[E. e. gdoutos(auth.),emmanuel_e._gdoutos,__ch(book_zz.org)

Problems of Fracture Mechanics and Fatigue

Problems of Fracture Mechanics
and Fatigue
A Solution Guide
Edited by
E.E. GDOUTOS
Democritus University ofThrace,
Xanthi, Greece
C.A. RODOPOULOS
Materials Research Institute,
Sheffield Hallam University,
Sheffield, United Kingdom
J.R. YATES
University ofSheffield,
SPRINGER-SCIENCE+BUSINESS MEDIA, B.V.

A C.I.P. Catalogue record for this book is available from the Library of Congress.
ISBN 978-90-481-6491-2 ISBN 978-94-017-2774-7 (eBook)
DOI 10.1007/978-94-017-2774-7
Printed on acid-free paper
Ali Rights Reserved
© 2003 Springer Science+Business Media Dordrecht
Originally published by Kluwer Academic Publishers in 2003
Softcover reprint ofthe hardcover 1st edition 2003
No part of this work rnay be reproduced, stored in a retrieval system, or transrnitted
in any form or by any means, electronic, rnechanical, photocopying, rnicrofilrning,
recording or otherwise, without written perrnission from the Publisher, with the
exception of any material supplied specifically for the purpose of being entered
and executed on a computer system, for exclusive use by the purchaser of the work.

A book dedicated to
those who can think,
observe and imagine

Table of Contents
Editor's Preface on Fracture Mechanics
Editors Preface on Fatigue
List of Contributors
PART A: FRACTURE MECHANICS
1. Linear Elastic Stress Field
Problem 1: Airy Stress Function Method
E.E. Gdoutos
Problem 2: Westergaard Method for a Crack Under Concentrated Forces
E.E. Gdoutos
Problem 3: Westergaard Method for a Periodic Array of Cracks Under
Concentrated Forces
E.E. Gdoutos
Problem 4: Westergaard Method for a Periodic Array of Cracks Under
xix
xxiii
XXV
3
11
17
Uniform Stress 21
E.E. Gdoutos
Problem 5: Calculation of Stress Intensity Factors by the Westergaard Method 25
E.E. Gdoutos
Problem 6: Westergaard Method for a Crack Under Distributed Forces
E.E. Gdoutos
Problem 7: Westergaard Method for a Crack Under Concentrated Forces
E.E. Gdoutos
Problem 8: Westergaard Method for a Crack Problem
E.E. Gdoutos
Problem 9: Westergaard Method for a Crack Subjected to Shear Forces
E.E. Gdoutos
31
33
39
41

Vlll Table of Contents
Problem 10: Calculation of Stress Intensity Factors by Superposition
M.S. Konsta-Gdoutos
Problem 11: Calculation of Stress Intensity Factors by Integration
E.E. Gdoutos
Problem 12: Stress Intensity Factors for a Linear Stress Distribution
E.E. Gdoutos
Problem 13: Mixed-Mode Stress Intensity Factors in Cylindrical Shells
E.E. Gdoutos
Problem 14: Photoelastic Determination of Stress Intensity Factor K1
E.E. Gdoutos
Problem 15: Photoelastic Determination of Mixed-Mode Stress Intensity
Factors K1 and Kn
M.S. Konsta-Gdoutos
Problem 16: Application ofthe Method of Weight Function for the
Determination of Stress Intensity Factors
L. Banks-Sills
2. Elastic-Plastic Stress Field
Problem 17: Approximate Determination of the Crack Tip Plastic Zone
for Mode-l and Mode-ll Loading
E.E. Gdoutos
for Mixed-Mode Loading
E.E. Gdoutos
According to the Tresca Yield Criterion
M.S. Konsta-Gdoutos
According to a Pressure Modified Mises Yield Criterion
E.E. Gdoutos
Problem 21: Crack Tip Plastic Zone According to Irwin's Model
E.E. Gdoutos
Problem 22: Effective Stress Intensity factor According to Irwin's Model
E.E. Gdoutos
45
49
53
57
63
65
69
75
81
83
91
95
99

Table of Contents
Problem 23: Plastic Zone at the Tip of a Semi-Infinite Crack According
to the Dugdale Model
E.E. Gdoutos
ix
103
Problem 24: Mode-III Crack Tip Plastic Zone According to the Dugdale Model 107
E.E. Gdoutos
Problem 25: Plastic Zone at the Tip of a Penny-Shaped Crack According
to the Dugdale Model
E.E. Gdoutos
3. Strain Energy Release Rate
Problem 26: Calculation of Strain Energy Release Rate from Load - Displacement -
113
Crack Area Equation 117
M.S. Konsta-Gdoutos
Problem 27: Calculation of Strain Energy Release Rate
for Deformation Modes I, II and III
E.E. Gdoutos
Problem 28: Compliance of a Plate with a Central Crack
E.E. Gdoutos
121
127
Problem 29: Strain Energy Release Rate for a Semi-Infinite Plate with a Crack 131
E.E. Gdoutos
Problem 30: Strain Energy Release Rate for the Short Rod Specimen
E.E. Gdoutos
Problem 31: Strain Energy Release Rate for the Blister Test
E.E. Gdoutos
Problem 32: Calculation of Stress Intensity Factors Based on Strain Energy
Release Rate
E.E. Gdoutos
Problem 33: Critical Strain Energy Release Rate
E.E. Gdoutos
4. Critical Stress Intensity Factor Fracture Criterion
135
139
143
147
Problem 34: Experimental Determination of Critical Stress Intensity Factor K1c 155
E.E. Gdoutos

X Table of Contents
Problem 35: Experimental Determination of K1c
E.E. Gdoutos
Problem 36: Crack Stability
E.E. Gdoutos
161
163
Problem 37: Stable Crack Growth Based on the Resistance Curve Method 169
M.S. Konsta-Gdoutos
Problem 38: Three-Point Bending Test in Brittle Materials
A. Carpinteri, B. Chiaia and P. Cometti
Problem 39: Three-Point Bending Test in Quasi Brittle Materials
Problem 40: Double-Cantilever Beam Test in Brittle Materials
Problem 41: Design of a Pressure Vessel
E.E. Gdoutos
Problem 42: Thermal Loads in a Pipe
E.E. Gdoutos
5. J-integral and Crack Opening Displacement Fracture Criteria
173
177
183
189
193
Problem 43: J-integral for an Elastic Beam Partly Bonded to a Half-Plane 197
E.E. Gdoutos
Problem 44: J-integral for a Strip with a Semi-Infinite Crack 201
E.E. Gdoutos
Problem 45: J-integral for Two Partly Bonded Layers
E.E. Gdoutos
Problem 46: J-integral for Mode-l
E.E. Gdoutos
Problem 47: J-integral for Mode III
L. Banks-Sills
Problem 48: Path Independent Integrals
E.E. Gdoutos
207
211
219
223
Problem 49: Stresses Around Notches 229
E.E. Gdoutos
Problem 50: Experimental Determination of J1c from J - Crack Growth Curves 233

Table of Contents Xl
E.E. Gdoutos
Problem 51: Experimental Determination of J from Potential Energy - Crack
Length Curves 239
E.E. Gdoutos
Problem 52: Experimental Determination of J from Load-Displacement Records 243
E.E. Gdoutos
Problem 53: Experimental Determination of J from a Compact Tension Specimen 247
E.E. Gdoutos
Problem 54: Validity of J1c and K1c Tests
E.E. Gdoutos
Problem 55: Critical Crack Opening Displacement
E.E. Gdoutos
Problem 56: Crack Opening Displacement Design Methodology
E.E. Gdoutos
6. Strain Energy Density Fracture Criterion and Mixed-Mode Crack Growth
Problem 57: Critical Fracture Stress of a Plate with an Inclined Crack
M.S. Konsta-Gdoutos
Problem 58: Critical Crack Length of a Plate with an Inclined Crack
E.E. Gdoutos
Problem 59: Failure of a Plate with an Inclined Crack
E.E. Gdoutos
251
253
257
263
269
273
Problem 60: Growth of a Plate with an Inclined Crack Under Biaxial Stresses 277
E.E. Gdoutos
Problem 61: Crack Growth Under Mode-ll Loading 283
E.E. Gdoutos
Problem 62: Growth of a Circular Crack Loaded Perpendicularly to its Cord
by Tensile Stress
E.E. Gdoutos
Problem 63: Growth of a Circular Crack Loaded Perpendicular to its
Cord by Compressive Stress
E.E. Gdoutos
287
291

xu Table of Contents
Problem 64: Growth of a Circular Crack Loaded Parallel to its Cord
E.E. Gdoutos
Problem 65: Growth of Radial Cracks Emanating from a Hole
E.E. Gdoutos
293
297
Problem 66: Strain Energy Density in Cuspidal Points of Rigid Inclusions 301
E.E. Gdoutos
Problem 67: Failure from Cuspidal Points of Rigid Inclusions 305
E.E. Gdoutos
Problem 68: Failure ofa Plate with a Hypocycloidal Inclusion 309
E.E. Gdoutos
Problem 69: Crack Growth From Rigid Rectilinear Inclusions 315
E.E. Gdoutos
Problem 70: Crack Growth Under Pure Shear 319
E.E. Gdoutos
Problem 71: Critical Stress in Mixed Mode Fracture
L Banks-Sills
Problem 72: Critical Stress for an Interface Crack
L Banks-Sills
Problem 73: Failure of a Pressure Vessel with an Inclined Crack
E.E. Gdoutos
Problem 74: Failure ofa Cylindrical bar with a Circular Crack
E.E. Gdoutos
327
333
339
343
Problem 75: Failure ofa Pressure Vessel Containing a Crack with Inclined Edges 347
E.E. Gdoutos
Problem 76: Failure ofa Cylindrical Bar with a Ring-Shaped Edge Crack 351
G.C. Sih
Problem 77: Stable and Unstable Crack Growth 355
E.E. Gdoutos
7. Dynamic Fracture
Problem 78: Dynamic Stress Intensity Factor
E.E. Gdoutos
Problem 79: Crack Speed During Dynamic Crack Propagation
359
365

Table of Contents
E.E. Gdoutos
Problem 80: Rayleigh Wave Speed
E.E. Gdoutos
Problem 81: Dilatational, Shear and Rayleigh Wave Speeds
E.E. Gdoutos
Problem 82: Speed and Acceleration of Crack Propagation
E.E. Gdoutos
8. Environment-Assisted Fracture
xiii
369
373
377
Problem 83: Stress Enhanced Concentration of Hydrogen around Crack Tips 385
D.J. Unger
Problem 84: Subcritical Crack Growth due to the Presence of a Deleterious Species 397
D.J. Unger
PARTB: FATIGUE
1. Life Estimates
Problem 1: Estimating the Lifetime of Aircraft Wing Stringers
J.R. Yates
Problem 2: Estimating Long Life Fatigue of Components
J.R. Yates
Problem 3: Strain Life Fatigue Estimation of Automotive Component
J.R. Yates
Problem 4: Lifetime Estimates Using LEFM
J.R. Yates
Problem 5: Lifetime of a Gas Pipe
A. Afagh and Y.-W. Mai
Problem 6: Pipe Failure and Lifetime Using LEFM
M.N.James
405
409
413
419
423
427
Problem 7: Strain Life Fatigue Analysis of Automotive Suspension Component 431
J. R. Yates

XIV Table of Contents
2. Fatigue Crack Growth
Problem 8: Fatigue Crack Growth in a Center-Cracked Thin Aluminium Plate 439
Sp. Pantelakis and P. Papanikos
Problem 9: Effect ofCrack Size on Fatigue Life 441
A. Afaghi and Y.-W. Mai
Problem 10: Effect of Fatigue Crack Length on Failure Mode of a Center-Cracked
Thin Aluminium Plate 445
Problem 11: Crack Propagation Under Combined Tension and Bending 449
J. R. Yates
Problem 12: Influence of Mean Stress on Fatigue Crack Growth for Thin and Thick
Plates 453
Problem 13: Critical Fatigue Crack Growth in a Rotor Disk
Problem 14: Applicability ofLEFM to Fatigue Crack Growth
C.A. Rodopoulos
455
457
Problem 15: Fatigue Crack Growth in the Presence of Residual Stress Field 461
3. Effect of Notches on Fatigue
Problem 16: Fatigue Crack Growth in a Plate Containing an Open Hole
Problem 17: Infinite Life for a Plate with a Semi-Circular Notch
C.A. Rodopoulos
Problem 18: Infinite Life for a Plate with a Central Hole
C.A. Rodopoulos
Problem 19: Crack Initiation in a Sheet Containing a Central Hole
C.A. Rodopoulos
467
469
473
477

Table of Contents
4. Fatigue and Safety Factors
Problem 20: Inspection Scheduling
C.A. Rodopoulos
Problem 21: Safety Factor of aU-Notched Plate
C.A. Rodopoulos
Problem 22: Safety Factor and Fatigue Life Estimates
C.A. Rodopoulos
Problem 23: Design of a Circular Bar for Safe Life
Problem 24: Threshold and LEFM
C.A. Rodopoulos
XV
483
487
491
495
497
Problem 25: Safety Factor and Residual Strength 501
C.A. Rodopoulos
Problem 26: Design ofa Rotating Circular Shaft for Safe Life 505
Problem 27: Safety Factor of a Notched Member Containing a Central Crack 509
C.A. Rodopoulos
Problem 28: Safety Factor ofa Disk Sander
C.A. Rodopoulos
S. Short Cracks
Problem 29: Short Cracks and LEFM Error
C.A. Rodopoulos
Problem 30: Stress Ratio effect on the Kitagawa-Takahashi diagram
C.A. Rodopoulos
Problem 31: Susceptibility of Materials to Short Cracks
C.A. Rodopoulos
Problem 32: The effect of the Stress Ratio on the Propagation of Short
Fatigue Cracks in 2024-T3
C.A. Rodopoulos
519
529
533
539
543

xvi Table of Contents
6. Variable Amplitude Loading
Problem 33: Crack Growth Rate During Irregular Loading
Problem 34: Fatigue Life Under two-stage Block Loading
Problem 35: The Application of Wheeler's Model
C.A. Rodopoulos
Problem 36: Fatigue Life Under Multiple-Stage Block Loading
Problem 37: Fatigue Life Under two-stage Block Loading Using Non-Linear
Damage Accumulation
Problem 38: Fatigue Crack Retardation Following a Single Overload
Problem 39: Fatigue Life of a Pipe Under Variable Internal Pressure
Problem 40: Fatigue Crack Growth Following a Single Overload Based
on Crack Closure
Problem 41: Fatigue Crack Growth Following a Single Overload Based on
551
553
555
559
563
565
569
573
Crack-Tip Plasticity 575
Problem 42: Fatigue Crack Growth and Residual Strength of a Double Edge
Cracked Panel Under Irregular Fatigue Loading 579
Problem 43: Fatigue Crack Growth Rate Under Irregular Fatigue Loading 583
Problem 44: Fatigue Life of a Pressure Vessel Under Variable Internal Pressure 585

Table of Contents
7. Complex Cases
Problem 45: Equibiaxial Low Cycle Fatigue
J.R. Yates
XVll
589
Problem 46: Mixed Mode Fatigue Crack Growth in a Center-Cracked Panel 593
Problem 47: Collapse Stress and the Dugdale's Model 597
C.A. Rodopoulos
Problem 48: Torsional Low Cycle Fatigue 601
J.R. Yates and M. W Brown
Problem 49: Fatigue Life Assessment ofa Plate Containing Multiple Cracks 607
Problem 50: Fatigue Crack Growth and Residual Strength in a Simple MSD
Problem 611
INDEX 615

Editor's Preface
On Fracture Mechanics
A major objective of engineering design is the determination of the geometry and
dimensions of machine or structural elements and the selection of material in such a
way that the elements perform their operating function in an efficient, safe and
economic manner. For this reason the results of stress analysis are coupled with an
appropriate failure criterion. Traditional failure criteria based on maximum stress, strain
or energy density cannot adequately explain many structural failures that occurred at
stress levels considerably lower than the ultimate strength of the material. On the other
hand, experiments performed by Griffith in 1921 on glass fibers led to the conclusion
that the strength of real materials is much smaller, typically by two orders of magnitude,
than the theoretical strength.
The discipline of fracture mechanics has been created in an effort to explain these
phenomena. It is based on the realistic assumption that all materials contain crack-like
defects from which failure initiates. Defects can exist in a material due to its
composition, as second-phase particles, debonds in composites, etc., they can be
introduced into a structure during fabrication, as welds, or can be created during the
service life of a component like fatigue, environment-assisted or creep cracks. Fracture
mechanics studies the loading-bearing capacity of structures in the presence of initial
defects. A dominant crack is usually assumed to exist. The safe design of structures
proceeds along two lines: either the safe operating load is determined when a crack of a
prescribed size exists in the structure, or given the operating load, the size of the crack
that is created in the structure is determined.
Design by fracture mechanics necessitates knowledge of a parameter that characterizes
the propensity of a crack to extend. Such a parameter should be able to relate laboratory
test results to structural performance, so that the response of a structure with cracks can
be predicted from laboratory test data. This is determined as function of material
behavior, crack size, structural geometry and loading conditions. On the other l}.and, the
critical value of this parameter, known as fracture toughness, is a property of the
material and is determined from laboratory tests. Fracture toughness is the ability of the
material to resist fracture in the presence of cracks. By equating this parameter to its
critical value we obtain a relation between applied load, crack size and structure
geometry, which gives the necessary information for structural design. Fracture
mechanics is used to rank the ability of a material to resist fracture within the
framework of fracture mechanics, in the same way that yield or ultimate strength is used
to rank the resistance of the material to yield or fracture in the conventional design
criteria. In selecting materials for structural applications we must choose between
materials with high yield strength, but comparatively low fracture toughness, or those
with a lower yield strength but higher fracture toughness.

XX Editor's Preface
The theory of fracture mechanics has been presented in many excellent books, like
those written by the editor of the first part of the book devoted to fracture mechanics
entitled: "Problems of Mixed Mode Crack Propagation," "Fracture Mechanics Criteria
and Applications," and "Fracture Mechanics-An Introduction." However, students,
scholars and practicing engineers are still reluctant to implement and exploit the
potential of fracture mechanics in their work. This is because fracture is characterized
by complexity, empiricism and conflicting viewpoints. It is the objective of this book to
build and increase engineering confidence through worked exercises. The first part of
the book referred to fracture mechanics contains 84 solved problems. They cover the
following areas:
• The Westergaard method for crack problems
• Stress intensity factors
• Mixed-mode crack problems
• Elastic-plastic crack problems
• Determination of strain energy release rate
• Determination of the compliance of crack problems
• The critical strain energy release rate criterion
• The critical stress intensity factor criterion
• Experimental determination of critical stress intensity factor. The !-integral and
its experimental determination
• The crack opening displacement criterion
• Strain energy density criterion
• Dynamic fracture problems
• Environment assisted crack growth problems
• Photoelastic determination of stress intensity factors
• Crack growth from rigid inclusions
• Design of plates, bars and pressure vessels
The problems are divided into three groups: novice (for undergraduate students),
intermediate (for graduate students and practicing engineers) and advanced (for
researchers and professional engineers). They are marked by one, two and three
asterisks, respectively. At the beginning of each problem there is a part of "useful
information," in which the basic theory for the solution of the problem is briefly
outlined. For more information on the theory the reader is referred to the books of the
editor: "Fracture Mechanics Criteria and Applications," "Fracture Mechanics-An
Introduction," "Problems of Mixed-Mode Crack Propagation." The solution of each
problem is divided into several easy to follow steps. At the end of each problem the
relevant bibliography is given.

Editor's Preface XXl
I wish to express my sincere gratitude and thanks to the leading experts in fracture
mechanics and good friends and colleagues who accepted my proposal and contributed
to this part of the book referred to fracture mechanics: Professor L. Banks-Sills of the
Tel Aviv University, Professor A. Carpinteri, Professor B. Chiaia and Professor P.
Cometti of the Politecnico di Torino, Dr. M. S. Konsta-Gdoutos of the Democritus
University of Thrace, Professor G. C. Sib of Lehigh University and Professor D. J.
Unger of the University of Evansville.
My deep appreciation and thanks go to Mrs Litsa Adamidou for her help in typing the
manuscript. Finally, a special word of thanks goes to Ms Nathalie Jacobs of Kluwer
Academic Publishers for her kind collaboration and support during the preparation of
the book.
April, 2003
Xanthi, Greece
Emmanuel E. Gdoutos
Editor

Editor's Preface
On Fatigue
The second part of this book is devoted to fatigue. The word refers to the damage
caused by the cyclic duty imposed on an engineering component. In most cases, fatigue
will result into the development of a crack which will propagate until either the
component is retired or the component experiences catastrophic failure. Even though
fatigue research dates back to the nineteenth century (A. Wohler1860, H. Gerber 1874
and J. Goodman 1899), it is within the last five decades that has emerged as a major
area of research. This was because of major developments in materials science and
fracture mechanics which help researchers to better understand the complicated
mechanisms of crack growth. Fatigue in its current form wouldn't have happened if it
wasn't for a handful of inspired people. The gold medal should be undoubtedly given to
G. Irwin for his 1957 paper Analysis of Stresses and Strains Near the End ofa Crack
Traversing a Plate. The silver medal should go to Paris, Gomez and Anderson for their
1961 paper A Rational Analytic Theory of Fatigue. There are a few candidates for the
bronze which makes the selection a bit more difficult. In our opinion the medal should
be shared by D.S. Dugdale for his 1960 paper Yielding ofSteel Sheets Containing Slits,
W. Biber for the 1960 paper Fatigue Crack Closure under Cyclic Tension and K.
Kitagawa and S. Takahashi for their 1976 paper Applicability ofFracture Mechanics to
Very Small Cracks or the Cracks in the Early Stage. Unquestionably, if there was a
fourth place, we would have to put a list of hundreds of names and exceptionally good
works.
To write and editor a book about solved problems in fatigue it is more difficult than it
seems. Due to ongoing research and scientific disputes we are compelled to present
solutions which are well established and generally accepted. This is especially the case
for those problems designated for novice and intermediate level. In the advanced level,
there are some solutions based on the author's own research.
In this second part, there are 50 solved problems. They cover the following areas:
• Life estimates
• Fatigue crack growth
• Effect of Notches on Fatigue
• Fatigue and Safety factors
• Short cracks
• Variable amplitude loading
• Complex cases
As before, the problems are divided into three groups: novice (for undergraduate
students), intermediate (for graduate students and practicing engineers) and advanced
(for researchers and professional engineers).
Both the editors have been privileged to scientifically mature in an department with a
long tradition in fatigue research. Our minds have been shaped by people including
Bruce Bilby, Keith Miller, Mike Brown, Rod Smith and Eduardo de los Rios. We thank
them.
We wish to express our appreciation to the leading experts in the field of fatigue who
contributed to this second part of the book: Professor M. W. Brown from the University
of Sheffield, Professor M. N. James from the University of Plymouth, Professor Y-M.

xxiv Editor's Preface
Mai from the University of Sydney, Dr. P. Papanikos from the Institute of Structures
and Advanced Materials, Dr. A. Afaghi-Khatibi from the University of Melbourne and
Professor Sp. Pantelakis from the University of Patras. Finally, we are indebted to Ms.
Nathalie Jacobs for immense patience that she showed during the preparation of this
manuscript.
April, 2003
Chris A. Rodopoulos
John R. Yates
Editors

List of Contributors
Afaghi-Khatibi, A., Department of Mechanical and Manufacturing Engineering. The University of
Melbourne, Victoria 3010, Australia.
Banks-Sills, L., Department of Solid Mechanics, Materials and Systems, Faculty of Engineering, Tel
Aviv University, Ramat Aviv, Tel Aviv 69978, Israel.
Brown, M. W., Department of Mechanical Engineering, The University of Sheffield, Sheffield, S1 3JD,
UK.
Carpinteri, A., Department of Structural Engineering and Geotechnics, Politecnico di Torino, Corso
Duca degli Abruzzi 24, 10129 Torino, Italy.
Chiaia, B., Department of Structural Engineering and Geotechnics, Politecnico di Torino, Corso Duca
degli Abruzzi 24, 10129 Torino, Italy.
Cometti, P., Department of Structural Engineering and Geotechnics, Politecnico di Torino, Corso Duca
degli Abruzzi 24, 10129 Torino, Italy.
Gdoutos, E. E., School of Engineering, Democritus University ofThrace, GR-671 00 Xanthi, Greece.
James, M. N., Department of Mechanical and Marine Engineering, University of Plymouth, Drake
Circus, Plymouth, Devon PL4 8AA, UK.
Konsta-Gdoutos, M., School of Engineering, Democritus University of Thrace, GR-671 00 Xanthi,
Greece.
Mai, Yiu-Wing, Centre for Advanced Materials Technology, School of Aerospace, Mechanical and
Mechatronic Engineering, The University of Sydney, NSW 2006, Australia.
Pantelakis, Sp., Department of Mechanical Engineering and Aeronautics, University of Patras, GR
26500, Patras, Greece.
Papanikos, P., ISTRAM, Institute of Structures & Advanced Materials, Patron-Athinon 57, Patras,
26441, Greece.
Rodopoulos, C. A., Structural Integrity Research Institute of the University of Sheffield, Department of
Mechanical Engineering, The University of Sheffield, Sheffield, S1 3JD, UK.
Unger, D. J., Department of Mechanical and Civil Engineering, University of Evansville, 1800 Lincoln
Avenue, Evansville, IN 47722, USA.
Yates, J. R, Department of Mechanical Engineering, The University of Sheffield, Sheffield, S1 3JD,
UK.

1. Linear Elastic Stress Field

Problem 1: Airy Stress Function Method ***
E.E. Gdoutos
1. Problem
In William's eigenfunction expansion method [I] the Airy stress function for a semi-
infinite crack in an infinite plate subjected to general loading is assumed in the form
(1)
where r, 9 are polar coordinates centered at the crack tip and). is real.
Using the boundary conditions along the crack faces, determine the function U and
find the expressions for the singular stress and displacement components for opening
mode and sliding mode loading.
Observe that negative values of A. are ignored since they produce infinite displacements
at the crack tip. Furthermore, use the result that the total strain energy contained in
any circular region surrounding the crack tip is bounded to show that the value ). = 0
should also be excluded from the solution.
2. Useful Information
In the Airy stress function method the solution of a plane elasticity problem in polar
coordinates is reduced to finding a function U = U(r, 9) (Airy function) which satisfies
the biharmonic equation in polar coordinates
and the appropriate boundary conditions [2]. The stress components are given by
(3)

4 E.E. Gdoutos
3. Solution
3.I GENERAL REMARKS
From Equation (I) we obtain
(4)
Thus Equation (2) becomes
(5)
The solution ofthis equation is
(6)
(7)
f 2 = C2 sin (A. -I) e+C4 sin (A.+ I) e
where the symmetric part f1 corresponds to opening-mode and the anti-symmetric part
f2 corresponds to sliding-mode.
The boundary conditions are
<Jo= 1:t9= 0, fore=± 1t (8)
We consider the two cases ofopening-mode and sliding-mode separately.
3.2. OPENING-MODE:
We have
(9)

Airy Stress Function Method
'tril =-~(.!.au)=-lr'--1df
Or r aa d9
The boundary conditions (Equation (8)) give
or
c1 cos (l-1) x + C3 cos (A.+ I) x = o
cI(A. -I) sin (A. -I) 7[ + c3(A.+ I) sin (A.+ I) 7[ = 0
[
cos (l-1) x
(l-1) sin (l-I) x
cos (A.+ I)x l[C1]
(l + 1) sin (A.+ 1) 7[ c3 =
0
For nontrivial solution the determinant ofthis equation should vanish. We get
sin 2xA. = 0
or
n
l =-, n =0, 1, 2, 3, ...
2
5
(10)
(II)
(12)
(13)
(14)
(15)
We will show later that nonpositive values of A. lead to unacceptable singularities and,
therefore, they are omitted.
The boundary conditions (Equations (II) and (12)) give
For n = 1, 3, 5, ... we have
cos ( ~- I)x =cos (~+I) x =0
sin ( ~- I) x = sin (~+ 1)x = 1 (18)

6 E.E. Gdoutos
and Equation (16) is satisfied automatically, while Equation (17) gives
For n = 2, 4, 6, ... we have
n-2
c3 =---c1
n n+2 n
and Equation (17) is satisfied, automatically, while Equation (16) gives
The function U = U1 becomes
~ I+n/2 ( n-2 n-2 n+2 )U1 = L..J r C10 cos--6---cos--6 +
2 n+2 2n = 1.3•...
~ I+n/2 ( n-2 n+2 )
~ r cln cos-2-6-cos-2-6
n -2.4....
For n = 1 we obtain the singular solution
3/2 ( 9 1 39)U1 =C11 r cos -+-cos-
2 3 2
(19)
(20)
(21)
(22)
(23)
The singular stresses corresponding to the Airy function U1 are obtained from Equa-
tions (3) as
a =-- 5cos--cos-C11 ( 6 36)
r 4rl/2 2 2
C11 ( 6 36)(J 9 =-----u-2 3 cos - + cos -
4r 2 2
(24)

Airy Stress Function Method 7
ell ( . 9 . 39)1:16 =-- SID-+SID-
4r1/2 2 2
3.3. SLIDING-MODE
Following the same procedure we obtain Equation (I5) for A, while the Airy function
u2 becomes
""' 1+n/2 (. n-2 . n+2 )U2 = ~ r C2n SID-2-9-sm-2-9 +
n -1,3,...
r 2 SID-- ---siD--L 1+nt2c (. n-2 9 n-2 . n+2 9)
n 2 n+2 2
n = 1,3,...
The singular stresses are
C21 ( 5 . 9 3 . 39)G r = ----.!2 - SID - + SID -
4r 2 2
C21 ( 3 . 9 3 . 39)Ge = --- SID-+ SID-
4r112 2 2
C21 ( 9 39)1:16 = - - cos-+3cos-
4r112 2 2
3.4. DETERMINATION OF DISPLACEMENTS
(25)
(26)
For the determination of the displacement components Ur and u9 the strain-
displacement equations in conjunction with Hooke's law are used. We have
and
for plane stress, and
Our ur I 8u8
Er=-, Ee=-+---
ar r r ae
I aur aee Ue
"Y.e=--+----
r ae Or r
(27)
(28)

8 E.E. Gdoutos
for plane strain, where E is Young' modulus, J1 is shear modulus and v is Poisson' ra-
tio.
The singular displacement u., and u9 are obtained as
C r112
[ 6 36]Ur = ~ (2K -})COS 2 - COS l
(30)
u9 =-11- - - (2K+I)sin -+sin-C r112
[ 6 36]
4J1 2 2
for opening-mode, and
ur = 21 - (2K - I) sin - + 3 sin -C r112
[ 6 36]
4J1 2 2
(3I)
u9 = 21 - (2K +I) cos-+ 3 cos-C r112
[ 6 36]
4J1 2 2
for sliding-mode, where K=(3-v)/(l+v) for plane stress, and K=3-4v for plane strain.
Equation (30) and (31) suggest that the displacements Ur, u9 for A. < 0 become infinite
and, therefore, these values of A. are unacceptable. For A. = 0 the stresses cr;i and strains
e;i take the form
(32)
where g(9) and h(9) are functions of9, and the strain energy density becomes
(33)
where ro(9) is function of9.
Then the total strain energy W contained in an circular area r < R enclosing the crack
tip is
2x R
W = JJcor drd6 (34)
0 r0

or
Airy Stress Function Method
2% R 2%
W = JJro~O) drdO= Jro(O)[logr]~ dO
0 ~ 0
2%
= (logR -logr0 ) Jro(9)d9
0
9
(35)
For r0 -+ 0, W--+ oc. Thus, the root A.= 0 ofEquation {14) is physically unacceptable.
4. References
[I] M.L. Williams (1975) On the Stress Distribution at the Base ofa Statiomuy Crack, J. Appl. Mech. Trans
ASME, 24, 109-114.
[2] S. Timoshenko and J.N. Goodier (1951) Theory ofElasticity, Second Ed., McGraw-Hill, New York, To-
ronto, London.

Problem 2: Westergaard Method for a Crack Under
Concentrated Forces ***
E.E. Gdoutos
1. Problem
Verify that the Westergaard function for an infinite plate with a crack of length 2a sub-
jected to a pair offorces at x = b (Figure la) is
( 2 2r2P a -b
(1)ZI=
x(z-b) z2-a2
y
p
b
X
1---0 •I•
01;
(a)
y
p p
b b
X
a a
p p
(b)
Figure I. An infinite plate with a crack oflength 2a subjected (a) to a pair offurces Pat x =band (b) to two
pair offurces at x = ±b.

12 E.E. Gdoutos
Then show that the stress intensity factor ofthe tip x =a is given by
K- ~
P ( b)l/2
I- (1ta)J/2 a-b
(2)
Use these results to show that for an additional pair offorces at x =-b (Figure 1b) the
Westergaard function is
(3)
and the stress intensity factor is
(4)
The Westergaard semi-inverse method constitutes a simple and versatile tool for solv-
ing crack problems. The Westergaard function for a crack problem is an analytic func-
tion that satisfies the boundary conditions of the problem. The stress field is obtained
from the Westergard function Z. For mode-l crack problems the stresses u,, uy, rxy are
obtained from Z1 as [1]
uY =ReZ1 +ylmz;
rxy =-yRez;
(5)
where Re and In denote the real and imaginary parts of a function and the prime de-
notes differentiation with respect to z.
3. Solution
3.1. WESTERGAARD FUNCTION FOR PROBLEM OF FIGURE la
To verify that the function Z1 given by Eq. (1) is the Westergaard function for an infi-
nite plate with a crack of length 2a subjected to a pair offorces at x = b (Figure 1a), we
have to show that it satisfies the boundary conditions ofthe problem. By differentiating
Eq. (I) we obtain

Westergaard Method for a Crack Under Concentrated Forces 13
At infinity we obtain from Eqs (1) and (6) for lzl~ ao:
(7)
Then Eq. (5) gives for the stresses at infinity
(8)
which indicates that the stress-free boundary condition at infinity is satisfied.
For the boundary conditions along the crack length, except point x = b, we obtain from
Eqs (1) and (6) for y = 0, z = x, jxj <a that
Re z;, lm z; are finite quantities
Under such circumstances, we obtain from the second and third Eq (5) that
which indicates that the crack lips except point x =bare stress-free.
At point x = b, y = 0 we obtain from the second Eq (5) that
which indicates the existence ofa concentrated force at that point.
At point x = b, we obtain for x ~ b that
iP
ZI=----
2[ (z-b)
The magnitude ofthe concentrated force at point x = b, y = 0 is calculated as
(9a)
(9b)
(10)
(II)
(12)

14 E.E. Gdoutos
b+£ b+£ .
PY =lim JGydx =lim J- 1 tP . dx
y--+0 y--+O x (x- b)+ 1y
b-£ b-£
I. R b+J£ l iP(x-b-i y) d
=•m e- x
y--+0 x (x _ b)2 + y2
b-£
[ ]
b+£ ( )p . -I X - b 2P . -1 8 2P 1[
=--hm tan - - =--hmtan - =---=-P(l3)
1[ y--+0 y b-£ 1[ y--+0 y 1[ 2
which indicates that at point x=b exists a pair of concentrated compressive forces of
magnitude P.
3.2. STRESS INTENSITY FACTOR FOR PROBLEM OF FIGURE la
The stress intensity factor can be calculated from the Westergaard function of a given
problem. For mode-l crack problems the stress intensity factor K1 is calculated by [1]
where the complex variable ~is measured from the crack tip.
We obtain
P a2-b2
K1 = lim J2x~ - - - -
1~1--+0 x(~+a-b) ~(~+2a)
p...{2; Ja2 -b2 P ~+b
= x(a-b) ~ = ..j;; a-b
which shows that K1 is given by Eq. (2).
(14)
(15)

3.3. WESTERGAARD FUNCTION FOR PROBLEM OF FIGURE 1b
The Westergaard function for a pair offorces at x =- b is obtained from Eq. (1) as
Z1(-b) = p
x (z+b)
(16)
Thus, the Westergaard function Z1 for the problem of Figure 1b is obtained by adding
the Westergaard function for a pair of concentrated forces at points x =band x =-b.
We have
(17)
which shows that Z1is given by Eq. (3).
3.4. STRESS INTENSITY FACTOR FOR PROBLEM OF FIGURE 1b
The stress intensity factor is calculated from the Westergaard function using Eq. (14).
We obtain
2P Ja
= ~a2 -b2 v-;;
which shows that K1 is given by Eq. (4).
4. References
(18)
[IJ E.E. Gdoutos (1993) Fracture Mechanics- An Introduction, K.luwer Academic Publishers, Dordrecht,
Boston, London.

Problem 3: Westergaard Method for a Periodic Array of
Cracks Under Concentrated Forces **
E.E. Gdoutos
1. Problem
Consider an infinite periodic array of equally spaced cracks along the x-axis with each
crack subjected to a pair of concentrated forces at the center of the crack (Figure I).
Verify that the Westergaard function is
z = Psin(xa/W) 1 _ sin(xa/W) 2
[ l-112
1 W (sin (xz/ W))2 (sin (xz/ W))
Then show that the stress intensity factor is given by
p
p
K, =(w . xa)I/2.
2 smW
y
(1)
(2)
X
Figure /. An infinite periodic array ofequally spaced cracks subjected to a pair ofconcentrated forces P at their
center in an infinite plate.

18 E.E. Gdoutos
See Problem 2.
3. Solution
From Equation (1) we have
(3)
For lzl ~ oo we have
Z1 =0, y Im Z~ = y Re Z~ =0 (4)
Then Equation (5) ofProblem 2 gives
(5)
Fory =0, lx-nWI <a, n =0, 1, 2, 3, ... we have
z=x
. (xx) . (xa)SID W <SID W (6)
Re Z1 = 0, Re Z~, Im Z~ = finite
Then, Equation (5) ofProblem 2 gives
(7)
At x = 0, ay becomes infinite, indicating the existence of a concentrated force at that
point. For y = 0, a< lx-nWI < (n+112)W, n =0, 1, 2, 3, ... the quantity [sin2
(xz/W) -sin 2 (xa/W)]"2 is real and according to Equation (5) of Problem 2 ay is
given by Equation (1) for z = x, y = 0. The magnitude ofthe concentrated force at x =

Westergaard Method for a Periodic Array ofCrack Under Concentrated 19
Forces
0, y = 0 is obtained by taking equilibrium equation along the x-axis ofthe half-plane y
> 0. As in Problem 2 we obtain that this force is equal toP.
K1 is calculated from [I]
(8)
We have
I
(9)
. 2[x(a+~)J . 2 (xa)SID -SID -
w w
and
sin [ x (~+~)J = sin ( ~) + ~ cos ( ~) (IO)
. 2 [x(a+~)J . 2 (xa) x2 ~2 2 (xa) 2x~ . (xa) (xa)SID -sm - =--cos - +--sm - cos -
w w w2 w w w w
(11)
Thus
. Psin ( ~)
K1 =hmJ2x~ Z1 =----''----'-
i~l-+0 w
[ x2
~2 2 (xa) 2x~ . (xa) (xa)]w2 cos w +-wsm w cos w
(I2)
We have for K1

20 E.E. Gdoutos
p
(13)
W. (2xa}-Stn -~
2 w
Note that for W/a ~ oc the above solution reduces to the case of a single crack (K1 =
PI ,J;;, Equation (2) ofProblem 2 with b = 0).
4. References
[l] E.E. Gdoutos (1993) Fracture Mechanics - An Introduction, Kluwer Academic Publishers, Dordrecht,
Boston, London.

Problem 4: Westergaard Method for a Periodic Array of
Cracks Under Uniform Stress**
E.E. Gdoutos
1. Problem
Consider an infinite periodic array of equally spaced cracks along the x-axis in an infi-
nite plate subjected to equal uniform stresses u along the x- and y-axes at infinity (Fig-
ure l). VerifY that the Westergaard function is
0 (n:z)USID W
(l)
Then show that the stress intensity factor is given by
( )
1/2
112 W n:aK 1 =u(n:a) -tan-
n:a W
(2)
r-------------1~------------,I I
I I
I Yf I
I W W I
a 1 1 a
...._.. - .--...I X I
I ~~~ ~~~ ~~~ I
I I
I I
I I
L-------------l~------------J
Figure lo An infinite periodic array of equally spaced cracks in an infinite plate subjected to equal unifunn
stresses a at infinityo

22 E.E. Gdoutos
See Problem 2.
3. Solution
From Equation (I) we have
For y = 0, lx-WI< awe have
z=x
Thus,
. (1tx) . (1ta)sm W <sm W, since
wa<-
2
ReZ1 = 0, Rez; , Imz; =finite
(3)
(4)
(5)
(6)
(7)
(8)
(9)
Since Z1 satisfies all boundary conditions it is the Westergaard function ofthe problem.
K1 is calculated from [1]

Westergaard Method for a Periodic Array ofCracks Under Uniform Stress 23
where
~=z-a (II)
We have
(12)
Since for ~~ ~ 0, cos x~ ~I and sin x~ =x~, we obtain
w w w
(13)
and
(14)
or
(15)
( xa xa) .Note that for W I a ~ ex> tan W =W the above solution reduces to the case of a
single crack (K 1 = a/ii).
4. References
[1] E.E. Gdoutos (1993) Fracture Mechanics- An Introduction, Kluwer Academic Publishers, Dordrecht,
Boston, London.

Problem 5: Calculation of Stress Intensity Factors by
the Westergaard Method**
E.E. Gdoutos
1. Problem
The Westergaard function Z for the concentrated forces P and Qapplied at the point x
= b(b <a) ofa crack AB oflength 2a in an infinite plate (Figure Ia) is
Z=-- - - +-- +IQ+iP [(K -I) I I [J%2-a2 ]]
2n: K +I ~z2 -a2 b-z z2 -a2
(I)
Show that the complex stress intensity factor K = K1- iKn at the tip B ofthe crack is
K = Q + iP ( K - 1 + ~a + b ) .
2..[;; K +I b-a
(2)
Then show that for equal and opposite distributed forces ay(x,O) and r"Y(x,O) on the
upper and lower crack faces (figures lb) K1 and Kn are given by
a~1 a+x
K 1 = r- Jay(x,O) - - dx
vn:a a-x
-a
(3a)
a ~I a+x
Kn = r- Jrxy(x,O) - - dx.
vn:a a-x-a
(3b)
Then determine the values ofK1 for uniform (Figure 2a) and triangular (Figure 2b, c)
equal and opposite distributed forces on the upper and lower faces of a crack of length
2a in an infinite plate.

26 E.E. Gdoutos
y
b
a
,........~--:
,/
!tor.IJf.O
p
A-=====~====-=-B--x
A
.___a a- ........
...............'-_
-a----a-
(a) (b)
Figure 1. A crack oflength 2a subjected (a) to concentrated furces P and Q and (b) to distributed forces ay(x,O)
and txy(x, 0) along the crack faces.
IIIII.,.__ 2a ----t
(a)
,__ 2a ---t
{b)
,___ 2a ---.t
(c)
Figure 2. A crack oflength 2a in an infinite plate subjected to (a) a uniform and (b, c) triangular opposite forces
on the upper and lower crack faces.
See Problem 2.
3. Solution
3. I STRESS INTENSITY FACTOR K AT TIP B
K is calculated as [1]
or
K = K 1 - iK 11 =lim .J'2« Z
!~l-+0
(4)

Calculation of Stress Intensity Factors by the Westergaard Method 27
K I. ~Y (Q+iP) [(JC-1) I I [ b2
-a2
Ill- IIDv.<.1t<, - - - -- + +
i~J-->o 21t JC+l ~ (l;+a)z -az b-(a+ I;) (a+ Q2 -a2
r .j21tf,(Q+iP)[(JC-l) 1 1 [ b2
-a2
1]]
=:~1~!1J 1t ~ K+l ~1;(1;+2a) + b-(a+ I;) 1;(1;+2a) +
= (Q+iP) [(~) ~+-I~ x (b 2
-a2
) l21t IC+I v-; b-a a
(5)
= (Q +iP) [(~) + r;;+b l21t IC+I V~
The mode-l and mode-II stress intensity factors K1 and Krr are given by
(6a)
p (IC-1) Q ra+b
Ku =- 2~ K+I + 2~ v;=t; (6b)
3.2. STRESS INTENSITY FACTOR FOR FIGURE Ib
Letting P = cry (x, 0) dx and Q= r,y (x, 0) dx integrating from x = -a to x = a the K1 and
Kn expressions for a crack subjected to arbitrary loads on the upper crack surface are
I af Jg[+x I (IC-1) 3
K 1 = r-- cry(x,O) --dx+ r-- -- Jrxy(x,O)dx
2-yxa a-x 2-vxa K+I-a -a
(7a)
I (IC-I) 3
1 a J§+xK 11 =- r-- -- Jay (x, 0) dx + r-- Jrxy (x, 0) - - dx
2-vxa K+l 2-vxa a-x-a -a
(7b)
For equal and opposite forces on the upper and lower crack faces using the symmetry
equations
(8a)

28 E.E. Gdoutos
(8b)
we obtain
a~I a+x
KI = ~ Jay(x, 0) - - dx
-vna a-x
-a
(9a)
a ~I a+x
K 11 = ,...- Jtxy(x, 0) - - dx
vna a-x
-a
(9b)
3.3. STRESS INTENSITY FACTOR FOR FIGURE 2a
For uniform stress distribution Cfy (x, 0) = cs0 we obtain from equation (9a)
a~I a+x
KI = ,...- Ja0 - - dx
vna a-x
-a
(10)
Putting u = x/a we have
a~ I~ Ia +x l+u . -I 2
J - dx =a J - du =a [sin u- ~]a-x I-u -1
-a -1
(11)
and
cso ,--
KI = ,...-an= cs0 -vna
vna
(I2)
which is identical to the value of K1 for a crack of length 2a subjected to a uniform re-
mote stress cs0•
3.4. STRESS INTENSITY FACTOR FOR FIGURE 2b
For the triangular stress distribution of Figure 2b, Cfy = x cs0/a. We obtain from Equa-
tion (9a)

Calculation of Stress Intensity Factors by the Westergaard Method 29
K, =_I_ aJ<rox ~a+x dx (13)
.,j;"; a a-x
-a
Putting u = x/a, and integrating by parts we have
I I
= a2 sin-1(1)- (-a2) sin-1(-l) -a2 Jsin-1 udu + a2 JPdu
-1 -1
=-a2[usin-1 u + ~]1
+ ~[u~+ sin-1 u]
1
_, 2 -1
2
(14)
and
(15)
3.5. STRESS INTENSITY FACTOR FOR FIGURE 2c:
For a triangular stress distribution with <ry (-a)= 0 and Oy (a)= o0 we obtain by super-
position ofthe stress intensity factors ofFigures 2a and 2b.
(16)
4. References
[I] E.E. Gdoutos (1993) Fracture Mechanics- An Introduction, Kluwer Academic Publishers, Dordrecht,
Boston, London.

Distributed Forces **
E.E. Gdoutos
1. Problem
Show that the Westergaard function for the configuration ofFigure 1 is
[ [ ( )1/2]]2cr z b b z2 - a 2
Z1=- 2 2112 arccos(-)- arccot - - 2--21t (z -a ) a z a - b
and the stress intensity factor is
y
a a
X
a a
...1·-- a--·~1~-"~•...,_- a
(1)
(2)
Figure /. A crack of length 2a in an infinite plate subjected to a unifunn stress distribution a along the interval
b:<;;lxl:<;;a.

32 E.E. Gdoutos
See Problem 2.
3. Solution
According to Problem 2 the Westergaard function Z for a pair of concentrated forces a
at the points ±x is given by
(3)
The function Z1 for the problem ofFigure 1 is
(4)
or
2a [ z (b) [bJ%2
-a
2
]]Z1 =- ,-:;----;;arccos - -arccot - - 2--2
x vz2 - a2 a z a - b
(5)
From Problem 2 we obtain that K, for a pair of concentrated forces a at points ± x is
given by
(6)
Thus, the stress intensity factor, K" for the Problem ofFigure 1 is
K1 =aJ 2a ~dx =-2a ~[arccos(~)]a=2a ~arcsin(~) (7)
b ~a2- x2 ~; ~; a b ~; a
4. References
[I] E.E. Gdoutos (1993) Fracture Mechanics -An Introduction, Kluwer Academic Publishers, Dordrecht,
Boston, London.

Concentrated Forces **
E.E. Gdoutos
1. Problem
Consider a crack of length 2a in an infinite plate subjected to the concentrated forces P
at a distance y0 from the crack (Figure 1). Verify that the Westergaard function is
(I)
where
(a2 +y~)I/2 z
f(z, Yo· a)= 2 2 2 2 112 ·
z +Yo (z -a )
(2)
Determine the stress intensity factor K1•
(
T
YI Yo
t
..X
1---0 ..,. 0 "'"1 Yo
lp
.i
Figure I. Acrack oflength 2a in an infinite plate subjected to concentrated forces P.

34 E.E. Gdoutos
See Problem 2.
3. Solution
From Equations (1) and (2) we obtain for the Westergaard function z,
(3)
(4)
(5)
For y = 0, z = x, we have
2(1- v)~a2 +y~
(6)
For y = 0, lxl < a, we have
(7)
(8)
For x = 0, y = y0, z = iy0, the stress ay calculated from Equation (5) of Problem 2 be-
comes infinite, indicating the existence of a concentrated force at that point. The
magnitude of the concentrated force is obtained by taking the equilibrium equation
along the x-axis ofthe half-plane y > 0. We have

<Xl
PY- 2 J<Jydx = 0 (9)
0
or
<Xl
P -2J<J dx=Oy y (10)
a
or
p ~a2 +y~ j d(x 2 +y~)
Jt 2 a2+y~(x2+y~)~(x2+y~)-(a2+y~)
p y~
Py~ ~a2 +y~ j d(x2 +y~)
+ 2x(l-v) 2 82+y~ (x2 +y~)~(x2 +y~)-(a2 +y~)

36
Thus
p
2n(l-v)
+ Py~Ja2 +y~
4n(l-v)
E.E. Gdoutos
which indicates that the concentrated load at point x = 0, y = y0 has magnitude P.
The stress intensity factor is calculated as [1]
or
(11)
(12)
(13)

p 1 Yo
~[ 2 l= ~a2 +y~ + 2(1-v)(a2 +y~)312
(14)
Note that for y0 = 0 the above solution reduces to the value of stress intensity factor of
case (a) ofProblem 2 (K, = PI ..r;;).
4. References
[l] E.E. Gdoutos (1993) Fracture Mechanics- An Introduction, K.luwer Academic Publishers, Dordrecht,
Boston, London.

Problem 8: Westergaard Method for a Crack Problem**
E.E. Gdoutos
1. Problem
Consider the Westergaard stress function
(1)
Find the loading that represents in an infinite plate with a crack of length 2a along the
x-axis and determine the stress intensity factor.
See Problem 2.
3. Solution
From Equation (1) we have
(2)
For y =0, lxl < a we have
ReZ1 = 0, y lm Z~ =y Re Z~ =y Re Z~ =0 (3)
(4)
Let put
(5)

40 E.E. Gdoutos
For lzl ~ ao, r = r1 = r2 and 9 = 91 = 92 we obtain
(6)
(7)
Equations (5) ofProblem (2) give [1]
. a a
ax =ReZ-ylmZi =-rcos8=-x (Sa)
a a
. a a
ay =ReZ+ylmZi =-rcos 9=-x
a a
(8b)
. a
't' =-yReZ1 =--yxy a
(8c)
From Equations (3) and (8) we conclude that the Westergaard function Z; corresponds
to an infinite plate with a crack of length 2a subjected to stresses ax= ay = (a/a) x and
't'xy = - (a/a) y at infinity.
The stress intensity factor is calculated as [1]
4. References
[I) E.E. Gdoutos (1993) Fracture Mechanics -An Introduction, Kluwer Academic Publishers, Dordrecht,
Boston, London.

Problem 9: Westergaard Method for a Crack Subjected to
Shear Forces **
E.E. Gdoutos
1. Problem
VerifY that the function z;11 for an infinite plate with a crack of length 2a subjected to
a pair ofshear forces Sat x = b (Figure 1) is
z;n =___s_~(-a2_-_b_2)"2
x(z-b) z2 -a2
Determine the stress intensity factor.
y
b~s
0
....t......_- a ----•-+t••-a ®
X
Figure 1. An infinite plate with a crack oflength 2a subjected to a pair ofshear forces S at x = b.
See Problem 2.
(1)

42 E.E. Gdoutos
3. Solution
From Equation (1) we have for the stress t yz along the crack surfaces (y =0, z =x *b,
!xl<a)[l]
tyz= Re Z~11 = 0 (2)
Furthermore, we have for the stresses tyz and tyz. at infinity (z --+ co)
(3)
At y = 0, x = b, the tyz stress becomes infinite, indicating the existence of a concen-
trated force at that point.
For x --+ b we have
S ~a2 -b2 iSZ~u = - - - =
x(z-b) b2 -a2 x(z-b)
(4)
The magnitude ofthe concentrated force at x = b is calculated as
b+t
T =lim r+tt dx =lim Re I- I iS dx
y-+0 -s yz y-+O x x(x-b)+iy
b-e
l . b+Is I iS(x-b-iy) d
= 1m -- x
y-+O x (x-b)2 +y2
b-s
b+s b+£
= lim J-..!._ SY dx = -~lim I y dx
y-+O x (x-b)2+y2 x (x-bi+y2
b-s b-s
[ ]
b+£ ( )s - -I X- b 2S . -I 6 - 2S 1t
=-- hm tan - - = --hm 1an - =--- =-S
X y-+0 y b-s 1t y-+0 y X 2
(5)

Westergaard Method for a Crack Subjected to Shear Forces 43
Thus, the Westergaard function defined by Equation (1) satisfies the boundary condi-
tions ofthe problem of an infinite plate with a crack of length 2a subjected to a pair of
shear forces S at x = b.
The mode-III stress intensity factor is calculated as [I]
We have
4. References
S a2 - b2
Km =lim J2x~ - - - - 1 - - -
ICI--+D 1t (~+a- b) ~(~ + 2a)
s..fi; ~a2 -b2 s /a+b
= x(a-b) 2a = ;;;v-;=t;
(6)
(7)
[I] E.E. Gdoutos (1993) Fracture Mechanics -An Introduction, Kluwer Academic Publishers, Dordrecht,
Boston, London.

Superposition *
MS. Konsta-Gdoutos
1. Problem
Consider a strip specimen of width b with an edge crack of length a loaded by a trian-
gular tensile stress as shown in Figure. 1. Determine the stress intensity factor at the
crack tip. Obtain numerical results for alb= 0.4
Figure I. A strip with an edge crack subjected to a triangular stress distribution perpendicular to the crack
along its upper and lower boundaries.
The stress intensity factor expresses the strength of the singular elastic stress field in
the vicinity of the crack tip. For opening-mode problems the stresses Gx, Gy and Txynear
the crack tip are given by [1]

46 M.S. Konsta-Gdoutos
K1 8 (1 . 8 . 38)
ox== ~21tr cos 2 -SID 2 SID 2
K, 8 ( 1 . 8 . 38)Oy == r::;::::: COS- +SID -SID -
v21tr 2 2 2
(1)
K, 8 . 8 38
T - - - COS - SID - COS -
xy - ~21tr 2 2 2
where K1 is the stress intensity factor, and r, e are the polar coordinates at the point
considered centered at the crack tip.
Equation (1) applies to all crack-tip stress fields independently of crack/body geometry
and loading conditions. The stress intensity factor depends linearly on the applied load
and is a function of the crack length and the geometrical configuration of the cracked
body. Results for stress intensity factors for a host ofcrack problems ofpractical impor-
tance are presented in relevant handbooks [2, 3].
3. Solution
3.1. SUPERPOSITION
The problem of Figure 1 can be considered as a superposition of the two problems
shown in Figure 2 for which the stress intensity factor is obtained from existing solu-
tions [2, 3].
3.2. STRESS INTENSITY FACTOR FOR UNIFORM AND BENDING LOADS
The stress intensity factor K: for a single-edge cracked plate under uniform tension o
is given by [2, 3]
KJ ~n,J,;";[112- 023 [~l+ 1055 [ ~r-21.72 [ ~)' + 3039 [~n(2)
a
-< 0.6.
b

Calculation of Stress Intensity Factors by Superposition 47
a~, 0/2 C/2 ~M
~
I 0/2
+
14-------- b bI
o/2
a 0/2
(a) (b) (c)
Figure 2. Superposition ofthe triangular load as sum ofa uniform tensile and a bending load.
The stress intensity factor K ~ for a finite width strip with an edge crack under bend-
ing is given by (2, 3]
3.3. STRESS INTENSITY FACTOR FOR TRIANGULAR LOAD
The stress intensity factor K1 for the triangular load of Figure l is obtained by adding
the stress intensity factors for the uniform and bending loads. We have
In our case we have
cr
cro =-
2
(4)
(5a)
(5b)

3.4. NUMERICAL RESULTS FOR alb= 0.4
For alb = 0.4 we obtain for K: and K ~ :
Kl = aJ;'; [1.12 -0.23x(0.4) +10.55x(0.4)2 - 21.72x(0.4)3 + 30.39x(0.4)4 ]
2
=l.052aJ;';
b Gb2 J;;[ 2 3 4]K1 = 6--- 1.12-1.40x(0.4)+7.33x(0.4) -13.08x(0.4) + 14.0x(0.4)
12 b2
aJ;'; c-=--xl.254 =0.627a.yxa
2
Thus, the stress intensity factor for the triangular load is obtained as
(6a)
(6b)
K1 =K; +K~ = 1.052 aJ;';+ 0.627 aJ;'; = 1.679 aJ;'; (7)
4. References
Boston, London.
[2] G.C. Sib (1973) Handbook ofStress Intensity Factors, Institute ofFracture and Solid Mechanics, Lehigh
University.
[3] Y. Murakami (ed.) (1987) Stress Intensity Factors Handbook, Pergamon Press.

Integration *
E.E. Gdoutos
1. Problem
Use Problem 5 to determine the stress intensity factor for a crack of length 2a in an
infinite plate subjected to uniform normal stress a and shear stress t along the upper
crack surface from x = b to x = c. Then determine the stress intensity factor when the
same stresses apply to the lower crack surface. Finally use Problem 2 to determine the
stress intensity factor when additional normal and shear stresses a and t apply along
the upper and lower crack surface from x =- c to x = - b (Figure I).
c •I
f--b-i
a a
7" T
a a
._1·-- a ----t·l~-·-- a ---~·I
Figure I. A crack oflength 2a in an infinite plate subjected to a unifurm normal stress a and shear stress T from x
=±btox=±a.
See Problems 5 and I0.
3. Solution
3.1 STRESS APPLIED ALONG THE UPPER CRACK SURFACE FROM x =b TO
x=c

50 E.E. Gdoutos
Using Problem 5 we obtain the values of stress intensity factors K1 and Kn for a uni-
form normal stress a and shear stress t along the upper crack surface
(J cJg+X t (K-1) C
K1 = - - --dx+-- - - dx
2~ b a-x 2~ K +1 I (1)
or
K _ oa [. _1 xgx2 lc t{c-b) (-K-1)1 - - - sm -- -- +_...:...-=,:..
2~ a a2 2~ K+1
b
(2)
or
K _ a.Ja [ . _1 c . _1 b gc2
Hb2
] t (c-b) (K -1)I - - - sm -- sm -- -- + - - + - -
2.{; a a a2 a2 2~ K+ 1
(3)
and
(J (K-1)CJ t CI§+XK 11 =--- - - dx+-- --dx
2~ K+1 2~ J a-x
b b
(4)
or
o(c-b)(K-1) t.Ja [. _1 c . _1 b g2 ~2 ]K11 =- - - +-- sm --sm -- 1--+ 1--
2~ K+ 1 2.{; a a a2 a2
(5)
3.2 STRESSES APPLIED ALONG THE LOWER CRACK SURFACES FROM x = b
TOx=c
From symmetry considerations we obtain the values ofK1 and Kn for a uniform normal
stress a and shear stress t along the lower crack surface
K a.Ja [ . -I c . -1 b /.71 c2 [b21 b2 ]
I= 2.{; Sin ;-Sin ;-vI-~ + f-~ - t (c-b) (~)
2~ K+1
o(c-b) (K-1) t.Ja [. _1 c . _1 b g 2 H 2 ]K11 = - - + -- SID --SID - - 1- - + 1- -
2& K+ 1 2.{; a a a 2 a2
(6)
(7)

Calculation ofStress Intensity Factors by Integration 51
By superposing the above solutions we obtain the values ofK1and Kn when normal and
shear stresses a and t apply along the upper and lower crack surface from x = b to x =
c
3.3 STRESSES APPLIED ALONG THE UPPER AND LOWER CRACK SURFACE
FROM x = b TO x = c
K o.Ja[. _1 c ._1 b~c2 gb2
]1 = - - SID - - SID - - - - + --
.;; a a a2 a2
(8)
K r:.Ja[. _1 c . _1 b g c2
gb2]II=-- SID --SID - - - - + --
.;; a a a2 a2
(9)
3.4 STRESSES APPLIED ALONG THE UPPER AND LOWER CRACK SURFACES
FROM x=b TO x=c AND FROM x = -c TO x = -b
Using Problem 2 we obtain the value of K, when an additional normal stress a applies
along the upper and lower crack surfaces from x = -c to x = -b
or
4. References
K 2o.Ja(.-1 c .-1 b)1 =--SID --SID -
.fi a a
(10)
(11)
Boston, London.

Problem 12: Stress Intensity Factors for a Linear Stress
Distribution *
E.E. Gdoutos
1. Problem
The stress intensity factor for an edge crack of length a in a semi-infinite plate sub-
jected to a pair of equal and opposite concentrated forces at a distance b from the plate
edge (Figure 1) is given by [I]
where
F(bl a)== [I- (b I a)2 ] [0.2945- 0.3912(b I a)2 + 0.7685(bl a)4
-0.9942(bla)6 +0.5094(bla)8].
(1)
(2)
Using this result show that the stress intensity factor for this crack subjected to a self-
balanced linear tensile stress distribution acting along the crack faces (Figure Ib) is
K1 == 0.683a.J;;.. (3)

54 E.E. Gdoutos
------------, ------------,
p a
p a
____________.J ____________.J
(a) (b)
Figure 1. An edge crack in an infinite plate subjected to (a) a pair ofconcentrated furces at a distance b from the
plate edge and (b) a self-balanced linear tensile stress distribution acting along the crack taces.
See Problem l0
3. Solution
The stress intensity factor K1 for the case of Figure lb is calculated by integration of
the stress intensity factor ofFigure I a with P =a (b/a)
K, =_3__ aJ l+F(b/a) ..Ja(a~)db = 2afa JI l+F(x) xdx (4)
..[; o ~a2-b2 a ..[; o ~
or
K1
= 2a../a Jl.2945x -0.6857x3 +1.1597x5 -1.7627x7 +1.5036x9 -0.5094x II dx (S)
..[; o J1-x 2
where x = b/a
We have

so that
We have
Stress Intensity Factors for a Linear Stress Distribution
1 xdx
J -1
o .JI-x2 -
1 x3dx
fp=0.6667
o 1-x
f1 x5dx 4
~ = -5 X 0.6667 = 0.5334
o -yt-x
f1 x7dx 6
~ = -7 X 0.5334 = 0.4572
o-yt-x
f1
x9dx 8
~ =-9 X 0.4572 =0.4063
o '/1- x-
1 x11dx 10
JJ0! =-11 X 0.4063 =0.3694
0 1-x
The stress intensity factor K1 is calculated as:
55
(6)
(7)
(8)
(9)

56 E.E. Gdoutos
2
K 1 =- cr.J;;(1.2945 x1- 0.6857 x0.6667 +1.1597 x0.5334
1t
- 1.7627 X 0.4572 +1.5036 X 0.4063 - 0.5094 X 0.3694)
=3_ x1.0729 cr.J;; =0.683 cr.J;;
1t
4. Referentes
(10)
[1] G.C. Sib (1973) Handbook ofStress-Intensity Factors, Institute of Fracture and Solid Mechanics, Lehigh
University.

Problem 13: Mixed-Mode Stress Intensity Factors in
Cylindrical Shells **
E.E. Gdoutos
1. Problem
A cylindrical pressure vessel of radius R and thickness t contains a through crack of
length 2a oriented at an angle Pwith the circumferential direction (Figure I). When
the vessel is subjected to an internal pressure p, determine the stress field in the vicin-
ity ofthe crack tip.
(a)
....--11~-,I , I
I I
I I
I Ll
a:4-t 0 ......
z I "V I Uz
I ./ I
I I
I I
L---l---~
aa
(b)
Figure I. (a) A cylindrical pressure vessel with an inclined though the thickness crack and (b) stresses acting
in a local element containing the crack.
Because the crack is oriented at an angle with the circumferential direction the stress
field in the vicinity of the crack tip is of mixed-mode, that is, a combination of open-

58 E.E. Gdoutos
ing-mode (mode-l) and sliding-mode (mode-11). The stress components crx, cry, txy for
opening-mode loading are given by [I]
K I a (1 . a . Je )CJ =--cos- -sm- sm-
x .J21tr 2 2 2
KI a (1 . a . 3a)Gy = r;;-:- COS - +SID -SID -
.y27tr 2 2 2
(1)
KI a . a 3at = - - cos - sm - cos -
xy .J21tr 2 2 2
where KI is the opening-mode stress intensity factor and r and aare the polar coordi-
nates ofthe point considered centred at the crack tip.
For sliding-mode we have [1]
CJ = - - - sm- 2+cos- cos-Kn . a ( a 3a )
x .J21tr 2 2 2
K11 • a a 3aCJ = - - sm - cos - cos -
Y .J21tr 2 2 2
(2)
K11 a (1 . a . 3a )t =--cos- -sm -sm-
xy .J21tr 2 2 2
where K11 is the sliding-mode stress intensity factor.
When the cracked plate is subjected to uniform stresses a and kCJ perpendicular and
along the crack axis, respectively, the CJx stress along the crack axis is given by
KI a (1 . a . 3a) (I k)CJx =--cos- -sm- sm- - - CJ
.J2u 2 2 2
(3)
3. Solution
We consider a local element containing the crack and calculate the stresses acting on
the element. Then we determine the stress field in the vicinity of the crack tip using
Equations (1) to (3).

Mixed-Mode Stress Intensity Factors in Cylindrical Shells 59
3.1. STRESSES IN THE VESSEL
The longitudinal a. and hoop a9 stresses in the cylindrical vessel are obtained from
equilibrium along the longitudinal and hoop directions, respectively.
Equilibrium along the longitudinal axis ofthe vessel (Figure 2b) gives
(a) (b)
Figure 2. Stress equilibrium along (a) the longitudinal and (b) hoop directions of the cylindrical vessel of
Figure I
or
pR
0' =-
z 2t
Equilibrium along the hoop direction (Figure 2b) gives
or
2ta9 =2Rp
pR
0' =-
z t
(4)
(5)
(6)
(7)

60 E.E. Gdoutos
3.2. STRESS TRANSFORMATION
Consider a local element containing a crack of length 2a that makes an angle ~ with
the y direction and subjected to stresses CJ and ko along the y and x directions, respec-
tively (Figure 3). By stress transformation we obtain the following stresses
, .
T~y in the system x'y' (Figure 3b).
Figure 3. An inclined crack (a) in a biaxial stress field and (b) stress transformation along and perpendicular
to the crack plane.
. k+l k-1
CJ =--CJ - - - CJ cos 2f.l.
X 2 2 I'
(8a)
. k+l k-1
CJY = --CJ +--ocos2~
2 2
(8b)
. k-1
T = ---osin 2~
xy 2
(8c)
The crack is subjected: (a) to a biaxial stress CJY, (b) to a normal stress
(o~ -o~)along the x-axis and (c) to a shear stress T~. Thus, the stress field at the
crack tip is obtained by superposing an opening-mode loading caused by the stress CJ~
and a sliding-mode loading caused by the stress t~ . The stress (CJ~ - o~) does not

Mixed-Mode Stress Intensity Factors in Cylindrical Shells 61
create singular stress but should be subtracted from the cr~ stress along the x -axis.
From Equations (6), (1 ), (2) and (3) we obtain for the stresses cr~, cr~, r~
where
. KI 9 (• . 9 . 39)
crx = ~21tr cos 2 -sm 2 sm T -
- - sm- 2+cos- cos- -(k-1) cos 29Kn . 9 ( 9 39 )
~ 2 2 2
. KI 9 ( . 9 . 39 ) Ku . 9 9 39
cry= ~21tr cos 2 l+sm 2 sm 2 + J2nr sm 2 cos 2 cosT
t - - - cos- sm - cos -+ - - cos - 1- sm - sm -
. KI 9 . 9 39 Ku 9 ( . 9 . 39 )
xy-~ 2 2 2 ~ 2 2 2
KI = .!_[k +I+ (k -1) cos 2~] crm
2
k-1 0 c-
Kn =---sm 2~ crvna.
2
3.3. STRESS FIELD AROUND THE CRACK TIP
(9a)
(9b)
(9c)
(lOa)
(lOb)
For the case of the cylindrical vessel the stress field around the crack tip is given by
Equations (7) where the stress intensity factors K1 and Krr are given by
(lla)
(lib)
4. References
Boston, London.

Problem 14: Photoelastic Determination of Stress Intensity
Factor K1 *
E.E. Gdoutos
1. Problem
The Westergaard function for the stress field near the tip of an opening-mode crack is
put in the form
(1)
where the parameter pmodels the effect ofnear field boundaries and boundary loading.
Determine the singular stresses o., oy and txy from Z1• According to photoelastic law,
the isochromatic fringe order N is related to the maximum shear stress tm by [1]
Nf
2t =-
m t
where f is the stress-optical constant and t is the plate thickness.
(2)
Show that this equation can be used to determine K1 from the isochromatic fringe pat-
tern in the neighborhood ofthe crack tip.
See Problem 2.
3. Solution
Introducing the value ofthe Westergaard function Z1 given by Equation (1) into Equa-
tions (5) ofProblem 2 we obtain for the stress components
o • = ~[cos~(1-sin~sin 39) + cos ~(1+sin 2 ~) p(rIa)+ a ..rrf;.]v2xr 2 2 2 2 2

64 E.E. Gdoutos
cry= ~[cos ~(I+sin ~sin 3e)+cos ~(I-sin2 ~) ~(r/a)] (3)
v2nr 2 2 2 2 2
r = - - sm - cos - cos -- ~ (rIa) cos -K1 • e e [ 3e e]
xr .J2n 2 2 2 2
The maximum shear stress tm is given by
(4)
We have
-2a Msin e [sin 3;- ~<rta)sin ~]+ arta>] <s>
Introducing this value of tm into the photoelastic law expressed by Equation (2) we
obtain for the distance of a point on the isochromatic fringe of order N from the crack
tip
(6)
where
(7)
This equation can be solved in a computer to give the polar distance r as a function of
the polar angle e withy, a, a and~ as parameters. Physically accepted solutions ofthis
equation should give real values of r, such that r/a < I. The four parameters y, a, a and
~ can be adjusted so that the analytical isochromatics match the experimental ones.
When a close fit is achieved the stress intensity factor K1 is determined. This analysis
permits the fringe loops to tilt, stretch and become unsymmetrical, so that they can be
used to determine K1 for a wide variety of specimen geometries and loading conditions
[I].
4. References
(I] J.W. Dally and W.F. Riley (1991) Experimental Stress Analysis, Third Ed., McGraw-Hill, New York.

Problem 15: Photoelastic Determination of Mixed-Mode
Stress Intensity Factors K1 and K11 **
MS. Konsta-Gdoutos
1. Problem
Consider a crack in a mixed-mode stress field governed by the values of the opening-
mode K1 and sliding-mode Krr stress intensity factors. Obtain the singular stress com-
ponents and subtract the constant term O"ox from the stress O"x to account for distant field
stresses. Determine the isochromatic fringe order N from equation [I]
Nf
2T =-
m t
(l)
where Tm is the maximum in-plane shear stress, f is the stress-optical constant and t is
the plate thickness. Obtain an expression for N. Consider the opening-mode. Ifrm and
9m are the polar coordinates of the point on an isochromatic loop, furthest from the
crack tip (Figure I), show that [2]
_ Nf~[ ( 2 )
2
]
112
( 2tan(39m /2)K 1 - I + 1 + ---'---"::._-
tsin9m 3tan9m 3tan9m
(2)
(3)
For the problem of a mixed-mode stress field if only the singular stresses are consid-
ered, show that the maximum in-plane shear stress Tm is given by
I r. 2 2 • • 2 2 f'2
Tm = ,;;--!_SID 9K1 + 2sm 29K 1Kn + (4-3sm 9)Kn .
2v2xr
(4)
Then show that the polar angle 9m ofthe point furthest from the crack tip on the curve
Tmax = constant (Figure I) satisfies the following equation

(5)
y
Ta•Constant
X
Figure 1. A crack-tip isochromatic fringe loop.
See Problem 13.
3. Solution
The stress field in the vicinity ofthe crack tip for mixed-mode conditions is given by
1 [ 6 (I . 6 . 36} K . 6 ( 2 6 36 }]crx = r;:;-- K1 cos- -SID- sm- - II sm- +cos- cos- - crox
-v2xr 2 2 2 2 2 2
1 [K 6 ( 1 . 6 . 36} K . 6 9 39]cry= ~ 1cos- +sm -sm- + IIsm- cos -cos-
-v2u 2 2 2 2 2 2
(6)
I [K . 6 6 36 K 6 ( 1 . 6 . 36 }]t xy = r;:;-- 1 sm - cos - cos - + II cos - - sm - SID -
-v2xr 2 2 2 2 2 2

Photoelastic Determination ofMixed-Mode Stress Intensity Factors K1 and Kn 67
where K1 and Kn are the mode-l and mode-II stress intensity factors and r, 9 the polar
coordinates referred to the crack tip.
The maximum in-plane shear stress 'tm is given by
We obtain from Equations (6) and (7)
(2•mY = - 1-[(K1 sinO+ 2Ku cos of +(Ku sin of]
21tr
(7)
+ ~sin! [K1sin0 (1 + 2cos0 )+ Ku(1 + 2cos2 0 +coso)]+ cr~x (8)
v21tr 2
For opening-mode (Kn = 0) we obtain
(9)
The position ofthe farthest point on a given loop is dictated by
(IO)
which gives
-K1 sin9m cos Om
cro =
X~( .39 3. 39)V£. JL lm COS e Slfl __tJI_ + - Slfl 0 COS ___IJI_
m 2 2 m 2
(II)
From Equation (9) and (II) using the photoelastic law ofEquation (I) we obtain for K1
and O"ox
Nf ~21trm 2 2 2 tan 2
t sin9m (3tanem) 3tan9m
[ l-1/2 [ 30m ]
K 1 = - 1+ 1+ (I2)

Nf cosem
O"ox =- ----------=------
t (36 ) ( 9 )112
cos 2m cos 2 6m+ 4sin 2 em
(13)
From Equation (8) we obtain for Oox = 0
1 r. 2 2 • • 2 2 ] 112
•m= r.:;--I_Sm 6K1 +2sm26K1Ku+(4-3sm 6)Ku
2v2 n:r
(14)
From Equations (5) and (14) we obtain
(15)
4. References
[1) J.W. Dally and W.F. Riley(l99l)Experimental Stress Analysis, Third Ed., McGraw-Hill, New York.
[2) G.R. Irwin (1958) Discussion ofpaper ''The Dynamic Stress Distribution Surrounding a Running Crack-A
Photoelastic Analysis", by A. Wells and D. Post, Proc. SESA, Vol. XVI, pp. 69-92, Proc. SESA, Vol. XVI,
pp. 93-96.

Problem 16: Application of the Method of Weight Function
for the Determination of Stress Intensity Factors **
L. Banks-Sills
1. Problem
(a) By means of the weight function, determine an integral expression for the stress
intensity factor of the geometry and loading shown in Fig. I. Assume plane strain
conditions.
(b) Carry out the integration to obtain an explicit expression for K1 •
- 2 a -
Figure. I A crack in an infinite plate subjected to a triangular stress distribution along the crack surfaces
The stress intensity factor K1 may be written as
(l)

70 L. Banks-Sills
where Ti is the traction vector on the boundary Sr along which the tractions are
applied, ds is differential arc length, and m; is the Bueckner-Rice weight function
given by [1, 2)
( n)- H au~(x,y,l)
mi x,y,<e - •
2K1 at
(2)
where £is crack length, H = E /(1- v2 ) or H = E for plane strain or generalized plane
stress conditions, respectively, K; and u; are the stress intensity factor and
displacement vector on Sr for another loading applied to the same geometry in Fig. 1,
respectively.
3. Solution
3.1 STRESS INTENSITY FACTOR AND DISPLACEMENT FIELDS FOR THE
AUXILIARY PROBLEM
The auxiliary or starred problem is chosen as illustrated in Fig. 2.
X
Figure 2 Auxiliaryproblem

Application ofthe Method ofWeight Function for the Determination ofSIFs 71
This is a Griffith crack of length l with remote applied tensile stress u..,. The stress
intensity factor is
(3)
The displacement vector required to solve the problem in Fig. 1 (along Sr) is along the
crack faces. Thus, for plane strain conditions
•( 0 n)= (l+v)(l-2v) ( _!:_)U1 X, , t: CJ.., X
E 2
<o ~x ~n.
3.2 CRACK FACE TRACTIONS
In Fig. I, the traction T; on the upper crack face is given as
T1 =0
T2 = {
- 2; (x -l/2)
2cr (x-l/2)
l
and on the lower crack face by
3.3 THE WEIGHT FUNCTION
O<x<l/2
ll2<x<f
O<x<l/2
f.l2<x<f
(4a)
(4b)
(5a)
(5b)
6a)
(6b)
The weight function is a universal function which depends only upon geometry. In
•order to obtain m; along the crack faces, only the derivative of u2 is required. It is
found to be

72 L. Banks-Sills
(7)
The weight function m2 is found by substituting (3) and (7) into (2) as
(8)
3.4 DETERMINATION OF THE STRESS INTENSITY FACTOR
Substituting (8) and the expressions for the tractions T2 in (Sb) and (6b) into (1) leads
to
K1 =-~ /2{ret2
(x-!:._) /xdx - re (x-!:._) /xdx} (9)
e1J;i Jo 2 v~ Jm 2 v~
To carry out the integration in (9), use is made of the transformation X =f sin 2 0 to
obtain
(lO)
which agrees with the value ofK1 found in [3].
4. References
[I] H.F. Bueckner (1970) A Novel Principle fur the Computation of Stress Intensity Factors. Zenschriftfor
Angewandte Mathematic und Mechanik, 50, 529-546.
[2] J.R. Rice (1972) Some Remarks on Elastic Crack Tip Stress Fields. International Journal ofSolids and
Structures, 8, 751-758.
[3] H. Tada, P. Paris and G. Irwin (1987) The Stress Analysis of Cracks Handbook, Del Research
Corporation, Missouri.

2. Elastic-Plastic Stress Field

Problem 17: Approximate Determination of the Crack Tip
Plastic Zone for Mode-l and Mode-ll Loading*
E.E. Gdoutos
1. Problem
Determine the crack tip plastic zone for mode-l and mode-11 loading according to the
Mises yield criterion.
A first estimate ofthe extent of the plastic zone attending the crack tip can be obtained
by determining the locus of points where the elastic stress field satisfies the yield crite-
rion. This calculation is very approximate, since yielding leads to stress redistribution
and modifies the size and shape of the plastic zone. Strictly speaking the plastic zone
should be determined from an elastic-plastic analysis of the stress field around the
crack tip. However, we can obtain some useful results regarding the shape of the plas-
tic zone from the approximate calculation.The most frequently used criteria for yield-
ing are the Tresca and von Mises criteria.
The Tresca criterion states that a material element under a multiaxial stress state enters
a state of yielding when the maximum shear stress becomes equal to the critical shear
stress in a pure shear test at the point of yielding. The latter is a material parameter.
Mathematically speaking, this criterion is expressed by [I]
(l)
where cr~o cr2, cr3 are the principal stresses and k is the yield stress in a pure shear test.
The von Mises criterion is based on the distortional energy, and states that a material
element initially yields when it absorbs a critical amount of distortional strain energy
which is equal to the distortional energy in uniaxial tension at the point of yield. The
yield condition is written in the form [I]
(2)

76 E.E. Gdoutos
where cry is the yield stress in uniaxial tension.
3. Solution
3.1 MODE-I:
The principal singular stresses are given by (2]
cr1 = ~cos ~(1+ sin~)..;21fT 2 2
(3)
K1 e( . e)0"2 = r;;::::: COS - 1-Stn-
...;27tr 2 2
Introducing these values of cr1 and cr2 into the von Mises yield criterion expressed by
Equation (2), we obtain the following expression for the radius of the plastic zone
rp(e) = _..!.__ (.!S_)2
(~sin2 e +I+ cos e)
47t O"y 2
(4)
(5)
for plane strain.
The extent ofthe plastic zone along the crack axis (e = 0) is given by
( )
2
1 KI
r (0)=- -
p 27t O"y
(6)
( )
2
1 KI
r (0)=--
p 187t O"y
(7)
for plane strain, with v = l/3.

Crack Tip Plastic Zone for Mode-l and Mode-ll Loading 77
Figure 1 shows the shapes ofthe plastic zones for plane stress and plane strain with v =
1/3. Observe that the plane stress zone is much larger than the plane strain zone be-
cause of the higher constraint for plane strain. Equations (6) and (7) show that the ex-
tent ofthe plastic zone along the crack axis for plane strain is 1/9 that of plane stress.
)
K
}. 0.7 plane stress
Figure I. Approximate estimation of the crack-tip plastic zones for mode-l loading under plane stress and
plane strain. v ~ 1/3.
3.2 MODE-II:
For mode-11 we have [2]
Thus, we obtain
2Krr . 9
cr +cr =---sm-
X y Fr 2
2K . 9 ( 9 39)cr,- cry=- r;;!!-: sm- I+ cos- cos-
v27tr 2 2 2
Krr 9 (I . 9 . 39)t,Y = Fr cos 2 - sm 2 sm T
(8)

78 E.E. Gdoutos
(cr -cr )2 +4-r2 =--1-1 I-3sm 2 -cos2 -
4K 2
( • e e)x Y xy 21tr 2 2
(9)
The principal stresses cr~> cr2 are given by
(10)
or
Kn . e Kn
cr1 2 =- r;;::: sm- ± r;;:::
· ...;2Jtr 2 ...;2Jtr
. z e 2 el-3sm -cos -
2 2
(II)
For conditions ofgeneralized plane stress (cr3 = 0) the Mises yield criterion becomes
(I2)
or
2
Kn (6 + 2sin 2 ~- 2.sin 2 e)= 2crt
21tr 2 2
(13)
0.4
r/(K,/a;r
0.6
Figure 2. Approximate estimation of the crack-tip plastic zones for mode-II loading under plane stress and
plane strain. v=I/3.
The radius ofthe plastic zone is given by

Crack Tip Plastic Zone for Mode-l and Mode-ll Loading
rp(9) =r =-1-(~)2
(14- 2cos9- 9sin 2 9)
87t Oy
For conditions ofplane strain (aJ = v(a1 + a2)) the Mises yield criterion yields
79
(14)
rP (9) =r =-1-(~)2
[12 + 2(1 - 2v)2 (1-cos 9)- 9sin 2 9] (15)
87t Oy
Figure 2 shows the shapes ofthe plastic zones for plane stress and plane strain with v =
1/3.
4. References
[I] A. Nadai (1950) Theory ofFlow and Fracture ofSolids, McGraw-Hill, New York.
Boston, London.

Plastic Zone for Mixed-Mode Loading *
E.E. Gdoutos
1. Problem
Determine the radius of the plastic zone accompanying the crack tip for mixed-mode
(opening-mode and sliding-mode) loading under plane strain conditions according to
the Mises Yield criterion. Plot the resulting elastic-plastic boundary for a crack of
length 2a in an infinite plate subtending an angle p= 30° with the direction ofapplied
uniaxial stress at infinity. v =0.3.
See Problem 17
3. Solution
By superimposing the stresses for opening-mode and sliding-mode loading and omit-
ting the constant term we obtain, after some algebra, for plane strain conditions (cr.= v
(crx+ cry), see Equations (7) of Problem 13) for the radius r ofthe plastic zone
r =- 1- 2-[Kf cos2 ~[<1-2v)2 + 3 sin2 ~] + K1K11 sin 9 [3 cos9
2xcry 2 2
(I)
Equation (1) for mode-l (Kn= 0) coincides with Equation (5) ofProblem 17, while for
mode-11 (K1= 0) coincides with Equation (15) ofProblem 17.

82 Approximate Determination ofthe Crack Tip Plastic Zone for Mixed-Mode
Loading
For a crack of length 2a in an infinite plate subtending an angle p=30° with the direc-
tion of applied uniaxial stress at infinity the stress intensity factors KJ. Ku are given by
[I]
(2)
Introducing these values into Equation (1) we obtain the radius of the plastic zone. For
p=30° and v =0.3 it is shown in Figure I.
r-----------1=----------1
I
I
I
I
I
I
r(2a.1)?0
L-----------l~----------~
Figure I. Elastic-plastic boundaly surrounding the tip ofan inclined crack in an infinite plate.
4. References
[I] E.E. Gdoutos (1993) Fracture Mechanics· An Introduction, Kluwer Academic Publishers, Dordrecht,
Boston, London.

Plastic Zone According to the Tresca Yield Criterion **
MS. Konsta-Gdoutos
1. Problem
Find the equation of the plastic zone ahead of a crack for mode-l and mode-II loading
under conditions of plane stress and plane strain for a material obeying the Tresca
yield criterion. Compare the plastic zones with those obtained with the Mises yield
criterion.
See Problem 17.
3. Solution
3.1 MODE-I
3.1. 1. Plane stress (u3 = 0): We have [I]
K1 e( . e)a1 = ~ cos - I + Sin -
..;'brr 2 2
(I)
K1 e( . e)a2 = ~ cos- I - sm -
..;'brr 2 2
Thus
KI . ea1 - a 2 = ,;;--- sm
.y2ltr
(2)
The Tresca yield criterion for plane stress is expressed by [2]

(3)
For our case we have
(4)
Thus, the Tresca criterion gives
(5)
or
KI 9( . 9)h::: cos - I + Stn - = cry
v2nr 2 2
(6)
The radius ofthe plastic zone is given by
K
2
[ e( 9)]
2
r (9) = r = - -1 - cos - I + sin -
P 2ncr~ 2 2
(7)
3.1.2. Plane strain (cr3 = v (cr1 + cr2)):
We have
KI . ecrl - cr2 = ,.-;:;----- Stn
-v2nr
K1 e( . e)CJ1 - cr3 = ,.-;:;----- cos - I - 2v + sm -
.y2nr 2 2
(8)
KI 9( . 9)cr2 - cr3 = .J2 nr cos 2 1- 2v -sm 2
The radius ofplastic zone is the larger of
(9)
and

Crack Tip Plastic Zone According to the Tresca Yield Criterion 85
K2
9( 9)2
r (9) =r2 =--1 - cos2 - I - 2v + sin -
P 21ta~ 2 2
(10)
For v = 113, it can be shown that r1 > r2 for 9 < 38.94°. Thus, for 9 < 38.94° the elastic-
plastic boundary is represented by Equation (8), while for 9 > 38.94° the elastic-plastic
boundary is represented by Equation (9).
The elastic-plastic boundary for v = 113 according to the Tresca yield criterion for con-
ditions of plane stress and plane strain is plotted in Figure I. Comparing Figure I with
Figure I ofProblem 17 we observe that the elastic-plastic boundaries according to Tre-
sca yield criterion are slightly different than the elastic-plastic boundaries according to
..0.4
Figure 1. Plastic zones around the crack tip for mode-l under plane stress and plane strain conditions accord-
ing to the Tresca yield criterion.
3.2 MODE II
3.2.1. Plane stress_(a3 = 0):

We have
(11)
The Tresca yield criterion is expressed by
(12)
Since
(13)
we have for the Trasca criterion
(14)
and the elastic-plastic boundary is determined by
2KnW·2r:;::: 1- -SID 9 = Oy
-v27rr 4
(15)
or
Kn ( . e ~1 3 · 2 e)r;;=: sm-+ - -sm =cry
-v2nr 2 4
(16)
(17)
and

Note that r2 > r1 for 1e1 >76.7°. Thus the elastic-plastic boundary is represented by
equation (17) for 1e1 < 76.7° and by equation (18) for lei > 76.7°.
3.2.2. Plane strain: (cr3 = v (cri + crz)):
We have
cr1-cr3 = -~[(1- 2v) sin!!__~I- ~sin2 e]~r 2 4
(19)
The Tresca yield criterion is expressed by
(20)
Since
(21)
we have for the Tresca criterion
(22)
and the elastic-plastic boundary is determined by
2Ku ~ 3 . 2
r;;::::: 1- -Sin 9 = Oy
v21tr 4
(23)
or

(24)
(25)
and
rp(9) =r3 =-1-(Ku )
2
[(1-2v) sin!+~1- ~sin2e]2
2n oy 2 4
(26)
Since r1 > r3 the elastic-plastic boundary is represented by Equation (25) and coincides
with the elastic-plastic boundary for conditions ofplane stress for 1e1 < 76.7".
The elastic-plastic boundary for v = 1/3 according to Tresca yield criterion for condi-
tions of plane stress and plane strain is plotted in Figure 2. Comparing Figure 2 with
Figure 2 of Problem 17 we observe that the elastic-plastic boundaries according to Tre-
sca yield criterion are slightly different than the elastic-plastic boundaries according to
Figure 2. Plastic zones around the crack tip for mode-11 under plane stress and plane strain conditions ac-
cording to the Tresca yield criterion.

4. References
Boston, London.
[2] A. Nadai (1950) Theory ofFlow and Fracture ofSolids, McGraw-Hill, New York.

Plastic Zone According to a Pressure Modified Mises Yield
Criterion**
E.E. Gdoutos
1. Problem
Determine the crack tip plastic zone for opening-mode loading for a pressure modified
von Mises yield criterion expressed by
where R = crcfcr, and cr, and crc are the yield stress of the material in tension and com-
pression, respectively. Plot the resulting elastic-plastic boundaries for plane stress and
plane strain conditions when R = 1.2 and 1.5. Compare the results with those obtained
by the von Mises criterion [I].
See Problem I7.
3. Solution
3.1. PLASTIC ZONE ACCORDING TO EQUATION (I)
Introducing the values of crh cr2(Problem 17) into the modified von Mises yield crite-
rion expressed by Equation (I) we obtain for the radius r of the elastic-plastic boundary
(3)

92 E.E. Gdoutos
{ 2;~~)=(~;1J[~[[2(l-2v)2 +3(1-cos e)](l+cos e)]112
]
2
R-t e
+ 2 (l+v)(--) cos-
R+l 2
(4)
for plane strain.
The elastic-plastic boundaries for conditions of plane stress (v =0) and plane strain for
v = 0.3 and v = 0.5 when R = l (von Mises yield criterion), R = 1.2 and R = 1.5 are
shown by the upper halves ofFigures l, 2 and 3.
Figure 1. Plastic zones for conditions of plane stress (v =0) and plane strain for v = 0, R = I, 1.2 and 1.5
according to the modified Mises yield criteria expressed by Equations I (upper half curves) and
Equation 2 (lower halfcurves).

Crack Tip Plastic Zone According to a Pressure Modified Mises Yield Criterion 93
Figure 2. As in Figure I for v = 0.3
Figure 3. As in Figure I for v =0.5

94 E.E. Gdoutos
3.2. PLASTIC ZONE ACCORDING TO EQUATION (2)
Working as in the previous case we obtain:
for plane stress and
r (21t~~) =~[2(1+v)(R -1)cos ~+[4(1+v)2 (R -1)2 cos2 ~K1 4R 2 2
+R(1+cos9) [2(1- 2v)2 + 3(1-cos 9)]]112 ]
2
(6)
for plane strain.
The elastic-plastic boundaries for conditions of plane stress (v = 0) and plane strain for
v = 0.3 and v = 0.5 when R = 1 (von Mises yield criterion), R = 1.2 and R = 1.5 are
shown by the lower halves ofFigures 1, 2 and 3.
4. References
[I] A. Nadai ( 1950) Theory ofFlow and Fracture ofSolids, McGraw-Hill, New York.

Problem 21: Crack Tip Plastic Zone According to
Irwin's Model *
E.E. Gdoutos
1. Problem
Consider a central crack of length 2a in an infinite plate subjected to uniaxial stress a
at infinity perpendicular to the crack plane. According to the Irwin model, the effective
crack is larger than the actual crack by the length of plastic zone. Show that the stress
intensity factor corresponding to the effective crack, called effective stress intensity
factor Keffi for conditions ofplane stress, is given by
(1)
Then, consider a large plate of steel that contains a crack of length 20 mm and is sub-
jected to a stress a = 500 MPa normal to the crack plane. Plot the ay stress distribution
directly ahead of the crack according to the Irwin model. The yield stress of the mate-
rial is 2000 MPa.
Irwin [I, 2] presented a simplified model for the determination of the plastic zone at-
tending the crack tip under small-scale yielding. He focused attention only on the ex-
tent along the crack axis and not on the shape of the plastic zone, for an elastic-
perfectly plastic material.
The model is based on the notion that as a result of crack tip plasticity the stiffness of
the plate is lower than in the elastic case. The length of the plastic zone c in front of
the crack is given by
c = _!_(.!S..)2
1t ay
(2)

96 E.E. Gdoutos
c =_I(.!S_)2
31t CJy
for plane strain, where K1 is the stress intensity factor and CJyis the yield stress.
3. Solution
The effective crack has a length 2(a+c/2) where for plane stress c/2 is (Equation (2))
~=-1(KerrJ2
2 21t CJy
(3)
(4)
The stress intensity factor Ketr for a crack of length 2(a+c/2) in an infinite plate sub-
jected to the stress cr is
(5)
or
(6)
This Equation leads to Equation (I).
Since the plate is large the effective stress intensity factor Keffis computed from Equa-
tion (1). We have
[ ] 1/2
K = cr 1t (O.Ol) =90MP r
err 1/2 a-vm
[~-0.5 ( ;oooooyl
The length ofthe plastic zone cis
c =_!_ (__2Q_)2
= 0.64mm
1t 2000
(7)
(8)

Crack Tip Plastic Zone According to Irwin's Model 97
The cry stress is constant along the length of plastic zone, while in the elastic region it
varies according to
Keff
cr =----
y (21tx)I/2
where xis measured from the tip of the effective crack (x > 0.32 mm).
The cry stress distribution is shown in Figure 1.
original crock
fictitious crock
}-- o=20 mm ----.i
T2000UPo
1--x
(2-nx)•l'l
Figure 1. Original and fictitious crack and cr, stress distribution according to the Irwin modeL
4. References
(9)
(I] G.R. Irwin (1960) Plastic Zone Near a Crack Tip and Fracture Toughness. Sagamore Ordnance Mate-
rial Conference. pp. IV63-N78.
(2] G.R. Irwin (1968) Linear Fracture Mechanics, Fracture Transition, and Fracture Control, Engineering.
Fracture Mechanics., 1, 241-257.

Problem 22: Effective Stress intensity Factor According to
Irwin' Model **
E.E. Gdoutos
1. Problem
Consider a crack in a finite width plate subjected to opening-mode loading. Establish
an iterative process for determining the effective stress intensity factor Keffaccording to
the Irwin model.
Then consider a thin steel plate of width 2b = 40 mm with a central crack of length 2a
= 20 mm that is subjected to a stress a = 500 MPa normal to the crack plane. Plot the
ay stress distribution directly ahead of the crack according to the Irwin model. The
yield stress ofthe material is 2000 MPa.
See Problem 21and references I and 2.
3. Solution
The effective crack has a length 2(a + c/2), where for conditions ofplane stress c/2 is
(Equation (2) ofProblem 21)
~=-1 (~)2
2 2n ay
(I)
and for conditions ofplane strain c/2 is (Equation (3) ofProblem 21)
~=-1 (~)2
2 6n ay
(2)
Kefffor a crack oflength 2(a + c/2) in a finite width plate is
[ ]
1/2
KelT =f((a+c/2)/b)a n(a+i) (3)

100 E.E. Gdoutos
where the function t((a + c/2)/b) depends on the ratio (a+ c/2)/b, where b is the plate
thickness.
A flow chart of a computer program for the solution ofequations (I) and (3) or (2) and
(3) is shown below:
START
I
Aaoume SlJIISs
Intensity factor K.
F1
Calculate length of Plane srrain
ptasUc zone c
I
c -~[~f ·-~r~r
I I
K, •roJn(a •i> K,•foJn(a•il
T ff (ABS(K.-KJJ < •
T
NO I YES
I I
K.•K, K,,=K.
-T I
Print K,,
I
END
From the computer program based on the above flow chart it is found
Kerr = 109.48 MParrn
The length ofplastic zone calculated from Equation (I) is
c= 0.954 mm
The cry stress distribution directly ahead ofthe crack is calculated from
Kerf
0 =---
y .J21tX
(4)
(5)
(6)

Effective Stress intensity Factor According to the Irwin Model
where xis measured from the tip ofthe fictitious crack. It is shown in Figure 1.
4. References
4000
I
I
I
a _ 109.48
~·-72RX
2 4 6
x(mm)
Figure/. Stress distribution ahead of the cmck tip
8 10
101
[I] G.R. Irwin (1960) Plastic Zone Near a Cmck Tip and Fmcture Toughness, Sagamore Ordnance Mate-
rial Conference, pp. IV63-IV78.
[2] G.R. Irwin (1968) Linear Fmcture Mechanics, Fmcture Transition, and Fracture Control, Engineering
Fracture. Mech., l, 241-257.

Problem 23: Plastic Zone at the Tip of a Semi-Infinite
Crack According to the Dugdale Model *
E.E. Gdoutos
1. Problem
The stress intensity factor for an infinite plate with a semi-infinite crack subjected to
concentrated loads Pat distance L from the crack tip (Figure I) is given by [I)
K _ 2P
I- (2nL)l/2
(I)
For this situation determine the length of the plastic zone according to the Dugdale
model.
p
p
t-- L--t
(a)
t
r llllll::p -x--1
1-- L ..,.. C----1
Figure 1. (a) A semi-infinite crack subjected to concentrated loads P and (b) calculation of the length of
plastic zone according to the Dugdale model.

104 E.E. Gdoutos
Calculation ofthe plastic state of stress around the crack tip and the extent of the plas-
tic zone is a difficult task. A simplified model for plane stress yielding which avoids
the complexities ofthe true elastic-plastic solution was introduced by Dugdale [2]. The
model applies to very thin plates in which plane stress conditions dominate, and to
materials with elastic-perfectly plastic behavior which obey the Tresca yield criterion.
According to the Dugdale model there is a fictitious crack equal to the real crack plus
the length of plastic zone (Figure 1b). This crack is loaded by the applied loads P and
an additional uniform compressive stress equal to the yield stress of the material, av.
along the plastic zone.
The length of plastic zone c is determined from the condition that the stresses should
remain bounded at the tip ofthe fictitious crack. This condition is expressed by zeroing
the stress intensity factor.
3. Solution
The stress intensity factor at the tip of the fictitious crack is obtained by adding the
stress intensity factors due to the applied loads P and the uniform compression stress
av along the plastic zone.
3.1. STRESS INTENSITY FACTOR DUE TO APPLIED LOADS P
The stress intensity factor K ~P) at the tip ofthe fictitious crack due to applied loads P
is according to Eq. (I)
(2)
3 2. STRESS INTENSITY FACTOR DUE TO THE STRESS ov
The stress intensity factor Kov l at the tip of the fictitious crack due to the uniform
compressive stress av along the length of plastic zone is calculated by integrating the
expression for the stress intensity factor due to a pair of concentrated loads along the
length ofthe plastic zone (Eq. (1)). We have
(3)
or

Plastic Zone at the Tip ofa Semi-Infinite Crack According to the Dugdale 105
Model
K (Gy)-
I -
3.3. SUPERPOSITION OF STRESS INTENSITY FACTORS
(4)
The stress intensity factor at the tip of the fictitious crack is obtained from Equations
(2) and (4) as
(5)
2P 40"yC112
[21t (c+L)]112 (21t)112
3.4. CONDITION OF ZEROING THE STRESS INTENSITY FACTOR
The condition that the stress intensity factor be zero at the tip of the fictitious crack
expressed as
(6)
leads to
(7)
Equation (7) expresses the length of plastic zone ahead of the crack tip according to
the Dugdale model.
4. References
[I] G.C. Sih (1973) Handbook of Stress Intensity Factors, Institute of Fracture and Solid Mechanics,
Lehigh University.
[2] D.S. Dugdale (1960) Yielding ofSteel Sheets Containing Slits, Journal. ofthe Mechanics and Physics
of. Solids, 8, 100-104.

Problem 24: Mode-III Crack Tip Plastic Zone According to
the Dugdale Model • •
E.E. Gdoutos
1. Problem
The stress intensity factor for an edge crack of length a in a semi-infinite solid sub-
jected to a pair of concentrated shear forces S applied to the crack at a distance b from
the solid edge (Figure I) is [I]
(I)
Determine the length of plastic zone according to the Dugdale model [2], and plot the
variation of cia versus SlaTy for different values of b/a, where Tv is the yield stress in
shear.
-------------..,
-------------~
Figure I. A crack of length a in a semi-infinite solid subjected to a pair of shear forces S.
Look in Problem 23.

108 E.E. Gdoutos
3. Solution
The stress intensity factor at the tip of the fictitious crack is obtained by adding the
stress intensity factors due to the applied shear forces Sand the uniform shear stress tv
along the plastic zone (2].
3.1. STRESS INTENSITY FACTOR DUE TO APPLIED SHEAR FORCES S
The mode-III stress intensity factor K~> at the tip of the fictitious crack of length
(a+c), where cis the length ofthe plastic zone, due to applied shear forces S, is accord-
ing to Equation (I)
K (S) _ 2S~7t(a+c)
III-
1t~ (a+c)2 -b2
3.2. STRESS INTENSITY FACTOR DUE TO THE STRESS tv
(2)
The stress intensity factor K j;t> at the tip of the fictitious crack due to the uniform
shear stress tv along the length of plastic zone is calculated by integrating the expres-
sion for the stress intensity factor due to the concentrated shear forces along the length
ofthe plastic zone. We have:
<~ ) a+Jc 2ty .J1t (a +c) 2ty ~ 1t (a +c) a-Jc dx
Kmv =- dx = __c:.....:____ -;=====
a 'It~ (a+c)2 -x 2 1t ~ (a+c)2-x 2
(3)
To calculate this integral we make the substitution
x =A sin t, A =a+ c (4)
We have
I dx = JAcost dt =t =sin_1 ~
~A2 -x2 Acost A
(5)
Thus, we obtain

Mode-III Crack Tip Plastic Zone According to the Dugdale Model 109
K (<y)-
III -
2ty~7t(a+c)
1t
2ty ~ 1t (a+ c)
1t
[ J
a+c
• -1 X
sm - -
a+c 1
( x . -J a )
2- sm a+c
(6)
The stress intensity factor at the tip ofthe fictitious crack is obtained from Eqs (2) and
(6) as
2S~ 1t (a +c)
- 7t~(a+c)2 -b2
2ty ~ 1t (a+ c)
1t
( 7t . -1 a )--sm - -
2 a+c
(7)
The condition that the stress intensity factor is zero at the tip of the fictitious crack
expressed as
Km= 0 (8)
leads to
(9)
or
(10)
or
( I+ ~)2
- (.!:)2
COS-I (-l)a a l+c/a
(11)
s
aty
Equation (II) expresses the length of plastic zone ahead of the crack tip according to
the Dugdale model.

110 E.E. Gdoutos
3.5. NUMERICAL RESULTS
Values of Slaty for various values of c/a for b/a = 0.1, 0.4, 0.7 and 1.0 are shown in the
following table:
c/a S/{_atv)
b/a=0.1 b/a=0.4 b/a=0.7 b/a=I.O
.01 0.141 0.131 0.102 0.020
.03 0.248 0.230 0.183 0.060
.05 0.324 0.300 0.242 0.099
.10 0.471 0.440 0.365 0.197
.15 0.592 0.557 0.471 0.293
.20 0.700 0.663 0.571 0.388
.25 0.802 0.762 0.666 0.483
.30 0.898 0.857 0.759 0.576
.35 0.992 0.950 0.850 0.668
.40 1.082 1.040 0.940 0.759
.45 - - 1.028 0.850
.50 - - - 0.940
.55 - - - 1.029
The variation of c/a versus S/(atv) for (b/a) = 0.1, 0.4, 0.7 and 1.0 is shown in Figure
2.
0.0 0.2 0.4 0.6 0.8 1.0
S{(aTv)
Figure 2. Plot ofc/a versus S/(aw) for various values ofb/a.

Mode-III Crack Tip Plastic Zone According to the Dugdale Model Ill
4. References
[I] G.C. Sib (1973) Handbook of Stress Intensity Factors, Institute of Fracture and Solid Mechanics,
Lehigh University, Bethlehem, Pennsylvania, USA
Boston, London.

Problem 25: Plastic Zone at the Tip of a Penny-Shaped
Crack According to the Dugdale Model *
E.E. Gdoutos
1. Problem
The stress intensity factor for a penny-shaped crack ofradius a in an infinite solid sub-
jected to a uniform stress a over a concentric circular area ofradius b (b <a) (Figure l)
is given by
(l)
Determine the length ofthe plastic zone according to the Dugdale model [l].
z
X
Figure I. A penny-shaped crack in an infinite plate subjected to a unifonn stress a over a concentric circular
area ofradius b.

114 E.E. Gdoutos
See Problem 23.
J. Solution
The stress intensity factor at the tip of the fictitious penny-shaped crack of radius (a+
c) is obtained by adding the stress intensity factors due to the applied uniform stress cr
over a concentric circular area of radius b and due to the uniform compression stress
crv along the plastic zone, where cry represents the yield stress of the material.
3.1. STRESS INTENSITY FACTOR KIA DUE TO APPLIED STRESS cr
K 1A = 2cr [a+c-~ (a+c)2 -b2 J
J1f(a+c)
(2)
3.2. STRESS INTENSITY FACTOR Km DUE TO UNIFORM STRESS crv IN THE
YIELD ZONE ENCLOSED BY TWO CIRCLES OF RADII A AND (a+ c)
Km is obtained by superimposing a uniform stress crv over a circular area of radius (a+
c) and a negative uniform stress crv over a circular area ofradius a. We obtain
(3)
+ 2cry [a+c-~(a+c)2 -a2 J
Jn(a+c)
The stress intensity factor along the circumference of the fictitious crack is obtained as
(4)
The condition that the stress intensity factor be zero along the circumference of the
fictitious crack expressed as

Plastic Zone at Tip ofa Penny-Shaped Crack According to the Dugdale Model 115
(5)
leads to equation
(6)
or
(7)
Equation (7) allows determination of the length c of the plastic zone as a function of
the ratio alav ofthe applied stress a to the yield stress CJy.
4. References
[I] D.S. Dugdale (1960) Yielding of Steel Sheets Containing Slits, Journal ofthe Mechanics and Physics
ofSolids, 8, I00-104.

Problem 26: Calculation of Strain Energy Release Rate
from Load - Displacement- Crack Area Equation **
MS. Konsta-Gdoutos
1. Problem
For a certain experiment, Gurney and Ngan (Proc. Roy Soc. Lond., Vol. A325, p. 207,
1971) expressed the load-displacement-crack area relation in the form
where Cm (m = 1, 2, 3, ...) are constants.
Use this expression to calculate U and then G. Show that
G=~
2A
(1)
(2)
For a specific test the load-displacement-crack area relation of the previous problem
takes the form
u u2 u3
p =350--1890-- + 5250-
A A312 A2
(3)
where Pis in Kgf (1 Kgf= 9.807 N), u is in em and A in cm2• In the test a crack of
area A == 50 cm2 starts to grow when u = 0.5 em. Determine R == Ge without resorting
to Equation (2). Then compare the value ofR with that obtained from Equation (2).
The strain energy release rate G represents the energy available for crack growth. It is
given the symbol G in honor of Griffith. G for an ideally brittle material where the
energy dissipated in plastic deformation is negligible is calculated as [1]

[E. e. gdoutos__(auth.),__emmanuel_e._gdoutos,__ch(book_zz.org)

[E. e. gdoutos(auth.),emmanuel_e._gdoutos,__ch(book_zz.org)

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[E. e. gdoutos(auth.),emmanuel_e._gdoutos,__ch(book_zz.org)