Preface
Introduction
Contents
1 Statements of Boundary Problems of Solid Mechanics
2 Basic Concepts of Stress–Strain State Theory
2.1 Defining Concepts of Continuum Mechanics
2.2 Stress Tensor
2.3 Stress Tensor Properties
2.4 Equilibrium Equations of Deformed Solid
2.5 Equilibrium Conditions at the Boundary
2.6 Principal Axes and Principal Values of Stress Tensor
2.7 Maximal Tangent Stresses
2.8 Stress Deviator and Spherical Part of Stress Tensor
2.9 Strains and Displacements
2.10 Principal Axes and Principal Values of Strain Tensor
2.11 Strain Deviator and Spherical Part of Strain Tensor
2.12 Strain Compatibility Equations
3 Material and Solid Mechanical Characteristics (Properties)
3.1 Main Mechanical Characteristics of Materials and Solids
3.2 Classification of Materials by the Nature of Deformation
3.3 Complete Diagrams of Deformation and Fracture
4 Construction of Mathematical Model Problems
4.1 The System of Equations to Describe the Medium Stress–Strain State
4.2 Physical Relationships Determining the Solid Behavior
5 Mathematical Models of the Theory of Elasticity
5.1 Basic Concepts and Definitions
5.2 Hooke’s Law
5.3 Clapeyron’s Formula and Clapeyron’s Theorem
5.4 Thermoelasticity
5.5 The Boundary Problems of Theory of Elasticity
5.6 The Theory of Elastic Boundary Problems in Displacements
5.7 Boundary Problems of the Theory of Elasticity in Stresses
5.8 Homogeneous Problem of the Theory of Elasticity in Stresses
5.9 The Problem of the Setting of Theory of Elasticity
5.10 2D-Problems of the Theory of Elasticity
6 Mathematical Models of Solid with Rheological Properties
6.1 Creep and Relaxation
6.2 Solid Mathematical Models
6.2.1 Structural Rheological Models
6.2.2 Creep Damage
6.3 Rheological Equations of Linear Viscoelasticity
6.3.1 General Concepts
6.3.2 Examples of Creep and Relaxation Kernels
6.3.3 Volterra Principle
6.4 Equations of Mechanics of Linear-Hereditary Media
7 Mathematical Models of Plasticity Theory
7.1 Solid-Plastic Behavior Models
7.2 Material Plasticity During Tension and Compression
7.3 Elastic–Plastic Models of Material Behavior
7.4 Viscoplasticity Models
7.5 Plasticity Conditions
7.7 Hypotheses of the Small Elastoplastic Deformation Theory
7.8 Theory of Problems of Small Elastic–Plastic Deformations
7.9 The Method of Elasticity Solutions
7.10 Plastic Flow Theory
7.11 Relationship Between Plastic Flow and Theory of Plasticity
7.12 Formula and General Methods
7.12.1 Formula of the Problems of Plasticity Theory
7.12.2 General Methods of Solving the Problems of Plasticity Theory
8 Fundamental Solutions
8.1 General Remarks
8.2 Boussinesq’s Problem
8.3 Hertz’s Formulas
8.4 Flamant’s Problem
8.4.1 Flamant’s Problem for a Half-Space
8.4.2 Flamant’s Problem for a Half-Plane
8.5 Kelvin’s Problem
8.6 Cerrutti’s Problem
8.7 General Case of Point Load Action on Surface of Falf-Space
8.8 Mindlin’s Problem
8.9 Some Generalizations
9 Dynamic Problems of Solid Mechanics
9.1 General Concepts and Definitions
9.2 Wave Equation
9.3 Model Problems of Dynamics of Elastic Body
9.3.2 Cauchy Problem and Boundary Problems
9.4 Stokes Problem
9.5 Natural and Forced Harmonic Oscillations
9.5.1 Natural Oscillations of Elastic Bodies
9.5.2 Forced Oscillations of Elastic Bodies
9.6 Propagation of Shock Waves in Unlimited Elastic Bodies
9.7 Progressive Waves
9.8 Rayleigh Waves
9.9 Progressive Waves in a Plane Layer
9.10 Semi-plane Under Action of Moving Surface Force
9.11 Model Problems on Perturbation Propagation in Elastic Space
9.12 Model Problems on Non-stationary Perturbation Propagation
9.13 Model Problems on Volume Perturbation Propagation
10 Mathematical Models of Special Classes of Solid Mechanics Problems
10.1 Solving Problems for Heterogeneous Nonlinear Elastic Media
10.1.1 Solving Problems for Physically Nonlinear Elastic Medium
10.1.2 Solution of Problems for Orthotropic Solids
10.2 Solving the Problems of Linear Theory of Viscoelasticity for Inhomogeneous Solids
10.3 The Theory of Creep of Isotropic Hardening Media
10.3.1 Consider the Situation of a Complex Stress State
10.4 Bilinear Theory of Elasticity
10.5 Nonlinear Viscoelastic Media
10.6 Method of Successive Approximations in Viscoplastic Problems

##### Citation preview

Michael Zhuravkov Yongtao Lyu Eduard Starovoitov

Mechanics of Solid Deformable Body

Mechanics of Solid Deformable Body

Michael Zhuravkov · Yongtao Lyu · Eduard Starovoitov

Mechanics of Solid Deformable Body

Michael Zhuravkov Theoretical and Applied Mechanics Belarusian State University Minsk, Belarus

Yongtao Lyu Department of Engineering Mechanics DUT-BSU Joint Institute Dalian University of Technology Dalian, Liaoning, China

Eduard Starovoitov Construction Mechanics Belarusian State University of Transport Gomel, Belarus

Preface

In this textbook, we consider the problems connecting with selecting and constructing mechanics-mathematical models that adequately describe the stress–strain state of deformed solids. We do not describe detailed approaches and methods for solving specific classes of boundary problems of deformable solids. These issues are set out in the relevant special publications. This publication is compiled on the basis of lecture courses for students of the joint institutes “DTU-BSU” and “BSU-DTU” as well as textbooks and lecture courses compiled by the authors for students of Belarussian State University, among them: Zhuravkov M. A. Mathematical modeling of deformation processes in solid deformable media (on the example of problems of rock mechanics). Minsk. Pub. by Belarusian State University, 2002. 456 p. (in Russian). Zhuravkov M. A. Fundamental solutions of the theory of elasticity and some of their applications in geomechanics, soil and base mechanics. Course of lectures. Minsk. Pub. by Belarusian State University, 2008. 247 p. (in Russian). Zhuravkov M. A., Starovoitov E. I. Continuum Mechanics. Theory of Elasticity and Plasticity. Coursebook. Minsk. Pub. by Belarusian State University, 2011. 543 p. (Classical University Edition). (in Russian). Zhuravkov M. A., Starovoitov E. I. Mathematical models of solid mechanics. Coursebook. Minsk. Pub. by Belarusian State University, 2021. 535 p. (Classical University Edition). (in Russian). Minsk, Belarus Dalian, China Gomel, Belarus

Michael Zhuravkov Yongtao Lyu Eduard Starovoitov

v

Introduction

Deformed Solid Body Mechanics or Solid Mechanics (SM) is a collection of disciplines that study the stress–strain state (SST) of deformed solids under the various physical laws of material behavior. The main “classical” sections of SM are the theories of elasticity, plasticity, and viscoelasticity. This can also include material strength and structural mechanics. The difference among these sections of mechanics lies in the objects under consideration, assumptions and physical equations defining material behavior, and so on. In the theories of elasticity, plasticity, and viscoelasticity, various physical laws that establish the connection between stresses and strains are used. Often these sections of SM are divided into mathematical and applied theories. Therefore, for example, the mathematical theory of elasticity does not use any deformation hypotheses, and the resulting equations are solved either by exact methods or by such approximate methods that allow us to limitlessly increase the degree of approximation to an exact solution. Therefore, one can consider the results obtained when solving problems by the mathematical theory of elasticity can be used as a standard for evaluating the accuracy of various approximate theories and methods for solving similar problems. The applied theory of elasticity differs from the mathematical one in that when solving problems, in addition to the Hooke’s1 law, some additional hypotheses are used (the hypothesis of flat sections for rods, straight normal for plates, and shells, etc.). In solving problems of the applied theory of elasticity, along with accurate methods for solving the corresponding equations, approximate methods are also used. Note that there is no “clear boundary” between the applied theory of elasticity and the resistance of materials. The first mathematician who was engaged in systematic studies of the resistance of solid bodies to destruction was Galileo.2 Although he considered solids to be inelastic and did not know the law linking displacements and forces, his work indicated the path that other researchers followed. 1

Robert Hooke (1635–1703), English scientist, architect, and polymath. Galileo di Vincenzo Bonaiuti de’ Galilei (1564–1642), Italian astronomer, physicist, and engineer, polymath.

2

vii

viii

Introduction

The discovery of Hooke’s law in 1660 and the establishment of Navier’s3 general equations in 1821 represent two important milestones in the further development of the theory that began with Galileo. Hooke’s law provided the necessary experimental justification for the theory. Finding common equations made it possible to reduce all problems related to small deformations of elastic bodies to mathematical calculations. In the period between the discovery of the Hooke’s law and the establishment of general differential equations of the theory of elasticity, the interest of researchers was focused on the problems of oscillations of rods and plates, as well as on the stability of columns. These were, first of all, the fundamental work of J. Bernoulli,4 devoted to the definition of the elastic curve shape, and the work of Euler,5 which laid the foundation for the research in the field of stability of elastic systems. Lagrange6 applied Euler’s theory to determine the most reliable shape of columns. The mathematical theory of elasticity as a science has been developed in the first half of the nineteenth century, mainly thanks to the work of French engineers and scientists. Therefore, for the first time, equilibrium and oscillation equations of elastic solids, assuming the body discrete molecular structure, were obtained by Navier. He deduced not only differential equations, but also boundary conditions that must be satisfied on the surface of the body. Lamé7 and Clapeyron8 developed Navier’s theory in relation to engineering. They wrote a special and scientific work on the internal equilibrium of solids, which solved the problem of stresses and strains of a thick-walled pipe under axisymmetric loading (Lamé’s problem). By the fall of 1822, Cauchy9 discovered most of the main elements of the pure theory of elasticity. He introduced the concept of stress and strain at a point. He showed that they (stress and strain) can be defined by six relevant components. Based on the hypothesis of a continuous and homogeneous solid structure, Cauchy obtained equations of motion (or equilibrium). He first introduced two elastic constants into the

3

Claude-Louis Navier (1785–1836), French mechanical engineer, a physicist whose work was specialized in continuum mechanics. First introduced the concept of stress. 4 Jacob Bernoulli (1655–1705), one of the many prominent mathematicians in the Bernoulli family. He was an early proponent of Leibnizian calculus and sided with Gottfried Wilhelm Leibniz during the Leibniz–Newton calculus controversy. 5 Leonhard Euler (1707–1783), Swiss mathematician, physicist, astronomer, geographer, logician, and engineer who made important and influential discoveries in many branches of mathematics. He is also known for his work in mechanics, fluid dynamics, optics, and astronomy. 6 Joseph-Louis Lagrange (1736–1813), Italian mathematician and astronomer, later naturalized French. He made significant contributions to the fields of analysis, number theory, and both classical and celestial mechanics. 7 Gabriel Lamé (1795–1870), French mathematician who contributed to the theory of partial differential equations by the use of curvilinear coordinates and the mathematical theory of elasticity (for which linear elasticity and finite strain theory elaborate the mathematical abstractions). 8 Benoît Paul Émile Clapeyron (1799–1864), French engineer and physicist, one of the founders of thermodynamics. 9 Baron Augustin-Louis Cauchy (1789–1857), French mathematician, engineer, and physicist who made pioneering contributions to several branches of mathematics, including mathematical analysis and continuum mechanics.

Introduction

ix

equations of elastic theory, while Navier’s equations contained only one. Relations that link small strains and displacements are named after him. A significant contribution to the development of the theory of elasticity belongs to Saint-Venan.10 He proposed a new approach for solving a number of applications (the semi-reverse method of Saint-Venan). Using this method, important problems were solved regarding the bending and torsion of a bar with a non-circular cross section. He also owns research on oscillation, shock, and plasticity theory. In the second half of the nineteenth century, Kirchhof11 formulated the main equations of thin rod theory, which laid the foundation for the development of methods for calculating elastic springs. In addition, he developed a consecutive theory of thin plates. The first attempt in this direction was made by Lagrange and Sophie Germain12 in 1814, but they were not able to correctly formulate the boundary conditions. At the end of the nineteenth century, Aron and Love13 gave the first version of the equations of shell theory of the model, based on the application of the hypothesis of a non-deformability of the normal rectilinear element. Bussinesk14 investigated the problem of determining stresses in an elastic body under the influence of a concentrated force. These studies allowed Hertz15 to formulate the problem about the interaction of two elastic bodies in contact. An important role in the development of the theory of elasticity was played by the work of Russian scientists. The fundamental results of the development of the principle of possible displacements, the theory of impact, as well as the integration of dynamics equations belong to Ostrogradsky.16 Gadolin17 studied the stresses in multilayer cylinders, thereby building the basis for the design of artillery barrels. Zhuravsky18 formulated the modern theory of bending beams. Significant progress in the construction of methods for solving 2D problems of the theory of elasticity is 10

Adhémar Jean Claude Barré de Saint-Venant (1797–1886), a mechanician and mathematician who contributed to early stress analysis and also developed the unsteady open channel flow shallow water equations, also known as the Saint-Venant equations. 11 Gustav Robert Kirchhoff (1824–1887), German physicist who contributed to the mechanics, mathematical physics, electricity, and spectral analysis. 12 Marie-Sophie Germain (1776–1831), French mathematician, physicist, and philosopher. One of the pioneers of elasticity theory. 13 Augustus Edward Hough Love (1863–1940), a mathematician famous for his work on the mathematical theory of elasticity. He also worked on wave propagation. 14 Joseph Valentin Boussinesq (1842–1929), French mathematician and physicist who made significant contributions to the theory of hydrodynamics, vibration, light, heat, and theory of elasticity. 15 Heinrich Rudolf Hertz (1857–1894), a German physicist who first conclusively proved the existence of the electromagnetic waves predicted by James Clerk Maxwell’s equations of electromagnetism. 16 Ostrogradsky Mikhail Vasilyevich (1801–1862), Russian mathematician, mechanician, and physicist. Performed important studies on integral calculus. 17 Gadolin Aksel Vilgelmovich (1828–1892), Russian scientist, developed the theory of fastening the barrels of artillery guns; works on physics, metal processing. 18 Zhuravsky Dmitriy Ivanovich (1821–1891), Russian scientist and engineer, founder of a school in the field of construction mechanics and bridge construction.

x

Introduction

associated with the names of Kolosov19 and Muskhelishvili,20 who first applied the method based on the use of complex variable functions. Bubnov21 solved a number of problems about a plate bending. Fundamental studies on the theory of plates and shells, vibrations of rods taking into account the influence of shear deformations were carried out by Tymoshenko.22 Subsequently, many problems were solved by the energy method proposed by him. Galerkin23 completed a series of studies on the theory of bending thin plates, thick plates, and the theory of shells. To derive the equations of shell theory, he apparently first applied the equations of the threedimensional theory of elasticity. Papkovich first proposed to build a solution to the problems of the theory of elasticity of displacements in the form of harmonic functions. He also performed studies of general stability theorems of elastic systems. In addition, he solved a large number of problems about bending plates under various boundary conditions. Vlasov,24 Novozhilov25 and Rabotnov26 made a great contribution to the development of the general theory of shells and other sections of the theory of elasticity. Therefore, Vlasov is the founder of a new scientific discipline—the construction mechanics of shells. The theory of plasticity, which studies irreversible deformations of solid bodies, as an independent section of mechanics, began “its history” about a hundred and fifty years ago. The foundations of the theory of plastic flow whose creation was aimed at describing metal forming processes were formulated in the first publication of Saint-Venan (1868–1871).

19 Kolosov Guriy Vasikyevich (1867–1936), Russian, Soviet mechanic; works on mathematics, theory of elasticity, machine theory, and mechanisms. 20 Muskhelishvili Nikolai Ivanovich (1891–1976), a renowned Soviet Georgian mathematician, physicist, and engineer. Muskhelisvili conducted fundamental research on the theories of physical elasticity, integral equations, boundary value problems, and others. 21 Bubnov Ivan Grigorievich (1872–1919), Russian engineer, developed the basics of the ship’s construction mechanics, the designer of two submarines. 22 Tymoshenko Stepan Prokofyevich (1878–1972), Ukrainian, Russian, and later, an American engineer and academician. He is considered to be the father of modern engineering mechanics. 23 Galerkin Boris Grigoryevich (1871–1945), Soviet mathematician and an engineer. One of the creators of the theory of bending of plates, his works contributed to the introduction of mathematical methods in engineering research. 24 Vlasov Vasiliy Zaharovich (1906–1958), Russian, Soviet scientist in the field of mechanics, works on the resistance of materials, construction mechanics, the theory of elasticity. 25 Novozhilov Valentin Valentinovich (1910–1987), Russian, Soviet mechanic; works on the theory of elasticity, plasticity, and calculation of shells and ship structures. 26 Rabotnov Uriy Nikolaevich (1914–1985), Russian and Soviet mechanical scientist; works on the theory of shells, the theory of plasticity, creep, and destruction of materials.

Introduction

xi

At the beginning of the twentieth century, Karman,27 Huber,28 Mises,29 Nadai,30 Henki,31 and others put forward new concepts and theories that, although they did not solve problems, expanded the range of ideas. In the 1920s and 1930s, in addition to propose different versions of plasticity theory, important fundamental experimental studies were carried out. It was an important stage in the development of plasticity theory. However, by the end of the 1930s–first years of the 1940s, a situation developed when experiments could confirm one theory and at the same time refute other theories. And the opposite, results from other experiments indicated otherwise. Clarity was introduced by Ilyushin32 (1943–1945), which pointed out the need for a clear distinction between the nature of deformation processes (simple and complex deformation). As a result of these researches, Ilyushin developed the theory of small elastoplastic deformations. In the early 1950s, various theories of plasticity were proposed under arbitrary complex loading. These approaches took the form of three theories: modern flow theory, sliding theory, and the general theory of elastoplastic deformations. The construction of the general mathematical deformation theory of plasticity was based on the isotropy postulate formulated by Ilyushin. The basis for the further development of the theory of the flow of elastoplastic bodies was the Drucker 33 hardening postulate on the non-negativity of the external forces work in a closed cycle of plastic loading. Models of viscoelastic behavior of materials began to actively develop in the second half of the last century. In many materials under operating conditions, the law of connection between force and displacement depends significantly on time. This dependence is exerted in the fact that, for example, under constant loading, the deformations do not remain constant, but increase. On the other hand, if the body is subject to deformation and some bonds keep the deformation unchanged, the reaction of the bonds decreases with time, “relaxed”.

27

Theodore von Kármán (1881–1963), Hungarian-American mathematician, aerospace engineer, and physicist who was active primarily in the fields of aeronautics and astronautics. 28 Hyber Maximilian Tytus (1872–1950), scientist in the field of mechanics; works in the field of material resistance, the theory of elasticity, the theory of plasticity, proposed a criterion for the plasticity of the potential energy of shape change. 29 Richard Edler von Mises (1883–1953), an Austrian scientist and mathematician who worked on solid mechanics, fluid mechanics, aerodynamics, aeronautics, statistics, and probability theory. 30 Nadai Arpad Ludwig (1883–1963), scientist mechanic, engineer; works in the field of the theory of elasticity, plasticity, together with Lode introduced the Nadai-Lode parameter. 31 Henki Heinrich (1885–1951), scientist mechanic, engineer; the most famous works in the field of sliding theory, plasticity, and rheology. 32 Ilyushin Alexey Antonovich (1911–1998), Russian and Soviet scientist mechanic, one of the founders of the theories of plasticity, viscoelasticity, and other sections of mechanics. 33 Daniel Charles Drucker (1918–2001) was an American civil and mechanical engineer. Drucker was known as an authority on the theory of plasticity in the field of applied mechanics. His key contributions to the field of plasticity include the concept of material stability described by the Drucker stability postulates and the Drucker–Prager yield criterion.

xii

Introduction

The initial development of the modern theory of viscoelasticity is associated with the names of L. Boltzmann,34 J. Maxwell,35 V. Kelvin, and36 W. Voigt.37 Subsequently, more complex models were created from elastic and viscous elements combined in various ways. Many achievements of the modern theory of viscoelasticity are associated with the works of A. A. Ilyushin, A. Yu. Ishlinsky,38 Yu. N. Rabotnov, and other scientists.

34

Ludwig Eduard Boltzmann (1844–1906), Austrian physicist and philosopher. His greatest achievements were the development of statistical mechanics and the statistical explanation of the second law of thermodynamics. 35 James Clerk Maxwell (1831–1879), Scottish scientist in the field of mathematical physics. 36 William Thomson, first Baron Kelvin (1824–1907), British mathematical physicist, and engineer. He did important work in the mathematical analysis of electricity and formulation of the first and second laws of thermodynamics and did much to unify the emerging discipline of physics in its modern form. 37 Woldemar Voigt (1850–1919), German physics. He worked on crystal physics, thermodynamics, and electro-optics. 38 Ishlinsky Alexander Yulievich (1913–2003), Russian and Soviet mechanical scientist, one of the founders of the theories of plasticity, viscoelasticity, and other sections of mechanics.

Contents

1

Statements of Boundary Problems of Solid Mechanics . . . . . . . . . . . .

1

2

Basic Concepts of Stress–Strain State Theory . . . . . . . . . . . . . . . . . . . . 2.1 Defining Concepts of Continuum Mechanics . . . . . . . . . . . . . . . . . 2.2 Stress Tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Stress Tensor Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Equilibrium Equations of Deformed Solid . . . . . . . . . . . . . . . . . . . 2.5 Equilibrium Conditions at the Boundary . . . . . . . . . . . . . . . . . . . . . 2.6 Principal Axes and Principal Values of Stress Tensor . . . . . . . . . . 2.7 Maximal Tangent Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8 Stress Deviator and Spherical Part of Stress Tensor . . . . . . . . . . . . 2.9 Strains and Displacements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.10 Principal Axes and Principal Values of Strain Tensor . . . . . . . . . . 2.11 Strain Deviator and Spherical Part of Strain Tensor . . . . . . . . . . . . 2.12 Strain Compatibility Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9 9 10 14 16 18 18 21 24 28 32 33 35

3

Material and Solid Mechanical Characteristics (Properties) . . . . . . . 3.1 Main Mechanical Characteristics of Materials and Solids . . . . . . . 3.2 Classification of Materials by the Nature of Deformation . . . . . . . 3.3 Complete Diagrams of Deformation and Fracture . . . . . . . . . . . . .

51 51 57 58

4

Construction of Mathematical Model Problems . . . . . . . . . . . . . . . . . . 4.1 The System of Equations to Describe the Medium Stress–Strain State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Physical Relationships Determining the Solid Behavior . . . . . . . .

63

Mathematical Models of the Theory of Elasticity . . . . . . . . . . . . . . . . . 5.1 Basic Concepts and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Hooke’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Clapeyron’s Formula and Clapeyron’s Theorem . . . . . . . . . . . . . . . 5.4 Thermoelasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 The Boundary Problems of Theory of Elasticity . . . . . . . . . . . . . . 5.6 The Theory of Elastic Boundary Problems in Displacements . . . .

69 69 72 76 78 79 80

5

63 64

xiii

xiv

Contents

5.7 5.8

Boundary Problems of the Theory of Elasticity in Stresses . . . . . Homogeneous Problem of the Theory of Elasticity in Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.9 The Problem of the Setting of Theory of Elasticity . . . . . . . . . . . . 5.10 2D-Problems of the Theory of Elasticity . . . . . . . . . . . . . . . . . . . . .

83

6

Mathematical Models of Solid with Rheological Properties . . . . . . . . 6.1 Creep and Relaxation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Solid Mathematical Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.1 Structural Rheological Models . . . . . . . . . . . . . . . . . . . . . . 6.2.2 Creep Damage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2.3 General Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Rheological Equations of Linear Viscoelasticity . . . . . . . . . . . . . . 6.3.1 General Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3.2 Examples of Creep and Relaxation Kernels . . . . . . . . . . . 6.3.3 Volterra Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Equations of Mechanics of Linear-Hereditary Media . . . . . . . . . .

101 101 103 103 120 123 128 128 129 134 137

7

Mathematical Models of Plasticity Theory . . . . . . . . . . . . . . . . . . . . . . . 7.1 Solid-Plastic Behavior Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Material Plasticity During Tension and Compression . . . . . . . . . . 7.3 Elastic–Plastic Models of Material Behavior . . . . . . . . . . . . . . . . . 7.4 Viscoplasticity Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Plasticity Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Simple and Complex Loading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 Hypotheses of the Small Elastoplastic Deformation Theory . . . . . 7.8 Theory of Problems of Small Elastic–Plastic Deformations . . . . . 7.9 The Method of Elasticity Solutions . . . . . . . . . . . . . . . . . . . . . . . . . 7.10 Plastic Flow Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.11 Relationship Between Plastic Flow and Theory of Plasticity . . . . 7.12 Formula and General Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.12.1 Formula of the Problems of Plasticity Theory . . . . . . . . . 7.12.2 General Methods of Solving the Problems of Plasticity Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

149 149 150 155 157 157 162 164 168 169 173 179 180 180

Fundamental Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 General Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Boussinesq’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Hertz’s Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Flamant’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4.1 Flamant’s Problem for a Half-Space . . . . . . . . . . . . . . . . . 8.4.2 Flamant’s Problem for a Half-Plane . . . . . . . . . . . . . . . . . 8.5 Kelvin’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.6 Cerrutti’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 General Case of Point Load Action on Surface of Falf-Space . . .

199 199 201 206 209 209 212 217 221 223

8

86 86 89

183

Contents

xv

8.8 8.9

Mindlin’s Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 Some Generalizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

9

Dynamic Problems of Solid Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 General Concepts and Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Wave Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 Model Problems of Dynamics of Elastic Body . . . . . . . . . . . . . . . . 9.3.1 General Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.3.2 Cauchy Problem and Boundary Problems . . . . . . . . . . . . 9.4 Stokes Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Natural and Forced Harmonic Oscillations . . . . . . . . . . . . . . . . . . . 9.5.1 Natural Oscillations of Elastic Bodies . . . . . . . . . . . . . . . . 9.5.2 Forced Oscillations of Elastic Bodies . . . . . . . . . . . . . . . . 9.6 Propagation of Shock Waves in Unlimited Elastic Bodies . . . . . . 9.7 Progressive Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.8 Rayleigh Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.9 Progressive Waves in a Plane Layer . . . . . . . . . . . . . . . . . . . . . . . . . 9.10 Semi-plane Under Action of Moving Surface Force . . . . . . . . . . . 9.11 Model Problems on Perturbation Propagation in Elastic Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.12 Model Problems on Non-stationary Perturbation Propagation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.13 Model Problems on Volume Perturbation Propagation . . . . . . . . .

10 Mathematical Models of Special Classes of Solid Mechanics Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Solving Problems for Heterogeneous Nonlinear Elastic Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.1 Solving Problems for Physically Nonlinear Elastic Medium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1.2 Solution of Problems for Orthotropic Solids . . . . . . . . . . 10.2 Solving the Problems of Linear Theory of Viscoelasticity for Inhomogeneous Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 The Theory of Creep of Isotropic Hardening Media . . . . . . . . . . . 10.3.1 Consider the Situation of a Complex Stress State . . . . . . 10.4 Bilinear Theory of Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.5 Nonlinear Viscoelastic Media . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Method of Successive Approximations in Viscoplastic Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

233 233 236 237 237 240 243 246 247 249 251 252 254 258 263 272 279 281 287 287 287 290 292 294 299 300 301 304

Chapter 1

Statements of Boundary Problems of Solid Mechanics

Solid Mechanics (Deformed Solid Body Mechanics) (SM), like the Continuum Mechanics (CM), is a phenomenological science. This means that it is based on an axiomatic approach, although not as complete as mathematics. The improvement and development of study methods for mechanical processes and phenomena are characterized by the aiming to find a solution in the most general form. This naturally requires a significant complication for the used mathematical methods, which in turn leads to additional requirements for functions describing the mechanical processes and phenomena being studied. Today, success in solving the problem of qualitative and quantitative correspondence of real mechanical processes and their mathematical description is impressive. Therefore, mathematical and computer modeling has become the basic elements for solving mechanics problems. Despite the fact that the general plan for solving the boundary problems of SM is quite clear, its implementation is accompanied by great difficulties. In general, it is impossible to solve the equations describing the state and behavior of deformable solids under various loads. Analytical solutions exist only for a limited number of applications. However, even these solutions are of great value, since there is a standard with which you can compare approximate solutions obtained as a result of the introduction of certain simplified hypotheses. Therefore, in mechanics, we’re dealing with modeling. In particular, in SM, we are dealing with modeling the deformation processes of solid bodies. SM operates models with elastic body, plastic body, etc., although real medium is described using these models only under certain assumptions. A continuous medium is introduced to describe discrete physical objects so that a known effective mathematical analysis apparatus can be used. Just as any set becomes a space only after introducing some structures, continuous medium needs to be introduced by axioms and postulates to become the object of constructing models.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Zhuravkov et al., Mechanics of Solid Deformable Body, https://doi.org/10.1007/978-981-19-8410-5_1

1

2

1 Statements of Boundary Problems of Solid Mechanics

As B.E. Pobedrya1 noted, the continuity principle used to introduce a continuum “does not solve all problems”. After setting the continuum mechanics problem, its solution is carried out using computational mathematics. To do this, the problem must be “transformed” into an algebraic problem, that is, discretization. Furthermore, in order to analyze the solution obtained and compare it with the experimental results, it is necessary to continue the problem again. Thus, the processes of discretization and continuity of CM problems and, in particular, SM problems are repeated several times at different levels. When we consider mathematical models for real complex mechanical processes, it is extremely difficult to achieve the high quantitative accuracy. One of the reasons for this situation is, for example, the discrepancy between the mechanical and strength characteristics used for calculations and the real characteristics of objects. The next reason is that it is impossible to construct a mechanical-mathematical model that accurately reflects all the features of the real process, etc. The main problem of the modeling is that, based on the results of model studies, it is possible to give the necessary answers about the qualitative and quantitative features and effects of the studied phenomenon in real conditions. Modeling in the general case can be divided into three types: physical, mathematical, and functional modeling. This publication relates to the mathematical modeling of the stress–strain state of solids under conditions of various forces and kinematic loadings. Mathematical modeling is used to study those processes that can be described mathematically (i.e., for which mathematical models can be built), in such a way that the mathematical problem of constructed model is solvable. Currently, due to the active introduction of computer tools in all areas of human research and production, the approaches and ideologies of scientific research and engineering work are radically changing. The computer today serves as an intelligent assistant. Now experts, in the majority, don’t put their attentions on the development of approaches to the solution of problems, but deal with a problem of adequacy of a model problem and real process. At the same time, the methods for solving the model problems can be very “bulky and heavy”. Since today’s tasks are solved using specialized software, this circumstance is not a determining factor. As a rule, solving a large number of applied mechanical problems requires passive or active experiments. The main disadvantage of a passive experiment (by this term, we mean studies or observations “in-situ”) is the impossibility of sufficient variation of the input parameters, which limits the use of the results obtained to the framework of the specific conditions in which the studies were performed. To eliminate this drawback, the real process is replaced by a model, and subsequent studies are carried out with a variation of input parameters with the help of this model (we will call it active experiment). The most widely used technologies of an active experiment are approaches based on the use of physical and mathematical models. 1

Pobedrya Boris Efimovich (1937–2016), Soviet and Russian scientist in the field of mechanics; works in the field of the theory of elasticity, viscoelasticity, plasticity, numerical methods.

1 Statements of Boundary Problems of Solid Mechanics

3

Before the active introduction of computer technologies, mathematical modeling methods were not widely used in the study of complex applied processes due to the laboriousness and unscheduled impossibility of conducting real calculations in accordance with the constructed mathematical model. In this regard, physical models were more preferred and common. However, physical models, in turn, in addition to the positive aspects, have a large number of negative factors that prevent the spread of physical modeling as a universal technology. Therefore, it is necessary to clearly represent the areas of approaches which are effectively used and limited to real-world process modeling. Today, there is a procedure for actively replacing physical modeling with mathematically based computer technologies in accordance with predetermined physical equations describing the behavior of the medium. Thus, today one of the most important problems is the development and adaptation of modern advanced approaches and methods of mathematical modeling to perform computer modeling for a wide range of applied mechanical processes. Real mechanical processes are very diverse and complex in terms of adequately qualitative and quantitative descriptions based on the mathematical modeling. Two approaches to the theoretical construction in fundamental and applied sciences are distinguished—phenomenology and structuralism. Phenomenological models are based on empirical data of object behavior. The structural approach consists of the development of models that allow you to describe and explain the phenomena based on the internal structure of the objects. Note that these two approaches are closely interrelated. Therefore, in addition to these two approaches, the approach called structural-phenomenological approach has become widespread. The essence of this approach is that the standard phenomenological equations and criteria in deformed solid mechanics are considered at two levels: microscopic and macroscopic, displaying behavior of non-homogeneous materials as homogeneous materials with effective properties. Theoretical modeling concerns two main points in the SM: a) Constructing models describing the physical state of the material of which the solid consists; b) Setting boundary problems taking into account constitutive (defining) equations of material behavior. The mathematical model of the original object is a system of equations in the general sense of this term. The same mathematical model can describe different physical processes (models). However, when studying the same real objects, fundamentally different models can be used. Therefore, the choice of model is very significant in the researches. The construction of different models of the same object is aimed at describing a different detailing of the processes or phenomena. Relative to the selected system of characteristics, the adequacy of the mathematical model to the studying object (process, phenomenon, etc.) belongs to the basic requirements for the mathematical model. At the same time, adequacy should be considered on two sides: the correct qualitative description of the object by the selected characteristics and the correct quantitative description of the object

4

1 Statements of Boundary Problems of Solid Mechanics

1 Statements of Boundary Problems of Solid Mechanics

5

Fig. 1.1 Qualitative diagram of the relationship of the adequacy and simplicity properties of the model

a problem, an infinite set can become finite and vice versa. Infinity can also result when each element loses its personality. In this case, discretization is replaced by continuity. The concept of “infinitesimal” is also important in the construction of mathematical models of many mechanical processes. It should be noted that there are no wellfounded general considerations and recommendations on the choice of the concepts of “infinitely large” and “infinitely small” and the related concepts of “significantly more”, “significantly less” still are not. Excessively large and excessively small numbers are also of auxiliary importance. An assessment of the accuracy of the solutions is associated with these concepts. For example, it is obvious that it makes no sense to spend effort to determine in the rock massif the area of disturbance with the additional stress state with an accuracy of 1 N. However, at the same time, the difference of 1 mm in displacements is important to estimate the plate stability. An important conclusion from the facts presented is: the choice of main characteristics and their values plays a very important role in the construction of the model. The main characteristics are the spatial and time variables that we use when constructing a model for an object of research. After defining the main characteristics (scales), all variables relative to them can be classified as variables with “normal”, “slow” and “rapid” rate of change. Slowly changing values can be considered parametrically (for example, taken as constant), and rapidly changing values are taken into account by averaging and introducing effective values. Therefore, for example, a variable value is the thickness of the layerlin a layered composite. The thickness of the layer has a characteristic range of change l∗ . The

6

1 Statements of Boundary Problems of Solid Mechanics

composite has a characteristic size d∗ . Then, as one of the main characteristics, it is advisable to take a value equal to the ratio l∗ to d∗ : δl∗ = l∗ /d∗ . In this case, when you build a model, all values are classified as follows: main if their respective parameters (defined similarly δl∗ ) are comparable to δl∗ ; slow if the values of these parameters are much smaller δl∗ ; fast if parameter values are much larger δl∗ . If you want to build more accurately than the original model, you enter clarifying values. The forming process of the main variables can be represented by the diagram shown in Fig. 1.2. Similar reasoning concerns the duration of the effects of disturbances on the mechanical system. Classification of disturbances when taking into account the time of external influence can be performed as follows: • Disturbances change throughout the entire time interval of the mechanical process study; • Influence of disturbance is felt only in the integral characteristic (e.g., dynamic impact, explosion); • Perturbation occurred so long ago that due to the dissipation of energy, its effect in the time interval under consideration is no longer felt. The mathematical modeling process of mechanical processes and phenomena can be structured as follows: • The ultimate aim of the problem is formulated; • The type of problem is set from initial geometric conditions; • Physical model of the object with indication of acting kinematic and force loads is accepted; • Quantitative estimates of mechanical properties of materials and structural features of the objects are determined; • Mathematical formula of the problem is performed, the calculation scheme is built with a setting of initial and boundary conditions; • The method of solving the formulated mathematical problem is chosen; • Direct solution of the model problem is performed; • Analysis and interpretation of the obtained results of the solution are carried out. It should be noted that if the formulas of the main aim of the study is usually not very difficult, since it is dictated by the requests of practice, then the determination of the necessary initial data sometimes requires special studies. Control Questions 1. 2. 3. 4.

Mechanics of deformed solids is a phenomenological science. What does it mean? What are the “principles of continuity and discretization” in SM? What are “active and passive experiments” in SM? How do you understand the terms “theoretical models of the mechanical behavior of material” and “setting a model boundary problem taking into account the model of material behavior”?

1 Statements of Boundary Problems of Solid Mechanics

7

Defining parameters

Defining variables

Defining constants

Main parts of defining variables

“Slow” changing variables

Constant values or values depending on t or x parametrically

“Fast” changing variables

Variables with a “normal” (basic) rate of change

Averaged values of “fast” changing variables

Main variables

a) Corrections that take into account changes in slow variables over time or space

b) Corrections that take into account rapid changes in variables relative to their averages

c) Small parts of defining variables

Clarifying variables

Fig. 1.2 Main variable generation process

8

1 Statements of Boundary Problems of Solid Mechanics

5. How do you understand the requirement of the adequacy of a mathematical model to a real object of research? 6. What are the concepts of “infinitely small” and “infinitely large” in SM? 7. What are the “basic characteristics” of concepts; variables with a “normal”, “slow” and “rapid” rate of change; “clarifying values” in SM? 8. What are the process steps of mathematical modeling of mechanical processes and phenomena?

Chapter 2

Basic Concepts of Stress–Strain State Theory

2.1 Defining Concepts of Continuum Mechanics The mechanical behavior of a solid can be very diverse and complex. In this course, the description of the deformed solid state and behavior is based on the theory of a continuous medium. We will ignore (but not always) the discrete structure of the material. Therefore, it is assumed that the volume occupied by the body is continuously filled with matter. We consider the infinitesimal volume of material as a “particle” of a continuous medium. At the same time, we accept that all individual particles of a continuous medium are absolutely the same and indistinguishable. This assumption is one of the basic concepts of the continuum mechanics. The choice of the continuum model is based on the concepts of elementary volume, the criteria of quasi-continuum, and quasi-heterogeneity. The choice of elementary volume size is more influenced by the structural defects of medium. Classification of volume as elementary volume depends on the size of the object. The question of the legality of using continuum mechanics methods should be solved for the specific volume of each body, taking into account its structural and mechanical features. The criterion of quasi-continuum can be formulated as follows: ΔA < ε

by

Δa < l0 ,

where ΔA is the difference among the values of stresses, strains, or displacements in neighboring points of the medium with an increment of coordinates; Δa, l0 are the characteristic linear sizes of the elementary volume; ε is the permissible error in the determination. The criterion for considering the medium as quasi-homogeneous materials can be taken to exceed the impact area over the size of the inhomogeneous medium by at least a value order in linear dimensions:

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Zhuravkov et al., Mechanics of Solid Deformable Body, https://doi.org/10.1007/978-981-19-8410-5_2

9

10

2 Basic Concepts of Stress–Strain State Theory

Vg / Vn = 103 , where Vg and Vn are the volume of the impact area and heterogeneity, respectively. Deformable solids change sizes and shapes (deformation) under the influence of external forces (loads). Internal forces arise in deformed solids. The magnitude and distribution of internal forces depend on the load and the body’s geometric shape. External loads can be divided into volume, surface, and concentrated loads. The concentrated loads can be considered as a limit case of applying surface loads on a small part of the body surface.

2.2 Stress Tensor We divide the body into two parts by an imaginary section: V 1 and V 2 , (Fig. 2.1a). The surface element of section ΔS with the focal point A has a unit normal vector ν directed to V 1 . The force vector ΔP and the moment vector ΔM represent the action of the part V 1 body at point A on part V 2 . In the limit, when the elementary section aims for point A, ΔS → 0, (with a fixed direction ν), the following physically reasonable assumptions can be made: dP ΔM ΔP = = σν , lim =0 ΔS→0 ΔS ΔS→0 ΔS dS lim

(2.1)

The vector σν (2.1) is called the stress vector at point A. It is important that vector σν acts on the surface element ΔS with the normal direction ν and changes when the direction of this normal to ΔS changes. For stress vectors acting at the same point of the section, but directed in opposite directions, the following equality is true: σv = −σ−v

Fig. 2.1 To the definition of “stress on the site”

(2.2)

2.2 Stress Tensor

11

Fig. 2.2 Normal and tangent components of the stress vector

Equality (2.2) describes the effect of the body part V 2 on the part V 1 and vice versa, (Fig. 2.1b). Remark. The relation (2.2) can be interpreted as a direct expression of the Newton1 third law (the principle of equality of action and opposition). The set of stress vectors σν (A) for all directions ν determines the stress state at point A. In the general case, the stress vector σν (A) is not directed along the normal ν. Its projection to an arbitrary direction (defined by a unit vector) is called the component of the stress vector in this direction. If we decompose the stress vector σν into normal and tangent to the site dS, then we get the normal stress σn and the tangent stress τn , (Fig. 2.2). For modulus of these values, the ratio is true: σ2ν = σn2 + τn2 . It should be noted that the stresses σn and τn are not components of the vector in the usual sense. We will use Cartesian coordinates x 1 , x 2 , x 3 , which we will denote with the symbol x i , while remembering that i and other Latin indices take values of 1, 2, 3. Vectors ei denote the basis vectors of the coordinate system. To describe the stress state at a point of solid, an elementary parallelepiped is cut around this point, the edges of which are parallel to the coordinate axes and have small lengths dx i , (Fig. 2.3a). Due to the small parallelepiped, it can be considered that the stress vectors σi on its edges coincide with the stress vectors on the parallel coordinate sites drawn through the point under consideration. The important assumption that deformed and undeformed solid elements are identical should be noted. This follows from the hypothesis of small strains. The stress vectors σi acting on the parallelepiped faces can be decomposed into normal σii (perpendicular to the faces) and tangent σij (i /= j) components (lying in the plane of the faces), (Fig. 2.3b). 1

Newton Isaac (1643–1727), English mathematician, mechanician; founder of classical mechanics.

12

2 Basic Concepts of Stress–Strain State Theory

Fig. 2.3 Decomposition of the stress vector in the general case

We accept that the first subscript in the symbols indicates the normal of the section on which this stress acts, and the second index indicates the axis on which it is parallel. Tensile normal stresses are positive and compressive stresses are negative. We emphasize that the values σij are not components of the vector in the usual sense. They are measured most often in pascals (1 Pa = 1 N/m2 ). For example, the decomposition of the stress vector σ3 (acting on the face x 3 = constant) along the coordinate axes x i has the form: σ3 = σ31 e1 + σ32 e2 + σ33 e3 =

3 Σ

σ3k ek ≡ σ3k ek .

k=1

We accept, as in the classical course of the continuum mechanics, that when a Latin index is found twice in a single member, a summation of this index from 1 to 3 occurs. We will not write the summation mark itself (this index is called mute). This index can be replaced by any other indices. For example, xi yi = xk yk . If the Greek index is repeated, then it is not summed. The index be found once in the term is called free. With the correct spelling of any formula, each of its terms must have the same free indices. In the general case, for the stress vector σi , the following equality is true:

2.2 Stress Tensor

13

σi = σi j e j ,

(2.3)

where ej is the unit vector of coordinate axis x j . We will actively use the concept of a metric tensor. Therefore, recall that the metric tensor δij is defined as the scalar product of the basis vectors. In Cartesian coordinate system ei · e j = 0, if i /= j, so: ⎧ δi j = ei · e j , δi j =

1, i = j, 0, i /= j.

(2.4)

The matrix of this tensor is diagonal. Generally, the components of the metric tensor (2.4) are called Kronecker 2 symbols. Scalar multiplication of the ratio (2.3) by the basis vector ek gives: σi · ek = σi j e j · ek = σi j δi j . It follows from (2.4) that δkj = 1 only with j = k, therefore, we get: σik = σi · ek .

(2.5)

The nine stress components σij represent, in aggregate, a physical quantity called the Cauchy stress tensor3 (T s ). This is a second-rank tensor (by the number of indices). Write it as a matrix, like this: ⎛

⎞ σ11 σ12 σ13 T s = (σi j ) = ⎝ σ21 σ22 σ23 ⎠. σ31 σ32 σ33 Matrix rows contain stress tensor components parallel to one axis, and columns contain tensor components acting on the same section. Components σi j mean stresses at point N acting in the direction of axis i on a section with an external normal towards axis j. In this case, the stress vector σ→i at point N in the direction of axis i is defined as: σ→i = σ→i {σi1 , σi2 , σi3 }. Recall that the tensor is an invariant object, that is, it does not change when transitioning from one coordinate system to another. Only its components change according to a certain “tensor” law with such a transition. However, in the future, we will sometimes call “tensor” the combination of its components. Different symbols can be used for stress tensor components. Therefore, for example: 2

Leopold Kronecker (1823–1891), a German mathematician who worked on number theory, algebra, and logic. 3 The concept of “tensor” was formed from the Latin “tension”—stress.

14

2 Basic Concepts of Stress–Strain State Theory

⎞ σx x σx y σx z Ts = ⎝ σ yx σ yy σ yz ⎠, σzx σzy σzz where x, y, z are the coordinate axes. The representation of the Karman has the form: ⎛

⎞ σx x τx y τx z Ts = ⎝ τ yx σ yy τ yz ⎠. τzx τzy σzz Thus, the stress state in the body is determined by the field of stress tensors, that is, by setting at each point of the body nine functions σi j (x, y, z, t) (t is time), which are related to the external normal to the site. The stress tensor is symmetrical, i.e., σi j = σ ji . A very important statement is the law of reciprocity: if there are two intersecting elementary sites at the body point, then the projection of stress on the first site to the normal to the second equals to the projection of stress on the second site to the normal to the first (this is a more general statement than the symmetry of the stress tensor).

2.3 Stress Tensor Properties The components of the stress vector σν on an arbitrary inclined site, are carried out through the observation point, and can be expressed through the stresses on the coordinate sites (faces of the elementary parallelepiped) at this point, (Fig. 2.4a).

Fig. 2.4 To define stress on an arbitrary site with normal ν

2.3 Stress Tensor Properties

15

To prove this statement, we consider an elementary pyramid formed by coordinate faces and an inclined site with the normal ν, (Fig. 2.4b), where σνi are projections of a stress vector σν to coordinate axes; S is the area of the inclined site; S i are areas of corresponding coordinate faces (subscript indicates normal to the site), which are related to S by the following relations: Si = Sli , (i = 1, 2, 3),

(2.6)

where l i = cos(ν, x i ) are the directing cosines of the normal ν to the inclined site. We compose equilibrium equations of forces on the faces of the selected pyramid, which are the products of stresses on the faces on the area of the corresponding faces, (Fig. 2.4b): σν1 S − σ11 S1 − σ21 S2 − σ31 S3 = 0, σν2 S − σ12 S1 − σ22 S2 − σ32 S3 = 0, σν3 S − σ13 S1 − σ23 S2 − σ33 S3 = 0. This system of equations takes the form in tensor symbols: σνi S − σ ji S j = 0.

(2.7)

Recall that the summation is made according to the repeated Latin index j, that is, in the case under the consideration:σ ji S j = σ1i S1 + σ2i S2 + σ3i S3 . We substitute the relationship (2.6) into Eq. (2.7). Reducing then by S, we obtain formulas expressing the components of the stress vector on an arbitrary inclined site through coordinate stresses: σνi = σi j l j .

(2.8)

Thus, the stress tensor components at the coordinate sites can fully describe the stress state at the point. The stress vector σν on an arbitrary inclined site drawn through a given point can be represented as a decomposition according to the basis vectors: σν = σνi ei ≡ σν1 e1 + σν2 e2 + σν3 e3 .

(2.9)

The vector modulus σν can be expressed through the sum of the squares of its projections using (2.9): 2 2 2 σν2 = σν1 + σν2 + σν3 .

The appropriate normal σn and tangent τn stresses on the same site are determined as follows, (Fig. 2.2):

16

2 Basic Concepts of Stress–Strain State Theory

σn = σi j li l j = σ11l12 + σ22 l22 + σ33l32 + 2σ12 l1l2 + 2σ23l2 l3 + 2σ31 l3l1 τ2n = σν2 − σn2 .

2.4 Equilibrium Equations of Deformed Solid If the deformable body is in equilibrium, then any of its part should also be in equilibrium. Let the body be loaded with given surface forces Rν and homogeneous volume forces ρF (ρ is the density of the material): ρF = ρFi ei ≡ ρF1 e1 + ρF2 e2 + ρF3 e3 . We assume that the components of stresses σij , as well as their first partial derivatives, are continuous coordinate functions. It is important that the strains are considered small, so equilibrium equations can be formulated for an undeformed body or parts of it. For the selected element with small edges dx i , loaded according to Fig. 2.5, the stress components change accordingly to dσij , at the increment of coordinates by infinitesimal value dxi change accordingly to dx i . For example, a change of the normal stresses σ11 acting along the axis x 1 by incrementing only dx 1 will be as follows: dσ11 = σ11 (x1 + dx1 , x2 , x3 ) − σ11 (x1 , x2 , x3 ) =

∂σ11 dx1 ≡ σ11 ,1 dx1 ∂ x1

The stress σ21 acting along the axis x 1 on the other face is incremented by changing the coordinate x 2 by the value dx 2 : Fig. 2.5 To construct equilibrium equations

2.4 Equilibrium Equations of Deformed Solid

dσ21 = σ11 (x1 , x2 + dx2 , x3 ) − σ11 (x1 , x2 , x3 ) =

17

∂σ21 dx2 ≡ σ21 ,2 dx2 ∂ x2

The stress σ31 is similarly changed: dσ31 = σ31 ,3 dx 3 . Remark. For brevity, the partial differential operation is indicated by a comma before the corresponding lower index:

∂ (...) ≡ (...),i . ∂ xi The remaining stress increments along the coordinate axes are as follows: dσαβ = σαβ ,α dxα , (α, β = 1, 2, 3; no summation by repeating Greek index). We obtain the known law of tangent stress parity σij = σji (i /= j)4 from the condition that the sum of moments of forces acting on the element relative to coordinate axes is zero. This law shows that the stress tensor has only six independent components and its matrix is symmetrical with respect to the main diagonal. Consider the equilibrium of the forces acting on the selected element in the direction of the x 1 axis. We project all the active forces, including volume ones, on this axis. For this, stresses parallel to the x 1 axis are multiplied by the area of the faces on which they act, and the force ρF 1 is multiplied by the volume of the parallelepiped and then summed up: ( ) ) ( σ11 + σ11,1 dx1 − σ11 dx2 dx3 + σ21 + σ21,2 dx2 − σ22 dx3 dx1 + +(σ31 + σ31 ,3 d x3 − σ31 )d x1 d x2 + ρ F1 d x1 d x2 d x3 = 0, from which: σ11 ,1 +σ21 ,2 +σ31 ,3 +ρF1 = 0 We take into account the symmetry of the stress tensor (σij = σji ) and obtain: σ11 ,1 +σ12 ,2 +σ13 ,3 +ρF1 = 0 The other two directions are similar: σ21 ,1 +σ22 ,2 +σ23 ,3 +ρF2 = 0, σ31 ,1 +σ32 ,2 +σ33 ,3 +ρF3 = 0 In general, these three equations can be written in the following tensor form: σi j , j +ρFi = 0 4

Cauchy first proved in 1822 when obtaining equilibrium equations.

(2.10)

18

2 Basic Concepts of Stress–Strain State Theory

Equation (2.10) is the equilibrium equation of deformed solids or static equilibrium equations. It should be noted that in the theory of generalized media, when moment stresses appear (for example, in the theory of the Cosser continuum5 ), the stress tensor is no longer symmetric.

2.5 Equilibrium Conditions at the Boundary Equilibrium Eq. (2.10) is valid everywhere within a deformable solid body. At the boundary, that is, on the surface of the body, the equilibrium conditions in stresses must be satisfied. This means that there must be a continuous transition of the stress tensor to the surface load at the boundary. From (2.2) follows: Rν = −σ−ν = σν , where Rν is the stress vector on the boundary specified on the surface side with the normal ν. We multiply this expression scalarly by ei and use the formulas (2.8). As a result, we obtain boundary conditions in stresses: Rνi = σi j l j

(2.11)

Therefore, stresses at the boundary (surface loads) are in equilibrium with stresses inside the body. Equality (2.11) in coordinate form has the following form: Rν1 = σ11 cos(ν, x1 ) + σ12 cos(ν, x2 ) + σ13 cos(ν, x3 ), Rν2 = σ21 cos(ν, x1 ) + σ22 cos(ν, x2 ) + σ23 cos(ν, x3 ), Rν3 = σ31 cos(ν, x1 ) + σ32 cos(ν, x2 ) + σ33 cos(ν, x3 ). Note that the components of the normal vector are guide cosines. In the future, we will not write the sign “ν” in the lower index.

2.6 Principal Axes and Principal Values of Stress Tensor We take the elementary parallelepiped of a solid in the Cartesian coordinate system and begin to rotate it around a point. The values of the stress tensor components on its faces will change. It has been proved that there is at least one parallelepiped 5

The Cosser brothers first introduced the concept of a generalized media and studied it.

2.6 Principal Axes and Principal Values of Stress Tensor

19

Fig. 2.6 To define principal axes and principal values of stress tensor

position at which tangent stresses on its faces turn to zero, and normal stresses become extreme, (Fig. 2.6). Coordinate axes at a given point in this position are called the principal axes of the stress tensor. Stress components in these axes denote σ1 , σ2 , σ3 and are called the principal values of the stress tensor. The main axes are numbered so that in the algebraic sense the following conditions are fulfilled for the principal values: σ1 ≥ σ2 ≥ σ3 . Matrix of the stress tensor in principal axes takes diagonal form. The components of the stress vector on the inclined site (2.8) in the principal axes are as follows (without summation by i): σνi = σi li .

(2.12)

Consider the procedure for determining the principal stresses values σ1 , σ2 , σ3 from the known values of the six-coordinate components of the stress tensor σij . We use representations (2.5) and Fig. 2.4. Let the inclined site be the principal one. Then the vector of total stress σν is directed normal ν to this site, since there are no tangent stresses, (Fig. 2.7). We denote the algebraic value of this principal stress as σ. Then the projections of vector σν on the coordinate axes will be as follows: σνi = σli . We substitute these expressions into relations (2.8) and transfer all the terms to the left. As a result, we get such a system of equations: Fig. 2.7 Direction of stress vector at principal site

20

2 Basic Concepts of Stress–Strain State Theory

σi j l j − σli = 0.

(2.13)

We represent the term σli in Eq. (2.13) in the form: σli ≡ σδi j l j Remark. This operation (li ≡ δij l j ) is called “juggling” by indices using a metric tensor. We check this expression, for example, with i = 1: σl1 = σδ1 j l j = σδ11l1 + σδ12 l2 + σδ13l3 = σδ11l1 = σl1 , where δ12 = δ13 = 0, δ11 = 1, due to (2.4). We now substitute these relations into Eq. (2.13). As a result, we obtain a system of three homogeneous linear algebraic relations relative to unknown cosines of the normal, which determine the orientation of the principal site in the original coordinate system: ( ) σi j − σ δi j l j = 0 (i, j = 1, 2, 3),

(2.14)

If the solution of the received system is unique, it will be zero: li = 0, (i = 1, 2, 3). But a zero solution does not make physical sense. For the existence of a non-zero solution, it is necessary and sufficient that the determinant of the system is zero: | | | σ13 || | | | σ11 − σ σ12 |σi j − σδi j | ≡ | σ21 σ22 − σ σ23 | = 0 | | | σ σ32 σ33 − σ | 31

(2.15)

This is achieved by the proper (correct) selection of the value σ. If the condition (2.15) is satisfied, then at least one of the three equations of the system (2.14) is not independent and is linearly expressed through the other two. Add this condition to independent relations: li li = 1, or l12 + l22 + l32 = 1 As a result, we get a new system of three independent equations. This system is sufficient to find cosines of the normal of the principal site, on which the stress equals to σ. The equation for determining the principal stresses values is the determiner (2.15). We disclose this equation and arrange the terms according to the power σ. As a result, we obtain a cubic equation, which is called the age-old equation: σ3 − σ2 J1 + σ J2 − J3 = 0,

2.7 Maximal Tangent Stresses

21

where: J1 = σii ≡ σ11 + σ22 + σ33 = σ1 + σ2 + σ3 , J2 = (σii σ j j − σi j σi j )/2 2 2 2 = σ11 σ22 + σ22 σ33 + σ33 σ11 − σ12 − σ23 − σ31 = σ1 σ2 + σ2 σ3 + σ3 σ1 , | | | | || || | σ11 σ12 σ13 | | || || J3 = σi j ≡ | σ21 σ22 σ23 || = σ1 σ2 σ3 |σ σ σ | 31 32 33

(2.16)

You can see that all the three roots of an age-old equation are real. They are the principal values of the stress tensor and are indicated by σ1 , σ2 , σ3 . Their values are determined by the nature of the external load and do not depend on the initial orientation of the coordinate system. That is, the principal stresses are invariant with respect to the transformation of the coordinate system. Therefore, the values of the coefficients J 1 , J 2 , J 3 in the age-old Eq. (2.16), must remain unchanged when the axes are rotated. In this regard, they are called the stress tensor invariants. Planes perpendicular to the principal axes of the stress tensor is called the principal planes. We will follow the generally accepted rules and consider stresses positive if they are tensile. In addition, we will number the principal stresses in descending order of their values, that is: σ1 ≥ σ2 ≥ σ3 . In the principal axis coordinate system, the normal stress vector σ→n = n , x) + σ→ y cos(→ n , y) + σ→z cos(→ n , z) is collinear with the external normal σ→x cos(→ vector n→ to the site. If the direction is the principal one, the tangent stress σ→nτ is zero on the site perpendicular to it. Stress tensor in principal axes has the form: ⎛ ⎞ σ1 0 0 [ ] σi j = ⎝ 0 σ2 0 ⎠, 0 0 σ3

(2.17)

The stress state is called three-dimensional (triaxial) stress state, if all principal stresses are not zero. The stress state is called two-dimensional (plane) stress state, if one of the principal stresses is zero. The one-dimensional (linear) stress state occurs if two principal stresses equal to zero at the same time.

2.7 Maximal Tangent Stresses Sites are called octahedron, if they are equally inclined to the principal axes of the stress tensor.

22

2 Basic Concepts of Stress–Strain State Theory

Fig. 2.8 Arbitrary site with stresses in main axes

Let’s build expressions for stresses on octahedral sites. However, first, we consider an arbitrary site in the principal axes, (Fig. 2.8). We denote the stress vector on this site as σν , which was decomposed into the normal σn and the tangent τn components: σν2 = σn2 + τ2n

(2.18)

The projections of vector σν on the principal axes, in accordance with formulas (2.12), take the form: σvα = σα lα , where l α are guiding cosines of the normal to the site. Then the modulus of the total stress vector on this site can be obtained from the sum of the squares of the projections, and the normal stress value σn is obtained as the sum of the projections of total stress components on the normal: 2 2 2 σv2 = σv1 + σv2 + σv3 = σi2 li2 ,

σn = σv li = σ1l12 + σ2 l22 + σ3l32 = σi li2

(2.19)

Substituting the resulting expressions for σν 2 and σn into Eq. (2.18) and considering that the sum of the cosine squares li l i = 1, we obtain the following expression for tangent stresses on an arbitrary site: τ2n = (σ1 − σ2 )2 l12 l22 + (σ2 − σ3 )2 l22 l32 + (σ3 − σ1 )2 l32 l12 .

(2.20)

As can be seen from (2.20), τn2 is the positive value and turning to zero at the principal sites. Indeed, if the normal ν coincides with one of the principal axes, then one of the guide cosines equals to a unit, and the other two equal to zero, and then τn2 = 0. Now let’s look at octahedral sites. Since they are equally inclined to the principal axes, all directional cosines for them are: li 2 = 1/3. We substitute these values l i 2 = 1/3 into formulas (2.19), (2.20) and obtain the values of normal and tangent stresses on octahedral sites:

2.7 Maximal Tangent Stresses

23

σoct = (σ11 + σ22 + σ33 )/3 = (σ1 + σ2 + σ3 )/3, τoct =

(σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 /3

(2.21)

The formula for τoct through arbitrary coordinate components of stresses on the basis of (2.21) is as follows: τoct =

/

2 2 2 (σ11 − σ22 )2 + (σ22 − σ33 )2 + (σ33 − σ11 )2 + 6(σ12 + σ23 + σ31 )/3.

Octahedral stresses are invariant values, which are expressed through stress tensor invariants: / σoct = J1 /3, τoct = 2J12 − 2J2 /3. These stresses were introduced by Nadai. They not only play an important role in the calculation of equivalent stresses for a complex stress state, but also are used to formulate plasticity criteria. We will determine the sites on which maximal stress occurs. The position of these sites can be determined by examining the extremum of the expression (2.20). Because: (σ1 − σ3 ) = (σ1 − σ2 ) + (σ2 − σ3 ), and the square of the number is more than the sum of squares of numbers of its components, so: (σ1 − σ3 )2 ≥ (σ1 − σ2 )2 + (σ2 − σ3 )2 Therefore, τn reaches the maximum on those sites on which / the third of the terms in formula (2.20) is maximal. Therefore, with l12 = l32 = 1 2, l22 = 0 we get that τn = τmax (index 2 shows the axis which the site is parallel): / τmax = τ2 = (σ1 − σ3 ) 2

(2.22)

Thus, the site with maximal tangent stress is parallel to the principal axis 2 and is equally inclined to the principal sites on which the maximal (σ1 ) and minimal (σ3 ) principal stresses act, (Fig. 2.9). If you consider two more sites that are parallel to the first and third principal axes, respectively, and to the other two are equally inclined, then we get two more extreme values of tangent stresses: / / τ1 = (σ2 − σ3 ) 2, τ3 = (σ1 − σ2 ) 2

24

2 Basic Concepts of Stress–Strain State Theory

Fig. 2.9 Site with maximal tangent stresses

The values of τ1 , τ2 , τ3 are called the principal tangent stresses. For them, the ratio is correct: τ1 + τ2 + τ3 = 0. It should be noted that the surfaces on which the principal tangent stresses are may not be mutually orthogonal. They form the sides of a regular dodecahedron. Its sides are not free from normal stresses. On sites where the principal tangent stresses act, normal stresses are: σn1 =

σ2 + σ3 σ3 + σ1 σ1 + σ2 , σn2 = , σn3 = 2 2 2

2.8 Stress Deviator and Spherical Part of Stress Tensor A stress tensor exists at each point in the solid. Therefore, there is a field of stress tensors in the body. The two properties of tensors that we will need in the future should be noted: • The sum of two tensors is a tensor, the components of which are the sum of the corresponding components of the term tensors; • The product of the tensor per scalar λ is a tensor whose components are λ times larger than the corresponding components of the multiplied tensor. Let’s consider the stress state, in which there are only three identical principal stresses σ, equal to the average stress at a given point of the body, on three mutually perpendicular sites: σ1 = σ2 = σ3 = σ = σii /3 ≡ (σ11 + σ22 + σ33 )/3. This stress state is described using the tensor: ⎛

⎞ σ00 Tσ = ⎝ 0 σ 0 ⎠ 00σ

(2.23)

2.8 Stress Deviator and Spherical Part of Stress Tensor

25

The tensor (2.23) is called a spherical stress tensor (the spherical part of a stress tensor). Subtract the spherical tensor from the stress tensor. As a result, we get a new tensor, called a stress deviator. ⎛ ⎞ σ11 − σ σ12 σ13 ( ) Dσ ≡ si j ≡ ⎝ σ21 σ22 − σ σ23 ⎠ σ31 σ32 σ33 − σ

(2.24)

Thus, the stress tensor at each point can be represented as the sum of the spherical stress tensor (σ) and the stress deviator (sij ). For their components, such ratio is performed: σi j = si j + σδi j (i, j = 1, 2, 3),

(2.25)

where δij is the Kronecker delta (2.4). The decomposition of the stress tensor into the spherical and deviated parts is of great importance when studying the behavior of elastic and plastic bodies under loads. The spherical part extracts a uniform tension or compression from the stress state, in which only the volume of this body element changes without changing its shape. The stress deviator characterizes the shear state in which the shape of the element changes without changing its volume. Therefore, the stress deviator indicates the deviation of the stress state from the all-around tension (compression) or the deviation of the deformed body shape from the original. Some examples of the simplest stress states. 1. Uniform all-round tension (compression) of the body. In this case, the stress deviator is zero and the spherical tensor is different from zero. Vectors σ→n and n→ are collinear. The principal normal stresses equal to each other. 2. Tension (compression) in one direction only. In this case, the stress tensor is uniaxial, i.e., for example, σx x /= 0, and the remaining components are all zero. If σx x > 0, then the stress tensor is the simple tension tensor. If σx x < 0, then the stress tensor is the simple compression tensor. If n→ is collinear with the x axis, then σ→n is also collinear with the same axis. If the vector n→ is normal to the axis x, then σ→n = 0. The other two principal normal stresses are identical to zero. 3. Plane stress state. Let σzx = σzy = σzz = 0. If the vector n→ is collinear with the axis z, then σ→n = 0. One of the principal stresses is zero, and the stress matrix has the form:

26

2 Basic Concepts of Stress–Strain State Theory

⎞ σx x σx y 0 Ts = ⎝ σ yx σ yy 0 ⎠ 0 0 0 Let’s build invariants for the spherical tensor and stress deviator. The first invariant of the spherical tensor is the same as the first invariant of the stress tensor: J1s = 3σ ≡ σ11 + σ22 + σ33 The first invariant of the stress deviator is zero, since: J1d = sii ≡ (σ11 − σ) + (σ22 − σ) + (σ33 − σ) = σ11 + σ22 + σ33 − 3σ = 0 We use the expression for the second invariant of the stress tensor (2.16), substituting instead of the stresses σii of the difference σii –σ, to determine the second invariant of the stress deviator. After simple transformations, we get: [ J2d = −si j si j /2 = − (σ11 − σ22 )2 + (σ22 − σ33 )2 + (σ33 − σ11 )2 )] [ ] ( 2 2 2 /6 = − (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 /6 +6 σ12 + σ23 + σ31 The third invariant of the stress deviator has the form: J3d = σi j σ jk σki /3 In the theory of plasticity, the concept of stress tensor intensity, which was introduced by Ilyushin, is widely used. Stress tensor intensity is formally defined through the second invariant of the stress deviator: ( / ) σu2 = −3J2d = 3 2 si j si j It follows that: √ / 2 2 2 2 σu = (σ11 − σ22 )2 + (σ22 − σ33 )2 + (σ33 − σ11 )2 + 6(σ12 + σ23 + σ31 ) 2 √ / 2 = (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 (2.26) 2 The stress tensor intensity is an invariant value, since it is expressed through the second invariant of the stress deviator. It is exactly the same as the octahedral tangent stress (2.21). The numerical coefficient in formula (2.26) is chosen so that in the case of simple tension or compression (σ11 = σ1 , all other components equal to zero), the condition σu = |σ1 | is fulfilled.

2.8 Stress Deviator and Spherical Part of Stress Tensor

27

Positive value: s=

√ si j si j =

√ / 3 (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 , 3

(2.27)

is called a stress deviator modulus. The guide tensor of the stress deviator is called a tensor, each component is divided into a modulus of the deviator (2.27): ⎛

⎞ (σ11 − σ )/s σ12 /s σ13 /s Dσ ⎠ = ⎝ σ21 /s Dσ ≡ (¯s i j ) = (σ22 − σ )/s σ23 /s s σ32 /s (σ33 − σ )/s σ31 /s

(2.28)

Since the first invariant of the deviator is zero, then: s ii = s 11 + s 22 + s 33 = 0. At the same time: s i j s i j = 1. Therefore, the guide tensor is completely characterized by four numbers, since its six components are already connected by the two given relationships. It should be noted that the principal axes of the guide tensor coincide with the principal axes of the stress tensor and the stress deviator tensor. Using the definition of the guide tensor, the components of the stress tensor can be represented in the following form: σi j = σδi j + s s i j Two modulus σ and s determine the scalar properties of the material. Its vector properties are defined by a guide tensor ( s i j ). Loading at a given point in the body is called simple, if all components of the stress tensor change it in proportion to one common parameter λ, that is σi j = λσi0j . In that case, the components of the guide tensor remained unchanged. In other cases, loading is called complex. A few words about an additional physical parameter—the Nadai-Lode parameter μσ , characterizing the stress state in three-dimensional deformable solids. As it is known, the three-dimensional stress state can be reduced to several types of the generalized stress state. This can be done using the Nadai-Lode parameter, which characterizes the volumetric stress state in deformable bodies. The Nadai-Lode parameter μσ is defined by this expression: μσ =

2σ2 − σ1 − σ3 , −1 ≤ μσ ≤ 1 σ1 − σ3

The Nadai-Lode parameter varies from “–1” to “1” and characterizes the type of volumetric stress state: • If μσ ∈ [−1; −0.5[—this corresponds to a generalized tension state; • If μσ ∈ [−0.5; +0.5]—this corresponds to a generalized shearing state; • If μσ ∈ ]0.5; +1]—this corresponds to a generalized compression state.

28

2 Basic Concepts of Stress–Strain State Theory

Using the parameter μσ , the stress tensor (2.17) can be decomposed into several components. For example, the stress tensor (2.17) can be written as a sum of the three simple stress states: ⎡ ⎡ ⎤ ⎤ ) 100 10 0 σ1 − σ3 ⎣ σ1 + σ3 σ1 − σ3 + μσ TH = (1 − μσ )⎣ 0 0 0 ⎦ 0 1 0⎦ + 2 2 2 0 0 −1 001 ⎡ ⎤ 00 0 σ1 − σ3 ⎣ + 2μσ 0 0 0 ⎦, 2 0 0 −1 (

where the first tensor characterizes hydrostatic compression, the second is the pure shear, and the third tensor is the uniaxial compression at μσ > 0 or uniaxial tension at μσ < 0.

2.9 Strains and Displacements Under the influence of external forces, all solids change their shapes (deformed). Solid points change the position in space continuously. Solid does not contain breaks and voids after deformation (that is, no cracks occur). First, we define what is meant by the terms, “deformation” and “strain”. Deformation indicates a change in solid geometry. This is a general concept for various types of deformations: stretching, compression, twisting, bending, crushing, etc. Strain means a deformation measure, or the degree of deformation. Consider, for example, a material sample with an initial length of L0. Stretch the sample, that is, increase its length by a value of ΔL to the length L 1 L 0 + ΔL. Strain is called dimensionless quantity ε: ε=

ΔL L1 − L0 = L0 L0

It should be noted that strain is not additive. That is, if we add up the strains, then the error accumulates. For example, the first experimenter stretched a sample with a length of 10 cm / by 1 cm. Strain equals to ε1 = 1 10. The sample length is now 11 cm. Another experimenter does not know about / the sample history and stretch it by another 1 cm. The stain in this case is ε2 = 1 11. Then the total strain is: / / ε = 1 10 + 1 11

2.9 Strains and Displacements

29

Fig. 2.10 To define a displacement vector

However,/this is less than the strain obtained by this sample from its original length ε = 2 10. Now introduce the concept of “displacement” for a deformed solid. A vector u→, having a start at point A of an undeformed body and an end at the corresponding point A’ of the deformed body is called the full displacement vector of the point, (Fig. 2.10). Its projections on the coordinate axis: u→ = (u 1 , u 2 , u 3 ), u i ≡ u i (x), x ≡ (x1 , x2 , x3 ) The displacements thus defined for most deformed solids are small, compared to the geometric sizes of the body. It should be noted that, as a rule, we consider kinematically unchanged systems in solid mechanics. That is, the movement of a body in space is not considered as a rigid whole. Based on the general statements of the theory of tensor field, the increment of the displacement vector d u→ can be represented as: [ ]/ ← d u→ = (E · d r→) + rot→ u×d r 2,

(2.29)

where E is the symmetric strain tensor, the components of which are associated with the components of the displacement vector u→ according to known formulas (see below). The second term in (2.29) describes the part of the point A’ displacement relative to A that occurs when the small element containing point A rotates as a rigid body around point A. Thus, the “pure deformation” (strain) is described using the first term in the expression (2.29), i.e., the symmetric tensor E. The method of strain determining is that we calculate changes in the lengths of linear elements, as well as changes in the angles between two linear elements by the point displacements. We get the formulas linking the point A displacement and the deformation in its vicinity.

30

2 Basic Concepts of Stress–Strain State Theory

Fig. 2.11 To define Solid strains

Let’s first consider a plane problem, that is, accept that: u 3 (x) ≡ 0, u 1 ≡ u 1 (x1 , x2 ), u 2 ≡ u 2 (x1 , x2 ) After the deformation, segment AB with projections dx 1 , dx 2 occupies position A’B’, (Fig. 2.11). The full displacement of point A along the axis x 1 is u1 . Point B will move along with the point A by u1 plus additional displacement du1 due to the deformation of segment AB along the axis x 1 . Since the displacement increment is small and du1 = du1 (x 1 , x 2 ), so with accuracy to terms of the higher order of small: du 1 =

∂u 1 ∂u 1 ∂u 1 dx1 + dx2 ≡ dx j ≡ u 1, j dx j ( j = 1, 2), ∂ x1 ∂ x2 ∂x j

where the first term corresponds to the component dx 1 extension along the axis x 1 . The second term (since the deformations are small) describes the displacement due to the rotation of the segment AB relative to the axis x 2 by an angle α1 : α1 ≈ tgα1 =

∂u 1 ≡ u 1 ,2 ∂ x2

We consider the point B displacement along the axis x 2 , by analogy. As a result, we get: du 2 =

∂u 2 ∂u 2 dx1 + dx2 ≡ u 2, j dx j , ∂ x1 ∂ x2

α2 ≈ tgα2 =

∂u 2 ≡ u 2,1 ( j = 1, 2). ∂ x1

Values describing linear extensions along axes are called linear strains and are denoted by: ε11 =

∂u 1 ≡ u 1,1 , ∂ x1

ε22 =

∂u 2 ≡ u 2 ,2 ∂ x2

2.9 Strains and Displacements

31

To describe shear strains, a value equal to half of the total change of right angle among coordinate axes is used: / / ε12 = (α1 + α2 ) 2 = (u 1 ,2 +u 2 ,1 ) 2 If the deformation of the neighborhood of point A is spatial (ui (x)/= 0, i = 1, 2, 3), then the increment of displacements will be: du i = u i , j dx j

(i, j = 1, 2, 3)

and such strains are added: ε33 = u 3,3 ,

) ( ε32 = u 2,β + u 3,2 /2,

) ( ε31 = u 3,1 + u 1,3 /2

As a result, it is obtained that in the general case the small strains (u i2 , j 3 2 k, i.e., when the shear modulus is more than 50% of the volume strain modulus. Remark. All known theoretical and experimental approaches to create composites with ν < 0 can be divided into two categories: either all components of the composite have ν > 0 or some of them are auxetic. Elastic properties together with material density ρ define dynamic properties of material—the ability of distribution of elastic strain waves or elastic vibrations in materials. Dynamic properties are characterized by the propagation velocity C, acoustic resistance Q and absorption coefficient α of elastic waves.

56

3 Material and Solid Mechanical Characteristics (Properties)

Among the various types of elastic oscillation in solids, the most interesting are longitudinal waves (waves of compaction-rarefaction, P-waves), transverse waves (shear strain propagation waves, S-waves) and surface waves (Rayleigh waves). In the longitudinal waves, the direction of particle oscillation coincides with the direction of wave propagation. In transverse waves, the direction of particle oscillation is perpendicular to the direction of wave propagation. Surface waves are oscillations of the solids’ surface. Longitudinal wave propagation velocity: / CP =

λ + 2μ = ρ

/

E(1 − ν) . ρ(1 + ν)(1 − 2ν)

Transverse wave propagation velocity: / CS =

μ = ρ

/ E . 2ρ(1 + ν)

Surface wave propagation velocity: C R = Kν CS , where K ν is the dimensionless coefficient depending on Poisson’s ratio (for example, at ν = 0.25, K ν = 0.9194; at ν = 0.5, K ν = 0.9553). The propagation velocities of these waves are characterized by the following inequality: C P > CS > C R . The product of the material density by the velocity of the corresponding wave is called the acoustic resistance or acoustic rigidity. Q = ρ · C. The acoustic resistance characterizes the effect of the medium properties on the intensity I (frequency) of oscillations in this medium. The velocity of longitudinal elastic waves is the most common characteristic. The velocity of elastic waves in materials increases with the increase of compressive loads. For materials having a small elastic limit, elastic waves are acoustic oscillations. Two types of waves are possible in a limitless medium: longitudinal and transverse waves. The absorption of mechanical energy is impossible in an ideal elastic body, only the dissipation of mechanical energy. The mechanical characteristics of most materials depend on the speed at which the load acts. The strength and yield limits of most materials are several times higher than that in static tests with a small number of dynamic loading cycles. At the same time,

3.2 Classification of Materials by the Nature of Deformation

57

with multiple loading with a load varying in value and in sign, destruction can occur at stresses much lower than static strength. With a high-speed dynamic loading (for example, blow-up), the failure mechanism has a different character from considering these processes from the point of view of “classical representations”. In this case, there are complex wave processes: first, there is the material hardening and then (with repeated periodic loads) there is the material fatigue destruction. It should be noted that the dynamic modulus of elasticity is less dependent on the experimental conditions than the static modulus of elasticity. Therefore, the dynamic modulus of elasticity more reflects the properties of the samples as a material, wherein the dynamic modulus of elasticity increases as the samples increase and approaches the calculated effective modulus.

3.2 Classification of Materials by the Nature of Deformation Materials are divided into several basic types depending on their response to external influences. First of all, a large group is composed of materials that behave as elastic media under the loading action. A perfectly elastic material is characterized by direct proportionality between the applied loads and the deformations. A large number of materials exhibit rheological properties that characterize the growth of deformations (absolute and relative deformations) in time under constant external loads (creep) or a gradual stress decrease under constant deformations supported by constant loads (stress relaxation). Stress relaxation is a reverse creep process. During relaxation, elastic strains in the material gradually change over time into plastic ones, but the total strains do not change with time. The stresses decrease in this case. The strength and elasticity of natural materials under the prolonged action of large loads are reduced, asymptotically approaching some limit values—the limit of long-term strength σ∞ and the long-term limit of modulus of elasticity of long-term elasticity E ∞ . For most materials: σ∞ = (0.7 − 0.8)σ; E ∞ = (0.65–0.95)E. When stresses are greater than the elastic limit, both elastic and plastic strains appear. The plastic strains remain after the body is not loaded. The plastic properties can be characterized by a plastic coefficient, which is the ratio of the work W pl spent on breaking a given material volume to the work Wel needed to break the same material volume with the same value of compressive strength value under the assumption of the material ideal elasticity: / K pl = W pl Wel .

58

3 Material and Solid Mechanical Characteristics (Properties)

Such characteristic as brittleness is important for a large class of materials. Brittleness is the material ability to destroy under the influence of applied loads without significant residual (plastic) strains. Brittleness can be characterized by a brittleness coefficient, which is the ratio of the work Wel taken to deform a material sample to the elastic limit to the total work W f r to destroy a sample: / K f r = Wel W f r . A characteristic property of most natural materials is their high degree of heterogeneity and anisotropy. A material is called homogeneous material if its physical properties are the same at all the points. A material is called isotropic material with respect to a property, if this property at a point is the same in all directions. If the properties of a material at a point depending on the direction, then it is called anisotropic material. Real materials can be considered quasi-isotropic in many applications.

3.3 Complete Diagrams of Deformation and Fracture

59

Fig. 3.3 Diagrams of specified loading and deformation modes. a—specified loading mode on sample (R = P, Δl = f 1 (P), Δl does not affect P); b—mutual deformation mode on sample (R almost does not affect Δl 1 ); c—interaction deformation mode (Δl = f 1 (Q, R)); d—combined deformation mode (Δl = f 1 (P, Q, R))

Fig. 3.4 Scheme of the complete rock deformation diagram

layer. But at the same time, the underlying layer exerts opposition (pressure) on the upper one. In practice, cases of a combined mode of “operation” are very common, when at the same time there are mutual deformation mode and specified loading mode (Fig. 3.3d). It should be noted that traditional tests of rock samples for the specified loading mode often do not correspond to the real mechanical state of rocks in the massif, where the given deformation mode mainly takes place. In traditional methods for testing building materials and in real building structures, as a rule, the material “works” in the given loading mode.

60

3 Material and Solid Mechanical Characteristics (Properties)

The methods of testing materials (especially natural materials), developed in recent years, allow you to build complete deformation diagrams. These diagrams characterize the mechanical state of the sample in all loading areas: up to strength (pre-limit branch of the diagram); at the level of strength; beyond the strength, including the section of residual strength (beyond the limit branch of the diagram). It should be noted that complete sample deformation diagrams reflect, as a rule, “instant” mechanical states and do not characterize the material rheological properties. Studies of the material behavior and properties at the pre-limit stage of deformation were carried out quite a lot. Similar studies for the “over-limit deformation section” are single. However, the practical importance and relevance of these studies are evident. For example, in an over-limit deformation branch, the solid takes on a block structure. Let’s consider the complete deformation diagram using the example of a rock sample deformation. The deformation diagram under the uniaxial compression becomes more informative when it is supplemented by the graphical dependence of compressive stresses not only on longitudinal strains, but also on transverse strains, (Fig. 3.4). We consider the case of uniaxial compression. We use the following symbols: σ1 is compressive normal stress; e1 is longitudinal principal linear compression strain; e3 is transverse principal linear strain, having negative values. With the increase of the external compressive load to the level σ1a , there are processes of closing microcracks and pores, which is accompanied by the increase of the effective modulus of elasticity E e f to the value of the modulus of the material elasticity E, i.e., in this section, the dependence σ1 (e1 ) is non-linear. The relationa lesser ship between σ1 (e3 ) and the strain level e3a is also nonlinear, but | to / | extent than σ1 (e1 ). At this stage, the transverse strain coefficient β = |Δe3 Δe1 | (where Δe | 1 , Δe / 3 are| increments of corresponding strains) increases to a constant value |Δe3a Δe1a |. The volume of the sample decreases as a result of its compression. From the level of compressive stresses σ1a to the level σ1b , there is an elastic compression of the sample with a constant modulus of elasticity E. Experimental relationships between σ|1 (e1 )/and σ|1 (e3 ) are linear at this stage. The transverse deformation coefficient β = |Δe3 Δe1 | < 0, 5 and remains constant up to the strain e1b and e3b , that is, it represents the Poisson’s ratio at this stage. The volume of the sample continues to decrease. At a stress level greater than σ1b , the formation of microcracks begins. Thus, for example, the first step of the sample breaking may begin to be realized. In this case, dependence σ1 (e1 ) remains linear, and dependence σ1 (e3 ) (due to the formation and opening of vertical microcracks) becomes non-linear. As compressive stress σ1 increases, the increase rate of transverse strain e3 is faster than that of longitudinal strain e1 , i.e., the coefficient of transverse strain β increases, and the decrease in sample volume slows down. If the external load does not increase in the future, then the formed microcracks will stop their growth, and when the load is removed, they will close.

3.3 Complete Diagrams of Deformation and Fracture

61

62

3 Material and Solid Mechanical Characteristics (Properties)

With a rigid loading mode, successive steps of the sample breaking can be possibly observed at this stage, which is practically impossible to do with a soft loading mode when the breaking is dynamic. At the beyond-limit deformation section de, the bearing capacity of the sample is accompanied by an increase in the sample transverse strain |e3 |, causing an increase in the transverse strain coefficient β to values greater than one. Control Questions 1.

What is the difference between the Mechanical properties of samples and mechanical properties of the material itself? 2. What are the main groups of mechanical properties of materials? 3. What are the basic strength properties of materials? 4. What are the basic elastic properties of materials? 5. What are auxetics (auxetic materials)? 6. What are the dynamic material properties? 7. What is the material classification according to the nature of the “response” to external mechanical loading? What are the terms of “rheology” and “plasticity”? 8. What are isotropy and anisotropy with respect to some mechanical properties? 9. What are the various modes of mechanical loading of samples (specified loading mode and specified deformation mode). Give some examples. 10. What is the complete diagrams of deformation and fracture at uniaxial compression and tension? 11. What are the definitions of the modulus of elasticity and modulus of deformation?

Chapter 4

Construction of Mathematical Model Problems

4.1 The System of Equations to Describe the Medium Stress–Strain State The mathematical condition of the physical (in-situ) existence of a solid is the continuity equation: ∂ρ + div(ρ→ v ) = 0, ∂t

(4.1)

where ρ is the medium density; v→ is the medium velocity. Remark. div(ρ→ v ) = ∂∂x ρvx + ∂∂y ρv y + ∂∂z ρvz . The differential equilibrium equations of a continuous medium in stresses σi j , j +ρ Fi = 0 are performed at any continuous strain of all continuous media. However, various real deformable bodies behave differently under the same external loads. Thus, these equations with the addition of appropriate boundary conditions are not sufficient to describe the specific solid motion. Such system of equations is not complete. Therefore, to determine the stresses, strains, and displacements of a solid under the external loads, equilibrium equations, boundary conditions, strain compatibility equations and Cauchy equations are not enough. To construct a closed and complete system of resolving equations, it is still necessary to have physical relations between stresses and strains, taking into account the behavior peculiarities of materials. It should be noted that for a large number of tasks, it seems appropriate and justified to use the hypothesis of small strains. This hypothesis allows you to limit yourself to the geometrically linear setting of problems, that is, to exclude second and higher degrees of strains from the equations, which greatly simplifies analytical studies. However, the hypothesis of small strains does not contradict reality, given that the displacements in most cases are negligible compared to the size of the bodies themselves. Saint-Venan principle is important for a wide class of solid mechanics

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Zhuravkov et al., Mechanics of Solid Deformable Body, https://doi.org/10.1007/978-981-19-8410-5_4

63

64

4 Construction of Mathematical Model Problems

problems. This principle is that the forces applied to the small surface of the body produced only local stresses and strains. Thus, a complete system of equations constructed by a mechanical-mathematical model describing the states of a solid mechanical object includes: • • • •

Differential equilibrium equations of a continuous medium; Boundary conditions; Physical relations between stress and strain states of solid; Strain compatibility equations.

4.2 Physical Relationships Determining the Solid Behavior Since a huge number of different materials are available, many physical laws describing their behavior under various loading conditions are also possible. Unfortunately, it was impossible to establish a universal physical law describing the behavior of all materials. Therefore, we seek to establish relationships that can describe the most important types and behaviors of materials under certain loadings and deformation conditions. As a result, a specific mathematical model is introduced. Model type is significantly determined by experimentally established relations between stress and strain states at points of deformable medium. The validity and correctness of the model describing the real material state and behavior are tested experimentally. Structural models are often used to describe the solid behavior and build mathematical models based on them. A large number of structural models are known. Examples of the most famous structural models are given in Fig. 4.1. Among them, for example: the elastic Hooke model (Fig. 4.1a), the viscous model (Fig. 4.1b), the viscous-elastic relaxing model (Fig. 4.1c), the viscous-elastic Kelvin-Foigt model (linear elastic-late) (Fig. 4.1d), the Hohenemser-Prager model (Fig. 4.1e), Point-Thompson model (Fig. 4.1i), rigid-plastic model (Fig. 4.1g), elastic–plastic model (Fig. 4.1j), viscoelastic model of Shvedova-Bingam (Fig. 4.1h), model with the connection of serial elements (Fig. 4.1k), model with elastic delay and stress relaxation (Fig. 4.1l) etc. Thus, the process of material loading behavior can be mathematically described using the abstract representation of deformable materials in the form of some structural models consisting of elementary structural units, each of these units is a certain element (elastic, plastic or viscous). Structural units with elastic properties are similar to springs with the Hooke deformation law: σ = E · ε, where σ is stress; ε is strain; E is modulus of elasticity (proportionality coefficient). Structural units with viscous properties correspond to the Newton’s deformation law:

4.2 Physical Relationships Determining the Solid Behavior b

а

d c

e

i g

j

h

Fig. 4.1 Examples of some structural models

65

66

4 Construction of Mathematical Model Problems k

l

Fig. 4.1 (continued)

σ = η · (dε/dt), where dε/dt is the strain rate; η is the viscosity coefficient. Remark. They are represented by a damper with holes in a cylinder containing viscous liquid. The plastic properties of structural units are modeled by dry friction. In this case, deformation can occur only at stresses exceeding a certain value σm called the yield strength. The rigid-plastic element can be characterized by cohesion (C) and internal friction angle (coefficient) (tgϕ). The deformation of materials in accordance with different laws can be reflected by an appropriate combination of different elements in the structural model. Typically, the views of structural model and parameters of model elements are set according to special experiments, obtaining a family of curves corresponding to different levels of active stresses. There are some principal differences among elastic, plastic, and rheological models of solid behavior. • The state of elastic body after the load change is uniquely determined by the new conditions and does not depend on the previous one. • Plastic models “remember” the history of previous loads. But since the change in states occurs instantly, in plastic models, this story is not connected with time in any way. • If the plasticity of materials characterizes their behavior at stresses exceeding the limit of elasticity, creep is also come through at stresses less than the limit of elasticity, but under the influence of loads for a long time. Consider some of the defining points regarding the description of the behavior of deformable solids within different approaches and models. • Elastic Medium Model

4.2 Physical Relationships Determining the Solid Behavior

67

Based on the elastic model, it is assumed that some so-called “natural” states of the medium can be distinguished, in which the stress tensor is identical to zero. For all other states of elastic medium at any point and at any time, the stress tensor is a one-to-one function of the strain tensor. It should be noted that the stress state of the elastic medium at a given time only depends on its deformed state, not on the way of transition from the natural state to the current state and the speed of this way. In the linear theory of elasticity, it is assumed that each point makes small displacements relative to some “natural” states. If the medium is homogeneous and isotropic, then there is a classical theory of elasticity which is based on the linear connection between stresses and strains. The relationship between stresses and strains is described using the Hooke’s law. • Elastic–plastic Model Common in applications is the theory of small elastoplastic deformations. This theory belongs to the number of deformation theories of plasticity. Plastic deformation occurs when stresses reach the required limit value and is irreversible. In case of elastic–plastic deformation, the strain increment is composed of two parts: elasticity and plasticity: dei j = de(i j) + de[i j] . Square brackets in the indices denote the values related to the plastic region, and parentheses refer to elastic. It is assumed that the elastic and plastic properties of the material are manifested independently. Plastic (residual) deformations occur when stresses reach a certain limit value. For ideal plastic bodies, there are (some ) fixed relationships among stress components called the plasticity condition: ϕ σi j = constant. Elastic–plastic problems usually include the cases where elastic and plastic strains occurring in the body are of the same order. The setting of elastic-plastic problems should take into account the ratios in the elastic and plastic zones and at the border between them: • Elastic region: – – – –

Relationship between stresses and strains in the form of Hooke’s law; Boundary conditions formulated with respect to stresses or/and displacements; Compatibility conditions; Equilibrium equations.

• Plasticity region: – – – – –

Relationship between stresses and strain increments; Compatibility conditions of complete deformations; Plasticity condition; Boundary conditions at the boundary of the plastic region; Equilibrium equations in stresses.

• Boundary between elastic and plastic regions: – Stresses and displacements change continuously, i.e., coupling conditions must be carrying out at the boundary of elastic and plastic zones.

68

4 Construction of Mathematical Model Problems

• Viscous-Elastic Model Viscoelastic media are solids whose deformation changes with time after applying constant external loadings. If in such medium, the relations between stresses and strains are described using linear differential operators (relative to time), then it is called a linear viscous-elastic medium. Rheological models are used to simulate viscoelastic processes. Rheological models have the memory of previous medium states. Rheological structural models can be represented by different connections of elastic springs and viscous Newtonian bodies (a piston moving in a cylinder with viscous liquid). Control Questions 1. 2.

What is the definition of the Saint-Venan principle of loading static equivalence? What is the complete system of equations for constructing a mechanicalmathematical model that describes the stress–strain state of a solid? 3. What are the structural models for describing the solid behavior? Give some examples. 4. What is the structural element of elastic medium? Explain the Hooke’s law. 5. What is the structural element with viscous properties? Explain the Newton’s law. 6. What is the structural element with plastic properties? Explain the element “dry friction”. 7. What is the principal difference among elastic, plastic, and rheological models of solid behavior? 8. What are the defining positions regarding the description of the deformable solid behavior within elasticity model? 9. What are the defining positions regarding the description of the deformable solid behavior within elastic–plastic model? 10. What are the defining positions regarding the description of the deformable solid behavior within viscous-elastic model?

Chapter 5

Mathematical Models of the Theory of Elasticity

5.1 Basic Concepts and Definitions The theory of elasticity is a part of the continuum mechanics and the solid mechanics. The theory of elasticity deals with the determination of a stress–strain state in elastic bodies at specified external loads. The following basic hypotheses and assumptions regarding material properties, loads and the nature of deformations are accepted in the theory of elasticity: • Homogeneity and continuity hypothesis—the assumption makes it possible to study the mechanical properties of bodies on samples with relatively small sizes and allows you to use a differential calculus apparatus to study strain states; • Assumption of small strains—at the points on the body, strains are taken so small that they do not significantly affect the mutual location of loads applied to the body, which allows us to consider the geometry of body unchanged when constructing the equilibrium equations; • The Euler and Cauchy stress principle—in each cross section, mentally drawn inside the body, there is a force interaction according to the type of loads distributed on the body surface, that is, using the method of sections, the effect of one part of the body on another can be replaced by the surface forces acting on the section. • The axiom of solidification (freezing)—at any fixed time t, the material body is considered as an absolute solid, and the laws of theoretical mechanics, including Newton’s laws, are valid for it, while the axiomatic concepts of force and mass are used. In establishing the constitutive relations of the theory of elasticity, in addition to the accepted ones, the following hypotheses are introduced: • The material ideal elasticity hypothesis—the ideal elasticity is the ability of body to restore its original shape and size after “eliminating the causes” that caused its deformation (removal of external loads, temperature, electromagnetic and radiation fields, etc.); © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Zhuravkov et al., Mechanics of Solid Deformable Body, https://doi.org/10.1007/978-981-19-8410-5_5

69

70

5 Mathematical Models of the Theory of Elasticity

• Assumption of linear dependence between deformations and loads (Hooke’s law)—it is assumed that for most materials, the displacements resulting from the body strains are directly proportional to the stresses that caused them. We assume that in its initial state, there is no stress in the body, the object has a constant temperature and is in thermodynamic equilibrium. Therefore, in the initial (“natural”) state of the elastic body, the stress tensor is identical to zero. For all other states of elastic body at any point and at any time, the stress tensor is a one-to-one function of the strain tensor. It should be noted that the stress state of the elastic medium at a given time depends only on its deformed state, not on the method of transition from the natural state to the studied one and the speed of this transition. In the linear theory of elasticity, it is assumed that each point makes small displacements relative to some natural states. If the medium is also homogeneous and isotropic, then we come to the classical theory of elasticity. The classical theory of elasticity is based on a linear connection between stresses and strains and is a linearized theory of the behavior of elastic media in the vicinity of a natural state, when for any point of media, the displacements, and their gradients relative to some natural states are small. Thus, the stress at each point of the elastic body is a unique strain function: σi j = ϕi j (εkm ).

(5.1)

The loading path/deformation path, respectively, is the process of the stress/strain tensor changing, depending on some monotonically increasing parameters (which we call “time”). We emphasize that, in fact, real time does not play a role in determining the model of elastic body. Using this term, we talk only about the sequence of events, not about the time of their action. The stress tensor or strain tensor can be represented by vectors whose components equals to the components of the corresponding tensors: σ = σ(σ11 , σ22 , σ33 , σ12 , σ23 , σ31 ), ε = ε(ε11 , ε22 , ε33 , ε12 , ε23 , ε31 ). Thus, the loading or deformation paths can be represented as curves described using the ends of the vectors σ and ε in the respective spaces. The law of elasticity [i.e., Eq. (5.1)] establishes, in particular, that the closed deformation path corresponds to the closed loading path, and vice versa. The specific potential energy of deformation is the value defined as the work performed during the deformation of unit volume body: ∫ εi j U=

σi j dεi j . εi j =0

5.1 Basic Concepts and Definitions

71

The internal energy change (due to the symmetry of the tensors εi j and σi j ) can be written in an expanded form: dU = σ11 dε11 + σ22 dε22 + σ33 dε33 + + 2(σ12 dε12 + σ23 dε23 + σ31 dε31 ).

(5.2)

For a perfectly elastic body, the work done on an elementary volume in a closed cycle by strain or stress is zero. The condition equals to zero work on an arbitrary closed cycle can be represented as: ∫

∫ dU =

σi j dεi j = 0.

In order to perform the calculated ratio, it is necessary that the sub-integral expression is a complete differential, i.e., σi j =

∂U . ∂εi j

(5.3)

( ) The specific potential strain energy U εi j is a unique strain function, also called the elastic potential energy. However, on the other hand, the expression: dU ∗ = ε11 dσ11 + ε22 dσ22 + ε33 dσ33 + 2(ε12 dσ12 + ε23 dσ23 + ε31 dσ31 ) is also a complete differential, since: dU + dU ∗ = d(σi j εi j ). For this reason: εi j =

∂U ∗ . ∂σi j

(5.4)

The value U ∗ (σi j ) is called the additional specific strain energy and is an elastic stress potential. For the introduced energies U (εi j ) and U ∗ (σi j ), the general ratios are correct: dU (ε) =

∂U ∂U ∗ dεi j , dU ∗ (σ) = dσi j . ∂εi j ∂σi j

The meaning of the U and U * values for a nonlinear elastic material can be clearly explained using the example of a uniaxial stress state, (Fig. 5.1). The figure shows that the values:

72

5 Mathematical Models of the Theory of Elasticity

Fig. 5.1 Visual representation for the concepts of additional specific strain energy and elastic stress potential

∫ ε U (ε) = 0

σdε, U ∗ (σ) =

∫ σ εdσ 0

complement each other to the rectangle “σε” under the curve and above curve of the “stress–strain” diagram. It should be emphasized that the assumption of the existence of specific potential strain energy does not necessarily mean a linear relationship between stresses and strains. Linearity is introduced only by Hooke’s law.

5.2 Hooke’s Law The relationship between stresses and deformations for an elastic body is described using the Hooke’s law.1 The generalized Hooke’s law in the tensor representation is as follows: e=

νΣ 1 1+ν TH − I, E E

(5.5)

where e is the principal strain tensor, corresponding to principal stresses σi ; T H is the stress tensor; Σ 1 is the first invariant of stress tensor T H ; I is the unit tensor. Remark The relationship (5.5) equals to three vectors: ei = 1

σi νΣ 1 . (1 + ν) − E E

The same law was independently opened in 1680 by the French physicist E. Mariott (Mariotte E., 1620–1684).

5.2 Hooke’s Law

73

The tensors in (5.5) can be written in any coordinate system. Constants E and v fully characterize ideal elastic medium. However, you can also specify two other constants. It is important that there is a relationship between each pair of constants (see Chap. 3). The generalized ratios of Hooke’s law for isotropic bodies are as follows: 1 1 [σ11 − ν(σ22 + σ33 )], ε22 = [σ22 − ν(σ33 + σ11 )], E E 1 = [σ33 − ν(σ11 + σ22 )], E 1 1 1 σ12 , ε23 = σ23 , ε31 = σ31 , = 2G 2G 2G

ε11 = ε33 ε12

(5.6)

where E and G are, respectively, modulus of elasticity and shear modulus, v is Poisson’s ratio. They are related by a known dependency: 2G = E/(1 + ν). In solving the problems of the theory of elasticity, there is a need for inverse relations when stresses are expressed through strains. Using the ratios (5.6), we get: σi j = 2μεi j + λθδi j ,

(5.7)

where: θ = ε11 + ε22 + ε33 =

1 − 2ν (σ11 + σ22 + σ33 ), E

(5.8)

θ is the volume strain; λ and μ are Lamé constants that are associated with the dependencies of G, v and E (see Chap. 3): λ=

E λ νE , μ=G= , ν= . (1 + ν)(1 − 2ν) 2(1 + ν) 2(λ + μ)

Recall that for an ideal elastic body from five elastic constants λ, μ, K, ν, E, only any two are independent. The dimensionalities of the values λ, μ, K, E correspond to the stress or pressure dimensions. These constants are positive. Poisson’s ratio is dimensionless. Its change interval is −1 ≤ ν ≤ 1/2. Value ν = 1/2 corresponds to incompressible material. For all known isotropic materials, we have ν > 0. Equation (5.7) with respect to εij is the inverse of Hooke’s law: ] [ 1 3ν σi j − σδ i j . εi j = 2G 1+ν

(5.9)

With ν → 0.5 Lamé constant λ → ∞, which, according to the expression (5.8), corresponds to incompressible material (θ = 0). In this case, it is difficult to use the ratio (5.7). Therefore, it is recommended to record two separate ratios in which the volume deformation would be explicitly written out. This is achieved, for example,

74

5 Mathematical Models of the Theory of Elasticity

by using the expression of the Hooke’s law through the spherical and deviatoric components of stress and strain tensors: si j = 2G C- i j , σ = K θ,

(5.10)

2 E K =λ+ μ= , 3 3(1 − 2ν) where K is the volume strain modulus. For incompressible materials, instead of the second in Eq. (5.10), this condition is used: θ = 0. The value θ is defined by (5.8). It follows from Hooke’s law in the form (5.10), that the specific potential strain energy can be divided into two independent parts: ( ) ε2 1 σ2 1 1 σi j σi j − , Uv = σθ = K , Ud = si j C- i j = 2 4G 3 2 2

(5.11)

where U d is called the specific energy of the form change; U v is called the specific energy of the volume change. In the expanded form, formula (5.11) takes the following form: 1 − 2ν (σ11 + σ22 + σ33 )2 , 6E ) 1+ν( Ud = (σ11 − σ22 )2 + (σ22 − σ33 )2 + (σ33 − σ11 )2 6E ) 1 ( 2 2 2 + σ + σ23 . + σ31 2G 12 Uν =

Both values U d and U v play the important roles, for example, in formulating strength criteria and flow laws during plastic deformation. The relation (5.7) can be generalized to the case of an arbitrary anisotropic material, assuming a linear relationship between the stress tensor and strain tensor components in the form of: σi j = E i jkl εkl .

(5.12)

The relation (5.12) contains, in general, nine equations having nine terms each. The components of E ijkl are coordinate independent values for homogeneous material. Since stresses and strains depend on the orientation of coordinate system, elastic constants E ijkl must also correspond to this dependence. They form 81 elastic constant matrices and are transformed into components of a fourth-rank tensor, which is called the elastic modulus tensor. Elastic constants do not depend on the orientation of coordinate system only for isotropic materials. In view of the symmetry of tensors σi j and εkl , the elastic modulus tensor is also symmetrical with respect to the indices i and j, k and l: E ijkl = E jikl = E ijlk = E jilk .

5.2 Hooke’s Law

75

Thus, the independent components remain 36. follows: ⎡ E 1111 E 1122 E 1133 E 1112 ⎢E ⎢ 2211 E 2222 E 2233 E 2212 ⎢ ⎢ E 3311 E 3322 E 3333 E 3312 ⎢ ⎢ E 1211 E 1222 E 1233 E 1212 ⎢ ⎣ E 1311 E 1322 E 1333 E 1312 E 2311 E 2322 E 2333 E 2312

The matrix of its components is as

E 1113 E 2213 E 3313 E 1213 E 1313 E 2313

⎤ E 1123 E 2223 ⎥ ⎥ ⎥ E 3323 ⎥ ⎥. E 1223 ⎥ ⎥ E 1323 ⎦ E 2323

A further decrease in the number of independent components is obtained from thermodynamic considerations, assuming the existence of a specific potential strain energy. Substituting the ratios (5.12) and (5.3) into the expression for the specific potential energy difference (5.2) in turn, we have: dU = σi j dεi j = E i jkl εkl dεi j , dU = σi j dεi j =

∂U dεi j , ∂εi j

where ∂U = E i jkl εkl . ∂εi j After re-differentiation and taking into account the possibility of changing the order of differentiation, we obtain: ( ) ( ) ∂U ∂U ∂ ∂ = E i jkl , = E kli j . ∂εkl ∂εi j ∂εi j ∂εkl Therefore, ∂ 2U = E i jkl = E kli j . ∂εi j ∂εkl Therefore, the number of independent constants for the anisotropic body becomes 21. If the elastic body has a plane of symmetrical elastic properties, then for it, the number of independent constants is reduced to 13. For a body having three mutually orthogonal symmetry planes (orthotropic body), the number of constants is reduced to 9. The number of independent constants for an isotropic body, as previously shown, is two. The generalized Hooke’s law can be reversed by expressing strains through stresses. Then: εi j = Di jkl σkl .

(5.13)

76

5 Mathematical Models of the Theory of Elasticity

The fourth-rank tensor Dijkl (elastic pliability tensor) has the same symmetric properties as the elastic modulus tensor E ijkl . It should be emphasized that the number of elastic constants in Hooke’s law is reduced only when the symmetric planes are taken as coordinate. Other coordinate systems would still contain 21 constants, expressed through nine independent constants.

5.3 Clapeyron’s Formula and Clapeyron’s Theorem Clapeyron’s formula We use the expression (5.2) to calculate the specific potential strain energy. We consider an arbitrary element of the volume of an elastic body, the stress–strain state of which is given by the values σi j and εi j . We introduce a local system of principal axes for stresses and strains. Let the principal stresses receive increments that are the same as their final values (this is proportional loading). Then the work performed, assigned to the unit of volume, is defined as: W = (σ1 ε1 + σ2 ε2 + σ3 ε3 )/2. It is independent for the loading path, and according to the law of conservation of energy, it equals to the specific potential strain energy U. In arbitrary coordinate axes, accordingly, we obtain: U=

1 σi j εi j . 2

(5.14)

This expression is called the Clapeyron’s formula. We can write it in an expanded form: U=

1 (σ11 ε11 + σ22 ε22 + σ33 ε33 ) + σ12 ε12 + σ23 ε23 + σ31 ε31 . 2

Taking into account the generalized Hooke’s law in form (5.12) or (5.13), the Clapeyron’s formula (5.14) takes the form: U=

1 1 E i jkl εi j εkl or U = Di jkl σi j σkl . 2 2

Hence: ∂U 1 ∂ 1 = (Dmnkl σmn σkl ) = Dmnkl (δim δ jn σkl + δik δ jl σmn ) ∂σi j 2 ∂σi j 2 1 = (Di jkl σkl + Dmni j σmn ) = Di jkl σkl , 2

5.3 Clapeyron’s Formula and Clapeyron’s Theorem

77

or ∂U = εi j . ∂σi j

(5.15)

It should be noted that this ratio should not be lumped together with a very similar ratio (5.4), which instead of U contains U * . At the same time, the ratio (5.15) is assumed to satisfy the Hooke’s law and therefore the linear-elastic behavior of the material. Meanwhile, the ratio (5.4) follows the general thermodynamic laws. Remark Formulas (5.4) and (5.15) are partial formulas of Castigliano’s2 theorem (see below). Clapeyron’s theorem Let the linearly elastic body under the influence of body forces F i and surface forces Ri be in an equilibrium state. Let’s scalarly multiply the equilibrium equations by the displacements ui and then integrate on the body volume V (with the surface S). As a result, we obtain: ∫ ∫ u i σi j , j dV + u i ρFi dV = 0. V

V

We transform the sub-integral function in the first integral using the formula: (u i σi j ), j = u i , j σi j + u i σi j , j , and, therefore: ∫

∫ (u i σi j ), j dV + V

∫ u i ρFi dV =

V

u i , j σi j dV . V

The first integral on the left side can be transformed into a surface integral using the known Gauss3 —Ostrogradsky formula. In the right part, due to the symmetric stress tensor, equality will be performed: u i , j σi j = εi j σi j . As a result, we get the ratio: ∫ ∫ ∫ u i σi j l j dS + u i ρFi dV = εi j σi j dV . S

V

V

The sub-integral expression in the right part according to the Clapeyron’s formula (5.14) equals to twice the specific potential strain energy. Therefore: 2

Castigliano C. A. (1847–1884), Italian scientist-mechanics. Gauss C. F. (1777–1855), German mathematician, physicist; works in the field of algebra, number theory, differential geometry.

3

78

5 Mathematical Models of the Theory of Elasticity

∫ u i Ri dS + S

∫ u i ρFi dV = 2

V

U dV .

(5.16)

V

Thus, the work of elastic strain equals to half of the work performed by the external surface and body forces on the displacements from the initial state of equilibrium to the final state (Clapeyron’s theorem).

5.4 Thermoelasticity We assume that the body is in a heterogeneous and non-stationary temperature field T (x i , t). Initial temperature value is T 0 . Complete linear strains εii consists of strains from the force load εii' and temperature expansion εii'' = αT (Neumann4 hypothesis): εii = εii' + εii'' . Temperature increments for shear strains are zeros. Therefore, the Hooke’s law, taking into account the temperature, has the form: εi j =

[ ] 3ν 1 σi j − σδ i j + αT δi j . 2G 1+ν

(5.17)

The value α is called the coefficient of linear thermal expansion of the material. By expressing stresses through strains from (5.17), we obtain: [ σi j = 2G εi j +

] 1+ν νθ δi j − αT δi j . 1 − 2ν 1 − 2ν

(5.18)

The ratio between the deviators in thermoelasticity remains the same as in the case of ideal elasticity (5.10). Only the spherical parts connected by stress and strain tensors take another form: si j = 2G C- i j , σ = K (θ − 3αT ). Strains in formulas (5.17) and (5.18) are related to displacements by Cauchy ratios (2.35). For complex thermoelastic problems, in general, it is also previously necessary to solve the heat distribution problem. The temperature T (xi , t) at each point of the body must satisfy the thermal conductivity equation

4

Franz Ernst Neumann (1798–1895), German physicist.

5.5 The Boundary Problems of Theory of Elasticity

Q ∂T = BΔ T + , ∂t cγ

79

(5.19)

where B is the coefficient of heat transfer of materials; Δ is the Laplacian; Q(xi , t) is the function showing the heat quantity produced by the thermal energy source in unit volume and per unit time; c is the specific heat capacity; γ is the specific weight. The initial condition of temperature distribution T (xi , 0) = f (xi ) should be added to Eq. (5.19). In addition, the conditions of heat exchange with the environment (three types of heat exchange conditions) shall be executed at the solid boundary. The Q value in Eq. (5.19) for solids generally depends on the stress–strain state. In this case, they talk about coupled thermoelastic problems. For an elastic anisotropic body, Hooke’s law (5.12) takes the form: σi j = E i jkl (εkl − αkl T ), where αkl is a coefficient of the thermal expansion tensor of materials. Constant E ijkl is determined under isothermal conditions at T = T 0 . If the temperature increment is not small, then E ijkl and αkl should be considered as functions of temperature.

5.5 The Boundary Problems of Theory of Elasticity For the solution of the mathematical theory of direct problems in elasticity, that is for definition of 15 unknown functions ui , σij , εij (i, j = 1, 2, 3), the following system of the equations is available: • Equilibrium equations σi j , j +ρFi = 0,

(5.20)

• Cauchy ratios 1 (u i , j +u j ,i ), 2

(5.21)

σi j = 2μεi j + λθ δ i j .

(5.22)

εi j = • Hooke’s Law

The system (5.20)–(5.22) includes 15 linear equations in partial derivatives. When displacements are not explicitly unknown, Eq. (5.21) is replaced by the conditions of strain compatibility (without summation by repeating indices): εαα ,ββ +εββ ,αα −2εαβ ,αβ = 0,

80

5 Mathematical Models of the Theory of Elasticity

(εαβ ,γ −εβγ ,α +εγα ,β ),α −εαα ,βγ = 0.

(5.23)

To close a resolving system of equations, you must add conditions on the boundary (boundary conditions). The most common situation is that the surface of the body S can be represented as consisting of two parts: S = S σ + S u . It is assumed that the boundary conditions at each point of surface are as follows: u i = u i0 (x) on Su ,

Rνi = σi j l j on Sσ .

(5.24)

This simplest case is not the only one and does not exhaust all possible options for boundary conditions. It will be appreciated that other combinations of boundary conditions are possible. Note that if the body contains an infinitely distant point, the requirement of the regularity of the solution at infinity should to be added to the boundary conditions, which, as a rule, is reduced to the constraint condition. Therefore, in the general case, the theory of elastic problem is to solve Eqs. (5.20)–(5.23) under boundary conditions (5.24). It is also possible to set a theory of elastic inverse problem. In this case, you specify stresses, strains, or displacements for all internal points of the body as coordinate functions. Stress–strain state at the body’s boundary (boundary conditions) must be defined.

5.6 The Theory of Elastic Boundary Problems in Displacements Depending on the setting of the boundary problem, the main defined functions are either stresses or displacements. If we substitute into equilibrium Eq. (5.20) the Hooke’s law (5.22) and exclude strains using Cauchy relations (5.21), we get a system of three differential equations with three unknown functions ui (Lamé equation): (λ + μ)θ,i +μΔ u i + ρFi = 0,

(5.25)

where θ = u1,1 + u2,2 + u3,3 = ui,i is the volume strain; Δ is the Laplacian. The system of Eq. (5.25) is elliptical by Petrovsky5 in the area V body with all values of the Poisson’s ratio, except for ν = 0.5 and ν = 1. To construct a complete system of resolving equations, boundary conditions should be attached to the three Eq. (5.25). On one part S u of the boundary surface, displacements u0i can be set as known coordinate functions: 5

Petrovsky Ivan Georgievich (1901–1973), Russian mathematician, rector of Moscow State University (1951), academician of the USSR Academy of Sciences.

5.6 The Theory of Elastic Boundary Problems in Displacements

u i = u i0 (x) on Su .

81

(5.26)

External surface forces Ri = Ri (x i ) can be set on the part Sσ of the boundary surface. Moreover, since the problem is solved in displacements, the values Ri should be associated with the values ui using the Hooke’s law and Cauchy ratios. As a result, the boundary conditions in displacements take the form on the part Sσ : λθδi j l j + μ(u i , j +u j ,i )l j = Ri .

(5.27)

In general, three types of boundary problems in displacements occur. The first type of boundary problems consists in determining the displacements and stresses within the elastic body if the displacements of the point are known on the body surface (S = S u ). The mathematical setting of the first boundary problem is as follows. Determine the elastostatic state [u, σ] of the medium D corresponding to the mass force X and satisfy the Lamé equation: μΔ u + (λ + μ)grad divu + ρX = 0 by the boundary condition: ∀y ∈ S : lim u ± (x) = f (y) x→y

(

) x ∈ D± ,

(5.28)

where μ, λ are Lamé constants; f = ( f 1 , f 2 f 3 ) is the specified displacement vector at the body boundary; S is the boundary of domain D. If the domain contains an infinitely point, then the solution must satisfy the conditions of regularity at infinity: ( ) lim |u(x)| = 0, lim ∂u j /∂ xi R = 0, (i, j = 1, ..., m), (m = 2, 3),

|x|→∞

R→∞

where R is the distance from current point x to the origin. In the second boundary problem, the distribution of forces on the surface (S = S σ ) is known. The mathematical setting of the second boundary problem is as follows. Determine the elastostatic state [u, σ] of the medium D corresponding to the mass force X and satisfy the Lamé Eq. (5.25) by boundary condition: ) ( ∀ y ∈ S lim (T(∂ x, n x )u(x))± = f (y) x ∈ D ± , x→y

where T(∂ x, n x ) is the stress operator defined by the formula: || || T(∂ x, n x ) = ||Ti j (∂ x, n(x))||, where, in turn,

(5.29)

82

5 Mathematical Models of the Theory of Elasticity

Ti j (∂ x, n(x)) = λn i (x)

∂ ∂ ∂ . + μn j (x) + μδ i j ∂x j ∂ xi ∂n(x)

In condition (5.29), the normal at point x is selected so that in the limit n(x) → n(y), where n(y) is the external normal to surface S. It should be noted that in the second boundary problem, the displacements are determined by integral equations which are accurate to the displacement of the body as a rigid. To eliminate this uncertainty, attention should be paid to the conditions for the existence of the single solution. For example, when constructing a system of resolution, one should take into account either the immobility of the vicinity of a fixed-point M, or the immobility of the plane in the case of symmetrically deformable bodies, etc. Setting the boundary problem with conditions (5.24) refers to the third or mixed boundary problem, that is, the mixed problem is characterized by setting the conditions (5.28) on a part of the boundary surface, and the conditions (5.29) on the rest of it. Remark In the second and mixed problems of the theory of elasticity for the regions with finite dimensions, the boundary conditions must be supplemented by the solvability conditions of these problems, which consist in the requirement of the selfequilibrium of the applied load. For a mixed task, unknown forces acting on Su must be added to the applied load on Sσ : ∫

∫ f (y)dS = 0; S

r × f (y)dS = 0. S

If there are several boundary surfaces, it is necessary to require the self-equilibrium of the applied load as a whole rather than each surface. Lamé equations are widely used in many methods for solving the problems of theory of elasticity, since it does not need the equations of the strain compatibility, which in this case are satisfied identity. Particularly important problem is the homogeneous problem of the theory of elasticity, when body forces can be zero. We get some properties of displacements ui , resulting from (5.25), in the absence of body forces, that is, F i = 0. We differentiate each of Eq. (5.25) by the corresponding coordinate x i and perform convolution (summation by index i). We get: (λ + μ)θ,ii +μΔ u i ,i = 0. Because: θ,ii = θ,11 +θ,22 +θ,33 = Δ θ, Δ u i ,i = Δ (u 1 ,1 +u 2 ,2 +u 3 ,3 ) = Δ θ, then from (5.28), we get

(5.30)

5.7 Boundary Problems of the Theory of Elasticity in Stresses

Δ θ = 0.

83

(5.31)

Thus, the volume strain in an elastic isotropic solid with the absence of mass forces is a harmonic function, which satisfies the Laplace’s Eq. (5.31). We take now the Laplace operator from the left side of the Lamé equation, assuming F i = 0: (λ + μ)Δ θ,i +μΔ Δ u i = 0. From which, taking into account (5.31), we get that Δ θ,i = (Δ θ),i = 0. Therefore: Δ Δ u i = 0.

(5.32)

Thus, each of the components of displacement vector is a biharmonic function of the coordinates (satisfying the double Laplace’s equation). However, it should not be thought, that the problem of the theory of elasticity can be reduced to the integration of the system (5.32) or that the value θ can be found using the known methods for solving the Laplace’s equation. To define a biharmonic function, two conditions must be specified at the domain boundary (for example, ui and ∂u i /∂n), while to solve the system (5.25), it is enough to set only the ui values at each point on the surface. In addition, it is relatively easy to construct three biharmonic functions with set values at the boundary, but they may not satisfy Eq. (5.25). Finally, the system (5.32) is of the twelfth order, whereas the initial system (5.25) is of the sixth order. In addition, the value θ is never set at the boundary. It cannot be determined by solving the Dirichlet6 problem. Remark The system order can be defined as the product of the maximal derivative order and the number of equations. It should be noted that the described boundary problem for incompressible materials (ν = 0.5) loses its ellipticity and its solution may not be unique.

5.7 Boundary Problems of the Theory of Elasticity in Stresses The theory of elastic problem can be set not only on displacements, but also on stresses. This is useful when the body boundary is loading rather than displacements and the stress state must be determined first. 6

Dirichlet Peter Gustav (1805–1859), German mathematician. The Dirichlet problem is the search for a function satisfying the Laplace equation inside a domain and given at its boundary. In the Neumann problem, the function derivative is given on the boundary.

84

5 Mathematical Models of the Theory of Elasticity

Stress problem as a boundary problem for an overridden system and its simplification. The resolving system of equations for the theoretical problems of elasticity in stresses consists of three equilibrium equations in the form of Cauchy, six strain compatibility equations and six ratios of the Hooke’s law. In total, these are 15 equations relative to 12 unknowns σi j , εi j , i, j = 1, 3 to which three boundary conditions in the forces are attached. Since this system obviously does not include displacements, it is natural to expect that in this formula the problem in stresses has the unique solution. In fact, the uniqueness theorem of the problem in stresses with this approach has been proved by A.N. Konovalov.7 Using Hooke’s law, we exclude strains from the compatibility relations. As a result, we obtain the following system to determine the unknown components of the stress tensor: 3 ∑

∂ j σi j + f i = 0, i = 1, 3,

(5.33)

j=1

∑ ∑ ν ν ∂i ∂ j δi j σkk = − ∂k f k 1+ν 1 − ν k=1 k=1 3

Δ σi j +

3

− (∂i f j + ∂ j f i ),

i, j=1, 3

(5.34)

The system of (5.33)–(5.34), as before, is a redefined system of equations (six unknowns σi j = σ ji , i, j = 1, 3 satisfy nine equations: three equilibrium Eq. (5.33) of the first order and six equations Beltrami8 -Michell9 (5.34) of the second order. Equilibrium Eq. (5.33) was used in obtaining the Beltrami-Michell’s equations. Therefore, Eq. (5.34) is always satisfied, if as σi j i, j = 1, 3, to take the system (5.33) solutions. The opposite is not always true, since the differentiation of Eq. (5.33) used in the construction of the Beltrami-Michell’s equations led to an expansion of the class of functions σi j , i, j = 1, 3. Therefore, equilibrium Eq. (5.33) must be involved in solving the theory of elasticity in stresses. This enables us to distinguish from the Beltrami-Michell’s equations which class of solutions that lead to physically justified stress–strain states of solids. The definition of (5.33)–(5.34) leads to two important problems for the theory of elasticity. The first problem is the study of the system (5.33)–(5.34) for compatibility.

7

Konovalov Anatoly Nikolaevich, Soviet and Russian scientist, Academician of the Russian Academy of Sciences; one of the main areas of research is mathematical models and numerical methods for problems of continua mechanics. 8 Eugenio Beltrami (1835–1900), Italian mathematician notable for his work concerning differential geometry and mathematical physics. 9 John Henry Michell (1863–1940), Australian mathematician, Professor of Mathematics at the University of Melbourne.

5.7 Boundary Problems of the Theory of Elasticity in Stresses

85

The second problem is the separation from the system (5.33)–(5.34) of the principal system of six equations, which leads to the solution of the problem in stresses. To separate a basic system of equations Consider the problem of separation from the (5.33)–(5.34) basic system of equations. Remark Detailed information on solving the problems of solid mechanics in stresses is available in monographs. First version. We select Eq. (5.34) as the basic system and satisfy Eq. (5.33) at the boundary. We come to this formula of the problem in stress: Determine the solution of such system in the area D occupied by the elastic body: ∑ ν ν ∂i ∂ j δi j σkk = − 1+ν 1−ν k=1 3

Δ σi j + 3 ∑

∂k f k − (∂i f j + ∂ j f i ); i, j = 1, 3,

k=1

when the following conditions are met at the boundary S of the elastic body: 3 ∑

n j σi j = Pni ,

j=1

3 ∑

∂ j σi j = − f i ; i = 1, 3.

j=1

In this formula, the equilibrium equations of stress problems are used to release a unique solution to the system (5.33)–(5.34). Second version. As the basic system of equations for the stress problem, in addition to the previously mentioned, a system consisting of three equilibrium equations and three Beltrami-Michell’s equations with respect to the tangents of the stress tensor components can be used: ⎧ 3 ∑ ⎪ ⎪ ⎪ ∂ j σi j + f i = 0 ⎪ ⎪ ⎨ j=1

⎪ ⎪ ⎪ ⎪ ⎪ ⎩ Δ σi j +

∑ 1 ∂i ∂ j σkk = −(∂ j f i + ∂i f i ); i /= j = 1, 3. 1+ν k=1 3

(5.35)

Therefore, for the problems of solid mechanics in stresses, the two formulas of boundary problems are physically reliable.

86

5 Mathematical Models of the Theory of Elasticity

5.8 Homogeneous Problem of the Theory of Elasticity in Stresses The differential equilibrium equations at F i = 0 take the form: σi j , j = 0.

(5.36)

These three equations are not sufficient to determine the stress state in the elastic body (six independent components of the stress tensor). As additional equations, we can use the Beltrami-Michell’s equations in the absence of mass forces (compatibility equations in stresses): (1 + ν)Δ σi j + 3σ,i j = 0.

(5.37)

Stresses, in the absence of mass forces, as well as components of the displacement vector, have the properties (5.31) and (5.32). Indeed, summing the Eq. (5.37) at i = j, we come to the equation: Δ σ = 0.

(5.38)

where σ = (σ11 + σ22 + σ33 )/3 is the average hydrostatic stress. Taking into account (5.38), after applying the Laplace operator to the Eq. (5.37), it follows that the stresses in static problems without mass forces are biharmonic functions: Δ Δ σi j = 0.

(5.39)

5.9 The Problem of the Setting of Theory of Elasticity A large class of the theory of elastic problems is convenient to solve not only in Cartesian, but also in polar, cylindric or spherical coordinate systems. Let’s give the main relations for the theory of elasticity in cylindrical and spherical coordinate systems. The cylindrical coordinates r, ϕ, and z are associated with Cartesian in the following way, (Fig. 5.2): x1 = x = r cos ϕ, x2 = y = r sin ϕ, x3 = z = z. The linear element is given by the following formula:

5.9 The Problem of the Setting of Theory of Elasticity

87

z

Fig. 5.2 Relationship of cylindrical and Cartesian coordinates

P(r, , z) r

x

z r

y

(ds)2 = (dr )2 + r 2 (dϕ)2 + (dz)2 . Vector components of displacements: u 1 = ur , u 2 = u ϕ , u 3 = u z . Strain tensor components: ε11 = εrr , ε22 = εϕϕ , ε33 = εzz , ε12 = εr ϕ , ε23 = εϕz , ε31 = εzr . The following kinematic relations are valid: 1 ∂u ϕ ∂u z ∂u r ur , εϕϕ = + , εzz = , ∂r r ∂ϕ r ∂z ( ) ( ) ( ) ∂u ϕ uϕ 1 1 ∂u r 1 ∂u ϕ 1 ∂u z 1 ∂u z ∂u r = + − , εϕz = + , εr z = + . 2 r ∂ϕ ∂r r 2 ∂z r ∂ϕ 2 ∂r ∂z εrr =

εr ϕ

The components of the stress tensor σrr , σϕϕ , σzz , τr ϕ , τϕz , τzr , are associated with strains of the Hooke’s law with Lamé constants: σrr = 2μεrr + λθ, . . . τr ϕ = 2μεr ϕ , . . . , where θ is the relative volumetric strain: θ = εrr + εϕϕ + εzz . Equilibrium equations in stresses have the form:

88

5 Mathematical Models of the Theory of Elasticity

z

Fig. 5.3 Relationship of spherical and Cartesian coordinates

P(R, , ) R z

x

r

σϕϕ 1 ∂ 1 ∂τrϕ ∂τrz (r σrr ) + + + ρFr = 0, r ∂r r ∂ϕ ∂z r ∂τϕz 1 ∂σϕϕ 1 ∂ 2 (r τrϕ ) + + + ρFϕ = 0, 2 r ∂r r ∂ϕ ∂z 1 ∂τϕz ∂σzz 1 ∂ (r τr z ) + + + ρFz = 0, r ∂r r ∂ϕ ∂z where ρF r , ρF ϕ , ρF z are vector components of the body force. In spherical coordinates, such conversion formulas are valid, (Fig. 5.3): x1 = x = R sin ϑ cos ϕ, x2 = y = R sin ϑ sin ϕ, x3 = z = R cos ϑ. Linear element: (ds)2 = (dR)2 + R 2 sin2 ϑ(dϕ)2 + R 2 (dϑ)2 . Displacements: u 1 = u R , u 2 = u ϕ , u 3 = u ϑ . Strain tensor components: ε11 = ε R R , ε22 = εϕϕ , ε33 = εϑϑ , ε12 = ε Rϕ , ε23 = εϕϑ , ε31 = ε Rϑ . The following kinematic relations are valid: ∂u R 1 ∂u ϕ uR , εϕϕ = + ∂r R sin ϑ ∂ϕ R ( ) ∂u ϕ uϕ 1 1 ∂u R = + − , 2 R sin ϑ ∂ϕ ∂R R ( 1 1 ∂u ϑ + εϕϑ = 2 R sin ϑ ∂ϕ

εR R = ε Rϕ

uϕ 1 ∂u ϕ uR , εϑϑ = + , R R ∂ϑ R ( ) ∂u ϕ uϕ 1 1 ∂u R ε Rϑ = + − , 2 R ∂ϑ ∂R R ) uϕ 1 ∂u ϕ − ctgϑ . R ∂ϑ R + ctgϑ

y

5.10 2D-Problems of the Theory of Elasticity

89

The components of the stress tensor σ R R , σϕϕ , σϑϑ , τ Rϕ , τϕϑ , τϑR are associated with strains of the Hooke’s law with Lamé constants: σ R R = 2με R R + λθ, . . . , τ Rϕ = 2με Rϕ , . . . , where θ = ε R R + εϕϕ + εϑϑ . Equilibrium equations: σϕϕ + σϑϑ 1 ∂(τ Rϕ sin ϕ) 1 ∂τ Rϑ 1 ∂(R 2 σ R R ) + + − + ρFR = 0, R ∂R R sin ϕ ∂ϕ R sin ϕ ∂ϑ R ∂(τϕϑ sin2 ϑ) 1 1 ∂σϕϕ 1 ∂(R 3 τ Rϕ ) + + + ρFϕ = 0, 2 3 R ∂R ∂ϕ R sin ϕ ∂ϕ R sin ϕ 1 ∂(σϑϑ sin ϑ) 1 ∂τϕϑ ctgϑ 1 ∂(R 3 τ Rϑ ) + + − σϕϕ + ρFϑ = 0. 3 R ∂R R sin ϕ ∂ϑ R sin ϑ ∂ϕ R

5.10 2D-Problems of the Theory of Elasticity The problem that the stress–strain state functions only depend on two coordinates is very important for applications. These are 2D-problems of the theory of elasticity. At the same time, two cases are distinguished: plane strain state and plane stress state. • Plane strain state. This case corresponds to the stress–strain state formation in a long prismatic body (with a longitudinal axis of coordinate z) loaded with surface forces independent of z and has no component along this axis. As an example, the cylinder loaded by forces p1 , p2 , p3 , which linearly distributed along its general axis is shown in the Fig. 5.4. The elastic body may be either infinitely long or of finite length, but its edges shall be appropriately fixed. In this case, the stress–strain state on the cross section of each body is plane strain state. In the future, simplified relationships will be used, so the tensor form of recording will only be used for compact writing of formulas. Displacements in Cartesian coordinate system x, y, z are defined through u, v, w, respectively: u = u(x, y), v = v(x, y), w = const or w = 0.

(5.40)

In addition, all the derivatives of displacements by z (by x 3 ) are zeros. The deformas, including the volume deformation, will be as follows: εx x =

( ) ∂v ∂v 1 ∂u ∂u , ε yy = , εx y = + , ∂x ∂y 2 ∂y ∂x

90

5 Mathematical Models of the Theory of Elasticity

p1

Fig. 5.4 Example of plane strain state

y p3

p

2

z

x

εx z = ε yz = εzz = 0, θ =

∂v ∂u + . ∂x ∂y

(5.41)

Only one strain compatibility condition remains: ∂ 2 ε yy ∂ 2 εx y ∂ 2 εx x . + =2 2 2 ∂y ∂x ∂ x∂ y

(5.42)

Non-zero stress tensor components are σx x , σx y , σ yy , σzz . The formulas of the generalized Hooke’s law take the form: σx x = 2με x x + λθ , σ yy = 2με yy + λθ , σx y = 2με x y , σzz = λθ .

(5.43)

Strains are expressed through stresses by the following formulas [obtained from (5.6)]: ] ] 1[ 1[ σx x − ν(σ yy + σzz ) , ε yy = σ yy − ν(σzz + σx x ) , E E 1+ν σx y . = E

εx x = εx y

It is easy to show that for a plane strain of a body, the number of independent components of the stress tensor is three. In fact, from the Hooke’s law, under the condition εzz = 0, it is as follows: σzz = ν(σx x + σ yy ).

(5.44)

Only the following equations remain from equilibrium equations: ∂σx y ∂σx x + = 0, ∂x ∂y

∂σ yy ∂σx y + = 0. ∂y ∂x

(5.45)

5.10 2D-Problems of the Theory of Elasticity

91

body forces are not taken into account here. The boundary conditions for stresses, if Rx and Ry are set at the cylinder edge, are: σx x cos(ν, x) + σx y cos(ν, y) = Rx , σx y cos(ν, x) + σ yy cos(ν, y) = R y . (5.46) For a prismatic body with a length of l, the end sections of which are supported, the boundary conditions have the form: w(x, y, 0) = w(x, y, l) = 0. Then the forces acting on the cross sections of the end equal to: ∫ Pz =

σzz dS, S

where stresses are integrated along the cross section of the prismatic body. Equilibrium equations in displacements (Lamé’s equations) (5.25) give the form: ∂ 1 Δ u + 1 − 2ν ∂ x

(

∂u ∂v + ∂x ∂y

)

( ) 1 ∂ ∂u ∂v = 0, Δ v + + = 0. 1 − 2ν ∂ y ∂ x ∂y

Accordingly, the compatibility equations in stresses (Beltrami-Michell’s equations) (5.37) become as follows: 3 ∂ 2σ = 0, 1 + ν ∂x2 3 ∂ 2σ + = 0, 1 + ν ∂ y2 3 ∂ 2σ = 0, + 1 + ν ∂ x∂ y

Δ σx x + Δ σ yy Δ σx y

where the average stress is (the expression (5.27) for σz is taken into account): σ=

1 1+ν (σx x + σ yy + σzz ) = (σx x + σ yy ). 3 3

If the prismatic body is finite in length and there is no load in its ends, then the conditions of plane strain are broken. However, even in this case, for sections sufficiently remote from the ends, on the basis of the Saint-Venan principle, the stress–strain state can be considered corresponding to the plane strain conditions. • Plane stress state. In this case, we are talking about a flat elastic body of small thickness (plate), which is only loaded by forces on its plane, and has no normal stresses in the thickness direction, (Fig. 5.5). The applied forces are

92

5 Mathematical Models of the Theory of Elasticity

Fig. 5.5 Example of plane stress state

P3

z y x

P1

h

P2

evenly distributed across the thickness and are independent of z (which is always done with good approximation for thin plates). Forces can also be distributed symmetrically with respect to the median plane (an imaginary surface that bisects the thickness) of the plate. In this case, their values of average plate thickness can be entered. With a plane stress state in the (x, y) plane, there are the following non-zero stress tensor components: σx x = σx x (x, y), σ yy = σ yy (x, y), σx y = σx y (x, y). The remaining components are zeros: σzz = σx z = σ yz = 0. The components of displacements u, v, and w in this case are independent of the z coordinate. In the case of a plane stress state, kinematic equations correspond to Eq. (5.41) for a plane strain state, compatibility conditions correspond to Eq. (5.42), equilibrium equations correspond to Eq. (5.45), and boundary conditions correspond to conditions (5.46). When considering strains, the difference between plane strain state and plane stress state is exerted, for example, in Hooke’s law. Since σzz = 0, it follows from Hooke’s law: ] ] 1[ 1[ σx x − νσ yy , ε yy = σ yy − νσ x x , E E σx y ν εzz = − (σx x + σ yy ), εx y = . E 2G

εx x =

Reversing (5.47), yield stress expressions through strains are: σx x =

E (εx x + νε yy ), 1 − ν2

(5.47)

5.10 2D-Problems of the Theory of Elasticity

σ yy =

93

E E εx y . (ε yy + νε x x ), σx y = 2 1−ν 1+ν

These formulas differ from the corresponding formulas (5.43) and (5.44). As already mentioned, the plane stress state is realized not precisely, but only approximately in thin plates. Filon10 showed that it is possible to consider a generalized case and modify the given ratios. Symmetric distribution of applied forces relative to median plane of plate is assumed. If it is assumed that the stresses σzz in the plate are absent, and on the outer surfaces of the plate, σzx and σzy are zeros, then for displacements and stresses, their average values are calculated for the thickness of the plate, for example: σ∗x x

1 = 2h

∫ h σx x dz, −h

where 2h is the plate thickness. As a result, the task is reduced to the previous task. Such a stress state is called a generalized plane stress state. Stress compatibility equations for plane stress problems To solve the plane problem in stresses, strain compatibility Eq. (5.42) must be written in stresses, using the Hooke’s law (5.47). As a result of this operation, we obtain: ] 1 ∂2 [ ] 1 ∂2 [ 1 ∂ 2 σx y = 0. σ + σ − 2 − νσ − νσ x x yy yy x x E ∂ y2 E ∂x2 2G ∂ x∂ y Next, we express the derivatives of tangent stresses using equilibrium Eq. (5.45): ∂ 2 σx y ∂ 2 σx x =− , ∂ x∂ y ∂x2

∂ 2 σ yy ∂ 2 σx y =− ∂ x∂ y ∂ y2

and substitute these ratios into the previous ratio. After replacing the shear modulus G with the expression 2G = E/(1 + v), we have: ( 2 ) ] ] ∂ 2 σ yy ∂ σx x ∂2 [ ∂2 [ σ + σ + (1 + ν) − νσ − νσ + = 0. xx yy yy xx ∂ y2 ∂x2 ∂x2 ∂ y2 After reductions: (

10

) ∂2 ∂2 + 2 (σx x + σ yy ) = 0, ∂x2 ∂y

Filon L. N. G., 1875–1937, English mathematician, physicist.

94

5 Mathematical Models of the Theory of Elasticity

or, using the Laplace operator in a planar coordinate system Δ ≡ ∂ 2 /∂ x 2 + ∂ 2 /∂ y 2 , we get: Δ (σx x + σ yy ) = 0.

(5.48)

Therefore, the sum of stresses is a harmonic function. • Airy11 stress function In the absence of mass forces, equilibrium Eq. (5.45) are usually satisfied by the introduction of a stress function in accordance with the formulas: σx x =

∂ 2ϕ ∂ 2ϕ ∂ 2ϕ . , σ = , σ = − yy xy ∂ y2 ∂x2 ∂ x∂ y

(5.49)

The function ϕ(x, y) is called the Airy function. Sum up normal stresses: σx x + σ yy =

∂ 2ϕ ∂ 2ϕ + = Δ ϕ. ∂x2 ∂ y2

Take the Laplace operator from both parts of the obtained equality. Since stress compatibility equations must be satisfied (5.48), this leads to a main differential equation to determine the Airy function: Δ Δ ϕ =

∂ 4ϕ ∂ 4ϕ ∂ 4ϕ + 2 + = 0. ∂x4 ∂ x 2∂ y2 ∂ y4

(5.50)

Equation (5.50) is a biharmonic equation and its solution is biharmonic functions. Many of its partial solutions are known, each of which corresponds to a certain stress state satisfying equilibrium equation and compatibility equation. For example: x 2 , y 2 , x y, x 3 , y 3 , x 2 y, x y 2 , x 4 − y 4 , cos(λx)ch(λy), cos(λy)ch(λx), . . . The main difficulty of the process of solution building is the selection of functions that satisfy boundary conditions. Numerous problems of the theory of elasticity, which are of great practical importance, have been solved taking into account specific boundary conditions. However, there are no general solutions to the biharmonic equation, and there are also no general methods for solving it. In order to close the mathematical boundary problem by defining the Airy function, boundary conditions and displacements must be expressed through it. 11

Sir George Biddell Airy (1801–1892) was an English mathematician and astronomer. His many achievements include work on planetary orbits, a method of solution of two-dimensional problems in solid mechanics.

5.10 2D-Problems of the Theory of Elasticity

95

Thus, in order to obtain an exact solution to the theory of elastic problem, it is necessary to find such functions that, in addition to satisfy the biharmonic Eq. (5.50), would also strictly satisfy the boundary conditions at each point of the solid surface. In detail, the plane problem of the theory of elasticity is considered in many textbooks. It should be noted that the classical work of N.I. Muskhelishvili. According to the theory of elasticity, much literature has been published. Control Questions 1.

Hooke’s generalized law for an isotropic body: 1) σi j = 2μεi j + λθδi j ; 2) U = σi j εi j /2; 3) εαα,ββ + εββ,αα − 2εαβ,αβ = 0.

2.

Hooke’s generalized law for an anisotropic body: 1) U = σi j εi j /2; 2) σi j = E i jkl εkl ; 3) σi j = 2μεi j + λθδi j .

3.

Clapeyron Formula: [ ] 3ν 1 σi j − 1+ν 1) εi j = 2G σδ i j ; 2) σi j = E i jkl εkl ; 3) U = σi j εi j /2.

4.

Total potential strain energy U: ( ) (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 ; 1) U = 1+ν 6E 2) U = 1−2ν + σ 2 + σ 3 )2 ; 6E( (σ1 ) 1 σ12 + σ22 + σ32 − 2ν(σ1 σ2 + σ2 σ3 + σ3 σ1 ) . 3) U = 2E

5.

Temperature is included in expressions linking stresses and strains: 1) deviators; 2) spherical parts; 3) intensities.

6.

In the absence of mass forces, displacements are: 1) harmonic functions; 2) biharmonic functions; 3) generalize functions.

7.

Potential Energy of shape change Uf : ( ) (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 ; 1) U f = 1+ν 6E + σ 2 + σ 3 )2 ; 2) U f = 1−2ν 6E( (σ1 ) 1 2 3) U f = 2E σ1 + σ22 + σ32 − 2ν(σ1 σ2 + σ2 σ3 + σ3 σ1 ) .

96

5 Mathematical Models of the Theory of Elasticity

8.

Potential energy of volume change UV : ( ) (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 ; 1) UV = 1+ν 6E + σ2 + σ3 )2 ; 2) UV = 1−2ν 6E( (σ1 ) 1 2 3) UV = 2E σ1 + σ22 + σ32 − 2ν(σ1 σ2 + σ2 σ3 + σ3 σ1 ) .

9.

Lamé’s equations are equilibrium equations in: 1) stresses; 2) strains; 3) displacements.

10. Beltrami-Michell equations are equations: 1) equilibrium; 2) compatibility; 3) expressions for elastic potential. 11. Lamé’s equations: [ ] 1 3ν σi j − 1+ν 1) εi j = 2G σδ i j ; 2) σi j = E i jkl εkl ; 3) (λ + μ)θ,i +μΔ u i + ρFi = 0 12. Beltrami-Michell equations: [ ] 3ν 1 σi j − 1+ν 1) εi j = 2G σδi j ; 2) σi j = E i jkl εkl ; Σ Σ ν ν ∂i ∂ j 3k=1 σkk = − 1−ν δi j 3k=1 ∂k f k − (∂i f j + ∂ j f i ), i, j = 3) Δ σi j + 1+ν 1, 3 13. In the absence of mass forces, the displacements satisfy the equations: 1) harmonic equations; 2) biharmonic equations; 3) Bessel equations. 14. Boundary conditions are specified on the surface: 1) deformed; 2) not deformed; 3) on both surfaces. 15. On the surface of the body carry out: 1) equilibrium equations; 2) initial conditions; 3) boundary conditions. 16. What are the boundary problems of the theory of elasticity in displacements? What are the first, second and mixed problems of the theory of elasticity in displacements?

5.10 2D-Problems of the Theory of Elasticity

97

17. What are the boundary problems of the theory of elasticity in stresses? What is the definition of an overridden system? 18. Stresses σzz = 0 in case of plane state: 1) deformed; 2) loading; 3) in both cases. 19. In case of plane strain state: 1) u = constant, v = constant, w = constant; 2) u = (x, y),v = (x, y),w = constant; 3) u = u(x, y, z), v = v(x, y, z), w = (x, y, z). 20. In case of plane strain state: 1) εzz = εx z = ε yz = 0; 2) εzz = constant, εx z = constant, ε yz = constant; 3) εzz = εzz (x, y), εx z = εx z (x, y), ε yz = ε yz (x, y). 21. Compatibility equations at plane stress state: ( ) 1 ∂u ∂v ; , ε = + 1) εx x = ∂∂ux , ε yy = ∂v xy ∂y 2 ∂y ∂x 2) θ =

∂u ∂x

∂ 2 εx x ∂ y2

+

3)

+

∂v ; ∂y

∂ 2 ε yy ∂x2

∂2ε

= 2 ∂ x ∂x yy .

22. For plane strain state, the number of independent stress components: 1) three; 2) four; 3) six. 23. Equilibrium equations at plane strain state: 1) 2) 3)

∂ 2 ε yy ∂2ε = 2 ∂ x ∂x yy ; ∂x2 ∂σ ∂σ + ∂ yx y = 0, ∂ yyy , εx x = ∂∂ux , ε yy = ∂v ∂y ∂ 2 εx x ∂ y2 ∂σx x ∂x

+

+

∂σx y ∂x

εx y =

= 0, ( 1 2

∂u ∂y

+

∂v ∂x

) .

24. For plane stress state: 1) σx x = constant, σ yy = constant, σzz = constant; 2) σzz = σx z = σ yz = 0; 3) σx x = σx x (x, y, z), σ yy = σ yy (x, y, z), σx y = σx y (x, y, z). 25. Airy stress function is defined: 1)

∂σx x ∂x

+

2) εx x = 3) σx x =

∂σx y ∂σ ∂σ = 0, ∂ yyy + ∂ xx y = 0; ∂y ( ) ∂u , ε yy = ∂v , εx y = 21 ∂u + ∂∂vx ; ∂x ∂y ∂y 2 2 ∂2ϕ , σ yy = ∂∂ xϕ2 , σx y = − ∂∂x∂ϕy . ∂ y2

98

5 Mathematical Models of the Theory of Elasticity

26. Airy stress function satisfies the equation: 1) harmonic; 2) biharmonic; 3) Laplace’s. Control Exercises 1.

2.

Show that for isotropic elastic media: 2μν ν λ 3kν 1 = 2(λ+μ) ; b) 1−ν = λ+2μ ; c) 1−2ν = 1+ν ; d) 2(1 + ν)μ = a) 1+ν 3λ+2μ 3k(1 − 2ν). Let the axis x 1 be the symmetry axis of order N = 2. Define the type of elastic constant matrix Cαβ , considering that Cαβ = Cβα . Answer: ⎤ ⎡ C11 C12 C13 C14 0 0 ⎥ ⎢C C C C ⎢ 12 22 23 24 0 0 ⎥ ⎥ [ ] ⎢ 0 0 ⎥ ⎢C C C C Cαβ = ⎢ 13 23 33 34 ⎥ ⎢ C14 C24 C34 C44 0 0 ⎥ ⎥ ⎢ ⎣ 0 0 0 0 C55 C56 ⎦ 0 0 0 0 C56 C66

3.

Let the axes that make up equal angles with coordinate axes be axes of elastic symmetry of order N = 3. Show that there are twelve independent elastic constants and the matrix of elastic constants has the form: ⎤ ⎡ C11 C12 C13 C14 C15 C16 ⎢C C C C C C ⎥ ⎢ 13 11 12 16 14 15 ⎥ ⎥ [ ] ⎢ ⎢C C C C C C ⎥ Cαβ = ⎢ 12 13 11 15 16 14 ⎥. ⎢ C41 C42 C43 C44 C45 C46 ⎥ ⎥ ⎢ ⎣ C43 C41 C42 C46 C44 C45 ⎦ C42 C43 C41 C45 C46 C44

4.

For an elastic body, let axis x 3 be the axis of elastic symmetry of order N = 6. Show that only the following elastic constants are non-zero: C11 = C22 , C33 , C55 = C44 , C66 = 0, 5(C11 − C12 ), C13 = C23 .

5.

Let the stress field for the medium be: ⎡ ⎤ x1 + x2 σ12 0 [ ] σi j = ⎣ σ12 x1 − x2 0 ⎦, 0 0 x2 where σ12 is function from x 1 and x 2 .

5.10 2D-Problems of the Theory of Elasticity

99

Let equilibrium equations be satisfied in the(absence of mass forces and let ) e1 the stress vector in the plane x1 = 1 be given as t = (1 + x2 )e1 +(6 − x2 )e2 . Define σ12 as a function from x 1 and x 2 . Answer: σ12 = x1 − x2 + 5. The displacement field is given through some vectors qi by the following equation: Λ

Λ

6.

Λ

) ( u i = 2(1 − ν)qi, j j − q j, ji /G. Show that the Lamé’s equations are satisfied if the mass forces are zeros and qi are biharmonic functions such that qi, j jkk = 0. Let q1 = x2 /r and q2 = −x1 /r (r 2 = xi xi ). Define the resulting stress field. Answer: ) ( σ11 = −σ22 = 6Qx1 x2 /r 5 ; σ33 = 0; σ12 = 3QG x22 − x12 /r 5 ; σ13 = −σ31 = 3Qx2 x3 /r 5 ; σ23 = 0, Q = 4(1 − ν)/G. 7.

Show what ϕ(x1 , x2 ) = x14 x2 +4x12 x23 −x25 is the Airy stress function. Calculate the stress tensor for this case under the assumption of a plane strain state with Poisson’s ratio ν = 0.25. Answer: ⎡ ⎤ 0 24x12 x2 − 20x23 −4x13 − 24x1 x22 [ ] ⎦. σi j = ⎣ −4x13 − 24x1 x22 12x12 x2 + 8x23 0 2 3 0 0 9x1 x2 − 3x2

8.

Airy stress function has the form: ϕ(x1 , x2 ) = Dx12 x23 + F x25 .

9.

Show that in this case F = −D/5. The( stress function ) has the form: ϕ = ϕ2 + ϕ3 + ϕ5 , where ϕ(x1 , x2 ) = D5 x12 x23 − 0.2x25 + 0.5B3 x12 x2 + 0.5A2 x12 x2 . Show that for this stress function the stress components are defined as: ( ) σ11 = D5 6x12 x2 − 4x23 ; σ22 = 2D5 x23 + B3 x2 + A2 ; σ12 = −6D5 x1 x22 − B3 x1 .

10. Beam with rectangular cross section of unit width and length 2L loaded with distributed load of intensity q (Fig. 5.6). Transverse forces act on both ends of beam V. List the six boundary conditions for this beam that the stresses must satisfy. Answer: 1) σ22 = −q, x2 = +c; 2) σ22 = 0, x2 = −c; 3) σ12 = 0, x2 = ±c;

100

5 Mathematical Models of the Theory of Elasticity

Fig. 5.6 Illustration for exercise 10

∫ c ∫ c 4) −c σ12 dx2 = q L , x1 = ±L; 5) −c σ11 dx2 = 0, x1 = ±L; 6) −c σ11 x 2 dx 2 = q L , x 1 = ±L. 11. Using the boundary conditions 1, 2, and 3 listed in exercise 10, show that for stress expressions in exercise 9, you need to: ∫ c

( ) A2 = −q/2; B3 = −3q/(4c); D5 = q/ 8c3 . In this case, the formulas for stresses have the form: ) ) ( ( 2 3 q 1 3 2 3 q 2 2 x x2 − x2 ; σ22 = x − c x2 − c ; σ11 = 2I 1 3 2I 3 2 3 ) q ( 2 x 1 x 2 + c2 x 1 , σ12 = − 2I where I = (2/3)c3 is plane moment of inertia of beam cross section. 12. Show that using the stresses calculated in exercise 11, the boundary conditions 4 and 5 of exercise 10 are satisfied and the boundary condition 6 is not satisfied. 13. Let’s continue exercises 11 and 12. We add an additional term to the stress function for implementation of boundary condition 6: ϕ = D3 x23 . Show that from the boundary condition 6 it follows that: D3 =

( ) L2 3q 1 − 2 . 4c 15 6c

Therefore, finally: σ11

) ( 2 2 1 2 1 3 q 2 x − x + c − L x2 . = 2I 1 3 2 15 6

Chapter 6

Mathematical Models of Solid with Rheological Properties

As mentioned repeatedly before, in order to construct a full system of equations for describing the solid behavior, it is necessary to attach the special relations describing the solid mechanical behavior (the relations between the solid stress–strain state) to the constitutive equations. In this section, we will talk about theories describing deformation processes in solids with rheological properties.

6.1 Creep and Relaxation In the theory of elasticity, it is assumed that external influences do not change over time, so the solid stress–strain state remains unchanged over time. However, many materials deform over time even under constant stresses. This property of materials is called creep. Creep can result in significant changes in the stress–strain state of solids over time. The mechanical properties of materials that are significantly time-dependent are called rheonomic properties. Creep can be experimentally examined by the stretching of cylindrical samples. Strain growth graphs over time at constant stresses are called creep curves. Test conditions and creep curve are shown schematically in Fig. 6.1. The upper end of the sample is fixed and the load is applied to the lower end. The strain change curve from time (creep curve) is constructed on the results of the length change of the sample. The strain increases from its initial value ε0 , which reflects the elastic properties of the material and corresponds to the “instantaneous” applied load P, that is, ε0 is the elastic initial strain. We can distinguish three characteristic sections on the creep curve. In the first section (AB), the creep speed is high, but decreases monotonically, starting from a certain value. This is a section of unsteady creep. The second curve section (BC) is almost straight, where the speed of creep is almost constant. This is a section of © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Zhuravkov et al., Mechanics of Solid Deformable Body, https://doi.org/10.1007/978-981-19-8410-5_6

101

102

6 Mathematical Models of Solid with Rheological Properties

Fig. 6.1 Creep curve

steady creep. The third section (CD) predates the destruction, Where the creep rate increases monotonically. It should be noted that creep is observed at any stresses. Therefore, it is impossible to specify any ultimate load at the beginning of creep. The phenomenon of reducing stresses in the body at a constant strain is called relaxation. Suppose that the sample was “instantaneously” stretched so that its “useful” length l 0 equal to l, (Fig. 6.2). To do this, the load P1 was required. Then we fixed the sample in a stretched state for a period of time. After that, the sample was released from fixation and a load was applied again, under the influence of which the sample “useful” part was stretched to the value of l. In this case, we need a load P2 less than the original load P1 . Therefore, the stress required to maintain a constant strain is reduced (relaxed). In solid deformable bodies, creep and relaxation, as rheonomic properties of the material, take place simultaneously and are interconnected. The relationship between creep and relaxation can be expressed analytically, introducing time into the relationship of solid stresses and strains. Viscoelasticity (relaxing) are solids with elastic and rheonomic properties. At the same time, a viscous element prevails in such bodies, and strains grow indefinitely Fig. 6.2 Relaxation curve

6.2 Solid Mathematical Models

103

under constant stresses. With constant strains, the initial stresses will decrease over time (relax) in such bodies. If the relationship between strain increments and stress increments at any moment is almost linear, then this medium behavior can be described using the linear creep model. In this case, the theory of linear hereditary media can be used to describe the medium deformation over time. The complete strain at any moment consists of two components: elastic strain at the loading and creep strain itself. Mathematically, this can be expressed in the following form: 1 1 ε(t) = σ(t) + E E

∫t L(t, τ)σ(τ)dτ, 0

where L(t, τ) is a function expressing hereditary material properties.

6.2 Solid Mathematical Models 6.2.1 Structural Rheological Models In the most general form, the differential equation linking stress and strain can be represented in the following form, while taking into account the development of deformation processes over time: dσ d2 σ dn σ + a2 2 + · · · + an n dt dt dt dε d2 ε dm ε b0 ε + b1 + b2 2 + · · · + bm m , dt dt dt

a0 σ + a1 =

(6.1)

where ai (i = 0, 1, . . . , n), b j ( j = 0, 1, . . . , m) are factors determined by material properties. The most common approach to obtain a rheological equation for a specific material today is to summarize a large number of experimental research data, construct the corresponding creep curves, and compare them with the curves with the same name based on theoretical models. In practice, a common approach is to use only a few terms in both parts of Eq. (6.1) to describe experimental data. This is equivalent to describe the viscoelastic behavior of materials based on rheological structural models consisting of elastic elements (their behavior is described using Hooke’s law) and viscous dampers (deformed according to Newton’s law), (Fig. 6.3). Structural units with elastic properties can be presented to springs with the Hooke’s deformation law:

104

6 Mathematical Models of Solid with Rheological Properties b

a

c

d

Fig. 6.3 Structural models of solids viscoelastic behavior: a viscoelastic (relaxing) model; b linear elastic model with lag; c linear solid model; d model with elastic lag and stress relaxation

σ = E ε,

(6.2)

where σ is the stress; ε is the strain; E is the modulus of elasticity (coefficient proportionality). A structural element that describes the material viscous behavior is called a Newton element. It is a piston with through holes in a cylinder containing viscous liquid (damper), (Fig. 6.3). This element corresponds to the simplest model of material linear viscous behaviour—Newton’s law of linear viscous deformation: dε or ε(t) = ε(0) + σ=η· dt

∫t 0

σ(τ) dτ, η

where dε/dt is the strain rate; η is the viscosity coefficient. Material has creep at a constant speed under a constant value of loading.

(6.3)

6.2 Solid Mathematical Models

105

Fig. 6.4 Parallel connection of elements

It should be noted that the Newton element cannot reproduce instantaneous elasticity. In addition, the concept of initial length cannot be defined for the Newton element, and it can accumulate unlimited deformation under constant loads. Remark In “pure form” the Newton element corresponds to the liquid behavior. Other linear viscoelastic models can be assembled as a parallel or serial connection of springs (Hooke’s model) and pistons (dampers) (Newton’s model): • When the elements are connected in parallel (Fig. 6.4), the total stress equals to the sum of the stresses in the elements, and the strains in the elements are the same and equal to the total structural strain: σ = σ1 + σ2 , ε = ε1 = ε2 .

(6.4)

• When connecting elements are consecutive (Fig. 6.5), the general rules are as follows: σ = σ1 = σ2 , ε = ε1 + ε2 .

(6.5)

Arbitrary rheological models can be compiled according to these rules. Note that there may be situations where different models describe the same material behavior (with selected appropriate material properties). Consider some “classic models”. Note that in this section, the models are described in the one-dimensional case. • Maxwell model Maxwell model is one of the simplest models to describe the viscoelastic behavior of solids. It represents a consecutive connection of a spring and the damper (Fig. 6.3a). In this case, if an external force is applied at the initial moment, then the spring will be instantly stretched (instantaneous elastic strains). If in the future the total stretch of the system remains constant (constant strains), then the spring will shrink and

Fig. 6.5 Consecutive (serial) connection of elements

106

6 Mathematical Models of Solid with Rheological Properties

gradually pull up the piston. At the same time, the external force should obviously decrease. Thus, Maxwell model illustrates the stress relaxation phenomenon. In a one-dimensional Maxwell model, following the rule (6.5), the stresses in the elements are the same, and the strains are summed up. If the equation of Hooke’s law is time differentiated, we get the connection between stress and strain, and we add a viscous component to the obtained strain rate: σ˙ E ση + . E η

ε = ε E + εη ⇒ ε˙ = ε˙ E + ε˙ η ⇒ ε˙ = Since σ = σ E = ση , we finally get: ε˙ =

σ˙ σ + . E η

(6.6)

Let’s consider the process of limited stress action and their corresponding strains in the period of time from 0 to t. Integrate the representation (6.6) and get the expression for deformations: 1 1 ε(t) = ε(0) + [σ(t) − σ(0)] + E η

∫t σ(τ)dτ.

(6.7)

0

We get expressions for stresses by multiplying (6.6) by E exp((E/η)t) [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge—Springer International Publishing Switzerland, 2015. 331 p.]. As a result, we obtain an expression that is a complete time derivative: ) ( ) E E t = σ˙ (t) exp t E ε˙ (t) exp η η ( ) ( ( )) E d E E t = σ(t) exp t . + σ(t) exp η η dt η (

Integrate this expression over time between 0 and t: ∫t 0

) ( ( ))|t | E E τ dτ = σ(τ) exp τ || E ε˙ (τ) exp η η τ=0 ( ) E t − σ(0). = σ(t) exp η (

Let’s rewrite this formula with respect to stresses:

6.2 Solid Mathematical Models

107

) ) ( ( ∫t E E σ(t) = σ(0) exp − t + E ε˙ (τ) exp − (τ − t) dτ. η η

(6.8)

0

The integral in the expression (6.9) can be rewritten as follows: ∫t 0

(

) ( )|t E E || ε˙ (τ) exp τ dτ = ε(τ) exp τ η η |τ=0 E − η

∫t 0

) E τ dτ. ε(τ) exp η (

As a result, we have: ) ( E σ(t) = σ(0) exp − t η ] )[ ( ) ( E E t − ε(0) + E exp − t ε(t) exp η μ ) ( ∫t E2 E − dτ, ε(τ) exp − t) (τ μ η 0

or: ) ( E σ(t) = Eε(t) + [σ(0) − E ε(0)] exp − t η t ( ) ∫ E E2 − ε(τ) exp (τ − t) dτ. μ η

(6.9)

0

The first term in (6.9) is the elastic component of the stresses. The second term expresses the influence of initial values, which decreases over time according to the exponential function. The third term is the convolution integral, which shows the contribution of strains: more remote events have less impact at the current time than more recent events (fading memory). Material behavior within Maxwell model demonstrates the results of creep test (Fig. 6.6, Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge—Springer International Publishing Switzerland, 2015. 331 p.). It is assumed that before t = 0 the stresses are zeros (σ = 0) and after that they are constant, i.e., σ = σk = constant, t > 0. We obtain it on the basis of (6.8), if ε(0) = 0:

108

6 Mathematical Models of Solid with Rheological Properties

Fig. 6.6 Tests for creep (a) and relaxation (b) within the Maxwell model

ε=

σk σk + t, t > 0. E η

(6.10)

From (6.11), it follows that there is an unlimited linear creep. In addition, it follows from (6.11) that strains from fast loads in this case are “extinguished” by elastic strains (instantaneous elasticity). In case there are constant strains εr (relaxation test), the damper will instantly operate and the spring elastically stretches so that the initial stresses equal to σ0 = Eεr . After that, according to (6.9), we have: ) ( E σ(t) = Eεr exp − t . η Thus, the stresses tend to zero until the spring is unloaded due to the piston displacements. The material constant tr = η/E (dimension “time”) is called the relaxation time. For t = tr , we get: σ(tr ) = Eεr e−1 ≈ 0.368σ0 . Using (6.6) (in this case ε˙ = 0) for t ≥ 0, we have: σ(t) = −tr σ˙ (t). .

For t = 0, the initial value σ˙ (t) is: σ(0) = −σ0 /tr . Therefore, the relaxation time is the inverse of the relaxation rate. The description of tensor of the Maxwell model can generally be presented as follows:

6.2 Solid Mathematical Models

109

[ ( ) ] ∂σi j ∂ei j 1 1 2 ∂θ + σi j = 2μ + δi j λ+ μ θ+λ . ∂t η ∂t η 3 ∂t • Kelvin-Foigt model Kelvin-Foigt model is a parallel connection between spring and damper. This model clearly explains the phenomenon of creep (Fig. 6.3b). If a constant external load (for example, tension) is applied to this model, then there will be no initial instantaneous strains, and the viscous piston will move as long as the spring allows. The Kelvin-Foigt body is an example of the simplest linear elastic-lag medium. In the Kelvin Foigt model, according to (6.4), the stresses in the elements are summed, and the strains in the elements are the same, i.e., ε = ε E = εη , σ E = Eε, σ = σ E + ση , .

.

ση = η ε, ⇒ σ = Eε + η ε .

(6.11)

To obtain the expression of strains, we apply to (6.11), the procedure we used to obtain expressions for stresses in the case of Maxwell model. We multiply (6.11) by η1 exp((E/η)t) [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge—Springer International Publishing Switzerland, 2015. 331 p.]. As a result, we get: ( ) ( ) ( ( )) ( ) E E E d E 1 E . σ(t) exp t = ε(t) exp t + ε(t) exp t = ε(t) exp t . η η η η η dt η Integrating over time in the range from 0 to t, we have: 1 η

(

∫t σ(τ) exp 0

) ( )|t ( ) E || E E τ dτ = ε(τ) exp τ | = ε(t) exp t − ε(0). η η η 0

We rewrite this expression regarding strains: ) ) ( ( ∫t 1 E E σ(τ) exp ε(t) = ε(0) exp − t + (τ − t) dτ. η η η

(6.12)

0

Thus, for the Kelvin model, the initial strain distribution fades exponentially over time, and stresses are introduced as an integral convolution with the exponential core. Let’s consider a creep test with the creep load σk from the time t = 0 and initial strains ε(0) = 0 (Fig. 6.7, Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge—Springer International Publishing Switzerland, 2015. 331 p.). In accordance with (6.12), we obtain the following

110

6 Mathematical Models of Solid with Rheological Properties

Fig. 6.7 Creep tests within the Kelvin model

expression: ) ( ) E E τ dτ exp − t exp η η 0 ) [ ( )] ( )|t ( σk E 1η E || E σk exp τ 1 − exp − t . = exp − t = ηE η |0 η E η

1 ε(t) = σk η

∫t

(

From this expression, we get: .

ε(t) =

( ) E σk exp − t . E η

Therefore, in this case there is no instantaneous elasticity. Creep speed slowly decreases and creep strains converge to elastic strains: ε∞ = σk /E. Relaxation time for the Kelvin-Foigt model tk = η/E is the time during which strains increase to the value: ε(tk ) =

) σk σk ( 1 − e−1 ≈ 0.632 . E E

The relation of the relaxation time to the value of initial strain rate ε˙ (t) is: ε˙ (0) = ε∞ /tk .

6.2 Solid Mathematical Models

111

Relaxation test is not feasible for this model because limited (final) loads cannot produce fracturing strains. The constitutive equations for the Kelvin-Foigt medium can be written in the following tensor representation: ( ) ∂ei j 2 ∂θ + δi j λθ − μtk , σi j = 2μei j + 2μtk ∂t 3 ∂t

(6.13)

where δi j is Kronecker delta (δi j = 1, δi j = 0 under i /= j); tk is stress relaxation time. Remark If tk = 0, from (6.13) follows the Hooke’s law. Therefore, it focused more on the shortcomings of the Maxwell and Kelvin-Foigt models. • The Maxwell element has properties that contradict the real material properties, namely: the invariability of strain rate during creep, the irreversibility of accumulated strain component and the complete stress relaxation. The irreversibility of strain component, which increases over time during creep and stress relaxation, is the characteristic of elastic–plastic but not viscoelastic bodies. The Kelvin-Foigt element does not have an elastic deformation component, so that the initial value of the modulus of elasticity tends to infinity: E → ∞ Due to the marked negative properties, both considered elements are rarely used as independent models. We can obtain media with a variety of properties by combining Maxwell and Kelvin-Foigt models (connecting mechanical analogues in series or in parallel). We’ll consider some of the most commonly used models. • Poynting model The Poynting model (another name is the “standard” linear solid model) is the “parallel connection” of an elastic body with a Maxwell body or the “consecutive” connection of a Kelvin body with an elastic body (Figs. 6.3c and 6.8). We perform a serial transformation in accordance with a similar procedure that used for previous models:

Fig. 6.8 The Poynting model or the “standard” linear solid model

112

6 Mathematical Models of Solid with Rheological Properties

σ = σ K + σC = σ D + σC ; σC = CεC , σ K = K ε K , σ D = D˙ε D ε = εC = ε K + ε D , ε˙ = ε˙C = ε˙ K + ε˙ D . σC C Therefore: ε˙ = σ˙CC = σ˙KK + σDD = σ˙ −˙ + σ−σ . K D After performing further transformations, we get:

ε˙ =

C σ C σ˙ − ε˙ + − ε. K K D D

(6.14)

C+K D σ˙ = Cε + D ε˙ . K K

(6.15)

And based on (6.14): σ+

As a result, we get the following exact expressions for the stress–strain state component with initial conditions at t = 0 σ(0) = σ0 , ε(0) = 0 [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge— Springer International Publishing Switzerland, 2015. 331 p.]: ) ] ( ) ( [ ∫ K K K2 t τ dτ exp − t , ε(τ) exp σ(t) = (C + K )ε(t) + σ0 − D 0 D D ) [ ( σ0 K KC σ(t) + − + − ε(t) = C+K C+K D(K + C) D(K + C)2 ⎤ ) ( ) ( ∫t KC KC ⎦ τ dτ × exp − t . (6.16) σ(τ) exp D(K + C) D(K + C) 0

The Poynting model demonstrates the properties of creep (Fig. 6.9a) and relaxation (Fig. 6.9b) and instantaneous elasticity (Fig. 6.9a) [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge—Springer International Publishing Switzerland, 2015. 331 p.]. To describe the Poynting model, the following representations are often used: ) ( ( ) ∂ Si j ∂ei j 1 1 1 1 ∂σ ∂e + Si j = 2μ + εi j ; a1 + σ = 3k + ε , a ∂t τ ∂t τ ∂t τ1 ∂t τ1 where εi j and Si j are the components of strain and stress deviators; μ and k are the initial elastic shear modulus and volume expansion modulus; a, a1 are rheological coefficients (a ≥ 0, a1 ≤ 1);τ, τ1 are shear and average stress relaxation times. • Burgers model Burgers model is a model formed by consecutively connecting a Maxwell body and a Kelvin body or a parallel connection of two Maxwell bodies. Burgers model

6.2 Solid Mathematical Models

113

Fig. 6.9 Material behavior corresponding to the Poynting model: a is the creep curve; b is the relaxation curve

describes a medium with both elastic-lag and stress relaxation and creep properties (Figs. 6.3d and 6.10). This model shows instantaneous elasticity due to the presence of spring C. Burgers model can be introduced as a parallel connection of two Maxwell models. In accordance with this, we have: σ = σC = σ D = σ K + σ R ; σC = CεC , σ K = K ε K , σ D = D˙ε D , σ R = R˙ε R ; ε R = ε K , ε = εC + ε D + ε K ,R . Therefore: σ˙ = K ε˙ K + R¨ε R = K (˙ε − ε˙ .C − ε˙ D ) + R(¨ε − ε¨ C − ε¨ D ) ) ( ) ( σ σ¨ σ˙ σ˙ + R ε¨ − − . = K ε˙ − − C D C D Finally, we have: σ¨ + C

(

Fig. 6.10 Structural element of the Burgers model

) 1 K 1 K K + + σ˙ + σ = ε¨ + ε˙ . D CR R DR R

(6.17)

114

6 Mathematical Models of Solid with Rheological Properties

The numerical integration allows us to transform the second-order differential Eq. (6.17) into a system of two first-order differential equations if we introduce the stress in the spring K as an internal variable: ε˙ =

σ σ σK K K σ˙ + + − and σ˙ K = K ε˙ − σ˙ − σ, C D R R C D

or in stress increments: ( σ K σ σK ) , σ˙ K = (σ − σ K ). σ˙ = C ε˙ − − + D R R R The creep curve for Burgers model under the creep load σ K (Fig. 6.11a) is the result of the superposition of creep curves of the Kelvin and Maxwell models [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge—Springer International Publishing Switzerland, 2015. 331 p.]: [ ( )] K σk σk σk 1 − exp − t ε(t) = + t+ . K R D C

(6.18)

At the beginning of the creep process, the damper R contributes to the creep strains (initial creep). The contribution of the damper R decreases more and more over time, and then the damper D begins to dominate with a steady creep speed σk /D. This is called the second or steady-state creep phase. For many real materials, this creep phase continues until the third creep phase, which cannot be described only by Burgers model. Fig. 6.11 Material behavior corresponding to the Burgers model: a creep curve; b relaxation curve

6.2 Solid Mathematical Models

115

Fig. 6.12 Structural diagram of the Burgers model

In the relaxation test (Fig. 6.11b), the stresses in spring C are relaxed due to the displacement of damper D and they tend to zero. In general, the Burgers model can be described using this equation: a

∂ 2 εi j ∂ 2 Si j ∂ei j ∂ Si j + b 2 = Si j + c + d 2 , a, b, c, d > 0. ∂t ∂t ∂t ∂t

Consider some of the Burgers model features in more detail [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge— Springer International Publishing Switzerland, 2015. 331 p.]. According to the Burgers structural scheme (Fig. 6.12), the relationship between strain and stress can be described using a second-order differential equation in this representation: η1 η2 η1 η2 ε¨ + η1 ε˙ = σ¨ + E2 E1 E2

(

) η1 η2 η1 η2 η1 η2 σ˙ + σ. + + E2 E1 E1

(6.19)

Direct integration of Eq. (6.19) is difficult due to the instability of the numerical method (second derivatives reach large values, which is caused by a strong oscillation of a system). Therefore, from the point of view of numerical implementation, it seems effective to implement the Burgers model at a uniaxial load, but the calculation is stable. The extension of the Burgers model at constant stress (creep) is expressed using the creep function in the following form (Fig. 6.13) [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge—Springer International Publishing Switzerland, 2015. 331 p.]: ( ( ( ))) E1 E2 σ E1 1+ 1 − exp − t ε= t+ . E1 η1 E2 η2

(6.20)

We show that the Burgers material creep at the beginning of the process corresponds to the Kelvin material (damping), at the end of the process corresponds to the Maxwell material (reaching the straight line of uniform extension). We decompose the function (6.20) into a Taylor row when t → 0. As a result, we obtain:

116

6 Mathematical Models of Solid with Rheological Properties

Fig. 6.13 Burgers model extension at constant stress

ε=

) ( ( ( ( ))) E1 E2 E1 σ E1 σ E1 1+ 1 − exp − t 1+ t+ = t+ t + o(t) . E1 η1 E2 η2 E1 η1 η2

For physical reasons, when t → 0, the “greater weight” has a viscous element η1 , that is, it can be considered that most often η1 >> η2 . For this reason: ( ) E1 σ 1+ t . ε≈ E1 η2 This expression corresponds to the Kelvin creep function: ( ( )) E2 σ 1 − exp t . ε= E2 η2 Remark In this case, E 1 = E 2 is accepted, but this is not essential for demonstration. Similarly, when t → ∞, we obtain: ε≈

( ) E1 σ 1+ t . E1 η1

That is, creep begins to correspond to Maxwell material. Therefore, in the Burgers model, two components can be distinguished: 1. The first component corresponds to the Kelvin model (Fig. 6.14) and it is described using the equation: ε˙ +

σ E2 ε= . η2 η2

It contributes significantly to the general expression only at high strain rates (˙ε). 2. The second component of the Burgers model corresponds to the Maxwell model (Fig. 6.15) and it is described using the equation:

6.2 Solid Mathematical Models

117

Fig. 6.14 Kelvin model extension graph

η1 ε˙ = σ +

η1 σ˙ . E

(6.21)

It makes a significant contribution at low strain rates (˙ε). Under constant load σ (t) = σ1 , converting Eq. (6.21), we obtain: σ=

E1 ε. 1 + (E 1 /η1 )t

(6.22)

The resulting expression is similar to the models of accumulated damage in the elements (in our case in springs). There are several approaches to construct algorithms for the numerical implementation of the “accumulated destruction” process. The first approach is as follows: we record the time that the spring was under loading, the value of which is greater than σ0 (we consider that with a lower load, there is no creep). When calculating the bonding forces, the spring is considered “degraded” and its stiffness is replaced as follows: k→

Fig. 6.15 Maxwell model extension graph

k . 1 + (k/η1 )t

118 Table 6.1 Various viscoelastic differential models

6 Mathematical Models of Solid with Rheological Properties Hooke model

P0 σ

= Q0ε

Newton model

P0 σ

= Q 1 ε˙

Maxwell model

P0 σ + P1 σ˙

= Q 1 ε˙

Kelvin model

P0 σ

= Q 0 ε + Q 1 ε˙

Poynting model

P0 σ + P1 σ˙

= Q 0 ε + Q 1 ε˙

Burgers model

P0 σ + P1 σ˙ + P2 σ¨

= Q 1 ε˙ + Q 2 ε¨

Model (p, q)-type P0 σ + · · · + Pp σ( p) = Q 0 ε + · · · + Q q ε(q)

Second approach (integral). We just integrate the expression (6.21). The result is the following formula for stress increment: ⎛ 1 σ˙ = E 1 ε⎝1 − εη1

∫t

⎞ σdt ⎠.

0

Then the spring stiffness is changed as follows: ⎛ 1 k → k ⎝1 − εη1

∫t

⎞ σdt ⎠.

0

• Viscoelastic Differential Models Table 6.1 summarizes the differential equations of the described rheological models. These equations are generally presented in the form as ordinary differential equations (ODE) of the order p with a time derivative of stresses and of the order q with a time derivative of strains. Constants P0 , . . . , Pp and Q 0 , . . . , Q q are determined by the properties of springs and dampers. In all cases, one of the non-zero constants must be normalized using the unit. These differential equations fully describe the behavior of their respective models. • Viscoelastic Models of Integral Type The following models describing linear viscoelastic processes are based on the Boltzmann superposition principle. The Boltzmann superposition principle can be formulated as follows: if the effects of deformation processes εi (t), i = 1, 2 are stresses σi (t), i = 1, 2, then the superposition ε1 (t) + ε2 (t) leads to stresses σ1 (t) + σ2 (t). We describe the procedure for obtaining a viscoelastic model of integral type.

6.2 Solid Mathematical Models

119

Let a constant load Δσi (t) act on the body during the period of time ti . This loading process can be determined through the Heaviside1 function H (t) = 0, t < 0; H (t) = 1, t ≥ 0 in the following form: σi (t) = Δσ(ti )H (t − ti ). That is, the stress Δσ(ti ) remains constant from the time ti to the new time t j . Let J (t − ti ) be the resulting deformation process from the action of a single load H (t − ti ) that began at the time ti . According to the superposition principle, the load Δσ(ti )H (t − ti ) initiates a deformation process of this type: εi (t) = Δσ(ti )J (t − ti ). The function J (t − ti ) is called the creep function. An arbitrary loading process σ(t) can be approximated as the sum of the following functions: σ(t) =

n Σ

Δσ(ti )H (t − ti ).

i=1

Then the corresponding strains have the form: ε(t) =

n Σ

Δσ(ti )J (t − ti ).

i=1

In the limit n → ∞, the strain function tends to the integral in the following form (Stieltjes2 integral): ∫σ(t) ε(t) =

∫t J (t − τ)dσ(τ) =

σ(0)

J (t − τ)σ˙ (τ)dτ.

(6.23)

0

By changing the role of stresses and strains, we get this resulting equation [the actions are the same as Eq. (6.23)]: ∫ε(t) σ(t) =

R(t − τ)dε(τ) = ε(0)

1

∫t R(t − τ)˙ε(τ)dτ,

(6.24)

0

Oliver Heaviside (1850–1925), an English mathematician and physicist who brought complex numbers to circuit analysis, invented a new technique for solving differential equations (equivalent to the Laplace transform), independently developed vector calculus. 2 Thomas Joannes Stieltjes (1856–1894), a Dutch mathematician. He was a pioneer in the field of moment problems and contributed to the study of continued fractions.

120

6 Mathematical Models of Solid with Rheological Properties

where R(t) is the relaxation function. The relaxation function and the creep function are related by the following relation [follows from Eqs. (6.23) and (6.24)]: ∫t t=

∫t R(t − τ)J (τ)dτ =

0

J (t − τ)R(τ)dτ. 0

Remark Since all rheological models of the (p,q)-type satisfy the superposition principle, all of them can be possibly represented with similar Stieltjes integrals. All viscoelastic models of this type have the following characteristic features: • They are linear. • Their behavior is temporarily dependent that is viscous (rheonomic). • They describe behavior with decaying memory. This means that if an event occurred long enough ago, then its effect in the present is negligible. • With very fast or very slow processes, these models behave elastically as long as all dampers become rigid or pliable, respectively. The rank of all these models is limited by their linearity. In general, linear differential operators of various orders are used to describe linear viscous-elastic media. The Volterra3 principle is very important, which allows the results of solving static problems in elasticity theory to be recalculated into the states of hereditary viscoelasticity.

6.2.2 Creep Damage A typical creep curve starts from the initial creep phase with high creep speeds. Then the creep curve goes into the second creep phase with a minimum and constant creep rate (steady creep phase). In the third stage, the creep rate increases again (nonlinear creep phase), (Fig. 6.16) [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge—Springer International Publishing Switzerland, 2015. 331 p.]. The tertiary creep stage ends in destruction. While the initial and second creep phases may be described using rheological models within linear viscoelasticity, such models are not applicable to the tertiary creep phase because they have non-linear behavior. Non-linear behavior in the third phase of creep is a consequence of various mechanisms. Among them are such as a reduction in the transverse area, which leads to an increase in real stresses; the appearance and growth of microcracks; changes of 3

Vito Volterra (1860–1940) was an Italian mathematician and physicist, known for his contributions to mathematical biology and integral equations, being one of the founders of functional analysis.

6.2 Solid Mathematical Models

121

typical creep stages

typical creep stages

Fig. 6.16 Typical creep curves with three creep phases

microstructure and others. The effect of these mechanisms is summed up into the “Creep Damage” (or “Creep Rupture”) phase as a phenomenological theory. An effective approach to describe the fracture of third creep stage was proposed by Kachanov.4 The main idea of the Kachanov approach is that only a part of the cross section of the sample S, namely a reduced cross section S l , withstands the force when creeping. This cross section is called the effective cross-section. This leads to the introduction of the concept of effective stresses: σl = F/Sl .

(6.25)

Effective stresses are greater than the true stresses σ = F/S. If the effective stresses are introduced into the creep law, then the creep speed increases in this case. We enter the fracture parameter d, defined as follows: d=

Sl S − Sl =1− or Sl = (1 − d)S. S S

(6.26)

Based on (6.26), it follows that the initial value of d is zero (d = 0) for the undisturbed material and grows monotonically during creep. If the value of parameter d approaches to one (d → 1), then effective stresses increase almost infinitely. In fact, the creep process stops and the destruction occurs when the critical value dkr is less than one: 0 < dkr < 1. The value dkr is considered as a material constant. The failure parameter d is considered as an internal variable for which an evolutionary equation is required. Therefore, for example, by analogy with the Burgers model, you can use such equation: 4

Kachanov Lazar Markovich (1914–1993), Russian mechanic, the largest specialist in the field of creep theory and strength theory.

122

6 Mathematical Models of Solid with Rheological Properties

d˙ = α˙σl + βσl + γd, where α, β, γ are constants. Yu. N. Rabotnov proposed the following equation for the parameter d: d˙ =

(

σl μ

)N ,

(6.27)

where μ, N are two positive constants. This equation gives good results in many applications. Equation (6.27) with (6.2) and (6.26) constitute the Kachanov-Rabotnov destruction theory. According to the definition of effective stresses (6.25), the speed of effective stresses is: ( ) F˙ F d F = − 2 S˙l . (6.28) σ˙ l = dt Sl Sl Sl In the case of a monotonous creep test, when the applied load F is constant, we obtain F˙ = 0. The growth of the effective cross-sectional area according to (6.26) is: S˙l = (1 − d) S˙ − d˙ S.

(6.29)

It is often assumed that the creep strains (but not fracture strains) occur isochorically (for example, this is a general assumption for metals). In addition, if we neglect the change in volume at the elastic stage, then for strains ˙ 0 and current volume V = l S, we ε, defined as ε = (l − l0 )/l0 , strain rate ε˙ = l/l obtain that the volume change of the prismatic sample is: ˙ V˙ = 0 = l˙S + l S˙ = ε˙l0 S + l S. From this, we obtain that: ε˙ ε˙l0 S =− S. S˙ = − l 1+ε Finally, the increase of effective stresses is: σ˙ l =

] ] [ [ ˙ F˙ ε˙ d F˙ F F ε˙ S = + . + 2 d˙ S + (1 − d) + Sl 1+ε Sl Sl 1 − d 1+ε Sl

(6.30)

Based on this, it can be concluded that the growth of effective stresses occurs due to: • Increments of applied load F;

6.2 Solid Mathematical Models

123

• Growth of failure parameter d; • Transverse contraction.

6.2.3 General Comments The “mechanical” setting of problems of creep theory is usually formulated as follows: we study the state of solid for a certain time interval 0 ≤ τ ≤ t when a given system of external loads and temperature acting on the body. The internal interaction between isotropic particles is characterized by stress, strain, and strain rate tensors. An equation or system of equations is used to connect the parameters of stress–strain state; specify a basic experiment to define material constants; find out the law of changing deformations over time. f p It is traditional to separate complete strains εi j into elastic εiej , plastic εi j and viscous (creep strain) εicj components: f

p

εi j = εiej + εi j + εicj (i, j = 1, 2, 3),

(6.31)

Remark In the case where these variables are used without indices, we assume that they correspond to the longitudinal component of the same strain under uniaxial loading. Currently, two alternative ideas are mainly used in constructing the defining creep equations. The first idea is based on the concept of a mechanical equation of state. The initial ratios for creep strain rates ε˙icj in this case are: ε˙icj = λ

∂f ∂f + binj , ∂σi j ∂qn

(6.32)

where f is creep potential given by the ratio: [ ] f = f I2 (Dσ ), I3 (Dσ ), ε˙ic , q1 , . . . , qr , T .

(6.33)

As follows from expression (6.33), the creep potential includes the invariants of the stress deviation I2 (Dσ ), I3 (Dσ ), the intensity of the creep strain rate ε˙ic and some internal parameters qn , the kinetics of which are described using the following evolutionary equation: dqn = ainj dσi j + binj dεicj + cn dt + d n dT ,

(6.34)

where a n , bn , cn , d n are some functions from σi j , εi j , t, T , as well as from parameters qn , (n = 1, r ), if the number of internal parameters is r.

124

6 Mathematical Models of Solid with Rheological Properties

The other designations in (6.32)–(6.34) are as follows: λ is proportionality coefficient; t is time: T is temperature. Summation is based on duplicate indices. Depending on the internal variable qn contained in the creep potential f, we obtain one or another creep theory. In particular, time t is∫ included in flow theory, and in hardening theory, the work ∫ of plastic deformation σi j dεicj , the Odqvist parameter dεic or the value of the accumulated creep strain εicj are included. In the theory of aging, instead of the potential of creep rates, the potential of creep strains is used, instead of the rate of creep strain intensity ε˙ic , the actual intensity of creep strains εic is used, and the time t is set as an internal parameter. To describe the third stage, internal parameters of scalar, vector, or tensor nature, reflecting the process of damage accumulation are additionally included in the creep potential f . In the framework of flow theory, for example, for the simplest case of onedimensional creep, it is accepted: ( )n σ dεc =B , dt 1−ω

(6.35)

where ω is a scalar damage parameter for which it is generally possible to be written as: [ ] σ(1 + εc ) m dω 1 =C ωβ , (6.36) r dt (1 − ω ) (1 − ω)q where C, m, r, q, β are experimentally determined coefficients (C > 0; m ≥ 1; q ≥ 0; β ≥ 0). Another approach to construct defining creep equations is based on the use of Volterra-Frechet5 integral representation: e = J1 ∗ dσ + J2 ∗ dσdσ + · · · ,

(6.37)

In (6.37), it is accepted that e = εe + εc and Ji are the weight functions of time. As simplified options (6.37), the following relationships are most often used: ∫t ε = ψ(σ) +

J0 (t1 , t2 , . . . , tn )σ(τ)dτ 0

and

5

Maurice Rene Frechet (1878–1973), French mathematician.

(6.38)

6.2 Solid Mathematical Models

125

∫t ϕ(ε) = σ +

J0 (t1 , t2 , . . . , tn )σ(τ)dτ,

(6.39)

0

where J0 is the Volterra integral operator with a special type of difference or nondifference core; ψ is the stress function. Equations (6.38) and (6.39) (unlike the first approach) take into account the loading history, but the third creep stage is also described using the damage parameter. The concretization of Eqs. (6.33), (6.38), and (6.39) is carried out on the basis of certain concepts (for example, aging or hardening theories) implemented in experiments (most often for uniaxial stretching). Generalization in the case of a complex stress state and non-stationary loading is carried out using the hypothesis of a “single creep curve”. Therefore, all the limitations and disadvantages of the one-dimensional models in this approach are completely preserved. The most principle is the inadequacy of the description of the third creep stage. For example, Eq. (6.35) together with (6.36) solve this problem approximately, since the true law of the damage effect on creep speed is unknown. When choosing a rheological equation for a specific material, the following procedure must be followed: 1. Static and dynamic experiments are carried out with samples from the examining material. For example, a diagram of uniaxial tension–compression is plotted at static and dynamic loading (with different loading rates). 2. Constructed experimental curves are compared with theoretical creep curves. By comparing the curves of the same name, a rheological body is selected, the theoretical curves of which are most “close” to the corresponding experimental curves. From the comparison of the same-name (theoretical and experimental) curves, the values of rheological parameters for the defined rheological equation are specified. This path selected by the rheological equation is the most common. Therefore, in the general case, the rheological equation describing the deformation process can be represented by a functionality in the following form: F(t, ε, ε˙ , ε¨, . . . , s, s˙ , s¨ ) = 0,

(6.40)

or in the form of functionals permitted relative to stress and strain components: [( ) ] [ ( ) ] t t s(t) = F1 ε τ , t , ε(t) = F2 s τ , t . 0

0

(6.41)

Functionals (6.41) are continuous in the region of functions s(t) and ε(t). If you designate the required functions as z(t) and y(t), these functionals can be represented in the following form:

126

6 Mathematical Models of Solid with Rheological Properties

∫t az(t) = y(t) + b

K 1 (t, τ1 )y(τ1 )dτ1 0

+

2

b 2!

∫t ∫t K 2 (t, τ1 , τ2 )y(τ1 )y(τ2 )dτ1 dτ2 0

,

(6.42)

0

bn ··· + n!

∫t

∫t ...

0

K n (t, τ1 , . . . τn )y(τ1 ) . . . y(τn )dτ1 . . . dτn 0

where a, b are constants; K n (t, τ1 , . . . τn ) are known functions specified independently from y(τ). Remark This conclusion follows from the Weierstrass6 theorem that any continuous function on the interval (c, d) can be approximated with a pre-set accuracy of the usual power series. Let’s limit in the right parts (6.42) to two terms and take into account that for “nonaging” linear-hereditary media, the kernels of integral equations can be different. Then they can be represented in the following form: ⎤ ⎡ ∫t 1 ⎣ ε(t) = s(t) + J (t − τ)s(τ)dτ⎦, 2μ ⎡ s(t) = 2μ⎣ε(t) −

0

∫t

(6.43)

⎤ R(t − τ)ε(τ)dτ⎦.

(6.44)

0

The representations for ε(t) and s(t) in the form (6.43) and (6.44) are very convenient, since they make it possible to move away from mechanical structural models and introduce new abstract models of rheological bodies—mathematical models of rheological bodies. Note that the expressions (6.43) and (6.44) are general, si j and εi j are components of stress and strain deviators, i.e., si j = σi j − σ · δi j , εi j = ei j − δi j e. By setting functions J (x, t) and R(x, t), and associating a new rheological body with them, an infinite set of new mathematical models of rheological bodies can be obtained. All these mathematical models will be linear models. They characterize the stress–strain state of hereditary elastic bodies. Representation (6.44) can be considered as the solution of (6.43). Therefore, R(t, τ) in (6.44) is resolvent, J (t, τ) in (6.43) and vice versa. The expression (6.43) can be divided into two parts: 6

Karl Theodor Wilhelm Weierstrass (1815–1897), German mathematician often cited as the "father of modern analysis".

6.2 Solid Mathematical Models

127

• The first term is instantaneous deformation (elastic) εel (t) = s(t)/2μ; • The second term is hereditary deformation (also elastic): 1 ε (t) = 2μ

∫t J (t − τ)s(τ)dτ,

sp

0

where the hereditary function J (t − τ) “remembers” the effect of loading at the time τ on the body deformation at the moment of observation t. According to the physical meaning J (t − τ) must be decreasing function and in case of μ = constant, it must be function of not t and τ, but (t − τ), since it depends only on the time passed from the moment τ of loading beginning to the moment t when we observe deformations from loading. If on the right side of the expression (6.42), we leave respectively 3, 4, 5 … first terms, then we get the non-linear integral equations respectively 3, 4, 5th … orders that can be mathematical models of the corresponding rheological bodies. Remark It should be emphasized that such a way to build new mathematical models of rheological bodies is not the only approach. For rheological equations in which the coefficients are constant, the kernels K (x, t) and ζ(x, t) in the corresponding integral equations can be represented as exponential functions, i.e., as follows: n Σ

( ) ) ( exp −k j (t − τ) , k j > 0 , ( j = 1, 2, . . . , n).

(6.45)

j

With the complication of integral Eq. (6.42) and the inclusion of a larger number of terms, there are great mathematical difficulties in solving them. Therefore, in this case, instead of (6.42), non-linear integral equations of the following form are more often considered: ∫t ϕ(ε) = s(t) +

J (t − τ)s(τ)dτ,

(6.46)

0

or such: ⎤ ⎡ ∫t 1 ⎣ ε(t) = s(t) + J (t − τ) f [s(τ)]dτ⎦, 2μ 0

(6.47)

128

6 Mathematical Models of Solid with Rheological Properties

where ϕ(ε) and f [s(t)] are defined functions characterizing non-linear deformations.

6.3 Rheological Equations of Linear Viscoelasticity 6.3.1 General Concepts First, recall some common basic concepts introduced earlier. Physical relationships between stress and strain deviators (si j = σi j − σ · δi j , εi j = ei j − δi j e) in the case of the linear viscoelasticity have the following form: ∫t J (t − τ)si j (τ)dτ,

2μεi j (t) = si j (t) +

K θ(t) = σ(t),

(6.48)

0

where θ = ekk is the relative volume change; σ = σkk /3 is the average (hydrostatic) stress; μ is the instantaneous elastic shear modulus; K is the instantaneous volume deformation modulus and function J(t) is the creep kernel. Physical interpretation of Eq. (6.48): the field of deformations εi j at a current moment of time is determined not only by the instantaneous stress si j (connect with strains by the generalized Hooke’s law), but also by previous stress values by means of some hereditary functions. The volumetric strain θ is taken to be elastic since the volumetric creep is small compared to the shear one. Equation (6.48) is invariant with respect to the beginning of the time reference, since the creep kernel J(t) has a difference (t – τ ) with its arguments. The kernel J(t) is a monotonically positive decreasing function of its arguments. At infinity, it asymptotically tends to zero. It is identical to zero, if the argument is negative. If Eq. (6.48) can be resolved with respect to si j and σ, then: ⎛ si j (t) = 2μ⎝εi j (t) −

∫t

⎞ R(t − τ)εi j (τ)dτ⎠,

K θ(t) = σ(t).

(6.49)

0

Function R(t) is the relaxation kernel. It is a kernel J(t) resolvent. Of course, the function J(t) in turn is the resolute kernel R(t) resolvent: ∫t R(t) − J (t) =

J (t − τ)R(τ)dτ. 0

Deviator ratios (6.48), (6.49) can be written in another form:

(6.50)

6.3 Rheological Equations of Linear Viscoelasticity

∫t εi j (t) =

129

∫t K (t − τ)dsi j (τ)dτ, si j (t) =

0

F(t − τ)dεi j (τ)dτ,

(6.51)

0

where K(t) is the creep function, F(t) is the relaxation function. Between these functions and the previously introduced kernels J(t), R(t), a connection can be established by integrating the ratios (6.51) in parts: |t εi j (t) = K (t − τ) si j (τ)|0 +

∫t

dK si j (τ)dτ. dτ

0

We accept sij (0) = 0 and K (0) = 1/(2μ), J (t) = 2μdK (t)/dt. As a result, we come to Eq. (6.48). Similarly, the identity of Eq. (6.49) and the second equation of (6.51) can be established.

6.3.2 Examples of Creep and Relaxation Kernels The functions J(t) and R(t) are defined most often as the results of experiments on pure shear stress test. Based on the experiments, creep curves are built ε12 (t)/σ12 0 ~ t. In this case, all other components of stress and strain tensors are zeros. Equation (6.48) will be reduced to one: ∫t J (t − τ)s12 (τ)dτ.

2με12 (t) = s12 (t) + 0

0 From this expression, in the case of creep at constant stress σ12 (t) = σ12 = const, we obtain:

ε12 (t) 2μ 0 = 1 + s12

∫t J (ξ)dξ. 0

The obtained expression allows you to determine the function J(t): J (t) = 2μ

ε12 (t) dK (t) , K (t) = 0 . dt s12

The K(t) is the pliability (creep function) in pure shear. Similarly, relaxation kernel R(t) can be determined by the pure shearing experiments. Various analytical

130

6 Mathematical Models of Solid with Rheological Properties

expressions for creep and relaxation kernels are used to approximate experimental data. The simplest kernel of the integral equation of the most commonly used linear hereditary creep theory is the whole exponent. However, we can describe real creep curves with only one exponential function and only in a short time interval. Therefore, kernels as the sum of exponent are more often used. The main disadvantage of exponential kernels is that a large number of material parameters are required to describe the mechanical behavior of real materials. In addition, for such kernels, the strain rate dε/dt at the initial time t = 0 is a finite value, but numerous experiments indicate that the strain rate at the initial time is large, and its concrete determination is very difficult. Therefore, the heredity kernels are usually given using various weakly singular functions that set an infinitely large creep rate at t = 0. 1. Exponential creep kernel: J (t) = Ae− pt (A > 0, p > 0). Unfortunately, there are only two material constants A and p in this representation, making it difficult to describe the experimental data over a wide time range, especially in the initial loading period. The resolvent kernel (relaxation kernel) has the form: R(t) = Be−qt . Substituting the exponential creep kernel and its resolvent into the ratio (6.50) and requiring that it be an identity by t, we obtain: B = A, q = A + p. A more accurate representation gives the kernel and its resolvent as a sum of exponential functions: J (t) =

m Σ i=1

Ai e− pi t ,

R(t) =

m Σ

Bi e−qi t .

i−1

Substituting these expressions into the integral relationship between creep and relaxation kernels (6.50) and requiring them to be identical, we obtain 2m equations connecting the constants of resolvents Bi and qi and the kernel constants Ai and pi : m Σ i−1

m Σ ( ) Bi Ai = 1, = 1, j = 1, m q j − pi q − p j i−1 i

6.3 Rheological Equations of Linear Viscoelasticity

131

These equations are useful in determining one of the kernels if the others are known. For example, if we determined the constants Ai and pi after processing experimental data. Then, from the first system of equations, the roots qi are determined. They must be real, unequal among themselves and different from pi . After that, constants Bi are determined by the second system of linear equations. 2. Power kernel Duffing7 :

J (t) =

C t (1−β)

(0 < β < 1).

The limitation on the constant β is due to the fact that the strain rate and the strain itself at the time of the loading application become infinitely large when β = 0. The Duffing kernel “works worse” with sufficiently long loading periods. The Duffing kernel resolvent is the following function: R(t) =

Σ

(−1)k+1

C k Γ∗k (β)t kβ−1 . Γ∗ (kβ)

where Γ∗ (β) is the gamma-function of the argument β. With limitations on the value of β (0 < β < 1) this series converges everywhere. This type of kernel is called kernel with a weak singularity or kernel with a weak feature. This feature is that at t → 0, J(t) → ∞. This corresponds to an infinitely high strain rate at the moment of the loading application. A weak desire for infinity is provided by the restriction 0 < β < 1. The Duffing kernel cannot satisfactorily describe the experimental data with a wide time interval due to the presence of a limited number of arbitrary constants. 3. Rzhanitsyn8 core:

R(t − τ) =

Ae− p(t−τ) (t − τ)(1−β)

, ( p > 0, 0 < β < 1).

(6.52)

This is a typical relaxation core in which exponential and weakly singular properties are combined. Kernel (6.52) resolvent is the following function: ∞ Ak Γ∗k (β)(t − τ)kβ e− p(t−τ) Σ , J (t − τ) = (−1)k+1 t − τ k=1 Γ∗ (kβ)

where Γ∗ (β) is the gamma-function. 7 8

Georg Wilhelm Christian Caspar Duffing (1861–1944), German physics-engineer. Aleksey R. Rzhanitsyn (1911–1987), Soviet famous scientist, mechanic.

132

6 Mathematical Models of Solid with Rheological Properties

It differs from the Duffing core resolvent only by the multiplier e−pt . The Rzhanitsyn core is often used in applied research. 4. Koltunov9 kernel. Koltunov kernel has a more general representation than the Rzhanitsyn kernel: R(t − τ) =

Ae− p(t−τ)

α

(t − τ)(1−β)

, ( p > 0, 0 < α < 1, 0 < β < 1)

(6.53)

The resolvent of this kernel is the function:

J (t − τ) =

− p(t−τ)α

e (t − τ)α

∞ Σ

[ (−1)k+1

( AΓ∗

k=1

α+β−1 α

)]k

(t − τ)k(α+β−1) ) ( . Γ∗ k α+β−1 α

5. Abel10 kernel:

Jα (t − τ) =

(t − τ)α (−1 < α < 0), Γ∗ (1 + α)

(6.54)

where Γ∗ (1 + α) = αΓ(α) is the gamma–function. Abel kernel allows the reversibility of integral ratio (6.48), wherein the constant α is determined by experimental creep or relaxation curves. Abel kernel corresponds to experimental data only in the initial stage of deformation, but then the unlimited growth of deformation is not confirmed by the experiment. The applicability of Abel kernel can be substantially extended by moving to the non-linear hereditary elasticity equations. 6. Rabotnov fractional-exponential kernel. The Rabotnov function is a generalization of the Abel kernel (for the first time for β = –1): .α (β, t − τ) = (t − τ)α

∞ Σ n=0

βn (t − τ)n(1+α) . Γ∗ [(n + 1)(1 + α)]

Series .α (β, t − τ) converges for all values t and β. Obviously: .α (β, t − τ) = Jα (t − τ).

9

Michail A. Koltunov, Soviet scientist-mechanic. Niels Henrik Abel (1802–1829), Norwegian mathematician who made pioneering contributions in a variety of fields. His most famous single result is the first complete proof demonstrating the impossibility of solving the general quintic equation in radicals.

10

6.3 Rheological Equations of Linear Viscoelasticity

133

The Yu. N. Rabotnov function .α (β, t − τ) is called a fractional-exponential kernel. This core has become widespread in the literature, due to the great possibility of describing the rheological properties of materials and the presence of a sufficiently developed special algebra of the corresponding operators. At the same time, it is very difficult to calculate the Rabotnov kernel, due to the presence of a large number of defined constants. 7. Power core of general form. In this case, the creep function is: J (t − τ) = (t − τ)α−1 e−β(t−τ) .

(6.55)

The physical meaning (6.55) follows from the following formula for the creep function: J (t) = (E/σ)e(t). ˙ Therefore, J (t − τ), with accuracy to a constant factor, is a strain rate function of a rod that is in a uniaxial tension (or compression) state by stresses of a constant value: σ = constant. 8. Wulfson11 -Koltunov kernel is a generalization of the Rzhanitsyn and Rabotnov kernels:

K (A, α, β, t − τ) = a exp[−β(t − τ)]

∞ Σ An (t − τ)n(1+α)+α . Γ [(n + 1)(1 + α)] n=0 ∗

(6.56)

9. The method of constructing linear defining equations, based on the use of the Gavrillac-Negami ratio for the complex modulus of elasticity is promising (for example, for composite, anisotropic materials): ]−γ [ E(i ω) = E 0 − (E 0 − E ∞ ) 1 + (i ωτ)(α+1) , 0 < γ ≤ 1, where E 0 is the instantaneous modulus of elasticity and E ∞ is the long-term modulus (about ω → 0). Expressions for relaxation functions include the corresponding GavrillacNegami kernel: )−γ ( Γ α,β,γ ( p) = p (α+1) + β . 11

Iosif I. Wulfson, Soviet scientist-mechanic.

134

6 Mathematical Models of Solid with Rheological Properties

In general, many types of different heredity kernels are proposed. The advantages of one or another kernel depend on the physical properties of the material and the type of task to be solved. The choice of linear kernel is explained primarily by the desire to simplify the integral equations of viscoelasticity, which in this case allow reversibility, that is, an analytical solution to the problem of resolvent can be obtained. However, it should be noted that the problem of obtaining new kernels continues to be urgent.

6.3.3 Volterra Principle Let’s describe more detail about the Volterra method related to the linear theory of solving viscoelastic problems of the type (6.49). Expressions (6.49) are rewritten in symbolic form by entering the following integral operator G* : si j = 2G ∗ .i j , σ = K θ,

(6.57)

where ) ( G = G 1 − R∗ , R∗ f = ∗

∫t R(t − τ) f (τ)dτ, G is shear modulus. 0

By comparing (6.57) with the relations of Hooke’s law for the linear theory of elasticity, it can be concluded that the physical relations of linear viscoelasticity have exactly the same form as the equations of generalized Hooke’s law, and only the shear modulus G is replaced by the operator G* . In the operator form, we also represent Eq. (6.48) in the following form: .i j =

1 si j , σ = K θ, 2G ∗

where the operator 1/G* is: ) 1 1( 1 + J∗ , J∗ f = = G∗ G

∫t J (t − τ) f (τ)dτ. 0

Substituting where the expression from (6.57) G ∗ = G(1 − R ∗ ), we get: 1 = 1 + J ∗. 1 − R∗

6.3 Rheological Equations of Linear Viscoelasticity

135

Thus, we determined the operation of dividing by some operators. From where, we come to the following formula of the Voltérra principle: the solution of linear problem of viscoelasticity can be obtained as solution of the corresponding problems of linear theory of elasticity, in which elastic constants (properties) are replaced by some operators. In this case, the shear modulus G should be replaced by operator G* . Solutions to the problems of elastic theory, in addition to the shear modulus, contain other elastic constants, for example, the modulus of elasticity E and the Poisson’s ratio ν. In accordance with the described approach, in the elastic solution, replace the modulus of elasticity E and Poisson’s ratio v with constants G and K, and then replace the constant G with an operator G* . Using the Voltérra principle, we get a solution including algebraic or transcendent functions of time operators. In addition, this solution must also be “decrypted”. In the general case, for example, the Laplace-Carson12 integral transform, the Ilyushin approximation method, the Rabotnov operators, and others are used. In particular, the Laplace-Carson image of the function f (t) is called the following function f * (p) of the real parameter p: ∫∞

f ( p) = p

f (t)e− pt dt.

0

If f (t) = f 0 , then f * (p) = f 0 . If the function f (t) is a convolution of the two functions F and ψ, that is: ∫t f (t) =

F(t − τ)dψ(τ),

(6.58)

0

then its image equals to the product of images of sub-integral functions: f ∗ ( p) = F ∗ ( p)ψ∗ ( p).

(6.59)

In turn, if the image of the function f (t) is determined by the relation (6.59), then the function itself is represented by the formula (6.58). The complete system of equations includes the equations of linear viscoelasticity (6.51) with elastic volumetric strain, equilibrium equations, Cauchy ratios, and boundary conditions: ∫t εi j (t) =

J (t − τ)dsi j (τ)dτ, σ = K θ; σi j, j + ρFi = 0; 0

) ( ei j (t) = u i, j + u j,i /2; σi j l j = Ri on Sσ , u i = u 0i on Su . 12

J. Carson Mark (1913–1997), Canadian-American mathematician.

(6.60)

136

6 Mathematical Models of Solid with Rheological Properties

We apply the Laplace-Carson transform to the ratios (6.60) and considering (6.59), we obtain: .i∗j = J ∗ si∗j , σ∗ = K θ∗ , σi∗j, j + ρFi∗ = 0, 2ei∗j = u i,∗ j + u ∗j,i , σi∗j l j = Ri∗ on Sσ , u i = u ∗0i on Su .

(6.61)

It follows that the linear viscoelastic problem in images is identical to the corresponding problem for an elastic body. Therefore, the solutions to these problems coincide with the only difference that the constant (1/2)G is replaced by the image G* . The modulus of volume compression in the solution is saved. If we find the solution of the linear problem of elasticity theory for images σi∗j , ei∗j , then we will find the originals from these images and determine the solutions σi j (x, t), ei j (x, t). Therefore, when solving problems for viscoelastic bodies, it is enough to first determine the solution of the same problem in the classical theory of elasticity, and then in the found solution replace the elastic constants E and ν, operators E and ν. Remark It should be noted that the Volterra principle is not the only method for solving viscoelastic problems. There is no exact classification of problems that can be resolved using the Volterra principle. Therefore, it has been more or less established that for problems with time-interchangeable boundaries, the Volterra principle is not suitable. According to the Volterra principle, it follows that in all problems whose final results do not include constant materials, the results of the classical theory of elasticity can be right for any rheological body with the general law of linear deformation: ⎡ si j (t) = 2μ⎣εi j (t) −

∫t

⎤ ϕ(t − τ)εi j (τ)dτ⎦.

(6.62)

0

Such problems include, for example: 1. The first main problem of the theory of elasticity in stresses for single-connected regions (the problem of the stress state of the elastic body at external forces set on its surface). In the formulas for stress components (rather than strains), the constants of elastic material do not enter. Therefore, this solution can be right for the rheological body with the general law of linear deformation (6.62). 2. Stress problems when examining the stress concentration around various holes whose contour is free from external forces, or when the principal vector of external forces applied to the hole contour is zero. Solutions of viscous-elastic non-linear problems on stress concentration around curvilinear holes show that the formulas for stress components near arbitrary holes include elastic constants of non-linear materials. Thus, for viscoelastic materials, not only the displacements, but also the stresses near the hole change over time. In order to adequately describe the changes in stresses near arbitrary holes in real

6.4 Equations of Mechanics of Linear-Hereditary Media

137

materials without leaving the area of linear problems, it is enough to refuse the ideal isotropic model and take as the main anisotropic model. In this case, it is possible to use appropriate solutions to the problems of the theory of elasticity of an anisotropic medium. Next, the Volterra principle is used, that is, instead of elastic constants, operators of the form (6.57) are introduced and we obtain the law of changing stresses near holes over time.

6.4 Equations of Mechanics of Linear-Hereditary Media The method of obtaining a resolving system of equations of mechanics of small viscoelastic strains for media subjects to the general law of linear deformation does not differ from the method of obtaining these equations used in the classical theory of elasticity. The only difference among these systems of equations is that the law describing the relations between stresses and strains in the theory of elasticity must be replaced by the corresponding rheological equations. The complete system of equations of the theory of linear viscoelasticity in the case of small strains has the form: (a) Equilibrium equations Σ ∂σi j j

∂x j

+ X i = 0, i = 1, 3

(6.63)

(b) Boundary conditions Σ

) ( σi j cos n, x j = X ni ,

(6.64)

(c) Saint-Venan strain compatibility equation ∂ 2 e11 ∂ 2 e22 ∂ 2 e12 + =2 ,··· 2 2 ∂ x1 ∂ x2 ∂ x2 ∂ x1

(6.65)

(d) Cauchy relation ei j =

) ( ∂u j 1 ∂u i , + 2 ∂x j ∂ xi

(e) Rheological equations

Ee11 = σ11 − ν(σ22 + σ33 ) Ee22 = σ22 − ν(σ11 + σ33 )

(6.66)

138

6 Mathematical Models of Solid with Rheological Properties

Ee33 = σ33 − ν(σ11 + σ22 ) σ23 =

E E E e12 , e13 , σ12 = e23 , σ13 = 1+ν 1+ν 1+ν

(6.67)

where ) ( ( − ∗) ∗ E = E 1− E , ν =ν 1+ν , ∗

∫t

∫t

ε(t − τ ) f (τ )dτ, ν f (t) =

E f (t) = 0

(6.68)

N (t − τ ) f (τ )dτ,

(6.69)

0

where E, ν are respectively, modulus of elasticity and Poisson’s ratio at elastic deformation of the material. ε(t − τ) and N (t − τ ) are heredity kernels. From expressions (6.67), it follows that: −

E=

( ) μ 3λ + 2μ λ+μ

λ ), ; ν= ( 2 λ+μ

or λ=

Eν E E 2 ; μ=G= ;χ = λ + μ = ; (6.70) 3 2(1 + ν) 3(1 − 2ν) (1 + ν)(1 − 2ν)

where χ is the modulus of material volume deformation. The introduction of two independent kernels ε(t − τ ) and N (t − τ ) [in accordance with (6.69)] gives more opportunities to select analytical dependencies to adequately describe the real material behavior. Many materials (especially metals) at volumetric deformation follow the linear E elastic law, i.e., the Hooke’s law. For such materials, the characteristic χ = 3(1−2ν) must be a constant, i.e., time independent. Therefore, for such materials χ = χ = constant or E E = . 3(1 − 2ν) 3(1 − 2ν)

(6.71)

Then (

] [ ] [ ( )) 1 − 2ν 1 − 2ν ∗ 1 ∗ 1− E =ν 1+ E = E 1− E ν= E . 2 E 2ν

(6.72)

In this case, we obtain [on the basis of (6.52), (6.53) and (6.60)] that only one operator will be included in the formulas for stresses and strains of isotropic materials whose volumetric deformation is linear elastic:

6.4 Equations of Mechanics of Linear-Hereditary Media

( ∗) E = E 1− E , ∗

where E ·1 =

∫t

139

(6.73)

ε(t − τ ) · 1 · dτ .

0

That is, in this case, the stress and strain components will have one function of influence ε(t − τ ) . The system (6.63)–(6.67) is a full system of equations of mechanics of linearhereditary media. The hereditary theory of viscoelasticity corresponds to the bodies in which mechanical structural models, the viscous element is connected in a pair parallel to the elasticity element. Therefore, for example, Maxwell’s elastic-viscous medium, Burger’s body and the like cannot be the basis of the viscoelastic hereditary theory, since they describe inelastic deformations. Therefore, functions R(t − τ) and J (t − τ) in mathematical models of the corresponding rheological bodies in the hereditary theory of viscoelasticity should reflect linear reversible processes. Control Questions 1.

With creep strains over time: 1) increase; 2) decrease; 3) do not change.

2.

Stresses during relaxation over time: 1) increase; 2) decrease; 3) do not change.

3.

Physical relationships of linear viscoelasticity: [ ] ∫t 1) si j (t) = 2μ εi j (t) − R(t − τ)εi j (τ)dτ ; 0

2) 2Gϕ(εu ) .i j (t) = si j (t); 3) 2G .i j (t) = si j (t). 4.

Volumetric deformation of viscoelastic bodies: 1) elastic; 2) plastic; 3) linearly viscoelastic.

5.

Rheological properties of the material characterize: 1) modulus of volume strains; 2) plasticity function; 3) creep and relaxation kernels.

140

6.

6 Mathematical Models of Solid with Rheological Properties

Creep and relaxation cores are connected by dependency: 1) linear; 2) integral; 3) differential.

7.

In the exponential creep kernel the material constants: 1) one; 2) two; 3) three.

8.

In the Duffing creep kernel, the material constants: 1) one; 2) two; 3) three.

9.

Constant of material in Rzhanitsyn creep kernel: 1) one; 2) two; 3) three.

10. With increasing strain rate, the yield strength: 1) does not change; 2) increases; 3) decreases. 11. Volterra principle: the solution of the viscoelastic problem can be obtained from the solution of the corresponding problem: 1) elasticity; 2) plasticity; 3) damage. 12. Maxwell model, Kelvin-Foigt model? 1) ε˙ = Eσ˙ + λσ ; 2) σ = Eε + λ˙ε. Control Exercises Remark Exercises are taken from the book [Mase, G. T. Continuum mechanics for engineers/G. T. Mase, G. E. Mase—CRC Press LLC, 1999. 380 p.]. We recommend solving other exercises given in the book. 1. A four-parametric model consists of a Kelvin structural model and a Maxwell model (Fig. E.1).

6.4 Equations of Mechanics of Linear-Hereditary Media

141

Fig. E.1 Four-parametric structural model

Knowing that γMODEL = γKELVIN + γMAXWELL , where γ is strain, and taking into account the operator equation (∂t ≡ ∂/∂t ) for Kelvin structural model (a) σ12 = (G + η∂t )γ12 and for Maxwell model (b) (∂t + 1/τ)σ12 = (G∂t )γ12

Construct a relation describing the behavior of this model. Answer: G 2 η1 γ¨ + G 1 G 2 γ˙ = η1 σ¨ + (G 1 + G 2 + η1 /τ2 )˙σ + (G 1 /τ2 )σ . 2. Construct defining equations for the three-parametric models shown in Fig. E.2. Answer: ˙ 1 = [(η1 + η2 )/η1 η2 ]˙σ + (1/τ1 η2 )σ; a) γ¨ + γ/τ b) σ˙ + σ/τ2 = (G 2 + G 1 )γ˙ + (G 1 /τ2 )γ; ˙ c) σ˙ + σ/τ2 = η1 γ¨ + (G 2 + η1 /τ2 )γ. 3. The model consists of a Kelvin unit connected in parallel to the Maxwell unit, (Fig. E.3). Construct a defining relation for such a model. Answer: σ˙ + σ/τ2 = η1 γ¨ + (G 1 + G 2 + η1 /τ2 )γ˙ + (G 1 /τ2 )γ. 4. For the four-parametric model shown in Fig. E.4, to define: a) Defining relation; b) Relaxation function G(t). Remark In this case G(t) is the sum of the functions G(t) of all parallel connected structural units. Answer: a) σ˙ + σ/τ2 = η3 γ¨ + (G 1 + G 2 + η3 /τ2 )γ˙ + (G 1 /τ2 )γ; b) G(t) = G 1 + G 2 exp(−t/τ2 ) + η3 δ(t).

142

6 Mathematical Models of Solid with Rheological Properties

Fig. E.2 Different types of three-parametric structural models

Fig. E.3 Example of a Kelvin unit connected in parallel to the Maxwell unit

Fig. E.4 Special type of four-parametric structural model

5. There is the history of the model stress over time in Fig. E.5. Determine strains γ(t) for this load in the time interval: a) 0 ≤ t/τ ≤ 2; b) 0 ≤ t/τ ≤ 4. To get an answer in (b), use the superposition principle. Answer:

6.4 Equations of Mechanics of Linear-Hereditary Media

143

Fig. E.5 One of a possible type of three-parametric structural models

a) γ(t) = σ0 J (2 − exp(−t/τ))U (t); b) γ(t) = σ0 J (2 − exp(−t/τ))U (t) − σ0 J (2 − exp(−(t − 2τ)/τ))U (t − 2τ). 6. For the model shown in Fig. E.6, define: a) Defining equation; b) Relaxation function G(t); c) Stresses σ(t) for the interval 0 ≤ t ≤ t1 , when strains are set by the corresponding graph Answer: a) σ˙ + σ/τ = η¨γ + 3G γ˙ + (G/τ)γ; b) G(t) = ηδ(t) + G(1 + exp(−t/τ)); c) σ(t) = λ(2η − η exp(−t/τ) + Gt)U (t). 7.

For the model from the previous exercise, define the stresses σ(t) for the case where the strains γ(t) are set by the graph shown in Fig. E.7. Answer: γ0 (ηδ(t) + G(1 + exp(−t/τ)))U (t) + σ(t) = λ(η(2 − exp(−t/τ)) + Gt)U (t).

Fig. E.6 Four-parametric structural model and its graph of strain mode

144

6 Mathematical Models of Solid with Rheological Properties

Fig. E.7 The graph of strains mode

8.

9.

The three-parameter model shown in Fig. E.8 undergoes the strain mode shown in the graph (Fig. E.8). Using the superposition principle, obtain an expression for the stress σ(t) for the interval t ≥ t1 based on the expression for σ(t) for the interval t ≤ t1 . Accept that γ0 /t1 = 2. Answer: for t ≤ t1 : σ(t) = λ(η1 (1 − exp(−t/τ1 )) + G 2 t)U (t); for t > t1 : σ(t) = λ(η1 (1 − exp(−t1 /τ1 )) + G 2 t1 )U (t1 ) + λ(η1 (1 − exp(−(t − t1 )/τ1 )) + G 2 (t − t1 ))U (t − t1 ). For the model shown in Fig. E.9, define the stresses σ(t) at: (a) t = t1 ; (b) t = 2t1 and (c) t = 3t1 , for the case where the applied strains are given by the corresponding diagram. For options (b) and (c) you use the superposition principle. Answer: a) σ(t1 ) = (γ0 η/t1 )(2 − exp(−t1 /τ)); b) σ(2t1 ) = (γ0 η/t1 )(−2 + 2 exp(−t1 /τ) − exp(−2t1 /τ)); c) σ(3t1 ) = (γ0 η/t1 )(− exp(−t1 /τ) + 2 exp(−2t1 /τ) − exp(−3t1 /τ)).

Fig. E.8 Three-parametric structural model and its graph of strain mode for exercise E.8

Fig. E.9 Three-parametric structural model and its graph of strain mode for exercise E.9

6.4 Equations of Mechanics of Linear-Hereditary Media

145

Fig. E.10 Three-parametric structural model and diagram of strains develop over time

10. Define the stress function σ(t) for the model in Fig. E.10 for the case where strains develop over time in accordance with the diagram γ(t) = (γ0 /2G)(2 − exp(−t/2τ))U (t). Answer: σ(t) = γ0 U (t). 11. The stress relaxation function is given by G(t) = a(b/t)m , where a, b and m are constant, t is time. ( )m Show that the creep function in this case is J (t) = a m1 π sin mπ bt under m σ y ). Therefore, when the sample is re-loaded after plastic deformation, the proportional breaking point of the material increases. This phenomenon is called hardening. Thus, for real materials, the yield stress is not constant, but develops during plastic deformations. In general, the yield stress may increase due to the deformation (this is hardening as previously stated), but the yield stress may decrease. This behavior is called softening. For many materials, hardening is first observed to some deformations and then softening occurs to the moment of destruction. When the sample is being softened, the material becomes unstable and disposed to localization like shear strips or the appearance of a “yield neck” (Fig. 7.3). Therefore, in the softening mode, the determination of the relations between stresses and strains is rather complicated. Determining the ultimate stiffness also becomes difficult. Therefore, it is usually limited to consideration of the hardening regime. In this case, the following hardening rules can be adopted: • Linear hardening law σ y = K |ε| + σ y0 , K is positive constant;

152

7 Mathematical Models of Plasticity Theory

Fig. 7.3 Hardening and softening phenomena of the material

• Or potential hardening law σ y = K |ε|n + σ y0 , K, n are positive constants. Thus, when the material is re-loaded from tension to compression or vice versa, there may be a difference between the yield stresses in both modes. That is, the material enters plastic area under repeated loading at other stresses than the initial limit of proportionality σ y . Usually, when the sample is loaded with the load of an opposite sign (compression), the material will enter the plastic area at stresses less than the initial limit of proportionality, i.e., σ''y < σ y (Fig. 7.2c). Such a phenomenon was studied in detail by I. Bauschinger and is called by his name—the Bauschinger1 effect, (Fig. 7.4). If equality is achieved (as far as the tensile yield stress is increased, by the same amount it decreases under compression), then the material is called cyclical ideal. If the equality σ'y + σ''y = 2σ y is true (how much the tensile yield stress increases, by the same value it decreases under compression), then the material is called cyclically ideal. Thus, the pre-stretched sample has an increased tensile proportional limit and a reduced compression limit due to hardening. Two types of hardening are described in the literature. The difference between them is observed when the load is reversed, (Fig. 7.5) [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems/Albrecht Bertram, Rainer Glüge. –Springer International Publishing Switzerland, 2015. 331 p.]: 1. Isotropic hardening, for which the yield limit under tension and compression changes in parallel; 2. Kinematic hardening, for which the yield limit under tension increases in a way other than its reduction under compression. 1

Johann Bauschinger (1834–1893), mathematician, builder, and professor of Engineering Mechanics at Munich Polytechnic from 1868 until his death. The Bauschinger effect in materials science is named after him.

7.2 Material Plasticity During Tension and Compression

153

Fig. 7.4 Saturation hardening (a) and Bauschinger effect (b)

Fig. 7.5 Two types of hardening

The softening behavior often lies between these two extremes, which can be described using a combination of the two hardening models. For that purpose, we need two hardening variables, namely one for the isotropic hardening σ y and a back stress σb for kinematic hardening. For both variables, the evolution equations are needed. As the experiments show, a certain effect on the deformation curve of the sample is the rate of loading increase or the strain rate. In conventional test machines, the strain rate of the samples varies within ε˙ = 10−5 − 10−2 c−1 . This deformation mode is called static. Static test diagram is independent of strain rate. This dependence is noticeably manifested, starting with order rate of 10−1 c−1 . Schematic diagrams of small-carbon steel tests obtained at three different levels of strain rates are shown in Fig. 7.6: curve 1 corresponds to the value ε˙ = 10−4 c−1 ; curve 2 corresponds to ε˙ = 0.5 c−1 , and curve 3 corresponds to ε˙ = 102 c−1 . These comparisons show that the modulus of elasticity at uniaxial tension is practically unchanged. Stress and strain yield limits increase with increasing tension rates. Such expansion of elastic deformation range is related to the inertia of plastic deformation mechanism.

154

7 Mathematical Models of Plasticity Theory

Fig. 7.6 Deformation modes at different strain rates

An analytical representation of the dependence of the yield limit value on the strain rate can be taken as linear: ·

σ y = σ y0 + μ0 ε, where σy0 is the static elasticity limit; μ0 is the experimentally determined constant of material. With an increase of the deformation rate, the material ultimate tensile stress σb (stress corresponding to the fracture) increases, and the corresponding strain (ultimate plastic elongation) decreases. This phenomenon is called material embrittlement. Brittle material is destroyed without noticeable plastic deformations preceding the destruction. Remark It should be noted that there are still no reliable systematic experimental studies on the effect of strain rates for many materials. This is due to the fact, that the limit rates at which the described effects are observed are not widely used in practice. Therefore, in most cases, the static diagram of the tensile test material is sufficient to characterize its mechanical properties. A common approach is that the plastic properties of the material are characterized by a plastic coefficient. The plastic coefficient is the ratio of the work W p spent on destroying a given volume of real material to the work W i.e required to destroy the same volume of material, with the same value of compressive strength under the assumption of material ideal elasticity: K pl = W p /Wel .

7.3 Elastic–Plastic Models of Material Behavior

155

Fig. 7.7 Prandtl or Genky structural model

7.3 Elastic–Plastic Models of Material Behavior Therefore, the behavior of many real materials is first elastic until the deformations reach plastic yield limit. Only upon reaching the yield limit σ y (often referred to as the elastic limit), the material deforms plastically (plastic yielding). This behavior can be modeled by placing in a series of Coulomb and Hooke’s elements. The constructed model is called the Prandtl2 model or the Genky3 model (Fig. 7.7). Consider a one-dimensional case. The total deformation in the Prandtl model can be represented as the sum of the elastic and plastic components: ε = εl + ε p , εl = σ/E.

(7.2)

In the hardening mode, only plastic deformations appear ε p . Stress power for the Prandtl model is: | | ) ( πi = σ˙ε = σ ε˙l + ε˙ p = Eεl ε˙l + σ y |ε˙ p | = w˙ + δ. That is, in this case, the stress power consists of elastic and dissipative parts. The elastic–plastic model with kinematic hardening is, for example, the Masing model (Fig. 7.8). For metals, the elastic regions in the strain space are often small enough to justifiably use only the laws of the linear theory of elasticity. 2 3

Ludwig Prandtl (1875–1953), German fluid dynamitist, physicist, and aerospace scientist. Heinrich Hencky (1885–1951), German engineer-mechanics.

156

7 Mathematical Models of Plasticity Theory

Fig. 7.8 Masing structural model

If one considers elastoplastic material behavior under monotonous loading without loading reversal, one cannot distinguish its behavior from a non-linear elastic behavior. This fact inspired Hencky to suggest a finite deformation rule, which allows for treating the material behavior in the context of non-linear elasticity as long as no loading reversal occurs. A similar approach is the Ramberg–Osgood law: ( )N σ σ + k2 , k1 , k2 , k3 , N − constants. ε= k1 k3 Plasticity theories establish a connection between stresses and strains (deformation theories) or strains rates (plastic flow theory) in the areas of the plasticity of materials. At the same time, stresses often depend not only on current strains, but also on the history (process) of deformation. The deformation theory of plasticity has a more limited area of application than the plastic flow theory, since it has a wider range of restrictions.

7.5 Plasticity Conditions

157

Fig. 7.9 Viscoplastic structural model

7.4 Viscoplasticity Models In addition to non-linearity, speed independence is a characteristic of all elastoplastic models, which exclude creep and relaxation. Viscoplasticity models combine plasticity and creep. A simple method to include viscous effects is adding viscous elements like dampers in series (V 1 ) or in parallel (V 2 ) (or both) to the plastic elements (Fig. 7.9) [Bertram, A. Solid Mechanics. Theory, Modeling, and Problems / Albrecht Ber-tram, Rainer Glüge. –Springer International Publishing Switzerland, 2015. 331 p.]. The effect of V 1 is that there are no more elastic ranges but rather viscoelastic ones. Under deformation cycles even below the yield limit, one will observe a hysteresis (be similar to the Maxwell model). What is even more important is the application of a viscous element V 2 in parallel with the Coulomb model. V 2 is only activated under inelastic deformations. Therefore, elastic ranges may still exist. During yielding, the stresses are composed by the yield stress σ y and a viscous overstress σ0 : σ = σ y + σ0 . As a consequence, the stresses can be arbitrarily large or small, which is perhaps more realistic than the restriction to ±σ y for a perfect plastic material. If one combines such models with complex hardening mechanisms, one can describe rather complex material behavior. Common of all such suggestions is the intention to describe the material behavior under essentially all kinds of loadings (creep, cyclic loads with or without holding time, relaxation, etc.) as realistically as possible.

7.5 Plasticity Conditions For the purpose of simplification, in the general case, non-linear relations between stresses and strains, the dependencies σ~ε for real materials are often approximated in the form of piece-broken lines (Fig. 7.10a–c).

158

7 Mathematical Models of Plasticity Theory

Fig. 7.10 Various approximations of the dependencies σ~ε for real materials

The simplest is the Prandtl diagram for a perfectly elastoplastic material, (Fig. 7.10a). A diagram with linear hardening is shown in Fig. 7.10c. These two approximations are most often used in solving problems of plasticity theory. The tension and compression curves “stress–strain” for most materials are very close, and we will consider them to be coincidental in the future. Remark More careful experiments show that the unloading law is not always linear. In existing plasticity theories, these minor deviations from Hooke’s law during unloading are neglected, as the difference between the proportional limit and the yield limit. Consider a body of arbitrary shape, and consider that there are no initial stresses and strains in it. At the initial stage of such solid loading, only elastic deformations occur and, therefore, the appearance of plastic deformations is uniquely determined by the current stresses. Therefore, the plasticity condition can be written as some functions of the stress tensor components. For an isotropic material, the plasticity condition cannot depend on the choice of coordinate system. Therefore, the corresponding function must be determined by three stress tensor invariants, which can be, for example, three principal stresses: f (σ1 ,σ2 ,σ3 , k) = 0,

(7.3)

where k is a constant value, characteristic of material (for example, yield limit σ y ). For example, with uniaxial tension, we have: σ1 = σ y . As experiments show, plastic deformations are associated with shear phenomena. Therefore, the basic plasticity criteria, which are widely used, compare some tangent stresses with yield stress limits. Therefore, according to the TrescaSaint-Venan condition, the transition from an elastic state to a plastic state occurs if the maximal tangent stress reaches a certain limit value for a given material: |τmax | =

|σ1 − σ3 | σy = or σ1 − σ3 = σ y , 2 2

(7.4)

where, as before, the principal stresses are numbered in decreasing order: σ1 ≥ σ2 ≥ σ3 ; σ y is the tensile yield limit. Solid is assumed to be initially isotropic.

7.5 Plasticity Conditions

159

Remark In the course of mechanics of materials, the Tresca-Saint-Venan criterion is used as the criterion of maximal tangent stresses: σred = σ1 − σ3 ≤ [σ]. Another plasticity criterion determines the transition from an elastic state to a plastic state of an initially isotropic material, if: σu = σ y ,

(7.5)

where σu is the stress intensity (see Chap. 2). σu is proportional to the octahedral tangent stress and its square is proportional to the second invariant of the stress deviation: ) ( σu2 = −3J2d = (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 /2. In this case, not the maximum, but the octahedral tangent stress reaches the limit value for this material. This criterion is used in the energy theory of strength and the name is the Mises (Huber4 —Mises—Genky) plasticity condition. Results obtained by plasticity criteria (7.4), (7.5) are very close. When solving problems, it is usually recommended to use the criterion that simplifies the solution. Since the solid deformed substantially and elastically during all-round tension or compression, it can therefore be assumed that the plasticity condition generally does not define all the stress tensor, but only its deviatoric part. As we have already said, the transition to the plastic state cannot depend on the choice of the coordinate system, so the plasticity condition is a function of the invariants of the stress deviator. The first invariant is zero (J 1d = 0), so in the general case the condition for the appearance of plastic deformations is determined by the second and third invariants of the stress deviator: f 1 (J2d , J3d ) = 0.

(7.6)

This equation in the coordinate system σ1 , σ2 , σ3 describes a surface that is called the yield surface. Condition (7.6) contains the invariants of the stress deviator and some material constants, e.g., yield limit. These plasticity criteria make it possible to fix the moment of the first appearance of plastic deformations. These criteria are sufficient to solve plastic problems when the material deformation in a uniaxial stress state respected to the Prandtl diagram (Fig. 7.10a). This is explained by the fact that during such materials repeated loading there is no change in the condition of plastic deformations. The situation changes if the material is hardened (Fig. 7.10b, c). Such re-loading materials are characterized by an increase in yield strength, the value of which 4

Huber Maximilian Tytus (1872–1950), a mechanical scientist, was president of the Academy of Technical Sciences of Poland.

160

7 Mathematical Models of Plasticity Theory

depends on the accumulated plastic deformation. In such cases, a hardening condition that looks like a plasticity condition, (7.6) should be introduced: f 1 (J2d , J3d ) = .(η).

(7.7)

Condition (7.7) includes a function .(η) that depends on the material hardening parameter η. This equation in the space of principal stresses also determines the yield surface, the change in position, shape, and size of which during loading characterizes the material deformation hardening. For example, if the hardening parameter coincides with the strain intensity, then one of the variants of the criterion (7.7) can be taken as a generalization of the Mises condition: σu = .(εu ),

(7.8)

which are examples of the yield surface (7.6), construction for the considered Tresca-Saint-Venan and Mises plasticity criteria. In the case of the Mises criterion (7.5), the yield surface equation (σu = σ y ) can be written as: (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 − 2σ2y = 0. This equation in axes σ1 , σ2 , σ3 describes a cylinder, the axis of which is equally inclined to coordinate axes (Fig. 7.11). If you cut the cylinder with a plane σ3 = 0, the section will be an ellipse whose equation is: σ12 − σ1 σ2 + σ22 = σ2y .

Fig. 7.11 Yield surface for Mises criterion

7.5 Plasticity Conditions

161

Fig. 7.12 Yield surface for Tresca-Saint-Venan criterion

Therefore, the Mises plasticity criterion corresponds to the yield surface in the form of a circular cylinder, the √ radius of which in the plane perpendicular to the axis of the cylinder equals to σ y / 2. If you accept the Tresca-Saint-Venan criterion (σ1 −σ3 = σ y ), taking into account the fact that the number of the principal stresses is not always known in advance, then the following equalities may occur: σ1 − σ3 = ±σ y , σ3 − σ2 = ±σ y , σ2 − σ1 = ±σ y . The yield surface in this case is represented as a hexagonal prism with an axis also equally inclined to the axes σ1 , σ2 , σ3 (Fig. 7.12a). This prism is called the Coulomb prism. It is inscribed in the Mises cylinder. Prism and cylinder axes coincide. The equation of this axis is σ1 = σ2 = σ3 . Figure 7.12b shows the cross section of the cylinder and prism with a plane corresponding to the plane stress state. Remark In order to take advantage of the Saint-Venan plasticity condition, it is necessary to know in advance which of the principal stresses is maximal and which is minimal. When using the Mises condition, you do not need to define the principal stresses at all. The condition σu = σ y can be written using the definition of the stress intensity: 2 2 2 (σ11 − σ22 )2 + (σ22 − σ33 )2 + (σ33 − σ11 )2 + 6(σ12 + σ23 + σ31 ) = 2σ2y .

Therefore, in many cases, the use of the Mises plasticity condition is more convenient than the Saint-Venan conditions. The maximal difference in calculations for these criteria does not exceed 13% (case of pure shearing).

162

7 Mathematical Models of Plasticity Theory

mz P , τ= . 2πRδ πR 2 δ

(7.9)

When the values of external loads P and mz change in proportion to one parameter λ, for example, time, simple loading is carried out, since the of the stress tensor components, according to (7.9), change in proportion to the same parameter. The loading trajectory is in the stress axes σx , τ is shown in Fig. 7.14. Fig. 7.13 Example of simple loading

Fig. 7.14 Diagrams of various loads

163

Simple loading corresponds to beam OA. Thus, for simple loading, the following equation is true: σi j = λ σi0j , where σi0j are some initial values of stress tensors at the loading application. Then the average stress σ, the stress intensity σu and the modulus of deviator stress s, due to formulas (2.27), (2.28), will also be linear with respect to the parameter λ: σ = λ σ0 , σu = λ σu0 , s =

Σ

3/2σu = λ s 0 .

In turn, the components of the stress guide tensor (2.28) do not depend on λ, since: si j =

λσi0j − λσ0 δi j σi j − σδi j = s i0j . = s λs 0

164

7 Mathematical Models of Plasticity Theory

With an active process, also called a loading increment of work performed by external loads over the body is positive. Negative increment of this work corresponds to passive process, or unloading. For elementary material volume, the active process condition can be written as the following inequality: σi j δεi j > 0, and passive process condition as follows: σi j δεi j < 0. Sometimes an active process is taken as such that the plastic deformation increases, and a passive process is taken as such that the plastic deformation remains unchanged. In plasticity theories, it is assumed that at any moment of a deformation process, the tensor of complete strain is represented as the sum of elastic and plastic strain tensors: p

εi j = εiej + εi j .

(7.10)

Moreover, the first tensor changes in both active and passive processes, while the plastic strain tensor changes only in the active process. Plastic deformation is defined as the collection of strain tensor components that remain at the solid point when all stress tensor components at that point turn to zero. Thus, plastic deformations are identified with residual deformations.

7.7 Hypotheses of the Small Elastoplastic Deformation Theory The theory of small elastoplastic deformations is the simplest theory of a hardening plastic material. It is very common in applications. This theory refers to deformation theories of plasticity. Consider the slow rod extension process, (Fig. 7.15). A rod is loaded along the section OAB, a line BC corresponds to unloading. The AB loading branch equation σ1 = f (ε1 ) can represent both plastic and non-linear elastic deformation of the rod. In connection with this remark, it is possible to attempt to construct plastic deformation equations in the form of finite relations between stresses and strains. Such equations would be significantly simpler than the theory of plastic flow equations. Therefore, in the theory of small elastoplastic deformations, it is assumed that for elastoplastic bodies, it is possible to establish relationships between stresses and strains, like Hooke’s law for elastic bodies.

7.7 Hypotheses of the Small Elastoplastic Deformation Theory

165

Fig. 7.15 Rod extension-compression diagram

В

А

О

С

The theory of small elastoplastic deformations is based on the hypotheses proposed by M.T. Huber, R. Mises, G. Genki and generalized in the case of a material with hardening by A. Nadai. The development and substantiation of the theory of small elastoplastic deformations was carried out in the works of A.A. Ilyushin. Therefore, often the theory of small elastoplastic deformations is called the Ilyushin plasticity theory. In the theory of small elastoplastic deformations, it is assumed that with a simple active deformation of an initially isotropic material, the properties of which do not depend on the third invariant of the stress tensor, the following three hypotheses are true. 1. Volume strain is elastic. Therefore, the volumetric strain of the body θ is considered elastic. It is directly proportional to the average normal stress σ and for it the Hooke’s law is valid: σ = K θ = 3K ε. This means that the volume changes only due to elastic deformations, and the material behaves as incompressible materials during plastic deformation. Thus, due to plastic deformation, the body volume does not change. 2. Coaxiality of stress and strain deviators. Strain deviator components .ij are proportional to stress deviator components sij . The relationship between them can be represented in the form proposed by A.A. Ilyushin:

si j =

2 σu .i j , 3 eu

(7.11)

where σu , εu are intensities of the stress and strain tensors. In stress and strain components, the expression (7.11) is: σi j − σδi j =

2 σu (εi j − eδi j ), (i, j = 1, 2, 3). 3 eu

) ) ( ( where σ = σx x + σ yy + σzz /3; e = ex x + e yy + ezz /3;

(7.12)

166

7 Mathematical Models of Plasticity Theory

√ /( )2 ( )2 ) ( 2 + σ 2 is σu = 22 σx x − σ yy + σ yy − σzz + (σzz − σx x )2 + 6 σx2y + σ yz zx the stress intensity; √ /( )2 ( )2 ) ( 2 is the eu = 32 ex x − e yy + e yy − ezz + (ezz − ex x )2 + 6 ex2 y + e2yz + ezx strain intensity. Similar to elastic deformation in the model of small elastoplastic deformations in the area of plastic deformations, the values of stresses and strains are connected to each other by unique dependence. It has been shown experimentally that the law of the relationship between stresses and strains (7.12) is fulfilled for the case of proportional or simple loading.

3. Hardening hypothesis. Regardless of the type of stress state for each material, there is a universal relationship between stress intensity and strain intensity:

σu = .(εu ).

(7.13)

Remark For an elastic material, this relationship is expressed by a linear relationship: σu = 3G εu . Experimental tests of these hypotheses gave good enough results for simple loading or little different from simple loading. The deformation process must be active without unloading. Remark The theory of small elastic–plastic deformations was proposed by Genki and is based on finite dependencies between stress components and strain components. Genki equations are written in the following form: ( ) ( ) εi j − δi j ε0 = . σi j − δi j σ0 ,

(7.14)

where ε0 = σ0 /3K , K = E/(1 − 2ν). The function . is defined as follows: .=

3 εi , 2 σi

(7.15)

where εi is the strain intensity; σi is the stress intensity. Then the Eq. (7.14) can be represented as: ( ) 3 εi ( ) εi j − δi j ε0 = σi j − δi j σ0 . 2 σi

(7.16)

The inverse dependencies between the stress deviator components and the strain deviator components are recorded in this way: ( ) 2 σi ( ) σi j − δi j σ0 = εi j − δi j ε0 . 3 εi

(7.17)

7.7 Hypotheses of the Small Elastoplastic Deformation Theory

167

As can be seen, the representation (7.17) is similar to (7.12). There is a certain relationship between stress intensity and strain intensity σi = .(εi ), with small elastoplastic deformations for each material, similar to the relationship between stress and strain in tension σ = f (ε). The relationship σi = .(εi ) can be plotted from the tension pattern. To do this, you must first replace σi with σ, and εi with (ε − ε0 ). Since the theory of small elastoplastic deformations assumes that the volume does not change during plastic deformations (θ = 0), it is necessary to put ε0 = 0 in Eqs. (7.16) and (7.17). In this case, the relationship between stress and strain intensities σi = .(εi ) beyond elasticity is determined by the tension diagram. Thus, the hypothesis is accepted that there is a “single deformation curve” for a material, regardless of the stress state type. Under conditions of material incompressibility: K → ∞; μ = 0.5; E = 3 μ. One of the varieties of the deformation theories of Genka, characterizing the relationships between the components of strains and stresses for elastoplastic deformations, has the form: ) ( 3ν ϕ + 1+ν 1+ϕ σ0 . εi j = (7.18) σi j − δi j 2μ 1+ϕ In these relationships, the value of shear modulus is reduced by 1/(1 + ϕ), i.e., the material becomes less rigid. The inverse dependencies between stress and strain components are recorded as follows: ) ( 3ν ϕ + 1+ν 2μ σi j = ε . (7.19) εi j + δi j 3ν 0 1+ϕ 1 − 1+ν If in Eq. ((7.19) we express strains through displacements along Cauchy formulas ) /2, u + u and then we substitute these formulas into equilibrium (i.e., εi j = i, j j,i Σ equation σi j, j + X i = 0, then we get three differential equations relative to four j

unknown functions (u i , ϕ). We add to these three differential equations a fourth, representing the Huber-Mises-Genky plasticity condition: ( ( )2 ( )2 ) σx − σ y + σ y − σz + (σz − σx )2 + 6 τ2x y + τ2yz + τ2zx = 2σ2y . In this equation, we replace stresses with representations through displacements u i and the function ϕ. As a result, we obtain four equations to determine four unknown coordinate functions. From the relationships (7.16) and (7.17), it follows the proportionality of the stress deviator components to the strain deviator components, as well as the proportionality of the principal angular strains to the principal tangent stresses.

168

7 Mathematical Models of Plasticity Theory

A.A. Ilyushin on the basis of experimental data showed that Genka equations are confirmed for simple loading processes or loading processes close to simple ones. The equations of deformation theory of plasticity are the equations of a nonlinear elastic solid. Therefore, using these equations to describe plastic deformations in complex loading paths can lead to unsatisfactory results. Therefore, we can assume that the equations of the deformation theory of plasticity are suitable if the plastic deformations develop in a certain direction.

7.8 Theory of Problems of Small Elastic–Plastic Deformations We consider an elastic–plastic body that is under the influence of mass forces ρF i and surface loads Ri . For the solution of the theory of problems of small elastic–plastic deformations (that is, for definition of unknown displacements, strains, and stresses (ui , σij , εij ; i, j = 1, 2, 3)), there are equilibrium equations, Cauchy ratios, strain compatibility equations, and boundary conditions: σi j , j + ρ Fi = 0; εi j = (u i , j +u j ,i )/2; εαα ,ββ + εββ ,αα − 2εαβ ,αβ = 0, (εαβ ,γ − εβγ ,α + εγ α ,β ),α − εαα ,βγ = 0; u i = u i0 (x) on Su , σi j l j = Ri on Sσ .

(7.20)

Constitutive equations are conveniently written as: si j = 2G ϕ(εu ) .i j , σ = 3K ε,

(7.21)

where si j = σi j − σ δi j ; .i j = εi j − εi j δi j ; i, j = 1, 2, 3. Comparing the dependencies (7.12) and (7.21), we can find the expression of the plastic function ϕ(εu ) through the previously introduced universal function F(εu ) (7.13): ϕ(εu ) ≡

.(εu ) . 3Gεu

Equations (7.21) are valid only in the loading process. In the case of elastic unloading from the generalized Hooke’s law, such ratios follow: σi j − σi' j = 2μ(εi j − εi' j ) + λ(θ − θ' )δi j ,

(7.22)

where σi' j and εi' j are stresses and strains existing before the start of unloading; λ and μ are Lamé constants.

7.9 The Method of Elasticity Solutions

169

Unloading equations in form (7.22) are maintained until new (secondary) plastic deformations appear during unloading. When we solve elastic–plastic problems, the obtained solution must satisfy not only forces and kinematic boundary conditions, but also additional conditions on the interface of the zones of elastic and plastic deformations. The problem of plasticity theory is non-linear, so the question arises with existence and uniqueness of the solution. A.A. Ilyushin ([Ilushin, A.A. Plasticity. Fundamentals of general mathematical theory / A.A. Ilushin. – Moscow. Pub. by Academy of Sciences USSR, 1963. 272 p. (in Russian)]) proved that the system of Eqs. (7.20), (7.21) is of the elliptical type, if the following condition is met: 3G ≥

d σu σu ≥ > 0. εu d εu

(7.23)

At the same time, a solution to this problem exists if there is a solution to the corresponding problem of the linear theory of elasticity. A.A. Ilyushin proved the following theorem regarding the uniqueness of the solution: “With given body forces ρF i and surface forces Ri on a part of the boundary surface S σ and displacements ui on a part of the boundary surface S u , the stress–strain state of the body (that is, ui , σij , εij ) are determined in the only way if the loading is simple”.

7.9 The Method of Elasticity Solutions The solution of the problems of plasticity theory is associated with the solution of a system of non-linear differential equations in partial derivatives, which is an extremely complex mathematical problem, which in analytical form is solved, as a rule, in exceptional cases. Therefore, most often used approximate methods. One of them is the method of successive approximations proposed by A.A. Ilyushin to solve the problems of the theory of small elastic–plastic deformations with active loading and called in the theory of plasticity the method of elasticity solutions. The essence of this method is to consider the sequence of linear problems of the theory of elasticity, the solutions of which with an increase in ordinal number converge to the solution of the problem of plasticity theory. ϕ(εu ) = 1 − ω(εu ).

(7.24)

The representation (7.24) is substituted into the expression (7.21), then the physical relations become as follows: si j = 2G(1 − ω(εu )).ij , σ = 3K ε, where 0 ≤ ω < 1; ω(εu ) = 0, if εu ≤ ε y .

(7.25)

170

7 Mathematical Models of Plasticity Theory

Thus, with ω = 0, Eq. (7.25) coincide with the relations of the linear theory of elasticity. We write the differential equilibrium equation and boundary conditions by decomposing the stress tensor into the deviatoric and spherical parts: si j , j +σ,i +ρ Fi = 0, si j l j + σli = Ri on Sσ , u i = u i0 (x) on Su .

(7.26)

We solve the problem of elastic-plasticity in displacements. Therefore, components (7.25) are substituted into equilibrium equations and force boundary conditions (7.26) and Cauchy ratios are taken into account. As a result, generalizations of Lamé equations are obtained: (λ + μ)θ,i +μ .u i + ρ Fi − ρ Fωi = 0,

(7.27)

λθ li + μ(u i , j +u j ,i )l j = Ri + Rωi , where Fωi = 2G(ω .ij ),j ; Rω i = 2G .i j l j ω . For zero approximation ω(0) = 0 is taken. Then, fictitious loads are Fωi = Rωi = 0. Therefore, to determine the first approximation u i(1) , we have the usual problem of the linear theory of elasticity. The following components are defined from the determined displacements u i(1) : (1) (1) (1) .i(1) ≡ ω(εu(1) ), Fωi , Rω(1)i . j , εu , ω

For any kth approximation, there are such equilibrium equations and boundary conditions: (k−1) (λ + μ)θ,i(k) + μ.u i(k) + ρ Fi − ρ Fωi = 0, (k−1) , λθ(k)li + μ(u i(k) , j −u (k) j ,i )l j = Ri + Rωi

(7.28)

where Fω(k−1) , Rω(k−1) determined by the preceding (k-1)th approximation. i i Modified Lamé equations and boundary conditions (7.28) are linear with respect to unknown displacements u i(k) . They differ from the corresponding equations of the theory of elasticity in those fictitious forces Fω(k−1) , Rω(k−1) are added to the external i i forces ρF i , Ri . This allows, according to the known solution of the corresponding problem of the theory of elasticity, to construct a solution of the elastic–plastic problem in a recurved form. Remark An example of building such an iterative solution is discussed in the Sect. 7.12. For convergence of the methods of elastic solutions, it is necessary that the parameter ω associated the relation (7.24) with the function ϕ(εu ) is small compared to the unit. At the same time, the following condition must be met:

7.9 The Method of Elasticity Solutions

171

1 > ω + εu

dω ≥ ω ≥ 0. dεu

Convergence of this method has been investigated by various authors. In practice, it has been found that the convergence rate of the methods of elastic solutions is very high, so that several approximations are enough to obtain the necessary accuracy. Let’s give an algorithm for solving the problems of elastic-plasticity by the methods of successive approximations, namely, by the method of elastic solutions. There are several ways to implement this method. Consider two of them: the method of additional loads and the method of additional deformations. Using them allows you to solve elastic–plastic problems using the approaches of the theory of elasticity to determine functions u i (x) and σ i j (x). Remark Note that we are talking about the theory of plasticity problems without unloading zones when using the deformation theory of plasticity. Let’s consider an approach based on the method of additional loads. Equilibrium equations obtained by the method of additional loads have the following form: ) ( ( ) μ.u j + (λ + μ) ∂ θ/∂ x j + X j + X 0j = 0, j = 1, m ,

(7.29)

where X 0j

∂ =− ∂x j

[( [( )] )( )] Σ )( m ∂u j ∂u j 1 1 ∂ 1 ∂u k 2μ − μ− . − θ − + ψ ∂x j 3 ∂ xk 2ψ ∂ xk ∂x j k,k/= j (7.30)

where ψ = (3/2)(εi /σi ); εi and σi are strain and stress intensities, respectively: [ | Σ m m Σ ( ) 2 | ] · eii − e j j + 6 ei2j ; εi = 3 j,i i/= j j,i i/= j [ | Σ m m Σ ( )2 1 | ] σii − σ j j + 6 σi2j . σi = √ 6 j,i i/= j j,i i/= j /

m Σ ∂u j (y) k=1

where

∂ xk

nk + μ

m Σ ∂u k k=1

∂x j

n k = X ν j + X ν(0)j ,

(7.31)

172

7 Mathematical Models of Plasticity Theory

X ν(0)j

) ] ( )[( )[ ] m ( Σ ∂u j ∂u j 1 1 ∂u k 1 μ− n k + 2μ − = + − θ n j. 2ψ ∂ xk ∂x j ψ ∂x j 3 k,k/= j (7.32)

According to (7.29) and (7.31), the initial elastic–plastic problems can be reduced to a sequence of “pseudo-problems” for solving the theory of elasticity. The procedure for solving the “pseudo-problems” of theory of elasticity is the implementation of the methods of successive approximations, taking in the first approximation that all additional volumetric and surface loads are zeros (that is, ψ = (1/2)μ). After solving the theory of elastic problem for given forces (k) (k) X (k) j , X ν j ( j = 1, . . . , m), we find displacements in the kth approximation u j (k) . Then we determine strains e(k) deforj and strain intensity ε j . And from the given ( ) (k) and mation diagram, taking into account hardening, we determine σik = . εi ( ) (k) (k) then ψ(k+1) = (3/2)εi /. εi . By formulas (7.30) and (7.32), we find additional

(0) loads X (0) j , X ν j and again solve the theory of elastic problem. The solution must be continued until two consecutive approximations differ among themselves by an infinitesimal predetermined value. Let’s consider an approach based on the method of additional deformations. The transformed continuity equations in the case of solving elastic–plastic problems by the method of additional deformations have the form:

)] [( m ∂Xj ∂ 2 σ0 ν Σ ∂ X k 1 ∂ Xi − δi j .σi j + =− + + X i(0) j , 1 + ν ∂ xi ∂ x j ∂x j ∂ xi 1 − ν k=1 ∂ xk (7.33) where ⎧

[( ] ) ∂2 1 ( σ0 ) ψ− σkk − = 2μ 2μ 3 ∂ x 2j [( [( ) ] ) ]⎫ 2 ( 1 ∂2 ∂2 σ0 ) ψ − 2ψ − σ − σk j ; (7.34) + − jj μ 3 ∂ xk ∂ x j μ ∂ xk2 [( ) ] ) ] ⎧ 2 [( 1 ∂2 1 ∂ 2ψ − + 2ψ − σ σk j X i(0) = μ i j j μ ∂ xk ∂ xi μ ∂ xk2 [( [( ) ] ) ]⎫ 1 ∂2 1 ( ∂2 σ0 ) 2ψ − ψ− + σki − 2 σkk − , ∂ xk ∂ x j μ ∂ xi ∂ x j 2μ 3

X ii(0)

(i, j, k = 1, . . . , m; )

i /= j /= k.

(7.35)

7.10 Plastic Flow Theory

173

As in the previous case, the initial problem is given to the sequence of the theory of elastic problems based on the method of successive approximations, in the first approximation believing ψ = (1/2)μ. The components of the strain tensor in the k th approximation εi(k) j are determined (k) th through the k approximate solution σi j by the following formulas: [ εii(k)

= (1/E) εi(k) j

]

) ( σll + (ψ − (1/2μ)) σii(k) − σ0 ; l,l/=i ) ( + (2ψ − (1/μ))σi(k) = (1/μ)σi(k) j j .

σii(k)

−ν

Σ

Then we calculate the function ψ, and by the formulas (7.34) and (7.35) we find X ii(0) and X i(0) j . Then we solve the theory of elastic problems with new body forces. The calculations continue until the difference between the two consecutive approximations is sufficiently small.

7.10 Plastic Flow Theory For calculations in the case of large plastic deformations, the theory of plastic flow is used. Its main difference from deformation theory is that it is accepted that there is no unambiguous connection between stresses and plastic strains under both simple and complex loadings. Stresses at plastic deformations in final state depend on deformation path. In this regard, the equations describing plastic deformation cannot in principle be finite relations linking the components of stresses and strains, but should be differential relationships. The equations of the plastic flow theory establish a connection between infinitesimal increments of strains and stresses, the stresses themselves and some parameters of the plastic state. Remark The physical equations corresponding to this theory for the flat problem were first obtained by Saint-Venan, and for the spatial problem—M.K. Levi5 and later Mises. In flow theory, the plastic deformation of material is likened to a viscous fluid flow. As noted earlier, the transition to the plastic state in the vicinity of the body point is determined by the equation f (σij ) = 0, which in the six-dimensional stress space describes the surface of yield (loading). If the material is hardened, the yield surface changes with increasing plastic strains, its equation contains the hardening parameter η. The following hypotheses underlie the theory of plastic flow. 5

Maurice Lévy (1838–1910), French mathematician, mechanic and engineer.

174

7 Mathematical Models of Plasticity Theory

1. Volumetric strains are elastic. As in the theory of small elastoplastic deformations (7.6), the ratio is fulfilled:

σ = 3K ε. That is, the volume relative change is an elastic strain proportional to the average stress: θ = σ0 /K ; d θ = 3(d ε0e + d ε0 p ) = d σ0 /K .

(7.36)

In this case, the proportional coefficient K is the same as in the ratios used when the behavior of the material is within the elastic range: K = E/3(1 − 2ν), where σ0 =

1 3

3

Σ σii is the average stress; ε0e =

i=1

1 3

Σ

εiie , ε0 p =

i

1 3

Σ

εii p ; εiie and

i

εii p are, respectively, elastic and plastic parts of general strains εii . We remind that during plastic deformations, the body volume practically does not change. Therefore, the plastic strain increment tensor is a deviator. Then: d εop = 0, d εoe = d σ0 /3K . That is, the material is incompressible in the plastic state: p

p

p

ε p = (ε11 + ε22 + ε33 )/3 = 0. Remark This incompressible condition can also be written as equal to zero the rate of volumetric plastic strain, i.e., d ε p /d t = 0. 2. Gradient hypothesis. Strain increment vector is directed perpendicular to yield surface. This is equivalent to the assumption of the proportionality of strain component increment and the gradient vector component to the yield surface (partial derivatives of the surface equation over the corresponding stress components):

p

d ε11 = d λ

∂f ∂f p , . . . , 2d ε13 = d λ . ∂ σ11 ∂ σ13

(7.37)

Ratios (7.37) express the law of flow associated with the accepted plasticity condition f = 0,where d λ is a differential-small multiplier. By definition, we have:

7.10 Plastic Flow Theory

175

Fig. 7.16 Yield surface

p

dU = σi j d εi j . Using the ratio (7.37), we obtain: dU = d λσi j

∂f = d λ σ grad f, ∂ σi j

(7.38)

where σ is the radius-vector of the point in the six-dimensional stress space corresponding to the components σij . The part of the yield surface and the vectors constituting the scalar product in the formula (7.38) is showed in Fig. 7.16. Therefore, the multiplier d λ is proportional to the stress working density on plastic strains. The function f is called the plastic potential, since the increment of plastic strains is determined through the derivative function f by the corresponding arguments (7.38). The yield surface, or plastic potential, is determined experimentally. One of the important requirements for the yield surface construction is the hardening criterion formulated by D. Drucker. The hardening criterion formulated by D. Drucker is as follows. Let there be a certain stress state σ* to which the yield surface S * corresponds (Fig. 7.17). The yield surface divides the six-dimensional stress space into two areas. Transition from a point σ* to a region of its positive values is accompanied by active plastic deformation and an increment of the vector of plastic deformation. Transition from the end of the vector σ* to the area of negative values corresponds to unloading subject to elastic laws; plastic deformation is unchanged. Fig. 7.17 Yield surfaces during loading

176

7 Mathematical Models of Plasticity Theory

Let’s now consider the state defined by vector σ, the end of which lies in a positive region. This state corresponds to the yield surface S. The Drucker postulate claims the non-negativity of the stress increment work on real deformation displacements of per cycle, when the state from the point σ* along some path passes to σ, then returns to σ* : ∫

(σi j − σi∗j )d εi j ≥ 0.

(7.39)

In the theory of plastic flow, it is assumed that for a given material the stress intensity is a function of the integral from the intensity of the plastic strain increments: ∫ σi = F(

d εi p ).

(7.40)

Function F is determined by the diagram of material tension. To do this, you must first convert the function σ = f (ε) to a function σ = f ∗ (ε p ), (Fig. 7.18). Indeed, for uniaxial tension, we have:

d εx p

σx = σ y = 0; σz = σ; τx y = τ yz = τzx = 0; dε = d ε yp = − 2 p ; d εx p + d ε yp + d εzp = 0; d εzp = d ε p .

Then, with due regard for: 1 Σ σi = √ (σ1 − σ2 )2 + (σ2 − σ3 )2 + (σ3 − σ1 )2 , 2

D1 C1 B1 А1

А

Fig. 7.18 Material tensile diagram

B

C

D

7.10 Plastic Flow Theory d εip =

177

√ / )2 ( )2 ( )2 ] 2 3 [( d γxyp + d γyzp + d γzxp , (d εxp − d εyp )2 + (d εyp − d εzp )2 + (d εzp − d εxp )2 + 3 2

we get: σi = σ , d εi p = d ε p .

(7.41)

The curve A1 B1 C 1 D1 expresses the relationship between stress intensity and the Odquist parameter, i.e., σi = F(q). ∫ Remark Odqvist parameter q characterizes accumulated plastic strain: q = d εi p . The increment of the plastic strain components can be expressed as follows: ) ( ) 3 d εi p ( σi j − δi j σ0 . d εi j p = 2 σi

(7.42)

Equation (7.42) shows that components of plastic strain increment are proportional to components of a stress deviator. We add to the plastic strain components defined by (7.42) the elastic strain components defined by the formulas: ( ) ( ) 3ν 1 σi j − σ0 δi j . εi j e = 2μ 1+ν As a result, we obtain formulas for determining the components of complete strain increments: ] [ ) 3 d εi p ( 3ν 1 d σi j − δi j d σ0 + σi j − δi j σ0 . (7.43) d εi j = 2μ 1+ν 2 σi Equation (7.43) is the basic equation of plastic flow theory and it is called Prandtl-Reis equation. The relationship between stress intensity and strain increment intensity is taken as (7.40). If in the Prandtl-Reis equations, we neglect the components of elastic strains (which are permissible for developed plastic deformation), then we get the equation of the Saint-Venan-Mises plasticity theory: d εi j =

) 3 d εi ( σi j − δi j σ0 . 2 σi

(7.44)

Dividing both parts of the Eq. (7.44) by dt, and taking into account d εi /dt = ξi , we obtain the physical equation of the relationship between the strain rate and the components of the stress deviator:

178

7 Mathematical Models of Plasticity Theory

ξi j =

) 3 ξi ( σi j − δi j σ0 , 2 σi

(7.45)

where ξi is the intensity of strain rates determined by the formula: √ / )2 ( )2 ) 2 ( 3( ξx − ξ y + ξ y − ξz + (ξz − ξx )2 + η2x y + η2yz + η2zx . ξi = 3 2 Obviously, strain rates ξi j are not unambiguously determined when setting the stresses. When setting the strain rates ξi j , the components of stress deviator si j are determined uniquely. It is easy to verify that the components si j defined by formulas (7.45) (because si j = σi j − δi j σ0 ) are identical to the Mises yield condition. It should also be noted that in the yield state, the uncertainty of the strain rate component is necessary to be able to meet the strain compatibility conditions. The Prandtl-Reis Eq. (7.43) associates stress components with infinitesimal increments of stress and strain components, i.e., they are not finite relations (unlike the deformation theory of plasticity). The ratios (7.43), generally speaking, are not integrated, i.e., in other words, are not reduced to finite ratios between stress and strain components. This mathematical fact reflects the O1 (with stresses σi j ) of two paths I and II, then the components of the strains at the point O1 according to the equations of the plastic flow theory will be different (Fig. 7.19). Equation (7.43) does not contain time. However, by dividing them into dt, you can formally switch from increments d εi j to strain rates ξi j . Then the resulting equations will look like viscous fluid flow equations. This analogy to some extent justifies the name of the theory of plastic flow. It should be emphasized that the variable t can be understood as a time or a monotonically increasing loading parameter, or a monotonically increasing value (For example, the characteristic size of the plastic zone). At the same time, the obtained equations of the theory of plastic flow are fundamentally different from the equations of viscous flow. In equations of plastic flow, unlike the latter, you can always discard dt and return to formula (7.43) that does not contain time. In case of hardening, it is possible to calculate strains when setting the

I

Fig. 7.19 Various deformation ways upon transition from a point O to point O1

О1 О

I

7.11 Relationship Between Plastic Flow and Theory of Plasticity

179

loading path (i.e., when specifying σi j = σi j (t), where t is a parameter (for example, time). You can also find stresses if you specify a strain path, i.e., εi j = εi j (t). The equations of the Saint-Venan-Mises plasticity theory have a much simpler structure and are finite dependencies between stress components and strain rates. The complete system of equations in solving the problems of plastic flow theory includes: equilibrium equations, Cauchy ratios, strain compatibility equations and boundary conditions. σi j , j +ρ Fi = 0; εi j = (u i , j +u j ,i )/2; εαα ,ββ + εββ ,αα − 2 εαβ ,αβ = 0, (εαβ ,γ − εβγ ,α + εγα ,β ),α − εαα ,βγ = 0; u i = u i0 (x) on Su , σi j l j = Ri on Sσ . In the theory of plastic flow, the theorem on the uniqueness of fields of increments of stresses, strains, and displacements in a strengthening body has been proved. It is impossible to guarantee the singular increments of strains and displacements in the case of non-hardening materials. Therefore, obviously, the equations of flow theory are more complex than the equations of the theory of small elastoplastic deformations. However, it has been shown that with simple loading, both theories give the same solution. In the case of complex loading, the results, obtained using the theory of plastic flow, are better aligned with experimental data. In more details, the theory and problems of the plastic flow are considered in numerous monographs and articles.

7.11 Relationship Between Plastic Flow and Theory of Plasticity Theory of plastic flow and deformation theory of plasticity coincide only in the case of simple loading. In the case of complex loading, these theories lead to different results. Note an important case for applications when the results obtained from two theories are close to each other. As it is known, in deformation spaces, the deformation path is depicted as a line (Fig. 7.20). Let, starting from a point, the deformation path approach to a straight line (dashed line). In this case, we will say that deformation develops in a certain direction. If this case occurs, then the stress states calculated according to both theories come closer. At the same time, the influence of the history of complex deformation rapidly Fig. 7.20 Deformation path

180

7 Mathematical Models of Plasticity Theory

weakens and a constant stress state is established, which determined by those fixed deformation rates that are specific to the rectilinear section. Experiments confirm the theory of plastic flow much more fully than the deformation theory. Experimental data also indicate the coincidence of the directions of the principal axes of the stress tensor and the plastic strain increment tensor. It should be noted, however, that under complex loading, especially with intermediate unloading, there is a marked influence of the anisotropy that the material acquires during the plastic deformation.

7.12 Formula and General Methods 7.12.1 Formula of the Problems of Plasticity Theory Summarizing the material presented in the previous sections concerning the setting of problems of plasticity theory. The problems of elastoplasticity usually include cases where the elastic and plastic deformations that occur in the body during loading have almost the same order. The setting of such problems should take into account ratios in elastic and plastic zones and at the boundary between them. We consider a solid deformable body under the influence of surface (X ν , Yν , Z ν ) (including reactions) and volumetric (X, Y, Z ) forces. Elastoplastic properties of materials are given by deformation diagram (σi − εi ). The task is( to) define 17 unknown coordinate( functions (x, y, z): six stress tensor components σi j ; six strain ) tensor components εi j ; three components of the deformation vector (u, v, w); stress intensity (σi ) (one); strain intensity (εi ) (one). To determine these functions in the framework of plasticity theory, there is the following system of equations. 1. Equations arising from the consideration of the static equilibrium state of the solid: (a) Differential equilibrium equations:

3 Σ ∂σi j j=1

∂x j

+ X i = 0,

(7.46)

(b) Boundary conditions:

3 Σ j=1

σi j n j = X νi .

(7.47)

7.12 Formula and General Methods

181

2. Equations arising from considering the geometry of a problem: (a) Dependencies between strain components and deformation vector components (Cauchy equations) ⎫ ∂v ⎪ γx y = ∂u + εx = ∂∂ux , ∂y ∂x ⎬ ..................................... , ⎪ ..................................... ⎭

(7.48)

(b) Saint-Venan compatibility equations (as a result of Eqs. (7.48)) ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ . ⎪ ⎪ ⎪ ⎪ ⎪ ......................................................... ⎪ ⎪ ⎪ ⎭ ......................................................... ∂ 2 εx ∂ y2

∂2ε

∂2γ

+ ∂ x 2y = ∂ x∂xyy ............................. ............................ ( ) ( 2 ) ∂γ yz ∂γx y ∂γzx ∂ εx ∂ − = 2 + + ∂x ∂x ∂y ∂z ∂ y∂ z

(7.49)

Remark As can be seen from the above equations, the static Eqs. (7.46) and (7.47) and the geometric relations (7.48) and (7.49) in the theory of plasticity have the same form as it in the theory of elasticity. The other equations are only the constitutive equations, i.e., the relations between stresses and strains. Therefore, when solving elastoplastic problems, it is necessary to write physical constitutive equations according to one of the plasticity theories instead of the Hooke’s law. 3. Equations arising from consideration of the mechanical behavior of materials, i.e., constitutive equations between stresses and strains: (a) On the theory of small elastoplastic deformations

τx y

σi σx − σ0 = 2 3ε (εx − ε0 ), i( ) 2 σi σ y − σ0 = 3 εi ε y − ε0 , σz − σ0 = 23 σεii (εz − ε0 ), = 3εσii γx y , τ yz = 3εσii γ yz , τzx =

γx y

3εi εx − ε0 = 2σ (σx − σ0 ), i( ) 3εi ε y − ε0 = 2σi σ y − σ0 , 3εi εz − ε0 = 2σ (σz − σ0 ), i 3εi = σi τx y , γ yz = 3εσii τ yz , γzx =

⎫ ⎪ ⎪ ⎪ ⎬ σi 3εi

γzx .

⎪ ⎪ ⎪ ⎭

(7.50)

or

3εi σi

τzx

⎫ ⎪ ⎪ ⎪ ⎬ , ⎪ ⎪ ⎪ ⎭

(7.51)

182

7 Mathematical Models of Plasticity Theory

where σi = .(εi ), σi , εi are stress and strain intensities. (b) On the theory of plastic flow ⎧ [ )] 3 d εi p ( 1 ⎪ ⎪ d εx = E [d σx − μ d σ y + d σz ] + 2 σi ((σx − σ0)) ⎪ ⎪ dε ⎪ ⎪ d ε y = E1 d σ y − μ(d σx + d σz ) + 23 σii p σ y − σ0 ⎪ ⎪ [ )] ( ⎨ dε d εz = E1 d σz − μ d σx + d σ y + 23 σii p (σz − σ0 ) , dε dτ ⎪ d γx y = Gx y + 3 σii p τx y ⎪ ⎪ ⎪ ⎪ ⎪ ....................................................................... ⎪ ⎪ ⎩ ....................................................................... where σi = .

(∫

(7.52)

) d εi p ,

[( )2 ( )2 ( )2 √ | | ⎫ d εxp 2 | d ε⎧xp[ − d εyp ] + [d εyp − d]εzp + d εzp − ( )2 2 ]2 ( ) 2 [ ] 2 . d ε ip = + 3 d γxy p + d γyz p + d(γzx ) p 3 The use of physical equations according to the theory of plastic flow in the form (7.52) in solving elastoplastic problems is associated with great mathematical difficulties, since they are non-linear and have a rather complex structure. Therefore, when solving problems in which significant plastic deformations are taken into account in comparison with elastic ones, the components of elastic strains are neglected and use the Saint-Venan-Mises equations, which for a rigid plastic body have the form: ξx = ξy = ξz =

3 ξi − σ0 ) ⎪ 2 σi (σ ( x )⎬ 3 ξi σ − σ , y 0 2 σi ⎪ ⎭ 3 ξi − σ0 ) 2 σi (σz

⎧ ⎪ ⎨ ηx y = η yz = ⎪ ⎩η = zx

3ξi τ σi x y 3ξi τ σi yz 3ξi τ σi zx

.

(7.53)

/( )2 ( )2 ξx (− ξ y + ξ y − ξ)z + (ξz − ξx )2 where σi = .(ξi ); ξi = . 2 + 23 ηx2 y + η2yz + ηzx When solving elastoplastic problems using physical Eq. (7.53) according to the theory of plastic flow, it is necessary to write Cauchy Eq. (7.48) and, therefore, strain compatibility equations respectively in the following form: √ 2 3

∂ 2ξy ∂ 2 ηx y ∂ 2 ξx , + = ∂ y2 ∂x2 ∂ x∂ y ∂ 2ξy ∂ 2 η yz ∂ 2 ξz , + = ∂ z2 ∂ y2 ∂ y∂z ∂ 2 ξz ∂ 2 ξx ∂ 2 ηzx , + = 2 2 ∂x ∂z ∂ x∂z

7.12 Formula and General Methods

183

( ) ∂η yz ∂ηx y ∂ 2 ξx ∂ηzx ∂ − + + =2 , ∂x ∂x ∂y ∂z ∂ y∂ z ( ) ∂ηx y ∂ 2ξy ∂ ∂η yz ∂ηzx − + =2 , ∂y ∂x ∂y ∂z ∂z∂ x ( ) ∂ηx y ∂ ∂η yz ∂ 2 ξz ∂ηzx =2 + − . ∂z ∂x ∂y ∂z ∂ x∂ y where ⎧

∂v

ηx y =

∂vx ∂y

z x , ξ y = ∂ yy , ξz = ∂v , ξx = ∂v ∂x ∂z ∂v y ∂v y ∂vz + ∂ x , η yz = ∂ y + ∂ z , ηzx =

∂vx ∂z

+

∂vz . ∂x

7.12.2 General Methods of Solving the Problems of Plasticity Theory As in the theory of elasticity, the problems of plasticity theory can be solved in displacements or stresses, as well as in a mixed way. For most practical problems, solving in a closed form is difficult due to the non-linearity of the governing partial differential equations. Therefore, to solve non-linear equations of plasticity theory, in most cases, various versions of the method of successive approximations are used. Solving the problems of plasticity theory is usually reduced to solve a sequence of linear problems, each of which can be interpreted as a certain problem of elasticity theory (a method of elasticity solutions). Such an idea was first applied by A.A. Ilyushin, and then developed by I.A. Birger6 and others. Summary of the most common methods for solving problems of plasticity theories are as follows. Method of Additional Loads We take the Genki-Nadai equations as basic.

γx y

⎫ εx − ε0 = .(σ ⎪ ⎪ ) ( x − σ0 ), ⎬ ε y − ε0 = . σ y − σ0 , , ⎪ εz − ε0 = .(σz − σ0 ), ⎪ ⎭ = 2.τx y , γ yz = 2.τ yz , γ yz = 2.τ yz ,

(7.54)

where in the elastic area: . = 1/2G; in the elastoplastic region: . = 1/2G + ϕ. 6

Birger Isaac Aronovich (1918–1993), Soviet mechanical scientist in the field of strength and dynamics of aircraft engines for aviation and space purposes, doctor of technical sciences, professor.

184

7 Mathematical Models of Plasticity Theory

Let’s accept that 1/. = 2G −(2G − 1/.). In addition, we will take into account ε0 = θ/3 , σ0 = 3K ε0 , K = E/(3(1 − 2μ)), λ = μ E/(1 − 2μ)(1 − μ). Then the Eq. (7.54) are rewritten in the following form: ⎫ ) ( ⎪ σx = 2G εx + λθ − (2G − .1 )((εx − ε0 ), ) ⎪ ⎪ 1 ⎪ σ y = 2G ε y + λθ − (2G − . ) ε y − ε0 , ⎪ ⎪ ⎪ ⎬ 1 σz = 2G εz + λθ − (2G − . )(εz − ε0 ), . 1 ⎪ γx y , τx y = G γx y − ( G − 2. ⎪ ) ⎪ ⎪ 1 ⎪ τ yz = G γ yz − (G − 2. ⎪ ⎪ )γ yz , ⎭ 1 τzx = G τzx − G − 2. γzx .

(7.55)

Equations (7.55) differ from the Hooke’s law by additional terms. With . = 1/2G, these equations express the generalized Hooke’s law and determine the stresses in the elastic body: ⎫ σ∗x = 2G εx + λθ, ⎬ σ∗y = 2G ε y + λθ, ⎭ σ∗z = 2G εz + λθ,

⎫ τ∗x y = G γx y , ⎪ ⎬ τ∗yz = G γ yz , . ⎪ τ∗zx = G γzx . ⎭

(7.56)

Enter such expressions for stresses: ⎫ σx = σ∗x − σ(0) x , ⎪ ⎪ ⎪ ⎪ ................... , ⎪ ⎪ ⎪ ⎬ ................... , ∗ (0) ⎪, τx y = τx y − τx y , ⎪ ⎪ ⎪ ................... , ⎪ ⎪ ⎪ ⎭ ................... .

(7.57)

(0) where σ(0) x , . . . , τx y , . . . are additional stresses:

⎫ ) ( 1 σ(0) x = 2G − . (εx − ε0 ), ⎪ ⎪ ⎪ ⎪ .................................... , ⎪ ⎪ ⎪ ⎬ ................................... , ) ( , 1 (0) τx y = G − 2. γx y , ⎪ ⎪ ⎪ ⎪ ................................. , ⎪ ⎪ ⎪ ⎭ .................................. .

(7.58)

There is such a relationship between stress intensities in the elastic body and additional stresses, as it shown in Fig. 7.21: σi = σi∗ − σi(0) .

(7.59)

7.12 Formula and General Methods

185

B

А

The physical meaning of equality (7.59) is clear from the generalized diagram (see Fig. 7.21). According to the Fig. 7.21: σi(0)

=

σi∗

( − σi = εi (tg β − tg α) = 3 εi

) 1 G− . 2.

(7.60)

Assuming that the material is elastic, using a generalized deformation diagram at a given strain intensity εi , we find a point B. The correction σi(0) “returns” the calculated point to the deformation curve (point A). The stresses σx , σ y , . . . ., τzx must satisfy the Eq. (7.46). These equations, taking into account (7.57), have the form: ⎫ ∂τ∗ ∂τ∗ ∂σ∗x + ∂ yx y + ∂ zx z + X + X 0 = 0, ⎪ ⎪ ∂x ⎬ ∂σ∗y ∂τ∗x y ∂τ∗yz 0 , (7.61) + + + Y + Y = 0, ∂y ∂x ∂z ⎪ ⎪ ∂τ∗yz ∂τ∗x z ∂σ∗z ⎭ 0 + + + Z + Z = 0, ∂x

∂y

∂z

where X, Y, Z are components of the body force; X 0 , Y 0 , Z 0 are components of additional body force: ( (0) ) ⎧ (0) (0) ⎪ ⎪ 0 = − ∂σx + ∂τx y + ∂τx z ⎪ = X ⎪ ⎪ ∂x ∂y ∂z ⎪ ⎪ ] ) [( ) [( ) ]⎫ ] ⎧ [( ⎪ ⎪ ⎪ ∂ 1 1 1 ∂ ∂ ⎪ ⎪ + 2G − G − γ G − γzx + − ε − (ε ) ⎪ x xy 0 ⎪ ∂x ( . ∂y 2. ∂z 2. ⎪ ⎪ ) ⎪ (0) (0) (0) ⎪ ⎪ ∂σ y ∂τ yz ∂τ yx ⎪ ⎪ ⎨ Y0 = − + + = ∂x ∂y ∂z ⎧ [( ] ) [( ) [( ) ]⎫ ] ⎪ ) ∂ ∂ 1 1 ( 1 ∂ ⎪ ⎪ ⎪ ⎪ − ∂ x G − 2. γ yx + ∂ y 2G − . ε y − ε0 + ∂z G − 2. γ yz ⎪ ⎪ ) ( (0) ⎪ ⎪ (0) (0) ⎪ ⎪ ⎪ 0 = − ∂τzx + ∂τzy + ∂σz ⎪ = Z ⎪ ⎪ ∂x ∂y ∂z ⎪ ⎪ ) ] [( ) ] [( ) ]⎫ ⎧ [( ⎪ ⎪ ⎪ 1 1 1 ∂ ∂ ∂ ⎪ ⎩− G− γzx + G− γzy + 2G − (εz − ε0 ) . ∂x 2. ∂y 2. ∂z .

(7.62)

186

7 Mathematical Models of Plasticity Theory

Stresses must satisfy the boundary conditions (7.47) on the body surface. Expression (7.47), taking into account (7.57), is written in the form: ⎫ σ∗x l + τ∗x y m + σ∗x z n = X v + X ν(0) ⎪ ⎬ τ∗yx l + σ∗y m + τ∗yz n = Yv + Yν(0) , ⎪ τ∗zx l + τ∗zy m + σ∗z n = Z v + Z ν(0) ⎭

(7.63)

where X v , Yv , Z v are surface load components; X v(0) , Yv(0) , Z v(0) are components of additional surface loads: ⎧ [( ) ) ] ( (0) (0) (0) (0) 1 (ε − ε ) l + G − 1 (γ m + γ n ), ⎪ X v = σx l + τx y m + τx z n = 2G − . ⎪ x xy xz 0 ⎪ 2. ⎨ ] ( [( ) ) (0) (0) (0) (0) 1 (γ l + γ n ) + 2G − 1 (ε − ε ) m, Yv = τ yx l + σ y m + τ yz n = G − 2. yx yz y 0 . ) ⎪ ( [( ) ] ⎪ ) ⎪ (0) (0) 1 1 ( ⎩ Z v(0) = τ(0) zx l + τzy m + σz n = G − 2. γzx l + γzy m + 2G − . (εz − ε0 ) n.

(7.64)

Using Eqs. (7.56), (7.57), Cauchy Eqs. (7.48), after some transformations, the differential equilibrium Eqs. (7.61), taking into account (7.62), are written in the following form: ⎧ {( ( )]} )[ ⎪ + + ∂w (λ + G) ∂∂θx + G.u + X = ∂∂x 2G − .1 ∂∂ux − 13 ∂∂ux + ∂v ⎪ ⎪ ∂ y ∂ z ⎪ [( )] ( ⎪ ) [( )( )] ⎪ 1 ∂u 1 ∂u ⎪ + ∂∂y G − 2. + ∂∂vx + ∂∂z G − 2. + ∂w ⎪ ⎪ ∂y ∂x ⎪ {( ( ∂z )]} [ ⎪ ) ⎪ ⎨ (λ + G) ∂θ + G.v + Y = ∂ 2G − 1 ∂v − 1 ∂u + ∂v + ∂w + ∂y [ ∂ y . ∂ y 3 ∂ x ∂ y ∂ z )] [( )] ( ( ( ) ) 1 ∂u ∂ 1 ∂v ⎪ + ∂∂x G − 2. G − 2. + ∂∂vx + ∂z + ∂w ⎪ ⎪ ∂y ∂y ⎪ {( ( ∂z )]} [ ⎪ ) ⎪ ∂ 1 ∂w 1 ∂u ∂v ∂w ∂θ ⎪ 2G − + + G.w + Z = − + + + G) (λ ⎪ ⎪ ∂z ∂z .[ ∂z 3 (∂ x ∂ y )] ∂z ⎪ ⎪ ( [( )( )] ) ⎪ 1 ∂u 1 ∂w ⎩ + ∂∂x G − 2. . + ∂w + ∂v + ∂∂y G − 2. ∂z ∂x ∂y ∂z (7.65) The system of Eqs. (7.65) is a synthesis of static, geometric, and physical equations of the original problem. The Eqs. (7.65) differ from the Lamé equations in the theory of elasticity in the presence of additional terms on the right side. Similarly, you can convert boundary conditions (7.63), taking into account (7.64), have the form: ) ( ) ∂u ∂u ∂u ∂v ∂w ∂u l+ m+ n +G l+ m+ n = Xν ∂x ∂y ∂z ∂x ∂x ∂x )[( ) ( ) ] ( )[ ( )] ( ∂u ∂v ∂u ∂w 1 ∂u 1 ∂u ∂v ∂w 1 + m+ + n + 2G − − + + l, + G− 2. ∂y ∂x ∂z ∂x . ∂x 3 ∂x ∂y ∂z (

λθl + G

) ( ) ( ∂u ∂u ∂v ∂w ∂u ∂u l+ m+ n +G l+ m+ n = Xν λθ l + G ∂x ∂y ∂z ∂x ∂x ∂x )[( ) ( ) ] ( ∂u ∂u 1 ∂v ∂w m+ n + G− + + 2. ∂y ∂x ∂z ∂x

7.12 Formula and General Methods

187

)[ ( )] ( ∂u 1 ∂u ∂v ∂w 1 − + + l, + 2G − . ∂x 3 ∂x ∂y ∂z ) ( ) ( ∂v ∂v ∂u ∂v ∂w ∂v l+ m+ n +G l+ m+ n = Yν λθ m + G ∂x ∂y ∂z ∂y ∂y ∂y )[( ) ( ) ] ( ∂u ∂v ∂v ∂w 1 + l+ + n + G− 2. ∂y ∂x ∂z ∂y )[ ( )] ( ∂v 1 ∂u ∂v ∂w 1 − + + m, + 2G − . ∂y 3 ∂x ∂y ∂z ) ( ) ( ∂w ∂w ∂u ∂v ∂w ∂w l+ m+ n +G l+ m+ n = Zν λθ n + G ∂x ∂y ∂z ∂z ∂z ∂z )[( ) ( ) ] ( ∂u ∂w ∂w ∂v 1 + l+ + m + G− 2. ∂z ∂x ∂y ∂z )[ ( )] ( ∂ w 1 ∂u ∂v ∂w 1 − + + n. (7.66) + 2G − . ∂z 3 ∂x ∂y ∂z Equations(7.65) together with Eq. (7.66) allow to solve the problem of plasticity theory in displacements. If we assume that the terms arising from the presence of additional terms in (7.55) and transferred to the right parts (7.65) and (7.66) are known, then we get a system of equations of(the theory of elasticity with respect to displacements, but with ) ( ) additional volumetric X (0) , Y (0) , Z (0) and surface X v(0) , Yv(0) , Z v(0) forces. In the first approximation, we assume that all additional volumetric and surface loads are zeros (i.e., . = 1/2G). Then we have the “usual” problem of the theory of elasticity in displacements. Equations (7.65) are converted to the Lamé equations of the theory of elasticity, and Eqs. (7.66) to the boundary conditions of the theory of elasticity in displacements. Let the specified problem of the theory of elasticity for given forces X, ( Y, Z ; X v , Yv , Z v be solved ) and we obtained displacements (u 0 , v0 , w0 ), strains εx0 , ε y0 , εz0 , γx y0 , γ yz0 , γzx0 , and intensity of strains (εi0 ). We determine σi0 = εi0 .(εi0 ), and then .1 = 23 .(ε from the given deformation diagram (Fig. 7.23), i0 ) taking into account the hardening. We know the displacements and value .1 , and now we obtain additional loads X (0) , Y (0) , Z (0) ; X v(0) , Yv(0) , Z v(0) (which in this case are different from zero), using (7.62) and (7.64). Again, we solve the problem of the theory of elasticity with additional loads and define u 1 , v1 , w1 , and, therefore, εx1 , ε y1 , εz1 , γx y1 , γ yz1 , γzx1 and εi1 . From the deformation diagram at the calculated value εi1 , taking into account hardening, we obtain σi1 = .(εi1 ). Then we determine εi1 again, etc. .2 = 23 .(ε i1 ) The solution must continue until the current approximation differs from the previous one by an infinitesimal predetermined value. Let’s consider the implementation of the method of elastic solutions in the form of the method of additional loads in a slightly different form We use the relationship between stress and strain intensities in the form proposed by A.A. Ilyushin. (i.e. σi = 3G εi [1 − ωi (εi )]). Then the Eq. (7.55) take the form:

188

7 Mathematical Models of Plasticity Theory

( σx = 2G ∂∂ux + 2G ωi( 3θ −

+ 2G ωi σ y = 2G ∂v ∂y

θ 3

)

∂u , ∂x ) ∂v , ∂y

. . . . . . . . . . . . . (. . . . . . . . ) τx y = G(1 − ωi ) ∂u + ∂∂vx , ∂y

(7.67)

..................... Now equilibrium Eq. (7.46), taking into account (7.67) and provided that all terms containing the function ωi are transferred to the right part, are written as: ⎫ (G + λ) ∂∂θx + G.u + X = X (0) ⎪ ⎬ (G + λ) ∂∂θy + G.v + Y = Y (0) , ⎪ + G.w + Z = Z (0) ⎭ (G + λ) ∂θ ∂z where ⎧ ( ) [ ∂v i ∂u i ⎪ + − 23 ∂ω + ∂w ⎪ X (0) = G ωi .u + 13 ωi ∂∂θx + 43 ∂ω ⎪ ∂ x ∂ x ∂ x ∂ y ∂ z ⎪ ( ) ] ⎪ ( ) ⎪ i ∂w ⎪ + ∂ωi ∂u + ∂∂vx + ∂ω ; + ∂u ⎪ ⎪ ∂z ∂x ∂z ⎪ [ ∂y ∂y ⎪ ( ) ⎪ ∂ω ∂ω ∂v 4 2 1 ∂θ i ⎨ Y (0) = G ωi .v + ωi + − 3 ∂ yi ∂w + ∂∂ux + 3 ∂ y 3 ∂ y ∂ y ∂z ( ) ( )] ∂ωi ∂u ∂ωi ∂v ∂v ∂w ⎪ + + ; + + ⎪ ⎪ ∂x ∂z ∂ z ∂y ( ⎪ ) [ ∂x ∂y ⎪ ⎪ ∂ω ∂ω 1 4 2 ∂θ ∂w ⎪ Z (0) = G ωi .w + 3 ωi ∂ z + 3 ∂zi ∂ z − 3 ∂zi ∂∂ux + ∂v + ⎪ ⎪ ∂ y ⎪ ( )] ⎪ ) ( ⎪ ∂v i ∂w i ⎩ + ∂ω . + ∂ω + ∂u + ∂w ∂z ∂y ∂x ∂x ∂y ∂z

(7.68)

(7.69)

To solve Eq. (7.68), we add boundary conditions in displacements in the following form: ⎧( ) ( ( ) ) ⎪ 2G ∂∂ux + λθ l + G ∂u n = X v + X v(0) , + ∂∂vx m + G ∂w + ∂u ⎪ ⎪ ∂ y ∂z ) ⎨( ) (∂ x ) ( 2G ∂v n = Yv + Yv(0) , + λθ m + G ∂u + ∂∂vx l + G ∂v + ∂w (7.70) ∂y ∂y ∂ y) ⎪ ( ∂z ⎪ ( ) ) ( ⎪ ∂w ∂u ∂v ∂w ∂w ⎩ 2G + λθ n + G + ∂ z l + G ∂z + ∂ y m = Z v + Z v(0) , ∂z ∂x where ⎧ [( ) ( ( ∂w ∂u ) ] ) 2 ∂u ∂v ∂u (0) ⎪ X 2 m + = G ω − θ l + + + ∂z n , ⎪ i ⎪ 3) ∂x ) ⎨ v ( ∂y ) ] [( ∂ x ( ∂x Yv(0) = G ωi ∂u n , + ∂∂vx l + 2 ∂v − 23 θ m + ∂v + ∂w ∂ y ∂ y ∂z ∂ y ⎪ [( ) ( ⎪ ( ∂w 2 ) ] ) ⎪ (0) ∂u ∂v ∂w ∂w ⎩ Z v = G ωi ∂ x + ∂ z l + ∂ z + ∂ y m + 2 ∂z − 3 θ n .

(7.71)

Thus, the solution of the problem of the theory of plasticity under active deformation and simple loading, taking into account the Ilyushin dependence σi = 3G εi (1 − ωi ), came down to solve the Eqs. (7.68) under boundary conditions (7.70).

7.12 Formula and General Methods

189

This method of elastic solutions of A.A. Ilyushin is also based on the principle of successive approximations (allocation procedure). In the first approximation, it to (7.69) and (7.71), all additional is accepted that ωi0 = 0. In this case, according ( ) volumetric and surface forces are zeros X v(0) = Yv(0) = Z v(0) = X (0) = Y (0) = Z (0) . Equations (7.68) are Lamé equations of the theory of elasticity. Therefore, the solution in the first approximation is to solve the problem of the theory of elasticity. Supposing that the solution to this problem at given forces X, Y, Z ; X v , Yv , Z v is u, v, w. Using (7.67), we obtain stresses as a function of coordinates σx , σ y , σz , τx y , τ yz , τzx , and, hence, stress intensity σi . From the deformation diagram, we determine εi , and from the analytical dependence of A.A. Ilyushin we obtain ωi as a function of coordinates. Now by formulas (7.69) and (7.71) we find additional loads X (0) , Y (0) , Z (0) ; X v(0) , Yv(0) , Z v(0) as coordinate functhat, we solve the) problem of the theory of elasticity for volumetric (tions. After X − X (0) , Y − Y (0) , Z − Z (0) and surface X v + X v(0) , Yv + Yv(0) , Z v + Z v(0) forces. The solution to the problem of the theory of elasticity in this case is displacements u 2 , v2 , w2 in the second approximation, etc. The calculations can be completed when the difference between the results of two consecutive approximations is sufficiently small, that is, it will be within the required accuracy. Method of additional deformations One of the types of elastic solutions is the method of additional deformations. If in Genki-Nadai Eqs. (7.54), we denote(. = (1/2G) )+ (. − (1/2G)), and we also take into account ε0 = [(1 − 2μ)/(3E)] σx + σ y + σz , then after small transformations of Eqs. (7.54), we can write these equations in the following form: ⎧ ⎪ εx = ⎪ ⎪ ⎪ ⎪ εy = ⎪ ⎪ ⎨ εz = ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

[

)] ( ) ( 1 − μ σ y + σz + . − 2G (σx − σ0)), )( ] ( 1 σ − σ0 , − μ(σx + σz ) + . − 2G )] ( ) y ( 1 − μ σx + σ y + . − 2G (σz − σ0 ), ) ( γx y = G1 τx y + 2. − G1 τx y , ) ( γ yz = G1 τ yz + 2. − G1 τ yz , ) ( τzx = G1 τzx + 2. − G1 τzx .

1 σ E[ x 1 σ E[ y 1 σz E

(7.72)

These equations differ from the generalized Hooke’s law by additional terms. If . = (1/2G) , then these equations express the generalized Hooke’s law and determine the strains (relative deformations) in the elastic body: ⎧ [ )] ( ⎪ εx∗ = E1 [σx − μ σy + σz ] , ⎪ ⎪ ⎨ ε∗ = 1 σ − μ(σ + σ ) , z y E[ y )] ( x 1 ∗ ⎪ σ , = − μ σ + σ ε z x y ⎪ z E ⎪ ⎩γ∗ = 1 τ , γ∗ = 1 τ , γ∗ = xy yz zx G xy G yz Then

(7.73) 1 τ . G zx

190

7 Mathematical Models of Plasticity Theory

⎧ γx y

∗ ∗ (0) (0) εx = ε∗x + ε(0) x , ε y = ε y + ε y , εz = εz + εz , ∗ ∗ ∗ (0) (0) = γx y + γx y , γ yz = γ yz + γ yz , γzx = γzx + γ(0) zx .

(7.74)

(0) (0) (0) (0) (0) where ε(0) x , ε y , εz , γx y , γ yz , γzx are plastic strains in elastic–plastic bodies:

⎧ (0) ( ) 1 εx = . − 2G (σx − σ0 ), ⎪ ⎪ )( ) ( ⎪ 1 ⎪ (0) ⎪ σ y − σ0 , = . − ε ⎪ y 2G ⎪ (0) ( ) ⎨ 1 εz = . − 2G (σz) − σ0 ), ( 1 (0) ⎪ τ , = 2. − γ xy ⎪ G ) xy ( ⎪ ⎪ 1 (0) ⎪ γ τ , = 2. − ⎪ yz G ) yz ⎪ ( ⎩ 1 (0) γzx = 2. − G τzx .

(7.75)

Strain intensity is defined as: εi = εi∗ + εi(0) ,

(7.76)

where εi∗ , εi(0) are intensity of elastic and plastic strains, respectively. The physical meaning of equality (7.76) is clear from Fig. 7.22, where: εi(0) = 3σi

(

) 1 −G . 2.

(7.77)

Consider the problem of plasticity theory in stresses Differential equilibrium Eqs. (7.46) and boundary conditions (7.47) remain the same. The strain compatibility Eqs. (7.49), due to the presence of additional (underlined) terms in Eq. (7.72), will contain additional terms:

Fig. 7.22 To the method of additional deformations

B

А

7.12 Formula and General Methods Fig. 7.23 To the difference in the methods of solving problems in additional loads and additional deformations

191

B

А2 А1

( ) { [( )( )] μ ∂ X + ∂Y + ∂ Z + 2G ∂ 2 1 = −2 ∂∂ Xx − 1−μ . − 2. σ y − 13 I1 ( Tσ ) 2 ∂ x ∂ y ∂z ∂z [( )( [( ) ]} )] 2 2 1 σz − 13 I1 ( Tσ ) − ∂ ∂y∂z 2. − G1 τ yz ; + ∂ 2 . − 2. ∂y ( ) { 2 [( )( )] μ ∂X ∂Y ∂Z ∂ 1 σx − 13 I1 ( Tσ ) = −2 ∂Y ∂ y − 1−μ ∂ x + ∂ y + ∂z + 2G ∂z 2 . − 2. [( )( [( )]} )] 2 2 1 σz − 13 I1 ( Tσ ) − ∂ ∂x∂ z 2. − G1 τzx ; + ∂ 2 . − 2G ∂x ( ) { 2 [( )( )] μ ∂ Z ∂ X ∂Y ∂ Z ∂ 1 = −2 ∂z − 1−μ ∂ x + ∂ y + ∂z + 2G . − 2G σx − 13 I1 ( Tσ ) ∂ y 2[( )] [( )( ) ]} 2 2 1 σ y − 13 I1 ( Tσ ) − ∂ ∂x∂ y 2. − G1 τx y ; + ∂ 2 . − 2G ∂x ( ) { [( ) ] [( ) ] 1 τ ∂2 1 τ 1 ∂ 2 I1 ( Tσ ) = − ∂ Z + ∂Y + G ∂ 2 2. − + 2. − .τ + x y yz zx ⎪ 2 1+μ ∂ y∂z ∂z ∂z G ∂ x∂ y G ⎪ )]} [( ) ] ∂ x 2 [( )( ⎪ ⎪ ⎪ ∂2 1 τ ∂ 1 1 I (T ) ⎪ ; − 2. − − 2 . − σ + ⎪ x σ 3 1 ∂ x∂ z G xy y∂z ⎪ ⎪ ) { ∂[( )2G ] [( ) ] ( ⎪ ⎪ 1 τ ∂2 1 ∂ 2 I1 ( Tσ ) = − ∂ Z + ∂ X + G ∂ 2 ⎪ ⎪ .τ 2. − + 2. − G1 τ yz + zx zx ⎪ 2 1+μ ∂ z∂ x ∂x ∂z G ∂ x∂ y ∂ y ⎪ [( ) ] [( )( )]} ⎪ ⎪ 2 ⎪ ∂2 1 ⎪ + ∂ ∂y∂ z 2. − G1 τx y − 2 ∂z∂ . − 2G σ y − 13 I1 ( Tσ ) ; ⎪ x ⎪ ⎪ ( ) { 2 [( ) ] [( ) ] ⎪ 2 ⎪ ∂ 2 I1 ( Tσ ) ∂ X ∂Y ∂ 1 ∂ 1 1 ⎪ ⎪ 2. − τx y + ∂ x∂z 2. − G τ yz ⎪ .τx y + 1+μ ∂ x∂ y = −[(∂ y + ∂ x ) + G ⎪ ] ∂z 2 2 [( G )( )]} ⎪ 2 ⎪ 1 ∂ 1 ∂ ⎪ ⎪ + ∂ y∂ z 2. − G τx z − 2 ∂ x∂ y . − 2G σz − 13 I1 ( Tσ ) . ⎪ ⎪ ⎩ ⎧ 1 ∂ 2 I 1 ( Tσ ) ⎪ .σx + μ+1 ⎪ ⎪ ∂x2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ∂ 2 I 1 ( Tσ ) ⎪ ⎪ .σ y + 1+μ ⎪ ⎪ ∂ y2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ∂ 2 I 1 ( Tσ ) ⎪ ⎪ .σz + 1+μ ⎪ ⎪ ∂z 2 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨

(7.78) These equations are similar to the Beltrami-Michell equations, but with additional terms. If we put . = (1/2G) in the Eq. (7.78), then all the additional terms on the right side will turn to zero, and we will get the usual equations of the theory of elasticity. Equations (7.78) together with boundary conditions (7.47) allow to solve the problem of plasticity theory in stresses. Remark Equations (7.78) do not integrate in closed form. To do this, various approximate methods of solving are used. The difference in the methods of solving problems in additional loads and additional deformations is shown in Fig. 7.23. After solving the problem of the theory of elasticity, we “get” to the point B. Further “movement” according to the method of additional loads is carried out in the direction to the point A1 , while according to the method of additional deformations—in the direction to the point A2 . The convergence

192

7 Mathematical Models of Plasticity Theory

criterion of these methods is the proximity of stresses in the previous and current approximations. Method of elastic parameters The physical Eqs. (7.50) and (7.51) of the theory of small elastoplastic deformations after some transformations can be written in the form of Hooke’s law: ⎧ [ ( )] ⎪ εx = E1∗ σx − μ∗ σ y + σz , ⎪ ⎪ ⎪ ⎪ ....................................... , ⎪ ⎪ ⎨ ...................................... , (7.79) ⎪ γx y = G1∗ τx y , ⎪ ⎪ ⎪ ⎪ ...................................... , ⎪ ⎪ ⎩ ....................................... where E∗ =

(σi /εi ) 1+

1−2μ /ε 3E (σi i )

;

μ∗ =

1 2

1+

1−2μ /ε 3E (σi i ) ; 1−2μ /ε 3E (σi i )

G ∗ = (σi /εi ).

(7.80)

Besides, σ0 =

E∗ ε0 . 1 − 2μ∗

(7.81)

As 3εi /2σi = ., therefore, formulas (7.80) can be rewritten as: G ∗ = 1/2.;

E∗ =

3E ; 2E. − 1 − 2μ

μ∗ =

1 2

1+

1−2μ E (1/2.) . 1−2μ (1/2.) E

(7.82)

Therefore, if the physical equations of small elastoplastic deformations (7.50) and (7.51) are replaced respectively by Eqs. (7.79) and (7.81), then the solution to the problem of plasticity theory is reduced to solving the problem of elastic theory with parameters of elasticity theory determined by the formulas (7.80) and (7.82). According to (7.80), the relationship between elastic parameters has the same form for elastic constants E, G, μ , namely G ∗ = E ∗ /(2(1 + μ∗ )). Remark This method was first proposed by prof. I.A. Birger. The algorithm for solving the problems of plasticity theory using the method of variable elasticity parameters is as follows. In the first approximation we accept that . = 1/(2G) . Therefore, the elastic parameters equal to the elastic constants: E ∗ = E, G ∗ = G, μ∗ = μ. The physical equations in this case turn to the generalized Hooke’s law and solving the problem at this iteration step is reduced to solve the usual problem of the theory of elasticity. As a result, we determine stresses σx1 , σ y1 , σz1 , τx y1 , τ yz1 , τzx1 and strains

7.12 Formula and General Methods

193

εx1 , ε y1 , εz1 , γx y1 , γ yz , γzx1 in the first approximation. Then we determine the stress intensities σi1 and strain intensities εi1 at each point of the body. On the plane in coordinates σi − εi (Fig. 7.22), the stress–strain state of some points of the body at the stage of the first approach is depicted as the point 1 lying on the beam OA. At the same time tg α ∼ = 3G is the angle tangent of beam OA. =E∼ In the second approximation, the 3G value must be corrected. In this case ∗ ∗ tg α1 = 3G ∗1 = σi1 /εi1 , where σi1 is stress intensity, which corresponds to strain ∗ and εi1 , we obtain paramintensity εi1 , taken from the strain diagram. By values σi1 ∗ ∗ ∗ eters E 1 , μ1 , G 1 that differ at different points in the body. Now, knowing these parameters, we solve the problem of the theory of elasticity and define stresses σx2 , σ y2 , σz2 , τx y2 , τ yz2 , τzx2 and strains εx2 , ε y2 , εz2 , γx y2 , γ yz2 , γzx2 in the second approximation. Then we determine the stress intensities σi2 and strain intensities εi2 at each point of the body. On the plane σi − εi (Fig. 7.22), the stress–strain state of some points of the body at the stage of the second approach is depicted as the point 2 lying on the beam OB with angle tangent tg α ∼ = 3G ∗1 . ∗ ∗ ∗ In the third approximation tg α2 = 3G 2 = σi2 /εi2 , where σi2 is stress intensity, which corresponds to strain intensity εi2 , taken from the strain diagram. By values ∗ σi2 and εi2 , we obtain parameters E 2∗ , μ∗2 , G ∗2 . Then we solve the elastic problem with the specified elastic parameters and determine stresses and strains in the third approximation. Then we determine the stress intensities σi3 and strain intensities εi3 . On the plane σi − εi (Fig. 7.22), the stress–strain state of some points of the body at the stage of the third approach is depicted as the point 3 lying on the beam OC with angle tangent tg α ∼ = 3G ∗2 and so on (Fig. 7.24). The calculations should be continued until the results of the calculations of the nth approximation differ from the results of the calculations of the (n-1)th approximation by a given value with the required accuracy. Note that the process is always convergent.

А

1

B

2

3

Fig. 7.24 To the method of elastic parameters

С

194

7 Mathematical Models of Plasticity Theory

The method of “steps” in the theory of plastic flow Solving the problem of plastic flow theory presents significant difficulties due to the fact that the physical equations of plastic flow theory (7.52) contain not only the components of stresses, but also their increments. Since these equations cannot be solved with respect to stresses, therefore, it is impossible to construct a system of equations in displacements. In many practical problems, numerical integration with the “step-by-step” procedure of the study of the development of plastic deformations is usually used. At each stage, the external load receives increments from which the corresponding increments of stresses and strains are calculated. In addition, at each stage, it is necessary to solve some problems for an elastic-anisotropic body with elastic parameters. The problem of integrating the equations of plastic flow theory is somewhat simplified if you can neglect the increments of the elastic strain components compared to the increments of the plastic strain components. Control Questions 1.

If the yield stress is reached, then yield strains in the direction of the applied forces: (1) remain constant; (2) can reach any value; (3) disappear.

2.

Conditional yield strength: (1) stress at ε = 0.2%; (2) stress at ε = εu ; (3) yield strength at repeated loading.

3.

Deformation theory of plasticity connects: (1) stresses and strains; (2) stresses and rate of strains; (3) rate of loading and strains.

4.

Theory of plastic flow connects: (1) stresses and strains; (2) stresses and rate of strains; (3) rate of loading and strains.

5.

At repeated loading of sample yield strength: (1) raises; (2) goes down; (3) does not change.

7.12 Formula and General Methods

6.

In case of alternating loading of sample yield stress: (1) raises; (2) goes down; (3) does not change.

7.

The theory of perfectly plasticity assumes constancy: (1) stresses and strains; (2) strains; (3) stresses.

8.

Tresca-Saint-Venan plasticity condition: (1) σu = σ y ; (2) σ1 − σ3 = σ y ; (3) f 1 (J2d , J3d ) = 0.

9.

Huber-Mises-Genky plasticity condition: (1) σu = σ y ; (2) σ1 − σ3 = σ y ; (3) f 1 (J2d , J3d ) = 0.

10. Mises yield surface in axes σ1 , σ2 , σ3 : (1) cylinder; (2) prism; (3) sphere. 11. Saint-Venan yield surface in axes σ1 , σ2 , σ3 : (1) cylinder; (2) prism; (3) sphere. 12. With simple loading, stress tensor components: (1) proportional to the general parameter; (2) are extreme; (3) do not exceed yield strength. 13. Simple loading trajectory in axes σ1 , σ2 , σ3 : (1) circle; (2) parabola; (3) straight line. 14. The dependency of Stress and strain intensity for simple loading is: (1) linear;

195

196

7 Mathematical Models of Plasticity Theory

(2) power; (3) exponential function. 15. In the deformation theory of plasticity, volumetric strain is: (1) elastic; (2) plastic; (3) elastoplastic. 16. In the theory of plastic flow, volumetric strain is: (1) elastic; (2) plastic; (3) elastoplastic. 17. In the deformation theory of plasticity, stress and strain deviators are: (1) proportional; (2) linearly dependent; (3) not linked. 18. In the theory of plastic flow, the stress deviator is connected to: (1) strain deviator; (2) deviator of plastic strains; (3) rate of strains. 19. In the deformation theory of plastic material is: (1) not hardening; (2) hardening; (3) perfectly plastic. 20. Hardening hypothesis of the deformation theory of plasticity: (1) σu = .(ε (∫u );p ) (2) σu = . εu ; (3) si j = 23 σξuu ξi j . 21. Hardening hypothesis of the theory of plastic flow: (1) σu = .(ε (∫u );p ) (2) σu = . εu ; (3) si j = 23 σξuu ξi j . 22. In the method of elastic solutions at the first step is zero: (1) strain rate; (2) volumetric strain; (3) plasticity function.

7.12 Formula and General Methods

197

23. With increasing of the strain rate, the yield strength: (1) does not change; (2) increases; (3) decreases. 24. Volterra principle: the solution of the viscoelastic problem can be obtained from the solution of the corresponding problem: (1) elasticity; (2) plasticity; (3) fracture theory. 25. Draker’s postulate refers to criteria: (1) hardening; (2) softening; (3) perfectly plasticity of the material. 26. Gradient hypothesis: strain increment vector is directed to the yield surface: (1) along the tangent; (2) normal; (3) at an angle 45° . 27. The flow surface separates the elastic region from the: (1) plastic area; (2) viscoelastic area; (3) destruction area.

Chapter 8

Fundamental Solutions

8.1 General Remarks Fundamental solutions relate to solve boundary problems in case of point loading action. This is a finite force applied at a point: a surface of zero area. The point load is a mathematical artifice because we cannot achieve it in reality. Intuitively, we expect one or more components of stress to become infinite at the point where the load is placed, and this is realized in the mathematical solutions. Because of the stress singularities, understanding point-load problems will involve limiting procedures, which, as we know, are a bit dubious in regards to solids. But this should not deter us, since we will show the applications where the stresses remain finite and the applied loads are not point loads. When constructing mathematical models of solid mechanics problems, in most cases, we are dealing with the solution of boundary problems for elliptic equations. Therefore, it seems logical to consider primarily fundamental solutions for elliptic equations. Fundamental solutions to elliptic equations play an important role in constructing the general theory of elliptic boundary problems. Therefore, for example, they allow you to obtain integral representations of solutions, investigate their properties and reduce the boundary problem to integral equations. A classic example of an elliptical equation is the Laplace equation: Δ x u ≡

n Σ ∂ 2u i=1

∂ xi2

= 0,

n = 2, 3.

(8.1)

The fundamental solution of the Laplace Eq. (8.1) is a function in the following form: [ | n ⎧ 1 |Σ − 4πr , n = 3, r =] K 1 (x, y) = 1 (xi − yi )2 . 1 ln , n = 2, 2π r i=1

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Zhuravkov et al., Mechanics of Solid Deformable Body, https://doi.org/10.1007/978-981-19-8410-5_8

199

200

8 Fundamental Solutions

The function K 1 (x, y) is a solution to such differential equation: Δ x K 1 (x, y) = δ(x, y),

(8.2)

where δ (x − y) is the Dirac1 function. Solution (8.2) has the property that for any infinitely differentiable function ϕ(x) with a compact carrier, the following identity takes place: ∫ ϕ(x) =

∫ (n)

∫ K 1 (x, y)Δ y ϕ(y)dy = Δ x

∫ (n)

K 1 (x, y)ϕ(y)dy.

or ∫

∫ (n)

ϕ(y)δ(x − y)dy = ϕ(x).

Denote via L differential operator: Lu =

n Σ

2m Σ

k=0 i 1 +...+i k =1

ai1 ...ik

∂ i1 +...+ik u ∂ x1i1 . . . ∂ xkik

.

(8.3)

The fundamental solution K (x, y) of the operator (8.3) is determined by the following equation: Lx K (x, y) = δ (x − y).

(8.4)

Formula (8.4) should be understood as follows: ∫ ∫ L (n) K (x, y)ϕ(y)dy = ϕ(x). ...

Note that for second-order elliptic equations with constant coefficients in the following form: n Σ i, j=1

ai j

∂ 2u = 0, ai j = const, i, j = 1, n ∂ xi ∂ x j

(8.5)

the fundamental solution is obtained by converting the corresponding quadratic form to a canonical form. As a result:

1

Paul Adrien Maurice Dirac (1902–1984), British theoretical physicist who is regarded as one of the most significant physicists of the twentieth century.

8.2 Boussinesq’s Problem

201

⎧ ( ) 2−n ⎪ n ) 2 ( Σ ⎪ ⎪ ⎪ ,n>2 Ai j (xi − yi ) x j − y j ⎨c i, j = 1 ( ) K (x, y) = ⎪ n ) ( Σ ⎪ ⎪ ⎪ Ai j (xi − yi ) x j − y j , n = 2, ⎩ c ln i, j = 1

|| ||n where Aij is the algebraic complement to aij in matrix ||ai j ||i, j=1 ; c is a constant defined by coefficients (8.5).

8.2 Boussinesq’s Problem Of all the point load problems, the most useful in solid mechanics is the problem of a point acting normal to the surface of an elastic half-space (Fig. 8.1). The Boussinesq’s problem is to determine stress–strain state in an elastic isotropic and homogeneous half-space x3 ≥ 0, loaded at the origin of coordinates by a pointload directed along the x 3 axis. It has magnitude P: p3 (x1 , x2 ) = Pδ(x1 )δ(x2 ). Because the load is symmetrical about the planes (x 1 x 3 ) and (x 2 x 3 ), cylindrical coordinates are most convenient here (Fig. 8.1). The boundary conditions for this problem are as follows. Everywhere on the surface x3 = 0, except at the origin (0,0,0) or r = 0, loads are specified to zero. At the origin, the stresses equilibrate the applied load P, that is: p1 = p2 = 0, p3 (x1 , x2 ) = Pδ(x1 )δ(x2 ) = −σ33 (x1 , x2 , 0), x3 = 0. Besides, for any point in the half-space infinitely distant form the origin, all the displacements must vanish: .

u i (→ x) → 0, x → ∞, i = 1, 2, 3. For solution building of this problem, we can introduce such guess: the solution must have radial symmetry. That means nothing can depend on the θ coordinate, that is, P

Fig. 8.1 Boussinesq’s problem

θ

x1 r

x2

ψ x3

a

202

8 Fundamental Solutions

whatever the angle θ we choose, the solution must be exactly the same. The second guess is that the component of the displacement u θ must be zero. If u θ were not zero everywhere, there would be torsion of the half-space, at least, at some points (as the point load could not cause any torsional motions). Therefore, displacement vector has the form: u→ = (u r , 0, u z ), where u r and u z are not functions of the angle θ. Now to get expressions for strains, we will use representation for strains in the form: e=

) 1( ∇ u→ + (∇ u) → T , 2

(8.6)

where e is the strain matrix, ∇ u→ is the displacement gradient matrix. In cylindrical coordinates, the displacement gradient matrix looks like this: ⎡ ⎢ ∇ u→ = ⎣

∂u r ∂r ∂u θ ∂r ∂u z ∂r

1 r 1 r

∂u r − ∂θ ∂u θ + ∂θ 1 ∂u z r ∂θ

u θ ∂u r r ∂z u r ∂u θ r ∂z ∂u z ∂z

⎤ ⎥ ⎦.

If u θ is zero, u r and u z do not depend upon θ, then we see that ∇ u→ will have only five non-zero components lying on the two diagonals of the matrix. The strain– displacement relationship (8.6) tells us that there will similarly be only five non-zero strain components. Finally, if we use Hooke’s law, we see the stress matrix must have this form: ⎡ ⎤ σrr 0 σr z σ = ⎣ 0 σθθ 0 ⎦, σr z 0 σzz where the stress σzz is the axial stress; σrr is the radial stress; σθθ is the hoop stress. Only one non-zero shear stress is present. The procedures for building up the Boussinesq’s solution are described in various textbooks. Here we just write out Boussinesq’s solution for displacements and stress components for the vertical point-load problem. In Cartesian coordinates: [ 2 ] x3 P 1 , u3 = + 2(1 − ν) 4πμ R 3 R [ ] P · xi x3 1 , i = 1, 2. (8.7) ui = − (1 − 2ν) 4πμ R 3 R(R + x3 )

8.2 Boussinesq’s Problem

203

3P x j x32 · , j = 1, 3 2π R 5 [ ( 2 R + R · x3 + x32 3P x12 x3 1 − =− + − 2ν) (1 2π R 5 3 R 3 (R + x3 ) [ ( 2 R + R · x3 + x32 3P x22 x3 1 − =− + − 2ν) (1 5 2π R 3 R 3 (R + x3 ) ] [ 1 3P x1 x2 x3 x1 x2 2R + x3 . − =− − 2ν) (1 2π R5 3 R 3 (R + x3 )2 σ3 j = −

σ11 σ22 σ12

)] x12 (2R + x3 ) , R 3 (R + x3 )2 )] x22 (2R + x3 ) , R 3 (R + x3 )2 (8.8)

In cylindrical coordinates: [ ] rz P r ur = − (1 − 2ν) 4πμR R 3 (R + z) uθ = 0 ] [ 2 z P uz = + 2(1 − ν) 4πμR R 2 [ ] P 3r 2 z (1 − 2ν) , σrr = − − 2π R 5 R(R + z) [ ] z 1 P(1 − 2ν) − 3+ , σθθ = − 2π R R(R + z) [ 3] P 3z σzz = − , 2π R 5 ] [ P 3r z 2 , σr z = σzr = − 2π R 5 σr θ = σθr = σθz = σzθ = 0.

(8.9)

(8.10)

In these equations R 2 = r 2 + z 2 . For any point in the half-space, R is the straightline distance to the origin. Considering the stress field in Eqs. (8.10), we note that as R becomes larger, all the stress components approach to zero. On the boundary z = 0, we see that σzz and σr z vanish at every point, except at the origin of coordinates. At the origin, the stresses become singular, just as we expect for a point load. The unit normal vector to the boundary is n→ = (0, 0, −1). If we use this with equation for traction vector T = σT n, → we find that the traction vector on the boundary is zero everywhere except at origin. We’ll see how the singular stresses exactly equilibrate the point load in a moment. Next, let’s consider the displacement field. If we let R become larger, both u r and u z approach to zero, as our boundary conditions specified. On the boundary x 3 = 0 (z = 0), we have:

204

8 Fundamental Solutions

)1 2 ( P · xi (1 − 2ν) (1 − ν)P , i = 1, 2, u 3 = , r = x12 + x22 / ; 2 4πμr 2πμr P(1 − 2ν) P(1 − ν) , u zz = . ur = − 4πμr 2πμr ui = −

These formulas show that displacements become singular at the origin. Now let’s find out how the stress field equilibrates the point load. For this purpose, we consider the hemispherical surface of radius a (Fig. 8.1). Let ψ be the angle between a radius of the hemisphere and the z axis (Fig. 8.1). For any point on this surface, we have R = a = constant. The unit normal vector of the surface at any point can be written as: ⎛ ⎞ sin ψ n→ = ⎝ 0 ⎠. (8.11) cos ψ Components of the point r and z are: z = a cos ψ, r = a sin ψ.

(8.12)

The components of stress at any point on the surface can be obtained by setting R = a and using (8.12) and (8.10): [ ] P 3 sin2 ψ cos ψ (1 − 2ν) − , 2π a2 a 2 (1 + cos ψ) [ ] cos ψ 1 P(1 − 2ν) − 2 + 2 , =− 2π a a (1 + cos ψ) 3P cos3 ψ =− , 2πa 2 3P sin ψ cos2 ψ = σzr = − , 2πa 2 = σθr = σθz = σzθ = 0.

σrr = − σθθ σzz σr z σr θ

(8.13)

Using (8.11), we can obtain the traction vector that acts on the hemispherical surface: ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎤ ⎡ Tr σrr 0 σr z sin ψ σrr sin ψ + σr z cos ψ ⎦. T = ⎣ Tθ ⎦ = ⎣ 0 σθθ 0 ⎦⎣ 0 ⎦ = ⎣ (8.14) 0 Tz σr z 0 σzz σr z sin ψ + σzz cos ψ cos ψ And, finally, we use (8.13) to completely specify T in terms of P, a, and ψ. Now some words about the overall equilibrium of that part of the half-space contained inside the hemispherical surface. The horizontal upper surface just supports

8.2 Boussinesq’s Problem

205

P

Fig. 8.2 Geometry for integrating to find resultant force on hemispherical surface

ψ

a

r

z

the vertical force P. The hemispherical surface is acted on the tractions given in Eq. (8.14). These tractions have one horizontal component T r and one vertical component T z . The horizontal components exactly cancel. For every T r there is an equal and opposite component on the opposing side of the hemisphere. The vertical components T z do not cancel. They have to equilibrate the applied load P. We can write out T z by using the equations of (8.13) and (8.14): Tz = −

] 3P [ 2 3P sin ψ cos2 ψ + cos4 ψ = − cos2 ψ. 2 2πa 2πa 2

(8.15)

If we integrate T z over the hemispherical surface, we will find the resultant of upward forces, which equal to P. Let’s show it. Considering the elementary site around the hemispherical surface (Fig. 8.2) with the width a dψ, the total length of the elementary site is 2πa sin ψ. For calculating the total upward force acting on the elementary site we use (8.15): Tz Sar ea = Tz 2πa 2 sin ψdψ = 3P sin ψ cos2 ψdψ,

(8.16)

where Sar ea is total area of the elementary site. From (8.16), we see that the upward force on the site depends upon ψ and P, but not on a, that is, if we hold ψ and P constant, then this site force will be the same on every hemispherical surface no matter how big or how small the size of a is. And finally, we can find the resultant of the upward force F s that acts on the hemispherical surface by integrating: ∫π/ 2 Fs = 3P sin ψ cos2 ψdψ = P. 0

F s exactly equals to P, which is what we expected to obtain. We emphasize again that this conclusion is true for any radius a. Even if we let a approach to zero, and hence have infinite stresses, the resultant force will exactly equilibrate P. Thus, we

206

8 Fundamental Solutions

have found that all the specified boundary conditions are satisfied. Now we verify that the equilibrium equations are also satisfied. For cylindrical coordinates, equilibrium equations have the form: 1 ∂σr θ ∂σr z 1 ∂σrr + + + (σrr − σθθ ) + fr = 0, ∂r r ∂θ ∂z r ∂σrθ 1 ∂σθθ ∂σθz 2 + + + σr θ + f θ = 0, ∂r r ∂θ ∂z r ∂σr z 1 ∂σθz ∂σzz 1 + + + σr z + f z = 0. ∂r r ∂θ ∂z r

(8.17)

If we substitute expressions for the stresses (8.10), we will find these Eqs. (8.17) are exactly satisfied for the case where the body forces ( fr , f θ , f z ) are zeros. Remark Boussinesq’s solution has a lot of applications, especially in geotechnical engineering.

8.3 Hertz’s Formulas Consider the following problem: in the plane x 3 = 0, which bounds the elastic half-space, within the region S p there is a normal load p3 = p(x1 , x2 ) (Fig. 8.3). For the correctness of the setting of the boundary problem, we will assume that ˜ the integral | p(x1 , x2 )|d x1 d x2 is the limited value. To find the stress–strain state caused by such a load, we use the Boussinesq’s solution for the point load acting at the point x = (x1 , x2 , 0). We will use the formulas (8.7), rewriting them in the following form: Fig. 8.3 Principle loading pattern for Hertz’s problem

x1

р3

x2

x3

8.3 Hertz’s Formulas

207

] [ ( − x3 P ∂ →) (1 − ν)Pδi3 − · + (1 − 2ν) ln(R(x, ξ) + x3 ) , → ξ = Ui x, π μ R(x, ξ) 4π μ ∂ xi R(x, ξ) (8.18) )1 2 ( where R(x, ξ) = (x1 − ξ1 )2 + (x2 − ξ2 )2 + x32 / is the distance between points x and ξ. Let’s integrate the expression (8.18) over the area S p in the plane x3 = 0 on which the distributed load acts: ¨ p(ξ1 , ξ2 ) (1 − ν)δi3 dξ1 dξ2 u i (→ x) = πμ R(x, ξ) Sp

1 ∂ − · 4π μ ∂ xi

¨ Sp

[

] x3 + (1 − 2ν) ln(R(x, ξ) + x3 ) dξ1 dξ. p(ξ1 , ξ2 ) R(x, ξ) (8.19)

Enter the following functions: ¨ .(→ x) =

p(ξ1 , ξ2 ) ln[R(x, ξ) + x3 ]dξ1 dξ2 , Sp

¨ .(→ x) = Sp

p(ξ1 , ξ2 ) ∂ .(→ x) dξ1 dξ2 = . R(x, ξ) ∂ x3

(8.20)

From the definition of the function .(→ x) (8.20), it follows that it satisfies the Laplace equation outside the region S p , it is a continuous function everywhere except x) for the region S p , and tends to zero at R → ∞. The derivative of the function .(→ in the direction of normal to the boundary of the half-plane has a break: when moving from point x→ = (x1 , x2 , x3 ) to point x→ 0 = (x1 , x2 , 0) on the side of positive values of x 3 , the following relation occurs: | ⎧ ∂ . || −2π p(x1 , x2 ), x ∈ S p , = | 0, x∈ / Sp. ∂ x3 x3 →+0

(8.21)

The function .(x) in half-space is harmonic. At R → ∞ this function increases indefinitely. Derivatives of .(x) by coordinates in the half-space x3 > 0 at R → ∞ tend to zero. Using the functions .(→ x) and .(x), relation (8.19) can be represented in the following form: u→ =

1 1 − ν → i3 . − grad[x3 . + (1 − 2ν).]. π·μ 4π μ

(8.22)

208

8 Fundamental Solutions

Rewrite (8.22) in the components of the displacement vector: ] ] [ [ ∂. ∂. 1 ∂. 1 ∂. , u2 = − , x3 x3 u1 = − + (1 − 2ν) + (1 − 2ν) 4π μ ∂ x1 ∂ x1 4π μ ∂ x2 ∂ x2 ∂. 1−ν 1 u3 = . (8.23) .− x3 2π · μ 4π μ ∂ x3 Based on (8.23), the displacement values on the plane x3 = 0 are defined by the following formulas: uβ = −

1 − 2ν ∂.(x1 , x2 , 0) 1−ν .(x1 , x2 , 0). , β = 1, 2 u 3 = 4π μ ∂ xβ 2π · μ

For the stress components, we get: σ13

) ( ∂ 2. x3 ∂ 2 . 1 ∂. x3 ∂ 2 . , σ23 = − , σ33 = − x3 2 . (8.24) =− 2π ∂ x1 ∂ x3 2π ∂ x2 ∂ x3 2π ∂ x3 ∂ x3

From (8.24), it is easy to obtain that at x 3 = 0 it is performed that: σ31 = 0, σ32 = 0, σ33 =

) ( 1 ∂. . 2π ∂ x3 x3 →+0

From (8.21), we obtain that: 1 σ33 (x1 , x2 , 0) == 2π

(

)| ⎧ ∂ . || − p(x1 , x2 ), x ∈ S p , = 0, x∈ / Sp. ∂ x3 |x3 →0

If you enter a new function: F =−

1 [x3 . + (1 − 2ν).], 4π μ

then the displacements (8.23) can be represented by a shorter record, namely: ui =

∂F 1−ν δi3 .. + ∂ xi πμ

(8.25)

Formulas (8.25) were first introduced by Hertz and were used by him in solving contact problems. Remarks The problems connected with determination of the stress–strain state in an elastic half-space in case of action on boundary x 3 = 0 loadings, perpendicular to the plane x 3 = 0, were solved by many researchers.

8.4 Flamant’s Problem

209

Fig. 8.4 Flamant’s problem for a half-space

p

dy

xa ba

Ra

y

za φ

A

x

z

8.4 Flamant’s2 Problem 8.4.1 Flamant’s Problem for a Half-Space In this section, we consider a line load acting on the surface of a half-space (Fig. 8.4). The line load acts on an infinitely long line. The intensity of this line load is p, and it has dimensions of force per unit length. We will use the Boussinesq’s solution together with the principle of superposition to solve the stress–strain state in the half-space. Consider the point A in Fig. 8.4. Note, the fact that this point lies beneath the x axis doesn’t mean any loss of generality. Since the load extends to infinity in both directions on the y axis, we can place the origin of coordinates wherever we like on the y axis. Let’s find the σzz components of stress at point A due to the action of the line load on the surface of a half-space. We consider an elementary length dy of the y axis (Fig. 8.4). For any point inside the half-space, including point A, the line load p acting on the element dy may be considered as a point load. There will be an increment of stress dσzz caused by this “point load” pdy. According to the Boussinesq’s solution (8.10), we have: dσzz = −

[ ] ( pdy) 3z 3 . 2π R5

(8.26)

The distances between z and R are shown in Fig. 8.4. To obtain the stress σzz , we must integrate both sides of (8.26): ∫∞ σzz = − −∞

2

3 pz 3 dy. 2πR 5

Flamant A. (1839–1915), French mechanical engineer.

(8.27)

210

8 Fundamental Solutions

)1 2 ( Note that (Fig. 8.4) b = x 2 + z 2 / , y = btgϕ. So dy = b sec2 ϕdϕ. We can rewrite (8.27) as: ∫π/ 2 σzz = − −π/ 2

3 pz 3 cos3 ϕdϕ 2πb4

(8.28)

Note that z and b in (8.28) are constants. Now we integrate (8.28) and obtain: σzz = −

2 pz 3 2 pz 3 = − )2 . ( πb4 π x 2 + z2

(8.29)

The other components of stress are obtained by similar methods: 2 px 2 z σx x = − ( )2 , π x 2 + z2 2 pyz ), σ yy = − ( 2 π x + z2 2 px z 2 σx z = − ( )2 = σzx , π x 2 + z2 σx y = σ yx = σzy = σ yz = 0.

(8.30)

Remark Obtaining all these components is not quite easy as obtaining σzz . The vertical stress component σzz doesn’t differentiate between cylindrical and rectangular coordinates, but the other stress components do, and we must be careful. Note that when we integrated (8.26) to get (8.27), we were using the principle of superposition. However, if we did not deal with linear elasticity, this might not be possible. Let’s consider a cylindrical surface of radius b whose axis coincides with direction of line load (Fig. 8.5). Let’s obtain the tractions that act on this surface by using the expressions for stresses (8.29) and (8.30). As a result of performing mathematical operations, we get: T=−

2 pz n, → πb2

(8.31)

where n→ is the unit normal to the cylindrical surface. If we integrate the vertical component of the traction (8.31) around the cylindrical surface, we can make sure that the resultant of upward forces exactly equilibrates the applied load p.

8.4 Flamant’s Problem

211

p

Fig. 8.5 Cylindrical surface whose axis coincides with direction of line load

x α

b n

z

From (8.31), we can conclude that the cylindrical surface itself is a principal surface. The minor principal stress acts on it: σ1 = −

2 pz . πb2

(8.32)

Remark The principal stress σ1 is the major by absolute value (by modulus). The intermediate principal surface is defined by n→ = (0, 1, 0) and the intermediate principal stress is σ2 = νσ 1 . The major principal surface is perpendicular to the cylindrical surface and to the intermediate principal surface. The major principal stress is exactly zero. Some words about the distribution of the principal stress in space. Consider the locus of points on which the major principal stress σ1 is a constant. From (8.32), we see this will be a surface for which: 1 z π|σ1 | = , = b2 2p 2c where c is a constant. From this ratio, it follows that: b2 = x 2 + z 2 = 2cz. This is the equation of a circle with the radius c centered on the z axis at a depth c beneath the origin. At each point on this circle, the major principal stress is the same. It points directly to the origin. If we were to consider a larger c and consequently a larger circle, the value of σ1 would be smaller in inverse proportion to c. This result allows us to formulate the idea that there is a pressure bubble under the foundation in the soil.

212

8 Fundamental Solutions

8.4.2 Flamant’s Problem for a Half-Plane Interesting point is when we are dealing with a plane strain problem. For this type of problem e yy = ex y = e yx = ezy = e yz = 0. Also, the non-zero strains are not functions of y. A particle that initially has coordinate y0 in the referred configuration will always have coordinate y0 in any deformed configuration unless a rigid translation in the y direction occurs. We will make some comments regarding the selection of a coordinate system when constructing model problems in a two-dimensional statement. This is important since the displacements of the points of the elastic body during its deformation are determined before additive displacements occur. Remarks Additive displacements are displacements of a deformed body with respect to some selected coordinate system as a rigid body. Since the components of strains and stresses in the elastic half-plane decrease, as / they are removed from the place of application of the load, with order 1 r , when they are integrated (when determining displacements), logarithmic features appear. Therefore, with a poorly selected coordinate system, when moving away from the point of application of the load (r → ∞), displacements may be infinite. Note that in the case of considering a three-dimensional elastic body, the decrease in strains / 2 and/stresses is in order 1 r . Accordingly, for displacements, the decreasing order is 1 r . Some explanations in this regard are given as below. Consider an elastic half-space, to the boundary of which an arbitrary load is applied on a bounded section with a finite resultant force and a finite principal moment. According to the above facts, the origin of the coordinate system can be associated with a half-space point, as far as you want from its boundary. One of the axes can be directed along an element that was perpendicular to the half-space boundary before deformation. The displacements of the elastic half-space points in such a coordinate system are always finite and disappear as they move away from the place of the boundary where the load is applied. As a result, this coordinate system can be replaced by another coordinate system, starting at one of the points of the body boundary before deformations, with two axes lying in the boundary plane and a third axis perpendicular to it. Displacements of a half-infinite solid with respect to this new coordinate system will be simultaneous displacements with respect to infinitely distant elements. That is, you can “fix” the body at infinitely distant points. Unfortunately, a similar procedure cannot be done in the case of an elastic halfplane, even if a load is applied at a section of the boundary of the finite length. Indeed, in this case, the change in distance (and therefore displacements) between two arbitrary points of a 2D body (located, for example, perpendicular to the undeformed half-plane boundary) grows according to the logarithm law depending on the distance itself. Therefore, if you take a point at infinity as the origin (i.e., “fix” half-plane in infinity), then the displacements of boundary are unlimited. That is, the origin of the coordinate system with respect to which the displacements are determined must be

8.4 Flamant’s Problem

213

a

b

x

h

z Fig. 8.6 Flamant’s problem for a half-plane

associated with a particular point of the elastic half-plane (and not be arbitrary, as in the case of half-space). Thus, it can be, for example, a point located on the boundary of the half-plane. A. Consider the problem of the action of a point load P→ on an elastic half-plane (Fig. 8.6). The load P→ is directed along the z axis and applied at the origin. In order to correctly set the problem, as mentioned above, it is necessary to clearly indicate the coordinate system with respect to which the displacements are calculated. We fix the half-plane element located on the z axis at the depth h. In this case, the following equalities are fulfilled: at x = 0, z = h: u = 0, v = 0, ∂z u = 0, where u = u(x, z) and v = v(x, z) are correspondingly the components of the displacement of the points of the elastic half-plane parallel to the axes x and z. The solution of the formulated problem on the effect of the point load on the elastic half-plane, taking into account the fixing conditions, can be presented in the following form: ] [ ] λ+μ xz r λ + μ x2 1 ϑ − ln + , v = A , 2(λ + 2μ) 2μ(λ + 2μ) r 2 2μ h 2μ(λ + 2μ) r 2 . λ + 2μ P (8.33) , r = x 2 + z 2 , ϑ = arctg(x / z). A=− λ+μ π [

u=A

For points of elastic half-plane boundary, formulas (8.33) take the following form (z = 0): A π sign x, 2(λ + 2μ) 2 [ ] |x| 1 λ+μ v(x, 0) = V (x) = A ln + . 2μ h 2μ(λ + 2μ)

u(x, 0) = U (x) =

(8.34)

214

8 Fundamental Solutions dξ

Fig. 8.7 To problem about the normally distributed load acts normally on horizontal boundary of elastic half-plane

p(ξ)

ξ

x

h

y

B. Now let’s consider the next problem. The normally distributed load of intensity p(ξ) acts normally on horizontal boundary of elastic half-plane in section S : ξ ∈ [a, b] (Fig. 8.6). Define the displacements of half-plane boundary points. If a load pdξ acts at the origin, then the displacements on the boundary element, remote from the point (0, 0), by the distance ξ (Fig. 8.7), in accordance with the formula (8.34), are: dVf = −

|ξ| λ + 2μ ln pdξ. 2 π μ(λ + μ) h

(8.35)

If the load p(ξ) is distributed in the section ξ ∈ [a, b], then the displacements at the origin are: λ + 2μ V f (x) = − 2 π μ(λ + μ)

∫b ln

|ξ| p(ξ)dξ. h

(8.36)

a

Formulas (8.35–8.36) for determining displacements at the origin are not accurate. Therefore, the expression (8.35) for displacements at the origin from the load pdξ is true in the assumption of fixing the half-plane element at the depth h, located directly below the point of application of the load, that is, at the distance ξ from the z axis (Fig. 8.7). Thus, in the expression (8.35), in fact, summation of displacements of an infinite number of deformed states with fixations at different points ξ is performed, which is fundamentally unacceptable. We describe the procedure for constructing the correct solution A. Let the point load act at the points of the boundary with the coordinate (ξ, 0). Formulas for displacements are obtained from formulas (8.33), in which the coordinate x is replaced by the difference (x − ξ). In addition, displacements corresponding to the additive displacement of the half-plane as a rigid body should be added to the values u and v. As a result, we get:

8.4 Flamant’s Problem

215

x −ξ A λ+μ (x − ξ)z arctg A 2 + α + ω z, − 2(λ + 2μ) z 2μ(λ + 2μ) z + (x − ξ)2 / A λ+μ (x − ξ)2 ln z 2 + (x − ξ)2 + A 2 v= + β − ω x. 2μ 2μ(λ + 2μ) z + (x − ξ)2 (8.37)

u=

To determine the additive displacement constants α, β, ω, we assume that at the point (0, h), the displacements u and v are turned to zero along with the derivative ∂ u/∂ z (thereby fixing the element of the axis z at the depth h). Solving the corresponding equations, we get: ξ h·ξ 2h 3 ξ A λ+μ A arctg + − , 2(λ + 2μ) h 2(λ + 2μ) h 2 + ξ2 2μ(λ + 2μ) (h 2 + ξ2 )2 ξ2 λ+μ A . 2 ln h + ξ2 − A 2 , β=− 2μ 2μ(λ + 2μ) h + ξ2 ) ( 2 ξ − h2 ξ ξ λ+μ A A( − ω=− ) . 2(λ + 2μ) h 2 + ξ2 2μ(λ + 2μ) h 2 + ξ2 2 α=

We substitute these values α, β, ω into expressions (8.37) for u and v, and after obvious transformations, we have: ] [ x −ξ ξ hξ A yξ arctg + arctg + 2 u(x, z, ξ) = − 2(λ + 2μ) z h h + ξ2 h 2 + ξ2 [ ) ] ( 2 ξ − h2 ξ z λ+μ 2h 3 ξ (x − ξ)z − A 2 +( , )2 + 2μ(λ + 2μ) h 2 + ξ2 z + (x − ξ)2 h 2 + ξ2 / ξx A A (x − ξ)2 + z 2 ln + v(x, z, ξ) = 2 2 2 2μ h +ξ 2(λ + 2μ) h + ξ2 [ ) ] ( 2 ξ − h2 ξ x λ+μ ξ2 (x − ξ)2 A 2 + − 2 + ( )2 . 2μ(λ + 2μ) h + ξ2 z + (x − ξ)2 h 2 + ξ2 (8.38) The formula for determining the vertical displacements of the boundary surface points (at z = 0), on the basis of (8.38), is as follows: |x − ξ| ξ· x A A ln . + 2 2 2 2μ 2(λ + 2μ) h + ξ2 h +ξ [ ] ) ( 2 ξ − h2 ξ · x ξ2 λ+μ A 1− 2 + ( + )2 2μ(λ + 2μ) h + ξ2 h 2 + ξ2

V (x, ξ) = v(x, 0, ξ) =

(8.39)

216

8 Fundamental Solutions

As a result, (8.39) is the displacement on the surface of the elastic half-plane as a result of the action at the point (ξ, 0) of the point load P→ perpendicular to the boundary of the elastic half-plane. B. Now let’s look at the case where the distributed load p(ξ) is acting on the site (a, b). Vertical displacements from such load at any point (x, 0) of the boundary according to (8.39) in this case have the following representation: ⎧ |x − ξ| 1 ξ· x 1 λ + 2μ 1 p(ξ) ln . + V (x) = 2 2 2 π λ+μ 2μ 2(λ + 2μ) h + ξ2 h +ξ a [ ]⎧ ( 2 ) ξ − h2 ξ · x ξ2 λ+μ A 1− 2 + ( + dξ. (8.40) )2 2μ(λ + 2μ) h + ξ2 h 2 + ξ2 ∫b

Compare the obtained formulas (8.40) with inaccurate expressions (8.36). Let x = 0, a = 0, b = c, λ = μ, p = p0 = const. According to the exact formula (8.40), we obtain: 3 · p0 V (0) = 4π μ

⎧ ∫c ⎧ . 2 2 ξ2 2 2 ln h + ξ − ln ξ − + dξ. 3 3 h 2 + ξ2 0

According to the imprecise formula (8.36), we have: 3 · p0 V f (0) = 4πμ

∫c (ln h − ln ξ)dξ. 0

After calculating the integrals, we finally get: ( ) √ 2 c h 2 + c2 3 · p0 + h arctg , V (0) = c ln 4π μ c 3 h ( ) h 3 · p0 c ln + c . V f (0) = 4πμ c It follows that when using the formula for V f , we get the wrong result, if we take in it that h = 0 (i.e., fix the half-plane element to the origin (0, 0)). In this case, the displacements become infinitely larger. The exact formula for V at (0, 0) gives the value V (0) = 0.

8.5 Kelvin’s Problem

217

8.5 Kelvin’s Problem In this section, we consider another point load problem. This is the problem of a point load acting in the interior of an infinite elastic body (Fig. 8.8). The procedure for building up the Kelvin’s solution base on Kelvin’s methods and it is described in detail, for example, in textbooks [Nowacki, W. Theory elasticity/W. Nowacki.—Moscow. Pub. by “Mir”, 1975. 872 p. (in Russian), Zhuravkov, M.A., Continua Mechanics. Theory of Elasticity and Plasticity. Coursebook/M.A. Zhuravkov, E.I. Starovoitov. Minsk. Pub. by Belarusian State University, 2011. 543 p. (Classical University Edition). (in Russian)]. The Kelvin’s solution, written in cylindrical coordinates has such form: Pzr , 16πμ(1 − ν)R 3 u θ = 0, [ 2 ] z P 1 2(1 − ν) uz = . + + 16πμ(1 − ν) R 3 R R ] [ 2 3r z P (1 − 2ν)z , σrr = − − 8π(1 − ν) R 5 R3 P(1 − 2ν)z σθθ = , 8π(1 − ν)R 3 ] [ 3 3z P (1 − 2ν)r , σzz = − + 8π(1 − ν) R 5 R3 [ ] 3r z 2 P (1 − 2ν)r σr z = σzr = − , + 8π(1 − ν) R 5 R3 ur =

(8.41)

Fig. 8.8 Kelvin’s problem

P

θ

x1 r

x2 x3/z

218

8 Fundamental Solutions

P

Fig. 8.9 Geometry for integrating vertical stress σzz

c

ψ

r

R dr

r

z

σr θ = σθr = σθz = σzθ = 0,

(8.42)

where R 2 = z 2 + r 2 . Solutions (8.41–8.42) have singularities at the origin, where the point load acts, and both displacements and stresses die out for large R. Note that on the plane z = 0, all the stress components except for σr z vanish at all points except for the origin. Now we investigate the solutions (8.41–8.42) by considering a planar surface defined by z = c. The vertical component of traction on this surface is σzz . Let’s now integrate σzz over this entire surface for calculating the resultant force. Note that the value of σzz will be a constant on any horizontal circle centered on the z axis. Therefore, the force acting on the annulus (Fig. 8.9) will equal to σzz multiplied by the area 2πr dr . The total resultant force Rsum on the surface z = c is given by the following formula: ∫∞ Rsum =

∫∞ σzz 2πr dr = −

0

0

] [ 3 3c P (1 − 2ν)c r dr . + 4(1 − ν) R 5 R3

For integration, we introduce the angle ψ (Fig. 8.9). We have: r = ctgψ and dr = c sec2 ψdψ, taking into account these relations, the integral takes the form: ∫π/ 2 Rsum = − 0

[ ] P (1 − 2ν) sin ψ + 3 cos2 ψ sin ψ dψ. 4(1 − ν)

After integral calculation, we find that the resultant force equals to –P/2, exactly one-half of the applied load. If now we consider a similar surface z = −c, we find tensile stresses of the same magnitude as the compressive stresses on the lower plane. The resultant force on the upper plane is P/2 (a tensile force). Together, the two resultant forces exactly equilibrate the applied load.

8.5 Kelvin’s Problem

219

As we noted earlier, on the plane z = 0, all the stress components except for σr z vanish / at all points except for the origin. Let’s consider the special case, where ν = 1 2 (an incompressible material). Then σr z will also be zero on this surface, and that part of body below the z = 0 plane becomes equivalent to the half-space of the Boussinesq’s problem. Comparing the Kelvin’s solution with Boussinesq’s solution / for the case ν = 1 2, we see they are identical for all z ≥ 0. For z ≤ 0 we also have the Boussinesq’s solution, but with a negative load –P. The two half-spaces, which together comprise the infinite body of Kelvin’s problem, act as if they are uncoupled on the plane z = 0 where they meet [Davis, R.O. Elasticity and geomechanics / R.O. Davis, A.P.S. Selvadurai. Cambridge University press. 1996. 201 p.]. If we consider a spherical surface centered on the origin, we find it is a principal surface, supporting the minor principal stress σ1 : σ1 = −

3Pz , 2πR 3

where R is the sphere radius. Note how σ1 change the sign from negative to positive values of z, giving tensile stresses above the median plane z = 0. Let’s write Kelvin’s solution in the Cartesian coordinate system. First, we consider the action of the point load applied at the origin and directed parallel to the x 3 axis (Fig. 8.8). The point load can be written as below: X i = Pδi3 · δ(→ x), where δ(→ x) is the Dirac delta function. Kelvin’s solution for this case has the form: ( ) xi · x3 λ + 3μ δi3 λ+μ (3) · , + ui = P 8π μ(λ + 2μ) R3 λ+μ R

(8.43)

where u i(3) are displacement components caused by the action of a point load directed along the x 3 axis and applied at the origin R(x, 0) = (xi · xi )1/ 2 . A solution of formula (8.43) is often called an elementary solution of the first kind. Formula (8.43) in the literature is often recorded as: ) ( δi3 − (R),i3 , u i(3) = P A B · R

(8.44)

λ+μ , B = 2(λ+2μ) . where A = 8π μ(λ+2μ) (λ+μ) Stresses and dilatations corresponding to the displacements defined by formula (8.44) are given by such formulas:

220

8 Fundamental Solutions

σ11

[ ( ] ) ∂ R−1 μ ∂R 2 = P2μA − 3 , ∂ x3 ∂ x1 λ+μ ∂ R ∂ R ∂ R−1 , ∂ x1 ∂ x2 ∂ x3 [ ( ) ∂ R−1 ∂R 2 = P2μA + 3 ∂ x1 ∂ x3 [ ( ) ∂ R−1 ∂R 2 = P2μA − 3 ∂ x3 ∂ x2 [ ( ) ∂ R−1 ∂R 2 = P2μA + 3 ∂ x2 ∂ x3 [ ( ) ∂ R−1 ∂R 2 + = P2μA 3 ∂ x3 ∂ x3

σ12 = P6μA σ13 σ22 σ23 σ33

θ = PA

] μ , λ+μ ] μ , λ+μ ] μ , λ+μ ] μ , λ+μ

(8.45)

2μ ∂ R−1 . λ + μ ∂ x3

Consider a sphere centered at the origin and calculate the value of the forces on that sphere pi = σi j · n j . We assume that the normal directs inside the sphere (that is, n j = −x j /R) and P = 1. Then in accordance with the formulas (8.45), we have: p1 =

6μ A x2 x3 2μ A 6μ A x1 x3 , p2 = , p3 = R4 R4 R2

(

) 3 · x32 μ . + R2 λ+μ

(8.46)

We investigate whether the ball cut from the body will be in equilibrium. To do this, check whether the equilibrium conditions are met: ∫

∫ p1 d S = 0, S

∫ p2 d S = 0,

S

p3 d S − 1 = 0.

(8.47)

S

We substitute (8.46) into expressions (8.47) and integrate along the surface of the ball r = a. As a result, we get that the equilibrium conditions are met. Now let’s summarize the results for the general case of the point load in unlimited space. − → ( j) Denote through displacement Ui (→ x, ξ ) caused by the action of a point load − → directed parallel to the x j axis and applied at a point ξ , magnitude is P = 1. By analogy with the function (8.43) and expression (8.44), you can write: ( j)

Ui

(

) ( ) ( B · δi j xi · x j λ + 3μ δi j − →) · = A − x, → ξ = A + (R) ,i j . R3 λ+μ R R

8.6 Cerrutti’s Problem

221

− → ( j) Functions Ui (→ x, ξ ) are form the Green’s displacement tensors. Recall that the Green’s displacement tensor is symmetrical, that is: (− → ) − → ( j) Ui (→ x, ξ ) = U (ij ) ξ , x→ , − → and at a point ξ , its components have a feature of order 1/R, and their values decrease with increasing values R. − → If a mass force X i (ξ) · d V (ξ) is acting at a point ξ , the displacements caused by that force at the point x→ will be: [ ] u i = X 1 (ξ)Ui(1) (x, ξ) + X 2 (ξ)Ui(2) (x, ξ)+ X 3 (ξ)Ui(3) (x, ξ) d V (ξ) ( j)

= X j (ξ)Ui (x, ξ)d V (ξ). We integrate in the area V ' in which mass forces operate. As a result, we get the following formula: (− ∫ →) ( j) ( − →) → ξ d V (ξ) u i (→ X j ξ Ui x, or u i (→ x) = x) = V' ( ) ( ) ∫ − → − → X j ξ U (ij ) ξ , x→ d V (ξ).

V'

8.6 Cerrutti’s3 Problem Cerrutti’s problem is the problem about a horizontal point load acting on the surface of an elastic half-space (Fig. 8.10). We consider the problem of loading the surface x 3 = 0, which bounds the elastic half-space, with tangent forces. Tangent load is the point load P · δ(x1 )δ(x2 ) and acts at the origin of coordinates, pointing in the x 1 direction. The point load is represented by P and acts at the origin of coordinates, pointing in the x 1 direction. This is a more complicated problem than either the Boussinesq’s or Kelvin’s problem due to the absence of radial symmetry enjoyed by those two problems. We use a rectangular coordinate system. Cerrutti’s solution is: ( )] [ 1 1 x12 P x12 , + 3 + (1 − 2ν) u1 = − 4π μ R R R + x3 R(R + x3 )2 ] [ x1 x2 P x1 x2 u2 = , − (1 − 2ν) 4π μ R 3 R(R + x3 )2 ] [ P x1 x3 x12 u3 = . + − 2ν) (1 4π μ R 3 R(R + x3 )2 3

Cerrutti V. (1850–1909), Italian scientist-mechanic.

(8.48)

222

8 Fundamental Solutions

Fig. 8.10 Cerrutti’s problem

P x1

x2

σ11 σ22 σ33 σ12 σ23

[ 3x12 P x1 − = + 2π R 3 R2 [ P x1 3x22 = + − 2π R 3 R2 3P x1 x32 , =− 2π R 5 [ 3x12 P x2 − + = 2π R 3 R2 3P x1 x2 x3 =− , σ31 2π R 5

x3

( )] 2Rx22 (1 − 2ν) 2 2 R − x2 − , (R + x3 ) (R + x3 )2 ( )] 2Rx12 (1 − 2ν) 2 2 − x − 3R , 1 (R + x3 ) (R + x3 )2 ( )] 2Rx12 (1 − 2ν) 2 2 −R + x1 + , (R + x3 ) (R + x3 )2 3P x12 x3 =− , 2π R 5

(8.49)

)1 2 ( where R = x12 + x22 + x32 / . Analysis of solutions (8.48–8.49) shows that displacements and stresses at the origin are singular, while for large R, all components approach to zero. The vertical plane that contains the x 1 axis is a plane of symmetry. Looking at the x 1 -component of the displacement field, we see that particles are displaced in the direction of the point load. The x 2 -component of displacement moves particles away from the x 1 axis for positive values of x 1 and toward the x 1 axis for negative x 1 . Vertical displacements take the sign of x 1 and hence particles move downward in front of the load and upward behind the load. Both the displacements and stresses for Cerrutti’s problem appear more complex than for either the Boussinesq’s or Kelvin’s problem, but this is largely an illusion. Cerrutti’s solution is embedded in the Kelvin’s solution for the special case where ν = 0.5. If we set ν equal to 0.5 in both solutions, we find the stresses and displacements in the left- or right-hand half-space of Kelvin’s problem are identical to the Cerrutti’s solution, provided we to designate the coordinate axes appropriately.

8.7 General Case of Point Load Action on Surface of Falf-Space

223

8.7 General Case of Point Load Action on Surface of Falf-Space − → Consider the general case of loading action: an arbitrary point load P acts at the origin. Force components are (P1 , P2 , P3 ) and they are directed along three respective coordinate axes (x 1 , x 2 , x 3 ). We build a total solution, adding together the Boussinesq’s solution and Cherutti’s solution. Therefore, the total displacements are obtained by adding the displacements expressed by the formulas (8.7) for the component of the load P3 , and the displacements caused by the forces P1 and P2 (formulas (8.48)): ( )] x12 x12 1 1 + 3 + (1 − 2ν) − 4 π μ u 1 = P1 R R R + x3 R(R + x3 )2 [ ] [ ] x1 x2 x1 x3 x1 x2 x1 + P2 + P − − 2ν) − − 2ν) , (1 (1 3 R3 R3 R(R + x3 ) R(R + x3 )2 ] [ x1 x2 x1 x2 − (1 − 2ν) 4π μ u 2 = P1 R3 R(R + x3 )2 [ )] ( [ ] x2 x22 x2 x3 1 1 x2 + 23 + (1 − 2ν) − − − 2ν) , + P2 + P (1 3 R R R + x3 R3 R(R + x3 ) R(R + x3 )2 [ ] x12 x1 x3 4 π μ u 3 = P1 + − 2ν) (1 R3 R(R + x3 )2 [ ] [ ] x32 x22 x2 x3 1 + (1 − 2ν) + 2(1 − ν) + P2 + P3 , R3 R3 R R(R + x3 )2 [

)1 2 ( where R = x12 + x22 + x32 / . If distributed loads p(x1 , x2 ) ≡ ( p1 (x1 , x2 ), p2 (x1 , x2 ), p3 (x1 , x2 )) act on the plane × 3 = 0 in the area D, then the displacements caused by them can be found using the superposition principle. Therefore, for example, the displacement component u 1 (→ x) is expressed by the following formula: 4 π μ u1 (→ x) = ¨ ⎧

( )] 1 1 x 1 − ξ1 (x1 − ξ1 )2 p1 (ξ1 , ξ2 ) + + (1 − 2ν) − R R3 R + x3 R(R + x3 )2 D [ ] (x1 − ξ1 )(x2 − ξ2 ) (x1 − ξ1 )(x2 − ξ2 ) + p2 (ξ1 , ξ2 ) − − 2ν) (1 R3 R(R + x3 )2 [ ]⎧ (x1 − ξ1 ) (x1 − ξ1 )x3 dξ1 dξ2 . − − 2ν) + p3 (ξ1 , ξ2 ) (1 R3 R(R + x3 ) [

224

8 Fundamental Solutions

x1 h

x2 P

x3 Fig. 8.11 Mindlin’s problem

8.8 Mindlin’s4 Problem Mindlin’s problem is the problem of a point load (arbitrary direction in space) acting in the interior of an elastic half-space. The load P acts at a point, a distance h beneath the half-space surface (Fig. 8.11). Plane x3 = 0 is free from external forces. Mindlin’s problem is a generalization of the Kelvin’s, Bussinesq’s, and Cherutti’s problems. As the first problem, consider a special case where the load P acts at the point (0, 0, h) and directed in the positive direction of the x 3 axis. We divide the solution of the formulated problem into two stages. First, we consider the problem of the action in unlimited space of two oppositely directed forces: the force P = +1 applied at a point (0, 0, h) and the force P = −1 acting at{a point}(0, 0, −h). We denote the stress–strain state corresponding to this load as u i0 , σi0j . To determine it, we use the Kelvin’s method. We use formulas (8.44) and apply the principle of superposition. As a result, for displacements u i0 we get the following formulas: ( u i0 = P A

) ) ( 1 xi · (x3 − h) xi (x3 + h) 1 δi3 , − + B − R1 R2 R13 R23

where λ + 3μ λ+μ , ,B = 8 π μ(λ + 2μ) λ+μ ]1 2 ]1 2 [ [ R1 = r 2 + (x3 − h)2 / , R2 = r 2 + (x3 + h)2 / , A=

r 2 = x12 + x22 . By defining the stress field σi0j , we can see that in the plane x3 = 0, only the stress 0 σ33 is not zero: 4

Raymond David Mindlin (1906–1987), American mechanical scientist.

8.8 Mindlin’s Problem 0 σ33

225

) ( )1 2 ( λ 1 − 2ν 3h 2 h , R0 = h 2 + x12 + x22 / . = + 5 ,ν = 4π(1 − ν) 2(λ + μ) R03 R0

Therefore, in the second stage, we determine the displacements u i' caused by the 0 action of a load p3 (x1 , x2 ) = −σ33 (x1 , x2 , 0) in the plane x3 = 0 of the elastic half-space. This load is applied to balance the loads in the plane x3 = 0. This is the problem discussed in Sects. 8.2 and 8.3 (the Bussinesq’s and Hertzs’ problems). Finally, formulas for complete displacements have the form: ) ( [ ] P x3 − h ∂. ∂. x3 + h Pr − x3 + (1 − 2ν) , − ur = 16 π μ (1 − ν) 4π μ ∂r ∂r R13 R23 ( )] [ 1 1 P (x3 + h)2 (x3 − h)2 u3 = − + (3 − 4ν) − 3 3 16π μ (1 − ν) R1 R2 R1 R2 1−ν x3 ∂ . + , .− 2π μ 4π μ ∂ x3 where u r = u 1 cos θ + u 2 sin θ; .(→ x) =

x3 + h h 1 + , x3 > 0. R2 2(1 − ν) R23

.(→ x) = ln(R2 + x3 + h) −

1 ∂. h , = .. 2(1 − ν) R2 ∂ x3

The components of stresses in cylindrical coordinate system (r, θ, z) have such representations: [ 2 3r z 1 P (1 − 2ν)z 1 (1 − 2ν)z − 12(1 − ν)h − + 3 5 8π(1 − ν) R3 R3 R3 ] 30hz 2 (z − h) 3r 2 z − 6(7 − 2ν)hz 2 + 24h 2 z B + σrr − , − 5 R R7

σrr = −

P σθθ = 8π (1 − ν) [ ] −6(1 − 2ν)hz 2 − 64h 2 z (1 − 2ν)z 1 (1 − 2ν)(z + 6h) B − + σθθ + − , R3 R5 R33 [ 3 3z 1 P (1 − 2ν)z 1 (1 − 2ν)(z − 2h) σzz = − + − 3 5 8π(1 − ν) R3 R3 R3 ] 30hz 2 (z − h) 3z 3 + 12(2 − ν)hz 2 − 18h 2 z B + σzz + , − R5 R7 [ 2 3z 1 Pr (1 − 2ν) (1 − 2ν) + − σr z = σzr = − 5 8π(1 − ν) R3 R3 R33

226

8 Fundamental Solutions

] 3z 2 + 6(3 − 2ν)hz − 6h 2 30hz(z − h) + σrBz , − + R5 R7 σr θ = σθr = σθz = σzθ = 0, ]1 2 ]1 2 [ [ where z 1 = x3 − 2h, z 1 = x3 − 2h, R = r 2 + x32 / , R3 = r 2 + (x3 − 2h)2 / , r 2 = x12 + x22 , σ B is the stress component of the Boussinesq’s solution (8.10). As the next problem, let’s consider the Mindlin’s solution for a horizontal point load P acting at the point (0, 0, h) and directed in the positive direction of the x 1 axis. As with the previous problem, in this case, the solution can be represented as a superposition of two parts, after we use the Kelvin’s solution and Bussinesq’s solution. But we can use the Cerruti’s solution together with Bussinesq’s solution. As a result, the additional displacements and stresses are: P u x1 = 16 π μ (1 − ν) ) ( 2 −x12 + 2h(x3 − h) 6hx12 (x3 − h) x1 3 − 4ν 3 − 4ν + + − − R3 R R3 R5 R33 ( ) x1 x2 P x1 x2 6hx1 x2 (x3 − h) , u x2 = − − 16 π μ (1 − ν) R33 R3 R5 ) ( x1 x3 x1 x3 + 2hx1 (3 − 4ν) 6hx1 x3 (x3 − h) P ; − − u x3 = 16 π μ (1 − ν) R33 R3 R5 [ 2 3x1 P x1 (1 − 2ν) (1 − 2ν) σx1 x1 = − + − 8π(1 − ν) R35 R3 R33 ] 3x12 − 6(3 − 2ν)hx3 + 18h 2 30hx12 (x3 − h) , − − R5 R7 [ 2 3x2 P x1 (1 − 2ν) (1 − 2ν) − + σx2 x2 = − 5 8π(1 − ν) R3 R3 R33 ] 3x22 − 6(1 − 2ν)hx3 + 6h 2 30hx22 (x3 − h) , − − R5 R7 [ 2 3z 1 P x1 (1 − 2ν) (1 − 2ν) σx3 x3 = − + − 5 8π(1 − ν) R3 R3 R33 ] 3x32 + 6(1 − 2ν)hx3 + 6h 2 30hx32 (x3 − h) , − − R5 R7 [ 2 3x1 P x2 (1 − 2ν) (1 − 2ν) + − σx1 x2 = σx2 x1 = − 5 8π(1 − ν) R3 R3 R33

8.9 Some Generalizations

227

] 3x12 − 6h(x3 − h) 30hx12 (x3 − h) , − − R5 R7 ] [ 3z 1 3x3 + 6(1 − 2ν)h + 18h 2 P x1 x2 30hx3 (x3 − h) , σx2 x3 = σx3 x2 = − − − 8π(1 − ν) R35 R5 R7 [ 2 3x1 z 1 P (1 − 2ν)z 1 (1 − 2ν)(x3 − 2h) σx1 x3 = σx3 x1 = − + − 8π(1 − ν) R35 R3 R33 ] 2 2 2 3x1 x3 + 6(1 − 2ν)hx1 − 6hx3 (x3 − h) 30hx1 (x3 − h) . − − R5 R7 For the complete solution, we must add the Boussinesq’s solutions (8.6) and (8.10) to them.

8.9 Some Generalizations We will make some generalizations regarding the classical fundamental solutions of the theory of elasticity, which were discussed in this chapter. Classic fundamental solutions can be grouped into two classes. The first class of fundamental solutions is the problem for an infinite elastic body. That is, in this case, the area under consideration .∗ is an infinite elastic medium and its corresponding boundary . ∗ is an infinite boundary. In general, this class corresponds to the fundamental Kelvin’s solution for an infinite elastic medium. Kelvin’s solution for 3D-problems has the form: u i∗j (ξ, x) =

[ ] 1 (3 − 4ν)δi j + r,i r, j . 16π(1 − ν)μ r

(8.50)

Kelvin’s solution for 2D-problems has the form: u i∗j (ξ, x) =

[ ] −1 (3 − 4ν) ln(r )δi j − r,i r, j , 8π(1 − ν)μ

(8.51)

where u i∗j (ξ, x) are displacements occurring at the point x in the jth direction and corresponding to a point load acting in the ith direction and applied at the point ξ. Expressions for stresses are written as: pi∗j (ξ, x) =

⎧ ⎧ [ ( ]∂ r ) −1 − 2ν)δ −(1 − 2ν) r , + β r r n − r n (1 i j ,i , j ,i j , j i 4α π (1 − ν)r α ∂n

where α = 2, 1, β = 3, 2 respectively, for 3D-problems and plane-deformed state problems.

228

8 Fundamental Solutions

Function r = r (ξ, x) is the distance between the point ξ to which the load is applied and the point x. The derivatives of function r are taken by the coordinates of the point x, i.e., r=

/ / √ ri ri , ri = xi (x) − xi (ξ), r,i = ∂r ∂ xi (x) = ri r .

The expressions for deformations ε∗jk at an arbitrary point x due to the point load applied at the point ξ and directed along the ith axis have the form: { ( ) −1 (1 − 2ν) r,k δi j + r, j δik α 8α π (1 − ν)μr ( )} − r,i δ jk + β r,i r, j r,k .

ε∗jki (ξ, x) =

Formulas for stresses can be written in the following form: σ∗jki (ξ, x) =

{ ( ) } −1 (1 − 2ν) r,k δi j + r, j δik − r,i δ jk + β r,i r, j r,k . α 4α π (1 − ν)r (8.52)

The expressions for the plane-deformed state are also valid for the plane-stressed state, if the coefficient ν is replaced by ν = ν/(1 + ν). The second class of fundamental solutions relates to problems for half-space. In this case, a semi-infinite medium is considered with a plane section of the boundary— an infinite horizontal plane .. Typically, a lower half-space is considered. This class of solutions can be represented in the following form: ()∗ = () K + ()C , where () K and ()C are, respectively, Kelvin’s solutions (expressions (8.50−8.52) for three- and two-dimensional solutions) and an additional solution. For example, expressions of additional solutions ()C for displacements due to a point load applied within a half-space, have the form [Brebbia, C.A. Boundary element techniques/C.A. Brebbia, J.C.F. Telles, L.C. Wrobel. Springer-Verlag Berlin, Heidelberg, 1984. 524 p.]: / ) (( )/ ( )/ u c11 = K d 8(1 − ν)2 − (3 − 4ν) R + (3 − 4ν)R12 − 2cx R 3 + 6cx R12 R 5 , / ) ( / / u c12 = K d r2 (3 − 4ν)r1 R 3 − 4(1 − ν)(1 − 2ν) (R(R + R1 )) + 6cx R1 R 5 , ( / / ) / u c21 = K d r2 (3 − 4ν)r1 R 3 + 4(1 − ν)(1 − 2ν) (R(R + R1 )) − 6cx R1 R 5 , / ) ( / / / )( ( u c22 = K d 1 R + (3 − 4ν)r22 R 3 + 2cx R 3 1 − 3r22 R 2

8.9 Some Generalizations

229

/ ( / )( )) + 4(1 − ν)(1 − 2ν) (R + R1 ) 1 − r22 (R(R + R1 )) , /( / ) ( / ) u c23 = K d r2 r3 (3 − 4ν) R 3 − 4(1 − ν)(1 − 2ν) R(R + R1 )2 − 6cx R 5 , / )( ( ( / / / ) u c33 = K d 1 R + (3 − 4ν)r32 R 3 + 2cx R 3 1 − 3r32 R 2 ( / / )( )) + 4(1 − ν)(1 − 2ν) (R + R1 ) 1 − r32 (R(R + R1 ) , / u c13 = u c12 r3 r2 ;

/ u c31 = u c21 r3 r2 ;

u c32 = u c23 ,

where i = 1, 2, 3; R( =) R(Ri Ri )1/2 , ri = xi (x) − xi (ξ); K d = 1/(16π(1 − ν)μ), Ri = xi (x) − xi ξ' ; c = x1 (ξ) ≥ 0 , x = x1 (x) ≥ 0. The geometry of problem about the point load (|P1 | = |P2 | = |P3 | = 1 ) applied within the half-space is shown in Fig. 8.12. An additional part of the displacements for the plane-deformed state can be written in the following form [Brebbia, C.A. Boundary element techniques/C.A. Brebbia, J.C.F. Telles, L.C. Wrobel. Springer-Verlag Berlin, Heidelberg, 1984. 524 p.]: ) ( )/ { ( / } u c11 = K d − 8(1 − ν)2 − (3 − 4ν) lnR + (3 − 4)R12 − 2cx R 2 + 4cx R12 R 4 ,

с ‘

x2 c r2 = R2 R1

x3

x

P2 P3 P1

r3 = R3

R r1 r x

x1 Fig. 8.12 The geometry of problem about the point load (|P1 | = |P2 | = |P3 | = 1) applied within the half-space

230

8 Fundamental Solutions

Fig. 8.13 The geometry of problem about the point load (|P1 | = |P2 | = 1) applied within the half-plane

с

R1

c

R P2 P1

r1 x1

x2

x

r

x

r2 = R2

} { / / u c12 = K d (3 − 4ν)r1r2 R 2 + 4cx R1r2 R 4 − 4(1 − ν)(1 − 2ν)θ , / { / } u c21 = K d (3 − 4ν)r1r2 R 2 + 4cx R1r2 R 4 − 4(1 − ν)(1 − 2ν)θ , { ( / } ) ( )/ u c22 = K d − 8(1 − ν)2 − (3 − 4ν) lnR + (3 − 4ν)r22 + 2cx R 2 − 4cxr22 R 4 , / / θ = arctgR2 R1 , K d = 1 (8π(1 − ν)μ), other symbols are where i = 1, 2, shown in Fig. 8.13. Expressions for additional stresses due to the point loads applied inside the halfplane for the plane-deformed state are given. / For a plane-stressed state, the coefficient ν is replaced by ν = ν (1 + ν) in the formulas. It follows from expressions for additional terms that they do not contain singularities at x1 ≥ 0 and c > 0 (that is, when the point of loading application is inside the area .∗ ). When the point of application of the load lies on the surface .(c → 0), then additional expressions, together with the Kelvin’s solution, give a complete solution to the three-dimensional Bussinesque-Cherruti problem or two-dimensional Flaman’s problem. In the case of the Flaman’s problem, ( for )example, the fundamental displacements and stresses on the surface equal to ξ ∈ . : { } ∗ { } 2 , u 12 = −K d' (1 − 2ν)θ − r,1r,2 u ∗11 = K d' 2(1 − ν)lnr − r,1 { } { } 2 u ∗21 = −K d' −(1 − 2ν)θ − r,2 r,1 , u ∗22 = −K d' 2(1 − ν)lnr − r,2 , ( ) ∂r 2 r,i r, j (8.53) pi∗j = − πr ∂n / where K d' = 1 (2π μ).

8.9 Some Generalizations

231

Once again, some comments on such an important question as the transition from a three-dimensional problem to a plane-deformed state problem. It should be noted that in Sect. 8.4 this problem was already discussed regarding the Flaman’s solution. In three-dimensional displacement problems, the Kelvin’s solution u i∗j tends to zero at r → ∞. In two-dimensional problems, this is not the case, since u i∗j → −∞ at r → ∞ due to the presence of a logarithmic function in the expression for a fundamental solution with a flat-deformed state. This behavior in the two-dimensional case does not represent anything unexpected. Therefore, from a physical point of view, one can consider the case of a semi-infinite rod, the x coordinate of which varies from x(A) = 0 to x(B) → ∞. We assume that the end B is rigidly fixed, and at the point A we apply an axial load in the positive direction. As a result, constant deformations are obtained in the rod. Then, after integration, we get that u( A) → ∞. On the other hand, if the displacement at the point A is counted from a point C at a finite distance from the point A, then the displacement u(A) will be finite and u(C) = 0, u(B) → −∞. This simple example clearly shows the physical nature of the expression (8.51) and can be used to explain the transition from a three-dimensional to a two-dimensional problem by integration of the coordinate x3 (ξ). Algorithms of integration of fundamental solutions under action of loads on arbitrary areas of influence are described, for example, in [Davis, R.O. Elasticity and geomechanics / R.O. Davis, A.P.S. Selvadurai.—Cambridge University press. 1996. 201 p.]. Control Questions 1.

The fundamental solution of the Laplace equation for 3D-problem: 1 (1) K (x, y) = − 4πr ; 1 (2) K (x, y) = 2π ln r1 ; ( ) (3) K (x, y) = exp r1 .

2. 3. 4. 5. 6.

What is the mathematical statement of the Boussinesq’s problem? Where the Bussinesq’s solution has a singular feature? What is the Hertz’s task? What is the mathematical statement of the Flamant’s problem for a half-space? What are the difficulties and features of building up the Flamant’s solution for a half-plane? 7. What is the mathematical statement of the Kelvin’s problem? 8. What is the definition of the Green’s displacement tensor? 9. What is the mathematical statement of the Cerrutti’s problem? 10. What is the difference in the mathematical statement of the Mindlin’s problem and Kelvin’s problem?

Chapter 9

Dynamic Problems of Solid Mechanics

9.1 General Concepts and Definitions In previous chapters, we considered the solid mechanics problems in a static formula. However, there is a large class of problems where the external loads on deformable bodies are clearly dynamic. In this chapter, we consider problems whose model analogues related to wave problems of the mechanics of deformable solids. In short, we give some basic concepts related to the description of wave processes in solids. If oscillations of particles are induced in some places of the elastic solid medium, then, due to the interaction among the particles, these oscillations propagate from the particle to the particle at the velocity v. The process of propagating fluctuations in space is called a wave. It is important that such disturbances, which occur in some parts of the elastic solid medium, are propagating in it at a finite velocity. This means that a point at a distance L from the source of disturbances, for the time τ, a rest will be maintained. Thus, the particles of the medium in which the wave propagates are not involved by the wave in the translational movements, and they only oscillate near their equilibrium positions. The characteristic property of the wave is the transfer of the energy without the transfer of the matter. The concepts of “oscillations” and “waves” are interconnected, but also different from each other. The motion of the body G is called the oscillatory motion (or the simply oscillations) at the finite time interval [t1 , t2 ] (at the semi-infinite interval t ≥ t1 ), if at this interval all points M ∈ G oscillate. The oscillations of the point M in the interval [t1 , t2 ] is meant such a law of change in time of at least one defining parameter u i (M, t) for which there exists a function u i∗ (M, t) with a trajectory more than two (counting set) points of intersection, that is, the equation u i (M, t) = u i∗ (M, t) has more than two (counting set) roots (Fig. 9.1). Usually, u i∗ (M, t) corresponds to the position of equilibrium of the body. From the point of view of such definition of the oscillatory motion, if there are points in the body G that oscillate with different laws of motion, then such a movement is called a wave process (waves). © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Zhuravkov et al., Mechanics of Solid Deformable Body, https://doi.org/10.1007/978-981-19-8410-5_9

233

234

9 Dynamic Problems of Solid Mechanics

Fig. 9.1 Point oscillation (counting set of roots with u i∗ )

Once again, it should be emphasized that the fundamental difference between waves and oscillations is the energy transfer, since only “local” energy transformations occur in oscillations. Waves, as a rule, are able to move long distances from their places to the origin. Remark Most waves are by nature not new physical phenomena, but only a conditional name for a certain type of collective movement of medium particles. Most waves are oscillations of some media. Outside this medium, waves of this type do not exist (for example, sound in a vacuum). Wavelength is the distance over which a wave propagates in one period of oscillation. Depending on the direction of oscillation of the particles with respect to the direction in which the wave propagates, longitudinal and transverse waves are distinguished. Longitudinal waves are waves in which the oscillations of the medium are parallel (“along”) to the direction of the wave traveling and the displacement of the medium is in the same (or opposite) direction of the wave propagation. Mechanical longitudinal waves are also called compressional or compression waves, because they produce the compression and rarefaction when traveling through a medium, and pressure waves, because they produce increases and decreases in pressure. The other main type of wave is the transverse wave, in which the displacements of the medium are at right angles to the direction of propagation. Transverse waves, for instance, describe some bulk sound waves in solid materials (but not in fluids); these are also called “shear waves” to differentiate them from the pressure (longitudinal) waves that these materials also support.

9.1 General Concepts and Definitions

235

If the relationship between the particles of the medium is made by the elastic forces, resulting from the deformation of the medium during the transfer of oscillations from one particle to another, then the waves are called elastic waves. Elastic transverse waves can only occur in a medium having shear resistance. Therefore, only longitudinal waves are possible in liquid and gaseous media. Both longitudinal and transverse waves can occur in the solid medium. In fact, not only particles along the x axis oscillate in solids, but also a collection of particles enclosed in a certain volume. Propagating from the source of oscillations, the wave process covers more and more parts of solid. The geometric location of the points to which the oscillations reach the time t is called the wave front. The wave front is the surface that separates the part of space already involved in the wave process from the area in which the oscillations have not yet occurred. The geometric location of points oscillating in the same phase is called the wave surface. The wave surface can be passed through any point of the space covered by the wave process. Therefore, there is an infinite set of wave surfaces, while the wave front at each point in time is only one. Wave surfaces remain stationary. Wave front displaces all the time. Wave surfaces can be any shape. If the wave surfaces are planes, then the waves are called plane waves. Spherical waves are waves whose wave surfaces are spheres. Waves that have breaks derived from displacements (i.e., velocity and stress breaks) are called strong breaking waves or shock waves. Shock waves occur during explosions, detonation, supersonic movements etc. Two and only two types of elastic waves can propagate in an unlimited isotropic solid: longitudinal and transverse waves. In addition to the above types of waves that can propagate within the limitless solids, elastic waves can propagate along the surface as well. At the same time, the displacements associated with these waves decrease with the depth according to the exponential law. These waves are called Rayleigh surface waves and play an important role, for example, in the study of seismic phenomena. Since these waves diverge in only two dimensions, they attenuate with a distance slower than other types of waves. The waves on the surface of the body are neither longitudinal nor transverse. Surface waves in solids are of two classes: with vertical polarization, in which the vector of oscillatory displacement of medium particles is located in a plane perpendicular to the boundary surface and with horizontal polarization, in which the vector of displacement of medium particles is parallel to the boundary surface. The most common particular cases of surface waves include the following. 1. Rayleigh waves propagating along the boundary of a solid with a vacuum or a sufficiently rarefied gas medium. The energy of these waves is localized in the surface layer with a thickness of l to 2l, where l is the wavelength. 2. Decaying waves of the Rayleigh type at the boundary of a solid with a liquid, provided that the phase velocity in the liquid is less than in the solid (which is true for almost all real media). This wave continuously emits energy into the liquid, forming in it an inhomogeneous wave extending from the boundary. The

236

9 Dynamic Problems of Solid Mechanics

depth distribution of displacements and stresses is the same as in the Rayleigh wave. 3. Stoney wave1 propagating along the flat boundary of two solid media, in the case where the modulus of elasticity and density of which do not differ much. Such a wave can be represented as the sum of two Rayleigh waves—one in each media. 4. Love waves are surface waves with horizontal polarization that propagate at the boundary of the solid half-space with the solid layer. These waves are purely transverse: they have only one displacement component, and elastic deformation in the Love wave is a pure shift. An important property of waves is the superposition property, according to which in any region of the medium any number of waves can propagate independently from each other. In the theory of elasticity, the body is considered to be in equilibrium under the influence of applied forces. It is assumed that elastic deformations have already taken their static values. Such an interpretation is exact enough for problems in which the period between the moment of loading application and the establishment of actual equilibrium is small compared to the time periods during which observations are made. However, when the effect of forces applied only for a short period of time or rapidly changing is examined, this phenomenon must be considered from the point of view of the propagation of stress waves. Real bodies are never completely elastic. Therefore, when disturbances propagate in solids, part of the mechanical energy turns into heat (several different mechanisms of these transformations are combined by the common name, namely, internal friction). The velocity gradients produced by the stress wave cause a second type of loss due to the viscosity of the material. The nature of the attenuation for these two types of internal friction is different. Experimental data show that both types occur in solids.

9.2 Wave Equation The one-dimensional wave equation can be represented in the following form: • In Cartesian coordinates: ∂ 2ϕ ∂ 2ϕ = a2 2 ; 2 ∂x ∂t

(9.1)

• In spherical coordinates (r is radial coordinate): ∂ 2ψ ∂ 2 ψ 2 ∂ψ = a2 2 . + 2 ∂r r ∂r ∂t 1

George Johnstone Stoney (1826–1911), British, Irish physicist and mathematician.

(9.2)

9.3 Model Problems of Dynamics of Elastic Body

237

Remark Equations of this type are considered in the course of the equations of mathematical physics. The main feature of the solutions of the wave equation is the description of a perturbation propagating with a finite velocity, i.e., a wave. Using the Dalamber method, one can consider the process of propagation of waves (perturbations) of the type: ϕ(x, t) = Φ 1 (x − t/a) + Φ 2 (x + t/a),

(9.3)

where Φ 1 , Φ 2 are twice continuously differentiable functions of their arguments. The expression (9.3) represents a general solution to Eq. (9.1). It can be shown that Eq. (9.1) does not have solutions other than those that can be represented by functions of the form (9.3) or by a superposition of such functions. If ax − t = const, , then there is a wave (disturbance) propagating towards the increase of the ordinate x (when increasing t). If ax + t = const, then there is a wave, spreading in the opposite direction. Using (9.3), the properties of the spherical symmetric wave described using the Eq. (9.2) can be investigated: r ψ(r, t) = Ψ1 (r − t/a) + Ψ2 (r + t/a).

(9.4)

In (9.4), the first term Ψ1 (r − t/a) describes a diverging wave, and the second term Ψ2 (r + t/a) describes a converging wave. Therefore, if a certain value of the function ϕ(x, 0) is specified at the beginning of time (t = 0), then it causes perturbations that can be detected when t > 0 in the vicinity of points x1 and x2 . That is, the feature of Eq. (9.1) is that it is possible to describe perturbations (waves) propagating at a finite velocity by this equation.

9.3 Model Problems of Dynamics of Elastic Body 9.3.1 General Comments Consider various representations of the equations of dynamics of homogeneous isotropic elastic body. The equations of motion of an elastic body can be obtained from equilibrium equations by adding inertia forces to the acting mass forces, according to the D’Alembert principle ρF i , i.e., ρFi → ρ(Fi − u¨ i ). Thus, the equations of motion can be represented in the following form: σi j, j + ρ(Fi − u¨ i ) = 0, (i, j = 1, 2, 3). In the expanded form, the Eq. (9.5) is written as:

(9.5)

238

9 Dynamic Problems of Solid Mechanics

∂τx y ∂τx z ∂ 2u ∂σx + + +X =ρ 2, ∂x ∂y ∂z ∂t ∂σ y ∂τ yz ∂τx y ∂ 2v + + +Y =ρ 2, ∂x ∂y ∂z ∂t ∂τ yz ∂σz ∂ 2w ∂τx z + + +Z =ρ 2 . ∂x ∂y ∂z ∂t

(9.6)

where X, Y, Z are components of mass forces in the Cartesian coordinate system x, y, z, and t is time. The remaining equations of the complete resolution system of equations (equation of the compatibility of deformations, Cauchy relations, boundary conditions) are preserved in the same form. In addition, initial conditions must be added to these equations: u i = u 0i , u˙ i = u˙ 0i , t = 0.

(9.7)

Remark We will not consider the question of proofing the existence and uniqueness of solving dynamic problems of the theory of elasticity. This question is considered in detail in the literature on the general course of the dynamic theory of elasticity. The relationship of stresses with strains when considering an elastic isotropic body for dynamic problems of the theory of elasticity is determined by the Hooke’s law: ∂u ∂v ∂w , σx = λε + 2μ , σ y = λε + 2μ , σz = λε + 2μ ∂x ∂y ∂z ) ) ) ( ( ( ∂v ∂w ∂u ∂w ∂v ∂u τx y = μ + , τx z = μ + , τx y = μ + , ∂y ∂x ∂x ∂z ∂y ∂z ∂v ∂w ∂u + + , (9.8) ε= ∂x ∂y ∂z where λ, μ are elastic Lamé constants. As in the construction of the Lamé equations for static problems of the theory of elasticity, from the system (9.6), we exclude the components σi j of the stress tensor using the relations (9.8). As a result, we obtain a system of three second-order equations from displacements u, v, w by coordinates, x, y, z and time t: ∂ 2u ∂ε + μΔu + X − ρ 2 = 0, ∂x ∂t ∂ 2v ∂ε + μΔv + Y − ρ 2 = 0, (λ + μ) ∂y ∂t ∂ 2w ∂ε (λ + μ) + μΔw + Z − ρ 2 = 0. ∂z ∂t

(λ + μ)

(9.9)

9.3 Model Problems of Dynamics of Elastic Body

239

Equation (9.9) can be written in compact form as follows: ρu¨ i = (λ + μ)θ,i + μΔu i + ρFi .

(9.9*)

The system of Eq. (9.9) is called the Lamé system of dynamic problems of the theory of elasticity. In most cases, when considering the dynamic problems of the theory of elasticity, the Eq. (9.9) are used as the basic ones. Therefore, approaches and methods for constructing general solutions for dynamic problems of elasticity theory are mostly developed for a system of resolving equations with basic Eq. (9.9). It should be noted that in addition to this statement, depending on the boundary and initial conditions, equations of motion (9.6) relative to stresses are often adopted as the basic equations of the resolving system of equations. In this case, the system (9.6) is supplemented in the general case by six Beltrami-Michell equations (compatibility conditions for stresses). The Lamé system (9.9) can be presented in several different forms. When constructing a solution to a system (9.9), it is often convenient, on the basis of the known theorem of vector analysis (Helmholtz theorem), to represent a vector of displacements u→ = u = (u, v, w) in the form of potential and solenoidal parts: → or u = gradϕ + rotψ, u→ = gradϕ + rotψ

(9.10)

} { → ψx , ψ y , ψz is the vector potential of where ϕ(x, y, z, t) is the scalar potential, ψ the displacement field. The “grad” and “rot” operators in a rectangular Cartesian coordinate system have the form: | | |e e e | 1 2 3 | | ∂ϕ | | gradϕ = ei , rotψ = | ∂∂x1 ∂∂x2 ∂∂x3 |. | | ∂ xi | ψ1 ψ2 ψ3 | Note that for uniqueness of motion potentials, an additional condition is necessary, which, as a rule, is accepted in the following form: divψ = ψi,i = 0.

(9.11)

The mass force field is also represented as a potential and solenoid component: F = gradΦ + rotΨ. . Note that when using the Eq. (9.10), four functions ϕ, ψx , ψ y , ψz are introduced instead of the three functions u, v, w. It follows, in particular, that there is a “arbitrariness” in choosing such representation of displacements through potentials.

240

9 Dynamic Problems of Solid Mechanics

Using the representation (9.10), the Lamé equations of motion (9.9) are satisfied if the displacements potentials are solutions to the following equations (note, that div grad = δ, div rot = 0): ¨ = c22 δψ + ψ, δψ = δψi ei . ϕ¨ = c12 δϕ + Φ , ψ

(9.12)

In (9.12), values c1 =

∫ ∫ (λ + 2μ)/ρ, c2 = μ/ρ, c1 > c2

have a velocity dimension and are accordingly called the velocity of longitudinal waves (tension–compression waves) and transverse waves (forming waves). Thus, for a homogeneous elastic isotropic medium, a closed system of motion equations in displacement potentials consist of Eqs. (9.12) and (9.11). In this case, the initial conditions (9.7) for it must be recorded in potentials: | ˙| = ψ ˙ 0, ϕ|t=0 = ϕ0 , ϕ| ˙ t=0 = ϕ˙ 0 , ψ|t=0 = ψ0 , ψ t=0 ˙ 0 , divψ0 = 0, divψ ˙ 0 = 0. where u 0i ei = gradϕ0 + rotψ0 , u˙ 0i ei = gradϕ˙ 0 + rotψ Note that, despite the independence of Eq. (9.12), as a rule, in initial-boundary problems they are connected by boundary conditions. In the absence of mass forces from (9.9), it follows four wave equations for ϕ → which can be written as: and ψ, Δϕ = a 2

2→ ∂ 2ϕ → = b2 ∂ ψ , , Δψ 2 ∂t ∂t 2

(9.13)

where Δ = ∂∂x 2 + ∂∂y 2 + ∂∂z 2 is the Laplacian, a and b represent, respectively, inverse √ values √of velocities of longitudinal and transverse waves: a = ρ/(λ + 2μ) = 1 c1 , b = ρ/μ = 1/c2 . A similar view is true in the presence of mass forces. In this case, it is necessary to also represent the mass force vector as a sum of scalar and vector potentials. 2

2

2

9.3.2 Cauchy Problem and Boundary Problems The fifth chapter has considered the boundary problems of the theory of elasticity and defined the main types of boundary conditions for the boundary value problems of the static and dynamic theory of elasticity. In the case of dynamic processes, boundary conditions must be supplemented by initial conditions that can be formulated for the elastic displacement vector:

9.3 Model Problems of Dynamics of Elastic Body

u→|t=t0 = u→0 (x, y, z),

| ∂u || = u→1 (x, y, z). ∂t |t=t0

241

(9.14)

Consider the applied and scientifically important Cauchy problem for a system of Eq. (9.9). The Cauchy problem for the system (9.9) is a problem with initial data for a limitless elastic medium. Initial conditions (9.14) are set in elastic medium or in some parts of it. It is necessary to build and investigate the solution of system (9.9) at t > t0 . From a physical point of view, this is a problem to spread perturbation in an elastic medium with velocities c1 and c2 . These perturbations are “induced” by the part of the elastic medium in which at t = t0 functions u→0 (x, y, z), u→1 (x, y, z) were determined. In the other parts of the medium these functions are zeros. Remark We will return to the solution of the formulated Cauchy problem further. Now we turn to the wording of the main boundary value problem in terms of elastic → displacements and in terms of scalar ϕ(x, y, z, t) and vector ψ(x, y, z, t) potentials. The boundary value problem in the case of boundary conditions in stresses is formulated as follows (first boundary value problem). Let the elastic medium occupy some parts of the space D which bounded by the smooth surface S on which the stresses are set. Mass forces X, Y, Z are defined in the region D. It is necessary to find a solution to the system of Eq. (9.9), satisfying the initial conditions (9.14) and boundary conditions on the surface S. The solution must have a specified smoothness (which is determined by the required smoothness of the boundary and initial conditions). The components of the stress vector on the surface, which are included in the formula of boundary conditions, are expressed by Cauchy formulas through the components of the stress tensor and the direction cosines of the outer normal line n→(x, y, z) to the surface S: Tnx = σx cos(n, x) + τx y cos(n, y) + τx z cos(n, z), Tny = τx y cos(n, x) + σ y cos(n, y) + τ yz cos(n, z), Tnz = τx z cos(n, x) + τ yz cos(n, z) + σz cos(n, z).

(9.15)

Note that the normal to the surface S is known because the surface S is set and boundary conditions at moments t ≥ t0 relate to the position of the surface S before deformations. Therefore, for example, in the case of an elastic half-space z ≥ 0 we have: cos(n, x) = cos(n, y) = 0, cos(n, z) = −1. Then from the conditions (9.15), it is as follows: Tnx = −τx z , Tny = −τ yz , Tnz = −σz .

(9.16)

242

9 Dynamic Problems of Solid Mechanics

That is, on the surface of the half-space in this case, a normal force Tnz = Tzz and two conditions with respect to tangent forces Tnx , Tny must be set. The construction of a complete system of resolving equations of the boundary value problem of the dynamics of an elastic body in the case of boundary conditions in stresses can be carried out in two ways. The first approach: in boundary conditions (9.15) we turn to displacements, replacing stresses with their expressions (9.8) in accordance with the Hooke’s law. In this case, the boundary conditions of type (9.16) for the half-space, in accordance with this solution construction, are written as follows: ) ) ( ( ∂w ∂w ∂v ∂u + , −Tny = μ + , −Tnx = μ ∂z ∂x ∂z ∂y ) ( ∂v ∂w ∂w ∂u + + + 2μ . (9.17) −Tzz = λ ∂x ∂y ∂z ∂z As a result, the boundary value problem is formulated entirely in terms of displacements: find a solution to the system (9.9) under initial conditions (9.14) and boundary conditions (9.17), assuming some smoothness of functions u→0 (x, y, z), u→1 (x, y, z), Tzx (x, y, 0, t),Tzy (x, y, 0, t) and Tzz (x, y, 0, t). The second approach is to supplement the dynamics Eq. (9.6) with BeltramiMichell ratios and formulate the initial conditions (9.14) in terms of stresses. Thus, in the second case, the problem is fully formulated in terms of stresses. Finally, we note another possible formula of boundary problems related to the transition from displacements (or stresses) to scalar and vector potentials, that is, to the transition from the system (9.9) to the system (9.13). Obviously, in order to construct a problem, as a basic system (9.13), it is necessary to write both boundary and initial conditions in terms of potentials. Initial conditions (9.14) in terms of potentials take the form: | → || → 1 (x, y, z), ϕ|t=t0 = ϕ0 (x, y, z), ψ =ψ t=t0 | | → || ∂ψ ∂ϕ || → 1 (x, y, z), = ϕ (x, y, z), =ψ | 1 ∂t |t=t0 ∂t |

(9.18)

t=t0

→0 , ψ → 1 are associated with u→0 and u→1 according to ratios (9.10), where ϕ0 , ϕ1 and ψ which are right with all t ≥ t0 . When formulating the boundary conditions of the first boundary value problem in terms of potentials, it is necessary in (9.15) to replace stresses through strains according to (9.8), and then switch from displacements to derivatives from potentials according to (9.10). In expanded form, formulas (9.10) have the form: u=

∂ψ y ∂ϕ ∂ψx ∂ψz ∂ϕ ∂ψ y ∂ψx ∂ϕ ∂ψz + − ,v = + − ,w = + − , ∂x ∂y ∂z ∂y ∂z ∂x ∂z ∂x ∂y

9.4 Stokes Problem

243

where ψx , ψ y , ψz are components of vector potentials. In accordance with the remark to formulas (9.10), in the general case, one of the components of the vector potential or some combinations of functions ϕ, ψx , ψ y , ψz can be set to zero. Usually, this fact is used in solving specific problems in which this arbitrariness is used for the wording (for example, symmetry considerations) and reduction of calculations. As an example, the boundary conditions of the first boundary value problem for the elastic half-space (9.17) are taken (bearing in mind this arbitrariness)ψz = 0: ∂ 2ψy ∂ 2ψy 1 ∂ 2ϕ ∂ 2 ψx − , − Tzx = 2 + − μ ∂ x∂z ∂z 2 ∂x2 ∂ y∂z ∂ 2ψy 1 ∂ 2 ψx ∂ 2ϕ ∂ 2 ψx − Tzy = 2 + , − + μ ∂ y∂ z ∂z 2 ∂ y2 ∂ x∂ y [ 2 ] 1 ∂ 2 ψx 1 − 2v ∂ ϕ ∂ 2 ψ y − Tzz = − + + μ 2(1 − v) ∂z 2 ∂ x∂z ∂ y∂z [ 2 ] ∂ 2ψy 2v ∂ ϕ ∂ 2 ϕ ∂ 2 ψx − . + + 1 − 2v ∂ x 2 ∂ y2 ∂ y∂z ∂z∂ x

(9.19)

Thus, in terms of potentials, the first boundary value problem is formulated as follows: it is necessary to construct a solution of the system of Eq. (9.13) under initial conditions (9.18) and boundary conditions (9.19). Remark The boundary conditions (9.19) are complex. They are combinations of second derivatives from potentials through three functions Tzx , Tzy and Tzz known on the surface of the elastic medium. The formulas of boundary value problems in the case of displacements set at the body boundary (the second boundary value problem) and the mixed problem are performed by analogy with the first boundary value problem.

9.4 Stokes Problem Dynamic Problems on the Action of Point Forces in Limitless Elastic Medium The problem of the action of a point force in a limitless elastic medium for problems of the static theory of elasticity was considered in the corresponding section on fundamental solutions of the theory of elasticity. Consider a similar problem, but in the case of the action of a point force, the modulus of which is a function of time. As with a static problem, the solution for a dynamic problem has a wide range of applications. Therefore, let’s at some points in the elastic medium, taken as the origin of coordinates, act a point force P(t), varying in time. For an infinite elastic medium, boundary conditions at infinity are conditions of no stresses and displacements. Therefore, the

244

9 Dynamic Problems of Solid Mechanics

solution must satisfy the conditions at infinity and have a feature at the origin of coordinates that leads to a finite point force in magnitude. As in the case of a static problem, if we cut a small neighborhood in the form of a ball around the point of application of forces, then stresses are equivalent to the applied force in magnitude and direction should be distributed over the surface of this ball. At the same time, it is easy to find (as in the static problem) that at the origin of the coordinates the stresses should turn to infinity of order r −2 . Then displacements (since they are expressed through derivatives of stresses and deformations) must have a peculiarity of order r −1 . In order to find a class of solutions with a feature of this type, consider the solutions of the wave equation: Δf =

1 ∂2 f . c2 ∂t 2

(9.20)

We investigate the solution (9.20) in the case of central symmetry, i.e., f = f (r, t), r is the radial coordinate, t is time. In this case, the Dalamber operator is simplified and Eq. (9.20) takes the form: (

∂2 2 ∂ + 2 ∂r r ∂r

) f =

1 ∂2 f 2 ∂2 1 ∂2 f or . f = (r ) c2 ∂t 2 r ∂r 2 c2 ∂t 2

Remark The Dalamber operator ◻ a is defined as follows: ( ◻ a

∂ϕ ∂x

)

(

∂ϕ =δ ∂x

)

( ) 1 ∂ 2 ∂ϕ − 2 2 . a ∂t ∂ x

From this equation it follows (using the Dalamber integral) the representation of the solution in the spherical symmetric case: f =

1 1 f 1 (t − r/c) + f 2 (t + r/c), r r

where f 1 and f 2 are arbitrary functions of specified arguments. The terms f 1 and f 2 describe divergent and convergent spherical waves. If the action of the force P(t) began at a point in time t = 0, then at t < 0, all the elastic media were at rest, so that there are no disturbances in the form of converging waves at t ≥ 0. Therefore, only divergent waves can be considered as the required solution with the desired feature: f =

1 f 1 (t − r/c). r

(9.21)

Indeed, the function f has the feature at the origin of coordinates exactly those displacements should have in the point force problem. Note that the derivatives of function f by coordinates are solutions to the wave equation, but already have terms

9.4 Stokes Problem

245

with a higher feature at the origin of coordinates. Therefore, the first derivatives have a feature of order r −2 . For example: ) ( ∂ (1/r ) = − 1/r 2 (x/r ), ∂x where x/r = cos(r, x). The solution to the Stokes problem has the form [Shemyakin, E.I. Dynamic problems of the theory of elasticity and plasticity / E.I. Shemyakin. – Moscow. Pub. by “NNCGP-IGD A.A. Skochinskiy”, 2007. 207 p. (in Russian)]: ( ) ∫τ/ b ⎧ ⎧ 1 ( ∂2 1 x2 1 ( r) r) − 2P t− , 4πρu = 2 τP(t − τ)dτ + 3 2 P t − ∂x r r a a b b τ/ a

⎧ ( ) ∫τ/ b ⎧ 1 1 ( ∂2 xy 1 ( r) r) − 4πρv = τP(t − τ)dτ + 3 P t − P t − , ∂ x∂ y r r a2 a b2 b τ/ a

⎧ ( ) ∫τ/ b ⎧ 1 1 ( xz 1 ( r) r) ∂2 − 2P t− τP(t − τ)dτ + 3 P t− 4πρw = ∂ x∂z r r a2 a b b τ/ a

where (u, v, w) are components of the displacement vector. Force P(t) acts in the axis x direction. Constants “a” and “b” are wave propagation velocities. Using Stokes formulas, you can reduce the general case of a problem with mass forces to the case when they are absent. To do this, we use the linearity of the equation and the boundary conditions of the problem and sum up two solutions: the general solution of homogeneous dynamic equations and the partial solution of a heterogeneous system of equations, constructed using Stokes formulas. Let mass forces (X, Y, Z ) depending on coordinates and time t act on the investigated elastic body in a volume V at points (ξ, η, ζ). Denote displacements caused in a limitless elastic medium at a point (x, y, z) by a force X (ξ, η, ζ, t) acting at a point (ξ, η, ζ) (the center of gravity of a small volume dω) in the x axis direction as: U (x) (ξ, η, ζ, x, y, z), V (x) , W (x) . Then the force X (ξ, η, ζ, t) causes a displacement equal to: ) ( (x) U dω,V (x) dω, W (x) dω . And finally summing up the action of all mass forces, we find:

246

9 Dynamic Problems of Solid Mechanics

U∗ =

˚

{ (x) } U + U (y) + U (z) dω, V ∗ = · · · , W ∗ = · · · .

V

These integrals represent elastic displacements in a limitless body due to the action of mass forces. If the body is bounded, then you need to construct total displacements in the following form: u = U ∗ + u ' , v = V ∗ + v ' , w = W ∗ + w' where the components of displacements u ' , v ' , w ' satisfy homogeneous equations and so that the boundary conditions on the surface and the initial conditions for u, v, w are satisfied accurately. Therefore, with given mass forces (and therefore with known U ∗ , V ∗ , W ∗ ), the problem of constructing auxiliary functions u ' , v ' , w ' is a well-defined boundary value problem of the dynamics of an elastic body. Thus, one of the important applications of the Stokes solution is the construction of general solutions to the dynamic equations of the theory of elasticity.

9.5 Natural and Forced Harmonic Oscillations Consider problems about free (natural) and forced harmonic oscillations, as well as problems about the propagation of progressive waves in an elastic body. Natural (free) oscillations. The free oscillation problem refers to the eigenvalue problem (natural frequency) for homogeneous equations of elasticity theory under homogeneous boundary conditions. In this case, the elastic body is free from the action of external forces, that is: Fi = 0, σi j l j = 0, on Sσ . Part of the surface S u can be fixed, that is, on it ui = 0. Initial conditions are given (9.7), as a result of which the body comes into motion. Forced harmonic oscillations. In this case, the body forces F i , the surface forces Ri , and the predetermined surface point displacements u0i are periodic time functions such that: Fi = Fi0 ϕ(t),

Ri = Ri0 ϕ(t), u 0i = u i0 ϕ(t).

Multipliers with the upper index “0” are time-independent, so as a typical representative of the function ϕ(t), you can take: ϕ(t) = exp(i pt) = cos pt + i sin pt.

9.5 Natural and Forced Harmonic Oscillations

247

Such representation is true, since any periodic function is represented by a Fourier series. By building a solution for one member of this series, you can use the superposition principle to build a complete solution.

9.5.1 Natural Oscillations of Elastic Bodies Consider the problem of natural oscillations of elastic bodies. We build the solution of the equations of motion in the form: u k = Uk exp(i ωt),

(9.22)

where U k is the function of coordinate only, not time. In the same way, all components of the strain tensor εij and stress tensor σij are presented. In the future, for convenience, through ui , σij , we will denote the amplitudes of displacements and stresses. That is, instead of U i , we will further write ui . Then, substituting the required solution (9.22) into the equations of motion (9.6), we obtain such a system of equations for amplitudes: σi j , j −ρω2 u i = 0.

(9.23)

The external forces and boundary conditions in the case of free oscillations must be zero, while the factor exp(iωt) is reduced. Remark We will consider the question of initial conditions later. Equations that establish a relationship between the amplitudes of stresses and strains retain the form of ordinary Hooke’s law equation: σi j = E i jkl u k ,l . The system of Eq. (9.23) under uniform boundary conditions can have an obvious trivial solution: u i ≡ 0, σi j ≡ 0. However, with some values of the parameter ω = ωk , a non-zero solution is also maybe: u i ≡ u ik , σi j ≡ σikj . The corresponding values of the parameter ωk are called the fundamental frequencies of the elastic body and the functions u ik determine the fundamental forms of oscillation. Note that (9.23) includes squares of fundamental frequency, so the root ωk will always correspond to the second, equal in magnitude and opposite in sign, root (−ωk ). We will not introduce special numbering for these negative roots, but it should be remembered that in addition to the solution u i exp(i ωt) there is always a second solution (−u i ) exp(−i ωt). This makes it possible to form real combinations from these roots, which only have a mechanical meaning. The equation linking the quantities u ik and σikj follows the from (9.23): σikj , j − ρω2k u ik = 0.

(9.24)

248

9 Dynamic Problems of Solid Mechanics

Obviously, due to the homogeneity of the system of equations and boundary conditions, the required functions included in (9.24) are determined with accuracy to an arbitrary factor. Note that Eq. (10.6) can be considered as equations of the static problem of the theory of elasticity with mass forces −ρω2 u i . Let ω = ωk be any fundamental frequency. Then u i = u ik is a displacement caused by the action of distributed forces Pik = −ρω2k u ik . Similarly, forces Pis = −ρω2s u is at frequency ωs cause displacements u is . By Betty’s theorem, we have: ∫

Pis u ik d V = V

Pik u is d V or ω2s V

∫ ρu is u ik d V = ω2k

V

ρu ik u is d V . V

Since ωs /= ωk , then the last equality is possible if the integral is zero. Therefore, ∫ ρu ik u is d V = 0 (k /= s). V

This equality expresses the orthogonality property of the fundamental forms of oscillation. From the condition of orthogonality, it follows, in particular, that frequencies ωk are always real number. Since u ik are defined only by the accuracy of an arbitrary constant factor, they can be normalized arbitrarily. Usually it is accepted: ∫ ρu ik u is d V = δks ,

(9.25)

V

where δks are Kronecker deltas. Relations (9.25) express normalization conditions and simultaneously repeat conditions of orthogonality of fundamental forms of oscillation. Let’s do some studies of natural oscillation now. We use the superposition principle (Fourier method). Then we can represent the general expression for displacements under natural oscillations of the elastic body as follows: ui =

∞ Σ

(Ak sin ωk t + Bk cos ωk t) u ik (xs ),

k=1

where Ak , Bk are undefined constants. Differentiating (9.26) by time and the following is obtained: u˙ i =

∞ Σ k=1

(Ak ωk cos ωk t − Bk ωk sin ωk t) u ik (xs ).

(9.26)

9.5 Natural and Forced Harmonic Oscillations

249

Equating at t = 0, the values of displacements and velocities are their initial values u 0i , v0i and we obtain: ∞ Σ

Bk u ik (xs ) = u 0i (xs ) ,

k=1

∞ Σ

Ak ωk u ik (xs ) = v0i (xs ) .

k=1

We multiply each of these equations by ρu ik and integrate by volume. Due to (9.25), only one member remains on the left side of each series: ∫ ρu 0i u ik d V ,

Bk = V

1 Ak = ωk

∫ ρv0i u ik d V .

(9.27)

V

It should be noted that the ratios (9.27) do not imply the possibility of decomposition of functions u 0i and v0i into series according to their fundamental forms of oscillations or fundamental functions u ik . The initial velocity distribution may not even be continuous at all. Therefore, if we talk about convergence, then we can only talk about convergence on average.

9.5.2 Forced Oscillations of Elastic Bodies Let periodic forces act on the body with a circular frequency p. For simplicity, we will assume that there are no mass forces (F i = 0), and on the body surface, the external forces are represented in the form: Ri = Ri0 exp(i pt). Remark Consideration of a more general case of additional difficulties does not occur. Let’s assume that the displacements and stresses are also proportional to exp(ipt).We will also use the designations ui and σij for the amplitudes of displacements and stresses. From Eq. (9.6), then we obtain the following equation: σi j , j − ρ p 2 u i = 0.

(9.28)

The following boundary conditions shall be performed on the surface S σ : σi j l j = Ri0 .

(9.29)

We present the solution in the form: /

/

u i = u i + u i0 , σi j = σi j + σi0j ,

(9.30)

250

9 Dynamic Problems of Solid Mechanics

where u i0 , σi0j are solutions to the static problem of the theory of elasticity. This solution satisfies the equilibrium equation σi0j , j = 0, Cauchy relations, and boundary conditions (9.29). We substitute (9.30) into (9.28). As a result, the first part of the solution in (9.30) satisfies the following equation of motion: σi'j , j −ρ p 2 u i' = ρ p 2 u i0

(9.31)

under uniform boundary conditions. / / Present u i , σi j as series: /

ui =

∞ Σ

/

ak u ik , σi j =

k=1

∞ Σ

ak σikj .

k=1

Now we substitute these expressions into (9.31) and exclude the stress amplitude derivatives by using (9.24). As a result, we get: ∞ Σ

ak ρ(ω2k − p 2 )u ik = ρ p 2 u i0 .

k=1

Multiply the obtained equality by u im and integrate by volume. Then we use the orthonormality condition of eigenfunction (9.25). As a result, we obtain: ∫ am (ω2m

−p )= p 2

ρu i0 u im d V .

2 V

From which it follows: p2 am = 2 ωm − p 2

∫ ρu i0 u im d V . V

If p = ωm , then the frequency of the perturbing force coincides with one of the fundamental frequencies of the elastic body, and the corresponding coefficient turns to infinity over time. This phenomenon is called resonance. In the case of non-periodic external forces to describe forced oscillations of the elastic body, the surface load and the solution are presented as decompositions in a series according to the system of eigenvalue functions. Substituting these series into equations of motion allows us to obtain equations for determining unknown time functions.

9.6 Propagation of Shock Waves in Unlimited Elastic Bodies

251

9.6 Propagation of Shock Waves in Unlimited Elastic Bodies Waves that have breaks of derivatives from displacements (that is, velocities and stresses) are called strong breaking waves or shock waves. The velocity of propagation of the longitudinal wave in the rod is: c0 =

/ / E ρ.

(9.32)

Remark When proving this statement, it is immediately assumed that at the front of the wave, the velocity and deformation and therefore the stress, change by a jump. According to the definition, the field of displacements corresponding to the plane longitudinal wave propagating in the isotropic medium has the form: u i = Ui (ct − n s xs ).

(9.33)

If you substitute (9.33) into the equations of motion (9.9) and solve it, then it can be established that in an isotropic body, there are two velocities of plane wave √ and c , having the “classical” expressions c = (λ + 2μ)/ρ, c2 = propagation c 1 2 1 / / μ ρ. As can be seen, the velocity of propagation of the longitudinal wave in the rod (9.32) is different from the velocities c1 and c2 . The possibility of propagating shock waves in an unlimited elastic medium with velocities c1 and c2 requires justification. Let’s show that the longitudinal waves of strong rupture in an unlimited elastic medium can propagate. Consider the longitudinal waves propagating along the x1 axis (Fig. 9.2). the section of medium of the length ct along x1 axis after time interval t will be uniformly deformed and the rest of medium remains not stressed, i.e., the section (mn), which is the boundary between the stressed and non-stressed parts of the medium, will move at a velocity c. We fix a section (pq) with a coordinate x1 . The distance from the section (pq) to the boundary (mn) is (ct − x1 ). Section with the length (ct − x1 ) uniformly compressed, its deformation is e11 . Therefore, the displacement of the cross section (pq) from the initial position is u 1 = e11 (ct − x1 ). Let the wave pass through section 1–1 at time t and through section 2-2 at time t + dt. The distance among these sections is d x1 = cdt. We differentiate by time and obtain the velocity of the section movement: / v = du 1 dt = e11 c. For a period of time dt, a force σF acts in the section 1-1 (F is the cross-sectional area), while the section 2-2 remains not stressed. Therefore, the impulse equals to σ11 Fdt. At the beginning of time, all the selected parts were at rest, and at the moment

252

9 Dynamic Problems of Solid Mechanics

Fig. 9.2 Longitudinal waves propagation

t + dt, it is all moving at velocity v. Therefore, there is a change in the momentum: vρFd x1 = vρFcdt, i.e., σ11 Fdt = vρFcdt, or σ11 = vρc = e11 ρc2 .

(9.34)

Based on the Hooke’s law for isotropic media: σ11 = (λ + 2μ)e11 + λ(e11 + e22 ). )/ ( From Cauchy ratios ei j = u i, j + u j,i 2 and conditions u 2 = u 3 = 0, it follows that e11 = e22 = 0, which means: σ11 = (λ + 2μ)e11 .

(9.35)

Substituting (9.35) into (9.34), √ we obtain the wave propagation velocity in an unlimited elastic medium: c = (λ + 2μ)/ρ. Thus, the equations of motion of the elastic medium allow solutions containing breaks of the first derivatives from displacements.

9.7 Progressive Waves Progressive waves mean partial solutions of the equations of the dynamic theory of elasticity, corresponding to waves propagating along a line in the absence of initial conditions. Suppose that the studied progressive waves propagate along a line, as which the axis Ox, x = x1 is selected. Then the corresponding partial solution of the equations of motion can be taken in the following form: u k (t, x1 , x2 , x3 ) = Vk (x2 , x3 )e−iq(x−ct) , (q > 0, c > 0).

(9.36)

9.7 Progressive Waves

253

Solutions (9.36) make sense for regions of the infinite size along the Ox axis, for example, infinite cylinder, in general, of variable section (waveguide), half-space, plane infinite layer. Included in (9.36), values have the following names: Vk is the wave amplitude; q is the wave number; ω = qc is the phase frequency; −q(x − ct) is the wave phase; value L = 2π/q is the wave length (period). In addition, since there is a point at which the phase keeps a constant value, that is −q(x − ct) = const, so the value c is called the phase velocity. As with harmonic oscillations, the complex solution (9.36) should be understood as two processes: its real part changes according to the law of cosine, imaginary part changes according to the law of sine. If there are no perturbations (zero mass forces and homogeneous boundary conditions), then the problem of progressive waves is reduced to the problem of eigenvalues. The eigenvalue is the phase velocity c, which in general implicitly determines the dependence c = c(q) and the corresponding system of eigenfunctions (modes) Vk (q, x2 , x3 ). Media (waves) have dispersion if the function c(q) /= const and do not have dispersion otherwise. In stationary wave theory, the concept of group velocity cg , introduced by Stokes and Rayleigh, is used, which is defined as the velocity of the point x = x∗ in which the phase of the wave is stationary, that is: d (−q(x∗ − ct)) = 0. dq

(9.37)

/ cg = x∗ t = c + qc' (q).

(9.38)

From which it follows:

From (9.38), it follows that group and phase velocities coincide only with nondispersed media. The importance of information about group velocity is confirmed by the following qualitative considerations. Consider several (group) waves of the form (9.36) with close phase frequencies. If at some points, wave phases coincide or are close, then the amplitudes of individual waves add up modulus. The smallest phase difference will be provided by (9.37). Therefore, the group velocity defined by formula (9.38) is the propagation velocity of the maximal perturbation formed by the group of waves. It can be shown that the concept of progressive waves, as for harmonic waves, has a direct connection with the complex Fourier transform over the x coordinate. Namely, if we are looking for a solution in the following form: u k (t, x, x2 , x3 ) = vk (x − ct, x2 , x3 ),

(9.39)

then there is an equality: vkF (q, x2 , x3 ) = Vk (q, x2 , x3 ),

(9.40)

254

9 Dynamic Problems of Solid Mechanics

where the index “F” denotes the image; q is the transformation parameter. Then the amplitude of the group wave is the Fourier image of the function vk (x, x2 , x3 ) in (9.40). At the same time, (9.36) is (with accuracy to the constant factor) a sub-integral expression in the inverse integral of the Fourier transform. Note that (9.40), there is a special case of auto-model solutions, that is, solutions that depend on a smaller number of variables than the original problem. If there is an infinitely distant point in the body, the limitation condition of the solution specified in Sect. 9.1 must be replaced by the emission condition introduced by Sommerfeld,2 which is the condition for the existence of a single solution ˜ k > 0, where ϕ˜ corresponding to the Helmholtz wave equation: Δϕ˜ + k 2 ϕ˜ = Φ , ˜ amplitudes or modes ϕ and Φ . and Φ The emission condition introduced by Sommerfeld has the form (r is the length of the radius vector; i is an imaginary unit): • Spatial and one-dimensional problems ( / ) ϕ,r + ikφ = o 1 r , r → ∞, • Plane problems ( /√ ) r , r → ∞. ϕ,r + ikϕ = o 1 The concept of progressive waves is widely used when considering waves propagating over the surface of a semi-space or flat layer.

9.8 Rayleigh Waves Consider the problem of propagation of a plane progressive wave (9.36) in the direction of the boundary of the half-plane occupied by a homogeneous linear elastic isotropic medium (plane-strain state condition). The half-plane boundary is free from stresses, there are no mass forces and disturbances at infinity. We introduce such a rectangular Cartesian coordinate system O x yz that the Oz axis is directed deep into the half-plane and the Ox axis coincides with the boundary z = 0. The motion of the medium is described using equations in potentials (9.12) (ϕ = ϕ(x, z),ψ1 = ψ2 ≡ 0, ψ3 = ψ(x, z)): ∂ 2ϕ = c12 Δϕ, ∂t 2

∂ 2ψ ∂2 ∂2 2 = c Δψ, Δ = + . 2 ∂t 2 ∂x2 ∂ z2

(9.41)

Displacements, strains, stresses, and potentials are related by known relationships: 2

Arnold Johannes Wilhelm Sommerfeld, 1868–1951, German theoretical physicist and mathematician.

9.8 Rayleigh Waves

255

∂ϕ ∂ψ ∂ϕ ∂ψ − , u3 = + , ∂x ∂z ∂z ∂x ( ) ∂u 1 ∂u 1 1 ∂u 3 ∂u 3 , ε13 = + , ε33 = , ε11 = ∂x 2 ∂x ∂z ∂z σ11 = (λ + 2μ)ε11 + λε 33 , σ13 = 2με 13 , σ33 = (λ + 2μ)ε33 + λε 11 .

u1 =

(9.42)

The conditions at the boundary of the half-plane and at infinity have the form: σ13 |z=0 = σ33 |z=0 = 0, ϕ = O(1), ψ = O(1)(z → +∞).

(9.43)

To find the solution of the problem (9.41)–(9.43) we will look in the form of a progressive wave: ϕ(t, x, z) = Φ (z)E(x, t),

ψ(t, x, z) = ψ(z)E(x, t), E(x, t) = e−iq(x−ct) . (9.44)

Substituting (9.44) into (9.41), we obtain equations regarding the functions Φ (z) and Ψ(z): / Φ '' (z) − q 2 β1 Φ (z) = 0, ψ'' (z) − q 2 β2 ψ(z) = 0, β j = 1 − c2 c2j .

(9.45)

With c ≥ c j (β j ≤ 0), solutions to these equations do not satisfy the boundary conditions at infinity (see 9.43). If c < c j , then β j > 0 (see inequality in (9.12)) and the solutions, satisfying the condition at infinity, have the form: Φ (z) = C1 E 1 (z), ψ(z) = C2 E 2 (z), E j (z) = e−qα j z , α j =

/ ∫ / β j = 1 − c2 c2j ,

where C1 and C2 are arbitrary constants. Therefore: ϕ(t, x, z) = C1 E 1 (z)E(x, t), ψ(t, x, z) = C2 E 2 (z)E(x, t). Then from (9.42), we obtain: • Displacements u 1 = q[−iC1 E 1 (z) + α2 C2 E 2 (z)]E(x, t), u 3 = −q[α1 C1 E 1 (z) + iC2 E 2 (z)]E(x, t); • Strains ε11 = −q 2 [C1 E 1 (z) + iα2 C2 E 2 (z)]E(x, t), ) ( ] q2 [ ε13 = 2i α1 C1 E 1 (z) − 1 + α22 C2 E 2 (z) E(x, t), 2[ ] ε33 = q 2 α21 C1 E 1 (z) + i α2 C2 E 2 (z) E(x, t),

256

9 Dynamic Problems of Solid Mechanics

θ = −q 2

c2 C1 E 1 (z)E(x, t); c12

• Stresses ) [( ] c2 2 2α1 + 2 C1 E 1 (z) + 2i α2 C2 E 2 (z) E(x, t), σ11 = −μq c2 [ ) ( ] 2 σ13 = μq 2i α1 C1 E 1 (z) − 1 + α22 C2 E 2 (z) E(x, t), [( ) ] σ33 = μq 2 1 + α22 C1 E 1 (z) + 2i α2 C2 E 2 (z) E(x, t). 2

(9.46)

Substituting Eq. (9.46) into the first two boundary conditions (9.43), we obtain a homogeneous system of linear algebraic equations with respect to constants C1 and C2 : ) ) ( ( 2i α1 C1 − 1 + α22 C2 = 0, 1 + α22 C1 + 2i α2 C2 = 0.

(9.47)

The condition for the existence of a non-trivial solution of the system (9.47) is the equality to zero of its determinants: (

R(α1 , α2 ) = 1 +

)2 α22

/

( / )∫ − 4α1 α2 = R0 (ξ) = (2 − ξ) − 4 1 − ξ η2 1 − ξ = 0, / / ξ = c2 c22 , η2 = c12 c22 . (9.48) 2

Multiply the left of this equation by the conjugate expression / / ∫ R˜ 0 (ξ) = (2 − ξ)2 + 4 1 − ξ η2 1 − ξ. As a result, we get a polynomial of the fourth degree P42 (ξ) = ξP32 (ξ), P32 (ξ) = ξ3 − 8ξ2 + 8(2 + κ)ξ − 8(1 + κ), where / / / κ = λ (λ + 2μ) = ν (1 − ν) = 1 − 2 η2 .

(9.49)

Since in the real domain R˜ 0 (ξ) > 0, so (9.48) is equivalent to the following cubic equation (multiplier ξ rejected because it gives zero root, which corresponds to no wave): P32 (ξ) = 0.

(9.50)

/ Due to the inequality 0 < ν < 1 2 for real solids, the parameters κ and η change in the following ranges:

9.8 Rayleigh Waves

257

0 < κ < 1,

√ 2 < η < ∞.

(9.51)

Limit case κ = 1 (η = ∞) corresponds to acoustic medium. In this case, the denominator in (9.48) in equality for ξ turns to zero (C 2 = 0). Therefore, in parallel with Eq. (9.50), we will consider the equation equivalent to it: ) ( P31 (ς) = ς3 − 4(1 − κ)ς2 + 2(2 + κ)(1 − κ)2 ς − 1 − κ2 (1 − κ)2 = 0, P32 (η2 ς) = η6 P31 (ς), R0 (η2 ς) R˜ 0 (η2 ς) = P42 (η2 ς) = P41 (ς) = η2 ςP31 (ς).

(9.52)

The discriminant of the polynomial P31 (ς) has the form: D3 (κ) =

( p )3 3

+

( q )2

1 8 8 Δ3 (κ), p = (3κ − 2), q = (45κ − 11), 54 3 27 11 3 5 Δ3 (κ) = κ3 + κ2 + κ − . 32 16 32

2

=

/

As Δ3 (0) < 0, Δ3 (1) > 0 and Δ3 (κ) > 0, then there is a single real root κ∗ of the polynomial Δ3 (κ). Moreover, for D3 (κ) we have (values κ∗ , ν∗ and η2∗ are related by ratios (9.49)): D3 (κ∗ ) = 0, ) D3 (κ) < 0 (0 ≤ κ < κ∗ , 0 ≤ ν < ν∗ , 2 ≤ η2 < η∗2 , / ) D3 (κ) > 0 (κ∗ < κ < 1 , ν∗ < ν < 1 2, η∗2 < η2 < +∞ , κ∗ ≈ 0, 357003205, ν∗ ≈ 0, 263082064, η∗2 = 3, 1104351. Thus, polynomials P31 (ς) and P32 (ξ) at 0 ≤ κ < κ∗ have three different real roots. One polynomial root is real and the other two are complex conjugates if κ∗ < κ < 1. Note that, for real roots and ς3 polynomials P31 (ς), the following inequalities / ς1 , ς2 / are valid: 0 < ς1 = c2R c12 < 1 η2 (0 ≤ κ < 1), ς2,3 > 1 (0 < κ ≤ κ∗ ). Thus, there is a single-phase velocity ∫ √ (9.53) c = c R = ξ R c2 = ς R c1 < c2 , ξ R = ξ1 , ς R = ς1 . The general solution of the system of algebraic equations is represented as follows (C is arbitrary constant): ∫ ) ( C1 = −2i α2R C, C2 = 1 + α22R C, α2R = 1 − ξ R . We substitute these constants into (9.46) and obtain displacements and stresses:

258

9 Dynamic Problems of Solid Mechanics

[ ) ( ] u 1 = Cqα2R −2E 1 (z) + 1 + α22R E 2 (z) E(x, t), [ ] ) 1 + α22R ( 2 E 1 (z) − E 2 (z) E(x, t); u 3 = Ciq 1 + α2R 2 [( / ) ) ( ] σ11 = 2iCμq 2 α2R 2α21R + c2 c22 E 1 (z) − 1 + α22R E 2 (z) E(x, t), ( )2 σ13 = Cμq 2 1 + α22R [E 1 (z) − E 2 (z)]E(x, t), / / ( ) σ33 = 2iCμq 2 α2R 1 + α22R [−E 1 (z) + E 2 (z)]E(x, t), α1R = 1 − ξ R η2 . (9.54) Note that the stress–strain state of the solid is not singly determined, since in formulas (9.54) there is an arbitrary constant factor C. This is because the original problem is homogeneous. Due to the linearity of the problem, either the real or imaginary part of the problem is real, respectively. The resulting waves are called Rayleigh waves (see Sect. 9.1) and the corresponding phase velocity c R is called the Rayleigh wave velocity. Note that, as follows from Eq. (9.48), the phase velocity in this case does not depend on the wavenumber q, therefore, the Rayleigh waves These waves, due to exponen( are dispersion-free. ) tial multipliers with indicators −qα j z , decay very quickly in the half-plane depth. The decaying velocity is determined by the phase velocity q, that is, the wavelength / L of the x coordinate L = 2π q. The greater the phase velocity (the shorter the wavelength), the faster the decay occurs. Therefore, Rayleigh waves are surface waves (their main energy is concentrated at the boundary of the half-plane). These waves are of great importance in seismology, since they are observed in earthquakes far from the epicenter, and they are the causes of destruction of objects on the earth surface. Rayleigh waves also play a large role in ultrasound, also in fault detection.

9.9 Progressive Waves in a Plane Layer Rayleigh surface waves can exist in half-space. Consider the problem of propagation of plane progressive waves in a plane elastic layer of the thickness 2 h. Evaluate the impact of the presence of the second boundary. We use the setting of the model problem and the approach to build its solution similar to the previous section. The geometry of the task and the selected rectangular Cartesian coordinate system are shown in Fig. 9.3. The motion of the layer is described using Eq. (9.41) and relations (9.42). Assume that the layer boundaries are planes z = ±h and they are free (see (9.43)). The boundary conditions are written as follows: σ13 |z=±h = σ33 |z=±h = 0.

(9.55)

9.9 Progressive Waves in a Plane Layer

259

Fig. 9.3 To the problem of the propagation of plane progressive waves in a plane elastic layer

Just as in the Rayleigh wave problem, we build a solution in the form (9.44) and as a result we come to Eq. (9.45). The general solution of these equations depends on the magnitude of the phase velocity: c < c2 , c = c2 , c2 < c < c1 , c = c1 or c > c1 . For the examining problem in ranges c < c2 , c2 < c < c1 and c > c1 it is convenient to write the solutions of Eq. (9.45) in the same form (C k and Dk are arbitrary constants): ∫ ∫ Φ (z) = C1 sh(qz β1 ) + C2 ch(qz β1 ), ∫ ∫ / Ψ(z) = D1 sh(qz β2 ) + D2 ch(qz β2 ), β j = 1 − c2 c2j .

(9.56)

Remark The results for the bounds of the ranges are obtained using the corresponding limit transitions. From (9.44) and (9.56), we get the following representations for potentials: ( ∫ ∫ ) ϕ(t, x, z) = C1 sh(qz β1 ) + C2 ch(qz β1 ) E(x, t), ( ∫ ∫ ) ψ(t, x, z) = D1 sh(qz β2 ) + D2 ch(qz β2 ) E(x, t). Then from (9.42), we find: {( ∫ ∫ ) u 1 = − q i C1 sh(qz β1 ) + C2 ch(qz β1 ) ∫ ( ∫ ∫ )} + β2 D1 ch(qz β2 ) + D2 sh(qz β2 ) E(x, t) , {∫ ( ∫ ∫ ) u 3 = q β1 C1 ch(qz β1 ) + C2 sh(qz β1 ) ( ∫ ∫ )} − i D1 sh(qz β2 ) + D2 ch(qz β2 ) E(x, t) { ∫ ∫ ε11 = −q 2 C1 sh(qz β1 ) + C2 ch(qz β1 ) ∫ ( ∫ ∫ )} − i β2 D1 ch(qz β2 ) + D2 sh(qz β2 ) E(x, t) ∫ ∫ ) q2 { ∫ ( 2i β1 C1 ch(qz β1 ) + C2 sh(qz β1 ) ε13 = − 2

260

9 Dynamic Problems of Solid Mechanics

)} ( ∫ ∫ + (1 + β2 ) D1 sh(qz β2 ) + D2 ch(qz β2 )) E(x, t) { ( ∫ ) ∫ ε33 = q 2 β1 C1 sh(qz β1 ) + C2 ch(qz β1 ) ∫ ( ∫ ∫ )} −i β2 D1 ch(qz β2 ) + D2 sh(qz β2 ) E(x, t) { ( ∫ ∫ ) σ11 = μq 2 −(1 + 2β1 − β2 ) C1 sh(qz β1 ) + C2 ch(qz β1 ) ∫ ( ∫ ∫ )} +2i β2 D1 ch(qz β2 ) + D2 sh(qz β2 ) E(x, t) { ∫ ( ∫ ∫ ) σ13 = −μq 2 2i β1 C1 ch(qz β1 ) + C2 sh(qz β1 ) )} ( ∫ ∫ +(1 + β2 ) D1 sh(qz β2 ) + D2 ch(qz β2 )) E(x, t) { ( ∫ ∫ ) σ33 = q 2 (1 + β2 ) C1 sh(qz β1 ) + C2 ch(qz β1 ) ∫ ( ∫ ∫ )} −2i β2 D1 ch(qz β2 ) + D2 sh(qz β2 ) E(x, t).

(9.57)

Substituting Eq. (9.57) into boundary conditions (9.55), after transformations, we obtain two independent homogeneous systems of linear algebraic equations with respect to C 2 , D1 and C 1 , D2 : ∫ ∫ ∫ 2i β1 C2 sh(qh β1 ) + (1 + β2 )D1 sh(qh β2 ) = 0, ∫ ∫ ∫ (1 + β2 )C2 ch(qh β1 ) + 2i β2 D1 ch(qh β2 ) = 0, ∫ ∫ ∫ 2i β1 C1 ch(qh β1 ) + (1 + β2 )D2 ch(qh β2 ) = 0, ∫ ∫ ∫ (1 + β2 )C1 sh(qh β1 ) + 2i β2 D1 sh(qh β2 ) = 0.

(9.58)

As follows from (9.57), with C1 = D2 = 0 the displacements u1 and the stress σ11 , σ33 are even functions and the displacements u3 and the stress σ13 are odd functions along the z coordinate. If C2 = D3 = 0, then on the contrary, u3 and σ13 are even functions and u1 and σ11 , σ33 are odd functions. We call these two types of waves symmetric and antisymmetric, respectively, and due to the linearity of the problem, consider them separately. 1. Symmetric waves. The condition for the existence of a non-trivial solution of the system (9.46) is the equality to zero of its determinants, which is equivalent to the transcendent equation (the equation is written at various values of the phase velocity taking into account the equality thi x = itgx):

9.9 Progressive Waves in a Plane Layer / / th(qh 1 − ξ η2 ) (2 − ξ)2 • At c < c2 (ξ < 1) = / √ / ; √ th(qh 1 − ξ) 4 1 − ξ 1 − ξ η2 / / 1 − ξ η2 ) th(qh ( ) (2 − ξ)2 • At c2 < c < c1 1 < ξ < η2 = / √ / ; √ tg(qh ξ − 1) 4 0 − 1 1 − ξ η2 / / ( ) tg(qh ξ η2 − 1) (2 − ξ)2 • At c > c1 ξ > η2 =− . / / √ √ tg(qh ξ − 1) 4 ξ − 1 ξ η2 − 1

261

(9.59)

Parameter designations are given in (9.48). Due to the periodicity of the tangent, Eq. (9.59) have a counting set of solutions ξ√s1 < ξs2 < ... < ξsk < ... that correspond to phase velocities csk = ςsk c2 , ςsk = ξsk . We investigate the behavior of phase velocities in two limit cases: a wavelength is much longer than the layer thickness (L >> 2h, qh → 0) and a wavelength is much smaller than the layer thickness (L c1 ξ > η then . =− √ tg(qh ξ − 1) (2 − ξ)2

(9.63)

Equation (9.63) have a counting set of solutions ξα1 0 and v = c in accordance with (9.40), the formulas (9.71) coincide with the equalities (9.46)–(9.47).

266

9 Dynamic Problems of Solid Mechanics

Substituting stresses from (9.71) into boundary conditions in (9.67), we obtain a system of linear algebraic equations with respect to constants C 1 and C 2 : ) ) ( ( / 2i|q|α1 C1 − q 1 + α22 C2 = 0, q 1 + α22 C1 + 2i|q|α2 C2 = −P (μq). (9.72) The determinant of this system equals to q 2 R(α1 , α2 ), that is, it is proportional to the determinant (9.47) at v = c. Its solution has the form: ( ) P K u 1 + α22 C1 = − , μq 2

C2 = −

2i P K u α1 signq 1 . , Ku = 2 μq R(α1 , α2 )

Substituting these constants into (9.71), we find an image of the displacements and stresses: ) ] i P K u [( 1 + α22 E 1 (z) − 2α1 α2 E 2 (z) , μq ) ] P K u α1 [( u 3F = 1 + α22 E 1 (z) − 2E 2 (z) , μ|q| [( )( ) ] F σ11 = P K u 1 + α22 2α21 + M22 E 1 (z) − 4α1 α2 E 2 (z) , ( ) F σ13 = −2i P K u α1 1 + α22 [E 1 (z) − E 2 (z)]signq, [( ] )2 F σ33 = −P K u 1 + α22 E 1 (z) − 4α1 α2 E 2 (z) .

u 1F =

Using the Fourier transform properties, we finally find displacements and stresses from these relations: [ ] ) x1 P Ku ( x1 2 1 + α2 arctg − 2α1 α2 arctg , u 1 (x1 , z) = πμ α1 z α2 z ] [ / ) / 2 ( P K u α1 2 2 2 2 2 2 2 C M2 − 1 + α2 ln x1 + α1 z + 2 ln x1 + α2 z , u3 = πμ ] [ ) 1 + α22 4α22 P K u α1 z ( 2 , 2α1 + M22 2 σ11 = − π x1 + α21 z 2 x12 + α22 z 2 ( )( ) 2P K u α1 1 + α22 M22 − M12 x1 z 2 )( ) ( σ13 = , π x12 + α21 z 2 x12 + α22 z 2 ] [( )2 4α22 P K u α1 z 1 + α22 − 2 . σ33 = − π x12 + α21 z 2 x1 + α22 z 2 It is obvious if the velocity √ of the force √ movement coincides with the velocity of Rayleigh waves cR (M2 = ξ R , M1 = ζ R , see (9.53)), then all components of the solid stress–strain state strive for infinity. That is, the solution to the problem does not exist (there is no solution to the system of Eq. (9.72)).

9.10 Semi-plane Under Action of Moving Surface Force

267

/ The limit value of the force velocity v = c2 (M2 = 1,M1 = 1 η, α2 = 0, / / α1 = (1 + κ) 2, K u = 1) is not critical, that is, the solution to the problem exists and has the form: √ x1 2 P arctg √ , u 1 (x1 , z) = πμ z 1+κ / ⎡ ⎤ √ 2x12 + (1 + κ)z 2 P 1 + κ⎣ ⎦, u3 = C − ln √ √ πμ 2 x12 2 √ √ P z 2 (1 + κ) 2(1 + κ) Pz(2 + κ) 2(1 + κ) ] , σ13 = − [ 2 ] , σ11 = [ 2 π 2x1 + (1 + κ)z 2 π 2x1 + (1 + κ)z 2 x1 √ Pz 2(1 + κ) ]. σ33 = − [ 2 (9.73) π 2x1 + (1 + κ)z 2 Analysis of formulas (9.73) shows that stresses are zeros at infinity, tangent displacements u1 are bounded in the vicinity of a point at infinity x1 = ∞, on the surface of the half-plane they are bounded but have a first kind of break. Normal displacements u3 have a logarithmic feature at the beginning of the moving coordinate system and at x1 = ∞: ) P K u α1 M22 P Ku ( 1 + α22 − 2α1 α2 signx1 , u 3 |z=0 = (C + ln|x1 |); πμ πμ ) P Ku ( P K u α1 M22 1 + α22 − 2α1 α2 , u 3 ∼ ln|x1 |(x1 → ±∞). lim u 1 = ± x1 →±∞ πμ πμ u 1 |z=0 =

Note that the presence of features in displacements does not contradict the medium continuity hypothesis, since an idealized concentrated load is considered. In the case of distributed load, these features are not present. 2. Transonic speed c2 < v < c1 (M1 < 1 < M2 ,β2 < 0 < β1 ). Equation (9.68) together with infinity conditions (9.69) are kept. Also in this case, the corresponding solution is retained (9.70). The second equation from (9.67) is written as follows: ∫ ∂ 2ψF + q 2 λ22 ψ F = 0, λ2 = −β2 . 2 ∂z

(9.74)

For this equation, the infinity condition (one-dimensional problem) is the radiation condition (9.41): ( ) ∂ψ F 1 + iqλ2 ψ F = o (z → +∞). ∂z z

(9.75)

268

9 Dynamic Problems of Solid Mechanics

The solution of Eq. (9.74) satisfying condition (9.75) is as follows: ψ F (q, z) = C2 E˜ 2 (z), E˜ 2 (z) = e−iλ2 qz . Similarly, in Eq. (9.71) we obtain the representations of transformants of displacements, deformations, stresses, and a system of linear algebraic equations relative to constants C 1 and C 2 : u 1F = −iqC1 E 1 (z) + i λ2 qC2 E˜ 2 (z), u 3F = −α1 |q|C1 E 1 (z) − iqC2 E˜ 2 (z), [ [ ] ] F F ε11 = q 2 −C1 E 1 (z) + λ2 C2 E˜ 2 (z) , ε33 = q 2 α21 C1 E 1 (z) − λ2 C2 E˜ 2 (z) , ] ) ( q[ F ε13 2i|q|α1 C1 E 1 (z) − q 1 − λ22 C2 E˜ 2 (z) , = 2 [ ] ( ) F σ11 = μq 2 − 2α21 + M22 C1 E 1 (z) + 2λ2 C2 E˜ 2 (z) , [ ] ) ( F σ13 = μq 2i|q|α1 C1 E 1 (z) − q 1 − λ22 C2 E˜ 2 (z) , [( ] ) F σ33 = μq 2 1 − λ22 C1 E 1 (z) − 2λ2 C2 E˜ 2 (z) . ) ) ) ( ( /( 2i|q|α1 C1 − q 1 − λ22 C2 = 0, 1 − λ22 C1 − 2λ2 C2 = −P μq 2 . (9.76) The solution of the system of Eq. (9.76) has the following form (the polynomial P31 (ς ) is defined in (9.52)): ( ) ] )2 P K t 1 − λ22 [( 1 − λ22 + 4i α1 λ2 signq , 2 μq ] ) 2i P K t α1 signq [( 2 2 C2 = − 1 − λ + 4i α λ signq , 1 2 2 μq 2 ( )−1 K t = η2 M12 P31 (M12 ) . C1 = −

(9.77)

Substituting these constants in (9.76) we obtain images of displacements and stresses: ] [( ] )2 ) P K t i [( 1 − λ22 + 4i α1 λ2 signq 1 − λ22 E 1 (z) − i2α1 λ2 E˜ 2 (z)signq , μq ] [( ] )2 ) P K t α1 [( u 3F = 1 − λ22 E 1 (z) − 2 E˜ 2 (z) , 1 − λ22 + 4i α1 λ2 signq μ|q| ] ( )[( )2 F σ11 = P K t 1 − λ22 1 − λ22 + 4i α1 λ2 signq [( ] )( ) 2α21 + M22 1 − λ22 E 1 (z) − 4iα1 λ2 E˜ 2 (z)signq ][ ] ( )[( )2 F σ13 = −2i P K t α1 1 − λ22 1 − λ22 + 4i α1 λ2 signq E 1 (z) − E˜ 2 (z) signq, u 1F =

9.10 Semi-plane Under Action of Moving Surface Force

269

[( ] [( ] )2 )2 F σ33 = −P K t 1 − λ22 + 4i α1 λ2 signq 1 − λ22 E 1 (z) − 4i α1 λ2 E˜ 2 (z)signq . (9.78) From (9.78), using the Fourier transform properties, we find displacements and stresses: [ )3 ( ) / P Kt ( x1 1 − λ22 arctg + 4α1 λ2 1 − λ22 ln x12 + α21 z 2 − πμ α1 z ( )2 ( ) ] 2 −2α1 λ2 1 − λ2 ln|x1 + λ2 z| + 4α21 λ22 sign(x1 + λ2 z) + 2α1 λ2 1 − λ42 C ,

u 1 (x1 , z) =

( ) )3 / x1 P K t α1 ( + [− 1 − λ22 ln x12 + α21 z 2 + 4α1 λ2 1 − λ22 arctg πμ α1 z ( )2 )( ) ] ( +2 1 − λ22 ln|x1 + λ2 z| − 4α1 λ2 sign(x1 + λ2 z) + 1 − λ22 1 − λ42 C ,

u 3 (x1 , z) =

⎡ ( )2 )( ) 1 − λ22 z + 4λ2 x1 ( ⎢ ) ( σ11 (x1 , z) = P K t α1 ⎣ 2α21 + M22 1 − λ22 π x12 + α21 z 2 ⎤ ( )2 4λ2 1 − λ22 ⎥ + 16α1 λ22 δ(x1 + λ2 z)⎦, − π(x1 + λ2 z) )( ) ⎧( )⎨ 1 − λ22 λ2 x1 − α21 z z ) ( σ13 (x1 , z) = −2P K t α1 1 − λ22 ⎩ π x 2 + α2 z 2 (x + λ z) 1 2 1 1 ⎡ ⎤⎫ ⎬ α1 z )⎦ , +4α1 λ2 ⎣δ(x1 + λ2 z) − ( π x 2 + α2 z 2 ⎭ (

1

1

⎡ ( )2 )2 1 − λ22 z + 4λ2 x1 ( ⎢ ) ( σ33 (x1 , z) = −P K t α1 ⎣ 1 − λ22 π x12 + α21 z 2 ⎤ ( )2 4λ2 1 − λ22 ⎥ + 16α1 λ22 δ(x1 + λ2 z)⎦. − π(x1 + λ2 z)

(9.79)

In the transonic case, the argument ς = M12 of polynomial P31 (ς) (see (9.77)) is in the range η−2 < ς < 1. There are no valid polynomial roots in this range. Therefore, the solution of the problem, unlike the subsonic case, is determined at any loading speed from the specified range. The limit values of the force speed corresponding to the ends of the range are also not critical in the transonic case. Therefore, displacements and stresses coincide with corresponding values defined by/formulas (9.73) for subsonic case if v = c2 / / (M1 = 1 η, M2 = 1, λ2 = 0, α1 = (1 + κ) 2, K t = 1).

270

9 Dynamic Problems of Solid Mechanics

( ( )2 ) If v = c1 M2 = η, M1 = 1, λ22 = η2 − 1, α1 = 0, K t−1 = η2 − 2 , then as follows from (9.79), we are dealing with one-dimensional deformation: ( / ) u 1 = P 2λ signx1 , u 3 ≡ 0; ( / ) σ11 = − P κ δ(x1 ), σ13 ≡ 0, σ33 = −Pδ(x1 ).

(9.80)

When obtaining the first and third equations in (9.80), the following property of the delta-set of functions f (x1 , z) at z → +0 was used: lim f (x1 , z) = δ(x1 ), f (x1 , z) =

z→+0

π(x12

α1 z . + α12 z 2 )

(9.81)

Remark The same result can be obtained, if you perform the limit transition in Eq. (9.78), and then the inverse Fourier transform. It follows from (9.79) that displacements have a logarithmic feature at the halfplane boundary and at x1 → ∞: } ] )3 ( ) P K t {[ π ( 1 − λ22 + 4α21 λ22 signx1 + 2α1 λ2 1 − λ42 (ln|x1 | + C) , πμ 2 )( ) [ ( ) ] } P K t α1 {( 1 − λ22 1 − λ42 (ln|x1 | + C) + 2α1 λ2 π 1 − λ22 − 2 signx1 , u 3 |z=0 = πμ ( ) )2 ( ) P K t α1 ( 2P K t α1 λ2 1 − λ42 ln|x1 |, u 3 ∼ 1 − λ22 1 − λ42 ln|x1 | (x1 → ∞). u1 ∼ πμ πμ

u 1 |z=0 =

From (9.79), it follows that stresses have a singular feature x1 + λ2 z = 0 in the front of the Mach wave. The boundary conditions in (9.67) are fulfilled, since the limit equality (9.81) is true. For the same reason, the stresses at the half-plane boundary are defined as follows: ⎧ ( ) ⎧ (( )( )3 )( )α λ 1 2 + 2α21 + M22 1 − λ22 + 16α21 λ22 δ(x1 ) . σ11 (x1 , z) = P K t 8 α21 + λ22 1 − λ22 πx1

3. Supersonic speed c2 < c1 < v (1 < M1 < M2 ,β2 < β1 < 0). In this case, the solution can also be constructed using the Fourier transform. In this case, the potential image ϕ must satisfy √ Eq. (9.74) and condition (9.75), in which λ2 must be replaced by a value λ1 = −β1 . However, since in this case both equations in (9.65) are hyperbolic, it is more convenient to use a different approach based on the functional-invariant solutions of D’Alembert: ϕ(x1 , z) = f (x1 + λ1 z), ψ(x1 , z) = g(x1 + λ2 z),

9.10 Semi-plane Under Action of Moving Surface Force

271

where f (x) and g(x) are arbitrary, required number of times of differentiable functions. Substituting these representations in (9.42) we find (x = x1 ): u 1 = f ' (x1 + λ1 z) − λ2 g ' (x1 + λ2 z), u 3 = λ1 f ' (x1 + λ1 z) + g ' (x1 + λ2 z), ε11 = f '' (x1 + λ1 z) − λ2 g '' (x1 + λ2 z), ( ) ] 1[ ε13 = 2λ1 f '' (x1 + λ1 z) − λ22 − 1 g '' (x1 + λ2 z) , 2 ε33 = λ21 f '' (x1 + λ1 z) + λ2 g '' (x1 + λ2 z), ) [( ] σ11 = μ M22 − 2λ21 f '' (x1 + λ1 z) − 2λ2 g '' (x1 + λ2 z) , [ ( ) ] σ13 = μ 2λ1 f '' (x1 + λ1 z) − λ22 − 1 g '' (x1 + λ2 z) , [( ) ] σ33 = μ λ22 − 1 f '' (x1 + λ1 z) + 2λ2 g '' (x1 + λ2 z) . (9.82) Substituting the last two equations from (9.82) into boundary conditions (9.66), we get a system of linear algebraic equations with respect to f '' (x1 ) and g '' (x1 ): ) ( 2λ1 f '' (x1 ) − λ22 − 1 g '' (x1 ) = 0, ) ( 2 ( / ) λ2 − 1 f '' (x1 ) + 2λ2 g '' (x1 ) = − P μ δ(x1 ). Its solution has the form (see (9.47)): ) P Ks ( 2 2P K s λ2 − 1 δ(x1 ), g '' (x1 ) = − λ1 δ(x1 ), μ μ ( ) 1 ˜˜ 1 , λ2 ) = R(iλ1 , iλ2 ) = λ22 − 1 2 + 4λ1 λ2 Ks = , R(λ ˜ ˜ 1 , λ2 ) R(λ f '' (x1 ) = −

(9.83)

By integrating these equations, we find the first derivatives of the functions: f ' (x1 ) = −

) 2P K s P Ks ( 2 λ2 − 1 H (x1 ), g ' (x1 ) = − λ1 H (x1 ). μ μ

Substituting these expressions for derivatives into (9.82) leads to final formulas for displacements and stresses: ) ] P Ks [ ( 2 − λ2 − 1 H (x1 + λ1 z) + 2λ1 λ2 H (x1 + λ2 z) , μ ) ] P K s λ1 [( 2 λ2 − 1 H (x1 + λ1 z) + 2H (x1 + λ2 z) ; u3 = − μ [ ( )( ) ] σ11 = P K s − M22 − 2λ21 λ22 − 1 δ(x1 + λ1 z) + 4λ1 λ2 δ(x1 + λ2 z) , ) ( σ13 = 2P K s λ1 λ22 − 1 [δ(x1 + λ2 z) − δ(x1 + λ1 z)], ] [( )2 σ33 = −P K s λ22 − 1 δ(x1 + λ1 z) + 4λ1 λ2 δ(x1 + λ2 z) ,

u1 =

(9.84)

272

9 Dynamic Problems of Solid Mechanics

where H (x) is Heaviside step function. ˜˜ , λ ) (see (9.83)) in the supersonic case is strictly positive, Since the function R(λ 1 2 the solution to the problem is determined at any loading velocities from the specified range. velocity v = c1 ( At the limit value of the supersonic force ( 2 )2 ) 2 2 −1 M2 = η, M1 = 1, λ2 = η − 1, λ1 = 0, K s = η − 2 , the stresses and normal displacements coincide with the corresponding values determined by the formulas (9.80) for the transonic case. Only tangent displacements differ: ( / ) u 1 = P λ H (x1 ). This is due to the hyperbolic type of both Eq. (9.65) at supersonic velocity. It follows from (9.84) that by the fronts x1 + λ1 z = 0 and x1 + λ2 z = 0 of Mach waves, the half-plane is divided into three regions in each of which the displacements are constant and they have breaks of the first kind on the fronts: • At x1 < −λ2 z, u 1 = u 3 ≡ 0; 2K s λ1 λ2 2K s λ1 , u 3 = −P ; μ μ ) P K s λ1 2 Ks ( 2 λ2 − 1 − 2λ1 λ2 , u 3 = − M2 . • At x1 > −λ1 z, u 1 = −P μ μ • At − λ2 z < x1 < −λ1 z, u 1 = P

Formulas in (9.84) show that stresses have a singular feature on the selected fronts.

9.11 Model Problems on Perturbation Propagation in Elastic Space Consider model problems about the propagation of perturbations in an elastic homogeneous isotropic space, taking the equations of motion relative to the displacement vector as the original equations u k (→ x , t): ) ∂θ ( u¨ k = c12 − c22 + c22 δu k + Fk , x→ = (x1 , x2 , x3 ) ∈ R 3 , t > 0, ∂ xk

(9.85)

where the points denote the time differentiation operation t; Δ is the Laplacian; c1 and c2 are, as before, / / expansion-compression and shear wave speeds: c1 = √ (λ + 2μ)/ρ, c2 = μ ρ. To the system of Eq. (9.85) we add initial conditions to close the problem: u k |t=0 = ϕk (x1 , x2 , x3 ),

u˙ k |t=0 = ψk (x1 , x2 , x3 ), (x1 , x2 , x3 ) ∈ R 3 .

9.11 Model Problems on Perturbation Propagation in Elastic Space

273

The solution of model problems is based on the use of the Green tensor concept. Let’s give some information about fundamental solutions and the Green tensor of non-stationary processes. The Green tensor (influence function) G km (→ x , t) of the wave equation is the solution to the following boundary problems [Gorshkov, A.G. Waves in continuous media: coursebook / A.G. Gorshkov, A.L. Medvedskiy, L.N. Rabinskiy, D.V. Tarlakovskiy. – Moscow. Pub by “FIZMATLIT”, 2004. 472 p. (in Russian)]: )∂ θ ( m + c22 δG km + δkm δ(t)δ(x1 , x2 , x3 ), (x1 , x2 , x3 ) ∈ R 3 , t > 0, G¨ km = c12 − c22 ∂ xk | G km |t=0 = 0, G˙ km |t=0 = 0.

⎧ 1 xk xm 1( xk xm ) δ δ(c2 t − r ) δ(c t − r + − ) 1 km 4πr c1r 2 c2 r2 ⎧ ( ) where t 3xk xm . − 2 δkm − t − r − H t − r [H ) (c )] (c 1 2 r r2 If the Green function is known, you can define displacements based on the following integral representations: G km =

u k (→ x , t) = G km (→ x , t) ∗ ∗Fm (→ x , t) + G˙ km (→ x , t) ∗ ϕm (→ x ) + G km (→ x , t) ∗ ψm (→ x ), (9.86) where the sign “* ” denotes an operation of function convolution. In (9.86), the convolution in the first term is calculated both by spatial coordinates and by time, and in the rest – only by spatial coordinates. The deformation G εklm and the volumetric expansion factor θm corresponding to the influence function G km are defined as follows: G εklm =

) ) ( ( 1 ∂G lm ∂G km ∂G 1m ∂G 2m ∂G 3m ; θm = G ε11m + G ε22m + G ε33m = . + + 2 ∂ xk ∂ xl ∂ x1 ∂ x2 ∂ x3

It should be noted that if the Green tensor G km (→ x , t) is a symmetric tensor of the second rank, then G εklm (→ x , t) is a tensor of the third rank. Representations for stresses with known tensors G km (→ x , t) and G εklm (→ x , t) have the form: σ σkl (→ x , t) = G σklm (→ x , t) ∗ ∗Fm (→ x , t) + G˙ lkm x , t) ∗ ϕm (→ x ) + G σklm (→ x , t) ∗ ψm (→ x ), (→ (9.87)

where G σklm (→ x , t) is the influence function for stresses. They are components of the third-rank tensor and they are defined as follows: G σklm (→ x , t) = λ θm δkl + 2μG εklm .

(9.88)

274

9 Dynamic Problems of Solid Mechanics

By performing the procedure for obtaining influence function G εklm , we get: G εklm

⎧ x k xl x m ' 1 δ (c1 t − r ) =− 2 4π r c1r 2 ( ) 1 6xk xl xm xk δlm + xl δkm + xm δkl − δ(c1 t − r ) − c1 r r2 ( ) 1 xk δlm + xl δkm x k xl x m ' + δ (c2 t − r ) − c2 2 r2 ) ( 1 xk δlm + xl δkm 6xk xl xm + δ(c2 t − r ) 3 + xm δkl − c2 r 2 r2 ⎧ ( ) 3t 5xk xl xm − 3 xk δlm + xl δkm + xm δkl − [H (c1 t − r ) − H (c2 t − r )] . r r2 (9.89)

The influence function for stresses G σklm (→ x , t) is determined by simply substituting Eq. (9.89) into (9.88). Based on the above formulas, we will construct several concrete types of influence functions. Example 1 Find displacements in elastic space under the action of external forces F1 = δ(t)δ(x1 ), F2 = F3 ≡ 0 and zero initial conditions. Displacements in this case are defined as: ¨ G k1 (→ x , t)d x2 d x3 .

u k (→ x , t) = G kl (→ x , t) ∗ ∗δ(t)δ(x1 ) =

(9.90)

R2

The corresponding influence function for displacements is: G 11 = − G k1 = +

( ) ⎧ 2 x12 x1 1 1 1 − 2 δ(c2 t − r ) δ(c1 t − r ) + 4π r c1r 2 c2 r ⎧ ( ) 2 3x1 t 1 − 2 [H (c1 t − r ) − H (c2 t − r )] ; r2 r ⎧ 1 x1 xk 1 δ(c1 t − r ) − δ(c2 t − r ) 3 4π r c1 c2 ⎧ 3t [H (c1 t − r ) − H (c2 t − r )] , (k = 2, 3). r2

(9.91)

Since the function G k1 is odd in coordinate xk , if k = 2, 3, so the corresponding integrals in (9.90) are zeros and, therefore, at a specified type of disturbance u 2 = u 3 ≡ 0. Expressions for displacements u 1 , based on (9.90) and (9.91), take the form:

9.11 Model Problems on Perturbation Propagation in Elastic Space

275

⎧ ] 1[ 1 x12 J1 (x1 , c2 t) − x12 J3 (x1 , c2 t) J3 (x1 , c1 t) + u1 = G 11 d x2 d x3 = 4 π c1 c2 R2 ( ( ))]} [ , −t I3 (x1 , c1 t) − I3 (x1 , c2 t) − 3x12 I5 (x1 , c1 t) − I5 x1 , c2 t (9.92) ¨

where: ¨ Jl (x1 , t) =

1 δ(t − r )d x2 d x3 , Il (x1 , t) = rl

¨

1 H (t − r )d x2 d x3 . rl

(9.93)

R2

R2

It is convenient to calculate integrals in / (9.93) in polar coordinates ξ and α: x2 = ξ cos α, x2 = ξ sin α, where ξ = x22 + x32 ≥ 0, −π < α ≤ π. At the same time, we take into account that the/equation f (ξ) = t − r = 0 and inequality

f (ξ) ≥ 0, have solutions ξ = ξ0 = t 2 − x12 and ξ ≤ ξ0 only if t ≥ |x1 |. In addition, according to the properties of the Dirac delta function: | 1 r || t δ(ξ − ξ0 ) = | δ(ξ − ξ0 ) = δ(ξ − ξ0 ), r |ξ=ξ0 = t. δ(t − r ) = ' | f (ξ0 )| ξ ξ=ξ0 ξ0 Then, the integrals I 3 and I 5 in (9.93) are equal: t J1 (x1 , t) = H (t − |x1 |)2π ξ0 J3 (x1 , t) = H (t − |x1 |)2π

t ξ0

∫∞

ξ δ(ξ − ξ0 )dξ = 2πH (t − |x1 |), r

0

∫∞

ξ 2π δ(ξ − ξ0 )dξ = 2 H (t − |x1 |). 3 r t

(9.94)

0

Integrals I 3 and I 5 are calculated taking into account the media of sub-integral functions and replacing the integral variable y = ξ2 : ∫ξ0 I3 (x1 , t) = H (t − |x1 |)2π

ξ dξ = π H (t − |x1 |) r3

0 2 ∫ξ0

( 0

dy

= 2π ) 2 3/ 2

y + x1

(t − |x1 |) H (t − |x1 |) t|x1 | ∫ξ0

I5 (x1 , t) = H (t − |x1 |)2π 0

ξ dξ = π H (t − |x1 |) r5

276

9 Dynamic Problems of Solid Mechanics

∫ξ0

2

( 0

dy

)5 2 = 2π y + x12 /

) (3 t − |x1 |3 3t 3 |x1 |3

H (t − |x1 |)

(9.95)

Substituting (9.94) and (9.95) into (9.92), we finally get: (. ) 1 u 1 x , t = u 1 (x1 , t) = H (c1 t − |x1 |). 2c1 Example 2 Find displacements in elastic space under the action of external forces Fk = δ(t)δ(x1 , x2 )δkm , (k, m = 1, 2), F3 ≡ 0 and zero initial conditions. In this case, the displacements are defined as: ∫∞ u k (→ x , t) = G kl (→ x , t) ∗ ∗δ(t)δ(x1 , x2 )δlm =

G km (→ x , t)d x3 , −∞ ∫∞

u 3 (→ x , t) = G 3l (→ x , t) ∗ ∗δ(t)δ(x1 , x2 )δlm =

(9.96) G 3m (→ x , t)d x3 .

−∞

Remark In this example, in (9.96) and beyond, all integer indices are 1 and 2. The function G 3m is odd in coordinate x3 at m = 1, 2. Therefore, the corresponding integrals in (9.96) are zeros and therefore u 3 ≡ 0 at this type of perturbation. For the remaining components of the displacement vector, we get: ⎧ 1 1 xk xm K 3 (x1 , x2 , c1 t) + [δkm K 1 (x1 , x2 , c2 t) − xk xm K 3 (x1 , x2 , c2 t)] 4π c1 c2 − t[δkm [L 3 (x1 , x2 , c1 t) − L 3 (x1 , x2 , c2 t)] ( ( ))]} −3xk xm L 5 (x1 , x2 , c1 t) − L 5 x1 , x2 , c2 t (9.97)

uk =

where ∫∞ K l (x1 , x2 , t) = −∞

1 δ(t − r )d x3 , L l (x1 , x2 , t) = rl

∫∞ −∞

1 H (t − r )d x3 . rl

(9.98)

When calculating integral (9.98), we take into account that the equation t − r = 0 and inequality t −r ≥ 0 have solutions only / when t ≥ r2 . These / solutions have these forms: x3 = ±ζ0 and |x3 | ≤ ζ0 , where ζ0 = t 2 − r22 and r2 = x12 + x22 . In addition, according to the properties of the Dirac delta function, we have: | r || δ(t − r ) = [δ(x3 + ζ0 ) + δ(x3 − ζ0 )] |x3 | |x3 =±ζ0

9.11 Model Problems on Perturbation Propagation in Elastic Space

=

277

t [δ(x3 + ζ0 ) + δ(x3 − ζ0 )], r |x3 =±ζ0 = t ζ0

Then for integrals K 1 and K 3 in (9.98), taking into account the even function of the sub-integral function, we get: ∫∞

1 δ(t − r )d x3 r

K 1 (x1 , x2 , t) = 2 0

t = 2H (t − r2 ) ζ0

∫∞

)−1 2 ( 1 δ(x3 − ζ0 )d x3 = 2 t 2 − r22 + / r

0

∫∞

1 δ(t − r )d x3 r3

K 3 (x1 , x2 , t) = 2 0

t = 2H (t − r2 ) ζ0

∫∞

)−1 2 1 2( δ(x3 − ζ0 )d x3 = 2 t 2 − r22 + / 3 r t

(9.99)

0

The integrals L 3 and L 5 are calculated taking into account the even function of sub-integral functions: ∫ζ0 1 d x3 L 3 (x1 , x2 , t) = 2 H (t − r )d x3 = 2H (t − r2 ) ( )3 2 = 3 2 r r2 + x32 / 0 0 |ζ0 | | ( ) x3 | = 2 t 2 − r 2 1/ 2 = 2H (t − r2 ) / 2 + | 2 t r2 r2 r2 + x2 | ∫∞

2

2

3 0

∫∞

∫ζ0 d x3 1 L 5 (x1 , x2 , t) = 2 H (t − r )d x3 = 2H (t − r2 ) ( )5 2 2 r5 r2 + x32 / 0 0 ( ) |ζ0 )1 2 x3 3r22 + x32 || r22 + 2t 2 ( 2 t − r22 +/ (9.100) = 2 = 2H (t − r2 ) | ( ) 4 3 2 3 3t r2 3r 4 r 2 + x 2 / | 2

2

3

0

Substituting (9.99) and (9.100) into (9.97), we finally get: ⎧ [ ] ) ( ) 1( 2 2 1 2 1/ 2 2 2 2 2 −1/ 2 δkm c t − r2 + − c2 t c2 t − r2 + u k = u k (x1 , x2 , t) = − c1 1 2π r22 ⎫ 2 ⎬ j( Σ )( ) xk xm (−1) −1 2 + 2 2c2j t 2 − r22 c2j t 2 − r22 + / . (9.101) ⎭ cj r2 j=1

278

9 Dynamic Problems of Solid Mechanics

Example 3 Find displacements in elastic space under the action of the concentrated force F applied at the origin of coordinates and directed along the axis Ox 1 and equal to F = ρ f (t)H (t) and zero initial conditions. Calculations are performed in dimensionless variables: c∗ t ∂ ϑ' xi ui ϑ Fi L ; u i' = ; τ = ; ϑ' = ; Fi' = 2 − .' ' ; L L L T0 ∂ xi c1 σi j ' σi j = (i, j = 1, 2, 3); λ + 2μ . T0 c∗ ν λ ; γm = = ; .' = χ= (m = 1, 2); 2 λ + 2μ 1−ν cm ρ c1 / 2 c1 γ2 ; η= = = c2 γ1 1−χ / / / γ = γ1 ; r ' = r L; ϕ' = ϕ L 2 ; Φ '1 = Φ 1 L.

xi' =

) ( / where u i , σi j , Fi , Fi − . ∂ϑ ∂ xi are, accordingly, components of displacement vector,the stress tensor, the mass force vector, and the fictitious mass force vector; cm are velocities of propagation of longitudinal and transverse waves in elastic medium; λ, μ are Lamé constants; ν is the Poisson’s ratio; ρ is the medium density; . is the coefficient of thermal expansion of the medium; T0 is the medium temperature in initial condition; c∗ is a parameter having velocity dimension; L is the characteristic linear dimension; ϕ and Φ 1 are respectively scalar potentials of displacement vector and mass forces. /( ) We accept that: f (t) = f 0 = const; f 0' = f c12 L = 1; γ1 = 1; γ2 = 1, 871. For the example under consideration, the coordinates of body force have the form: F1 = f + (t)δ(→ x ), F2 = F3 ≡ 0;

f + (t) = f (t)H (t).

The expression for the components of the displacement vector, taking into account the conditions of the task, takes this form: u k (→ x , t) = G k1 (→ x , t) ∗ ∗ f + (t)δ(x1 , x2 , x3 ) = G k1 (→ x , t) ∗ f + (t). We substitute the influence function G k1 from (9.91) and we come to the following equalities: ( ) ( ) ( ) [ 2 r 1 x12 r x1 1 + 2 1 − 2 f+ t − + f+ t − u 1 (x1 , x2 , x3 , t) = 4π c12 r 2 c1 r c2 c2 [ ( ) ( )] 2 r 3 (2) r x1 xk Σ j 1 + 2f t− u k (x1 , x2 , x3 , t) = − f+ t − , (−1) 4π r 3 j=1 cj r cj c2j

9.12 Model Problems on Non-stationary Perturbation Propagation

279

where f

(2)

∫t (t) = t+ ∗ f + (t) =

∫t (t − τ)H (t − τ) f (τ)dτ = H (t)

0

(t − τ) f (τ)dτ. 0

(9.102) The function f (2) (t) defined by formula (9.102) is a primitive function (the integral of the function) f + (t). From (9.102), for the function f (t), given by the conditions of the example, we obtain: f

(2)

∫t (t) = f 0 H (t)

(t − τ) f (τ)dτ =

f0 2 t . 2 +

0

9.12 Model Problems on Non-stationary Perturbation Propagation In the case of problems for plane, taking into account the setting of problems described in the previous section, we have: u k = u k (x1 , x2 , t), u 3 ≡ 0; ε13 = ε23 = ε33 ≡ 0; σ13 = σ23 ≡ 0, λ σ33 = (σ11 + σ22 ). 2(λ + μ) The equations of motion are written as: ) ∂θ ( + c22 δu k + Fk (x1 , x2 , t), x→ = (x1 , x2 ) ∈ R 2 , t > 0. u¨ k = c12 − c22 ∂ xk (9.103) To construct a full resolution system of equations, we must add initial conditions to Eq. (9.103): u k |t=0 = ϕk (x1 , x2 ),

u˙ k |t=0 = ψk (x1 , x2 ), (x1 , x2 ) ∈ R 2 .

Integral representations for displacements and stresses have the same form as expressions (9.86) and (9.87). The influence function G km (x1 , x2 , t) for two-dimensional problems was actually constructed in Example 2 previously discussed (see expression (9.101)):

280

9 Dynamic Problems of Solid Mechanics

⎧ [ ] ) ( ) 1( 2 2 1 2 1/ 2 2 2 2 2 −1/ 2 δ t − r − c t t − r c c km 2 2 + + 2πr 2 c1 1 ⎫ 2 / )( )−1 2 ⎬ xk xm Σ (−1) j ( 2 2 2c j t − r 2 c2j t 2 − r 2 + / , r = x12 + x22 . + 2 ⎭ r cj j=1 G km = −

(9.104)

Example 4 Find displacements in elastic plane under the action of the concentrated force F applied at the origin of coordinates and directed along the axis Ox 1 and equal to: F = ρ f (t)H (t) and zero initial conditions. Calculations are performed in dimensionless /( ) variables (see example 3) and we accept that: f (t) = f 0 = const; f 0' = f c12 L = 1; γ1 = 1; γ2 = 1, 871. For the example under consideration, the body force coordinates have the form: F1 = f + (t)δ(x1 , x2 ), F2 ≡ 0;

f + (t) = f (t)H (t).

The expression for the components of the displacement vector, taking into account the conditions of the example, takes the form: u k (x1 , x2 , t) = G k1 (x1 , x2 , t) ∗ ∗ f + (t)δ(x1 , x2 ) = G k1 (x1 , x2 , t) ∗ f + (t). (9.105) We obtain the influence function G k1 from (9.104): ⎧[

)1 2 ( )−1 2 1( 2 2 c1 t − r 2 +/ − c2 t 2 c22 t 2 − r 2 + / c1 ⎫ 2 ⎬ )( ) xα2 Σ (−1) j ( 2 2 −1 2 / 2c j t − r 2 c2j t 2 − r 2 + + 2 ; ⎭ r j=1 c j

G αα = −

1 2π r 2

G 12 = −

2 )( )−1 2 x1 x2 Σ (−1) j ( 2 2 2c j t − r 2 c2j t 2 − r 2 + / 4 2π r j=1 c j

]

We substitute these equations into (9.105) and we obtain that the displacements are determined by the following formulas: ) ( [ 1 1 1 1 N11 (r, t) − N (r, t) + N−1,2 (r, t) u 1 (x1 , x2 , t) = − 2π c1 r 2 c2 r 2 12 ⎤ ) ( 2 x 2 Σ (−1) j 2 ⎦; + 12 N t) + N t) (r, (r, 1j −1, j r j=1 c j r2 ) ( 2 x1 x2 Σ (−1) j 2 u 2 (x1 , x2 , t) = − N1 j (r, t) + N−1, j (r, t) , 2π r 2 j=1 c j r2

9.13 Model Problems on Volume Perturbation Propagation

where N−1, j (r, t) =

∫t (

c2j τ2 − r 2

0

)−1/ 2 +

281

( ) ∫t f (t−τ)dτ √ 2 2 2, f (t − τ)dτ = H c j t − r / r cj

∫t ( ∫t )−1/ 2 ( ) 2 2 2 cjτ − r N1, j (r, t) = f (t − τ)dτ = H c j t − r +

/ r cj

0

c j τ −r

/ f (t − τ) c2j τ2 − r 2 dτ.

We use a convolution operation to determine the displacements, then: [ 2 )1 / 2 x2 − x2 ( f0 Σ 1 (−1) j 2 4 1 t c2j t 2 − r 2 + 2π c 2r j=1 j ] / )/ ) ( (( ) 1 r H cjt − r , + ln c j t + c2j t 2 − r 2 2c j

u 1 (x1 , x2 , t) = f 0 G 11 (x1 , x2 , t) ∗ H (t) =

) f 0 x1 x2 Σ (−1) j ( 2 2 2 1/ 2 . c t t − r + j + cj 2π r 4 2

u 2 (x1 , x2 , t) = f 0 G 21 (x1 , x2 , t) ∗ H (t) =

j=1

(9.106) Remark Construct formulas (9.106) yourself.

9.13 Model Problems on Volume Perturbation Propagation Consider the problem of spreading perturbations in a semi-space x1 ≥ 0, that are uniformly distributed over the boundary plane x1 = 0. Half-space is an elastic homogeneous isotropic medium. The initial-boundary-value problem includes [Gorshkov, A.G. Waves in continuous media: coursebook/A.G. Gorshkov, A.L. Medvedskiy, L.N. Rabinskiy, D.V. Tarlakovskiy. – Moscow. Pub by “FIZMATLIT”, 2004. 472 p. (in Russian)]: • Equation of elastic medium motion

γα2 u¨ α = u α,11 + ηα Fα (α = 1, 2, 3), • Initial conditions

u α |τ=0 = ϕα (x1 ),

u˙ α |τ=0 = ψα (x1 ),

282

9 Dynamic Problems of Solid Mechanics

• Boundary condition ( )| α u + βu ,x |x=0 = q(τ). Remark It can be shown that all types of solid mechanics boundary conditions can be written in one general form: ( )| )| ( α0 u i + β0 u i,x |x=0 = q0i (τ), α1 u˙ j + β1 u j,x |x=0 = q1 j (τ), (i, j = 1, 2, 3; i / = j )

where αk , βk ∈ R; k = 0, 1;

1 Σ l=0

α2k /= 0,

1 Σ l=0

β2k /= 0

Therefore, if α0 = 1, α1 = 0, β0 = 0, β1 = γ2j , and functions q0i (τ) = u 0i (τ), q1 j (τ) = p0 j (τ), then there are mixed boundary conditions: ( )| ( u i )|x=0 = u 0i (τ), σ1 j |x=0 = p0 j (τ), (i, j = 1, 2, 3; i /= j). Since the components of the displacement vector in this case are independent, it is sufficient to construct a solution to the following problem: γ2 u¨ = u ,11 + F, u|τ=0 = ϕ(x),

u| ˙ τ=0 = ψ(x),

( )| α u + β u ,x |x=0 = 0, u(x, τ) = O(1), x → +∞. Remark The components of the displacement vector are independent if they can be found independently of each other and are not connected to each other through boundary conditions. We will use the definition of fundamental solutions. Then the displacements and stresses for the problem under consideration can be represented as follows: ∫∞ u(x, τ) =

G(x, τ; ξ) ∗ F(ξ, τ)dξ 0

+ γ2 ⎣

∫∞ 0

˙ G(x, τ; ξ)ϕ(ξ)dξ +

∫∞ 0

∫∞ σ(x, τ) =

G σ (x, τ; ξ) ∗ F(ξ, τ)dξ 0

⎤ G(x, τ; ξ)ψ(ξ)dξ⎦;

(9.107)

9.13 Model Problems on Volume Perturbation Propagation

283

⎡∞ ⎤ ∫ ∫∞ + γ2 ⎣ G˙ σ (x, τ; ξ)ϕ(ξ)dξ + G σ (x, τ; ξ)ψ(ξ)dξ⎦, 0

0

where G(x, τ ; ξ) and G σ (x, τ ; ξ) are volumetric influence functions. The first of these functions is the solution to the boundary-value problem: γ2 G¨ = G ,x x + δ(τ)δ(x − ξ), G|τ=0 = 0,

| G˙ |τ=0 = 0,

(9.108)

( )| α G + β G ,x |x=0 = 0, G = O(1), x → +∞. The second function is written on the basis of the relation determining the stresses through the displacements: / G σ (x, τ; ξ) = G ,x (x, τ; ξ) ηα .. We apply the Laplace time transform to the problem (9.108) and solve the boundary-value problem with respect to images G L (x, s; ξ). As a result, we get the following formulas for images of influence functions: 1 {exp(γ s (x − ξ )) 2s γ ⎧ [ ] α + βγ s − exp(−γ s (x + ξ )) − 2sh γ s (x − ξ ) H (x − ξ ) ; α−βγ s 1 G σL (x, s; ξ ) = exp(γ s (x − ξ ))} 2ηα ⎧ [ ] α + βγ s + exp(−γ s (x + ξ )) − 2ch γ s (x − ξ ) H (x − ξ ) . α−βγ s

G L (x, s; ξ ) =

The originals of these functions are easily defined using Laplace transform properties. It is more convenient to do this for each specific case of boundary conditions on the surface x = 0. Example 5 Determine the stress–strain state of the elastic half-space, in which at the initial moment of time, the displacements equal to ϕ(x) = x e−x / 3 , and the velocities equal to zero. The half-space boundary is fixed, there are no mass forces. Put in γ = ηα = 1. For this problem, the formulas for the stress–strain state component (9.107) have the form: ∫∞ u(x, τ) = 0

˙ G(x, τ; ξ)ϕ(ξ)dξ,

(9.109)

284

9 Dynamic Problems of Solid Mechanics

∫∞ σ(x, τ) =

G˙ σ (x, τ; ξ)φ(ξ)dξ.

(9.110)

0

Time derived from influence functions is the differentiation of influence functions of the following form: ] [ ] 1 { [ H τ + γ (x − ξ ) − H τ − γ (x + ξ ) 2γ ( [ ] [ ])} −H (x − ξ ) H τ + γ (x − ξ ) − H τ − γ (x − ξ ) ] [ ] 1 { [ G α (x, τ ; ξ ) = δ τ + γ (x − ξ ) + δ τ − γ (x + ξ ) 2 γ ηα ( [ ] [ ]) − H (x − ξ ) δ τ + γ (x − ξ ) + δ τ − γ (x − ξ ) G(x, τ ; ξ ) =

Then ] [ ] 1{ [ ˙ δ τ + γ (x − ξ ) − δ τ − γ (x + ξ ) G(x, τ; ξ) = 2 ( [ ] [ ])} −H (x − ξ ) δ τ + γ (x − ξ ) − δ τ − γ (x − ξ ) ] [ ] 1 { '[ G˙ α (x, τ ; ξ ) = δ τ + γ (x − ξ ) + δ ' τ − γ (x + ξ ) 2γ ηα ] [ ])} ( [ −H (x − ξ ) δ ' τ + γ (x − ξ ) + δ ' τ − γ (x − ξ ) We substitute these functions into (9.109) and (9.110), and take into account their media and initial conditions of the problem. As a result, we determine the displacements and stresses: ( / ) 1{ (τ + x) exp −(τ + x) 3 H (τ ) [2 ( / ) ( / )]} + (x − τ )+ exp −(x − τ ) 3 − (τ − x)+ exp −(τ − x) 3 , ⎧( ) ( / ) τ +x 1 1− exp −(τ + x) 3 H (τ ) σ (x, τ ) = 2 3 ( ) ( / ) τ −x + 1− exp −(τ − x) 3 H (τ − x) 3 ⎧ ( ) ( / ) x −τ + 1− exp −(x − τ ) 3 H (x − τ ) . 3 u(x, τ ) =

Control Questions 1. 2.

What is called a “wave in solid deformable media”? Wave “carries” in solid: 1) energy; 2) material;

9.13 Model Problems on Volume Perturbation Propagation

3. 4. 5. 6. 7. 8.

3) both. What is the difference between the concepts of “wave” and “oscillation”? What are the main characteristics of the wave? What are the definitions of longitudinal and transverse surface waves? What are the definitions of plane waves and spherical waves? What is the definition of Shock waves? Wave equation: 1) ∂∂ xϕ2 = a 2 ∂∂tϕ2 ; [ 1 σi j − 2) εi j = 2G 2

3) 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

285

∂2ψ ∂r 2

2

+

2 ∂ψ r ∂r

=

3ν σδi j 1+ν 2 2∂ ψ a ∂t 2 .

] ;

What is the complete system of resolving equations of the dynamic theory of elasticity? What is the Lamé equations of dynamic elasticity theory? / / √ Propagation velocities of which waves are: c1 = (λ + 2μ)/ρ, c2 = μ ρ? What is the boundary-value problem of dynamic theory of elasticity in case of boundary conditions in stresses? What is the Cauchy boundary problem of dynamic elasticity theory? What are the Kinematic compatibility conditions of Lamé system? What are the dynamic compatibility conditions of Lamé system? What is the definition of Discontinuity surface? What is the definition of Stokes problem? D’Alembertian: ( ) ( ) ( ) 2 = Δ ∂ϕ − a12 ∂t∂ 2 ∂ϕ ; 1) ◻ a ∂ϕ ∂x ∂x ∂x 2) ϕ(t) = exp(i pt) = cos( pt) + i sin( pt); 3) σikj, j − ρωk2 u ik = 0.

19. What are the natural oscillations of elastic bodies? What is the fundamental (natural) frequencies of elastic body? 20. What are the definitions of forced oscillations of elastic bodies? 21. What are the main features of plane wave propagation in unlimited elastic bodies? 22. What are the main features of shock wave propagation in unlimited elastic bodies? 23. What are the definitions of progressive waves? 24. What are the definitions of Rayleigh waves? 25. What are the definitions of Mach numbers?

Chapter 10

Mathematical Models of Special Classes of Solid Mechanics Problems

One of the most important points in constructing mathematical models of a solid deformable body state and behavior is the choice of constitutive equations between components of stress–strain state (which determines the medium behavior: elasticity, viscoelasticity, plasticity, etc.). Therefore, it is natural to wish to have an “Simulation Software” that allows one to consider as wide a range of physical relationships as possible to model the behavior of the real media. Unfortunately, there are no universal, “life-friendly” methods and technologies. Due to these circumstances, it is desirable to have several “application software” implementing different laws of the behavior of solid media. At the same time, another way is possible: the development of approaches that allow modeling mechanical processes and phenomena to reduce real media to some “effective media”. As an “effective medium” it is convenient to accept a model of elastic medium, since it is obvious that to study the mechanical behavior of the elastic medium, there is the largest number of well-studied and tested methods and approaches. Consider some of the approaches that allow you to perform this procedure.

10.1 Solving Problems for Heterogeneous Nonlinear Elastic Media 10.1.1 Solving Problems for Physically Nonlinear Elastic Medium We consider the problems of the theory of elasticity, which are non-linear physic, but linear geometry. In this case, the strains of extensions and shifts are small compared to the unit.

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 M. Zhuravkov et al., Mechanics of Solid Deformable Body, https://doi.org/10.1007/978-981-19-8410-5_10

287

288

10 Mathematical Models of Special Classes of Solid Mechanics Problems

The relations between stresses σi j and small elastic strains ei j for a homogeneous isotropic body according to a physically nonlinear theory can be written as follows: ( ) ( ) σ ii = 3K χ(e0 )e0 + 2μγ ψ20 (eii − e0 ) ; σ i j = μγ ψ20 ei j , (i /= j ) ,

(10.1)

where K = λ + 2μ/3 is the bulk modulus; e0 = (1/3)(e11 + e22 + e33 ) = eii /3 is the average volumetric strain; ψ0 is shear strain intensity: ) [( / )( 2 2 2 ψ20 = (4/3) 2 3 e11 + e22 + e33 − e11 e22 − e11 e33 − e22 e33 ( / )( 2 )] 2 2 + 1 2 e12 ; + e13 + e23 ( ) where χ(e0 ) and γ ψ20 are respectively, extension and shift functions, which can be represented by power series in the following form: χ(e0 ) = 1 + χ1 e0 + χ2 e02 + χ3 e03 + · · · , ( ) γ ψ20 = 1 + γ2 ψ20 + γ4 ψ40 + γ6 ψ60 + · · · .

(10.2)

Constants χi and γ2 j can be determined experimentally. Remark For a large range of materials, the extension and shift functions with a sufficient degree of accuracy can be expressed in the form: ( ) χ(e0 ) = 1 ; γ ψ20 = 1 + γ2 ψ20 . We present the ratios (10.1) as: / ) ( / )) (( σ i j = λθδi j + μ ∂ u i ∂ x j + ∂ u j ∂ xi + 2μδηi j ,

(10.3)

where 0 ≤ δ ≤ 1; if δ = 0 then Eq. (10.3) moves to linear, and if δs = 1, move to non-linear Eq. (10.1). The perturbation function ηi j characterizes the deviation of physically nonlinear relation (10.3) from the equations of the Hooke’s law. When the extension and shift functions are represented in rows (10.2), perturbation function is as follows: η ij =

∞ [ Σ

( γ2n ψ2n 0 ei j + δi j

n=1

1+ν χn e0n+1 − γ2n ψ2n 0 e0 1 − 2ν

)] .

(10.4)

We substitute (10.3) into the Cauchy equilibrium equations. We obtain: Δu i + where Fi = 2

m Σ j=1

∂θ Xi 1 + δFi = 0 , (i = 1, ..., m), + 1 − 2ν ∂ xi μ

/ ∂ηi j ∂ x j .

(10.5)

10.1 Solving Problems for Heterogeneous Nonlinear Elastic Media

289

The procedure for solving problems according to the described approach is as follows. Let it be necessary to determine the stress–strain state of a physically non-linear body. On one part of the boundary surface of the body Su , displacements u i0 are specified, and on the rest part Sσ , external forces Ti0 are specified, S = Su ∪ Sσ : u i | Su

| | | = u i0 , σi j n j || | j=1 m Σ

= Ti0 .

(10.6)

Represent u i , ei j , σi j , ψ20 , ηi j , Fi in rows by degree of parameter δ: {

u i , ei j , σi j , ψ20 , ηi j , Fi

}

=

∞ Σ

{ } (n) (2n) (n) (n) . δn u i(n) , ei(n) , σ , ψ , η , F 0 j ij ij i

(10.7)

n=0

We add the corresponding series (10.7) to the Eq. (10.5) and equate the expressions with the same degrees of the parameter δ. We get: Δu i(0) +

Xi 1 ∂θ(0) =− , 1 − 2ν ∂ xi μ

Δu i(n) +

(i = 1, ..., m);

1 ∂θ(n) = −Fi(n−1) , (n ≥ 1). 1 − 2ν ∂ xi

(10.8)

(10.9)

Thus, in zero approximation we have a system of Lamé equations corresponding to linear theory of elasticity. And in each subsequent approximation (n ≥ 1), the right parts of the system (10.9) are known functions depending on the solution of the problem in the previous approximations. They are defined by the formula: Fi(n−1) = 2

m Σ

∂ηi(n−1) j

/

∂x j.

j=1

Boundary conditions (10.6) in the nth approximation will have the following form: | | u i(n) |

Su

=

u i0(n) ;

| | | σi j · n j || | j=1

m Σ

= Ti(n) .

(10.10)

To solve the problem (10.6), (10.8) in zero approximation and (10.9), (10.10) at n ≥ 1, you can use some approaches to solve the problems of the theory of elasticity. After the determination of u i(n) , strain components ei(n) j are determined from Cauchy ratios

290

10 Mathematical Models of Special Classes of Solid Mechanics Problems

ei(n) j

) ( ∂u (n) 1 ∂u i(n) j = + . 2 ∂x j ∂ xi

Stresses σi(n) j are calculated by the formula: (n) (n−1) (n) σi(n) . j = λ θ δi j + 2μ ei j + 2μ ηi j

Thus, the physically non-linear boundary problem of the theory of elasticity was reduced to a sequence of corresponding linear boundary problems.

10.1.2 Solution of Problems for Orthotropic Solids We consider an elastic homogeneous solid, for which the connection between stresses σi j and small strains ei j is described using the following relations: σ11 = c11 e11 + c12 e22 + c13 e33 ; σ22 = c21 e11 + c22 e22 + c23 e33 ; σ33 = c13 e11 + c23 e22 + c33 e33 ; σ12 = c66 e12 ; σ23 = c44 e23 ; σ13 = c55 e13 .

(10.11)

We select a part in these ratios in the form corresponding to the Hooke’s law for an isotropic solid: σi j = λθδi j + 2μei j . Then we present (10.11) in the following form: σ11 = (c11 − c12 )e11 + c12 (e11 + e22 + e33 ) + (c13 − c12 )e33 ; σ22 = (c11 − c12 )e11 + c12 (e11 + e22 + e33 ) + (c22 − c11 )e22 + (c23 − c12 )e33 ; σ33 = (c11 − c12 )e11 + c12 (e11 + e22 + e33 ) + (c13 − c12 )e11 + (c23 − c12 )e22 + (c33 − c11 )e33 ; σ23 = μ e23 + (c44 − μ)e23 ; σ13 = μ e13 + (c55 − μ)e13 ; σ12 = μ e12 + (c66 − μ)e12 .

(10.12)

Acceptance for an orthotropic body is as follows: μ = (1/2)(c11 − c12 ),

λ = c12 .

(10.13)

Then the ratios (10.12) can be written as: σ i j = 2μei j + λθδi j + 2μδηi j (0 ≤ δ ≤ 1),

(10.14)

10.1 Solving Problems for Heterogeneous Nonlinear Elastic Media

291

where perturbation functions ηi j characterize deviation of elastic properties of an orthotropic solid from those of an isotropic solid. Functions ηi j according to (10.12) have the form: ( / ) η11 = 1 2 (c13 − c12 )e33 ; η22 = (1/2μ)[(c22 − c11 )e22 + (c23 − c12 )e33 ] ; η33 = (1/2μ)[(c13 − c12 )e11 + (c23 − c12 )e22 + (c33 − c11 )e33 ] ; η23 = (1/2μ)(c44 − μ)e23 ; η13 = (1/2μ)(c55 − μ)e13 ; η12 = (1/2μ)(c66 − μ)e12 . (10.15) At δ = 0, the ratio (10.14) turns into the Hooke’s law equations for an isotropic medium, and at the δ = 1, in the Eq. (10.11) of the generalized Hooke’s law for an orthotropic medium. Characteristics μ and λ are determined according to (10.13). Expressions (10.13) can also be written using “technical constants”: μ = (E 1 /2)[1 − ν12 − ν23 (ν13 + ν23 )(E 2 /E 3 )]· ) [ ( 2 ]−1 2 2 1 − ν12 + 2ν12 ν13 ν23 (E 1 /E 3 ) ; (E 1 /E 2 ) − ν23 (E 2 /E 3 ) − ν13 [ ] 2 2 2 −1 λ = (ν21 + ν31 ν23 )E 2 1 − 2ν12 ν23 ν13 − ν13 − ν12 − ν23 . Let’s describe the task-solving procedure based on this approach. Let it be necessary to determine the stress–strain state of the orthotropic solid under given boundary conditions. We represent unknown components of the stress tensor σi j and displacement vector u i in rows: {

∞ { } } Σ σi j , u i = δk u i(k) , σi(k) j .

(10.16)

k=0

Note that the type of Eqs. (10.3) and (10.14) is identical. In addition, the selection of unknowns u i and σi j in form (10.16) and (10.7) is also identical. Therefore, the method of finding the components u i(k) and σi(k) j in this case is similar to the method described in the previous Sect. 10.1.1. The difference is that instead of the ratio (10.4), perturbations ηi j should be determined on the basis of the formulas (10.15). Remark This method of perturbing isotropic elastic properties allows you to find solutions with the highest degree of accuracy for orthotropic bodies with elastic properties that deviate slightly from the properties of the isotropic medium.

292

10 Mathematical Models of Special Classes of Solid Mechanics Problems

10.2 Solving the Problems of Linear Theory of Viscoelasticity for Inhomogeneous Solids This section describes a possible approach to account for rheological effects for inhomogeneous solids. We show that the problems of the linear theory of viscoelasticity for inhomogeneous bodies can be reduced to solve the problems of the theory of elasticity of inhomogeneous bodies. The relationship between stresses and strains for an anisotropic inhomogeneous linear viscoelastic body is determined by the relationships in the following form: ∫ t σi j =

λi jkl (t − τ, xs )dεkl (τ).

(10.17)

0

Applying to (10.17), Laplace transform is as follows: ∫ ∞ f ( p) =

f (t) exp(− pt)dt.

(10.18)

0

As a result, we have: σi j = C i jkl (xs )εkl , C i jkl = pλijkl (xs ). In this case, for the displacement transformants u k , the equilibrium equation in the differential form and the boundary conditions are written as:

C i jkl (xs )

( ) ∂ ∂u k + ρX i = 0; C i jkl ∂x j ∂ xl

(10.19)

| ∂ u k || n j Sσ = q i (xs ) ; u i | Su = ϕi (xs ). ∂ xl

(10.20)

Note that the relations (10.19) and (10.20) are valid if the boundary of the body S and the boundary between the parts Sσ and Su , the surface S does not change over time. Thus, in these assumptions, the problem of deformation of a viscoelastic inhomogeneous solid is reduced for Laplace-transformed values to the boundary problems (10.19) and (10.20) of the theory of elasticity of inhomogeneous solid. Consider, for example, the problem of deforming an isotropic inhomogeneous viscoelastic body. Let the volumetric properties of the material be perfectly elastic, and the shear properties are viscoelastic. Then:

10.2 Solving the Problems of Linear Theory of Viscoelasticity …

293

∫ t si j = 2Gei j −

(t − τ)ei j (τ)dτ;

θ = σ/K ,

(10.21)

0

where si j , ei j are stress and strain tensor deviators, i.e., si j = σi j − σ σi j ; ei j = εi j − θδi j /3.

(10.22)

From (10.21) and (10.22), the stress tensor can be obtained: ( ) σi j = K θδi j + 2G εi j − θδi j /3 −

∫ t

) ( (t − τ) εi j − θδi j /3 dτ.

(10.23)

0

Apply to (10.23) Laplace transform (10.18), and determine the values σi j = λθδi j + 2μεi j , where λ = K − (2/3)G + (1/3)∂( p); μ = G − (1/2)∂( p) . Therefore, a boundary-value problem for determining displacement transformants using Laplace-transformed equations of equilibrium and boundary conditions is brought to such boundary-value problem of elasticity theory of inhomogeneous media relative to displacement transformants. ( ) ( ) ∂θ ∂u j ∂μ ∂u i ∂λ + ρX i = 0; + + λ+μ + μΔu i + θ ∂ xi ∂ xi ∂ x j ∂ xi ∂ xi [ )] ( | | ∂ uj ∂ ui n j | Sσ = q i (xs ); u i | Su = ϕi (xs ). λθδi j + μ + ∂ xj ∂ xi In turn, we give one of the possible options for constructing resolving systems of equations of problems of the theory of elasticity of inhomogeneous media. The boundary-value problem of the theory of elasticity for an inhomogeneous body is formulated as follows. You want to define a vector u(x) that satisfies in the volume D = D ∪ S, the equilibrium equation is in the following form: ∂μ ∂θ + μΔu i + (λ + μ) ∂ xi ∂x j

(

∂u j ∂u i + ∂x j ∂ xi

) +θ

) ( ∂λ + ρX i = 0, i = 1, m ∂ xi (10.24)

under such boundary conditions at the S: )] [ ( | | ) ( ∂u j ∂u k ∂u i n j | Sσ = qi (xs ); u i | Su = ϕi (xs ), i = 1, m , λ δi j + μ + ∂ xk ∂x j ∂ xi where λ = λ(xs ) and μ = μ(xs ) are Lamé constants; n j are components of the outer normal vector restored at the point xs of the surface Sσ ; Sσ and Su are parts of surface

294

10 Mathematical Models of Special Classes of Solid Mechanics Problems

S with specified external loads qi and displacements ϕi , respectively; S = Sσ + Su ; m = 2 or m = 3 is space dimension. Differential equilibrium Eq. (10.24) can be written as: θ ∂λ ∂θ Xi 1 ∂μ 1 − = −ρ − Δu i + 1 − 2ν(x) ∂ xi μ(x) μ(x) ∂ xi μ(x) ∂ x j

(

) ∂u j ∂u i . + ∂x j ∂ xi (10.25)

The representation of Eq. (10.25) corresponds to the Lamé equilibrium equations of the theory of elasticity. It should be noted that in “pseudo-elastic” problems, the vector u(x ) means either the displacement vector transformant u(x ) or the vector of elastic fictitious displacements u y (x ) . Accordingly, in this case, the components of either the stress tensor transformant σi j or the stress tensor itself σi j are determined based on the Hooke’s law.

10.3 The Theory of Creep of Isotropic Hardening Media Remark The material in this section is based on work [Golub, V.P. To the creep theory of isotropic initially hardening media / V.P. Golub // Applied mechanics, 1989. V.25. N2. P.90–100. (in Russian)]. The properties of the real medium in the constitutive creep equations are usually given by one-dimensional models of various structures. Therefore, consider the solution of the problem for the one-dimensional case when stresses and strains are not tensors, but scalars. The initial state of the isotropic medium with simple loading for one-dimensional problems can be represented by the following equation: σ = ϕ0 (ε) = g(ε)ε.

(10.26)

We consider hardening media that are stable in the sense of M. Drucker’s postulate and satisfy such condition: dσ dε p > 0,

(10.27)

where σ is the true tensile stress value; ε is the strain corresponding to this stress; ϕ0 (ε) is the strain function, generally non-linear; g(ε) is the plastic modulus; dε p is the increment of the plastic component of strain. For the function g(ε), we can write it in the following expression: ⎧ g(ε) =

) ( E 0 ≤ ε ≤ εy ; ) ( / ϕ0 (ε) ε y < ε ≤ εb ,

(10.28)

10.3 The Theory of Creep of Isotropic Hardening Media

295 t=0

Fig. 10.1 Creep curves

C t=constant

B A

D

/

where ϕ0 (ε) > 0 is the derivative from ϕ0 (ε) to ε; E is the modulus of elasticity; ε y , εb are strain values corresponding to yield strength σ y and temporary resistance σb . To build a creep model, the principle of similarity of isochronous creep diagrams, formulated by Yu. N. Rabotnov, is used. Isochronous creep diagrams where mean creep diagrams rebuilt in coordinates “σ − ε” by time parameters (Fig. 10.1). The whole collection of isochrones is quite well approximated by the expression in the following form: )−1 ( σ = ϕ(ε)θ(t) = ϕ(ε) 1 + a t b ,

(10.29)

where a, b are factors determined experimentally; ε = ε0 + εc . In the general case, the explicit form of the function θ (t) is given by the weight function of the linear-hereditary equation so that at t = 0 the function (10.29) describes the elastic part ϕ0 (ε) of the instantaneous deformation diagram. Accordingly, the similarity in the frame (10.29) is defined as the possibility of shifting from diagram ϕ0 (ε) to isochrones along the abscissa ε at a given stress σ0 (Fig. 10.2, dash-and-dot curve). Equation (10.29) is essentially a relaxation equation. Another formula of the similarity condition of isochronous creep diagrams may be as follows. With a fixed value ε, the instantaneous values of the functions of the scleronomic ϕ0 (εs ) and rheonomic ϕ0 (εr ) components of the complete strain are equivalent and the following condition is met: ( )| ( ) ( ) ϕt εr |ε=const = ϕ0 εs = σ 1 + K ∗ ,

(10.30)

/ where εr = ε0 + εc ; ε0 = σ E is the elastic strain; εs = ε0 + εe if εs < ε y and εs = ε y + ε p if εs > ε y ; K ∗ is the integral operator.

296

10 Mathematical Models of Special Classes of Solid Mechanics Problems

Fig. 10.2 Principle of similarity of creep curves

C

t=0

A

t1

А A

t2

D

A

From the condition of similarity (10.30), after simple transformations, taking into account (10.26), a creep equation is obtained, relative to εc : ( ) ( ) ϕt (εr ) − ϕ0 ε y + ε y − ε0 . ε = g(εs ) c

(10.31)

In Eq. (10.31), the creep strain is given by the following relation: ⎧ εc =

( ) εe − ε0 ε0 ≤(εe ≤ ε y ; ) ε y + ε p − ε0 0 ≤ ε p ≤ εb − ε y .

(10.32)

Determining the full creep strain rate as a partial time derivative of the expression (10.31) leads to the following equation: ( )] [ ∂ ϕt (εr ) − ϕ0 ε y 1 . ε˙ = ∂t g(εs ) c

(10.33)

The Eq. (10.33) in the general case of variable loading, taking into account the history of loading and the ratio (10.28), is converted to the type: ( )] [ [ ] ∂ ψ(σ(τ), (t − τ)) − ϕ0 ε y ∂ ϕ0 (εs ) , ε˙ = ∂t ∂ εs c

(10.34)

where ψ is the creep function, defined by relation (10.30). The practical use of the governing Eqs. (10.33) and (10.34) assumes the specificity of weight functions and function ϕ0 (s) . This corresponds to the choice of the core structure of the integral operator (the selection should satisfy the best agreement with the creep experiment). In the simplest case, the integral operator core can be

10.3 The Theory of Creep of Isotropic Hardening Media

297

selected in the form (10.29). The structure of the function ϕ0 (εs ) is determined from the results of processing experimental diagrams of tension curves. The similarity condition (10.30) is fundamentally different from (10.29), since the equivalence of the isochronous diagram and the diagram of instantaneous strain is not given by abscissa ε, but by ordinate ϕ(ε) and linear displacement is replaced by a non-linear function (Fig. 10.2, dash-and-dot curve). For real hardening media, differentiate (10.26) by ε and consider (10.27). As a result, we can write: / dσ dε − σ/ ε = g ' (ε)ε ≤ 0. / / Therefore, dσ dε > 0 or dσ dε = const. As a result, the general case of hardening can be reduced to linear and non-linear laws. The non-linear law is most often given by a power function. Let’s consider features of medium creep with various nature of hardening deformation. As a creep function, we choose the simplest version of the kernel in the form similar to (10.29). For the creep rate ε˙ c at constant stress σ from (10.33), we obtain: σ 1 dεc =ab , dt g(ε) t 1−b

(10.35)

where a > 0; 0 < b < 1. For the isotropic medium with linear hardening in the elastic region g(εs ) = E, and plastic region g(εs ) = E ∗ = const, and E ∗ < E. In this case, the creep rate according to (10.35), in the whole variation range t, is a decreasing function of time (at t → ∞, ε˙ c → 0) and the creep curve has no inflection because the second derivative equals to d 2 εc 1 σ = a b(b − 1) s 2−b dt 2 g(ε ) t and always less than zero (b − 1 < 0). Therefore, the creep of the linearly hardening medium is unsteady (Fig. 10.3, curve 1). In solids with a clearly defined transition from the elastic section to the linear hardening section, the effect of a hopping increase in creep rate occurs at a constant stress (Fig. 10.3, curve 2). The effect is caused by a hopping change in the plasticity modulus g(ε) at the moment of this transition to the value Δ g(ε) = E − E ∗ and it is realized when the creep strain εc reaches the value εT − ε0 . It can also be seen from (10.35) that the smaller the hardening modulus E ∗ , the higher the creep rate at this point and at E ∗ → 0 (non-hardening media) ε˙ c → ∞ (curve 3). In the case where there is a transition region in the instantaneous deformation diagram ϕ0 (εs ) between the elastic section and the hardening section, there is no effect of the hopping creep rate (curve 4).

298

10 Mathematical Models of Special Classes of Solid Mechanics Problems

3

2

4

1

0

t

Fig. 10.3 Creep curves of linearly hardening medium

For an isotropic medium with power hardening, the creep determined by the value g(εs ) = E ∗ = const in the region, where εc < ε y − ε0 , as in the previous case, is unsteady (Fig. 10.4, curve 1). In the area of strains where εc > ε y − ε0 , for creep rate from (10.33), we obtain: ( ) m1 ) ] 1−m 1 σ[ ( 1 dεc = ab σ 1 + at b − EεT m 1−b . dt B m t

(10.36)

For the second derivative: ( ) m1 ) ] 1−m d 2 εc σ[ ( 1 b m = ab − Eε σ 1 + at T dt 2 B m ⎫ ⎧ 1 abσ t b (1 − m) ( ) − (1 − b) 2−b . m t σ 1 + at b − EεT

2 3

Х

Х 1

0

Fig. 10.4 Creep curves of isotropic medium with power hardening

t

10.3 The Theory of Creep of Isotropic Hardening Media

299

This second derivative at a time t ∗ equals to [ ∗

t =

]1 / ) b ( (1 − b) ( / ) · 1 − σy σ , a b m−1

(10.37)

and takes a zero value. In (10.36)–(10.37) B > 0, m < 1 are parameters of hardening in power law. Therefore, there is an inflection point on the creep curve, the creep rate after the inflection point becomes increasing and the system of Eqs. (10.35) and (10.36) describe all three creep stages (Fig. 10.4, curve 2). For non-linear elastic media (σ y = 0) [ ∗

t =

(1 − b) ( / ) a b m−1

] b1

and inflection points on the creep curve are independent for stress. An additional condition for the existence of inflection points on the creep curve according to (10.37) is the realization of inequality b > m, since σ < σ y (the initial state is elastic). With b < m, Eq. (10.37) loses meaning and the inflection point is determined by the moment of the creep strain εc reaches a value ε y − ε0 (curve 3). Initial deformation hardening of medium is taken into account by introduction of / / hardening speed ϕ0 (εs ) into governing Eqs. (10.33) and (10.34). As the value ϕ0 (εs ) c c decreases, the creep speed ε˙ increases (˙ε → ∞). Then the condition: | ∂ ϕ0 (εs ) ∂ ϕt (εr ) || = →0 ∂ εs ∂ εr |t=const

(10.38)

can be considered as a criterion for creep failure. The condition (10.38) corresponds to the moment of the loss of deformation stability and is implemented in medium with power hardening.

10.3.1 Consider the Situation of a Complex Stress State The solution of creep problems in a complex stress state within the framework of the approaches considered in this paragraph is based on the hypothesis of a “uniform creep curve”. An obvious condition for the validity of such and similar hypotheses is the implementation of the principle of similarity. When the theory is being constructing, it is effective to use the similarity of isochronous creep diagrams and diagrams of instantaneous strains, which makes it possible to generalize a onedimensional creep model (10.33) to a complex stress state based on the hypothesis of “uniform deformation curve”.

300

10 Mathematical Models of Special Classes of Solid Mechanics Problems

The hardening medium Eq. (10.26) in this case is written as: σ = ϕ(ε) = g(ε)ε,

(10.39)

where σ , ∈ are the generalized stress and strain; ϕ(ε) is the known function of material properties; g(ε) is the function independent of the type of stress state and g(ε) ≡ g(ε) . Remark Stress and strain intensities can be used as σ and ∈, respectively, as well as the octahedral tangent stress and octahedral shift. When constructing the constitutive creep equations for three-dimensional stress and strain fields, we use the following assumptions: • Medium is considered incompressible up to destruction; • The law of plastic flow (10.27) is assumed to be fair, but does not contain hidden variables; • The generalized creep rate is associated with the generalized stress of the same relationship as in the one-dimensional case. Thus, for the creep rate ε˙icj in the general case of the stress state, based on (10.27), (10.33), and (10.39), we obtain: ( ) 1 3 ε˙ic (σi ) σi j − δi j σi j , = − 2 σi 3 ] [ ( r) ε − σ ∂ ϕ 1 t i i , ε˙ic = ∂t g(εi )

ε˙icj

(10.40)

where ε˙ic is the intensity of strain rates; σi is the stress intensity; δi j is the Kronecker delta. In (10.40), the Huber-Mises plasticity condition was used. Therefore, the procedure for constructing the theory of creep is described, based on the refined principle of similarity of isochronous diagrams. This theory can be seen as an attempt to mechanically generalize the concepts of equations of state and nonlinear heredity. The theory allows you to take into account the initial deformation hardening of the medium, describe the third stage of creep, take into account the history and cycle of loading.

10.4 Bilinear Theory of Elasticity Determining constitutive equations of bilinear theory of elasticity in case of complex stress state takes into account plastic compressibility of material so that relations between strains and stresses for plastic bodies are bilinear [Nowacki, W.K. Stress waves in non-elastic Solids \ W.K. Nowacki. – Pergamon Press Ltd, 1978. 247 p.].

10.5 Nonlinear Viscoelastic Media

301

/ These ratios for the active plastic loading process (when d J2 dt > 0, where J2 = 0.5 si j si j is the second invariant of the stress tensor deviator) have the form: ) ( 1 1 1 (si j − si0j ), ei j = si j + − 2μ1 2μ2 2μ1 ) ( 1 1 1 (σi j − σi0j ), σi j + − εi j = 3K 1 2K 2 2K 1 where σi0j is the initial stress tensor corresponding to material transition points in plastic state; μ1 ,K 1 are shape and volumetric deformation modulus respectively; μ2 ,K 2 are plastic material constants: / / μ2 = E 2 (2(1 + ν2 )), 3K 2 = E 2 (1 − 2ν2 ).

(10.41)

As it is known, the coefficient ν2 is determined from the experience of uniaxial p/ p tension as the ratio of transverse plastic strain to longitudinal strain:ν2 = −ε2 ε1 . Thus, ν2 and μ2 , respectively, in the ratios (10.41), play the role of Poisson’s ratio and shear modulus in the plastic region. The equations of bilinear theory in the case of a uniaxial stress state pass to the relations of the deformation theory of plasticity. The application of bilinear theory in complex stress state problems has the advantage with respect to other plasticity theories that its equations are equally integrated in both elastic and plastic regions (due to the same linear relationships between strain and stress deviators and spherical components of tensors in both elastic and plastic deformation regions). This is the theoretical convenience. At the same time, this theory is associated with some simplifications of the physical nature of problems.

10.5 Nonlinear Viscoelastic Media Currently, the mathematical apparatus of linear viscoelasticity has been developed quite fully. Therefore, linear viscoelastic models are widely used in practical applications. Attempting to describe the deformation behavior of bodies with non-linear viscoelasticity led to the need to generalize the principle of Boltzmann superposition to the superposition of responses, taking into account the history of deformation. Thus, in a number of works for materials having similar stress relaxation curves, it is proposed to use an equation in the following form [Ferry, John D. Viscoelastic properties of polymers / John D. Ferry. – New York–London, 1961. 535 p.]: ∫ t σ(t) = ϕ(ε(t)) −

R(t − Θ)ϕ(ε(Θ))dΘ 0

302

10 Mathematical Models of Special Classes of Solid Mechanics Problems

where ϕ(ε(t)) is the empirical strain function. For materials that have similar isochronous curves (i.e., dependencies built from relaxation curves at fixed t), the hereditary non-linear theory of viscoelasticity, based on the Rabotnov’s equation, has become very widespread. General view of the Rabotnov’s equation: ∫ t ψ[σ(t)] = ε(t) −

R(t − Θ)ε(Θ)dΘ, 0

where ψ[σ] is the empirical stress function. In general, the non-linear theory of viscoelasticity can be represented by the Fréchet-Volterre’s equations, a special case of which is the non-linear principle of superposition developed by Persaud, who generalized Boltzmann’s postulates: ∫ t σ(t) = Eε(t)−

R(t − Θ, ε(Θ))dΘ, 0

where R(t, ε) is the non-linear relaxation core. Another equation of the non-linear theory of viscoelasticity is the Moskvitin’s equation. This equation takes into account the accumulation of damages using a linear Voltaire’s equation with time modified from the accumulated damages: ∫ t ϕ(εu )K = f (σu , σ)σ +

U (t − τ) f (σu , σ)σ(τ)dτ, 0

where ϕ(εu ) is the required function from strains. The presence of nonlinearity in viscoelasticity equations as a function of strain or stress significantly complicates the problem compared to the linear version of viscoelasticity equations. Consider one of the most common and rather simple approaches to considering the nonlinear viscoelastic properties of materials. Constitutive equations between stresses and strains are taken as: ∫ t si j = 2G( f (εu ) C- i j −

R(t − τ) f (εu )i j (τ)dτ), σ = 3K ε,

(10.42)

0

where si j C- i j , are deviators of stress and strain tensors; σ, ε are respectively, the average stress and strain; G, K are instantaneous elastic moduli of shear and volumetric strain; R(t) is the relaxation function. Physical non-linearity of a material describes a universal function f (εu ), wherein f (εu ) = 1, if the strain intensity εu does not exceed a certain threshold value εs : εu ≤ εs .

10.5 Nonlinear Viscoelastic Media

303

Fig. 10.5 Results of creep tests at pure shear

Physical Eq. (10.42) express the following: after the strain intensity at the point reaches a certain threshold value εs , the strain field ∋ij at a given moment of time is determined not only by instantaneous and preceding stress values sij , but also depends on the type of the deformed state itself. Volume change is assumed to be elastic. By resolving (10.42) with respect to sij , we obtain: ∫ t 2G f (εu ) C- i j = si j +

Γ(t − τ )si j (τ)dτ, 3K ε = σ,

(10.43)

0

where Γ(t) is the creep kernel. Consider the technique for determining the non-linear function f (εu ) from experimental creep curves. Let, for example, the results of creep experiments at pure shear be known (Fig. 10.5). In this case, six Eqs. (10.43) will be reduced to one, since ∋ij = 0 for all values i and j, except for one component ∋12 = ε12 , similar to sij = 0 except for s12 = σ12 . ε12 is)the relative shear; σ12 is the corresponding shearing stress. In our case, Where( 2/ √ εu = 2 3 ε12 . The lower curve in Fig. 10.5 corresponds to the linear behavior of the material (0) (εu ≤ εs ) at constant stress σ12 . The ratio is true along it: ⎛ 2Gε(0) 12 (t)

=

(0) ⎝ σ12 1

∫ t +

⎞ Γ(t − τ)dτ⎠.

0

This curve can also be used to determine the creep kernel: ∫ t Γ(t − τ)dτ = 0

For other curves from (10.43) follow:

2Gε(0) 12 (t) (0) σ12

− 1.

(10.44)

304

10 Mathematical Models of Special Classes of Solid Mechanics Problems

⎛ (k) (k) ⎝ 2G f (ε(k) u )ε12 (t) = σ12 1 +

∫ t

⎞ Γ(t − τ)dτ⎠,

(10.45)

0 (k) where σ12 is constant stress along the k th curve (k = 1, 2, 3). The left and right parts of Eq. (10.45) are divided into the corresponding parts of the relation (10.44). After elementary transformations, we get experimental values of the non-linear function at different points in time:

/( ) (k) (0) (0) (k) f (t) = σ12 ε12 (t) σ12 ε12 (t) . Since the strain intensity εu (t) and the non-linear function f (t) are known at each time, it is possible to construct an experimental curve f ∼ εu by comparing their values for the same t. Then constants are determined in accepted approximate formula for non-linear function. Note that the non-linear constitutive equations between stresses and strains can be taken in another form: ∫ t 2G C- i j = g(σu )si j +

Γ(t − τ)g(σu )si j (τ)dτ, 3K ε = σ,

(10.46)

0

where g(σu ) is the universal stress intensity function determined experimentally. Equations (10.42) and (10.46) are not equivalent to each other. One or another method of accounting for the material physical non-linearity is selected depending on the properties of the material and the available experimental data.

10.6 Method of Successive Approximations in Viscoplastic Problems Many materials, especially at high temperatures, show clearly rheonomic properties in addition to plasticity. Such materials are called viscoplasticity. Let’s consider one model that allows you to describe this behavior of materials. Constitutive equations in the presence of temperature field T (x,t) assume the following: ⎛ si j = 2G(T )⎝ϕ(εu , T ) C- i j −

∫ t

⎞ R(t − τ) f (εu , T ) C- i j (τ)dτ⎠, σ = 3K (T )(ε − αT ),

0

(10.47)

10.6 Method of Successive Approximations in Viscoplastic Problems

305

where ϕ(εu , T ) is the plastic function; f (εu , T ) is the universal function of rheonomic non-linearity of materials. Inverse to (10.47), they are expressions of strains through stresses: ∫ t 2G(T ) C- i j = ϕ1 (σu ,T )si j +

Γ(t − τ, T )g(σu , T )si j (τ)dτ, 0

(10.48)

3K (T )(ε − αT ) = σ. In (10.48), it is noted that instantaneous modulus of elasticity G(T ) and K(T ), creep kernel Γ(t,T ), and relaxation kernel R(t,T ), as well as plastic functions ϕ(εu , T ), ϕ1 (σu , T ) and universal functions of physical non-linearity f (εu , T ), g(σu , T ) can depend on temperature. At the same time ϕ(εu , T ) = 1 , if εu ≤ ε y (T ); ϕ1 (σu , T ) = 1 , if σu ≤ σ y (T ); f (εu , T ) = 1, if εu ≤ εs (T ); g(σu , T ) = 1, if σu ≤ σs (T ). For a general statement of the problem, it is necessary to attach to equilibrium Eqs. (10.47) or (10.48) (which are valid only with active loading), Cauchy relations, and boundary conditions: ⎛ si j = 2G(T )⎝ϕ(εu , T ) C- i j −

∫ t

⎞ R(t − τ) f (εu , T ) C- i j (τ)dτ⎠, σ = 3K (T )(ε − αT ),

0

/ σi j , j +ρFi = 0, εi j = (u i , j +u j ,i ) 2,

(10.49)

u i = u i0 (x) on Su , σi j l j = Ri on Sσ . The solution of viscoplastic problems is associated with the solution of a system of non-linear integra-differential partial differential Eqs. (10.49), which is an extremely complex mathematical problem. Therefore, it is almost impossible to build an analytical solution to this problem. Therefore, you can use the method of successive approximations (allocation procedure), which is based on the method of Ilyushin elastic solutions. We present the functions of plastic and physical non-linearity in the following form: ϕ(εu , T ) = 1 − ω(εu , T ), f (εu , T ) = 1 − ω1 (εu , T ),

(10.50)

where 0 ≤ ω < 1, 0 ≤ ω1 < 1; and ω(εu ) = 0, if εu ≤ ε y . Substitute (10.50) into (10.49). The physical relationships then become the following:

306

10 Mathematical Models of Special Classes of Solid Mechanics Problems

⎛ si j = 2G(T )⎝(1 − ω(εu , T )) C- i j −

∫ t

⎞ R(t − τ)(1 − ω1 (εu , T )) C- i j (τ)dτ⎠,

0

σ = 3K (T )(ε − αT ). (10.51) Thus, with ω = ω1 = 0, Eqs. (10.51) describe linear viscoelastic properties of materials. We write the differential equilibrium equations and boundary conditions by decomposing the stress tensor into the deviatoric and spherical parts: si j , j +σ,i +ρFi = 0,

(10.52)

u i = u i0 (x) on the Su , si j l j + σli = Ri on the Sσ

(10.53)

The problem of viscoplasticity is solved in displacements. We substitute the components of the deviatoric and spherical parts of the stress tensor (10.51) into the equilibrium Eqs. (10.52) and force boundary conditions (10.53), and take into account the Cauchy relations. As a result, we get the following generalized Lamé equations and boundary conditions: (λ + μ)θ,i +μΔu i + ρFi − Fω i − 3K αT,i = 0, λθli + μ(u i , j +u j ,i )l j = Ri + Rωi + 3K αT li , where ⎛ Fω i = 2G ⎝ω C- i j + ⎛ Rω i = 2G ⎝ω C- i j +

∫ t 0

∫ t

⎞ R(t − τ, T ) f (εu , T ) C- i j (τ)dτ⎠, j ; ⎞ R(t − τ, T ) f (εu , T ) C- i j (τ)dτ⎠l j .

0

We take zero approximation as ω(0) = f (0) = 0. Then F ωi = Rωi = 0 and to determine the first approximation ui (1) , we have the usual problem of linear thermoelasticity. Other values are determined by displacements ui (1) : (1) (1) (1) (1) (1) C- i(1) ≡ ω(ε(1) ≡ f (ε(1) u , T ), f u , T ), Fωi , Rω i . j , εu , ω

For any kth approximation, the following equilibrium equations and boundary conditions occur:

10.6 Method of Successive Approximations in Viscoplastic Problems

307

(k−1) (λ + μ)θ,i(k) +μΔu i(k) + ρFi − Fωi − 3K αT,i = 0,

(10.54)

(k−1) λθ(k)li + μ(u i(k) , j +u (k) + 3K αT li , j ,i )l j = Ri + Rωi

(10.55)

where Fω(k−1) , Rω(k−1) are known since they are determined by the previous i i approximation (k − 1). Equation (10.54) and boundary condition (10.55) are linear with respect to unknown displacements ui (k) . They differ from the corresponding equations of the , Rω(k−1) are added to the theory of thermoelasticity in those fictitious forces Fω(k−1) i i external forces F i , Ri . This allows, according to the known solution of the corresponding problem of elasticity theory, to construct a solution to the problem of viscoplasticity in the recurrent form. To ensure the convergence of the considered method of successive approximations, it is necessary that the ω parameter associated with the ϕ(εu ) function of the relation (10.50) and the non-linear function f (εu ) are small compared to the unit. If we additionally assume that ϕ (εu )≡ f (εu ), then the specified process of constructing approximations in images will not differ from the process of constructing approximations in the method of Ilyushin elastic solutions discussed earlier in the theory of small elastoplastic deformations. Remark When constructing solutions to a problem in images, it is convenient for any (k−1) and surface Ri + Rω(k−1) forces approximation to decompose volumetric Fi + Fωi i into a row according to known functions. In this case, solutions in any approximation will differ only in constant values. The method of successive approximations in a form other than the considered variant is obtained if we take the zero approximation as ω(0) = 0 and ω(0) 1 = 1− f (0) = 0. In this case, the first approximation ui (1) is defined as a solution to the usual problem of linear viscoelasticity. For the kth approximation, there is a linear (k−1) , Rω(k−1) , viscoelastic problem with some other additional “external” loads Fωi i which are calculated from the results of the previous (k-1)th approximation and are corrections for the plastic and physical nonlinearity of the material. Control Questions 1. What is the essence of the method of perturbation of linear elastic properties? What is the algorithm for solving problems for physically nonlinear elastic medium? 2. What is the essence of the method of perturbation of linear elastic properties? What is the algorithm for solving problems for orthotropic media? 3. What is the algorithm for solving the problems of linear theory of viscoelasticity for inhomogeneous bodies by reducing to solving the problems of the theory of elasticity of inhomogeneous bodies? 4. What are hardening media? 5. What is the essence of bilinear theory of elasticity?

308

6. 7. 8. 9.

10 Mathematical Models of Special Classes of Solid Mechanics Problems

What are the constitutive relations of nonlinear viscoelastic media? What are the Fréchet-Volterre equations of nonlinear viscoelasticity theory? What are the constitutive equations for viscoplastic media? What is the method of successive approximations in viscoplastic problems?