Stress Concentrators in Continuous Deformable Bodies (Advanced Structured Materials, 181) 3031160223, 9783031160226

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Advanced Structured Materials

Vahram N. Hakobyan

Stress Concentrators in Continuous Deformable Bodies

Advanced Structured Materials Volume 181

Series Editors Andreas Öchsner, Faculty of Mechanical Engineering, Esslingen University of Applied Sciences, Esslingen, Germany Lucas F. M. da Silva, Department of Mechanical Engineering, Faculty of Engineering, University of Porto, Porto, Portugal Holm Altenbach , Faculty of Mechanical Engineering, Otto von Guericke University Magdeburg, Magdeburg, Sachsen-Anhalt, Germany

Common engineering materials are reaching their limits in many applications, and new developments are required to meet the increasing demands on engineering materials. The performance of materials can be improved by combining different materials to achieve better properties than with a single constituent, or by shaping the material or constituents into a specific structure. The interaction between material and structure can occur at different length scales, such as the micro, meso, or macro scale, and offers potential applications in very different fields. This book series addresses the fundamental relationships between materials and their structure on overall properties (e.g., mechanical, thermal, chemical, electrical, or magnetic properties, etc.). Experimental data and procedures are presented, as well as methods for modeling structures and materials using numerical and analytical approaches. In addition, the series shows how these materials engineering and design processes are implemented and how new technologies can be used to optimize materials and processes. Advanced Structured Materials is indexed in Google Scholar and Scopus.

Vahram N. Hakobyan

Stress Concentrators in Continuous Deformable Bodies

Vahram N. Hakobyan Institute of Mechanics National Academy of Sciences Yerevan Yerevan, Armenia

ISSN 1869-8433 ISSN 1869-8441 (electronic) Advanced Structured Materials ISBN 978-3-031-16022-6 ISBN 978-3-031-16023-3 (eBook) https://doi.org/10.1007/978-3-031-16023-3 © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Preface

In engineering practice, structures or their parts are often found, which, according to the structural conditions of manufacture, contain stress concentrators such as cracks (cuts, cracks), inclusions, patches, holes and corner points. Some of these concentrators can also arise during the exploitation of various structures and their parts. A characteristic feature of the stress state of such structures and parts is that local stress fields with large and intensively changing gradients are formed around these concentrators, which lead to the destruction of these structures and parts. Therefore, the study of the stress state of massive bodies containing stress concentrators, in general, as well as local stress fields arising around the concentrators, is a topical problem both from scientific and engineering perspectives. On the other hand, the study of these issues, in most cases, is reduced to contact and mixed problems in the mechanics of a deformable solid. On the basis of powerful analytical methods of mathematical physics and algorithms of computational mathematics, exact and effective solutions of a number of complicated contact and mixed problems are constructed, and simple formulas are obtained for their main mechanical characteristics [1–6, 14, 33, 36, 37, 43, 44, 50, 55, 59, 60, 66–68, 74, 75, 79, 81–84, 96, 99, 111, 114, 115, 117–119, 128, 130, 132– 134, 137–141, 147, 148]. Many results on the study of this kind of problems have found important application in matters of strength and rigidity of machines and their parts in mechanical engineering, in calculations of various industrial and hydraulic structures, in seismology, in the design of flying vehicles, in the practice of welding joints and road coatings, in other branches of applied mechanics and practice. At present, the influence of defects such as cracks and completely adhered to the matrix of thin absolutely rigid or deformable inclusions and patches on the stress state of homogeneous and composite massive isotropic and anisotropic elastic bodies has been sufficiently studied. Many fundamental results in this direction are studied in the monographs of V. V. Panasyuk [111], V. V. Panasyuk, M. P. Savruk, A. P. Datsyshin [108], G. P. Cherepanov [43, 45], V. Z. Parton, E. M. Morozov [112], N. F. Morozov [93, 94], G. Ya. Popov [116], Ya. S. Uflyand [143], L. I. Slepyan [135], L. T. Berezhnitsky, V. V. v

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Panasyuk, N. G. Stashchuk [31], N. I. Muskhelishvili [100], M. P. Savruk [129], G. P. Cherepanova [43, 44], L. A. Tolokonnikov and V. B. Penkov [141], G. A. Morar [92], V. M. Alexandrov and S. M. Mkhitaryan [14], also in books and articles [16– 18, 22, 23, 28, 29, 50, 53, 54, 56–58, 76, 85, 86, 92, 95, 97, 98, 101–103, 109, 110, 125–127, 142, 144, 150, 152]. Let us turn to some works related to problems for massive elastic homogeneous and composite bodies, with given mixed-type conditions on the banks, which, in particular, can be interpreted as the presence of a rigid inclusion, one side of which is adhered to the matrix, and the other side wholly or partially has moved away from the matrix, or is in smooth contact with the matrix. The first work in this direction belongs to D I Sherman [131], where, using the method of singular integral equations, a closed solution of the problem was constructed for a homogeneous plane containing several collinear cracks with given stresses on one banks, and displacements on the other banks. In his monograph [100] N I Muskhelishvili obtained a closed solution of the same problem by the method of potential theory. Later, G. P. Cherepanov in his well-known work [42] showed by the method of potential theory that the solution of the problem for a homogeneous plane containing several finite cracks, on arbitrary sections of the edges of which displacements are given, and on the rest—stresses, can be reduced to the solution the Riemann problem for two functions admitting a closed solution. In this regard, we also acknowledge the works of E. I. Zverovich [153], S. M. Mkhitaryan [87–90], B. M. Nuller [105, 106] and G. Ya. Popov [116], where closed solutions of some problems for a homogeneous plane containing a crack and for the axis-symmetric problem were constructed for a homogeneous space containing a coin-shaped crack, on the edges of which mixed-type conditions are given. For composite bodies containing a crack at the junction line of heterogeneous materials, on the banks of which mixed conditions are given, most probably, the first work in this direction was the paper of G. P. Cherepanov [41], where the stress state of a composite plane of two heterogeneous half-planes containing a crack, on the edges of which mixed-type conditions are given. It is shown that when displacements are given on one bank of the crack, and stresses on the other, then using the complex potentials for a composite plane proposed by G. P. Cherepanov, the solution of the problem can be reduced to the solution of the Riemann problem. The works of X. Markenscoff and L. Ni [84], D. Elata [51], T. C. T. Ting [140], as well as research [12, 13, 15–18, 20, 38–40, 46–49, 70–73, 102, 103, 136] are devoted to the problems concerning a thin rigid interfacial inclusion partially detached from isotropic or anisotropic media in various formulations. In recent decades, a new scientific direction of the theory of contact and mixed problems has been formed to study the interaction of various types of stress concentrators with each other and with homogeneous and composite massive deformable bodies [30–32, 61–65, 145, 146]. The symbiosis of the principles of classical contact problems and crack mechanics served as the basis for the theory of contact fracture [78]. The issues of preventing the propagation of cracks in solids led to the study of problems on the interaction of stamps and thin-walled elements in the form of stringers, thin-walled inclusions,

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plates and shells with massive deformable bodies containing cracks [7–11, 34, 72, 146]. The interaction of different types of stress concentrators leads to new formulations of contact and mixed problems, qualitatively changes the nature of stress concentration, significantly affects the stress singularity index and stress distribution in concentration zones [104–107]. It should also be noted that both for constructing effective solutions for contact and mixed boundary value problems, and for revealing the effects of mutual influence of various concentrators, regardless of the chosen research methodology, an important role is played by exact solutions of the corresponding model problems. In this aspect, we note the works [55, 108, 114, 115, 132], where closed solutions of a number of classical and nonstandard contact and mixed problems are constructed, which are fundamental for the further development of this branch of science. Along with static contact and mixed problems the dynamic contact and mixed problems of the theory of elasticity also have practical importance. A huge number of works are devoted. In particular, many works consider the problems of forced stationary vibrations of multilayer systems with interfacial defects such as cracks and absolutely rigid inclusions that are relevant from a practical point of view. They have developed and proposed effective methods for solving dynamic problems for layered media with interfacial defects [24–27, 124]. We also point out the works [120–123], where solutions of some contact and mixed boundary value problems of the dynamic theory of elasticity are given. At the same time, it is worth to acknowledge that few works have been devoted to dynamic problems on the stress state of layered structures that simultaneously contain different interfacial defects. On the other hand, from the point of view of seismology, seismic-resistant construction, seismic exploration and defectoscopy, one of the important problems is to identify the patterns of mutual influence of these stress concentrators depending on the physical–mechanical and geometric parameters of the problems, as well as the frequency of forced vibrations. This monograph is devoted to the development of a scientific direction in the study of issues of the simultaneous interaction of various types of stress concentrators with massive deformable bodies, both in the direction of considering composite, uniformly layered and anisotropic bodies, and in the direction of considering various options for combining concentrators of stress. On the basis of the methods of discontinuous solutions of the equations of the theory of elasticity, singular integral equations and orthogonal polynomials, the apparatus for solving the Riemann problem for two functions in combination with the numerical–analytical method of mechanical quadratures, a wide class of new static and dynamic contact and mixed problems was researched. The closed or effective solutions were obtained. The behavior of the local fields of breaking and contact stresses near the concentrators is studied, and the regularities of their mutual influence are revealed.

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This monograph is based on the research by the author with his students and Prof. Avetik Sahakyan. The author is very grateful to Ph.D. Lilit Dashtoyan, Ph.D. Harutyun Amirjanyan and Ph.D. Asatour Khurshudyan for their invaluable help in preparing the manuscript for printing the book. Yerevan, Armenia

Vahram N. Hakobyan

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122. Popov V (2020) A dynamic contact problem which reduces to a singular integral equation with two fixed singularities. Proc Third Int Conf Theor, Appl Exp Mech 187–192 123. Popov VG (1999) Interaction of plane elastic waves with systems of radial defects Mech Solids 44(4):98–107 124. Pryakhina OD, Smirnova AV (2004) An effective method of solving dynamic problems for layered media with discontinuous boundary conditions Appl Math Mech 68(3):500–507 125. Reissner E (1940) Note in the problem of the distribution of stress in a thin stiffened elastic sheet. Proc Nat Acad Sci USA 26:300–305 126. Rice JR (1988) Elastic fracture mechanics concepts for interfacial cracks. Trans ASME. J Appl Mech V 55(1):98–103 127. Rice JR, Sih GC (1965) Plane problems of cracks in dissimilar media. Trans ASME. J Appl Mech 32(2):418–423 128. Rvachev VL (1967) Researches of scientists of ukraine in the field of contact problems of the theory of elasticity. Appl Mech 3(10):109–116 (in Russian) 129. Savruk MP (1988) Stress intensity factors in solids with cracks. Fract Mech Strength Mater. Manual Edited by V V Panasyuk. 2. Kiev. Naukova Dumka 620 (in Russian) 130. Selvadurai APS, Singh BM (1984) On the expansion of pennyshaped crack by a rigid circular disc inclusion. Int J Fract 25 69–77 131. Sherman DI (1940) Mixed problem of the theory of potential and theory of elasticity for plane with finite number of rectilinear Slits. Rep Acad Sci USSR 4:330–334 (in Russian) 132. Shtaerman IYa (1949) Contact problem of elasticity theory. Gostekhteorizdat 270 (in Russian) 133. Sih GC, Chen EP (1981) Craks in composite materials: a compilation of stress solutions for composite systems with craks. London etc., Martinus Nijhoff publishes. 581 (Mechanics of fracture, 6) 134. Simonov IV (1990) An interface crack in an inhomogeneous stress field. Int J Fract 46 223– 235 135. Slepian LI (1981) Mechanics of cracks. Shipbuilding. 295 136. Tamate O, Sekine H, Ozawa Y (1983) Crack initiation from the ends of partially debonded surfaces of a flat inclusion Acta Mech 50:59–70 137. Tepliy MI (1983) Contact problems for regions with circular boundaries. Lvov Visha Shkola. 176 (in Russian) 138. The development of the theory of contact problems in USSR (1979) Moscow, Nauka, 493 (in Russian) 139. The mechanics of the contact between deformable bodies (1975) Proc Symp IUTAM. Delft University Press 414 140. Ting TCT (1986) Explicit solution and invariance of the singularities at an interface crack in anisotropic composites. Int J Solids Struct 22(9):965–983 141. Tolokonnikov LA, Penkov VB (1980) Application of the riemann boundary value problem with discontinuous matrix coefficients. Proc High Educ Inst Math 2:55–59 (in Russian) 142. Tonoyan VS (1963) Plane contact problem for elastic quarter-plane with unmovable vertical edge. Rep NAS RA 37(5):249–258 (in Russian) 143. Uflyand Ya (1968) Integral transformations in problems of the elasticity theory. Leningrad. Nauka 402 (in Russian) 144. Ulitko AF (1979) Eigenfunction method in spatial problems of the theory of elasticity. Kiev Naukova Dumka 261 (in Russian) 145. YuO Vasil’eva VV Sil’vestrov (2011) The problem of an interface crack with a rigid patch plate on part of its edge J Appl Math Mech 75(6):716–730 146. Vasil’eva YuO, Sil’vestrov VV (2011) Concentrated force acting near the tip of an interface crack with a rigid overlay on its side. J Samara State Tech Univ, Ser Phys Math Sci 1(22):191– 195 147. Vorovich II, Aleksandrov VM, Babeshko VA, (1974) Non-classical mixed problems of elasticity. Moscow: Nauka Publishing, 456 (in Russian) 148. Vorovich II, Babeshko VA (1979) Dynamic mixed boundary of the theory of elasticity for non-classical regions. Nauka 320 (in Russian)

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149. Wang ZY, Zhang HT, Chou YT (1985) Characteristics of the elastic field of a rigid line in homogeneity. Tran ASME J Appl Mech 52:818–822 150. Williams WE (1971) A star-shaped crack deformed by an arbitrary internal pressure. Int J Enj Sci 9(4):303–315 151. Willis JR (1972) The penny-shaped crack on an interface. Q J Mech Appl Math 25(3):367– 385 152. Wu KC (1990) Line inclusion at anisotropic bimaterial interface Mech Mater 10:173–182 153. Zverovich EI (1973) Mixed boundary value problems of theory of elasticity for plane with slits on real axis. In: Proceedings of symposium on continuum mechanics and related problems of analysis, vol 1, pp 103–114

About This Book

This monograph is devoted to the study of topical issues of simultaneous interaction of various types of stress concentrators with massive homogeneous and composite deformable bodies. A wide class of new contact and mixed problems has been studied, and their closed or efficient solutions have been constructed. The features of both static and dynamic mutual influence of various stress concentrators simultaneously located in homogeneous or piecewise homogeneous massive bodies are studied. The monograph is intended for researchers and specialists in the field of mixed and contact problems, as well as for research engineers, graduate students, young scientists, senior students and undergraduates of universities.

xv

Contents

1 Stresses Near the Absolutely Rigid Inclusion in an Orthotropic Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Mixed Problem for an Orthotropic Plane with a Crack . . . . . . . . . . . 1.3 Some Special Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Mixed Boundary Value Problem for Orthotropic Plane with Cut, on the Banks of Which Identical Absolutely Rigid Stamps with a Flat Base Are Pressed . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Mixed Boundary Value Problem for an Orthotropic Plane with a Cut, into One of the Edges of Which an Absolutely Rigid Stamp with a Flat Base Is Pressed . . . . . . . . . . . . . . . . . . . . . . . 1.6 Orthotropic Plane with Cuts and Absolutely Rigid Inclusion Under the Influence of a Pair of Concentrated Loads . . . . . . . . . . . . . 1.7 Plane Stress State of an Orthotropic Plane with Two Cuts Connected by an Absolutely Rigid Inclusion . . . . . . . . . . . . . . . . . . . 1.8 Stress State of Orthotropic Plane, Containing the Absolutely Rigid Inclusion, When One Edge of Inclusion Is Rigidly Coupled and the Other Is in the Smooth Contact with Plane . . . . . . . 1.9 Load Transfer from Elastic Inclusions to an Infinite Elastic Orthotropic Plane with Slits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Some Mixed Boundary Value Problems for Homogeneous Elastic Plane and Half-Plane, Weakened by Cracks . . . . . . . . . . . . . . . . 2.1 Stress State of Elastic Plane with Semi-infinite Slit Perpendicularly Positioned in the Absolutely Rigid Inclusion of Finite Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Stress State of an Elastic Plane with a Semi-infinite Slit Extending Perpendicular to a Crack of Finite Length . . . . . . . . . . . .

1 1 1 7

22

30 39 47

51 54 64 67

67 81

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2.3 Stress State of an Elastic Plane with a Semi-infinite Slit, Perpendicularly Situated on an Absolutely Rigid Inclusion Partially Detached from the Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 2.4 Stress State of Elastic Half-Plane, Containing Thin Rigid Inclusion on Bound, One Edge of Which Is Separated from Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 2.5 Stress State of Homogeneous Elastic Plane, Containing Opposite Cracks with Mixed Boundary Conditions on Banks of Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 2.6 Stress State of a Semi-plane with an Absolutely Rigid Inclusion and Crack Rising on the Boundary . . . . . . . . . . . . . . . . . . . 116 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 3 Plane Strain State of Piecewise Homogeneous Elastic Plane With Interfacial Absolutely Rigid Inclusions . . . . . . . . . . . . . . . . . . . . . . 3.1 Mixed Boundary Value Problem for a Compound Plane Weakened by Crack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Mixed Problem for Compound Plane Weakened by Periodic System of Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Stress State of Compound Plane with Absolutely Rigid Inclusion on Boundary Line of Materials . . . . . . . . . . . . . . . . . . . . . . . 3.4 Contact Problem for a Piecewise Homogeneous Plane with an Interfacial Crack Under Dry Friction . . . . . . . . . . . . . . . . . . . 3.5 Stress State of a Compound Plane with Interface Absolutely Rigid Inclusion and Crack Having Common Tip . . . . . . . . . . . . . . . . 3.6 The Stress State of Compound Plane, Weakened by Arc-Shaped Cracks with Stresses Given on One Bank and Displacements on the Other . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Plane Stress State of Half-Plane Consisting of Interfacial Crack . . . . 4.1 Discontinuous Solutions for a Coated Half-Plane with Interfacial Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Contact Problem for Piecewise Homogeneous Half-Plane . . . . . . . . 4.3 On the Load Transfer from a Deformed Patch to the Compound Half-Plane with Interfacial Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Mixed Boundary Value Problem For Compound Space With Interphase Defects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Axis-Symmetrical Mixed Boundary Value Problem for Compound Space with Coin-Shaped Crack . . . . . . . . . . . . . . . . . . 5.3 Axis-Symmetrical Mixed Problem for Compound Space, Weakened by Semi-Infinite Ring-Shaped Crack . . . . . . . . . . . . . . . . .

127 127 139 151 157 162

171 183 185 185 190 199 205 207 207 208 227

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xix

5.4 Stresses Near the Absolutely Rigid Coin-Shaped Inclusion in Piecewise Homogeneous Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 5.5 Axisymmetric Smooth Contact Problem for a Composite Space with a Circular Disk-Shaped Crack . . . . . . . . . . . . . . . . . . . . . . 243 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 6 Periodic and Doubly Periodic Problem for Piecewise Homogeneous Plane with Defects . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Statement of the Problem and Derivation of Discontinuous Solutions of Elasticity Theory Equations . . . . . . . . . . . . . . . . . . . . . . . 6.3 Piecewise Homogeneous Plane with a Doubly Periodic System of Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Doubly Periodic Problem for Piecewise Homogeneous Plane with Absolutely Rigid Inclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.5 The Stress State of a Piecewise Uniform Layered Space with Doubly Periodic Internal Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 The Stress State of an Uniformly Piecewise Homogeneous Layered Plane with a System of Periodic Parallel Internal Inclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.7 Periodic Problem for a Plane Composed of Two-Layer Strips with a System of Longitudinal Internal Inclusions and Cracks . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Axis-Symmetrical Stress State of Piecewise Homogeneous, Uniformly Layered Space with Periodical Disk-Shaped Defects . . . . . 7.1 The Discontinuous Solutions of Axis-Symmetrical Theory of Elasticity for Piecewise Homogeneous, Layered Space with Periodical Interphase Disk-Shaped Defects . . . . . . . . . . . . . . . . 7.2 Piecewise Homogeneous Layered Space with Parallel Disk-Shaped Absolutely Rigid Inclusions . . . . . . . . . . . . . . . . . . . . . . 7.3 Piecewise Homogeneous Layered Space with Parallel Disk-Shaped Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 On Axial Symmetric Stress State of Uniformly Layered Space with System of Periodical Inner Disk-Shaped Cracks . . . . . . 7.5 Axisymmetric Stressed State of Uniformly Layered Space with Periodic Systems of Internal Disk-Shaped Cracks and Inclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

253 253 253 260 266 271

278 288 296 299

299 305 309 316

325 335

8 Dynamic Mixed Boundary Value Problems for a Compound Half-Space and a Compound Plane with Interphase Defects . . . . . . . . 337 8.1 Forced Shear Vibrations of a Stamp at the Boundary of a Composite Half-Space with Interphase Defects . . . . . . . . . . . . . 337

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8.2 Forced Shift Vibrations of Stamp on the Border of Composite Half-Space with Interfacial Partially Detached Thin Inclusions from the Matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 8.3 Plane Stress State of an Elastic Compound Plane with Interfacial Cracks and Inclusions Under the Influence of Dynamic Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380

Chapter 1

Stresses Near the Absolutely Rigid Inclusion in an Orthotropic Plane

Abstract Based on the discontinuous solutions of equations of the elasticity theory both for an orthotropic plane and for an isotropic plane, the complex functions reducing the two-dimensional problem for an orthotropic plane with defects on the one of the principal directions of orthotropy to the Riemann problem for one or two functions are considered in this section. Some particular cases are being of special attention. Keywords Orthotropic plane · Absolutely rigid stamp · Absolutely rigid inclusion · The cracks · Mixed boundary value problem

1.1 Introduction It is well known that the two-dimensional problem of the elasticity theory for isotropic plane containing defects can be reduced to the Riemann problem for one or two functions using Kolosov–Muskhelishvili complex potentials under certain conditions. Such problems can be solved in closed form [3, 21, 22]. In this regard the method of discontinuous solutions of the elasticity theory equations [13, 18] is very effective. The solution of mentioned class of problems is reduced to the solution of the system of singular integral equations allowing the closed solutions. Complex potentials are ineffective for anisotropic plane, in particular for orthotropic plane.

1.2 Mixed Problem for an Orthotropic Plane with a Crack Let the orthotropic elastic plane contains the slit on line y = 0 along line L, consisting of finite number of nonintersecting intervals. The orthotropic elastic plane is considered in the Cartesian coordinate system O x y. The directions of axes coincide with principal directions of material’s orthotropy. Suppose that normal P0(±) (x) and shear τ0(±) (x) stresses are given on the intervals L (+) and L (−) for the upper and lower edges of slit respectively. The vertical v± (x) and horizontal u ± (x) displacements, as © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 V. N. Hakobyan, Stress Concentrators in Continuous Deformable Bodies, Advanced Structured Materials 181, https://doi.org/10.1007/978-3-031-16023-3_1

1

2

1 Stresses Near the Absolutely Rigid Inclusion …

well as the resultant of stresses are given at the upper and lower edges of slit on the intervals S (+) and S (−) correspondingly. We must build the closed solution of stated problem [9]. We divide the plane into upper and lower half-planes and denote the indices “+” and “−” for the components of stress and displacements tensors in corresponding half-planes. Then the stated problem can be mathematically formulated as the following boundary value problem: σ y(+) (x, +0) = σ y(−) (x, −0) (−) τx(+) y (x, +0) = τx y (x, −0)

U+ (x, +0) = U− (x, −0)

/ L) (x ∈

(1.1a)

V+ (x, +0) = V− (x, −0) U± (x, ±0) = u ± (x) V± (x, ±0) = v± (x)

(x ∈ S± ) (x ∈ S± )

σ y(±) (x, ±0) = −P0 (±) (x) (±) τx(±) y (x, ±0) = τ0 (x)

(1.1b)

(x ∈ L ± ) (x ∈ L ± )

U± (x, y) and V± (x, y) are the horizontal and normal components of displacements for points of corresponding half-planes. These components (each of its domains) satisfy the following equations [6, 12]: ∂ 2 U± ∂ 2 U± ∂ 2 V± + + + a =0 (1 ) 12 ∂x2 ∂ y2 ∂ x∂ y ∂ 2 V± ∂ 2 V± ∂ 2 U± + a + + a =0 (1 ) 22 12 ∂x2 ∂ y2 ∂ x∂ y

a11

(1.2)

where σ y(±) (x, y) and τx(±) y (x, y) are the components of stress tensor. The stress and displacement components are connected by formulas:   ∂U± ∂ V± , + a22 σ y(±) = μ12 a12 ∂x ∂y

τx(±) y = μ12



∂U± ∂ V± + ∂y ∂x

 (1.3)

Moreover, u ± (x), v± (x), P (±) (x) and τ0(±) (x) are smooth functions in mentioned domains and ai j (i, j = 1, 2)—are combinations of elastic constants of plane’s material, which are connected with elastic modulus E j , shear modulus μi j and Poisson’s ratio νi j (i, j = 1, 2) by formulas: a11 = E 1 [μ12 (1 − ν12 ν21 )]−1 , ν12 E 2 = ν21 E 1 ; a22 = a11 E 2 /E 1 , a12 = ν12 a22 = ν21 a11 .

1.2 Mixed Problem for an Orthotropic Plane with a Crack

3

To solve the stated problem let us introduce the jump functions of displacements and stresses U (x), V (x), σ (x), τ (x). Then the conditions (1.1a) are represented in the following form: σ y(+) (x, +0) − σ y(−) (x, −0) = σ (x) (−) τx(+) y (x, +0) − τx y (x, −0) = τ (x)

(1.4)

U+ (x, +0) − U− (x, −0) = U (x) V+ (x, +0) − V− (x, −0) = V (x)

(−∞ < x < ∞)

Here U (x) = V (x) = σ (x) = τ (x) = 0 (x ∈ / L) First we construct the solution of the system (1.2) and express the components of stresses and displacements by jump functions. Note that the solution satisfies the conditions (1.1a) and (1.4). Using the Fourier generalized transformation [2, 20] with respect to relations (1.2) and (1.3) ∞ [ϕ(x)] = ϕ¯ (s) =

ϕ(x)eisx dx,

 (n)  ϕ (x) = (−is)n [ϕ(x)]

−∞

ϕ(x) = we get

1 2π

∞

ϕ(s)e−isx ds

−∞

d2 U¯ ± dV¯ ± − is (1 + a12 ) − s 2 a11 U¯ ± = 0 2 dy dy d2 V¯ ± dU¯ ± a22 − is + a − s 2 V¯ = 0 (1 ) 12 dy 2 dy

(1.5)

  dV¯ ± ¯ − isa12 U ± (s, y) = μ12 a22 dy  ¯  dU ± ¯ τ¯ ± − is V y) = μ (s, 12 ± xy dy

(1.6)

σ¯ ± y

The solutions of (1.5), vanishing at infinity for corresponding half-planes, will be determined as ∓μ1 |s|y ∓μ2 |s|y V¯ ± (s, y) = A± + A± 1 (s)e 2 (s)e

U¯ ± (s, y) = ∓

2  iμ j (1 + a12 ) j=1

μ2j − a11

sgn(s)A±j (s)e∓μ j |s|y

(1.7)

4

1 Stresses Near the Absolutely Rigid Inclusion …

Here  μj =

b0 + (−1) j+1 b0 2 − 4a11 a22 , b0 = a11 a22 − 2a12 − a12 2 ( j = 1, 2) 2a22

i is an imaginary unit and the coefficients A±j (s) ( j = 1, 2) are unknown. Now let us apply the Fourier transformation to the coefficients A±j (s) ( j = 1, 2) and express them by jump functions, using formulas (1.6) and (1.7) and satisfying the conditions (1.4). The following is obtained p2 σ¯ (s) τ¯ (s) + 2μ12 ( p1 μ2 − p2 μ1 ) |s| 2isμ12 ( p1 μ1 − p2 μ2 ) i (a22 μ2 + p2 a12 ) sgn(s)U¯ (s) ( p2 μ2 − 1) V¯ (s) + − 2a22 ( p1 μ2 − p2 μ1 ) 2 ( p1 μ1 − p2 μ2 )

A+ 1 (s) =

p2 σ¯ (s) τ¯ (s) − 2μ12 ( p1 μ2 − p2 μ1 ) |s| 2isμ12 ( p1 μ1 − p2 μ2 ) i (a22 μ2 + p2 a12 ) sgn(s)U¯ (s) ( p2 μ2 − 1) V¯ (s) + + 2a22 ( p1 μ2 − p2 μ1 ) 2 ( p1 μ1 − p2 μ2 )

A− 1 (s) =

p1 σ¯ (s) τ¯ (s) − 2a22 μ12 ( p1 μ2 − p2 μ1 ) |s| 2isμ12 ( p1 μ1 − p2 μ2 ) i (a22 μ1 + p1 a12 ) sgn(s)U¯ (s) ( p1 μ1 − 1) V¯ (s) − + 2a22 ( p1 μ2 − p2 μ1 ) 2 ( p1 μ1 − p2 μ2 )

A+ 2 (s) = −

p1 σ¯ (s) τ¯ (s) + 2a22 μ12 ( p1 μ2 − p2 μ1 ) |s| 2isμ12 ( p1 μ1 − p2 μ2 ) i (a22 μ1 + p1 a12 ) sgn(s)U¯ (s) ( p1 μ1 − 1) V¯ (s) − − 2a22 ( p1 μ2 − p2 μ1 ) 2 ( p1 μ1 − p2 μ2 )

A− 2 (s) = −



μ j (1 + a12 ) pj = μ2j − a11



Substituting the obtained values of coefficients A±j (s) in (1.6) and (1.7) and using the Fourier inverse transformation the following expressions are correct for components of stresses and derivatives of displacements on line y = 0:

1.2 Mixed Problem for an Orthotropic Plane with a Crack

dU± (x, 0) a1 =− dx π dV± (x, 0) a2 = dx π c1 σ y(±) (x, 0) = π c2 τx(±) y (x, 0) = π

 L

 L

 L

 L

V  (s) b1 ds − s−x π 

U  (s) b2 ds − s−x π V  (s) a1 ds − s−x π U  (s) a2 ds + s−x π

L



 L

5

τ (s) 1 ds ± U  (x) s−x 2

σ (s) 1 ds ± V  (x) s−x 2 (−∞ < x < ∞)

σ (s) 1 ds ± σ (x) s−x 2

L



σ (s) 1 ds ± τ (x) s−x 2

L

(1.8) Here √ √ (a12 − a11 a22 ) (1 + a11 a22 ) a11 a1 = √ , a2 = a1 , b1 = √ 2 a11 a22 (μ1 + μ2 ) a22 2μ12 a11 a22 (μ1 + μ2 ) a11 μ12 (a11 a22 − a12 2 ) a11 b2 = b1 , c1 = √ , c2 = c1 a22 2 a11 a22 (μ1 + μ2 ) a22 It is easy to see that after introducing complex functions W  (x) = U  (x) + iαV  (x),

χ (x) = σ (x) − iατ (x),

(α =

4 a22 /a11 )

(1.9)

the relations (1.8) can be represented in the following form: dU± (x, 0) dV± (x, 0) a1 + iα =− dx dx απi c1 σ y(±) (x, 0) − iατx(±) y (x, 0) = απi

 L

 L

W  (s) b1 ds + s−x απi W  (s) a1 ds + s−x απi

 L

 L

χ (s) 1 ds ± W  (x) s−x 2 (1.10) χ (s) 1 ds ± χ (x) s−x 2 (1.11)

(−∞ < x < ∞) At this stage, we are applying the defined equations satisfying (1.1b) and stated initially as: dU± (x) dV± (x) + iα = u ± (x) + iαv± (x) = w  ± (x) dx dx (±) (±) (±) σ y(±) (x) − iατx(±) y (x) = −P0 (x) − iατ0 (x) = χ0 (x)

6

1 Stresses Near the Absolutely Rigid Inclusion …

As a result the following system of singular integral equations is obtained: ⎧   1  a1 W  (s) b1 χ (s) ⎪ ⎪ ± W (s) − ds + ds = w  ± (x) ⎪ ⎪ ⎪ 2 απi s−x απi s−x ⎨ L L    ⎪ 1 c1 W (s) a1 χ (s) ⎪ ⎪ ⎪ ± χ (s) + ds + ds = χ0± (x) ⎪ ⎩ 2 απi s−x απi s−x L

 

x ∈ S (±)



x ∈ L (±)



L

(1.12) This system should be considered under the conditions of type 

( j)

χ (x)dx = P0 , W (ak ) = W (bk ) = 0

(1.13)

Lj

which are secured the equilibrium in zones, where the displacements are given, and continuity of displacements at the end points of intervals of slit L. Now we pass on to solution of governing system of singular integral equations (1.12). For this purpose let us consider the functions [4, 15], which are analytical in whole complex plane, besides the line L: 1

j (z) = 2πi

 L

χ (s) + k j W  (s) ds, s−z

  2(−1) j+1 c1 kj = α − 2(−1) j a1

( j = 1, 2) (1.14)

Applying the Plemelj–Sokhotski formulas [15] χ (x) + k j W  (x) 1

j (x) = ± + 2 2πi ±

 L

χ (s) + k j W  (s) ds s−x

( j = 1, 2)

it is easy to prove that the system of integral equations (1.9) is represented as the Riemann problem for two functions by using functions j (z) ( j = 1, 2): 

−

+ 1 (x) = ν1 (x) 2 (x) + F1 (x) −

+ 2 (x) = ν2 (x) 1 (x) + F2 (x)

(x ∈ L)

(1.15)

Here ± (x) are the values of analytical functions (z) on the upper and lower edges of line L. The following notation is introduced for functions ν j (x) and F j (x) ( j = 1, 2):  ν1 (x) =

ν

x ∈ L (+)

−1 x ∈ S

(+)

 , ν2 (x) =

1/ν −1

x ∈ L (−) x∈S

(−)

, ν=

α − 2a1 α + 2a1

1.3 Some Special Problems

⎧ ⎨ (1 + ν) χ1 (x) x ∈ L (+) F1 (x) = , α ⎩ W  1 (x) x ∈ S (+) b1

7

⎧ 1+ν ⎪ ⎨− χ2 (x) x ∈ L (−) ν F2 (x) = α ⎪ ⎩ W  2 (x) x ∈ S (−) b1

So that, the solution of stated problem is reduced to the Riemann problem for two functions (1.15). Taking into account the results of paper [3], the solution of this problem always can be constructed in closed form, reducing it to the combined solution of linear and nonlinear Riemann’s problem. However, since the coefficients ν j (x) ( j = 1, 2) are nonzero constants at the edges of line L, the solution of the problem can be reduced to combined solution of two linear Riemann problems. In fact, the following two linear Riemann √ equations are obtained, multiplying second equation of (1.12) by λ j = (−1) j+1 ν1 /ν2 , then summing with first equation and introducing new functions by formulas: j (z) = 1 (z) + λ j 2 (z) +j (x) = G j −j (x) + Q j (x) (x ∈ L)   √ Q j (x) = F1 (x) + λ j F2 (x); G j = (−1) j+1 ν1 ν2 ; j = 1, 2

(1.16)

1.3 Some Special Problems For illustration we consider some special cases of stated problem. (a) First discuss the problem for orthotropic plane, containing the absolutely rigid thin inclusion on the interval (−a, a), which is positioned on the main direction of orthotropy y = 0. Suppose the plane is deformed under action of concentrated load P0 applied at middle point of the inclusion (Fig. 1.1). In this case ν1 (x) = ν2 (x) = −1, F1 (x) = F2 (x) = 0 and the system (1.12) is represented in the following form:  −

+ 1 (x) = − 2 (x) (1.17) (−a < x < a) −

+ 2 (x) = − 1 (x) Then 

 −  + −

+ 1 (x) + 2 (x) = − 1 (x) + 2 (x) + − −

+ 1 (x) − 2 (x) = 1 (x) − 2 (x)

(−a < x < a)

(1.18)

Since the functions j (z) ( j = 1, 2) tend to zero at infinity in this case we obtain that 1 (z) = 2 (z) = (z) from second Eq. (1.18). Hence from first equation of system (1.17) we have

8

1 Stresses Near the Absolutely Rigid Inclusion …

Fig. 1.1 Orthotropic plane with absolutely rigid thin inclusion

+ (x) = − − (x) (−a < x < a)

(1.19)

The general solution of Eq. (1.19) vanishing at infinity is the following [15]:

(z) = √

C z2

(1.20)

− a2

Here we consider the branch of radical having the same behavior as z. Unknown constant C will be determined from condition (1.13). In this case the condition is represented in following form: a χ (x)dx = P0

(1.21)

−a

Using formulas (1.21) and taking into consideration that in this case W  (x) = 0, as well as Plemelj–Sokhotski formulas, we get 2C χ (x) = + (x) − − (x) = √ i a2 − x 2

(−a < x < a)

Finally, satisfying the condition (1.21) for jump function of stress we obtain χ (x) =

P0 √ π a2 − x 2

(−a < x < a)

The formulas for contact stresses under inclusion are the following: 1 σ y(±) (x) = ± χ (s), 2

τ (±) (x) = 0

(−a < x < a)

(1.22)

1.3 Some Special Problems

9

Fig. 1.2 Elastic orthotropic plane with absolutely rigid thin inclusion

(b) Now let us consider the plane-deformed state of elastic orthotropic plane. There is an absolutely rigid thin inclusion of length 2a on one of the principal direction of orthotropy on the interval (−a, a). The lower edge of inclusion wholly joined with plane and the edge is separated from matrix, forming the crack. Suppose the plane is deformed under action of concentrated load Q, imposed at middle point of inclusion (Fig. 1.2). In this case ν1 (x) = ν, ν2 (x) = −1, F1 (x) = F2 (x) = 0 and the system (1.15) is the following:  −

+ 1 (x) = ν 2 (x) (1.23) (−a < x < a) −

+ 2 (x) = − 1 (x) It is not difficult to show that the system (1.23) and the following two Riemann equations for one function are equivalent: +j (x) = −λ j −j (x)

(−a < x < a)

Here 1 + λj j (z) = 1 (z) + λ j 2 (z) = 2πi ϕ j (x) = χ (x) − β j W  (x),  β = μ12

a −a

√ λ j = (−1) j+1 i ν,

(a11 a22 − a12 2 ) √ (1 + a11 a22 )

ϕ j (s) ds s−z β j = (−1) j iβ

( j = 1, 2)

(1.24)

10

1 Stresses Near the Absolutely Rigid Inclusion …

The general solution of Eq. (1.24), vanishing at infinity is the following: j (z) = C j χ j (z) = C j (z + a)−γ j (z − a)γ j −1 The functions ϕ j (x) are defined by the Plemelj–Sokhotski formulas. Then satisfying the equilibrium equation for inclusion we get ϕ j (x) = −

Q sin π γ j + X j (x) π Gj

( j = 1, 2)

(1.25)

    ln G j  θj ia  0 ≤ θ j ≤ 2π G j = (−1)  , γj = + b +1 2πi 2π  √  √ √ ( a11 a22 + a12 )(1 + a11 a22 ) ( a11 a22 − a12 )   a = , b =− √ √ √ √ ( a11 a22 + a12 + 2) a11 a22 ( a11 a22 + a12 + 2) a11 a22

where

j

It can be shown that b + 1 > 0, θ1 = 3π/2 and θ2 = π/2. Hence, γ1 = 3/4 − iβ  , γ2 = 1/4 − iβ  ,

     β = ln G j  /2π

Taking into account that X +j (x) = − G j ω j (x), ω2 (x) = ω¯ 1 (−x) ,

ω j (x) = (x + a)−γ j (a − x)1−γ j γ2 = 1 − γ¯ 1 ( j = 1, 2)

for (−a < x < a), the formula (1.25) will be written in form ϕ j (x) =

Q sin π γ j ω j (x) π

(1.26)

W  (x) = (ϕ2 (x) − ϕ1 (x)) /2β

(1.27)

Then using formulas χ (x) = (ϕ1 (x) + ϕ2 (x)) /2,

for complex combination of jump functions of stresses and derivatives of displacements we obtain χ (x) = W  (x) =

 Q  sin π γ1 ω1 (x) + sin π γ¯ 1 ω¯ 1 (−x) 2π

(1.28)

 Q  sin π γ1 ω1 (x) − sin π γ¯ 1 ω¯ 1 (−x) 2πiβ

(1.29)

1.3 Some Special Problems

11

Fig. 1.3 Elastic orthotropic plane with absolutely rigid inclusion and semi-infinite crack

After all it is easy to determine the jump functions of stresses and derivatives of displacements:   Q Re sin π γ1 (ω1 (x) + ω1 (−x)) 2π

(1.30)

τ (x) = −

  Q Im sin π γ1 (ω1 (x) − ω1 (−x)) 2π α

(1.31)

U  (x) =

  Q Im sin π γ1 (ω1 (x) + ω1 (−x)) πβ

(1.32)

  Q Re sin π γ1 (ω1 (x) − ω1 (−x)) 2π αβ

(1.33)

σ (x) =

V  (x) =

(c) Now we consider the plane stress state of elastic orthotropic plane. There is absolutely rigid thin inclusion of length a on interval (0, a) and semi-infinite crack on interval (−∞, 0) in one of the main direction of material orthotropy y = 0. Suppose that the plane is deformed under action of shear load T0 , imposed at the end point x = 0 of inclusion (Fig. 1.3). For this case  ν1 (x) =

ν x 1. This parameter shows the distance of a crack with constant length (b − a = const) from the tip of the quarter-plane. Hence, the crack is moving away from the tip of quarter-plane when b/a → 1.

2.5 Stress State of Homogeneous Elastic Plane, Containing … Table 2.5 Intensity stress factors K1 , K2 and γ (case (a)) b/a Angle γ in radians K1 1.05 1.2 1.5 1.75 2 2.5 3

− 0.000036 − 0.00646 − 0.01371 − 0.01796 − 0.02100 − 0.02457 − 0.025996

0.19947 0.19624 0.18575 0.17537 0.16518 0.14739 0.13349

Table 2.6 Intensity stress factors K1 , K2 and γ (case (b)) b/a Angle γ in radians K1 1.05 1.5 1.75 2 2.5 3

0.00031 0.01223 0.01447 0.01506 0.01343 0.01002

0.12218 0.12273 0.12035 0.11829 0.11662 0.1181

Table 2.7 Intensity stress factors K1 , K2 and γ (case (c)) b/a Angle γ in radians K1 1.05 1.2 1.5 1.75 2 2.5 3

− 0.0000012 − 0.0001495 − 0.001483 − 0.003499 − 0.005949 − 0.01114 − 0.01597

0.31876 0.31404 0.29693 0.28071 0.26588 0.24306 0.22841

113

K2 0.19947 0.19776 0.19065 0.18408 0.17815 0.16889 0.16237

K2 0.12218 0.10977 0.10517 0.10234 0.10056 0.10171

K2 0.31876 0.31440 0.30041 0.28872 0.27909 0.26611 0.25899

  − The graphs of dimensionless normal −σϕ− /μ and shear τrϕ /μ stresses, acting under the inclusion for above described cases of loading are shown in Figs. 2.20, 2.21, 2.22, 2.23, 2.24 and 2.25 for different values of parameter ab correspondingly. The values of rotation angle ν = 0.3 of inclusion and modulus of the intensity coefficients ν = 0.1 and ν = 0.49 are shown in (Tables 2.5, 2.6 and 2.7).

114

2 Some Mixed Boundary Value Problems for Homogeneous Elastic …

Fig. 2.20 Distribution of normal contact stresses

Fig. 2.21 Distribution of shear contact stresses

Fig. 2.22 Distribution of normal contact stresses

From the obtained results we can draw the following conclusions: (1) In all three cases the values of ν = 0.3 and ν = 0.1 are decreasing as the stress concentrators are approaching each other. This is the result of increasing of the absolute values of the fracture stresses at the middle zone of contact. In case (b) there are tensile stresses, which can be cause for the separation of the lower banks of crack from inclusion, at the middle of the contact zone.

2.5 Stress State of Homogeneous Elastic Plane, Containing …

Fig. 2.23 Distribution of shear contact stresses

Fig. 2.24 Distribution of normal contact stresses

Fig. 2.25 Distribution of shear contact stresses

115

116

2 Some Mixed Boundary Value Problems for Homogeneous Elastic …

(2) Note that in the case (a) we have obtained the maximum angular span ν = 0.49. In the vicinity of the end of inclusion, moving downwards the shear stresses are changing their signs, i.e. the distribution of contact shear stresses are changing characteristics. If this distribution is strictly antisymmetrical with respect to the center of the contact zone for the infinitely far inclusion, the distribution is similar to symmetrical distribution approaching to the apex of considering half-plane.

2.6 Stress State of a Semi-plane with an Absolutely Rigid Inclusion and Crack Rising on the Boundary In this section the problem of transferring the load from a rigid thin inclusion perpendicularly outgoing to the boundary to an elastic half-plane is discussed and solved, when ahead the inclusion there is a finite crack that immediately continues it. This problem is closely related to the issues of preventing the propagation of cracks in solid deformable bodies, which is an urgent problem not only from the points of view of the mechanics of a deformed solid and fracture mechanics, but is also of great practical importance in calculating various structures and their parts for strength and durability. Let the elastic half-plane, referred to the polar coordinate system rϕ and occupying the region {0 ≤ r < ∞, −π/2 ≤ ϕ ≤ π/2 }, at intervals (0, a) and (a, b) of line ϕ = 0, respectively, be strengthened by an absolutely rigid thin inclusion and weakened by a finite crack [7]. It is assumed that the half-plane is deformed under the influence of a concentrated load P0 applied to the inclusion at a point r = 0 and making an angle α with the boundary of the half-plane (Fig. 2.26).

Fig. 2.26 Elastic plane with absolutely rigid thin inclusion and finite crack

2.6 Stress State of a Semi-plane with an Absolutely …

117

It is required to determine the contact stresses acting on the contact zones of the inclusion with the half-plane, the opening of the crack, the intensity factors of the fracture stresses on the line ϕ = 0, as well as the angle of rotation of the inclusion. Let us divide the half-plane along the line ϕ = 0 into two quarter-planes and supply the stress and displacement components related to the points of the quarterplanes D± = {0 ≤ ±ϕ ≤ π/2; 0 ≤ r < ∞} with ( + ) and ( − ), respectively. Then the problem can be formulated as the following mixed boundary value problem: ⎧ ± τ (r, ±π/2) = 0; σϕ± (r, ±π/2) = 0; ⎪ ⎨ rϕ ur+ (r, 0) + iuϕ+ (r, 0) = ur− (r, 0) + iuϕ− (r, 0) ; ⎪ ⎩ + + − σϕ (r, 0) − iτrϕ (r, 0) = σϕ− (r, 0) − iτrϕ (r, 0) ; 

(0 < r < ∞) (b < r < ∞) (b < r < ∞)

ur± (r, 0) + iuϕ± (r, 0) = c + iγ r;

(0 < r < a)

σϕ±

(a < r < b)

(r, 0) = −σ0 (r) ;

± τrϕ

(r, 0) = 0;

(2.91a)

(2.91b)

Here, c and γ are the constants that determine the rigid shit and rotation of the inclusion, respectively. First construct discontinuous solutions of the Lamé equations for the half-plane. For this purpose, we will introduce into consideration the jump function χ (r) of stresses acting on the inclusion and the difference in the displacements W (r) of the points of the crack edges  +    + − σϕ (r, 0) − iτrϕ (r, 0) − σϕ− (r, 0) − iτrϕ (r, 0)  χ (r) (0 < r < a) ; = 0 (a < r < b)  +    ur (r, 0) + iuϕ+ (r, 0) − ur− (r, 0) + iuϕ− (r, 0)  0 (0 < r < a) , = W (r) (a < r < b)

(2.92)

The auxiliary boundary value problem determined by conditions 2.91a and 2.92 is solved using the formulas 2.13 and 2.14. The stress components and derivatives of displacements are expressed in terms of the introduced unknown functions χ (r) and W (r). The following is obtained ⎧ a  W  (r) i ϑ22 − ϑ12 ⎨ =± − R11 (r, r0 ) χ (r0 ) dr0 ⎩ dr 2 2ϑ2 0 ⎫ b b ⎬ + R12 (r, r0 ) W  (r0 ) dr0 + R13 (r, r0 ) W¯  (r0 ) dr0 ; (0 < r < ∞) ⎭

  d ur± (r, 0) + iuϕ± (r, 0)

a

a

(2.93)

118

2 Some Mixed Boundary Value Problems for Homogeneous Elastic …

⎧ a ⎨ χ iϑ (r) 1 ± ± σϕ (r, 0) − iτrϕ (r, 0) = ± + R21 (r, r0 ) χ (r0 ) dr0 2 2ϑ2 ⎩ 0 (2.94) ⎫ b a ⎬ + R22 (r, r0 ) W  (r0 ) dr0 + R23 (r, r0 ) χ¯ (r0 ) dr0 , (0 < r < ∞) ⎭ a

0

where   1 1 A11 r0 (r0 − r) ; R11 (r, r0 ) = − + B11 π r0 − r r0 + r (r0 + r)3   C12 2r0 r0 (r0 − r) ; R12 (r, r0 ) = − B12 2 2 π r0 − r (r0 + r)3 2ϑ2 r0 R13 (r, r0 ) =  2 : R21 (r, r0 ) = R12 (r, r0 ) /C12 ; 2  π ϑ2 − ϑ1 r0 + r 2   1 2r0 2r0 (r0 − r) 2ϑ2 r0 ; R23 (r, r0 ) = R22 (r, r0 ) = − −  ; 2 3 2 π ϑ1 r0 − r π ϑ1 r + r 2 (r0 + r) 0 

2 ϑ2 − ϑ1 ϑ22 + ϑ12 ϑ2 ϑ ; C12 = 2 1 2 ; A11 = 2 ; B = ; B = 2 1 − 11 12 2 ϑ2 + ϑ1 ϑ1 ϑ2 − ϑ1 ϑ2 − ϑ1 1 λ + 2μ ϑ1 = ; ϑ2 = . 2 (λ + μ) 2μ (λ + μ) Using formulas 2.93 and 2.94 and satisfying the conditions 2.91b, initially differentiating the first of them, the following system of defining singular integral equations is obtained: ⎧ a b ⎪ ⎪ ⎪  ⎪ ⎪ ⎪ ⎪ R11 (r, r0 ) χ (r0 ) dr0 + R12 (r, r0 ) W (r0 ) dr0 ⎪ ⎪ ⎪ a 0 ⎪ ⎪ ⎪ ⎪ b  ⎪ ⎪ ⎪ ⎪ ⎪ + R13 (r, r0 ) W¯  (r0 ) dr0 = −γ ∗ (0 < r < a) ⎪ ⎪ ⎨ a (2.95) ⎪ a b ⎪ ⎪ ⎪ ⎪ ⎪ R21 (r, r0 ) χ (r0 ) dr0 + R22 (r, r0 ) W  (r0 ) dr0 ⎪ ⎪ ⎪ ⎪ a ⎪ ⎪ 0 ⎪ ⎪ a  ⎪ ⎪ ⎪ ⎪ ⎪ + R23 (r, r0 ) χ¯ (r0 ) dr0 = σ0∗ (r) , (a < r < b) ⎪ ⎪ ⎩ 0



2iϑ2 2ϑ2 γ σ0∗ (r) = . σ0 (r) ; γ ∗ = 2 ϑ1 ϑ2 − ϑ12

2.6 Stress State of a Semi-plane with an Absolutely …

119

System 2.95 should be considered under the conditions of equilibrium of inclusion and continuity of displacements at the end points of the crack: a χ (r) dr =

b

P0∗ ;

a rχ (r) dr = 0;

W (r) dr = 0; Re a

0



 ∗ P0 = P0 eiα .

(2.96)

0

By changing the variables r0 = a (s + 1) /2, r = a (x + 1) /2 on interval (0, a) and r0 = ps + q, r = px + q (p = (b − a) /2, q = (b + a) /2) on interval (a, b), the system 2.95 will be formulated on interval (−1, 1) and using dimensionless functions ϕ1 (x) = aχ (a (x + 1) /2) /P0 ; ϕ2 (x) = W  (px + q) ; ϕ3 (x) = ϕ¯ 1 (x) ; ϕ4 (x) = ϕ¯ 2 (x) , The following system of singular integral equations will be obtained: 1 1 4 " ϕj (s) 1 1 ds + Kjk (s, x) ϕk (s)ds = fj (x) π s−x π k=1 −1 −1  j, k = 1 − 4; −1 < x < 1

.

(2.97)

Conditions 2.96 will be represented in the following form 1

1 ϕ1 (x) dx = e ; Re iα

−1

−1

1 ϕ2 (x) dx = 0.

(x + 1) ϕ1 (x) dx = 0; −1

Here 

 4(1 − ν)2 + (1 − 2ν)2 2 (s + 1) (s − x) K11 (x, s) = − + ; æ (s + x + 2) æ(s + x + 2)3 4 (1 − ν) E0 (s + λ3 ) K14 (x, s) = − ; æ (s + λ4 x + 2λ2 )2 K12 (x, s) =

 1 1 (1 − 2ν) E0 − s + λ4 x + λ2 (1 + ν) æ s − λ4 x + λ1  2 (s + λ3 ) (s − λ4 x + λ1 ) ; + (1 − 2ν) (s + λ4 x + λ2 )3

(2.98)

120

2 Some Mixed Boundary Value Problems for Homogeneous Elastic …

K21 (x, s) = −

 1 1 (1 + ν) (1 − 2ν) + + E0 λ 4 s − x − λ1 λ 4 s + x + λ2  2λ4 (s + 1) (λ4 s − x − λ1 ) ; (1 − 2ν) (λ4 s + x + λ2 )3

1 2 (s + λ3 ) (s − x) + ; s + x + 2λ3 (s + x + 2λ3 )3  4 1 − ν2 λ4 (s + 1) K23 (x, s) = − ; E0 (λ4 s + x + 2λ2 )2

K22 (x, s) =

K13 (x, s) = K31 (x, s) = K24 (x, s) = K42 (x, s) = 0; K32 (x, s) = K14 (x, s) ; K33 (x, s) = K11 (x, s) ; K34 (x, s) = K12 (x, s) ; K41 (x, s) = K23 (x, s) ; K43 (x, s) = K21 (x, s) ; K44 (x, s) = K22 (x, s) ; 4 (1 − ν) γ f1 (x) = f3 (x) = −γ∗ = − ; f2 (x) = −f4 (x) æ (1 + ν)  4 1 − ν2 ∗ aσ0 ((x + 1)/2) =− σ0 (x) ; ; σ0∗ (x) = E0 P0 E0 =

aE λ λ+2 λ+1 1 b ; λ1 = ; λ2 = ; λ3 = ; λ4 = ; λ = > 1; P0 λ−1 λ−1 λ−1 λ−1 a

The bar above the functions denotes their complex conjugate values, E is the modulus of elasticity of the half-plane material, and ν is the Poisson’s ratio. Note that the solution of system 2.97 under conditions 2.98 can be represented as: ϕj (x) = ϕj(1) (x) cos α + ϕj(2) (x) sin α

 j =1−4 ,

where the functions ϕj() (x) ( j = 1 − 4) at k = 1, 2 are, respectively, solutions of system 2.97 under conditions 2.98 when α = 0 and α = π/2. Therefore, taking into account that the rotation angle of the inclusion in the case α = π/2 is equal to zero due to the symmetry of the problem with respect to the Oy axis, for the angle of rotation of the inclusions at any value α the following formula is true γ∗ (α) = γ∗ (0) cos α. Note that the dimensionless contact stresses under inclusion given by formula 2.94 for (0 < r < a), through functions ϕj (x) (j = 1 − 4), will be given by the formula:

2.6 Stress State of a Semi-plane with an Absolutely …

χ∗±

(x) =

121

  ± a σϕ± (a (x + 1) /2, 0) − iτrϕ (a (x + 1) /2, 0)

ϕ1 (x) i (1 − 2ν) =± + 2 π (1 − ν)

P0 1 −1

" ϕ1 (s) ds + s−x j=1 3

1 Qj (x, s) ϕk (s)ds

(2.99)

−1

(−1 < x < 1) , where  1 (s + 1) (s − x) + ; 3 (s + x + 2) (1 − 2ν) (s + x + 2)   iE 1 1 2 (s + λ3 ) (s − λ4 x + λ1 )  0 ; Q2 (x, s) = − − + 2 3 s − λ4 x + λ1 s + λ4 x + λ2 4π 1 − ν (s + λ4 x + λ2 ) Q1 (x, s) = −

Q3 (x, s) =

2i (1 − 2ν) π (1 − ν)



2i (s + 1) , π (s + x + 2)2

The dimensionless fracture stresses on the line ϕ = 0 for r > b are determined by the formula χ∗±

(x) =

  ± a σϕ± (a (x + 1) /2, 0) − iτrϕ (a (x + 1) /2, 0)

P0 ⎫ ⎧ 1  4 1 ⎬ ⎨ ϕ (s) " iE0 2 =−  ds + K (s)ds . x) ϕ (s, 2k k ⎭ 4π 1 − ν 2 ⎩ s − x

(2.100)

k=1 −1

−1

(x > 1) Using this formula, it is also easy to determine the reduced fracture stress intensity factors at the end point r = b of the crack, which corresponds to the point x = 1: KI (1) − iKII (1) = lim

x→1+0

√ x − 1χ∗± (x)

(2.101)

The following formula will be used to determine the reduced crack opening: 2 v∗ (x) = Im W (a (x + 1) /2) = Im a

x ϕ2 (s) ds.

(2.102)

−1

The solution of the governing system of Eq. 2.97 under conditions 2.98 is constructed by the method of mechanical quadratures [17, 18]. First, define the behavior of the unknown functions at the end points of the intervals of integration. Note that functions Kij (x, s) (i, j = 1 − 4) are regular functions almost everywhere in a square (−1 ≤ x, s ≤ 1), except perhaps for points x + s = 0 where they can have a unmoved singularity.

122

2 Some Mixed Boundary Value Problems for Homogeneous Elastic …

Using the Muskhelishvili method [14], it is easy to establish that the function ϕj (x) (j = 1, 3), as expected [4], at the end point r = 0 (x = −1) has a singularity (x + 1)β−1 (0 ≤ β < 1), where the exponent β is the root of the transcendental equation βπ (1 + æ)2 β2 sin2 = − ; (æ = 3 − 4ν) . 2 4æ æ At a point r = a (x = ±1), functions ϕj (x) (j = 1 − 4) have singularity 1 (1 ± x)− 2 ±iβ1 (β1 = ln æ/2π ). At the point r = b (x = 1), the functions ϕj (x) (j = 2, 4) have the root singularity. Using the behavior of the unknown functions at the end points of the intervals of integration, we represent them in the following form: ϕ1 (x) =

ϕ1∗ (x) (1 − x)0.5+iβ1 (1 + x)1−β

; ϕ2 (x) =

ϕ2∗ (x) (1 − x)0.5 (1 + x)0.5+iβ1

,

(2.103)

where ϕi∗ (x) (i = 1, 2) are continuous smooth functions bounded up to the ends of the interval (−1, 1). Note that in this case, according to the formula 2.101, to determine the dimensionless fracture stress intensity factors at the end point of the crack r = b, the following formula is obtained: √ i 2−2−iβ1 π E0 ∗  ϕ2 (1) . KI (1) − iKII (1) = 1 − υ2 Substituting representations 2.102 into 2.97 and 2.98, based on the method of mechanical quadratures [18], we get the following system of algebraic equations: n " i=1 n " i=1

 wik

 4 " n "  (1 − qik (xk )) ∗ ∗ ϕm (sik ) + wip Kmp (xk , si ) ϕp∗ sip = π fm (xk ) sik − xk p=1 i=1

wi1 ϕ1∗ (si1 ) = eiα

n "

wi1 ϕ3∗ (si3 ) = e−iα

i=1

n "

 wij ϕ3∗ sij = 0; (j = 2, 4) .

i=1

 (γ ,σ ) Here sik i = 1, n; k = 1, 2 are the roots of the Jacobi polynomial Pn k k (x) ,  (γk ,σk ) xk k = 1, n − 1 and are the roots of the Jacobi polynomial Pn−1 (x) , (γ1 , σ1 ) = (β − 1, −0.5 + iβ1 ); (γ2 , σ2 ) = (−0.5 − iβ1 , β − 1) (γ ,σ )

wik =

2 Qn k k (sik ) ; (γk ,σk ) n + γk + σk Pn−1 (sik )

2.6 Stress State of a Semi-plane with an Absolutely …

123

(γ ,σk )

qik (z) =

Qn k

(γ ,σ ) Qn k k

(z)

(sik )

;

⎧

n+1 2  (n + γk )  (n + σk ) ⎪ ⎪ ⎪ 22+γk +σk ⎪ ⎪ z−1  (2n + 2 + γk + σk ) ⎪ ⎪

⎪ ⎨ 2 (γk ,σk ) z∈ / [−1, 1] F n + 1, n + γk ; 2n + 2 + γk + σk ; Qn (z) = 1−z ⎪ ⎪ ⎪ ⎪ ⎪ ⎪   ⎪ ⎪ ⎩ Qn(γk ,σk ) (z + i0) + Qn(γk ,σk ) (z − i0) /2 (−1 < z < 1). ∗  Thus, a closed system has been obtained for determining the values ϕk (sik ) i = 1, n; k = 1, 2 by means of which it is possible to calculate the values of the regular parts of the unknown jumps of dimensionless stresses ϕj∗ (x) (j = 1, 2) at any point of the segment [−1, 1]. A numerical analysis of the problem is carried out, and the regularities of changes in the distribution of contact stresses under the inclusion, the angle of rotation of the inclusion, opening of the crack and the intensity factor of fracture stresses at the end point of the crack are studied, depending on the ratio λ = a/b and angle α in the case when the crack edges are free of stresses, i.e. σ0∗ (x) ≡ 0. The calculation results are presented in tables and graphs. Table 2.8 lists the values of the reduced angle of rotation γ∗ (0) of the inclusion depending on the change in Poisson’s ratio ν in the case λ = 2; E0 = 10. As shown in Table 2.8 that with an increase in Poisson’s ratio, the angle of rotation of the inclusions decreases. Table 2.9 shows the values of the reduced intensity factor KI (1) depending on λ in the case when, α = π/2, E0 = 10 and ν = 0.3. It is clear that with increasing λ, which at a constant length of the inclusion can be interpreted as an increase in the length of the crack, the reduced intensity factor decreases, i.e. the shorter the crack, the more likely it is to propagate. Table 2.10 shows the values of the reduced intensity factor KI (1) and, depending on the angle α, in the case λ = 2; E0 = 10 and ν = 0.3. It can be seen that as the angle α increases, KI (1) increases and KII (1) decreases.

Table 2.8 Rotation angle γ∗ (0) ν 0.1 0.2 γ (0)

0.2965

0.2936

0.3

0.4

0.45

0.2845

0.2683

0.2571

Table 2.9 Intensity factor KI (1) λ

1.1

1.5

2

2.5

3.

4.

KI (1)

0.7617

0.3292

0.2162

0.1636

0.1319

0.0946

124

2 Some Mixed Boundary Value Problems for Homogeneous Elastic …

Table 2.10 Intensity factors KI (1) and KII (1) (for ν = 0.3) α

0

π/6

π/4

π/3

π/2

KI (1) KII (1)

0 0.0258

0.1080 0.0224

0.1528 0.0183

0.11872 0.0130

0.2162 0

Fig. 2.27 Crack opening depending on the parameter λ

Fig. 2.28 Crack opening depending on the Poisson’s ratio

The graphs of normal components of the reduced crack opening for case α = π/2, E0 = 10, ν = 0.4, and for case α = π/2, λ = 2; E0 = 10 are shown in Figs. 2.27 and 2.28, respectively. In the first case, the regularity of the change in the reduced crack opening was studied depending on the parameter λ, and in the second case, depending on the Poisson’s ratio. Figures demonstrate that with an increase λ, the reduced crack opening decreases, and with an increase in Poisson’s ratio, it increases. The graphs of the reduced normal and shear contact stresses in the case when α = π/2, b/a = 2, and ν = 0.4 are shown in Figs. 2.29 and 2.30, respectively. The calculations show that the contact stresses are practically independent of both the parameter b/a and the Poisson’s ratio.

References

125

Fig. 2.29 Reduced normal stresses

Fig. 2.30 Reduced shear stresses

References 1. Abrahamyan BL (1969) Contact (mixed) boundary value problems of the theory of elasticity. Proc Acad Sci USSR Mechan Solid 4:181–197 (in Russian) 2. Aghayan KL (1972) Some contact problems for infinite plate, strengthened by elastic patches. Proc Russ Acad Sci Mechan Solid 5:34–45 (in Russian) 3. Arutyunyan NKh, Mkhitaryan SM (1969) Periodic contact problem for half-plane with coverings. J Appl Math Mechan 33(5):813–820 4. Beitmen G, Erdeyi A (1966) Higher transcendental functions. Nauka 2:295 (in Russian) 5. Elata D (1999) On the problem of rigid inclusions between two dissimilar elastic half-spaces with smooth surfaces. Int J Solids Struct 36:2633–2636 6. Grigoryan EKh (1981) A problem for elastic infinite plate, armed by cross-formed infinite stringer. Proc NAS RA Mechan 47(1–2):3–13 (in Russian) 7. Hakobyan VN, Amirjanyan HA (2015) Stressed state of semi-infinite plane with absolutely rigid inclusion and crack. Proc NAS RA Mechan 68(1):25–36 (in Russian) 8. Hakobyan VN, Dashtoyan LL, Hakobyan LV (2009) On the stressed state of an elastic plane with a semi-infinite cut perpendicular to a crack of finite length. In: Proceedings of International School—conference of young scientists 28 September–1 October. Aghavnadzor Armenia 118– 124 (in Russian) 9. Hakobyan VN, Dashtoyan LL, Hakobyan LV (2012) On stress state of elastic plane with semiinfinite slit, perpendicularly situated at the absolutely rigid inclusion, which is partially detached from matrix. In: Proceedings of international conference dedicated to the 100th anniversary of academician Nagush Kh. Arutyunyan “topical problems of continuum mechanics”. vol 1, pp 57–61 (in Russian) 10. Hakobyan VN, Sahakyan AV (1999) A mixed problem for an elastic wedge weakened by a crack. Mechan Solid J Russ Acad Sci 6:56–65

126

2 Some Mixed Boundary Value Problems for Homogeneous Elastic …

11. Hakobyan VN, Sahakyan AV (2007) On a stress state of elastic plane with semi-infinite slit perpendicularly arising to rigid inclusion of finite length. In: Proceedings on dedicated to the 85 Anniversary of Academician S. A. Ambartsumyan, pp 16–26 (in Russian) 12. Hakobyan VN, Sahakyan AV (2002) Stress state of an elastic half-plane containing a thin rigid inclusion. Mechan Solids J Russ Acad Sci 6:64–69 13. Muskhelishvili NI (1966) Some problems of mathematical theory of elasticity. Nauka 708 (in Russian) 14. Muskhelishvili NI (1968) Singular integral equations. Nauka 511 (in Russian) 15. Parton VZ, Perlin PI (1977) Integral equations of the theory of elasticity. Nauka, Moscow, p 311 (in Russian) 16. Popov GYa (1966) Two dimensional contact problem of the theory of elasticity taking into account of cohesion forces or friction. J Appl Math Mechan 30:551–563 (in Russian) 17. Sahakyan AV (2000) Method of discrete singularities in application towards solving of singular integral equations with unmovable singularity. Proc NAS RA Mechan 53(N3):12–19 (in Russian) 18. Sahakyan AV, Amirjanyan HA (2018) Method of mechanical quadrature’s for solving singular integral equations of various types. IOP Conf Ser J Phys Conf Ser 991:012070. https://doi.org/ 10.1088/1742-6596/991/1/01207 19. Uflyand Y (1968) Integral transformations in problems of the elasticity theory. Nauka, Leningrad, p 402 (in Russian)

Chapter 3

Plane Strain State of Piecewise Homogeneous Elastic Plane With Interfacial Absolutely Rigid Inclusions

Abstract In the present section based on the principles of the linear theory of elasticity the plane strain state of piecewise homogeneous planes containing rectilinear or arch-shaped interfacial absolutely rigid thin inclusions under different conditions of contact between the inclusion and the matrix is considered. We assume that the short sides of inclusion are not contacted with the matrix. For some cases the exact solutions of general problems are constructed, when the piecewise homogeneous planes contain interfacial cracks and the mixed conditions are given on the banks, the special cases of which can be considered as a presence of absolutely rigid inclusions. Keywords Piecewise homogeneous plane · Interfacial absolutely rigid thin inclusion · Crack · Arc-shaped interfacial absolutely rigid thin · Mixed boundary value problem

3.1 Mixed Boundary Value Problem for a Compound Plane Weakened by Crack Let us consider the strain state of compound elastic plane, consisting of two homogeneous half-planes with Lamé coefficients μ1 , λ1 and μ2 , λ2 , respectively. The plane is weakened by crack of length 2a on join-line of the half-planes. The components of stresses σ1 (x) − iτ1 (x) are given at the upper bank, the components of displacements u 2 (x) + iv2 (x) and the resultant vector of acting stresses are given at the lower bank of crack (Fig. 3.1). Now, we are inserting the upper indices 1 and 2 for all values describing the stressed state of upper and lower half-planes, correspondingly. This problem mathematically can be stated as the following boundary value problem [4]: σ y(1) (x, 0) = σ y(2) (x, 0) ; u

(1)

(x, 0) = u

(2)

(x, 0) ;

(2) τx(1) y (x, 0) = τx y (x, 0)

v

(1)

(x, 0) = v

(2)

(x, 0)

(|x| > a) (|x| > a)

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 V. N. Hakobyan, Stress Concentrators in Continuous Deformable Bodies, Advanced Structured Materials 181, https://doi.org/10.1007/978-3-031-16023-3_3

(3.1a)

127

128

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

σ y(1) (x, 0) = σ1 (x) ;

τx(1) y (x, 0) = τ1 (x) (|x| < a)

u (2) (x, 0) = u 2 (x) ;

v (2) (x, 0) = v2 (x) (|x| < a)

(3.1b)

for components of displacements u ( j) (x, y) , v ( j) (x, y) ( j = 1, 2), satisfying the Lamé equations in corresponding half-planes:   ∂θ j μ j u ( j) + λ j + μ j =0 ∂x   ∂θ j =0 μ j v ( j) + λ j + μ j ∂x ∂u ( j) ∂v ( j) + θj = ∂x ∂x ( j)

( j)

and components of stresses σ y (x, y) , τx y (x, y) ( j = 1, 2), which are connected with the components of displacements by Hook’s law [12, 18]: ∂u ( j) ( j) , σx = λ j θ j + 2μ j ∂x

∂v ( j) ( j) σ y = λ j θ j + 2μ j , ∂y

 ( j) τx y = μ j

∂v ( j) ∂u ( j) + ∂x ∂y



To construct the solutions of described mixed boundary value problem the solutions of Lamé equations for upper and lower half-planes are represented as the Fourier integrals

Fig. 3.1 Compound plane with inclusion partially detached from matrix

3.1 Mixed Boundary Value Problem for a Compound Plane Weakened by Crack

129

  ∞  y (−1) j |s| B j (s) + isC j (s) 1 u (x, y) = B j (s) + 2π ae j −∞

× exp −isx + (−1) j |s| y ds ( j = 1, 2)  ∞    y is B j (s) − (−1) j |s| C j (s) 1 ( j) v (x, y) = C j (s) + 2π ae j −∞

× exp −isx + (−1) j |s| y ds ( j = 1, 2)

(3.2)

( j)

Stresses acting in the heterogeneous half-planes are given by the following formulas: ( j) σ y (x,

( j)

    ∞  (−1) j ( j) j λj + μj |s| y × 1 + (−1) μ j |s| C j (s) − ϑ1 y) = π μj −∞



× |s| C j (s) − (−1) j is B j (s) exp −iλx + (−1) j |s| y ds ( j = 1, 2)  ∞  (1 + 2 |s| y) 1− sC j (s) + ae j −∞   

(1 + 2 |s| y) + 1+ i |s| B j (s) × exp −isx + (−1) j |s| y ds ae j

τx y (x, y) = −

iμ j 2π

Here ae j =

μ2j λ j + 3μ j ( j)  ; θ1 =  λj + μj λ j + 3μ j

( j = 1, 2)

( j = 1, 2) ,

B j (s) and C j (s) ( j = 1, 2) are an unknown coefficients. Introducing the functions σ (x) , τ (x) , u (x) and v (x), we will have equations defining the jump functions for stresses and the and displacements on the banks of the crack σ y(1) (x, 0) − σ y(2) (x, 0) = σ (x) ; (2) τx(1) y (x, 0) − τx y (x, 0) = τ (x) ;

u (1) (x, 0) − u (2) (x, 0) = u (x) /ϑ2(2) ;

(3.3)

v (1) (x, 0) − v (2) (x, 0) = v (x) /ϑ2(2) ; First, using the representations for displacements and stresses in Fourier integrals, we solve the auxiliary problem for compound plane under conditions 3.1a, 3.1b and 3.3 and stresses and displacements of upper and lower banks via unknown functions σ (x) , τ (x) , u (x) and v (x).

130

3 Plane Strain State of Piecewise Homogeneous Elastic Plane … σ y(1) (x, 0) =

τx(1) y (x, 0) =

l0 l2 l1 σ (x) + u  (x) +   π l0 l2 l1 τ (x) − v  (x) −   π

a −a

a −a

τ (s) l3 ds − s−x π σ (s) l3 ds − s−x π a

l0

d0 d1 v  (x) − τ (x) − v  2 (x, 0) = − (2)  π  ϑ 2

u  2 (x, 0) = −

l0 (2) ϑ2

u  (x) +

−a a

d0 d1 σ (x) −  π

−a

a

v  (s) ds; s−x

−a

a −a

u  (s) ds; s−x

σ (s) l2 ds − s−x π ϑ2(2)

a −a a

τ (s) l2 ds + (2) s−x π ϑ2

−a

u  (s) ds, s−x

(3.4)

v  (s) ds s−x

combinations of stresses and displacements are given by the formulas: ⎧ ⎫ a a  ⎨ ⎬ 1 il il χ ds w ds (s) (s) 2 3 + σ y(1) (x, 0) − iτx(1) l0 χ (x) + l1 w  (x) + y (x, 0) = ⎩ π s−x π s−x ⎭  1 d  (2) u (x, 0) + iv (2) (x, 0) = − dx 



−a

l0 w  (x) (2)

θ2

−a

− d0 χ (x) +

l2 i

a

(2)

πθ2

−a

d1 i w  (s) ds+ s−x π

a −a

⎫ χ (s) ds ⎬ s−x ⎭

(−∞ < x < ∞)

(3.5) where



  (2) ; χ (x) = σ (x) − iτ (x) = σ y(1) (x, 0) − σ y(2) (x, 0) − i τx(1) y (x, 0) − iτx y (x, 0) 

 (2) (1) (2) (1) (2) w (x) = u (x) + iv (x) = θ2 u (x, 0) − u (x, 0) + i v (x, 0) − v (x, 0) .

Deriving these equations, we are applying the value of integral [1] ∞ sign (s) e −∞

isx

2i ds = , x

∞ eisx ds = 2π δ (x) −∞

and the following notation: (1)

(2)

(1)

(2)

2

μj θ + θ2 − θ1 μj ( j)  = , d1 = 2 , θ1 =  , 2 2 2 + αj λ j + 3μ j     1 + αj μj μ j λ j + 2μ j Ej νj Ej ( j)  = , λj =   , , μj =  θ2 =  2 + αj 2 1 + νj λ j + 3μ j 1 + ν j 1 − 2ν j



1 (1) (2) 2 (1) (2) 2  = θ2 + θ 2 − θ1 − θ1 , αj = ( j = 1, 2) 1 − 2ν j

θ d0 = 1





l0 = θ2(1) θ2(1) + θ2(2) − θ1(1) θ1(1) − θ1(2) ; l2 = θ1(1) θ2(2) + θ2(1) θ1(2) ;

3.1 Mixed Boundary Value Problem for a Compound Plane Weakened by Crack

131









 (2) −1 (1) (2) (1) (2) (1) (2) (1) (2) (1) (2) (1) (2) θ2 + θ2 θ2 θ1 − θ1 θ2 + θ1 − θ1 θ2 θ2 − θ1 θ1 ; l1 = 2 θ2

−1 







 θ2(1) + θ2(2) θ1(1) θ1(2) − θ2(1) θ2(2) + θ1(1) − θ1(2) θ1(1) θ2(2) − θ2(1) θ1(2) ; l3 = 2 θ2(2)

or

−1

−1

ϑ1(2)l0 − ϑ2(2)l2 ; l3 = 2 ϑ2(2) ϑ1(2)l2 − ϑ2(2)l0 , l1 = 2 ϑ2(2) Here ν j , E j ( j = 1, 2) are Poisson’s ratio and Young’s modulus of different halfplanes, respectively. These notations will keep, besides special cases. Satisfying conditions 3.1b by Eq. 3.5, the system of singular integral equations of the second kind can be get for the determination of functions χ (x) and w (x): ⎧ a a  ⎪ il3 χ (s) ds w (s) ds il2 ⎪  ⎪ l0 χ (x) + l1 w (x) + + =  [σ1 (x) − iτ (x)] ⎪ ⎪ ⎪ π s−x π s−x ⎨ −a

−a

a  a ⎪ ⎪ ⎪ w  (x) d1 i w (s) χ (s) ds l2 i ⎪ ⎪ ds+ = − [u 1 (x) + iv (x)] ⎪ ⎩ θ 2 − d0 χ (x) + π θ 2 s−x π s−x 2 2 −a

−a

Reducing the obtained system of singular integral equations to the canonical form, we get ⎧ a  a ⎪ ia2 w (s) χ (s) ia1 ⎪  ⎪ w (x) + ds + ds = F1 (x) ⎪ ⎪ ⎪ π s − x π s −x ⎨ −a

−a

a  a ⎪ ⎪ ⎪ ib1 ib2 w (s) χ (s) ⎪ ⎪ ds + ds = F2 (x). ⎪ ⎩ χ (x) − π s−x π s−x −a

(|x| < a)

(3.6)

−a

Meanwhile, the solutions of this system should satisfy the conditions of equilibrium and the boundedness of jump of the displacements at points ±a, i.e. a χ (x) dx = T0 , w (±a) = 0

(3.7)

−a

Constants a j , b j and functions F j (x)

F1 (x) =

( j = 1, 2) are given by formulas:

 

  d θ2(2) (1) θ1 − θ1(2) σ1 (x) − iτ1 (x) − l0 (u 2 (x) + iv2 (x)) θ dx

132

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …  2 l0 F2 (x) = (σ1 (x) − iτ1 (x)) θ 2    



2  d (2) (1) 2 (1) 2 (1) (2) 2 θ2 − θ1 θ2 − θ1 − θ12 + θ1 (u 2 (x) + iv2 (x)) , dx

2



2  2 θ2(1) − θ1(1)

θ2(1) θ1(2) θ (1) θ (2) θ (1) θ (2) , a2 = 2 2 , b1 = , b2 = 1 2 , θ 2θ θ θ a 2 2 θ = θ2(1) − θ1(1) + θ1(1) θ1(2) , T0 = [σ1 (x) − iτ1 (x)] dx − (σ2 − iτ2 ).

a1 =

−a

Now let us to solve the system of singular integral Eq. 3.6 under conditions 3.7. Multiplying the first equation from 3.6 by λ∗ = 0 and summing up with the second equation, we get: λ∗ a2 + b2 χ (x) + λ∗ w (x) + i π 

a χ (s) + −a

λ∗ a1 −b1  w λ∗ a2 +b2

(s)

s−x

ds = λ∗ F1 (x) + F2 (x) (3.8)

It is required that λ∗ should be the root of quadratic equation a2 λ2∗ − (a1 − b2 ) λ∗ + b1 = 0

(3.9)

The discriminant of Eq. 3.9 can be represented in the following form: D=

  4μ μ(ν1 − ν2 )2 − 4 (1 − ν1 ) (1 − ν2 ) (3 − 4ν2 ) [μ (1 − 2ν1 ) (1 − 2ν2 ) + 2 (1 − 2ν2 )]2

(μ = μ2 /μ1 )

It is clear that Eq. 3.9 has one real root as μ = 4 (1 − ν1 ) (1 − ν2 ) (3 − 4ν2 ) / (ν1 − ν2 )2 . In other cases this equation has either two real roots or complex conjugate roots. In Table 3.1 for different Poisson’s ratio some numerical values for parameter μ are shown when D = 0. Let us consider two possible cases: (a) Eq. 3.9 has two different roots

Table 3.1 Numerical values for parameter μ ν1 ν2 0.1 0.2 0.1 0.2 0.3 0.4

633.6 748.8 163.8 62.4

492.8 105.6

0.3

0.4

113.4 403.2

33.6 67.2 233.2

302.4

3.1 Mixed Boundary Value Problem for a Compound Plane Weakened by Crack

λ(∗j)

 a1 − b2 + (−1) j+1 (a1 − b2 )2 − 4a2 b1 = , 2a2

133

( j = 1, 2) .

(2) Substituting by turns λ = λ(1) ∗ and λ = λ∗ into 3.8 we get two independent equations of second kind

iq j ϕ j (x) + π Here

a −a

ϕ j (s) ds = Q j (x) s−x

(−a < x < a,

j = 1, 2)

(3.10)

ϕ j (x) = χ (x) + λ(∗j) w  (x) , Q j (x) = F2 (x) + λ(∗j) F1 (x)  (a1 + b2 ) − (−1) j (a1 − b2 )2 − 4b1 a2 qj = ( j = 1, 2) 2

To solve Eq. 3.10 let us consider the analytical functions over the whole plane, besides points of interval (−a, a). 1  j (z) = 2πi

a −a

ϕ j (s) ds s−z

( j = 1, 2)

(3.11)

The values of these functions are defined by Plemelj–Sokhotski formulas on banks of interval (−a, a) [12, 13] ±j

1 1 = ± ϕ j (x) + 2 2πi

a −a

ϕ j (s) ds s−x

Therefore, ϕ j (x) =

+j

(x) −

−j

(x) ,

1 πi

a −a

ϕ j (s) ds = +j (x) + −j (x) s−x

( j = 1, 2)

(3.12) Substituting these relations in Eq. 3.10 we obtain the following two Riemann problems for determination of functions  j (z)   +j (x) − g j −j (x) = Q j (x) / 1 − q j ( j = 1, 2)  gj =

1 + qj 1 − qj

(3.13)



The solutions of this problem vanishing at infinity and unbounded at both ends of interval (−a, a) will be [2, 13]

134

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

⎧ ⎨

1    j (z) = X j (z) ⎩ 2πi 1 − q j

a −a

⎫ ⎬ Q j (s) ( j) ds + c 0 ⎭ X +j (s) (s − z)

( j = 1, 2) (3.14)

Here X j (z) = (z + a)−γ j (z − a)γ j −1 , γ j =

ln g j θj + , 2πi 2π

  0 < θ j = arg g j < 2π

X +j (s)—are boundary values of analytical functions in whole plane X j (z) on the ( j)

upper bank of the interval (−a, a) and c0 ( j = 1, 2) are unknown constants that should be defined. Furthermore, applying both relations 3.14 and Plemelj–Sokhotski formula, we can find the values of functions ±j (x). After substituting into the first relation 3.12 and introducing notation √ X +j (x) = − g j ω j (x) ,

! ω j (x) = (x + a)−γ j (a − x)γ j −1 , d j = −2q j 1 − q 2j

we get ϕ j (x) =

⎧ ⎨

q j ω j (x) 1 Q j (x) + πi 1 − q 2j ⎩

a −a

(−a < x < a,

⎫ ⎬

Q j (s) ds + d j ω j (x) ⎭ ω j (s) (s − x)

(3.15)

j = 1, 2)

Note that g j < 0

( j = 1, 2) in the case of real roots of Eq. 3.9. Actually, it is easy 2 2 ( j) ( j) − θ1 > 0 ( j = 1, 2). On the other to check that q1 > q2 > 0 and θ2 hand





 θ2(2) − θ1(2) + θ2(1) − θ1(1) θ2(1) − θ1(1) q22 − 1 q2 − 1 = >0 = q2 + 1 θ (1 + q2 )

Hence, g j < 0 and θ j = π,

γj =

1 1 − iβ j , β j = ln g j , ( j = 1, 2) 2 2π

In the case of complex roots of Eq. 3.9, we have λ1 = λ¯ 2 , quently,  γ1 = α − iβ, γ2 = 1 − α − iβ

θ2 = 2π − θ1 . Conse-

1 ln |g1 | α = θ1 /2π, β = 2π



The unknown jump function χ (x) of stresses and derivative of crack’s opening w (x) will be defined via functions ϕ j (x) ( j = 1, 2)

3.1 Mixed Boundary Value Problem for a Compound Plane Weakened by Crack

w  (x) =

ϕ1 (x) − ϕ2 (x) (2) λ(1) ∗ − λ∗

,

χ (x) =

(1) λ(2) ∗ ϕ1 (x) − λ∗ ϕ2 (x) (1) λ(2) ∗ − λ∗

135

(3.16)

Constants d j ( j = 1, 2) in ϕ j (x) can be determined from conditions 3.7, which have the following form in ϕ j (x) ( j = 1, 2): a ϕ j (x) dx = T0

(3.17)

−a

Applying formula 3.4 for |x| > a and taking into consideration that χ (x) = w (x) = 0, it is not difficult to find the fracture stresses out of crack on junction line of halfplanes. The stresses out of crack are expressed in functions ϕ j (x) by formula ( j) ( j) σ y (x, 0) − iτx y (x, 0)

(|x| > a,



⎧ ⎫ (2) a a i l3 − λ(1) ∗ l2 1 ⎨ i l 3 − λ∗ l 2 ϕ1 (s) ds ϕ2 (s) ds ⎬



= − (1) (2)  ⎩ π λ(1) − λ(2) s−x π λ∗ − λ∗ −a s − x ⎭ ∗ ∗ −a

j = 1, 2)

(3.18) For instance, let us consider a case, when the upper bank of crack is free of stresses, i.e. σ1 (x) − iτ1 (x) =  thin inclusion welded with  0. There is an absolutely rigid lower bank of crack u (2) (x) = 0, v (2) (x) = const . 

ϕ j (x) = A j ω j (x)

  A j = d j / 1 − q 2j ,

j = 1, 2



(3.19)

From 3.12 we can find w  (x) = χ (x) = −

1 λ(1) ∗

λ(2) ∗

−

w (x) =

1 λ(1) ∗

− λ(2) ∗

[A1 ω1 (x) − A2 ω2 (x)]

 (2)  λ∗ A1 ω1 (x) − λ(1) ∗ A2 ω2 (x) 1

(2) λ(1) ∗ − λ∗

(|x| < a)

(3.20)

x [A1 ω1 (x) − A2 ω2 (x)] dx + c0 −a

Taking into account the value of integral [15] a

(x + a)−γ (a − x)γ −1 dx =

−a

and satisfying the conditions 3.17 we get c0 ≡ 0,

π sin π γ

(3.21)

136

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

Aj =

sin π γ j T0 ( j = 1, 2) π

(3.22)

Hence   T0

λ(2) χ (x) = − sin (π γ1 ) ω1 (x) − λ(1) sin (π γ2 ) ω2 (x) (3.23) ∗ ∗ (2) π λ(1) ∗ − λ∗ 

  ∞ sin (π γ1 ) "  (k + 1 − γ1 ) a + x k+1−γ1  (1 − γ1 ) k=0 k + 1 − γ1 2    ∞ sin (π γ2 ) "  (R + 1 − γ2 ) a + x k+1−γ2 − (−a < x < a)  (1 − γ2 ) k=0 k + 1 − γ2 2 (3.24) We derived the 3.19 applying the formula [15] w (x) =

T0

(1) π λ(2) ∗ − λ∗

x (x + a)

−γ

(a − x)

γ −1

−a

  ∞ "  (k − γ + 1) a + x k+1−γ dx = k+1−γ 2 k=0

(|x| < a) (3.25)

Substituting the derived expressions for functions ϕ j (x) ( j = 1, 2) in 3.18 and recall that [15] a −a

a+x π (s + a)−γ (a − s)γ −1 = s−x (a − x) sin π (1 − γ ) a − x

−γ

(|x| > a) (3.26)

for fracture stresses out of crack on join-line of two half-planes the following formula is obtained ( j) ( j) σ y (x, 0) − iτx y (x, 0)

=



i T0 (1)

(2)

π  λ∗ − λ∗





(a − x)

a+x A a−x

−γ1

a+x − A¯ a−x

−γ2



(3.27)

(|x| > a) , A = (l3 − λ2 l2 )

Now we are calculating the intensity coefficients of fracture stresses at the end points of the crack x = ±a. It is obvious from 3.27 that the stresses are defined as a sum of two summands, having different power singularities at the end points of the crack. If materials of half-planes are chosen such as Re γ1 > Re γ2 , then the main singularity of stresses at point x = −a is determined by the first summand. Second summand is dominating at a point x = a.

3.1 Mixed Boundary Value Problem for a Compound Plane Weakened by Crack

137

Therefore,     √ K I (−a) j − i K I I (−a) j = 2π

lim

x→(−1) j (a+0)

|a + x|γ j |a − x|γ j

 i √2π T0 1 − (−1) j  A + 1 + (−1) j  A¯   ( j) ( j)

× σ y (x, 0) − iτx y (x, 0) = (2) 2π  λ(1) ∗ − λ∗

( j = 1, 2)

(3.28) It is not difficult to calculate the intensity coefficients of fracture stresses in the case Re γ1 < Re γ2 . (b) Let us consider the case, when Eq. 3.9 has two equal roots, i.e. (2) λ(1) ∗ = λ∗ = (a1 − b2 ) /2a2 . Substituting λ = λ1 into 3.8 the only one singular integral equation is obtained for determination of the function ϕ1 (x) = χ (x) +  λ(1) ∗ w (x). This equation is similar to 3.10 and in this case q1 = (a1 + b1 ) /2 > 1. The solution has the form 3.15 for j = 1. γ1 =

1 1 − iβ, β = ln |g1 | , g1 = (1 + q1 ) / (1 − q1 ) 2 2π

Let us define the function χ (x) through the functions ϕ1 (x) and w  (x) and substitute the obtained expression in first equation of 3.6. Then, taking into account the equality i π

a −a

1 ϕ1 (s) ds = [Q 1 (x) − ϕ1 (x)] s−x q1

we get the singular integral equation just like for function ϕ1 (x) with only difference that right part of this equation has the following form for determination of function w  (x) Q 0 (x) = F1 (x) +

a2 [ϕ1 (x) − Q 1 (x)] q1

Therefore, we are able to define the function w (x) in the way similar to the determination of function ϕ1 (x). Afterward, the function χ (x) can be obtained by equation  χ (x) = ϕ1 (x) − λ(1) ∗ w (x). The unknown constants and the fracture stresses out of crack can be found from conditions 3.7 and formula 3.4 for |x| > a. For an instance, we consider case of the absolutely rigid inclusion. The upper edge of inclusion is free of loads, i.e. F1 (x) = F2 (x) = Q 1 (x) ≡ 0 . Using integral value  a  a−x shπβ 1 1 ln (a + x)− 2 +iβ (a − x)− 2 −iβ dx = −π 2 i 2 a+x ch πβ −a

and integral relation

138

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …



a ln −a

a−s a+s



1 1  π π (a + s)− 2 +iβ (a − s)− 2 −iβ ds = s−x (a − x) ch 2 (πβ)

−ith (πβ) ln

a−x a+x



1 a + x − 2 +iβ a−x

(|x| > a)

what we can get from the known integrals [15], substituting in the following way u = ln [(a + x) / (a − x)] we find a2 1 1 ϕ1 (x) = e0 ω (x) ω (x) = (x + a)− 2 +iβ (a − x)− 2 −iβ Q 0 (x) = ϕ1 (x) q1     a2 λ(1) a−x ∗ e0 (1)   ln + λ∗ e1 − e0 ω (x) , χ (x) = − a+x πi 1 − q12     a − x e a 2 0  ln  w  (x) = + e1 ω (x) a+x πi 1 − q12 iβ   sgnx x+a x +a x−a +i K 2 σ y( j) (x, 0) − iτx(yj) (x, 0) = K 1 ln √ (|x| > a) x −a ch (πβ) x 2 − a 2 Here

a2 e0 l3 − λ(1) ∗ l2  ,  K1 = π 1 − q12 e0 =

    e th a (πβ) 2 0  e1 +  K 2 = e0 l2 + l3 − λ(1) ∗ l2 1 − q12 chβ a2 sh (πβ)  T0 T0 , e1 =  π π 1 − q12

So that, it is obvious that contact stresses at the end points of the crack have power singularity x −ν (0.5 ≤ ν < 1) for D (μ) = 0 and singularity x −0.5 ln x for D (μ) = 0 without taking into consideration the oscillating part. Note that D (μ) < 0 for any value of μ, if Poisson’s ratios of half-planes are equal. This means that Eq. 3.9 has two complex conjugate roots, and, consequently, the contact stresses at the end points of the crack have x −ν singularity. As a conclusion, let us consider one more special case of the above-mentioned problem: μ1 = μ2 ;

ν1 = ν2 = ν;

u (2) (x) = 0; v (2) (x) = const;

σ1 (x) − iτ1 (x) = 0

. i.e. homogeneous plane contains crack with given vertical displacements on lower

bank and upper bank is free of stresses. It is easy to prove that

3.2 Mixed Problem for Compound Plane Weakened by Periodic System of Cracks

139

√ √ (−1) j+1 i ae (1 − 2ν) + (−1) j+1 i ae , qj = = 1−ν 2 (1 − ν) √ 1 i ln ae 3 i ln ae g j = (−1) j+1 i ae, γ1 = − , γ2 = − 4 4π 4 4π

λ(∗j)

here ae = 3 − 4ν is the Muskhelishvili constant. Substituting these expressions into 3.23 and taking into account that √    √  2  1 − (−1) j 1 + (−1) j i ae sin π γ j = √ 4 4 ae For the contact stresses we get √

χ (x) =

   √  √  2T0  √ (1 + i) 1 − ae ω1 (x) + (1 − i) 1 + i ae ω2 (x) (3.29) 4 8π ae

ω1 (x) = (x + a)− 4 +iβ (a − x)− 4 −iβ , (β = ln (ae/4π )) 1

3

ω2 (x) = (x + a)− 4 +iβ (a − x)− 4 −iβ 3

1

This result coincides with the result, obtained by Muskhelishvili for this problem [12].

3.2 Mixed Problem for Compound Plane Weakened by Periodic System of Cracks In this subsection we consider the stress state of compound plane weakened by periodic system of cracks. On one bank of cracks the stress components and on the other bank displacements are given. Let the elastic compound plane consisting of two half-planes with different shear modulus μ1 , μ2 and Poisson’s ratio ν1 , ν2 is weakened by cracks the system of length 2a and period 2l on junction line. The stresses σ1 (x) − iτ1 (x) are given at the upper bank and displacement components u (2) (x) + iv (2) (x) and stress resultant σ2 − iτ2 are given at the lower bank of cracks (Fig. 3.2). For definiteness, as before, all values describing the stress–strain state of the upper and lower half-planes are labeled by indices 1 and 2 correspondingly. Taking into account the periodicity of stated problem let us consider the stress state of compound infinite strip with width 2l, which is weakened by central crack with length 2a on junction line of two semi-infinite strips in domain D± = {−l ≤ x ≤ l; 0 ≤ ±y < ∞}. The stress components are given at the upper bank and displacements at the lower bank of the central crack. The symmetry conditions exist on bounds x = ±l of strip. Saving notation of previous subsection the stated problem can be formulated as the following boundary problem [5]:

140

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

(2) (2) σ y(1) (x, 0) − iτx(1) y (x, 0) = σ y (x, 0) − iτx y (x, 0) (l > |x| > a)

u (1) (x, 0) + iv (1) (x, 0) = u (2) (x, 0) + iv (2) (x, 0)

(l > |x| > a)

σ y(1) (x, 0) − iτx(1) y (x, 0) = σ1 (x) − iτ1 (x) (|x| < a) u (2) (x, 0) + iv (2) (x, 0) = u 2 (x) + iv2 (x) (|x| < a) (1) τx(1) (±l, y) = 0 (0 < y < ∞) y (±l, y) = u (2) τx(2) (±l, y) = 0 (−∞ < y < 0) y (±l, y) = u

(3.30a)

(3.30b)

(3.30c)

To construct the solution of mixed boundary value problem let us introduce the jump functions χ (x) and w (x) by formulas 3.13. First we solve the auxiliary problem similar with 3.30. The only difference is that conditions 3.30b are replaced by 3.3. Let us define the stress components of the upper bank of the crack and displacements of lower bank by functions χ (x) and w (x). Applying the biharmonic function of stresses (Airy function) for domains D± and represent them in series form: F j (x, y) =

∞  "

( j)

( j)

Ak − (−1) j αk y Bk



  exp (−1) j+1 αk y cos αk x

(3.31)

k=1 ( j)

( j)

here αk = π k/l and Ak , Bk ( j = 1, 2; k = 1, 2, ...) are the unknown coefficients. Stresses and displacements are determined using stress functions according to the formulas [12]: Fig. 3.2 Compound plane with periodic system with interfacial inclusions partially detached from the matrix

3.2 Mixed Problem for Compound Plane Weakened by Periodic System of Cracks

∂ 2 F j (x, y) ∂ 2 F j (x, y) ( j) , τ = − x y ∂x2 ∂ x∂ y  2 ∂ F j (x, y) Ej νj ∂ F j (x, y) ( j) ( j)  u ( j) (x, y) = dx −  + a0 y + c0 ; ∂ y2 ∂x 1 − ν 2j 1 − νj  2 Ej ∂ F j (x, y) νj ∂ F j (x, y) ( j) ( j) ( j)  − a0 x + b0 v dy −  y) = (x, ∂x2 ∂y 1 − ν 2j 1 − νj (3.32) It is easy to prove that the last four conditions of 3.30 are satisfied by such choice of functions F j (x, y) ( j = 1, 2). Satisfying the rest conditions of auxiliary problem ( j) ( j) and expressing Ak , Bk ( j = 1, 2) by Fourier coefficients of functions χ (x) and w (x) the following is obtained for determination of stress components at the upper bank and displacements on lower bank of crack: σx( j) =

∂ 2 F j (x, y) , ∂ y2

141

σ y( j) =

⎧ a 1 ⎨ il2 π (s − x) l0 χ (x) + l1 w  (x) + = ctg χ (s) ds+ ⎩ 2l 2l −a ⎫ a ⎬ π (s − x)  il3 ctg + w (s) ds ⎭ 2l 2l

σ y(1) (x, 0) − iτx(1) y (x, 0)

(3.33)

−a

⎧    a d u (2) (x, 0) + iv (2) (x, 0) il2 π (s − x)  1 ⎨ l0  − d w χ + ctg w (s) ds+ =− (x) (x) 0 dx  ⎩ θ22 2l 2lθ22 −a ⎫ a ⎬ id1 π (s − x) + ctg χ (s) ds (−l < x < l) ⎭ 2l 2l −a

(3.34)

It should be noted that relations 3.33 and 3.34 could be obtained from 3.5. For this purpose functions χ (x) and# w  (x) are considered 2l-periodical, integrals in 3.5 extended over the interval L = ∞ n=−∞ (−a + 2nl; a + 2nl) and take into account the well-known formula [14, 17] ∞ " n=−∞

π πx 1 = ctg x − 2nl 2l 2l

Note that the relations 3.33 and 3.34 also can be obtained from 3.5. For this purpose now let us use the conditions 3.30b. After some operations the following system of singular integral equations of second kind with Hilbert kernel is obtained:

142

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

⎧ a a ⎪ π (s − x)  π (s − x) a1 i a2 i ⎪  ⎪ w (x) + ctg ctg w (s) ds + χ (s) ds = F1 (x) ⎪ ⎪ ⎪ 2l 2l 2l 2l ⎨ −a

−a

a a ⎪ ⎪ ⎪ b1 i π (s − x)  π (s − x) b2 i ⎪ ⎪ w (s) ds + χ (s) ds = F2 (x) ctg ctg ⎪ ⎩ χ (x) − 2l 2l 2l 2l −a

−a

(3.35) this system should be considered together with conditions a χ (x) dx = T0 , w (±a) = 0

(3.36)

−a

Now let us solve the system 3.35. It is clear that the system 3.35 almost coincides with 3.6 and the only difference is that Hilbert kernel is instead of Cauchy kernel. Therefore, the solution of this system can be reduced to solution of two integral equations with Hilbert kernel in case D = 0 and successive solving of two integral equations in case D = 0. (1) Let us consider case when D = 0, i.e. Eq. 3.9 has two different nonzero roots (2) (1) (2) λ(1) ∗ and λ∗ . Multiplying the first equation by λ∗ and λ∗ alternately and summing with second equation we get two independent integral equations: iq j ϕ j (x) + 2l

a ctg −a

π (s − x) ϕ j (s) ds = Q j (x) 2l

( j = 1, 2;

(3.37)

|x| < a)

( j)

The functions ϕ j (x) = χ (x) + λ∗ w  (x) should be also satisfied the conditions a ϕ j (x) dx = T0 ( j = 1, 2)

(3.38)

−a

These conditions follow from 3.36. To construct the solution of Eq. 3.37 let us pass to new variables by formula: τ = tg

πs , 2l

Now introduce the following notation:

t = tg

πx 2l

(3.39)

3.2 Mixed Problem for Compound Plane Weakened by Periodic System of Cracks

   2l 2l arctg t , Q ∗j (t) = Q j arctg t π π α τ ϕ ∗j (τ ) iq j πa , Aj = α = tg dτ 2l π 1 + τ2 ϕ ∗j (t) = ϕ j

143



(3.40)

−α

Then, Eq. 3.37 and conditions 3.38 will be represented in form: ϕ ∗j

iq j (t) + π

α −α

ϕ ∗j (τ ) dτ τ −t

= Q ∗j (t) + A j α −α

ϕ ∗j (τ ) 1 + τ2

dτ =

( j = 1, 2;

|t| < α)

π T0 2l

(3.41)

(3.42)

We see that Eq. 3.41 coincide with Eq. 3.10. Therefore, solutions of equations are obtained by formula 3.15. The only difference is that instead of the functions Q j (x) will be functions Q ∗j (t) + A j . For further calculations it is convenient to represent these formulas in the following form: ϕ ∗j (t) =

⎧ ⎨

1 Q ∗ (t) + A j + 1 − q 2j ⎩ j ( j = 1, 2;

q j X +j (t) πi

α −α

⎫ ⎬

Q j (τ ) + A j dτ + d ∗j (t) X +j (t) X +j (τ ) (τ − t) ⎭

−α < t < α)

(3.43) Here X +j (t) ( j = 1, 2) are boundary values of analytical functions in whole plane, cut-to-interval (−α, α) at the upper bank of slit X j (z) = (z + α)−γ j (z − α)γ j −1 ( j = 1, 2)

(3.44)

The coefficients d ∗j ( j = 1, 2) are unknown. It is easy to show that X +j (t) = eiπ (γ j −1) ω j (t) X +j (t) = g j X −j (t)       g j = 1 + q j / 1 − q j ; j = 1, 2 Here X −j (t) are boundary values of the functions X j (z) ( j = 1, 2) at lower bank of slit. Satisfying the conditions 3.42 and using the representations for constants A j ( j = 1, 2) in functions ϕ ∗j (t) ( j = 1, 2) from 3.40 the system of the algebraic equations is obtained for determination of an unknown constants d ∗j and A j ( j = 1, 2). Likewise previous subsection for illustration the results let us consider the case when the upper banks of cracks are stress-free and the punches with straight bases

144

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

are acting on the lower banks (u 2 (x) = 0, v2 (x) = const). It is not difficult to see that Q ∗j (t) ≡ 0 ( j = 1, 2). Therefore, the and functions ϕ ∗j (t) ( j = 1, 2) will be ⎡ ⎤ α + X q (t) A dτ j j j ⎣1 + ⎦ + d ∗j X +j (t) ϕ ∗j (t) = πi 1 − q 2j X +j (τ ) (τ − t)

(3.45)

−α

Applying formula [12] α

dτ πi = 1 − gj X +j (τ ) (τ − t)

−α



     1 + gj − 2 t − 1 − 2γ j α X +j (t)

(3.46)

for functions ϕ ∗j (t) we get  ϕ ∗j

(t) =

   t − α 1 − 2γ j ∗ A j + d j X +j (t) 1 + qj

( j = 1, 2)

(3.47)

Substituting ϕ ∗j (t) from 3.47 in expression for A j ( j = 1, 2) and taking into account that α −α

τ m X +j (t) 1 + t2

G (mj) =

dt = −(i)

m+1 π

  1 + q j ( j) G m (m = 0, 1 ; j = 1, 2) qj

(3.48)

 i  (i + α)−γ j (i − α)γ j −1 + (−1)m (i + α)γ j −1 (i − α)−γ j 2

we find A j = νi∗ d ∗j ( j = 1, 2). Then satisfying conditions 3.42 the following expression is obtained for d ∗j ( j = 1, 2) d ∗j =

q j T0     ( j) ( j) ( j) ( j) 2l 1 + q j G 1 a1 − i G 0 a0

( j = 1, 2) ,

Here 

ν ∗j =

 ( j) 1 + q j i G1   ( j) , ( j) G 0 + iα 1 − 2γ j G 1

( j)

a0

=

( j) ν ∗j G0 ( j)   ( j) , a1 = 1 + q ( j) j G 0 + iα 1 − 2γ j G 1

( j = 1, 2)

Now find contact stresses acting under punch and out of crack on junction line of half-planes, as well as the jump functions of displacement on banks of crack. Using the connection of functions ϕ ∗j (x) ( j = 1, 2) with functions χ (x) and w  (x) and passing on previous variables we can find

3.2 Mixed Problem for Compound Plane Weakened by Periodic System of Cracks

  i 1 + α 2 T0 cos πa

2l χ (x) = − (1) 2l λ(2) − λ ∗ ∗

145

⎧ ⎪ (1) πx πx ⎨ λ(2) q G (1) ∗ 1 0 cos 2l + i G 1 sin 2l !     γ 1−γ1 ⎪ ⎩ 1 − q 2 sin π(a+x) 1 sin π(a−x) 1 2l 2l

⎫ ⎪ (2) πx πx ⎬ G (2) cos + i G sin 0 1 2l 2l −!  γ2  1−γ2 ⎪ ; ⎭ 1 − q22 sin π(a+x) sin π(a−x) 2l 2l λ(1) ∗ q2

(3.49)

⎧   ⎪ (1) πa ⎨ πx πx 2 G (1) i 1 + α T0 cos 2l q1 0 cos 2l + i G 1 sin 2l

! w  (x) = −  γ  1−γ1 (2) ⎪ ⎩ 1 − q 2 sin π(a+x) 1 sin π(a−x) 2l λ(1) ∗ − λ∗ 1 2l 2l ⎫ ⎪ (2) (2) πx πx G 0 cos 2l + i G 1 sin 2l ⎬ q2 −! (−a < x < a)  γ2  1−γ2 ⎪ π(a−x) ⎭ 1 − q22 sin π(a+x) sin 2l 2l (3.50) To find fracture stresses out of the crack we use the expression 3.33 for a < |x| < l, which will be represented in form 3.51 by functions ϕ ∗j (x)



⎫ ⎧ (2) (1) α ∗ α ∗ ⎨ l λ l − l − λ l 2 3 3 2 ∗ ∗ ϕ ϕ2 (s) ⎬ (s) i ∗ ∗ 1



ds+ ds + E 0 σ y (t, 0) − iτx y (t, 0) = (2) (1) π  ⎩ λ(2) − λ(1) s−τ λ∗ − λ∗ −α s − t ⎭ ∗ ∗ −α (a < |x| < l)

(3.51) E0 = −





1

(1)  λ(2) ∗ − λ∗



λ(2) l3 − λ(1) ∗ l2 − l3 ∗ l2 A1 + A2 q1 q2



After substitution of expressions 3.47 for functions ϕ ∗j (t) and taking into consideration

⎧ ⎫ ( j) ( j) α ∗ ( j) ⎬ −γ j ⎨ + a t π a ϕ j (s) 0 1 α + t πa √ 1   ds = − g j d ∗j g + ⎩ sin π 1 − γ j (α − t) α − t s−t sin π γ j ⎭ −α

(α < |t| < ∞)

146

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

we get σ y± (x, 0) − iτx±y (x, 0) = ( j)

×



( j)

G 0 cos π2lx + i G 1 sin sin

π(x+a) 2l

γj

sin

π(x−a) 2l



λ(2) ∗

πx 2l 1−γ j

T0 sgnx

2l cos πa − λ(1) ∗ 2l



2 q j λ(∗j) l 2 − l3 (−1) j " !

1 − q 2j sin π γ j j=1

+ E 1∗ (a < |x| < l) (3.52)



√ 2 (−1) j+1 λ(∗j) l 2 − l3 g j d ∗j " i ∗

  E1 = E0 + (1) sin π γ j  λ(2) j=1 ∗ − λ∗

Here

From the obtained expressions it is easy to define the intensity coefficients at points x = ±a. If material of half-planes such that Re γ1 > Re γ2 , then

( j) √ j q j λ∗ l 2 − l 3     2π (−1) ! K I (−a) j − i K I I (−a) j = − ×  2l cos πa 2l 1 − q 2j sin π γ j  πa πa  ( j) + i(−1) j sin × G 0 cos ( j = 1, 2) . 2l 2l

(3.53)

2) Now let us discuss the case when the materials of half-plane materials such that (2) D = 0, i.e. Eq. 3.9 has two identical roots λ(1) ∗ = λ∗ = λ∗ = (a1 − b2 ) /2a2 . Then as in previous subsection we successively solve two integral equations and get the solution of stated problem. Without dwelling on this, let us show the solution of the problem when the upper banks of the cracks are stress-free, and there are absolutely rigid punches with flat bases on the lower bank. In this case Q ∗j ≡ 0 and as in similar problem of (1) for functions ϕ1∗ (t) we obtain   t − α (1 − 2γ1 ) d1∗ X 1+ (t) ϕ1∗ (t) = 1 + ν1∗ 1 + q1

(3.54)

here q1 = ν1∗ =

a1 + b2 , 2 G (1) 0

γ1 =

1 − iβ ; 2

(1 + q1 ) i G (1) 1 + iα (1 −

2γ1 ) G (1) 1

β=

, d1∗ =

Then, we take into account the relation

1 1 + q1 ln |g| , g = a) (3.66a) (1) ⎪ u (x, 0) = u (2) (x, 0) ⎪ ⎪ ⎪ ⎩ (1) v (x, 0) = v (2) (x, 0) 

v (1) (x, +0) = v (2) (x, −0) = δ (2) τx(1) y (x, +0) = u (x, −0) = 0

|x| < a

(3.66b)

Here δ is an unknown constant characterizing the rigid displacement of inclusion. To solve this problem the discontinuous solutions of equations of the elasticity theory for piecewise homogeneous represented by formula 3.4. In these formulas, we turn to the jumps of displacements u (1) (x, 0) − u (2) (x, 0) = u (x) (|x| < a) v (1) (x, 0) − v (2) (x, 0) = v (x) ≡ 0 (|x| < a)

Fig. 3.3 Compound plane with absolutely rigid inclusion on boundary line of materials

152

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

Then, after differentiation of first equation and satisfying the conditions 3.66b we get the governing system of singular integral equations: ⎧ a ⎪ d0 l0  τ (s) d1 ⎪ ⎪ ⎪ σ (x) − u (x) − ds = 0 ⎪ ⎪   π s−x ⎪ ⎪ ⎪ −a ⎪ ⎪ ⎪ ⎪ a a  ⎨d d1 l2 σ (s) u (s) 0 τ (x) + ds + ds = 0  π s−x π s−x ⎪ ⎪ ⎪ −a −a ⎪ ⎪ ⎪ a ⎪  a  ⎪ (2) ⎪ l0 l2 ϑ2 l3 σ (s) u (s) ⎪ ⎪ ⎪ τ (x) − ds − ds = 0 ⎪ ⎩ π s−x π s−x −a

(3.67)

−a

The unknown functions should be satisfied the conditions of equilibrium of inclusion and continuity of displacements at the end points of inclusion, i.e. a

a σ (x)dx = Q 0 ,

−a

a τ (x)dx = 0,

−a

u  (x)dx = 0

(3.68)

−a

Note, that the last problem was partially researched by the method of complex potentials in work [11]. Now let us construct the solution of system 3.67 under conditions 3.68. Eliminating the function τ (x) from Eq. 3.67 we obtain 1 π

a −a

Mσ (s) + N u  (s) ds = 0 s−x

M = l0 d1 + l2 d0 = ϑ2(1) /2,



N = ϑ1(2) ϑ2(1) 

Therefore [12] Mσ (x) + N u  (x) =

C0 √ π a2 − x 2

(3.69)

Here C0 in an unknown constant. After integrating Eq. 3.69 from −a to a and taking into account the conditions 3.68 the following value in obtained for unknown constant C0 : C0 = M Q 0

(3.70)

Then, we determine the function u  (x) from 3.69 by function σ (x). After substitution of function u  (x) into first two Eq. 3.67 we get the system of two singular integral equations for finding the jump functions of stresses.

3.3 Stress State of Compound Plane with Absolutely Rigid …

153

⎧ a ⎪ α τ (s)ds Q ∗0 ⎪ ⎪ σ (x) − = √ ⎪ ⎪ ⎪ π s−x π a2 − x 2 ⎨ −a

(3.71)

a ⎪ ⎪ ⎪ σ (s) β ⎪ ⎪ ds = 0 ⎪ ⎩ τ (x) − π s−x −a

Here



2 ˜ = ϑ2(1) ϑ2(1) + ϑ2(2) − ϑ1(1) − ϑ1(2) , Q ∗0 = l0 Q 0 / ˜ 

˜ β = ϑ2(2) /ϑ1(2) α = ϑ1(2) ϑ2(1) + ϑ2(2) /, ˜ = 0. In this case let us introduce Note also that the system 3.71 is true only for  functions ϕ j (x) = σ (x) + λ∗j τ (x)

 (λ∗j = (−1) j+1 α/β,

j = 1, 2)

Then the system of Eq. 3.71 will be represented in the form of two singular integral equations a ϕ j (s) qj Q∗ ds = √ 0 (3.72) ϕ j (x) + π s−x π a2 − x 2 −a



 −a < x < a, q j = (−1) j αβ, j = 1, 2 At the same time

a ϕ j (x)dx = Q 0 ( j = 1, 2)

(3.73)

−a

˜ > 0. Recalling First we consider the case when numbers q j ( j = 1, 2) are real, i.e.  this fact we can write ⎤ ⎡ a + X (x) q g(s)ds 1 ⎣ j j ⎦ + C j X +j (x) (3.74) g(x) − ϕ j (x) = π 1 + q 2j X +j (s)(s − x) −a

ϑj ϑj 1 ln G j + = 2πi 2π 2π 1 − iq j Q∗ Gj = , G j = 1, g(x) = √ 0 1 + iq j π a2 − x 2

X j (z) = (z + a)−γ j (z − a)γ j −1 , γ j = 0 < ϑ j = arg G j < 2π,

154

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

Taking into account the condition 3.73 we find C j = −Q ∗0

sin π α j  π Gj

( j = 1, 2)

Let us define normal and shear components of jumping functions of stresses on inclusions. Using σ (x) =

ϕ1 (x) − ϕ2 (x) ϕ1 (x) + ϕ2 (x) , τ (x) = 2 2λ∗

and substituting the value of functions ϕ j (x) ( j = 1, 2) from 3.74, as well as taking into account X 1+ (x) X 1+ (s)

=

X 2+ (x)

ω(x) ; ω(s)

X 2+ (s)

ϑ2 = 2π − ϑ1 ,

=

ω(−x) , q j = (−1) j q, ω(−s)

q=

 αβ,

G¯ 2 = G 1

γ2 = 1 − γ1 , ω(x) = (a + x)−γ1 (a − x)γ1 −1

we get ⎧ −1/2 ⎫  a  ds ⎬ ω(x) ω(−x) a 2 − s 2 1 ⎨ Q ∗0 2Q ∗0 − σ (x) = + √ ⎭ 1 + q 2 ⎩ π a2 − x 2 2π 2 ω(s) ω(−s) (s − x) −a

Q ∗ sin π γ1 + 0 [ω(x) + ω(−x)] π τ (x) =

q Q ∗0 2π 2 λ∗ (1 +

a  q 2)

−a



ds ω(x) ω(−x) + √ 2 ω(s) ω(−s) a − s 2 (s − x)

(3.75)

(3.76)

Q 0 sin π γ1 + [ω(x) − ω(−x)] π Now using the value of integral [15] b  a

x −a b−x

α−1

    −π dx y − a α−1 = +1 , cos π α x−y sin π α b−y

(0 < Re α < 2,

(3.77)

a < y < b)

finally we have KI σ (x) = √ + K I I [ω(x) + ω(−x)] 2 a − x2

(3.78)

3.3 Stress State of Compound Plane with Absolutely Rigid …

155

τ (x) = K I I I [ω(x) − ω(−x)]

(3.79)

  ˜ sin π γ1   Q ∗0  q KI = + 1 + qtgπ γ1 , K I I = − ˜ cos π γ1 2π l0 π(1 + q 2 ) (1 + q 2 )   ˜ sin π γ1 Q∗  q K I I I = − ∗0 + ˜ cos π γ1 2λ π l0 (1 + q 2 ) Q ∗0

To determine the last unknown function u (x) we find the following expression for derivative of this function from 3.69 u  (x) = −

1 2ϑ1(2)

σ (x) +

2ϑ1(2) π

Q0 √ a2 − x 2

(3.80)

Integrating 3.69 with respect to x from −a to x and taking into account 3.68, as well as using the value of integral 3.24 the following is obtained u(x) =

1 2ϑ1(2)



  Q0 x − K I arcsin + K I I [ω1 (x) − ω1 (−x) π a

(3.81)

Thus, all unknown functions are determined. Using these functions and the aboveobtained formulas we can determine the stress–strain state of compound plane completely. ˜ < 0, i.e. numbers q j ( j = 1, 2) are pure imagiNow let us discuss the case  nary. In this case the solution of Eq. 3.72 will also be given by formula 3.74. However, 1 − iβ1 , 2  q = −αβ, γ1 =

γ2 = β1 =

1 + iβ1 , 2

1 ln |G 1 | , 2π

q j = (−1) j iq G1 =

1−q a (3.86a) (1) ⎪ u (x, 0) = u (2) (x, 0) ⎪ ⎪ ⎪ ⎩ (1) v (x, 0) = v (2) (x, 0)

Fig. 3.6 Compound plane with interfacial crack with rigid stamp

158

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

⎧ (1) σ (x, 0) = τx(1) ⎪ y (x, 0) = 0 ⎨ y (2) τx y (x, 0) = ρσ y(2) (x, 0) ⎪ ⎩ (2) v (x, 0) = δ

|x| < a

(3.86b)

where δ is the rigid displacement of the stamp along the direction O y -axis, and ( j) τx y (x, y) ( j = 1, 2) are components of the stress tensor acting in the upper and lower half-planes, respectively, u ( j) (x, y) and v ( j) (x, y) ( j = 1, 2) are horizontal and normal components of the displacement vector in the corresponding half-planes. To construct an exact solution of the boundary value problem 3.86, we use discontinuous solutions of the elasticity theory equations for a piecewise homogeneous plane 3.4. σ (x) = σ y(1) (x, 0) − σ y(2) (x, 0) = −σ y(2) (x, 0) ; (2) (2) (2) τ (x) = τx(1) y (x, 0) − τx y (x, 0) = −τx y (x, 0) = −ρσ y (x, 0) = ρσ (x) ;

u

(1)

(x, 0) − u

(2)

(x, 0) =

(2) u (x) /θ2 ;

v

(1)

(x, 0) − v

(2)

(x, 0) =

(3.87)

(2) v (x) /θ2 ;

We use representations 3.4, which satisfies the conditions 3.86b on the crack edges, previously differentiated the last of them. As a result, to determine the unknown contact pressure σ (x), shear stress τ (x) and dislocation displacement functions u  (x) and v  (x), the following governing system of singular integral equations is obtained: l0 l1 l2 σ (x) + u  (x) +   π l0 l1 l2 τ (x) − v  (x) −   π l0

a −a a

−a

l3 τ (s) ds − s−x π l3 σ (s) ds − s−x π

d1 d0 v (x) + τ (x) + (2)  π ϑ2 

a −a

a −a a

−a

v  (s) ds = 0; s−x u  (s) ds = 0; s−x

l2 σ (s) ds + s−x π ϑ2(2)

a −a

(3.88)

u  (s) ds = 0; s−x

τ (x) = ρσ (x) . To construct an unambiguous solution for system 3.88, there is need to use the conditions for the equilibrium of the stamp and the continuity of displacements at the end points of the crack: a

a σ (x)dx = P0 ;

−a



a

v (x)dx = 0; −a

u  (x)dx = 0

(3.89)

−a

Thus, the solution of the problem is reduced to the solution of the system of singular integral Eq. 3.88 under conditions 3.89.

3.4 Contact Problem for a Piecewise Homogeneous Plane …

159

To construct a solution of system 3.88 under conditions 3.89, we substitute the value τ (x) from the last Eq. 3.88 into the first three equations and bring them to the canonical form. a1 σ (x) + π v  (x) +

u  (x) +

b1 π c1 π

a −a a

−a a

−a

σ (s) a2 ds + s−x π b2 σ (s) ds + s−x π c2 σ (s) ds + s−x π

a −a a

−a a

−a

u  (s) ds = 0; s−x u  (s) ds = 0; s−x c3 u  (s) ds + s−x π

(3.90) a

−a

v  (s) ds = 0. s−x

here 2



2

2  ϑ2(1) − ϑ1(1)

ϑ1(1) ϑ2(2) ϑ (1) ϑ (2) ϑ (1) ϑ (2) ; a2 = ; b1 = 2 2 ; b2 = 2 1 ; ρϑ ρϑ 2ϑ ϑ



2 2 ρl2 − a1 l0 a2 l 0 l3 c1 = ; c2 = − ; c3 = − ; ϑ = ϑ2(1) − ϑ1(1) + ϑ1(1) ϑ1(2) . l1 l1 l1

a1 = −

Let us reduce the solution of system 3.90 to the solution of three independent singular integral equations. For this purpose, we multiply the second and third of Eq. 3.90, respectively, by the unknown constants n = 0 and m = 0 and sum up with the first Eq. 3.90. σ (x) + nv  (x) + mu  (x) a σ (s) + a1 + nb1 + mc1 + π

mc3 v  (s) a1 +nb1 +mc1

−a

+

(a2 +nb2 +mc2 )u  (s) a1 +nb1 +mc1

s−x

ds = 0

(3.91)

Further, the following equalities must hold: a2 + nb2 + mc2 mc3 = n; = m. a1 + nb1 + mc1 a1 + nb1 + mc1

(3.92)

Hence, introducing new variables by formulas m/n = x, n = y and excluding y from 3.92, to determine x the following third-order equation is obtained: c32 x 3 − c3 (a1 + c2 ) x 2 − (c3 b2 − a1 c2 + a2 c1 ) x + (a1 b2 − a2 b1 ) = 0. where n=y=

c3 x − a1 c3 x − a1 ; m = nx = x . c1 x + b1 c1 x + b1

(3.93)

(3.94)

160

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

It is not hard to check that a1 b2 − a2 b1 = −

ϑ2(1) ϑ2(2) = 0. ρϑ

Hence 3.93 has no trivial roots. Consider the case when 3.93 has three different roots. In this case, introducing the functions ϕ j (x) = σ (x) + n j v  (x) + m j u  (x) ( j = 1, 2, 3) ,

(3.95)

where n j and m j ( j = 1, 2, 3) for each solution x j of Eq. 3.93 are determined by formulas 3.94, the solution of system 3.90 is reduced to the solution of the following three independent singular integral equations: qj ϕ j (x) + π

a −a

ϕ j (s) ds = 0 s−x

(3.96)

  −a < x < a; q j = c3 x j ; j = 1, 2, 3 The equations should be considered under condition a ϕ j (x) dx = P0

( j = 1, 2, 3) .

(3.97)

−a

The solutions of 3.96 have the following form [12] ϕ j (x) = C j X +j (x); X j (z) = (z + a)−γ j (z − a)γ j −1 ; 0 < ϑ j = arg G j < 2π ;

Gj =

γj = 1 − iq j , 1 + iq j

(3.98)

ϑj 1 ln G j + = α j − iβ j ; 2πi 2π

where function   X +j (x) = − G j (a + x)−γ j (a − x)γ j −1 = − G j ω j (x) is the value of function X j (z) on the upper edge of interval (−a, a). Using 3.98, considering conditions 3.97 and the value of the integral 3.21 we get sin π γ j C j = −P0  π Gj

( j = 1, 2, 3) .

3.4 Contact Problem for a Piecewise Homogeneous Plane …

Therefore ϕ j (x) =

P0 sin π γ j . π (a + x)γ j (a − x)1−γ j

161

(3.99)

To determine the normal contact pressure and dislocation displacement components, substituting the value of functions ϕ j (x) from 3.98 into relation 3.94, a system of algebraic equations is obtained. After solving of system, we get σ (x) =

3 " j=1

u  (x) =

" P0 A j sin π γ j P0 B j sin π γ j ; v  (x) = ; γj 1−γ j π (x + a) (a − x) π + a)γ j (a − x)1−γ j (x j=1

3 " j=1

3

P0 C j sin π γ j . π (x + a)γ j (a − x)1−γ j

Here m2 − m3 (m 2 n 3 − m 3 n 2 ) (m 3 n 1 − m 1 n 3 ) (m 2 n 1 − m 1 n 2 ) ; A2 = − ; A3 = ; B1 = ;     m1 − m3 m1 − m3 n3 − n2 n1 − n3 n2 − n1 ; B3 = ; C1 = ; C2 = ; C1 = ; B2 =       = (m 2 n 3 − m 3 n 2 ) − (m 3 n 1 − m 1 n 3 ) + (m 2 n 1 − m 1 n 2 ) . A1 =

It is clear that in the case when 3.93 has three different real roots, then both G j =1 ( j = 1, 2, 3) the contact pressure σ (x) and the components of the dislocation of the points of the crack edges v  (x) and u  (x) at the end points of the crack will have an exponential singularity without oscillations with three different exponents γ j = ϑ j /2π . In the case when 3.93 has one real and two complex conjugate roots, these functions at the end points of the crack will have one exponential singularity without oscillation and two with oscillation. Let us also determine the contact stresses outside the crack. For this purpose, we use the first two Eq. 3.87. Substituting into these equations the values of the contact pressure and dislocation components of the points of the crack edges, and using the values of the integral 3.25 after some calculations we get:

(1)

(2)

σ y (x, 0) = σ y (x, 0) = −

 3  P0 " ρl2 A j − l3 B j sgnx π |a + x|γ j |a − x|1−γ j j=1

 3  P0 " l2 A j + l3 C j sgnx (1) (2) τx y (x, 0) = τx y (x, 0) = π |a + x|γ j |a − x|1−γ j j=1

(|x| > a, j = 1, 2) ;

(|x| > a, j = 1, 2) .

Hence, it can be seen that the contact stresses outside the crack at the junction line of the heterogeneous half-planes at the end points of the crack have the same

162

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

singularities as the contact pressure. Without referring to the case when Eq. 3.93 has multiple roots, we only note that in this case the number of Eq. 3.96 decreases. This does not allow us to determine the unknown functions immediately. However, it is possible to determine one of the unknown functions through the other two, if 3.93 has a threefold root, or two of the unknown functions through one of them, if 3.93 has one simple and one double root. Then, using, respectively, two or one of Eq. 3.90, the solution of the problem is reduced to the solution of a system of two heterogeneous singular integral equations, in the first case, and one heterogeneous singular integral equation in the second case. Numerical calculations show that when the value of the friction coefficient ρ is less than 0.6, Eq. 3.93 has three real roots for all values of the elastic constants of heterogeneous half-planes. Thus, using the method of singular integral equations, an exact solution for the problem of contact interaction of a piecewise homogeneous elastic plane with a finite interfacial crack and an absolutely rigid stamp is constructed taking into account dry friction. Simple formulas are obtained both for determining the contact stresses under the stamp and outside the crack, and for the components of the dislocation of the points of the crack edges. It is shown that in the case when the characteristic cubic equation has three real roots, both the contact pressure and the normal and tangential contact stresses outside the crack at the end points of the crack have an exponential singularity without oscillation with three different exponents. The case when the characteristic cubic equation has one real and two complex conjugate roots is also studied. It is shown that, in this case, the contact stresses have both the exponential singularity without oscillation and an exponential singularity with oscillation.

3.5 Stress State of a Compound Plane with Interface Absolutely Rigid Inclusion and Crack Having Common Tip This section discusses the plane stress state of homogeneous plane with two heterogeneous half-planes, containing a finite crack and an absolute rigid inclusion at one of the end points of the crack. Let the composite elastic plane consisting of two homogeneous half-planes with shear moduli μ j and Poisson’s ratios ν j ( j = 1, 2) on the segments (−b, 0) and (0, a) of the joining line of the half-planes coinciding with the O x-axis of the rectangular Cartesian coordinate system O x y contain, respectively, a crack and a thin absolutely rigid inclusion. It’s assumed that the compound plane is deformed by the distributed normal q j (x) and tangential τ j (x) ( j = 1, 2) loads applied to the crack faces, and also the concentrated force P0 applied to the inclusion at a point x = x0 at an angle α relative to the axis (Fig. 3.7). For clarity, tangential loads τ j (x) are not shown in the figure. It is required to determine the contact stresses acting on the long sides of the inclusion, the crack opening and the Cherepanov–Rice J -integral at the end point of the crack x = −b, and the distance x0 of the point of application of the force P0

3.5 Stress State of a Compound Plane with Interface Absolutely Rigid

163

Fig. 3.7 Schematic representation of the problem

from the origin, at which the resultant moment of the forces acting on the inclusion turns to zero, such that, the possibility of rotating the entire system as a whole is excluded [9]. The stated problem is mathematically formulated in the form of the following boundary value problem: the Lamé equations for each half-plane and the boundary conditions on the joining line of the half-planes ⎧ ( j) σ y (x, 0) = −q j (x) (−b < x < 0), ⎪ ⎪ ⎪ ⎪ ⎨ τ ( j) (x, 0) = τ (x) (−b < x < 0), j xy ⎪ V1 (x, 0) = V2 (x, 0) = δ1 ⎪ (0 < x < a) , ⎪ ⎪ ⎩ U1 (x, 0) = U2 (x, 0) = δ2 (0 < x < a) , ⎧ (1) σ y (x, 0) = σ y(2) (x, 0) ⎪ ⎪ ⎪ ⎪ ⎨ τ (1) (x, 0) = τ (2) (x, 0) xy xy ⎪ ⎪ U1 (x, 0) = U2 (x, 0) ⎪ ⎪ ⎩ V1 (x, 0) = V2 (x, 0) ( j)

( j)

(3.100)

(−∞ < x < −b; x > a).

Here σ y (x, y) and τx y (x, y) ( j = 1, 2) are the normal and tangential components of the stresses, U j (x, y) and V j (x, y) are the horizontal and vertical displacements of the points of the corresponding half-plane, and δ1 , δ2 are the unknown constants expressing a rigid displacement not only of the inclusion, but also of the system as a whole. In order to write down the governing system of equations for the stated problem, we use discontinuous solutions of the equations of plane elasticity theory for a compound plane with interface defects 3.4. In this case, we write the equations in more convenient form:

164

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

d∗0 μl ∗ d∗1 U 2 (x, 0) = σ (x) − 0 U  (x) − ∗ μ2 ∗ π ∗ μ2

a



−b

d∗ μl ∗ d∗1 V 2 (x, 0) = − 0 τ (x) − 0 V  (x) − ∗ μ2 ∗ π ∗ μ2

a



σ y(1) (x, 0) =

τx(1) y (x, 0) =

μl0∗ ∗

σ (x) +

μ l1∗ ∗

μ2 U  (x) +

π ∗

μl0∗ μl ∗ μl2∗ τ (x) − 1 μ2 V  (x) − ∗ ∗ π ∗

−b

a

μl2∗

−b

a −b

τ (ξ )dξ μl2∗ + ξ −x π ∗

a −b

μl2∗ σ (ξ )dξ − ξ −x π ∗

τ (s)ds − μ2 s−x π ∗

σ (s)ds μl ∗ − μ2 3 s−x π ∗

a −b

a

μl3∗

V  (ξ )dξ , ξ −x

−b

a −b

U  (ξ )dξ , ξ −x V  (s)ds , s−x

U  (s)ds , s−x (3.101)

Here we use the difference functions  U1 (x, 0) − U2 (x, 0) (−b ≤ x ≤ 0) , U (x) = 0 (0 ≤ x ≤ a) ,  V (x) =  σ (x) =

V1 (x, 0) − V2 (x, 0) (−b ≤ x ≤ 0) , 0 (0 ≤ x ≤ a) ,

σ y(1) (x, 0) − σ y(2) (x, 0) ( 0 ≤ x ≤ a) ,

(3.102)

σ0 (x) = −q1 (x) + q2 (x) ( −b ≤ x ≤ 0) , 

τ (x) =

(2) τx(1) y (x, 0) − τx y (x, 0)

( 0 ≤ x ≤ a) ,

τ0 (x) = τ1 (x) − τ2 (x) (−b ≤ x ≤ 0) ,

which are finite functions that differ from zero only on the interval (−b, a) occupied by the crack and inclusion, and the following notation: l0∗ = ϑ2(1) (μϑ2(1) + ϑ2(2) ) − ϑ1(1) (μϑ1(1) − ϑ1(2) ),

l3∗ =2(ϑ1(2)l2 ∗ − ϑ2(2)l0 ∗ ),

l2∗ = ϑ1(1) ϑ2(2) + ϑ2(1) ϑ1(2) ,

d0∗ = μϑ1(1) − ϑ1(2) /2, 2

l1∗ = 2(ϑ1(2) l0 ∗ − ϑ2(2) l2 ∗ ),

d1∗ = μϑ2(1) + ϑ2(2) /2, 2

∗ = (μϑ2(1) + ϑ2(2) ) − (μϑ1(1) − ϑ1(2) ) ,   2 1 − νj 1 − 2ν j ( j) ( j) , ϑ2 = , ϑ1 = ae j ae j

μ=

μ1 . μ2

As the unknown functions, we choose the complex combinations of jumps of stresses and displacements at the corresponding intervals:

3.5 Stress State of a Compound Plane with Interface Absolutely Rigid

165

    (2) χ (x) = σ y(1) (x, 0) − σ y(2) (x, 0) − i τx(1) (0 < x < a) , y (x, 0) − τx y (x, 0) W (x) = [U1 (x, 0) − U2 (x, 0)] + i [V1 (x, 0) − V2 (x, 0)] (−b < x < 0) . (3.103) Taking into account that in the representations 3.102 the difference of boundary conditions 3.100 has already been used, we write down the sum of these conditions in the form of complex combinations: ⎧    ⎨ σ (1) (x, 0) + σ (2) (x, 0) − i τ (1) (x, 0) + τ (2) (x, 0) = P+ (x) − i T+ (x) y y xy xy ⎩ U  (x, 0) + U  (x, 0) + i V  (x, 0) + V  (x, 0) = 0 1 2 1 2

(−b < x < 0), (0 < x < a) ,

(3.104) P+ (x) = −q1 (x) − q2 (x) ,

T+ (x) = τ1 (x) + τ2 (x) .

Substituting in 3.104 the representation 3.101 and using 3.102, to determine the unknown functions 3.103 we obtain the following system of governing equations: ⎧ 0  a ⎪ ⎪ il3∗ il2∗ W X (ξ ) (ξ ) ⎪ ∗  ⎪ l1 W (x) + dξ + dξ = ∗1 (x) (−b < x < 0) , ⎪ ⎪ ⎪ π ξ − x π μ ξ − x 2 ⎨ −b

0

⎪ a 0  ⎪ ⎪ iμ2 l2∗ id1∗ X (ξ ) W (ξ ) ⎪ ∗ ⎪ dξ − dξ = ∗2 (x) (0 < x < a) . ⎪ ⎪ d0 X (x) − π ⎩ ξ −x π ξ −x −b

0

(3.105) This system must be considered together with the conditions of equilibrium of inclusion and continuity of displacements at the end points of the crack: a

0 X (ξ ) dξ = P0 (sin α − i cos α) ,

W  (ξ ) dξ = 0.

(3.106)

−b

0

Here ∗1 = −

l0∗ μ2

(σ0 (x) − iτ0 (x)) −

∗2

l2∗ π μ2

=

0 −b

d∗1 π

0 −b

τ0 (ξ ) + iσ0 (ξ ) ∗ dξ + (P+ (x) − i T+ (x)) , ξ −x 2μμ2 τ0 (ξ ) + iσ0 (ξ ) dξ . ξ −x

166

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

Algorithm for Solving the Governing System of Equations We switch to the dimensionless quantities, by dividing the stress components by the shear moduli μ2 , and the linear values by the length of the inclusion a, and bringing the integration intervals to the interval (−1, 1):   a

b W  (x) = W  − (1 − t) = ϕ (t) , X (t) = X (1 + t) = μ2 ψ (t) , 2 2   a

b ∗1 (x) = ∗1 − (1 − t) = g1 (t) , ∗2 (t) = ∗2 (t) (1 + t) = μ2 g2 (t) , 2 2

   2x 2x ∗ 1+ , τ0 (x) = μ2 τ0 1 + , σ0 (x) = b b     2x 2x ∗ ∗ , T+ (x) = μ2 T+ 1 + , P+ (x) = μ2 P+ 1 + b b μ2 σ0∗



  l∗ g1 (t) = −l0∗ σ0∗ (t) − iτ0∗ (t) − 2 π

d∗ g2 (t) = 1 π

1 −1

1 ∗  τ0 (ξ ) + iσ0∗ (ξ ) ∗  ∗ dξ + P+ (t) − i T+∗ (t) , ξ −t 2μ

−1

τ0∗ (ξ ) + iσ0∗ (ξ ) dξ , ξ −t

P0 = aμ2 P0∗ .

The system of governing Eq. 3.105 and condition 3.106 takes the form: ⎧ ⎪ ⎪ ∗ ⎪ ⎨ l1 ϕ (t) +

il3∗ π

⎪ ⎪ ∗ ⎪ ⎩d0 ψ (t) −

id1 π

)1

−1 ∗ )1 −1

ϕ(s) ds s−t

+

ψ(s) ds s−t

il2∗ π

−

)1 −1 ∗ )1

ii 2 π

−1

ψ(s) ds s+1+ ab (1−t)

= g1 (t) ,

ϕ(s) ds s−1− ab (1+t)

(3.107) = g2 (t) ,

1 ϕ (s) ds = 0, −1

1

(3.108) ψ (s) ds = 2P0∗ (sin α − i cos α) = −2i P0∗ eiα .

−1

The solution of the last system will be constructed using the method of mechanical quadratures (MMQ) [16]. The required functions will be sought in the following form:

3.5 Stress State of a Compound Plane with Interface Absolutely Rigid

ϕ (t) = ϕ∗ (t) (1 − t)α (1 + t)β , ψ (t) = ψ∗ (t) (1 − t)γ (1 + t)δ ,

    −1 < Re α, β, γ , δ < 1 .

167

(3.109)

According to the standard MMQ procedure, it is required to study the behavior of the equations of the system in the vicinity of the ends of the integration interval using the formulas 1.118 in order to determine the exponents in the representations 3.109, the following system of trigonometric equations are obtained: l1∗ − il3∗ ctg πβ = 0, tg2 π α + i

d0∗ − id1∗ ctg π γ = 0, α = δ,

4 (1 − ν1 ) (1 − ν2 ) 2 (ν1 − ν2 ) tg π α + = 0. (1 − 2ν1 ) (1 − 2ν2 ) (1 − 2ν1 ) (1 − 2ν2 )

(3.110)

The exponents β and γ are uniquely determined, but for α we have two different values:   1 1 ae1 1 + μ ae2 1 1 ln ln , θ= , β = − − iθ, γ = − + i θ + 2 2 2π ae2 2π μ + ae1 1 i α1 = − − ln ae1 , 2 2π

1 i α2 = − + ln ae2 . 2 2π

(3.111)

Therefore, we seek the solution of the defining system 3.107 in the following form: 1 − 2ν1  (t) (1 − t)α1 (1 + t)β +  (t) (1 − t)α2 (1 + t)β 2μ cos (π α1 ) (3.112) 2μ cos (π α2 )1 ψ (t) =  (t) (1 − t)γ (1 + t)α1 +  (t) (1 − t)γ (1 + t)α2 1 − 2ν2 ϕ (t) =

and the new unknown functions will be  (t) and  (t) which satisfy the Hölder continuity condition on the interval [−1, 1]. Continuing the MMQ procedure, we can represent the new unknown functions in the form of interpolation polynomials in the roots of the Jacobi polynomial corresponding to the weight function

 (t) =

 (t) =

⎧ n * 2 ⎪ ⎪ ⎨ n+α1 +β+1

in expression of ϕ (t) ,

⎪ ⎪ ⎩

in expression of ψ (t),

(α ,β ) Pn 1 (t) X (ϕ)

j α +1,β+1 α ,β

α ,β ) ( 1 ) ( 1 ) ( 1 Pn−1 ξj j=1 t−ξ j (ψ) (γ ,α1 ) n * X j Pn (t) 2



(γ ,α1 ) (γ +1,α1 +1) (γ ,α1 ) n+γ +α1 +1 t−ξ P ξj n−1 j j=1

⎧ n * 2 ⎪ ⎪ ⎨ n+α +β+1

in expression of ϕ (t) ,

⎪ ⎪ ⎩

in expression of ψ (t) .

(α ,β ) Y j(ϕ) Pn 2 (t)

α +1,β+1 α ,β

α ,β ( 2 ) ( 2 ) ( 2 ) 2 Pn−1 ξj j=1 t−ξ j γ ,α2 ) ( (ψ) n * Y j Pn (t) 2



(γ ,α2 ) (γ +1,α2 +1) (γ ,α2 ) n+α2 +γ +1 Pn−1 ξj j=1 t−ξ j

(3.113)

168

3 Plane Strain State of Piecewise Homogeneous Elastic Plane … (α,β)

(α,β)

Here ξ j

are the roots of the Jacobi polynomial Pn (t), the coefficients  (ψ) (ψ)  (ϕ) (ϕ) X j , X j , Yj , Yj j = 1, n are to be determined. We substitute 3.113 into the equations of system 3.107 and use the standard MMQ procedure to replace the integrals by finite sums with respect to each term in 3.112. Taking into account the fact that the real part of the singularity exponent at both ends of the integration interval is equal to −0.5, as the collocation points for each equation of the system, it is necessary to choose a system of n − 1 points. In the numerical analysis, the roots of the Chebyshev polynomial Un−1 (t) were chosen. Adding two conditions 3.108 to 2 (n − 1) equations and equating of different representations of the same functions 3.113 in the roots of the Chebyshev polynomial Tn (t), we obtain a closed system of linear algebraic equations with respect to 4n unknown coefficients. After determining the unknown coefficients, the derivatives of the jump of the crack face displacements and the jump of the stresses on the edges of the inclusion can be represented as interpolation polynomials, and using the corresponding quadrature formula [16], one can write out a representation for the jump of the displacements and construct graphs for crack opening. It is also possible to determine other values related to the stress state in the vicinity of stress concentrators. For example, from the condition that the resultant moment of the loads acting on the inclusion be zero, we find the point of application of the concentrated force x0∗ = x0 /a , when the rotation of the inclusion is excluded x0∗ =

M0 1 − . ∗ 2 4P0 sin π α

(3.114)

Here M0 is the resultant moment of contact stresses in the case when the force is applied at the center of inclusion, defined by the formula 1 M0 = −1

 n  " 2 cos π α2 (γ ,α2 ) (γ ,α2 ) (ψ) (γ ,α1 ) (γ ,α1 ) (ψ) wi s Re [ψ (s)] ds = Re ξi Xi + w ξi Yi , 1 − 2ν2 i i=1

(3.115)

(α,β)

wj

⎡ ⎤2 1 B (α, β + n + 1) ⎣ 2α+β+3

⎦ . =

(α+1,β+1) (α,β) B (α, n + 1) (α,β) 2 ξj (α + β + n + 1) Pn−1 1 − ξj

In the crack tip disconnected with inclusion, we define such a characteristic for cracks, as the dimensionless Cherepanov–Rice J-integral [10] J∗ =

1 μ+1 K (−1) K¯ (−1) , J (−1) = μ2 4μ

(3.116)

3.5 Stress State of a Compound Plane with Interface Absolutely Rigid

169

Fig. 3.8 Crack opening w (t) for ν1 = 0.25, ν2 = 0.3

where K (−1) is the dimensionless complex coefficient of stress concentration at the tip of the crack, defined by the formula √ 

4 2π 1 − ν1 + μ (1 − ν2 ) 1 − 2ν1 α1 α2 , K (−1) = i  (−1) 2 + μ (−1) 2 sin πβ (μ + κ1 ) (1 + μκ2 ) 2 cos π α1 (α ,β)

 (−1) =

n 1 " X (ϕ) (−1) 2 j Pn



, ,β) +1,β+1) (α (α (α ,β) n + α1 + β + 1 j=1 −1 − ξ 1 P 1 ξ 1 j

n−1

j

(α ,β)

 (−1) =

n " Y j(ϕ) Pn 2 (−1) 2



. n + α2 + β + 1 j=1 −1 − ξ (α2 ,β) P (α2 +1,β+1) ξ (α2 ,β) j

n−1

j

For definiteness, suppose that the concentrated load P0∗ = 1 acts on the inclusion at an angle α = π/3 , and normal uniformly distributed load q0 /μ2 = 0.01 acts on the crack faces, hence, σ0∗ (t) = τ0∗ (t) = 0; P+∗ = 0.02. The computational process analysis shows that the accuracy of calculations is sufficient for constructing the corresponding graphs and analyzing the dependence of the jump of the contact stresses )t ψ (t), the opening of the crack w (t) = −1 ϕ (s) ds, the J -integral and the point of application of the concentrated force, when the rotation of the inclusion is excluded, depended on the mechanical characteristics of the half-planes and the ratio of the lengths of the stress concentrators. Figures 3.8, 3.9 and 3.10 show graphs of crack opening, jumps of normal and tangential contact stresses for ν1 = 0.25, ν2 = 0.3 and different values of shear moduli ratio μ = 0.5, 1, 2. Similar graphs are constructed for different values of Poisson’s ratio of both the upper and the lower half-planes for ν1 , ν2 = 0.15, 0.3, 0.45 at a constant value of the ratio μ = 2. These graphs are not shown, because they qualitatively repeat the curves given in the figures. It should be noted that the order of the curves is similar: larger values μ correspond to a larger value ν.

170

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

Fig. 3.9 Distribution of the normal contact pressure

Fig. 3.10 Distribution of tangential contact stresses

Fig. 3.11 Dependence J -integral of the ratio μ for different ν1 and ν2

Figures 3.11 and 3.12 show curves of the dependence of J∗ and x0∗ on the ratio μ for different values of Poisson’s ratios. The graphs in Fig. 3.11 indicate that for any values of Poisson’s coefficients ν1 and ν2 J -integral takes on a maximum value for values μ close to 0.5. This means that for these values μ the probability of crack propagation is much higher. An interesting result can be seen by comparing the graphs in Fig. 3.13. Comparing the elastic characteristics for the curves 1 and 5, which are very close to each other, with the characteristics for the curve 2, we note that in both cases one of the parameters decreases by two times. In the first case, this is Poisson’s ratio ν2 , and in the second case it is the ratio of the shear moduli μ. The regulation is maintained

3.6 The Stress State of Compound Plane, Weakened by Arc-Shaped Cracks …

171

Fig. 3.12 Dependence of the point of application of force x0∗ on the ratio μ for different ν1 and ν2

Fig. 3.13 Dependence J -integral of the ratio of the lengths of the crack and the inclusion b/a

even for increasing of the last parameters, which corresponds to the curves 3 and 4. The difference between them is slightly larger, since the ratio of the shear moduli μ is doubled, but Poisson’s ratio ν2 is increased by one and a half time.

3.6 The Stress State of Compound Plane, Weakened by Arc-Shaped Cracks with Stresses Given on One Bank and Displacements on the Other In this subsection we argue the problem on stress state of compound elastic plane, formed by junction of elastic circular disk with radius R and Lamé coefficients μ1 , λ1 and elastic plane with circular hole of the same radius and Lamé coefficients μ2 , λ2 . There are disjoint arcs (αk , βk ) (k = 1, 2, ...N ) along line L. The elastic plane is weakened by arc-shaped cracks with given complex combinations of stress components χk(1) (θ ) on one bank and complex combinations of displacements Wk(2) (θ ), as well as the resultant of contact stresses Pk (k = 1, 2, ..., N ) on the other bank (Fig. 3.14). Further assume that self-balancing system of forces is acting on each crack, i.e.

172

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

βk

χ (1) k (θ ) dθ = Pk (k = 1, 2, ...N )

αk

We state the problem to determine the contact stresses acting on banks of cracks, where the displacements are given, as well as on junction line of circular disk and plane with hole [6]. The described problem can be mathematically stated as the following boundary value problem: ⎧ (1) X n + iYn(1) |r =R = X n(2) + iYn(2) |r =R ⎪ ⎪ ⎪ ⎨ u 1 + iv1 |r =R = u 2 + iv2 |r =R for θ ∈ /L (1) (1) ⎪ X + iYn |r =R = χ1 (θ ) ⎪ ⎪ ⎩ n u 2 + iv2 |r =R = W0 (θ ) for θ ∈ L

W0 (θ ) = Wk(2) (θ ) , χ1 (θ ) = χk(1) (θ ) ,

αk < θ < βk ,

(3.117)

k = 1, 2, ..., N



Here u j + iv j ( j = 1, 2) are complex combination of displacements of circular disk and plane with hole in Cartesian coordinate system. These combinations satisfy the Lamé equations in corresponding domains and connect with stress components ( j) ( j) X n + iYn ( j = 1, 2) by well-known formulas [12]. To construct the solution of mixed boundary value problem 3.117 we take into account the following formulas [12] for elastic displacements and stresses   2μ j u j + iv j = aej ϕ j (z) − zϕ  j (z) − ψ j (z) ( j)

( j)

f1 + i f2

Fig. 3.14 Elastic plane is weakened by arc-shaped cracks

= ϕ j (z) + zϕ  j (z) + ψ j (z)

( j = 1, 2)

3.6 The Stress State of Compound Plane, Weakened by Arc-Shaped Cracks …

173

here ϕ j (z) and ψ j (z) are Kolosov–Muskhelishvili potentials for disk and plane with hole, ae j = 3 − 4ν j ( ν j is Poisson’s ratio) are Muskhelishvili constants and ( j) ( j) f 1 + i f 2 are complex integrals of stresses, i.e. ( j) f1

+

( j) i f2

θ = iR



 X n( j) + iYn( j) dθ

−π

To solve the boundary value problem 3.117 we continue the first two conditions and introduce the jump functions of stresses, acting on banks of cracks, and difference of displacements of banks of cracks by formulas:

χ˜ (θ ) =







f 1(1) + i f 2(1) − f 1(2) + i f 2(2)

=Reiθ

=

⎧ 0, ⎪ ⎪ ⎪ ⎨

θ∈ /L

if

⎪ iR ⎪ ⎪ ⎩

θ χ (θ ) dθ,

if

−π

 W˜ (θ ) = (u 1 + iv1 )

z=Reiθ

− (u 2 + iv2 )

z=Reiθ

=

θ∈L (3.118)

0, if θ ∈ /L W (θ ) , if θ ∈ L

(3.119)

(χ (θ ) = χ1 (θ ) − χ2 (θ )) Here χ2 (θ ) are an unknown contact stresses, acting under inclusion. The complex functions ϕ j (z) , ψ j (z) , W (θ ) , χ (θ ) are represented as the following series: ϕ1 (z) = ϕ2 (z) =

∞ "

an z n ,

ψ1 (z) =

n=1

n=1

∞ "

∞ "

bn z −n ,

ψ2 (z) =

n=1

θ iR −π

∞ "

χ (θ )dθ =

an zn b n z −n

n=1 ∞ " n=−∞

An einθ ,

W (θ ) =

∞ "

Bn einθ

n=−∞

Substituting these representations in 3.118 and 3.119 and taking into account that z = Reiθ on junction circle, after equaling the coefficients by same powers einθ (n = 0, ±1, ±2...) we get the following system of equations for determination the coefficients an , a  n , bn , b n :

174

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

⎧ b 1 ⎪ ⎪ ⎪ a1 R + a¯ 1 R − = A1 ⎪ ⎪ R ⎪ ⎪ ⎨  1  2 2R a¯ 2 + a  0 = B0 − 2μ1 ⎪ ⎪ ⎪ ⎪ ⎪ ae b 1 a¯ 1 R ⎪ 1 ⎪ ⎩ = B1 a1 R − + 2μ1 2μ1 2μ2 R

(3.120)

⎧ bn ⎪ n+2 ⎪ + a  n R n − n = A−n ⎨ (n + 2) an+2 R R   ae2 bn 1 ⎪ n+2 ⎪− + an Rn − · = B−n (n + 2) an+2 R ⎩ 2μ1 2μ2 R n

(n ≥ 1) (3.121)

⎧ b n (n − 2) bn−2 ⎪ n ⎪ a R + − = An ⎪ n ⎨ R n−2  Rn  ⎪ ae1 b n 1 (n − 2) bn−2 n ⎪ ⎪ an R − − n = Bn ⎩ 2μ1 2μ2 R n−2 R

(n ≥ 2)

(3.122)

From the obtained system 3.120 the unknown coefficients in expansion of complex potentials are expressed by Fourier coefficients An , Bn (n ∈ Z) of functions χ (θ ) and W (θ ) by following formulas: an =

Rn

1 (μAn + 2μ1 Bn ) (μ + ae1 )

bn = −

Rn (A−n + 2μ1 B−n ) (n ≥ 1) 1 + ae2 μ

a  n = − (n + 2) an+2 R 2 +  b n = R n

(n ≥ 2)

Rn

  1 μae2 A−n − 2μ1 B−n (n ≥ 1) (1 + ae2 μ)

(n − 2) bn−2 − An R n an + R n−2

 (n ≥ 2) , a  0 = −2μ1 B0 − 2R 2 a2 ,

    2μ1 B1 (μ + ae1 ) − B1 (μ − 1) μ A1 (μ + ae1 ) − A1 (μ − 1) a1 = + R (ae1 + 1) (ae1 + 2μ − 1) R (ae1 + 1) (ae1 + 2μ − 1)       μ A1 + A1 + 2μ1 B1 + B1 − A1 , b1= R ae1 + 2μ − 1 

here

  μ1 , μ= μ2

3.6 The Stress State of Compound Plane, Weakened by Arc-Shaped Cracks …

iR An = 2π

1 Bn = 2π

π

⎡ ⎣

−π

π W (θ )e −π

θ

⎤ χ (θ ) dθ ⎦e

−inθ

−π

−inθ

1 dθ = − 2πin

R dθ = 2π n

π W (θ )de −π

−inθ

π

175

χ (θ ) e−inθ dθ

−π

1 = 2πin

π

W  (θ )e−inθ dθ

−π

Hence all unknown coefficients in expansion of complex potentials are expressed by Fourier coefficients of functions χ (θ ) and W (θ ). Now using the obtained expressions let us calculate the components of stresses, acting on the junction circle of disk and displacements of points of junction circle with hole. Thus we have   l0 l1 i R X n(1) + iYn(1) |r =R = − χ∗ (θ ) − W  (θ )   π  il3 ϕ−θ ϕ−θ il2 dϕ − dϕ + iRe2 · eiθ χ∗ (ϕ) ctg W  (ϕ)ctg − 2π  2 2π  2 −π

L

d d0 l0 (u 2 + iv2 ) |r =R = − χ∗ (θ ) + W  (θ ) dθ   π  il2 id1 ϕ−θ ϕ−θ dϕ + dϕ + e3 · eiθ + χ∗ (ϕ)ctg W  (ϕ)ctg 2π  2 2π  2 −π

L

Note that we use the relation for derivation of the above written formulas ∞

θ 1 " ±inθ i + e = π δ (θ ) ± ctg 2 n=1 2 2

(−π < θ < π)

and introduce the following notation: e2 = a1 + a1 − e3 = −

iae1 2μ2 (μ + ae1 )

 A1 −

μA1 + 2μ1 B1 R (μ + ae1 )

 b 1 2μ1 , B1 − ae1 2μ2 R

χ∗ (θ ) = i Rχ (θ )

Now let us satisfy the last two conditions 3.117 differentiate beforehand the second one with respect to θ ; i.e. the following conditions are satisfied: ⎧  (1)  ⎨ i R X n + iYn(1) |r =R = χ1 (θ ) ⎩ d (u 2 + iv2 ) r =R = W  0 (θ ) dθ

176

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

As a result we get the system of singular integral equations of the second kind with Gilbert kernel:  ⎧ l0 l1  il2 ϕ−θ ⎪ ⎪ χ∗ (ϕ)ctg χ∗ (θ ) + W (θ ) + dϕ ⎪ ⎪   2π  2 ⎪ ⎪ ⎪ L ⎪ ⎪ ⎪ π ⎪  ⎪ ⎪ ϕ−θ il3 ⎪ ⎪ dϕ = F2 (θ ) W  (ϕ) ctg + ⎪ ⎪ ⎪ 2π  2 ⎨ −π  (3.123) d0 l0  id1 ϕ−θ ⎪ ⎪ ⎪ χ − − χ W dϕ (θ ) (θ ) (ϕ)ctg ∗ ∗ ⎪ ⎪   2π  2 ⎪ ⎪ ⎪ L ⎪ ⎪ ⎪ ⎪ π ⎪ ⎪ ϕ−θ il2 ⎪ ⎪ dϕ = F1 (θ ) W  (ϕ) ctg ⎪− ⎪ 2π  2 ⎩ −π

This system should be considered together with conditions, first one of which is equilibrium and the second one is the condition of continuity at the end points of cracks. β j

β j χ∗ (θ ) dθ = 0,

αj

W  (θ ) dθ = 0

(3.124)

αj

here F2 (θ ) = −χ1 (θ ) + i Re2 eiθ ,

F1 (θ ) = −W  0 (θ ) + e3 · eiθ

First the system 3.123 should be reduced to normal form: ⎧  ⎪ θ2(1) θ1(1) ϕ−θ ⎪ ⎪ χ dϕ χ∗ (ϕ) ctg + i · (θ ) ⎪ ∗ ⎪ ⎪ 2π θ 2 ⎪ ⎪ L ⎪ ⎪ 

⎪ 2  ⎪ 2 ⎪ (2) (1) ⎪ ⎪ −2θ θ − θ1(1)  ⎪ 2 2 ⎪ ϕ−θ ⎪ ⎪ + i · dϕ = F2∗ (θ ) W  (ϕ)ctg ⎪ ⎨ 2π θ 2 L ⎪  ⎪ (1) ⎪ θ2 ϕ−θ ⎪ ⎪ ⎪ W  (θ ) + i · dϕ χ∗ (ϕ) ctg ⎪ ⎪ 2π · 2θ 2 ⎪ ⎪ ⎪ L ⎪ ⎪  ⎪ ⎪ ⎪ θ2(1) θ1(2) ϕ−θ ⎪ ⎪ +i · W  (ϕ) ctg dϕ = F1∗ (θ ) ⎪ ⎪ 2π θ 2 ⎩ L

here

3.6 The Stress State of Compound Plane, Weakened by Arc-Shaped Cracks …

177

  d0 l0 F2 (θ ) − F1∗ (θ ) = 2 F1 (θ) d0 l0 + d0 l1   l0 l1 F2 (θ ) + F1 (θ) F2∗ (θ ) = 2 l0 l0 + d0 l1

Then using notation W  ∗ (θ ) = θ2(2) W  (θ ) the governing system is written in the following form: ⎧   ib1 ib2 ϕ−θ ϕ−θ ⎪ ⎪ χ dϕ − dϕ = F2∗ (θ) χ W  ∗ (ϕ) ctg + ctg (θ ) (ϕ) ⎪ ∗ ∗ ⎪ 2π 2 2π 2 ⎪ ⎨ L L   ia ia ϕ − θ ϕ−θ ⎪ (2) 2 1  (θ ) + ⎪ ⎪ dϕ + dϕ = θ2 F1∗ (θ) χ W  ∗ (ϕ) ctg ctg W ∗ ∗ (ϕ) ⎪ ⎪ 2π 2 2π 2 ⎩ L

L

(3.125) Now let us solve the system of singular integral Eq. 3.125 under conditions 3.124. (2) First we consider the case when Eq. 3.9 has two different roots λ(1) ∗ and λ∗ . In ( j) this case, as above multiplying the second equation by turns λ∗ ( j = 1, 2) and summing with first equation we obtain two independent singular integral equations of the second kind with Gilbert kernel:  iq j ϕ−θ dϕ = Q ∗j (θ ) η∗j (ϕ) ctg (3.126) η∗j (θ ) + ( j = 1, 2) 2π 2 L

where Q ∗j (θ ) = λ(∗j) θ2(2) F1∗ (θ ) + F2∗ (θ ) , η∗j (θ ) = χ∗ (θ ) + λ(∗j) W∗  (θ ) Now we introduce new variables by formulas θ x = tg , 2

ϕ s = tg , 2

ak = tg

αk , 2

bk = tg

βk , 2

L=

N +

(ak , bk )

k=1

Denoted by η j (x) and Q j (x) the following expressions: η j (x) =

χ∗ (2 arctg x) + λ j W∗  (2 arctg x) , 1 + x2

Q j (x) =

Q ∗j (2 arctg x) 1 + x2

Then we have η j (x) +

iq j π

 L

η j (s) x ds = Q j (x) − s−x 1 + x2

 η j (s)ds L

178

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

On the other hand 



  χ∗ (s) + λ(∗j) W∗  (s) ds = 0,

η j (s)ds = L

L

Finally we get η j (x) +

iq j π



η j (s) ds = Q j (x) s−x

( j = 1, 2)

(3.127)

( j = 1, 2; k = 1, 2, ..., N )

(3.128)

L

The conditions 3.124 will have the following form: bk η j (x) dx = 0; ak

The solution of Eq. 3.127 ) is following [3, 13] ⎧ ⎪ ⎨

( j)+

q j X0 (x) Q j (x) + η j (x) = 2 ⎪ πi 1 − qj ⎩ 1



Q j (s) ds ( j)+

L

X0

⎫ ⎪ ⎬

⎭ (s) (s − x) ⎪

( j)+

+ X0

( j)

(x) PN −1 (x)

(3.129) Here ( j)+

X0

, √ |x − ak |−γ j ·|x − bk |γ j −1 (x) = i g j (−1) N −l+1 N

k=1

γj =

θj 1 ln g j + , 2πi 2π

gj =

1 + qj , 1 − qj

x ∈ (al , bl )

  0 < θ = arg g j < 2π

¯ function the value of analytical in whole plane, besides L, ( j)

X 0 (z) =

N ,

(z − ak )−γ j ·(z − bk )γ j −1

k=1 ( j)

j

PN −1 (x) is the polynomial of N − 1 degree with unknown coefficients Ck (k = 1, 2, ..., N ), which will be found . j Note that except unknown constants Ck the expression of function Q j (x) includes constants e2 , e3 , which will be found. For this purpose we represent the constants e2 , e3 through the unknown functions.

3.6 The Stress State of Compound Plane, Weakened by Arc-Shaped Cracks … e2 =

θ1 i πR



θ2 i s +i ψ (s) ds + s −i πR

L

θ e3 = 5 π





s −i θ3 i ψ (s)ds + s +i πR

L

s +i θ6 ψ (s) ds − s −i π

L





θ4 i s +i ϕ (s) ds − s −i πR

L

179 

s −i ϕ (s)ds s +i

L

s +i ϕ (s) ds s −i

L

Here θ1 = θ4 =

μ (1 − μ) μ , θ2 = , ae1 + 2μ − 1 (μ + ae1 ) (ae1 + 2μ − 1) 2μ1 θ2(2) (ae1

+ 2μ − 1)

,

ψ (s) =

2μ1 (1 − μ)

θ3 =

θ2(2) (ae1 + 1) (ae1 + 2μ − 1) ae1 μ , θ6 = (2) , θ5 = μ2 (μ + ae1 ) θ2 (μ + ae1 )

χ∗ (2 arctg s) , 1 + s2

ϕ (s) =

,

W  ∗ (2 arctg s) . 1 + s2

Now taking into account that right parts of integral Eq. 3.127 can be written in form ( j)

( j)

l0 + λ∗ θ2(2) d0 χ1 (2 arctg x) l1 − λ∗ θ2(2)l0 · + Q j (x) = θ 1 + x2 θ

( j) (2) ( j) i R l0 + λ∗ θ2 d0 1 l1 − λ∗ θ2(2)l0 · · + e + 2 θ θ (x + i)2 (x

·

W0  (2 arctg x) 1 + x2

1 · e3 + i)2

The function ηi (x) is represented as following ηi (x) = η1(i) (x) + e2 · η2(i) (x) + e3 · η3(i) (x) + X 0(i) (x) PN −1 (x) Here

⎡

qi X 0(i)+ (x) 1 ⎢ (i) = + η(i) Q (x) (x) ⎣ j j πi 1 − q 2j ( j) (2)



Q (i) j (s) ds (i)+

L

X0

(s) (s − x) ( j) (2)

(3.130)

⎤ ⎥ ⎦

(i = 1, 2

j = 1, 2, 3)

(3.131)

l0 + λ∗ θ2 d0 χ1 (2 arctg x) l1 − λ∗ θ2 l0 W0  (2 arctg x) = + · · θ 1 + x2 θ 1 + x2

( j) (2) ( j) i R l0 + λ∗ θ2 d0 l1 − λ∗ θ2(2) l0 1 1 Q (i) , Q (i) · · 2 (x) = 3 (x) = 2 θ θ (x + i) (x + i)2 (i) Q 1 (x)

To determine Cki (k = 1, ..., N ) the following system of algebraic equations is obtained after substitution the expression for η j (x) from 3.130 in conditions 3.128:

180

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

   η2(1) (x) − η2(2) (x) dx +

e2 (2) λ(1) ∗ − λ∗

L

+

 

1 (2) λ(1) ∗ − λ∗

=−

L

1 (2) λ(1) ∗ − λ∗

X 0(1)+

(x)

 

e3 (2) λ(1) ∗ − λ∗

N "

 η3(1) (x) − η3(2) (x) dx

L

Ck(1) x k−1

−

X 0(2)+

(x)

k=1

N "

 Ck(2) x k−1

dx

k=1

   η2(1) (x) − η2(2) (x) dx L

(3.132a)       e3 (1) (2) (2) (1) (1) (2) (2) (1) −λ dx+ · η · η λ λ∗ · η3 (x) − λ∗ · η3 (x) dx (x) (x) ∗ ∗ 2 2 (1) (2) (1) (2) λ∗ −λ∗ λ∗ − λ∗ L L ⎡ ⎤  N N " (2) " 1 (1) (2)+ (2) (1)+ (1) k−1 ⎦ k−1 ⎣ + (1) Ck x − λ∗ · X 0 Ck x λ∗ · X 0 dx (x) (x) (2) λ∗ − λ∗ k=1 k=1 L    1 (1) (2) (2) (1) λ∗ · η1 (x) − λ∗ · η1 (x) dx = p0 − (1) (2) λ∗ − λ∗ e2

L

(3.132b) Then, we write functions ϕ (x) and ψ (x) using η j (x) ϕ (x) =

η1 (x) − η2 (x) (2) λ(1) ∗ − λ∗

;

ψ (x) =

j = 1, 2 by formulas

(2) λ(1) ∗ η2 (x) − λ∗ η1 (x) (2) λ(1) ∗ − λ∗

and substitute them in expressions for constants e2 and e3 . After all we get two algebraic equations e2 =

θ1 i

(1) (2) π R λ∗ − λ∗



   s + i / (1) (2) (1) (1) (2) (2) (1) λ∗ η1 (s) − λ(2) ∗ η1 (s) + e2 λ∗ η2 (s) − λ∗ η2 (s) s −i

L

( N N   " " (2) (2) k−1 (1) k−1 (2) (1) (1) (2)+ (2) (1)+ + [λ +e3 λ(1) η η X C s − λ X C s ] ds − λ (s) (s) (s) (s) ∗ 3 ∗ 3 ∗ ∗ 0 0 k k θ2 i + πR

k=1



s −i s +i

k=1



   (1) (2) (2) (1) (1) (2) (2) (1) λ∗ η1 (s) − λ∗ η1 (s) + e2 λ∗ η2 (s) − λ∗ η2 (s)

L

(   N N " " (1) (2) (2) (1) (1) (2)+ (2) (2) (1)+ (1) Ck s k−1 − λ∗ X 0 (s) Ck s k−1 ] ds + e3 λ∗ η3 (s) − λ∗ η3 (s) + [λ∗ X 0 (s) +



θ3 i

(1) (2) π R λ∗ − λ∗ (1)+

+[X 0

(s)

N " k=1

(1)

k=1

k=1

     s + i / (1) (2) (1) (2) (1) (2) η1 (s) − η1 (s) + e2 η2 (s) − η2 (s) + e3 η3 (s) − η3 (s) s −i

L (2)+

Ck s k−1 − X 0

(s)

N "

( (2)

Ck s k−1 ] ds

k=1

(3.132c)

3.6 The Stress State of Compound Plane, Weakened by Arc-Shaped Cracks … −

θ4 i   (1) (2) π R λ∗ − λ∗

(1)+ + [X 0

(s)

N "



s −i s +i



     (1) (2) (1) (2) (1) (2) η1 (s) − η1 (s) + e2 η2 (s) − η2 (s) + e3 η3 (s) − η3 (s)

L

(1) Ck s k−1

−

(2)+ X0

(s)

k=1

e3 =

(2) +e3 λ(1) ∗ η3

N "

( (2) Ck s k−1 ]

ds

k=1

θ5

(1) (2) π λ∗ − λ∗ 

181



   s + i / (1) (2) (1) (1) (2) (2) (1) λ∗ η1 (s) − λ(2) ∗ η1 (s) + e2 λ∗ η2 (s) − λ∗ η2 (s) s −i

L

 ( N N  " " (1) (2) k−1 (1) k−1 (1) (2)+ (2) (1)+ + λ η X C s − λ X C s ds (s) − λ(2) (s) ∗ 3 ∗ ∗ 0 0 k k 

θ6

k=1

k=1

     s + i / (1) (2) (1) (2) (1) (2) η1 (s) − η1 (s) + e2 η2 (s) − η2 (s) + e3 η3 (s) − η3 (s) s −i

(1) (2) π λ∗ − λ∗ L  ( N N " " (1)+ (1) (2)+ (2) + X0 Ck s k−1 − X 0 Ck s k−1 ds

−

k=1

k=1

(3.132d) Hence, we obtain the system of linear algebraic Eqs. 3.132a, 3.132b, 3.132c and 3.132d, in which the coefficients Ck(i) (k = 1, ..., N ) and constants e2 , e3 are unknown. After determination of the constants the opening of crack and stress components can be found from system ⎧  χ∗ (2 arctg x) ⎪ (1) W ∗ (2 arctg x) ⎪ + λ ⎨ η1 (x) = ∗ 1 + x2 1 + x2  ⎪ W ∗ (2 arctg x) χ (2 arctg x) ⎪ ⎩ η2 (x) = ∗ + λ(2) ∗ 1 + x2 1 + x2 It follows that χ∗ (θ) =

W∗(k) (θ) =

θ αk

 θ  θ (2) λ(1) ∗ η2 tg 2 − λ∗ η1 tg 2

, (1) (2) λ∗ − λ∗ cos2 θ2

η1 (tg (θ/2)) − η2 (tg (θ/2))

dθ, (1) (2) λ∗ − λ∗ cos2 (θ/2) θ ∈ (αk , βk )

    η1 tg θ2 − η2 tg θ2

W  ∗ (θ) = (1) (2) λ∗ − λ∗ cos2 θ2 θ W (θ) = αk

η1 (tg (θ/2)) − η2 (tg (θ/2))

dθ (1) (2) (2) λ∗ − λ∗ θ2 cos2 (θ/2)

(k = 1, ..., N )

(2) Now discuss the case when the quadratic Eq. 3.9 has two equal roots λ(1) ∗ = λ∗ = λ∗ = (a1 − b2 ) /2a2 . Denote by

η1 (x) =

χ∗ (2 arctg x) + λ∗ W∗  (2 arctg x) 1 + x2

the following singular equation is obtained

182

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …



iq1 π

η1 (x) +

η1 (s) ds = Q 1 (x) s−x

(3.133)

L

The solution of this equation is written in form ⎡ η1 (x) =

(1)+ q1 X 0 (x)

1 ⎢ ⎣ Q 1 (x) + 1 − q12

Here PN(1)−1 (x) =

Q 1 (s) ds (1)+

πi L

N "

⎤

 X0

Ck(1) x k−1 ,

(s) (s − x)

⎥ (1)+ (1) ⎦ + X 0 (x) PN −1 (x)

q1 = (a1 + b2 ) /2

k=1

Let us express the function χ∗ (x) by the functions η1 (x) and W∗  (x). χ∗ (x) = η1 (x) − λ∗ W∗  (x) Substituting the obtained expression in second Eq. 3.125 and take into consideration that  iq1 η1 (s) ds = Q 1 (x) − η1 (x) π s−x L

we get



iq1 ϕ (x) + π

ϕ (s) ds = Q 0 (x) s−x

L

Q 0 (x) = θ2(2) F1∗ (x) −

x a2 ia2 · p1 + (η1 (x) − Q 1 (x)) π 1 + x2 q1

We obtain the same singular integral equation with Cauchy kernel with respect to ϕ (x) as for η1 (x) with only deference of right part, containing Q 0 (x). Hence the solution of obtained equation will be ⎡ ϕ (x) =

1 ⎢ ⎣ Q 0 (x) + 1 − q12

q1 X 0+

(x)



πi

⎤ Q 0 (s) ds ⎥ + ⎦ + X 0 (x) PN −1 (x) (s − x) X 0+ (s)

L

PN −1 (x) =

N " k=1

Ck x k−1

(k = 1, ..., N )

Refrences

183

Thus after determination of the function ϕ (x) we can determine function ψ (x) by formula ψ (x) = η1 (x) − λ1 ϕ (x). As far as the unknown constants e2 , e3 , Ck , Ck1 (k = 1, ..., N ) are concerned they are determined in the same way as for case of different roots. Using the expression for Q 1 (x) and representing the function Q 0 (x) in the following form:

⎞ ⎛ (2) (2) a l q + λ θ d 2 1 0 ∗ 2 0 θ d0 ⎠ · χ1 (2 arctg x)   + Q 0 (x) = ⎝ 2 2 θ 1 + x2 1 − q1 θ

⎛ ⎞ (2) a2 q1 l1 − λ∗ θ2(2)l0 θ l0 W0  (2 arctg x)   +⎝ − 2 ⎠· 2 θ 1 + x2 1 − q1 θ

⎞

⎞ ⎛ a2 q1 l0 + λ∗ θ2(2) d0 a2 q1 l1 − λ∗ θ2(2) l0 θ2(2) l0 θ2(2) d0 i R 1 ⎠ ⎠· ⎝ ⎝     · − + + e2 + e3 θ θ 1 − q12 θ 1 − q12 θ (x + i)2 (x + i)2 ⎛

a2 X (1)+ (x) +  0 2 1 − q1 πi +

 L

1 (1)+ X0



(2) i R l0 + λ∗ θ2 d0 θ

(s) (s − x)



l1 − λ∗ θ2(2) l0 W0  (2 arctg s) l0 + λ∗ θ2(2) d0 χ1 (2 arctg s) · · + + 2 θ 1+s θ 1 + s2

⎞ (2) l1 − λ∗ θ2 l0 e3 ⎠ (1)+ (1) ds + a2 X 0 (x) PN −1 (x) · · + θ (s + i)2 (s + i)2 e2

As above the constants e2 and e3 are written as the sum of partial solutions of obtained integral equations. Joining the conditions 3.128 to obtained two equations we get 2N + 2 algebraic equations for determination of constants e2 , e3 , Ck , Ck1 (k = 1, ..., N ).

References 1. Brychkov A, Prudnikov A (1977) Integral Transformations of Generalized Functions. Nauka. 287 (in Russian) 2. Gakhov FD (1977) Boundary problems. Nauka 640 (in Russian) 3. Gradshtein IS, Ryzhik IM (1962) Tables of integrals, series and products. Moscow. FizMatLit. 1110 (in Russian) 4. Hakobyan VN (1995) On one mixed problem for composite plate, weakened by a crack. Proc NAS RA Mech 48(4):57–65 (in Russian) 5. Hakobyan VN (2002) Mixed problem for a composite plane weakened by a periodic system of cracks. NAS RA Rep 102(1):29–34 (in Russian) 6. Hakobyan VN, Dashtoyan LL (2004) On one mixed problem for composite plane weakened by Arc-type cracks. Topics in analysis and its applications. NATO Sci Ser Math Phys Chem 47:377–383 7. Hakobyan VN, Sargsyan AH (2010) Stress concentration near the absolutely rigid inclusion in compound elastic half-plane. Proc NAS RA Mech 63(4):12–22

184

3 Plane Strain State of Piecewise Homogeneous Elastic Plane …

8. Hakobyan VN, Hakobyan LV, Dashtoyan LL (2021) Contact problem for a piecewisehomogeneous plane with an interfacial crack under dry friction. J Phys Conf Ser 2231 (VII International Conference: Topical Problems of Continuum Mechanics (TPCM 2021) 04/10/2021– 08/10/2021 , Published online: 25 April 2022) 9. Hakobyan VN, Sahakyan AA (2022) Stress state of a compound plane with interface absolutely rigid inclusion and crack having common tip. In: Altenbach H, Eremeyev VA, Galybin A, Vasiliev A (eds) Advanced materials modelling for mechanical, medical and biological applications. Advanced structured materials, vol 155. Springer, Cham, pp 225–237. https:// doi.org/10.1007/978-3-030-81705-3_13 10. Handbook of stress intensity factors, vol 1, vol 2. Edited by Murakami Y. Mir.1014 11. Il’ina II, Sil’vestrov VV (2005) The problem of a thin interfacial inclusion detached from the medium along one side. J Mech Solids 40(3):123–133 12. Muskhelishvili NI (1966) Some problems of mathematical theory of elasticity. Nauka 708 (in Russian) 13. Muskhelishvili NI (1968) Singular integral equations. Nauka 511 (in Russian) 14. Panasiuk V, Savruk M (1976) Distribution of Stresses Near the Cracks in Plates and Shells. Kiev. Naukova Dumka 443 (in Russian) 15. Prudnikov AP, Brychkov YuA, Marichev OI (1981) Integrals and series (elementary functions), vol 800. Nauka, Moscow (in Russian) 16. Sahakyan AV (2010) Quadrature formulas for calculation of the integral with a variable upper limit. In: Proceedings of II international conference topical problems of continuum mechanics, pp 107–111. http://www.mechins.sci.am/publ/avetik_sahakyan/Dilijan2010 doc (in Russian) 17. Shilov GY (1965) Mathematical analysis. Second course, vol 327. Nauka, Moscow (in Russian) 18. The development of the theory of contact problems in USSR. 1979. Nauka, Moscow, vol 493 (in Russian)

Chapter 4

Plane Stress State of Half-Plane Consisting of Interfacial Crack

Abstract In this section, discontinuous solutions of the equations of the theory of elasticity are constructed for a piecewise homogeneous elastic half-plane made by connecting an infinite elastic strip and a half-plane of various materials. The halfplane is in a plane stress state and contains finite interfacial cracks on the line of connection of various materials. Based on these solutions, two contact problems are considered. In the first problem, it is assumed that the half-plane is deformed under the influence of an absolutely rigid stamp with a flat base, which is pressed into the composite half-plane, taking into account dry friction. The mutual influence of an interfacial crack and an absolutely rigid stamp has been studied. In the second problem, it is assumed that the half-plane is deformed by the finite, thin deformable patch acting on the boundary of the half-plane. The mutual influence of an interfacial crack and an elastic patch has been studied. Keywords Piecewise homogeneous half-plane · Interfacial cracks · Discontinuous solutions · Rigid stamp · Stringer · Mixed boundary value problem

4.1 Discontinuous Solutions for a Coated Half-Plane with Interfacial Cracks In all the main problems, there are cracks at the line of the junction of the strip with the half-plane, i.e. lines with discontinuous displacements. Therefore, first of all, we construct Lamé discontinuous solutions for the compound half-plane consisting of strip with thickness h and Lamé coefficients λ1 , μ1 and a half-plane with Lamé coefficients λ2 , μ2 on the line of connections y = 0 on the interval L consisting of mutually nonintersecting intervals (ak , bk ) , (k = 1, 2, . . . , N ), There are cracks which are subjected to equal normal stresses pk (x) , (k = 1, 2, . . . , N ) and zero tangential stress at the line of the junction [2, 4]. It is assumed that the composite half-plane is also deformed under action of normal and shear stresses of intensity P0 (x) and τ0 (x), respectively, applied on the section L 1 of the upper bank of the strip y = h. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 V. N. Hakobyan, Stress Concentrators in Continuous Deformable Bodies, Advanced Structured Materials 181, https://doi.org/10.1007/978-3-031-16023-3_4

185

186

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

Let us introduce the following functions of crack surface displacement jumps  U1 (x, 0) − U2 (x, 0) =  V1 (x, 0) − V2 (x, 0) =

U (x) 0

V (x) 0

( x ∈ L) , / L) (x ∈

(4.1)

( x ∈ L) , / L) (x ∈

Then using the Fourier transform we obtain the solutions of the Lamé equations [6] for the coated half-plane assuming that

 σ y(1) (x, h) =

σ y(1) (x, 0) − σ y(2) (x, 0) = 0

(|x| < ∞) ,

τx(1) y (x, 0)

(|x| < ∞) ,

−

τx(2) y (x, 0)

=0

−P0 (x) (x ∈ L 1 ) , τx(1) y (x, h) = 0 / L 1) (x ∈



τ0 (x) 0

(4.2) (x ∈ L 1 ) , / L 1) , (x ∈

(4.3)

Let us introduce the relationships (i) J j,k (s) =

θj=

 −θi−

 θi− schs + (θi + k) shs j chs − θi− sshs ; − schs + (θi + k) shs j chs + θi sshs

μj 1 − 2ν j Ej ; μj =  ; θ± =  j = 1 ± θj; λ j + 2μ j 2 1−ν j 2 1 + νj

M

( j)

 (s) =

   ( j) ( j) J1,0 (s) −J0,−1 (s) M (1) (s) M (2) (s) = , ( j) ( j) M (3) (s) M (4) (s) J0,1 (s) −J−1,0 (s)

where E j and ν j ( j = 1, 2) Young modulus and Poisson’s coefficient of layer and half-plane, respectively. Then the solutions of the Lamé equations in the general case can be represented by the following Fourier integrals: 1 U j (x, y) = 2π V j (x, y) =

1 2π

∞ i sgn s −∞ ∞

4

( j)

( j)

M3,k (|s| y) Ak (s) e−isx ds;

k=1 4

−∞ k=1

(4.4a) ( j)

( j)

M4,k (|s| y) Ak (s) e−isx ds,

4.1 Discontinuous Solutions for a Coated Half-Plane with Interfacial Cracks

187

while the stress components can be written in the form: σ y( j) (x,

μj y) = π

μj τx(yj) (x, y) = π

∞

4

|s|

( j)

( j)

M1,k (|s| y) Ak (s) e−isx ds,

k=1

−∞ ∞

is −∞

4

(4.4b) ( j)

( j)

M2,k (|s| y) Ak (s) e−isx ds,

k=1

( j)

where Ak (s) (k = 1 − 4, j = 1, 2) are unknown coefficients. On the line y = 0 we will have 1 U j (x, 0) = 2π

∞ −∞

1 V j (x, 0) = 2π

σ y( j) (x, 0)

μj = π

τx(yj) (x, y) =

( j)

i sgn s A3 (s) e−isx ds,

μj π

∞

( j)

A4 (s) e−isx ds,

(4.5)

−∞

∞

|s| A(1j) (s) e−isx ds,

−∞

∞

( j)

i s A2 (s) e−isx ds.

−∞

Taking into account that for y → −∞ the stresses in the substrate vanish from (4.4a) and (4.4b) we obtain A(2) 3 (s) =

1  (2) (2) − −θ2 A1 (s) + A2 (s) , θ2

A(2) 4 (s) =

1  (2) (2) − A1 (s) − θ2 A2 (s) , θ2

while from the conditions (4.2) and (4.3) we find (2) A(1) 3 (s) = U (s) + A3 (s) ,

A(1) 1 (s) =

μ2 (2) A (s) , μ1 1

(2) A(1) 4 (s) = V (s) + A4 (s) ,

A(1) 2 (s) =

μ2 (2) A (s) . μ1 2

Now, let us satisfy the conditions (4.3) at y = h. As a result, for determining of functions A(2) ( j = 1, 2) we obtain a system of linear algebraic equations. The j (s) solution of this system has the form:

188

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack



       1 g7,1 g6,1 1 g10 g11,1 U (s) A(2) − P¯ 0 (s) 1 . = + τ¯ 0 (s) V (s)  g6,2 g7,2  g11,2 g10 A(2) 2

Here we introduced the following notation μ= =

θ1− 2

 d0 − 2μ2

μ2 , μ1

d0 = 2μθ1 θ2 + θ1− θ2+ + μ2 θ1+ θ2− ,

   − θ2− + 2 − − 2μsh(2s) − d + 2s θ + θ ch(2s) , − μ) μθ (1 0 1 2 2 θ1−

   2 1 1 + (−1)i 2s + 2s 2 2θ2 μ + μ2 θ2− − θ2+ θ1− 2    1  − μ θ2 θ1− − θ1 θ2− μ + θ1− (1 − μ) θ1− θ2+ − μθ2− θ1+ e−2s , 2      = −(−1)i s θ1− − θ2− μ ch s − θ1− 1 + (−1)i s θ2− μ + θ2 sh s, g2,i =

g5,i

(4.6)

    g6,i = −(−1)i θ1− θ2− μ + θ2 s ch s + −(−1)i s θ1− − θ2 θ1− + θ1 θ2− μ sh s,      g7,i = − μθ2− − (−1)i s θ1− ch s − θ1− 1 − (−1)i s θ2 + μθ2− sh s, g10 =

      1 −  θ1 − 1 + 2s 2 θ2 θ1− + θ1 − 2s 2 θ1− θ2− μ + θ2 θ1− − θ1 θ2− μ ch(2s) , 2

   1  g11,i = − θ1− θ1− + 2s s θ1− + θ2− μ − θ1− cosh(2s) − θ2− μ sinh(2s)  2  + 1 + (−1)i s θ1− θ2− μ,      d = θ1− + μθ1+ θ2+ + μθ2− , l = 2μ μθ1 θ2− − θ2 θ1− , m = −2μ(θ1− − μθ2− ). ( j)

Further, using the found representations of functions Ak (s) we obtain the expressions for stresses and displacements in the coated half-plane through the functions of the displacement jumps. For the further analysis we will need the stresses acting on the crack edges as well as derivatives of the normal displacements of the coating at y = h. These quantities are represented by the following formulas: 1 (1) l m V  (s)ds σ y (x, 0) = − U  (x) + + R11 (x − s)U  (s) ds μ1 d πd s−x L L P0 (s) τ0 (s)  + R12 (x − s)V (s) ds + R13 (x − s) ds + R14 (x − s) ds, μ1 μ1 L

L1

L1

4.1 Discontinuous Solutions for a Coated Half-Plane with Interfacial Cracks

189

1 (1) l m U  (s)ds τx y (x, 0) = V  (x) + + R21 (x − s)U  (s) ds μ1 d πd s−x L L P0 (s) τ0 (s)  ds + R24 (x − s) ds, + R22 (x − s)V (s) ds + R23 (x − s) μ1 μ1 L

V  1 (x, h) = −

L1

θ1

1

P0 (s)ds + s−x

(4.7)





R31 (x − s)U (s)ds 2π θ1− μ1 L1 L  P0 (s) τ0 (s) + R32 (x − s)V (s)ds + R33 (x − s) ds + R34 (x − s) ds, μ1 μ1 2θ1− μ1

τ0 (x) +

L1



L

L1

(2)

(1)

σ y (x, 0) = σ y (x, 0),

L1

(2)

(1)

τx y (x, 0) = τx y (x, 0),

where ∞

μ R13 (t) = − π μ R23 (t) = π

0

∞

g6,2 (s) μ sin st ds, R24 (t) =  (s) π

0

R11 (t) = −

R12 (t) =

1 π

1 π

0

0

∞ 0

∞ 0

1 R33 (t) = 2π

∞

∞

1 R21 (t) = π μ R31 (t) = π

g7,1 (s) μ cos st ds, R14 (t) =  (s) π

0

0

∞ 0

g6,1 (s) sin st ds,  (s) (4.8)

g7,2 (s) cos st ds,  (s)

 2 μ g10 (s) l cos st ds, −  (s) d

 2 μ g11,1 (s) m sin st ds, −  (s) d  2 μg11,2 (s) m sin st ds, R22 (t) = −R11 (t), −  (s) d

g6,2 (s) μ sin st ds, R32 (t) =  (s) π

∞

∞

∞ 0

g2,2 (s) 1 sin st ds, R34 (t) = − 2π θ1  (s)

g5,2 (s) cos st ds,  (s) ∞ 0

g1 (s) cos st ds. θ1−  (s)

190

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

The values of these integrals are calculated by making a transition from the integration interval [0, +∞) to [0, 1) after which the interpolation and analytical integration are used.

4.2 Contact Problem for Piecewise Homogeneous Half-Plane This section considers the plane stress state of a piecewise homogeneous half-plane with interfacial cracks under the influence of absolutely rigid stamp acting on the boundary of the half-plane with dry friction. Let us consider a plane problem for a coated half-plane made of homogeneous elastic material with Lamé coefficients μ2 , λ2 and a coating of thickness h also made of homogeneous elastic material with Lamé coefficients μ1 , λ1 . The coordinate system is introduced in such a way that the x-axis is directed along the interface of the half-plane and coating while the y-axis is directed upward. A rigid stamp with the bottom shape y = f (x) and fixe boundaries is indented in the coating. The stamp is subjected to a normal P and tangential T forces. We will assume that the contact region L 1 is known and fixed. There is a dry friction between the stamp and the coating characterized by the coefficient of friction λ. The interface has a set L of colinear cracks occupying intervals (ak , bk ) , (k = 1, 2, . . . , N ) the edges of which are subjected to equal normal stresses pk (x) , (k = 1, 2, . . . , N ) and zero tangential stress. The problem is to determine the contact pressure under the stamp and stress intensity factors at the crack tips as well as crack opening patterns and to analyze them as functions of various problem parameters. Taking into account the absence of stresses outside the contact region, the continuity of all stresses and displacements at the coating/substrate interface, the presence of normal and absence of the tangential stresses at the crack edges the mathematical formulation of the problem is as follows: ⎧ V (x, h) = f (x) + δ (x ∈ L 1 ) ⎪ ⎨ 1 (1) τx y (x, h) = −λP0 (x) (x ∈ L 1 ) ⎪ ⎩ (1) (1) / L 1) σ y (x, h) = τx y (x, h) = 0 (x ∈ ⎧ (1) σ y (x, 0) = σ y(2) (x, 0), ⎪ ⎪ ⎪ ⎪ ⎨ τ (1) (x, 0) = τ (2) (x, 0), xy xy ⎪ U1 (x, 0) = U2 (x, 0), ⎪ ⎪ ⎪ ⎩ V1 (x, 0) = V2 (x, 0), 

(x ∈ / L)

σ y(1) (x, 0) + σ y(2) (x, 0) = 2 pk (x) = (+) (x) k (2) τx(1) y (x, 0) + τx y (x, 0) = 0 (ak < x < bk ; k = 1 − N )

(4.9a)

(4.9b)

(4.9c)

4.2 Contact Problem for Piecewise Homogeneous Half-Plane ( j)

( j)

191

( j)

where σx (x, y), σ y (x, y), and τx y (x, y) ( j = 1, 2) are normal and tangential stress components in the coating and half-plane ( j = 1 for coating and j = 2 for substrate), U j (x, y) and V j (x, y) are the horizontal and vertical displacements of the points of the coating and half-plane, δ is the unknown vertical stamp displacement, P0 (x) is contact pressure, τ0 (x) is the contact tangential stress, and λ is the friction coefficient. Assuming that λ ≥ 0, we consider that the frictional stress applied to the coating is directed from the right to the left. Using the obtained relationships (4.7), we will satisfy the first condition in (4.9a) and conditions (4.9c). As a result, using the second condition (4.9a), we derive the following system of governing equations:     l m V  (s)ds x −s x −s U  (s) ds + R12 V  (s) ds − U  (x) + + R11 d πd s−x h h L L L   x − s P p (s) (x) 0 k ∗ + R13 ds = (ak < x < bk , k = 1, . . . , N ) , h μ1 μ1 L

    l  m U  (s)ds x −s x −s V (x) + + R21 U  (s) + R22 V  (s) ds d πd s−x h h L L L   x − s P0 (s) ∗ + R23 ds = 0 (ak < x < bk , k = 1, . . . , N ) , h μ1 L1

−

λθ1 P0 (x) 1 + 2μ1 θ1− 2π μ1 θ1−

+

∗ R33

L1

where



x −s h



L1

P0 (s)ds + s−x

P0 (s) ds = f  (x) , μ1



R31

x −s h



L







U (s)ds +

R32

x −s h

(4.10) 



V (s)ds

L

(x ∈ L 1 ) .

∗ Ri3 (x) = Ri3 (x) − λRi4 (x) (i = 1, 2, 3) .

The system (4.10) needs to be considered together with the continuity conditions of the crack displacement jumps at the crack tips U (ak ) = V (ak ) = U (bk ) = V (bk ) = 0,

(4.11)

as well as with the stamp equilibrium condition P0 (x) dx = P.

(4.12)

L1

The second set of inequalities in (4.11) represents the requirement that the crack faces do not overlap, and there is a normal stress where the crack faces are in contact.

192

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

By introducing the new function V∗ (s) = U  (s) + i V  (s), the first two equations from system (4.10) can be reduced to the following equation

l m − V∗ (s) + d πid +



∗ [R11

L

+

∗ R13

x −s h



V∗ (s) ds s−x

L



x −s h

∗ V∗ (s) + R12

 −i

∗ R23





x −s h

x −s h



L1



V¯ ∗ (s)]ds

(4.13)

P0 (s) pk (x) ds = . μ1 μ1

where V∗ (s) + V¯ ∗ (s) V∗ (s) − V¯ ∗ (s) ; V  (s) = −i ; 2 2 R11 − R22 − i(R12 + R21 ) R11 + R22 + i(R12 − R21 ) ∗ = ; R12 = . 2 2

U  (s) = ∗ R11

Therefore, the solution of the problem was reduced to solution of the system of Eqs. (4.13) and the last equation from (4.10). For the case of just one crack occupying (−a, a) and the contact region occupying L 1 = (b − c, b + c) this system is reduced to m πid

a −a

b+c +

V∗ (s) ds l − V∗ (s) + s−x d ∗ R13



x −s h



∗ − i R23

a

∗ R11



−a



x −s h



b−c

−

λθ1 1 P0 (x) + 2μ1 θ1− 2π μ1 θ1− a

+ −a

∗ [R31



x −s h



b+c b−c

∗ V∗ (s) + R32

x −s h



∗ V∗ (s) + R12



x −s h

 V¯ ∗ (s) ds

P0 (s) p0 (x) ds = , μ1 μ1

P0 (s)ds + s−x

b+c

∗ R33



x −s h



b−c





x −s h



(4.14) P0 (s) ds μ1

V¯ ∗ (s)]ds = f  (x) ,

∗ ∗ R31 (t) = [R31 (t) − i R32 (t)]/2, R32 (t) = [R31 (t) + i R32 (t)]/2.

The solution of this system will be determined by the method of mechanical quadratures [7]. To do that we will use the following substitutions for the independent

4.2 Contact Problem for Piecewise Homogeneous Half-Plane

193

variables s = aξ , x = aη and x = cη + b, s = cξ + b, and reformulate the system on the interval (-1,1) by introducing the following dependent variables 2 P 1 , ϕ (η) = P0 (c η) , P˜ = π c P˜ Q 2 j (η, ξ ) = a

4 ∗ R μ 3j

∗ Q 23 (η) = c R32



 cη h

 cη − aξ + b , h

ψ (η) =

Q 1 j (η) = a

μ2 4 P˜

V∗ (ηa) ,

μ ∗  aη R 4 1j h

( j = 1, 2) ,

∗ ( aη − cξ − b ) − i R ∗ ( aη − cξ − b ) Q 1,3 (η, ξ ) = a R13 23 h h

,

p1 (η) =

μ p0 (aη) , 4 P˜

f ∗ (η) =



μ2 1  f (cη + b) P˜ μ

in the form ⎧ 1 1 ⎪ ⎪ l m ψ(ξ ) ⎪ ⎪ − ψ(η) + dξ + Q 11 (ξ − η) ψ(ξ )dξ ⎪ ⎪ ⎪ d πi d ξ −η ⎪ ⎪ −1 −1 ⎪ ⎪ ⎪ ⎪ 1 1 ⎪ ⎪ ⎪ ⎪ ⎪ ¯ )dξ + Q 13 (η, ξ ) ϕ(ξ )dξ = p1 (η) Q ψ(ξ − η) + (ξ ⎪ 12 ⎪ ⎪ ⎨ −1

−1

⎪ 1 1 ⎪ ⎪ λθ1 1 1 ϕ(ξ ) ⎪ ⎪ − ϕ(η) + dξ + Q 21 (ξ − η) ψ(ξ )dξ ⎪ ⎪ ⎪ π 2θ1− ξ −η 2θ1− ⎪ ⎪ −1 −1 ⎪ ⎪ ⎪ ⎪ 1 1 ⎪ ⎪ ⎪ ⎪ ⎪ ¯ )dξ + Q 23 (η, ξ ) ϕ(ξ )dξ = f ∗ (η) Q ψ(ξ − η) + (ξ ⎪ 22 ⎪ ⎩ −1

(4.15)

−1

Using these variables, equations (4.11) and (4.12) take the form 1

1 ψ(ξ )dξ = 0,

−1

ϕ (x) dx = −1

π , 2

(4.16)

Taking into account the fact that Ri∗j (η, ξ ) (i = 1, 2, j = 1, 2, 3), are regular functions it is easy to determine the behavior of functions ψ(η) and ϕ(η) in the vicinity of the points η = ±1; i.e. these functions can be represented in the form ψ (η) = ψ ∗ (η) (1 − η)α1 (1 + η)β1 , ϕ (η) = ϕ ∗ (η) (1 − η)α2 (1 + η)β2 ,

(4.17)



 1 1 1 μ1 + æ1 μ2 α1 = − − iγ ; β1 = − + iγ ; γ = ln ; 2 2 2π μ2 + æ2 μ1

 j = 1, 2 ,

194

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

   1 1 1 2a2 α2 = − − γ∗ ; β2 = − + γ∗ ; γ∗ = , ar ctg 2 2 2π a22 − 1 where ψ ∗ (η) and ϕ ∗ (η) are continuous smooth functions bounded on [−1, 1]. Using the following quadrature formula 1.125 and formula [7] 1 −1

  n (−α,−β)

Pn+k (y) ψ (x) dx wi ψ ∗ (ξi ) + π λψ (y) = 1 − (−α,−β) , x−y ξi − y Pn−1 (ξi ) i=1

(4.18)

where z = ±1, Re α, Re β > −1, and ξi are the roots of the Jacobi polynomials (α,β) Pn (ξi ) = 0, qi (z) and wi are given by formulas 1.126–1.127 in Eq. (4.14) are ∗ reduced to a system of 3n linear algebraic equations for ψ ∗ (ξi ), ϕ ∗ (ξi ) and ψ¯ (ξi ) , i = 1, . . . , n. After variables ψ ∗ (ξi ) and ϕ ∗ (ξi ) are determined it is easy to obtain the approximations for functions ψ(η) and ϕ(η) by using Lagrange’s interpolation polynomials as follows n (α ,β )

ψ ∗ (ξi ) Pn 1 1 (x) ψn (x) = , 1 ,β1 ) (x − ξi ) P  (α (ξi ) n i=1

n (α ,β )

ϕ ∗ (ξi ) Pn 2 2 (x) ϕn (x) = . (x − ξi ) P  n(α2 ,β2 ) (ξi ) i=1

That allows for finding all necessary variable characterizing the deformed state of the coated half-plane with an interfacial crack. In particular, the complex stress intensity factor at the crack tips is determined from Eq. (4.13) the complex combination of which for y = 0 and |η| > 1 in new variables has form σ y( j) (aη, 0)

−

iτx(yj) (aη, 0)

mμ1 = πid

1 −1

ψ(ξ ) dξ +R (η) ξ −η

(|η| > 1; j = 1, 2) ,

where R (η) is a regular function bounded at η = ±1. Substituting here the expression for ψn (η) from (4.17) and using the relationship (3.26) we find mμ1 sign (η) + R1 (η) (|η| > 1) πid ch (π γ ) |1 − η|1/2−iγ |1 + η|1/2+iγ ⎛ ⎞  1  ∗ ψ (ξ ) − ψ ∗ (±1) (1 − ξ )α1 (1 + ξ )β1 mμ 1 ⎝ R1 (η) = R (η) + dξ ⎠ . πid ξ −η ( j)

( j)

σ y (aη, 0) − iτx y (aη, 0) = −

−1

From here, for the complex stress intensity factor at the crack tips we get K (±1) = K I (±1) − i K I I (±1) = √ i(2)±iγ π μ1 m ∗ =∓ ψ (±1) . dch (π γ )

√ 2π

lim

x→±a±0

  ( j) ( j) |η ∓ 1|1/2±iγ σ y (aη, 0) − iτ y (aη, 0)

(4.19)

4.2 Contact Problem for Piecewise Homogeneous Half-Plane

195

Using the obtained expressions for the stress intensity factors and formulas from [3] it is easy to determine the Cherepanov–Rice J integral π μμ ˜ 2 m2 ∗ ∗ J (±1) = μK ˜ (±1) K¯ (±1) = 2 2 1 ψ (±1) ψ¯ (±1) , d ch (π γ )   1 1 − ν12 1 − ν22 . + μ˜ = 2 E1 E2

(4.20)

To determine the crack opening we can use the formula 1 V∗ (η) = V (aη) /2a = 2

η Im ψ (ξ ) dξ .

(4.21)

−1

Let us consider some numerical results. It is assumed that the bottom of the stamp is flat (i.e. f (x) = 0) and the crack surfaces are free of stresses (i.e. p0 (x) = 0), the frictional stress applied to the coating is acting in the direction of the negative x-axis. Also, we suppose that the dimensionless crack half-length a/ h = 0.1 while the dimensionless half-width of the contact region c/ h = 0.5. Two sets of coating and substrate elastic parameters were chosen: for relatively soft coating we take μ∗ = E 2 /E 1 = 5 while for relatively hard coating we take μ∗ = 0.2. In these cases, the material Poisson’s ratios wereν1 = 0.2, and ν2 = 0.3. Also, we consider the case of a homogeneous material with μ∗ = 1 and ν1 = ν2 = 0.3. The dimensionless applied load was chosen to be P/ (cμ1 ) = 0.1. The behavior of the dimensionless pressure, crack opening, and the dimensionless Cherepanov–Rice integral J∗ (±1) = 104 J (±1) /μ1 as functions of the friction coefficient λ and the relative distance between the stamp and crack b∗ = b/ h (the projection of the distance between the centers of the contact region and the crack; the distance is considered to be negative if the crack is to the left of the stamp and positive if it is to the right of the stamp) were analyzed. In the case of a homogeneous material (in the absence of a coating) in [5] it is shown that the crack is closed (and the normal stress intensity factors K I = 0 at both crack tips) while being to the left of the stamp and right below it. At some point to the right of the stamp the first opens up just partially, then it opens completely (and the normal stress intensity factors K I become positive at both crack tips). As the crack moves away from the stamp further to the right it opens wider (the normal stress intensity factors K I grows at both crack tips), reaches its maximally open state and then becomes less and less open. At the same time, the normal stress intensity factors K I reach their maximum and then diminish to zero. The further analysis pertains to the case of a coated elastic solid and one interfacial crack. For the case of soft coating μ∗ = 5 the graphs of the dimensionless contact pressure ϕ (η) are presented for two values of the friction coefficient λ = 0 and λ = 0.3 in Fig. 4.1. These data show that for small cracks located relatively far away from the contact region the contact pressure ϕ (η) is almost independent from the friction coefficient λ and the relative distance λ1 between the stamp and the crack.

196

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

Fig. 4.1 Dimensionless contact pressure under the stamp for λ = 0 and λ = 0.3

Fig. 4.2 a Subsurface stress σ y (x, 0) for the case of (λ = 0.3, μ∗ = 0.2; 1; 5) b Subsurface stress τx y (x, 0) for the case of(λ = 0.3, μ∗ = 0.2; 1; 5)

The graphs of the subsurface stress σ y (x, 0) and τx y (x, 0) for the cases of hard (μ∗ = 5) and soft (μ∗ = 0.2) coatings as well as for the case of homogeneous material (μ∗ = 1) are presented in Fig. 4.2a–b. The behavior of σ y (x, y) in each of these cases gives us an idea what to expect from the crack behavior. Specifically, in locations where the stress σ y (x, 0) is tensile we can expect the crack to be partially or fully open while at the places where σ y (x, 0) is compressive the crack is supposed to be closed. The normal displacements of the crack surfaces are presented in Fig. 4.3a–c for different relative stamp and crack locations b∗ . The values of the dimensionless Cherepanov–Rice integral J∗ (±1) at the corresponding crack tips are presented in Fig. 4.4a and 4.4b as functions of the relative stamp and crack location b∗ . We assume that the value of J∗ = 0 for those values of b∗ for which the crack is closed at least at one of its tips. It can be seen from Fig. 4.4a–b that, at the right end, the cracks open only when the stamp is located to the left of the origin, and the left end of the crack opens both in the case when the stamp is to the right of the origin and when the stamp is to the left of the origin.

4.2 Contact Problem for Piecewise Homogeneous Half-Plane

197

Fig. 4.3 Crack surface normal displacements for λ = 0.3 and different relative stamp and crack locations b∗ for μ∗ = 0.2; 1; 5, respectively

Fig. 4.4 Dimensionless Cherepanov–Rice integral J∗ as a function of the friction coefficient λ obtained for soft coating with μ∗ = 5

In the case when μ∗ = 0.2 (Fig. 4.5a, b) it is clear from the graphs that the right end of the crack opens when the stamp is to the right of the origin, while the left end of the crack opens when the stamp is to the left of the origin. At the same time, shear contact stresses in the case under consideration prevent the propagation of a crack from the left end and increase the probability of crack propagation from the right end.

198

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

Fig. 4.5 Dimensionless Cherepanov–Rice integral J∗ as a function of the friction coefficient λ obtained for hard coating with μ∗ = 0.2

Fig. 4.6 Dimensionless Cherepanov–Rice integral J∗ as a function of the friction coefficient λ obtained for the case of a homogeneous material with μ∗ = 1

In the case of a homogeneous half-plane Fig. 4.6a–b, both the left and right ends of the crack open only when the stamp is located to the left of the origin. We also present the graphs of the reduced integral J∗ in the case when a/ h = c/ h = 1 and μ∗ = 0.2, 5 (Figs. 4.7 and 4.8). As it is clear from Figs. 4.4, 4.5, 4.6, 4.7, and 4.8 the behavior of the Cherepanov– Rice integral J∗ different depends on the elastic parameters of the system coating/substrate. Specifically, for the cases of hard coatings the increase in the friction coefficient λ leads to a decrease in the value of the integral J∗ while for soft coatings the situation is opposite, i.e. J∗ increases as λ increases.

4.3 On the Load Transfer from a Deformed Patch to the Compound …

199

Fig. 4.7 Dimensionless Cherepanov–Rice integral J∗ as a function of the friction coefficient λ obtained for hard coating with a/ h = c/ h = 1; μ∗ = 0.2

Fig. 4.8 Dimensionless Cherepanov–Rice integral J∗ as a function of the friction coefficient λ obtained for hard coating with a/ h = c/ h = 1; μ∗ = 5

4.3 On the Load Transfer from a Deformed Patch to the Compound Half-Plane with Interfacial Cracks This section discusses the plane stress state of a piecewise homogeneous half-plane with interfacial cracks under the influence of a finite elastic patch acting on the boundary of the half-plane. Consider the piecewise homogeneous half-plane that includes an infinite elastic strip h in thickness and an elastic half-plane with elastic moduli and Poisson’s ratios, respectively, E 1 , ν1 and E 2 , ν2 and is referred to a Cartesian coordinate system O x y, where the axis O x is directed along the butt line between the strip and the half-plane. Let the compound half-plane be in a plane strained state. In this case, we assume that the compound half-plane is weakened by interphase cracks along the butt of the strip and the half-plane on the segment L including a finite number of disjoint intervals (ak , bk ), k = 1, 2, . . . , N and is strengthened on the section L 1 of the free boundary of the strip y = h by an elastic thin patch hs in thickness with the reduced modulus of strains E s . Let the compound half-plane be deformed under the actions of identical normal distributed loads with intensities pk (x), (k = 1, 2, . . . , N ), acting

200

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

on the edges of cracks and under the action of a concentrated horizontal load T0 applied to the patch at the point x0 ∈ L 1 . We will determine the contact stresses under the patch, values of the Cherepanov– Rice J -integral at the end points of cracks, and the openings of cracks and will study the regularities of variations of these quantities depending on the physical and geometric characteristics of the posed problem. We write the boundary and contact conditions of the problem of elasticity theory in the following form:



⎧ du 1 (x, h) ⎪ ⎪ = εs (x, t) (x ∈ L 1 ) ⎪ ⎨ dx τx(1) / L 1) (x ∈ ⎪ y (x, h) = 0 ⎪ ⎪ ⎩ (1) σ y (x, h) = 0 (−∞ < x < ∞)

(4.22a)

⎧ (1) σ y (x, 0) = σ y(2) (x, 0) ⎪ ⎪ ⎪ ⎪ ⎨ τ (1) (x, 0) = τ (2) (x, 0) xy xy ⎪ u 1 (x, 0) = u 2 (x, 0) ⎪ ⎪ ⎪ ⎩ v1 (x, 0) = v2 (x, 0)

(4.22b)

σ y(1) (x, 0) + σ y(2) (x, 0) = 2 pk (x) (2) τx(1) y (x, 0) + τx y (x, 0) = 0

(x ∈ / L)

(ak < x < bk ; k = 1 − N )

(4.22c)

Here εs (x, t) are the axial strains of inclusions. Using the relations (4.7) and taking into account that P0 (x) = 0 in the case under consideration, we satisfy the first condition in (4.22a) and conditions (4.22c). Let a horizontal concentrated force T0 be applied at the point x0 of the patch. Then the axial strains in the patch are described by the formula [1] ⎡ ⎤ x 1 ⎣ εs (x) = T0 H (x − x0 ) − τ0 (s) ds ⎦ hs Es

(x, x0 ∈ L 1 ) ,

(4.23)

x1

where x1 is the initial point of the interval L 1 , and H (x) is the Heaviside function. Introducing the dislocation function for points of the edges of cracks V∗ (s) = U  (s) + i V  (s), we get the following defining system of N + 1 singular integral and integro-differential equations: l m − V∗ (s) + d πid + L1

L

∗ R13 (x − s)

V∗ (s) ds + s−x





 ∗ ∗ R11 (x − s) V∗ (s) + R12 (x − s) V¯ ∗ (s) ds

L

τ0 (s) pk (x) ds = μ1 μ1

(ak < x < bk , k = 1 − N )

(4.24)

4.3 On the Load Transfer from a Deformed Patch to the Compound …

−

1 2π θ1− μ1

+

L1

τ0 (s)ds + s−x

∗ R23 (x − s)

L1

L

∗ R21 (x − s)V∗ (s)ds +

⎡



∗ R22 (x − s)V¯ ∗ (s)ds

L

τ0 (s) 1 ⎣ ds = T0 H (x − x0 ) − μ1 hs Es

201

x

⎤ τ0 (s) ds ⎦ ; (x ∈ L 1 ) .

x1

System (4.24) should be considered together with the conditions of continuity of displacements at the end points of cracks and the conditions of equilibrium of the patch: bk V∗ (x) dx = 0 (k = 1 − N ) ; τ (x) dx = T0 . (4.25) ak

L1

Here, ∗ = [R − R − i(R + R )] /2, R ∗ = [R + R + i(R − R )] /2, R ∗ (x) = R (x) , R11 11 22 12 21 11 22 12 21 44 12 23 1 1 ∗ ∗ ∗ R13 = R14 − i R24 , R21 (x) = [R41 (x) − i R42 (x)] , R22 (x) = [R41 (x) + i R42 (x)] . 2 2

Hence, the solution of the formulated problem is reduced to that of the system of governing Eq. (4.24) under conditions (4.25). Solutions of the system of governing Eq. (4.24) under conditions (4.25) can be constructed by the method of orthogonal polynomials or by the method of mechanical quadratures. In our opinion, the latter is more efficient for considered case. Without any loss of generality, we consider the case where the compound half-plane (Fig. 4.9) is weakened only by a single interfacial crack occupying the interval (−a, a) and is strengthened on the line y = h by a patch on the section (b − c, b + c), i.e. L = (−a, a)) and L 1 = (b − c, b + c) . If we set that the concentrated load T0 is applied to the left end of the patch, and the crack edges undergo the action of a normal load with intensity p0 (x), the defining system of equations takes the form

Fig. 4.9 Compound half-plane with interfacial crack and patch

202

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

l m − V∗ (s) + d iπ d b+c +

a −a

V∗ (s) ds + s−x

∗ R13 (x − s)

b−c

1 − 2π θ1− μ1 b+c +

b+c b−c



 ∗ ∗ R11 (x − s) V∗ (s) + R12 (x − s) V¯ ∗ (s) ds+

−a

τ0 (s) p0 (x) ds = . μ1 μ1

τ0 (s)ds + s−x

a

∗ R21 (x

−a

∗ R23 (x − s)

b−c

a

(−a < x < a) a

− s)V∗ (s)ds + −a

⎡

τ0 (s) 1 ⎣ ds = T0 − μ1 hs Es

x

∗ R22 (x − s)V¯ ∗ (s)ds+

⎤

τ0 (s) ds ⎦ ; (x ∈ L 1 ) .

b−c

(4.26) We change the variables s = aξ, x = aη, in the first equation in (4.26) and x = cη + b , s = cξ + b in the second one and introduce the notations cτ0 (cη + b) V∗ (ηa) = ψ1 (η) , V¯ ∗ (ηa) = ψ2 (η) , = ψ3 (η) , T0 Q 1 j (η, ξ ) = −

ad ∗ T0 d ∗ R (a (η − ξ )) ( j = 1, 2) ; Q 1,3 (η, ξ ) = − R (aη − cξ − b); l 1j lμ1 13

Q 2 j (η, ξ ) = −

2acμ1 ϑ1− ∗ R2 j (cη − aξ + b) ; ( j = 1, 2) T0

∗ Q 23 (η, ξ ) = −2acϑ1− R23 (c (η − ξ )) ; f 1 (η) = −

2cμ1 ϑ1− d p0 (aη) ; a1 = m/l; λ = , lμ1 h s Es

Then system (4.26) on the interval (−1, 1) can be written as follows: ⎧ 1 3 1 ⎪

⎪ a1 ψ1 (ξ ) ⎪ ⎪ ψ1 (η) − i dξ + Q 1i (η, ξ ) ψi (ξ )dξ = f 1 (η) ⎪ ⎪ ⎪ π ξ −η ⎨ i=1 −1

⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ ⎩π

1 −1

ψ3 (ξ ) dξ + ξ −η

−1

3 1

⎡

Q 2i (ξ − η) ψi (ξ )dξ = −λ ⎣1 −

i=1 −1

η −1

⎤

ψ3 (ξ )dξ ⎦ (4.27)

In this case, conditions (4.25) take the form 1

1 ψ j (ξ )dξ = 0,

−1

ψ3 (x) dx = 1 −1

( j = 1, 2) .

(4.28)

4.3 On the Load Transfer from a Deformed Patch to the Compound …

203

In view of the regularity of the functions Rm∗ j (η, ξ ) (m = 1, 2, j = 1, 2, 3), it is easy to establish [7] that the required functions have an exponential singularity at the end points of the interval of integration and can be presented β

ψ1 (η) = ψ1∗ (η) (1 − x)α1 (1 + η)1 ; ψ2 (η) = ψ¯ 1 (η) ; 1 μ1 + æ1 μ2 ψ3 (η) = ψ3∗ (η) (1 − x)−1/2 (1 + η)−1/2 ; γ = ln ; 2π μ2 + æ2 μ1 

1 1 α1 = − − iγ ; β1 = − + iγ ; 2 2

 Ej  ; j = 1, 2 , μj =  2 1 + νj

where ψ j ∗ (η) are continuous smooth functions bounded up to the ends of the interval [−1, 1], j = 1, 2, 3. Substituting the values of the functions ψ j (η) , j = 1, 2, 3, in (4.27) and (4.27) and using the relations given in [7] and the standard procedure, we get a system of 3n algebraic equations for ψ j ∗ (ξt ) , t = 1, 2, . . . , n, where ξt (α,β) are the roots of the Jacobi polynomial Pn (ξi ) = 0. When the functions ψ j ∗ (ξt ) are determined, it is easy, with the use of the Lagrange interpolation polynomials to restore the functions ψ ∗ (η) , ϕ ∗ (η) (−1 < η < 1), and to find all necessary quantities characterizing the stress–strain state of the compound half-plane. In particular, the dimensionless crack opening and the well-known Cherepanov–Rice J -integrals, as above, can be determined by formulas (4.20) and (4.21), replacing the function ψ (η) in them by ψ1 (η) . We analyze the problem in the case where the edges of the interphase crack are free from stresses. Setting h s / h = 0.05, T0 / (cμ1 ) = 0.1, E s /E 1 = 10, ν1 = 0.2,, and ν2 = 0.3, we studied the regularities of changes in the dimensionless tangential contact stresses under the patch, crack opening, and the Cherepanov–Rice reduced integral J∗ (±1) = 104 J (±1) /μ1 depending on the distance of the patch from the coordinate origin b∗ = b/ h, ratio of the elasticity moduli of the strip and the halfplane μ∗ = E 2 /E 1 , and the ratios of the lengths of the crack and the patch to the strip height a/ h and c/ h. The calculations indicate that, at the constant physicomechanical parameters, when the crack length is much less than the layer height, the tangential contact stresses under the patch depend slightly on μ∗ at fixed b∗ . In Fig. 4.10, we show the plots of the crack opening displacement V∗ (η) for different distances b∗ of the patch from the coordinate origin for a/ h = c/ h = 0.5 and μ∗ = 0.5, 2.5. It is seen that if the patch moves to the left from the coordinate origin, then the crack becomes to open, starting from b∗ = 0.3, from the right end and opens completely for b∗ = −0.3. At the further displacement of the patch to the left, the difference of the displacements of points of the crack edges decreases and becomes zero, starting from b∗ = −5.5 for μ∗ = 0.5 and from b∗ = −7.5 for μ∗ = 2.5, which can be interpreted as the crack closure. In Fig. 4.11, we present the plots of the crack opening displacement which depends on the ratios a∗ = a/ h = c/ h for b∗ = −1 and μ∗ = 0.5, 2.5. It is seen that the

204

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

Fig. 4.10 Crack opening displacements for μ∗ = 0.5 (a) and μ∗ = 2.5 (b)

Fig. 4.11 Crack opening displacements for μ∗ = 0.5 (a) and μ∗ = 2.5 (b)

maximum crack opening increases with a∗ , which can be interpreted as a decrease in the strip height at the same lengths of the crack and the patch. In Fig. 4.12, we present the plots of the Cherepanov–Rice reduced integral J∗ (±1) at the end points of a crack as a function of the parameter b∗ in the cases where a/ h = c/ h = 0.5 and μ∗ = 0.5, 1, 2.5, 5. In this case, for the values of the parameter b∗ at which at least one of the ends of the crack is closed, the value of the integral J∗ at this end is considered to be zero. The plots show that the behaviors of the Cherepanov–Rice reduced integral J∗ at the end points of a crack for all values of the reduced parameter μ∗ as a function of the parameter b∗ are practically identical. However, the stiffer the half-plane relative to the strip, the larger the maximum value of the integral J∗ and, hence, the higher the probability for the crack to propagate and for the strip to be separated from the half-plane. In Fig. 4.13, we give the plots of the Cherepanov–Rice reduced integral J∗ (±1) for different values of μ∗ at the end points of a crack as a function of the parameter h ∗ = h/a = h/c, when b∗ = −1. As the parameter h ∗ increases, which can be interpreted as an increase in the strip height at the constant lengths of the crack and the patch, it is seen that the Cherepanov–Rice reduced integral J∗ tends to zero. The softer the halfplane, the larger the maximum value of the Cherepanov–Rice reduced integral J∗ .

References

205

Fig. 4.12 Values of the Cherepanov–Rice integral J∗ (1) (a) and J∗ (−1) (b)

Fig. 4.13 Values of the Cherepanov–Rice integral J∗ (1) (a) and J∗ (−1) (b)

Thus, in the case of one interfacial crack, when its edges are free from loads, the crack partially opens from the left end and opens completely when the center of the patch is to the left of the crack center. It turns out that a decrease in the height of the strip, at constant lengths of the crack and patch, leads to an increase in the crack opening, while the stiffer the half-plane, the smaller the crack opening. An increase in the rigidity of the half-plane leads to an increase in the maximum value of the Cherepanov–Rice integral.

References 1. Alexandrov VM, Mkhitaryan SM (1983) Contact problems for bodies with thin coverings and interlayer, vol 488. Nauka, Moscow (in Russian) 2. Hakobyan VN, Hakobyan LV, Amirjanyan HA (2019) On load transfer from a deformable patch to a compound half-plane with interfacial cracks. Math Methods Phys Mech Fields 62(3):26–37 (in Russian) 3. Murakami Y (ed) (1988) Handbook of Stress Intensity Factors, vols. 1 and 2 (Mir.1014) 4. Kudish II, Amirjanyan HA, Hakobyan VN (2019) Probabilistic modeling of coating delamination: FFW 2018, 9-10 July 2018, Ghent University, Belgium. In: Proceedings of the 7th

206

4 Plane Stress State of Half-Plane Consisting of Interfacial Crack

international conference on fracture fatigue and wear January 2019, pp 359–370. https://doi. org/10.1007/978-981-13-0411-8_32 5. Kudish II, Covitch MJ (2010) Modeling and analytical methods in tribology. CRC Press. Taylor & Francis Group, Boca Raton 6. Muskhelishvili NI (1966) Some problems of mathematical theory of elasticity, vol 708. Nauka (in Russian) 7. Sahakyan AV (2000) Method of discrete singularities in application towards solving of singular integral equations with unmovable singularity. Proceedings of NAS RA Mech 53(N3):12–19 (in Russian)

Chapter 5

Mixed Boundary Value Problem For Compound Space With Interphase Defects

Abstract In present section we consider the axis-symmetrical stress state of compound elastic space, consisting of two different half-spaces with coin-shaped or semi-infinite ring-shaped cracks on their junction plane. There are mixed boundary conditions on junction line of half-spaces. Closed solutions of two problems for compound space consisting of different half-spaces with absolutely rigid thin coinshaped inclusion are obtained. In mentioned problems various models of contact conditions for inclusion with matrix are considered. Keywords Compound space · Interfacial coin-shaped crack · Axisymmetric smooth contact problem · Rigid coin-shaped inclusion

5.1 Introduction Investigation of the axisymmetric stress state of an elastic homogeneous compound space consisting of two dissimilar half-spaces with interfacial disk- or ring-shaped defects has been the subject of many studies. Among them, works [2, 10, 12–14, 19], which are directly connected with the present paper, should be noted. In [1, 10, 19], the axisymmetric stress distribution in bonded dissimilar materials containing disk- and ring-shaped cracks on the joining plane was studied. A sufficiently detailed review of the studies performed in this direction is also presented in [2]. In [14], the axisymmetric stress state was considered for a homogeneous elastic space with a disk-shaped crack on whose one face stresses are specified and a smooth stamp is pressed into the other one. With the help of rotation operators, a closed solution of the problem was constructed. By the same method, closed solutions of similar problems were constructed in [5–8] for a two-component space with disk-shaped and semiinfinite ring-shaped cracks on whose faces mixed-type conditions were specified. We should also point to papers [12, 13], where many basic results on axisymmetric contact and mixed elasticity theory problems were obtained.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2023 V. N. Hakobyan, Stress Concentrators in Continuous Deformable Bodies, Advanced Structured Materials 181, https://doi.org/10.1007/978-3-031-16023-3_5

207

208

5 Mixed Boundary Value Problem for Compound Space With Interphase …

5.2 Axis-Symmetrical Mixed Boundary Value Problem for Compound Space with Coin-Shaped Crack Let us consider the elastic compound space consists of two heterogeneous half-spaces with Lamé coefficients λ1 , μ1 and λ2 , μ2 . This space occupies the upper (z ≥ 0) and lower (z ≤ 0) half-spaces in cylindrical system of coordinates. There is coin-shaped crack with radius a on junction plane (z = 0) of half-spaces. The normal stresses P0(1) (r ) and shear stresses τ0(1) (r ) are given on the upper bank and the radial and normal components of displacements u 0 (r ) and w0 (r ), as well as the resultant of forces P0(2) are given on the lower bank of crack. The problem is to determine contact stresses, acting both on junction plane and on lower bank of crack, and their intensity coefficients. The stated problem can be mathematically formulated as the following boundary value problem: ⎧ u 1 (r, 0) = u 2 (r, 0) ⎪ ⎪ ⎪ ⎪ ⎨ w1 (r, 0) = w2 (r, 0) a 0 Taking into account that numbers a ∗j , b∗j ( j = 1, 2) are positive, as well as the above obtained inequality we can assert that the required inequality is proved. Since g j > 0 ( j = 1, 2) it follows points ±a are the end points of automatic boundedness. There are two unknown constants c∗ and rigid displacement δ in expression of function F j (x) ( j = 1, 2). The rigid displacement δ enters in expression of given function w (r ) in implicit form. These constants can be determined from conditions (5.21). In case of equal roots of Eq. (5.19) the solution of system (5.17) as in planar case can be reduced to successive integration of two singular integral equations. Then the solution is constructed in the same way. After determination of functions ϕ j (x) ( j = 1, 2) and using formulas V  (x) =

ϕ1 (x) − ϕ2 (x) (2) λ(1) ∗ − λ∗

,

χ (x) =

(1) λ(2) ∗ ϕ1 (x) − λ∗ ϕ2 (x) (2) λ(1) ∗ − λ∗

σ∗ (x) = Re χ (x) , τ∗ (x) = Im χ (x) 1 1 u ∗ (x) = 2 Re V  (x) , w∗ (x) = − 2 Im V  (x) θ2 θ2

220

5 Mixed Boundary Value Problem for Compound Space With Interphase …

x V (x) =

ϕ1 (x) − ϕ2 (x) (2) λ(1) ∗ − λ∗

−a

dx

we find the following by applying inverse formulas of Abel integral operator 1 d σ (r ) = − r dr

a

s Re [χ (s)] 1 d ds = − √ 2 2 r dr s −r

r

τ (r ) = −

1 d r dr

a r

a s Re

!

√ s2 − r 2

r

a Im Im [χ (s)] 1 d ds = − √ r dr s2 − r 2

!

(1) λ(2) ∗ ϕ1 (s)−λ∗ ϕ2 (s) (2) λ∗ −λ(1) ∗

(1) λ(2) ∗ ϕ1 (s)−λ∗ ϕ2 (s) (1) λ(2) ∗ −λ∗

√

r

 u (r ) = −

d dr

a

#s

Re

−a

r

w (r ) =

1 d r dr

a

s Im

ϕ1 (x)−ϕ2 (x) (1) dx λ(2) ∗ −λ∗

√ s2 − r 2  s # ϕ1 (x)−ϕ2 (x) −a

√

r

(1) λ(2) ∗ −λ∗

s2 − r 2

" ds;

" ds;

ds;

dx ds.

s2 − r 2

Let us consider the special case of stated problem when u 0 (r ) = P0(1) (r ) ≡ 0, w0 (r ) = δ = const, i.e. the rigid disk (inclusion) is intended in lower bank of coinshaped crack under action of normal concentrated load P. The rigid disk (inclusion) is fully coupled with coin-shaped crack. In this case we have x V2 (x) = I1 [u 2 (r )] − i I [w2 (r )] = −i I [w2 (r )] = −i 0

x = −i

√ 0

δr dr x2

− r2

V2  (x) = −

iδ , 2

=−

iδ 2

x √ 0

dr 2 x2

− r2

=−

iδx 2

! " χ0 (x) = I P0(1) (r ) = 0,

l2 δ f 1 (x) = − (1) + d1 c∗ = c1 2 π θ2  2 b3 δ f 2 (x) = + l c 2 ∗ = c2 π θ2(1) θ2(2) 2 2



w2 (r ) r dr √ x2 − r2

P0 =

P π

5.2 Axis-Symmetrical Mixed Boundary Value Problem …

221

F j (x) = λ j f 1 (x) + f 2 (x) = λ j c1 + c2 = K j = const

( j = 1, 2)

Substituting the values of F j (x) ( j = 1, 2) in formula (5.20) and having the value of integral [15] b

(x − a)α−1 (b − x)β−1 dx = (y − a)α−1 (b − y)β−1 π · ctgπ α x−y a   b−y α+β−2 − (b − a) B(α; β − 1)2 F 1, 2 − α − β; 2 − β; b−a (Re α > 0; Re β > 0;

a < y < b)

After some simple operations we obtain K j q j ω j (x)  ϕ j (x) =  i 1 − q 2j sin π γ j

(−a < x < a;

j = 1, 2)

(5.24)

To determine the coefficients K j ( j = 1, 2) we use the conditions (5.21). Substituting the values of functions ϕ j (x) ( j = 1, 2) from (5.24) and formulas a 

a ω j (x) dx = −a

−a

sin2 π γ j =

a−x a+x

γ j

dx =

2πaγ j sin π γ j

q 2j 1 − cos 2π γ j =− 2 1 − q 2j

we obtain for coefficients K j ( j = 1, 2) Kj = −

i P0 · 2πa

From which we get ϕ j (x) = −



qj γj

 ( j = 1, 2)

(5.25)

P0 q 2j ω j (x)   2πaγ j 1 − q 2j sin π γ j

(5.26)

After finding functions ϕ j (x) ( j = 1, 2) by formula (5.23) the contact stresses, acting under disk, and opening of crack can be determined. However besides of these values one of the most important mechanical characteristics in stated problem is the rigid displacement, which will be easy determined from relation (5.25). In fact considering (5.25) as a system of algebraic equations we find

222

5 Mixed Boundary Value Problem for Compound Space With Interphase …

c1 =

i P0 (γ1 q2 − γ2 q1 )   , (2) γ2 γ1 2πa λ(1) ∗ − λ∗

c2 =

i P0 (γ1 λ1 q2 − γ2 λ2 q1 )   (1) γ2 γ1 2πa λ(2) ∗ − λ∗

Taking into consideration the representations for constants c j ( j = 1, 2) in δ and c∗ we get the following formula for determination of rigid displacement of disk   π θ2(1) l2 c1 + θ2(2) d1 c2

δ=

d1 b3 − l22

(5.27)

Let us remark that in case of homogeneous space the following expression for rigid displacement of disk is obtained from formula (5.27) √ P (1 + ν) 3 − 4ν   δ= πa E 1 + 16β 2 here β=

(5.28)

√ 1 ln 3 − 4ν 2π

The constants ν, E are Poisson’s ratio and modulus of elasticity for homogeneous space, respectively. The expression for δ coincides with expression obtained by G.Ya. Popov in paper [14]. Finally we determine the coefficients of stress concentration on circle r = a. For this purpose we consider the last two Eq. (5.8) out of interval (0, a) and introduce the functions u ∗ (t), w∗ (t), σ∗ (t) and τ∗ (t). These equations will be written in following explicit form (1)

σz

+

+

(r, 0) =

l2 b3

a 0

a

l0

+

b2

0

0

a 0

0

0

0

∞ w∗ (t) dt s 2 J0 (sr ) cos tsds;

0

a

∞ s J0 (sr ) cos tsds

0

0



l0

σ∗ (t) dt

∞ a ∞ b2 τ∗ (t) dt s J0 (sr ) sin tsds + u ∗ (t) dt s 2 J0 (sr ) sin tsds

l2 (1) τr z (r, 0) =

+

a

a

(5.29)

∞ σ∗ (t) dt s J1 (r s) cos tsds

0

0

∞ a ∞ b3 τ∗ (t) dt s J1 (r s) sin tsds + u ∗ (t) dt s 2 J1 (r s) cos tsds 0

0

∞ w∗ (t) dt s 2 J1 (r s) cos tsds 0

0

(r > a)

5.2 Axis-Symmetrical Mixed Boundary Value Problem …

223

The last two summands are integrated by parts and recalling the formulas [19] s J1 (sr ) = −

d J0 (sr ) , dr

s J0 (sr ) =

1 d [r J1 (sr )] . r dr

(5.30)

Then (5.29) will be written in following form σz(1) (r, 0)

⎧ a ∞ 1 d ⎨ l0 = r σ∗ (t) dt J1 (sr ) cos tsds r dr ⎩ 0

+

l2

b2 +

a τ∗ (t) dt

0

a

∞

u  ∗ (t)dt

0

0

τr z (r, 0) = −

+

l0

b3 +

J1 (sr ) sin tsds

0

(1)

d dr

b3 J1 (sr ) cos tsds −

l2

w  ∗ (t) dt

0

∞ J1 (sr ) sin tsds

⎫ ⎬

(r > a)

⎭

0

∞

a σ∗ (t) dt 0

a

J0 (sr ) cos tsds 0

∞

a τ∗ (t) dt

J0 (sr ) sin tsds

0

0

a

∞

0

0

∞

u  ∗ (t)dt

0

b2 J0 (sr ) cos tsds −

a

w ∗ (t) dt

0

∞ J0 (sr ) sin tsds

⎫ ⎬ ⎭

(r > a)

0

Using the well-known values of Weber integrals [4, 19] ⎧ 1 ⎪ ⎪ t r ⎩ − √2 0 r r t − r2 ⎧ t >r ∞ ⎨0 J1 (sr ) sin tsds = t 1 ⎩ √ t r t a)

Substituting the expressions for τ∗ (t), w  ∗ (t), σ∗ (t) and u  ∗ (t), written in functions ϕ j (x) ( j = 1, 2), after some elementary operations we get ⎧ ⎨

⎫ 1 d tϕ1 (t) dt tϕ2 (t) dt ⎬ σz(1) (r, 0) = − Im A1 √ − B1 √ , ⎩ r dr r2 − t2 r2 − t2 ⎭ 0 0 ⎧ ⎫ a ⎨ t ϕ (t) dt ⎬ d ϕ dt (t) 1 2 Re A1 √ − B1 √ τr(1) z (r, 0) = ⎩ dr r2 − t2 r2 − t2 ⎭ a

0

a

(r > a) (5.34)

0

here A1 = −

θ2(2) λ(2) ∗ l 2 + b3   , (2) (1) θ2 λ∗ − λ(2) ∗

B1 = −

θ2(2) λ(1) ∗ l 2 + b3   (2) (1) θ2 λ∗ − λ(2) ∗

Without loss of generality we discuss the case, when quadratic Eq. (5.18) has different ¯ (1) complex roots, i.e. λ(2) ∗ =λ ∗ . In this case it is obvious that g2 = g1 and γ2 = −γ1 . Let 0 < Re γ1 < 1/2, then we have 

   a−x γ a + x γ¯ ϕ1 (x) = D1 , ϕ2 (x) = D2 a+x a−x 2 P0 q j     Dj = − , γ = γ1 = −γ¯ 2 2πaγ j 1 − q 2j sin π γ j

( j = 1, 2)

It is obvious that ϕ1 (a) ≡ 0 and therefore the first summands, entering in representation (5.34), have lower order at end point r = a, than second summands. It means that the singularity of stresses in circle r = a is determined by second summand. Let us investigate the behavior of these summands at r = a. We have d dr

a 0

ϕ2 (t) dt d = D2 √ 2 2 dr r −t

γ a  a+t γ a  dt d t dt a−t = D2 ψ (a, r ) √ 2 2 dr a−t r −t r −t 0

0

5.2 Axis-Symmetrical Mixed Boundary Value Problem …

d +D2 dr

a 0



ψ (t, r ) − ψ (a, r ) r −t

t a−t

γ

225

a 



dt = D2 ψ (a, r ) 0

t a−t

γ

dt + r −t

γ  a dt d ψ (t, r ) − ψ (a, r ) t + D2 dt r −t dr r −t a−t 0 0 ( ( ' ' γ γ  π D2 ψ (a, r ) r π D2 ψ (a, r ) d r −1 + −1 = sin π γ r −a sin π γ dr r −a

d + D2 ψ (a, r ) dr

a 

t a−t

d +D2 dr

γ

a

ψ (t, r ) − ψ (a, r ) r −t

0



t a−t

γ dt

It is clear that d dr

a 0

( γ r −1 r −a ( ' γ π D2 ψ (a, r ) d r + − 1 + 1 (r ) sin π γ dr r −a

ϕ2 (t) dt π D2 ψ  (a, r ) = √ sin π γ r2 − t2

'

Notice that, for obtaining the formula (5.35) we use the value of integral [15] b  a

x −a b−x

α−1

dx π = x−y sin απ

   a − y α−1   −1 b − y 

(0 < Re α < 2; y < a < b or y > b > a) Here the following notation is introduced  ψ (t, r ) = d 1 (r ) = D2 dr

a 0

a+t t

γ )

r −t ; r +t

 γ ψ (t, r ) − ψ (a, r ) t dt. r −t a−t

(5.35)

226

5 Mixed Boundary Value Problem for Compound Space With Interphase …

Substituting the value of function ψ (t, r ) at point a in (5.35) we get d dr

a 0

 γ¯ ϕ2 (t) dt 2γ¯ πa D2 r − (r − a)γ¯ γ¯ −1 = − γ¯ r √ √ r +a sin π γ¯ r + a (r − a)γ¯ +1/2 r2 − t2 + 1 (r )

(5.36) It is not difficult to prove that the singularity of function 1 (r ) at point r = a has lower order than the singularity of first summand. On the other hand 1 d r dr

a 0

tϕ2 (t) dt d = √ 2 2 dr r −t

1 2 (r ) = r

a 0

a 0

ϕ2 (t) dt + 2 (r ) √ r2 − t2

ϕ2 (t) dt 1 d − √ r dr r2 − t2

a ) 0

r −t ϕ2 (t) dt. r +t

Therefore formulas (5.34) can be rewritten as:  (1) σz (r, 0) = Im

2γ¯ πa B1 D2 √ sin π γ¯ r + a (r − a)γ¯ +1/2

+ ∗1 (r ) (1)



τr z (r, 0) = − Re

'

r γ¯ − (r − a)γ¯ − γ¯ r γ¯ −1 r +a

2γ¯ πa B1 D2 √ sin π γ¯ r + a (r − a)γ¯ +1/2

'

(*

r γ¯ − (r − a)γ¯ − γ¯ r γ¯ −1 r +a

+ ∗2 (r )

(*

(5.37) ⎧ ⎨

⎫ a ⎬ 1 d tϕ dt (t) 1 ∗1 (r ) = − Im A1 √ + Im {B1 1 (r )} ⎩ r dr r2 − t2 ⎭ 0 ⎧ ⎫ a ⎨ d ϕ1 (t) dt ⎬ ∗2 (r ) = Re A1 √ − Re {B1 2 (r )} ⎩ dr r2 − t2 ⎭ 0

Multiplying the second equation from (5.36) by i and summing with first equation we find (1)

σz

(1)

(r, 0) + iτr z (r, 0) = −

' ( 2γ¯ πa B1 D2 r γ¯ − (r − a)γ¯ γ¯ −1 − γ ¯ r √ r +a sin π γ¯ r + a (r − a)γ¯ +1/2   ∗ ∗ r >a  (r ) = 1 (r ) + i2 (r ) ,

+ (r )

(5.38)

5.3 Axis-Symmetrical Mixed Problem for Compound Space, Weakened …

227

It follows from (5.36) that contact stresses, acting on junction plane of half-spaces out of crack, are unbounded at the end point and have the singularity γ + 1/2 with Re (γ + 1/2) > 1/2. Then the intensity coefficient of fracture stresses, acting on line r = a is determined by formula K I + i K I I = lim

r →a+0

r

γ¯ +1/2   σz(1) (r, 0) + iτr(1) −1 z (r, 0)

a πi (1 − 2γ¯ ) 2γ −3/2 B1 D2 =− a sin π γ¯

(5.39)

In the same way the intensity coefficient is obtained for other values of γ1 .

5.3 Axis-Symmetrical Mixed Problem for Compound Space, Weakened by Semi-Infinite Ring-Shaped Crack In this subsection we consider the axis-symmetrical stress state of compound space, consisting of two heterogeneous half-spaces, weakened by semi-infinite ring-shaped crack on junction plane. The stresses are given on the one bank of crack and the displacements on the other. Let the elastic compound space, consisting of two heterogeneous half-spaces with Lamé coefficients λ1 , μ1 and λ2 , μ2 . These half-spaces occupy the upper (z ≥ 0) and the lower (z ≤ 0) half-spaces, respectively, in cylindrical system of coordinates with semi-infinite ring-shaped crack on junction line (z = 0). The crack is situated in the domain a < r < ∞, one bank of which are rigidly clamped and there are axissymmetrically distributed normal P0 (r ) and shear τ0 (r ) loads on the other bank of crack. We state the problem to determine the contact stresses on junction line of halfspaces and their intensity coefficients. The stated problem can be mathematically formulated in cylindrical system of coordinates, the origin of which is at the center of contact circle as the following boundary value problem: ⎧ (1) σz (r, 0) = P0 (r ) ⎪ ⎪ ⎪ (1) ⎨ τr z (r, 0) = τ0 (r ) ⎪ u 2 (r, 0) = 0 ⎪ ⎪ ⎩ w2 (r, 0) = 0 ⎧ (1) σz (r, 0) = σz(2) (r, 0) ⎪ ⎪ ⎪ ⎨ τ (1) (r, 0) = τ (2) (r, 0) rz rz ⎪ u 0) = u (r, ⎪ 1 2 (r, 0) ⎪ ⎩ w1 (r, 0) = w2 (r, 0)

(a < r < ∞)

(5.40a)

r ∈ (0, a)

(5.40b)

228

5 Mixed Boundary Value Problem for Compound Space With Interphase …

To solve the stated problem we denote by σ (r ) , τ (r ) , u (r ) and w (r ) components of stresses and displacements, acting in contact zone and consider two auxiliary problems for upper and lower half-spaces, which will be determined by the following boundary conditions:  σz(1)

(r, 0) =

 u 2 (r, 0) =

P0 (r ) ,

r >a

σ (r ) ,

0 1) . =− π ξ −η

j (p2 η + k2 ) =

(8.53)

k=1 −1

−1

It is obvious that the functions under the sum sign are bounded at the end points of the interval [−1, 1]. Therefore, substituting the value of the function ψ3 (ξ ) in (8.53) and using the value of the integral (6.2.9), (8.53) is represented in the form: j (p2 η + k2 ) = −

il4 ψ3∗ (±1) +  (η) (|η| > 1) , ch (π γ3 ) |1 + η|1/2−iγ3 |1 − η|1/2+iγ3

where il4  (η) = − π

1 −1

 ∗ ψ3 (ξ ) − ψ3∗ (±1) dξ (ξ − η) (1 + η)α3 (1 − η)β3

−

4  

1

l3 R∗3k (ξ − η) ψk (ξ ) dη,

k=1 −1

the upper sign corresponds to the ray η > 1, and the lower ray to the ray η < −1. Then the complex factor of the intensity of the fracture stresses will be given by the formula: √ KI (±1) − iKII (±1) = 2π lim |1 + η|1/2−iγ3 |1 − η|1/2+iγ3 χj∗ (η) = x→±1±0 √ ∗ 2π i l4 ψ3 (±1) . =− ch (π γ3 )

8.3 Plane Stress State of an Elastic Compound Plane with Interfacial …

377

Fig. 8.22 Jump of normal stresses α = 0, μ = 2

Fig. 8.23 Jump of shear stresses α = 0, μ = 2

A numerical calculation was carried out and the regularities of the change in the real parts of the amplitudes of the normal and tangential contact stresses acting on the long sides of the inclusions, crack opening, the

absolute value of the complex coef-

ficient of fracture stress intensity K∗ (±1) = KI2 (±1) + KII2 (±1) and the reduced rotation angle δ∗ = δ/l6 of inclusions were determined depending on the reduced frequency ω∗ = ω/ac2(1) of forced vibrations in two cases when concentrated load is applied to the center of the inclusion and is directed along the inclusion (α = π/2) or perpendicular to it (α = 0). It is assumed that the inclusion is in the interval (−a, a), the crack is in the interval (3a, 5a), and the elastic constants of the heterogeneous half-planes have the following values ν1 = ν2 = 0, 3, μ = μ2 /μ1 = 2, P0∗ = 1. Figures 8.22 and 8.23 show graphs of the real parts of the jump amplitudes, respectively, of normal and tangential contact stresses, depending on the reduced frequency of forced vibrations ω∗ , when the load is perpendicular to inclusion (α = 0). It can be seen from them that with an increase in the frequency of forced oscillations, the amplitude of the jump of normal contact stresses in absolute value in some part of the contact zone increases, and decreases in the rest of the part and becomes oscillating. In this case, the amplitude of the shear contact stresses increases in absolute value and reverses sign. Table 8.9 shows the values of the reduced angle of rotation of the inclusion and the absolute value of the complex factor of the intensity of fracture stresses in this case. It can be seen from them that both the real part of the angle of rotation of the inclusion

378

8 Dynamic Mixed Boundary Value Problems …

Table 8.9 Values of the reduced angle of rotation of the inclusion and the absolute value of the complex factor of the intensity of fracture stresses in the case α = 0, μ = 2 ω∗ 0.2 0.5 1 1.5 2.5 Re δ∗  ∗  K (−1)  ∗  K (1)

−0.00055 0.0622 0.0739

−0.0038 0.3180 0.3466

−0.0039 0.4893 0.8637

−0.0334 0.5939 0.9143

−0.0214 0.2857 0.2736

Fig. 8.24 Jump of normal stresses α = π/2, μ = 2

Fig. 8.25 Jump of shear stresses α = π/2, μ = 2

and the absolute value of the complex factor of the intensity of the fracture stresses with an increase in the frequency of forced vibrations first increase in absolute value and then decrease. Figures 8.24 and 8.25 show graphs of the real parts of the jump amplitudes, respectively, of normal and shear contact stresses, depending on the reduced frequency ω∗ of forced vibrations, when the load is directed along the inclusion (α = π/2). Tables display that an increase in the frequency of forced vibrations has little effect on the distribution of the real part of the amplitudes of the shear contact stress jumps. The real parts of the amplitudes for the jumps of normal contact stresses, in this case, are an order of magnitude small. However, their change leads to a change in the sign of the real part of the amplitude of the angle of rotation of the inclusion, which is confirmed by the results of calculations given in Table 8.10. This table also shows the values of the absolute value of the complex intensity factor of fracture

8.3 Plane Stress State of an Elastic Compound Plane with Interfacial …

379

Table 8.10 Absolute value of the complex intensity factor of fracture stresses in the case α = π/2, μ = 2 ω∗ 0.2 0.5 1 1.5 2.5 Re δ∗  ∗  K (−1)  ∗  K (1)

−0.0039 0.0404 0.0351

−0.0039 0.1220 0.1319

−0.0038 0.2234 0.3029

−0.0008 0.0673 0.2773

−0.00009 0.0143 0.3190

Fig. 8.26 Jump of normal displacements α = 0, μ = 2

Fig. 8.27 Jump of shear displacements α = 0, μ = 2

stresses. As in the first case, with an increase in the frequency of forced oscillations, first they increase and then decrease. Figures 8.26 and 8.27 show graphs of the real parts of the amplitudes of the jumps of the normal component of the displacement of the crack edges at the moments of time tk = 2π k/ω (k = 1, 2, . . .), in the cases α = 0 and α = π/2, respectively, when, together with the dynamic concentrated load P0∗ = 1 acting on the inclusion, a static distributed load with intensity q0 is acting on the crack edges and preventing crack closure. In this case, q0∗ = q0 /μ2 l3 = −0.1. It is clear that for the selected values of the parameters, an increase in the frequency of forced vibrations leads to a loss of symmetry in the shape of the crack opening. Thus, it turned out that, with the selected physical–mechanical and geometric parameters, in the case when the concentrated load is directed along the inclusion, an increase in the frequency of forced vibrations leads to a redistribution of the amplitudes of the jumps of normal contact stresses in

380

8 Dynamic Mixed Boundary Value Problems …

the contact zone, as a result of which the real part of the amplitude of the rotation angle of the inclusion changes. Therefore, it can be argued that by choosing the frequency of forced vibrations, it is possible to control the real part of the rotation angle of the inclusion, in particular to achieve the elimination of its rotation.

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