingredients (input) for the analysis must be obtained. Apart from the general
stress distribution the load spectrum (and environmental spectrum) must be
known. These lead to the basic input for the analysis, namely
(a) Stress history, Chapter 6.
(b) Materials data (dajdN, Kin Kn JR , F;y, F;,o' etc), Chapter 7.
(c) Geometry factor(s), Chapter 8.
Material data are generally obtained from data handbooks. Care must be
taken that they are for the proper material direction. Assumptions are often
necessary, as discussed extensively in Chapter 7.
The stress history is obtained on the basis of the load spectrum and the stress
analysis. The input may be in tabular form. If an exceedance diagram is
available the stress history can be generated using procedures as discussed in
Chapter 6. Decisions with regard to sequencing, clipping and truncation must
be made. Some computer codes perform the stress history generation automati-
cally from the exceedance diagram [4] input and provide options for truncation
and clipping. The many issues involved were discussed in Chapter 6.
Geometry factors must be obtained using one of the methods reviewed in

332
 Tahle 10.1. Analysis (and fracture control) complications for various types of structures (subjective rating)

Item Complexity due to: Ships Offshore Pipelines Nuclear Airplanes Chemical Heavy
processing machinery
Load and Spectrum 8 8 2-7' 2 9b 2 7-8
spectrum Load response 7-8 6-7 1-4' 3-4 8-9' 2 6-7
Load and weight 6-9 2 2 2 8 2 3
configurations
Environment 6 8 3-8' 9 4 9 7-9
Stresses Structural geometry 7 6 3 5 9 3-5 8
Stress analysis 7 6 4 6 8-9 5 8
Cracks and Detail design 8 6 4 5 9 4-5 6
fracture Residual stress 6 8 5 4 2 7 7-8
Welded joints 8 8 7 6 1 8 8
Fastened joints 3 8 2
Active fracture Inspection 6 8 4-7" 9 6 6-7 5
control Total complexity 70-74 67-68 36-52 54-55 74-76 50-54 66-71
a High numbers for submarine pipeline.
h Maneuvers, gusts, and taxi-loads.
C Including control surfaces.

w
w
w
334
Chapter 8. Libraries of geometry factors are sometimes included in computer
codes. This is useful mostly for standard geometries. When compounding,
superposition and other methods are necessary, a pre-processor performing all
of the procedures discussed in Chapter 8 is more versatile [4].
After the input is complete residual strength analysis (Chapters 3, 4, 9) is
performed to determine the permissible crack size, ap • Crack growth analysis
(Chapters· 5, 9) then provides the life to ap (Chapter I), upon which fracture
control decisions can be made (Chapter II, 12). The latter computations are
usually performed by employing proven computer software. As such they are
essentially the easiest part of the analysis. Nevertheless, the analyst should be
aware of the procedures used in order to appreciate the possible consequences
of· approximations and judgements, which largely determine the accuracy of the
analysis. Preparation of the input is the most difficult and most crucial task.
Every type of structure presents its own special difficulties (Chapter 9), and the
difficulties are more or less of the same level regardless of the application. This
can be appreciated from Table 10.1, where the various complicating factors are
subjectively rated on a scale from 1 to 10. With few exceptions the total rating
is of the same order of magnitude. Naturally, these ratings are somewhat argu-
able, but they were based mostly upon actual experience with damage tolerance
analysis for these structures, so that they certainly provide a good indication
of complexities.

10.3. Critical locations and flaw assumptions

Perhaps the most important step in the analysis is the very first (Figure 10.1),
namely the selection and identification of the critical location. It is also the most
difficult to discuss. Analysis must be performed for those structural elements
and details liable to develop cracks. The assessment of these critical locations
requires sound judgement and should take into account the following factors:
(a) Areas of high stress (stress calculations and strain measurements).
(b) Stress concentrations due to discontinuities and eccentricities, particularly
in the areas mentioned sub a.
(c) Location where stresses would be high if adjacent elements failed (load
transfer, Chapter 9)
(d) Eccentricities and stress concentrations resulting from fracture of one
member of a back-to-back or multiple load path structure.
(e) Details prone to cracking according to service experience or tests.
(f) Areas where fretting may occur.
(g) Areas of secondary stress due to displacements of other parts.
Rather than relying on computer stress analysis the damage tolerance analyst
should develop a second nature for identifying stress concentrations. Especially
in complex structures the load path should be 'envisioned'; where the load must
 335

Critical location Damage tolerance

requirement chapter 12

t t
Environment load spec- Stress Permissible minimum
spectrum trum distribution ~ residual strengt, 0',
Chapter 7 Chapter 6 Chapters 1 and 12

t
da/dN. da/dt Stress Geometry
t
Toughness
retardation
parameters fly r- history factors Ke , K1c. • JR
also ~DI.f1y
Chapter 7 Chapter 6 Chapter 8 Chapter 7

t t
t
Crack growth Resi dual strenghth
cal culati on diagram calculation
Chapter 5 (and 9) Chapters 3, 4,9

t
I
Permissi ble
crack Qp

Chapter 1. 12

t
llifo to a,
Chapters 1. 12

i
I
I Frature control

I
measures
Chapters 11,12

Figure 10.1. Damage tolerance analysis.

go and where it cannot go depends upon stiffness distributions and location of

reaction forces. For design details the load-flow lines discussed in Chapter 2 are
useful.
'Hidden' stress concentrations should be recognized on the basis of stiffness
distributions and eccentricities. For example the bolted joint in Figure 1O.2a
may be perfect from a static design point of view. At the design load ample
plastic deformation occurs at the bolt holes for the load to be evenly distributed
over the bolts, each bolt then carrying Ij4P. However, the design is poor from
the point of view of fatigue. This can be readily seen from a comparison of
stiffness as shown below.
First note that during normal service loading the stresses are elastic. Now
336
c o

IlIa rJ.Lif.. I
p p

I
A 04- ~
8 0/ /
LlIt, 04- rJ.Li!!. I
(a )
c o
c o

P 3/4 P
+0-
-
1/2 P
.114-P
o II P

K l M
(b)

-L

(e)
D
~

/ '-
--L ( )
(d)
" /'
Figure 10.2. Hidden stress concentration in bolted joint. (a) Poor design; (b) Improved design; (c)
Full body equivalent of a; (d) Full body equivalent of h.

consider the area of the joint between bolts C and D. Parts IIa and IIb together
have the same stiffness (equal thickness) as part I. Since the parts are attached
at C and D they must stretch equally: they undergo equal strain. But for equal
strain they must have equal stress. Therefore IIa and lIb together carry half the
load, and part I carries half the load. This means that half of the load is
transferred at bolt C, the other half at bolt D; the two other bolts carry no load
at all. Equal strain in the parts between C and D, causes the holes of the centre
bolts in the three parts to remain lined up: hence the bolts cannot transfer load.
A much better design is shown in Figure 10.2b. If bolt C is to transfer 1/4P
then there will be 3/4P between GH, while there will be 1/8P between CE and
KL each. The parts being attached, they must have equal strain (stress).
Therefore the thickness of CE and KL each must be 1/8 x 3/4 = 3/32 of the
center part, and so on, in order to carry loads of iP, tP and iP.
The design of Figure IO.2b provides a gradual change in stiffness. The
 337
equivalent in full body design is shown is Figure 10.2d. No designer would
conceive the configuration of Figure 1O.2c, but the design of Figure 1O.2a is
quite common. Apart from the stress concentration of the bolt holes per se,
there is a 'hidden' stress concentration of a factor of 2 (the bolt carrying 1/2P
instead of 1/4P). Cracking will occur at C and D, and the analysis must account
for a bolt load of ~ P. (Somewhat less ifbolt hole deformation occurs)
Once the critical areas have been established, it is still not trivial how the
analysis should proceed. One must also know (or assume) how cracks will
develop. This assumption is very influential on the results of the life calcula-
tion and may overshadow other errors (Chapter 12).
In the case of surface cracks or corner cracks there is usually little choice but
to assume that the flaw is elliptical, otherwise geometry factors are not available.
Yet many surface flaws are not elliptical. A circular flaw is often assumed. As
discussed in Chapter 9 this assumption may have more effect on the results of
the analysis than all acclaimed analysis shortcomings together (see also Figure
7.8). One might select to go through the costly exercise of generating geometry
factors by means of 3-D finite element analysis, but if the wrong flaw shape
assumption is made also this may be in vain.
For example, nuclear pressure vessels tend to develop cracks at the nozzle
(Figure 10.3) where cyclic (thermal) stresses occur due to the 'cold' water
returning from heat exchanger or turbine. The stress distribution is as shown in
Figure lO.4a. Stress intensity factors were obtained [I] by 3-D finite element
analysis (Figure lO.4b) on the basis of the flaw assumptions of Figure lO.4c.
Consider the case of crack front 3. The stress intensity is higher at the free
surface. This means that crack growth will be faster there, so that crack front
4 will no longer be a circle (as was assumed). Indeed, model tests confirmed this
faster growth at the surface, as shown in Figure 1O.5b. Subsequent fracture tests
overpredicted (unconservative) the fracture stress as shown in Figure 10.5.
It is strongly emphasized that this is not intended to suggest that the des-
crepancies in Figure 10.5 are due to the flaw assumptions. On the other hand,
if the tests had been done first, proper flaw assumptions could have been made,
and at least one area of uncertainty could have been eliminated. The example
shows that answers do not lie in costly analysis. Unfortunately, engineering
insight and foresight tend to be lost in the computer world. Any of the methods
discussed in Chapter 8 (possibly combined with the FEM stress distribution for
the uncracked model) would have given equally useful results, the (flaw)
assumptions being the crucial point.
Similarly, flaw development presents a problem, in particular when the
structure consists of multiple elements. It would be hard to decide which
assumptions should be made with regard to the sequencing and interaction of
cracks I through 8 in a case such as in Figure 10.6: Of the many possible
assumptions none will occur in reality, and no useful answer will be obtained
338

WELD WALL THICKNESS

TYPICALLY 10-12 INCHES
VESSEL
STAINLESS STEEL CLADDING

Figure 10.3. Nozzle crack in nuclear pressure vessel..

(c)

. ..... ,1.

(a) (b) D~'------*".'----~'.~

-0

Figure 10.4. Analysis of nozzle corner cracks [I]. (a) Stress distribution at nozzle corner; (b)
Calculated stress intensity on basis of a and c. (c) Flaw assumption.
 339

o Kave
*
• Kmax

* K45

*/
"e.
/ 0
/
/
/

(b)

TEST
(a)
Figure 10.5. Possible effect offiaw assumption [1]. (a) Predicted fracture stress and test results; (b)
Actual crack shapes in test.

unless the effect of different assumptions is analysed and the worst scenario
adopted. The latter must follow from multiple analyses; it cannot be foreseen.
Naturally, if the structure is really damage tolerant (can sustain large damage)
no assumption may be necessary. The analysis could start by noting that
eventually all of the area in Figure 10.6b will be cracked, and the inspection
interval could be based upon the time for this damage to grow to ap (Chapter
11,12).

10.4. LEFM versus EPFM

The concepts of residual strength analysis were discussed in Chapters 3 and 4;

special situations were reviewed in Chapter 9. Collapse must always be
considered as a competing condition; the lower result of fracture and collapse
analysis is the residual strength. In principle, the fracture analysis is not essen-
tially different from conventional design analysis as demonstrated in Figure
10.7.
A question may arise as to whether to use LEFM or EPFM. In this respect
Figure 10.8 provides some guidelines. Most cases can be treated with LEFM
340

. ' Crack.d ~-
I Rear view
I
.

f1:f-,J
Side view
Crack stopper
Joggi. Stringer strap
I I
---~--------------

7 . B

a.

• ' Crackod

-------------

b.

Figure 10.6. Crack development and sequencing in multiple element structure (example: aircraft
fuselage) (a) Eight different cracks; (b) total assumed damage.

and collapse, be it that some approximations may have to be made for small
cracks (fracture at high stress), but such approximations are still necessary with
EPFM, as shown later in this chapter. The significance of EPFM is exaggarated
in the literature. Judicious use ofLEFM in conjuction with collapse analysis will
provide the residual strength with good engineering accuracy in most cases. The
intrinsic inaccuracy of the toughness measured in terms of J R (Chapter 7) usually
causes EPFM analysis to be no more accurate than LEFM even for very ductile
materials, in particular because collapse must still be separately evaluated.
Besides, fracture control usually depends only slightly on permissible crack size.
(Chapter II, 12) and is more influenced by crack growth.
In this respect the so-called 'failure analysis diagram' developed [2] in Great
Britain is of great interest, because it provides much insight. The whole gamut
of fractures from brittle to fully plastic can be represented in the failure analysis
diagram as shown in Figure 1O.9a. The stress intensity is plotted along the
 341
TEST MATE RIALS STRUCTURE HANDBOOK OR FRACTURE
HANDBOOK FINITE ELEMENTS STRESS

P=~t
R

tJ ". ~ 0--
Figure 10.7. Fracture analysis (bottom lines) as compared to conventional design analysis (to line).

ordinate and stress along the abscissa. The stress is limited by collapse and the
stress intensity by the toughness.
The limits of Kc at the one end and collapse at the other end, require that there
is a limiting line going from Kc to arc. The end portions of this contour must be
straight so that there is only a relatively small curved part. The top horizontal
part is governed by LEFM and the vertical portion by collapse. The curved part
is the regime ofEPFM. For some applications it may be permissible to approxi-
mate the diagram by two straight lines (Figure 10.9b). The failure analysis
diagram is usually presented in normalized form by plotting K/Kc and a/arc so
that the intercepts with the axes are at 1 (Figure 1O.9c). This normalization
makes the diagram universal.
The use of the failure analysis diagram can best be demonstrated by an
example. Consider a material with Klc = 50 ksi Fn and Fcol = 60 ksi. Assume
that the 'structure' is a center cracked panel, 12 in wide with a crack 2a = 2 inch
subjected to a stress of 10 ksi. The nominal stress at collapse would be:

w - 2a
W F col , (10.1)

which for the given case becomes: af<. = 60(12 - 2) = 50 ksi. Thus,
a/arc = 10/50 = 0.20. The stress intensity is K = /3aFa, so that with /3::::; 1
itsvalueisK = 1O~ = 17.7ksiFn. Thus,K/Kc = 0.35. Now the point
342

~ Plasticity
C " COLLAPSE
L " LEFM
P " EPFM

Higher
Stfess

EQUAL
OR
Ftv
SMALLER
iHIGHER TOUGHNESS
STRUCTURE
..
Figure 10.8. Regimes of LEFM, EPFM and collapse.

a/ajc = 0.20 with K/Kc = 0.35 can be plotted in the diagram as point A in
Figure 10.10. If the stress is raised to 20 ksi, the stress intensity becomes
K = 20FXT = 35.5ksiy'lrl and K/Ke = 35.5/50 = 0.71. Further,
a/a!e = 20/50 = 0.40. This produces point B in the diagram
Clearly, K/Ke and a/aj( increase proportionally. Therefore, a straight line
through the origin provides all combinations of K and a. Fracture occurs where
this line intersects the fracture locus. Judgement of the proximity offracture can
be made from the distance between a point and the contour. Extension of the
line will also show whether fracture occurs by LEFM or collapse. In the
present example LEFM applies (point C).
Consider another case with a crack of 2a = 0.6 in. The stress at collapse is
arc = 60(12 - 0.6)/12 = 57 ksi. For example, with a = 20 ksi one obtains
a/a}e = 20/57 = 0.35, and K = 20)n x 0.3 = 19.4ksiy'lrl or K/Ke =
19.4/50 = 0.39, plotted as point D. The extension of the line OD predicts that
 343
K
K
Ker---------------___
LEFM Ke 1------- - -- .,
I
EPFM I
I
I

COLLAPSE

a a
ale .!L
(a) (b) (c) ale

Figure 10.9. Principle of failure analysis diagram[2]. (a) Regimes of fracture failure; (b) Question-
able area in transition region; (c) Normalized diagram.

fracture will occur at ulUfe = 0.84 so that the fracture stress is

0.84 x 57 = 48 ksi. This point is not on the LEFM part in accordance with the
fact that the fracture stress is close to ufe. The tangent to the LEFM curve
discussed in Chapter 3 predicts that for a crack of2a = 0.6 the residual strength
is 47 ksi, as shown in Figure 1O.1Ob, which is almost exactly the same number
as obtained above. Hence, the failure analysis diagram can be constructed on
the basis of the tangent. The residual strength diagram for the above case was
calculated on the basis of ufr = KelP Fa,
and the tangent constructed. For
crack sizes of e.g. 2a = 0.25,0.5 and 1 inch the fracture stresses following from
the tangent are 55, 50 and 40 ksi respectively. With these stresses, the following
values for KIKc and ulufe are obtained.

Point E F G On tangent
2a = 0.25 0.5 in
(Jres = 55 48 40 ksi (from tangent)
K = {3uFa = 34.5 42.5 50 ksi~
K/K/c = 0.69 0.85
af' = (W - 2a)FcodW 58.8 57.5 55 ksi
ures/af , = 0.94 0.80 0.73

Then the three points F, and G and E in Figures 10.10 a can be constructed
so that the curved part of in Figure 1O.1Oa can be drawn. Clearly in this case the
curved part can still be treated with the LEFM tangent approximation.
In its normalized form the failure analysis diagram is the same for all
materials and structures, regardless of K,. and ~ol' Given the scatter in material
behavior, an approximation of the curved part will suffice for many purposes.
A more precise diagram (curved part) can be drawn based on J, but obviously
this cannot be too different.
344
\
Uc 60 \
(kSi) \
K G \
50 E
Kc F
~
/
.6 40
B/
- - - - - - -<I
G

6 /1 / 30
. 1:/
.4 ---,Lll.t 20
---~~ II
/1 J: 10
.2 //1 II
I J I
I
(a) .2 .4 .6 .6 .5 2 3 4
U
2a (in)
Ufc

(b)

Figure 10.10. Approximation of curved part by LEFM. (a) Failure analysis diagram; (b) Tangent.

The use of the diagram for a particular application requires the calculation of
the stress at collapse and the stress intensity at a given stress and crack length.
This permits calculation of and K/ K, at that stress, which can be plotted on the
diagram.
The diagram presents a means for a judgement of the proximity of fracture
and it shows what kind of fracture to expect, putting the three areas of fracture
analysis, LEFM, EPFM and collapse in perspective. It is useful in conjunction
with (not insteasJ of) the residual strength diagram, as it can be derived from the
latter as shown above. Its significance is possibly that it illustrates that from a
technical point of view the EPFM fracture criterion is not very sensitive, and
that also in EPFM collapse must treated separately as a competing condition.

Figure 10.11. Conditions to be considered for LEFM residual strength analysis.

 345

Fi nd Kc or KJc
Find bela
Find yield stress F,y

yes

no

Cres is Low,r of
u'r and Cife

Figure 10.12. Procedure for LEFM residual strength analysis.

This discussion would not be complete without mentioning of the so-called

R-6 diagram [3]. The latter is an extension of the failure analysis diagram to
account for stable fracture, but it opens no new avenues, as the residual strength
whether in LEFM or EPFM can be readily calculated as shown in the following
sections.

10.5. Residual strength analysis

The concepts and conditions for residual strength analysis with LEFM were
discussed in Chapter 3. Figure 10.11 summarizes the conditions, while Figure
346

10.12 shows the procedure. Note again that for small cracks unconservative
results would be obtained, so that a tangent approximation to the calculated
curve must be made. Similarly, for high toughness, or in the general case for
smaller structures, collapse may prevail.
Generally the residual strength analysis (LEFM, EPFM and Collapse), will
be performed using some sort of computer software but hand calculations are
well possible. It should be noted once again, that the analysis will be the same
for any kind of structure, provided one finds P (or H) for the structural
geometry as described in Chapter 8. Since the center crack is easy to envision,
examples here will be based on the center crack.
The stress intensity is K = pa Fa. Fracture occurs when the stress intensity
is equal to the toughness, Klc or Kc whichever is applicable. Fracture occurs if:
pa Fa = Klc (or Kc). Thus for given crack size the fracture stress is found as:
air = Klc(or KJ (10.2)
pFa
Table 10.2. Critical crack size calculation (not advisable; see test)

K 1c = 60ksiJln, FlY = 120ksi.

What is the critical crack size for an applied stress of 30 ksi in a center cracked panel of 30 inch
width?

Solution
According to Figure 3.3, the stress intensity is:

na/ W in radians!

- -KJc
-- First assume fJ I, so that a
n na'
(J2 sec W

~G~Y = 1.27 in This means fJ

J sec
n x 1.27
12 = 1.029;

I( 60
n 1.029 x 30
)2 1.20 in This means fJ J nxl.2
sec-I- 2- 1.025;

Then ac = -I (
n 1.025 x 30
60 )2 = 1.21 in.

The last ac differs only slightly from the previous one so that iteration complete: ac = 1.21 in, and
2ac = 2.42 in.
 347
Table 10.3. Calculation of residual strength center crack; Fry = 120ksi, KJc = 60ksi.JTn" and
W= 12.

2a a p=
H
sec W Fa (J/r

pFa
= KIc/ W - 2a
(iI' = --W- x Fry (Ires

0.5 0.25 0.89 67 115 67

I 0.50 1.25 48 110 48
2 1.02 1.77 33 100 33
4 2 1.07 2.51 22 80 22
6 3 1.19 3.07 16 60 16
8 4 1.41 3.54 12 40 12
10 5 1.97 3.96 7.7 24 7.7
11 5.5 2.77 4.16 5.2 10 5.2

This fracture stress is the residual strength unless collapse occurs at a lower
stress, or unless af' is close to F;;ol so that the tangent must be used (Figures 10.11
and 10.l2.
It is possible to calculate the crack size at fracture directly, or rather, the
permissible crack size ap for the minimum permissible residual strength, ap • It
followed from Equation (3.25a) that

a
p
= .!.(K/c)2
n f3a
(10.3)

Unfortunately, this equation cannot be solved directly, because f3 depends

upon a; hence the value of f3 is not known before a is known and vice versa.
Often f3 is given in graphical form, or at best in terms of a complicated
polynomial. Therefore Equation (10.3) can be solved only by iteration. A
numerical example is shown in Table 10.2. Note that it would still be necessary
to check for collapse.
Similar calculations would show that for a stress of 30 ksi the permissible
crack size would be 1.27 in, and for a stress of 33 ksi, it would be 1.05 inch. The
chance of knowing the acting stress to an accuracy of 10% is small, but 33 ksi
as opposed to 30 ksi makes a difference of almost 20% in the critical crack size.
Thus, one calculation may be inadequate. If the complete residual strength.
diagram is calculated instead, one would immediately see from the slope of the
curve how inaccuracies in stress would affect the permissible crack size. Cal-
culating the entire residual strength diagram is no more work than calculating
crack size with Equation (10.3) because no iteration is necessary, as shown in the
example in Table 10.3, which is for the same case as the previous. Besides the
calculation of Table 10.3 provides much more information.
Consider a situation where f3 is a long polynomial in a/w (as e.g. for an edge
crack, see Figure 3.3). In that case the iteration procedure would be more
cU!flbersome. Last but not least, the iteration does not always converge.
348

Improper choice of the starting value of f3 may cause diversion instead of

conversion. Then the procedure must be started all over again. Finally, if the
actual result would fall on the tangent, the whole procedure would be erroneous,
as fracture would occur below Kc.
These examples illustrate that calculating a permissible crack size directly by
iteration is inefficient. It is advisable under all circumstances to calculate a
complete residual strength diagram. The calculation is no more work, while the
complete diagram provides much more information. The results of Table
10.3 were plotted in Figure 10.13. It is now immediately obvious in which regime
the fastest drop in strength occurs. Similarly, it can be seen in which regime
errors in stress have the largest effect on permissible crack size. It is a simple
matter to sketch operating stress levels in the diagram which facilitates a
judgement of the criticality of a certain crack or loading situation.
The variability of toughness within an apparently homogeneous material can
be significant, and the heat-to-heat variation of the toughness may even be
larger. A scatter in fracture toughness on the order of 15% is not unusal. One
can account for the effect of scatter in the residual strength diagram (Figure
10.13). The figure demonstrates the usefulness of constructing complete residual
strength diagram when evaluating the criticality of a structure with respect to
cracks.
A problem occurs in the calculation of the residual strength for small cracks
because Equation (10.2) predicts that a/i. approaches infmity if a approaches
zero. It was shown in Chapter 3 that a tangent to the curve is a good approxi-
mation. Also collapse must be considered. Assuming (Chapter 3) that in LEFM
the collapse strength, Fool, is equal to the yield strength, Fr" the stress for
collapse of a center cracked panel follows from Equation' (10.1) Collapse
conditions for other cases were shown in Chapter 2 in Table 2.1.

28
.........

12
40 (J = 60 ~(10%)
f31/i8
30

20

10

2 3 4 5 9 10 11 12
28 (jn)

Figure 10.13. Residual strength diagram for case of Table 10.3.

Table 10.4. Residual strength calculation for 3 materials with different toughness and same yield. Fcol = F" ..

(al 12-inch wide center cracked panel.

2a a f3 = ,jsec (nal W Fa aF = K"If3Fa a" = (I - 2a1WlF" a,,, lower of a"

and (Jt,
K,< = 40 K" = 80 K" = 160ksi,jin F" = 60 ksi; all cases
A B C A B C

0.5 0.25 0.89 44 88 176 57.5 57.5

0.50 I 1.25 32 64 128 55 32 * 55
2 1.02 1.77 22 44 88 50 22 50
4 2 1.07 2.51 14 28 56 40 14 28 40
6 3 l.l9 3.07 II 22 44 30 II 22 30
8 4 1.41 3.54 8 16 32 20 8 16 20
10 5 1.97 3.96 5.2 10.4 20.8 10 5 10 10
11 5.5 2.77 4.16 3.4 6.8 13.6 5 3 5 5

(b) 60 inch-wide center cracked panel.

2a a f3 Fa (fIr a" a

K,< = 40 K" = 80 K" = 160ksi-Jln F". = 60ksi A B C

0.5 0.25 0.89 44 88 176 59.5 * *
1 0.50 1.25 32 64 128 59 32 *
2 1.77 22 44 88 58 22 *
5 2.50 1.80 14 29 57 55 14 29
10 5 1.02 3.96 10 20 40 50 10 20 *
20 10 1.07 5.60 7 13 27 40 7 13 27
30 15 l.l9 6.86 5 10 20 30 5 10 20
40 20 1.41 7.93 4 7 14 20 4 7 14
50 25 1.97 8.86 2 5 9 10 2 5 9
VJ
* On tangent from F" (see figure 10.14).
'-0
"""
 350

O"res(k,i)
0",., (k'i)
70r-----~----------------_.r_--~ 70r------------------------.r---,
\
\
60 \
\ 60
\
'I~
" '- ,, 12

.................K{C=160 ksi~
30
'-
'- ,,
20
, \
10
" \
"~
(a)
6 7 9 10 11 12 (b) L---"--,--=======:::2~
10 15 20 25 30 35 40 45 50 55 60
~~ ~~

Figure 10.14. Residual strength diagrrn for cases of Table 10.4. (a) 12 inch panel; (b) 60 inch panel.

Accounting for all of the above, residual strength analysis can be performed
as in Table 10.4. In order to show that collapse may occur even if the toughness
is low, two panel sizes were considered. Also some of the results may fall on the
tangent regardless of how low the toughness. This becomes clear from the
residual strength diagrams obtained by plotting the data of table 10.4 as in
Figure 10.14 (Note: in order to avoid clutter and to better demonstrate the
effect, F;y was assumed the same for all three materials; naturally, the principle
would not change if the materials had different F;y.
Figure 10.14 illustrates one more reason for which it is important to construct
a complete residual strength diagram. The line for collapse can be constructed
and it can be seen immediately whether or not the residual strength should be
found from collapse, LEFM or from the tangent to the curve. Also, note that
the whole curve has to be determined before the tangent can be constructed.
In the case of plane stress or transitional fracture the question arises as to
whether one should use Keff instead of Kc (Chapter 3). From a technical point
of view Keff is the relevant toughness. Naturally, the calculation procedure
remains the same regardless of the toughness used.
The limitations of LEFM are by no means as severe as sometimes suggested,
if sensible approximations are made where necessary, as shown. Besides EPFM
is no panacea either.Calculations are more involved and often not more
accurate than the approximations used in LEFM border cases.
The concept and equations for EPFM residual strength analysis were
discussed in Chapter 4. The calculation is simplest when stable fracture is
ignored (as it is in LEFM). In many cases this will lead to slightly conservative
results. The calculation procedure for that case is shown in Figure 10.15. The
question of constraint (Chapters 4, 7) has not been resolved for EPFM, so that
 351
G.om.try

Find J. for appropiat. thickn.ss

(through crack) plain. strain for
P.T. crack no cn t.rion availai(abl.

Calculat. ~,from EO. (10.5)

solv. by it.ration
If ," tffm n.gligibl. us. proc.-
duro of tabl. 10.5
If 2nd ' t.rm n'gligibl. us. proc.-
duro of tabl. , 0.3

Calculat. str.ss
0;. for collaps.

Find G, from U.

Figure 10.15. Simplified procedure for EPFM residual strength analysis.

'toughness' values must be used for the appropriate thickness in the case of
through cracks. For part-through cracks (plane strain) there is no obvious
answer; the use of conservative values is recommendable.
Generally the fracture equation (Chapter 4):
np2 aa F Hri'+la
~+ F = JR (10.4)

must be solved for al" the fracture stress. Solution is possible only by iteration
which is best done by means of appropriate computer software [4]. If either of
352
Table 10.5. Residual strength analysis using EPFM for four cases, ignoring stable fracture.
a (in) a/W H (fl' = (FJ/ Ha)I/(n+ 1) (fli = (1 - 2a/W) Fe,1
Feol = 56ksi
JR = 0.83 kips/in JR = 5.3 kip/in
W = 8inch
0.5 0.063 8.72 38.9 53.3 49
1.0 0.125 13.5 32.4 44.1 42
1.5 0.188 24.6 27.4 37.3 35
2.0 0.250 57.9 22.6 30.8 28
2.5 0.313 190 17.9 24.4 21
3.0 0.375 1239 12.7 17.3 14
3.5 0.438 35389 7.1 9.6 7
W = 24inch
1.5 0.063 8.72 32.6 44.4 49
3.0 0.125 13.5 27.1 86.9 42
4.5 0.188 24.6 22.8 29.6 35
6.0 0.250 57.9 18.8 25.7 28
7.5 0.313 190 14.9 20.3 21
9.0 0.375 1239 10.5 14.4 14
10.5 0.438 35389 5.9 8.0 7

n = 5;F= 1.9ElOksi 5 .

the terms in Equation (l0.4) is small with respect to the other, hand calculations
are a simple matter. The case of the second term being negligible is a trivial one,
because then the case reverts to LEFM and it is solved as discussed above
(toughness .jEJR ). If the first term is negligible solution proceeds as in the
numerical example in Table 10.5.
The results of this example are plotted in Figure 10.16. Note again that the
same problems exist as in LEFM. An approximation must still be made for
small crack sizes (because for a approaching zero Equation (10.4) leads to
infinite stress); also collapse must be evaluated separately. A complete residual
strength diagram is again necessary for a complete evaluation. Almost all
remarks in this section made regarding LEFM apply to EPFM residual strength
analysis as well, as can be judged easily from a comparison of Figures 10.16 and
10.14.
There is an additional nuisance of EPFM. The criterion is expressed in terms
of strain energy and therefore the 'toughness' or fracture resistance (JR ) is a
meaningless number for direct comparison of alloys. This can best be demon-
strated on the basis of the LEFM expressions. For example, consider a steel and
an aluminum alloy both with the same toughness (e.g. K, = 50 ksi v'In). This
means that the fracture resistance in terms of G (which may be called Gi, or R)
irrespective of its denotation is G, = R (= J R ) = KJ,/E. For the aluminum
with E = 10 000 ksi one obtains G, = 502 /10.000 = 0.25 kips/in, and for the
steel with E 30000 ksi; Gc = 502 /30000 = 0.083 kips/in.
 353
cr~ r-------------------~ crresr-------------------~

.
rksi] [kSi] ,
W=8 in. w =24 in.

\ , ""
"
';
501 1-"
, "
~
COLLAPSE
r """ '- '- ;COLLAPSE

t:"::
"

10
~
(a) --,'----~-----'3:----~ (b) L---~----~----~9----~,2
a[i.]

Figure 10./6. Residual strength analysis with EPFM (Tble 10.5) .• Calculated points from EPFM
(Table 10.5). (a) Panel of 8 inch width; (b) Panel of 24 inch width.

If the fracture resistance of these two materials is quoted as 0.25 and

0.083 kips/in respectively, one is inclined to believe that the aluminum alloy is
a better material. Yet, as both materials have same toughness of 50 ksi.Jill their
residual strength curves are identical. From the engineering point view, the
important issue is the stress that can be sustained at a certain crack size. Clearly,
this is the same for both materials as can be.readily seen because both materials
have the same toughness. The above numbers of 0.25 and 0.083 are meaningless
without knowledge of E.
In the case of EPFM the situation is worse. Not only is the 'stiffness' F
involved, but also the strain hardening exponent n. Thus comparing J R values
gives no indication at all which is the better material from the point of view of
residual strength. The material with the lower J R may well have the higher
residual strength. This is demonstrated in Table 10.6.

10.6. Use of R-curve and JR-curve

Whether the R-curve (LEFM) or the JR-curve (EPFM) is used, the procedure is
essentially the same. The solution is the easiest when done graphically, using

Table 10.6. Examples of lower fracture stress for higher JR'

JR n F aa H«a/W), n) (iF = (FJR/Ha)'/lh+l)

(kips/in) ksi" in (a/w) = 0.125 (ksi)
3 5 lEIl 2 13.5 47.2
5 7 3El3 2 23.3 36.6
5 10 4El4 2 50.8 22.4
1 5 lEll 2 13.5 39.3
3 7 3El3 2 23.3 34.3

a Same configuration and same a in all case.

354
Table 10.7. Calculation of G lines for graphical solution.

atotala pb G = np2(f2 a/E E = 10000

(a + ~a)
= = =
(f = 25 (f 30 (f 33 (f 35
o 0 0 0 0
0.5 0.031 0.045 0.171 0.192
0.062 0.090 0.342 0.384
1.5 0.093 0.135 0.256 0.577
2 0.124 0.180 0.342 0.780
Note: Plot calculated G-lines and R-curve as in Figure 10.17. The G-line for u= 35 ksi appears
about tangent; hence ui' ~ 35 ksi.
a More intermediate crack sizes should be taken if P#-I, because then lines will be curved.
b fJ should be calculated for the case at hand; fJ = I assumed here.

G,R
kips/'n 1

.1

8 .1 ·5
68
.5
I ,
1.5

o .5 1.5 total a

Figure 10.17. Graphical solution (see Table 10.7).

plots such as in Figures 3.17 and 4.2. The G or J lines are represented by the
equations:

(10.5)

First the R or JR curve (material data) is plotted. Then G(or 1) are calculated
in tabular form for a number of crack sizes and five or six values of the stress.
The fracture stress is the one associated with that curve which is tangent to the
R or J R curve. An example is shown in Table 10.7 and Figure 10.17. The
procedure works the same for EPFM using the second of Equations (10.5).
 355
Table 10.8. Logic for iterative solution

if either term is negligible solve directly, otherwise solve by iteration (software).

(a) Solve equation for commencement of fracture to obtain CT, (JR or R for !la = 0).
(b) Increment by small ba.
(c) anew = aold + ba; !lanew = !laold + ba.
J -J G -G
(d) Calculate dJ/da or dG/da as a+b~a a or aH~a a.

(e) Obtain J R or R for !lanew from JR-curve or R-curve.

JRIM< - J R!>.a R&anew - RIM<
(I) Calculate dJR/da or dR/da as new!la or !la

(g) if dJ/da ;;: dJR/da or dG/da ;;: dR/da then done (tangency).
(h) Otherwide go back to b.

Table 10.9. LEFM; R-curve analysis.

Crack. L !la Stable growth and instability dG/da dR/da
a inches
Stress G R
inches
ksi kips/in kips/in
1.500 0.000 14.002 0.100 0.100 0.000 0.000
1.530 0.030 14.115 0.100 0.104 0.078 0.133
1.561 0.061 14.223 0.104 0.108 0.080 0.133
1.592 0.092 14.326 0.108 0.112 0.082 0.133
1.624 0.124 14.424 0.112 0.116 0.084 0.133
1.656 0.156 14.516 0.116 0.121 0.086 0.133
1.689 0.189 14.604 0.121 0.125 0.088 0.133
1.723 0.223 14.687 0.125 0.130 0.090 0.133
1.757 0.257 14.765 0.130 0.134 0.092 0.133
1.793 0.293 14.838 0.134 0.139 0.094 0.133
1.828 0.328 14.857 0.139 0.143 0.097 0.107
1.865 0.365 14.860 0.143 0.147 0.098 0.100
1.902 0.402 14.861 0.147 0.150 0.100 0.100

Fracture instability, dG/da > dR/da, max load

1.940 0.440 14.859 0.150 0.154 0.101 0.100
1.979 0.479 14.854 0.154 0.158 0.000 0.100
2.058 0.558 14.836 0.158 0.166 0.00 0.100

Center cracked panel; plane stress.

Calculation of initiation and instability in load control Note:
Yield strength = 50 ksi in each step G is
Vlt. tens. st. = 70 ksi increased to match R of
Collapse str. = 50ksi previous step
Modulus = 10000 ksi
Crack length = 1.5 inches
2a = 3 inches
Width = 12 inches
Thickness = 0.2 inches
356

Numerical calculations by iteration, according to the logic of Table 10.8, are

shown in Table 10.9 for LEFM (R), and in Table 10.10 for EPFM (JR). SO
many iterations are necessary both in LEFM and EPFM that the procedure is
best done by computer [4]. Tables 10.9 and 10.10 are result of computer
analysis.

10.7. Crack growth analysis

The crack growth analysis procedure is shown in Figure 10.18. Extensive details,
algorithms and examples were shown in Chapters 5 and 6, specific problems
illustrated with examples in Chapter 9. Crack growth analysis, it was pointed
out, requires appropriate computer software [e.g. 4]; in few cases an analysis by
hand is possible (constant amplitude). The following example may serve as a
general illustration.
It is anticipated that cracks might occur at the edge of the reinforcement plate
of a nozzle in a pressure vessel Figure 10.19). The pressure fluctuates approxi-

special analysis
may be Establish stress
necessary history
(clipping, truncationl

yes

no

Draw crack propagation curve

Establish detectable crock size
Establish inspection interval
Sat isfy requirements

Figure 10.18. Crack growth analysis.

 357

l.ocation
I
I

/(f)
R

Lpossible I
cracks

crack

Figure 10.19. Examples of nozzle weld crack.

mately once per two hours between p = 1500 and 3000 psi. Under certain
conditions if the process stagnates the pressure may reach 3600 psi, but this is
an unlikely event. Failure of the vessel would cause explosion damage and
involve loss oflife. The fracture control objective is to reduce the probability of
failure to essentially zero even if this would involve high fracture control costs.
A conservative analysis is indicated.
Cracks are expected to occur in a plane perpendicular to the hoop stress. As
a first approximation, the vessel is treated as a thin-wall cylinder. The hoop
stress (Figure 10.19) is (Jh = pRJB = P x 20/2 = lOp. Hence, (Jh = 30.000 psi
(30 ksi) at maximum operating pressure (and 36 ksi at overpressure) with fluc-
tuations of 15 ksi with a stress ratio of R = 15/30 = 0.5. The yield strength of
the steel is }~y = 60 ksi. The toughness is 80 ksi JITi at operating temperature.
Fatigue crack growth data appear to be reasonably well described by the Walker
equation: da/dN = 10- 10 11[(38/(l-R)2.
Cracks will initate as surface flaws. Several of these may initiate simultaneous-
358
Table 10.10. Computational solution in EPFM
Load/mom Crack. Length J-appl Stress
inkips inches kips/in ksi
0.000 1.000 0.000 0.000
124.386 1.000 0.035 11.330
248.772 1.000 0.358 22.659
373.159 1.000 1.796 33.989
497.545 1.000 6.009 45.318
621.931 1.000 15.641 56.648
621.933 1.002 15.690 56.648
621.933 1.004 15.739 56.648
Max load; instability dJ/da > dJR/da
621.933 1.006 15.789 56.648
621.932 1.008 15.838 56.648
621.930 1.010 15.888 56.648

Crack ~a Stress J-appl JR dJ/dA dJR/da

length inches ksi kips/in kips/in
inches
1.0000 0.0000 56.65 15.64 15.63 24.49 24.5
1.0020 0.0020 56.65 15.69 15.68 24.53 24.5
1.0040 0.0040 56.65 15.74 15.73 24.58 24.5
Max load; instability dJ/da > dJR/da
1.0060 0.0060 56.65 15.79 15.78 0.00 24.5
1.0080 0.0080 56.65 15.84 15.83 0.00 24.5
1.0100 0.0100 56.65 15.89 15.88 0.00 24.5

Cylindrical container
Bending; through crack; circumferential; bending or tension
Outside diameter = 6 inches
Wall thickness = 0.5 inches
Crack size 2a = 2 inches
Yield strength F,y = 30 ksi
Ultimate tensile strength FlU = 60 ksi
Collapse stress F,ol = 40 ksi
Reference stress 0'0 = 40 ksi
Reference strain 6 0 = 1.73913E-03
(X = 15.99831
Strain hardening exponent, n = 3.4
Young's modulus = 23000 ksi
Plastic modulus = 1.006 E + 07 ksi /\ 3.4
Nominal stress at collapse = 44.39756 ksi
Table 10.11. Approximate analysis of case of Figure 10.19.
(a) Residual strength.
B = 2inch; K/c = 80ksiFn; a/2c = 0.1; F,y = 60ksip = 1.12pFFSk,/l/J; l/J ;:::: 1 (Figure 8.3); k, = 1.3;13 ;:::: 1.46pFFS
Crack Fa a/ B pFFS P u, = (K,)pFa)
depth from
a (in) Figure 8.3
0.1 0.56 0.05 1.02 1.49 96
0.3 0.97 0.15 1.07 1.56 53
0.5 1.25 0.25 1.15 1.68 38
0.7 1.48 0.35 1.28 1.87 29
1 1.77 0.50 1.60 2.34 19
1.2 1.94 0.60 2.00 2.86 14

(b) Crack growth.

10- 10
(da/dN) = I!!.KJ.8. 13 as above
(1 - R)2 '
Assumed: constant flaw shape; R = 0.1; 1 cycie/2 hours I!!.u = 15ksi.
a I!!.a Fa alB 13 I!!.K (da/dN) = (10- 10 I!!.K 38/(I _ R)2 I!!.N = l!!.a/(da/dN) N Hours
0.05 0 0
0.075 0.025 0.49 0.025 1.47 10.8 1.13 x 10- 6 22181 22181 44362
0.1 0.025 0.56 0.05 1.49 12.5 1.96 x 10- 6 12727 34908 69816
0.15 0.05 0.69 0.075 1.50 15.6 4.55 x 10- 6 10968 45876 91752
0.20 0.05 0.79 0.1 1.53 18.2 8.19 x 10- 6 6106 51982 103964
0.25 0.05 0.89 0.125 1.55 20.7 1.34 x 10- 5 3744 55726 111452
0.30 0.05 0.97 0.15 1.56 22.7 1.89 x 10- 5 2637 58363 116726
0.40 0.10 1.12 0.20 1.61 27.0 3.67 x 10- 5 2728 61091 122182
0.50 0.10 1.23 0.25 1.66 31.2 6.34 x 10- 5 1574 62665 125330 Vol
VI
0.60 0.10 1.37 0.30 1.75 36.0 1.09 x 10- 4 914 63579 127158 '-0
 360
6 res r---------,
crack depth a (in)

-.
( ksi)
60 \ .6
\ tangent ___ .... _ ~r~is~bl!.
crack
- - - - - - - - - - -
50 .5
\
40 .4
- - over pressure

30 _max operating
pressure

20 .2

tpermiSSible
10
I crack .1

.2 .4 .6 .8 1.9 10 50
crack depth a (in)

Figure 10.20. Results of Table !O.l! for case of Figure 10.19.

ly and join up to form a long slender flaw, or a slender flaw may form im-
mediately. For flaws with large aspect ratios, ¢ will approach I (Figure 8.3). A
stress concentration factor, k t = 1.3 is included to account for the stress con-
centration due to the weld. A hand analysis is shown in Table 10.11; the results
are plotted in Figure 10.20.
Cracks mayor may not occur, but if they do, they will cause catastrophic
failure. Therefore, the fracture control is to be based upon inspection (Chapter
II). Since the maximum permissible crack size (for the case of overpressure) is
0.54 inch, much -smaller cracks must be detected. Ultrasonic inspection (Chapter
II) is probably best suited for the purpose. Considering the circumstances
(height, surrounding piping) the probability of detecting cracks with a depth less
than 0.15 in is considered low regardless of their length. In that case, the
remaining life to failure would be 126000-92 000 = 34000 hours or 3.9 years
(Figure 10.20). The calculations were generally conservative, but nevertheless a
factor of 4 on life is considered advisable. Thus, the inspection interval is taken
as 3.9/4 = I year. Further analysis should be performed to evaluate cracks of
various depth-length ratios. Clearly, more refined analysis is possible with
sophisticated computer software [4]. It should be realized however, that this
would not necessary lead to better answers. (Chapter 12; Sources of error).
Most practical problems are more complicated than the above example.
Questions of stress history, sequencing, clipping and truncation, retardation
parameters and so on, must be considered, as discussed in Chapters 5 and 6. The
above merely shows the principles. Analysis is best performed using reliable
software, but the reliability of software is largely determined by the user (input);
with a sound knowledge of the principles behind the software the effects of
assumptions can be assessed (Chapter 12).
 361

10.S. Exercises
1. Given that K/c = 55 ksiFo and that (Jp = 16 ksi calculate ap using iteration,
and Equation (10.3). The configuration is a single edge crack (Figure 3.3)
with W = 10 inches; Fry = 100 ksi; B = 0.8 inch
2. Calculate the complete residual strength diagram for the case of Exercise 1
and determine ap • How much time did you save and how much more
information did you get, comparing Exercises 1 and 2?
3. Calculate and plot the residual strength diagrams for three panels with center
cracks; W = 4, 10 and 30 inches respectively. F;ol = F" = 60 ksi,
Kc = 65 ksiFo. In each case determine the permissible crack· size if the
permissible minimum residual strength is 20 ksi, and if it is 35 ksi, and if it is
51 ksi, use 10 crack sizes in each case.
4. Repeat Exercise 3 for the case that J c = 0.75 kips/in; Fcol = 60 ksi. Use Has
in Table 10.5. Assume F = 1.2 E 11, n = 5, and neglect elastic term of J.
Take crack sizes so that a/W in each case comes out at values for which H
is known, so that you do not need to interpolate.
5. Calculate graphically the fracture stress for a fatigue crack of a = 2 inch,
using the information of Table 10.7 and the R curve of Figure 10.18; How
much stable fracture will occur?
6. Repeat Exercise 5 using the procedure of Table 10.9, assuming f3 = 1.
Compare with results of Exercise 4.
7. Repeat Exercises 5 and 6 using F and n of Exercise 4. Assume that JR curve
is same as R-curve in Exercise 5, and assuming W = 16inch.
8. On the basis of Exercise 7 estimate K. from J R (E = 10 000 ksi) Then repeat
Exercise 4 using LEFM procedures.
9. Compare results of Exercise 3-8 above and draw your own conclusions.

References
[I] M.J. Broekhoven and M.O. Ruytenbeek, Fatigue crack extension in nozz/ejunctions. 3rd SMIRT
ConY., paper 04.7 (1975)
[2] 0.0. Chell, A procedure for incorporating thermal and residual stresses into the concept of a
failure analysis diagram. ASTM STP 668 (1979)
[3] J .M. Bloom, Prediction of ductile tearing using a proposed strain hardening failure assessment
diagram, Int. J. Fract. 16 (1980) pp. R 163-167.
[4] D. Broek, Fracture mechanics software, FractuREsearch (1987).
 CHAPTER 11

Fracture control

11.1. Scope
Chapters 2 through 5 provided the concepts of damage tolerance analysis, while
Chapter 6 through 10 were concerned with input and analysis practice. This
chapter considers the use of the analysis for fracture control. Crack growth and
fracture analysis is not an end by itself. Its sole purpose is to provide a basis for
fracture control.
Fracture control can be exercised in many different ways. Apart from a review
of fracture control options, this chapter provides procedures for the use of
analysis results in scheduling inspections, repair and replacements, proof tests
and so on. In view of the nature of the problem, the discussions do not provide
clear-cut recipes. Even more so than the analysis, fracture control measures
require engineering judgement and pragmatism. The considerations upon which
such judgement may be based are reviewed. Some damage tolerance require-
ments already specify the fracture control procedure, as discussed in Chapter 12.
These can be understood in the light of the possible fracture control measures
presented here.
After a summary of fracture control options, the selection of inspection
intervals on the basis of analysis results is disussed. Fracture control by
inspection is probably the most universal; safety depends upon the timely
detection and repair of cracks. The sole purpose of the damage tolerance
analysis is then to establish the inspection procedure and the inspection interval.
In view of the cost of analysis, it is important that this be done rationally. The
analysis efforts are futile if the inspection interval is still determined haphaz-
ardly. The chapter is concluded with a survey of itemized fracture control plans,
and a discussion of the cost offracture and fracture control.

11.2. Fracture control options

Structural strength is affected by cracks. The residual strength as a function of
crack size can be calculated, using fracture mechanics concepts. A condition has

362
 363

DESIGN
STRENGTH
o•"maxi

j .. safety factor

CRACK H (TOTAL LIFE)

SIZE

"max
{MAX SERVICE LOADI

RANGEOFj
NORMAL
SERVICE
LOADS

CRACK SIZE
a;b::======-__4--____!--__ TIME

Figure 11.1. Time available for fracture control. (a) Residual strength diagram providing ap; (b)
Crack growth curve providing H.

to be set (generally by offical rules, regulations or requirements) as to the lowest

acceptable strength in the case of cracks, i.e. the minimum permissible residual
strength, up. When the residual strength diagram has been calculated (Figure
ll.la) the maximum permissible crack size, ap ' follows from the minimum
permissible residual strength.
The other information from analysis is the crack propagation curve. It shows
how a crack develops by fatigue or stress corrosion as a function of time. The
maximum permissible crack, ap ' following from the residual strength analysis of
Figure II. la, can be plotted on the calculated crack growth curve as in Figure
11.1 b.
There are several ways in which this information can be used to exercise
fracture control. In all cases, the time period, H, to reach ap (Figure 11.1 b) is the
essential information needed. As no crack is allowed to grow beyond ap ' repair
or replacement is dictated by H. The following options are available for the
implementation of fracture control.
(a) Periodic inspection; repair upon crack detection.
(b) Fail safe design; repair upon occurrence of partial failure.
(c) Durability desig':l; replacement or retirement after time H.
(d) Periodic proof testing; repair after failure in proof test.
(e) Stripping; periodic removal of crack.
Damage tolerance requirements sometimes prescribe the fracture control
procedure. For example military airplane requirements prescribe methods (a)
and (c), commercial airplane requirements prescribe methods (a) and by their
intent promote (b). Requirements will be discussed in Chapter 12.
The above fracture control options are discussed below.
364
Table 11.1. Inspection methods
Method Principle Comments
Visual Naked eye, assisted by Only at places easily
magnifying glass, lamps accessible.
and mirrors.
Penetrant Coloured liquid is Only at places easily
brushed on to accessible.
penetrate into crack,
then washed off.
Quickly-drying sus-
pension of chalk is
applied (Developer).
Penetrant in crack is
extracted by developer
to give coloured line.
Magnetic Liquid containing iron Only for magnetic
particles powder. Part placed materials. Parts
in magnetic field and must be dismounted
observed under ultra- and inspected in
violet light. Magnetic special cabin.
field lines indicate
cracks.
X-ray X-rays pass through Method with
structure and are versatility and
caught on film. sensitivity. Small
Cracks are delineated surface flaws
by black line on film. difficult.
Ultrasonic Probe (piezo-electric Universal method;
crystal) transmits variety of probes
high frequency wave and input pulses.
into material. The
wave is reflected by
crack. Time between
pulse and reflection
indicates position of
crack.
Eddy current Coil induces eddy Cheap method (no
current in the metal. expensive equipment)
In turn this induces and easy to apply.
a current in the coil. Coils can be made
Under the presence of small enough to fit
a crack the induction into holes. Sensitive.
changes.
Acoustic Measurement of the Inspection while
emission intensity of waves structure is under
emitted in the material load. Continuous
due to plastic deform- surveillance possible.
ation at crack tip. Interpretation difficult.
 365

(a) Periodic inspection

Safety is insured when cracks are eliminated before they impair the strength
more than acceptable: they must be repaired before reaching ap • Therefore, any
cracks must be discovered before that point by means of periodic inspection.
Various inspection methods are possible, viz. visual inspection (including loupes
and magnifying glasses), penetrant, eddy current, ultrasonic, X-ray, and
acoustic emission. These methods are summarized in Table 11.1; for details on
these procedures the reader is referred to the extensive literature on the subject.
Whatever the inspection method, there is a certain crack of size, ao, detection
of which is questionable; during inspections before ao, the crack is unlikely to
be discovered. This implies that discov(!ry and repair must occur in the time
interval H between ao and ap as shown in Figure 11.1. Should an inspection take
place at time t 1 , the crack will be missed, and should the next inspection be at
t2 , after an interval H, the crack would already be too long (having reached ap ).
Hence, the inspection interval, I, must be shorter than H. It is often taken as
I = Hj2, but a more rational procedure to determine inspection intervals can
be employed, as discussed later in this chapter.
Damage tolerance analysis, to obtain residual strength and crack growth
curves, is performed solely to determine H and from this the inspection interval.
Safety is maintained by providing a sufficient number of inspections (at least 2)
during H, to ensure crack discovery before ap • Naturally, a crack once
discovered must be repaired at the operator's earliest convenience. Since ap is a
permissible and not a critical crack, and since detection will commonly occur at
sizes less than ap ' immediate repair may not be necessary, but any complacency
will defy all analysis and inspection efforts.
Regardless of how long or short H (the inspection interval) or the inspection
procedures used, safety is maintained, with some reservations as discussed later
in this chapter. Whether inspections must be performed every day (e.g. H = 2
days) or every year (H = 2 years), there will always be two inspections between
aoand ap • Although a daily inspection might be cumbersome, the achieved safety
is not really different in the cases of daily or yearly inspections. If a crack is
missed in daily inspections a potential fracture will occur sooner, but if a crack
is missed in yearly inspections fracture will occur nonetheless before the year is
over. If short inspection intervals are undesirable, one has the option of selecting
a more difficult but more refined inspection procedure with a smaller ao. Then
H, and hence the inspection intervals, will be longer. Fewer inspections are
necessary, but the cost of individual inspections may be higher.
It does not matter either at which time the crack initiates, as illustrated in
Figure 11.2. Inspections scheduled at e.g. HI2 interval will always give two
opportunities for detection regardless of when crack growth begins, provided
that inspections are scheduled at H/2 interval starting from hour zero (even if
initially the chance of a crack is small). Similarly, if the interval is chosen as H13,
366

I: H/2 .~ ~ INSPECTIONS

ap --------:--

TIME

Figure 11.2. Two detection possibilities with interval H/2, no matter when crack starts.

there will always be three inspections between ao and ap ' whether the crack
occurs early or late, or whether simple visual or other inspections are used
(different ao and H).

(b) Fail safety

Providing fail safety by means of crack arresters or multiple load paths, is
essentially a variation of method (a): cracks or failed members must be detected
and repaired. The only difference is that the structure is designed for tolerance
of large damage which is more readily apparent.
For example, crack arresters can be designed specifically for such a large
permissible crack size (Chapter 9) that cracks will be obvious in superficial (but
frequent) inspections. Alternatively, a pipeline or pressure vessel can be
designed so that cracks will cause leaks rather than breaks. As a leak is
presumably obvious, no special inspections are necessary other than frequent
checks for leaks (leak-before-break is also discussed in Chapter 9). In either case,
the structure must be designed to provide for a large and obvious permissible
crack or leak, otherwise the effort is in vain and one of the other fracture control
methods must be employed.
The same can be accomplished by providing multiple load paths (parallel
members), as discussed in Chapter 9. When properly designed, the structure can
 367

still sustain (f p when one member fails. Inspection for cracks would not be
required, but regular checks for failed members would be. Of course member
failure must be obvious, otherwise there is no advantage; a second member will
soon develop cracks when it must carry additional load.

(c) Durability
Ifno inspections can or will be done, a small crack, ai' could be assumed to exist
initially in the new structure (Figure ILl). The time H, for the crack to grow
from ai to ap is then the available safe life. In that case the structure or
component must be retired or replaced after e.g. H/2 hours. This is called the
durability approach.
This may prove a wasteful method. Since H must be very long, heavier and
more costly design may be necessary. If no inspections are performed, there is
no other choice than replacement after H/2 hours. In the case of inspections, the
structure can essentially be operated forever (Figure 11.2). If no cracks should
occur, this would be evident; if they do occur they are detected and repaired and
so eliminated. In the durability approach without any inspection, cracks of size
ap could be present after H, but there would be no way of knowing their
presence.
A major problem with the durability approach is the necessary assumption
for the size of the initial crack. This is an odd problem for structures and
components that are essentially defect free. In that case the initial flaw may just
represent an equivalent crack; at best it is the size of a flaw that can pass initial
quality control. In welded structures the assumption of an initial flaw is more
realistic. Welds often contain defects such as porosity or lack of fusion. In
particular the latter is a sharp defect equivalent to a crack of about equal size.

(d) Proof testing

If the toughness is very low the maximum permissible flaw, ap ' may be smaller
than the detectable crack, ad (Figure 11.3). This can also be the case when the
structure is so large that inspections for cracks are impractical (e~g. a '1000 miles
pipeline), so that the 'detectable' crack size is effectively infinite. In such
situations proof testing is another fracture control option.
Let at a certain time a component be subjected to a proof stress, (fproof.
Fracture would occur if a crack aproof were present, as shown in Figure 11.3b.
Conversely, if no fracture occurs, the maximum possible crack is aproof, so that
a safe operational period H (for growth from aproof to a p ) is ensured (Figure
11.3a). If the proof test is repeated every H hours, a period of at least H hours
of safe operation is available after each successful proof test. Should failure
occur during the proof test (aproof present) then a repair or replacement is made.
The life can be extended forever if no proof test failures occur, provided proof
tests are always conducted at the proper interval, H.
368

ad - - - - - - - - - - - - - - - - -
ap _ _ _ _ _ _ _ _ _ _ _ _ _

CRACK
GROWTH

H
.. I
N
(a)

RESIDUAL STRENGTH
----NORMAL
"" -DURING PROOF
" TEST
RESIDUAL STRENGTH
, RT

-- ---'", '. ;
I '

'--

(b) a (e) a

Figure 11.3. Proof testing. (a) Crack growth; (b) Proof test; no crack larger than aproof; e.g.
hydrostatic test of pipeline or pressure vessel; (c) Lower proof test load after cooling; example USAF
F-ili.

Pipelines and pressure vessels are eminently suitable for the proof test
approach. A line or vessel normally filled with gas or dangerous chemicals can
be proof tested (hydro-tested) with water. A failure during the proof-test would
happen under controlled circumstances, causing a water leak only. In many
cases hydro-tests are already performed anyway. Selecting the proof stress level
and interval on the basis of fracture mechanics analysis, H, would give these a
rational foundation.
Proof tests on structures other than pressurized containers are often hard to
perform. However, if the component can be removed and easily loaded, the
option is available and has been exercised (wing hinges of F-Ill aircraft).
Cooling the structure or component during the proof test causes a drop in
toughness. This permits the use of lower proof stresses to 'detect' the same aproof,
as shown in Figure 11.3c. After the test and warm up, the original toughness and
residual strength are automatically restored.

(e) Stripping
Stripping is another option for fracture control in components with permissible
crack sizes so small that they defy detection. If at a certain time the crack has
reached the permissible size ap , machining (stripping) away a surface layer b
would reduce the crack size to a, = ap - b (Figure 11.4b). As it would take H
 369

ad - - - - _ _ _ _ _ _ _ _ _

ap ___ - - _ _ ._--

(b)

(a)

!I. HOLE I::STRIP6

.OVERSIZE HOLE.
(d)
(e)
Figure 11.4.. Stripping for fracture control. (a) Crack growth; (B) Stripping; (c) Practice at fillet; (d)
Practice in oversizing holes.

hours for the crack to grow from as to ap (Figure II.4a), H hours of safe
operation would be available after stripping. After these H hours the crack
could be of size ap again (note that cracks are too small for detection), hence the
stripping of (i would have to be repeated every H hours.
It would seem that stripping cannot be repeated too many times, but it should
be realized that the stripping layer {) is very small indeed, the cracks not being
detectable. Furthermore, it should be known where the cracks occur, as for
example in a fillet radius (Figure 11.4c, d). Machining the fillet radius by a small
amount is repeatable many times without affecting the general stress level. Shot
peening after the operation, to introduce residual compressive stresses, would be
beneficial. Another case where stripping is feasible and often performed is a
fastener hole. Oversizing the hole and using an oversize fastener can be repeated
several times. A safe period H is available after each such operation.

11.3. The probability of missing the crack

All information needed to determine H and the inspection interval can be

calculated, except for the detectable crack size. The latter has to be obained from
inspection experience, which is not very well documented in the open literature.
Yet, H, or the length of the inspection interval is very sensitive to the (choice of)
detectable crack size, because the slope of the crack growth curve is small for
370
probabilily'-----------,
~ ~
detection 1 ______________ _

Pd

(a) O.o1'-----'-_ _ _ _ _ _ _ _ _--J

O.01L-----L_.L-_'--_ _ _ _ _- '
crack size a
(b)
Figure 11.5. Probability of crack detection in one inspection as a function of crack size. (a)
Probability of detection in one inspection; (b) Effect of circumstances on detectability.

small cracks. Consequently, an ASSUMPTION with regard to the detectable

crack size, may have more weight in the determination of the inspection interval
than the painstaking and costly damage tolerance analysis. This is unsatisfac-
tory. A more rational procedure for establishing inspection intervals is
desirable, as discussed in this and the following two sections.
Detection of cracks larger than the 'detectable size' is not a certainty. It is
affected by many factors: (a) the skill of the inspector; (b) the specificity of the
assignment: e.g. one sn~dfic location, as opposed to a whole wing or bridge; (c)
the accessibility and Viewing angles; (d) exposure; part of a crack may be hidden
behind other structural elements; (e) possible corrosion products inside the
crack, and so on.
Typically, the probability of crack detection depends upon crack size in the
manner shown in Figure 11.5. There is a certain crack size, ao, below which
detection is physically impossible. For example, for visual inspection this would
be determined by the resolution of the eye, for ultrasonic inspection by the wave
length, and so on. In reality, ao is larger than these physical limits.
The probability of detection is never equal to 1 even for large cracks; any
crack may be missed. It follows that the probability curve must have the general
form as shown in Figure 11.5, which can be described by the equation:
(11.1 )
where ao is the crack size for which detection is absolutely impossible (zero
probability of detection), C( and A. are parameters determining the shape of the
curve. The equation gives the probability, p, that a crack of size a will be
detected in one inspection by one inspector. This inspector may not detect the
crack. The probability of non-detection is 1 - p. Note that Equation (11.1) is a
curve-fitting equation; it does not make any statement about the statistics
i:wolved, despite the fact that a similar equation is used for certain statistical
procedures. Equation (1Ll) is merely a convenient format for describing the
general shape of the curve. Any other equation providing a similar curve shape
would be equally useful.
 371

A crack is subject to inspection several times before it reaches the permissible

size. At each inspection there is a chance that it will be missed. At successive
inspections, the crack will be longer, and the probability of detection higher, but
there is still a chance that it goes undetected.
Consider 100 cracks growing at equal rates, (same population), all in the same
stage of growth (same size). Let the probability of detection at a certain
inspection be P = 0.2. The probability that a crack will be missed is then
q = 1 - P = 0.8. That means that 80 cracks will go undetected. At the next
inspection the cracks are longer; let the probabilty of detection then be P = 0.6,
so that q = 0.4. Thus of the remaining 80 cracks 0.4 x 80 = 32 cracks will go
undetected, etc. Apparently the cumulative probability that a crack will be
missed in successive inspections is Q = ql X q2 X •.• X qn. In the above
example Q = 0.8 x 0.4 = 0.32: of the 100 cracks 32 cracks remain undetected
after two inspections. The cumulative probability of detection is P = 1- Q. In
the above example P = 0.68: of the 100 cracks 68 were detected after two
inspections, but 32 were missed.
The cumulative probability of detection is then:

n (1 -
n
P = 1- Pi) (11.2)
i=1

where P at each crack size follows from a curve such as in Figure 11.5 or from
an equation such as Equation (11.1).
Figure 11.6 shows what happens if inspection intervals are determined as
I = H/2, where H is the time required for crack growth from ad to ap- The

,i ,
p

IJ".~
a Hn/2

r" H
I

ap

as

(a)

Figure 11.6. Inspection intervals following from H/2. (a) Probability of detection in one inspection
of crack size a, with two methods; (b) Crack growth curve and inspection intervals for two
inspection methods.
372

detectable crack size, ad' is often taken more or less arbitrarily, but it might be
selected as a crack with a certain probability of detection. For example, the
detectable crack size could be defined as aso, a crack with 50% probabilty of
detection. Such a criterion certainly has appeal, because it seems consistent; yet
it still leads to inconsistencies.
For the case of Figure 11.6 either method lor method II could be prescribed.
The detectable crack sizes, a so , lead to different inspection intervals, Hd2 and
H1/12. Inspections would take place as indicated by arrows in Figure 11.6b. The
cracks would be inspected for the first time when they still have a size smaller
than aso . At this first inspection, the probability of detection is not zero (unless
a < ao). There a distinct probability, PI' that the crack is already detected
during that first inspection; the probability that it is missed is ql = 1 - PI' At the
next inspection the crack is larger, and the probabilty of detection is Pl' etc. By
the time the crack reaches ap ' it has been inspected n times. The cumulative
probability of the crack having been detected at anyone of these inspections
follows from Equation (11.2).
The probability-of-detection curves are different for the two procedures in
Figure 11.6, but either method would be satisfactory on the basis of the criterion
I = H12, where I is the inspection interval. The more involved method II (for
example ultrasonic) inspection with a conveniently long interval, and the easier
method I (for example visual) inspection with the more cumbersome shorter

CASE 2 CASE 1

H =4800 hrs H/2 = 2400 h,.

W 40
N
lJ)

'"-<
u
I
<r
u 30

I
INSPECTION 2 i
20 i
INSPECTION 1
I
10
. _Hj2 )
._.--.--.-'"
ad L-~=._=. _ _. _ . _ . _ ' _ - - ' - ' - '

0.5 1.0 1.5 2.0 2.5 3. 0 3.5 4. 0 4. 5

LlFE (1000 HOU,~S)

Figure 11.7. Alternative crack growth curves.

 373
interval are equally acceptable under the criterion I = Hj2. However, the safety
level certainly would not be the same in both cases: the cumulative probability
of detection certainly would be different, as can be readily appreciated from
Figure 11.6 and Equation (11.2).
The effect is even more striking for the case of different crack growth curves,
as illustrated in Figure 11.7. Such different growth curves are e.g. the result of
different f3 (different geometries). Let the method of inspection envisaged be one
with a 'detectable crack size aso' of 5 mm, and let in both cases the maximum
permissible crack size be ap = 33 mm. Then H would be 4800 hours for both
cracks. This would lead to an inspection interval of I = Hj2 = 2400 hours in
both cases. Consequently, both cracks would be inspected equally often. But
crack I would have a much better chance of being detected than crack 2, the
latter being much shorter at any of the inspections. Although the same criterion
was applied for both cracks, not the same level of safety is achieved. A better
rationale for determining inspection intervals [1] is discussed in the following
sections.

11.4. The physics and statistics of crack detection

A sizeable body of inspection data was generated by Lewis et al. [2] for several
large and small structural components inspected by many inspectors. An
example of the results is shown in Figure 11.8. The figure shows the detection
ratio, defined as the number of detections of a crack of a certain size divided by

P
,9 00
8 o o
o
c
,8
o ag °0
oco o o o
o 0
.7
00 0 0

.5
g 0
.4
o
8 0 o
o
.3
EDDY CURRENT
.2 96 INSPECTORS

52CRACKS
.1 '

5 10 15 a (mm)

Figure 11,8, Detection ratio, jJ; from Lewis et al (2), as a function of crack size a,
374

the number of inspectors. If the number of inspectors is large, as it was, the

detection ratio will approach the true probability of detection.
The number of inspectors covered in Figrue 11.8 is 96. For convenience,
consider a situation where 100 inspectors would look at the same crack for
which the probability of detection is known to be p. The results of these 100
inspectors are known beforehand for any probability p, because the distribution
is binomial, as if these 100 inspectors were picking marbles out of a vase with
white (detect) and red (miss) marbles. They either pick a white or a red one. For
a certain probability, p, the band in which e.g. 90% of the observations must fall
can be constructed a priori.
For example, for p = 0.8 and q = 0.2, the standard deviation of this
oinomial distribution is:
s = .J(pqn) = .J(0.8 x 0.2 x 100) = 4, (11.3)

where p is the probability of detection, q = 1 - p the probability to miss, and

n the number of inspectors. Since the 90% band is given by
1.64 s = 1.64 x 4 = 6.4, the results are pre-determined. The expectation is for
80 inspectors to detect and for 20 to miss. There is a five percent chance that only
8-6.4 = 73 inspectors would detect, and a five percent chance that
80 + 6.4 = 86 inspectors would detect the crack. Conversely, if 73 inspectors
out of the 100 detect a certain crack, the probability of detection could still be
as high as 0.8; if 86 inspectors would detect, the probability of detection could
still be as low as 0.8. The 90% band for the detection ratio of 100 inspectors
MUST be between 0.73 and 0.86 if the actual probability of detection is
p = 0.8. Hence, if only 20 out of 100 inspectors detect a crack, the probability
of detection is definitely less than 0.8.
It follows that if 100 inspectors 'look' at two different cracks of the same size,
and report 75 detects (25 misses) for one crack, and 30 detects (70 misses) for
the other, then the probability of detection of these two cracks is different:
although they may have the same size, they belong to different categories with
different probability-of-detection curves. For any such curve, the 90% band for
100 inspectors is pre-established by Equation (11.3) as shown in Figure 11.9. If
some data obtained by 100 inspectors fall outside this band, the cracks involved
belonged to different categories (populations).
For example, the data in Figure 11.8 show that certain cracks of approxi-
mately 9 mm size were detected with a ratio of 0.4, while others of the same size
were detected with a ratio of 0.9. The above discussion shows that the spread
for 96 inspectors cannot be that large, so that these cracks of approximately
9 mm were not of the same population. Ergo, the data set in Figure 11.8 contains
more than one population and more than one probability curve.
The 'probability-of-detection' curve for cracks of any type in any structure is
 375

o. 9
o. B
0.7

0.6

0.5

0.4

0.3

0.2

0.1

O. 5 1. 0 1. 5 2. 0 2. 5 3. 0 3. 5 4. 0 4. 5
CRACK LENGTH (lNCH)

Figure 11.9. Probability of detection in one inspection; 90% band for 100 inspectors.

low and follows the lower bound of the data in Figure 11.8. Indeed, this low
curve should be used if inspectors were assigned just to look for cracks in a large
structure somewhere. That is not the way inspections are specified in practice.
The data of Figure 11.8 are repeated in Figure 11.10 to show categories of
difficulty of inspection. Figure 11.10 also contains data [3] for a case where

("'""
ii
I /
.9
I

.8

.5

EDDY CURRENT

96 INSPECTOR':'

52CRACKS
.1

5 10 15 a
Figure 11.10. Data from Figuf(~ 11.8 split into different families.
376
inspectors were assigned one particular location in different aircraft: they knew
where to inspect and what type of crack to look for (high specificity), as is
normally the case; their results are much better. Had all inspectors represented
in Figure 11.10 been given specific assignments their results would have been
better as well. As the assignment was for any cracks in large areas the results are
varied, namely the way they came out.
The 90 percent bands, known a priori as discussed on the basis of Equation
(11.3) and Figure 11.9, can be drawn for the 96 inspectors involved for certain
probability curves. Three such bands are shown in Figure 11.10. They indicate
that the data cover at least three different populations, determined by specificity
(some components were small, others very large) and accessibility (some cracks
were easily accessible, others were not).
A category or population can be defined qualitatively only. An 'ideal'
population is formed by the increasing crack sizes during successive stages of
growth of a single crack. These cracks of different length are indeed of the same
class of difficulty (access). Similarly, cracks of different sizes at the same location
in a number of identical components would be in the same category. But two
such crack types would be in different populations if one were located in a niche
with a difficult viewing angle, the other in a smooth flat surface. The data
reported in [3], shown in Figure 11.10, belong to more than one population.
A comprehensive damage tolerance assessment will identify types of cracks
and their locations. Inspections will be prescribed for specific locations with
known access. The probability-of-detection curve can be defined for the specific
circumstances oJ the inspection envisaged.
Clearly, probability-of-detection curves obtained in the laboratory are not
very relevant. In the first place the inspectors know that there are cracks,
otherwise the experiment would not be conducted; this introduces bias. In the
second place laboratory specimens are ideally accessible under comfortable
circumstances. Third, the assignment is very specific: small specimens and
usually one specific location. Results of such investigations can provide data for
the most ideal circumstances only. Cracks in the tension bars of a suspension
bridge will be in a different category. In each case a different probability of
detection curve applies.
Available data obtained under realistic circumstances for structures [2] are
useful, provided it is realized that they cover more than one population. For
inspections of high specificity and/or easy access their upper bound applies,
while for general inspections and/or poor access their lower bound applies. As
long as inspection assignments are accounted for, the relevant probability curve
can be determined from those data. Thus a re-evaluation of the data would be
highly worthwhile. In the mean time, the data set is still useful, because of its
extent. Not only eddy current inspection was covered, but also X-ray, penetrant
and ultrasonic inspection.
 377

11.5. Determining the inspection interval

Inspection must be prescribed with due account of accessibility and specificity.

This can be done if the critical locations are properly identified. Specificity and
accessibility determine the applicable probability-of-detection curve. Categories
of accessibility and specificity must be established first [1], and for each of these
the probability curve established on the basis of available data [1, 4].
The length of the inspection interval should be established such as to provide
a consistent safety level (cumulative probability of detection), independent of
the shape of the crack growth curve, the accessibility, and the specificity of the
inspection. The aimed for cumulative probability of detection could be set for
example at 95 or 98%, and be specified in damage tolerance requirements.
Given the calculated crack growth curve and permissible crack size, and the
probability-of-detection for the relevant specificity and accessibility, the
cumulative probability of detection can be calculated for different lengths of the
inspection interval by means of Equation (11.2). When the results are plotted, the
interval for the desirable probabilty of detection can be obtained from the curve
[1, 5]. The interval will be different for different inspection methods, different
crack growth curves, accessibility and specificity, but the cumulative probability
of detection is always the same (equal safety). The problems discussed in Section
11.3 are then eliminated automatically.
A computer [5] can perform the calculation for different interval lengths,
provided the crack growth curve calculated in the damage tolerance analysis and
the applicable parameters to Equation (11.1) are provided as input. For a
certain inspection interval it finds the crack sizes, at which the inspection will
take place, from the crack propagation curve. At each inspection (crack size) the
probability of detection follows from the probability curve with the parameters
appropriate for the inspection method and category. Equation (11.2) is then
applied to obtain the cumulative probability. An example of a hand calculation
for 2 inspection interval-lengths is shown in Table 11.2. For a complete analysis
more inspection intervals-lengths must be considered. In a computer program
the procedure can be further refined by accounting for the fact that the crack
size at the first inspection may vary in accordance with the time of crack
initiation (Figure 11.2).
Typical computer results [1, 5] are shown in Figures 11.11 and 11.12 for the
two crack propagation curves in Figure 11.7. As discussed in Section 11.3 the
criterion I = H/2 would assign the same inspection interval to both cracks.
According to Figures 11.11 and 11.12 this would lead to different cumulative
probabilities of detection. In order to ensure the same probability in both cases
the intervals must be shorter in case 1. Figure 11.13 shows how the probability
of detection of the case 1 crack would be affected by accessibility and specificity.
Although refinements can be made, the above provides a rational procedure
 w
-...)
00

Table 11.2. Hand calculation of cumulative probability of detection for two inspection intervals.
Inspection interval = 500 hours Inspection interval = 1000 hours

10 11 12 13 14

Inspection Hours a from Pd from Pmiss Pmiss Pdetect Inspection Hours a from Pd from Pmiss Pmiss Pdetect
Fig. 11.7 Equation (1 - Pd) cumulative cumulative Fig. 11.7 Equation (1 - Pd) cumulative cumulative
(or figure) n x Pmiss (I - 6) x 100% n x Pmiss (I - 13) x 100%

5 5
500 5.5 0 1 1000 6.5 0.27 0.73 0.7300 27
1000 6.5 0.27 0.73 0.7300 27 2000 0.54 0.46 0.3358 66
1500 7.5 0.42 0.58 0.4234 58 3 3000 13 0.69 0.31 0.1041 90
4 2000 9 0.54 0.46 0.1948 80 4 4000 19 0.80 0.20 0.D208 98
5 2500 11 0.63 0.37 0.0721 93 Total at ap : 98
6 6000 13 0.69 0.31 0.0223 97
3500 15.5 0.75 0.25 0.0056 99.4
4000 19 0.80 0.20 0.0011 99.9
4500 28 0.88 0.12 0.0001 99.99
Total at 0p: 99.99

Notes: (I.) Crack growth curve (calculated) must be known (Case I of Figure 11.7 assumed). (2.) Columns 3 and 10 follow from crack growth curve. (3.) Pd in columns 4 and II follow from curve
as in Figure t 1.10 or from Equation 01.1). In this case Equation (11.1) assumed with ao = 6.), = I L:x. = 0.5. (4.) Columns 6 and 13 [rom Equation (11.2): Column 5 or 12 times previous number
in same column.
 379
z
8
~
u
~
WJ
Cl

"-
CJ
>-
I- \
;; 0.9
aJ
\
-<
gJ 0.8 \
'"
0- \
~ 0.7 \

""
_ULTRASONIC
;::
-<
:5
a """
0.6 . _ _ PENETRANT

0.5 _ _ EDDY CURRENT

0.4 " "" _X-RAY

0.3 "" _ _ VISUo\L

0.2
" '-----
0.1

D. 5 1. 0 1. 5 2. 0 2. 5 3. 0 3. 5 4. 0 4. 5
~ENGTH iNS?ECTJON INTERVAL (1000 HOURS)

Figure 11.11. Cumulative probability of detection; Case I crack growth curve of Figure 11.7 (small
area; good access).

to establish inspection intervals for which the probability of detection is in-

dependent of the inspection technique, the crack growth curve, and the
assignment. The procedure is finding acceptance in the aircraft industry. As
calculation of the crack growth curve and ap requires expensive damage
tolerance analysis, using the results for determination of inspection intervals as
I = H/2 is unsatisfactory indeed. The above procedure is a much better
approach.

11.6. Fracture control plans

The optimum fracture control plan depends upon the consequences of a
fracture. If the number of fractures experienced is considerd to be at an
acceptable level with a certain fracture control plan at acceptable costs, the plan
is close to optimum. Before implementation of a fracture control program the
objectives must be identified. If a structure can sustain assumed damage under
an assumed loading condition, it is not necessarily safe despite all analysis.
Before defining the permissible residual strength or permissible crack size, the
desired level of safety should be established, even if only qualitatively. It will
appear that every component and structure sets different fracture control re-
quirements.
380
~
,-
~
'-"
n

'"
~
>-
_J 0.9
aJ
< I
aJ
co 0.8
'"w
n.
I
:> 0.7 \ _ _ lA...TRASONIC

\
1-
<
..J
:::J 0.6
\
:>:
:::J
w
0.5
\ _~_ EDDY CURRENT

O. 4
\ _._X-R.\Y
\
0.3 \ _ _ VISUAL

0.2

\
"\" '-...
0.1 ............"'""----._._.-
\ ...............
" 0.5 1. 0 1. 5 2. 0 2.5 3.0 3. 5 4.0 4.5
LENGTH INSPECTION iNTERVAL (1000 HOURS)

Figure 11.12. Cumulative probability of detection; Case 2 crack growth curve of Figure I 1.7 (small
area; good access).
~
~
u
~
'"
o
"-
o

\\
>-
~

0.9
\ \
..J
;;;
<
~ 0.8 \ \.
'"n. \ \.
~ 0.7

:s
~ \
\ \.
\ _ _ ULiRASONIC

a
:::J 0.6 ___ P::~Ei"RANT

\ "-.
0.5 \ "-. _ _ EDDY CURR::~T

\~
o. 4
CAT E
_ _ X-RAY

0.3 __

.......... J--
VISU,~l

0.2 '....... ~
0.1
------ - - - ---

0.51.01.52.02.53.03.54.04.5
LENGTH INSPECT ION INTERVAL (1000 HOURS)
Figure 11.13. Cumulative probability of detection Case 1 crack growth curve of Figure I 1.7. Cat.
A; One location: Good accessibility Cat. E: Large area; access not easy.
 381
The consequences of a fracture must be acceptable, the fracture control
measures in accordance with the acceptable risk. The structure must have
adequate damage tolerance to meet this risk. Designer or manufacturer
prescribe the details of the fracture control plan, the operator implements this
plan through maintenance, inspection., repair, replacement, proof testing, and
possibly load monitoring. The plan must be suitable for a particular structure,
component or part; it also must be suitable for the potential operators. Pro-
fessional operators of pressure vessels, airplanes and the like, can implement
more complex fracture control measures than the general public operating
automobiles. When fractures can be adequately controlled by selecting
materials of sufficient toughness, the fracture control plan is indeed simple. But
here the concern is with those cases where fractures can have serious conse-
quences and where material selection alone does not provide adequate
safeguards against such fractures.

(a) Detectable cracks

Table 11.3 shows the ingredients of fracture control plans for structures in which
cracks are detectable by inspection. If initial defects will not grow during service,
Plan I is applicable. If defects, whether initial or developing later, may grow
under service loading, a crack eventually will become critical, unless it is timely
discovered and repaired. If the permissible crack is large and readily apparent
Plan IIa is applicable, otherwise inspections should be scheduled in accordance
with Plan IIb.

(b) Cracks not detectable by inspection

Cracks may not be detectable, either because their permissible size is so small
that it defies inspection, the location not accessible, or the structure so large that
inspections are not feasible. Plans IIIa and IIIb are applicable in such cases
(Table 11.4). If stripping (Section 11.2) is possible, Plan IIIc may present an
alternative. This could be for cracks at fastener holes (oversizing of holes and
use of oversize fasteners), or for cracks at fillets if the component can be easily
removed. Plan IV in Table 11.5 involves proof testing to show that no cracks
larger than aproof are present (Section 11.2). If larger cracks are present, a failure
will occur during the proof test, but this failure must not be catastrophic in its
consequences. The latter should be ensured by the use of water instead of gas
for proof testing pressurized containers, testing small sections of a pipeline at a
time, evacuation of surroundings, and if possible, cooling so that low pressures
are sufficient.

(c) After crack detection

All fracture control calls for immediate repair or replacement when a crack is
discovered. This is not always convenient. Large savings may be realized if
382
Table 11.3. Fracture control plans for anticipated cracks that are detectable by inspection

Plan I For initial defect not expected to grow by fatigue.

- Calculate permissible size of defect.
- If stress corrosion Can occur, calculate which size of defect
can be sustained indefinitely given the K/sec of the material.
- Inspect once using a technique that can reliably detect defects
of above sizes.
- Eliminate all detected defects larger than above.

Plan 1I
For all defects (initial or initiating later) that will grow during service - these will reach critical size.

Alternative lIa
- Show by analysis (or tests) that the structure can sustain without failure such large defects that
the damage will be obvious (e.g., readily apparent leak or failed component; fail safety).
- Repair when damage is discovered.

Alternative lIb
- If above cannot be shown, calculate permissible crack size.
- Establish crack size that can be detected reliably with inspection technique envisaged.
- Calculate time for crack growth.
- Implement periodic inspection based on crack growth calculation, using adequate factor or
procedure of Section 1I.S.
- Start inspection immediately as time of crack initiation is not known.
- Repair or replace when crack is detected.

Table 11.4. Fracture control plans for anticipated cracks not detectable by inspection because they
are too small.
Plan 1I1 For parts where a is so small that it defies inspection.

Alternative lIla
- Calculate life.
- Replace/retire after calculated life expires (using adequate factor).

Alternative 1IIb
- Make best estimate of possible initial defects.
- Calculate permissible crack size ap •
- Calculate crack growth life from initial defect size to ap •
- Replace/retire after calculated life expires (using adequate factor).

Alternative lIIc If stripping is possible.

- Calculate permissible crack ap •
- Establish feasible stripping depth, fJ (see Section 11.2).
- Calculate crack growth life, B, from (ap - fJ) to ap •
- Repeat stripping of fJ at intervals (B/2.)
 383

Table 11.5. Fracture control plan for anticipated cracks not detectable by inspection because
inspection is not feasible, but proof testing is possible.
Plan IV For components or structures that can be proof tested and where failure during proof
testing is not a catastrophy.
- Determine feasible proof test pressure or load.
- Calculate maximum crack size a pmof that could be present after proof test (see Section 11.2).
- Calculate maximum permissible crack ap '
- Calculate crack growth time, H, from apmof to ap •
- Repeat proof test before H has expired (using adequate
factor).

Table 11.6. Fracture control plans for cracks discovered in service

Plan V For detected cracks for which no analysis is done.
- Repair or replace unconditionally.

Plan VI
For detected cracks for which analysis is done (if immediate replacement is impractical).

Alternative VIa
Show that larger defect can be sustained.
- Check growth daily; drill stop hole if possible.
- Prepare for repair or replacement at earliest convenience.

Alternative VIb
Determine exact size and shape.
- Find materials data; if possible cut test specimens from structure.
- Obtain reliable load and stress information.
- Calculate ap •
- Calculate time, H, for growth to ap '
- Prepare for repair or replacement before H (with adequate factor) expires.
- Check growth daily; drill stop hole if possible.
- If crack grows faster than calculated, update prognosis and speed up replacement or repair
actions.
- If possible reduce operational loads.
- Repair or replace as soon as possible.

Plan VII
For structures identical to those in which a crack was detected.
- Use parts of cracked or failed structure to obtain material properties.
- Implement one of Plans II a-b, III a-b, IV.

remedial action can be scheduled for the next major overhaul or shut down, or
when at least operations can continue until a new part or component has been
manufactured and received. Whether or not this is possible depends upon the
fracture control plan in force. A well-conceived Plan IIa already contains
information on crack growth and residual strength. Using this information as
an initial safeguard, operation can be continued but the analysis should be
384

updated and Plans VIa or VIb (Table 11.6) be put into action. As it is often
difficult to measure the exact size and shape of the crack, the more stringent plan
VIb may be indicated.
A crack may be discovered accidentally in a structure not subject to a fracture
control plan. When no analysis is to be done, Plan V is the only possible course.
Otherwise, Plans VIa or b can be used. Recurrence of the incident can be
prevented using Plan VII.

11.7. Repairs

The sole objective of damage tolerance analysis is to establish fracture control

measures so that cracks can be eliminated before they become dangerous, by
either repair or replacement of the component. The objective is not to determine
whether a crack appearing in service can be sustained. If a crack appears it must
be repaired; there is no excuse for a fracture resulting from known cracks
regardless of what analysis predicts. Naturally, replacement or repair is not
always convenient immediately upon crack discovery, but damage tolerance
analysis is not intended to show how long 'one can live with' cracks. The above
are repetitious statements of the same issue, but repetition is justified as the
objective of damage tolerance analysis is too often misinterpreted: it is to prevent
fractures, not to evaluate how long discovered cracks can be sustained.
When crack discovery demands repair a new damage tolerance analysis
problem arises. Not only must the repair be adequate to restore strength, it must
be analyzed for damage tolerance again. A simple cover plate usually does
not suffice (Figure 11.14). In view of Figure 10.2 such a repair may rather
aggravate the situation and cause new cracks in due time. The stiffness of the
cover plates may introduce a stress concentration, the solution may be 'worse
than the disease'. (Consider the fact that the increased stiffness will attract
loads to the bolt holes; attached parts must undergo the same displacement -
strain - so that the stiffer part will take most of the load/stress.)
Repairs must be designed to cause gradual transfer of loads and stresses as
discussed on the basis of Figure 10.2. A new damage tolerance analysis must be
performed for the repair. It should not be assumed that the repair is a permanent
solution. Fracture control measures must be reeinstated for the repair.
The above may seem trivial at first sight. However, severe accidents have
occurred as a result of inappropriate repairs; in a recent case more than 400
people lost their lives owing to inadequacy of a 'so-called repair'. Upon dis-
covery cracks must be repaired at the earliest possible convenience. Repairs
are not a 'final solution'. Damage tolerance analysis of the repair must be
performed, and fracture control measures taken for the repaired structure; all
issues discussed in Chapters II and 12 must be accounted for. Efforts to reduce
stresses by including a (stitT) load bypass or second elements often make the
 385
CRACK BOLTS OR WELD

===~~~~~~*I~=======
4 ' ....
I B
NEW PROBLEMS
AT A OR B

Figure 1J .14. Unsatisfactory repair (for better solution see Figure IO.2b).

situation worse than it was, because the load will be attracted to the stiff element
(see Figure 10.2).

II.S. Statistical Aspects

Many of the parameters and variables playing a role in fracture control vary
beyond control of human beings. Usually, the statistical variabilty is dealt with
in a deterministic way by assuming that estimates of the average values provide
adequate answers to engineering problems. The answers are factored to account
for variability. Sometimes variability is accounted for by taking 90 or 95 percent
exceedance values.
All material properties, including ultimate tensile strength and yield strength,
show variability (scatter). Fracture toughness and crack growth properties do
too. A scatter in fracture toughness of 10 to 15% is not unusual; variability by
a factor of about 2 of fatigue crack growth rates is normal. In most cases the
structural loads are statistical variables. The pressure in a vessel may be well
controlled, but random fluctuations may occur. The loads on bridges vary
widely depending upon traffic; they can be estimated but cannot be known until
after the fact. Finally, crack detection is governed by statistical variables. There
is a non-zero probability that a crack will be missed. In spite of sophisticated
386
fracture control, the probability of fracture will never be zero. Ideally the
fracture control plan should be based upon the acceptable probability offailure.
Because of the variability described, a safety factor is necessary if a deterministic
analysis is performed.
In addition, there are errors due to shortcomings and limitations of the
analysis, due to the limited accuracy of loads and stress history, and due to
simplifying assumptions (Chapter 12). The magnitude of the necessary safety
factor then depends upon the 'total uncertainty'. There is a natural tendency to
cover every uncertainty when it appears by taking conservative numbers:
highest estimates for loads and stresses, low estimates for toughness, upper
bound growth rates, worst crack configurations, and so on. This amounts to a
compounding of 'safety factors' of unknown magnitude which may lead to
conservative answers, but the final conservatism is unknown. For the effects of
all these assumptions see the discussion on accuracy in Chapter 12. It is
preferable to use best estimates and average data and to apply a factor of known
magnitude at the end of the calculation. Ideally, regulating societies or authories
should establish rules and recommendations for safety factors, as they do for
general design. Otherwise safety factors must be decided upon on a case-by-case
basis.
As an example, consider a crack growth analysis to determine an inspection
interval. The rate of crack propagation may depend upon 11K to the 4th power.
If there is a possible uncertainty of 10% in the loads, 10% in the stresses
following from these loads, and 10% in p, the potential error in 11K may
approach 30%. The effect on da/dN will be a factor of (1.3)4 = 2.86. If growth
rates can vary by a factor of 2, the calculated life might be off by a factor of 5.72.
One could then apply a factor of 5.72 on the calculated life, by scheduling 6
inspections: I = H/6.
Statistical fracture mechanics have been developed, in which all variables are
accounted for by the rules of statistics. Such procedures are of great interest,
provided the statistical distribution parameters could be known. If these have
to be estimated the more complicated technique may not lead to more reliable
answers. The simple way of applying statistical fracture mechanics is to
determine the statistical distributions of all input variables. By employing a
Monte Carlo technique a value for each input parameter can be selected and a
deterministic analysis performed. Subsequently, new input values are selected,
again with the Monte Carlo technique, and another deterministic analysis
performed. This process is repeated many times, so that eventually a distri-
bution of answers is obtained. The latter can then be analyzed statistically to
determine the probability offailure, given that certain fracture control measures
are implemented.
The problem with the latter procedure is to establish the statistical distri-
bution of the input. This can be done only if assumptions are made with regard
 387

to interdependence. For example, the statistical distributions of such input as

F;y, KIn and da/dN are dependent, because these properties are intrinsically
related to the material. Both the toughness and F;y for a certain alloy may show
variations by 15%. However, if F;y falls at the low end of its range, it is more
than likely that K lc will fall at the high end of its range, and vice versa. By
assuming that these properties are independently variable, the physics of the
problem are violated no matter how elegant the subsequent statistics and
mathematics. The problem could be analyzed if the physics of the dependence
were known, but they are not. Determining this dependence on the basis of data
would require many more test data than are usually available. Although statisti-
cal fracture mechanics are of interest, it would seem that much more develop-
ment is needed for general engineering applications.

11.9. The cost of fracture and fracture control

The acceptable consequences of failure form the basis for the fracture control
philosophy. These consequences must be weighed against the probability of
failures other than by fracture. Establishing the acceptable consequences of
fracture is an economic as well as an ethical problem; they must be considered
in the light of other circumstances endangering life. From a technical point of
view, the problem can be dealt with only if the consequences ofa failure in terms
of economic, ecological, and human loss can be quantified (expressed in cost)
and compared to the cost of fracture control. Then the cost effectiveness of
fracture control measures can be compared with their effect. If the probability
of fracture is low and the consequential cost of fracture manageable, costly
analysis and a costly fracture control plan cannot be justified. It is morally
difficult to assign a cost to a human life, but practically it is not. An individual
buying life insurance, in principle assigns a value to life, although courts of law
may ignore this personal assessment and appropriate higher values. Be that as
it may, a monetary value is assigned.
Let the total cost of a single fracture be S and probability of fracture P, then
the expected cost of fracture is PS. Obviously, if P were equal to e.g. 10- 5, it
would not be wise to use a fracture control measure costing lOS. This would be
insurance against a loss of 10- 5 S at a premium of lOS.
The potential costs of fracture include;
(a) Loss of human lives.
(b) Impact on environment, including natural habitat.
(c) Litigation expenses.
(d) Replacement of structure.
(e) Damage to buildings and surrounding structures.
(f) Down time (loss of production).
(g) Goodwill loss of sales and contracts.
388

The total potential cost of fracture is the sum, S, of the above. The anticipated
cost is P*S.
The costs of fracture control include:
(a) Damage tolerance analysis (20000-50000 man hours for an airplane).
(b) Coupon tests and verification tests.
(c) Inspections (or stripping or proof tests).
(d) Repairs or periodic replacements.
Some of these are incurred by the manufacturer, some by the operator, but
the manfuacturer's cost (including those of fracture) are obviously calculated in
the price, so that eventually all costs are incurred by the operator.
The costs of fracture control as listed above, can be easily assessed, but
determining those of fracture is more difficult. Some items can be estimated,
others can be 'guessed' only. Besides, the anticipated cost of fracture depends
upon the probability of fracture, which is the most difficult to estimate. Never-
theless, the principle applies, whether the numbers are 'hard' or 'soft'.
To facilitate the discussion, consider a qualitative Fracture Control Index
(FCI), a higher FCI signifying more extensive fracture control measures. The
probability of failure decreases within increasing FCI (Figure 11.l5a). The

...o
a:
!Zo
u
w
a:
'"
~
u.
o
Iii
8

FRACTURE CONTROL INDEX _ _ (b) FRACTURE CONTROL INDEX _

{8J

FRACTURE CONTROL INDEX_ (d) FRACTURE CONTROL INDEX_

Figure 11.15. Cost of fracture and fracture control. (a) Probability of fracture; (b) Cost offracture
control; (c, d) Total cost.
 389
decrease is faster for high strength materials, because low strength (high
toughness) materials have a lower probability of failure in general. The cost of
fracture control increases with FCI, both for design and operation (Figure
IU5b; the curves show trends only).
The probability offailure can be translated into an expenditure, and the costs
plotted versus the FCI (Figure Il.15c). The minimum of the total-cost curve
indicates the most economic fracture control. For high-cost structures and
high-strength materials, the minimum shifts to the right so that more extensive
fracture control is warranted (Figure Il.15d). If only parts of the structure are
fracture critical, the cost of fracture control would pertain mostly to those parts,
and fracture-control costs could be much lower. (If fracture of a given
component would cause loss of structure a higher FCI is warranted only for that
specific part). Should the cost lines be different than assumed, a minimum may
not be achievable. The probability of fracture of various components of a
system may be different; then the probability of fracture can be made the lowest
for those components for which fracture control is the easiest. This permits
acceptance of a somewhat higher probability for components for which fracture
control is more difficult, while the total probability could remain the same.
The above is but a qualitative assessment of the problem. Nevertheless, it
touches upon the relevant issues. When the logical process of decision-making
leads to a fracture control plan involving analysis, information on loads must
be available. The cost of obtaining load data must be expended. Any analysis
without detailed information on loads, load history and stresses is wasteful. The
decision maker should be aware of the obtainable accuracy in analysis (Chapter
12) and of the statistical aspects of fracture control as discussed. The decision
maker, if aware of the above considerations and of the sources of inaccuracy,
will not embark on finite element analysis to obtain geometry factors when loads
and load history are not known accurately. Cheaper, approximate analysis will
suffice in such cases; uncertainties should be covered by safety factors. Where
fracture control calls for inspection, the decision maker should appreciate that
even detectable cracks may be missed. Inspection intervals should be determined
rationally as discussed in Section 11.5, otherwise all analysis, regardless of
accuracy is futile. If the cost of fracture control (including analysis) far exceeds
the cost of fracture, a simple fracture control plan should be selected. Analysis
then may serve as a guideline; it may bound the problem. But in such cases
rough assessments should suffice.

11.10. Exercises

1. Determine the inspection interval on the basis of the criterion I = H/6 for
structures with crack growth curves as in Figure 11.7, assuming the 'detectable
crack size' is 5 mm, and the permissible crack size 33 mm.
390
2. Using the results of Exercise I, determine the cumulative probability of
detection for the two cases, assuming that the middle probability of detection
curve in Figure 11.10 is applicable.
3. Repeat Exercise 2 for the case that IY. = 0.5, a o = 5 mm, )0 = 8 mm.
4. Select three inspection intervals, 500, 1000, and 1500 hours. Determine the
cumulative probability of detection for each of the cases of Exercise 2 using the
upper curve of Figure 11.10; then estimate the required inspection interval for
a cumulative probability of detection of95%. Compare the results with those
obtained in Exercise 2.
5. Assuming the crack propagation curves of Figure 11.7, determine a proof test
. interval. Assume that Kc = 50 MPa Jill, f3 = I, and select proof test
conditions that would eliminate cracks larger than 15 mm. What is the
required proof stress?
6. A large component made of a material with K[c = 30 ksi Jill and
F,y = 200 ksi is subjected to service stresses of 100 ksi. Crack growth from
a = 0.01 inch to ap has been calculated to cover two years of operation given
that (Jp = 150 ksi. The crack occurs at a fillet. Determine a stripping depth;
assume f3 = 1.

7. Fracture of a certain structure is assessed at Sh' Replacement of the critical part

would cost S. No other costs are anticipated. Analysis costs are Sa per hour.
Load data are available. Extensive analysis including finite element evaluation
would require h hours. A total T of these structures are anticipated to be in
operation. The probability of a fracture is estimated to be P. Which course of
action would you recommend?

References
[I) D. Broek, Fracture control by periodic inspection with fixed cumulative probabilty of crack
detections, Structural failure, product liability and technical insurance, Rossmanith Ed. pp.
238-358, Interscience Enterprises, Ltd (1987).
(2) W.H. Lewis et aI., Reliability of non-destructive inspections, SA-ALCjMME 76-6-38-1 (1978).
(3) E. Knorr, Reliability of the detection of flaws and of the determination of flaw size, AGARDo-
graph 176, pp. 396-412 (1974).
[4] U. Gorenson, Paper presented at ICAF meeting, Toulouse (1983).
(5) D. Broek, IPOCRE, Software FractuREsearch Inc. (1985).
 CHAPTER 12

Damage tolerance substantiation

12.1. Scope

Previous chapters dealt with analysis procedures, (Chapters 2-5), the

ingredients needed for the analysis (Chapters 7-10) and with the use of the
results for fracture control (Chapter 11). This Chapter concentrates on the
general scope of the analysis, its relationship to tests (verification and substan-
tiation), the assumptions and sources of error, and the design options for
improvement of damage tolerance. In short, it considers the analysis in the
framework of damage tolerance provisions.
Damage tolerance analysis substantiation is governed by damage tolerance
requirements if any are in effect. Indubitably, the damage tolerance require-
ments for commercial and military aircraft are the most widely enforced, and
presently there is considerable experience with their use. Thus, a discussion in
some detail of the aircraft requirements is certainly worthwhile, even for readers
not concerned with aircraft, as it will bring out the good and bad aspects of
requirements in general. Although incidental rules may be in effect here and
there, the only other requirements addressing damage tolerance directly are
embedded in the ASME boiler and pressure vessel code. These will be reviewed
as well. Requirements for the use of arrester strakes in ships are discmised in one
of the examples in Chapter 14.
Compliance with requirements is an issue in this chapter, but the discussions
concentrate on the damage tolerance substantiation in general. In particular the
effects of the assumptions on the accuracy of the analysis - and thus on fracture
control and safety - will be discussed. Damage tolerance requirements may
enforce certain assumptions, and so be of greater effect on accuracy than the
analysis itself (Section 12.8).

12.2. Objectives

The objectives of damage tolerance provisions have been discussed at various

391
392

Residual
Strength
tI
- - - - - - - Design Strength a __________ __

,
---1"----

tI
I

(a) -
c rack size (b) - time

Figure 12.1. The engineering problem. (a) Residual strength curve; (b) Crack growth curve.

places in this book. They are briefly summarized here, in the context of damage
tolerance requirements.
A new structure can sustain the design load, which is higher than the
maximum expected service load, because of safety factors on loads or allowable
stresses. The probability of occurrence of the design load is small, but finite for
many structures, so that the probability of failure is not zero. When cracks are
present, the strength is less than the design strength so that fracture may occur
during extreme( or even normal operation. A fracture control plan is established
to prevent such fractures.

Fracture control is not always completely effective. The adoption of fracture

control measures can be rationalized only if they achieve the goals of the
fracture control philosophy. The latter forms the basis of rules and regulations,
which may also prescribe the specific fracture control measures to be used. For
example, inspection is essentially the only fracture control option for
commercial airplanes. The fracture control philosophy will also lead to a
decision with respect to the lowest strength that will ever be permitted. As the
strength decreases during crack growth, the safety factor against fracture is
reduced. A damage tolerance requirement sets a limit to the remaining safety
factor, by specifying the minimum permissible residual strength, (Jp (Figure
12.la). After the residual strength curve has been calculated, the specified
minimum residual strength, will enable determination of permissible crack size,
ap • Larger cracks will cause the strength to be less than (Jp. By implication cracks
larger than ap are not permitted.
Next, analysis must provide the time of crack growth by fatigue or stress
corrosion to a size ap (Figure 12.1 b). When this curve and the largest permissible
crack size are known, the time, H, for a crack to develop to size ap is obtained;
 393
it is the time available for fracture control. As the crack may not grow beyond
ap ' the structure or component must either be replaced or the crack must be
discovered (by inspection or proof testing) and repaired before H expires. In
either case, fracture control is based upon H (Chapter 11).
The damage tolerance analysis must provide:
(a) The residual strength as a function of crack size.
(b) The permissible crack size.
(c) The crack growth time H.
(d) The size of a pre-existing flaw that can be permitted in a new structure (in
some cases).
(e) The interval for inspection, proof testing, or stripping, or the replacement
time.
Many low-stress fractures occurred during the early years of the industrial
era. Better materials and detail design reduced their number to acceptable
proportions. With the introduction of all-welded structures, the number
increased again and once more was controlled by better materials (higher
transition temperature) and detail design.
The modern era brought about a new generation of fracture-prone structures,
often operating in hostile environments and at extreme temperatures where
material behavior is less predictable. Among these are offshore platforms,
certain chemical plants, nuclear plants, aircraft, etc. These can be realized only
if weight and costs are controllable. This drives the design to high quality
materials and high operating stress. Refined stress analysis has improved
confidence so that the high quality materials are operating closer to their limits
than the materials of the past, which increases the risk of crack formation.
The oldest remedy, material improvement, may still have potential. But
material improvements often are immediately exploited by increasing stresses to
reduce weight and costs. Another conventional remedy, improved detail design,
can be exploited by means of modern stress analysis, again to increase allowable
stresses, rather than to reduce fracture risks. This creates a vicious circle and the
danger of mal performance remains.
It is a compelling necessity to exploit new developments for further progress
and higher performance. Thus, low-stress fractures must be prevented by
fracture control. But fracture control technology (e.g. inspection techniques) is
also subject to further development. Improvements making it more efficient will
be exploited by designing closer to the limits. Fracture control is also part of the
vicious circle, and hence must be based on a more or less time-independent
philosophy. Damage tolerance requirements should reflect this philosophy.

12.3. Analysis and damage tolerance substantiation

Damage tolerance analysis provides the information needed to exercise fracture

394

control; it is only one link in the chain of fracture prevention. The manner in
which the analysis results are used to implement fracture control was discussed
extensively in Chapter 11. The damage tolerance substantiation consists of the
proof that the damage tolerance requirements can be met (these may be self-
imposed). Although this proof is provided by analysis, uncertainties and engin-
eering judgements often require tests to verify that the analysis is adequate (a
full-scale test will often be part of the damage tolerance substantiation of
airplanes).
Figure 12.2 shows the elements of the damage tolerance substantiation
program. The small center box pertains to the fracture mechanics analysis.
Material data handbooks may be useful, but some tests are often necessary to

Update

DISUbst~ Verification
tlotlon
I---

~I cr,itica(
POints
l
- ..... -I
I
I
I

I
g.om.try I
- ..... _1
damage

--
I

+ i component
and

....-- I d.t.ctabl.
do mao.
full
scale
load
meosu-
rem.nts I
""'
spe-ctrum
str.ss his!.
tests

Calibration

l7/'I .,
L--
t
anal b.ta
I chock
mod.l
tests
on
Ir.nv,roni r l coupons

18~
T

B
ment
t
tests c.K r K" K,1- t.sts
t
Iranalyzol
-t
r analyze I
II a·NIFl I r a, l
I inspection
thr.shold
J

I !nspection
Interval
I

t
rservice
experience r
Figure 12.2. Damage tolerance substantiation.
 395
substantiate and adjust the analysis. These may include (semi-random loading)
crack growth tests, arrest tests, and tests for calibration of retardation models,
effects of clipping and truncation. Such tests establish confidence that the
analysis procedure is adequate for the substantiation of all other critical
locations, loading cases and crack cases. Component tests (and/or full-scale
tests) are used to substantiate the analysis, at least for some of the most critical
locations. They provide guidelines for the engineering judgements involved.
This phase is not a check of the models per-se, but of the analysis capability in
general, including the basic stress analysis.
The ultimate check of damage tolerance is in the service experience. Feed
backs on crack detection and (perhaps most important) measured load/stress
histories are extremely useful for analysis updates and refinement of inspections,
inspection intervals, and replacement schedules. In view of the many
.assumptions and judgements, expenditures can be saved (and safety improved)
by monitoring service loads, and by updating analysis and fracture control plans
during operation.

12.4. Options to improve damage tolerance

The time available for fracture control is H, which is governed by the residual
strength (ap ) and the crack growth curve. In essence, fracture safety is not
affected by the length of H (Chapter 11). If H is short, frequent inspections must
be scheduled, or the component replaced soon. As long as all fracture control
decisions are indeed based on H, safety will be maintained. But long inspection
intervals or replacement times are desirable from an economic point of view.
The question then is which measures can be taken to improve the situation
when H is too small to be economically acceptable. The following avenues are
open (Figure 12.3).

(a) Use of a material with better properties (Figure 12.3b).

A higher toughness will provide a somewhat larger ap ' but generally speaking,
is not of great influence on H; most of the life is in the early phase of crack
growth. Increasing toughness (ap ) only affects the steeper part of the curve which
has only a small effect in general. Should ap be non-detectable then the effect of
increased toughness is more significant. An average reduction in rates by a
factor of 2, immediately increases H by a factor of 2. Protective coatings may
also help, but it should be noted that surface layers only protect the free surface
while the crack surface is still unprotected.

(b) Selection of a better inspection procedure (Figure 12.3c)

Improving the inspection technique, i.e. by selecting a more sophisticated
inspection procedure, reduces detectable crack sizes. This usually has a very
significant effect on H, because of the small slope of the initial part of the crack
 396
CRACK
SIZE,_

HRS
(a)

, , ,
-~/.
a NEW H a a a H
I NEW H ,NEW H I I NEW

;;; ---p
ap /

----~)J
~, ~, ,
I
I ap
, I

~---P'
ap

ad - ---~
I
I
I
I
ad - -- ad
- :: :~I
ad
---:/1
... : I t I

HRS HRS HRS HAS

(b) (c) (d) (e)

Figure 12.3. Options to increase inspection interval. (a) Crack growth time from ad to ap; (b) Better
material; (c) Other inspection method; (d) Redesign or lower stress; (e) Redundancy or arresters.

growth curve. Note that inspection intervals (Chapter 11) on the basis of
probability of detection are still governed by H. The penalty will be a more
difficult inspection, but the inspection interval is longer; fewer inspections
are needed.
If the structure is not inspected and the component replaced after H hours
better quality control can be used to reduce ao and so increase the replacement
life H. In the case of proof testing, higher proof loads or lower temperatures
would reduce a proof and hence increase the proof test interval (H).

(c) Redesign or lower stress (Figure 12.3d)

The crack growth curve is governed by the stress intensity. Reducing the stress
by e.g. 15% will reduce K by 15%. As crack growth rates are roughly propor-
tional to K4 a 15% reduction in stress will increase H by a factor 1.154 = 1.75.
Such a stress reduction seldom requires a general 'beef up' of the structure;
cracks occur where the local stresses are high, and the stress reductions are
needed only locally: reduction of stress concentrations, larger fillet radii, less
eccentricity will hardly add material, cost, or weight.
Redesign may also affect K; a reduction in f3 is just as effective as a reduction
 397

in (T. In the above example of the reduced stress concentration, the effect is
actually in {3(k,) instead of in (T (the nominal stress does not change). Last but
not least, the redesign may be in the production procedure so that cracks occur
in cross-grain direction instead of along exposed grain boundaries (Figure 7.8).

(d) Providing redundance and arresters (Figure 12.3e)

Building the structure out of more than one element provides multiple load
paths (Chapter 9). In a well-designed multiple load path structure, only
inspections for a failed member might be necessary, provided the fasteners can
transfer the load of the failed member by shear. (Chapter 9). Similarly, stringers
and arresters (Chapter 9) can improve the design and increase H.

All of the above options can be exercised during design. It is crucial, therefore
that damage tolerance assessments commence in the early design phase when
modifications are still possible. Once the design is finalized, the options for
improvement are drastically reduced. Essentially for finalized designs and
existing structures, only option b remains (Figure 12.3c), although doublers or
arresters sometimes can be added later .

12.5. Aircraft damage tolerance requirements

Damage tolerance is the ability of the structure to sustain damage in the form
of cracks, without catastrophic consequences, until such time that the damaged
component can be repaired (Commercial Aircraft Requirement), or (Military
Aircraft Requirements) until the economic service life is expired and the airplane
or component retired. Damage tolerance can be achieved more easily by incor-
porating fail-safety features, such as redundancy, multiple load paths and crack
arresters. Fail-safe structures can sustain larger damage, but if unattended this
damage will eventually still cause a catastrophic failure. Hence, fail-safety
features by themselves do not prevent fracture: the partial failure (e.g. the failed
load path) still must be detected and repaired; even if the structure is fail-safe,
inspection is essential to achieve safety. Without fail-safety features the
structure can still be damage tolerant, provided cracks are detected and repaired
before they impair the safety. Fail-safety features merely alleviate the inspection
problem. The Military Requirements also use the damage tolerance analysis for
a durability requirement (Chapter II). Cracks growing from a presumed initial
size must be sustainable throughout the economic service life.
The requirement that damage can be safely sustained should be interpreted to
mean that the probability of failure must remain acceptably low. If cracks are
left unattended, the probability of fracture will eventually become equal to I
(fracture will occur). Thus, a criterion for lowest strength permitted must be
based upon the acceptable probability of fracture.
398

(a) Requirements for commercial airplanes [1]

The U.S. Federal Aviation Requirements [FAR. 25b], enforced in a similar way
in other countries, stipulate that the residual strength shall not fall below limit
load PL, so that ~ = PL. The so called limit load is, generally speaking, the load
anticipated to occur once in the aircraft life. Given ~, the residual strength
diagram provides the maximum permissible crack size, ap- It should be noted
again that ap is not a critical crack, but is the maximum permissible size under
the regulations. It would be critical only if the load ~ would occur; the probabil-
ity of ~ coinciding with the occurrence of ap is extremely small, so that an
acceptably low probability of fracture is indeed achieved.
In essence the above is the complete requirement; indeed, little more is
necessary. To satisfy the requirement, the manufacturer is obliged to design in
such a manner that cracks can be detected before they reach ap and to prescribe
to the operator where and how often to inspect. Similarly, the operator is
obliged to follow the manufacturer's inspection instructions. Fracture control
by FAR-rules must be exercised by inspection. The excuse that some cracks are
non-inspectable is not maintainable. Every crack will become detectable if large
enough. Thus, the requirement forces tolerance of damage large enough for
detection (Figures 10.6 and 12.9), which promotes fail safe design with multiple
load path and crack arrest features.
In a competitive field, it is in the manufacturer's best interest to ensure easy
inspection; designs with high residual strength and large ap (Figure 9.9) will
ensure long inspection intervals. If too heavy a burden is put on the operator,
the latter will prefer a competitive airplane requiring less and easier inspections.
Hence, the requirements do not have to prescribe the inspection intervals and
detectable crack sizes. If the design requires an unacceptably small inspection
interval, options for improvement as discussed in the previous section can be
exercised. These drive the design to fail-safety, and ease of fracture control for
operators. The requirement, as simple as it is, accomplishes its objectives, a safe
highly damage tolerant structure, at the lowest cost.
Although there is a problem in the definition of detectable cracks, an
(arbitrary) specification of detectable size would not improve the requirement,
because detectability depends upon the type of structure, its location and
accessibility. The best way to determine inspection intervals is based on the
cumulative probability of detecton discussed in Chapter 11. A useful improve-
ment of the requirements would specify the desirable cumulative probability of
inspection.

(b) Military aircraft requirements [2]

The U.S. Air Force requirements (adopted by some other forces as well)
distinguish three types of structures, namely Slow Crack Growth (SCG),
Multiple Load Path (MLP) and the Crack Arrest Fail Safe (CAFS) structure
 399
(Chapter 9). When a crack in SCG structure would cause fracture instability the
airplane would be lost. In both MLP and CAFS structure large damage is
permissible (Chapter 9). Therefore, the requirements for MLP and CAFS
structures are less stringent than for SCG structures. The term SCG is a
misnomer; slow crack growth is always desirable. As opposed to MLP and
CAFS, SCG is rather a Non-Fail-Safe structure (NFS). The commercial re-
quirements do not need to make these distinctions because they automatically
promote fail-safety features (see above). Since the primary Military Require-
ments are less stringent for MLP and CAFS than for NFS structures, they also
should promote fail-safe structures.
The minimum permissible residual strength is somewhat higher than in the
commercial requirements. This is certainly necessary for fighter airplanes and
trainers which experience the 'limit' load more often than once in their life, but
since the requirements cover all military airplanes, one would have expected a
different residual strength requirement for transport airplanes.
Up to this point, the military requirements differ only slightly from the
commercial requirements. As argued, no further rules are needed for
commercial aircraft, because it is in the manufacturer's best interest to build an
easily inspectable airplane. This does not quite hold for U.S.A.F. procurements:
in that case, the regulator is also the operator. Accordingly, the inspection
interval is specified as 1/4 life, for which there is a compelling reason.
Although conceptual designs are competitive, the final design and production
usually reside at a sole contractor, under which monopoly the operator must be
protected. Yet, a fixed inspection interval, regardless of location and accessibil-
ity, may promote convenience more than safety. For NFS structure the
inspection interval (/ = 1/4 life), requires a factor of two, so that H must be
equal to 1/2 life. For MLP and CAFS on the other hand, no factor is required
so that the H need be only 1/4 life.
The requirements also consider durability. A small initial crack must be
assumed present in the new structure; it must then be shown that this initial
crack will not grow to ap within the economic service life for CAFS and MLP
structure or within twice the life for NFS structure. Formerly, these intial crack
sizes were prescribed to be 0.02 inch for MLP and CAFS and 0.05 inch for NFS;
they have since become negotiable. Note that these flaw sizes are not based upon
quality control (the latter is no more difficult for NFS than for CAFS, but the
initial flaws are different). These primary requirements are contrasted with the
commercial requirements in Table 12.1 and Figure 12.4. Apart from these, there
are several secondary requirements.
A fracture instability occurring in CAFS and MLP structure results in a large
(arrested) crack. While this large damage is more easily detectable, it is not
necessarily obvious. A large crack in a stringer-stiffened wing will cause fuel
leakage and might be detected during a cursory visual inspection. However, a
400
Table 12.1. Comparison of Federal and Military Requirements for aircraft.
Damage tolerance Federal Military
MLP and CAFS NFS
Minimum residual PLL [I + :xlPu [I +:xlPLL
strength ~
Detectable crack ad ad or fixed ad or fixed
size
Growth period H H = ±Iifc'" H ~ life'"

= !!.h, =:t life'

H I

'"
Inspection interval H - life'"
2 2 4

Post arrest 2 bays orh , 2 bays or

insta bility damage failed member failed member

Post instability
growth: if
Ground evident I flight
Walk around visual 50 flights
Special visu 2 years
Depot level ~ life I

Safe Life
Non-attended
hypothetical crack
~ [I + :x + {I)PIL [I + :x + ~]Pu
Initial crack size 0.02 inch 0.05 inch
Growth period I life" 2 lives'"

", Life is economical service life (or design goal)

h, By implication

similar crack in a pressure bulkhead may still not be discovered until the next
major overhaul, while further growth by fatigue of the large crack will be fast.
To cover this there are secondary requirements for post-arrest, the severity of
which depends upon the detectability of the post-arrest damage. It can be
so-called ground evident, walk around visual, or detectable only at the next
major overhaul. These secondary requirements are also shown in Table 12.1. In
the commercial requirements this post-instability is not covered explicitly.
Two other secondary requirements are of interest. These concern so-called
'continuing damage' and 'dependent damage'. Continuing damage is a conser-
vative 'invention' to facilitate crack growth calculations. Its effect is illustrated
in Figure 12.5. When a crack grows into a hole it is effectively terminated; a
certain time is required for reinitiation of a crack at the other side of the hole
(Chapter 9). Crack growth analysis is powerless in calculating the reinitiation
time. Continuing damage conveniently provides for immediate reinitiation
through a mandatory assumption of a pre-existing crack of 0.005 inch at
every hole in the structure. Crack growth thus calculated will follow curve C in
 401

RESIDUAL
STRENGTH RESIDUAL
STRENG
ILL

a
I LIFE
/
a /
MLP
CAFS
OR
NOT
ATTEPlDED
ALL
TYPES HRS HRS

a
2 LIVES
, /

INSPECTED
.!f ORLUS NFS

HRS
INSPECTED NOT
1 LIFE ATTENDED

HRS HRS

(a) (b)
Figure 12.4. Comparison of requirements for aircraft damage tolerance. (a) Federal; (b) Military.

a
w
'"a:
~

0~
:;) wJ:
!;;:
Z
Bo
J:Z

HRS

Figure 12.5. Crack running into hole.

402

Figure 12.5. Further artificiality is introduced because the 0.005 inch continuing
damage cracks in some cases must be assumed to grow simultaneously with the
main crack, in other cases they are assumed dormant until reached by the main
crack. It is questionable whether natural cracks will comply with these PRE-
SCRIPTIONS (assumptions).
When the crack approaches the hole it accelerates. After reinitiation it is
longer by the diameter of the hole (curve B in Figure 12.5). It has been shown
experimentally for through-the thickness cracks [3] that regardless of distance
and diameter of the hole these two effects approximately cancel the gain due to
reinitiation. A crack growth calculation ignoring the hole (curve A) would
provide a 'good' answer within the accuracy of analysis. Admittedly, examples
can be given where this simple solution is not so obvious. Besides, in a row of
fasteners several holes may crack simultaneouly, so that continuing damage is
indeed present. Some assumption may be necessary, but whether that
assumption should be prescribed quantitatively in an official requirement is
questionable.
Another secondary requirement concerns so-called dependent damage. This
is based on the notion that if a crack initiates in one of multiple parts, joined
together by one fastener, the holes for which are drilled in a 'stack', cracks in
the other parts will follow soon. The cracked part loses stiffness and, therefore,
sheds its load to the other parts which then will crack as well. However, by the
nature of the load shedding through adjacent fasteners, cracking of the other
parts should typically occur at the next fastener. Because of the assumption of
pre-existing cracks - supposedly due to manufacturing - parts joined by one
fastener must be assumed to have the same initial damage if the fastener hole
in all parts is drilled in one operation in a stack.
Non-fail-safe structure (NFS) is penalized because an initial crack of 0.05 inch
has to be assumed instead of 0.02 inch for fail safe structure, and because growth
period H must cover two lives (durability) or two-inspection intervals (inspect-
able), as opposed to one life or one inspection interval for fail-safe structures
(Figure 12.4). Accepting the assumption of initially cracked structure, an initial
0.05 inch crack is more conservative than an 0.02 inch crack. But, whether or
not the longer crack has a much shorter life depends upon its location. For
example, five types of cracks are compared in Figure 12.6. Crack type B, starting
at 0.02 inch, would have a life of 34900 flights; starting at 0.05 inch the life
would be (34000 - 19000) = 15000 flights. Thus the assumption of the larger
initial crack implies an additional safety factor of more than two.
In the case of crack type D on the other hand, the life of an 0.02 inch crack
would be 9000 flights, whereas an 0.05 inch crack would have a life of
(9000 - 2000) = 7000 flights. The difference is much smaller here, so that for
cracks of this type the additional safety factor is much less. For many types of
cracks, it can be readily foreseen how much NSF structure is penalized, and as
Figure 12.6 shows, the extra conservatism is small for some types of cracks. A
numerical example is shown in the solution to Exercises 1-3. Besides, the
 403

_ wwn • WI}; Ia ~
O.l25--j I-- --jO.25
D
I-- f-05-j
E
C
_ 0.125-ln. thickness

1- in. thickness
E C B
1.0

0.9 IC
I
0,8 I
0.7
I
I
0.6 I

0.5
I
/
0.4 I
/

.... '"
--
0.3 "',;
0.2

0.1
-,----
.05
.02
0 2 4 6 8 10 12 14 16 18 20 22 24 26 28

Figure 12.6. Differences in crack growth life for O.02-in and O.05-in initial crack.

question arises whether NSF structures are built so differently that they have
larger initial damage than fail-safe structures.
Although the requirements for non-inspectable structure (durability) may be
defendable, commercial requirements are based on the fact that all cracks are
detectable sooner or later (Figures 10.6 and 12.9). If a hidden crack becomes
long enough it will eventually run into inspectable area. For such large damage
to be sustainable, only very damage tolerant designs can satisfy the require-
ments, so that the design must be made more damage tolerant.
The assumed initial cracks are small. This means that crack growth is largely
influenced by adjoining structural elements. Due to the dependent damage
assumptions, neighboring holes are cracked as well. Finally all cracks are
prescribed to be circular. But natural cracks have no obligation to the require-
ments; natural crack growth will be different from the calculated growth. Thus,
the calculation becomes hypothetical due to too many assumptions.
404

A lug such as in Figure 12.7 would be considered well designed, because it

provides multiple load paths. However, because the three holes in the fork will
be commonly drilled they must be assumed to have the same initial crack. Since
the three prongs are equally stressed, the three cracks will grow at the same rate
and will reach a p at the same time. Hence, the design has (artificially) lost its
fail-safety (all three prongs will break at the same time). A one-prong lug (worse
design) could more easily pass the requirements (one crack only). The latter
would be cheaper to make. Thus the requirements may accomplish the opposite
of what was intended due to the assumptions.
Ifit is demonstrated on paper that a structure can sustain postulated damage,
there is still no proof of damage tolerance. The question remains whether
realistic damage can be safely sustained. The military requirements prescribe a
set of rigid assumptions concerning initial, detectable, continuing and
dependent cracks. As these are unrealistic, damage tolerance analysis provides
'numbers' only.

12.6. Other requirements

Other damage tolerance requirements exist for ships and for nuclear pressure
vessels. Requirements for ships are issued by Shipping Bureaus, such as Lloyds
of London, Veritas (Norway) and ABS (American Bureau of Shipping). Similar
requirements exist for military ships. Essentially these are preventive require-
ments; no analysis is necessary. Ships of a certain size and over must be
equipped with so called arrest strakes, which are located at the gunwale and at
the bilge and sometimes mid decks. They are longitudinal strakes of a higher
quality (higher toughness) material than the normal hull plating. The strakes are
essentially of the same thickness as the plating. For a more detailed discussion
of these see the example in Chapter 14, Section 3.
The damage tolerance requirements for nuclear pressure vessels are contained
in the ASME boiler and pressure vessel code [5], Section XI and its Appendix

-----*"

---- .........

Figure 12.7. Fail-safe (MLP) three-prong lug.

 405

A. Essentially the requirements provide acceptance limits for cracks detected in

service. A great variety of possible crack configurations and locations are
identified. The requirements then provide the crack sizes for each case that may
be left unattended. Should a detected crack exceed the prescribed limits then one
has two options
(a) Unconditional repair (repair weld).
(b) Perform analysis.
Since damage tolerance analysis is not mandatory; one has the option to
conduct analysis if a detected crack exceeds a pre-set limit (the requirements give
no justification for these pre-set limits).
If the option (b) above is selected, the following damage tolerance require-
ments apply:
K < Arrest Toughness/v'IO for upset conditions, and K < Toughness/.y1
for emergency and faulty conditions, where K is the stress intensity at these
conditions. Strangely enough, these requirements are expressed in terms of the
stress intensity and toughness. However, realizing that the fracture condition is
K = Toughness, in the upset condition, at the stress a eu , fracture would occur
if ppapFa; = Toughness. If the actual stress intensity must be smaller by v'IO,
it follows that ap/a cu :::::: v'IO = 3.16, assuming Pp :::::: Peu. This means that a
safety factor of 3.16 must remain with regard to upset conditions. Thus the
requirement can be stated in terms of the minimum permissible residual strength
ap.1t follows from the above that ap = 3.16 aeu , in accordance with the previous
discussions in this book. This case is displayed in Figure 12.8a using the same
nomenclature as before.
At the same time the stress intensity must be less than the toughness divided
by.y1 for emergency conditions (stress ace). Using the same arguments as above
the minimum permissible residual strength, ap ' must provide a safety factor of
.y1 = 1.41 with regard to upset conditions. This is shown in Figure 12.8c.
A different toughness is used in the two cases, namely the arrest toughness
(Chapter 9) and the regular toughness. Since the former is less than the latter,
there are effectively two residual strength curves in playas shown in Figure
12.8a, c. Naturally, this does not change the principle of the analysis. In either
case the permissible crack size ap ' follows from the residual strength diagram as
shown. Obviously, it is impossible to satisfy both requirements exactly at the
the time. If it can be foreseen which of the two generally is the severest, the
requirement can be simplified, because then the other is superfluous.
The rule presents an alternative. Instead of the above one may satisfy the
following requirements:
ap :::;; aeu/IO upset and operating conditions;
ap :::;; aee /2 emergency conditions.
Where aeu is the critical crack (causing fracture) at upset conditions and ace the
406

-~

+I +, +,'
I ,
Ocu -4 - - .. - - - ---,;---
I , , I

(a) (e)

a cu - - - - - - - - - - - - - - - - -

ace

ad-=-_ _ _ _ _ _ _ _ _ __

time time
(b) (d)
Figure 12.8. ASME requirements. Subscript cu = critical at upset conditions; Subscript ce =
critical at emergency conditions; I = time till next inspection (shut down). (a-b) Upset conditions;
(c-d) emergency conditions.

critical crack in emergency conditions. In the ASME code ap is denoted as at.

This provides with acu = 10ap: K = {Jcuacu.Jn x 10ap = {Jpap.J1Ul;, so that
{Jpap/{J,uacu = )TO = 3.16. This requirement will be identical to the one stated
previously (ap/a,u :::::; 3.16) only if {Jp = {Jcu' Since acu is a longer crack than ap,
in general /3ell will be larger than /3 r so that the requirement leads to a safety
factor somewhat smaller than 3.16. The same arguments hold for ai' and ace.
Both sets of requirements apparently attempt to cover the same conditions. and
only one set is necessary.
The rules are specifically for a case where a crack is detected in service.
One may then prove by analysis that this crack is not dangerous during further
operation until the next shut-down. Essentially, this is contradictory to the
general idea of damage tolerance presented in this book. The analysis is
intended to provide the information to ensure timely crack detection (e.g.
inspection interval) and repair. For this reason a crack then must be repaired
when detected. Instead in the ASME requirements the damage tolerance
analysis is used to decide whether a structure with a KNOWN crack can be left
in service without repair. Besides, analysis is not used to determine the
 407
inspection interval. This is an important difference with the approach in aircraft
where cracks must be repaired and where the analysis is used to ensure detection
and repair, not to determine whether it is 'safe' to fly with a known crack. The
above is not meant as a critique but to point out the difference in approach.
Once a crack is detected and analysis is preferred above immediate repair,
crack growth must be analyzed as well. Fatigue crack growth must be calculated
starting at ad, which is the crack actually present and discovered, and continuing
over the period until the next inspection (shut down), and using a load history
as e.g. in Figure 6.2. Over this period the crack may not grow beyond ap as
determined by the criteria discussed above. This condition is shown in Figure
12.8b, d (Note again that ap is denoted as af in the ASME code).
In most structures (bridges, ships, offshore, airplanes, cranes, etc.) inspections
can be scheduled at almost any time and be dictated by the damage tolerance
analysis. These inspection schedules may be bothersome, but in essence this
bother is only a consideration of cost. A nuclear reactor can be inspected only
during shut down periods which are dictated by many other considerations as
well as cost. Thus the inspection interval necessarily is determined more by
criteria other than damage tolerance.
The time to the next inspection thus being predetermined (Figure 12.8) the
normal process is more or less reversed. Crack growth is calculated as it will
occur during the service period until the next inspection. The growth may not
exceed ap as determined above. If the calculated growth does not take the crack
beyond ap ' no repair is required and the crack may remain 'in service' until the
next shut down. Should the calculated growth go beyond ap then the crack may
not be left in the structure and a repair be made immediately.
The requirements do not leave anything to chance; in Appendix A to the
requirements [4], the analysis procedure is fully prescribed and even the
toughness and rate data are prescribed. (Use of other analysis and data is subject
to approval by authorities.) The prescribed analysis procedure is mostly in
agreement with the discussions in this book. The stress distribution at the crack
location must be obtained. If it has large gradients it may be approximated by
procedures as shown in Figure 8.22: a uniform stress and (a number of) bending
moment(s). The stress intensity is to be obtained by superposition as described
here in Chapter 8. Surprisingly however, superposition of the stress intensity due
to pressure inside the crack for flaws at the inner surface (Figure 8.7) is not
required.
The mentioned appendix also prescribes that flaws be assumed elliptical (for
effect see Chapter 9) and it provides the geometry factors for elliptical cracks in
the same way as in Figure 8.3. However, it uses the obsolete procedure of
modifying Q by accounting for a plastic zone corrrection to the crack size
(Chapter 9). Plastic zone corrections were once thought [5] to be a way to
account for e.g. the tangent to the residual strength diagram (Chapters 3 and
408

10). The procedure has long been abandoned as impractical and inadequate. If
a plastic zone correction is applied to K, it should be done so generally, and
therefore also in the determination of K[c in toughness tests. Although the
procedure is obsolete it will provide somewhat conservative results, and as such
is not objectionable in a requirement. However, the requirement ignores the fact
that for a small a the elastic fracture mechanics approach is unconservative and
that a tangent must be used (Chapters 3 and 10) or possibly a collapse approach.
As mentioned, the appendix to the requirements even prescribes the
toughness and rate data to be used. The given data are very conservative, so that
the actual safety factors are likely to be much higher than those discussed earlier
in this section. Operators wanting to use more realisitic data must seek approval
from the authorities.

12.7. Flaw assumptions

In most cases fracture control is to be planned for anticipated (i.e. postulated)

cracks. It is not known in advance where cracks will occur, only where they
might occur. Establishing the potential crack sites requires a diligent review of
critical locations, stress concentrations, eccentricities and so on (Chapter 10).
Also crack shapes or damage configurations must be postulated. As the shape
and configuration are of great influence on crack growth it is this
ASSUMPTION that may overshadow all efforts for preciseness.
If fracture control is to be based upon inspections, the problem may be
somewhat simpler, especially if larger damage is covered. This is shown by the
examples in Figure 12.9. Since only detectable cracks are of interest no
assumptions would be necessary with regard to damage development. The
configurations of detectable damage for visual inspection are rather obvious.

DETECTABLE VISUALL Y
• TOTAL DAMAGE

Figure 12.9. Detectable (from top) cracks.

 409

However, if for these same cases inspections would be by, e.g., X-ray, some
ASSUMPTIONS on flaw shape development would be necessary, since X-
raying can potentially reveal cracks in hidden layers and components. In such
a case several scenarios of damage development may be postulated, to assess
which of these would lead to the shortest life (inspection interval).
If initial flaws are postulated, the problem is of a different nature. Initial
defects in welded structures can be defined. Weld defects such as porosity,
undercut, lack of fusion, lack of penetration can be identified. On the basis of
the weld quality control criteria it can be concluded which size defects might
pass quality control inspection, and be present in the new structure. Such defects
can be treated as initial cracks (Chapter 14).
In other cases initial flaw assumptions are a more delicate problem. Indeed
sometimes they may be based on quality control experience (or criteria) if they
are of sufficient size to represent a crack, as for example in the case of castings.
But it is questionable what would be the size of an initial flaw in a crankshaft
built under stringent quality control.
In the case of the military airplane requirements this problem was addressed
as follows [6]. A total of 2000 holes in a wing that had been subjected to a fatigue
test (known loads), were broken open to reveal any cracks. Of these, 119 holes
were found cracked; the crack sizes were established. Crack growth analysis was
then perfonned for each crack starting at a very small size, using the appropriate
{J and stress history for each location. The calculated crack growth curves were
shifted so that the final crack size matched the one in the test (Figure 12.10). It
followed which size of crack should have been assumed present at the start of

CRACK SIZE AT END OF LIFE

COMPUTED CRACK ~ROWTH CURVES

MATCHING END OF LIFE
WITH PROPER BET,

oL---------------------------------tim-e~-

ElF IS FLAW THAT SUPPOSEDLY EXISTED AT BEGINNING

Figure 12.10. ElF (equivalent initial flaw) determination.

410
the test to produce the final crack size as observed. These initial cracks can be
considered Equivalent Initial Flaws (ElF). The statistical distribution of
these 119 ElF's was obtained and extrapolated to extreme values, which led to
the flaw sizes of 0.02 and 0.05 inch (Section 12.5). There were also 1881
uncracked holes. Clearly, the ElF for these holes was much smaller than for the
other 119. Eliminating these from the statistical distribution has biased the
results.
In subsequent efforts many specimens with holes were tested and analyzed [6]
in the same manner as above, to determine the ElF. Attempts were made to
correlate the ElF with hole quality (roundness, scratches, reaming, burrs, etc.).
Most calculated ElF's were on the order of 0.001 to 0.002 inch; significant
correlations with hole quality failed to emerge. Other attempts to correlate the
ElF with e.g. inclusions and second phase particles [7] were inconclusive as
well. In view of the above, the 0.02 and 0.05 inch initial flaws assumptions are
arbitrary and have no bearing upon initial quality control.
From a practical point of view, there is no objection to assumed flaw sizes and
shapes if they lead to safe structures. The only danger is that the numbers have
tended to become a standard, which is extremely unfortunate and objectionable
as they are ARBITRARY. Specification of initial cracks and detectable sizes in
requirements and regulations are tied to the present state of technology.

12.8. Sources of error and safety factors

Contrary to common beliefs, the short-comings of fracture mechanics methods
are NOT the important error drivers. Rather, these are the input and the
assumptions. In this section the various sources of error will be identified, and
estimates made of their maximum possible effects; actual errors are often
smaller.
Error sources can be classified in six main categories:
(a) Intrinsic shortcomings of fracture mechanics.
(b) Uncertainty and assumptions in data input.
(c) Uncertainty due to flaw assumptions.
(d) Interpretations of and assumptions in stress history.
(e) Inaccuracies in stress intensity.
(f) Intrinsic shortcomings of computer software.
In each of these categories, there are a number of factors contributing to
inaccuracies in the analysis. They are listed in Table 12.2. Many of these already
received attention previously. Therefore Table 12.2. and the following
discussion make references to other chapters for details and illustrations; they
are intended only to provide an overview of the error sources and their effects.
As the calculated life H is decisive for fracture control and safety, Table 12.2.
provides rough estimates of how particular errors may affect the calculated life
H. This estimate is given as a factor on life (not as a percentage error). The
 411

numbers are somewhat subjective but of the right order of magnitude (from
analysis experience).
The error due to the use of LEFM for the residual strength analysis occurs
mainly at small crack sizes (tangent approximation) and for small structures
(collapse) as discussed in Chapter 3 and 10, but accounts for these make this
error quite acceptable. For longer cracks and larger structures the error is very
small (Figure 5.29). Note that this is the error due to the procedure alone, and
does not include the one due to data scatter. The error will hardly be less if
EPFM is used; collapse will still be a problem especially for small cracks and
components. However, errors in IIp have only a small effect on the life H, as was
shown already in Figure 5.29; most of the life is in the early stages of growth.
Retardation models are not ideal, but as shown in Figures 5.19 and 5.22 well
calibrated models provide results in which the error in life is generally only
around 10% (1.1) with few exceptions running as high as 30% (1.3). This is
under conditions where p, da/dN, stress history and calibration factors are
known accurately; i.e. it is the intrinsic error of the models. The fact that
retarded crack growth analysis is generally less accurate is due to other factors
which will be considered separately.
The errors due to data input are larger. Misinterpretation of scatter and
force-fitting by unsuitable equations may introduce a factor of 2 to 3. But even
careful assumptions may well cause a factor of 1.5-2. The situation is worse for
mixed environments where the data for the separate environments must be used
to obtain a weighted average. By itself this may be an acceptable engineering
approach, but estimating the mixture and sequence of environment requires
judgement and assumptions, and is a delicate matter. These issues were discussed
extensively in Chapter 7.
Flaw assumptions are other big drivers of errors. By assuming a 'conser-
vative' circular instead of elliptical flaw, one may 'casually' introduce factors of
2 or 3. As many flaws are not elliptical, the assumption of ellipticity by itself
causes errors (Chapter 9). Assumptions for initial flaw size may introduce
equally large factors (Figure 12.6 and Exercises 1-5). Flaw development
assumptions, continued cracking assumptions when cracks run into holes
(Figure 12.5) and so on, are equally influential.
Every load history is an approximation (Chapter 6). Loads and number of
occurrences must be approximated (number of levels). Decisions have to be
made about clipping and truncation. Since the simple clipping of a few loads can
have dramatic effects on H (Figure 6.23), the decision on clipping should not be
made by load experts but by damage tolerance experts. Improper sequencing is
another error driver. Randomizing the history while in reality it is semi-random
(mild weather-storms), can cause great differences. All these are introduced by
assumptions.
Errors in stress intensity are drivers of intermediate importance. Crack
 Table 12.2. Error sources +>-
N
Category Cause of error Comment Possible
factor on
calculated H
(a) Intrinsic I. LEFM approximation Small error in ap for 1.1-2 Figure 5.29
shortcomings small cracks or small
of fracture parts
mechanics 2. Retardation model Small error if well 1.1-1.3 Figure 5.19
calibrated

(b) Data input 3. Kk or K" JR Small error in ap 1.1-1.2 Figure 5.29

(20%)
4. daJdN-data Normal scatter 1.1-1.5 Figure 7.16
5. Assumption 90% band Erroneous apparent 1-2' Figure 7.17
scatter Figure 7.18
6. Assumption variable environment; E.g. weighted averages 1-2' Table 7.3
7. Equations for daJdN Unnecessary force fits 1-2a Section 7.7

(c) Assumptions 8. Direction (e.g. LT versus SL) Wrong data applied 1-2a Figure 7.8
9. Size Important for initial 1-3" Figure 12.6
flaw only
10. Shape Surface flaws 1-3' Figure 9.3
II. Development E.g. multiple cracks, 1.1-2' Figure 12.5
load transfer etc.,
continuing damage

(d) Interpretation 12. Sequence Semi-random vs random 1-2a Figure 6.19

of stress 13. Truncation Improper truncation 1.1-1.3 Figure 6.25
history 14. Clipping Assumptions 1.5-3' Figure 6.22
Figure 6.23
 Table 12.2. (Continued)

Category Cause of error Comment Possible

factor on
calculated H

(e) Stress intensity 15. Actual load values Measurement, analysis 1-1.75 Chapter 6
15% (1.15)4
16 . Stresses Assumptions, boundary 1-1.5"
conditions, load
transfer 10% (1.1t
17. f3 10% (1.1)4 1-1.5 Chapter 8

(f) Computer 18. Integration scheme Minor if small steps l-l.l Chapter 12
software 19. Double precision Usually minor; Large 1-(2) Chapter 12
possible
20. State of stress for retardation No large error if 1-1.5 Chapter 12
calibrated
Total possible: 2.7-1.37000
a Fully, or partly due to assumptions.

\;.l
"""
414

growth is roughly proportional to the 3rd or 4th power of K. Since K = f3(J J1W,
all errors in life are proportional to the errors in f3 and (J to approximately the
t
4th power. A 10% error in stress causes a factor of (1.1 = 1.46 on life. Errors
in stress stem from errors in loads and stress analysis.
The calculated loads contain an error, a 10% error being quite acceptable.
Subsequently these loads are used for stress analysis. No matter how sophis-
ticated the latter, the error is unlikely to be much less than 10%, especially in
places of importance (stress concentrations, eccentricities, load transfer). In
finite element analysis complex structures are often crudely modeled at such
places, boundary conditions are assumed, fasteners represented by assumed
springs, three-dimensional cases approximated in 2-dimensions, etc.
Admittedly, this situation can be improved, but the cost may be prohibitive for
analysis covering hundreds of potential crack locations, or even for one crack
in a common hammer. FEM has a potential for good accuracy, but in general
applications an accuracy of 10% is all that may be expected.
The stresses may be obtained within 10% for the given load, but also the load
contains an error. Hence, the final stress may have an error larger than 10%,
possibly 15%. This causes a factor between (0.85)-(1.15) or between 0.52 and
1. 75 on life, the expected life being 1.
Also the error in f3 is included in the stress intensity. If this error can be
reduced from e.g. five to three percent, the gain is only from a factor on life of
(1.05)4 = 1.2 to (1.03)4 = 1.13, a small improvement indeed in comparison
with other factors. If the inaccuracies in loads and stresses together account for
a factor of 1.5 or more, while a simple assumption on flaw shape may cause a
factor of 2, it is hardly a worthwhile effort to obtain f3 for this assumed flaw
within three percent through a costly analysis if a simple procedure (Chapter 8)
can yield a 5% accuracy. And if the assumption of a circular surface flaw (for
conservatism or otherwise) introduces a factor of 2 to 3, it is not realistic to
demand a high accuracy for f3.
Finally there are errors due to specific computer modelling. These may be due
to (1) the integration scheme, (2) rounding errors, and (3) equations for retar-
dation and state of stress.
A crack growth calculation per se is but a simple numerical integration which
does not give rise to large errors. Integration scheme errors can be introduced
only in the case of constant amplitude where integration is performed in steps,
because for variable amplitude loading integration is performed cycle-by-cycle
anyway. If integration in constant amplitude is done in large steps, the accuracy
is less. This was demonstrated by the hand-calculation and other examples in
Chapter 5. But it was also shown there that integration is an intrinsically
accurate process, as opposed to differentiation. Numerical integration
procedures such as the Runge-Kutta and Simpson rules were devised in the
pre-computer era when hand-calculations forced large steps. With the introduc-
 415
tion of the computer, the step-size does approach zero (as it should), because the
computer can perform many steps in a short time. Since integration is very
forgiving in the first place (Chapter 5), these small step sizes are adequate and
produce negligible errors, in particular when the result is seen in the context of
the other errors discussed above.
The use of single instead of double precision can sometimes cause significant
errors, especially in variable amplitude loading where da (one cycle) is very
small, but also in constant amplitude loading with very small steps (in that case
smaller steps give a LESS accurate answer than larger steps). This is an intrinsic
problem of numerical computers. Personal computers provide eight significant
figures in single precision and 16 in double precision, while mainframe
computers generally work with 16 significant figures in single and 32 in double
precision. The following examples are for personal computers; they apply
equally to mainframes if one just changes the numbers.
An output given as e.g. 831 259 cycles is erroneous, if the accuracy is a factor
of 2. The number should be 830000, the error being much larger than 1259/
830000 = 0.1 %. Similarly an input for da/dN = CpKmp as 9.4327 E-IO K 3.7234 is
unrealistic. Considering the accuracy of the data an input of 9.45 E-lO K 3.7 is
more than adequate. Giving m = 3.7234 is implying that
3.7233 < m < 3.7235. Clearly m is not known that accurately: at best
3.6 < m < 3.8.
However, double precision has nothing to do with the accuracy of input and
output; it defines the number of significant figures carried in the computations,
not in input and output. If the computer must evaluate 1.79 E-l 0 x 2.73 E-ll,
the result is 4.8867 E-21. Note that these numbers provide 12, 13 and 25
decimals respectively, and the result is evaluated properly. The number of
decimals is not important. In single precision the product 1879.43284 x 3.83
will be evaluated as 1879.4328 x 3.83 because the first number has nine signifi-
cant figures of which only eight are carried. The difference is insignificant for
engineering calculations. Therefore, throughout most of the crack growth
analysis, single precision is MORE than adequate for multiplications, divisions,
power, logs, etc.
However, double precision may become important in addition of large and
small numbers and in subtractions oflarge numbers. This situation occurs when
the small crack growth in one cycle is added to a large crack (a + da). For
example, in a particular cycle da is evaluated as 7.45 E-8 =
0.000000074500000. This occurs PROPERLY in eight significant figures;
leading zeros do not count. If the crack size is 12, the results will be
a + da = 12.000000 + 0.0000000745 = 12.000000. As 12.000000 has eight
significant figures, da will be rounded off and not be counted. It will appear as
if there is no growth. This might occur in a similar way in 10000000 successive
cycles. The total growth would then have been 10000000 x 7.45 E-8 = 0.745,
416
so that a + da = 12.745. However, in each cycle the growth was rounded off
and after the 10 000 000 cycles a is still 12.
Double precision will mend this problem, but only to a degree:
a + da = 12.00000000000000 + 0.0000000745 = 12.00000007450000,
and indeed after a lO-million cycles the size will be 12.745. However if da
appears to be 7.45E-16, this crack growth will still be ignored. The problem
occurs in mainframes also, but it is less important because 32 significant figures
are carried in double precision. Hence, double precision is useful, but there is a
limit to accuracy. Fortunately, the above problem seldom arises, but the use of
double precision is recommendable at one place in the software, namely where
a + da is evaluated.
Other inaccuracies are introduced when the computer model cannot cope
with the simultaneous growth of two axis of a surface flaw. Then the user is
forced to make the circular flaw assumption, the inaccuracy of which was
discussed above and at other places. (Chapter 9 shows how this problem can be
circumvented, at least partially.) Serious errors may occur if the code can
perform random loading only, thus ignoring semi-randomness as discussed in
Chapter 6.
The way retardation is treated may affect the results considerably. All models
use F;y, but F;y is but an arbitrarily defined number. One can argue whether the
model requires the use of the 0.01 % yield strength, the 0.02% yield strength, or
the cyclic yield strength, where the latter can be defined qualitatively only. All
models make use of a plastic zone equation which contains arbitrary numbers
for plane stress and plane strain. Different computer codes use different
numbers and equations. A proper code will check the state of stress in every
cycle. The latter depends upon thickness, Kmax and F;v.
Even if using the most sophisticated retardation models, all computer codes
contain assumptions, with regard to retardation. Consequently, retardation
calibration parameters are not transferable between codes. Calibration must be
performed using the same code as used for subsequent analysis. If the models
are thus calibrated it does not matter what the code's assumptions are. The same
assumptions used in calibration will be used in analysis, and inaccuracies due to
assumptions are compensated for by the same assumptions in the analysis. But
if these calibration factors are used with other computer codes the results will
be different. As long as codes are used in a consistent manner, only small errors
occur. Naturally, if the computer code provides more options, it is more
versatile and can provide somewhat better results.
In the case that all errors discussed are active (which depends upon the
complexity of the problem), the total factor on life (Table 12.2) would be
between 2.7 and 137000, with a logarithmic average of 600. This can hardly
be called an error; it is a total misrepresentation. Naturally, errors generally will
not operate in the same direction, and some will compensate others. However,
 417

it can be seen readily that the reliability of the result is affected much more by
assumptions than by shortcomings of fracture mechanics or computer software.
It is not worthwhile to improve the strong links in a chain; the weak link must
be improved. The geometry factors, fracture mechanics concepts and calibrated
retardation models are not the weak links. Improving these will hardly improve
the result. The weak links are the assumptions involved in rate data, clipping,
flaw size, flaw shape and so on.
There is only one way in which the magnitude of the inaccuracies due to
assumptions can be assessed, namely, by repeating the analysis using different
assumptions. It should be second nature to a damage tolerance analyst to
perform calculations a number of times to evaluate the effects of assumptions
with regard to stresses, loads, stress history, clipping levels and so on. Once the
analysis is set up, such evaluations amount to no more than a number of similar
computer runs.
'Garbage in, garbage out' is a worn phrase, but it needs more repetition. No
answer to engineering problems is more suspect than the one generated by a
computer. Although the computer is perfect, and good computer programs are
nearly perfect, the result is still dependent upon input and assumptions. The
effects of assumptions should be assessed. Only then can the problem be
bounded and an impression of the 'true answer' obtained. A single analysis is
never adequate.
The common practice of making 'conservative' assumptions everywhere is
ASSUMING that all errors work in the same direction. Table 12.2. shows that
the answer could be off by a mere factor of 137000. Realism and sound
judgement are necessary and it is better to use best estimates, than conservative
estimates. Even with the best estimates the answer will be in error, but it will be
closer to the truth. Analysis to assess the sensitivity to assumptions is required.
In the end, one must admit ignorance; the result is dubious, but this holds for
any other engineering analysis. The magnitude of the safety factor should
depend upon the total 'uncertainty', as in conventional design. Regulating
societies and/or authorities would have to establish rules and recommendations.
Where such information is lacking, engineers will have to decide on a case-by-
case basis.

12.9. Misconceptions

A number of misconceptions have crept into the engineering world about

fracture mechanics and damage tolerance analysis. These have all been proven
wrong in this book, but it seems worthwhile to briefly review a number of them.
The most persistent and most damaging misconception is that fracture
mechanics is inaccurate, almost to the point of being useless. Some of the
concepts are indeed less than ideal when considered from a fundamental point
418

of view. However, fracture mechanics is an engineering tool for damage

tolerance analysis. Almost no engineering method is ideal, but if it provides
needed answers, the method is useful.
Fracture mechanics can provide useful anwers with reasonable accuracy when
it is used judiciously in the manner described in this book. Anyone can use
household tools such as hammer, saw and screw driver, but it requires expertise
to produce a piece of furniture with these tools. Similarly, it requires expertise
to obtain useful answers with fracture mechanics. This book shows that the
acclaimed inaccuracy of fracture mechanics is mostly due to unknowns,
assumptions, and inaccuracy in input: the tool (fracture mechanics) is not the
cause. Hammer, saw and screw driver cannot be blamed for poor results if used
on third grade knotted wood. Fracture mechanics is oflittle help if the user does
not have (or refuses to obtain) basic information on loads, stresses and material
data.
Another persistent misconception is that linear elastic fracture mechanics can
be used only if there is plane strain. This is probably brought about by the facts
that (1) only plane-strain toughness tests have been standardized, and (2) short
popular summaries of LEFM emphasize plane strain. As shown in Chapters 3
and 10, the procedure for using LEFM is the same whether there is plane stress,
plane strain or a transitional state of stress. If there is no plane strain the
toughness is usually high, and the higher the toughness the sooner LEFM leads
to errors. But, as shown, in such cases reliable approximations can be made. For
very short cracks the fracture strength will tend to infinity, or at least will be
close to the yield strength. In such cases the tangent from Fool or F;y provides a
good approximation as was clearly demonstrated. Also the possibility that
failure occurs by net section yield or collapse must be considered. But for this
case good engineering solutions are available as well.
Often forgotten is the fact that the same problem exists in plane strain. For
small cracks the fracture strength will still tend to infinity no matter how low the
toughness. In small structural components, failure will still be by collapse. Thus,
the problems of small cracks and collapse must be faced in plane strain as well.
As a matter offact, they even must be faced in elastic-plastic fracture mechanics:
the fracture stress still tends to infinity for a --+ zero, and collapse will still occur
in small components. Whether there is plane stress or plane strain, whether
LEFM or EPFM is used, approximations for small cracks and net section
collapse are necessary. In all cases these conditions MUST be evaluated
together with the fracture strength on the basis of K or J. The condition
first satisfied (at the lowest stress) is the failure strength. If collapse or net section
yield prevails, the fracture strength based on K or J (also in EPFM) is too high.
This leads to the third misconception, namely that the use of LEFM is always
conservative. If for a --+ 0 the calculated fracture strength is infinite; one can
hardly maintain that this answer is conservative. If the calculated fracture
 419

strength is much higher than the stress for collapse, failure occurs by collapse
and not at the calculated LEFM fracture strength (the latter's result is unconser-
vative). Again, the same problem exist in EPFM.
In many practical cases, the cracks of interest are surface flaws or corner
cracks, which indeed are in plane strain and should be treated as such (that
thickness has no relevance here was discussed extensively in Chapters 3 and 7).
Thus the plane strain toughness is needed in such cases. But if the toughness is
high, the ASTM standard of B > 2.5(K/c/F;y)2 may require such a large
specimen that K/c cannot be measured. Ergo, fracture mechanics cannot be
applied; another misconception.
In the first place, the number 2.5 used in the standard is rather arbitrary and
not rigorous as it is often considered (Chapter 7). But apart from that, the value
of K[c can be reasonably well estimated from a specimen that is too thin, as was
discussed in Chapter 7 and in the solution to Exercise 6 of Chapter 3. Insight
and ingenuity go a long way in obtaining engineering solutions. Certainly, in
such cases the ASTM standard is not satisfied, but the standard is there for
convenience, not to make engineering impossible. Of course, one MUST check
whether collapse occurred in the test; but if it did the 'apparent' toughness
following from the test is too low (conservative). Actually a check for collapse
should always be made, whether mentioned in the standard or not.
In EPFM and JR-tests the problem of constraint still exists. The LEFM test
for K[c puts much emphasis on thickness and state of stress, but the EPFM test
for JR is unrealistic (Chapter 4). This raises the impression that constraint is of
less importance in EPFM. But as shown above, whether there is plane stress or
plane strain the 'procedures' still apply provided collapse is recognized as a
failure criterion.
The present standard for the JR-test analysis is based on a collapse condition.
In that case the obtained J is too low; as a matter of fact it is only an 'apparent'
JR' Collapse and fracture are competing conditions, and the one satisfied first
will prevail. The true JR-test should be on larger specimens (no collapse) and be
evaluated with J = HcrD + 1 a/F. Indeed, a new standard for the EPFM test is
badly needed (Chapters 4, 7).
In some cases, where approximations must be used in LEFM, it is better to
use EPFM. This is not a misconception, but certainly a statement reflecting
more academic than pragmatic wisdom. JR-curves are difficult to measure so
that the data are inaccurate. Thus, it is questionable whether the answers EPFM
provides, are any more accurate than those obtained with LEFM using the
appropriate approximations as discussed. If the fracture stress can be calculated
within 10% using a simple procedure, one would not want to use a 'sophis-
ticated' and complicated procedure, to produce results to the same or poorer
accuracy. This may not always be the case, but from an engineering point of
view, fundamental rigor does not count; only results do.
420
As far as crack growth analysis is concerned, a few persistent misconceptions
exist as well. Should crack growth be based on K or J? In fatigue crack growth
most of the life is at small 11K (Chapter 5), i.e. from K = 5-20 ksi JUl, and
most of that part at K = 5-10 ksi JUl. Even in a material with a yield strength
as low as F;y = 50 ksi, the plastic zone size at 11K = 15 ksi JUl (R = 0), is only
0.004 inch; during most of the crack growth it is even smaller. This is a very
small plastic zone indeed. The data clearly show that it is not necessary to
question the use of K. The high growth rate regime affects only a small portion
of the life; it does not change the life H to a significant degree. Considering the
general accuracy, using J in this regime amounts to a third order correction and
is of no practical interest.
Retardation models are inaccurate and hence crack growth analysis is useless;
another misconception. Certainly, the first part of the statement is true. Re-
tardation models are inaccurate, but they can be calibrated empirically, and
then they work satisfactorily. This is based upon empiricism, but also the da/dN
data are empirical, F;y is empirical, and even E is empirical. A calibrated
retardation model provides useful results. Any errors due to other unkowns in
load and stress input, and clipping, overshadow those due to the models.
A final misconception is that test data can be obtained only from standard
specimens. If that were true they could be applied only to standard specimens
and not to structures. If the data can be applied to configurations other than test
specimens, they can be obtained from any non-standard specimen, as long as the
f3 for the specimen is known (Chapter 7). Some specimens have been standard-
ized for which a very accurate f3 is available. This is a matter of convenience
only. it is not a restriction.

12.10. Outlook

Speculations about the outlook for the future must take due account of the
points discussed in the previous two sections. At present it is possible with
judicious use offracture mechanics, combined with small crack approximations
and collapse analysis, to predict the failure strength of a structure in most cases
within about 10%. This is as good as is desirable for engineering analysis.
Buckling strength, or for that matter the strength of uncracked structures,
cannot be predicted with greater accuracy. If a more rigorous fracture theory
emerges in the future, the resulting engineering analysis will not be better than
it is now. Scatter in material data will not be less. It will be equally difficult to
predict the actual loads on a structure. Hence, the predicted fracture case will
still be within about 10% only.
Most likely more refined crack growth and retardation models will be
developed in the future. But these will not improve crack growth analysis much.
The inherent large scatter in da/dN will remain; it is 'in the nature of the animal'
 421

(Chapter 7). Loads and stresses in complex structures will still contain errors,
and since crack growth rates are proportional to some power of stress/load, small
errors in the latter are magnifed by this power: (1.1.)4 = l.46. Thus, the
accuracy of crack growth analysis as presently obtained in engineering appli-
cation will not be much improved. It will remain equally difficult to make
projections in the future of stress histories and changing environments.
Predictions will remain predictions.
From a fundamental point of view it is desirable that research continue and
more is learned, better hypotheses and procedures developed. From a practical
point of view, the present situation cannot be much improved upon. The
exponential growth of the number of researchers in the field has come at an
inopportune time; their efforts could be better spent in other, new areas. After
a period of slow and consistent growth, subsequent exponential growth in any
bull market signifies that collapse is near. Many agencies have discovered this,
and research funds are decreasing. In 1987 more papers were published on
fracture mechanics than in the entire decade of 1960-1970. However, the results
were less worthwhile. Research should certainly be continued, but should
concentrate on problems of real engineering interest (dynamic fracture,
composites, etc.).
Fracture mechanics has become an established tool. It is not perfect, but it
provides engineering answers previously unobtainable. Engineers feel uncom-
fortable because of the possible errors (again, largely due to input), but this
can be cured by experience. Compare the situation as it was 100 years ago. A
bend member was designed by calculating the bending stress with (1 = Mh/I
(elastic) and by sizing the member with a safety factor. Obviously, the elastic
analysis was in error, and there was a great deal of uncertainty with respect to
the (inaccurate) results. But structures were designed on this basis, because there
was no alternative. With the years came experience. Certainly, there were
misphaps, but in the end the procedures were made to work, and eventually the
necessary safety factors were established. Today nobody questions this
procedure; 100 years of engineering experience has shown that it works. Even
today, virtually all structues are designed on this very basis: an 'inadequate'
elastic analysis and a safety factor. Plastic analysis is certainly possible but it is
also more complicated; it is more fundamental, but does not lead to better
results in the general design of load-bearing structures. Compare LEFM and
EPFM in this light.
Fracture mechanics should be considered against this background. Once a
century of experience is obtained, it will be as common as present day design
analysis. Mishaps will occur, but become fewer when experience accumulates.
Experience will be obtained only through application. Naturally, safety factors
are necessary. As before, experience must show how large these should be.
422
Fracture mechanics is useful now. Waiting for further technical improvement
will be in vain. Better methods will evolve, but the engineering results will
improve only marginally. (Elastic-plastic analysis has not displaced regular
elastic design analysis).
Inexperience and unjustified fears are no excuse. If they had been 100 years ago
with elastic design analysis, there would have been little progress in enginneer-
ing. But a century ago, engineers were willing 'to stick their neck out'. This is
true, now as much as a century ago. Fracture mechanics is a new tool to prevent
fractures. As such it can only lead to improvements. Damage tolerance analysis
may not prevent all fractures, but it can prevent many, which is sufficient
justification for its use.
Fracture prevention is not glamorous. It can never be proven that a fracture
was prevented. If the fracture does not happen, some may think that the effort
was 'money down the drain', but in today's litigious society a fracture may cost
much more than fracture prevention. If fracture mechanics is not used and a
fracture occurs, lawyers will be quick in pointing out that 'the best available'
techniques were not used. Even though fracture mechanics is not perfect, it is
the best available. The time for its application is now.

12.11. Exercises

1. A damage tolerance analysis is performed by assuming an initial edge crack

of 0.1 inch. Kc = 60 ksiJill and da/dN = 2 E-9 flK3.2 , F;y = 70 ksi, B = 0.2
inch. Assume constant amplitude loading to 15 ksi at R = O. The minimum
permissible residual strength must provide a safety factor of 2. The item will
be replaced afterH; inspections cannot be performed. Ifinitial quality control
can assure discovery of an 0.04 inch flaw, what is the extra factor on life
obtained. Assume f3 = 1.12 throughout (W = 100 inch).
2. Repeat Exercise 1 for a through-crack at a hole of one inch diameter in a wide
plate.
3. Compare the two factors on life obtained in Exercises I and 2. How would
you change the damage tolerance specification (minimum permissible
residual strength) or the assumptions to provide more consistent safety
regardless of the type of crack?
4. The rate data in Exercise 1 are the average of a scatter band that covers a
factor of 1.5 on da/dN. The toughness is an average number from tests with
a data scatter of 13%. The stresses as given are expected to have an accuracy
of five percent. Estimate the upper and lower bound of the life in Exercises
1 and 2 (estimate error in f3 yourself). Where do the largest inaccuracies come
from?
 423

5. Using the information obtained in Exercises 1 through 4 do you have reason

to change the damage tolerance specification or the assumptions? Given there
are 10000 cycles per year what would be the replacement times? Would you
consider inspection?

References
[I] Airworthiness requirements FAR 25b. U.S. Federal Aviation Administration.
[2] Damage tolerance requirements for military aircraft, MIL-A-83444.
[3] G.L. van Oosten and D. Broek, Fatigue cracks approaching circular holes, Delft Un. rept (1973).
[4] Anon. ASME boiler and pressure vessel code; Section XI; In service inspection of nuclear power
plant components,plus Appendix A, Analysis offlaw indications, ANSI/ASME, American Society
of Mechanical Engineers, New York, Issued annually.
[5] D. Broek, Elementary engineering fracture mechanics, 4th ed. Nijhoff (1985).
[6] J.P. Gallagher et aI., USAF damage tolerant design handbook, AFWAL-TR 82-3073.
[7] R.C. Rice and D. Broek, Evaluation of equivalent initial flaws for damage tolerance analysis,
Naval Air Dev. Center NADC-77250-30 (1978).
 CHAPTER 13

After the fact: fracture mechanics and failure analysis

13.1 Scope

Despite careful fracture control, service failures will continue to occur; but
without it more fractures would be experienced. Engineering journals from the
turn of the century discuss the large numbers of failures then occurring.
Presently, with many more structures in service, the number of structural
failures is relatively low (although these few get much more publicity). This is
due to better design, but not in the least due to fracture control and quality
control. Fracture mechanics and damage tolerance analysis can further improve
the situation, but cannot eliminate all failures.
When a service fracture occurs, a failure analysis is usually performed. Every
fracture, in principle, contains all the evidence about its cause, although this
information is iiometimes hard to extract. The broader the scope of the failure
analysis, the greater the likelihood that the scenario and cause can be recon-
structed. In this respect fracture mechanics can provide a great deal of infor-
mation as discussed in this chapter.
Fractography is an indispensable part of the failure analysis as it is usually the
only means by which the failure mechanism can be established, but the
knowledge that failure was caused by fatigue or stress corrosion does not solve
the problem. The purpose of the failure analysis is to arrive at 'solutions' that
will prevent subsequent failures. In the author's experience the majority of
failures is due to design and production deficiencies; few are due to material
defects. The remedies often lie in design and structural changes. Quantitative
fractographic analysis and fracture mechanics can be of help. This chapter does
not review fracto graphic features as excellent texts on the subject are available
[e.g. 1,2,3,4]. Instead, its purpose is to review the structural and design aspects
of failure analysis, and to show the role fracture mechanics can play. In this
respect some quantitive measurements of the microscopic features are reviewed;
also a few fracto graphic features observable with the naked eye will be
discussed, as they may be of help to the damage tolerance analyst involved in
failure analysis.

424
 425

13.2 The cause of service fractures

A load-bearing structure is designed to sustain the maximum anticipated service
loads with a safety factor between 1.5 and 3, depending upon the type of
structure. There is usually uncertainty about the maximum anticipated service
loads; the safety factor covers these, as well as inacurracies in stresses, possible
below-average material strength, unknown residual stresses, dimensional
tolerances, and - to some extent - small defects escaping quality control.
Because of the safety factor, defect free structures should never fail below or at
the maximum service load. Indeed they almost never do; a true 'overload failure'
is rare.
The causes of fracture are: (a) The remote possibility of a true overload
failure; (b) Development of cracks during service either due to a material defect
not detected during quality control, or (more often) due to poor detail design
(notches and eccentricities) so that conventional design analysis was inadequate;
(c) Crack development due to extreme circumstances (e.g. temperature, and
residual stresses) not accounted for in the design. As a rule, fractures are
precipitated by cracks. A crack may be considered a partial failure. The final
failure - the complete separation - is caused by fracture (Chapter 1). The
propagating fracture is often referred to as a fast propagating CRACK, but the
word crack is used here (and throughout this book) to pertain to partial failures
developing slowly in time.
Crack growth can occur by a variety of mechanisms; the most prominent of
which are, fatigue, stress corrosion and creep, or combinations of these (Chapter
1). By themselves these do not cause a fracture. The latter is a consequence of
the crack, and occurs by rupture, cleavage, or intergranular separation. Cracks
impair the strength, so that a fracture eventually may occur at the operating
stress (service failure).
Fractographers sometimes use the term 'overload fracture' to distinguish
between 'crack' and 'fracture'. This is confusing because an overload can be
interpreted as a load higher than the (maximum) service load. The vast majority
of fractures occur at service loads: these may be the high loads in the spectrum,
but in general, they are not overloads from the design point of view. When the
verdict from a fractographic analysis is "overload fracture' was the cause, it
usually signifies that the crack or defect that precipitated the fracture could not
be identified, for example because it was very small.
A true overload fracture can occur only due to (1) Extreme abuse by the user,
causing stresses higher than ja ma" where j is the safety factor and a max the
maximum service stress anticipated during design (Figure 11.1); (2) Gross
underestimate of the maximum service load so that the structure was under-
designed; or (3) Poor design with sharp notches, misfits etc, so that conventional
design analysis was inadequate.
With few exceptions service fractures are brittle from the engineering point of
426
view (Chapter 2). Yet, the great majority occurs by ductile rupture. The fracto-
grapher's definition of a brittle fracture pertains to whether or not plastic
deformation is required for fracture mechanism. Cleavage (Chapter 1) does not
require plastic deformation, although some plasticity may occur: the fracto-
grapher calls this brittle fracture. Rupture is the result of plastic defor-
mation: the fractographer calls it ductile, although virtually no plastic defor-
mation may occur. This dichotomy was already discussed in Chapter 2. In an
unnotched bar pulled to fracture ample plastic deformation will occur
throughout; the fracture is ductile. In the case of a crack, plastic deformation
is confined to the fracture path (Figure 2.12). The fracture is brittle, because
there is little overall plasticity; yet the fracture mechanism may be ductile
rupture. Since most service fractures are due to cracks almost all are brittle from
the engineering point of view, regardless of the fracture mechanism.
Fracture is the direct cause of the failure, but the actual culprit is usually a
crack or defect without which the fracture would not have occurred. Hence, a
failure analysis should determine the cause of the CRACKING (fatigue, stress
corrosion, etc). Each mechanism has certain characteristic features by which it
can be recognized [1-5]. However, the mechanism by itself does not explain the
failure; the real question is why this mechanism could become operative.
There are five fundamental causes for the start of cracking, namely
(a) Material defects.
(b) Manufacturing defects.
(c) Poor choice of material or heat treatment.
(d) Poor choice of production technique.
(e) Poor (detail) design.
One might add 'poor quality control', but this is not a fundamental cause: it
is the secondary reason by which others become possible.
If the failure analysis can identify one of the above to have been operative,
the remedy is at hand. Manufacturing defects can be introduced by blunt tools,
overheating during machining, welding, etc. Material selection is an obvious
culprit. The material might be perfect for fatigue, but propensity for stress
corrosion cracking due to the particular heat treatment might have been
overlooked. Local heat treatments such as carburizing, nitriding and surface
hardening, almost always cause a volume change of the surface layer, so that
they introduce residual stresses not accounted for in analysis. As for the
production technique, the problem of grain orientation was discussed in
Chapter 7, on the basis of Figure 7.8. It is often a cause of 'unexpected' failures.
Poor detail design is a major cause of service failures. That sharp notches
should be avoided is commonly understood, but hidden stress concentrations
are sometimes not recognized. A classical example was discussed in Chapter 10
on the basis of Figure 10.2. Statically the design is adequate: when loaded to
failure, plastic deformation will ensure even distribution of the load over the
 427

SHRI~K c
FIT

TORSION
II II

DJ5"LACEMENi J!FFE~;:NCES

(a)

(c)

(b)
Figure 13.1. Cracks due to relative displacements (fretting; a, b) or consequent secondary stresses (c).

bolts. However, during elastic service loading, the center bolts transfer no load,
causing cracking at the highly loaded outer bolts.
Other design details may cause secondary displacements, not accounted for
in design or fatigue analysis. Consider the shrink fitting in Figure 13.la.
Eventually, the load must be transferred from the shaft to the shrunk-on part.
However, at A the shrunk-on part is still relatively stress-free (no stress and no
strain). The shaft is strained, so that there will be relative movement between A
and B, which may cause fretting and subsequent fatigue cracks. The same
happens at bolt shafts, under bolt heads, and so on.
Secondary stresses due to displacements may cause problems as well. The
loading of parts I and II in Figure 13.1 b causes (small) upward displacements
of A and therefore of B. As B cannot undergo vertical displacements because of
the bolts, secondary bending stresses in the bolted flanges may cause cracking.
Elimination of some bolts may solve the problem, because there will be some
freedom for vertical displacements if bolt 1 and 2 are omitted.
Determining the cracking mechanism is a significant part of the failure
analysis, but the starting point of the crack must be found in order to establish
the reason why it became operative. All fundamental causes listed above must
then be considered. At this time there is an essential task for the damage
428

tolerance analyst. Questions regarding the origin of stresses, loads, load path,
primary and secondary displacements/stresses, must be answered. Answers to
these are often found in general area deformations (occurring during the
fracture process); deformation of adjacent parts will reveal directions of acting
loads, and stresses. It should be checked whether these are compatible with the
design assumptions and with the cracking mechanism. Analysis of stress fields
for 'unanticipated' stress concentrations, secondary stresses and displacements
(Figure 13.1) is necessary.

13.3. Fractography

The tools for fractography are loupe and stereo microscope, optical micros-
copes, electron microscopes, X-ray analysers, and image analysers. The non-
fractographer should be aware that an electron fracto graph (high magnification
photograph) shows only a very small area. This may be adequate, but it is
sometimes deceiving because it may not be representative of the whole.
The main cracking mechanisms were discussed in Chapter 1. Fatigue damage
and fatigue cracking in service take place under nominal elastic stresses, but a
fatigue crack cannot initate without plastic deformation, however local and
minute (Chapter I). Generally, such plastic deformation will occur at the tip of
a notch or at a stress raiser (including particles in the material). Once a crack
is initiated it grows by a mechanism similar to the example in Figure 1.4. A
regular repetition of blunting and sharpening causes the formation of distinct
lines on the fracture surface, the fatigue striations. One striation is formed
during each cycle as shown in the electron fracto graph in Figure 1.5. This opens
the possibility to measure the rate of growth by measurement of the striations
spacing.
Striations as regular as those in Figue 1.5 can be formed only when the
material can accomodate the mechanism of Figure 1.4 by opening and closing
in a uniform manner over some distance along the crack front, as shown in
Figure 13.2. If the material's deformation possibilities are insufficient to open

Figure 13.2. Formation of regular and ill-defined striations Left: uniform opening and closing over
some distance creates lines on crack surface (regular striation as in figure 1.5) Right: Non-uniform
opening and closing (ill-defined striations as in Figure 13.3).
 429

Figure 13.3. Patch of rather well-defined striations (center) and 'chopped' striations (top left and
bottom right); 4340 steel;3500 x.

and close the crack uniformly, the striations become ill-defined as illustrated in
Figure \3 .2 (right). This is the case in many steels (Figure \3 .3). Nevertheless
striations (even if ill-defined) can provide the rate of growth by their average
spacing.
Fractographic features of other cracking mechanisms are more variable and
differ from material to material. A stress corrosion crack often follows the grain
boundaries because the chemical composition at the grain boundaries is
different from that inside the grains. The fractographer will recognize the
features by which to identify a stress corrosion crack.
Fracture mechanisms were discussed in Chapter I as well. To a degree the
shape of dimples depends upon the (local) stress gradient. Thus, the dimple
shape can be used occasionally for a qualitative assessment of the local stress
field, but dimple shape can be deceiving as it depends upon the angle of view [4].
The dimple rupture surface is irregular and (in contrast to the glittering cleavage
430
surface) diffuses light, so that the fracture looks dull grey to the naked eye; as
such it often can be recognized without microscopic aid.
Whether fracture occurs hy cleavage or rupture depends upon rate ofloading,
temperature, and state of stress. Roughly speaking, if sufficient plastic deforma-
tion can occur to relieve stresses, cleavage will not occur. At low temperatures
and/or high-loading rates the yield strength is higher. If the state of stress is one
with high hydrostatic tension, yielding is postponed to stresses higher than the
uniaxial yield strength (Chapter 2). Hence, the above conditions tend to confine
plastic deformation and promote cleavage if the stress peaks at or above the
cleavage strength. In other cases local plasticity will ameliorate conditions
through lower stresses and larger plastic strains to set the rupture process in
motion. In many alloys it is virtually impossible to induce a cleavage fracture.
Of particular interest for the damage tolerance interpretation of the fracture
surface are transverse cracks, such as shown in Figure 13.4. They are an
indication of high hydrostatic tension (plane strain), or high (Jz as shown in
Figure 13.4, possibily combined with a low fracture toughness in Z-direction
(ST or SL).

13.4. Features of use in fracture mechanics analysis

Fracture surface topography can be measured [6, 7] from fractographs in much

the same way as terrestial topography from aerial photography. The projection-
al displacement of identical features in two photographs taken at different
angles, permit the determination of the height of the feature due to the viewing
angle difference, as shown in Figure 13.5. The larger the viewing angle
difference, the larger the relative displacement, and the more accurate the
procedure.
The height is determined from (Figure 13.5):

p = CD - CE = bcos() + hsin()
- (b cos () - h sin () = 2h sin ()
(13.1)
p
h =
2 sin ()

For fracture mechanics analysis the most fruitful application of this technique
is in the area of transition between crack and fracture, as in the example given
in Figure 13.6. A topographic measurement [8] provides the size of the Crack
Tip Opening Displacement, CTOD. Stereo pictures show the blunting in a
different projection (different size; Figure 13.7), so that CTOD can be measured
by Equation (13.1). Using fracture mechanics equations, the CTOD can be
 ay
z
a../ TRANSVERSE CRACKING

../
I.. I
../../~' /
../ ~/
../ .,,-'
../ '--/../
~z
../
../

~~x
z

Figure 13.4. Transverse cracks due to high <12 Left: crack surface at 50 x (optical microscope).

-""
W
432

related to the applied stress as discussed in next section; if the CTOD is

measured from the topography, the toughness can be calculated.
In the case of a fatigue crack, striation counts are a source of valuable
information. The technique is acclaimed inaccurate, but this is generally due to
insufficient numbers of measurements, and impatience of the fractographer.
Fractographs display an extremely small part of the crack, while crack growth
rates depend strongly upon local circumstances. In a crack growth test, growth
is measured as an average over hundreds of grains. The same averaging must be
done in striation counts, so that the procedure literally requires hundreds of
fractographs. The slow photographing process in many scanning microscopes is
the reason why striation counts are often limited to a few dozen fractographs;
this does not provide useful data. Therefore the use of replicas and transmission
microscopes is preferable, because photographing is semi-automatic; a modern
transmission microscope can easily produce a hundred fractographs per hour,
provided film is used instead of plates (which is adequate for the purpose).
There is basically one striation per load cycle. Whether striations occur only
here or there, are chopped up, or regular (Figures 1.5 and 13.3), their average
spacing does provide the rate of crack propagation. A typical result is shown in
Figure 13.8. (Each measurement of striation spacing is an average of many
measurements.) Numerical integration of the rates provides the crack propaga-
tion curve as a function of the number of cycles, as shown in Figure 13.8 and
Table 13.1. Such a curve can be compared with the one obtained by the fracture
mechanics analysis.

I
I I
, 'E
CI ,'0
,
I

'8

, ~/}-
~_(_ h

Figure 13.5. Height measurement from stereo micrographs. Features A and B project at different
distances on photographic plane (horizontal).
 433

/
STRIATED
FATIGUE
/
STRETCHED DIMPLE RUPTURE

CRAr crOD
ZONE DUE
TO FINAL
BLUNTING

\
Figure 13.6. Blunting of crack tip before fracture (slip) forming stretched zone characterized by
wavy slip lines.

Crack growth often is suspended for long periods because the structure is not
being loaded (not in use). The already present part of the crack will be subject
to environmental influences and may undergo slight discoloration. Then a ring
shaped mark (beach mark) will be visible on the crack surface, delineating the
shape of the crack (front) at a given point in time (see Figure 13.9). Changes in
load will change the rate of crack growth. Higher growth rates (higher stress)
result in a rougher crack surface, than low rates (low stress). A smoother crack
surface has more luster because it is more reflective. Also this will result in a
delineation on the crack surface, creating another beach mark . Beach marks
may occur on any crack surface, whether fatigue or stress corrosion. They
always signify a change in circumstances, either load or environment. An
absence of beach marks is an indication of uniform circumstances throughout
the cracking phase. As beach marks signify the positions of the crack front at
certain times, they are useful in correlations with the stress-time-environment
history.
Fracture is a fast process. Its macroscopic features are reflections of fracture
speed and hence of stress level and fracture toughness. Many fracture surfaces
(both cleavage and rupture) show chevron marks as depicted in Figure 13. 10:
434

PHOTO

PHOTO

Figure 13.7. Measurement ofh (1 /2 CTOD) in accordance with Figure 13.5. Stereo photographs at
+ and - 6 degrees.
they point to the origin of the fracture. The sharper the points of the chevrons,
the faster the fracture (low toughness; high stress). The chevrons are not
indicative of the shape of the fracture front ; instead they are more or less a
mirror image of the fracture front. They show increasing height (roughness)
towards the outer surface: in the interior the state of stress is often triaxial, but
at the surface (Jz is always zero, so that there is a gradual decrease of the
hydrostatic tension towards the surface. The higher the hydrostatic tension the
lower the fracture resistance (plane strain versus plane stress). Consequently the
fracture tries to run faster in the center giving a smoother fracture surface, but
naturally the fracture at the free surface must move along with that in the
interior.
If the stresses are high and/or the material's fracture resistance low, the energy
released during fracture is high. When sufficient energy is available, two or more
fractures can be maintained simultaneously (a branch is formed [5]). A higher
energy surplus causes more branching. High energy surplus occurs in the case
of low fracture resistance (windows shatter when fracturing) . Branching
provides another indication of the fracture origin. This is particularly helpful
when shattering has occurred.
Much information can be gained from the general condition of the broken
 o path A
a *B
j
10

o
~ ~9
6.5 t/
I /8
5
5.5 / /
,# ARROWS INDICATE DIRECTION
A" 5.5 /OF PROPAGATION
/
B/
/
//:
o

-g/
+
13 I ~/
9-- 40'
5.8- 50-
o 20,000 N

Figure J3.B. Calculation of crack growth curve from measurement of striation spacings. Numbers
in bottom left drawing are measured striation spacings in units of 0.1 microns (see Table 13.1). ~
w
v.
436
Table 13.1. Calculation of crack growth curve along path A in Figure 13.8
a Striation ~a aaverage da/dN da/dN ~a N = N + ~N
(mm) spacing (mm) over average average ~N = da/dN
p/c increment over increment mm/c
average
p/c
0 0 0
0.8 0.4 0.05 5.0 x 10- 5 16000
0.8 0.1 16000
0.93 1.27 0.13 1.3 x 10- 4 7150
1.73 0.16 23150
1.20 2.33 0.26 2.6 x 10- 4 5150
2.93 0.35 28300
3.27 4.57 0.45 4.5 x 10- 4 7270
6.20 0.55 35570
1.13 6.77 0.75 7.5 x 10- 4 1510
7.33 0.94 37080

component and surrounding structure. Deformations that occurred AFTER

fracture, secondary cracks and fractures, branches, etc., can be used to recon-
struct the fracture event and provide information on the direction or magnitude
of the loads. In general, as much of the surrounding structure should be
preserved for the analysis as practical. Considering the magnifications, fracto-
graphic features are very small. The slightest damage to the fracture surface will
obliterate them: fitting the pieces together will accomplish this; acid deposits of
fingerprints do the same.
One of the most important preliminaries to a failure analysis is the protection
of the fracture surface by means of clear acrylic lacquer spray or acid-free oil;
these can be removed later with organic solvents. If a reconstruction is to be
made on the spot, the pieces should be laid out (without the fracture surfaces
touching) in the position from whence they came. Photographs from many
different angles should be made for documentation. If it is not strictly necessary
to perform the reconstruction in the field, it should be done in the laboratory.

13.5. Use of fracture mechanics

Failure analysis as discussed, will show the mechanism of cracking and the
mechanism of fracture, and whether or not material defects or manufacturing
defects were involved. Fractography generally will show the crack size at which
final fracture occurred because cracks have a different surface roughness than
fracture. Examples are shown in Figure 13.9.
Fracture occurs in accordance with the concepts discussed in Chapter 3 and
4. If operational stresses are known, the equations in Chapters 3 and 4 can be
used to calculate the toughness from the crack size at fracture, which, if
compared with the anticipated toughness, will show whether or not the material
 437

Figure 13.9. Beach marks on various crack surfaces.

438
had a toughness higher or lower than expected and whether the material was in
accordance with the specification.
If the stresses are not known on the other hand, the toughness may be
obtained from data handbooks. The operating stresses then can be calculated.
Depending upon the material and toughness either LEFM (Chapter 3) or
EPFM (Chapter 4) must be used. The best way to proceed is to calculate the
entire residual strength diagram in the manner discussed in Chapter 10; collapse
should naturally be accounted for. Given the crack size at fracture as
determined from fractography, the fracture stess can be obtained from the
residual strength diagram.
If neither fracture stress nor toughness is known, an estimated toughness
value may be used, e.g. for comparative materials or estimated on the basis of
a Charpy-value (Chapter 7). Instead stretched zone measurements may be used
to obtain the toughness; at the least these may be of help to corroborate the
toughness estimate. The size of the stretched zone (Figures 13.6 and 13.7) is
related to the crack-tip opening displacement (CTOD) at the time of fracture;
it can be measured using the techniques described in the previous section on the
basis of Figures 13.5-13.7 and Equation (13.1). The CTOD at fracture is related
to the toughness as (Chapter 4):
K2
G = i = 1';y CTOD (13.2)

so that the toughness can be calculated from the measured stretched zone size.
Careful measurement can give reliable toughness values [8]. For example,
consider the case of Figure 13.7, which is for a material with 1';y = 65 ksi and
E = 10000 ksi. The distance between the two features indicated by the large
arrows can be measured from which the value of p in Equation 13.1 is obtained.
Using the micron-scale shown in Figure 13.7, the value of p in Equation (13.1)
is found as 2.7 microns = 0.0027 mm. Since (J = 6 degrees = 0.10 radians, the
height h is found from Equation (13.1) as h = 0.0027/0.2 = 0.0135mm. As
CTOD = 2h(Figure13.6),itfoliowsthatCTOD = 0.027mm = 0.00106 inch.
Then with the aid of Equation (13.2) the toughness is found as
,JIOOOO x 65 x 0.00106 = 26ksi.Jill. It was pointed out already that
micrographs exhibit only a small area, so that several measurements of different
areas are required to obtain an average. If more measurements are made using
the fractographs of Figure 13.7 an average toughness value of 28 ksi.Jill
emerges.
With the toughness known, fracture stresses can be calculated from (J = KI
f3 J1W;, where ac is the crack length at fracture; f3 is determined by one of
the procedures discussed in Chapter 8. It then can be determined whether the
fracture took place due to an excessively high load or whether the local stresses
were too high.
Fatigue-striation counts (Figure 13.8) generally provide a reasonable account
 439

Figure 13.10. Chevrons on surface of high speed fracture (high stress/low toughness).

of the crack-growth rates and crack-growth curve. If the crack-growth rate data
of the material and the stresses are known, the stress intensity can be inferred,
and a comparison made between actual and anticipated properties for a
conclusion about the adequacy of the material. Conversely, if the stresses are
not known, the measured rates and the rate properties can be used to estimate
the acting stresses. From the amount of crack growth (crack size at fracture),
known stresses, and growth-rate properties, a reasonable insight can be
obtained regarding the question of misuse (e.g. continual high loading). The
time to failure and final crack size are determined using fracture mechanics, as
discussed above. When the results are not in accordance with the observations,
the analysis can be repeated to determine how much higher (or lower) the
stresses would have had to be to produce the cracking time and crack size at
fracture as observed.
The most direct way to use the information is as a basis for determining
inspection intervals for similar components, using the procedures discussed in
Chapter 11. Indirect use can be made by comparing the curve with a calculated
one using the concepts described in Chapters 5, 6 and 10. This can provide
information regarding the adequacy of the analysis (e.g. load history, (J, dajdN).
If no corroboration is found the analysis can be performed for different load
histories or da jdN data, to evaluate the cause of the problem. Other indirect use
can be made by determining I1K-da jdN at various crack sizes from the striation
measurements. Comparison of these results with the anticipated rate properties
can reveal whether the material might have been below standard. (For an
example see Exercises 3, 4, 5.)
A change in loading or environment during the cracking process is likely to
leave a beach mark. If information on the nature of these changes is available,
the crack size at which they occurred can be used to obtain information on rate
properties or stresses at that time. Conversely, without knowledge of the nature
of the changes, information (a change in loading in particular) can be obtained
from known growth-rate properties: the time for growth between beach marks
can be calculated, and the stress required to produce the observed crack sizes
estimated.
If fractography shows the presence of initial defects - whether mechanical or
metallurgical - crack-growth analysis can be used, starting from the initial
440
defect. Again, depending upon the objective, information can be obtained on
stresses (known material properties) or material properties (known stresses).
The stress intensities due to different stress systems are additive as long as all
stress systems are of the same mode (Chapter 8). Residual stresses due to
welding, cold deformation, surface treatments, or heat treatments can thus be
accounted for. Although this is useful in principle, the magnitude of the residual
stresses is often unknown. It may be assumed that yielding at the crack tip is so
extensive that residual stresses can be ignored for the fracture analysis. For
fatigue-crack growth, the residual stress (static) causes only a change in R-ratio
(Chapter 9). In the case of sustained-load stress corrosion, the stress intensities
are additive. These guidelines can be used to evaluate the effect of residual stress.
The procedures can at least provide an estimate of the effect, but engineering
judgement is required. Residual stresses can be accounted for (Chapter 9), but
the real problem is to know their magnitude.

13.6. Possible actions based on failure analysis

The purpose of a failure analysis is to determine how other failures can be

prevented. If the fracture appears to be an incidental case only, no action may
be required at all. However, if it appears symptomatic, one or more of the
following measures may be taken:
(a) Relief of residual stress.
(b) Relief of secondary displacement.
(c) Selection of other material.
(d) Redesign.
(e) Change in manufacturing procedures.
(f) Replacement of similar components (or life limitation).
(g) Inspection on the basis of detectable crack size and maximum permissible
crack size (based on crack growth analysis); repair when crack is detected.
(h) Retrofit of load bypass (second member; Chapter 9).
These actions are in accordance with those discussed in Chapters II and 12.
Some of these will put part of the burden on the operator. A life limitation
may be preferable, if the costs of the part or component and of replacement are
low. For costly parts or components a retrofit may be a good alternative. If the
operator is the 'general public' only replacement may be possible, as inspection
is only an alternative for professional equipment. The possiblility of taking no
action at all should not be overlooked: the cost of fracture may be less than the
cost of fracture control.

13.7. Exercises
1. Using the same procedure as in the text calculate the toughness by making
 441
measurements at two or more locations in Figure 13.7 and estimate the
average toughness.
2. If the crack at fracture in Exercise I was 20 mm long and 13 = 0.7, what was
the nominal stress at fracture.
3. Measurements of striation spacings at 8000 x provide spacings of 0.03, 0.08
and 0.15 inch respectively at crack sizes 0.1, 0.2, 0.3, and striation spacings
of 0.06 and 0.09 inch at 2000 x at crack sizes of 0.4 and 0.5 inch resp.
Determine the growth rates and estimate the crack growth curve.
4. Assuming that 13 = 1 throughout, and assuming that a materials handbook
provides da/dN = 2£-8 IlK 3 for the material, estimate the stress range in
Exercise 3 if the loading was of constant amplitude. R = O.
5. Using the rate data quoted in Exercise 4, calculate the crack growth curve,
and compare the result with that of Exercise 4.
6. Suppose that the stress range in the case of Exercise 4 is known to be 15 ksi,
could there be a problem with the material?
7. It appears that in the component concerned in Exercise 4 a stress concentra-
tion exists with kr = 1.5. Re-evaluate the results of Exercises 5 and 6. Which
remedial action would you take.
8. A fracture occurs in a single lap joint with three bolts due to a crack. The
joint transfers a load of 100000 Ibs. The material has a yield strength of
F;y = 50 ksi, the bolts a shear strength of 100 ksi. The joint was designed
with a safety factor of 3 against the above numbers. No action was taken in
the original design for equal load distribution during elastic loading. The
width of the joined plates is two inches. What is the configuration of the
original design. Redesign the joint for optimal crack resistance still using
three bolts. Would four bolts in the original design provide a good alter-
native?
9. A service fracture appears to have been caused by a stress corrosion crack.
The crack size at fracture was I inch (13 = 0.7), the material's toughness is
50 ksi,Jlli, while K[scc = 7 ksi,Jlli. What was the operating stress and which
size of defect would have to be eliminated in quality control to prevent
recurrence.
10. A failure analysis leads to the hypothesis that the crack was caused by a
material deficiency. No crack is apparent. A crack larger than 0.05 inch
would have been identified. The specified toughness of the material is
60 ksi,Jlli, F;y = 80 ksi. At the location of the crack there is a stress con-
centration with k, = 2. The nominal stress was calculated to be 25 ksi. Is the
hypothesis tenable?
442
11. A material's yield strength is 90 ksi and its crack propagation properties at
R = 0 can be described by da/dN = 8 x 10- 9 (K?5 inch/cycle if K in ksi Jill.
Failure analysis of a very large failed part made of this material shows that
fracture occurred due to a fatigue crack penetrating 0.6 inches from the
surface (/3 = 0.64). At a magnification of 2000 x the striation spacing at the
very front of the fatigue crack is 0.12 inches. The loading was of constant
amplitude with occasional overloads of twice the maximum of the constant
amplitude loading (R = 0 in both cases). Calculate the material's toughness
and the maximum stress during the constant amplitude loading.

References
[I] D.A. Ryder, Elements oJJractography, AGARDograph 155-71 (1971).
[2] A. Phillips et aI., Electron Jractography handbook, AFML-TDR-64-416 (1965).
[3] Various authors, Failure analysis and prevention. Metals Handbook, Vol. II, 9th ed. ASM
(1986).
[4] D. Broek, Some contributions of electron fractography to the theory of fracture, Int. Met.
Reviews, Review 185, Vol. 9 (1974) pp. 135-181.
[5] D. Broek, Elementary engineering Jracture mechanics, Nyhoff, 1985.
[6] I.F. Nankivell, Minimum differences in height details in electron stereomicroscopy, Brit. J. Appl.
Phys. 13 (1962) pp. 126-128.
[7] D.C. Wells, Correction of errors in electron stereomicrography, Brit. J. Appl. Phys. 11 (1960)
pp. 199-200.
[8] D. Broek, Corelation between stretched zone size andJracture toughness, ICF conference, Munich
(1973).
 CHAPTER 14

Applications

14.1. Scope
Many examples of practical damage tolerance problems and analysis were given
throughout this book, especially in the Exercises and their solutions. This
chapter provides a number of examples in which application of damage
tolerance analysis and fracture control can be seen in the larger context of
structural design and operation. In some cases use is made of fictitious situations
and numbers, but they are derived from actual cases. Sometimes insufficient
input data are available, which is common in engineering. These conditions were
not changed artificially. Instead, the problems are represented as they might
appear in engineering, so that judgement and approximations can be illustrated.
Only scant details are provided on the actual analysis since the analysis
procedures were amply discussed in the forgoing chapters. Instead, the problems
are treated in their general context, so that the reader may appreciate how input,
assumptions, and judgement affect results, accuracy and subsequent fracture
control decisions. It would have been easy to fill this whole book with examples
from the author's experience concerning aircraft, but in order to avoid the
suggestion that damage tolerance analysis only works for aircraft, all of the
examples are for other structures. It stands to reason that not all areas of
technology can be covered. However, the examples are of such variety that
similarities to other problems may be found and ideas be transferred.

14.2. Storage tank (fictitious example)

(a) Problem definition

A storage tank for non-flammable chemicals (Figure 14.1) developed a complete
fracture, spilling its contents. Failure analysis revealed a pre-existing defect in
a vertical weld with dimensions as shown in the figure. At the time of the
incident the recorded temperature was - 20 F. Many other tanks of the same

443
444

longItudinal 7
weld defect
weld failure

(a)
weld d"'f~ct

~ \ ._____---L.-__.../ i
2a =0.1
/ /
0.4

(b) / / _.. _-,2=C=-,;',-_ '" ~~

Figure 14.1. Failure of storage tank. (a) Tank; (b) Fracture and weld defect.

design operate in areas with different climates. The cost of the incident,
including lost revenues, clean-up operation, and replacement is estimated at 30
man-year equivalents. A fracture control plan costing less than a fracture, would
be implemented.

(b) Loads, environment, stresses

Loads due to wind, snow and sleet are negligible; the only loading is the
hydrostatic load due to the contents. The chemical being non-corrosive, stress
corrosion is unlikely. The tanks are emptied approximately 12 times per year;
they are expected to have a remaining economical life of 50 years. The lowest
anticipated service temperature is - 40 F in the northern regions.
The chemical has a specific weight of 0.029Ib/cu in. Since the height of the
tank is 300 in (25 ft) the hydrostatic pressure at the bottom is
p = 300 x 0.029 = 8.7 psi. With the dimensions given in Figure 14.1 the hoop
stress is: (J = pR/B = 8.7 x 216/0.4 = 4698 psi = 4.7 ksi. The chemical can
expand freely in the empty top part so that no pressurization due to expansion
occurs. Due to the fixity at the base, thermal stresses can occur due to shrinkage
at low temperature. Erection took place at 70 F; hence there was a temperature
differential of 90 degrees at the time of failure. Assuming full constraint, this
would cause a thermal stress of (J = E(X~ T = 30.000 x

7 X 10- 6 X 90 = 18.9 ksi. Hence, the total stress could have been as high as
4.7 + 18.9 = 23.6ksi.
 445
(c) Material data
The material's yield strength at - 20 F is FII' = 55 ksi. Toughness data not being
available, Charpy specimens containing longitudinal welds were cut from the
failed tank (weld crosswise in the specimen at the location of the notch). These
tests showed the transition temperature at 10 F, and a Charpy energy of 2 ft/lbs
at - 20 F. Using the conversion equations discussed in Chapter 7, the toughness
is estimated to be K/c 12y'2 = 17ksi~in. Naturally, the weld defect is in plane
strain, but fracture would first cause a through-the-thickness crack. For
through-cracks plane strain occurs at a thickness larger than 2.5 x (17/
55)2 = 0.24 inch, so that with the given wall thickness the toughness estimated
above for fast loading, it is considered a useful number because at break-
through due to fracture the high fracture speed must be accounted for in
considering leak-before-break (Chapter 9).

(d) Analysis
The weld-defect here is a lack of fusion defect, with dimensions as in Figure 14.1.
It can be considered as an elliptical flaw with a/2e = 0.05. The geometry factor
is f3 = f3FFS/</J (Figure 8.3). Since this is a buried defect the back free surface
correction of 1.12 is not required. In determining f3FFS it must be realized that
the distance from crack center to the wall is equal to half the wall thickness, so
that the value of alB used in Figure 8.3c must be 0.05/0.2 = 0.25. This leads to
f3 = 1.15/1.05 = 1.1.
After breakthrough there is a through-the-thickness crack of 2a = 1 in. For
such a crack in a pressurized container f3 = .jl + 1.6la2 / RB =
.jI + 1.61 x 0.5 2 /(18 x 12 x 0.4) = 1.002. Since this is approximately
equal to I superposition to the thermal stress can be accomplished by directly
adding the stresses (same geometry factor).
With a total stress of 23.6 ksi the stress intensity at fracture of the embedded
flaw was K = 1.1 x 23.6 x .jnO.05 = 10.3 ksiFn, which is considerably
lower than the estimated toughness. The latter would require a stress of 17/
(1.1.jn x 0.05) = 39 ksi so that either an additional residual stress of about
15 ksi was present or the toughness was lower than estimated. The stress
intensity of the one inch through crack was 23.6.jnO.5 = 29.6 ksiFn, which is
larger than the toughness, so that fracture should indeed occur after break-
through of the internal defect.
It may be assumed that the residual stresses were largely relieved after
break-through. Using the stress intensity for the through crack, the stress
intensity and residual strength can be calculated for various locations in the
tank. For example at half of the height of the tank the hoop stress is reduced
to 4.7/2 = 2.35 ksi. Assuming (very conservatively) that the thermal stress
decreases linearly to the top, because fixation is at the bottom, the correspond-
ing thermal stresses is 9.3 ksi. If the toughness is 17 ksi Fn a through cracks of
446

2a = 2(l7/12.6)2/n = 1.6 inches can be sustained (leaks).

Fatigue crack growth may occur during emptying and refilling. The fatigue
crack growth properties are estimated to be da/dN = 10- 9 LlK'. With refill
cycles of 12 times per year, there will be 600 cycles in the remaining service life
of 50 years. Assuming that these will coincide with the thermal cycles (extremely
conservative) the stress intensity of the internal defect is 10.3 ksiJm as
calculated above. Using the above, crack growth in 50 years'
Lla = 10- 9 X (10.3)4 X 600 = 0.007 in, which is negligible. However, many
more thermal cycles of smaller magnitude may occur, so that fatigue crack
growth should be calculated more accurately. Assuming daily thermal cycles of
30 F, the thermal stress cycle is 18.9/3 = 6.3 ksi. The crack growth curve for this
case is shown in Figure 14.2.

(e) Fracture control plan

A toughness of 40 ksiJID or more is estimated for temperatures above 10 F
(transition temperature). With a toughness that high, through cracks of several
inches can be sustained (leak). On this basis the following fracture control plan
is established.
No action to be taken for tanks in those climates where the lowest anticipated
temperature is above 10 F. In tanks in colder climates all longitudinal welds in
the lower half are to be X-rayed. Any defects longer than 0.1 inch are to be
repaired (ground out and repair welded).

,4 21 28 35 42 49 56 63
C. COl ]NCH)

180

Q !50 150
D
~.

...LI 140 140

N

'" 120 ]20

u

g 100 100

80 80

50
L_--- 50

40 40

20 20

20 40 50 80 100 120 ]40 150 180

L] FE (1000 CYCLES)
Figure 14.2. Crack growth due to daily thermal cycle as assumed.
 447

The costs for inspection and repairs are estimated at an equivalent of three
man years. The fracture control plan is implemented because the cost is
negligible with respect to the anticipated cost of fracture.
NOTE: The above is a fictitious example. Numbers may bear no relation to
reality. Certain aspects of the problem may have been overlooked.

14.3. Fracture arrest in ships

(a) Problem definition

Of some 2000 Liberty ships built during World War II, more than 100 broke in
two; many others had serious cracks. Since then, ship design requirements have
been changed to include mandatory fracture arrest strakes. These strakes are
currently applied [1, 2] at the gunwale in deck and side plating (Figure 14.3) and
at the turn of the bilge. In very large ships additional strakes are required. In
general, the strake material is of a higher grade than the normal plating material.
The effectiveness of the strakes is assessed in this section. Also the feasibility of
hybrid arresters is assessed.
The purpose of an arrester is to stop a running fracture within the continuous
structure, so that the structure with the arrested crack is still capable of
sustaining appreciable loads until remedial action can be taken. This may be
accomplished by means of heavy stingers, strakes or hybrid arresters.
The first type is a heavy stringer or girder. Load transfer from the cracked
plate to the girder reduces the crack tip stresses and thus K (Chapter 9). An
arrester of the second type is a welded-in strake of a material of higher toughness
than the base plate. When the fracture enters the strake, it meets with a higher

~Deck plating

I
I
i
I
ct.

,I

Figure 14.3. Cross section of ship hull.

448
toughness which may cause arrest. A hybrid arrester is a combination of the two
other types, i.e. a heavy girder plus a high toughness strake.
Early research on arresters was conducted by the naval industry [3-7]. Then
the aircraft industry began developing crack arrest technology [8-12], carrying
it to a state of near perfection, and accounting for such things as stringer and
fastener plasticity (Chapter 9). Modern airplane designs take full advantage of
the arrest capability of skin-stringer structure. The present assessment of hybrid
arresters draws upon aircraft technology.

(b) Background
A Navy ship is designed [13, 14] for a Standard Design Wave with a wave length
equal to the Length Between Perpendiculars (LBP) and a wave height equal to
1.1 LBP, the LBP being roughly equal to the ship length. Two cases are
considered, one where the top of the wave is at midship (hogging) and one where
the trough of the wave is at midship (sagging), as shown earlier in Figure 6.10.
The Standard Design Wave provides the distribution of the upward forces on
the ship; the downward forces can be obtained from the estimated weight
distribution. These provide the bending moment.
The material allowables used are as follows (requirements [13, 14]):

Material F,u F,y Max. allowable Max. allowable Toughness

(ksi) (ksi) total stress bending stress at LAST
(ksi) (ksi) (estimated)
(ksi~)
Mild steel 60 34 27 19 50
ABS-Grade E 90
HY-80 100 80 55 23.5 175
HY-130 130 220

If the ship is built of mild steel the maximum allowable live stress (due to
applied loads) is 19 ksi. The maximum allowable for the 'total stress' is 27 ksi,
which means that there is an allowance of 8 ksi for secondary stresses, due to
dead weights, misfits, slamming, etc. These secondary stresses are not evaluated
in the design. Their existence is acknowledged and a margin is provided by
limiting the bending stress to 19ksi. Essentially, this provides a safety factor of
34/27 = 1.26 against yield and of 60/27 = 2.22 against F,u (compare I and 1.5
respectively for airplane structures).
Shell plating, and the size and spacing of longitudinals (stringers) are selected
to produce the section modulus required to sustain the bending at the Standard
Design Wave with stresses less than or equal to the allowable; the hull is then
roughly sized. The design, in particular the bottom, is checked for hydrostatic
pressure and elastic stability to determine the final size and spacing of the
 449
longitudinals. The loop is closed by calculating the section modulus of the final
design and by checking whether the bending stress is not in excess of the
allowable. Naturally this summary is simplifying matters. Merchant ship hulls
are designed in essentially the same manner, but the Standard Design Wave is
different.

(c) Arrest criterion

Before the usefulness of arresters can be assessed, a criterion for arrest must be
set. No arrester will be effective under all circumstances. Experience shows that
ships may incur damage during severe weather [15] and that even vessels fitted
with special steels may experience fractures. Yet, requiring arrest under the
worst conceivable circumstances leads to inordinately heavy, closely spaced
arresters, if arrest is at all feasible. The requirement would be unrealistic,
because of the low probability that the worst circumstances concur with the
presence of a critical defect. The toughness of ship steels depends strongly upon
temperature. Hence, the fracture arrest capacility at the Lowest Anticipated
Service Temperature (LAST) is the most relevant. Normally, a ship spends only
a few percent of its life at the LAST, whereas the Design Wave is expected to
occur only once in the ship life [13].
Measured and calculated stress spectra of commercial ships [16,17] show that
the once per lifetime (109 waves) peak-to-peak-stress range is on the order of
18-25ksi. For one year of operation, the maximum range would be 14-19ksi.
If approximately two-thirds of the range is in tension, the secondary stress about
5 ksi, then the once-per-year tension stress would be 15-17 ksi. Hence, a good
criterion would be that arrest must be possible at the LAST at stresses between
14 and 17 ksi.

(d) Effectiveness of arrester strakes

Linear elastic fracture mechanics apply at the LAST, especially because large
cracks are concerned (Chapters 3 and 12). Dynamic effects are ignored initially,
but their possible consequences will be discussed later. Four materials are con-
sidered, namely mild steel (MS), ABS Grade E steel, HY-80 and HY-130. The
toughness of these materials at LAST were estimated as shown in the table above.
Using p = 1 for wide plates (Chapter 3), the residual strength curves can be
calculated as in Figure 14.4. At the time a fracture in MS runs into an arrester
strake, the problem becomes essentially identical to the case in which the whole
plate is of the arrester material (both have the same modulus and thickness).
Then consider a MS deck plate with arrester strakes. Both operate at the same
stress. A crack of 4-inch length (a = 2 inches) in MS would be critical at 20 ksi
(point A in Figure 14.4). It could be arrested by an ABS Grade E strake if the
strake was no more than three inches away from the original crack tip, because
at 20ksi the strake is critical at a = 5-inch (point D). Similarly, HY-80 and
450
30r.-~~-- ________________________________________- .

25
:t== I QUlzmm I I
2a

20

Crock Size, 0 , inch

o 50 100 150 200 250 300 350 400 450

Crock Size, 20, inch

Figure 14.4. Arrest capability of various strake materials.

HY-130 strakes would provide arrest if the strakes were 23 and 28 inches away
respectively (points B and C). If arrest were required at a stress of 12 ksi, a
somewhat longer crack (point E) could be tolerated in the MS. Arrest would
occur in HY-80 arrester strakes at a distance of approximately 70 inches or in
HY-130 strakes at a distance of about 115 inches (points F and G).
On this basis residual strength diagrams for plates with arresters can be
constructed. If arrest were required for example at a stress of 14 ksi, the diagram
of Figure 14.5 would apply. For this situation, HY-80 arrester strakes at
100-inch spacing would be required. The width of the strakes was taken here as
25 inches, but the necessary strake width depends upon the dynamic effects (see
later). Figure 14.5 shows that any crack smaller than four inches (2a = 8
inches) will not be arrested because it will be critical at a stress higher than 14 ksi
(e.g. point C). When a fracture precipitated by a smaller crack (higher stress) has
propagated across the MS bay and reaches the arrester, the arrester cannot
sustain the damage at this higher stress (point D). Cracks longer than 4 inches
will be critical at stresses below 14 ksi. For example, a fracture starting at point
A runs to B, where it is arrested. In the above it was implicitly assumed that the
crack in the MS would occur in the center of the bay, but in the absence of
dynamic effects, the location of the crack (if it has two tips) is immaterial. A
fracture starting at both tips would still propagate through the whole bay.
 451
30
1 .
Ii=== • HY-BO •
X---IOO-ln.--I-25-in.-l

Il=
i
HY-I30
•• 160-ln. 25-in.-1

20

,
, , .D"" ,,
" ....

10
----

----------------
Crack Size, 0, inch

o 50 100 150 200 250 300 350 400 450

Crock Size, 20, Inch

Figure 14.5. Required spacing of two types of strakes for equal arrest capability.

The examples apply to plates, but a hull consists of plate and stringers.
(Figures 14.3). A running fracture in the plate can and will enter the welded
stringers, severing them in the process. The load originally carried by these
longitudinals also has to bypass the crack, which will increase the stress intensity
factor, and therefore, reduce the arrest capability of the strakes. In other words
Figures 14.4 and 14.5 may draw too optimistic a picture of the effect of strakes.
The figures show that crack arrester strakes as used presently have little effect.
If a crack would occur anywhere in the deck plating in Figure 14.3, it could
essentially run through the entire deck (e.g. 60 ft) if arresters were present only
at the gunwale.
For a 60 foot crack size, the residual strength of HY-80 is 5.2 ksi; arrest in
general then would be possible only at stresses below 5.2 ksi.

(e) Principle of hybrid arresters

For the discussion of load transfer arresters the reader is referred to Chapter 9.
A summary follows below. Consider a plate with arrester stringers and a central
crack shown in Figure 14.6. Since there are only small stress gradients, the
nominal stress in the arrester is equal to that in the plate. As such the arresters
are effective in sharing the load on the uncraked structure and they can be
accounted for in the sizing of the structure for the Standard Design Wave. In
452

(a) ,

Crack Size, a

(c) Crock Size, a

~

StrInger: a=~
fs L

Stiffened plate:
K

I
',, /unstiffened plate:
............. ~ K

(d)
,-----__
, -- "f=-:J;O

Crack Size, a

Figure 14.6. Principle of stringer arrester (see Chapter 9).

other words, they are not just an addition for the mere purpose of fracture
arrest. In the absence of arresters, /1 = I. The load originally carried by the
cracked material has to bypass the crack. Some of this load will be transferred
to the arresters which will cause a reduction of the stress intensity factor: /1 < I.
(Chapter 9). This causes the arrester stress to increase to af = La, where L
increases with crack size.
If the plate material has a toughness K" the residual strength of the unstif-
fened plate follows from aF = Kc//1.fiW. with /1 = I (dashed line in Figure
14.6). The residual strength of the plate with arresters follows from the same
equation with /1(a) < 1. Since /1 decreases with crack size, the residual strength
curve shows the typical relative maximum. The nominal stress at which arrester
fracture occurs is obtained as afa = Flu/L, where FlU is tensile strength of the
arrester material. See Chapter 9 for more details and a discussion of the arrester
critical case.
The residual strength diagram is determined by /1, L, Kn and FlU (of the
 453

stringer material), where 13 and L depend upon arrester spacing and stiffness,
and upon the fastening system. They are also affected by plastic deformation in
arrester and fastener holes. All these effects can be accounted for in the calcu-
lation. If the fastener spacing is smaller, load transfer is more effective, which
results in larger L and lower 13. Indeed, 13 is lower for welded arresters, but in
that case the fracture can readily enter the arrester and a failed arrester is worse
than none at all (Chapter 9).
A hybrid arrester is a combination of a strake and a load transfer arrester
(Figure 14.7). The residual strength diagram of a strake arrester is shown
schematically in Figure 14.7a, that of a stringer arrester in Figure 14.7b.
Combining these leads to the residual strength diagram of the hybrid as in
Figure 14.7c. The effect of the strake is a further increase of the relative
maximum.
For a mere assessment of their effectiveness the use of handbook solutions
[18,8] for 13 is satisfactory (Figure 14.8). This case is for a fastener spacing equal
to 1/12 of the bay width; smaller fastener spacings would give somewhat better
results. The discussion will be for riveted hybrids; welding, if performed in a
certain way, is feasible as will be discussed. The value of 13 depends upon the
stiffening ratio.
Ship plating is already stiffened by stringers. These longitudinals are relatively
light, but more important, will be penetrated by the fracture because they are
welded to the plate. Unlike riveted stringers they are not effective as arresters.
Special heavy stringers or girders are required. In order to avoid confusion
about the use of the word stringer, they are called arrester in previous paragraph
and in the following discussion. When the arrester must come into action, the
'hatched' plate and stringers in figure 14.9 will all be broken. The broken
stringers cause an increase in 13, depending upon their size and spacing. As a first
approximation, an increase of 13 by a factor 1.2 was assumed on the basis of
calculations for broken stringers [18]. This factor may be somewhat low, but the
13 from Figure 14.8 is too high, because the fastener spacing for the large bay
widths considered here will certainly be less than 1/12th of the bay width. Thus,

\
Max \
---\---------

__ ....
arrest O"amox
, I

I ' ......... :
£: stress,

---
CT
amox
I
I I
I ... I
I ......... I I
I . . -t----- --+-
O'min 0amal °amOl DornaK

Crack Size, a Crack Size, a

(a) (b)
Crack Size, a
(c)

Figure 14.7. Principle of hybrid arrester. (a) Tough strake; (b) Stringer or Doubler; (c) Hybrid.
454

5=0.05

0.11

.5
0.43

00

.1

o .1 .5 a
Ii'
Figure 14.B. Geometry factors.

errors in the above assumptions will compensate one another. As the broken
stringers in Figure 14.9 are part of the total load path interruption, they
must be accounted for in arrester stiffening ratio. In order to avoid complica-
tions due to curvature and hydrostatic pressure flat deck plates were considered
only. For an assessment as aimed at here, it is sufficient if f3 is within about 10%
of the actual value; for actual design more rigorous procedures [9, 12] can be
used. Plate critical conditions were assessed only; because of the high F tu of
HY-80 and HY-130 arrester critical cases will be rare.
(1) Effectiveness of hybrid arresters
The performance of two hybrid arrester designs is shown in Figure 14.10.
Design B has a crack arrest capacity of 14 ksi. In order to obtain the same crack
arrest stress by means of HY-80 strakes only, the strake spacing would have
to be 50 inches (Figure 14.5), whereas the hybrid arresters could be spaced at
160 inches (Figure 14.10). If the same design (B) were built of HY-130, the arrest
capability would go up to 18 ksi (Figure 14.11).
An example may show how the design of arresters would proceed. Let the
problem be that of designing arresters with a spacing of 300 inches for a required
arrest stress of 15 ksi. The two material candidates are HY-80 and Hy-130.
Figure 14.8 is assumed applicable, but for an actual design more specific curves
are needed [9, 12]. It is assumed that the general design specifies the deck plating
to be 20.41bs/sq ft with deck stringers of 1O.21bs/ft spaced at 30 inches (ship
plating is specified in weight per foot). According to Figures 14.8 and 14.9, the
minimum value of f3 occurs at alb = 0.6. Therefore, the arrested crack size
should be a = 0.6 x 300 = 180 inches. It is assumed that the broken stringers
 455
A- Due 10 broken slringer accounled for by 0=1.2
B - Due to intact arrester accounted for by 13

I!iI"M''''J:'''''IIIIWc:;:;=o*
\
, 12ZZZZZ211 = broken material
\
\
\\ Arrester with stringers
lca
\\\\ / "c: a13KFa

\'t:--.
\~A,
T ; ' : I a : e ;;:: stringers
c a~
'~>:t" I //ArreS~~:QWlthout stnngers
" . . .: 1-~-
............ ---- u c = Jr;O

Crack Size, 0

b-----------I

\
\
\
\ Arrester

--:>',~::~~~/
ilegular stiffening ratio: S: ~:
AQ + Wi
Arrester (stiffening) rolio: So: nAs + Bb
°amax
Crack Size, a

Figure 14.9. Effect of deck stringers. Top: increased K due to broken stringers. Bottom; redefinition
of arrester ratio.

account for a factor of 1.2 as before. In order for this crack to be arrested, the
stress intensity K = 1.2{3(JJ1Ui, must be less than the toughness
(Kc = 175 ksiFu for HY-80). Since' arrest must occur at (J = 15 ksi, the
required {3 follows from:
175
{3 == 0.41.
1.2(JJ1(i 1.2 x 15 x ~n x 180
For {3 = 0.41 at alb = 0.6, the required stiffening ratio is S 0.43, according
to Figure 14.8.
456

---Arrester spacing, 160-in

----- -------jJ---,I "
_____~~~-~:~A-L--~~~---
--
,::........ I .................................
.......... '1--. _: Design B

: i----
I I
I I
I I
I I
I I

CroCk Size, a, Inch

o 40 BO 120 160 200 240 2BO 320 360 400

Crack Size, 20, inch

Figure 14.10. Comparison of two arrester designs in HY-80 .

.--j~-------Arrester spacing, IGO-in - - - - - - - - - - - -

301r-------~--------------

25
~9F==~==~~~~~~==~~
'I
23.4#HY-130
20
i 25 -in--!
r-----'-'A"'rr.:::es"'~e:;;o"~'!:~=~C.,in,,g"-,

-~=~~~~~~~~~~~~~~~:~~~~~-------
. ."':...----I---~I --- -- -- - ...... _ I I Design C
1---_-1
I 1------
I I
I I
I I
I I
I I
I I

Crock Size, a, Inch

I
o 40 BO 120 160 200 240 2BO 320 360 400
Crock Size, 20, Inch

Figure 14.11. Comparison of two arrester designs in HY-130.

 457
30

25
20# 20.4 # HY-130

- - - Arrester spacing, 300-in.

20
_=-L---I_ i~trl±J1~:
20#
R4::HY-130

30-in.

Welded Doubler

10
r
Equal arrest capability
HY-130 __ -
HY-BO--
;;;
...:'

O~--~----~----~----~--~----~----~--~~--~
o ~ ~ ~ ~
Creck Size, c, inch

o 50 100 150 200 250 300 350 400 450

Crock Size.. 20. inch

Figure 14.12. Two designs with equal arrest capabilities.

The deck plating being 20.41bs/sq ft and the spacing between arresters 300
inches, the total weight of the plating is 20.4 x 300/12 = 510 lbs/ft. Attached
to this are nine stringers of 1O.21bs/ft with a total weight of 921bs/ft. Hence, the
weight of the cracked bay is 510 + 92 = 6021bs/ft. With a required arrester
stiffening ratio of S = 0.43, the required arrester weight is 0.43 x 602 =
2591bs/ft.
The arrester width is taken as e.g. 30 inches. The arrester strake is of equal
weight (equal thickness) as the deck plating and the arrester doubler has the
same width as the strake. If the arrester is provided with two stringers of 20 Ibs/ft
(40Ibs/ft total), the doublers have to provide 259 - 40 = 2191bs/ft. Then the
doubler plating should be 219/30/12 = 881bs/sq ft. Since such heavy plating
might be impractical, two doublers each of 44lbs/sq ft could be used. Given that
S = 0.43, the p-values for other crack sizes can be derived from Figure 14.8 and
the residual strength calculated from (fIr = K e /(1.2PJ1W,). The residual strength
diagram is so obtained (Figure 14.12). Also the design for HY-130 is shown
(Ke = 220 ksiJffi).
The example shows that the design procedure is straightforward if the
required arrest stress is specified. Almost any arrest requirements can be met,
but the resulting design may be expensive and impractical.
The analysis is somewhat crude, but is meant only as an assessment of
feasibility. As such it shows that hybrid arresters are feasible. Required spacings
458

would be on the order of 200-300 inches for an arrest capability of about 15 ksi,
in accordance with the criterion discussed (once per year wave at LAST). At
temperatures above LAST the arrest stress would be higher. In the examples, the
load tranfer part of the arresters is riveted to the strake. This would prevent the
fracture from entering into the arrester and from rendering it useless. However,
welding the doubler plates to the strakes along the edges of the doubler maya
viable solution, as the filled weld (Figure 14.12 insert) would provide relatively
little opportunity for the fracture to enter the doubler. In addition, the weld
would provide small (zero) 'fastener' spacing which would reduce p, which can
be exploited to increase arrester spacing or to increase the arrest stress.
However, the fillet weld must have sufficient shear strength to permit load
tranfer from the cracked plate to the doubler (Chapter 9). A doubler plate is
necessary. The required stiffening should not be effected by using just a thicker
strake because the running fracture would then sever the reinforcement.

(g) Dynamic effects

Dynamic effects may be due to (1) the material properties, (2) the stress
intensity, and (3) the kinetic energy. The material properties (including
toughness) are affected by the rate of deformation. If the dynamic toughness
(fast moving fracture) is known, its effect can be accounted for. Also the stress
intensity is different than in the quasi-static case, but at fracture velocities
observed in steels, the effect is small. However, the contribution of kinetic
energy could change the conclusions reached.
The rate of displacement, v, of the material at each side of the fracture is very
high and proportional to the fracture speed d: the material has a kinetic energy
112 mv2 = I12Cmd2, where m is the specific mass, and C a proportionality
factor depending upon location. The elastic energy release rate during fracture
is G, the energy required for fracture R (Chapter 3). The surplus (G - R) is
converted into kinetic energy, so that
a2 1
J (G - R) da = J "2Cmd2 d V,
~ v
where V is the volume of material involved. Since G and R are functions of a, the
equation can be solved [12] to obtain the fracture speed d.
When the fracture is to be arrested, this kinetic energy has to be dissipated
(the displacements MUST come to a halt). Since the kinetic energy enters the
energy conservation equation, fracture must be analyzed in terms of the energy
balance. It can be shown (Chapter 4) that the elastic energy release per unit
fracture extension, da, under constant stress is equal to G = K2 IE = np2 (J2 alE.
The fracture energy per unit fracture extension is R = K~ IE. It will be assumed
here that R is approximately constant, and it will be shown later that this
 459

".5TRAKE...
WIDTH

(a) f--f!,.a

G.R

---
Rms
G

f-~f!,.a
(b)

1lOF
G.R r--
PLASTIC
ENERGY
ARRESTER

I-- R.r

~
'--Rms

--
G

--- ---
(e) r-~-I\a

Figure ]4.13. Dynamic effect. (a) Arrester strake of insufficient width; (b) Strake of sufficient width;
(c) Hybrid.

assumption does not affect the conclusions for the case of hybrid arresters.
If the structure contains a crack arrester strake of a high toughness material,
the fracture energy (K~ar/ E) suddenly increases as shown in Figure 14.13a.
Assuming that f3 = I, the energy release G is represented by a straight line. The
kinetic energy, (G - R) da, is represented by the area between the G and R
lines. When the fracture enters the arrester this kinetic energy is given by the
area ABC (Figure 14.13a). At that time the energy release rate (driving energy)
is less than the fracture energy. In other terms: G < Rar , which means K2/
E < K~ar / E, or K < Kcar . The latter criterion was used in the foregoing analysis,
460
but this condition may not be sufficient for arrest.
The material around the crack is still displacing at a high rate and 'contains'
a kinetic energy represented by the area ABC. This energy may no'Y drive the
fracture. While fracture continues, G increases to point D where G > Ran so
that fracture is driven again by G; there is no arrest. While the fracture
propagates from C to D, the energy release is smaller than the fracture energy,
(G < Rar)' The difference, Rar - G, has to come from the kinetic energy.
Therefore, the area CDF represents the part of the kinetic energy used for
fracture from C to D. Since CDF < ABC, arrest does not occur; the fracture
slows down only, because of the loss of kinetic energy (CDF), but it continues
because G > Rar at point D. For the fracture to be arrested, all kinetic energy
must be used (displacements must stop). This means that the arrester strake
must have sufficient width, as illustrated in Figure 14.13b. Because area
CHLM = area ABC, all kinetic energy would be used (all displacement motion
stopped) and arrest would occur at H.
The case of a hybrid arrester is depicted in Figure 14.13c. Due to the load
transfer to the arrester f3 decreases with increasing a. Therefore the G-line
(G = nf32(12 a /E) slopes down as shown. Without dynamic effects arrest would
occur at C. This was the arrest condition used in the foregoing analysis (G < Rar
or K < Kcar ), but if kinetic energy can be used the fracture may run through the
strake as before. However, in the case of a hybrid arrester there is a much more
effective energy dissipator than the high toughness trake, namely the doubler
plate with attached stiffeners (both uncracked). If fracture were to continue
(large local displacements), the doubler and stiffeners would have to undergo
large strains in the region crossing the fracture path. The plastic deformation
energy of this uncracked material constitutes an effective increase in the fracture
resistance R, as shown in Figure 14.13c.
As a result, the hybrid arrester can easily absorb all kinetic energy. Consider
for example, arrester design B of Figure 14.11. The energy diagram for this
arrester is shown in Figure 14.14. Using Kcar = 174ksiFu for HY-80 as in the
foregoing, provides Rar = K~ar/E = 175 2 /30000 = 10201bs/in, as shown in
Figure 14.14. According to Figure 14.11 the maximum static arrest stress for this
design was 15 ksi; the G-line for this stress is shown in Figure 14.14, based upon
the appropriate values of f3 and with the factor 1.2. The static solution predicts
arrest at A as discussed. For the estimate of the plastic deformation energy that
can be absorbed by the doubler and stringers, it is conservatively assumed that
only one inch (length) of these takes part in the deformation. Also assuming,
again conservatively, that the fracture strain is 30% (uncracked) and that there
is no strain hardening, the 0.5 inch (thick) doubler can absorb 110001bs/(inch
width), which is the effective R of the doubler. The effective R of the stringer can
be estimated in the same manner, bringing the total effective R of the hybrid
arrester to at least 150001bs/(inch width). Thus, the kinetic energy can be
 461

15.--------Energy absorption --~---,---,

of 10101 arrester
ppr Inch width
14

13

12 Energy absorption ---

per inch width
of doubler

10

At arrest stress of 15 ksi.

3
G ~0I2i32 .. a!t
2 E
A

110

Crock Size, 0, In. Crock Extension,

60, In.
Figure 14.14. Energy diagram for arrester design B of figure 14.10.

absorbed; only about 40% of the energy absorption capacity is used when arrest
occurs. Since the estimate of the energy absorption capacity was low, the above
numbers are conservative.
The example shows that the dynamic effect in the case of hybrid arresters is
negligeable. In fact, the statically calculated arrest stress of 15 ksi still stands. It
also is obvious now that the assumption of constant R has little effect upon the
conclusion. Whether R is increasing or decreasing, it remains small in
comparison to the effective R of the load tranfer arrester. The example also
illustrates why quasi-static crack arrester analysis is in excellent agreement with
experimental data [9-12], and why it is used so successfully in aircraft design.

(h) Fracture control considerations

Strake arresters as presently used have a limited arrest capability, but hybrid
arresters can be effective. The cost of the latter may be a drawback. Considering
costs of material and additional welding, it is estimated that hybrid arresters
might increase cost of a ship by 1 to 1.5%. Iffewer than lout of 100 ships break
in two or have severe fractures (cost of fracture less than one percent of that of
462
a ship), it may not be economical to spend 1-1.5% on fracture control.
Design of arresters and arrest capability must be based upon an arrest
criterion to be specified in the design rules. As discussed it might not be rational
to require arrest at the standard design wave at LAST, because of the extremely
low probability of the simultaneous occurrence. A more realistic criterion may
be similar to the one used here.

14.4. Piping in chemical plant (fictitious example)

(a) Problem definition and fracture control objective

Two vessels operating at low pressure are connected by an annealed austenitic
stainless steel pipe with an elbow (Figure 14.15). The plant consists of a series
of such arrangements. The temperature varies from 350F to 200F in the last one.
Stress corrosion cracking may occur at the welds in the piping.
A leaking pipe can be tolerated. A pipe break would cause interruption of the
flow and could result in dangerous overheating further down. The fracture
control objective is to ensure that pipes will never be completely severed.

(b) Loads, stresses and data

The pressure is low; there is no significant stress d.ue to pressurization. Essen-
tially, the loading is due to thermal expansion. As the process is continuous, the
temperature of the pipe is constant and hence the thermal stress is constant.
Thermal expansion of one of the legs of the elbow has to be accomodated by
bending of the other leg (Figure 14.16). This introduces a load P and a bending
moment M. Because the legs are symmetric the angle of deflection is zero at the
elbow. The loads, moments and stresses are as shown in Figure 14.16. The
highest moment occurs at the welds closest to the vessels. If it is assumed that
the pipes were installed at room temperature (70 F) then the maximum
AT = 350 - 70 = 280 F. Substiting this and other relevant numbers in Figure

l \leak

Figure 14.15. Piping between vessels.

 463

-J>l.... i
------, At
!It ' l

/!J.t="'~Tf
a = 10 x 10-'/' F

- p

PI 2 Mel 1
rp = 2EI - Ei = 0 or Me = 2 PI (rp = 0 at x = f)

The deflectionf at x = I is equal to the thermal expansion of the other leg. which is I'll = a/lT/.

f = PI] _ Me /2 = rx/lTI
3EI 2EI
at Weld I:
I
6rx/lT EIjl - 12rx/lT EI 6/ 2 4rx/lT EIjl

at Weld 2:
111
Mw2 = PI- Me 12rx/lT EI 12/2 - 6rx/lT EIjl 5rx/lT EIjl

elastic bending stress

Mw D 5 5
(Jb = 12 2 rx/lT EDjl = 2 x 10 X 10- 6 x 280 x 30 x JOl x 6j36 35ksi

longitudinal compressive stress

- 12rx/lTEI
- 3.0ksi
~ (D 2 _ d2 ) 12
4

Figure 14.16. Stresses in piping.

14.16 provides Mw = 5 x 10 X 10- 6 x 280 x 30 x 106 X n/64(64 - 54)/

36 = 384000 in-lb, so that the acting stress is

(J = Mx ~o/[:iDri - Di)] = 64 x 384 x 3/[nW - 54)] = 35ksi

Austenitic stainless steel operating at 200 F to 350 F is very ductile so that

collapse may be the govering failure mode. Collapse conditions are evaluated in
Figure 14.17 (see also Chapters 2, 3 and 10).
464
Collapse strength data for the steel are provided in Figure 14.IS. The data are
for SIS-in plate, which is close enough to the 0.5-in wall thickness of the pipe.
Obviously, the plate fractures occurred at constant net section stress, with a
collapse strength of Feol = 50 ksi at 200 F and Feol = 30 ksi at 500 F. Therefore
it is estimated that Feol = 40 ksi at the operating temperature. Also, the J R curve
and stress strain data are given in Figure 14. I S.

(c) Analysis
The question is whether leaking pipes could cause complete fracture. Therefore,
the crack of concern is a circumferential through-the-thickness crack (Figure
14. I 7). The stresses at the crack location are due to the bending moment M and
axial compressive load P. The longitudinal compressive stresses due to Pare
small as compared to the bending stresses. Neglecting this compressive stress

crack 20:'

horizontal equilibrium
2f3 BFcol = (2n: - 2:x - 2f3) BF;ol or Ii (n: - :x)/2
(collapse) moment Meal around the line (i - ).)

f f
{I

Mcol = 2 BF,ol (R cos <p - R cos Ii)R d<p + 2 BF;ol (R cos f3 - R cos <p) R d<p
() {I

R (D + d)/4 = (6 + 5)/4 = 2.75in M,ol = 303(2 cos ~ sin :x)

Figure 14.17. Collapse conditions.
 465

will be conservative and simplify the problem. The collapse analysis is shown in
Figure 14.17. The residual strength can now be calculated for EPFM fracture
and collapse. Results are shown in the residual strength diagram of Figure
14.19. Collapse is prevalent (lowest). Through cracks up to 2a = 4 inches can
be sustained, without fracture, at the acting stress of 35 ksi.

(d) Fracture control plan

A through crack of a certain angle will form by penetration of a stress-corrosion
crack of equal angle. Stress-corrosion cracks will grow in the area of highest
tension stress during the normal (elastic) loadings; hence it is unlikely that a
stress-corrosion crack grows over an angle larger than 70 degrees before break
through. Thus it can be concluded that leaks and no breaks will occur. (Note,
however, that the bending moments should be determined at which surface flaws
would break through. These moments should be lower than the moments for
propagation of the through crack for leak-before-break under all circumstances,
see Chapter 9.)
Since leaks are expected to be readily apparent, no fracture control plan would
be necessary. Nevertheless, the following measures are implemented. (1) In-
stallation of leak detectors at all pipes. (2) Daily walk-around visual inspection
for leaks. Note: The above serves as an example. All numbers are fictitious.
Some aspects of the problem may have been overlooked.

14.5. Fatigue cracks in railroad rails

(a) Definition of the problem

Fracture due to fatigue cracks in railroad rails is a common cause of derailment
accidents in the USA. Reduction of the number of such fractures may be
f

;0
Z
E,O
Ul
Q

'S cr:
_:u

ac
.,'"
(ksi)
F _
I
'0
50
col
::G
40 I
Feor- 30
• 200 F
?G
20
;0
10

2 5

Figure 14.18. Material data (fictitious).

466

-- gr-
,-
2J
7

, -0
'"
E~

,CL
5:=:

40

3C

20

:C

0.5 1.0 1.5 2.0 2.5 3. 0 ~ 5 4.0 4.5

CRACK SIZE a (INCHES)

Figure 14.19. Residual strength diagram.

achieved by track maintenance, reduction of traffic (loads), or replacement of

rails after timely detection of fatigue cracks through periodic inspection. Such
fracture control measures can be effective only if adequate methods exist to
predict the rate of crack growth, which req uires knowledge of service loads, rail
stresses, and of crack growth properties of rail material.
A comprehensive program to address this problem funded by the US Federal
Railroad Administration [19-26], is outlined in Figure 14.20. Samples of rail
obtained from track of various railroads were used to measure material data,
including toughness and da/dN - AK,R, as well as residual stresses. Tracks were
straingaged to obtain loads and spectra. Tests, using simulated service loading
(train-by-train loading) were performed to verify crack growth integration
procedures. The objective was to arrive at a suitable damage tolerance require-
ment for railroad rails.
Rails are inspected for cracks by means of a car equipped with ultrasonic and
magnetic inspection devices running along the rail. Cracks in different directions
may initiate at various locations in the rail head; these are internal flaws of
quasi-elliptical shape. Some are in a longitudinal plane, either vertical or
horizontal, some in a transverse plane (commonly called detail cracks). An
example of the latter is shown in Figure 14.21 [23, 25].

(b) Loads and stresses

Track was straingaged (Figure 14.20) at several locations to measure load-spec-
tra for various kinds ofusuage [19]. Finite element analysis was performed [20]
 467

"'"rCIU4L CR:4CK GJb:tW'TW "M"CRTIf,.,S

RC41J)uA.L STRC~.$LS

Figure 14.20. Ingredients of analysis of cracks in railroad rails.

to obtain the stresses in the rail from the loads, and subsequently the stress
exceedance diagrams for different usage and locations in the rail head. A typical
exceedance diagram [21], shown in Figure 14.22, provides stress exceedances
during passage of 1 MGT (1 Million Gross Tons) of traffic. The diagram is
non-linear and asymmetric, which is typical for man-induced exceedance
diagrams (Chapter 6).
The stresses are given as ranges, /l.u, but since there is no live stress without
a load, these are excursions from zero. When a wheel passes the rail, there will
be downward bending, so that the stresses shown in the diagram are essentially
compressive stresses (some tension occurs when the wheel is further away due
to a tendency of the rail to be 'lifted' from the roadbed).
The question arises why fatigue cracks occur when the stresses in the rail head
are compressive. The answer lies in residual stresses (Figure 14.23). Continuous
'pounding' by the wheels causes plastic deformation of the upper layers (3 mm)
of the railhead, so that this layer is stretched (rolled out). However this layer,
468

Figure 14.21. Fracture of railroad rail as a result of detail crack. Courtesy TSC and Battelle.

being attached to the head, must remain of the same length, which requires a
(residual) compressive stress. Residual stresses must be in internal equilibrium,
so that, there will be residual tensile stresses in the rest of the rail head as shown
in Figure 14.23. The residual tension stresses are very high. Compressive stresses
 469

c
"-
::;:
b
<l
.,;
g-
o
0:

Exceedonces

Figure 14.22. Stress exceedance diagram for I million gross tons (IMGTM) of traffic.

COMPRESS I ON L! VE STRESS (CeMPRESS ION)

STRESS
RA1L HEAD

RESIDUAL SfRESS ~
ReSiDUAL - _ - - -
ENSjON

CYCLIC STRESS AT POINT A

OUE TO PASSAGE OF WHEELS

TIME

Figure 14.23. Residual stress and cyclic stress in rail head.

470
due to passing wheels will be superposed on the residual tension. Consequently,
the cyclic stresses are actually in tension as shown.
Analysis requires knowledge of the residual stresses. These were measured
[22] on samples cut from tracks with different traffic experience, which were
straingaged (Figure 14.24) and subsequently sectioned. Residual stress fields can
be reconstituted from the output of the gages after sectioning [22]. Various other
techniques were used for the same purpose. The residual stress field is a very
complicated tri-axial field with large gradients. Resulting tensile stresses are
shown schematically in Figure 14.25.
The detail crack (transverse) usually starts in a horizontal plane, with high
residual tension (Figure 14.26). This phase of cracking is called a 'shell'. If the
shell crack perturbates into the vertical plane it experiences increasing longitudi-
nal tension, so that the perturbation persists and the shell turns into a 'detail
crack' [23]. The stages of development are shown schematically in Figure 14.26.
(Other hypotheses have been advanced to explain the turning of the shell.)
The analysis problem concerns cracks of complex shape (13) in a stress field
with large gradients. By using a reference stress, the stress gradients can be
accounted for in 13 (Chapter 8), but the variation of 13 along the crack front must
be accounted for by dependent crack gro~th analysis for more than one
direction, because growth depends upon the changing crack shape (see surface
flaws; Chapter 8). The case is more complicated than a normal surface flaw as
there is no symmetry, and the analysis should be performed for several directions
simultaneously (instead of two for common surface flaws).

(c) Material data

Almost 70 rail samples were collected from different locations, and charac-
terized according to weight, year of production, chemical composition, and
mechanical properties. Crack growth data were obtained from all samples, for
various R-ratios, temperatures, and orientations.
Data for some individual rails and the scatter band of all samples are shown
in Figure 14.27. Data trends for room temperature crack-growth in LT orien-
tation are shown in Figure 14.28. The variability is substantial due to differences
among rails. This is true variability as opposed to apparent scatter as discussed
in Chapter 7. It will affect general purpose predictions; either statistical methods
must be used to deal with this problem, or predictions must be made for specific
(individual) rails.

(d) Model calibration and analysis

The objective called for a general procedure to predict crack growth in rails
under service loading. This is a complex problem of a quasi-elliptical embedded
flaw in a nonuniform stress field and a variable-amplitude stress history. The
effects of variable-amplitude loading (retardation) were evaluated for through-
 471

Figure 14.24. Strain gages on rail for measurement of residual stress. Courtesy TSC and Battelle.

the-thickness cracks subjected to simulated service stress histories. If crack

growth under these circumstances can be predicted, the model can be
generalized to include other complexities.
Stress exceedance diagrams (Figure 14.22) were used to generate semi-
random service stress histories, employing the procedures discussed in Chapter
6. A stress history with six different trains (Figure 14.29) was established. The
cycle content simulated actual trains according to length and weight. A mixture
of2-A I, 6-A2, 12-A3, 120-B, 20-C, and 10-0 trains was used. Cyclic stresses due
to passage of a car are shown in Figure 14.30, but simpler stress histories
(elimination of small cycle) are desirable for analysis, although not necessary
(for truncation see Chapter 6).
472

Rail
Head longitudinal _____-E::===aI
stresses

A. SHELL
B. DETAIL CRACK

Figure 14.25. Cyclic stresses on shell crack, A, and detail crack (transverse), B.

Contact
of
Wheels

1l-h1-H-\-- SHELL

---f-tI-i'r-t-.DETAIL

Figure 14.26. Schematic growth of shell and detail crack.

Load interaction (retardation) was negligible, and in view of the large varia-
bility in data, retardation was not included in the analysis. Predictions were
made using average rate data, and the actual rate data of the particular rail used
in a service simulation test. Some results and data are shown in Figure 14.31.
 473

IO·r-------------------------------------------~~----_,
o

~
~
~ 10-'
z Scatterband of
~ baseline data
c
"C

10 20 3Q 40 50 70
Stress Intensity Factor Range, L'lK, MPa.jm

Figure 14.27. Variability of rate data among rail steels.

'"
~
~
E
E
Z
~
.g

Stress Intensity, L'lK, MPa..rm

Figure 14.28. True scatter of rate data among rail steels.

474
Train A (extremely heavy)

II I

Itrr=J____
~

~,MI MInt nn
Train B (very heavy)

CYCLES

n_amrC_(_he_aVy_)t-~_ _
.,
i n-t-J Train 0 (medium I

R
ltD
Train E (medium)

~F(lf9ht)

CYCLES--_

Figure 14.29. Simulated train loading histories.

cor I truck

I truck I wheel

2 r- I wheel .J

If--~
o- I ,A

I-

-2 I-

f--

-
- A A
6 ,-
A
V
-7 f--

V
Stiff Rood Bed
Sf--

9f-- V
Soft Rood Bed
-I 0

Figure 14.30. Cycles due to passage of wheels, trucks, and cars.

Predictions were generally within a factor of 2. This may not be very accurate,
but the test data showed a similar variability. Obviously, if discrepancies are
caused by material variability, they cannot be blamed on the prediction method.
Examples of analysis results for actual rail cracks are shown in Figure 14.32.
 475
42.5 ....-----------=-Ra-n"""'d,...om----"S,...pe-c..,.tr-u-m-------------,
12 level o
o

E
E

25.0l.-_~ _ _..I.__.....l_ _...L_ _L__--L._ _.l-_.....l_ _..J....._ __.J

o 2 4 6 8 10 12 14 16 18 20
Million Gross Tons

Figure 14.31. Predictions of crack growth (solid lines) and test data under simulated train-by-train
loading.

0.8

..
.s
.!:!
(/)
0.7

0.&
High residual
stress

""ug
0 0.5

0.4

D.3

0.2

OJ

0
0 10 20 lSO 40 50 60 '70 80 90 100 160
MillIon Gross Tons
I I I I I
0 2 3 4 5 6 7 B
Years

Figure 14.32. Predicted crack growth in rails.

The significant effect of residual stresses, is understandable because all cyclic

stresses would be in compression in the absence of residual stresses (Figure
14.23).
476
(e) Fracture control
The use of the model (even with limited accuracy) does permit rational fracture
control. As an example, consider a railroad in which the number of failures is
considered unacceptable. The model can be exercised for the particular circum-
stances of that railroad, with a load spectrum for the prevailing traffic and track
conditions. The analysis can be run for different conditions, such as reduced
speed (loads), reduced traffic (load spectrum), and/or upgraded track and road
red (stresses). Each of these measures will increase (reduce) the life by a certain
factor. It then can be decided which measure is the most cost effective to obtain
a certain factor of life improvement. Since the actual failure rate is known, the
reduction factor can be applied to estimate the actual expected failure rate if one
of the above measures is implemented.
The absolute accuracy of the model is not important if the ratios are predicted
properly; the relative accuracy is significant; variability of parameters influenc-
ing crack growth is of no issue in relative assessments. Naturally, the absolute
life calculations can be used to determine inspection intervals. With procedures
as described in Chapter 11 to determine the interval, the variability can be
accounted for by using a safety factor, which will reduce the inspection interval.
Even though predicted crack growth has limited accuracy, the permissible
increase (or reduction) of inspection intervals upon certain maintenance efforts
could still be reliably analyzed, as they would be based on relative instead of
absolute changes in predicted life, academic objections notwithstanding.

14.6. Underwater pipeline (Courtesy SNAM) [27, 28]

(a) Definitions of the problem

A submarine pipeline is laid from a barge as shown schematically in Figure
14.33. At A a new section of pipe is welded on, previous welds are X-rayed at
B, protective coating applied at C and, to reduce buoyancy, a 1-2 inch concrete
coating is applied as well. Then the barge is moved so that the next section can
be built, the previous section being expelled. The line hangs down from the
barge (prevention of buckling in the sharp bend being an important issue) until
it hits bottom.
Free hanging spans may occur where the line encounters uneven bottom
(Figure 14.33). Current flow (streams or tides), around the pipe causes
turbulence (vorteces) as shown in the insert in Figure 14.33. Should the
frequency of the vortex shedding be about the same as the natural bending
frequency of a free span, bending excitation will occur, giving rise to cyclic
stresses. The latter may cause fatigue; cracks growing from defects in the
circumferential welds will eventually penetrate the wall. Repairs being
impossible, any leak constitutes a failure. Consequently, the life of the line is
governed by the time required for the growth of the largest possible weld defects
 477

ABC

,,- ---------

CURRENT

O ') .,:;)
VORTECES

SPAN SUPPORT

Figure 14.33. Laying of underwater pipiine.

to wall penetration. Analysis of the problem requires detailed information on

the amplitudes of vibration and consequent cyclic stresses, weld defect sizes,
crack growth rates in a sea water environment, and feasible fracture control
measures.

(b) Amplitudes and excitation times

First sonar measurements of bottom profiles were made to determine which
span lengths might occur. Next came the calculation of the natural bending
frequencies of all possible span sizes. Although this is trivial in principle, in
practice it is not. The bending frequency depends strongly upon the end-fixity
(Figure 14.33 insert). Depending upon the end support,and upon the (random)
length of adjacent spans, the effective fixity can vary dramatically. As the future
configuration is unknown, assumptions must be made.
VERIT AS, the Norwegian shipping and offshore bureau, provides a
procedure to determine vortex shedding frequencies as a functon of current
velocity. From these, and from the natural frequencies of the spans, the current
velocities causing excitation can be calculated. A typical result is shown in
Figure 14.34. This information is necessary but insufficient. For the calculation
478

SPAN
LENGTH \
\
\
\
\

r \
\ \
\ \

EXCITATION
HINGeD ENDS \
CLAMPED ENDS \
'\
'\
'\

10 100 1000
U CM/SEC

Figure 14.34. Excitation of span as a function of current velocity.

of stresses, the amplitude of the vibrations must be known. The response curve
can be calculated, provided the damping is known as well. Hydrodynamic and
material damping must be accounted for; many assumptions must be made. As
this is not a 'fracture mechanics' problem, further details are beyond the scope
of this book.
The above is a fine illustration of the arguments in Chapter 6 and 12 with
regard to accuracy of the loads analysis. So many assumptions are involved in
the solutions that the anticipated accuracy of the loads (amplitudes in this case)
is certainly not better than 10 to 15%. It also illustrates how engineering
depends upon assumptions. Vibration analysis is well founded, its rigor hardly
questionable. Yet, practical results and accuracy are governed by assumptions.
The same holds for fracture mechanics analysis. Deficiencies in the underlying
theory do not determine its accuracy, but the engineering assumptions and
judgements do.

(c) Stresses
The amplitude now being 'known', based upon reasonable assumptions, the
stresses can be calculated; but the stress history has yet to be determined from
the current spectrum. As spans of any size may occur anywhere, the current
spectrum had to be measured at many points along the projected line. Only then
was it be possible to determine the vibration response. With the information of
Figure 14.34, the current spectra can be used to establish the excitation spectrum
 479

for any span. Using estimated fixity and damping, the critical current velocity
and the maximum amplitude can be calculated. Vibration response curves show
that excitation will still occur at frequencies (current velocities) close to the
natural frequency, but the amplitudes are smaller (Figure 14.35). When this is
accounted for, the stress spectrum can be obtained as in Figure 14.36. As it is
not known which combinations will occur, this stress spectrum must be
obtained for a range of span sizes and locations, so that a total of at least
100-400 cases must be analyzed.
A simple formula for bending stress permits calculation of the cyclic stresses,
but crack growth is also affected by the mean stress (R-ratio; Chapter 5). The
mean stress can be calculated from the pre-tension in the line, which depends
upon the depth oflaying (vertical suspension from barge), bottom friction, and
so on. Again the problem is simple in principle, but assumptions must be made
to obtain a solution. The results are strongly dependent upon this assumption.

(d) Defect sizes

Weld defect sizes could be estimated on the basis of weld quality control
specifications (X-ray). But since the project was costly and the cost ofa fracture
beyond imagination, a more rigorous assessment of defect sizes was indicated.
A rational procedure was to cut many specimens from a large number of welds
produced on the barge under the most adverse (waves) conditions, but these
should have passed X-ray quality control. The specimens were fatigue tested to
failure. Examination of the fracture surfaces after failure revealed the defect that
initated the cracking.
A large number of such tests established the defect size distribution as in
Figure 14.37. In view of the severe consequences of a service failure, it was

U/Ucrit

L:::----- A/Arne.'

-............J 0.25
......... 0.50
1.1 ~ 0.75
1.0
0.9
------ - - -- -~ 1.0
"J 0.75
'\j 0.50
"- 0.25

Log
'\
(PERCENT TIME
U IS EXCEEDED)

Figure 14.35. Amplitudes of vibration as a function of current. velocity.

480

.6 1.0 1.4

Figure 14.36. Stress spectrum.

prudent to base the analysis on the largest possible defect size, but a sensitivity
analysis to assess the conservatism introduced by this assumption, is advisable.

(c) Material data

Defects may occur at various locations in the welds: at the outer surface
(exposure to seawater), close to the inner surface (exposure to gas), or in the
interior (vacuum). Crack growth rate data (da/dN-AK, R) were obtained for
all these cases. As the sea-water environment is likely to produce the fastest
growth, emphasis was on tests in sea-water, but the others were not discarded
a priori, because life also depends upon crack location (fJ).
Fatigue crack growth tests to obtain da/dN data were performed on welded
specimens (the cracks growing in the welds or HAZ), as cracks in practice will
start in the welds; the specimens were machined from pipe welded on the barge.
Typical data for crack growth in a seawater environment were shown in Figure
6.10. Schematically, the data can be represented as in Figure 14.38. A complica-
tion is that the rates depend upon cyclic frequency. Because the cyclic frequency
varies with span length, the data used in the analysis must be appropriate for
the frequency of the relevant span size.
 481

NUMBER

/' "-
I
I
I

,
I
\

I
I
,,
,

DEFECT SIZE

Figure 14.37. Weld defect sizes appearing from fatigue tests; Courtesy SNAM .

.!!!
dN
RI
I I
; 1/ R2
I /1
;
/ ;1
/ /1
r-~--r'7'/ /

tJ.K

Figure 14.38. Schematic representation of rate data (fixed frequency).

(f) Analysis and fracture control

Crack growth was calculated for a range of span sizes, span locations, and crack
locations, using a reliable computer program. In this case the crack growth
curves were not of great interest; only the growth life until penetration was
important. Examples of results are shown in Figure 4.39. Short spans are not
excited and have infinite life. Very long spans are excited but only part of the
482

LIFE

Nol
EXCITATION
- - INTERNAL DEFECT
- - - - SURFACE DEFECT
t
(SEAWATER)

\
\
\
\
\
\
\
\
,-
\ ,-
/
\ /
,~

SPAN LENGTH

Figure 14.39. Life of various spans. Courtesy SNAM.

time, and their frequency is so low that few cycles are accumulated. Thus, their
life (years; not cycles) is long. Note that the results may be quite different for
other current spectra, line diameters, coatings etc.
In this case intermediate span sizes show the shortest lives. Should this
shortest life (with adequate safety factors) be enough to cover the economic
service life of the line, no specific fracture control measures would be necessary.
Failing that, measures might be taken to reduce stresses, improve weldments,
and so on. A sensitivity study might indicate that not much can be gained by
such measures, so that other fracture control measures are indicated. Because
inspection for cracks is impossible, one alternative is to inspect the line (by
submarine) to find span sizes in the critical range. If any are discovered, they can
be eliminated by insterting an artificial support. New span sizes in the critical
range may develop (Figure 14.33) if the current washes away sand; should this
occur then the life would still be limited. Hence, future inspections (submarine)
would have to be scheduled at intervals equal to the life of the most critical span
(with safety factor).
It needs no emphasis that, owing to the assumptions, the results of this
damage tolerance analysis are not very accurate. Without damage tolerance
analysis no information at all would be available. If responsible for the line, in
which situation would be reader like to be: without any information, or with
that provided by an imperfect damage tolerance analysis with appropriate safety
factors? Obviously, the damage tolerance assessment is costly, because of the
necessary data generation, but is far outreached by the cost of fracture. It needs
 483

no further arguments that extensive finite element analysis would be out of place
in view of the many assumptions involved. Stresses in a pipe in bending and
tension can be determined accurately enough with conventional procedures;
errors due to the simple assumptions with regard to fixity will be much larger
than those introduced by stress analysis.

14.7. Closure

The foregoing examples demonstrate the power of fracture mechanics and

damage tolerance analysis, despite the fact that not all problems could be solved
rigorously. Further development of fracture mechanics within the realm of
engineering is worthwhile. So many assumptions are necessary, that concern
with the shortcomings of fracture mechanics per se seem naive.
The last example in particular shows, that all engineering analysis needs
assumptions; even 'rigorous' vibration analysis yields approximate answers
only. The situation is no different for fracture mechanics and damage tolerance
analysis.
Finally, claims that applications of fracture mechanics are more difficult for
certain structures are negated by Table 10.1.

References
[I] I. Fioriti, Fracture control of a naval destroyer, Private Communication.
[2] Anon., Rules for building and classing steel vessels, American Bureau of Shipping (1978).
[3] R.J. Mosberg, Behaviour of riveted and welded crack arresters, Final Report on SR-134 to Ship
Structures Committee, SSC-122 (1960).
[4] P. Romualdi et aI., Crack extension force near a riveted stringer, Naval Research Lab. Rept No.
4956 (1957).
[5] T. Kanazawa et aI., Study on Brittle Crack Arrester, Selected Papers from G of Soc., Naval
Architects of Japan 11 (1973) pp. 135-147.
[6] T. Kanazawa et aI., Fracture mechanics analysis and design of stiffener-type crack arrester,
Japan Shipbuilding and Marine Eng. (JSME) 3,6 (1968) pp. 10-19.
[7] M. Yoshiki et aI., Fracture mechanics analysis of stiffener-type crack arrester, JSNA 118
(1965); JSNA 122 (1967) and JSNA 124 (1968).
[8] c.c. Poe, The effect of riveted and uniformly spaced stringers on the stress intensity factor of a
cracked sheet AFFDL-TR-79-144, (1970) pp. 207-216.
[9] H. Vlieger, Residual strength of cracked stiffened panels, Engineer Fracture Mechanics, 5 (1973)
pp. 447-478.
[10] T. Swift, Development ofthe fail-safe design features of the DC-IO, ASTM STP486 (1974) pp.
164-214.
[II] T. Swift, Design of redundant structures, AGARD LSP 97, (1978) pp. 9/1-23.
[12] D. Broek, Elementary engineering fracture mechanics, Nijhoff, Fourth Edition (1986).
[13] I. Fioriti, Abbreviated hull structural design, Paper for Committee on Toughness Require-
ments for Materials in Weapon Systems (1978).
[14] T.C. Gillmer, Modern ship design, Second Edition, Naval Institute Press. Annapolis, Maryland
(1975).
484
[15] G. Buchanan et aI., Lloyd's register of shipping's approach to the control of the incidence of
brittle fracture in ship structures, Lloyd's Paper, No. 56 (1969).
[16] E.V. Lewis, Dynamic loadings due to waves and ship motions, Paper presented at Ship Structure
Symposium, Washington D.C. (1975).
[17] B.H. Barber et aI., Structural considerations in the design of the Solar class of coast guard
icebreakers, Paper presented at the Ship Structure Symposium, Washington, D.C. (1975).
[18] D.P. Rooke and D.J. Cartwright, Compendium of stress intensity factors, Her Majesty's
Stationery Office (1976).
[19] D.R. Ahlbeck et aI., Evaluation of analytical and experimental methodologies for the charac-
terization of a wheel/rail loads, Fed. Railroad Adm. Rept. 76-276 (1976).
[20] T.G. Johns, and K.B. Davis, A preliminary description of stress in railroad rails, Fed. Railroad
Adm. FRA-ORD-76-294 (1976).
[21] J.J. Groom, Residual stress determination in railroad rails. Dept of Transport. DOT-TSC-1426
(1979).
[22] D. Broek and R.C. Rice, Prediction of fatigue crack growth in railroad rails, SAMPE Nat.
Techn. Con! 9 (1977) pp. 392-408.
[23] R.C. Rice et aI., Post Service rail defect analysis, Battelle Rept. to Dept of Transp. (1983).
[24] O. Orringer et aI., Applied research on rail fatigue and fracture in the U.S., Theor. Appl. Fract.
Mech. I (1984), pp 23-49.
[25] O. Orringer et aI., Detail fracture growth in rails: test results Theor. Appl. Fract. Mech. 5 (1986)
pp 63-95
[26] O. Orringer, Rail testing: strategies for safe and economical rail quality assurance, TRB
symposium (1987).
[27] M. Celant et al. Fatigue characterization for probabilistic design of submarine pipelines, Corr.
Science 23,6 (1983) pp 621-636.
[28] M. Celant et al. Fatigue analysis for submarine pipelines, OTC paper 4233 (1982).
 CHAPTER 15

Solutions to exercises

15.1. Solutions to exercises of Chapter I

I. Material A: Safety factor of 3 against F tu ; allowable stress is 60/3 = 20 ksi.

Safety factor of 2 against F ty ; allowable stress is 50/2 = 25 ksi. Lowest is
applicable. Required cross section A = 1000/20 = 50 in 2 • A = nD2/4.
Required diameter D = 7.98 inch (8 inch). Ultimate design load:
P u = AFtu = 50 x 60 = 3000 kips.
Material B: Allowable stress lower of 50/2 and 80/3, i.e. allowable stress
25 ksi. Required area: A = 1000/25 = 40in 2 ; D = 7.l4inch;
P = 40 x 80 = 3200 kips.

2. Material A: Maximum service load is 1000 kip. Design load, Pu = 2500 kip.
Required area A = Pu/Ftu = 2500/60 = 41.7in2 • At 1.3 x 1000 = 1300kip
no yielding is allowed: Check (J = PIA = 1300/41.7 = 31 ksi, is indeed less
than F,y. D = 7.29 inch.
Material B: Area A = 2500/80 = 31.25 in 2 • (J = 1300/31.25 = 41.6 is less
than F ty (OK); D = 6.31 inch.

3. Material A, problem 1: Minimum permissible residual strength

Pp = 1.2 x 1000 = 1200 kip. In terms of stress (Jp = Pp/A = 1200/
50 = 24 ksi. Remaining safety factor j = 1200/1000 = 1.2.
Material B, problem I: Pp = 1200kip; up = 1200/40 = 30ksi,j = 1.2.
Material A, problem 2: Pp = 1200 kip, (Jp = 1200/41.7 = 28.8 ksij = 1.2.
Material B, problem 2: Pp = 1200 kip; (Jp = 1200/31.25 = 38.4ksij = 1.2.

4. Fracture is final event. Cracking by fatigue, stress corrosion, creep, etc.

5. Known points: a = 0, Ures ~ F tu = 60 ksi;

a = 2, (Jres = (Jp = 24 ksi (see Exercise 3);
a = D = 8, (Jres = 0 (D = 8, Exercise I).
Fair approximate curve through three points.

485
4~6

6. At 1500 kips stress is P/A = 1500/50 = 30ksi. Fracture at about a ~ 1.4

inch (estimated from curve in exercise 5). At 1000 kips stress is P/A = 1000/
50 = 20 ksi. Fracture at a ~ 3 inch (estimated from curve).

15.2. Solutions to exercises Chapter 2

I. k[ = I + 2-./2/0.4 = I + 2)5 = 5.47;

ka = 5.47, ke = 5.47 (elastic).
Full yield no strain hardening; ka = I; kake = k~; ke = 5.472 /1 = 29.94.

2. Since ka = I net stress has become approximately uniform and equal to F,y,
i.e. 50ksi; fracture load: (W - 2a)BF,r = (12 - 4)0.5 x 50 = 200 kips. In
second case at collapse uniform -net stress approximately 75 ksi; fracture load
300 kips. Nominal stress in the two cases: (I) 200/(12 x 0.5) = 33.3 ksi (2)
300/(12 x 0.5) = 50 ksi.

3. Circular hole k, = 3.
Elliptical notch k[ = I + 2-./0.8/0.1 = 6.66.

4. Before hole drilling k[ = I + 2JfifO = 00. After hole drilling

k[ = I + 2-./2a/d (p = 1/2d).

5. Plane strain: yielding starts at about 3F,y , i.e. stress at yielding is

3 x 30 = 90 ksi, a little away from notch. At free surface of notch root there
is plane stress «(T, = 0); yielding starts at (T = F,y = 30 ksi.

6. With k[ = 4, elastic stress (just before yielding) at crack tip is 30 ksi. Nominal
stress = 30/4 = 7.5 ksi, i.e. yielding at notch tip begins at nominal stress
7.5ksi.

7. No numerical answer.

8. No numerical answer.

9. (a) Center crack.

Linear curve from (a = 0, (Tfc = Fcol = 350 MPa) to (a = W/2 = 300 mm,
(Tfc = 0). With crack of 150mm; (Tfc = (W - 2a)Fcol/

W = (600 - 300) x 350/600 = 175 MPa.

h. Edge crack.
Linear curve from (a = 0, (Tfc = 350 MPa) to (a = W = 600, (Tfc = 0) (con-
strained edges). With crack of 150mm: (Tfc = (W - a)Fcol/

W = (600 - 150)350/600 = 262.5 MPa.

 487

15.3. Solutions to exercises of Chapter 3

1. Stress at fracture 300/(20 x 0.75) = 20 ksi; f3 ~ I (Figure 3.3). Tough-

ness = Kfacture = I x 20 x FX1= 35.4ksi.JID. With Fry = 70ksi for
collapse, the nominal stress at collapse would be (20 - 2)70/20 = 63 ksi; since
fracture at 20 ksi collapse did not occur. Hence, calculated toughness is useful.
For plane strain B > 2.5 (35.4/7W = 0.64inch. Since B > 0.64in, the above
toughness is plane strain by ASTM criterion. Plastic zone rp ~ (35.4/70)2/
6n ~ 0.014 inch.

2. W = 5 inch; a = 2 inch; f3 = 2.1. (Jfr = 35.4/2.1J1CX2 = 6.73 ksi.

(Jic= (5 - 2) 70/5 = 42ksi > (Jfr; (Jres = 6.73ksi. W = 6 inch, a = 0.5inch;
f3 = 1.16; (Jfr = 24.4 ksi; (Ife = 64 ksi; (Ires = 24.4 ksi.
3. W = 20; 2a = 3; a = 1.5; f3 = 1.01; (Ifr = 31.9ksi; (Ife = 63.7ksi > (Ifr;
(Ires = 31.9ksi. W = 2; 2a = 1; a = 0.5; f3 = 1.19; (Ifr = 46.9ksi;
(Ife = 37.5 ksi < (Ifr; (Jres = 37.5 ksi (collapse).

4. (Ires = 34.2 ksi; f3 = 1.07; calculated K at fracture:

1.07 x 34.2",1nX"l = 64.9 ksi.JID. Collapse would occur at:
(W - 2a)BFry = (6 - 2) 0.2 x 50 = 40 kip. Hence, fracture occurred by
collapse (net section yield). Above K value is not the toughness. Equation (3.25)
cannot be used because whole section is yielding; plastic zone is equal to
remaining ligament.

5. With Equation (3.28): KQ = 33.97ksi.JID.

With Equation (3.29): KQ = 34.18 ksi.JID. Same answer as it should be;
KQ = 34 ksi",lln. Valid because B > 2.5 (34/80)2 = 0.45 inch (actual 1 inch);
KIc = 34 ksi.JID.

6. For plane strain B > 2.5 (50/100)2 = 0.625 inch. Thickness is less; therefore
no plane strain. Toughness for 0.5 inch thickness is Ke = 50 ksi.JID. Plane strain
toughness is less. Specimen measures plane strain if 0.5 > 2.5 (Kle (100)2, i.e. if
KIc ~ 45 ksi.JID. Hence, actual toughness is larger than 45 ksi.JID otherwise test
would have been valid. Hence, 45 < K le < 50. Estimate KIc = 47.5 ksi.JID.
(accurate within about five to six percent even if toughness is as low as 45, which
is lowest possible). Highest possible 50 kSi.JID; required specimen thickness
0.625 in or larger.

7. Center crack f3 = 1.1; (Ifr = Kc/1.1.jn x 0.75 = Ke/1.69. Edge cracks

f3 = 1.13 (Figure 3.3); (Ifr = Ke/1.13.jn x 0.6 = Ke/1.55. Whatever Ke (same
488

for both cracks) fracture stress at center crack is lower; failure should occur at
center crack.

8. Case a: O"p = 58 ksi; ap = 0.6 inch (on tangent);

O"p = 20 ksi; ap = 5.1 inch (O"/r from Kc);
O"p = 10 ksi; ap = 8.8 inch (O"/r from Kc);
Case b: O"p = 58 ksi; ap = 0.5 inch (collapse);
O"p = 20ksi; ap = 2.14inch (collapse);
O"T = 10 ksi; ap = 2.57 inch (collapse);
Case c: O"p = 58 ksi; ap = 0.48 inch (O"/r);
O"p = 20ksi; ap = 2.7 inch (O"/r);
O"p = 10ksi; ap = 4.4 inch (O"/r)'

15.4. Solutions to exercises in Chapter 4

la. A = nd2 /4 = n(0.4?/4 = 0.126inch 2• 0" = PIA. Draw straight line SI to

determine E. For P = 9.5 kip, c5 = 0.01 inch. Hence, 0" = 9.5/0.126 =
75.4ksi. c5 = sL, so that s = c5/L = 0.01/4 = 0.0025, hence, E = 75.4/
0.0025 = 30000 ksi.
FlY is defined as stress for plastic strain of 0.2%, i.e. sp = 0.002. This means
c5p = 0.008. Draw line S2 parallel to SI starting at 0.008 and intersect with curve;
read load P = 5.67 kip; FI , = 5.67/0.126 = 45 ksi. P max = 7.56 kip; FlU =
7.56/0.126 = 60 ksi. From p'oints A through H read from curve and work out
(Stot = c5/L):

Point P Pd (J Btot 0" = (JIE epf = Btot - eel

A 5.67 0.014 45.0 0.0035 0.00150 0.00200

B 6.60 0.05 52.4 0.0125 0.00175 0.01075
C 6.8 0.1 53.9 0.0250 0.00180 0.02320
D 7.0 0.2 55.6 0.0500 0.00185 0.04815
E 7.3 0.3 57.9 0.0750 0.00193 0.07307
F 7.4 0.4 58.7 0.1000 0.00196 0.09804
G 7.5 0.5 59.5 0.1250 0.00198 0.12302
H 7.56 0.6 60.0 0.1500 0.00200 0.14800

1b. Plot 0" - Stot to obtain stress-strain curve. Plot log 0" - log Sp to obtain F
and n by determining slope and intercept from straight line through data points.
E = 30,000 ksi, n = 16.2, F = 3.9 X 1029 ksi A 16.2, depending upon how
you draw line.

lc. 0"0 = 100 ksi, So = 100/30000 = 3.33E - 3;

(3.33E-3 x 3.9 x 1029 ) = 193222
 489

Id. CTO = 50ksi, eo = 1.67E-3, IX = 5.12.

Ie. CT = 50ksi for CTo = 100ksi:

e = 50/30000 + 50 16.2/3.9 X 1029 = 0.0102
e/eo = 50/100 + 193222(50/100)162 = 3.067
e = eo x 3.067 = 3.33E-3 x 3.067 = 0.0102
CT = 55 ksi for CTo = 100 ksi:
e = 55/30000 + 55 162 /3.9 X 1029 = 0.042
e/eo = 55/100 + 193222(55/100)16.2 = 12.57
e = 12.57 x 3.33E-3 = 0.042
CT = 50 ksi for CTo = 50 ksi:
e/eo = 6.12
e = 6.12 x 1.67E-3 = 0.0102
CT = 55 ksi for CTo = 50 ksi:
e/eo = 25.08
e = 25.08 x 1.67E-3 = 0.042
Note that results for CT = 50 and 55 ksi are identical (CTo can be selected arbitrari-
ly). Check results in your CT - e diagram.

2. Reminder: CT true = CTeng (1 + eeng) and etrue = In (1 + eeng); Same procedure as

exercise I with different results (except for modulus E).

3. At P = 7.4kips the plastic strain is 0.009804. Hence, the bar has become
thinner. Constant volume requires that (L + .1.L)A = LAo or Ao = A(1 + e)
so that A = Ao/(1 + ep ) = 0.126/1.09804 = 0.1l5inch2 •
This new bar will yield at P = 7.4 kips, the cold worked material's Fry =
7.4/0.115 = 64.3ksi. The fracture load is 7.56 kips so that F ru = 7.56/0.115 =
65.7 ksi.
This is the common way to strengthen materials by cold work (e.g. cold
drawing).

4. J R = 8.72 X 5017.2 x 2/3.9E29 = 7.46 kips/inch.

CTtc = 28 x 55/32 = 48.1 ksi « 50 ksi; apparently collapse. J R value not
reliable).

5. CTtr = [(3.9E29 x 7.46)/(13.5 x 3)]1.'17.2 = 46.6 ksi.

CTtc = 18 x 55/24 = 41.3 ksi ( < 46.6 ksi; collapse at 41.3 ksi).

6. a/W = 0.125; a = 20 x 0.125 = 2.5 inch.

Case I: J = 193222 x 100 x 3.33E-3 x 0.75 x 2.5 x 4.13[CT/0.75 x
100]'7.2 = 2.77E-27CT 17.2 = 2; hence CT = 36.4ksi.
Case 2: J = 5.12 x 50 x 1.67E-3 x 0.75 x 2.5 x 4.13[CT/0.75 x
490

50]17.2 = 2.80E - 27 (117.2 = 2; hence (1 = 36.4 ksi.

Same result in both cases, independent of choice of (1o, because Co = (1o/E and
()( is adjusted accordingly.

7. Note that J = 2A/bB, b = W - a, P-!5-curve horizontal after rising part.

a t'la P <5 oM t'lA A tot b JR

0 5 0.15 0.15 t x 5 x 0.15 = 0.38 0.38 1.00 1.52

1.05 0.05 5 0.20 0.05 5 x 0.05 = 0.25 0.63 0.95 2.65
1.15 0.15 5 0.27 0.07 5 x 0.07 = 0.35 0.98 0.85 4.61

Plot J R versus t'la for J R curve.

15.5. Solutions to exercises of Chapter 5

1. da/dN ~ l1a/I1N = (1.05 - 1)/7000 = 7.14 x 10- 6 inch/cycle.

2. Kmax = 12.Jn x 0.5 = 15 ksi.jffi; 11K = (I-R)Kmax = 0.8 x 15 =

12ksi.jffi. da/dN = lE-8(12)2(15)15 = 8.4E-5inch/cycle.
Assuming K does not change over smalll1a = 0.51 - 0.5 = 0.01 inch then
da/dN remains constant. Then I1N = 0.01/8.4E-5 = 120 cycles.

3. C = 7.83E - 13; m = 4.54 MPa.jill, m/cycle;

C = 4.28E - 11; m = 4.54 ksi.jffi, in/cycle.

4. For 11(1 = I6ksi; R = 0 (f3 = .Jsec na/w):

a N t'la t'lN (da/dN) = (t'la/t'lN) aa"

0.1 0
0.005 1100 4.55E - 6 0.1025 9.08 9.08
0.105 1100
1.5
0.05 100 5.00E - 4 1.525 35.53 35.53
1.55 i + 100

Make a similar table for the case of R = 0.5, and plot da/dN versus 11K on
log-log.

5. From table in solution to exercise 4:

log 4.55E-6 = mplog 9.08 + log Cp;
log 5.00E-4 = mplog 35.53 + log Cp-
Solution: Cp = 2.21 E - 9; mp = 3.45. Same for other case with R = 0.5; 11(1 = 10
ksi; R = 0.5: Cp = 6.49E-9; mp = 3.45.
 491

6. Lines in Exercise 5 are parallel (Same m), so that Walker equation is certainly
applicable C R = CR~O/(1 - R)"w where CR~O = Cw (Equation 5.14), so that
(1 - R)"K = CR/CR~O' Using the C values from Exercise 5 for R = 0.5 and
R = 0 one obtains (1 - 0.5)"" = 2.21/6.49, so that OS, = 0.34 and
n.. = log 0.34/log 0.5 = 1.56. As flK/(l - R) = Kmax: mw = 3.45 - 1.56 =
1.89. Hence, the Walker equation reads:
da 2.21 E - 9 flK3.45 da
or 2.21E - 9flKL89 K~~~.
dN (1 + R)1.56 dN

7. Forman Equation:

2 3 4 5
.1.0- R .1.K da/dN {(l - R)Kc - .1.K}da/dN
16 0 9.08 4.55£-6 3.23£-4
35.53 5.00£-4 2.22£-2
10 0.5 5.68 2.50£-6 8.58£-5
22.21 2.94£-4 5.23£-3

Columns 1 through 4 follow from results of Exercise 4. Plot columns 3 versus

5 on double log scale. ALL data must fall on a SINGLE straight line as can be
deduced from Equaton 5.16. If they do not Forman equation does not apply.
Indeed they do approximately. Determine CF and mF from line.
log 8.58E-5 mFlog 5.68 + log C F;
log 2.22E-2 mFlog 35.53 + log CF.
From which: CF = 4.58E - 7; mF = 3.03. Equation:
da flK3.03
dN = 4.58E-7(1 - R)80 - flK

8. Naturally, all rate diagrams should be the same, otherwise equations would
not fit the data. Example: for flK = 8 ksi.Jffi and R = 0.5 the Walker equation
of Exercise 6 provides

:~ = 2.21E-9 x 81.89 x 161.56 = 8.5E-6inch/cycle.

The Forman equation of Exercise 7 provides:

da 83.03 .
dN = 4.58E-7(1 _ 0.5)80 _ 8 = 7.8E-6mch/cycle.
492

9. Walker equation; p = ..}sec na/W; Urnax = l5ksi; I1u = 15 x 0.7

10.5 ksi; da/dN from Exercise 6:

a l1.a aave I1.K = f3l1.aJna avo Kmax da/dN I1.N = l1.a/da/dN N

0.5 0
0.2 0.60 14.4 20.6 3.83£-5 5220
0.7 5220
0.3 0.85 17.2 24.6 7.07£-5 4243
1.0 9463
0.5 1.25 20.8 29.7 1.36£-4 3676
1.5 13139
0.5 1.75 24.6 35.1 2.42£-4 2066
2.0 15205
1.0 2.50 29.4 42.0 4.49£-5 2227
3.0 17432

Computer result: Ntot = 17717 cycles.

10. Forman Equation; same f3 as in Exercise 9. I1u = 15 x 0.7 10.5ksi;

da/dN from Exercise 7.

a l1.a Qave I1.K da/dN I1.N N

0.5 0
0.2 0.60 14.4 3.56£-5 5618
0.7 5618
0.3 0.85 17.2 6.84£-5 4587
1.0 10205
0.5 1.25 20.8 1.28£-4 3966
1.5 14111
0.5 1.75 24.6 2.39£-4 2092
2.0 16203
1.0 2.50 29.4 4.84£-5 2066
3.0 18269

Computer result: N tot = 18538 cycles.

11. u rn•x = 11 ksi; R = 0.2; I1u = 0.8 x 11 = 8.8 ksi 11K = 3 X 8.8 x
..}nx 0.1 = 14.8ksiJ1i1;Krnax = 18.5ksiJ1i1.
With Walker equation: da/dN 3.41E - 5 inch/cycle.
With Forman equation: da/dN 3.27 E - 5 inch/cycle.

12. Overload Krnax = 3 x 17 n x 0.1 = 28.6 ksiJ1i1.

Current Krnax = 3 x 11 n x 0.1 = 18.5 ksiJ1i1; 11K = 14.8 ksiJ1i1.
rpoverload = 0.00886 inch (p = rpoverload)'
rpcurren! = 0.00371 inch.
 493

da/dN = 2.2IE-9~KLR9K~~~ = 3.4IE-5inch/cycle(unretarded).

da/dN = (371/886)12 x 3.4IE-5inch/cycle = 1.20E-5inch/cycle
(retarded).
After 0.005/inch growth:
Current Kmax = 3 x II x JII x 0,103 = 18.8ksi,Jill; ~K = 15.0ksi,Jill.
p = 0.00886 - 0.003 = 0.00586 inch; rpcurrcnt = 0.00383 inch.
da/dN = 2.21 E - 9~KL89 K~~~ = 3.59E - 5 inch/cycle (unretarded).
da/dN = (383/586)12 x 3.59E - 5 inch/cycle = 2.16E - 5 inch/cycle (retarded)
For y = 0 unretarded rates.

13. For prevention K ~ KIscc

(Jp < 15/(3 x In x 0.1) =, 8.92 ksi.

14. Results (a, = 0.5 inch in alI cases; ac is crack size at fracture); (J = P/ BW.

p = j3uFa, Growth Time

K/C Y
K,
a, = ~ 1.I2u
1[
ac - G1

18 12 16.84 1.58 1.08 5000

24 16 22.46 0,89 0,39 100
30 20 28.10 0.57 0.07 3
32 21 30.00 0.50 0.00 0

Plot K,(linear) versus time (log). Draw asymptote for KIsC(. Graph shows Klscc ~
16 ksi,Jill. Amount of growth shown in fifth column.

15.6. Solutions to exercises of Chapter 6

1. For results see Table 15.1 and Figure 15.1.

2. For results see Table 15.2 and Figure 15.2. The exceedance diagram is for 300
periods (days). In Exercise 1 it was represented by a total of 200 000 cycles, i.e.
200000/300 = 667 cycles per period (day). In Exercise 2 this was represented by
100,000 cycles, i.e. 333 cycles per day (but with cycles of different magnitude).
Computer crack growth analysis, for each case are shown in Figures 15.3 and
15.4. The data used were the same in each case. The results are different in
numbers of cycles, (as they should) but not much in number of periods (days).
Hence, both stress history representations of the same spectrum give about the
same answer. Better agreement would be obtained if more levels were chosen
(10-12).
494

1.2

300 DAYS

.8
~------------~~~

.6
Truncation
/ a t level 7
.4~----------------------- ----'---'>~__rL_1 (Exercise 4)
I

.2t----------------------------------------+--~~----

-.2 4.SX10'

-.4

-.6

Figure 15.1. Results of Exercise I of Chapter 6.

3. Solutions in Figure 15.5 and Table 15.3 first 8 lines: 2 periods (days) type A,
5 types B, 34 types C, 130 types D, 129 E types E. Average total exceedances/day
= 4.5 x 103/300 = 1500, which provides 'pivot' point, P, for exceedance
diagram of periods in Figure 15.5.

4. Solution Table 15.3, bottom lines.

Table 15.1. Results of Exercise I of Chapter 6 (see Figure 15.1).

Level Relative Relative (Jmin (fmax .1.<1 Exceedances Occurences

(J'min (Jmax (cycles)
I -0.4 1.1 -6.0 16.5 22.5 2 2
2 -0.325 0.988 -4.9 14.8 19.7 II 9
3 -0.25 0.875 -3.8 13.1 16.9 63 52
4 -0.175 0.763 -2.6 11.4 14.0 316 253
5 -0.1 0.65 -1.5 9.8 11.3 1679 1363
6 -0.025 0.538 -0.38 8.1 8.48 7830 6151
7 0.05 0.425 0.75 6.4 5.65 43300 35470
8 0.125 0.313 1.9 4.7 2.80 200000 156700
 495
Table 15.2. Results of exercise 2 of Chapter 6 (see Figure 15.2).

Level Relative Relative (Jrnin Umax Exceedances Occurrences

O"min (Jmax (cycles)
I -0.4 l.l -6.00 16.50 2 2
2 -0.37 1.05 -5.55 15.75 10 8
3 -0.24 0.83 -3.60 12.45 100 90
4 -0.17 0.74 -2.55 1l.l0 1000 900
5 -0.01 0.51 -0.15 7.65 10000 9000
6 0.05 0.40 0.75 6.00 100000 90000

5. Instead of one cycle of level I and two of level 2, there will be three cycles
of level 2 in period A.

6. Similar as Tables 15.1 and 15.2, but all stress levels, exceedances and cycle
numbers in accordance, see Figure 15.6. Pivot point, P, at 1500 Exceedances.

7. All stresses will be multiplied by 21.5/15 = 1.43. Everything else remains the
same as in Tables 15.1 and 15.2.

1.2

1.0
300 DAYS

.8~------~--~~~3

r---------------~~~--_44

.6
r------ ---~ ----- ------
.41'-__~ ___ ~~_~ _ _ _ _ _ _ _ _ _ _'___~::____,

.2t-----------------------+-~~--

0~~--·----------------------------.-~~~-J6

105
I
I
-.2 4.5xl0 s
~----.~.-~~~

-.4

-.6

Figure 15.2. Results of Exercise 2 of Chapter 6

496
(1000 CYCLES)
650 1300 1950 2600 3250 3900 4550 5200 5850
(INCH)

4.5 4.5
ti1
w
I
u 4.0 4.0
z

w 3.5 3. 5
N
(JJ
3.0 3.0
'"u
'"u"'" 2.5 2.5

2.0 2.0

1.5 1.5

1.0 1.0

0.5 0.5

2 3 5 6 7 8 9
LIFE (l000 DAYS)

Figure 15.3. Same configuration and data as in next figure. Crack growth as in Exercise 1 of
Chapter 6.

(1000 CYCLES)
350 700 1050 1400 1750 2100 2450 2800 3150
(INCH)

4.5 4.5
ti1
w
I
u 4.0 4.0
z

w 3.5 3.5
N
(JJ
3.0 3.0
'"u
'""'"
L.J 2.5 2.5

2.0 2.0

1.5 1.5

1.0 1.0

O. 5 0.5

2 3 5 6 9
LHt: (lUOO C,~YS)

Figure 15.4. Same configuration and data as in previous figure. Crack growth in Exercise 2 of
Chapter 6.
 497

300 DAYS

Truncation
/ a t level 7
1-------.-::::-....=-=~~~VM"--------""LL>..."<777TL-1 (Exercise 4)
.4 I

.2+------------~~------_+-~~---

I
I

-.2 4.sx10'

-.4

-.6

Figure 15.5. Solution to Exercise 3 in Chapter 6.

8. If there is no retardation the highest load only contributes a small amount

of crack growth. Since there are only a few of these, their total contribution is
extremely small. Without retardation they have no effect on the growth at lower
loads.

15.7. Solutions to exercises of Chapter 7

I. Plane strain at B = 2.5 (40/8W = 0.625 inch.

Plane stress at B ~ (40/80? /2n = 0.04 inch. Assuming straight line between
(0.04, 90) and (0.65, 40) as in Figure 7.4, toughness for 0.3 inch thickness is
estimated as 68 ksiJm.

2. K, = {Ja-Jn x 4.3; Keff = {JaJ1CX4. Assuming {J is about the same for

a = 4.3 and 4 inches, Keff = -J4/4.3 x 75 = n.3ksi.Jll1;a!c = (16-8) x 80/
16 = 40ksi; aF = n.3/1.19J1CX4 = 17.1 ksi. Fracture was not by collapse,
i.e. Keff = 72.3 ksi-Jln.
498

1.2

300 DAYS

.2t-------------------------------------------4-~----

-.2 4.sx10'

-.6
Figure 15.6. Solution to Exercise 6 of Chapter 6.

3. (It, = (16-8) 35/16 = 17.5 ksi; (Irr = 17.1 ksi (see Exercise 2). Hence,
fracture was probably by collapse; therefore K, ~ 75 ksiyTil, and
Kerr ~ 72.3 ksiyTil.

4. K 1, = 12J'i6 = 48 ksiyTil.
!1T = 215 - 1.5 x 70 110 F, so that toughness might be useful for slow
loading at 70 - 110 = - 40 F.

5. Estimate KIc = 50 x 60/75 = 40 ksiyTil for B. If yield strength values are

reversed, estimate toughness B = toughness A.

6. Plane strain at B = 2.5 (40/70)2 = 0.82 inch.

Draw a straight line through (0.82, 40) and (0.5, 60) as In Figure 7.4 and
extrapolate to 0.3 inch to obtain: K, = 72 ksiyTil.

7. Should be about the same as in Table 7.1.

8. Should be about the same as in Table 7.2.

 Table 15.3. Results of Exercises 3, 4,5; See Figure 15.5
2 4 6 8 9 10 11 12 13 14
Level Exceed- Occur- A- A- Occur- Remain- B- B- Occur- Remain- C- Occur- Occur-
ances fences Exceed- Occur- rences der Exceed- Occur- rences der Exceed- fences fences
Table Table ances fences in 2A (3)-(6) ances fences in 5B (7)-(10) ances in C in 34C
15.1 15.1 Fig. 15.5 2 x (5) Fig. 15.5 5 x (9) Fig. 15.5 34x(13)

2 2 1 2 0
2 11 9 2 4 5 0
63 52 4 8 44 10 34 1 34
4 316 253 17 10 20 233 10 35 198 2 68
1680 1364 46 29 58 1306 29 19 95 1211 i4 11 374
6 7830 6150 124 78 156 5994 80 51 255 5739 56 42 1428
7 43300 35470 365 241 482 34988 254 174 870 34118 190 134 4556
200000 156700 1000 635 1270 155430 866 612 3060 152370 750 560 19040

7 43300 35470 365 241 482 34988 254 174 870 34118 190 134 45556
improper
truncation

81000 73170 487 363 726 72444 365 285 1425 71019 273 217 7378
correct
truncation

.j:.
\0
\0
 VI
:3

Table 15.3. (Continued)

15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
Remain- D- O- Occur- Remain- aFor Occur- Exceed· Actual Difference Add to New 8 New 8 Extra 'Remain-
der Exceed- Occur- rences der 129E fences ances Occur- Remainder 8 occurr. Exceed. in 58 der
(1IHI4) ances fences in 1300 (l5H18) Remains in E in E rences (19H23) (24)/5 b (9)+(25) Fig. 15.5 6(26)-(10) (24)-(28)
Fig. 15.5 130 x (17) (19) (20)/129 in E
(21) x 129

1 0
0 2 0
130 1 130 0 7 10 0
837 4 520 317 317 2 258 59 12 31 41 60 -1
4311 22 17 2210 2101 2101 16 18 2064 37 58 99 35 2
29562 124 102 13260 16302 16302 126 144 16254 48 10 184 283 50 -2
133330 650 526 68380 64950 64950 503 647 64887 63 13 625 908 65 -2

29562 124 102 13260 16302 16302 126 144 16254 48 10 184 283 50 -2

63641 177 155 20150 43491 43491 337 355 43473 18 4 289 388 20 -2

"Total of 2A + 58 + 34C + 1300 = 171 periods used; Remain: 300 - 171 129 periods E
h Or io other 'period'
'Differences negligeable (lower stresses: and consider log readings)
 501

9. Walker: m w , m R , nw remain the same (dimensionless).

Conversion of C..{mR = mw + nw):

I in = (1000)-mR in .
(ksiv'\nt R (psiv'\ntR
Cw must be multiplied by (1000)-307 if K in psiv'\n, i.e. C = 3.88E-19.

10. See solution to Exercise 9.

II. C = 0.05 x IE-8 + 0.25 x 3E-9 + 0.7 x 5E-1O = 1.6 E-9.

15.8. Solutions to exercises of Chapter 8

1. K = 1.481TJ1W; f3 = 1.48; ITfr = 11.9 ksi (max bending stress at fracture);

Collapse ITfc = 1.5FtAW - a)2/W2 = 1.5 x 60 x (5/10)2 = 22.5ksi (Figure
2.15); ITres = 11.9 ksi.

2. ITten = P/BW; ITben(maX) = MW/2I = 2PW/(BW3 /6) = 12P/BW2. ITben/

ITten = 18/W = 1.2; ITtat = 2.2ITten . For f3 see Table 15.4, columns I through 4.

3. See Table 15.4 columns 1 through 5.

4. See Table 15.4 columns 6 through 9 (note same results in columns 7 and 9).

5. See Figure 8.17; Compound with f3w = ..}sec 1raeff/W' =

..}sec n(a/2 + d/2)/W (example see problem 6).

6. For aiD = 0.3 we have f3hale = 1.47.

f3w = ..}sec n x 0.65/6 = 1.03. Total f3 = 1.47 x 1.03 = 1.52;
ilK = 1.52 x 10 x ..}n x 0.3 = 14.76ksiv'\n; Kmax = 18.44ksiv'\n.
da/dN = 2.4 E - 5 inch/cycle.

7. Case 1: For a ~ 0 stress at hole IS ITh = 3ITL - ITT = (3 - 0.33)ITL =

2.67IT L •
Hence f3 = 1.12 x 2.67 = 2.99.
Case 2: For a ~ 0 stress at hole is 3ITL - ITT = (3 + 0.5)ITL = 3.5ITL.
Hence f3 = 1.12 x 3.5 = 3.92.
In both cases f3 = ..}0.5 + D/2a for large a. Draw curve for latter and fair into
values for a ~ O. Case I: f3h = 2.26 (on faired) line; f3w = 1.02, total f3 = 2.31;
ilK = 2.31 x lO..}n x 0.1 = 12.9ksiv'\n;Kmax = 16.2ksiv'\n;da/dN = 1.53
E-5inch/cycle. Case 2: f3 = 2.35 x 1.02 = 2.39; da/dN = I.72E-5inch/
cycle.
 V1
o
N

Tahle 15.4. Solution to Exercise 2 (columns I through 4), Exercise 3 (columns I through 5), and Exercise 4 (columns 6 through 9) of Chapter 8.

(J"n/(Jtot = 1/2.2 = OA55; (J"n/(Jtot = 1.2/2.2 = 0.545; (Jho"/(J,," = 1.2.

2 3 4 5 6 7 8 9
a/w f3ten fib," (Jtot reference (JtO" reference a (ftot (lIen a lOI =
Fig. 8Aa Fig. SAb 2.2 (J"n
K,. K,
((JIe" (JI =-- (Ji = ---
fil = - fibon
fi,," + -(Jhe") fi2 = fihon
( fi"n + -(Jhm)
0"\01 0"101 O"ten ' filfo ' fi2fo
0.1 1.15 1.06 1.10 2A2 I 25.6 11.6 25.5
0.2 1.20 1.03 1.11 2A4 2 18.0 8.2 18.0
0.3 1.26 1.12 I.IS 2.60 3 13.8 6.3 13.9
OA 1.33 1.25 1.29 2.83 4 10.9 5.0 11.0
0.5 1.44 1.48 1.46 3.22 5 S.6 3.9 8.6
0.6 1.60 1.86 1.74 3.83 6 6.6 3.0 6.6
 503

8. For large cracks and tension: Phote = .JI + Dj2a. Due to fastener force:
Kp = Pj-j-rc(2a + D), where P is per unit thickness; so that we have K = aWj
.In(2a + D); Ku = .JI + Dj2a a.jiW. Ktot = 0.8Ku + 0.5 x 0.2(Kp +
Ku) = Pta.jiW. Substitute expressions and work out Pt. Compound P with
.Jsec n(2a + D)j2 W. Result in Table 15.5 first 2 columns.

9. See Table 15.5.

10. For a ~ 0: P= 1.12(1 + 2Jd!Y) 6.13.

Table 15.5. Beta(a) for exercises 8 and 9/Chapter 8.

Loaded hole crack at both sides.

Through-the-thickness crack.
Loading: Bypass ratio = 0.8.
Stress intensity defined as K = beta x sigma x sqr(pi x a)
Sigma defined as: Nominal tension stress.
All dimensions in inches.
Thickness = 0.5
Width = 8
Hole diameter = I
Crack.L Beta K, After arrest
inches
PFa
ksi ksi
0.100 3.022 29.517 29.517
0.200 2.455 25.694 26.768
0.300 1.924 26.768" 26.768
0.400 1.793 24.874 24.874
0.500 1.662 24.001 24.001
0.600 1.590 22.903 22.903
0.700 1.538 21.926 21.926
0.800 1.485 21.234 21.234
0.900 1.441 20.639 20.639
1.000 1.419 19.875 19.875
1.100 1.399 19.227 19.227
1.200 1.391 18.512 18.512
1.300 1.383 17.887 17.887
1.400 1.384 17.221 17.221
1.500 1.387 16.606 16.606
1.600 1.396 15.975 15.975
1.700 1.408 15.366 15.366
1.800 1.425 14.757 14.757
"Increasing residual strength (arrest conditions).
504

'-."
\
72 ----------i----------------_______ _
\
64 \"

56 '" "
"~,
_ ACTUAL RES. STRENGTH

______ £LASTle

~ ____ NSY

_ . _ TANGENT

24

16

50 100 150 200 250 300 350 400 450

CR,\CK LENGTH COOl INCH)

Figure 151 Residual strength diagram; solution to Exercise I of Chapter 9; ale = I.

For large a: 13 is f3e for edge crack (Figure 8.5) for (a + d)/W.
For example fora = d = 0.5 and W = 5:aetrlW = 0.1;13 = 1.15fromFigure
8.5a.

II. Same solution as in Exercise 10 with r = D/2 and d = e + D/2.

12. For small cracks 13 ~ 1.5 x 1.12 = 1.68.

(For surface flaws multiply by f3FFS/¢; Figure 8.3).

13. Apparent stress intensity in elements is:

first element K = 48.9,J2n x 0.05 = 27.4ksi,Jill;
second element 29.1 ksi,Jill, third element 31.6 ksi,Jill.
Plot versus r and extrapolate to r = 0 to final K ~ 27 ksi,Jill. Taking highest
stress as the reference stress: 13 = 27/5FX2 = 2.15.

14. Va = 1/2PfJ = 1/2 x 10000 x 0.0020 = 100 in Ibs/inch thickness.

V aHa = 1/2 x 10000 x 0.021 = 105 in Ibs/inch thickness.
dV/da = AV/Aa = (105 - 100)/(1.01 - 1) = 500lbs/in = 0.5 kip/inch.
K = ,JEdV/da = ,J5000 = 71 ksi,Jill.
15. Since it is assumed that the case of Figure 8.25 is for J1 = 0.4 it follows that
As = 0.4 x 8 x 0.2 = 0.64inch 2 •
Example for alb = 0.5: 13 = 0.5 from Figure 8.25d. a = 4 inch.
 505

'-.

72
\,
_. __ ~ _____________ .. ________ _

64 \
\.
\
_ ACTUAL RES. STRENGTH
56

48

40 f ____ NS)

EXERCISE 3 _._!"flNGENT
32

24

16

0.2 O. 4 0.6 C.8 1. 0 1. 2 1. 4 1. 6 1. 8

CRAcK LENGTH ( ]f~CHES)

Figure 15.8. Residual strength diagram; solution to Exercise 1 of Chapter 9; ale = 0.3.

From Equation (S.42): L = I + O.sJ2'(l - 0.4)/0.64 = 2.06 (compare with

Figure S.25e). Fastener load is according to Equation (S.43):
0.6(2.06 - I)O.64u = O.4lu (in kips if u is in ksi).

15.9. Solutions to exercises of Chapter 9

1. For external flaws: K = PUhooP..[iW, (P from Figure 8.3).

For internal flaws: K = Kpress + Pp p..[iW, + PhUhooP..[iW, = P(UhooP + p)..[iW,
(since Pp ~ Ph; P form Figure 8.3). Results shown in Figures 15.7 and 15.S.
(Subtract p from the given residual strength to get hoop stress at fracture for the
internal flaws).

2. Hoop stress at the given pressure is 3 x 5/0.5 = 30 ksi. For through-the-

thickness cracks there is still plane strain, because B > 2.5 (35/70)2 =
0.625 inch (B = 0.5 inch), so that K lc can be used.
(a) External cracks.
The circular crack will cause fracture at the hoop stress of 30 ksi when
a = e = 0.4 inch (Figure 15.7). A circular crack will cause a through-the-thick-
ness crack of length 2a = 2B, i.e. a = 0.5 inch. Its stress intensity is K =
30.jrc x 0.5 = 37.6 ksi,Jm. This is larger than KIn so that fracture continues
(break). The flaw with a/e = 0.3 causes fracture at e = 0.7 inch. The through-
crack will then have a length of 2a = 1.4 inch. Then K = 30.jrc x 0.7 =
44.5 ksi,Jm, which is larger than K/c(break).
506

67.5

60.0
2'
t.:l
~ 52.5 _RES.STR.AFT£RA~RESI

t;;
-"~.,:-,,",.,....- -
45.0

----.
...J
<
::J
- - - -- - - - -- - - - - - -- - -
8
<J) 37.5 ........... ~.. ____ STRINGER
ii2
30.0 _ _ TANGENT

22.5

15. Of
7.5

L-~ __~__~__~~___ ~.~I_ _~_ _~~

O. 5 1. 0 l. 5 2.0 2. 5 3. C 3. 5 4. 0 4. 5
CRACK LENGTH ( INCHES)

Figure 15.9. Residual strength diagram; solution to Exercise 5 of Chapter 9.

(b) Internal flaws

The same figures can be used, but as the figures provide (Jh + p at fracture for
internal flaws (see stress intensity superposition in Exercise 1), the critical crack
size must be found for a stress of (Jhoop + P = 35 ksi. The circular flaw breaks
through at a = 0.36 inch. It still gives a through crack 2a = 2B. The stress
intensity after break-through is K = (JhooPJ1W, = 30-/n x 0.5 = 37.6ksiFo
(break). The flaw with al c = 0.3 is critical at c = 0.59 inch. Break through
length is c, i.e. K = 30-/n x 0.59 = 40.8 ksiFo (break).

3. For plane strain B > 2.5 (35/50? = 1.225 inch, which is not the case (B =
0.5). Hence, the given Kc = 80 ksiFo applies. Note that a new tangent from
50ksi must be drawn in Figure 15.7 and 15.8 because of the lower yield.
The part through cracks are still in plane strain, so that otherwise Figure 15.7
and 15.8 apply. Following Exercise 3, the break-through crack sizes are:
External: alc = la = 0.40 inch. Through crack a = B; K =
30.jn x 0.5 = 37 ksiFo ( < Kc), fracture arrests (leak). For alc = 0.3: c =
0.7 inch. Through crack a = 0.7 inch; K = 30-/n x 0.7 = 44.5 ksiFo < K
(leak).
Internal: Through crack a = B for alc = 1: K = 37.6ksiFo (leak); alc =
0.3: K = 30.jn x 0.59 = 40.8 ksiFo (leak).

4. For alc = 0.3 about same result as in Table 9.1 for alc = 0.33. For alc =
flaw remains circular.
 507

80
i':
'-"
1'5 70 _ ACTUAL RES. STRE"lGTH
g:
U1
____ ELASTIC
-' 60
<:
::>
8U1 so _____ NSy
UJ

'" ____
40 TA~GENT

30

20

10

5
CRACK LENGTH ( I':CHES)

Figure 15.10. Residual strength diagram; solution to Exercise 2 of Chapter 10.

5. See Figure 15.9. Stringer critical. To make it stringer critical use stringers of
another material; F tu = 75 x 40/37.5 = 80 ksi. Fastener load at arrest (if skin
critical): P = 0.41 x 40 = 16.4 kip (Exercise 15 of Chapter 8). Fastener
diameter is D = .,)4 x 16.4/(100 x n) = 0.45 inch. Bearing stress: (fb =
P/DB = 16.4/(0.45 x 0.2) = l82ksi; elastic; redistribution will occur.

6. From Figure 15.9:

(a) Stress for fracture at a = 2 inch is (ffr = 34 ksi.
(b) Arrest at a = 3.5 inch.
(c) In Exercise 5 you calculated fj = 0.82 for a = 2 inch; 11K = 0.82 x
15 x FX2 = 30.8 ksi0ii; Kmax = 30.8/0.8 = 38.5 ksi0ii; da/dN =
3E-9 x 30.8 2.1 x 38.5°·9 = 1.07E-4in/cyde.
(d, e) Same rate as in C, because residual strength curve is line of constant K
(K = KJ, the stress intensities at a = 2 (fracture) and a = 3.5 (arrest) are
equal (larger a at arrest but smaller fj (fj = 0.62 for a = 3.5 inch).

7. Small a: fj ~ 1.12kt = 1.12(1 + 2.,)0.875/0.125 = 7.05. Larger a: fj =

1.12.,)1 + 0.875/a. Fair line from 7.05 at a = 0 to curve.

8. Before stophole (ffr = 50/1.12.,)n x 0.75 = 29.l ksi. If e.g. crack of 0.02
inch has developed from hole:
(ffr = 50/7.48.,)n x 0.02 = 26.7ksi.

9. Same answer as for Exercise 7.

508

10. Max stress is 30 ksi; minimum stress = 20 ksi.

= lOksi; R = 0.67; /).K = 1O~ = 17.7ksiJli1
/).(J

da/dN = 2 x 10- 7 x 17.72 /[(1 - 0.6)80 - 17.7] = 7.2E-6inch/cycle.

11. Stress intensity due to residual stress:

K = 20~ = 35.4ksiJli1, which is larger than Klscc • Crack will grow by
stress corrosion.

12. Assuming K2c = Kle = 80 ksiJli1 (circle):

K2 = 20.fiCX2 = 50 ksiJli1. Allowable KI is then KI = ../802 - 502 =
62.5 ksiJli1, hence (J = 62.5/.fiCX2 = 24.9 ksi.
Assuming K2e = 0.8Kle = 64 ksiJli1 (ellipse):
KI = 80../1 - 502 /642 = 49.9, hence (J = 49.9/.fiCX2 = 19.9 ksi.

15.10. Solutions to exercises of Chapter 10

1. ap = 2 inch.

2. See Figure 15.10.

3. W = 30 inch; ap = 0.38 inch for (Jp = 51 ksi (tangent);

ap = 1.1 inch for (Jp = 35 ksi (LEFM);
ap = 3.2inch for (Jp = 20ksi (LEFM).
W = lOinch ap =
0.38 inch for (Jp = 51 ksi (tangent);
ap = 1.04 inch for (Jp = 35 ksi (LEFM);
ap = 2.4 inch for (Jp = 20 ksi (LEFM).
W = 4inch ap = 0.30 inch for (Jp = 51 ksi (collapse);
ap = 0.83 inch for (Jp = 35 ksi (collapse);
ap = 1.33 inch for (Jp = 20ksi (collapse).

4. See Table 10.5 for example. In present case the numbers will be different. Plot
(Jlr and (Jle and draw tangent. Lowest of three prevails. Read ap from curves for

three values of (Jp.

5. (Je = 32 ksi; stable fracture /).a = 0.4 inch.

6. Same answer as in Exercise 5.

7. (Je ::::; 36ksi; /).a ::::; 0.3 inch; J at fracture::::; 0.68 kips/in.

8. K = ../6800 = 82 ksiJli1.
 509

15.11. Solutions to exercises of Chapter 11

1. Case I: H = 4800 hrs; I = 4800/6 = 800 hrs.

Case 2: H = 4800 hrs; I = 4800/6 = 800 hrs.

2. Solution in Table 15.6.

3. Solution in Table 15.7.

4. Computer results (may differ slightly from hand-calculations, because

computer varies time of first inspection):
Case I: I = 500 hrs, P det = I; I = 1000 hrs, P det = I; I = 1500 hrs, P det =
0.99. For P det = 95%: I = 1600hrs.
Case 2. 1= 500hrs, Pdet = I; 1= 1000hrs, Pdet = 0.98; 1= 1500 hrs,
Pdet = 0.91. For Pdet = 95%: I = 1360hrs.

5. Critical stress at ap = 33 mm is U c = 50/-Jn x 0.033 155 MPa. For

aproof= 15mm: Uproof = 50/-Jn x 0.015 = 230 MPa.
Interval Nac - Nproof = 4800-3400 = 1400 hrs (Case I).
Interval = 4800 - 4700 = 100 hrs (Case 2).
Both without safety factor

Table 15.6. Solution to Exercise 2 of Chapter II.

Insp. Hours Case I Case 2

#
a P de! P miss Cum. Cum. a Pde! P mlss Cum. Cum.
Fig. 11.7 Fig. ll.8 P miss P de, Fig. 11.7 Fig. 11.8 Pmiss P de!

I 800 6 0.6 004 004 0.6 5.2 0.5 0.5 0.5 0.5
2 1600 7.5 0.69 0.31 0.12 0.88 5.5 0.52 0048 0.24 0.76
3 2400 10.0 0.74 0.26 0.03 0.97 6.0 0.60 0.40 0.096 0.904
4 3200 13.0 0.80 0.20 0.006 0.994 6.5 0.63 0.37 0.036 0.974
5 4000 20.0 0.87 0.13 0.0008 0.9992 7.5 0.69 0.31 0.011 0.989

Table 15.7. Solution to Exercise 3 of Chapter 11

Equation: p = I _ e-{(a-5)!(8-5)}O.s

Insp. Hours Case 1 Case 2

#
a P de! Pmiss Cum. Cum. a P de! PUliS'S Cum. Cum.
Fig. 11.7 Eq. Pmiss P de! Fig. 11.7 Eq. Pmiss P de!
1 800 6 0.44 0.56 0.56 0.44 5.2 0.23 0.77 0.77 0.23
2 1600 7.5 0.60 0040 0.22 0.78 5.5 0.33 0.67 0.52 0048
3 2400 10.0 0.72 0.28 0.06 0.94 6.0 0043 0.57 0.29 0.71
4 3200 13.0 0.80 0.20 0.01 0.99 6.5 0.51 0.49 0.14 0.86
5 4000 20.0 0.89 0.11 0.001 0.999 7.5 0.60 0040 0.06 0.94
510

6. At operating stress ac = (30/IOW /n = 0.029 inch.

For (Jp = 150ksi: ap = (30/150)2/n = 0.013 inch.
Stripping depth: b = 0.029 - 0.013 = 0.016 inch every two years.

7. Cost of fracture is Cfr = P x T X Sh. Cost of fracture control:

Ceon = Soh TS, assuming one replacement of critical part during economic life
(durability). Do fracture control Ceon < Cf " otherwise fracture costs might be
acceptable.

15.12. Solutions to exercises of Chapter 12

1. With K, = 70 ksiFn and (Jp = 2 x 15 = 30 ksi, we have ap = 1.01 inch.

Life from a = 0.04 to 1.01 inch is 94,000 cycles.
Life from a = 0.04 to 0.1 inch is 47,000 cycles.
Hence life from 0.1 to 1.01 inch is 94,000 - 47,000 = 47,000 cycles. Safety
factor on life by assuming a = 0.01 inch is 94,000/47,000 = 2.

2. Permissible crack size a = 1.30 inch.

Life from a = 0.04 to 1.30 inch is 22,000 cycles.
Life from a = 0.04 to 0.1 inch is 3000 cycles.
hence, life from 0.1 to 1.30 inch is 19,000 cycles. Safety factor on life by assuming
a = 0.1 inch is 22000/19000 = 1.16.

3. By assuming a larger initial crack size, the extra safety for the crack at the
hole (1.16) is much less than for edge crack (2), showing that assuming larger
initial cracks for 'safety' is dubious; sometimes the extra safety is very small. For
a consistent safety factor, it is better to assume the correct crack size (0.04 inch)
in both cases, and then apply the same factor to both, e.g. factor of 2. In Exercise
1 life from 0.04 to 1.30 is 94,000 cycles. Factor of 2 gives H = 47,000 cycles. In
Exercise 2, the same factor on 22,000 cycles gives H = 11,000 cycles. Now both
cases are equally 'safe'.

4. Toughness is 60 ± 0.065 x 60 = 60 ± 4ksiFn. The actual toughness

may be 56 ksiFn. Estimate error in Pis two percent. If the stress is five percent
too high, we must take (Jp = 33 ksi, so that with Kc = 56 ksiFn, and P = 1.14
we have ap = 0.70 inch.
Life from 0.04 to 1.01 inch is 94,000 cycles.
Life from 0.04 to 0.70 inch is 90,000 cycles.
Factor La = 90/94 = 0.96.
Factor on rate data is 1.5 total; from average to upper bound is 1.25. This could
give a shorter life by a factor Lr = 1/1.25 = 0.8. If error in stress is 5 per cent,
actual stress could be 1.05 x 15 = 15.75 ksi. Hence, possible factor on life due
 511

to a and {3 is L, = [15 x 1.12/(15.75 x 1.14W 2 = 0.81. Hence, life may be

shorter by La X Lp X L, = 0.96 X 0.8 X 0.81 = 0.62, (i.e. 0.62 X
94,000 = 58,280 cycles). Factor of 2 will cover this (see Exercise 3), and in this
case the factor of 2 obtained by assuming initial a = 0.1 instead of 0.04 inch will
cover uncertainty as well.
For crack at hole toughness could be 56 ksi.jffi. Estimate error in {3 is 5
percent. For ap = 33ksi this leads to ap = 0.76 inch.
Life from 0.04 to 1.30 inch = 22,000 cycles.
Life from 0.04 to 0.76 inch = 16,000 cycles.
La = 16/22 = 0.73; L, is same as in previous case;
L, = [15{3/(15.75 x 1.05{3)]3.2 = 0.73.
Total uncertainty factor: 0.73 x 0.8 x 0.73 = 0.43.
A factor of 2 will not even cover this. In order to get the same 'safety' as in
previous case a factor of 2 x 0.62/0.43 = 2.9 would be needed; i.e. H =
22,000/2.9 = 7600 cycles. Note however that the factor of 1.16 obtained by
assuming initial a = 0.1 instead of 0.04 inch will certainly not cover uncer-
tainty.

5. Replacement times: 47,000/10,000 = 4.7 years for case 1 and 7600/10,000 =

0.76 years = 6 months for case 2.

15.13. Solutions for exercises of Chapter 13

1. Average toughness: K lc ~ 28 ksi.jffi

2. 20 mm = 0.8 inch; Nominal fracture stress aI' = 30/0.7.Jn x 0.8 17ksi.

3. See columns 1 through 8 of Table 15.8.

4. See columns 9 and 10 of Table 15.8.

5. See columns 11 through 15 of Table 15.8.

6. If the actual stress range was 15 ksi, the growth rates were too high by a
factor of (15/9.97)3 = 3.40 (third power of AK). If the handbook data were
correct then the material behaved as if C = 6.8E - 8 instead of C = 2E - 8
from the handbook. The material could be a 'bad lot', but may be the stress
analysis was incorrect and the stresses were indeed 9.97 ksi. Possibly there was
a hidden stress concentration, so that {3 = 1.50 instead of {3 = 1.

7. See solution to 6.
 Ul

tv

Table 15.8. Solutions to Exercises 3, 4 and 5 of Chapter 13.

2 3 4 5 6 7 8 9 10 II 12 13 14 15
a Spacing Magni- da/dN da/dN 8a 8N N 8K" 80- a 8K' da/dN 8N N
fication 2/3 average 6/5 8K average Equation 6/13

Fa
0.1 0.03 8000 3.75£-6 0 5.62 10.0 0
6.83£ -6 0.1 15000 0.15 6.84 6.4£-6 15625 15625
0.2 0.08 8000 1.00£ - 5 15000 7.77 9.80
1.44£-5 0.1 7000 0.25 8.84 1.4£-5 7143 22768
0.3 0.15 8000 1.88£ -5 22000 9.57 9.86
2.44£-5 0.1 4100 0.35 10.45 2.3£-5 4348 27116
0.4 0.09 2000 3.00£-5 26100 11.2 9.99
3.75£-5 0.1 2700 0.45 11.85 3.3£-5 3030 30146
0.5 0.09 2000 4.50£-5 28800 12.8 10.21

Average 80- = 9.97

'daldN = 3£-88KJ, so that 8K = (da/dN/2£-8)IJ; use da/dN from column 4.

h 11K = ,1,udverdge.vnaaverage'
 513

8. See Figure 12.8.

9. Operating stress: 50/0.7J1i. = 40ksi.

A defect of a = (7/0.7 x 40)2/n = 0.02 inch would not grow by stress
corrosion at the given stress and the given Kiser. Hence, quality control should
prevent parts with larger defects from entering into service.

10. The stress intensity at a crack of 0.05 inch is K = 2 x 25.Jn x 0.05 =

20 ksi.Jlll. This is much lower than the toughness; the material could not be that
much worse (specified toughness 60 kSi.Jlll. It might be somewhat below
standard, but more than likely the stress concentration was higher than
expected, and probably the stresses s well. May be the loads were higher than
anticipated, but also stress analysis and stress concentration factor are suspect.

11. If fracture would have been approached gradually at lower stress, the
striation spacing would have increased toward fracture. If striations suddenly
end (Figure 13.7), the fracture was due to one of the higher loads. The crack
growth rate just before fracture was: da/dN = 0.12/2000 = 6£ - 5. It follows
from the crack growth equation that 11K = (6£- 5/8£ _9Y/3.5 = 12.8 ksi.Jlll.
Hence, l1a = 12.8/0.64.Jn x 0.6 = 14.6 ksi. The occasional overloads
(R = 0) are a max = 29.2 ksi, and Kmax = 25.6 ksi.Jlll. Iffracture occurred at the
peak of the overload KIc = 25.6 ksi.Jlll. It could have occurred before the load
reached its maximum, so that K]c ~ 25.6 ksi.Jlll.
 Subject Index

Accessibility 376 Calibration 148

Accuracy (see also Errors) 243-280, 410, 418 Candidate fracture toughness (see Fracture
Acoustic emission 364 toughness)
Adhesive bonding 298, 299 Center notch 26
Airplane 5,17,18,149,155,172,173,177, Charpy 216, 241
179, 193, 203, 206, 237, 295, 302, Chevrons (on fracture) 434
327, 333, 340, 376, 391, 397, 398 Chevron notch 67
Analysis 332 Circular flaw (see Surface flaw or Corner
Anisotropy 221 cracks)
Apparent toughness 212 Cleavage
Arrest 292, 293, 308, 329, 330, 447, 502 -facets 12
Arrester 297, 397, 449, 454 -fracture 8, 12
Arrester failure (see Stringer failure) Clip gage 69
Arrest toughness 292, 405 Clipping 192, 201, 204, 206, 412
ASME Code 404 Closure 142, 145
ASTM standards (see Testing) Clustering 173
Asymmetric loading 255 Coalescence of voids 13, 14
COD 68,118,119,217
COD criterion 119
Back free surface correction (see Geometry COD tests 118
factor) Cold work 311, 313, 489
Beach marks 437 Collapse 22, 35,41, 46, 61, 68, 70, 86, 97,
Bending 100, 107, 121, 213, 340,418, 464,
collapse in- 39 486, 487, 507
Bend specimen 99 Collapse strength 39, 44
Beta (see Geometry factors) Combined bending and tension 39, 40, 250,
Biaxial loading 263 279
Biaxial stress field 25, 28, 48 Combined mode 17,319,330
Blunting 13,44, 101,430 Compact tension specimen 44, 100, 122
Blunting line 101 Composites 327
Bolt (see Fastener) Compounding 247, 257, 500
Branching 434 Compressive stress 140
Break-through 290 Computer 133, 135, 154,415
Bridges 5, 17, 237, 326, 376 Computer Software (Code) 136, 146, 150,
Brittle 152,154,158,170,185,197,205,
--cleavage (see Cleavage) 231, 241,244,268,286,301,332,
-fracture 12,41,43,425 412, 417

515
516
Concentration (Stress) (see Stress concentra- -opening displacement (see CTOD)
tion) -plastic zone (see Plastic zone)
Concept -stress field (see Stress field)
-of collapse 35 Crane 5, 172
-of EPFM 89 Creep 5, 8
-of fatigue crack growth 123 Criterion
-of LEFM 55 crack growth- 124
Conservation of energy (see Strain energy) energy- 73
Constant amplitude 126, 133 fracture- for EPFM 90, 92, 95
Constraint 28, 30, 58, 62, 64, 65, 10f, 284 fracture- for LEFM 56, 75, 78, 82
Contour integral 112, 114 instability- 82, 95
Contraction 28, 30, 58, 62, 102 plane strain- 64, 69, 102
Control (see Fracture control) Critical crack 7, 22, 94, 346
Convention (crack size) 50 Critical location 332, 334
Conversion of units 55, 229, 242 Critical stress intensity 56, 62, 82
Corner Crack 66, 214, 260, 282, 414 CTOD 118, 119,217,430,433,438
Correction factor (see Geometry factor) Cycle counting 174, 184, 202, 240
Cost 387, 390, 461 Cycle ratio (see Stress ratio)
Counting (loads or stresses) 174, 184, 202, Cycle-by-cycle 150
240
Counting (striations) 435, 439, 441, 442 Damage tolerance 1,391, 393, 395
Crack -analysis 1,3, 15,335,391
-arrest (see Arrest) -requirements 6, 17, 18,397,404
-at hole (see Holes) Data 208, 223, 412
-blunting (see Blunting) Defects 426
-closure 142, 145 Definition of
corner- (see Corner crack) -a 50
detail- 472 -C 131
-detection 369, 373, 381 -F 103
-growth (see Fatigue and/or Stress -Feo' 39
corrosion) -Fty 20
-growth curve (see Fatigue and/or Stress -F tu 20
corrosion) -G 78
-opening displacement (see COD) -H 89, 363, 366
permissible- I, 48 -h, 107
-propagation rate (see Fatigue or Rate -1365
equation) -J 89, 90
-resistance (force) (see Fracture -J R 90
resistance or R-curve) -K 49
-resistance curve (see R-curve) -Kapp 212
-size convention 50 -Ke 55, 62
small- 162 -Kelf 85
surface- (see Surface flaw) -K" K", Kill 50
shell- 472 -K" 62
Cracking mechanisms -K,= 165
of SCC 8, 10 -Km" 125
of fatigue 8, 9, 124 -Kmin 125
of hydrogen 8 -kt 26
Crack tip -kc 37
-blunting 13,44, 101 -ko 37
-closure 142, 145 -~K 125
 517
-L 53 Effect of- on fatigue crack
-m 131 anisotropy 221
-n 103 direction 221
-P 371 environment 157,204,222,236
-Pd" 370 frequency 157,222
-Pcurn 371 mean stress 124
-q 374 overloads (see Retardation)
-R(-curve) 80 stress ratio 126, 129,218
-R (stress ratio) 123, 125 temperature 157
-r 60 thickness 157
-Trn •t 97 Effect of- on toughness
-8T 216 anisotropy 213
-u 73 direction 213
-w 75 grain direction 213
-IX 106 loading rate 216
-f3 53 processing 213
-8124 specimen size 211
-60 106 strain rate 216
-/1 276 temperature 216
-Uk 36 thickness 61, 63,210,217
-u[,56 yield strength 215
-Urnin 123 ElF 409
-Urn., 123 Elastic energy (see Strain energy)
-U o 106 Elastic energy release rate (see Strain energy
-U,", 345, 351 release)
-8u 123 Elastic stress field (see Stress field)
Design Load (strength) 4, 20, 485 Elliptical cracks (see Corner crack, Surface
Detail crack 472 flaw, Geometry factor)
Detection of cracks 208, 369, 373, 381,412, Elliptical notch 26, 27
508 Embedded crack 260
Deterministic load 171, 202 Energy consumption (see R-curve and
Dimple 13, 14 Fracture energy)
Displacement Energy criterion 73, 79
constant- 77 Energy release rate (see Strain energy
--control 95, 196, 462 release)
Double shear 159 Environment 137, 204, 222, 236, 242
Ductile Equivalent initial flaw 409
-fracture 8, 13, 14 EPFM 22, 48, 62, 88, 93, 97, 102, 108, 116,
-rupture 8, 13, 43 119, 162, 214, 278, 339, 342, 350,
Durability 367, 399, 403 351,418
Dynamic Errors 208, 232, 410
-effect 292, 450, 458 Estimate of
-fracture 292 geometry factors (see Geometry factor)
toughness 87, 215, 241
Exceedance diagram 168, 176, 188, 205, 469,
Eddy current inspection 364, 375 493
Edge crack bending 246
Edge crack tension 54, 259
Edge notch 26, 36 Fail safety 366, 397, 398
Effective crack size 261, 304 Failure analysis 424, 441
Effective toughness 86, 213, 350 Failure analysis diagram 340, 343
518

Fastener 161,257,275,277,280,308,311, Fracture toughness

329,336,441,453,503 apparent- 212
Fatigue crack arrest- 292, 293, 308, 405
-growth 5, 123 candidate- 69, 72, 87
-growth analysis 123, 149, 355 definition of- 55, 92
-growth curve 7, 15, 126 effective- 85, 2 I 3, 350
-growth data 218, 223, 412 (see also estimate of- 215
Rate equation) plane strain- 62, 209, 419
-growth rate 125, 126, 130,218,228 plane strain- test 67
-initiation 366 plane stress- 62, 210, 212, 419
-mechanism 8, 9, 124 thickness effect on- 61, 63, 210, 217
-prediction 133, 147, 149, 152,223 transitional- 62, 212
-propagation (see Fatigue crack growth) unit of- 55
-retardation (see Retardation) valid- 68, 87, 210
-surface I I, 159 Free surface correction (see Geometry
-tests 126 factor)
-threshold 158 Freq uency effect 137, 204, 222
Finite elements 271, 280, 503 Frequency of inspection (see Inspection
Fitting data (see Rate data and Rate equation) interval)
Flaw (shape) (see Surface flaw or Corner Front free surface correction (see Geometry
cracks) factor)
Flow stress 105, 106, 488
Fractography 21, 424, 428 G.a.g 173,202
Fracture Geometry factor
-at notch 41 -for arrester structure 275
-analysis diagram 424, 441 -for asymmetric loading 255
-arrest (see Arrest) -for center crack 52, 54
-beyond yield (see Collapse) -for compact specimen 68
brittle- 12 compounding of- 247,257,501
cleavage- 8, 12 -for corner cracks 260
-control 1,2,362,379,387,392,408, -for crack at stress concentration 260,
446,461,465,481 266
ductile- 12 -for crack from hole 260, 267, 279, 280
elastic- (see LEFM) definition of- 53
elastic plastic- (see EPFM) -for edge crack bending 246
-energy 75, 79 -for edge crack tension 54, 259
-energy unit 79 -for embedded crack 260
-instability 5, 80, 82, 94, 353, 361, 400 -in EPFM 89, 93, 106, 122, 278
-mechanics 2 -in FEM 271, 280
-mechanisms 8, 12, 13 -from Green's functions 268, 3 I 7
overioad- 12, 425 -from handbooks 53, 87, 122,280
-probability 5, 389, 390, 392, 397, 398 -for I, L, U beams 301
-processes 8 -for irregular stress distribution 267
-resistance 77, 78, 79, 93 -for lug 255
-resistance curve (see R-curve or l-curve) -by simple methods 243, 260, 267
-speed II -for stiffened panel 275, 454
stable- 5, 80, 94, 353, 36 I -for surface flaws 247
-surface 12, 13, 14 -by using Green's functions 268, 3 I 7
-strength (see Residual strength) Grain direction 213, 221, 426
-stress 56 Green's function 268, 3 I 7
-velocity I I Ground-air-ground cycle 173
 519

Handbooks Leak-before-break 290, 329, 504

-for geometry factors 53, 87, 122, 243, LEFM 22, 48, 93, 97, 108, 119,339,342,
280 344,412,418
-for material data 209, 219 Limitations of EPFM 116
Helicopter 169 Load
Hinge, plastic 99 -COD diagram 68
History (see Stress history) -concentration factor (see Stringer stress
Hole concentration)
arrest at- 402 -control 95, 96
cold worked- 311, 313 deterministic- 171
crack approaching- 401 -displacement diagram 68, 73, 76, 99,
crack at- 163, 260, 279, 330, 422 121
--expansion 311 -flow lines 24, 25, 48
loaded- 256, 257, 502 -history (see Stress history)
oversizing- 369 -interaction (see Retardation)
stop- 46, 311, 330, 486, 506 over- (see Overload)
stress at- 26 -path 24
Hybrid arresters 451, 459 service- I, 4, 168
Hydrogen cracking 8 -spectrum (see Spectrum)
Loading
Initial flaw 367, 400, 402, 409, 479, 509 biaxial- 263
Initiation of fatigue crack 366 constant amplitude- 126, 133
Inspection 2, 290, 360, 363, 365, 393, 395, cyclic- 124
408, 476, 479 -mode 15, 16
Inspection accessibility 376 random- 149, 168
Inspection interval 365, 372, 377, 389, 390, semi-random- 168, 173, 190
508 service- (see Load)
Inspection specificity 376 sustained- 165
Inspection techniques 364 variable amplitude- 149,168
Instability Lug 255, 309, 404
fracture- (see Fracture instability)
post- 400 Magnetic particle inspection 364
Integral stringer 298 Material data 208
Integration of crack growth (see Fatigue Material data handbooks 209
crack growth analysis) Maximum shear stress 32, 58
Interaction effect (see Retardation) Mean stress 123
Intergranular fracture 8, 13 Measurement (see Testing)
Interference fit 311, 314 Mechanism of
Internal pressure 255 -cleavage 12
-dimple rupture 13
-ductile rupture 13
1 as energy release rate 88, 162, 420
-fatigue cracking 8, 9, 124
l-definition 88
-fracture 8, 12, 13, 425
l-integral 88, 112, 122
-stress corrosion 10, 425
lR (-curve) 93, 94, 101, 110, 121, 214, 353,
-void formation 13
361,419,489
Misconceptions 417
l-geometry factors (see Geometry factor)
Mixed mode 319
Mixed mode fatigue 324, 326
K (see Stress intensity and Definitions or Mode
Fracture toughness) I, II, III, definition 16, 50
Kinetic energy 460 I stress field 49
I stress intensity 49, 319
520
II stress field 318 Plane stress
II stress intensity 50 -conditions 28
III stress field 3 I 8 -cracking (see Plane stress fracture)
III stress intensity 50, 3 I 9 -fracture toughness 62
combined- 17, 319 -plastic zone 57, 60
combined-fatigue 324 -stress field 35, 59, 6 I
mixed- 17, 319 -toughness (see Plane stress fracture
opening- 16 toughness)
shear- 16 Plank 307
tearing- 16 Plastic collapse (see Collapse)
Mohr's circle 33 Plastic hinge 99
Multiple elements 299, 300, 306, 337, 340 Plastic modulus 103
MUltiple load path 258, 274, 307, 366 Plastic zone
-correction 284, 408
Net section collapse (see Collapse) fatigue- 139, 147
Net section stress: (see Collapse or -size 57, 60, 86
Reference stress) -size in plane strain 57, 60
Net section yield (see Collapse) -size in plane stress 57, 60
Notch (see Stress concentration) Post instability 400
Nozzle 338, 357 Prediction of crack growth (see Fatigue
Nuclear 17, 170, 206, 214, 333, 338,407 crack growth)
Pressure vessel 18, 290, 329, 333, 338, 357,
366, 391
Off-shore 5,171,177,178,333
Probability of
Opening mode 16
-crack detection 369, 373, 378, 390, 508
Outlook 420
-fracture 5, 389, 390, 392, 397, 398
Overload in fatigue (see Retardation) Proof testing 2, 367, 390, 393
Overload fracture 12, 425
Propeller 169

Paper 46, 87, 322, 325, 328, 329 Rail(road) 154, 155, 171, 172, 177, 180,206,
Panicles 13 465
Part-through-crack (see Surface flaw or Rainflow (see Cycle counting)
Corner crack) Random loading 168,171,201
Path independence 113, 115 Range K 125
Penetrant inspection 364, 376 Rate equation (function) 126, 130,222,226,
Permissible 232, 242, 490, 498
-crack I, 6, 20, 48, 208, 363, 392 Rate data 218, 232, 234
-residual strength 6, 48, 295, 347, 363, Ratio (stress ratio) (see Stress ratio)
392, 397 R-curve 79, 211,212,353,361
Piping 170, 462 Reference stress 244, 271, 279
Pipeline 169,222,298,333,338,357,366, Repair 2, 365, 384, 405
391 Regulations (see Damage tolerance require-
Plane strain ments)
-conditions 29, 30, 58, 64, 66, 102, 284, Replacement 2, 367, 396,423,440
487 Requirements (see Damage tolerance re-
-cracking (see Plane strain fracture) quirements)
-fracture toughness 62, 66, 67 Residual strength 4, 56, 62, 70, 82, 84, 86,
-plastic zone 57, 60 252
-stress field 34, 59, 6 I permissible- 6, 7
-testing 67 Residual strength diagram 4, 7, 15, 16,48,
-toughness (see Plane strain fracture 56, 71, 93, 96, 97, 294, 305, 345,
toughness) 361, 392
 521
Residual stress 15, 139, 145,312,316,440, arrest in- 293
467 design of- 303
Resistance (see Fracture resistance) fatigue crack growth in- 300
Retardation 136, 185,238,420,492 residual strength diagram of- 293
Retardation calibration 148, 154, 238, 416 Stiffening ratio 276, 455
Retardation models 143, 145, 154,239,412. Stiffener- (see stringer)
416,420 Stop hole - (see hole)
Rivets (see Fasteners) Storage tank 443
Rotating machinery 169 Storm 4, 171
Rules and Regulations (see Damage Strain concentration 37
tolerance requirements) Strain concentration factor 37
Rupture 12, 13,43 Strain distribution 38
Strain energy 73, 76, 88, 112
Safety factor 4, 5, 392, 402, 422, 485, 509 Strain energy release 22, 75, 77, 79, 88, 99,
Scatter 128, 222, 232, 412, 422, 470 117, 273, 322
Scc (see Stress corrosion) Strain hardening exponent 103
Secondary stresses and displacements 427 Stress concentration 22, 28, 37, 47, 336
Semi-random 168, 173, 186, 190,202,412 Stress concentration factor 25, 37, 47, 58,
Sequencing 183, 186, 202,412 276, 311
Service Stress corrosion cracking 5, 8, 15, 163,441,
-failure analysis 424 506
-loading I, 4, 168 Stress distribution (see Stress field)
-life 400 Stress field
Shear -at crack tip 48, 57, 59
double- 159 -at notch 24, 27, 34, 36, 38
-loading 324, 325 Stress at fracture (see Fracture stress,
-mode 16,319 Residual strength)
-stress maximum 32, 58 Stress history (see also Structures) 6, 151, 169,
single- 159 180,198,206,332,412,474,479,
Shell crack 472 498
Ship 5, 171, 172, 177, 178,203,206,333, Stress intensity
391,447 calculation of- (see Geometry factor)
Single shear 159 compounding of- 247, 257, 500
Size correction (see Geometry factor) definition of- 49, 51
Size requirements (see Testing and/or general form of- 52
Criteria) -range 125
Slip deformation 32, 33 simple solutions to- 243-280
Slow crack growth (see Fracture-stable) Superposition of- 249, 256, 500
Software (see Computer software) Stress range 123
Specifity 376 Stress ratio 124, 125, 220, 228
Spectrum 168, 175,205,206,240 Stress spectrum (see Spectrum)
Speed of fracture II Stress-strain curve 90, 102, 107, 121, 488
Stable crack growth (see Fracture-stable) Stress-strain equation 90, 98, 121
Stable fracture (see Fracture-stable) Stress-strain loop 141, 144
Standard spectra 205, 206, 207 Stretched zone 433
standard tests (see Testing) Striations 9, 11. 428, 432, 512
Starter notch 67 Striation count 435, 439, 441, 442
State of stress 28, 30, 58, 62 Stringer 275, 293, 448
Stereography 430 Stringer failure 296, 329
Stiffened structures Stringer stress concentration 276, 277, 280,
analysis of- 275, 293, 329, 330 294, 329
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Stripping 368, 390 -effect on toughness 61, 63, 69, 101,212
Structures Threshold
airplane 5, 17, 18, 149, 155, 172, 173, 177 fatigue cracking- 158
179, 193,203,206,237,295,302, stress corrosion- 165
327, 333, 340, 376, 391, 397, 398 Topography 430
bridge 5, 17, 237, 326, 376 Toughness (see Fracture toughness)
crane 5, 172 Transition
nuclear 17, 170,206,214,333,338,407 fatigue crack- 159
off-shore 5, 171, 177, 178, 333 temperature- 216
pipeline 169, 222, 298, 333, 366, 476 Transitional fracture toughness 62
piping 170, 462 Tresca yield criterion 32, 58, 59
pressure vessel 18, 290, 329, 333, 338, Triaxial stress field 28, 48
357, 366, 391 True stress-strain 103, 121
rail(road) 154,155,171,172,177,180, Truncation 192, 194, 195, 206, 494
206, 465
rotating machinery 169 Ultrasonic Inspection 360, 364, 376
ship 5, 171, 172, 177, 178, 203, 206, 333, Uncracked stress distribution rule 255, 267
391,447 Unstable fracture (see Fracture instability)
storage tank 443 Unit
Subcritical growth 94, (see Fatigue and -conversion 55, 229, 242
Stress corrosion) -of toughness 55
Superposition 249, 256, 317, 320, 550 -of fracture energy
Surface flaw 66, 214, 247,282, 291, 414 -of fracture resistance 79, 89
-of J 89, 90
Tangent approximation 71, 72, 350, 487, 507
Tearing mode 16 Valid toughness (see Fracture toughness)
Tearing modulus 97 Variable amplitude 136, 147
Temperature (see Effect of - on) Velocity of fracture II
Testing Visual inspection 364, 408
COD- 118, 119 Void coalescence 13
collapse strength- 44
fatigue- 126 Welding 24, 257, 298, 299, 318, 326, 357,
J R and J- 98, 100,419 367, 426, 444, 447, 476, 480
plane strain fracture toughness- 67, 87
proof- 2, 367, 390, 393 X-ray inspection 364, 376, 476, 479
stress corrosion- 164
Theoretical stress concentration factor 37, Yield (see Plastic or Collapse)
46 Yield at notch 31, 34
Thermal stress 462 Yield criterion
Thickness Von Mises- 33
-effect on fatigue crack 157 Tresca- 32, 58, 59
-effect on state of stress 30, 58, 65