Problems of Fracture Mechanics and Fatigue: A Solution Guide

Problems of Fracture Mechanics and Fatigue A Solution Guide
Edited by
C.A. RODOPOULOS Materials Research Institute, Sheffield Hallam University, Sheffield, United Kingdom
J.R. YATES University of Sheffield, Sheffield, United Kingdom
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 of the hardcover 1 st 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
Editors Preface on Fatigue
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
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
E.E. Gdoutos
Problem 9: Westergaard Method for a Crack Subjected to Shear Forces
E.E. Gdoutos
M.S. Konsta-Gdoutos
E.E. Gdoutos
Problem 12: Stress Intensity Factors for a Linear Stress Distribution
E.E. Gdoutos
E.E. Gdoutos
E.E. Gdoutos
Factors K1 and Kn
M.S. Konsta-Gdoutos
Problem 16: Application of the Method of Weight Function for the
Determination of Stress Intensity Factors
L. Banks-Sills
Problem 17: Approximate Determination of the Crack Tip Plastic Zone
for Mode-l and Mode-ll Loading
E.E. Gdoutos
for Mixed-Mode Loading
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
Table of Contents
Problem 23: Plastic Zone at the Tip of a Semi-Infinite Crack According
to the Dugdale Model
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
3. Strain Energy Release Rate
Problem 26: Calculation of Strain Energy Release Rate from Load - Displacement -
113
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
E.E. Gdoutos
135
139
143
147
Problem 34: Experimental Determination of Critical Stress Intensity Factor K1c 155
E.E. Gdoutos
E.E. Gdoutos
161
163
Problem 37: Stable Crack Growth Based on the Resistance Curve Method 169
M.S. Konsta-Gdoutos
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
E.E. Gdoutos
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
E.E. Gdoutos
E.E. Gdoutos
L. Banks-Sills
E.E. Gdoutos
E.E. Gdoutos
Problem 50: Experimental Determination of J1c from J - Crack Growth Curves 233
Table of Contents Xl
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
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 of a 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 of a Cylindrical bar with a Circular Crack E.E. Gdoutos
327
333
339
343
Problem 75: Failure of a Pressure Vessel Containing a Crack with Inclined Edges 347 E.E. Gdoutos
Problem 76: Failure of a 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
359
365
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
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
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 of Crack 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
Sp. Pantelakis and P. Papanikos
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 Sp. Pantelakis and P. Papanikos
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 Sp. Pantelakis and P. Papanikos
3. Effect of Notches on Fatigue
Problem 16: Fatigue Crack Growth in a Plate Containing an Open Hole Sp. Pantelakis and P. Papanikos
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
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 Sp. Pantelakis and P. Papanikos
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 of a Rotating Circular Shaft for Safe Life 505 Sp. Pantelakis and P. Papanikos
Problem 27: Safety Factor of a Notched Member Containing a Central Crack 509 C.A. Rodopoulos
Problem 28: Safety Factor of a Disk Sander C.A. Rodopoulos
S. Short Cracks
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
xvi Table of Contents
6. Variable Amplitude Loading
Problem 33: Crack Growth Rate During Irregular Loading Sp. Pantelakis and P. Papanikos
Problem 34: Fatigue Life Under two-stage Block Loading Sp. Pantelakis and P. Papanikos
Problem 35: The Application of Wheeler's Model C.A. Rodopoulos
Problem 36: Fatigue Life Under Multiple-Stage Block Loading Sp. Pantelakis and P. Papanikos
Problem 37: Fatigue Life Under two-stage Block Loading Using Non-Linear Damage Accumulation
Problem 38: Fatigue Crack Retardation Following a Single Overload Sp. Pantelakis and P. Papanikos
Problem 39: Fatigue Life of a Pipe Under Variable Internal Pressure Sp. Pantelakis and P. Papanikos
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 Sp. Pantelakis and P. Papanikos
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 Sp. Pantelakis and P. Papanikos
Problem 44: Fatigue Life of a Pressure Vessel Under Variable Internal Pressure 585 Sp. Pantelakis and P. Papanikos
Table of Contents
7. Complex Cases
XVll
589
Problem 46: Mixed Mode Fatigue Crack Growth in a Center-Cracked Panel 593 Sp. Pantelakis and P. Papanikos
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 of a Plate Containing Multiple Cracks 607 Sp. Pantelakis and P. Papanikos
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 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
• Strain energy density criterion
• 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 of a 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 of Steel 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 of Fracture 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 Sheffield, United Kingdom
Chris A. Rodopoulos John R. Yates
Editors
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.
PART A: FRACTURE MECHANICS
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)
(6)
(7) f 2 = C 2 sin (A. -I) e +C 4 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
We consider the two cases of opening-mode and sliding-mode separately.
3.2. OPENING-MODE:
We have
The boundary conditions (Equation (8)) give
or
c 1 cos (l-1) x + C3 cos (A.+ I) x = o
c I (A. -I) sin (A. -I) 7[ + c3 (A.+ I) sin (A.+ I) 7[ = 0
[ cos (l-1) x
(l + 1) sin (A.+ 1) 7[ c3 = 0
For nontrivial solution the determinant of this equation should vanish. We get
sin 2xA. = 0
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 2
n = 1.3 •...
n -2.4 ....
For n = 1 we obtain the singular solution
3/2 ( 9 1 39) U1 = C 11 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)
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, ...
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 8u 8 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
u r = 21 - (2K - I) sin - + 3 sin -C r 112 [ 6 36]
4J1 2 2 (3I)
u 9 = 21 - (2K +I) cos-+ 3 cos-C r 112 [ 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 of 9, 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
0 r0
2% R 2%
W = J J ro~O) drdO= J ro(O)[logr]~ dO 0 ~ 0
2%
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 of a Statiomuy Crack, J. Appl. Mech. Trans ASME, 24, 109-114.
[2] S. Timoshenko and J.N. Goodier (1951) Theory of Elasticity, 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 2 r2 P a -b (1) ZI=
x(z-b) z2 -a2
(b)
Figure I. An infinite plate with a crack of length 2a subjected (a) to a pair offurces Pat x =band (b) to two pair of furces at x = ± b.
12 E.E. Gdoutos
Then show that the stress intensity factor of the 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)
(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 =ReZ 1 +ylmz;
(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 of forces at x = b (Figure 1a), we have to show that it satisfies the boundary conditions of the 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)
(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 of a concentrated force at that point.
At point x = b, we obtain for x ~ b that
iP ZI=----
2[ (z-b)
The magnitude of the concentrated force at point x = b, y = 0 is calculated as
(9a)
(9b)
(10)
(II)
(12)
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
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 1 b
The Westergaard function for a pair offorces at x =- b is obtained from Eq. (1) as
Z 1(-b) = p x (z+b)
(16)
Thus, the Westergaard function Z1 for the problem of Figure 1 b is obtained by adding the Westergaard function for a pair of concentrated forces at points x =band x =-b. We have
(17)
3.4. STRESS INTENSITY FACTOR FOR PROBLEM OF FIGURE 1 b
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
[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
y
(1)
(2)
X
Figure /. An infinite periodic array of equally spaced cracks subjected to a pair of concentrated forces P at their center in an infinite plate.
18 E.E. Gdoutos
See Problem 2.
(3)
For lzl ~ oo we have
Z 1 = 0, y Im Z~ = y Re Z~ = 0 (4)
Then Equation (5) of Problem 2 gives
(5)
Fory = 0, lx-nWI <a, n = 0, 1, 2, 3, ... we have
z=x
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 of the concentrated force at x =
Westergaard Method for a Periodic Array of Crack Under Concentrated 19 Forces
0, y = 0 is obtained by taking equilibrium equation along the x-axis of the half-plane y > 0. As in Problem 2 we obtain that this force is equal toP.
K1 is calculated from [I]
(8)
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
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) of Problem 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
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 infinity o
22 E.E. Gdoutos
See Problem 2.
For y = 0, lx- WI< awe have
z=x Thus,
a<- 2
Then, Equation (5) ofProblem 2 gives
For lzl ~ oo we have
Then, Equation (5) of Problem 2 gives
(3)
(4)
(5)
(6)
(7)
(8)
(9)
Since Z1 satisfies all boundary conditions it is the Westergaard function of the problem.
K1 is calculated from [1]
Westergaard Method for a Periodic Array of Cracks Under Uniform Stress 23
where ~=z-a (II)
(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 Q applied at the point x = b(b <a) of a crack AB oflength 2a in an infinite plate (Figure Ia) is
Z=-- -- +-- +I Q+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 of the 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
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
(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.
I I I I I .,.__ 2a ----t
(a) ,__ 2a ---t
{b) ,___ 2a ---.t
(c)
Figure 2. A crack of length 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.
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 -a 2 ) l
21t IC+I v-; b-a a
(5)
= (Q + iP) [(~) + r;;+b l 21t 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§+x K 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)
-vna a-x -a
vna a-x -a
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 = ,...- Ja 0 -- dx
vna a-x -a
Putting u = x/a we have
a~ I~ I a +x l+u . -I 2 J - dx =a J - du =a [sin u-~]
a-x I-u -1 -a -1
(11)
and
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
I I
= a2 sin-1(1)- (-a 2 ) sin-1(-l) -a2 Jsin-1 udu + a 2 JP du -1 -1
=-a2[usin-1 u + ~]1 + ~[u~+ sin-1 u]
1
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 of the stress intensity factors of Figures 2a and 2b.
(16)
[I] E.E. Gdoutos (1993) Fracture Mechanics- An Introduction, Kluwer Academic Publishers, Dordrecht, Boston, London.
Problem 6: Westergaard Method for a Crack Under Distributed Forces **
E.E. Gdoutos
1. Problem
Show that the Westergaard function for the configuration of Figure 1 is
[ [ ( )1/2]] 2cr z b b z2 - a 2 Z1 =- 2 2 112 arccos(-)- arccot - - 2--2
1t (z -a ) a z a - b
and the stress intensity factor is
y
(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.
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 of Figure 1 is
(4)
or
2a [ z (b) [b J%2 -a2 ]] Z1 =- ,-:;----;;arccos - -arccot - - 2 --2
x v z2 - a 2 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 of Figure 1 is
K 1 = aJ 2a ~ dx = -2a ~[arccos(~)] a = 2a ~arc sin(~) (7) b ~a2- x2 ~; ~; a b ~; a
4. References
[I] E.E. Gdoutos (1993) Fracture Mechanics -An Introduction, Kluwer Academic Publishers, Dordrecht, Boston, London.
Problem 7: Westergaard Method for a Crack Under 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
lp .i
Figure I. A crack 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)
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
or <Xl
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(x 2 +y~) + 2x(l-v) 2 82 +y~ (x2 +y~)~(x2 +y~)-(a2 +y~)
36
Thus
p
2n(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) of Problem 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
(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.
(2)
For y = 0, lxl < a we have
ReZ1 = 0, y lm Z~ = y Re Z~ = y Re Z~ =0 (3)
Then Equation (5) ofProblem 2 gives
(4)
40 E.E. Gdoutos
For lzl ~ ao, r = r1 = r2 and 9 = 91 = 92 we obtain
(6)
(7)
. a a ax =ReZ-ylmZi =-rcos8=-x (Sa)
a a
a a (8b)
xy 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 of shear 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
.... t ... ..._- a ----•-+t••-a ® X
Figure 1. An infinite plate with a crack oflength 2a subjected to a pair of shear forces S at x = b.
See Problem 2.
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 iS Z~u = --- = x(z-b) b2 -a2 x(z-b)
(4)
The magnitude of the 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
y-+O x (x-b)2 +y2 b-s
b+s b+£
= lim J -..!._ S Y 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 of the 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
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.
Problem 10: Calculation of Stress Intensity Factors by 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)
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 of crack problems of practical 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
a~, 0/2 C/2 ~M
(a) (b) (c)
Figure 2. Superposition of the triangular load as sum of a 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 =-
For alb = 0.4 we obtain for K: and K ~ :
K l = 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 Gb 2 J;;[ 2 3 4] K 1 = 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)
K 1 =K; +K~ = 1.052 aJ;'; + 0.627 aJ;'; = 1.679 aJ;'; (7)
4. References
[2] G.C. Sib (1973) Handbook of Stress Intensity Factors, Institute of Fracture and Solid Mechanics, Lehigh University.
[3] Y. Murakami (ed.) (1987) Stress Intensity Factors Handbook, Pergamon Press.
Problem 11: Calculation of Stress Intensity Factors by 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
._1·-- a ----t·l~-· -- a ---~·I
Figure I. A crack of length 2a in an infinite plate subjected to a unifurm normal stress a and shear stress T from x =±btox=±a.
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 x gx2 lc t{c-b) (-K-1) 1 --- sm -- -- +_...:...-=,:.. 2~ a a 2 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 a 2 a 2 2~ K + 1
(3)
and
(J (K-1)CJ t CI§+X K 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 g 2 ~2 ] K 11 =- -- +-- sm --sm -- 1--+ 1-- 2~ K + 1 2 .{; a a a 2 a 2
(5)
3.2 STRESSES APPLIED ALONG THE LOWER CRACK SURF ACES 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 b 2 ] 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 ] K 11 = -- + -- SID --SID -- 1 - - + 1 - - 2& K + 1 2.{; a a a 2 a 2
(6)
(7)
Calculation of Stress 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 gc2 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
.fi a a
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 I b) 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 of concentrated furces at a distance b from the plate edge and (b) a self-balanced linear tensile stress distribution acting along the crack taces.
3. Solution
The stress intensity factor K1 for the case of Figure l b is calculated by integration of the stress intensity factor of Figure 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
K 1 = 2a../a Jl.2945x -0.6857x3 +1.1597x 5 -1.7627x 7 +1.5036x 9 -0.5094x II dx (S)
..[; o J1-x 2
where x = b/a
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
o-yt-x
o '\/1- x-
1 x11dx 10 J J0! = -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 x 1- 0.6857 x 0.6667 + 1.1597 x 0.5334
1t
- 1. 7627 X 0.4572 + 1.5036 X 0.4063 - 0.5094 X 0.3694)
= 3_ x 1.0729 cr.J;; = 0.683 cr.J;; 1t
4. Referentes
(10)
[1] G.C. Sib (1973) Handbook of Stress-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 P with the circumferential direction (Figure I). When the vessel is subjected to an internal pressure p, determine the stress field in the vicin ity of the crack tip.
(a)
a:4-t 0 ...... z I "V I Uz
I ./ I I I I I L---l---~
a a
(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)
xy .J21tr 2 2 2
where KI is the opening-mode stress intensity factor and r and a are the polar coordi nates of the 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 3a CJ = -- 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
K I 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 of the 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
or
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)
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
. K I 9 (• . 9 . 39) crx = ~21tr cos 2 -sm 2 sm T -
-- sm- 2+cos- cos- -(k-1) cos 29 Kn . 9 ( 9 39 ) ~ 2 2 2
. K I 9 ( . 9 . 39 ) K u . 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
(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)
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 p models the effect of near 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 of the crack tip.
See Problem 2.
3. Solution
Introducing the value of the Westergaard function Z1 given by Equation (1) into Equa tions (5) of Problem 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-sin 2 ~) ~(r/a)] (3) v2nr 2 2 2 2 2
r = -- sm - cos - cos -- ~ (rIa) cos -K 1 • e e [ 3e e] xr .J2n 2 2 2 2
The maximum shear stress tm is given by
(4)
-2a Msin e [sin 3;- ~<rta)sin ~]+ a\rta>] <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 of this 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 ]
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 9K 1 + 2sm 29K 1Kn + (4-3sm 9)Kn .
2v2xr (4)
Then show that the polar angle 9m of the point furthest from the crack tip on the curve Tmax = constant (Figure I) satisfies the following equation
See Problem 13.
3. Solution
The stress field in the vicinity of the crack tip for mixed-mode conditions is given by
1 [ 6 (I . 6 . 36} K . 6 ( 2 6 36 }] crx = r;:;-- K 1 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 of Mixed-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! [K 1sin0 (1 + 2cos0 )+ Ku( 1 + 2cos 2 0 +coso)]+ cr~x (8) v21tr 2
For opening-mode (Kn = 0) we obtain
(9)
The position of the farthest point on a given loop is dictated by
(IO)
which gives
-K 1 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 of Equation (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)
(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 of paper ''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 K 1 •
-2a-
Figure. I A crack in an infinite plate subjected to a triangular stress distribution along the crack surfaces
(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)
2K 1 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
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
3.2 CRACK FACE TRACTIONS
In Fig. I, the traction T; on the upper crack face is given as
T1 =0
3.3 THE WEIGHT FUNCTION
(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
K 1 =-~ /2{ ret 2(x-!:._) /x dx - re (x-!:._) /x dx} (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)
4. References
[I] H.F. Bueckner (1970) A Novel Principle fur the Computation of Stress Intensity Factors. Zenschrift for Angewandte Mathematic und Mechanik, 50, 529-546.
[2] J.R. Rice ( 1972) Some Remarks on Elastic Crack Tip Stress Fields. International Journal of Solids 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 of the 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)
3. Solution
3.1 MODE-I:
cr1 = ~cos ~(1+ sin~) ..;21fT 2 2
(3)
K 1 e ( . e) 0"2 = r;;::::: COS - 1-Stn- ...;27tr 2 2
Introducing these values of cr 1 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 (~sin 2 e +I+ cos e) 47t O"y 2
(4)
for plane strain.
( )
(6)
(7)
Crack Tip Plastic Zone for Mode-l and Mode-ll Loading 77
Figure 1 shows the shapes of the 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 of the 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:
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-r 2 =--1-1 I-3sm 2 -cos 2 - 4K 2
( • e e) x Y xy 21tr 2 2
(9)
(10)
or
· ...;2Jtr 2 ...;2Jtr
(II)
For conditions of generalized 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 of the 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 of plane 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 of the plastic zones for plane stress and plane strain with v = 1/3.
4. References
[I] A. Nadai ( 1950) Theory of Flow and Fracture of Solids, McGraw-Hill, New York. [2] E.E. Gdoutos (1993) Fracture Mechanics- An Introduction, Kluwer Academic Publishers, Dordrecht,
Boston, London.
Problem 18: Approximate Determination of the Crack Tip 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 of applied 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 of the plastic zone
r = - 1- 2-[Kf cos2 ~ [<1- 2v)2 + 3 sin 2 ~] + 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) of Problem 17.
82 Approximate Determination of the 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
r(2a.1) ?0
L-----------l~----------~
Figure I. Elastic-plastic boundaly surrounding the tip of an inclined crack in an infinite plate.
4. References
[I] E. E. Gdoutos (1993) Fracture Mechanics· An Introduction, Kluwer Academic Publishers, Dordrecht, Boston, London.
Problem 19: Approximate Determination of the Crack Tip 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.
3.1. 1. Plane stress (u3 = 0): We have [I]
K 1 e ( . e) a1 = ~ cos - I + Sin - ..;'brr 2 2
(I)
K1 e ( . e) a2 = ~ cos- I - sm - ..;'brr 2 2
Thus
(2)
The Tresca yield criterion for plane stress is expressed by [2]
(4)
(5)
or
KI 9 ( . 9) h::: cos - I + Stn - = cry v2nr 2 2
(6)
K 2
[ e ( 9)] 2
r (9) = r = --1 - cos - I + sin - P 2ncr~ 2 2
(7)
We have
KI . e crl - 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 of plastic 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 - cos 2 - 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 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 Mises yield criterion .
..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
(12)
Since
(13)
(14)
2KnW·2 r:;::: 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 > r 1 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- ~sin 2 e] ~r 2 4
(19)
(20)
Since
(21)
(22)
2Ku ~ 3 . 2 r;;::::: 1- -Sin 9 = Oy v21tr 4
(23)
or
(25)
and
rp(9) = r3 = -1-(Ku ) 2[(1- 2v) sin!+~ 1- ~sin 2 e]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 of plane 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 Mises yield criterion.
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
[2] A. Nadai ( 1950) Theory of Flow and Fracture of Solids, McGraw-Hill, New York.
Problem 20: Approximate Determination of the Crack Tip 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.
3.1. 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)
] 2
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 of Figures 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 half curves).
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
Working as in the previous case we obtain:
for plane stress and
r ( 2 1t~~) = ~[2(1 +v)(R -1)cos ~+[4(1 +v)2 (R -1)2 cos 2 ~ 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 of Figures 1, 2 and 3.
4. References
[I] A. Nadai ( 1950) Theory of Flow and Fracture of Solids, 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 of plane 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
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
(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 Keff is computed from Equa tion (1). We have
[ ] 1 /2 K = cr 1t (O.Ol) = 90MP r err 1/2 a-vm
[~- 0.5 ( ;ooooo y l The length of the plastic zone cis
c = _!_ (__2Q_)2 = 0.64mm
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
T 2000UPo
(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 of the material is 2000 MPa.
3. Solution
The effective crack has a length 2(a + c/2), where for conditions of plane stress c/2 is (Equation (2) of Problem 21)
~=-1 (~)2 2 2n ay
(I)
and for conditions of plane strain c/2 is (Equation (3) of Problem 21)
~=-1 (~)2 2 6n ay
(2)
Kefffor a crack oflength 2(a + c/2) in a finite width plate is
[ ] 1/2
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 of equations (I) and (3) or (2) and (3) is shown below:
START
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 < •
Print K,,
I END
From the computer program based on the above flow chart it is found
Kerr = 109.48 MParrn
The length of plastic zone calculated from Equation (I) is
c= 0.954 mm The cry stress distribution directly ahead of the crack is calculated from
Kerf 0 =---
y .J21tX
Effective Stress intensity Factor According to the Irwin Model
where xis measured from the tip of the fictitious crack. It is shown in Figure 1.
4. References
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
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 of the 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 of the 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 1 b). 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

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Problems of Fracture Mechanics and Fatigue: A Solution Guide