piaras Part I

Solid Mechanics Part I An Introduction to Solid Mechanics Piaras Kelly Version 4 Date Printed: 10/09/2008 Copyright ©2008 Piaras Kelly. All rights reserved. References Material, and many ideas and explanations for this book were taken from many sources, including the following (i n approximate order of usage) Mechanics of Materials , Roy R. Craig Jr., 2nd Edition, pub. John Wiley and Sons, 2000. Strength of Materials , William A. Nash, Schaum’s Outline Series, pub. McGraw- Hill, 1977. Vector Mechanics for Engineers, Vol 1: Statics , Joseph F. ShelleySchaum’s Solved Problems Series, pub. McGraw-Hill, 1980. Viscoelastic Solids , Roderic S. Lakes, 2 nd Edition, pub. CRC Press, 1998. Creep and Relaxation of Nonlinear Viscoelastic Materials , William Findley, James Lai and Kasif Onaran, pub. Dover, 1976. Advanced Engineering Mathematics , Erwin Kreyszig, 7 th Edition, pub. John Wiley & Sons, 1993. iTable of Contents 1. Introduction 1 1.1 What is Solid Mechanics 3 1.2 What is in this book? 6 2. Statics of Rigid Bodies 7 2.1 The Fundamental Concepts and Principles of Mechanics 9 2.1.1 The Fundamental Concepts 9 2.1.2 The Fundamental Principles 9 2.2 The Statics of Particles 12 2.2.1 Equilibrium of a Particle 12 2.2.2 Rough and Smooth Surfaces 14 2.2.3 Problems 15 2.3 The Statics of Rigid Bodies 17 2.3.1 Moments, Couples and Equivalent Forces 17 2.3.2 Equilibrium of Rigid Bodies 21 2.3.3 Joints and Connections 22 2.3.4 Problems 25 3. Stress and Strain 27 3.1 Surface and Contact Stress 29 3.1.1 Stress Distributions 29 3.1.2 Equivalent Forces and Moments 31 3.1.3 Shear Stress 33 3.1.4 Combined Normal and Shear Stress 34 3.1.5 Problems 34 3.2 Body Forces 37 3.2.1 Weight 37 3.2.2 Problems 39 3.3 Internal Stress 40 3.3.1 Cauchy’s Concept of Stress 40 3.3.2 Real Problems and Saint-Venant’s Principle 46 3.3.3 Problems 47 3.4 Equilibrium of Stress 49 3.4.1 Symmetry of the Shear Stress 49 3.4.2 Three Dimensional Stress 50 3.4.3 Stress Transformation Equations 51 3.4.4 Problems 54 3.5 Plane Stress 55 3.5.1 Stress Boundary Conditions 55 3.5.2 Description of Boundary Conditions 58 3.5.3 Plane Stress 61 3.5.4 Analysis of Plane Stress 63 3.5.5 Mohr’s Circle 68 3.5.6 Problems 70 3.5.7 Appendix to §3.5 72 3.6 Strain 74 3.6.1 Strain at a Point 74 3.6.2 Two Dimensional Strain 75 3.6.3 Sign Convention for Strain 77 3.6.4 Geometrical Interpretation of the Engineering Strain 78 3.6.5 Large Rotations 79 3.6.6 Three Dimensional States of Strain 80 ii 3.6.7 Problems 80 3.7 Plane Strain 82 3.7.1 Thick Components 82 3.8 Properties of the Strain 85 3.8.1 Strain Transformation Formula 85 3.8.2 Problems 88 3.8.3 Appendix to §3.8 88 4. Linear Elasticity I 91 4.1 The Continuum 93 4.1.1 Stress and Scale 93 4.1.2 Example: Metal 94 4.1.3 Problems 96 4.2 The Linear Elastic Model 97 4.2.1 The Response of Real Materials 97 4.2.2 Homogeneity and Isotropy 100 4.2.3 Stress-Strain Law 101 4.2.4 Volumetric Strain 105 4.2.5 Two-dimensional Elasticity 106 4.2.6 Problems 108 4.3 One-dimensional Axial Deformations 111 4.3.1 Basic Relations for Axial Deformations 111 4.3.2 Structures with Uniform Members 112 4.3.3 Structures with Non-uniform Members 118 4.3.4 Resultant Force and Moment 118 4.3.5 Problems 119 4.4 Torsion 121 4.4.1 Basic Relations for Torsion of Circular Members 121 4.4.2 Stress Distribution in Torsion Members 125 4.4.3 Problems 126 4.5 The Thin-walled Pressure Vessel Theory 127 4.5.1 Thin-walled Spheres 128 4.5.2 Thin-walled Cylinders 131 4.5.3 External Pressure 132 4.5.4 Problems 133 4.5.5 Appendix to Section §4.5 134 4.6 The Elementary Beam Theory 136 4.6.1 The Beam 136 4.6.2 Moments and Forces in a Beam 137 4.6.3 The Relationship between Loads, Shear Forces and Bending Moments 143 4.6.4 Deformation and Flexural Stresses in Beams 145 4.6.5 Shear Stresses in Beams 150 4.6.6 Approximate Nature of the Beam Theory 152 4.6.7 Beam Deflection 153 4.6.8 Statically Indeterminate Beams 156 4.6.9 Problems 158 4.6.10 Appendix to Section §4.6 160 4.7 Failure of Elastic Materials 163 4.7.1 Failure Theories 163 5. Energy and Virtual Work 167 5.1 Energy in Deforming Materials 169 5.1.1 Work and Energy in Particle Mechanics 169 5.1.2 The Principle of Work and Kinetic Energy 173 iii 5.1.3 The Principle of Conservation of Mechanical Energy 175 5.1.4 Deforming Materials 176 5.1.5 Energy Methods 178 5.1.6 Problems 178 5.2 Elastic Strain Energy 180 5.2.1 Strain Energy in Deformed Components 180 5.2.2 The Work-Energy Principle 183 5.2.3 The Strain Energy Density 184 5.2.4 Castigliano’s Second Theorem 187 5.2.5 Dynamic Elasticity 190 5.2.6 Problems 192 5.3 Complementary Energy 194 5.3.1 Complementary Energy 194 5.3.2 The Crotti-Engesser Theorem 195 5.3.3 Problems 195 5.4 Strain Energy Potentials 196 5.4.1 The Linear Elastic Strain Energy Potential 196 5.4.2 The Strain Energy Potential 197 5.4.3 The Complementary Energy Potential 198 5.4.4 Problems 199 5.5 Virtual Work 200 5.5.1 Principle of Virtual Work: a Single Particle 201 5.5.2 Principle of Virtual Work: Deformable Bodies 203 5.5.3 Virtual Strain Energy for a Uniaxially Loaded Bar 204 5.5.4 Virtual Strain Energy for a Beam 206 5.5.5 Problems 207 5.6 The Principle of Minimum Potential Energy 208 5.6.1 The Principle of Minimum Potential Energy 208 5.6.2 The Rayleigh-Ritz Method 209 5.6.3 Problems 212 6. Linear Elasticity II 213 6.1 Elastic Buckling 215 6.1.1 Columns and Buckling 215 6.1.2 A General Approach to Buckling 218 6.2 Anisotropic Elasticity 222 6.2.1 Material Constants 222 6.2.2 Orthotropic Linear Elasticity 223 6.2.3 Transversely Isotropic Linear Elasticity 227 6.2.4 Isotropic Linear Elasticity 229 6.2.5 Problems 230 7. Viscoelasticity 233 7.1 The Response of Viscoelastic Materials 235 7.1.1 Viscoelastic Materials 235 7.1.2 Testing of Viscoelastic Materials 236 7.2 Examples and Applications of Viscoelastic Materials 239 7.2.1 Creep and Recovery 239 7.2.2 Stress Relaxation 240 7.2.3 Energy Absorption 240 7.3 Rheological Models 242 7.3.1 Mechanical (Rheological) Models 242 7.3.2 The Maxwell Model 243 7.3.3 The Kelvin (Voigt) Model 246 7.3.4 Three-Element Models 248 iv 7.3.5 The Creep Compliance and the Relaxation Modulus 249 7.3.6 Generalised Models 249 7.3.7 Retardation and Relaxation Spectra 251 7.3.8 Problems 253 7.4 The Hereditary Integral 255 7.4.1 An Example: the Maxwell Model 255 7.4.2 Hereditary Integrals: General Formulation 258 7.4.3 Non-Linear Hereditary Integrals 262 7.4.4 Problems 262 7.4.5 Appendix to §7.4 263 7.5 Linear Viscoelasticity and the Laplace Transform 267 7.5.1 The Laplace Transform 267 7.5.2 Mechanical Models Revisited 267 7.5.3 Relationship between Creep and Relaxation 272 7.5.4 Problems 273 7.6 Oscillatory Stress, Dynamic Loading and Vibrations 274 7.6.1 Oscillatory Stress 274 7.6.2 Example: The Maxwell Model 278 7.6.3 Energy Dissipation 279 7.6.4 Impact 282 7.6.5 Damping of Vibrations 283 7.6.6 Problems 285 7.7 Temperature-dependant Viscoelastic Materials 286 7.7.1 Example: The Maxwell Model 286 7.7.2 Temporheological Materials 290 11 Introduction This brief chapter introduces the subject of Solid Mechanics and the contents of this book (Part I) and the books whic h follow (Parts II-IV). 2 Section 1.1 Solid Mechanics Part I Kelly 31.1 What is Solid Mechanics? Solid mechanics is the study of th e deformation and motion of solid materials under the action of forces. It is one of the fundamental applied engi neering sciences, in the sense that it is used to de scribe, explain and pred ict many of the physica l phenomena around us. Here are some of the wide-ra nging questions whic h solid mechanics tries to answer: Solid mechanics is a vast subj ect. One reason for this is the wide range of materials which falls under its ambit: st eel, wood, foam, plastic, f oodstuffs, concrete, biological materials, and so on. Anothe r reason is the wide range of applications in which these materials occur. For example, a metal compone nt can be manufactured by forgin g slowly at high temperature; the metal of an automobile which crashe s into a wall at high speed on a cold day will behave very differently. When will this cliff collapse? How does the heart deform when pumped? When will these gears wear out? How fast will a tuning fork vibrate? How will the San Andreas fault in California progress? How will the ground move during an earthquake? why does nature use the materials it does? how does one build a bridge which will not collapse? American Plate Pacific Plate Section 1.1 Solid Mechanics Part I Kelly 4Here are some examples of Solid Mechan ics of the cold, hot, slow and fast … Here are some examples of So lid Mechanics of the small, large, fragile and strong … 1.1.1 Aspects of Solid Mechanics At the foundation of Solid M echanics is the study of the rigid body , an ideal material in which the distance between an y two particle s remains fixed, a good approximation in some applications. Rigid body mech anics is usually subdivided into • statics , the mechanics of material s at rest, for example of a road taking the weight of a car • dynamics , the study of bod ies which are changing speed , for example of a pulley rotating at ever increasing speed Following on from statics and dynamics usually comes the topic of Mechanics of Materials (or Strength of Materials ). This is the study of so me elementary but relevant deformable materials and structures, for example beams and pressure vessels. Elasticity theory is used, in which a materi al is assumed to undergo small deformations when loaded and, when unloaded, returns to its original shape . The theo ry well approximates the behaviour of most real solid material s at low loads, and the behaviour of the “engineering materials”, for example steel and concrete, right up to fairly high loads. More advanced theories of defo rmable solid materials include • plasticity theory , which is used to mo del the behaviou r of materials which undergo permanent deformations, for example metals and soils why did this Antarctic ice fracture? what materials can withstand extreme heat? how much will this glacier move in one year? what damage will occur to a car in a car crash? what affects the quality of paper? (shown are fibers 0.02mm thick) how will a ship withstand wave slamming? how strong is an eggshell and what prevents it from cracking? how thick should a dam be to withstand the water pressure? Section 1.1 Solid Mechanics Part I Kelly 5• viscoelasticity theory , which models well material s which exhibit many “fluid- like” properties, fo r example plastics • viscoplasticity theory , which is a combination of viscoelasticity and plasticity Some other topics embraced by Solid Mechan ics, some of which could be described as “advanced”, are • rods, beams, shells and membranes , the study of material components which can be approximated by va rious model geometries, such as “very thin” • vibrations of solids and structures • composite materials , the study of co mponents made up of mo re than one material, for example fibre-glass reinforced plastics • geomechanics , the study of materials su ch as rock, soil and ice • contact mechanics , the study of materi als in contact, for ex ample a set of gears • fracture and damage mechanics , the mechanics of crack-growth and damage in materials • stability of structures • large deformation mechanics , the study of ma terials such as rubber and muscle tissue, which stretch fairly easily • biomechanics , the study of biological materials, such as bone an d heart tissue • variational formulations a nd computational mechanics , the study of the numerical (approximate ) solution of the mathematical equations which arise in the various branches of soli d mechanics, including th e Finite Element Method • dynamical systems and chaos , the study of mechanical systems which are highly sensitive to thei r initial position • experimental mechanics • thermomechanics , the analysis of materials using a formulation based on the principles of thermodynamics Section 1.2 Solid Mechanics Part I Kelly 61.2 What is in this Book? This book is divided into four parts: I. An Introduction to Solid Mechanics II. Engineering So ld Mechanics III. Large Strain Solid Mechanics IV. Thermomechanics The aim of Part I is to cover the essential concepts involved in solid mechanics, and the basic material models. It is primarily aimed at the Engineering or Science undergraduate student who has, perh aps, though not necessa rily, completed some introductory courses on mechanics and streng th of materials. Apart from gi ving a student a good grounding in the fundamentals, it should act as a stepping stone for further study into parts II to IV and into some of the more specialis ed topics ment ioned in §1.1. The outline of Part I is as fo llows: Chapter 2 covers the esse ntial material from a typical introductory course on m echanics; it serves as a brief review for th ose who have seen the material before, and serves as an introduction for those wh o are new to th e subject. Chapters 3, 4 and the first two sections of Chapter 5 cover much of the material typical of that included in a Strength of Materials or Mechanic s of Materials c ourse, and includes the elementary beam th eory and energy methods. The la tter part of th e book, Chapter 5, §5.3-5.6 and Chapters 6-7 cove r more advanced topics, for ex ample anisotropic elasticity, which is essential for the und erstanding of composite mate rials and biom aterials, and viscoelasticity, which has applications in plastics, biomaterials and many synthetic materials. In Part II, differential equili brium and strain is introduced, allowing for more complex problems to be tackled, includ ing problems of contact mech anics and elastodynamics, the study of wave propagation and vibrations, plasticity theory an d viscoplasticity. In Part III, tensor mathematic s is introduced, allo wing one to analys e the mechanics of large deformations, in particular the large-st rain elasticity of rubber-like materials and biomaterials, and the large defo rmation plasticity theories. Finally, in Part IV, a thermo mechanical approach to Solid Mechanics is taken, wherein much of the (essential) materi al from Parts I-III is re-formu lated beginning with the laws of thermodynamics. 72 Statics of Rigid Bodies Statics is the study of materials at rest. The action of all ex ternal forces acting on such materials are exactly counterbalanced and there is a zero net force effect on the material: such materials are said to be in a state of static equilibrium . In much of this book (Chapters 4-6), static elasticity will be examined. This is the study of materials which, when loaded by external fo rces, deform by a small amount from some initial configuration, and which then take up the state of sta tic equilibrium. An example might be that of floor boards deforming to take the weight of furniture. In this chapter, as an introduction to this subject, rigid bodies are considered. These are ideal materials which do not deform at all. The chapter begins with the fundamental concepts and principles of mechanics – Newton’s laws of motion . Then the mechanics of the particle , that is, of a very small amount of matter which is assumed to occ upy a single point in space, is examined. Finally, an analysis is made of the mechanics of the rigid body. The material in this chapter covers the essential material from a typical introductory course on statics. Although the concepts pres ented in this chapter serve mainly as an introduction for the later chapters, the ideas are very useful and important in themselves, for example in the design of machinery and in structural engineering. 8 Section 2.1 Solid Mechanics Part I Kelly 92.1 The Fundamental Concepts and Principles of Mechanics 2.1.1 The Fundamental Concepts The four fundamental concepts 1 used in mechanics are space , time , mass and force . It is not easy to define what these concepts are. Rather, one “knows” what they are, and they take on precise meaning when they appear in the principles and equations of mechanics discussed further below. The concept of space is associated with th e idea of the position of a point, which is described using coordinates ),,( zyx relative to an origin o as illustrated in Fig. 2.1.1. Figure 2.1.1: a particle in space The time at which events occur must be recorded if a material is in motion. The concept of mass enters Newton’s laws (see below) and in that way is used to characterize the relationship between the accel eration of a body and the fo rces acting on that body. Finally, a force is something that changes th e acceleration of matte r; it represents the action of one body on another. 2.1.2 The Fundamental Principles The fundamental laws of mechanics are Newt on’s three laws of motion. These are: Newton’s First Law : if the resultant force acting on a particle is zero, the particle remains at rest (if originally at rest) or will move with constant speed in a straight line (if originally in motion) By resultant force , one means the sum of the individual forces which act; the resultant is obtained by drawing the indivi dual forces end-to-end, in what is known as the vector 1 or at least the only ones needed outside “advanced topics” xy z•particle p o Section 2.1 Solid Mechanics Part I Kelly 10polygon law ; this is illustrated in Fig. 2.1.2, in which three forces 3 2 1,, FFF act on a single particle, leading to a non-zero resultant force2 F. Figure 2.1.2: the resultant of a system of forces acting on a particle; (a) three forces acting on a particle, (b) co nstruction of the resultant F, (c) an alternative construction, showing that the order in which the forces are drawn is immaterial, (d) the resultant force acting on the particle Example (illustrating Newton’s First Law) In Fig. 2.1.3 is shown a floating boat. It can be assumed that there are two forces acting on the boat. The first is the boat’s weight gF, that is its mass times the acceleration due to gravity g. There is also an upward buoyancy force bF exerted by the water on the boat. Assuming the boat to be still, these two forces must be equal and opposite, so that their resultant is zero. Figure 2.1.3: a zero resultant force acting on a boat ■ The resultant force acting on the particle of Fig. 2.1.2 is non-zero, and in that case one applies Newton’s second law: 2 the construction of the resultant force can be regarded also as a principle of mechanics, in that it is not proved or derived, but is taken as “given” and is borne out by experiment bFgF•1F 2F 3FF 2F 3F1F (a) (b)•F (d) (c)2F1F 3F F Section 2.1 Solid Mechanics Part I Kelly 11 Newton’s Second Law : if the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of the resultant fo rce and in the direction of this resultant force: a Fm= (2.1.1) where3 F is the resultant force, a is the acceleration and m is the mass of the particle. The units of the force are the Newton (N), the units of acceleration are metres per second squared (m/s2), and those of mass are the kilogram (kg); a force of 1 N gives a mass of 1 kg an acceleration of 1 m/s2. If the water were removed from beneath the boa t of Fig. 2.1.3, a non-zero resultant force would act, and the boa t would accelerate at g m/s2 in the direction of gF. Newton’s Third Law : each force (of “action”) has an equa l and opposite force (of “reaction”) Again, considering the boat of Fig. 2.1. 3, the water exerts an upward buoyancy force on the boat, and the boat exerts an equal and opposite force on the water. This is illustrated in Fig. 2.1.4. Figure 2.1.4: Newton’s third law; (a) the water exerts a force on the boat, (b) the boat exerts an equal and opposite force on the water Newton’s laws are used in the analysis of the most basic problems and in the analysis of the most advanced, complex, problems. They appear in many guises and sometimes they appear hidden, but they are always there in a Solid Mechanics problem. 3 vector quantities, that is, quantities which have both a magn itude and a direction associated with them, are represented by bold letters, like F here; scalars are represented by italics, like m here. The magnitude and direction of vectors are illustrated using arrows as in Fig. 2.1.2 bFbF− (a) (b)surface of water Section 2.2 Solid Mechanics Part I Kelly 122.2 The Statics of Particles 2.2.1 Equilibrium of a Particle The statics of particles is the study of particles at rest under the action of forces. Such particles can be analys ed using Newton’s first law only. Th is situation is referred to as equilibrium , which is defined as follows: Equilibrium of a Particle A particle is in equilibrium when the resultan t of all the forces acting on that particle is zero In practical problems, one w ill want to introduce a coordinate system to describe the action of forces on a particle. It is important to note that a force exists independently of any coordinate system one might use to desc ribe it. For example, consider the force F in Fig. 2.2.1. Using the vector polygon law, this force can be decomposed into combinations of any number of different indi vidual forces; these individual forces are referred to as components of F. In particular, shown in Fi g 2.2.1 are three cases in which F is decomposed into two rectangular (p erpendicular) components , the components of F in “direction x” and in “direction y”, xF and yF. Figure 2.2.1: A force F decomposed into components F x and F y using three different coordinate systems Resolving forces into rectangular component s, one can obtain analytic solutions to problems, rather than relying on graphical so lutions to problems, for example as done in Fig. 2.1.2. In order that the resultant force F on a body be zero, one must have that the resultant force in the x and y directions are zero individually1, as illustrated in the following example. 1 and in the z direction if one is consider ing a three dimensional problem xy •F • •xy xy xFF FyF xF xFyFyF (c) (a) (b) Section 2.2 Solid Mechanics Part I Kelly 13Example Consider the particle in Fig. 2.2.2, subjected to forces 3 2 1,, FFF . The particle is in equilibrium and so by definition the resultant force is zero, 0F=. The forces are decomposed into horizontal and vertical components x x x 3 2 1 ,, FFF and y y y 3 2 1 ,, FFF . The horizontal forces may be added togeth er to get a single horizontal force xF, which must equal zero. This force xF should be evaluated using the vector polygon law but, since the individual forces x x x 3 2 1 ,, FFF all lie along the same line, one need only add together the magnitudes of these vectors, which i nvolves simply an addition of scalars : 03 2 1 =++x x x F F F . Similarly, one has 03 2 1 =++y y y F F F . These equations could be used to evaluate, for example, the force 1F, if only 2F and 3F were known. Figure 2.2.2: Calculating the resultant of three forces by decomposing them into horizont al and vertical components ■ In general then, if a set of forces nF FF ,,,2 1L act on a particle, the particle is in equilibrium if and only if 0 ,0 ,0 = = = ∑∑∑ z y x F F F (2.2.1) These are known as the equations of equilibrium for a particle . They are three equations and so can be used to solve pr oblems involving three “unknowns”, for example the three components of one of the forces. In two-dimensiona l problems (as in the next example), they are a set of two equations. Example Consider the system of two cables attached to a wall shown in Fig. 2.2.3. The cables meet at C, and this point is subjected to the two forces shown. To evaluate the forces of tension arising in the cables AC and BC, one can draw a free body diagram . This is a diagram in which one body is isolated, and all the forces acting on that body are considered. A free body diagram of the particle C is shown in Fig 2.2.3b. xy •1F x1Fy1F 2F3Fx2F y2Fy3Fx3F Section 2.2 Solid Mechanics Part I Kelly 14 Figure 2.2.3: Calculating the tensio n in cables; (a) the cable system, (b) a free-body diagram of particle C, (c) cable AC in equilibrium The equations of equilibrium for particle C are 0 120 sin10030cos,0 cos10060cos ACAC BC =− + == + −−= ∑∑ θθ F FF F F yx leading to N 9.36 ,N2.46BC AC = = F F . From Newton’s third law, the cable exerts a tension force on particle C and so C must exert an equal and opposite force on the cable, as illustrated in Fig. 2.2.3c. ■ The concept of the free body is essential to So lid Mechanics. Again and again, problems will be solved by considering only a portion of the complete system, and analysing the forces acting on that portion only. 2.2.2 Rough and Smooth Surfaces Fig 2.2.4a shows a particle in equilibrium , sitting on a rough surface and subjected to a force F. Such a surface is one where frictio nal forces are large enough to prevent tangential motion. The free body diagram of the particle is shown in Fig. 2.2.4b. The friction reaction force is fR and the normal reaction force is N and these lead to the resultant reaction force R which, by Newton’s first law, must balance F. When a particle meets a smooth surface, there is no resistance to tangential movement. The particle is subjected to onl y a normal reaction force, and t hus a particle in equilibrium can only sustain a purely normal force. This is illustrated in Fig. 2.2.4c. xy (a) (b)•A BCN100 N120 N120BCFACF •N100 Cθ4 3o60 (c)ACF−C Section 2.2 Solid Mechanics Part I Kelly 15 Figure 2.2.4: a particle sitting on a surface; (a) a rough surface, (b) a free-body diagram of (a ), (c) a smooth surface 2.2.3 Problems 1. A 2000kg crate is being unloaded from a ship . A rope BC is pulled to position the crate correctly on the wharf. Use the equations of equilibrium to evaluate the tensions in the crane-cable AB and rope. [Hint: create a free body for particle B] 2. A metal ring sits over a stat ionary post. Two forces act on the ring, in opposite directions, as shown. Draw a free body diagram of the ring including the reaction force of the post on the ring. Evaluate this reac tion force. Draw a free body diagram of the post and show also the forces acting on it. 3. Two cylindrical barrels of radius mm400 are placed inside a container, a cross section of which is shown below. The mass of each barrel is 10kg. All surfaces are smooth . Draw free body diagrams of each ba rrel, including the reaction forces exerted by the container walls on the barrels, and the weight of each barrel, which acts through the barrel centres. Ev aluate all forces. What forces act on the container walls? • o15o10 A B Ccable rope N100 N200F F F fR NR (a) (b) (c) Section 2.2 Solid Mechanics Part I Kelly 16 .2m1 Section 2.3 Solid Mechanics Part I Kelly 172.3 The Statics of Rigid Bodies A material body can be considered to consist of a very large number of particles. A rigid body is one which does not deform, in other words the distance between the individual particles making up the rigid body remains unchange d under the action of external forces. A new aspect of mechanics to be considered here is that a rigid body under the action of a force has a tendency to rotate about some axis. Thus, in or der that a body be at rest, one not only needs to ensure that the resultant for ce is zero, but one must now also ensure that the forces acting on a body do not tend to make it rotate. This issue is addressed in what follows. 2.3.1 Moments, Couples and Equivalent Forces When one swings a door on its hinges, it will m ove more easily if (i) one pushes hard, i.e. if the force is large, and (ii) if one pushes furthest from th e hinges, near the edge of the door. It makes sense therefore to measure the rotational effect of a force on an object as follows: The tendency of a force to make a rigid body rotate is measured by the moment of that force about an axis. The moment of a force F about an axis through a point o is defined as the product of the magnitude of F times the perpendicular distance d from the line of action of F and the axis o. This is illustrated in Fig. 2.3.1. Figure 2.3.1: The moment of a forc e F about an axis o (the axis goes “into” the page) The moment oM of a force F can be written as Fd M=0 (2.3.1) Not only must the axis be specified (by the subscript o) when evaluating a moment, but the sense of that moment must be give n; by convention, a tendency to rotate counterclockwise is taken to be a positive moment. Thus the moment in Fig. 2.3.1 is positive. The units of moment are the Newton metre (Nm) Note that when the line of action of a fo rce goes through the axis, the moment is zero. d•Rigid body •oFline of action of force axis point of application of force Section 2.3 Solid Mechanics Part I Kelly 18 It should be emphasized that th ere is not actually a physical axis, such as a rod, at the point o of Fig. 2.3.1. In this discussion, it is imagined that an axis is there. Two forces of equal magnitude and acting along the same line of action have not only the same components y xFF, , but have equal moments about any axis. They are called equivalent forces since they have the same effect on a rigid body. This is illustrated in Fig. 2.3.2. Figure 2.3.2: Two equivalent forces Consider next the case of two forces of e qual magnitude, parallel lines of action separated by distance d, and opposite sense. Any two such forces are said to form a couple . The only motion that a couple can impart is a rota tion; unlike the forces of Fig. 2.3.2, the couple has no tendency to translate a rigid body. The moment of the couple of Fig. 2.3.3 about o is Fd Fd Fd M =−=1 2 o (2.3.2) Figure 2.3.3: A couple The sign convention which will be followed in mo st of what follows is that a couple is positive when it acts in a countercloc kwise sense, as in Fig. 2.3.3. It is straight forward to show the following: (a) the moment of Fig. 2.3.3 is also Fd about any axis in the rigid body, and so can be represented by M, without the subscript. In other wo rds, this moment of the couple is independent of the choice of axis. {see ▲Problem 1} (b) any two different couples having the same moment M are equivalent, in the sense that they tend to rotate the body in precisely the same way; it does not matter that the ••o F• d F1d 2dd•Rigid body •o1F line of action of force •2F Section 2.3 Solid Mechanics Part I Kelly 19forces forming these couples might have different magnitudes and act in different directions. (c) any two couples may be replaced by a single couple of moment equal to the algebraic sum of the moments of the individual couples. Example Consider the two couples shown in Fig. 2. 3.4a. These couples can conveniently be represented schematically by semi-circular arro ws, as shown in Fig. 2.3.4b. They can also be denoted by the letter M, the magnitude of their moment. In this example, the couples are taken to be equal and opposite, 1 2 M M−= , in which case the sum of the moments is zero and the net effect is to impart zero rotation on the body. Note that the curved arrow for 2M has been drawn countercl ockwise, even though it is negative. It could have been illustrated as in Fig. 2.3.4c, but the version of 2.3.4b is preferable as it is more consistent and reduces the likelihood of making errors when solving problems (see later). Figure 2.3.4: Two couples acting on a rigid body ■ Any force is equivalent to (i) a force acting at any (other) point and (ii) a couple. This is illustrated in Fig. 2.3.5. Referring to Fig. 2.3.5, a force F acts at position A. This for ce tends to translate the rigid body along its line of action and also to rotate it about any given axis. The system of forces in Fig. 2.3.5b are equivalent to thos e in Fig. 2.3.5a: a set of equal and opposite forces have simply been added at position B. Now the force at A and one of the forces at B form a couple, of moment M say. As in the previous example, the couple can conveniently be represented by a curved arrow, and the letter M. For illustrative purposes, the curved arrow is usually grouped with the force F at B, as shown in Fig. 2.3.5c. However, note that, representing the moment of a couple, which can be placed anywhere and have the same effect, the curved arrow is not associated with any particular point in the rigid body . ••1F 1d2d 1F•• (a)(b)11 1 dF M=22 2 dF M−= 2F2F (c)22 2 dF M+= Section 2.3 Solid Mechanics Part I Kelly 20 Figure 2.3.5: Equivalents force/moment systems; (a) a force F, (b) an equivalent system to (a), (c) an equivalent system involving a force and a couple M Note that if the force at A were moved to another position other than B, the moment M of Fig. 2.3.5c would be different. Example Consider the spanner and bolt system shown in Fig. 2.3.6. A downward force of 200N is applied at the point shown. This force can be replaced by a force acting somewhere else, together with a moment. For the case of the force moved to the bolt-centre, the moment has the magnitude shown in Fig. 2.3.6b. Figure 2.3.6: Equivalent force a nd force/moment acting on a spanner and bolt system As mentioned, it is best to draw th e semi-circle representing the moment counterclockwise (positive) and given a value of 40− as in Fig. 2.3.6b; rather than as in Fig. 2.3.6c. ■ Example Consider the plate subjected to the four external loads shown in Fig. 2.3.7a. An equivalent force-couple system F-M , with the force acting at the centre of the plate, can be calculated through •dF A•F A•B FF •BF Fd M= (a)(b) (c) (a) (b)N200cm20 N200mN40−=M (c)mN40 Section 2.3 Solid Mechanics Part I Kelly 21Nmm07. 7071 )50)(200()2/100)(50()2/100)(50()100)(100( MN100 N,200 o −= + − − −== = ∑∑ ∑ y x F F and is shown in Fig. 2.3.7b. A resultant force R can also be derived, that is, an equivalent force positioned so that a couple is not necessary, as shown in Fig. 2.3.7.c. Figure 2.3.7: Forces acting on a plate; (a) individual forces, (b) an equivalent force-couple system at th e plate-centre, (c) the resultant force The force systems in the three figures are equiva lent in the sense that they tend to impart (a) the same translation in the x direction, (b) the same translation in the y direction, and (c) the same rotation about any given point in the plate. For example, the moment about the upper left corner is Fig 2.3.7a: )100)(200()2/150)(50()2/50)(50()0)(100( + − − − Fig 2.3.7b: 7071 )44.89)(61.223( − + Fig 2.3.7c: ) 82.57)(61.223(+ all leading to Nmm 93. 12928=M about that point. ■ 2.3.2 Equilibrium of Rigid Bodies The concept of equilibrium encountered earlie r in the context of particles can now be generalized to the case of the rigid body: Equilibrium of a Rigid Body A rigid body is in equilibrium when the exte rnal forces acting on it form a system of forces equivalent to zero The necessary and sufficient conditions that a (two dimensional) rigid body is in equilibrium are then 0 ,0 ,0o= = = ∑∑∑ M F Fy x , (2.3.3) •N61.223=F Nmm07. 7071−=M (a) (b) (c)mm100 mm200N100 N200N50N50 mm50 o45o45 •N61.223=R mm.6213=doo Section 2.3 Solid Mechanics Part I Kelly 22 that is, there is no resultant force a nd no resultant moment. Note that the yx− axes and the axis of rotation o can be chosen arbitrar ily: if the resultant force is zero, and the resultant moment about one axis is zero, then the resultant moment about any other axis in the body will be zero also. 2.3.3 Joints and Connections Components in machinery, buildings, etc., conn ect with each other and are supported in a number of different ways. In order to solve for the forces acting in such assemblies, one must be able to analyse the forces acting at such connections/supports. One of the most commonly occurring supports can be idealised as a roller support , Fig. 2.3.8a. Here, the contacting surfaces ar e smooth and the roller offers only a normal reaction force (see §2.2.2). This reaction force is labelled yR, according to the conventional yx− coordinate system shown. This is shown in the free-body diagram of the component. Another commonly occurri ng connection is the pin joint , Fig. 2.3.8b. Here, the component is connected to a fixed hinge by a pin (going “int o the page”). The component is thus constrained to move in one plane, and the joint does not provide resistance to this turning movement. Th e underlying support transmits a reaction force through the hinge pin to the compon ent, which can have both normal (yR) and tangential (xR) components. Figure 2.3.8: Supports and connectio ns; (a) roller support, (b) pin joint, (c) clamped Finally, in Fig. 2.3.8c is shown a fixed (clamped) joint . Here the component is welded or glued and cannot move at th e base. It is said to be cantilevered . The support in this case reacts with normal and tangential forces , but also with a couple of moment M, roller yRpin hinge xR yR (a) (b) (c)xR yR Mxy Section 2.3 Solid Mechanics Part I Kelly 23which resists any bending/turning. For exampl e, consider such a component loaded with a force F a distance L from the base, as shown in Fig. 2.3.9a. A free-body diagram of the component is shown in Fig. 2.3.9b. The known force F acts on the body and so do two unknown forces xR, yR, and a couple of moment M. The unknown forces and moments will be called reactions henceforth. If the component is static, the equilibrium equations 2.3.3 apply. One has, taking moment s about the base of the component, 0 ,0 ,0o =+−= == =+= ∑ ∑ ∑ M FL M R F RF Fy y x x and so FL M R F Ry x ==−= ,0 , The moment is positive and so acts in the direction shown in the Figure. Figure 2.3.9: A loaded cantile vered component; (a) loaded component, (b) free body diagram of the component For ease of discussion, from now on, “couples” su ch as that encountered in Fig. 2.3.9 will simply be called “moments”. All the elements are now in pl ace to tackle fairly comple x static rigid body problems. Example Consider the plate subjected to the three external loads shown in Fig. 2.3.10a. The plate is supported by a roller at A and a pin-joint at B. The weight of the pl ate is assumed to be small relative to the applied loads and is neglected. A fr ee body diagram of the plate is shown in Fig 2.3.10b. This shows all the forces acting on the plate. Reactions act at A and B: these forces represent the action of the base on the plate, preventing it from moving downward and horizontally. The equilibrium equations can be used to find the reactions: N15 0N150 00 0 A =→==+→==→= ∑∑∑ yByB yA yxB x F MF F FF F , (b)xR yR M (a)F LF L Section 2.3 Solid Mechanics Part I Kelly 24The resultant moment was calculated by taking the moment about point A. One could have taken the moment about any other point in the plate. For example, taking moments about point B leads to N135 0B =→=∑ yAF M , which is the same result. There is usually a “most convenient” point about which to take the moments. In this example it would be point A or B, since in that case only one of the r eactions will appear in the moment equilibrium equation. Figure 2.3.10: Equilibrium of a plate; (a) forces acting on the plate, (b) free-body diagram of the plate ■ In the above example there were three unknown reactions and three equilibrium equations with which to find them. If the roller was replaced with a pin, there would be four unknown reactions, and now there would not be enough equations with which to find the reactions. When this situation arises, the system is called statically indeterminate . To find the unknown reactions, one must relax th e assumption of rigidi ty, and take into account the fact that all materi als deform. By calculating de formations within the plate, the reactions can be evaluated. The deformation of materials is studied in the following chapters. To end this Chapter, note the following: (i) the equilibrium equations 2.3.3 result fr om Newton’s first law, a universal mechanics principle, and are thus as valid for a body of water as they are for a body of hard steel; the external forces acti ng on a body of still water form a system of forces equivalent to zero. (ii) as mentioned already, Newton’s laws apply not only to a complete body or structure, but to any portion of a body. The external forces acting on any free-body portion of static material form a syst em of forces equivalent to zero. (iii) there is no such thing as a rigid body. Me tals and other engineering materials can be considered to be “nearl y rigid” as they do not defo rm by much under even fairly large loads. The analysis carried out in this Chapter is particularly relevant to these materials and in answering questions like: wh at forces act in the steel members of a suspension bridge under the load of self-weight a nd traffic? (which is just a more complicated version of the problem of Fig. 2.2.3 or Problem 3 below). (iv) if the loads on the plate of Fig. 2.3.10a ar e too large, the plate will “break”. The analysis carried out in this Chapter cannot answer where it will break or when it will (a) (b)mm150 mm200N100 mm100N100 N50 •mm30 mm100mm70N100 N100 N50 yAFyBFxBF A B Section 2.3 Solid Mechanics Part I Kelly 25break. The more sophisticated analysis ca rried out in the following Chapters is necessary to deal with th is and many other questions of material response. 2.3.4 Problems 1. A plate is subjected to a couple Fd, with cm20=d , as shown below left. Verify that the couple can be moved to the positi on shown below right, and the effect on the plate is the same, by evaluating moments about point o. 2. Consider the beam below left, of length 10m , supported by two pins and loaded at its two ends as shown. What couple (moment) M must be applied to the beam shown below right so that both beams are subjected to equivalent loadings? Where can the moment be applied? 3. What force F must be applied to the following sta tic component such that the tension in the cable, T, is 1kN? What are the reac tions at the pin support C? 4. A machine part is hinged at A and subjected to two forces through cables as shown. What couple M needs to be applied to the machine part for equilibrium to be maintained? N100 N 100 N 200 M •F CT mm150mm150 mm250cm100cm100 F F cm30cm30 cm20 cm100cm100F Fcm20 oo Section 2.3 Solid Mechanics Part I Kelly 26 N100 N50mm100 • •• Mmm75A 273 Stress & Strain Forces acting at the surfaces of components were considered in the previous chapter. The task now is to examine forces arising inside materials, internal forces . Internal forces are described using the idea of stress . There is a lot more to stre ss than “force over area”, as will become clear from §3.1-3.5. First, the id ea of surface (contact) stress distributions will be examined, together with their relations hip to resultant forces and moments. Then internal stress and traction will be discussed. The means by which internal forces are described is through the stress components , for exampleyy zxσσ, , and this “language” of sigmas and subscripts needs to be mastered in order to model sensibly the internal forces in real materials. Stress analysis involves representing the actual internal forces in a real physical component mathematically. Some of the limitations of th is are discussed in §3.3.2. Newton’s laws are used to derive the stress transformation equations , and these are then used to derive expressions for the principal stresses , stress invariants , principal directions and maximum shear stresses acting at a material particle. The practical case of two dimensional plane stress is discussed. Strain is a measure of deformation in a mate rial. The relationship between stress and strain depends on the particular material under study, and will be discussed in Chapter 4. The concept of strain is introduced in §3.6. The approximation to the true strain of the engineering strain is discussed. The practical case of two dimensional plane strain is then introduced, along with the strain transformation formulae , principal strains , principal strain directions and the maximum shear strain . 28 Section 3.1 Solid Mechanics Part I Kelly 293.1 Surface and Contact Stress The concept of the force is fundamental to mechanics and many important problems can be cast in terms of forces only, for exam ple the problems considered in Chapter 2. However, more sophisticated problems require that the action of forces be described in terms of stress , that is, force divided by area. Fo r example, if one hangs an object from a rope, it is not the wei ght of the object which determ ines whether the rope will break, but the weight divided by the cross-sectional area of the rope, a fact noted by Galileo in 1638. 3.1.1 Stress Distributions As an introduction to the idea of stress, consider the situation shown in Fig. 3.1.1a: a block of mass m and cross sectional area A sits on a bench. Following the methodology of Chapter 2, an analysis of a free-body of the block shows that a force equal to the weight mg acts upward on the block, Fig. 3.1.1b. Allowing for more detail now, this force will actually be di stributed over the surface of the block, as indicated in Fig. 3.1.1c. Defi ning the stress to be for ce divided by area, the stress acting on the block is Amg=σ (3.1.1) The unit of stress is the Pascal (Pa): 1Pa is equivalent to a force of 1 Newton acting over an area of 1 metre squared. Typical un its used in engineer ing applications are the kilopascal, kPa ( Pa 103), the megapascal, MPa ( Pa 106) and the gigapascal (P a109). Figure 3.1.1: a block resting on a bench; (a) weight of the block, (b) reaction of the bench on the block, (c) stress distribution acting on the block The stress distribution of Fig. 3.1.1c acts on the block. By Newton’s third law, an equal and opposite stress distri bution is exerted by the block on the bench; one says that the weight force of the block is transmitted to the underlying bench. The stress distribution of Fig. 3.1.1 is uniform , i.e. constant everywhere over the surface. In more complex and interesting situations in which materials contact, one is more likely to obtain a non-uniform distribution of stress. For example, consider the case of a metal ball being pushed into a similarly stiff object by a force F, as mg (a) (c)σmg (b) Section 3.1 Solid Mechanics Part I Kelly 30illustrated in Fig. 3.1.2.1 Again, an equal force F acts on the underside of the ball, Fig. 3.1.2b. As with the block, the force will actually be distributed over a contact region . It will be shown in Part II that a ci rcular contact region will arise where the ball and object meet2, and that the stress is largest at the centre of the contact surface, dying away to zero at the e dges of contact, Fig. 3.1.2c (2 1σσ> in Fig. 3.1.2c). In this case, with stress σ not constant, one can only write dA F A∫=σ (3.1.2) The stress varies from point to point over the surface but the sum (or integral) of the stresses (times areas) equals the total force applied to the ball. Figure 3.1.2: a ball being forced into a large object, (a) force applied to ball, (b) reaction of object on ball, (c) a no n-uniform stress distribution over the contacting surface A given stress distribution gives rise to a resultant force, which is obtained by integration, Eqn. 3.1.2. It will al so give rise to a resultant moment. This is examined in the following example. Example Consider the surface shown in Fig. 3.1.3, of length 2m and depth 2m (into the page). The stress over the surface is given by x=σ kPa, with x measured in m from the left- hand side of the surface. The force acting on an element of length dx at position x is ()()m m2 kPa dx x dA dF ××==σ The resultant force is then, from Eqn. 3.1.2 1 the weight of the ball is neglected here 2 the radius of which depends on the force applied and the materials in contact 1σ2σF (a) (c)F (b)Fcontact region Section 3.1 Solid Mechanics Part I Kelly 31() kN4 mkPa 222 0= ==∫∫xdx dF F A The moment of the stress distribution is given by ∫∫×== A AdAl dM M σ0 (3.1.3) where l is the length of the moment -arm from the chosen axis. Taking the axis to be at 0=x , the moment-arm is xl=, Fig. 3.1.3b, and () mkN316mkPa 232 00 = ×==∫∫= dxxx dM M Ax Taking moments about the right-hand end, 2=x , one has ()() mkN38mkPa 2 232 02 −= −×−==∫∫= dxx x dM M Ax Figure 3.1.3: a non-uniform stress acting over a surface; (a) the stress distribution, (b) stress acting on an element of size d x ■ 3.1.2 Equivalent Forces and Moments Stress distributions can be replaced by equiva lent forces, i.e. forces equal to the resultant force of the distribution and whic h also give the same moment about any axis as the distribution. Fo rmulae for equivalent forces are derived in what follows for triangular and arbitrary linear stress distributions. Triangular Stress Distribution x m2)(xσ dx)(xσ x (a) (b) Section 3.1 Solid Mechanics Part I Kelly 32Consider the triangular stress distribution s hown in Fig. 3.1.4. The stress at the end is 0σ, the length of the distribution is L and the thickness “into the page” is t. The equivalent force is , from Eqn. 3.1.2, Lt dxLxtFL 0 0021σ σ= =∫ (3.1.4) which is just the average stress times area. The point of action of this force should be such that the moment of the force is equivalent to the moment of the stress distribution. Taking moments about the left hand end, for the distribution one has, from 3.1.3, tL dxxxt ML 2 0 0o31)(σ σ= =∫ Placing the force at position cxx=, the moment of the force is ()cx Lt M 2/0 oσ= . Equating these expressions leads to the posit ion at which the equivalent force acts. L xc32= (3.1.5) Figure 3.1.4: triangular stress distribution and equivalent force ■ Arbitrary Linear Stress Distribution Consider the linear stress dist ribution shown in Fig. 3.1.5. The stress at the ends are 1σ and 2σ and this time the equivalent force is [] () 2/ )/)( (2 1 01 2 1 σσ σσσ += −+=∫Lt dxLx tFL (3.1.6) Taking moments about the left hand end, for the distribution one has 0σ Loequivalent force )(xσ cxx Section 3.1 Solid Mechanics Part I Kelly 33() 6/ 2 )(2 12 0o σσ σ += =∫tL dxxxt ML The moment of the force is ()2/2 1 o cx Lt M σσ+= . Equating these expressions leads to () ()2 12 1 32 σσσσ ++=Lxc (3.1.7) Eqn. 3.1.5 follows from 3.1.7 by setting 01=σ . Figure 3.1.5: a non-uniform stress di stribution and equivalent force ■ The Centroid Generalising the above cases, the line of ac tion of the resultant force for any arbitrary stress distribution )(xσ is FdFx dxx tdxxxt xc∫ ∫∫= = )()( σσ Centroid (3.1.8) This location is known as the centroid of the distribution. The forces considered thus far are normal forces, where the force acts perpendicular to a surface, and they give rise to normal stresses . Normal stresses are also called pressures when they are compressive as in Figd. 3.1.1-2. Note also that the most of the discussion above was for two-dimensional cases, i.e. the stress was assumed constant “into the pag e”. Three dimensional problems can be tackled in the same way, only now one must integrate two-dimensionally over a surface rather than one-di mensionally over a line. 3.1.3 Shear Stress 1σ2σ Loequivalent force )(xσ cx Section 3.1 Solid Mechanics Part I Kelly 34Consider now the case of shear forces, that is, forces which act tangentially to surfaces. A normal force F acts on the block of Fig. 3.1.6a . The block does not move and, to maintain equilibrium, the force is resisted by a friction force mg F μ= , where μ is the coefficient of friction. A free body diagram of the block is shown in Fig. 3.1.6.b. Assuming a uniform distribution of stress, th e stress and resultant force arising on the surfaces of the block and underlying object are as shown. The stresses are in this case called shear stresses . Figure 3.1.6: shear stress; (a) a force acting on a bloc k, (b) shear stresses arising on the contacting surfaces 3.1.4 Combined Normal and Shear Stress Forces acting inclined to a surface are most conveniently described by decomposing the force into components normal and tangent ial to the surface unde r consideration. Then one has both normal stress Nσ and shear stress Sσ, as in Fig. 3.1.7. Figure 3.1.7: a force F giving rise to norm al and shear stress over the contacting surfaces The stresses considered in th is section are examples of surface stresses or contact stresses . They arise when materials meet at a common surface. Other examples would be sea-water pressurising a material in deep water and the stress exerted by a train wheel on a train track. 3.1.5 Problems F (a) (b)F F F Nσθ Sσ Section 3.1 Solid Mechanics Part I Kelly 351. Consider the surface shown below, of length 4cm and unit depth (1cm into the page). The stress over the surface is given by x+=2σ kPa, with x measured in cm from the surface centre . (a) Evaluate the resultant force acting on the surface (in Newtons). (b) What is the moment about an axis (i nto the page) through the left-hand end of the surface? (c) What is the moment about an axis (into the page) through the centre of the surface? 2. Consider the surface shown below, of length 4mm and unit depth (1mm into the page). The stress over the surface is given by x=σ MPa, with x measured from the surface centre. What is the total fo rce acting on the surfa ce, and the moment acting about the centr e of the surface? 3. Find the reaction forces (per unit length) at the pin and roller for the following beam, which is subjected to a varyin g pressure distribu tion, the maximum pressure being kPa 20)(=xσ (all lengths are in cm – give answer in N/m) [Hint: first replace the stress distribu tion with three equivalent forces] 4. A block of material of width 10cm and length 100cm is pushed into an underlying substrate by a normal force of 100 N. It is found that a unif orm triangular normal x 4)(xσy x 4y • 4 16 441 2 8 Section 3.1 Solid Mechanics Part I Kelly 36stress distribution arises at the contacting surfaces, that is, the stress is maximum at the centre and dies off linearly to zero at the block edges, as sketched below right. What is the maximum pressure acting on the surface? σN100 cm10cm100typical cross- section stress distribution Section 3.2 Solid Mechanics Part I Kelly 373.2 Body Forces Surface forces act on surfaces. As discussed in the previous section, these are the forces which arise when bodies are in c ontact and which give rise to stress distributions. Surface forces also arise inside materials, acting on internal surfaces, Fig. 3.2.1a, as will be discussed in the following section. To complete the description of forces acting on real materials, one needs to deal with forces which act at a distance , for example the force of gravity. For these, one can define the body force , which acts on volume elements of material. Fig. 3.2.1b shows a sketch of a volume element subjected to a magnetic body force and a gravitational body force gF. Figure 3.2.1: forces acting on a body; (a) su rface forces acting on surfaces, (b) body forces acting on a material volume element 3.2.1 Weight The most important body force is the force due to gravity, i.e. the weight force. In Chapter 2 there were examples involving the wei ght of components. In those cases it was simply stated that the weight could be taken to be a single force acting at the component centre (for example, Problem 3 in §2.2.3). This is true when the component is symmetrical, for example, in the shape of a circ le or a square. However, it is not true in general for a component of arbitrary shape. The weight of a small volume element VΔ of material of density ρ is Vg dFΔ=ρ and the total weight is dVg F V∫=ρ (3.2.1) Consider the general two-dimensional case, Fig. 3.2.2, where material elements of area iAΔ (and constant thickness t) are subjected to forces i i Agt FΔ=Δρ . gFF e.g. air pressure internal surface contact force (a) (b) F Fvolume elemen t Section 3.2 Solid Mechanics Part I Kelly 38 Figure 3.2.2: Resultan t Weight on a body The resultant weight force due to all elements , for a component with uniform density, is gtA dAgt dF F ρρ= ==∫∫, where A is the cross-sectional area. The resultant moments about the x and y axes, which can be positioned anywhere in the body, are ∫= ydAgt Mxρ and ∫= xdAgt Myρ ; the moment xMΔ is shown in Fig. 3.2.3. The equivalent weight force is thus positioned as ) ,(c cyx , Fig. 3.2.2, where AydA yAxdA xc c∫∫= = , Centroid of Area (3.2.2) The position ) ,(c cyx is called the centroid of the area . The quantities ∫xdA, ∫ydA, are called the first moments of area about, respectively, the y and x axes. Figure 3.2.3: The moment Mx; (a) full view, (b) plane view iAΔ y iAgtΔρxy yz oiAΔ iAgtΔρyxMΔ (a) (b)i i AgtFΔ=Δρxy •()c cyx, gFiAΔ tz Section 3.2 Solid Mechanics Part I Kelly 393.2.2 Problems 1. Where does the resultant force due to grav ity act in the component shown below? (Gravity acts downward in the direction of the arrow shown, perpendicular to the component’s surface) o90 m1m1 Section 3.3 Solid Mechanics Part I Kelly 403.3 Internal Stress The idea of stress considered in §3.1 is not difficult to conceptualise since objects inte racting with other objects are encountered all around us. A more difficult concept is the idea of forces and stresses acting inside a material, “within the interior where neither eye nor experiment can reach” as Euler put it. It took many great minds working for centuries on this question to arrive at the concept of stress we use today, an idea finally brought to us by Augustin Cauchy, who presented a paper on the subject to the Academy of Sciences in Paris, in 1822. Augustin Cauchy 3.3.1 Cauchy’s Concept of Stress Uniform Internal Stress Consider first a long slender block of ma terial subject to equilibrating forces F at its ends, Fig. 3.3.1a. If the complete block is in equili brium, then any sub-division of the block must be in equilibrium also. By imagining the block to be cut in two, and considering free-body diagrams of each half, as in Fig. 3.3.1b, one can see that forces F must be acting within the block so that each half is in equilibrium. Thus external loads create internal forces ; internal forces represent the action of one part of a material on another part of the same material. If the material out of which the block is made is uniform over this cut, one can take it that a uniform stress AF/=σ acts over this interior surface, Fig. 3.3.1b. Figure 3.3.1: a slender block of material; (a) under the action of external forces F, (b) internal normal stress σ, (c) internal normal and shear stress FFFF )a() b( )c(F F FF AF=σNσ SσFimaginary cut F FF FF Section 3.3 Solid Mechanics Part I Kelly 41Note that, if the internal forces were not act ing over the internal surfaces, the two half- blocks of Fig. 3.3.1b would fly apart; one can th us regard the intern al forces as those required to maintain material in an un-cut state. If the internal surface is at an incline, as in Fig. 3.3.1c, then the internal force required for equilibrium will not act normal to the surface. There will be components of the force normal and tangential to the surface, and thus both normal ( Nσ) and shear (Sσ) stresses must arise. Thus, even though the material is subjected to a purely normal load, shear stresses develop. From Fig. 3.3.2a, the normal and shear stresses ar ising on an interior surface inclined at angle θ to the horizontal are { ▲Problem 1} θθ σθ σ cos sin , cos2 AF AF S N = = (3.3.1) Figure 3.3.2: stress on inclined surface; (a) decomposing the force into normal and shear forces, (b) stress at an internal point Although stress is associated w ith surfaces, one can speak of the stress “at a point”. For example, consider some point interior to the block, Fig 3.3.2b. The stress there evidently depends on which surface through that point is under consideration. From Eqn. 3.3.1a, the normal stress at the point is a maximum AF/ when 0=θ and a minimum of zero when o90=θ . The maximum normal stress arising at a point within a material is of special significance, for example it is this st ress value which often determines whether a material will fail (“break”) there. It has a special name: the maximum principal stress . From Eqn. 3.3.1b, the maximum shear stress at the point is A F2/± and arises on surfaces inclined at o45± . Non-Uniform Internal Stress The example illustrated above is straight forward. One can imagine a more complex geometry under a more complex loading, as in Fig. 3.3.3. Again, using equilibrium arguments, there will be some stress distributi on acting over any given internal surface. F SFNF θθ A FF internal point • internal surface θ )a() b( Section 3.3 Solid Mechanics Part I Kelly 42To evaluate these stresses is not an easy matte r, and much of Part II is devoted to doing just that. Suffice to say here that they will invariably be non-uniform over a surface, that is, the stress at some particle will differ from the stress at a neighbouring particle. Figure 3.3.3: a component subjected to a co mplex loading, giving rise to a non- uniform stress distribution over an internal surface Traction and the Physical M eaning of Internal Stress All materials have a complex molecular micros tructure and each molecule exerts a force on each of its neighbours. The complex in teraction of countless molecular forces maintains a body in equilibrium in its unstre ssed state. When th e body is disturbed and deformed into a new equilibrium position, net forces act, Fig. 3.3.4a. An imaginary plane can be drawn through the material, Fig. 3.3.4b. Unlike some of his predecessors, who attempted the impossible of accounting for a ll the molecular forces, Cauchy discounted the molecular structure of ma tter and replaced the molecular forces acting on the plane by a single force F, Fig 3.3.4c. This is the force ex erted by the molecules above the plane on the material below the plane and can be attrac tive or repulsive. Different planes can be taken through the same portion of material and, in general, a different force will act on the plane, Fig 3.3.4d. Figure 3.3.4: a multitude of molecular fo rces represented by a single force The definition of stress will now be made more precise. First, define the traction at some particular point in a material as follows: take a plane of surface area S through the point, on which acts a force F. Next shrink the plane – as it shrinks in size both S and F get smaller, and the direction in which the for ce acts may change, but eventually the ratio SF/ will remain constant and the force will act in a particular direction, Fig. 3.3.5. The limiting value of this ratio of for ce over surface area is defined as the traction vector (or stress vector ) t: )a() b() c() d(F F1F2F3F 4F Nσ Sσ Section 3.3 Solid Mechanics Part I Kelly 43 SF SΔΔ= →Δ 0limt ( 3 . 3 . 2 ) Figure 3.3.5: the traction vector - the limit ing value of force over area, as the surface area of the element on whic h the force acts is shrunk An infinite number of traction vectors act at any single point, since an infinite number of different planes pass through a point. Thus the notation S FSΔΔ→Δ / lim0 is ambiguous. For this reason the plane on which the traction vector acts must be specified; this can be done by specifying the normal n to the surface on which the traction acts, Fig 3.3.6. The traction is thus a special vect or – associated with it is no t only the direction in which it acts but also a second dir ection, the normal to the plane upon which it acts. Figure 3.3.6: two different traction vectors acting at the same point Stress Components The traction vector can be decomposed into components which act normal and parallel to the surface upon which it acts. These components are called the stress components , or simply stresses , and are denoted by the symbol σ; subscripts are added to signify the surface on which the stresses act and the directions in which the stresses act. same point with different planes passing through it (defined by different normals) 1n2n SSΔΔ= →ΔFtn 0)(lim1different forces act on different planes through the same point SΔFΔ 1n SSΔΔ= →ΔFtn 0)(lim2SΔFΔ 2nSFSF FΔ SΔa plane passing through some point in the material Section 3.3 Solid Mechanics Part I Kelly 44 Consider a particular traction vector acti ng on a surface element. Introduce a Cartesian coordinate system with base vectors kji,, so that one of the base vectors is a normal to the surface, and the origin of th e coordinate system is positio ned at the point at which the traction acts. For example, in Fig. 3.3.7, the k direction is taken to be normal to the plane, and kj i tk z y x t tt++=)(. Figure 3.3.7: stress components – the components of the traction vector Each of these components it is represented by ijσ where the first subscript denotes the direction of the normal to the plane and the second denotes the direction of the component . Thus, in Fig. 3.3.7, k j i tk zz zy zx σσσ ++=)(. The first two stresses, the components acting tangential to the su rface, are shear stresses, whereas zzσ, acting normal to the plane, is a normal stress1. Sign Convention for Stress Components The following convention is used: The stress is positive when the direction of the normal and the direction of the stress component are both positive or both negative The stress is negative when one of the directions is positive and the other is negative According to this convention, the three stresses in Fig. 3.3.7 are all positive. The sign convention makes sense for this reason : the traction vector shown in Fig. 3.3.7 represents the force (per unit area) ex erted by the material above the surface on the material below the surface. By Newton’s th ird law, an equal and opposite traction must be exerted by the material below the surface on the material above the surface, as shown in Fig. 3.3.8. If )(kt has positive stress components, then so should )(kt−. 1 this convention for the subscripts is not universa lly followed. Many authors, particularly in the mathematical community, use the exact opposite convention, the first subscript to denote the direction and the second to denote the normal. It turns out that both conventions are equivalent , since, as will be shown later, the stress is symmetric, i.e. ji ijσσ= )(kt y x)ˆ(nt ijk zyσ zxσzzσz Section 3.3 Solid Mechanics Part I Kelly 45 Figure 3.3.8: equal and opposite traction vectors – each with the same stress components Looking at the two-dimensional case for ease of visualisation, the (positive and negative) normal stresses and shear stresses on either side of a surface are as shown in Fig. 3.3.9. Note that the shear stresses go in opposite directions. Figure 3.3.9: stresses acting on either side of a material surface: (a) positive stresses, (b) negative stresses Examples of negative stresses are shown in Fig. 3.3.10 { ▲Problem 4}. Figure 3.3.10: examples of negative stress components )(kt y xz ijk zyσzxσzzσ )(kt−zzσzyσzxσ )(jt yyσ yxσ yzσ2e )(jt− yyσyxσyzσ k j i tj yz yy yx σσσ ++=−)(k j i tj yz yy yx σσσ ++=)(yx zij k xy zij k )a() b()a() b(xyyyσ yyσyyσ yyσyxσ yxσyxσ yxσ Section 3.3 Solid Mechanics Part I Kelly 46 3.3.2 Real Problems and Saint-Venant’s Principle Some examples have been given earlier of exte rnal forces acting on materials. In reality, an external force will be applied to a real material component in a complex way. For example, suppose that a block of material, welded to a large object at one end, is pulled at its other end by a rope attached to a metal hoop, which is itself attached to the block by a number of bolts, Fig. 3.3.11a. The block can be idealised as in Fig 3.3.11b; here, the precise details of the region in which the external force is applied are neglected. Figure 3.3.11: a block subjected to an extern al force: (a) real case, (b) ideal model, (c) stress in ideal model, (d) stress in ac tual material, (e) stress in real material modelled well by either (f) or (g) According to the earlier discussion, the stress in the ideal model is as in Fig. 3.3.11c. One will find that, in the real material, the stress is indeed (approximately) as predicted, but only at an appreciable distance from the right hand end. Near where the rope is attached, the force will differ considerably , as sketched in Fig.3.3.11d. Thus the ideal models of the type discussed in this section, and in much of this book, are useful only in predicting the stress field in real components in regions away from points of application of loads. Th is does not present too much of a problem, since the stresses internal to a structure in such regions are often of most interest. If one wants to know what happens near the bolted connection, then one will have to create a complex model incorporating all the details and the problem will be much more difficult to solve. It is an experimental fact that if two different force sy stems are applied to a material, but they are equivalent force systems, as in Fig. 3.3.11(f,g), then the stress fields in regions away from where the loads are applied will be the same. This is known as Saint- Venant’s Principle . Typically, one needs to move a distance away from where the loads are applied equal to the distance over which the loads are applied. )a( )b( )c(F F F AF/=σ )d( F)e( )f(F F )g(2/F2/Fstress differs here stress the same Section 3.3 Solid Mechanics Part I Kelly 473.3.3 Problems 1. Derive Eqns. 3.3.1. 2. The four sides of a square block are subjected to equal forces S, as illustrated. The length of each side is l and the block has unit depth (into the page). What normal and shear stresses act along the (dotted) diagonal? 3. A shaft is dug into the ground. A thick st eel rope is looped around the shaft and a force is applied normal to the shaft, as show n. The shaft is in static equilibrium. Draw a free body diagram of the shaft (fro m the top down to ground level) showing the forces/moments acting on th e shaft (ignore the weight of the shaft). Draw a free body diagram of the section of shaft from the top down to the cross section at A. Draw a free body diagram of the section of shaft from the top down to the cross section at B. Roughly sketch the stresses acting over the (horizontal) internal surfaces of the shaft at A and B. 4. In Fig. 3.3.10, which of the stress components is/are negative? 5. Label the following stress component acting on an internal material surface. Is it positive or negative? SS S S x y z acting parallel to surfacegroundF AB Section 3.3 Solid Mechanics Part I Kelly 486. Label the following shear stresses. Are they positive or negative? 7. Label the following normal stresses. Are they positive or negative? y xzy xz Section 3.4 Solid Mechanics Part I Kelly 493.4 Equilibrium of Stress Consider two perpendicular planes passing through a point p. The stress components acting on these planes are as shown in Fig. 3.4.1a. These stresses are usually shown together acting on a small material element of finite size, Fig. 3.4.1b. It has been seen that the stress may vary from point to point in a material but, if the element is very small, the stresses on one side can be taken to be equa l to the stresses acti ng on the other side. Figure 3.4.1: stress components acting on tw o perpendicular planes through a point; (a) two perpendicular surfaces at a point, (b) small material element at the point It will be shown below that the stress components acting on any other plane through p can be evaluated from a knowledge of only these stress components. 3.4.1 Symmetry of the Shear Stress Consider the material element shown in Fi g. 3.4.1b, reproduced in Fig. 3.4.2a below. The element has dimensions is y xΔ×Δ and is subjected to arbitrary uniform stresses over its sides. The resultant forces of the stresses acting on each side of the element act through the side-centres, and are shown in Fig. 3.4.2b. The stresses shown are positive, but note how positive stresses can lead to negative forces, depending on the definition of the yx− axes used. The resultant force on the co mplete element is seen to be zero. yyσyyσ xxσxxσyxσ yxσxyσ xyσ )a() b(xy •p xxσyxσ xxσ yxσ yyσxyσyyσ xyσ Section 3.4 Solid Mechanics Part I Kelly 50 Figure 3.4.2: stress components acting on a material element; (a) stresses, (b) resultant forces on each side By taking moments about any point in the block, one finds that { ▲Problem 1} yx xyσσ= (3.4.1) Thus the shear stresses acting on the elemen t are all equal, and for this reason the yxσ stresses are usually labelled xyσ, Fig. 3.4.3a, or simply labelled τ, Fig. 3.4.3b. Figure 3.4.3: shear stress acting on a material element 3.4.2 Three Dimensional Stress The three-dimensional counterpart to the tw o-dimensional element of Fig. 3.4.2 is shown in Fig. 3.4.4. xxσyxσ xxσ yxσ yyσxyσyyσ xyσ xΔyΔy Fxx xΔ−=σx Fyx xΔ+=σx Fyy yΔ+=σ )a() b(y Fxy yΔ−=σ x Fyx xΔ−=σ x Fyy yΔ−=σy Fxx xΔ+=σy Fxy yΔ+=σ ττ τ τ xyσxyσxyσ xyσ )a() b( Section 3.4 Solid Mechanics Part I Kelly 51 Figure 3.4.4: a three dimensional material element Moment equilibrium in this case requires that zy yz zx xz yx xy σσσσσσ = = = , , (3.4.2) The nine stress components, six of which are independent, can be conveniently written in the matrix form [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = zz zy zxyz yy yxxz xy xx ij σσσσσσσσσ σ (3.4.3) A vector F has one direction associated with it and is characterised by three components ),,(z y x FFF . The stress is a quantity which has tw o directions associated with it (the direction of a force and the normal to th e plane on which the force acts) and is characterised by the nine components of Eqn. 3.4.3. Such a mathematical object is called a tensor . Just as the thre e components of a vector change with a change of coordinate axes (for example, as in Fig. 2. 2.1), so the nine components of the stress tensor change with a change of axes. This is discussed in the next section for th e two-dimensional case. 3.4.3 Stress Transformation Equations Assume that the stress components of Fig. 3.4. 5a are known. It is required to find the stresses arising on ot her planes through p. Consider the perpendicular planes shown in Fig. 3.4.5b, obtained by rotating the original element through a positive (counterclockwise) angle θ. The new surfaces are defined by the axes yx′−′ . xyσ xxσyzσyxσyyσ zxσzyσ zzσxzσ xy z Section 3.4 Solid Mechanics Part I Kelly 52 Figure 3.4.5: stress components acting on two different sets of perpendicular surfaces, i.e. in two different coordinate systems; (a) original system, (b) rotated system To evaluate these new stress components, consid er a triangular element of material at the point, Fig. 3.4.6. Carrying out fo rce equilibrium in the direction x′, one has θτθτθ σθ σσ cos sin sin cos : OA OB OA OB AB Fyy xx xx x + + + =′∑′ (3.4.4) Since θ θ sin , cos AB OA AB OB = = , and dividing through by AB, θτθσθσσ 2sin sin cos2 2+ + =′yy xx xx (3.4.5) Figure 3.4.6: a free body diagram of a triangular element of material The forces can also be resolved in the y′ direction and one obtains the relation )a() b(xyxxστ xxσ yyσyyσ x′y′xxσ′yyσ′ xxσ′ yyσ′θττ′ τ′ x′y τ′ xxσ′y′ xθ θB Axxσ τ τ yyσO Section 3.4 Solid Mechanics Part I Kelly 53 θτθθσστ 2cos cos sin) ( + −=′xx yy (3.4.6) Body force terms can be included in the above calculation, but they tend to zero as the element is shrunk in size down to the vertex o. Finally, consideration of the element in Fig. 3.4.7 yields two furt her relations, one of which is the same as Eqn. 3.4.6. Figure 3.4.7: a free body diagram of a triangular element of material In summary, one obtains the stress transformation equations : xy xx yy xyxy yy xx yyxy yy xx xx θσ σσθθσθσ θσ θσ σθσ θσ θσ σ 2cos) ( cos sin2sin cos sin2sin sin cos 2 22 2 +− =′− + =′+ + =′ 2D Stress Transformation Equations (3.4.7) These equations have many uses, as will be seen in the next section. In matrix form, ⎥⎦⎤ ⎢⎣⎡− ⎥ ⎦⎤ ⎢ ⎣⎡ ⎥⎦⎤ ⎢⎣⎡ −=⎥ ⎦⎤ ⎢ ⎣⎡ ′′′′ θθθθ σσσσ θθθθ σσσσ cos sinsin cos cos sinsin cos yy yxxy xx yy yxxy xx (3.4.8) Transformation equations can also be derived for three-dimensional stress states, but the expressions are lengthy – they will be discussed in Parts II and III. θ θ ox′y y′ xτ′yyσ′ xxσ τ yyστ Section 3.4 Solid Mechanics Part I Kelly 543.4.4 Problems 1. Derive Eqns. 3.4.1 by taking moments about the lower left corner of the block in Fig. 3.4.2. 2. Suppose that the stresses acting on two pe rpendicular planes through a point are []⎥⎦⎤ ⎢⎣⎡ −−=⎥ ⎦⎤ ⎢ ⎣⎡=111 2 yy yxxy xx ijσσσσ σ Use the stress transformation formulae to evaluate the stre sses acting on two new perpendicular planes through the point, obt ained from the first set by a positive rotation of 30 ¡. Use the conventional notation yx′−′ to represent the coordinate axes parallel to these new planes. Section 3.5 Solid Mechanics Part I Kelly 553.5 Plane Stress This section is concerned with a special two-dimensional state of stress called plane stress . It is important for two reasons: (1) it has practical application in the analysis of thin components and (2) it is a two dimensional state of stress, and thus serves as an excellent introduction to more complicated three dimensional stress states. As discussed further below, plane stress arises in many situations, but particularly in thin materials. First, consid er stress boundary conditions. 3.5.1 Stress Boundary Conditions When solving problems, information is usually available on what is happening at the boundaries of materials. This information is called the boundary conditions . Information is usually not available on what is happening in th e interior of the material – information there is obtained by solving the equations of mechanics. A number of different conditions can be kn own at a boundary, for example it might be known that a certain part of the boundary is fixe d so that the displacements there are zero. This is known as a displacement boundary condition On the other hand the stresses over a certain part of the material boundary might be known. These are known as stress boundary conditions – this case will be examined here. General Stress Boundary Conditions It has been seen already that, when one mate rial contacts a second material, a force, or distribution of stress arises. This force F will have arbitrary direction, Fig. 3.5.1a, and can be decomposed into the sum of a normal stress distribution Nσ and a shear distribution Sσ, Fig. 3.5.1b. One can introduce a coordinate system to describe the applied stresses, for example the yx− axes shown in Fig. 3.5.1c. Figure 3.5.1: Stress boundary conditions; (a ) force acting on material due to contact with a second material, (b) the resulting normal and shear stress distributions, (c) applied stresses as stress componen ts in a given coordinate system SσF Nσ contact region (a) (b) (c)xyσ yyσy x Section 3.5 Solid Mechanics Part I Kelly 56Figure 3.5.2 shows the same component as Fig. 3.5.1. Shown in detail is a small material element at the boundary. From equilibrium of the element, stresses yy xyσσ, , equal to the applied stresses, must be acting inside the material, Fig. 3.5.2a. Note that the tangential stresses , which are the xxσ stresses in this example, can take on any value and the element will still be in equilibrium with the applied stresses, Fig. 3.5.2b. Figure 3.5.2: Stresses acting on a material element at the boundary, (a) normal and shear stresses, (b) tangential stresses Thus, if the applied stresses are known , then so also are the normal and shear stresses acting at the boundary of the material. This can be summarised as follows: Stress boundary conditions: Stress boundary conditions involve the normal and shear stresses acting on a surface Stress Boundary (Interface) Conditions between Two Materials Consider now in more detail a surface between tw o different materials, Fi g. 3.5.3. One says that the normal and shear stresses are continuous across the surface, as illustrated. Figure 3.5.3: normal and shear stress cont inuous across an interface between two different materials, material ‘1’ and material ‘2’ xyσ yyσ yyσ xyσxxσ xxσ (a) (b) 2 1 )2( xyσ)2( yyσ )1( xyσxy )2( )1()2( )1( xy xyyy yy σσσσ == )1( yyσ Section 3.5 Solid Mechanics Part I Kelly 57 Note also that, since the shear stress xyσ is the same on both sides of the surface, the shear stresses acting on both sides of a perpendicular plane passing through the interface between the materials, by the symmetry of stress, must also be the same, Fig. 3.5.4a. Figure 3.5.4: stresses at an interface; (a) shear stresses continuous across the interface, (b) tangential stresses not necessarily continuous However, again, the tangential stresses, th ose acting parallel to the interface, do not have to be equal. For example, shown in Fig. 3.5.4b are the tangential stresses acting in the upper material, )2( xxσ - they balance no matter what the magnitude of the stresses )1( xxσ. Stress Boundary Conditions at a Free Surface A free surface is a surface that has “nothing” on one side and so there is nothing to provide reaction forces. Thus there must also be no normal or shear stress on the other side (the inside). This leads to the following, Fig. 3.5.5: Stress boundary conditions at a free surface : the normal and shear stress at a free surface are zero This simple fact is used again and again to solve practical problems. Again, the stresses acting normal to any other plane at the surface do not have to be zero – they can be balanced as, for example, the tangential stresses Tσ and the stress σ in Fig. 3.5.5. xyσ)2( xxσ )1( xxσ )a( )b(xyσ Section 3.5 Solid Mechanics Part I Kelly 58 Figure 3.5.5: A free surface - the norm al and shear stresses there are zero Atmospheric Pressure There is something acting on the outside “free” surfaces of materials – the atmospheric pressure. This is a type of stress which is hydrostatic , that is, it acts normal at all points, as shown in Fig. 3.5.6. Also, it does not vary much. This pressure is present when one characterises a material, that is, when its material properties are determined from tests and so on, for example, its Young’s Modulus (see Chapte r 4). The atmospheric pressure is therefore a datum – stresses are really measured relative to this value, and so the atmospheric pressure is ignored. Figure 3.5.6: a material subjected to atmospheric pressure 3.5.2 Description of Boundary Conditions The following example brings together the notions of stress boundary conditions, stress components, equilibrium and equivalent forces. Tσ 0==S NσσTσ Tσ Sσ Nσ σ σ Section 3.5 Solid Mechanics Part I Kelly 59Example Consider the plate shown in Fig. 3.5.7. It is of width a2, height b and depth t. It is subjected to a tensile stress r, pressure p and shear stresses s. The applied stresses are uniform through the thickness of the plate. It is welded to a rigid base. Figure 3.5.7: a plate subjected to stress distributions Using the yx− axes shown, the stress boundary conditions can be expressed as: Left-hand surface: ⎩⎨⎧ −=−−=− s yap ya xyxx ),(),( σσ, by<<0 Top surface: ⎩⎨⎧ −=+= s bxr bx xyyy ),(),( σσ , a xa+<<− Right-hand surface: ⎩⎨⎧ −=+=+ s yaya xyxx ),(0),( σσ, by<<0 Note carefully the description of the normal and shear stresses over each side and the signs of the stress components. The stresses at the lower edge are unknown (there is a displacement boundary condition there: zero displacement). They will in general not be uniform. Using the given yx− axes, these unknown reaction stresses, exerted by the base on the plate, are (see Fig 3.5.8) Lower surface: ⎩⎨⎧)0,()0,( xx xyyy σσ, a xa+<<− xy b a2s ps sr Section 3.5 Solid Mechanics Part I Kelly 60 Note the directions of the arrows in Fig. 3.5. 8, they have been drawn in the direction of positive )0,( ),0,( x xxy yyσσ . Figure 3.5.8: unknown reaction stresses acting on the lower edge For force equilibrium of the complete plate, c onsider the free-body diagram 3.5.9; shown are the resultant forces of the stress distri butions. Force equilibrium requires that 0 )0,( 20 )0,( 2 = −== −−= ∫ ∑∫ ∑ + −+ − dxx t art Fdxx t ast bpt F a ayy ya axy x σσ Figure 3.5.9: a free-body diagram of the plat e in Fig. 3.5.7 showing the known resultant forces (forces on the lower boundary are not shown) For moment equilibrium, consider the moment s about, for example, the lower left-hand corner. One has () 0 )0,( )2( )(2)(2)2/(0 =+× −−++ −= ∫ ∑+ −dxxa x ta bstaart bast bbpt Ma ayyσ )0,(xxyσ )0,(xyyσ xy ast2art2 bst bstbpt Section 3.5 Solid Mechanics Part I Kelly 61If one had taken moments about the top-left corner, the equation would read () 0 )0,( )0,()2( )(2)2/(0 =+× −× −−+ += ∫ ∫∑ + −+ −dxxa x t dxb x ta bstaart bbpt M a ayya axy σ σ ■ 3.5.3 Plane Stress The state of plane stress is defined as follows: Plane Stress : If the stress state at a material particle is such that the only non-zero stress components act in one plane only, the particle is said to be in plane stress. The axes are usually chosen such that the yx− plane is the plane in which the stresses act, Fig. 3.5.10. Figure 3.5.10: non-zero stre ss components acting in the x – y plane An example of a state of plane stress is that in the previous example. Thin Components Consider a thin components as shown in Fig. 3.5.11. With the coordinate axes aligned as shown, one has 0===zz zy zxσσσ . Strictly speaking, these stresses are zero only at the free surfaces of the material but, b ecause it is thin, the stresses can be assumed not to vary much from zero within, and in fact are taken to be identically zero throughout the material1. In this 1 it will be shown in Part II that, when the applied stresses xy yy xxσσσ ,, vary only linearly over the thickness of the component, the stresses zy zx zzσσσ ,, are exactly zero throughout the component, otherwise they are only approximately zero yyσ xy xxσxyσ xyσyyσ Section 3.5 Solid Mechanics Part I Kelly 62sense, plane stress conditions are only approxima tely true. On the other hand, were the sheet not so thin the stress components that were zero at the free-surfaces might well deviate significantly from zero deep within the material. Figure 3.5.11: a thin material loaded in-plane, leading to a state of plane stress When analysing plane stress states, only one cross section of the material need be considered. This is illustrated in Fig. 3.5.12. Figure 3.5.12: one two-dimensional cross-section of material 3 – dimensional material 2 – dimensional cross-section of material xy zxy•xyσyyσ xxσxy Section 3.5 Solid Mechanics Part I Kelly 63In plane stress, the matrix of stress components reduces from 33× to 22×: ⎥⎦⎤ ⎢⎣⎡→⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ yy yxxy xx yy yxxy xx σσσσσσσσ 0 0 000 (3.5.1) Note that, although the stress normal to the plane, zzσ, is zero, the three dimensional sheet of material is deforming in this direction – it will obviously be getting thinner under the tensile loading shown in Fig. 3.5.12. Note also that plane stress arises in all thin materials, no matter what they are made of. Example (of a thick component in a state of plane stress) A material does not have to be thin to be in a state of plane stress. For example, the thick block of uniform material shown in Fig. 3.5.13, loaded by a constant stress oσ in the x direction, will have 0σσ=xx and all other components zero everywhere. Figure 3.5.13: a thick block of material in plane stress 3.5.4 Analysis of Plane Stress Next are discussed the stress invariants , principal stresses and maximum shear stresses for the two-dimensional plane state of stress, a nd tools for evaluating them. These quantities are useful because they tell us the complete state of stress at a point in simple terms. Further, these quantities are directly related to the strength and response of materials. For example, the way in which a material plastically (per manently) deforms is often related to the maximum shear stress, the directions in which fl aws/cracks grow in materials is often related to the principal stresses, and the energy stored in materials is often a function of the stress invariants. oσ xy z Section 3.5 Solid Mechanics Part I Kelly 64Stress Invariants A stress invariant is some function of the stress which is independent of the coordinate system being used. In a two dimensional space there are two stress invariants, labelled 1I and2I. These are 2 21 xy yy xxyy xx II σσσσσ −=+= Stress Invariants (3.5.2) These quantities can be proved to be invariant directly from the stress transformation equations, Eqns. 3.4.7 { ▲Problem 4}. Physically, invariance of 1I and2I means that they are the same for any chosen perpendicula r planes through a material particle. Combinations of the stress invariants are also invariant, for example the important quantity ()2 2 41 22 141 121 2xy yy xxyy xxI I I σσσσσ+−±+=−± (3.5.3) Principal Stresses Consider a material particle for wh ich the stress, with respect to an yx− coordinate system, is ⎥⎦⎤ ⎢⎣⎡ −−=⎥⎦⎤ ⎢⎣⎡ 111 2 yy yxxy xx σσσσ (3.5.4) The stress acting on different planes throug h the point can be evaluated using the stress transformation equations, Eqns. 3.4.7, and the resu lts are plotted in Fig. 3.5.14. The original planes are re-visited after rotating 180 ¡. Section 3.5 Solid Mechanics Part I Kelly 65 Figure 3.5.14: stresses on different planes through a point It can be seen that there are two perpendicular planes for which the shear stress is zero, for θ ≈ 58¡ and θ ≈ (58 + 90) ¡. In fact it can be proved that fo r every point in a material there are two (and only two) perpendicular planes on which the shear stress is zero (see below). These planes are called the principal planes . It will also be noted from the figure that the normal stresses acting on the planes of zero shear stress are either a maximum or minimum. Again, this can be proved (see below) –. These normal stresses are called principal stresses. The principal stresses are labelled 1σ and 2σ, Fig. 3.5.15. Figure 3.5.15: principal stresses The principal stresses can be obtained by setting 0=′xyσ in the stress transformation equations, Eqns. 3.4.7, which leads to the value of θ for which the planes have zero shear stress: yy xxxy σσσθ−=22tan Location of Principal Planes (3.5.5) For the example stress state, Eqn. 3.5.4, this leads to ()2 arctan21− =θ Stress at Point -2-10123 0 30 60 90 120 150 180 DegreesStress xyσ′yyσ′xxσ′ 1σ 2σPrincipal stresses zero shear stress on these planes (principal planes) 1σ2σ Section 3.5 Solid Mechanics Part I Kelly 66 and so the perpendicular planes are at ()°°−= 28.14872.31θ and °= 3.58θ . Explicit expressions for the principal stresses can be obtained by substituting the value of θ from Eqn. 3.5.5 into the stress transformation equations, leading to (see the Appendix to this section, §3.5.7) 2 2 41 21 22 2 41 21 1 ) ( ) () ( ) ( xy yy xx yy xxxy yy xx yy xx σσσ σσσσσσ σσσ +−−+=+−++= Principal Stresses (3.5.6) For the example stress state Eqn.3.5.4, one has 38.025 3,62.225 3 2 1 ≈−= ≈+= σ σ Note here that one uses the symbol 1σ to represent the maximum principal stress and 2σ to represent the minimum principal stress. By maximum, it is meant the algebraically largest stress so that, for example, 3 1−>+ . From Eqns. 3.5.3, 3.5.6, the principal stresses are invariant. This is as expected; they are intrinsic features of th e stress state at a point and cannot depend on the coordinate system used to describe the stress state. The question now arises: why are the principal stre sses so important? On e part of the answer is that the maximum principal stress is the la rgest normal stress acting on any plane through a material particle. This can be proved by diff erentiating the stress transformation formulae with respect to θ, xy yy xxxyxy yy xxyyxy yy xxxx dddddd θσ σσθθσθσ σσθθσθσ σσθθσ 2sin2) (2cos2cos2) (2sin2cos2) (2sin −− −=′−− +=′+− −=′ (3.5.7) The maximum/minimum values can now be obtained by setting these expressions to zero. One finds that the normal stresses are a maximum/minimum at the very value of θ in Eqn. 3.5.5 – the value of θ for which the shear stresses are zero – the principal planes. Very often the only thing one knows about the stress state at a point are the principal stresses. In that case one can derive a very useful formula as follows: align the coordinate axes in the principal directions, so Section 3.5 Solid Mechanics Part I Kelly 670 , ,2 1 = = =xy yy xx σσσσσ (3.5.8) Using the transformation formulae with the relations )2cos1( sin21 2θ θ−= and )2cos1( cos21 2θ θ+= then leads to θσσ σθσσσσσθσσσσσ 2sin) (212cos) (21) (212cos) (21) (21 2 12 1 2 12 1 2 1 −−=′−−+=′−++=′ xyyyxx (3.5.9) Here, θ is measured from the principal directions, as illustrated in Fig. 3.5.16. Figure 3.5.16: principal stresses and principal directions Maximum Shear Stress Eqns. 3.5.9 can be used to derive an expression for the maximum shear stress. Differentiating the expression for shear stress with respect to θ, setting to zero and solving, shows that the maximum/minimum occurs at 45±=θ , in which case () ()2 1452 145 21,21σσ σσσ σ θ θ−+= −−= −= +=xy xy or ()2 121max σσσ −=xy Maximum Shear Stress (3.5.10) Thus the shear stress reaches a maximu m on planes which are oriented at °±45 to the principal planes, and the value of the shear stre ss acting on these planes is as given above. Note that the formula Eqn. 3.5.10 does not let one know in which direction the shear stresses θ 2σ1σ2σ 1σprincipal directions Section 3.5 Solid Mechanics Part I Kelly 68are acting but this is not usually an important issue. Many materials respond in certain ways when the maximum shear stress reaches a critical value, and the actual direction of shear stress is unimportant. The direction of the maximum principal stress is, on the other hand, important – a material will respond differently according to whether the normal stress is compressive or tensile. The normal stress acting on the planes of maximum shear stress can be obtained by substituting 45±=θ back into the formulae for normal stress in Eqn. 3.5.9, and one sees that 2/) (2 1σσσσ +=′=′yy xx (3.5.11) The results of this section ar e summarised in Fig. 3.5.17. Figure 3.5.17: principal stress es and maximum shear stress 3.5.5 Mohr’s Circle Otto Mohr devised a way of describing the state of stress at a point using a single diagram, called the Mohr's circle . To construct the Mohr ci rcle, first introduce the stress coordinates ()τσ,, Fig. 3.5.18; the abscissae (horizontal) are the normal stresses σ and the ordinates (vertical) are the shear stresses τ. On the horizontal axis, locate the principal stresses 2 1,σσ , with 2 1σσ> . Next, draw a circle, centred at the average principal stress ()()( )0,2/ ,2 1σστσ += , having radius () 2/2 1σσ− . 1σ2σ o45−=θ 1σ 2σ() 2/2 1σσ+2/) max( 2 1σσσ −=xy()2/2 1σσ+ ()2/2 1σσ+ () 2/2 1σσ+ Section 3.5 Solid Mechanics Part I Kelly 69 The normal and shear stresses acting on a single plane are represented by a single point on the Mohr circle. The normal and shear stre sses acting on two perpendicular planes are represented by two points, one at each end of a diameter on the Mohr circle. Two such diameters are shown in the figure. The first is horizontal. Here, the stresses acting on two perpendicular planes are () () 0, ,1στσ= and ()()0, ,2στσ= and so this diameter represents the principal planes/stresses. Figure 3.5.18: Mohr’s Circle The stresses on planes rotated by an amount θ from the principal planes are given by Eqn. 3.5.9. Using elementary trigonometry, these st resses are represented by the points A and B in Fig. 3.5.18. Note that a rotation of θ in the physical plane corresponds to a rotation of θ2 in the Mohr diagram. Note also that the conventional labeling of shear stress has to be altered when using the Mohr diagram. On the Mohr circle, a shear stress is positive if it yields a clockwise moment about the centre of the element, and is "negative" when it yields a negative moment. For example, at point A the shear stress is "positive" ( 0>τ ), which means the direction of shear on face A of the element is actually opposite to that shown. This agrees with the formula 1σ2 σ θ2 1σ2σ θ στ A BA B θσσ σ 2sin) (21 2 1−−=′xy θ σσσσ σ 2cos) (21) (21 2 1 2 1 −−+=′yyθ σσσσσ 2cos) (21) (21 2 1 2 1 −++=′xx ) (21) (21 2 1σσ σσ +=′+′yy xx) (21) ( ) (41 2 12 2σσ σσσ −=′+′−′xy yy xxxxσ′yyσ′ xyσ′ Section 3.5 Solid Mechanics Part I Kelly 70θσσ σ 2sin) (21 2 1−−=′xy , which is less than zero for 2 1σσ> and o90≤θ . At point B the shear stress is "negative" ( 0 <τ ), which again agrees with formula. 3.5.6 Problems 1. Consider the point shown below, at the boundary between a wall and a dissimilar material. Label the stress components displa yed using the coordinate system shown. Which stress components are continuous acr oss the wall/material boundary? (Add a superscript ‘w’ for the stresses in the wall.) 2. A thin metal plate of width b2, height h and depth t is loaded by a pressure distribution )(xp along a xa+<<− and welded at its base to the ground, as shown in the figure below. Write down expressions for the stress boundary conditions (two on each of the three edges). Write down expressions for the force equilibrium of the plate and moment equilibrium of the plate about the corner A. 3. (a) Is a trampoline (the material you jump on) in a state of plane stress? When someone is actually jumping on it? (b) Is a picture hanging on a wall in a state of plane stress? xy )(xp h Aa2 b2xy Section 3.5 Solid Mechanics Part I Kelly 71(c) Is a glass window in a state of plane stress? On a windy day? (d) A piece of rabbit skin is stretched in a test ing machine – is it in a state of plane stress? 4. Prove that the function y xσσ+ , i.e. the sum of the normal stresses acting at a point, is a stress invariant. 5. Consider a material in plane stress conditions. An element at a free surface of this material is shown below left. Taking the coor dinate axes to be orthogonal to the surface as shown (so that the tangential stress is xxσ), one has ⎥⎦⎤ ⎢⎣⎡=⎥⎦⎤ ⎢⎣⎡ 000xx yy yxxy xx σ σσσσ (a) what are the principal stresses at the point? Which is the maximum and which is the minimum? (b) examine planes inclined at 45o to the free surface, as shown below right. What are the stresses acting on these planes and what have they got to do with maximum shear stress? 6. The stresses at a point in a state of plane stress are given by ⎥⎦⎤ ⎢⎣⎡−=⎥⎦⎤ ⎢⎣⎡ 2331 yy yxxy xx σσσσ (a) Draw a little box to represent the point and draw some arrows to indicate the magnitude and direction of the stresses acting at the point. (b) What relationship exists between Oxy and a second coordinate set yxO′′ , such that the shear stresses are zero in yxO′′? (c) Find the principal stresses. (d) Draw another box whose sides are aligned to the principal directions and draw some arrows to indicate the magnitude and direct ion of the principal stresses acting at the point. (e) Check that the sum of the normal stresses at the point is an invariant. 7. A material particle is subjecte d to a state of stress given by [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ = 00000 αααα σij xxσ y xxσ y o45 xyσ′ xyσ′ xxσ′yyσ′y′ x′ Section 3.5 Solid Mechanics Part I Kelly 72Find the principal stresses, maximum shear stresses, and the direction of the planes on which these stresses act. 8. Consider the two dimensional stress state []⎥⎦⎤ ⎢⎣⎡=αασ00 ij Show that this is an isotropic state of stress , that is, the stress components are the same on all planes through a material particle. 3.5.7 Appendix to §3.5 A Note on the Formulae for Principal Stresses To derive Eqns. 3.5.6, first rewrite the transformation equations in terms of θ2 using )2cos1( sin21 2θ θ−= and )2cos1( cos21 2θ θ+= to get xy xx yy xyxy yy xx yyxy yy xx xx θσ σσθ σθσ σθ σθ σθσ σθ σθ σ 2cos) (2sin2sin )2cos1( )2cos1(2sin )2cos1( )2cos1( 2121 2121 21 +− =′− ++ −=′+ −+ +=′ Next, from Eqn. 3.5.5, 2 2 2 24) (2cos , 4) (22sin xy yy xxyy xx xy yy xxxy σσσσσθ σσσσθ +−−= +−= Substituting into the rewritten transformation formulae then leads to 0) ( ) () ( ) ( 2 2 41 212 2 41 21 =′+−−+=′+−++=′ xyxy yy xx yy xx yyxy yy xx yy xx xx σσσσ σσσσσσ σσσ Here yy xxσσ′>′ so that the maximum principal stress is xxσσ′=1 and the minimum principal stress is yyσσ′=′2 . Here it is implic itly assumed that 0 2tan>θ , i.e. that 90 20<<θ or 270 2 180<<θ . On the other hand one could assume that 0 2tan<θ , i.e. that 180 2 90<<θ or 360 2 270<<θ , in which case one arrives at the formulae 2 2 41 212 2 41 21 ) ( ) () ( ) ( xy yy xx yy xx yyxy yy xx yy xx xx σσσ σσσσσσ σσσ +−++=′+−−+=′ The results can be summar ised as Eqn. 3.5.6, Section 3.5 Solid Mechanics Part I Kelly 73 2 2 41 21 22 2 41 21 1 ) ( ) () ( ) ( xy yy xx yy xxxy yy xx yy xx σσσ σσσσσσ σσσ +−−+=+−++= These formulae do not tell one on which of the two principal planes the maximum principal stress acts. This might not be an important issu e, but if this information is required one needs to go directly to the stress transformation equati ons. In the example stress state, Eqn. 3.5.4, one has )1(2sin)1( cos)2( sin)1(2sin)1( sin)2( cos 2 22 2 −− + =′−+ + =′ θ θ θ σθ θ θ σ yyxx For ()°°−= 28.14872.31θ , 62.2=′xxσ and 38.0=′yyσ . So one has the situation shown below. If one takes the other angle, °= 3.58θ , one has 38.0=′xxσ and 62.2=′yyσ , and the situation below θ=-31.72¡ yyσ′ xxσ′ θ=+ 58.3¡ x′ xxσ′ yyσ′ Section 3.6 Solid Mechanics Part I Kelly 743.6 Strain If an object is placed on a table and then the table is moved, each material particle moves in space. The particles are said to undergo a displacement . The particles have moved in space as a rigid body . The material remains unstressed. When a material is acted upon by a set of forces, it changes size and/or shape , it deforms . In this section, the kinematics of materials is studied, th at is, a number of tools and techniques are developed to describe the deformation of a material. 3.6.1 Strain at a Point Material deformation can be described by imag ining the material to be a collection of small line elements. As the material is deform ed, the line elements stretch, or get shorter, and rotate in space relative to each other. This movement of line elements is encompassed in the idea of strain : the “strain at a point” is all the stretching, contracting and rotating of all line elements emanating from that point, with all the line elements together making up the continuous ma terial, as illustrated in Fig. 3.6.1. Figure 3.6.1: a deforming material element; or iginal state of line elements and their final position after straining It turns out that the strain at a point is completely ch aracterised by the movement of any three mutually perpendicular line-segments . If it is known how these perpendicular line- segments are stretching, contra cting and rotating, it will be possible to determine how any other line element at the point is behaving, by using a strain transf ormation rule (see later). This is analogous to the way the stre ss at a point is charact erised by the stress acting on three perpendicular planes through a point, and the stress components on other planes can be obtained using th e stress transformation formulae. before deformation after deformation Section 3.6 Solid Mechanics Part I Kelly 753.6.2 Two Dimensional Strain Consider the two-dimensional case: two perpe ndicular line-elements emanate from a point and the material that contains the point is deformed. Then two things (can) happen: (1) the line segments will change length and (2) the perpendicular angle between the line-segments changes . The change in length of line-elements is called normal strain and the change in angle between initially perpendicu lar line-segments is called shear strain . The strains are now defined as follows: Normal strain in direction x: (denoted by xxε) change in length (per unit length) of a line element originally lying in the x−direction Normal strain in direction y: (denoted by yyε) change in length (per unit length) of a line element originally lying in the y−direction Shear strain : (denoted by xyε) (half) the change in the original right angle between the two perpendicular line elements Referring to Fig. 3.6.2, the strains are () ,21, , λθε ε ε +=−′′=−′′=xy yy xxACAC CA ABAB BA (3.6.1) Figure 3.6.2: strain at a point A Note that the point A in Fig. 3.6.2 has also undergone a displacement u(A). This displacement has two components, xu and yu, as shown in Fig. 3.6.3 (and similarly for the points B and C). C′ A BA′B′ θλ xy Cdeforms Section 3.6 Solid Mechanics Part I Kelly 76 Figure 3.6.3: displacement of a point A The line elements not only change length and the angle between them change – they can also move in space as rigid-bodies. Thus, for example, the normal and shear strain in the three examples shown in Fig. 3.6.4 are the same, even though the displacements occurring in each case are different – strain is independent of rigid body motions . Figure 3.6.4: rigid body motions The Engineering Strain Suppose now that the deformation is very small, so that, in Fig. 3.6.5, *BA BA′≈′′ - here, *BA′ is the projection of BA′′ in the x − direction. In that case, ABAB BA xx−′≈* ε (3.6.2) Similarly, one can make the approximations ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛′+′≈−′≈ABCC ABBB ACAC CA xy yy* * * 21,ε ε (3.6.3) the expression for shear strain fo llowing from the fact that, for a small angle , the angle (measured in radians) is approximate ly equal to the tan of the angle. )(Au xuyuA′ A B Section 3.6 Solid Mechanics Part I Kelly 77 Figure 3.6.5: small deformation This approximation for the normal strains is called the engineering strain or small strain or infinitesimal strain and is valid when the deformations are small . The advantage of the small strain approximation is that the mathematics is simplified greatly. Small strain is useful in characterising the small deform ations that take place in, for example, (1) engineering materials such as concrete, me tals, stiff plastics and so on, (2) linear viscoelastic materials such as many polymeric materials, (3) some porous media such as soils and clays at moderate loads, (4) almost any material if the loading is not too high. Small strain is inadequate for describing large deformations that occur, for example, in many rubbery materials, engineeri ng materials at large loads, et c. In these cases the more precise definition 3.6.1 as developed and used in Part III is required. Engineering Shear Strain and Tensorial Shear Strain The definition of shear stra in introduced above is the tensorial shear strain xyε. The engineering shear strain xyγ is defined as twice this angle, i.e. as λθ+, and is often used in Strength of Materials and elementary Solid Mechanics analyses. 3.6.3 Sign Convention for Strain A positive normal strain means that the line element is lengthening. A negative normal strain means the line element is shortening. For shear strain, one has the following conve ntion: when the two perpendicular line elements are both directed in the positive directions ( x and y), or both directed in the negative directions, then a positiv e shear strain corresponds to a decrease in right angle. Conversely, if one line segment is directed in a positive direction whilst the other is directed in a negative dir ection, then a positive shear strain corre sponds to an increase in angle. The four possible cases of shear stra in are shown in Fig. 3.6.6 (all four shear strains are positive). C′ A BA′B′ θλ xy Cdeforms *B*C Section 3.6 Solid Mechanics Part I Kelly 78 Figure 3.6.6: positive shear strain Thus if a “box” element of material u ndergoes a constant pos itive shear strain xyε, it will deform as shown in Fig. 3.6.7. Figure 3.6.7: a material element under going a positive shear strain By definition, the strain, as with the stress, is symmetrical: yz xyεε= (3.6.4) 3.6.4 Geometrical Interpretation of the Engineering Strain Consider a small “box” element and suppose it to be so small that the strain is constant throughout - one says th at the strain is homogeneous . This implies that straight lines remain straight after straining and parall el lines remain parallel. A few simple deformations are examined below and these are related to the strains. A positive normal strain 0 >xxε is shown in Fig. 3.6.8a. Here the undeformed box element (dashed) has elongated. Knowledge of the strain alone is not enough to determine the position of the strained element, since it is free to move in space as a rigid body. The displacement over some part of th e box is usually specified, for example the left hand end has been fixed in Fig. 3.6.8b. A negative normal strain acts in Fig. 3.6.8c and the element has contracted. xy Section 3.6 Solid Mechanics Part I Kelly 79 Figure 3.6.8: normal strain; (a) positive norm al strain, (b) positive normal strain with the left-hand end fixed in sp ace, (c) negative normal strain A case known as simple shear is shown in Fig. 3.6.9a, and that of pure shear is shown in Fig. 3.6.9b. In both illustrations, 0>xyε . Indeed, if the shear strain is the same in both, then one can be obtained from the other by a pure (rigid body) rotation, of the type shown in Fig. 3.6.9c. Figure 3.6.9: (a) simple shear, (b ) pure shear, (c) pure rotation Indeed, any shear strain can be decomposed into a pure shear and a pure rotation, as illustrated in Fig. 3.6.10. Figure 3.6.10: shear strain decomposed into a pure shear and a pure rotation 3.6.5 Large Rotations Note that if a material element undergoes a rigid body rotation, it can only be rotated by a small amount ; otherwise the small strain approximation will be invalid. This effect is illustrated in Fig. 3.6.11, which shows a shear ed material element, Fig. 3.6.11a. The xy )a() b() c(y x xyfixed )a() b() c(x x xy y y arbitrar y shear strain + pure shea r pure rotation = Section 3.6 Solid Mechanics Part I Kelly 80deformed element is translated (Fig. 3. 6.11b) and rotated by a small amount (Fig. 3.6.11c). In both cases, it can be seen that the normal strains xxε and yyε remain zero. In Fig. 3.6.11d, however, a large ro tation is seen to induce non-ze ro normal strains, which is contradictory, since, by defin ition, the strain should be zero for rigid body motions. Figure 3.6.11: an element und er shear strain; (a) the st rained element, (b) the element with an additional translation, (c ) with a small rotation, (d) with a large rotation As an example, consider a cantilevered beam which undergoes large bending, Fig. 3.6.12. The shaded element shown might well undergo small normal and shear strains. However, because of the large rotation of the element, additional spurious engineering strains are induced. Use of the precise defi nition, Eqn. 3.6.1, is requir ed in cases such as this. Figure 3.6.12: Large rotations of an element in a bent beam 3.6.6 Three Dimensional States of Strain The above can be generalized to three dimens ions. In the general case, there are three normal strains, zz yy xxεεε ,, , and three shear strains, zx yz xyεεε ,,. T h e zzε strain corresponds to a change in length of a line element initially lying along the z axis. The yzε strain corresponds to half the change in th e originally right angl e of two perpendicular line elements aligned with the y and z axes, and similarly for the zxε strain. 3.6.7 Problems 1. Under what conditions are the small strains a good approximation to the actual finite strains at a point? translation small rotation lar ge rotation strained elemen t Section 3.6 Solid Mechanics Part I Kelly 81 2. An element undergoes a homogeneous strain, as shown. There is no normal strain in the element. The angles are given by 001.0=λ and 002.0=θ radians. What is the (tensorial) shear strain in the elem ent? The engineering shear strain? 3. In a fixed yx− reference system established for the test of a large machine member, three points A, B and C on the member have the followi ng coordinates before and after loading: ) 0000.0, 0045.2(: ) 0000.0, 0000.2(:) 0000.0, 0000.0(: ) 0000.0, 0000.0(:) 5030.1, 0025.0(: ) 5000.1, 0000.0(: B BA AC C ′′−′ Determine the actual strains and the small strains (at/near point A). What is the error in the small strain compared to the actual strains? 4. Sketch the deformed shape for the materi al shown below under the following strains (A, B constant): (i) 0>=Axxε (taking 0 ==xy yyεε ) – assume that the right-hand edge is fixed (ii) 0<=Byyε (with 0==xy xxεε ) – assume that the lower edge is fixed (iii) 0<=Bxyε (with 0 ==yy xxεε ) – assume that the left-hand edge is fixed y xxy θλ Section 3.7 Solid Mechanics Part I Kelly 823.7 Plane Strain A state of plane strain is defined as follows: Plane Strain : If the strain state at a material particle is such that the only non- zero strain components act in one plane only, the particle is said to be in plane strain. The axes are usually chosen such that the yx− plane is the plane in which the strains are non-zero, Fig. 3.7.1. Figure 3.7.1: non-zero stress co mponents acting in the x – y plane It turns out that, just as the st ate of plane stress arises in th in components, a state of plane strain arises in very thick components. 3.7.1 Thick Components Consider the three dimensional block of material in Fig. 3.7.2. The material is constrained from undergoing normal strain in the z direction, for example by preventing movement with rigid immovable walls – and so 0=zzε . Figure 3.7.2: A block of material constrained by rigid walls If, in addition, the loading is as shown in Fig. 3.7.2, i.e. it is the same on all cross sections parallel to the zy− plane (or zx− plane) – then the line elements shown in Fig. 3.7.3 will remain perpendicular (although th ey might move out of plane). xy xy yy xxεεε ,, Rigid Walls xy z Section 3.7 Solid Mechanics Part I Kelly 83 Figure 3.7.3: Line elements etched in a block of material – they remain perpendicular in a state of plane strain Then 0==yz xzεε . Thus the fully three dimensiona l strain field reduces to a two dimensional one: ⎥ ⎦⎤ ⎢ ⎣⎡→ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ yy yxxy xx zz zy zxyz yy yxxz xy xx εεεε εεεεεεεεε The problem can now be analysed using the three independent strains, which simplifies matters considerable. Once a solution is found for the defo rmation of one plane, the solution has been found for the deformation of the whole body, Fig. 3.7.4. Figure 3.7.4: three dimensional problem redu ces to a two dimensional one for the case of plane strain The rigid walls (and particular loading) of Fig. 3.7.2 ensure s a state of plane strain. A state of plane strain will also exist in thick structures without such end walls. Material towards the centre is constrained by the mass of material on either side and will be (approximately) in a state of plane strain, Fig. 3.7.5. 0=zxe 0=yzε0=xzε Section 3.7 Solid Mechanics Part I Kelly 84 Figure 3.7.5: material in an appr oximate state of plane strain material in a state of plane strain Section 3.8 Solid Mechanics Part I Kelly 853.8 Properties of the Strain Stress transformation formulae, principal stre sses, stress invariants and formulae for maximum shear stress were presented in §3.4-§ 3.5. The strain is very similar to the stress. They are both tensor s, having nine components, and all the formulae for stress hold also for the strain. Formulae for a two-di mensional state of stra in are given in what follows. 3.8.1 Strain Transformation Formula Consider two perpendicular line-elements lying in the coordinate directions x and y, and suppose that it is known that the strains are xy yy xxεεε ,, , Fig. 3.8.1. Consider now a second coordinate system, with axes yx′′, , oriented at angle θ to the first system, and consider line-elements lying along these axes. It can be shown that the line-elements in the second system undergo strains accordi ng to the following (two dimensional) strain transformation equations : xy xx yy xyxy yy xx yyxy yy xx xx θε εεθθεθε θε θεεθε θε θε ε 2cos) ( cos sin2sin cos sin2sin sin cos 2 22 2 +− =′− + =′+ + =′ Strain Transformation Formulae (3.8.1) Figure 3.8.1: A rotated coordinate system Note the similarity between these equations and the stress transformation formulae, Eqns. 3.4.7. Although they have the same structur e, whereas the stress transformation equations were derived using Newton’s laws, no physical law is used to derive the strain transformation equations (see th e Appendix to this section, §3.8.3, for their derivation). Principal Strains Using exactly the same arguments as used to derive the expressions for principal stress, there is always at least one set of perpe ndicular line elements which stretch and/or contract, but which do not undergo angle changes. The strains in this special coordinate system are called principal strains , and are given by (compare with Eqns. 3.5.6) xy x′y′ θ Section 3.8 Solid Mechanics Part I Kelly 862 2 22 2 1 ) (41 2) (41 2 xy yy xxyy xxxy yy xxyy xx εεεεεεεεεεεε +−−+=+−++= Principal Strains (3.8.2) Further, it can be shown that 1e is the maximum normal strain occurring at the point, and that 2e is the minimum normal stra in occurring at the point. The principal directions , that is, the directions of th e line elements which undergo the principal strains, can be obtaine d from (compare with Eqns. 3.5.5) yy xxxy εεεθ−=22t a n ( 3 . 8 . 3 ) Here, θ is the angle at which the principal dir ections are oriented with respect to the 2 1,xx axes. Maximum Shear Strain Analogous to Eqn. 3.5.10, the maximum shear strain occurring at a point is ()2 1max 21eexy −=ε ( 3 . 8 . 4 ) and the perpendicular line elements undergoi ng this maximum angle change are oriented at o45 to the principal directions. Example (of Strain Transformation) Consider the block of material in Fig. 3.8.2a. Two sets of perpendicular lines are etched on its surface. The block is then stretched, Fig. 3.8.2b. Figure 3.8.2: A block with strain measured in two different coordinate systems • •θ • •fixed fixed xy x′y′ θ )b()a( Section 3.8 Solid Mechanics Part I Kelly 87 This is a homogeneous deformation, that is, the strain is the same at all points. However, in the yx− description, 0 >xxε and 0 ==xy yyεε , but in the yx′−′ description, none of the strains is zero. The two sets of strains are related th rough the strain transformation equations. ■ Example (of Strain Transformation) As another example, consider a square material which undergoes a pure shear, as illustrated in Fig. 3.8.3, with 01.0 ,0===xy yy xx εεε Figure 3.8.3: A block under pure shear From Eqn. 3.8.2, the principal strains are 01.0 ,01.02 1 −=+= e e and the principal directions are obtained from Eqn. 3.8.3 as °±=45θ . To find the direction in which the maximum normal strain occurs, put °+=45θ in the strain transformation formulae to find that 01 .01+=′=xxee , so the deformation occurring in a piece of material whose sides are aligned in these princi pal directions is as shown in Fig. 3.8.4. Figure 3.8.4: Principal strains for the block in pure shear The strain and principal strain are summarised in Fig. 3.8.5. o xy o 45¡ a a 0.01a x′ y′ Section 3.8 Solid Mechanics Part I Kelly 88Note that, since the original yx− axes were oriented at °±45 to the principal directions, these axes are those of maximum shear strain – the original 01 .0=xyε is the maximum shear strain occurring in the material. Figure 3.8.5: Strain viewed from two different coordinate systems ■ 3.8.2 Problems 1. In Fig. 3.8.2, take o30=θ and 02 .0=xxε . Calculate the strains xy yy xxεεε′′′ ,, . What are the principal strains? What is the maximu m shear strain? Of all the line elements which could be etched in the block, at what angle θ to the x axis are the perpendicular line elements which undergo the largest angl e change from the initial right angle? 3.8.3 Appendix to §3.8 Derivation of the Strain Transformation Formulae Consider an element ABCD undergoing a strain xxε with 0 ==xy yyεε to DCBA′′ as shown in the figure below. x′y A BxθEC′ CD B′E′*E θagrees with after before Section 3.8 Solid Mechanics Part I Kelly 89In the yx− coordinate system, by definition, ABBBxx /′=ε . In the yx′−′ system, one has ABBB ABEE AEEE xx′=′==′ θθθε2* coscos/cos which is the first term of the first of Eqn. 3.8.1. The other transformation formulae can be derived in a similar manner. Section 3.8 Solid Mechanics Part I Kelly 90 814 Linear Elasticity I The concepts of stress and strain were in troduced in the previous chapter. The relationship between stress and strain at any material par ticle depends on the material under consideration. For example the deformation in a rubbery material will be very different to that of a ceramic when both are subjected to the same stress. This relationship between stress and strain is called a constitutive law . The simplest constitutive law for solid materials is the linear elastic law, which assumes a linear relationship between stress and strain. This assumption turn s out to be an excellent predictor of the response of components which undergo small deformations, for example steel and concrete structures under large loads, and also works well for practically any material at a sufficiently small load. The linear elastic model is introduced in th is chapter and some elementary problems involving elastic materials are solv ed. Elastic materials undergoing homogeneous stress and strain are analysed in §4.2, that is, th e stress and strain are constant throughout. Sections 4.3-6 then deal with four important, practical, theori es. They are concerned with certain geometries under the action of particul ar types of load. In §4.3, the geometry is that of a long slender bar and the load is one which acts along the le ngth of the bar; in §4.4, the geometry is that of a long slender circular bar and th e load is one which twists the bar; in §4.5 the geometry is that of a thin-walled cylindrical or spherical component, and the load is normal to these walls; in §4.6 the geometry is that of a long and slender beam, and the load is transverse to the beam length. These four particular situations allow for simplifications (or approximations) to be made to the full three-dimensional linear elastic stress-strain relations; this allows one to write down simple expressions for the stress and strain and so solve some important practical problems analytically. Finally, in §4.7, there is a short discussion on the failure of elastic materials. 82 Section 4.1 Solid Mechanics Part I Kelly 934.1 The Continuum The Linear Elastic Model to be discussed in this Chapter is a continuum model . This type of model is discussed briefly in this section. 4.1.1 Stress and Scale In the definition of the traction vect or, §3.3.1, it was assumed that the ratio S FΔΔ/ would indeed reach some definite limit as the area SΔ of the surface upon which the force FΔ acts was shrunk to zero. This issue can be explored further by considering Fig. 4.1.1. Assume first that the plane upon which th e force acts is fairly large; it is then shrunk and the ratio SF/ tracked. A schematic of this ratio is shown in Fig. 4.1.2. At first (to the right of Fig. 4.1.2) the ratio SF/ undergoes change, assuming the stress to vary within the material, as it invariably will if a material is loaded in some complex way. Eventually the plane will be so small that th e ratio changes very litt le, perhaps with some small variability ε. If the plane is allowed to ge t too small, however, down towards the atomic level, where one might encounter “i ntermolecular space”, there will be large changes in the ratio and the whole concept of a force acing on a single surface breaks down. Figure 4.1.1: A force acting on an internal surface In a continuum model, it is assumed that the ratio SF/ follows the dotted path shown in Fig. 4.1.2. It should be kept in mind, then, that the traction in a real material should be evaluated through SF h SΔΔ= →Δ2*)(limt (4.1.1) point force plane force point plane Section 4.1 Solid Mechanics Part I Kelly 94 where *h is some minimum dimension below which there is no acceptable limit. It is necessary to take the limit to zero in the math ematical modelling of materials since that is the basis of calculus. Figure 4.1.2: the change in traction as th e plane upon which a force acts is reduced in size In a continuum model, then, there is a mini mum sized element one can consider, say of size 3*)(h V=Δ . When one talks about the stress on this element, the mass of this element, the density, velocity and acceleration of this element, one means the average of these quantities throughout or over the surf ace of the element – the discrete atomic structure within the element is ignored and is “smeared” out into a continuum element . One does not have any information about what is happening inside the continuum element - it is like a “black box”. The scal e of the element (and higher) is called the macroscale – continuum mechanics is mechanics on th e macroscale. The scale of entities within the element is termed the microscale – continuum models cannot give any information about what happens on the microscale. 4.1.2 Example: Metal A metal component, from a distance, appears fairly uniform. With the help of a microscope, however, it will be seen to consis t of many individual grains of metal - each grain is of the order 0.01mm across and each one has very individual properties (the crystals in each grain are aligned in different directions), Fig. 4.1.3. Molecular level more or less constant F/S – with some variability ε F/S changing as one moves away from “point” getting closer to the “point” SF εcontinuum approximation *h Section 4.1 Solid Mechanics Part I Kelly 95 Figure 4.1.3: metal grains If one is interested in the gross deformation of a m oderately sized metal component it would be sufficient to consider deformations that are averaged over volumes which are large compared to individual grains, but sm all compared to the whole component. A minimum dimension of, say, mm5.0*=h for the metal of Fig. 4.1.3 would suffice, and this would be the macro/micro-scale boundary, with a minimum surface area of dimension 2*)(h for the definition of stress. When one measures physical properties of the metal “at a point”, for example the density, one need only measure an average quantity over an element of the order 3mm)5.0( or higher. It is no t necessary to consider the individual grains of metal – these are inside the “black box”. The model will return valuable information about the deformation of th e gross material, but it will not be able to furnish any information about m ovement of individual grains. What if the response of individual grains to applied loads is required? In that case a model would have to be constructed whic h accounted for the different mechanical properties of each grain. The metal could no longer be considered to be a uniform material, but rather a complex one with ma ny interfaces between individual grains. The macro/micro boundary could be set at about m 0.1 *μ=h . Experiments would have to be conducted on pieces of material considerably smaller than the grain size in order to provide data for any model. Apart from the experimental difficulties, much more computational power would be required to obtain results for this model than for the former, simpler, model. The continuum element is also called a representative volume element (RVE), an element of material large enough for the hete rogeneities to be replaced by homogenised mean values of their properties. The order of the dimensions of RVE’s for some common engineering materials would be approximately Metal: 0.1mm Polymers/composites: 1mm Wood: 10mm Concrete: 100mm Section 4.1 Solid Mechanics Part I Kelly 964.1.3 Micromechanical Models As mentioned, micromechanical models whic h account for detail on the very small scale require large computer resources to account for the mechanics of the many elements which make up a “macro” sample of ma terial. Whereas this simply made micromechanics modelling impossible in the past , this is not now the case. With the improved power of computers, especially since the 1990s, micromechanical models are becoming more and more popular. Usually, one will have a micromechanical model of a small RVE of material. This then provides information regarding the properties of the RVE to be included in a continuum model (ra ther than having a micromechanical model of the complete material, which is in most cases still not practical). 4.1.4 Problems 1. You want to evaluate the stiffness of a metal for inclusion in a solid mechanics model. What minimum size specimen would you use in your test - m10μ, 1mmmm,1.0 o r 1cm ? 2. Rice flowing in a grain chute of a silo can be considered to be a fl uid, and the flow of rice can be solved using the equations of mechanics. What minimum dimension *h should be employed for measurements in th is case to ensure the validity of a continuum model of flowing rice? Section 4.2 Solid Mechanics Part I Kelly 974.2 The Linear Elastic Model 4.2.1 The Response of Real Materials The Tension Test Consider the following key experiment, the tensile test , in which a small, usually cylindrical, specimen is gripped and stretched, us ually at some given ra te of stretching. The force required to hold the specimen at a given displacem ent/stretch is recorded, Fig. 4.2.1. For many engineering materials, for example st eel, rocks and concrete, it is found that the force is proportional to displacement as with the portion OA in Fig. 4.2.1. The following observations will also be made: (1) if the material is unloaded, the force/ displacement curve wi ll trace back along the line OA down to zero force and zero displ acement; further loading and unloading will again be up and down OA (2) the force-displacement curve will be more or less the same regardless of the rate at which the specimen is stretched (at least at moderate temperatures). (3) the loading curve remains linear up to a certain force level, the yield point or elastic limit of the material (point A). Beyond this point, permanent deformations are induced; on unloading to zero force, the specimen will have a permanent elongation. (4) the strains up to the elastic limit are small Figure 4.2.1: force/displacement curve for the tension test Stress-Strain Curve There are two definitions of stress used to describe the tension test. First, there is the force divided by the original cross sectional area of the specimen 0A; this is the nominal stress or engineering stress , 0AF n=σ (4.2.1) 0AForce Displacement elastic limit Section 4.2 Solid Mechanics Part I Kelly 98Alternatively, one can evaluate the force divided by the (smaller) current cross-sectional area A, leading to the true stress AF=σ (4.2.2) in which F and A are both changing with time. For small elongations, within the linear range OA, the cross-sectional area of the material undergoes negligib le change and both definitions of stress are more or less equivalent. Similarly, one can describe the deformati on in two alternative ways. Denoting the original specimen length by 0l and the current length by l, one has the engineering strain 00 lll−=ε (4.2.3) Alternatively, the true strain accounts for the fact that the “original length” is continually changing; a small change in length dl leads to a strain increment ldl d /=ε and the true strain is defined as the accumulation of these increments: ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛==∫ 0ln 0ll ldll ltε (4.2.4) The true strain is also called the logarithmic strain . Again, at small deformations, the difference between these two strain measures is negligible. The stress-strain diagram for a tension te st can now be descri bed using the true stress/strain or nominal stress/strain definitio ns, as in Fig. 4.2.2. The stress value at the elastic limit is called the yield stress Y. The shape of the nominal stress/strain diagram, on the left, is of course the same as the graph of force versus displacement. C here denotes the point at which the maximum force the specimen can withstand has been reached. The nominal stress at C is called the Ultimate Tensile Strength (UTS) of the material. After this point, the specimen “necks”, with a very rapid reduction in cro ss-sectional area somewhere about the centre of the specimen until the specimen rupt ures, as indicated by the asterisk. A compression test will lead to similar results as the tensile stress. The yield stress in compression will be approximately the same as (the negative of) the yield stress in tension. If one plots the tr ue stress versus true strain curve for both tension and compression (absolute values for the compre ssion), the two curves will more or less coincide. This would indicate that the be haviour of the material under compression is broadly similar to that under tension. If one were to use the nominal stress and strain, then the two curves would not coincide (although they would of course in the small-strain linear region). Section 4.2 Solid Mechanics Part I Kelly 99 Figure 4.2.2: typical stress-strain curve for an engineering material; (a) engineering stress and strain, (b) true stress and strain Plasticity and Br ittle Fracture One says that the material fails once the yield stress is r eached (see §4.7). For most metals, soils and rocks, plasticity theory , to be discussed in Part II, describes well events after the yield stress is reached. Brittle materials, such as ceramics and glass, do not become plastic. Instead, they fail through brittle fracture (crack growth) before any yield stress is reached; brittle fracture can be modelled using elasticity theory (see Part II). The Linear Elastic Model The Linear Elastic model is used to de scribe materials which respond as follows: (i) the strains in the material are small 1 (ii) the stress is proportional to the strain, εσ∝ (linear ) (iii) the material returns to its original sh ape when the loads are removed, and the unloading path is the same as the loading path ( elastic ) (iv) there is no dependence on the ra te of loading or straining From the earlier discussion, this model well re presents the engineering materials up to their yield stress. It also models well almost any materi al provided the stresses are sufficiently small. The stress-strain (loading and unloading) curve for the Linear Elastic solid is shown in Fig. 4.2.3a. Other possible responses are shown in Figs. 4.2.3b,c. Fig. 4.2.3b shows the typical response of a rubbery -type material and many biol ogical tissues; these are non- linear elastic materials (see Part III). Fig. 4.2.3c shows the typical response of viscoelastic materials (see Chapter 7 and Part II) and that of many plastically and viscoplastically deforming materials (see Part II and Part III); the dissimilar loading/unloading curves here is called a hysteresis loop . 1 if the small-strain approximation is not made, the st ress-strain relationship will be inherently non-linear; the actual strain, Eqn. 3.6.2 involves (non-linear) squares and square-roots of lengths Y Ynσ ε∗∗ σ tε )a() b(C C Section 4.2 Solid Mechanics Part I Kelly 100 Figure 4.2.3: Different stress-stra in relationships; (a) linear elastic, (b) non-linear elastic, (c) viscoelastic/plastic/viscoplastic 4.2.2 Homogeneity and Isotropy In this Chapter, not only will it be assume d that the material responds according to the Linear Elastic Solid model, but that the material is homogeneous and isotropic . The term homogeneous means that the mechanical properties are th e same at each point throughout the material . In other words, the relationship between stress and strain is the same for all material particles. The term isotropic means that the mechanical properties are the same in all directions . In other words, the relationship between stress and strain at any single point is the same in all directions. This implies that if a specimen is cut from a material and subjected to a load, it would not matter in which orientation the specimen was cut, the resulting deformation w ould be the same – as illustrated in Fig. 4.2.4.2 Figure 4.2.4: Illustration of Isotropy Most metals and ceramics can be considered to be isotropic, on the macroscopic scale considered in a continuum model. Wood is an example of a material which is not isotropic (it is anisotropic ) – the fibres in the wood are a ligned in a preferred direction, along the grain, and the stiffness in this direct ion is much greater than in the transverse direction. Anisotropic elasti city will be examined in §6.2. 2 a material can be homogeneous and not isotropic, and vice versa σ εσ εσ εload unload )a() b() c(hysteresis loop oσ oσoσ oσ Section 4.2 Solid Mechanics Part I Kelly 101 4.2.3 Stress-Strain Law Consider a cube of material subjected to a uniaxial tensile stress xxσ, Fig. 4.2.5a. One would expect it respond by extending in the x direction, 0 >xxε , and to contract laterally, so 0<=zz yyεε , these last two being equal because of the isotropy of the material. With stress proportional to strain, one can write xx zz yy xx xxE Eσνεεσε −== = ,1 (4.2.1) Figure 4.2.5: an element of mate rial subjected to a uniaxial stress; (a) normal strain, (b) shear strain The constant of proportionality between th e normal stress and st rain is called the Young’s Modulus , and is a measure of the stiffness of the material. The material parameter ν is called the Poisson’s ratio . Since xx zz yy νεεε −== , it is a measure of the contraction relative to the normal extension. Because of the isotropy/symmetry of the materi al, the shear strains are zero, and so the deformation of Fig. 4.2.5b, which shows a non-zero xyε, is not possible – shear strain can arise if the material is not isotropic. One can write down similar expressions for the strains which result from a uniaxial tensile yyσ stress and a uniaxial zzσ stress: zz yy xx zz zzyy zz xx yy yy E EE E σνεεσεσνεεσε −== =−== = ,1,1 (4.2.2) Similar arguments can be used to write down the shear strains which result from the application of a shear stress: xz xz yz yz xy xy σμεσμεσμε21,21,21= = = (4.2.3) xxσxxσ )a(xxσxxσ )b( Section 4.2 Solid Mechanics Part I Kelly 102 The constant of proportionality introduced here is the Shear Modulus3 μ, and is a measure of the resistance to shear deformation. The strain which results from a combination of all six stresses is simply the sum of the strains which result from each 4: ()[] ()[] ()[]yy xx zz zzyz yz zz xx yy yyxz xz xy xy zz yy xx xx EEE σσνσεσμεσσνσεσμεσμεσσνσε +−== +−== = +−= 121,1,21,21,1 (4.2.4) These equations involve three material paramete rs. It will be proved in §6.2 that an isotropic linear elastic material can have onl y two independent material parameters and that in fact ()νμ+=12E ( 4 . 2 . 5 ) It will be verified in the following example. Example Consider the simple shear deformation shown in Fig. 4.2.6, with 0 >xyε and all other strains zero. With the material linear elastic, th e only non-zero stress is xy xyμεσ 2= . Figure 4.2.6: a simple shear deformation Using the strain transformation equations, E qns. 3.8.1, the only non-zero strains in a second coordinate system yx′−′ , with x′ at o45=θ from the x axis, are xy xxεε+=′ and xy yyεε−=′ . Because the material is isotropic, Eqns 4.2.4 hold also in this second 3 and is often denoted by the G 4 this is called the principle of linear superposition : the "effect" of a sum of "causes" is equal to the sum of the individual "effects" of each "cause" x′y xy′ Section 4.2 Solid Mechanics Part I Kelly 103coordinate system and so the stresses in the new coordinate system can be determined by solving the equations ()[] ()[] ()[]yy xx zz zzyz yz zz xx yy xy yyxz xz xy xy zz yy xx xy xx EEE σσνσ εσμεσσνσεεσμεσμεσσνσεε ′+′−′==′′==′′+′−′=−=′′==′′==′′+′−′=+=′ 10210 ,1,210 ,210 ,1 (4.2.6) which results in xy yy xy xxvE vEε σε σ+−=′ ++=′ 1,1 (4.2.7) But the stress transformation equations, with xy xyμεσ 2= , give xy xxμεσ 2+=′ and xy yyμεσ 2−=′ and so Eqn. 4.2.5 is verified. ■ Relation 4.2.5 allows the Linear Elastic Solid st ress-strain law, Eqn. 4.2.4, to be written as ()[] ()[] ()[] yz yzxz xzxy xyyy xx zz zzzz xx yy yyzz yy xx xx EEEEEE σνεσνεσνεσσνσεσσνσεσσνσε +=+=+=+−=+−=+−= 111111 Stress-Strain Relations (4.2.8) These equations can be solved for the stresses to get Section 4.2 Solid Mechanics Part I Kelly 104() [] () [] () [] yz yzxz xzxy xyyy xx zz zzzz xx yy yyzz yy xx xx EEEEEE ενσενσενσεενενννσεενενννσεενενννσ +=+=+=++−−+=++−−+=++−−+= 111) 1()21)( 1() 1()21)( 1() 1()21)( 1( Stress-Strain Relations (4.2.9) Values of E and ν for a number of materials are given in Table 4.2.1 below. Table 4.2.1: Young’s Modulus E and Poisson’s Ratio ν for a selection of materials at 20oC The stress-strain relations 4.2.8-9, also known as Hooke’s Law , can be used to solve some simple problems involving homogeneous stress and deformati on, that is, stress and strains which are the same at all points in a ma terial. If three of the six normal stresses and strains are known, the other three can be determined from the relations. Example Consider the block of linear elas tic material shown in Fig. 4.2. 7. It is subjected to an equi-biaxial stress of 0>==σσσyy xx . The surfaces parallel to the yx− plane are free surfaces and so 0=zzσ . From Eqn. 4.2.8 then, the strains are 0 , 2 ,) 1( === −= −==yz xz xy zz yy xxE Eεεεσνεσνεε Material E (GPa) ν Grey Cast Iron 100 0.29 A316 Stainless Steel 196 0.3 A5 Aluminium 68 0.33 Bronze 130 0.34 Plexiglass 2.9 0.4 Rubber 0.0017 0.4 Concrete 23-30 0.2 Granite 53-60 0.27 Wood (pinewood) fibre direction transverse direction 17 10.45 0.79 Section 4.2 Solid Mechanics Part I Kelly 105 As expected, yy xxεε= and 0<zzε . Figure 4.2.7: A block of linear elastic material subjected to an equi-biaxial stress ■ 4.2.4 Volumetric Strain Consider an element of ma terial under a general st ress system, and undergoing consequent strains xy xxεε, , etc., Fig. 4.2.8a. The same deformation is viewed along the principal directions in Fig. 4.2.8b, for which only normal strains arise. The volumetric strain is defined to be the unit change in volume: () 3 2 12 3 2 13 2 1 1) 1)( 1)( 1() )( )( ( εεεεεεεεεε ++≈+++=−+++=−Δ+Δ+Δ+=Δ iOabcabcc cb ba a VV (4.2.10) and the squared and cubed terms can be neglec ted because of the small-strain assumption. Since any elemental volume such as that in Fig. 4.2.8a can be constructed out of an infinite number of the elemental cubes show n in Fig. 4.2.8b, this result holds for any elemental volume irrespective of shape. Another way of saying this is that the sum of the normal strains zz yy xxεεε++ is an invariant and so also is the volumetric strain5. Thus no matter the coordinate system or strain field, the volumetric strain is alwa ys given by the sum of the normal strains zz yy xxεεε++ . From Hooke’s law, normal stresses cause norm al strain and shear stresses cause shear strain. It follows that normal stresses produce volume changes and shear stresses produce distortion (change in shape), but no volume change. 5 as with the sum of the normal stresses, Eqn. 3.5.2 xy zσ σσσ Section 4.2 Solid Mechanics Part I Kelly 106 Figure 4.2.8: A block of material subjected to a general strain 4.2.5 Two Dimensional Elasticity The above three-dimensional stress-strain relations reduce in the case of a two- dimensional stress state or a two-dimensional strain state. Plane Stress In plane stress, 0 ===zz yz xzσσσ , Fig. 4.2.9, so the stress- strain relations reduce to [] [] [] [] xy xyyy xx yyyy xx xxxy xyxx yy yyyy xx xx EEEEEE ενσενενσνεενσσνενσσενσσε +=+−=+−=+=−=−= 111111 22 Stress-Strain Relations (Plane Stress) (4.2.11) with [] 00 , ===== +−= yz xz zzyz xz yy xx zzE σσσεεσσνε (4.2.12) xy z )3(x)2(x )1(x principal directions )a(abc aΔbΔcΔ )b( Section 4.2 Solid Mechanics Part I Kelly 107 Figure 4.2.9: Plane stress - a thin component loaded in-plane Note that the zzε strain is not zero. Physically, zzε corresponds to a change in thickness of the thin sheet of material. Plane Strain In plane strain, 0 ===zz yz xzεεε , Fig. 4.2.10, so the stress-strain relations reduce to [ ] [] [] [] xy xyxx yy yyyy xx xxxy xyyy xx yyyy xx xx EEEEEE ενσνεενννσνεενννσσνεσν νσνενσσννε +=+−−+=+−−+=+=−+−+=−−+= 1) 1()21)( 1() 1()21)( 1(1) 1(1) 1(1 Stress-Strain Relations (Plane Strain) (4.2.13) with [] 0 ,0 == +==== yz xz yy xx zzyz xz zz σσσσνσεεε (4.2.14) Again, note here that the stress component zzσ is not zero. Physically, this stress corresponds to the forces pr eventing movement in the z direction (the loading on the component is not shown). y x z Section 4.2 Solid Mechanics Part I Kelly 108 Figure 4.2.10 Plane strain - a thick component constr ained in one direction Similar Solutions The expressions for plane stress and plane strain are very similar. For example, the plane strain constitutive law 4.2.13 can be derived from the corresponding plane stress expressions 4.2.11 by making the substitutions νννν ′−′=′−′=1,12EE (4.2.15) in 4.2.11 and then dropping the primes. The plane stress expressions can be derived from the plane strain expressions by making the substitutions () ννν νν ′+′= ′+′+′=1, 121 2EE (4.2.16) in 4.2.13 and then dropping the primes. Thus , if one solves a plane stress problem, one has automatically solved the corres ponding plane strain problem, and vice versa . 4.2.6 Problems 1. Steel and aluminium can be considered to be isotropic and homogene ous materials. Is the composite material shown here isotropi c and/or homogeneous? Everywhere in the sandwich? 2. Consider a very thin sheet of material subjected to a normal pressure p on one of its large surfaces. It is fixed along its edges. This is an example of a plate problem, an important branch of elasticity with applica tions to boat hulls, aircraft fuselage, etc. (a) write out the complete three dimensional stress-strain relations for this case (both cases, stress in terms of strain, strain in terms of stress) - simplify the relations using the fact that the sheet is thin, the stress boundary condition on the large face and the coordinate system shown (just substitute in appropriate values for yz xzσσ, and zzσ) aluminium steel zzσ Section 4.2 Solid Mechanics Part I Kelly 109(b) assuming that the through thickness change in the sheet can be neglected, show that ()yy xx p σσν+−= 3. A strain gauge at a certain point on the surface of a thin aluminium component (loaded in-plane) records strains of μm15 μm,30 μm,60 = = =xy yy xx ε ε ε . Determine the principal stresses. 4. A block of isotropic linear elastic material is subjected to a compressive normal stress oσ over two opposing faces. The material is constrained (prevented from moving) in one of the direction normal to thes e faces. The other faces are free. (a) What are the stresses and strain s in the block, in terms of Eo,,νσ ? (b) Calculate three maximum shear stresses, one for each plane (parallel to the faces of the block). Which of these is the overall maximum shear stress acting in the block? 5. Repeat problem 4a, only with the free faces now fixed also. 6. Use the stress-strain relations to prov e that, for a linear elastic solid, yy xxxy yy xxxy εεε σσσ −=−2 2 and, indeed, zz yyyz zz yyyz zz xxxz zz xxxz εεε σσσ εεε σσσ −=− −=−2 2,2 2 Note: from Eqns. 3.5.5 and 3.8.3, these s how that the principal axes of stress and strain coincide for an isotropic elastic material 7. Consider the case of hydrostatic pres sure in a linearly elastic solid: [] ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−− = ppp ij 0 00 00 0 σ as might occur, for example, when a s pherical component is surrounded by a fluid under high pressure, as illustrated in the figure below. Show that the volume tric strain is equal to () Epν213−− [Note: The factor relating the change in volum e and constant pressure is known as the Bulk modulus K: p xy z Section 4.2 Solid Mechanics Part I Kelly 110()ν213,/ /−= −=ΔEK Kp VV When ∞→= K,2/1ν , so that there is no change in volume, and hence no change in density (for constant mass element), and the material is said to be incompressible . For small deformations of rubber, 48.0≈ν .] 8. Consider again Problem 4 from §3.5. (a) Assuming the material to be linearly elas tic, what are the strains? Draw a second material element (superimposed on the one shown below) to show the deformed shape of the square element – assume th e displacement of the box-centre to be zero and that there is no rotation. No te how the free surface moves, even though there is no stress acting on it. (b) What are the principal strains 1ε and 2ε? You will see that the principal directions of stress and strain coincide (see Problem 6) – the largest normal stress and strain occur in the same direction 9. What is the volume change for a block of material which undergoes a simple shear deformation, as in Fig. 4.2.6? 10. A thin linear elastic plate is subjected to a uniform compressive stress 0σ as shown below. Show that the slope of the plate diagonal shown after deformation is given by ()⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ −+=+EE ab / 1/ 1tan 00 σνσδαα What is the magnitude of δα for a steel plate ( GPa210=E , 3.0=ν ) of dimensions 2cm2020× with MPa 10=σ ? 0σ0σ αxxσy xxσ Section 4.3 Solid Mechanics Part I Kelly 1114.3 One Dimensional Axial Deformations In this section, a specific simple geometry is considered, that of a long and thin straight component loaded in such a way that it de forms in the axial direction only. The x-axis is taken as the longitudinal axis, with the cross-section lying in the yx− plane, Fig. 4.3.1. Figure 4.3.1: A slender straig ht component; (a) longitudi nal axis, (b) cross-section 4.3.1 Basic relations for Axial Deformations Any static analysis of a stru ctural component involves the following three considerations: (1) constitutive response (2) kinematics (3) equilibrium In this Chapter, it is taken for (1) that the material responds as an isotropic linear elastic solid. It is assumed that the only significant stresses and strains occur in the axial direction, and so the stress- strain relations 4.2.8 reduce to the one-dimensional equation xx xxEεσ= or, dropping the subscripts, εσE= (4.3.1) Kinematics (2) is the study of deformati on, the subject of §3.6-3.8. In the theory developed here, known as axial deformation , it is assumed that the axis of the component remains straight and that cross-sect ions that are initially perpendicular to the axis remain perpendicular after deformation. This implies that, although the strain might vary along the axis, it remains constant over any cross section . As defined in the previous Chapter, the axial strain occurring over any section is given by 00 LLL−=ε (4.3.2) This is illustrated in Fig. 4.3.2, which shows a (shaded) re gion undergoing a compressive (negative) strain. Note that individual particles/points undergo displacements whereas regions/line- elements undergo strain. In Fig. 4.3.2, the particle originally at A has undergone a displacement )(Au whereas the particle originally at B has undergone a displacement )(Bu . From Fig. 4.3.2, another way of expressing the strain in the shaded region is xy z Section 4.3 Solid Mechanics Part I Kelly 1120)()( LAu Bu−=ε (4.3.3) Figure 4.3.2: axial strain; (a) before deformation, (b) after deformation Both displacements )(Au and ) (Bu of Fig. 4.3.2 are positive , since the particles displace in the positive x direction – if they move d to the left, for consistency, one would say they underwent negative displacements. Further, positive st resses are as shown in Fig. 4.3.3a and negative stresses are as shown in Fig. 4.3.3b. From Eqn. 4.3.1, a positive stress implies a positive strain (lengthening) and a compressive stress implies a negative strain (contracting) Figure 4.3.3: Stresses arising in the slender component; (a) positive (tensile) stress, (b) negative (compressive) stress Equilibrium, (3), will be considered in the individual examples below. Note that, in the previous section, problems were solved using only the stress-strain law (1). Kinematics (2) and equilibrium (3) were not considered, the reason being the problems were so simple, with uniform (homogene ous) stress and strain (as indeed also in the first example which follows). Whenever more complex problems encountered, with non-uniform stress and strains, (2) and (3) need to be considered to solve for the stress and strain. 4.3.2 Structures with Uniform Members A uniform axial member is one with cross-section A and modulus E constant along its length, and loaded with axial forces at its ends only. x x(a) (b)0L L•• ••)(Au )(BuA B x0>σ 0>σx0<σ 0<σ (a) (b) Section 4.3 Solid Mechanics Part I Kelly 113Example Consider the bar of initial length L shown in Fig. 4.3.4, subjected to equal and opposite end-forces F. The free-body (equilibrium) diagram of a section of the bar shown in Fig. 4.3.4b shows that the internal force is also F everywhere along the ba r. The stress is thus everywhere AF/=σ and the strain is everywhere EAF=ε ( 4 . 3 . 4 ) and, from Eq. 4.3.2, the bar extends in length by an amount EAFL=Δ (4.3.5) Note that although the force acting on the left-hand end is nega tive (acting in the x− direction), the stress there is positive (see Fig. 4.3.3). Figure 4.3.4: A uniform axial member; (a) subjected to axial forces F, (b) free-body diagram Displacements need to be calculated relative to some datum displacement. For example, suppose that the displacement at the centre of the bar is zero, 0 )(=Bu , Fig. 4.3.4. Then, from Eqn. 4.3.3, 2) ()( )(2) ()( )(4) ()( )( L EAFBA Bu AuL EAFBD Bu DuL EAFBC Bu Cu −=−+==−+==−+= εεε (4.3.6) ■ Example Consider the two-element structure shown in Fi g. 4.3.5. The first element is built-in at end A, is of length 1L, cross-sectional area 1A and Young’s modulus 1E. The second F FL F(a) (b)• B• •• A C D F Section 4.3 Solid Mechanics Part I Kelly 114element is attached at B and has properties 2 2 2 ,, EAL . External loads F and P are applied at B and C as shown. An unknown reaction force R acts at A. This can be determined from the force equilibrium equation for the structure: 0=+− PFR (4.3.7) As usual, the reaction is first assumed to act in the positive ( x) direction. With R known, the stress )1(σ in the first element can be evalua ted using the free-body diagram 4.3.5b, and )1(σ using 4.3.5c: 2)2( 1)1(,AP AFP=−= σ σ (4.3.8) and so the strain is 22)2( 11)1(,AEP AEFP=−= ε ε (4.3.9) Note that the stre ss and strain are discontinuous at B1. Figure 4.3.5: A two-element structure (a) subjected to axial forces F and P, (b,c) free-body diagrams For each element, the total elongations iΔ are () 222 2111 1 )()()()( AEPLBu CuAELFPAu Bu =−=Δ−=−=Δ (4.3.10) If FP>, then 01>Δ as expected, with 0<R and 0>σ . 1 this result, which can be viewed as a violation of equilibrium at B, is a result of the one-dimensional approximation of what is really a two-dimensional problem P1L (a) (b)B• • •A CF2L P •P (c)FPR−=−R FP− Section 4.3 Solid Mechanics Part I Kelly 115 Thus far, the stress and strain (and elongati ons) have been obtained. If one wants to evaluate displacements, then one needs to en sure that the strain s in each element are compatible , that is, that the elements fit together after deformation ju st like they did before deformation. In this example, the displacements at B and C are 2 1 )( )(, )( )( Δ+= Δ+= Bu Cu Au Bu (4.3.11) A compatibility condition , bringing together the separate relations in 4.3.11, is then () 222 111)( )(AEPL AELFPAu Cu +−+= (4.3.12) ensuring that ) (Bu is unique. As in the previous ex ample, the displacements can now be calculated if the displace ment at any one (datum) point is known. Indeed, it is known that 0)(=Au . ■ Example Consider next the similar situ ation shown in Fig. 4.3.6. Here, both ends of the two- element structure are built-in and there is only one applied force, F, at B. There are now two reaction forces, at ends A and C, but there is only one equilibrium equation to determine them: 0 =++C A RF R (4.3.13) Any structure for which there are more unknowns than equations of equilibrium, so that the stresses cannot be determined without considering the deform ation of the structure, is called a statically indeterminate structure2. Figure 4.3.6: A two-element structure built-i n and both ends; (a) subjected to an axial force F, (b,c) free-body diagrams 2 See the end of §2.3.3 1L (a) (b)B• • •A C F2L (c)• CR•CRC A RF R+=−CRF+AR CR Section 4.3 Solid Mechanics Part I Kelly 116 In terms of the unknown r eactions, the strains are 22 2)2( )2( 11 11 1)1( )1(,AER E AERF AER EC C A==+=−==σεσε (4.3.14) and, for each element, the total elongations are 222 2 111 1 ,AELR AELRC A=Δ =Δ (4.3.15) Finally, compatibility of both elemen ts implies that the total elongation 02 1=Δ+Δ . Using this relation with Eqn. 4.3.13-14 then gives 112 221221 112 221112,AEL AELAELF RAEL AELAELF RC A+−=++= (4.3.16) The displacements can now be evaluated, for example, 2 22 1 11 / /1)(LAELAEF Bu++= (4.3.17) so that a positive F displaces B to the right and a negative F displaces B to the left. ■ Note the general solution procedure in this last example, known as the basic force method : Equilibrium + Compatibility of St rain in terms of unknown Forces Æ Solve equations for unknown Forces The Stiffness Method The stiffness method (also known as the displacement method ) is a slight modification of the above solution procedure, where the final equations to be solved involve known forces and unknown displacements only: Equilibrium in terms of Displacement Æ Solve equations for unknown Displacements Example (The Stiffness Method) Consider a series of three ba rs of cross-sectional areas 3 2 1,, AAA , Young’s moduli 3 2 1,, EEE and lengths 3 2 1,,LLL , Fig. 4.3.7. The first and th ird bars are bui lt-in at points Section 4.3 Solid Mechanics Part I Kelly 117A and D, bars one and two meet at B and bars two and three meet at C. Forces BP and CP act at B and C respectively. The force is constant in each bar, and for each bar there is a rela tion between the force iF, and elongation, iΔ, Eqn. 4.3.5: ii ik FΔ= where iii iLEAk= (4.3.18) Here, ik is the effective stiffness of each bar. The elongations are related to the displacements, A Bu u−=Δ1 etc., so that, with 0==D Au u , ()C B C B uk F u uk F uk F3 3 2 2 1 1 , , −= −= = (4.3.19) There are two degrees of freedom in this problem, that is , two nodes are free to move. One therefore needs two equilibrium e quations. One could use any two of 0 ,0 ,03 2 2 1 3 1 =++−=++−=+++− F P F F P F F P P FC B C B (4.3.20) In the stiffness method, one uses the second and third of these; the second is the “node B” equation and the third is the “node C” equation. Substituting Eqns. 4.3.19 into 4.3.20 leads to the system of two equations () ()C C BB C BP uk k ukP uk ukk −=+−+−=++− 3 2 22 2 1 ( 4 . 3 . 2 1 ) which can be solved for the two unknown nodal displacements Figure 4.3.7: three bars in series; (a) subjected to external loads, (b,c,d) free-body diagrams (a) (b)B • • •A C (c) (d)BP •CPD •1F1F3F3F ••3F CP2F •BP •2F 1F • Section 4.3 Solid Mechanics Part I Kelly 118Note that it was not necessary to evaluate th e reactions to obtain a solution. Once the forces have been found, the reactions can be found using the free-body diagram of Fig. 4.3.7d. The stiffness method is a very systematic procedur e. It can be used to solve for structures with many elements, with the two equations 4.3.21 replaced by a large system of equations which can be solved numerically using a computer. 4.3.3 Structures with Non-uniform Members Consider the structure shown in Fig. 4.3.8, an axial bar consis ting of two separate components bonded together. The components have Young’s moduli 2 1,EE and cross- sectional areas 2 1,AA . The bar is subjected to equal and opposite forces F as shown, in such a way that axial deformations occur, that is, the cross-sections remain perpendicular to the x axis throughout the deformation. Since there are only axial deformations, the st rain is constant over a cross-section. However, the stress is not uniform, with εσ1 1E= and εσ2 2E= ; on any cross-section, the stress is higher in the stiffer compone nt. The resultant force acting on each component is ε11 1 AE F= and ε22 2 AE F= . Since F F F=+2 1 , the total elongation is 22 11 AE AEFL +=Δ (4.3.22) Figure 4.3.8: A bar consisting of two se parate materials bonded together 4.3.4 Resultant Force and Moment Consider the force and moments acting over any cross-section, Fig. 4.3.9. The resultant force is the integral of the stress times elemen tal area over the cross section, Eqn. 3.1.2, dA F A∫=σ (4.3.23) There are two moments; the moment yM about the y axis is the sum of the moments of the stresses () zdAσ , and similarly for zM, dAy M dAz M Az Ay ∫ ∫−= = σ σ, (4.3.24) F FL • • A B1E 2E1F1F 2F2F Section 4.3 Solid Mechanics Part I Kelly 119 Positive moments are defined through the right hand rule , i.e. with the thumb of the right hand pointing in the positive y direction, the closing of the fingers indicates the positive yM; the negative sign in Eqn. 4.3.24b is due to the fact that a positive stress with 0>y would lead to a negative moment zM. Figure 4.3.9: Resultants on a cross-sectio n; (a) resultant force, (b) resultant moments Consider now the case where the stress is constant over a cross-section . Since it is assumed that the strain is constant over the cr oss-section, from Eqn. 4.3.1 this will occur when the Young’s modulus is constant. In that case, Eqns. 4.3.23-24 can be re-written as dAy M dAz MA F Az Ay ∫ ∫−= = = σ σ σ , , (4.3.5) The quantities dAz A∫ and dAy A∫ are the first moments of area about, respectively, the y and z axes. These are equal to Az and Ay, where ),(zy are the coordinates of the centroid of the section (s ee Eqn. 3.2.2). Taking the x axis to run through the centroid, 0==zy results in 0==z yM M . Thus, a resultant axial force which acts through the centroid of the cross-section ensures that there is no moment/rotation of that cr oss-section, the main assumption of this section. For the non-uniform member of Fig. 4.3.8, sin ce the resultant of a c onstant stress over an area is a force acting through the cen troid of that area, the forces 2 1,FF act through the centroids of the respective areas 2 1,AA . The precise location of the total resultant force F can be determined by taking the moments of the forces 2 1,FF about the y and z axes, and equating this to the moment of the force F about these axes. 4.3.5 Problems 1. Consider the rigid beam supported by two deformable bars shown below. The bars have properties 1 1,AL and 1 2,AL and have the same Young’s modulus E. They are separated by a distance L. The beam supports an arbitrary load at position x, as shown. What is x if the beam is to remain horizontal after deformation. y zσy zσy F zM(a) (b)dA zyM Section 4.3 Solid Mechanics Part I Kelly 120 x1L2LL Section 4.4 Solid Mechanics Part I Kelly 1214.4 Torsion In this section, the geometry to be considered is that of a l ong slender circul ar bar and the load is one which twists the ba r. Such problems are important in the analysis of twisting components, for example lug wrenches and transmission shafts. 4.4.1 Basic relations for Torsion of Circular Members The theory of torsion presented here concerns torques 1 which twist the members but which do not induce any warping , that is, cross sections which are perpendicular to the axis of the member remain so after twisting. Further, radial line s remain straight and radial as the cross-sectio n rotates – they merely rotate with the section. For example, consider the member shown in Fig. 4.4.1, built-in at one end and subject to a torque T at the other. The x axis is drawn along its axis. The torque shown is positive, following the right-hand rule (see §4.3.4). The member twis ts under the action of the torque and the radial plane ABCD moves to DCAB′. Figure 4.4.1: A cylindrical member under the action of a torque Whereas in the last section the measure of deformation was elongation of the axial members, here an appropriate measure is the amount by which the member twists, the rotation angle φ. The rotation angle will vary along the member – the sign convention is that φ is positive in the same direction as positive T as indicated by the arrow in Fig. 4.4.1. Further, whereas the measure of strain used in the previous section was the normal strain xxε, here it will be the e ngineering shear strain xyγ (twice the tensorial shear strain xyε). A relationship between γ (dropping the subscript) and φ will next be established. As the line BC deforms into CB′, Fig. 4.4.1, it undergoes an angle change α. As defined in §3.6.2, the shear strain γ is the change in the orig inal right angle formed by BC and a tangent at B (indicated by the dotted line – this is the y axis to be used in xyγ). If α is small, then LLR BCCC )(tanφααγ ≈′=≈= (4.4.1) 1 the term torque is usually used instead of moment in the context of twisting shafts such as those considered in this section TxC B A DC′αφ L Section 4.4 Solid Mechanics Part I Kelly 122where L is the length, R the radius of the member and )(Lφ means the magnitude of φ at L. Note that the strain is constant along the length of the member although φ is not. Considering a general cro ss-section within the member, as in Fig. 4.4.2, one has xxR )(φαγ≈= (4.4.2) Figure 4.4.2: A section of a twisting cylindrical member The shear strain at an arbitrary radial location r, Rr<<0 , is evidently xxrr)()(φγ= (4.4.3) showing that the shear strain varies fr om zero at the centre of the shaft to a maximum () xxR LLR /)( /)( φ φ= on the outer surface of the shaft. Considering a free-body diagram of any portion of the shaft of Fig. 4.4.1, a torque T acts on all cross-sections. This torque must equal the resultant of the shear stresses acting over the section, as schematically i llustrated in Fig. 4.4.3a. The elemental force acting over an element with sides dr and θrd is θττ rdrd dA= , Fig. 4.4.3b, and so the resultant moment about 0=r is drrr drdrr TR R )( 2 )( 022 002∫ ∫∫= = τπθτπ (4.4.4) Hooke’s law is γτG= (4.4.5) where G is the shear modulus (the μ of Eqn. 4.2.5). But r/γ is a constant and so therefore also is r/τ (provided G is) and Eqn. 4.4.4 can be re-written as rJrdrrrrTR)(2)( 03τπτ=⎥ ⎦⎤ ⎢ ⎣⎡=∫ (4.4.6) The quantity in square brackets is called the polar moment of inertia of the cross- section (also called the polar second moment of area ) and is denoted by J. For this circular cross-sec tion it is given by TE B A FE′αφ x Section 4.4 Solid Mechanics Part I Kelly 123 32 224 4 03 D Rdrr JRπππ == =∫ (4.4.7) where D is the diameter. In general, for a cross-section of arbitrary shape, dAr J A∫=2 Polar Moment of Area (4.4.8) where dA is an element of area and the integra tion is over the complete cross-section. Figure 4.4.3: Shear stresses acting over a cross-section; (a) shear stress, (b) moment for an elemental area From Eqn. 4.4.6, the shear stress at any radial location is given by JrTr=)(τ (4.4.9) From Eqn. 4.4.1, 4.4.5 and 4.4.9, the angle of twist at the end of the member – or the twist at one end relative to th at at the other end – is GJTL=φ (4.4.10) Example Consider the problem shown in Fig.4.4.4, two torsion members of lengths 2 1,LL , diameters 2 1,dd and shear moduli 2 1,GG , built-in at A and subjected to torques BT and CT. Equilibrium of moments can be used to determine the unknown torques acting in each member: 0 ,02 1 =+−=++−C C B TT T TT (4.4.11) so that C BT TT+=1 and CT T=2 . τ )(rτ r (a) (b)θrdrd dA= Section 4.4 Solid Mechanics Part I Kelly 124 Figure 4.4.4: A structure consisting of two to rsion members; (a) subjected to torques BT and CT, (b,c) free-body diagrams The shear stresses in each member are therefore () 22 11 ,JrT JT TrC C B=+= τ τ (4.4.12) where 32 /4 1 1 d Jπ= and 32 /4 2 2 d Jπ= . From Eqn. 4.4.10, th e angle of twist at B is given by 11 11/JGLTB=φ . The angle of twist at C is then B CJGLTφ φ −= 2222 (4.4.13) ■ Statically indeterminate problems can be solv ed using methods analogous to those used in the previous section for uniaxial members. Example Consider the structure in Fig. 4.4.5, similar to that in Fig. 4.4.4 only now both ends are built-in and there is only a single applied torque, BT. AxDφCTBT B C 1L2L 1T(a) 2T(b) (c) Section 4.4 Solid Mechanics Part I Kelly 125 Figure 4.4.5: A structure consisting of tw o torsion members; (a) subjected to a Torque BT, (b) free-body diagram, (c) separate elements Referring to the free-body diagram of Fig. 4.4.5b, there is only one equation of equilibrium with which to determin e the two unknown member torques: 02 1 =++− T TTB (4.4.14) and so the deformation of the structure need s to be considered. A systematic way of dealing with this situation is to consider each element separately, as in Fig. 4.4.5c. The twist in each element is 2222 2 1111 1 ,JGLT JGLT= =φ φ (4.4.15) The total twist is zero and so 02 1=+φφ which, with Eqn. 4.4.14, can be solved to obtain B B TJGL JGLJGLT TJGL JGLJGLT 112 221221 2 112 221112 1 ,+−=++= (4.4.16) The rotation at B can now be determined, 2 1φφφ−==B . ■ 4.4.2 Stress Distribution in Torsion Members The shear stress in Eqn. 4.4.9 is acting over a cross-section of a torsion member. It follows from §3.4.1 that shear stresses act also along the length of the member, as illustrated to the left of Fig. 4.4.6. Shear stresses do not act on the surface of the element shown, as it is a free surface. AxDφBT B C 1L2L 1T(a) (b) 2T BT (c)1T1T2T2T Section 4.4 Solid Mechanics Part I Kelly 126Any element of material not aligned with the axis of the cylinder will undergo a complex stress state, as shown to the right of Fig. 4.4.6. The stresses acting on an element are given by the stress tran sformation equations, Eqns. 3.4.7: θτ σθτ σθτ σ 2cos ,2sin ,2sin +=′ −=′ +=′xy yy xx (4.4.17) Figure 4.4.6: Stress distribution in a torsion member From Eqns. 3.5.5-6, the maximum normal (principal) stresses arise on planes at o45±=θ and are τσ+=1 and τσ−=2 . Thus the maximum tensile stress in the member occurs at o45 to the axis and arises at the surface. The maximum shear stress is simply τ, with 0=θ . 4.4.3 Problems 1. A shaft of length L and built-in at both ends is subj ected to two external torques, T at A and T2 at B, as shown below. The shaft is of diameter d and shear modulus G. Determine the maximum (absolute value of) shear stress in the shaft and determine the angle of twist at B. T2 4/LT 4/L 2/LA B BφTτ τ 0=τx′ y′x yθ xyσ′xxσ′yyσ′ Section 4.5 Solid Mechanics Part I Kelly 1274.5 The Thin-walled Pressure Vessel Theory An important practical problem is that of a container subjected to an internal pressure p. Such a container is called a pressure vessel , Fig. 4.5.1. In many applications it is convenient and valid to assume that (i) the material is isotropic (ii) the strains resulting from the pressures are small (iii) the wall thickness t of the pressure vessel is much smaller than some characteristic radius: i o i o rr rrt ,<<−= Figure 4.5.1: A pressurised container Because of (i,ii), the isotropic linear Elastic model is used. Because of (iii), it will be assumed that there is negligible variation in the stress field across the thickness of the vessel, Fig. 4.5.2. Figure 4.5.2: Approximation to the st ress arising in a pressure vessel As a rule of thumb, if the thickness is less than a tenth of the vessel ra dius, then the actual stress will vary by less than about 5% through the thickness, and in these cases the constant stress assumption is valid. Note that a pressure i zz yy xx p−===σσσ means that the stress on any plane drawn inside the vessel is subjected to a normal stress ip− and zero shear stress. ir2 or2p t p tσσσσ tactual stress approximate stress p Section 4.5 Solid Mechanics Part I Kelly 1284.5.1 Thin Walled Spheres A thin-walled spherical shell is shown in Fig. 4.5.3. Because of the symmetry of the sphere and of the pressure loading, the circumferential (or tangential or hoop) stress tσ at any location and in any tangentia l orientation must be the same. Figure 4.5.3: a thin-walled spherical pressure vessel Considering a free-body diagram of one half of the sphere, Fig. 4.5.4, force equilibrium requires that () 02 2 2=−+−t i o i r r pr σππ (4.5.1) and so () trrpr ii t+= 02 σ ( 4 . 5 . 2 ) Figure 4.5.4: a free body diagram of one half of the spherical pressure vessel One can now take as a characte ristic radius the dimension r. This could be the inner radius, the outer radius, or the average of th e two – results for all three should be close: tpr t2=σ Tangential stress in a thin-wa lled spherical pressure vessel (4.5.3) This tangential stress accounts for the stress in the plane of the surface of the sphere. The stress normal to the walls of the sphere is called the radial stress , rσ. The radial stress is zero on the outer wall since that is a free surf ace. On the inner wall, the normal stress is pr−=σ , Fig. 4.5.5. From Eqn. 4.5.3, since 1 /<<rt , t pσ<<, and it is reasonable to tσptσ tσ Section 4.5 Solid Mechanics Part I Kelly 129take 0=rσ not only on the outer wall, but on the i nner wall also. The stress state in the spherical wall is then one of plane stress. Figure 4.5.5: An element at the surf ace of a spherical pressure vessel There are no in-plane shear stresses in the s pherical pressure vessel and so the tangential and radial stresses are the principal stresses: tσσσ==2 1 , and the minimum principal stress is 03==rσσ . Thus the radial direction is one principal direction, and any two perpendicular directions in the plane of the sphere’s wall can be taken as the other two principal directions. Strain in the Thin-walled Sphere The thin-walled pressure vessel expands when it is internally pressurised. This results in three principal strains, the circumferential strain cε (or tangential strain tε) in two perpendicular in-plane directions, and the radial strain rε. Referring to Fig. 4.5.6, these strains are ABAB BA CDCD DC ACAC CA r c−′′=−′′=−′′= ε ε , (4.5.4) Figure 4.5.6: Strain of an element at the surface of a spherical pressure vessel p ABC A′B′C′ before afte rDD′tσ tσtσ 0≈−=prσ 0=rσ Section 4.5 Solid Mechanics Part I Kelly 130From Hooke’s law (Eqns. 4.2.8 with z the radial direction, with 0=rσ ), ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ −−− −−− = ⎥⎥⎥ ⎦⎤ ⎢⎢⎢ ⎣⎡ ννν σσσ ννν ννν εεε 211 21 /1 / // /1 // / /1 tpr EE E EE E EE E E rtt rcc (4.5.5) To determine the amount by which the vessel expands, consider a circumference at average radius r which moves out with a displacement rδ, Fig. 4.5.7. From the definition of normal strain () r rr rr r cδ θθθδε =ΔΔ−Δ+= (4.5.6) This is the circumferential strain for points on the mid-radius. The strain at other points in the vessel can be approximated by this value. The expansion of the sphere is thus tpr Erc r212νεδ−== (4.5.7) Figure 4.5.7: Deformation in the thin -walled sphere as it expands To determine the amount by which the circumfe rence increases in size, consider Fig. 4.5.8, which shows the origin al circumference at radius r of length c increase in size by an amount cδ. One has tpr Er cc c c212 22νπεπεδ−=== (4.5.8) It follows from Eqn. 4.5.7-8 that the circ umference and radius increases are related through r cπδδ 2= (4.5.9) ir orrrδ θΔ()θδΔ+rr Section 4.5 Solid Mechanics Part I Kelly 131 Figure 4.5.8: Increase in circumferen ce length as the vessel expands Note that the circumferential strain is positive , since the circumference is increasing in size, but the radial strain is negative , since, as the vessel expands, the thickness decreases. 4.5.2 Thin Walled Cylinders The analysis of a thin-walled internally-pressuri sed cylindrical vessel is similar to that of the spherical vessel. The main difference is that the cylinder has th ree different principal stress values, the circumferential stress, the radial stress, and the longitudinal stress lσ, which acts in the direction of the cylinder axis, Fig. 4.5.9. Figure 4.5.9: free body diagram of a cylindrical pressure vessel Again taking a free-body diagram of the cylinder and carrying out an equilibrium analysis, one finds that, as for the spherical vessel, tpr l2=σ Longitudinal stress in a thin-walled cylindrical pressure vessel (4.5.10) Note that this analysis is only valid at pos itions sufficiently far away from the cylinder ends, where it might be closed in by caps – a more complex stress field would arise there. The circumferential stress can be evaluated from an equilibrium analysis of the free body diagram in Fig. 4.5.10: 0 2 2 =+− Lpr tLi cσ (4.5.11) and so r r cπ2=r rδ+ ()r c r c δπδ+=+ 2 lσp Section 4.5 Solid Mechanics Part I Kelly 132 tpr c=σ Circumferential stress in a thin-w alled cylindrical pressure vessel (4.5.12) Figure 4.5.10: free body diagram of a cylindrical pressure vessel As with the sphere, the ra dial stress varies from p− at the inner surface to zero at the outer surface, but again is small compared with the other two stresses, and so is taken to be 0=rσ . Strain in the Thin-walled cylinder The analysis of strain in the cylindrical pressure vessel is very similar to that of the spherical vessel. Eqns. 4.5.6 and 4.5.9 hold also here. Eqn. 4.5.5 would need to be amended to account for the three different principal stresses in the cylinder. 4.5.3 External Pressure The analysis given above can be extended to the case where there is also an external pressure acting on the vessel. The in ternal pressure is now denoted by ip and the external pressure is denoted by op, Fig. 4.5.11. Figure 4.5.11: A pressure vessel subjected to internal and external pressure In this case, the pressure p in formulae derived above can simply be replaced by ) (o ip p− , which is known as the gage pressure (see the Appendix to this section, §4.5.5). op ir2 or2ip tcσLp Section 4.5 Solid Mechanics Part I Kelly 1334.5.4 Problems 1. A 20m diameter spherical tank is to be used to store gas. The shell plating is 10 mm thick and the working stress of the material , that is, the maximum stress to which the material should be subjected, is 125 MPa. What is the maximum permissible gas pressure? 2. There are no shear stresses in the tangential plane of the spherica l pressure vessel. However, there are shear stresses acting on planes through the thic kness of the wall. A cross-section through the thickness is shown below. Take it that the radial stresses are zero. What are the maximum shear stre sses occurring on this cross section? 3. A steel propane tank for a BBQ grill has a 25cm diameter1 and a wall thickness of 5mm (see figure). The tank is pressurised to 1.2 MPa. (a) determine the longitudinal and circumfere ntial stresses in the cylindrical body of the tank (b) determine the absolute maximum shear stress in the cylindrical portion of the tank (c) determine the tensile force per cm length of the weld between the upper and lower sections of the tank. 4. What are the strains in the BBQ tank of ques tion 3? What is the radial displacement? [take the steel to be isotropic with 0.3 GPa,200 = = ν E ] 5. What are the strains in the cylindric al pressure vesse l, in terms of E, ν, p, t and r? 6. The three perpendicular planes in the cylin drical pressure vessel are the in-plane, through the thickness and longitudinal sec tions, as shown below. The non-zero (principal) stresses acting on these planes are also shown. Evaluate the maximum shear stresses on each of these three planes. Which of these three maxima is the overall maximum shear stre ss acting in the vessel? 1 this is an average diameter – the inside is 250-5mm and the outside is 250+5mm Weld tσtσ Section 4.5 Solid Mechanics Part I Kelly 134 4.5.5 Appendix to §4.5 Equilibrium of a Pressure Vessel with both internal and external pressure Consider the spherical pressure vessel. An external pressure op is distributed around its outer surface. Consider a fr ee-body diagram of one half of the vessel, as shown below. The force due to the external pressure acting in the horizontal direction can be evaluated using the spherical coordinates shown below. An element of surface area upon which the pres sure acts, swept out when the angles change by θd and φd, has sides θrd and φθd rsin . The force acting on this area is then φθθdd rposin2. Force equilibrium in the horizontal ( y) direction then leads to () 0 sin sin2 2 2 0 02 2=+−− − ∫∫ i i t i o o o pr r r d d pr πσπφφθθπ π xyz •()()φθ,, ,, r zyx≡ φrθ • φθd rsinθrdtσipopcσcσcσ cσlσlσ lσlσ in-plane through the thickness longitudinal section Section 4.5 Solid Mechanics Part I Kelly 135and so, () trrprpr io o i i t+−= 02 2 σ or trp po i t 2/) (−≈σ - see Eqn. 4.5.3. Section 4.6 Solid Mechanics Part I Kelly 1364.6 The Elementary Beam Theory In this section, problems i nvolving long and slender beams are addressed. As with pressure vessels, the geometry of the beam, and the specific type of loading which will be considered, allows for approximations to be made to the full three-dimensional linear elastic stress-strain relations. 4.6.1 The Beam The term beam has a very specific meaning in engine ering mechanics: it is a component that is designed to support transverse loads , that is, loads that act perpendicular to the longitudinal axis of the beam, Fig. 4.6.1. The beam supports the load by bending only. Other mechanisms, for example twisting of the beam, are not allowed for in this theory. Figure 4.6.1: A supported beam loaded by a force and a distribution of pressure It is convenient to show a two-dimensional cross-section of the three-dimensional beam together with the beam cross section, as in Fig. 4.6.1. The cross section of this beam happens to be rectangular but it can be any of many possible shapes. It will assumed that the beam has a longitudinal plane of symmetry , with the cross section symmetric about this plane, as shown in Fig. 4.6.2. Further, it will be assumed th at the loading and supports are also symmetric about this plan e. With these conditions, the beam has no tendency to twist and will undergo bending only 1. Figure 4.6.2: The longitudinal plane of symmetry of a beam Imagine now that the beam consists of ma ny fibres aligned longitudinally, as in Fig. 4.6.3. When the beam is bent by the action of do wnward transverse load s, the fibres near the top of the beam contract in length wher eas the fibres near the bottom of the beam extend. Somewhere in between, there will be a plane where the fibres do not change length. This is called the neutral surface . The intersection of the longitudinal plane of symmetry and the neutral surface is called the axis of the beam , and the deformed axis is called the deflection curve . 1 certain special cases, where there is not a plane of symmetry for geometry and/or loading, can lead also to bending with no twist, but th ese are not considered here longitudinal plane of symmetr y roller support pin support applied force applied pressure cross section Section 4.6 Solid Mechanics Part I Kelly 137 Figure 4.6.3: the neutral surface of a beam A conventional coordinate system is a ttached to the beam in Fig. 4.6.3. The x axis coincides with the (longitudina l) axis of the beam, the y axis is in the transverse direction and the longitudinal plane of symmetry is in the yx− plane, also called the plane of bending . 4.6.2 Moments and Forces in a Beam Normal and shear stresses act over any cross section of a beam, as shown in Fig. 4.6.4. The normal and shear stresses acting on each si de of the cross section are equal and opposite for equilibrium, Fig. 4.6.4b. The normal stresses σ will vary over a section during bending. Referring again to Fig. 4.6.3, ove r one part of the section the stress will be tensile, leading to extension of material fibres, whereas over the other part the stresses will be compressive, leading to contraction of material fibres. This distribution of normal stress results in a moment M acting on the section, as illu strated in Fig. 4.6.4c. Similarly, shear stresses τ act over a section and thes e result in a shear force V. The beams of Fig. 4.6.3 and Fig. 4.6.4 show the normal stress and deflection one would expect when a beam bends downward. There ar e situations when parts of a beam bend upwards, and in these cases the signs of the normal stresses will be opposite to those shown in Fig. 4.6.4. However, the moments (and shear forces) shown in Fig. 4.6.4 will be regarded as positive . This sign convention to be used is shown in Fig. 4.6.5. x y z fibres extending fibres contracting neutral surface Section 4.6 Solid Mechanics Part I Kelly 138 Figure 4.6.4: stresses and moments acti ng over a cross-section of a beam Figure 4.6.5: sign convention for moments and shear forces Note that the sign convention for the shear stre ss in the beam theory conflicts with the sign convention for shear stress used in the re st of mechanics, introduced in Chapter 3. This is shown in Fig. 4.6.6. Figure 4.6.6: sign convention for shear stress in beam theory The moments and forces acting within a beam can in many simple problems be evaluated from equilibrium considerations alone. Some examples are given next. cross-section in beam VV M Mσ σ ττ)a( )b( )c( positive bending positive shearing )a() b( )c(VV M M Mechanics (in general) Beam Theory ττ xy Section 4.6 Solid Mechanics Part I Kelly 139Example 1 Consider the simply supported beam in Fig. 4.6.7. The reaction at the roller support, end A, and the vertical reac tion at the pin support2, end B, can be evaluated from the equations of equilibrium, Eqns. 2.3.3: 3/2 ,3/ P R P RBy Ay = = (4.6.1) Figure 4.6.7: a simply supported beam The moments and forces acting within the beam can be evaluated by taking free-body diagrams of sections of the beam. There are clearly two distinct regions in this beam, to the left and right of the load. Fig. 4.6.8a shows an arbitrary portion of beam representing the left-hand side. A coordinate system has been introduced, with x measured from A3. An unknown moment M and shear force V act at the end. A positive moment and force have been drawn in Fig. 4.6.8a. From the equilibrium equations, one finds that the shear force is constant but that the mo ment varies linearly along the beam: xPMPV3,3= = )320(lx<< (4.6.2) Figure 4.6.8: free body diagram s of sections of a beam Cutting the beam to the right of the load, Fig. 4.6.8b, leads to 2 the horizontal reaction at the pin is zero since there are no applied forces in this direction; the beam theory does not consider such types of load 3 the coordinate x can be measured from any point in the beam; in this example it is convenient to measure it from point A xA 3/PVM )a() b(xA 3/PVM3/2l P3/2l Pdeflection curve lA B AyRByR Section 4.6 Solid Mechanics Part I Kelly 140 ()xlPMPV −= −=32,32 )32( lxl<< (4.6.3) The shear force is negative, so acts in the di rection opposite to that initially assumed in Fig. 4.6.8b. The results of the analysis can be displayed in what are known as a shear force diagram and a bending moment diagram , Fig. 4.6.9. Note that there is a “jump” in the shear force at 3/2l x= equal to the applied force, and in this example the bending moment is everywhere positive. Figure 4.6.9: results of anal ysis; (a) shear force diagram, (b) bending moment diagram ■ Example 2 Fig. 4.6.10 shows a cantilever , that is, a beam supported by clamping one end (refer to Fig. 2.3.8), and loaded by a force at its mi d-point and a (negative) moment at its end. Figure 4.6.10: a cantilevered beam loaded by a force and moment Again, positive unknown reactions AM and AV are considered at the support A. From the equilibrium equations, one finds that kN5 , kNm11 −= =A A V M (4.6.4) )a( )b(V M 3/2l92Pl 3P 32P− l l kN5 m3m 3 AV AMkNm4 A Section 4.6 Solid Mechanics Part I Kelly 141As in the previous example, there are two dist inct regions along the beam, to the left and to the right of the applied concentr ated force. Again, a coordinate x is introduced and the beam is sectioned as in Fig. 4.6.11. The unknown moment M and shear force V can then be evaluated from the equilibrium equations: () () 6x3 kNm4 ,03x0 kNm511 ,kN5 << −= =<< −= −= M Vx M V (4.6.5) Figure 4.6.11: free body diagrams of sections of a beam The results are summarized in the shear fo rce and bending moment diagrams of Fig. 4.6.12. Figure 4.6.12: results of an alysis; (a) shear force diagram, (b) bending moment diagram In this example the beam experiences nega tive bending moment over most of its length. ■ Example 3 Fig. 4.6.13 shows a simply supported beam subj ected to a distributed load (force per unit length). The load is uniformly distributed over half the length of the beam, with a triangular distribution over the remainder. )a( )b(V M m311 5− m64−x5− VM )a() b(xAVMm3kN5 11 5− 11 A Section 4.6 Solid Mechanics Part I Kelly 142 Figure 4.6.13: a beam subject ed to a distributed load The unknown reactions can be determined by repl acing the distributed lo ad with statically equivalent forces as in Fig. 4.6.14 (refer to §3.1.2). The equilibrium equations then give N140 ,N220 = =C A R R (4.6.6) Figure 4.6.14: equivalent forces acti ng on the beam of Fig. 4.6.13 Referring again to Fig. 4.6.13, there are two distinct regions in th e beam, that under the uniform load and that under the triangular distribution of load. The first case is considered in Fig. 4.6.15. Figure 4.6.15: free body diagram of a section of a beam The equilibrium equations give ()6x0 20 220 ,40 2202<< −= −= x x M x V (4.6.7) The region beneath the triangular distribution is shown in Fig. 4.6.16. Two possible approaches are illustrated: in Fig. 4.6.16a, the free body diagram consists of the complete length of beam to the left of the cross-sec tion under consideration; in Fig. 4.6.16b, only N/m40 x 220VMN240 m4 m3 m3 m2N120 ARCRN/m40 m6m 6A B C Section 4.6 Solid Mechanics Part I Kelly 143the portion to the right is considered, with di stance measured from the right hand end, as x−12 . The problem is easier to solve using th e second option. From Fig. 4.6.16b then, with the equilibrium equations, one finds that () 12x69/) 12(10) 12(140 ,3/) 12(10 1403 2<< −−−= −+−= x x M x V (4.6.8) Figure 4.6.16: free body diagrams of sections of a beam The results are summarized in the shear fo rce and bending moment diagrams of Fig. 4.6.17. Figure 4.6.17: results of an alysis; (a) shear force diagram, (b) bending moment diagram ■ 4.6.3 The Relationship betw een Loads, Shear Forces and Bending Moments Relationships between the applied loads and the internal shear fo rce and bending moment in a beam can be established by considering a small beam element, of width xΔ, and subjected to a distributed load ) (xp which varies along the sec tion of beam, and which is positive upward , Fig. 4.6.18. )a( )b(V M 600 140− m6 m6220 m6 m6x x−12 220 140VM VM )a() b( Section 4.6 Solid Mechanics Part I Kelly 144 Figure 4.6.18: forces and moments ac ting on a small element of beam At the left-hand end of the free body, at position x, the shear force, moment and distributed load have values ) (xF , )(xM and ) (xp respectively. On the right-hand end, at position x xΔ+ , their values are sl ightly different: ) ( x xFΔ+ , )( x xMΔ+ and ) ( x xpΔ+ . Since the element is very small, the di stributed load, even if it is varying, can be approximated by a linear variation over the element. The distributed load can therefore be considered to be a uniform distribution of intensity ) (xp over the length xΔ together with a triangular distribution, 0 at x and pΔ say, a small value, at x xΔ+ . Equilibrium of vertical forces then gives p xpxxVx xVx xVxp xxp xV Δ+=Δ−Δ+→=Δ+−ΔΔ+Δ+ 21)()( ) (0) (21)( )( (4.6.9) Now let the size of the element decrease towards zero. The left-hand side of Eqn. 4.6.9 is then the definition of the derivative, and the second term on the right-hand side tends to zero, so )(xpdxdV= (4.6.10) This relation can be seen to hold in Eqn. 4.6.7 and Fig. 4.6.17a, where the shear force over 6 0<<x has a slope of 40− and the pressure distributi on is uniform, of intensity N/m40− . Similarly, over 12 6<<x , the pressure decreases linearly and so does the slope in the shear force diagram, reachi ng zero slope at th e end of the beam. It also follows from 4.6.10 that the change in shear along a beam is equal to the area under the distributed load curve: () dxxp xV xVx x∫=−2 1)( )(1 2 (4.6.11) Consider now moment equilibrium, by taking moments about the point A in Fig. 4.6.18: x)(xV xΔ)(xM ) ( x xVΔ+) ( x xMΔ+ x xΔ+)(xp pΔ •A Section 4.6 Solid Mechanics Part I Kelly 1456 2)( )()( ) (03 21 2)( ) ( )( )( xpxxp xVxxMx xMxxpxxxpx xMxxVxM ΔΔ+Δ+=Δ−Δ+→=ΔΔΔ−ΔΔ−Δ++Δ−− (4.6.12) Again, as the size of the element decreases towards zero, the left-hand side becomes a derivative and the second and third terms on the right-hand side tend to zero, so that )(xVdxdM= (4.6.13) This relation can be seen to hold in Eqns. 4.6. 2-3, 4.6.5 and 4.6.7-8. It also follows from Eqn. 4.6.13 that the change in moment along a beam is equal to the area under the shear force curve: () dxxV xM xMx x∫=−2 1)( )(1 2 (4.6.14) 4.6.4 Deformation and Flexural Stresses in Beams The moment at any given cross-section of a b eam is due to a distribution of normal stress, or flexural stress (or bending stress ) across the section (see Fig. 4.6.4). As mentioned, the stresses to one side of the neutral axis are tensile whereas on th e other side of the neutral axis they are compressive. To determ ine the distribution of normal stress over the section, one must determine the precise loca tion of the neutral axis, and to do this one must consider the deformation of the beam. Apart from the assumption of there being a l ongitudinal plane of symmetry and a neutral axis along which material fibres do not extend, the following two assumptions will be made concerning the deformation of a beam: 1. cross sections which are plane and are perp endicular to the axis of the undeformed beam remain plane and remain perpendicula r to the deflection curve of the deformed beam. In short: “plane sections remain plane” . This is illustrated in Fig. 4.6.19. It will be seen later that this assumption is a valid one provided the beam is sufficiently long and slender. 2. deformation in the vertical direct ion, i.e. the transverse strain yyε, may be neglected in deriving an expression for the longitudinal strain xxε. This assumption is summarised in the deformation shown in Fig. 4.6.20, which shows an element of length l and height h undergoing transverse a nd longitudinal strain. Section 4.6 Solid Mechanics Part I Kelly 146 Figure 4.6.19: plane sections remain pl ane in the elementary beam theory Figure 4.6.20: transverse strain is negl ected in the elementary beam theory With these assumptions, consider now the element of beam shown in Fig. 4.6.21. Here, two material fibres ab and pq, of length xΔ in the undeformed beam, deform to ba′′ and qp′′. The deflection curve has a radius of curvature R. The above two assumptions imply that, referring to the figure: 2/π=′′′∠=′′′∠ qba bap (assumption 1) qb bq pa ap ′′=′′= , (assumption 2) (4.6.15) Since the fibre ab is on the neutral axis, by definition ab ba=′′ . However the fibre pq, a distance y from the neutral axis, extends in length from xΔ to length x′Δ. The longitudinal strain for this fibre is Ry RR yR xx x xx −=ΔΔ−Δ−=ΔΔ−′Δ=θθθε) ( (4.6.16) As one would expect, this relation implies that a small R (large curvature) is related to a large strain and a large R (small curvature) is related to a small strain. Further, for 0>y (above the neutral axis), the st rain is negative, whereas if 0<y (below the neutral axis), the strain is positive4, and the variation across th e cross-section is linear. 4 this is under the assumption that R is positive, which means that the beam is concave up; a negative R implies that the centre of cu rvature is below the beam plane in deformed beam remains perpendicular to the deflection curve deflection curve xy h ldl0 , ≈−= =hdh ldl yy xxε εdh Section 4.6 Solid Mechanics Part I Kelly 147 Figure 4.6.21: deformation of material fibres in an element of beam To relate this deformation to the stresses ar ising in the beam, it is necessary to postulate the stress-strain law for the material out of which the beam is made. Here, it is assumed that the beam is isotropic linear elastic 5. Since there are no forces acting in the z direction, the beam is in a state of plane stress, and the stress-strain equations are (see Eqns. 4.2.11) [] [] [] 0 ,111 ==+=+−=−=−= yz xz xy xyyy xx zzxx yy yyyy xx xx EEEE εεσνεσσνενσσενσσε (4.6.17) Yet another assumption is now made, th at the transverse normal stresses, yyσ, may be neglected in comparison with the flexural stresses xxσ. This is similar to the above assumption #2 concerning the deformation, where the transverse normal strain was neglected in comparison with the longitudinal strain. It might seem strange at first that the transverse stress is neglected , since all loads are in the tran sverse direction. However, just as the tangential stresses are much larger than the radial stresses in the pressure vessel, it is found that the longi tudinal stresses in a beam are very much greater than the transverse stresses. With this assumption, the first of Eqn. 4.6.17 reduces to a one- dimensional equation: E xx xx /σε= ( 4 . 6 . 1 8 ) 5 the beam theory can be extend ed to incorporate more complex material models (see Part II) before after •R b a′b′p p′q′ q •••xΔ yθΔ y x′Δ ••• • ax Section 4.6 Solid Mechanics Part I Kelly 148 and, from Eqn. 4.6.16, dropping the subscripts on σ, yRE−=σ (4.6.19) Finally, the resultant force of the normal stress distribution over the cr oss-section must be zero, and the resultant moment of the distribution must be M, leading to the conditions dAyydAyREdAy MdAyREdA A A AA A ∫∫∫∫∫ −= =−=−== 2 20 σσσ (4.6.20) and the integration is over the complete cross-sectional area A. The minus sign in the second of these equations arises becau se a positive moment and a positive y imply a compressive (negative) stress (see Fig. 4.6.4). The quantity dAy A∫ is the first moment of area about the neutral axis, and is equal to Ay, where y is the centroid of the section (see, for example, §3.2.1). Note that the horizontal component of the centroid will always be at the centre of the beam due to the symmetry of the beam about the plane of bending. Since the first moment of area is zero, it follows that the neutral axis passes through th e centroid of the cross-section . The quantity dAy A∫2 is called the second moment of area or the moment of inertia about the neutral axis, and is denoted by the symbol I. It follows that the flexural stress is related to the moment through IMy−=σ Flexural stress in a beam (4.6.21) The Moment of Inertia The moment of inertia depends on the shape of a beam’s cross-section. Consider the important case of a rectangular cross section. Before determining the moment of inertia one must locate the centroid (neutral axis). Due to symmetry, the neutral axis runs through the centre of the cro ss-section. To evaluate I for a rectangle of height h and width b, consider a small strip of height dy at location y, Fig. 4.6.22. Then 123 2/ 2/2 2 bhdyyb dAy Ih h A= ==∫∫+ − (4.6.22) This relation shows that the “t aller” the cross-section, the la rger the moment of inertia, something which holds generally for I. Further, the larger is I, the smaller is the flexural stress, which is always desirable. Section 4.6 Solid Mechanics Part I Kelly 149 Figure 4.6.22: Evaluation of the moment of inertia for a rect angular cross-section Similarly, it can be shown that the moment of inertia of a circular cross-section with radius r is given by 44rIπ= Example Consider the beam shown in Fig. 4.6.23. It is loaded symmetrically by two concentrated forces and has a circular cr oss-section of radius 100mm. The reactions at the two supports are found to be 100N. Sectioning the beam to the left of the forces, and then to the right of the first force, one finds that ( ) () 2 250 25000 ,0250 0 100 ,100 l/x M Vx x M V << = =<< = = (4.6.23) where l is the length of the beam. Figure 4.6.23: a loaded beam with circular cross-section The maximum tensile stress is then MPa8.31 4/25000) ( 4max max max = =−−= rr Iy M πσ (4.6.24) and occurs at all sections between the two loads. ■ mm250 mm250 mm100=rb0=y h • centroid ydy Section 4.6 Solid Mechanics Part I Kelly 1504.6.5 Shear Stresses in Beams In the derivation of the flexural stress form ula, Eqn. 4.6.21, it was assumed that plane sections remain plane. This implies that there is no shear strain and, for an isotropic elastic material, no shear stress, as indicated in Fig. 4.6.24. Figure 4.6.24: a section of beam before and after deformation This fact will now be ignored, a nd an expression for the shear stress τ within a beam will be developed. It is implicitly assumed th at this shear stress ha s little effect on the calculation of the flexural stress. As in Fig. 4.6.18, consider the equilibrium of a thin section of beam, as shown in Fig. 4.6.25. The beam has rectangular cross-section; although the theory developed here is strictly for rectangular cross se ctions only, it can be used to give approximate shear stress values in any beam with a plane of symmet ry. Consider the equilibrium of a section of this section, at the upper su rface of the beam, shown hatched in Fig. 4.6.25. The stresses acting on this section are as shown. Again, the normal stress is compressive at the surface, consistent with the sign convention fo r a positive moment. Note that there are no shear stresses acting at the surface – there may be distributed normal loads or forces acting at the surface but, for clarity, these are not shown, and they are not necessary for the following calculation. From equilibrium of forces in the hori zontal direction of the surface section: 0=Δ+⎥ ⎦⎤ ⎢ ⎣⎡+⎥ ⎦⎤ ⎢ ⎣⎡− Δ+∫∫xb dA dA xx A x Aτ σ σ (4.6.25) The third term on the left here assumes that the shear stress is uniform over the section – this is similar to the calculations of §4.6.3 – for a very small section, the variation in stress is a small term and may be neglected . Using the bending stress formula, Eqn. 4.6.21, 0)( ) (=+Δ−Δ+−∫b dAIy xxMx xM Aτ (4.6.26) and, with Eqn. 4.6.13, as 0→Δx , IbVQ=τ Shear stress in a beam (4.6.27) shear stresses would produce an angle change before deformation Section 4.6 Solid Mechanics Part I Kelly 151where Q is the first moment of area dAy A∫ of the surface section of the cross-section. Figure 4.6.25: stresses and forces acting on a small section of material at the surface of a beam As mentioned, this formula 4.6.27 can be used as an approximation of the shear stress in a beam of arbitrary cross-section, in which case b can be regarded as the depth of the beam at that section. For the rectangular beam, one has ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−==∫22 2/ 42yhbdyybQh y ( 4 . 6 . 2 8 ) so that ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− =22 346yh bhVτ (4.6.29) The maximum shear stress in the cross-se ction arises at the neutral surface: AV bhV 23 23 max ==τ (4.6.30) and the shear stress dies away towards the upper and lower surfaces. Note that the average shear stress over the cross-section is AV/ and the maximum shear stress is 150% of this value. Finally, since the shear stress on a vertical cr oss-section has been evaluated, the shear stress on a longitudinal section has been eval uated, since the shear stresses on all four sides of an element are the same, as in Fig.4.6.6. Example Consider the simply supported beam loaded by a concentrated force shown in Fig. 4.6.26. The cross-section is rect angular with height mm 100 and width mm 50 . The reactions at x)(xV xΔ)(xM ) ( x xVΔ+) ( x xMΔ+ x xΔ+h b σ σ τ ττ Section 4.6 Solid Mechanics Part I Kelly 152the supports are kN 5 and kN 15 . To the left of the load, one has kN 5=V and kNm5x M= . To the right of the load, one has kN 15−=V and kNm 1530 x M−= . The maximum shear stress will occur along the neutral axis and will clearly occur where V is largest, so anywhere to the right of the load: MPa5.423max max ==AVτ (4.6.31) Figure 4.6.26: a simply supported beam As an example of general shea r stress evaluation, th e shear stress at a point 25 mm below the top surface and 1 m in from the left-hand end is, from Eqn 4.6.29, MPa125.1+=τ . The shear stresses acting on an element at this location are shown in Fig. 4.6.27. Figure 4.6.27: shear stresses acti ng at a point in the beam ■ 4.6.6 Approximate nature of the beam theory The beam theory is only an approximate theo ry, with a number of simplifications made to the full equations of elasticity. The accuracy of the theory is briefly explored in this section. When a beam is in pure bending , that is when the shear for ce is everywhere zero, the full elasticity solution shows that plane sections do actually remain plane and the beam theory is exact. For more complex loadings, plane sections do actually deform. For example, it will be shown in Part II that th e initially plane sections of a cantilever subjected to an end force, Fig. 4.6.28, do not remain plane. Nevertheless, the beam theory prediction for normal and shear stress is exact in this simple case. MPa125.1=τm1kN20 m5.1m 5.0 Section 4.6 Solid Mechanics Part I Kelly 153 Figure 4.6.28: a cantilevered beam loaded by a force and moment Consider next a cantilevered beam of length l and rectangular cross section, height h and width b, subjected to a unifor mly distributed load p. With x measured from the cantilevered end, the shear fo rce and moment are given by ) (xlp V−= and ( )2 2)/(2/21)2/ ( lx lx pl M −+ = . The shear stress is ()xlyh bhp−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− =22 346τ (4.6.32) which turns out to be exact. The flexural st resses at the cantilevered end, at the upper surface, are 2 43⎟ ⎠⎞⎜ ⎝⎛=hl pσ ( 4 . 6 . 3 3 ) The exact solution is, however, 51 432 −⎟ ⎠⎞⎜ ⎝⎛=hl pσ ( 4 . 6 . 3 4 ) It can be seen that the beam theory is a good approximation for the case when hl/ is large, in which case the term 1/5 is negligible. In summary, for most configurations, the el ementary beam theory formulae for flexural stress and transverse shear stress are accurate to within about 3% for beams whose length- to-height ratio is greater than about 4. 4.6.7 Beam Deflection Consider the deflection curve of a beam. The displacement of the neut ral axis is denoted by v, positive upwards, as in Fig. 4.6.29. The sl ope at any point is then given by the first derivative, dxdv/. For any type of material, provi ded the displacement is small, it can be shown that the radius of curvature R is related to the second derivative 2 2/dxvd through (see the Appendix to this section, §4.6.10) deformed section not plane Section 4.6 Solid Mechanics Part I Kelly 154221 dxvd R= ( 4 . 6 . 3 5 ) and for this reason 2 2/dxvd is called the curvature of the beam. Using Eqn. 4.6.19, REy/−=σ , and the flexural stress expression, Eqn. 4.6.21, I My/−=σ , one has the moment-curvature equation 22 )(dxvdEI xM= moment-curvature equation (4.6.36) Figure 4.6.29: the deflection of a beam With the moment known, this differential equa tion can be integrated twice to obtain the deflection. Boundary conditions must be s upplied to obtain constants of integration. Example Consider the cantileve red beam of length L shown in Fig. 4.6.30, subjected to an end- force F and end-moment 0M. The moment is found to be 0 ) ( )( MxLF xM +−= , with x measured from the clamped end. The moment-curvature equation is then 2 13 2 012 00 22 61) (2121) () ( CxC Fx xM FL EIvC Fx xM FLdxdvEIFx M FL dxvdEI ++−+=→+−+= →−+= (4.6.37) The boundary conditions are that the displacement and slope ar e both zero at the clamped end, from which the two constant of integration can be obtained: 0 0)0(0 0)0( 12 =→=′=→= C vC v ( 4 . 6 . 3 8 ) Figure 4.6.30: a cantilevered beam loaded by an end-force and moment v F L 0M Section 4.6 Solid Mechanics Part I Kelly 155 The slope and deflection are therefore ⎥⎦⎤ ⎢⎣⎡−+ =⎥⎦⎤ ⎢⎣⎡−+ =2 03 2 021) (1,61) (211Fx xM FLEI dxdvFx xM FLEIv (4.6.39) The maximum deflection occurs at the end, where ⎥⎦⎤ ⎢⎣⎡+ =3 2 031 211)( FL LMEILv (4.6.40) ■ The term E I in Eqns. 4.6.39-40 is called the flexural rigidity , since it is a measure of the resistance of the beam to deflection. Example Consider the simply supported beam of length L shown in Fig. 4.6.31, subjected to a uniformly distributed load p over half its length. In this case, the moment is given by ⎪⎪ ⎩⎪⎪ ⎨⎧ << −<< − = LxLxLpLLx px pLx xM 2) (812021 83 )(2 (4.6.41) Figure 4.6.31: a simply supported beam subj ected to a uniformly distributed load over half its length It is necessary to apply the moment-curvature equation to each of the two regions 2/ 0 Lx<< and Lx L<<2/ separately, since the expressions for the moment in these regions differ. Thus there will be four constants of integration: 2 13 2212 22 22 2 14 313 22 22 481 161161 8181 81 241 48361 16321 83 DxD pLx xpL EIvD pLx xpLdxdvEIpLx pLdxvdEI CxC px pLx EIvC px pLxdxdvEIpx pLxdxvdEI ++ − =→+ −=→−= ++− =→+− =→−= (4.6.42) p 2/L 2/L Section 4.6 Solid Mechanics Part I Kelly 156 The boundary conditions are: (i) no deflection at pin support, 0 )0(= v and (ii) no deflection at roller support, 0)(=Lv , from which one finds that 02=C and LD pL D14 2 24/−−= . The other two necessa ry conditions are the continuity conditions where the two solutions meet. These are that (i) the deflection of bot h solutions agree at 2/Lx= and (ii) the slope of both solutions agree at 2/Lx= . Using these conditions, one finds that 38417,38493 23 1pLCpLC −= −= (4.6.43) so that ⎪⎪ ⎩⎪⎪ ⎨⎧ << ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−⎟ ⎠⎞⎜ ⎝⎛+⎟ ⎠⎞⎜ ⎝⎛−<< ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−⎟ ⎠⎞⎜ ⎝⎛+⎟ ⎠⎞⎜ ⎝⎛− = LxL Lx Lx Lx EIwLLxLx Lx Lx EIwL v 28 24 17138420 16 24 9384 3 2 44 3 4 (4.6.44) The deflection is shown in Fig. 4.6.32. Note that the maximum deflection occurs in 2/ 0 Lx<< ; it can be located by setting 0 /=dxdv there and solving. Figure 4.6.32: deflection of a beam ■ 4.6.8 Statically Indeterminate Beams Consider the beam shown in Fig. 4.6.33. It is cantilevered at one end and supported by a roller at its other end. A moment is app lied at its centre. There are three unknown reactions in this problem, the reaction forc e at the roller and the reaction force and moment at the built-in end. There are only two equilibrium equations with which to 4 3 16 24 9 ⎟ ⎠⎞⎜ ⎝⎛−⎟ ⎠⎞⎜ ⎝⎛+⎟ ⎠⎞⎜ ⎝⎛−Lx Lx LxLx/ vpLEI 43841− 2− 3 2 8 24 171 ⎟ ⎠⎞⎜ ⎝⎛−⎟ ⎠⎞⎜ ⎝⎛+⎟ ⎠⎞⎜ ⎝⎛−Lx Lx Lx Section 4.6 Solid Mechanics Part I Kelly 157determine these three unknowns and so it is not possible to solve the problem from equilibrium considerations alone. Such a beam is called statically indeterminat e. Figure 4.4.33: a cantilevered beam supported also by a roller More examples of statically indeterminate beam problem s are shown in Fig. 4.6.34. To solve such problems, one must consider th e deformation of the beam. The following example illustrates how this can be achieved. Figure 4.6.34: examples of stat ically indeterminate beams Example Consider the beam of length L shown in Fig. 4.6.35, cantilevered at end A and supported by a roller at end B. A moment 0M is applied at B. Figure 4.6.35: a statical ly indeterminate beam The moment along the beam can be expressed in terms of the unknown reaction force at end B: 0 ) ( )( MxLR xMB+−= . As before, one can integrate the moment-curvature equation: 0M0M A B0M Section 4.6 Solid Mechanics Part I Kelly 158() ()2 13 2 012 00 22 61 2121) ( CxCxR xMLR EIvC xR xMLRdxdvEIMxLR dxvdEI B BB BB ++−+=→+−+=→+−= (4.6.45) There are three boundary conditions, two to de termine the constants of integration and one can be used to determine the unknown reaction BR. The boundary conditions are (i) 0 0)0(2=→= C v , (ii) 0 0)0(/1=→= C dxdv and (iii) 0 )(=Lv from which one finds that L M RB 2/ 30−= . The slope and defl ection are therefore ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−⎟ ⎠⎞⎜ ⎝⎛=⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−⎟ ⎠⎞⎜ ⎝⎛= Lx Lx EILM dxdvLx Lx EILMv BB 2 344 22 3 2 (4.6.46) One can now return to the equilibrium equati ons to find the remaining reactions acting on the beam, which are B A R R−= and B A LR M M +=0 ■ 4.6.9 Problems 1. The simply supported beam shown below car ries a vertical load that increases uniformly from zero at the left end to a ma ximum value of 9 kN/m at the right end. Draw the shearing force and bending moment diagrams 2. The beam shown below is imply supported at two points and overhangs the supports at each end. It is subjected to a uniformly distributed load of 4 kN/m as well as a couple of magnitude 8 kN m applied to th e centre. Draw the shearing force and bending moment diagrams m6kN/m9 Section 4.6 Solid Mechanics Part I Kelly 159 3. Evaluate the centroid of the beam cross-se ction shown below (all measurements in mm) 4. Determine the maximum tensile and compressive stresses in the following beam (it has a rectangular cross-section wi th height 75 mm and depth 50 mm) 5. Consider the cantilever beam shown below. Determine the maximum shearing stress in the beam and determine the shearing stre ss 25 mm from the top surface of the beam at a section adjacent to the supporting wall. The cross-section is the “T” shape shown, for which 4 6mm1040×=I . [note: use the shear stress formula deri ved for rectangular cross-sections – as mentioned above, in this formula, b is the thickness of the beam at the point where the shear stress is being evaluated ] 75 50 225 50 100 5075kN/m4mkN8 m1m 2m 2m 1 m5kN1 Section 4.6 Solid Mechanics Part I Kelly 160 6. Obtain an expression for the maximum de flection of the simply supported beam shown here, subject to a unifo rmly distributed load of N/mw . 7. Determine the equation of the deflection cu rve for the cantilever beam loaded by a concentrated force P as shown below. 8. Determine the reactions for the following uniformly loaded beam clamped at both ends. 4.6.10 Appendix to §4.6 Curvature of the deflection curve Consider a deflection curve with deflection ) (xv and radius of curvature ) (xR , as shown in the figure below. Here, deflection is the transverse displacement (in the y direction) of m2kN50 125 50 20050 N/mw L LP a N/mw L Section 4.6 Solid Mechanics Part I Kelly 161the points that lie along the axis of the beam. A relationship between ) (xv and ) (xR is derived in what follows. First, consider a curve (arc) s. The tangent to some point p makes an angle ψ with the x – axis, as shown below. As one move along the arc, ψ changes. Define the curvature κ of the curve to be the rate at which ψ increases relative to s, dsdψκ= Thus if the curve is very “curved”, ψ is changing rapidly as one moves along the curve (as one increase s) and the curvature will be large. From the above figure, 22 2 )/(1)()(, tan dxdydxdy dx dxds dxdy+=+= =ψ , so that () ( ) () ()[]2/322 222 2 / 1// 11 / arctan dxdydxyddsdx dxyd dxdy dsdx dxdxdy d dsdx dxd dsd +=+= ===ψψκ x)(,xvy )(xv)(xR ψp xy dxdyds Section 4.6 Solid Mechanics Part I Kelly 162Finally, it will be shown that the curvature is simply the reciprocal of the radius of curvature. Draw a circle to the point p with radius R. Arbitrarily measure the arc length s from the point c, which is a point on the circle such that ψ=∠cop . Then arc length ψRs= , so that R dsd 1==ψκ Thus 23 222 11 ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛+= dxdvdxvd R If one assumes now that the slopes of the deflection curve are small, then 1 /<<dxdv and 221 dxvd R≈ xψψ pco xo ψdψd dsR R Section 4.7 Solid Mechanics Part I Kelly 1634.7 Failure of Elastic Materials In terms of material behavior, failure means a change in the normal constitutive behavior of a material, usually in response to excessive loads or deformations that cause irreparable changes of the microstructure. For exampl e, compressed rock will respond elastically up to a certain point but, if th e load is high enough, the rock will crush with permanent deformations. A model of crushing rock will involve a non-elastic constitutive law and is hence beyond the scope of elasticity theory. However, at issue here is the attempt to predict when the material first ceases to re spond elastically, not what happens after it does so. The failure of a specimen of rock under uniaxial tension1 can be predicted if a tension test has been carried out on a similar rock – it will fail when the applied tension reaches the yield stress Y (see §4.2.1). However, the question to be addressed here is how to predict the failure of a component which is loaded in a complex way, with a consequent complex stress state at any material particle. The theory of stress modulated failure assumes that failure occurs once some function of the stresses reaches some critical value. This function of the stress, or stress metric , might be the maximum principal stress, the maximum shear stress or some more complicated function of the stre ss components. Once the stress metric exceeds the critical value, the material no lo nger behaves elastically. 4.7.1 Failure Theories Three theories of material failure will be discussed in what follows. They are used principally in predicting failure in metals. 1. Maximum Principal Stress Theory The Maximum Principal Stress theory predicts that failure of a mate rial subjected to any state of stress occurs when the normal st ress of largest magnitude, i.e. the maximum principal stress, reaches th e appropriate failure stress fσ of the material: ( )fσσσσ =3 2 1 ,, max (4.7.1) This theory works reasonably well for predicting fracture of brittle materials and fσ here is the fracture stress recorded for a specimen in a tension test. 2. The Maximum Shear Stress (Tresca) Theory The Maximum Shear Stress (or Tresca) theory pred icts that failure of a material subjected to any state of stress occurs when the ma ximum shear stress reach es the appropriate failure stress fτ of the material, 1 i.e. acting along one axis, so one-dimensional Section 4.7 Solid Mechanics Part I Kelly 164fτσσσσσσ=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ −−− 2,2,2max3 2 3 1 2 1 (4.7.2) This theory predicts well the onset of plasticity in many ductile metals2. fτ here is the maximum shear stress at failure of a tensio n-test specimen. Since the maximum shear stress in the tension test at the yield stress Y is 2/Y (see Eqns. 3.3.1), the failure criterion 4.7.2 can be expressed as ( )Y=−−−3 2 3 1 2 1 , , max σσσσσσ (4.7.3) 3. The Von Mises Theory The Von Mised theory predicts th at failure of a material subj ected to any state of stress occurs when the following expression, involvi ng the sum of the squares of the differences between the principal stresses, is satisfied () () () Y=−+−+−2 3 22 3 12 2 121σσσσσσ (4.7.4) where Y is the yield stress in a tension test. Again, this theory is a good predictor, somewhat better than the Tresca theory, of th e onset of plasticity in ductile metals. Graphical Interpretation of the Failure Theories The three failure theories can conveniently be displayed on a si ngle graph. Assuming plane stress conditions, with 0 3=σ , the two principal stresses 1σ and 2σ can be plotted against each other in what is known as stress space , Fig. 4.7.1. Each closed curve is called the failure locus of the associated failure theory. When the stress state is such that the point ()2 1,σσ lies inside the locus, then the mate rial remains elastic. If the stress state is such that ()2 1,σσ reaches the locus, then the failure criterion is satisfied and failure occurs. As mentioned, what happens once failure o ccurs is beyond elasticity theory. “Beyond failure” of ductile metals and other materials which undergo plasticity is studied in Part II. 2 note that plastic flow in metals is often desirable, as in the forging and extrusion of metals in the manufacture of metal goods; in theses cases, satisfaction of the failure condition 8.1.2 is not considered to be “failure”, but rather a neccessity Section 4.7 Solid Mechanics Part I Kelly 165 Figure 4.7.1: Failure theori es in stress space Y1σY2σ Tresca Von Mises Maximum Principal Stress Section 4.7 Solid Mechanics Part I Kelly 166 1675 Energy and Virtual Work Thus far in this book, problems have been solved by using a combination of force- equilibrium and kinematics. Here, another appr oach is explored, in which expressions for work and energy are derived and utilised. Two important topics are discussed in this Chapter. The first is Energy Methods, which are techniques for solving problems involving el astic materials. Some of these methods, for example Castigliano’s second theorem, apply only to linear elastic materials, but most apply to generally non-lin ear elastic materials. The second topic is that of Vi rtual Work. The virtual work approach leads to powerful methods which can be used to solve static or dynamic problems involving any material model. 168 Section 5.1 Solid Mechanics Part I Kelly 1695.1 Energy in Deforming Materials There are many different types of energy: mechanical, chemical, nuclear, electrical, magnetic, etc. Energies can be grouped into kinetic energies (which are due to movement) and potential energies (which are stored energies - energy that a piece of matter has because of its position or becau se of the arrangement of its parts). A rubber ball held at some height above th e ground has (gravitationa l) potential energy. When dropped, this energy is progressively conv erted into kinetic en ergy as the ball’s speed increases until it reaches the ground wher e all its energy is kinetic. When the ball hits the ground it begins to deform elastical ly and, in so doing, the kinetic energy is progressively converted into elastic strain energy , which is stored inside the ball. This elastic energy is due to the re-arrangement of molecules in the ball – one can imagine this to be very like numerous springs being co mpressed inside the ball. The ball reaches maximum deformation when the kinetic energy has been completely converted into strain energy. The strain energy is then converte d back into kinetic en ergy, “pushing” the ball back up for the rebound. Elastic strain energy is a potential energy – el astically deforming a material is in many ways similar to raising a weight off the ground; in both cases th e potential energy is increased. Similarly, work is done in stretching a rubber band. This work is converted into elastic strain energy within the rubber. If the applied stretching forc e is then slowly reduced, the rubber band will use this energy to “pull” back. If the rubbe r band is stretched and then released suddenly, the band will retract quickly; the strain energy in this case is converted into kinetic energy – and sound energy (the “snap”). When a small weight is placed on a large meta l slab, the slab will undergo minute strains, too small to be noticed visually. Neverthe less, the metal behaves like the rubber ball and when the weight is removed the slab uses the in ternally stored strain energy to return to its initial state. On the othe r hand, a metal bar which is bent considerably, and then laid upon the ground, will not nearly recover its or iginal un-bent shape. It has undergone permanent deformation. Most of the energy supplie d has been lost; it has been converted into heat energy, which results in a very sli ght temperature rise in the bar. Permanent deformations of this type are accounted for by plasticity theory , which is treated in Part II. In any real material undergoing deformation, at least some of the supplied energy will be converted into heat. However, with the ideal elastic material under study in this chapter, it is assumed that all the ener gy supplied is converted into st rain energy. When the loads are removed, the material returns to its prec ise initial shape and there is no energy loss; for example, a purely elastic ball droppe d onto a purely elastic surface would bounce back up to the precise height from which it was released. As a prelude to a discussion of the energy of elastic materials, some important concepts from elementary particle mechanics are review ed in the following sections. It is shown that Newton’s second law, the principle of work and kinetic energy and the principle of conservation of mechanical energy are equivalent statements; each can be derived from the other. These concepts are then used to study the energetics of elastic materials. Section 5.1 Solid Mechanics Part I Kelly 170 5.1.1 Work and Energy in Particle Mechanics Work Consider a force F which acts on a particle , causing it to move through a displacement s, the directions in which they act being repr esented by the arrows in Fig. 5.1.1a. The work W done by F is defined to be θcosFs where θ is the angle formed by positioning the start of the F and s arrows at the same location with 180 0≤≤θ . Work can be positive or negative: when the force and displ acement are in the same direction, then 90 0≤≤θ and the work done is positive; when the for ce and displacement are in opposite directions, then 180 90≤≤θ and the work done is negative. Figure 5.1.1: (a) force acting on a partic le, which moves through a displacement s; (b) a varying force moving a particle along a path Consider next a particle moving alon g a certain path between the points 2 1,pp by the action of some force F, Fig. 5.1.1b. The work done is ∫=2 1cosp pds F W θ (5.1.1) where s is the displacement. For motion along a straight line, so that 0=θ , the work is dsF Wp p∫=2 1; if F here is constant then the work is simply F times the distance between 1p to 2p but, in most applications, the force w ill vary and an integral needs to be evaluated. Conservative Forces From Eqn. 5.1.1, the work done by a force in moving a particle through a displacement will in general depend on the path taken. There are many important practical cases, however, when the work is independent of the path taken, and simply depends on the initial and final positions, for example the work done in deforming elastic materials (see later) – these lead to the notion of a conservative (or potential ) force . Looking at the one-dimensional case, a conservative force conF is one which can always be written as the derivative of a function U, •sθ ds 1p2p •• F )a() b(F Section 5.1 Solid Mechanics Part I Kelly 171dxdUF−=con (5.1.2) since in that case () U pU pU dU dxdxdUdxF Wp pp pp pΔ−=−−=−= −= = ∫∫∫)( )(1 2 con2 12 12 1 (5.1.3) In this context, the function U is called the potential energy and UΔ is the change in potential energy of the part icle as it moves from 1p to 2p. If the particle is moved from 1p to 2p and then back to 1p, the net work done is ze ro and the potential energy U of the particle is that with which it started. Potential Energy The potential energy of a particle/s ystem can be defined as follows: Potential Energy : the work done in moving a system from some standard configuration to the current configuration Potential energy has the following characteristics: (1) The existence of a force field (2) To move something in the force field, work must be done (3) The force field is conservative (4) There is some reference configuration (5) The force field itself does negative work when another force is moving something against it (6) It is recoverable energy These six features are evident in the following example: a body attached to the coil of a spring is extended slowly by a force F, overcoming the spring (restoring) force sprF (so that there are no accelerations and sprF F−= at all times), Fig. 5.1.2. Figure 5.1.2: a force extending an elastic spring Let the initial position of the block be 0x (relative to the reference configuration, 0=x ). Assuming the force to be proportional to deflection, kxF= , the work done by F in extending the spring to a distance x is sprF F x Section 5.1 Solid Mechanics Part I Kelly 172U xUxU kx kx kxdx Fdx Wx xx xΔ=−≡−===∫∫)( )(02 0 21 2 21 0 0 (5.1.4) This is the work done to move something in the elastic spring “force field” and by definition is the potential en ergy (change in the body). The energy supplied in moving the body is said to be recoverable because the spring is read y to pull back and do the same amount of work. The corresponding work done by th e conservative spring force sprF is () U kx kx Fdx Wx xΔ−≡−−=−=∫2 0 21 2 21 spr 0 (5.1.5) This work can be seen from the area of the tria ngles in Fig. 5.1.3: th e spring force is zero at the equilibrium/reference position ( 0=x ) and increases linearly as x increases. Figure 5.1.3: force-extensio n curve for a spring ■ The forces in this example depend on the amount by which the spring is stretched. This is similar to the potential energy stored in ma terials – the potential force will depend in some way on the separation between material particles (see below). Also, from the example, it can be seen that an alternative definition for the potential energy U of a system is the negative of the work done by a conservative force in moving the system from some standard configur ation to the current configuration . In general then, the work done by a conservative force is related to the potential energy through U WconΔ−= (5.1.6) Dissipative (Non-Conservative) Forces When the forces are not conservative, that is, they are dissipative , one cannot find a universal function U such that the work done is the difference between the values of U at the beginning and end points – one has to consider the path taken by the particle and the work done will be different in each case. A ge neral feature of non-conservative forces is that if one moves a particle and then returns it to its orig inal position the net work done will not be zero. For example, consider a block being dragged across a rough surface, 0x xkx 0kxsprF F−= Section 5.1 Solid Mechanics Part I Kelly 173Fig. 5.1.4. In this case, if the block slides over and back a number of times, the work done by the pulling force F keeps increasing, and the work d one is not simply determined by the final position of the block, but by its comp lete path history. The energy used up in moving the block is dissipated as heat (the energy is irrecoverable ). Figure 5.1.4: Dragging a block over a frictional surface 5.1.2 The Principle of Work and Kinetic Energy In general, a mechanics problem can be solv ed using either Newt on’s second law or the principle of work and energy (w hich is discussed here). Th ese are two different equations which basically say the same thing, but one mi ght be preferable to the other depending on the problem under consideration. Wher eas Newton’s second law deals with forces , the work – energy principle casts problems in terms of energy . The kinetic energy of a particle of mass m and velocity v is defined to be 2 21mv K= . The rate of change of kinetic en ergy is, using Newton’s second law maF= , Fvvmadtdvmv mvdtdK ===⎟ ⎠⎞⎜ ⎝⎛= )(212 & (5.1.7) The change in kinetic energy over a time interval ) ,(10tt is then dtFv dtdtdKK KKt tt t∫∫==−=Δ1 01 00 1 (5.1.8) where K0 and K1 are the initial and final kinetic en ergies. The work done over this time interval is dtFv dxF dW W t ttx txtW tW∫∫∫===1 01 01 0)( )()( )( (5.1.9) and it follows that K WΔ= Work – Energy Principle (5.1.10) One has the followng: The principle of work and kinetic energy : F frF Section 5.1 Solid Mechanics Part I Kelly 174the total work done by the external forces acting on a particle equals the change in kinetic energy of the particle It is not a new principle of mechanics, rath er a rearrangement of Newton’s second law of motion (or one could have started with this principle, and derived Newton’s second law). The following example shows how the principl e holds for conserva tive, dissipative and applied forces. Example A block of mass m is attached to a spring and dragged along a rough surface. It is dragged from left to right, Fig. 5.1.5. Th ree forces act on the block, the applied force aplF (taken to be constant), the spring force sprF and the friction force friF (assumed constant). Figure 5.1.5: a block attached to a spring and dragged along a rough surface Newton’s second law, with kx Fspr= , leads to the non-homogeneous second order differential equation spr fri apl F F Fdtxdm −−=22 (5.1.11) Taking the initial position of the block to be 0x and the initial velocity to be 0x&, the solution is txtkF FxkF Ftxfri apl fri aplωωω sin cos )(0 0&+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ −−+−= (5.1.12) where mk/=ω . The total work done W is the sum of the work done by the applied force aplW, the work done by the spring force sprW and that done by the friction force friW: () ()()02 02 021xxF x xk xxF W W W Wfri apl fri spr apl −−−−−=++= (5.1.13) The change in kinetic energy of the block is ()2 02 21x xm K &&−=Δ ( 5 . 1 . 1 4 ) friFaplFsprF Section 5.1 Solid Mechanics Part I Kelly 175 Substituting Eqn. 5.1.12 into 5.1.13-14 and carrying out the algebra, one indeed finds that K WΔ= : K W W W Wfri spr apl Δ=++= (5.1.15) Now the work done by the spring force is equi valent to the negative of the potential energy change, so the work-energy equation (5 .1.15) can be written in the alternative form1 K U W Wspr fri apl Δ+Δ=+ (5.1.16) The friction force is dissipative – it leads to energy loss. In fact, the work done by the friction force is converted into heat which mani fests itself as a temperature change in the block. Denoting this energy loss by ) (0xxF Hfri fri −= , one has K U H Wspr fri apl Δ+Δ=− (5.1.17) ■ 5.1.3 The Principle of Conservation of Mechanical Energy In what follows, it is assumed that there is no energy loss , so that no dissipative forces act. Define the total mechanical energy of a body to be the sum of the kinetic and potential energies of the body. The work-ene rgy principle can then be expressed in two different ways: 1. The total work done by the external for ces acting on a body equals the change in kinetic energy of the body: K W W Wapl con Δ=+= (5.1.18) 2. The total work done by the external fo rces acting on a body, exclusive of the conservative forces, equals the change in the total mechani cal energy of the body K U W Δ+Δ=apl (5.1.19) The special case where all the external fo rces are conservative/potential leads to K UΔ+Δ=0 , so that the mechanical energy is constant . This situation occurs, for example, for a body in free-fall { ▲Problem 3} and for a freely oscillating spring {▲Problem 4}. Both forms of the work-energy principle can also be seen to apply for a spring subjected to an external force { ▲Problem 5}. 1 it is conventional to keep work terms on the left and energy terms on the right Section 5.1 Solid Mechanics Part I Kelly 176The Principle of Conservati on of Mechanical Energy The principle of conservation of energy states that the total en ergy of a system remains constant – energy cannot be created or destr oyed, it can only be changed from one form of energy to another. The principle of conservation of energy in the case where there is no energy dissipation is called the principle of conservation of mechanical energy and states that, if a system is subject only to conservati ve forces, its m echanical energy remains constant ; any system in which non-conservative forces act will in evitable involve non-mechanical energy (heat transfer). So, when there are only conservative forces acting, one has K UΔ+Δ=0 (5.1.20) or, equivalently, i i f f U K U K +=+ ( 5 . 1 . 2 1 ) where f iKK, are the initial and fina l kinetic energies and f iUU, are the initial and final potential energies. Note that the principle of mechanical ener gy conservation is not a new separate law of mechanics, it is merely a re-expression of the work-energy principle (or of Newton’s second law). 5.1.4 Deforming Materials The discussion above which concerned particle mechanics is now generalized to that of a deforming material. Any material consists of many molecules and particles, all interacting in some complex way. There will be a complex system of internal forces acting between the molecules, even when the material is in a natura l (undeformed) equilib rium state. If external forces are applied, the material will deform and the molecules will move, and hence not only will work be done by the external forces, but work will be done by the internal forces . The work-energy principle in this case states that the total work done by the external and internal forces equals the change in kinetic energy, K W W Δ=+int ext (5.1.22) In the special case where no external forces act on the system, one has K WΔ=int (5.1.23) which is a situation known as free vibration . The case where the kinetic energy is unchanging is Section 5.1 Solid Mechanics Part I Kelly 177 0int ext=+W W (5.1.24) and this situation is known as quasi-static (the quantities here can still depend on time). The force interaction between the molecules can be grouped into: (1) conservative internal force systems (2) non-conservative internal force systems (or at least partly non-conservative) Conservative Internal Forces First, assuming a conservative internal force system, one can imagine that the molecules interact with each other in the manner of elastic springs. Suppose one could apply an external force to pull two of these mol ecules apart, as shown in Fig. 5.1.6. Figure 5.1.6: external force pulling two molecules/particles apart In this ideal situati on one can say that the work done by the external forces equals the change in potential energy plus the change in kinetic energy, K U W Δ+Δ=ext (5.1.25) The energy U in this case of deforming materials is called the elastic strain energy , the energy due to the molecular arrangement relative to some equilibrium position. The free vibration case is now K UΔ+Δ=0 and the quasi-static situation is U WΔ=ext . Non-Conservative Internal Forces Consider now another example of internal fo rces acting within materials, that of a polymer with long-chain molecules. If one could somehow apply an external force to a pair of these molecules, as shown in Fig. 5.1.7, the molecules would slide over each other. Frictional (viscous) forces would act between the molecules, very much like the frictional force between the block and rough su rface of Fig. 5.1.4. This is called internal friction . Assuming that the internal forces are dissipative, the exte rnal work cannot be written in terms of a potential energy, K U W Δ+Δ≠ext , since the work done depends on the path taken . One would have to calculate the work done by evaluating an integral. internal forces external force external force internal forces external force external force Section 5.1 Solid Mechanics Part I Kelly 178 Figure 5.1.7: external force pulling two molecules/particles apart Similar to Eqn. 5.1.17, however, th e energy balance can be written as U K H W Δ+Δ=−ext (5.1.26) where H is the energy dissipated during the deformation and will depend on the precise deformation process . This energy is dissipated thr ough heat transfer and is conducted away through the material. 5.1.5 Energy Methods The work-energy principle pr ovides a method for obtaining solutions to conservative static problems and will be pursued in the next section. The principle is one of a number of tools which can be grouped under the heading Energy Methods , such as Castigliano’s theorems and the Crotti-Engesser theorem (see later). These methods can be used to solve a wide range of problems involving el astic (linear or no n-linear) materials. Virtual work methods are clos ely related to energy methods and provide powerful means for solving problems whether they involve elas tic materials or not; for the case of elastic materials, they lead naturally to the principle of minimum potential energy discussed in a later section. These virtual work methods will be discussed in sections 5.5-5.6. Finally, the concept of energy and the laws of thermodynamics provide a powerful means of deriving families of constitutive laws for material behaviour; this topic will be pursued in Part IV. 5.1.6 Problems 1. Consider the conservative force field 2 41 1 x xF−= What is the potential energy of a particle at some position 1xx= (define the point at infinity to be the reference point)? What work is done by F as the particle moves from the reference point to 1xx=? What is the work done by the applied force which moves the particle from the reference point to 1xx=? 2. Consider the gravitational force field mg. Consider a body acted upon by its weight mg w= and by an equal and opposite upward force F (arising, for instance, in a string). Suppose the weight to be moved at slow speed from one position to another one (so that there is no acceleration and w F−= ). Calculate the work done by F and Section 5.1 Solid Mechanics Part I Kelly 179show that it is independent of the path taken. What is the potential energy of the body? What is the work done by the gravitational force? 3. Show that both forms of the work-energy principle, Eqns. 5.1.18, 5.1.19, hold for a body in free-fall and that th e total mechanical energy is constant. (Use Newton’s second law with x positive up, initial height h and zero initial velocity.) 4. Consider a mass m attached to a freely oscillating spring, at initial position 0x and with initial velocity 0x&. Use Newton’s second law to show that () t xt x xt xt xx ωωω ωωωω cos)/( sinsin)/( cos 0 00 0 & && + −=+ = where mk/=ω . Show that both forms of the work-energy principle, Eqns. 5.1.18, 5.1.19, hold for the mass and that the total mechanical energy is constant. 5. Consider the case of an oscillating mass m attached to a spring with a constant force F applied to the mass. From Newton’s second law, one has F kx xm+−=&& which can solved to obtain ⎭⎬⎫ ⎩⎨⎧+⎟ ⎠⎞⎜ ⎝⎛−−=+ +⎟ ⎠⎞⎜ ⎝⎛−= t xtkFx xkFt xtkFx x ωωω ωωωω cos)/( sinsin)/( cos 0 00 0 & && Evaluate the change in kinetic energy and the total work done by the applied force to show that K WΔ= . Show also that the total work done by the applied force, exclusive of the conservative sp ring force, is equivalent to K UΔ+Δ . 6. Consider a body dragged a distance s along a rough horizontal surface by a force F, Fig. 5.1.4. By Newton’s second law, xm FFfr&&=− . By directly integrating this equation twice and lettin g the initial position and velocity of the body be 0x and 0x& respectively, show that the work done and the change in kinetic energy of the block are both given by ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−− txtmFFFFfr fr 02 21) ( & so that the principle of work and kinetic energy holds. How much energy is dissipated? Section 5.2 Solid Mechanics Part I Kelly 1805.2 Elastic Strain Energy The strain energy stored in an elastic materi al upon deformation is calculated below for a number of different geometries and loadi ng conditions. These expressions for stored energy will then be used to solve some elasticity problems using the energy methods mentioned in the previous section. 5.2.1 Strain energy in deformed Components Bar under axial load Consider a bar of elastic material fixed at one end and subjected to a steadily increasing force P, Fig. 5.2.1. The force is applied slowly so that kinetic energies are negligible. The initial length of the bar is L. The work dW done in extending the bar a small amount Δd is1 Δ=Pd dW (5.2.1) Figure 5.2.1: a bar loaded by a constant stress It was shown in §4.3.2 that the force and extension Δ are linearly related through EAPL/=Δ , Eqn. 4.3.5, where E is the Young’s modulus and A is the cross sectional area. This linear relationship is plotted in Fig. 5.2.2. The work expressed by Eqn. 5.2.1 is the white region under the force-extension curv e (line). The total work done during the complete extension up to a final force P and final extension Δ is the total area beneath the curve. The work done is stored as elastic strain energy U and so EALPP U2 212 =Δ= (5.2.2) If the axial force (and/or th e cross-sectional area and Y oung’s modulus) varies along the bar, then the above calculation can be done for a small element of length dx. The energy stored in this element would be EA dxP 2/2 and the total strain energy stored in the bar would be 1 the small change in force during this small extension may be neglected P LΔd Δ Section 5.2 Solid Mechanics Part I Kelly 181∫=L dxEAPU 02 2 (5.2.3) Figure 5.2.2: force-displacemen t curve for uniaxial load The strain energy is always positiv e, due to the square on the force P, regardless of whether the bar is being compressed or elongated. Note the factor of one half in Eqn. 5.2.2. The energy stored is not simply force times displacement because the force is changing during the deformation. Circular Bar in Torsion Consider a circular bar subjected to a torque T. The torque is equiva lent to a couple: two forces of magnitude F acting in opposite directions and separated by a distance r2 as in Fig. 5.2.3; Fr T2= . As the bar twists through a small angle φΔ, the forces each move through a distance φΔ=rs . The work done is therefore ()φΔ==Δ T Fs W 2 . Figure 5.2.3: torque acting on a circular bar It was shown in §4.4 that the torque and angl e of twist are linearly related through Eqn. 4.4.10, GJTL/=φ , where L is the length of the bar, G is the shear modulus and J is the polar moment of inertia. The angle of twist can be plotted ag ainst the torque as in Fig. 5.2.4. The total strain energy stored in the cylinde r during the straining up to a final angle of twist φ is the work done, equal to the shad ed area in Fig. 5.2.4, leading to GJLTT U2 212 ==φ (5.2.4) ΔdPforce-extension curve dW Δ rF φΔ Section 5.2 Solid Mechanics Part I Kelly 182 Figure 5.2.4: torque – angle of twist plot for torsion Again, if the various quantities are varying along the length of the bar, then the total strain energy can be expressed as dxGJTUL ∫= 02 2 (5.2.5) Beam subjected to a Pure Moment As with the bar under torsion, the work done by a moment M as it moves through an angle θd is θMd . The moment is related to the radius of curvature R through Eqns. 4.6.35-36, REI M /= , where E is the Young’s modulus and I is the moment of inertia. The length L of a beam and the angle subtended θ are related to R through θRL= , Fig. 5.2.5, and so moment and angle θ are linearly related through EI ML/=θ . Figure 5.2.5: beam of length L under pure bending The total strain energy stored in a bending beam is then EILMM U2 212 ==θ (5.2.6) and if the moment and other quantities vary along the beam, Rθ M M2/θ 2/θφΔT WΔ Section 5.2 Solid Mechanics Part I Kelly 183dxEIMUL ∫= 02 2 (5.2.7) This expression is due to the flexural stress xσ. A beam can also store energy due to shear stress τ; this latter energy is usually much less than that due to the flexural stresses provided the beam is slender – th is is discussed further below. Example Consider the bar with varying circular cro ss-section shown in Fig. 5.2.6. The Young’s modulus is GPa200 . Figure 5.2.6: a loaded bar The strain energy stored in the bar when a force of kN2 is applied at the free end is ()() ()() ()Nm 1062.9 1031 1051 10222 102 23 22221123 02 − − −×=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ×+ × ××= =∫πL dxEAPU (5.2.8) ■ 5.2.2 The Work-Energy Principle The work-energy principle for elastic material s, that is, the fact that the work done by external forces is stored as elastic energy, can be used directly to solve some simple problems. To be precise, it can be used to solve problems involving a single force and for solving for the displacement in the direction of that force. By force and displacement here it is meant generalised force and generalised displacement , that is a force/displacement pair, a torque/angle of twist pair or a moment/bending angle pair. More complex problems need to be solved using more sophisticated energy methods, such as Castiglianos’ method discussed further below. Example Consider the beam of length L shown in Fig. 5.2.7, pinned at one end (A) and simply supported at the other (C). A moment 0M acts at B, a distance 1L from the left-hand end. The cross-section is rectangular with depth b and height h. The work-energy principle can be used to calculate the angle Bθ through which the moment at B rotates. m2 m2 kN2 cm5=rcm3=r Section 5.2 Solid Mechanics Part I Kelly 184 Figure 5.2.7: a beam subjected to a moment at B The moment along the beam can be calcula ted from force and moment equilibrium, ()⎪⎩⎪⎨⎧ << −<< − = Lx L Lx MLx LxM M 1 01 0 ,/ 10 ,/ (5.2.9) The strain energy stored in the bar (due to the flexural stresses only) is 233 22 02 02 32 0 02216 22 11 L EbhLMdxLxdxLx EbhMdxEIMUL LL L = ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ ⎟ ⎠⎞⎜ ⎝⎛−+⎟ ⎠⎞⎜ ⎝⎛= = ∫∫ ∫ (5.2.10) The work done by the applied moment is 2 /0B Mθ and so 233 204 L EbhLM B=θ (5.2.11) ■ 5.2.3 Strain Energy Density The strain energy will in general vary thr oughout a body and for this reason it is useful to introduce the concept of strain energy density , which is a measure of how much energy is stored in small volume elements throughout a material. Consider again a bar of s ubjected to a uniaxial force P. A small volume element with edges aligned with the zyx,, axes as shown in Fig. 5.2.8 will then be subjected to a stress xxσ only. The volume of the element is dxdydz dV= . From Eqn. 5.2.2, the strain energy in the element is () Edydzdx dydzUxx 22σ= (5.2.12) L 0MA C B 1L2L Section 5.2 Solid Mechanics Part I Kelly 185 Figure 5.2.8: a volume element under stress The strain energy density u is defined as the strain energy per unit volume : Euxx 22σ= (5.2.13) The total strain energy in the bar may now be expressed as this quantity integrated over the whole volume, dVu U V∫= (5.2.14) which, for a constant cross-section A and length L reads dxuA UL∫= 0. From Hooke’s law, the strain energy density of Eqn. 5.2.13 can also be expressed as xx xx uεσ21= (5.2.15) As can be seen from Fig. 5.2.9, this is th e area under the uniaxial stress-strain curve. Figure 5.2.9: stress-strain cu rve for elastic material Note that the element does deform in the y and z directions but no work is associated with those displacements since there is no force acting in those directions. The strain energy density for an element subjected to a yyσ stress only is, by the same arguments, 2/yy yyεσ , and that due to a zzσ stress is 2/zz zzεσ . Consider next a shear stress xyσ acting on the volume element to produce a shear strain xyε as illustrated in Fig. σ εuxxσ volume element xxσ dxdzdyP Pxy z Section 5.2 Solid Mechanics Part I Kelly 1865.2.10. The element deforms with small angles θ and λ as illustrated. Only the stresses on the upper and right-hand surfaces are show n, since the stresses on the other two surfaces do no work. The force acting on the upper surface is dxdzxyσ and moves through a displacement dyλ. The force acting on the right-hand surface is dydzxyσ and moves through a displacement dxθ. The work done when the element moves through angles θd and λd is then, using the definition of shear strain, Eqn. 3.6.1, ()()()()()()xy xy xy xy d dxdydz dxd dydz dyd dxdz dW εσ θ σλ σ 2 = + = (5.2.16) and, with shear stress proportional to shear strain as in Fig. 5.2.9, the strain energy density is xy xy xy xyd u εσεσ= =∫2 (5.2.17) Figure 5.2.10: a volume element under shear stress The strain energy can be similarly calculated for the other shear stresses and, in summary, the strain energy density for a volume elem ent subjected to arbitrary stresses is () ( )zx zx yz yz xy xy zz zz yy yy xx xx u εσεσεσεσεσεσ +++++ =21 (5.2.18) Using Hooke’s law, Eqns. 4.2.9, with Eqn. 4. 2.5, the strain energy density can also be written in the alternative and useful forms { ▲Problem 4} () () () () () [] () () () ()2 2 2 2 2 2 22 2 2 2 2 22 2 2 2 2 2 2212 2 ) 1(2121 21 zx yz xy zz yy xx zz yy xxzx yz xy xx zz zz yy yy xx zz yy xxzx yz xy xx zz zz yy yy xx zz yy xxE Eu εεεμεεεμεεεννμεεεμεεεεεενεεεννμσσσμσσσσσσνσσσ ++++++++−=+++++ +++−−=+++++ −++ = (5.2.19) Strain Energy in a Beam due to Shear Stress dxdyxyσ xdyλy θλ xyσ dxθ Section 5.2 Solid Mechanics Part I Kelly 187The shear stresses arising in a beam at location y from the neutral axis are given by Eqn. 4.6.27, ) (/)( )( yIbVyQy=τ , where Q is the first moment of area of the section of beam from y to the outer surface, V is the shear force, I is the moment of inertia of the complete cross-section and b is the thickness of the beam at y. From Eqns. 5.2.19a and 5.2.14 then, the total strain energy in a beam of length L due to shear stress is dxdAbQ IVdV U AL V⎥ ⎦⎤ ⎢ ⎣⎡= = ∫∫∫ 22 022 2 21 2 μ μτ (5.2.20) Here V,μ and I are taken to be constant for any given cross-section but may vary along the beam; Q varies and b may vary over any given cross- section. Expression 5.2.20 can be simplified by introducing the form factor for shear sf, defined as dAbQ IAxf As∫=22 2)( (5.2.21) so that dxAVfUL s∫= 02 21 μ (5.2.22) The form factor depends only on the shape of the cross-section. For example, for a rectangular cross-section, using Eqn. 4.6.28, () 56 421 12/)(2/ 2/2/ 2/2 22 2 23= ⎥ ⎦⎤ ⎢ ⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛− = ∫ ∫+ −+ −b bh hs dz dy yhb b bhbhxf (5.2.23) In a similar manner, the form factor for a circular cross-sect ion is found to be 9/10 and that of a very thin tube is 2. 5.2.4 Castigliano’s Second Theorem The work-energy method is the simplest of energy methods. A more powerful method is that based on Castigliano’s second theorem2, which can be used to solve problems involving linear elastic materials. As an introduction to Cas tigliano’s second theorem, consider the case of uniaxial tension, where EALP U 2/2= . The displacement through which the force moves can be obtained by a diffe rentiation of this expression with respect to that force, Δ==EAPL dPdU ( 5 . 2 . 2 4 ) 2 Casigliano’s first theorem will be discussed in a later section Section 5.2 Solid Mechanics Part I Kelly 188Similarly, for torsion of a circular bar, GJLTU 2/2= , and a differentiation gives φ== GJTL dT dU / / . Further, for bending of a beam it is also seen that θ=dM dU/ . These are examples of Castiglia no’s theorem, which states that, provided the body is in equilibrium, the derivative of the strain energy with respect to the force gives the displacement corresponding to that for ce, in the direction of that force . When there is more than one force applied, then one take s the partial derivative . For example, if n independent forces nP PP ,,,2 1K act on a body, the displacement corresponding to the ith force is iiPU ∂∂=Δ (5.2.25) Before proving this theorem, here follow some examples. Example The beam shown in Fig. 5.2.11 is pinned at A, simply supported half-way along the beam at B and loaded at the end C by a force P and a moment 0M. Figure 5.2.11: a beam subjected to a force and moment at C The moment along the beam can be calcula ted from force and moment equilibrium, ⎪⎩⎪⎨⎧ << −−−<< −− = Lx L xLP MLx LxM Px M 2/ ), (2/ 0 ,/ 2 00 (5.2.26) The strain energy stored in the bar (due to the flexural stresses only) is () EILM EIL PM EILPdxxLP M dxxLMPEIUL LL 3 245 24) (2 21 2 02 0322/2 02/ 022 0 + +=⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ −++⎟ ⎠⎞⎜ ⎝⎛+ = ∫∫ (5.2.27) In order to apply Castigliano’s theorem, the st rain energy is considered to be a function of the two external loads, ()0,MPUU= . The displacement associated with the force P is then L 0M 2/L 2/LAC BP Section 5.2 Solid Mechanics Part I Kelly 189EILM EIPL PU C245 122 03 +=∂∂=Δ (5.2.28) The rotation associated with the moment is EILM EIPL MU C32 24502 0+=∂∂=θ (5.2.29) ■ Example Consider next the beam of length L shown in Fig. 5.2.12, built in at both ends and loaded centrally by a force P. This is a statically indeterminat e problem. In this case, the strain energy can be written as a function of the applied load and one of the unknown reactions. Figure 5.2.12: a statical ly indeterminate beam First, the moment in the beam is found from equilibrium considerations to be 2/ 0 ,2Lx xPM MA << += (5.2.30) where AM is the unknown reaction at the left-hand end. Then the strain energy in the left-hand half of the beam is EILM EIL PM EILPdxxPMEIUA AL A4 16 192 2 212 2 32 2/ 02 + +=⎟ ⎠⎞⎜ ⎝⎛+ =∫ (5.2.31) The strain energy in the complete beam is double this: EILM EIL PM EILPUA A 2 8 962 2 32 + += (5.2.32) Writing the strain energy as ()AMPUU ,= , the rotation at A is EILM EIPL MUA AA +=∂∂=82 θ (5.2.33) CM 2/L 2/LAC BP AM Section 5.2 Solid Mechanics Part I Kelly 190But 0=Aθ and so Eqn. 5.2.33 can be solved to get 8/PL MA−= . Then the displacement at the cen tre of the beam is EIPL EILM EIPL PUA B192 8 483 2 3 =+=∂∂=Δ (5.2.34) This is positive in the dir ection in which the associated force is acting, and so is downward. ■ Proof of Castigliano’s Theorem A proof of Castiligliano’s theorem will be give n here for a structure subjected to a single load. The load P produces a displacement Δ and the strain energy is 2/Δ=P U , Fig. 5.2.13. If an additional force dP is applied giving an additional deformation Δd, the additional strain energy is Δ+Δ= dPd Pd dU21 (5.2.35) If the load dPP+ is applied in one st ep, the work done is ()()2/Δ+Δ+ d dPP . Equating this to the strain energy dUU+ given by Eqn. 5.2.35 then gives dP PdΔ=Δ . Substituting into Eqn. 5.2.35 leads to Δ+Δ= dPd dP dU21 (5.2.36) Dividing through by dP and taking the limit as 0→dP results in Castigliano’s second theorem, Δ=dP dU/ . Figure 5.2.13: force-displacement curve In fact, dividing Eqn. 5.2.35 through by Δd and taking the limit as 0→Δd results in Castigliano’s first theorem , P d dU=Δ/ . It will be shown later that this first theorem, unlike the second, in fact holds also for the case when the elastic material is non-linear . 5.2.5 Dynamic Elasticiy P ΔdPP+ Δ+Δd Section 5.2 Solid Mechanics Part I Kelly 191Impact and Dynamic Loading Consider the case of a weight P dropped instantaneously onto the end of an elastic bar. If the weight P had been applied gradually from zero, the strain energy stored at the final force P and final displacement 0Δ would be 0 21ΔP. However, the instantaneously applied load is constant throughout the defo rmation and work done up to a displacement 0Δ is 0ΔP, Fig. 5.2.14. The difference between the two implies that the bar acquires a kinetic energy (see Eqn. 5.1.19); the material pa rticles accelerate from their equilibrium positions during the compression. As deformation proceeds beyond 0Δ, it is clear from Fig. 5.2.14 th at the strain energy is increasing faster than the work being done by the weight and so there must be a drop in kinetic energy; the particles begin to decelerate. Eventually, at 0 max 2Δ=Δ , the work done by the weight exactly equals the strain en ergy stored and the ma terial is at rest. However, the material is not in equili brium – the equilibrium position for a load P is 0Δ – and so the material begins to accelerate back to where 0Δ. Figure 5.2.14: non-equilibrium loading The bar and weight will conti nue to oscillate between 0 and maxΔ indefinitely. In a real material, internal friction will cause the vibration to decay. Thus the maximum compression of a bar under impact loading is twice that of a bar subjected to the same load gradually. Example Consider a weight w dropped from a height h. If one is interested in the final, maximum, displacement of the bar, maxΔ , one does not need to know about the detailed and complex transfer of energies during the impact; the en ergy lost by the weight equals the strain energy stored in the bar: ()max max21Δ=Δ+ P hw (5.2.37) where P is the force acting on the bar at its ma ximum compression. For an elastic bar, LEA P /maxΔ= , or, introducing the stiffness k so that maxΔ=kP , P 0ΔmaxΔ Section 5.2 Solid Mechanics Part I Kelly 192 ()LEAk k hw = Δ=Δ+ ,212 max max (5.2.38) which is a quadratic equation in maxΔ and can be solved to get ⎭⎬⎫ ⎩⎨⎧++=Δwhk kw 21 1max (5.2.39) If the force w had been applied gradually, then the displacement would have been kw/st=Δ , the “st” standing for “static”, and Eqn. 5.1.39 can be re-written as ⎭⎬⎫ ⎩⎨⎧ Δ++Δ=Δstst max21 1h (5.2.40) If 0=h , so that the weight is just touching the bar when released, then st max 2Δ=Δ . ■ 5.2.6 Problems 1. Show that the strain energy in a bar of length L and cross sectional area A hanging from a ceiling and subjected to its own wei ght is given by (at any section, the force acting is the weight of the material below that section) ELgAU6322ρ= 2. Consider the circular bar shown below subject to torques at the free end and where the cross-sectional area changes. The shear modulus is GPa80=G . Calculate the strain energy in the bar(s). 3. Two bars of equal length L and cross-sectional area A are pin-supported and loaded by a force F as shown below. Derive an expression for the vertical displacement at point A using the direct work-energy method, in terms of L, F, A and the Young’s modulus E. m1 m1 kNm4 cm5=rcm3=rkNm6 Section 5.2 Solid Mechanics Part I Kelly 193 4. Derive the strain energy density equations 5.2.19. 5. For the beam shown in Fig. 5.2.7, use th e expression 5.2.22 to calculate the strain energy due to the shear stresses. Take the shear modulus to be GPa80=G . Compare this with the strain energy due to flexural stress given by Eqn. 5.2.10. 6. Consider a simply supported beam of length L subjected to a uniform load w N/m. Calculate the strain energy due to both fl exural stress and shear stress for (a) a rectangular cross-section of depth times height hb×, (b) a circular cross-section with radius r. What is the ratio of the shear-to-fl exural strain energies in each case? 7. Consider the tapered bar of length L and square cross-section shown below, built-in at one end and subjected to a uniaxial force F at its free end. The thickness is h at the built-in end. Evaluate the displacement in terms of the (constant) Young’s modulus E at the free end using (i) the work-energy theorem, (ii) Castigliano’s theorem 8. Consider a cantilevered beam of length L and constant cross-s ection subjected to a uniform load w N/m. The beam is built-in at 0=x and has a Young’s modulus E. Use Castigliano’s theorem to calculate the deflection at Lx=. Consider only the flexural strain energy. [Hint: place a fictitious “dummy” load F at Lx= and set to zero once Castigliano’s theorem has been applied] 9. Consider the statically indeterminate uniax ial problem shown below, two bars joined at Lx=, built in at 0=x and L x2= , and subjected to a force F at the join. The cross-sectional area of the bar on the left is 2 A and that on the right is A. Use (i) the work-energy theorem and (ii) Castigliano’s theorem to evaluate the displacement at Lx=. F L LF L Lho45 Fo45 A Section 5.3 Solid Mechanics Part I Kelly 1945.3 Complementary Energy The linear elastic solid was considered in the previous section, with the characteristic straight force-deflection curve for axial defo rmations, Fig. 5.2.2. Here, consider the more general case of a bar of non-linear elastic material, of length L, fixed at one end and subjected to a steadily increasing force P. The work dW done in extending the bar a small amount Δd is Δ=Pd dW (5.3.1) Force is now no longer proportional to extension Δ, Fig. 5.3.1. However, the total work done during the complete exte nsion up to a final force P and final extension Δ is once again the total area beneath the force-extensi on curve. The work done is equal to the stored elastic strain energy which must now be expressed as an integral, Δ=∫Δ dP U 0 (5.3.2) The strain energy can be calcu lated if the precise force-de flection relationship is known. Figure 5.3.1: force-displacement cu rve for a non-linear material 5.3.1 Complementary Energy The force-deflection curve is naturally divided into two re gions, beneath the curve and above the curve, Fig. 5.3.2. The area of the regi on under the curve is th e strain energy. It is helpful to introduce a new concept, the complementary energy C, which is the area above the curve; this can be seen to be given by dP CP ∫Δ= 0 (5.3.3) For a linear elastic material, UC=. Although C has units of energy, it has no real physical meaning. ΔdPforce-extension curve dW Δ Section 5.3 Solid Mechanics Part I Kelly 195 Figure 5.3.2: strain energy and complement ary energy for an elastic material 5.3.2 The Crotti-Engesser Theorem Suppose an elastic body is loaded by n independent loads nP PP ,,,2 1K . The strain energy is then the work done by these loads, n ndP dP dP Un Δ++Δ+Δ= ∫∫∫Δ Δ Δ 02 02 1 012 1 L (5.3.4) It follows that j jPU=Δ∂∂ (5.3.5) which is known as Castigliano’s first theorem. Similarly, the total complementary energy is nP nP P dP dP dP Cn ∫∫∫Δ++Δ+Δ= 02 02 1 012 1 L (5.3.6) and it follows that j jPCΔ=∂∂ (5.3.7) which is known as the Crotti-Engesser theorem . For a linear elastic material, UC=, and the Crotti-Engesser theorem reduces to Castigliano’s second theorem, j j P U∂∂=Δ /, Eqn. 5.2.25. 5.3.3 Problems 1. The force-deflection equati on for a non-linear elastic material is given by 3Δ=αP . Find expressions for the strain energy and the complementary energy in terms of (i) P only, (ii) Δ only. Check that Δ=+ PCU . What is the ratio UC/? P UΔC Section 5.4 Solid Mechanics Part I Kelly 1965.4 Strain Energy Potentials 5.4.1 The Linear Elastic Strain Energy Potential The strain energy u was introduced in §5.2 1. From Eqn 5.2.19, the strain energy can be regarded as a functi on of the strains: () () () [] ()2 2 2 2 2 22 2 ) 1(21zx yz xy xx zz zz yy yy xx zz yy xxijuu εεεμεεεεεενεεεννμε +++++ +++−−== (5.4.1) Differentiating with respect to xxε (holding the other strains constant), ()() [ ]zz yy xx xxuεενεννμ ε++−−=∂∂) 1(212 (5.4.2) From Hooke’s law, Eqn 4.2.9, with Eqn 4.2.5, ()[]ν μ += 12/E , the expression on the right is simply xxσ. The strain energy can also be diffe rentiated with respect to the other normal strain components and one has zz zzyy yyxx xxu u uσεσεσε=∂∂=∂∂=∂∂, , (5.4.3) The strain energy is a potential , meaning that it provides information through a differentiation. Note the similarity between these equations and the equation relating a conservative force and the poten tial energy seen in §5.1: F dx dU=/ . Differentiating Eqn. 5.4.1 with respec t to the shear st resses results in zx zxyz yzxy xyu u uσεσεσε2 , 2 , 2 =∂∂=∂∂=∂∂ (5.4.4) The fact that Eqns. 5.4.4 has the factor of 2 on the right hands side but Eqns. 5.4.3 do not is not ideal. There are two common ways of viewing the strain energy potential to overcome this lack of symmetry. First, the st rain energy can be taken to be a function of the six independent strains, zx yz xy zz yy xx γγγεεε ,,,,, , the latter three being the engineering shear strains, xy xyεγ 2= , etc. Re-writing Eqn. 5.4.1 in terms of the engineering shear strains then leads to the set of equations { ▲Problem 1} zx zxyz yzxy xyzz zzyy yyxx xxu u u u u uσγσγσγσεσεσε=∂∂=∂∂=∂∂=∂∂=∂∂=∂∂, , , , , (5.4.5) 1 strictly speaking, this is the strain energy density , but it should be clear from the context whether it is energy per unit volume or not; the word density will often be omitted henceforth for brevity Section 5.4 Solid Mechanics Part I Kelly 197 The second method is to treat the strain energy as a function of nine independent strains, the three normal strains and xz zx zy yz yx xy εεεεεε ,,,,, . In other words the fact that the strains xyε and yxε are the same is ignored and the st rain energy is differentiated with respect to these as though they were independent. In order to implement this approach, the strain energy needs to be derived anew treating xyσ and yxσ as independent quantities. This simply means that Fig. 5.2. 10 is re-drawn as Fig. 5.4.1 below, and Eqn. 5.2.16 is re-expressed using 2 /) (yx xy xy εεε+= as ()()()()()[ ]yx yx xy xy xy yx d d dxdydz dxd dydz dyd dxdz dW εσεσ θ σλ σ + = + = (5.4.6) so that yx yx xy xy u εσεσ21 21+ = (5.4.7) Figure 5.4.1: a volume elem ent under shear stress Re-writing Eqn. 5.4.1 and diffe rentiation then leads to { ▲Problem 2} xz xzzy zyyx yxzx zxyz yzxy xyzz zzyy yyxx xx u u u u u uu u u σεσεσεσεσεσεσεσεσε =∂∂=∂∂=∂∂=∂∂=∂∂=∂∂=∂∂=∂∂=∂∂ , , , , ,, , , (5.4.8) These equations can be expressed in the succinct form ij ijuσε=∂∂ (5.4.9) 5.4.2 The Strain Energy Potential Generalising the above discussion now to non- linear elasticity, reca ll that the strain energy is the area beneath the stress -strain curve, Fig. 5.4.2, and dxdyyxσ xdyλy θλ xyσ dxθ Section 5.4 Solid Mechanics Part I Kelly 198 εσd du dW== (5.4.10) Figure 5.4.2: stress-strain curv e for a non-linear material When the material undergoes increments in strain xxdε, xydε, etc., the increment in strain energy is L+++=xy xy yy yy xx xx d d d du εσεσεσ (5.4.11) If the strain energy is a f unction of the nine strains ijε its increment can also be expressed as L+∂∂+∂∂+∂∂=xy xyyy yyxx xxdudududu εεεεεε (5.4.12) Subtracting Eqns 5.4.12 from 5.4.11 then gives L+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−=xy xyxy yy yyyy xx xxxx dududuεεσεεσεεσ 0 (5.4.13) Because the strains are indepe ndent, that is, any one of them can be adjusted without changing the others, one again arrives at Eqns . 5.4.8-9, only now it has been shown that this result holds generally for non-linear elastic materials. Note that in the case of an incompressible material, 0 =++zz yy xxεεε , so that the strains are not independent., and Eqns. 5.4.8-9 are not valid. 5.4.3 The Complementar y Energy Potential Analogous to Eqns. 5.4.10-13, an increment in complementary energy density can be expressed as σεd dc= (5.4.14) with εdσdW ε Section 5.4 Solid Mechanics Part I Kelly 199L++ +=xy xy yy yy xx xx d d d dc σεσεσε (5.4.15) and L+∂∂+∂∂+∂∂=xy xyyy yyxx xxdcdcdcdc σσσσσσ (5.4.16) so that L+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂−xy xyxy yy yyyy xx xxxx dududuσσεσσεσσε (5.4.17) With the stresses independent, one has an expression analogous to 5.4.9, ij ijcεσ=∂∂ (5.4.18) 5.4.4 Problems 1. Derive equations 5.4.5 2. Derive equations 5.4.8 Section 5.5 Solid Mechanics Part I Kelly 2005.5 Virtual Work Consider a mass attached to a spri ng and pulled by an applied force aplF, Fig. 5.5.1a. When the mass is in equilibrium, 0=+apl spr F F , where kx Fspr−= is the spring force with x the distance from the spring reference position. Figure 5.5.1: a force extending an elastic sp ring; (a) block in equilibrium, (b) block not at its equilibrium position In order to develop a number of powerful techniques based on a concept known as virtual work , imagine that the mass is not in fact at its equilibrium position but at an (incorrect) non-equilibrium position xxδ+, Fig. 5.5.1b. The imaginary displacement xδ is called a virtual displacement . Define the virtual work Wδ done by a force to be the equilibrium force times this small imaginary displacement xδ. It should be emphasized that virtual work is not real work – no work has been performed since xδ is not a real displacement which has taken place; this is mo re like a “thought experiment”. The virtual work of the spring force is then xkx xF Wspr spr δδ δ −== . The virtual work of the applied force is xF Wapl aplδ δ= . The total virtual work is ()x F kx W W Wapl apl spr δ δδδ +−=+= (5.5.1) There are two ways of viewing this expression. First, if the system is in equilibrium (0 =+−aplF kx ) then the virtual work is zero, 0=Wδ . Alternatively, if the virtual work is zero then, since xδ is arbitrary, the system must be in equilibrium. Thus the virtual work idea gives one an alternative means of determining whether a system is in equilibrium. The symbol δ is called a variation so that, for example, xδ is a variation in the displacement (from equilibrium). Virtual work is explored furt her in the following section. sprFaplF sprFaplFex xδ)a( )b( Section 5.5 Solid Mechanics Part I Kelly 2015.5.1 Principle of Virtual Work: a single particle A particle of mass m is acted upon by a number of forces, Nf ff ,,,2 1K , Fig. 5.5.2. Suppose the particle undergoe s a virtual displacement uδ; to reiterate, these impressed forces if do not cause the particle to move, one imagin es it to be incorrectly positioned a little away from the true equilibrium position. Figure 5.5.2: a particle in equilibrium und er the action of a number of forces If the particle is moving with an acceleration a, the quantity am− is treated as an inertial force. The total virtual work is then (each term here is the dot product of two vectors) u a fδ δ ⋅⎟ ⎠⎞⎜ ⎝⎛−=∑ =m WN ii 1 (5.5.2) Now if the particle is in equilibrium by the actio n of the effective (impressed plus inertial) force, then 0=Wδ (5.5.3) This can be expressed as follows: The principle of virtual work (or principle of virtual displacements ) I: if a particle is in e quilibrium under the action of a number of for ces (including the inertial force) the total work done by the forc es for a virtual displacement is zero Alternatively, one can define the external virtual work ∑⋅= ufδ δi Wext and the virtual kinetic energy uaδδ⋅=mK in which case the principle takes the form K Wδδ=ext (compare with the work-energy principle, Eqn. 5.1.10). In the above, the principle of virtual work was derived usi ng Newton’s second law. One could just as well regard the principle of vi rtual work as the fundamental principle and from it derive the conditions for equili brium. In this case one can say that 1 1 note the word any here: this must hold for all possible virtual displacements, for it will always be possible to find one virtual displacement which is perpendi cular to the resultant of the forces, so that () 0=⋅∑ ufδ even though ∑f is not necessarily zero uδ1f 2f3f Section 5.5 Solid Mechanics Part I Kelly 202The principle of virtual work (or principle of virtual displacements ) II: a particle is in equilibrium under the action of a system of forces (including the inertial force) if the total work done by the forces is zero for any virtua l displacement of the particle . Constraints In many practical problems, the particle will usually be constrained to move in only certain directions. For example consider a ball rolling over a table, Fig. 5.5.3. If the ball is in equilibrium then all the forces sum to zero, 0=−+∑ a f R m , where one distinguishes between the non-reaction forces if and the reaction force R. If the virtual displacement uδ is such that the constraint is not vi olated, that is the ball is not allowed to go “through” the table, then uδ and R are perpendicular, the virtual work done by the reaction force is zero and () 0=⋅−=∑ ua fδ δ m W . This is one of the benefits of the principle of virtual work; one does not need to calculate the forces of constraint R in order to determine the forces if which maintain the particle in equilibrium. Figure 5.5.3: a particle constrai ned to move over a surface The term kinematically admissible displacement is used to mean one that does not violate the constraints, and hen ce one arrives at the version of the principle which is often used in practice: The principle of virtual work (or principle of virtual displacements ) III: a particle is in equilibrium under the action of a system of forces (including the inertial force) if the total work done by the forces (excluding reaction forces) is zero for any kinematically admissible virtual displacement of the particle Whether one uses a kinematically admissibl e virtual displacement and so disregard reaction forces, or permit a virtual displacem ent that violates the constraint conditions will usually depend on the problem at hand. For example, in this next example use is made of a kinematically inadmissible virtual displacement. Example Consider a rigid bar of length L supported at its ends and loaded by a force F a distance a from the left hand end, Fig. 5.5.3a. Reaction forces C ARR, act at the ends. Let point C undergo a virtual displacement uδ. From similar triangles, the displacement at B is uLaδ)/( . End A does not move and so no virtual work is performed there. The total virtual work is R1f2f 3f Section 5.5 Solid Mechanics Part I Kelly 203uLaFuR WC δδδ −= (5.5.4) Note the minus sign here – the displacement at B is in a direction opposite to that of the action of the load and hence the work is negative. The beam is in equilibrium when 0=Wδ and hence LaF RC /= . Figure 5.5.3: a loaded rigid bar; (a) bar geom etry, (b) a virtual displacement at end C ■ 5.5.2 Principle of Virtual Work: deformable bodies A deformable body can be imagined to undergo virtual displacements (not necessarily the same throughout the body). Virtual work is done by the externally applied forces – external virtual work – and by the internal forces – internal virtual work . Looking again at the spring problem of Fig. 5. 5.1, the external virtual work is xF Wapl extδ δ= and, considering the spring force to be an “inter nal” force, the internal virtual work is xkx W δ δ−=int . This latter virtual work can be re-written as U Wδδ−=int where Uδ is the virtual potential energy change which o ccurs when the spring is moved a distance xδ (keeping the spring force constant). In the same way, the internal virtual work of an elastic body is the (negative of the) virtual strain energy and the principle of virtual work can be expressed as U Wextδδ= Principle of Virtual Work for an Elastic Body (5.5.4) The principle can be extended to accommodate dissipation (see Parts III and IV), but only elastic materials will be examined here. The virtual strain energy for a uniaxial rod is derived next. A B CF L aCRARuδ CRF Section 5.5 Solid Mechanics Part I Kelly 2045.5.3 Virtual Strain Energy for a Uniaxially Loaded Bar In what follows, to distinguish between the strain energy and the displacement, the former will now be denoted by w and the latter by u. Consider a uniaxial bar which undergoes strains ε. The strain is the unit change in length and, considering an element of length dx, Fig. 5.5.4a, the strain is [] dxdu xx xux xux=ΔΔ−−Δ++Δ=)() (ε (5.5.5) in the limit as 0→Δx . With εσd dw= , the strain energy density is then 2 2 21 21 21⎟ ⎠⎞⎜ ⎝⎛===dxduE E w εσε (5.5.6) and the strain energy is dxdxdu EAdVdxduE UL v∫ ∫ ⎟ ⎠⎞⎜ ⎝⎛=⎟ ⎠⎞⎜ ⎝⎛= 02 2 2 21 (5.5.7) This is the actual strain energy chan ge when the bar undergoes actual strains ε. For the simple case of constant A and L and constant strain L dxdu / /Δ= where Δ is the elongation of the bar, Eqn. 5.5.7 reduces to L AE U 2/2Δ= (equivalent to Eqn. 5.2.2). Figure 5.5.4: element undergoing actual and virtual displacem ents; (a) actual displacements, (b) virtual displacements It will now be shown that th e internal virtual work done as material particles undergo virtual displacements uδ is indeed given by Uδ, with U given by Eqn. 5.5.7. Consider an element to “unde rgo” virtual displacements uδ, Fig. 5.5.4b, which are, by definition, measured from the actual displacements . The virtual displacements give rise to virtual strains : xΔ)(xu ) ( x xuΔ+ x x xΔ+ )(xuδ ) ( x xuΔ+δ(a) (b) Section 5.5 Solid Mechanics Part I Kelly 205() dxud xxu x xu δ δ δδε =Δ−Δ+=)( ) ( (5.5.8) again in the limit as 0→Δx . Since ()dxdu/δδε= , it follows that () dxud dxduδδ=⎟ ⎠⎞⎜ ⎝⎛ (5.5.9) In other words, the variation of the derivativ e is equal to the derivative of the variation2. One other result is needed before calculati ng the internal virtual work. Consider a function of the displacement ) ( uf . The variation of f when u undergoes a virtual displacement is by definition ududfuuuf u ufuf u uff δδδδδ δ =−+=−+≡)( ) ()( ) ( (5.5.10) now in the limit as the virtual displacement 0→uδ . From this one can write ⎟ ⎠⎞⎜ ⎝⎛⎟ ⎠⎞⎜ ⎝⎛= ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛ dxdu dxdu dxduδ δ 22 (5.5.11) The stress σ applied to the surface of the elemen t under consideration is an “external force”. The internal force is the equal and opposite stress on the other side of the surface inside the element. The internal vi rtual work (per unit volume) is then σδε δ−=W . Since σ is the actual stress, unaffected by th e virtual straining, ⎥⎥ ⎦⎤ ⎢⎢ ⎣⎡ ⎟ ⎠⎞⎜ ⎝⎛−=⎟ ⎠⎞⎜ ⎝⎛−=⎟ ⎠⎞⎜ ⎝⎛⎟ ⎠⎞⎜ ⎝⎛−=−=2 2 21 21 dxduEdxduEdxdu dxduE E W δ δ δ εδεδ (5.5.12) since the Young’s modulus is unaffected by a ny virtual displacement. The total work done is then dVdxduE W v∫⎟ ⎠⎞⎜ ⎝⎛−=2 int21δδ (5.5.13) which, comparing with Eqn. 5.5.7, is the desired result, U Wδδ−=int . Example Two rods with cross sectional areas 2 1,AA , lengths 2 1,LL and Young’s moduli 2 1,EE and joined together with the other ends fixed, as shown in Fig. 5.5.5. The rods are 2 this holds in general for any function; manipulations with variations form a part of a branch of mathematics known as the Calculus of Variations , which is concerned in the main with minima/maxima problems Section 5.5 Solid Mechanics Part I Kelly 206subjected to a force P where they meet. As the rods el ongate/contract, the strain is simply LuB/=ε , where Bu is the displacement of the point at which the force is applied. The total elastic strain energy is, from Eqn. 5.5.7, 2 222 2 111 2 2B B uLAEuLAEU + = (5.5.14) Introduce now a virtual displacement Buδ at B. The external virtual work is BuP Wδδ=ext . The principle of virtual work, Eqn. 5.5.4, states that ⎭⎬⎫ ⎩⎨⎧ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ =2 222 111 2 2B B uLAE LAEuPδδ (5.5.15) Figure 5.5.5: two rods subjected to a force P From relation 5.5.10, B B B uuLAE LAEuP δ δ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+= 222 111 (5.5.16) The virtual displacement Buδ is arbitrary and so can be cancelled out, givi ng the result 1 222 111− ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ =LAE LAEP uB (5.5.17) from which the strains and hence stresses can be evaluated. Note that the reaction forces were not involved in this solution method. ■ 5.5.4 Virtual Strain Energy for a Beam The strain energy in a beam is given by Eqn. 5.2.7, viz. dxEIMUL ∫= 02 2 (5.5.18) Papplied force fixed fixed 1L2LB Section 5.5 Solid Mechanics Part I Kelly 207Using the moment-curvature relation 4.6.36, ()2 2/dxvdEI M= , where v is the deflection of the beam, dxdxvdEIUL ∫ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛= 02 22 2 (5.5.19) and the virtual strain energy is dxdxvdEIUL ∫ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛= 02 22 2δδ (5.5.20) It is not easy to analyse problems using this expression and the prin ciple of virtual work directly, but this expression will be used in the next section in conjunction with the related principle of minimum potential energy. 5.5.5 Problems 1. Consider a uniaxial bar of length L with constant cross section A and Young’s modulus E, fixed at one end and subjected to a force P at the other. Use the principle of virtual work to show that the displacement at the loaded end is EAPLu /= . 2. Consider a uniaxial bar of length L, cross sectional area A and Young’s modulus E. What factor of EAL is the strain energy when th e displacements in the bar are x u310−= , with x measured from one end of the bar? What is the internal virtual work for a virtual displacement x u510−=δ ? For a constant virtual displacement along the bar? 3. A rigid bar rests upon three columns, a central column with Young’s modulus GPa100 and two equidistant outer columns with Young’s moduli GPa200 . The columns are of equal length 1m and cross-sectional area 2cm1 . The rigid bar is subjected to a downward force of kN10 . Use the principle of virtual work to evaluate the vertical displacement dow nward of the rigid bar. 4. Re-solve problem 3 from §5.2.6 using th e principle of virtual displacements. Section 5.6 Solid Mechanics Part I Kelly 2085.6 The Principle of Minimum Potential Energy The principle of minimum potential energy follows directly from the principle of virtual work (for elastic materials). 5.6.1 The Principle of Minimum Potential Energy Consider again the example give n in the last section; in pa rticular re-write Eqn. 5.5.15 as 02 22 222 111= ⎭⎬⎫ ⎩⎨⎧ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+−B B uLAE LAEPuδ (5.6.1) The quantity inside the curly br ackets is defined to be the total potential energy of the system, Π, and the equation states that the variation of Π is zero – that this quantity does not vary when a virtual displacement is imposed: 0=Πδ (5.6.2) The total potential energy as a function of displacement u is sketched in Fig. 5.6.1. With reference to the figure, Eqn. 5.6.2 can be in terpreted as follows: the total potential energy attains a stationary value (m aximum or minimum) at the actual displacement (1u); for example, 0≠Πδ for an incorrect displacement 2u. Thus the solution for displacement can be obtained by finding a stationary value of the total potential energy. Indeed, it can be seen that the quantity inside the curly brackets in Fig. 5.6.1 attains a minimum for the solution already derived, Eqn. 5.5.17. Figure 5.6.1: the total pote ntial energy of a system To generalise, define the “potential energy” of the applied loads to be extW Vδδ−= so that V Uδδδ+=Π (5.6.3) The external loads must be conservative, precluding for example any sliding frictional loading. Taking the total potential en ergy to be a function of displacement u, one has 0)(=Π=Π uduudδ δ (5.6.4) )(uΠ 1u1uδ 2u2uδ Section 5.6 Solid Mechanics Part I Kelly 209 Thus of all possible displacements u satisfying the loading and boundary conditions, the actual displacement is that which gives rise to a stationary point 0 /=Πdu d and the problem reduces to finding a stationary value of the total potential energy VU+=Π . Stability To be precise, Eqn. 5.6.2 only demands that the total potential energy has a stationary point, and in that sense it is called the principle of stationary potential energy . One can have a number of stationary points as sketched in Fig. 5.6.2. The true displacement is one of the stationary values 3 2 1,,uuu . Figure 5.6.2: the total pote ntial energy of a system Consider the system with displacement 2u. If an external force acts to give the particles of the system some small initial velocity and hence kinetic energy, one has KΔ+ΔΠ=0 . The particles will now move and so the displacement 2u changes. Since Π is a minimum there it must increase and so the kinetic energy must decrease, and so the particles remain close to the equilibrium position. For this reason 2u is defined as a stable equilibrium point of the system. If on the other hand the particles of the body were given small initial velocities from an initial displacement 1u or 3u, the kinetic energy would increase dramatically; these points are called unstable equilibrium points. Only the state of stable equilibrium is of interest here and the principle of stationary potential energy in this case becomes the prin ciple of minimum potential energy. 5.6.2 The Rayleigh-Ritz Method In applications, the principle of mini mum potential energy is used to obtain approximate solutions to problems which are otherwise di fficult or, more usually, impossible to solve exactly. It forms one basis of the Finite Element Method (FEM), a general technique for solving systems of equations which ar ise in complex solid mechanics problems. Example Consider a uniaxial bar of length L, young’s modulus E and varying cross-section )/ 1( 0 Lx AA+= , fixed at one end and subjected to a force F at the other. The true Π 1u2u3u Section 5.6 Solid Mechanics Part I Kelly 210solution for displacement to this problem is ()()Lx EAFL u / 1ln /0+ = . To see how this might be approximated using the principle, one writes LxL uF dxdxduEA VU=−⎟ ⎠⎞⎜ ⎝⎛=+=Π∫2 021 (5.6.5) First, substituting in the exact solution leads to 02 02 0 00 22ln2ln/ 11)/ 1(2 EALF EAFLF dxLx EAFLxEAL −= −⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ++ =Π∫ (5.6.6) According to the principle, any other di splacement solution (which satisfies the displacement boundary condition 0 )0(= u ) will lead to a gr eater potential energy Π. Suppose now that the solution wa s unknown. In that case an estimate of the solution can be made in terms of some unknown parameter(s), substituted into Eqn. 5.6.5, and then minimised to find the parameters. This procedure is known as the Rayleigh Ritz method . For example, let the guess, or trial function , be the linear function x uβα+= . The boundary condition leads to 0=α . Substituting x uβ= into Eqn. 5.6.5 leads to LF LEA LF dxLx EAL ββ β β − =−+ =Π∫2 0 02 043)/ 1(21 (5.6.7) The principle states that () 0 /= Π=Π δββ δ dd , so that 0 0032 32023 EAFxuEAFFL LEAdd=→=→=− =Πβ ββ (5.6.8) The exact and approximate Ritz solu tion are plotted in Fig. 5.6.3. Figure 5.6.3: exact and (Ritz) approxi mate solution for axial problem 00.10.20.30.40.50.60.7 0.2 0.4 0.6 0.8 1x0EAFL Lx/exact approximate Section 5.6 Solid Mechanics Part I Kelly 211The total potential energy due to this approximate solution 03/2 EA Fx is, from Eqn. 5.6.5, 02 31 EALF−=Π (5.6.9) which is indeed greater than the minimum value Eqn. 5.6.6 (02/ 347.0 EALF−≈ ). ■ The accuracy of the solution 5.6.9 can be improved by using as the trial function a quadratic instead of a linear one, say 2xx u γβα++= . Again the boundary condition leads to 0=α . Then 2xx uγβ+= and there are now two unknowns to determine. Since Π is a function of two variables, () 0 , =∂Π∂+∂Π∂=Π δγγδββγβδ (5.6.10) and the two unknowns can be obtained from the two conditions 0 ,0 =∂Π∂=∂Π∂ γ β (5.6.11) Example A beam of length L and constant Young’s modulus E and moment of inertia I is supported at its ends and subjected to a uni form distributed force per length f. Let the beam undergo deflection ) (xv. The potential energy of the applied loads is dxxfv VL ∫−= 0)( (5.6.12) and, with Eqn. 5.5.19, the total potential energy is dxvf dxdxvd EIL L ∫ ∫−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛=Π 0 02 22 2 (5.6.13) Choose a quadratic trial function 2xx v γβα++= . The boundary conditions lead to ) (Lxxv−=γ . Substituting into 5.6.13 leads to 6/ 23 2Lf EILγγ−=Π (5.6.14) With () 0 /=Π=Π δγγ δ dd , one finds that Section 5.6 Solid Mechanics Part I Kelly 21222 3 2 24 24)(24xEIfLxEIfLxvEIfL+−=→=γ (5.6.15) which compares with the exact solution 4 33 24 12 24)( xEIfxEIfLxEIfLxv − +−= (5.6.16) ■ 5.6.3 Problems 1. Consider the statically indeterminate uniax ial problem shown below, two bars joined at Lx=, built in at 0=x and L x2= , and subjected to a force F at the join. The cross-sectional area of the bar on the left is 2 A and that on the right is A. Use the principle of minimum potential energy in conjunction with the Rayleigh-Ritz method with a trial displacement function of the form 2xx u γβα++= to approximate the exact displacement and in particular the displacement at Lx=. 2. A beam of length L and constant Young’s modulus E and moment of inertia I is supported at its ends and subjected to a uniform distributed force per length f and a concentrated force P at its centre. Use the principle of minimum potential energy in conjunction with the Rayleigh-Ritz method with a trial deflection ()Lx v / sinπα= , to approximate the exact deflection. 3. Use the principle of minimum potential ener gy in conjunction with the Rayleigh-Ritz method with a trial solution x uα= to approximately solve the problem of axial deformation of an elastic rod of varying cro ss section, built in at one end and loaded by a uniform distributed force/length f, and a force P at the free end, as shown below. The cross sectional are is ) / 2( )(0 Lx A xA −= and the length of the rod is L. F L L fP 2136 Linear Elasticity II Some further topics of elasti city are treated in this Chap ter. First, there is the buckling of elastic co lumns in section 6.1. In secti on 6.2, anisotropic elasticity is discussed. 214 Section 6.1 Solid Mechanics Part I Kelly 2156.1 Elastic Buckling The initial theory of the buc kling of columns was worked out by Euler in 1757, a nice example of a theory precedi ng the application, the appli cation mainly being for the later “invented” metal columns in modern structures. 6.1.1 Columns and Buckling A column is a long slender bar under axial compression, Fig. 6.1.1. A column can be horizontal, vertical or inclined; in the latter cases it is termed a strut . The column under axial compression responds el astically in exactly the same way as the axial bar of §4.3. For example, it decr eases in length unde r a compressive force P by an amount given by Eqn. 4.3.5, EAPL/=Δ . However, when the compressive force is large enough, the column will buckle with lateral deflection. This possibility is the subject of this section. Euler’s Theory of Buckling Consider an elastic column of length L, pin-ended so free to rotate at its ends, subjected to an axial load P, Fig. 6.1.1. Assume that it undergoes a lateral deflection denoted by v. Moment equilibrium of a section of the deflected column cut at a typical point x, and using the moment-curvature Eqn. 4.6.36, results in 22 )( )( dxvdEIxMxPv ==− (6.1.1) Hence the deflection v satisfies the differential equation 0)(2 22 =+ xvk dxvd (6.1.2) where EIPk=2 (6.1.3) Fig. 6.1.1: a column with deflection v P P xy )(xv L Section 6.1 Solid Mechanics Part I Kelly 216 The ordinary differential equation 6.1.2 is linear, homogeneous and with constant coefficients. Its solution can be found in any standard text on differential equations and is given by ()()kx B kx Axv sin cos )( + = (6.1.4) where A and B are as yet unknown constants. The boundary conditions for pinned- ends are 0)( ,0)0( = = Lv v (6.1.5) The first condition requires A to be zero and the second leads to ()0 sin=kL B (6.1.6) It follows that either: (a) 0=B , in which case 0)(=xv for all x and the column is not deflected or (b) () 0 sin=kL , which holds when kL is an integer number of π’s, i.e. K,3,2,1 ,= = nLnkπ, (6.1.7) As mentioned, the solution (a ) is governed by the axial deformation theory discussed in §4.3. Concentrating on (b), the co rresponding solution for the deflection is K,3,2,1 , sin )( =⎟ ⎠⎞⎜ ⎝⎛= nLxnBxvnπ (6.1.8) The parameter k is defined by Eqn. 1.6.3, so that, using 1.6.7, K,3,2,1 ,2 =⎟ ⎠⎞⎜ ⎝⎛= nLnEI Pnπ, (6.1.9) It has hence been show n that buckling, i.e. 0≠v , can only occur at a discreet set of applied loads - the buckling loads - given by 6.1.9. In practice the most important buckling load is the fi rst, corresponding to 1=n , since this will be the first of the loads reached as the applied load P is increased from zero; this is called the critical buckling load : 2 ⎟ ⎠⎞⎜ ⎝⎛=LEIPcπ (6.1.10) with associated deflection Section 6.1 Solid Mechanics Part I Kelly 217 ⎟ ⎠⎞⎜ ⎝⎛=LxBxvπsin )(1 (6.1.11) The column hence deforms into a single sine wave, which is termed the mode or mode shape of the deflected column. Note that B, the amplitude of the deflection, can not determined by this model. This is a consequence of assu ming the deflection is small; of linearising the problem. A more exact fini te deformation theory has been worked out and is called the theory of the elastica , but this is not pursued here. This mathematical structure, where one fi nds one can only get non-zero solutions of an equation for certain values of a parame ter is very common in engineering and theoretical physics. The critical values of the parameter, in this case k, are termed the eigenvalues of the problem, and the co rresponding non-zero solutions, ) (xv, are the eigenfunctions . The second moment of area I has dimensions of ()4length , and for columns is often written in the form 2Ar I= where A is the cross-sectional area of the column and the length r is called the radius of gyration . For example in the case of a circular shaft of radius a, 4 /4aIπ= (see §4.6.4) so 2/ar= . Failure of the Column The expression 6.1.10 for the critical buckli ng load can be written in terms of the radius of gyration: 2 2⎟ ⎠⎞⎜ ⎝⎛=LEAr Pcrπ or ()22 /rL Ecrπσ= (6.1.12) where crσ is the mean compressive stress on the loaded end of the column. The second equation in 6.1.12 is the mo st convenient non-dimensional form of presenting theoretical and experimental results for buckling problems. The ratio rL/ is called the slenderness ratio . Failure of the column will occur in pure ly axial compression if the stress in the column reaches the yield stress of the material (see §4.7, in particular 4.7.1). On the other hand, if the cr itical buckling stress crσ is less than the yield stress, then the column will fail by buckling before the yield stress is reached. Eqn. 6.1.12 is plotted in Fig. 6.1.2. Th e yield stress of the ma terial is denoted by Y. A critical slenderne ss ratio is denoted by ()crrL/ . For slenderness ratios less than the critical value, that is, for relatively squat columns, the stress in the column will reach the yield stress before buckling occurs. For example, consider a steel column for which GPa210=E and MPa210=Y . The critical value of the slenderness ratio is then 35.99 /=rL , which is a length to diameter ratio of about 25 for a circular column. Buckling will then occur in such Section 6.1 Solid Mechanics Part I Kelly 218columns which have 35.99 />rL , for sufficiently high applied axial compressive force. Fig. 6.1.2: critical values of the slenderness ratio 6.1.2 A General Approach to Buckling The model developed above only applies to columns simply supported at each end. To discuss the more general case one can retu rn to the formulati on of the bending of a beam discussed in §4.6.4, but include also axial forces. Fig. 4.6.18 is reproduced as Fig. 6.1.3 but now with compressive axia l forces, the forces offset by a small increment in deflection vΔ. Figure 6.1.3: forces and moments acting on a column Resolving vertically, one agai n arrives at Eqn. 4.6.10: )(xpdxdV= (6.1.13) Resolving horizontally, one simply gets ()x xPxP Δ+=)( , so that P is constant. Taking moments, one then gets, instead of 4.6.13, VdxdvPdxdM=+ (6.1.14) x)(xV xΔ)(xM ) ( xxVΔ+) ( xxMΔ+ xxΔ+)(xp •A )(xP) ( xxPΔ+ vΔ()crrL/crσ Y rLbuckling no buckling Section 6.1 Solid Mechanics Part I Kelly 219Note the extra term involving P, which is not present in pure bending theory. Eliminating M between 6.1.14 and the moment-curva ture equation 4.6.36 then leads to an expression for the shear force: dxdv EIP dxvd EIV+=33 (6.1.15) Note that, in the beam theory, where 0=P , the third derivative of the deflection is zero whenever the shear force is zero, in particular at a free, i.e. unsupported, end. Here, however, it is no longer true that the third derivative is zero. The final differential equation is now obt ained by differentiating 6.1.15 and using 6.1.13: EIp dxvd EIP dxvd=+22 44 (6.1.16) Concentrating on the buckling behaviour a nd so neglecting the transverse load ) (xp1, one arrives at the differential equation 022 2 44 =+dxvdkdxvd (6.1.17) where again EIP k /2= (Eqn. 6.1.3). Eqn. 6.1.17 is a homogeneous fourth-order differential equation and its solution is ()() D Cx kx B kx Axv ++ + = sin cos )( (6.1.18) The four constants are determined by the end conditions on ) (xv, two conditions at each end. There are three cases: (1) Pinned end: boundary conditions are 0=v and 0=M ; from the moment-curvature equation, 0=M can be replaced with 0 /2 2=dxvd (2) Fixed end: boundary conditions are 0 / ,0= = dxdv v (3) Free end: Boundary conditions are 0=M and 0=V ; again, this implies that 0 /2 2=dxvd and, from Eqn. 6.1.15, 0=V can be replaced with () 0 / /2 3 3= + dxdvk dxvd The case of pinned-pinned results again in the Euler solution given above. Consider now the case where one end is clamped and the other, loaded, end, is unrestrained (“fixed-free”), Fig. 6.1.4. 1 bars subjected to both axial compressive loads and transverse loads are called beam-columns Section 6.1 Solid Mechanics Part I Kelly 220 Fig. 6.1.4: a fixed-free column At the clamped end, 0)0( )0( =′=v v , giving 0 ,0 =+ =+ kBC DA (6.1.19) At the free end, 0)(=′′Lv and 0 )( )(2=′+′′′ LvkLv , leading to ()() 0 ,0 sin cos = = + C kx B kx A (6.1.20) Thus, from 6.1.19, B too is zero and A satisfies ()0 cos=kL A (6.1.21) Buckling hence can only occur when ()0 cos=kL , i.e. when K,2,1,0 ,21=⎟ ⎠⎞⎜ ⎝⎛+= n n kLπ (6.1.22) Using the definition of the parameter k the buckling loads are given by ()K,2,1,0 ,2 21 =⎥⎦⎤ ⎢⎣⎡+= nLnEIPπ (6.1.23) with the critical buckling load now P M Section 6.1 Solid Mechanics Part I Kelly 221 2 2⎟ ⎠⎞⎜ ⎝⎛=LEI Pcrπ (6.1.24) which is one quarter of the value for a pinne d strut, Eqn. 6.1.10. The buckling modes are given by 6.1.18: () K,2,1,0 , cos1 )(21 = ⎭⎬⎫ ⎩⎨⎧ ⎥⎦⎤ ⎢⎣⎡+−= nLxn Dxv π (6.1.25) The first three modes are sketched in Fig. 6.1.5; again, the amplitude is unknown, only the shape. Figure 6.1.5: mode shapes for the fixed-free column Other cases of end-support can be treated in the same way. Results for the critical buckling stress for various cases are sketched in Fig. 6.6.6. Fig. 6.1.6: critical values of the slen derness ratio for different end-cases crσ Y rL/free-fixedpinned- pinned (Euler ) fixed-fixed pinned-fixedLx/0=n1=n 2=n Section 6.2 Solid Mechanics Part I Kelly 2226.2 Anisotropic Elasticity There are many materials which, although well m odelled using the linear elastic model, are not nearly isotropic. Examples are wood and bone. The mechanical properties of these materials differ in different directions. Materials with this direction dependence are called anisotropic . 6.2.1 Material Constants The most general form of Hooke’s la w for a linear elastic material is ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ====== ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ = ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ======xyxzyzzzyyxx xyxzyzzzyyxx C C C C C CC C C C C CC C C C C CC C C C C CC C C C C CC C C C C C εεεεεεεεεεεε σσσσσσσσσσσσ 654321 66 65 64 63 62 6156 55 54 53 52 5146 45 44 43 42 4136 35 34 33 32 3126 25 24 23 22 2116 15 14 13 12 11 654321 (6.2.1) where each stress component depends on all st rain components. This new notation, with only one subscript for the stress and strain, numbered from 1…6, is helpful as it allows the equations of anisotropic elasticity to be written in matrix form. The 36 s'ijC are material constants called the stiffnesses , and in principle are to be obtained from experiment. The matrix of stiffnesses is called the stiffness matrix . Note that these equations imply that a normal stress xxσ will induce a material element to not only stretch in the x direction and contract laterally, but to undergo shear strain too, as illustrated schematically in Fig. 6.2.1. Figure 6.2.1: an element undergoing shear stra in when subjected to a normal stress only Strain Energy Considerations In section 5.4 it was shown that the elastic st rain energy, when differentiated with respect to the strains, gives the stresses, Eqn. 5.4.91: ij ijuσε=∂∂ (6.2.2) 1 this expression was derived for isotropic elastic materials, but it can be shown to hold for the more general anisotropic Hooke’s law; it is proved more rigorously in Part III xxσxxσ Section 6.2 Solid Mechanics Part I Kelly 223Using also 6.2.1, it follows that () ()61 16, CuCu xy xx xy xxxx xy xx xy=∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂=∂∂=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛ ∂∂ ∂∂σεεεσεεε (6.2.3) Since the order of partial differentiation fo r these second partial derivatives should be immaterial, it follows that 61 16C C= . Following the same procedure for the rest of the stresses and strains, it can be s een that the stiffness matrix is symmetric and so there are only 21 independent elastic constants in the mo st general case of anisotropic elasticity. Eqns. 6.2.1 can be inverted so that the strains are given explicitly in terms of the stresses: ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ = ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡654321 6656 5546 45 4436 35 34 3326 25 24 23 2216 15 14 13 12 11 654321 σσσσσσ εεεεεε SS SS S SS S S SS S S S SS S S S S S (6.2.4) The s 'ijS here are called compliances , and the matrix of compliances is called the compliance matrix . The bottom half of the compliance matrix has been omitted since it too is symmetric. It is difficult to model fully anisotropic materials due to the great number of elastic constants. Fortunately many materials which are not fully isotropic still have certain material symmetries which simplify the above equatio ns. These material types are considered next. 6.2.2 Orthotropic Linear Elasticity An orthotropic material is one which has three orthogonal planes of microstructural symmetry. An example is shown in Fig. 6.2.2a, which shows a glass-fibre composite material. The material consists of thousands of very slender, long, glass fibres bound together in bundles with oval cross-sections. These bundles are then surrounded by a plastic binder material. The c ontinuum model of this composite material is shown in Fig. 6.2.2b wherein the fine microstructural deta ils of the bundles and surrounding matrix are “smeared out”. Three mutually perpendicula r planes of symmetry can be passed through each point in the continuum model. The zyx,, axes forming these planes are called the material directions . Section 6.2 Solid Mechanics Part I Kelly 224 Figure 6.2.2: an orthotropic material; (a) micr ostructural detail, (b) continuum model The material symmetry inherent in the orth otropic material reduces the number of independent elastic constants. To see this, consider an element of orthotropic material subjected to a shear strain ()xyεε=6 and also a strain ()xyεε−=−6 , as in Fig. 6.2.3. Figure 6.2.3: an element of orthotropi c material undergoing shear strain From Eqns. 6.2.1, the st resses induced by a strain 6ε only are 6 66 6 6 56 5 6 46 46 36 3 6 26 2 6 16 1 , ,, , εσεσεσεσεσεσ C C CC C C = = == = = (6.2.5) The stresses induced by a strain 6ε− only are (the prime is added to distinguish these stresses from those of Eqn. 6.2.5) 6 66 6 6 56 5 6 46 46 36 3 6 26 2 6 16 1 , ,, , ε σεσε σεσε σεσ C C CC C C −=′ −=′ −=′−=′ −=′ −=′ (6.2.6) These stresses, together with the strain, are show n in Fig. 6.2.4 (the microstructure is also indicated) y zxy zx )a() b(binder fibre bundles 6ε 6ε− zxy Section 6.2 Solid Mechanics Part I Kelly 225 Figure 6.2.4: an element of orthotropic material undergoing shear strain; (a) positive strain, (b) negative strain Because of the symmetry of the material, one would expect the normal stresses in Fig. 6.2.4 to be the same, 1 1σσ′= , 2 2σσ′= , but the shear stresses to be of opposite sign, 6 6σσ′−= . Eqns. 6.2.5-6 then imply that 056 46 36 26 16 ===== C C C C C (6.2.7) Similar conclusions follow from consideri ng shear strains in th e other two planes: 0 :0 : 34 24 14 445 35 25 15 5 ======= C C CC C C C εε (6.2.8) The stiffness matrix is thus reduced, and there are only nine independent elastic constants: ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ = ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ 654321 6655443323 2213 12 11 65432100 00 0 00 0 00 0 0 εεεεεε σσσσσσ CCCCC CC C C (6.2.9) These equations can be inverted to get, introducing elastic constants E, ν and G in place of the sSij': 1σ2σ 6σ 1σ′2σ′ 6σ′ )a() b( Section 6.2 Solid Mechanics Part I Kelly 226⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ −−− −−− = ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡654321 1213233 223 113332 2 112331 221 1 654321 210 0 0 0 00210 0 0 00 0210 0 00 0 010 0 010 0 01 σσσσσσ ννν ννν εεεεεε GGGE E EE E EE E E (6.2.10) The nine independent constants he re have the following meanings: iE is the Young’s modulus (stiffness) of the material in direction 3,2,1=i ; for example, 11 1εσ E= for uniaxial tension in the direction 1. ijν is the Poisson’s ratio representing the ratio of a transverse strain to the applied strain in uniaxial tension; for example, 1 2 12 /εεν−= for uniaxial tension in the direction 1. ijG are the shear moduli representing the shear stiffness in the corresponding plane; for example, 12G is the shear stiffness for shearing in the 1-2 plane. 12G and 13G are the axial shear moduli and 23G is the transverse (out-of-plane ) shear modulus . Note that, from symmetry of the stiffness matrix, 1 21 2 12 1 31 3 13 2 32 3 23 , , E E E E E E νννννν = = = (6.2.11) An important feature of the orthotropic material is that there is no shear coupling with respect to the material axes. In other words, normal stresses result in normal strains only and shear stresses result in shear strains only. Note that there will in general be shear coupling when the reference axes used, zyx,, , are not aligned with the material directions 3,2,1 . For example, suppose that the yx− axes were oriented to the material axes as shown in Fig. 6.2.5. Section 6.2 Solid Mechanics Part I Kelly 227 Figure 6.2.5: reference axes not aligne d with the material directions Assuming that the material constants were known, the stresses and strains in the constitutive equations 6.2.10 can be transformed into xy xxεε, , etc. and xy xxσσ, , etc. using the strain and stress transformation equations. The resulting matrix equations relating the strains xy xxεε, to the stresses xy xxσσ, will then not contain zero entries in the stiffness matrix, and normal stresses, e.g. xxσ, will induce shear strain, e.g. xyε, and shear stress will induce normal strain. 6.2.3 Transversely Isotropic Linear Elasticity A transversely isotropic material is one which has a sing le material direction and whose response in the plane orthogonal to this directio n is isotropic. An example is shown in Fig. 6.2.6, which again shows a glass-fibre co mposite material with aligned fibres, only now the cross-sectional shapes of the fibres are circular. The characteristic material direction is z and the material is isotropic in any plane parallel to the yx− plane. The material properties are the same in all dire ctions transverse to the fibre direction. Figure 6.2.6: a transversely isotropic material This extra symmetry over that inherent in th e orthotropic material reduces the number of independent elastic constants further. To s ee this, consider an el ement of transversely isotropic material subjected to a normal strain ()xxεε=1 only of magnitude ε, Fig. 6.2.7a, and also a normal strain ()yyεε=2 of the same magnitude, ε, Fig. 6.2.7b. The yx− plane is the plane of isotropy. y zxxy 12 θ Section 6.2 Solid Mechanics Part I Kelly 228 Figure 6.2.7: elements of a transversely isot ropic material undergoing normal strain in the plane of isotropy From Eqns. 6.2.9, the st resses induced by a strain εε=1 only are 0 ,0 ,0, , 6 5 431 3 21 2 11 1 ==== = = σσσεσεσεσ C C C (6.2.12) The stresses induced by the strain εε=2 only are (the prime is a dded to distinguish these stresses from those of Eqn. 6.2.12) 0 ,0 ,0, , 6 5 432 3 22 2 12 1 =′=′=′=′ =′ =′ σσσεσεσεσ C C C (6.2.13) Because of the isotropy, the ) (1 xxσσ= due to the 1ε should be the same as the )(2 yyσσ= due to the 2ε, and it follows that 22 11C C= . Further, the ) (3 zzσσ= should be the same for both, and so 32 31C C= . Further simplifications arise from consider ation of shear deformations, and rotations about the material axis, and one finds that 55 44C C= and 12 11 66 C C C−= . The stiffness matrix is thus reduced, and there are only five independent elastic constants: ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ −= ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡654321 12 1144443313 1113 12 11 654321 00 00 0 00 0 00 0 0 εεεεεε σσσσσσ C CCCCC CC C C (6.2.14) with ‘3’ being the material dire ction. These equations can be inverted to get, introducing elastic constants E, ν and G in place of the sSij': xy xy )a() b( Section 6.2 Solid Mechanics Part I Kelly 229⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ −−− −−− = ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡654321 654321 210 0 0 0 00210 0 0 00 0210 0 00 0 010 0 010 0 01 σσσσσσ ννν ννν εεεεεε tfff ff ffff t ttff tt t GGGE E EE E EE E E (6.2.15) with tt t E Gν+=1 21 ( 6 . 2 . 1 6 ) and again ‘3’ is the material direction (given subscript ‘f’ for fibre) and ‘1’ and ‘2’ are the transverse directions, given subscript ‘t’). 6.2.4 Isotropic Linear Elasticity An isotropic material is one for which the ma terial response is independent of orientation. The symmetry he re further reduces the number of elastic constants to two, and the stiffness matrix reads ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ −−−= ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡654321 12 1112 1112 111112 1112 12 11 654321 00 00 0 00 0 00 0 0 εεεεεε σσσσσσ C CC CC CCC CC C C (6.2.17) These equations can be inverted to get, introducing elastic constants E, ν and G, Section 6.2 Solid Mechanics Part I Kelly 230⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ −−− −−− = ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡654321 654321 210 0 0 0 00210 0 0 00 0210 0 00 0 010 0 010 0 01 σσσσσσ ννν ννν εεεεεε GGGE E EE E EE E E (6.2.18) with E Gν+=1 21 ( 6 . 2 . 1 9 ) which are Eqns. 4.2.8 and 4.2.5. Eqns. 6.2.17 can also be written in terms of the engineering constants E, ν and G with the help of the Lamé constants , λ and μ: ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ +++ = ⎥⎥⎥⎥⎥⎥⎥⎥ ⎦⎤ ⎢⎢⎢⎢⎢⎢⎢⎢ ⎣⎡ 654321 654321 20 20 0 20 0 0 20 0 0 20 0 0 2 εεεεεε μμμμλλμλλλμλ σσσσσσ (6.2.20) with () ( ) ()νμνννλ+=−+=12,21 1E E (6.2.21) 6.2.5 Problems 1. A very thin piece of orthotropic material is loaded (in-plane) by a uniaxial stress 1σ (aligned with the material direction ‘1’). Wh at are the strains in the material, in terms of the engineering constants? Section 6.2 Solid Mechanics Part I Kelly 2312. A specimen of bone in the shape of a cube is fixed and loaded by a compressive stress MPa1=σ as shown below. The bone can be c onsidered to be orthotropic, with material properties 31.0 ,32.0 ,62.0GPa91.4 ,GPa56.3 ,GPa41.2GPa4.18 ,GPa51.8 ,GPa91.6 32 31 2123 13 123 2 1 = = == = == = = ν ν νG G GE E E What are the stresses and strains which arise from the test according to this model (the bone is compressed along the ‘1’ direction)? 3. Consider a block of transversely isotropic material subjected to a compressive stress p−=1σ (perpendicular to the material direct ion) and constrained from moving in the other two perpendicular dire ctions (as in Problem 2). Evaluate the stresses 2σ and 3σ in terms of the engineering constants f tEE, and f tνν,. 4. A strip of skin is tested in biaxial tension as shown below. The measured stresses and strains are as given in the figure. The orientation of the fibres in the material is later measured to be o20=θ . (a) Calculate the normal stresses along a nd transverse to the fibres, and the corresponding shear stress. (b) Calculate the normal strains along and transverse to the fibres, and the corresponding shear strain. (c) Assuming the material to be orthotropic determine the elastic constants of the material (assume the stiffness in the fibre direction to be five times greater than the stiffness in the transverse direction). (d) Calculate the magnitude and orientations of the principal normal stresses and strains. (e) Do the principal directions of stress and strain coincide? surfaces are fixed Pax50=σPay35=σ 0420.0147.00470.0 −=== xyyx εεεFibre orientation θ Section 6.2 Solid Mechanics Part I Kelly 232 5. A biaxial test is performed on a roughly plan ar section of skin (thickness 1mm) from the back of a rat. The test axes ( x and y) are aligned such that deformation is induced in the skin along the spinal direction and transverse to this direction, under the assumption that the fibres are oriented principally in these directions. However, it is found during the experiment that shear stress es are necessary to maintain a biaxial deformation state. Measured stresses are kPa1 ,kPa2 ,kPa5 = = =xy yy xx σ σ σ Determine the in-plane orientation of the fibres given the data kPa 10001=E , kPa5002=E , kPa 5006=G , 2.021=ν . [Hint: derive an expression for xyε involving θ only, where θ is the inclination of the material axes from the yx− axes] 2337 Viscoelasticity The Linear Elastic Solid has been the main mate rial model analysed in this book thus far. It has a long history and is still the most widely used model in applications today. Viscoelasticity is the study of ma terials which have a time-dependence . Vicat, a French engineer from the Department of Road Cons truction, noticed in the 1830’s that bridge- cables continued to elongate over time even though under cons tant load, a viscoelastic phenomenon known as creep . Many other investigators, such as Weber and Boltzmann, studied viscoelasticity throughout the nineteenth century, but the real driving force for its study came later – the increased demand fo r power and the associated demand for materials which would stand up to temperatur es and pressures that went beyond previous experience. By then it had been recognised th at significant creep occurred in metals at high temperatures. The theory developed further with the emergence of synthetic polymer plastics, which exhibit strong viscoelastic proper ties. The study of viscoelasticity is also important in Bi omechanics, since many biomaterials respond viscoelastically, for example, heart tissue, muscle tissue and cartilage. Viscoelastic materials are defi ned in section 7.1 and some ev eryday viscoelastic materials and phenomena are discussed in section 7.2. The basic mechanical models of viscoelasticity, the Maxwell and Kelvin mode ls, are introduced in section 7.3, as is the general differential equation form of the linear viscoelastic law. Th e hereditary integral form of the constitutive equation is disc ussed in section 7.4 a nd it is shown how the Laplace transform can be used to solve linear viscoelastic problems in section 7.5. In section 7.6, dynamic loading, impact and vibr ations of viscoelastic materials are considered. Finally, in the last section, temperature e ffects are briefly discussed, including the important concept of th ermorheologically simple materials. 234 Section 7.1 Solid Mechanics Part I Kelly 2357.1 The Response of Viscoelastic Materials 7.1.1 Viscoelastic Materials The response of the linear elastic model to loading and unloading in a tension test was discussed in §4.2.1. The typical response of a viscoelastic ma terial is sketched in Fig. 7.1.1. The following will be noted: (i) there is a dependence on the rate of straining dtd/ε , Fig. 7.1.1a; the faster the stretching, the larger th e stress required (ii) the loading and unloading curves do not coincide, Fig. 7.1.1b, but form a hysteresis loop (iii) there may or may not be some permanent deformation upon complete unloading, Fig. 7.1.1b Figure 7.1.1: Response of a Viscoelastic ma terial in the Tensio n test; (a) different rates of stretching, (b) loading and unload ing with possible permanent deformation (non-zero strain at zero stress) The effect of rate of stretching shows that the viscoelastic material depends on time . This contrasts with the elastic material, whose cons titutive equation is independent of time, for example it makes no difference whether an elastic material is loaded to some given stress level for one second or one day, the resulting strain will be the same. It was shown in Chapter 5 that the area bene ath the stress-strain curve is the energy per unit volume; during loading, it is the energy stor ed in a material, during unloading it is the energy recovered. There is a difference betw een the two for the viscoelastic material, indicated by the shaded region in Fig. 7.1.1b. This shaded region is a measure of the energy lost through heat transfer mechanisms during the deformation. Most engineering materials undergo plasticity , meaning permanent deformations occur once the stress goes above the elastic limit. Th e stress-strain curve for these materials can look very similar to that of Fig. 7.1.1, but, in contrast to viscoelas ticity, plasticity is rate independent . Plasticity will be discussed in detail in Part II. possible permanent deformation σ εσ εunload )a() b(1ε& 2ε& 2 1εε&&> load Section 7.1 Solid Mechanics Part I Kelly 236Linear Viscoelasticity Linear viscoelastic materials are those for which there is a linear relationship between stress and strain (at any given time), εσ∝. As mentioned in §4.2.1, this requires also that the strains are small. Strain-time curves for a linear viscoelastic mate rial subjected to vari ous constant stresses are shown in Fig. 7.1.2. At any given time, say 1t, the strain is proportional to stress, so that the strain there due to oσ3 is three times the strain due to oσ. Figure 7.1.2: Strain as a function of time at different loads Linear viscoelasticity is a reasonable appr oximation to the time-dependent behaviour of metals and ceramics at relatively low temp eratures and under relatively low stress. However, its most widespread applica tion is in the modelling of polymers. 7.1.2 Testing of Viscoelastic Materials The tension test is the standard materials test. A number of other tests which are especially useful for the char acterisation of viscoelastic ma terials have been developed, and these are discussed next. The Creep and Recovery Test The creep-recovery test involves loading a material at c onstant stress, holding that stress for some length of time and then removi ng the load. The response of a typical viscoelastic material to this test is show in Fig. 7.1.3. First there is an instantaneous elastic straini ng, followed by an ever-increasing strain over time known as creep strain . The creep strain usually increases with an ever decreasing strain rate so that eventually a more-or-le ss constant-strain steady state is reached, but many materials whose response is linear often do not reach such a noticeable steady-state. When unloaded, the elastic strain is r ecovered immediately. There is then anelastic recovery – strain recovered over time; this anelastic strain is usually very small for 0σ02σ03σ ε 1tt Section 7.1 Solid Mechanics Part I Kelly 237metals, but may be significant in polymeric ma terials. A permanent strain may then be left in the material1. A test which focuses on the loading phase only is simply called the creep test . Figure 7.1.3: Strain response to the creep-recovery test Stress Relaxation Test The stress relaxation test involve s straining a material at cons tant strain and then holding that strain, Fig. 7.1.4. The stress required to hold the viscoelastic material at the constant strain will be found to decrease over time. This phenomenon is called stress relaxation ; it is due to a re-arrangement of the material on the molecular or micro-scale. 1 if the load is above the yield stress, then some of the permanent deformation will be instantaneous plastic (rate-independent) strain; the subject of this chapter is confined to materials which are loaded up to a stress below any definable yield stress; rate–dependent ma terials with a yield stress above which permanent deformation take place are called viscoplastic (see Part II) timestress 0σ strain σ t tε instantaneous strain creep strain elastic recovery anelastic recovery permanent strain Section 7.1 Solid Mechanics Part I Kelly 238 Figure 7.1.4: Stress response to the stress-relaxation test The Cyclic Test The cyclic tests involves a repeating pattern of loading-unloading, Fig. 7.1.5. It can be strain-controlled (with the resulting stress observe d), as in Fig. 7.1.5, or stress-controlled (with the resulting strain observed). The results of a cyclic test can be quite complex, due to the creep, stress-relaxation and permanent deformations. Figure 7.1.5: Typical stress respo nse to the cyclic test strain stress strain σt εε 0ε 0εtimetimestress strain σt tε 0ε stress relaxation 0σ Section 7.2 Solid Mechanics Part I Kelly 2397.2 Examples and Applications of Viscoelastic Materials Some of the properties of visc oelastic materials ar e their ability to creep, recover, undergo stress relaxation and absorb energy. Some ex amples of these phenomena are discussed in this section. 7.2.1 Creep and Recovery The disks in the human spine are viscoela stic. Under normal body weight, the disks creep, that is they get shorter with time. Ly ing down allows the spinal disks to recover and this means that most people are taller in the morning than in the evening. Astronauts have gained up to 5cm in height under near-zero gravity conditions. Skin tissue is viscoelastic. This can be seen by pinching the skin at the back of the hand; it takes time to recover back to its original fl at position. The longer th e skin is held in the pinched position, the longer it takes to recove r. The more rapidly it is pinched, the less time it takes to recover – it behaves “more elastically”. Skin is an ageing material , that is, its physical properties change over time. Younger skin recovers more rapidly than older skin. Wood is viscoelastic. The beams of old woode n houses can often be seen to sag, but this creeping under the weight of the roof and gr avity can take many decades or centuries to be noticeable. Concrete and soils are other materials which creep, as is ice, which has consequences for glacial movements. Materials which behave elastically at r oom temperature often attain significant viscoelastic properties when heated. Such is the case with metal turbine blades in jet engines, which reach very high temperatures and need to withstand very high tensile stresses. Conventional metals can creep significantly at high temperatures and this has led to the development of creep -resistant alloys; turbine blad es are now often made of so- called superalloys which contai n some or all of nickel, c obalt, chromium, aluminium, titanium, tungsten and molybdenum. Creep is also one of the principal causes of failu re in the electric light bulb. The filaments in light bulbs are made of tungsten, a metal with a very high melting point (>3300 oC); this is essential because the filament needs to be electrically heated to a temperature high enough for light emission ( ≈2000oC). If the filament creeps to o much it sags and its coils touch each other, leading to a localised shor t circuit. Light bulbs last longer if the temperature is reduced, as in dimmed light s. Creep can also be reduced by adding potassium bubbles to the tungsten. Polymer foams used in seat cushions cree p, allowing progressive conformation of the cushion to the body shape. These cushions help reduce the pressu res on the body and are very helpful for people confined to wheelchai rs or hospital beds for lengthy periods. They often have to be repl aced after about 6 months becau se creep causes them to become more dense and stiff. A newly born baby’s head is viscoelastic and it s ability to creep and recover helps in the birthing process. Also, if a baby lies in one specific position for long, for example the Section 7.2 Solid Mechanics Part I Kelly 240same way of sleeping all the time, its h ead can become misshapen due to creep deformation. A baby's skull becomes more solid after about a year. Viscoelasticity is also involved in the moveme nt and behaviour of the tectonic plates, the plates which float on and travel independently over the mantle of the earth, and which are responsible for earthquakes, volcanoes, etc. 7.2.2 Stress Relaxation Guitar strings are viscoelastic. When tightened they take up a tensile stress. However, when fixed at constant length (strain), stress relaxation occurs. The speed of sound in a string 1 is ρσ/=c , where σ is the stress and ρ the density. The frequency is λ/cf= , where λ is the wavelength. The length of the string L is equal to half the wavelength: )2/(/ L fρσ= . The reduction in stress thus implies a reduction in frequency and a lowering of pitch – the guitar go es out of tune. The strings of a Classical guitar are made of Nylon, a synthetic polymer . The great classica l guitarists of the 19th Century did not have Nylon, i nvented in 1938, but used Catgut strings, usually made from the intestines of sheep; Catgut is a natural polymer . Metal guitar stri ngs do not go out of tune so easily since metals are less viscoelastic than polymers. 7.2.3 Energy Absorption Tall buildings vibrate when dynamically load ed by wind or earthquake s. Viscoelastic materials have the property of ab sorbing such vibrational energy – damping the vibrations. Viscoelastic damp ers are used in some tall buildings, for example in the Columbia Center in Seattle, in which the damp ers consist of steel pl ates coated with a viscoelastic polymer compound - the dampers are fixed to some of the diagonal bracing members. Sometimes it is necessary to control vibrations but the use of a polymer is inappropriate - in this case it is necessary to use some other material with good vibration-control properties. A good example is the use of c opper-manganese alloy to reduce vibration and noise from naval ship propelle rs. This alloy has also b een used in pneumatic rock crushers. Zinc is also relatively viscoelas tic for a metal and zinc-aluminium alloys are used in pneumatic drills - the alloy damps the vibrations and makes it a little less uncomfortable for anyone holding a pneumatic dril l. Viscoelastic materials are also used to line the gloves worn by people working with pneumatic drills and jackhammers. Helicopters make a lot of noise, which comes mainly from the turbine (rotary engine) and gears, but it is usually exacerbated by resonanc e of the fuselage skin. Acoustic blankets consisting of a layer of fibreglass sandwiched be tween layers of vinyl cloth, placed inside the fuselage, can reduce the noise. Sikorsky, in their HH-53C rescue helicopter, coated a small portion of the fuselage skin with da mping treatments, which helped reduce the high-frequency noise in the cabin by 10 dB. 1 stress waves and wave propagation in solid materi als will be discussed in detail in Part II, under elastodynamics Section 7.2 Solid Mechanics Part I Kelly 241 In quartz watches, vibrations are set up in quartz crystal at ultras onic frequency (32.768 kHz). The vibrations are then used to ge nerate periodic signals, which may be divided into intervals of time, like the second. Quartz (SiO 2) is a very low loss material, meaning that it is very un-viscoelastic. This ensures that the vibrations are not dampened and the watch keeps good time. Tuning forks are often made of aluminium as it is also a low-loss material. An aluminium tuning fork will continue vibrating for quite a long time after being struck – the vibrations eventually die down because of sound-energy loss, but also because of the small energy loss due to viscoelasticity in the aluminium fork. Viscoelastic materials are excellent impact absorbers. A peak impact force can be reduced by a factor of two if an impact buffer is made of viscoelastic, rather than elastic, material. Elastomers are highly viscoelastic and make good impact absorbers; these are any of various substances resembling rubber - they have trade names like Sorbothane, Implus and Noene. Viscoelastic materials are used in automob ile bumpers, on computer drives to protect from mechanical shock, in helmets (the foam padding inside), in wrestling mats, etc. Viscoelastic materials are also used in shoe insoles to reduce impact transmitted to a person's skeleton. The cartilage at the ends of the femur and ti bia, in the knee joint, is a natural shock absorber. In an osteoarthritic knee, th e cartilage has degraded - sometimes the bones grind against each other causing great pain. Synthetic viscoelastic materials can be injected directly into an osteoarthritic kn ee, enveloping cartilage- deficient joints and acting as a lubricant and shock absorber. Section 7.3 Solid Mechanics Part I Kelly 2427.3 Rheological Models In this section, a number of one-dimensional linear viscoelastic models are discussed. 7.3.1 Mechanical (rheological) models The word viscoelastic is derived from the words "viscous" + "elastic"; a viscoelastic material exhibits both viscous and elastic be haviour – a bit like a fluid and a bit like a solid. One can build up a model of linear visc oelasticity by considering combinations of the linear elastic spring and the linear viscous dash-pot 1. These are known as rheological models or mechanical models . The Linear Elastic Spring The constitutive equation for a material whic h responds as a linear elastic spring of stiffness E is (see Fig. 7.3.1) σεE1= (7.3.1) The response of this material to a creep-re covery test is to undergo an instantaneous elastic strain upon loading, to ma intain that strain so long as the load is applied, and then to undergo an instantaneous de-s training upon removal of the load. Figure 7.3.1: the linear elastic spring The Linear Viscous Dash-pot Imagine next a material which responds like a viscous dash-pot; the dash-pot is a piston- cylinder arrangement, filled with a viscous fl uid, Fig. 7.3.2 – a strain is achieved by dragging the piston through the fl uid. By definition, the dash-pot responds with a strain- rate proportional to stress: σηε1=& (7.3.2) where η is the viscosity of the material. This is the typical response of many fluids ; the larger the stress, the faster the strainin g (as can be seen by pushing your hand through water at different speeds). 1 a non-linear theory can be developed by including non-linear springs and dash pots σE σ Section 7.3 Solid Mechanics Part I Kelly 243 Figure 7.3.2: the linear dash-pot The strain due to a suddenly applied load oσ may be obtained by integrating the constitutive equation 7.3.2. A ssuming zero initial strain, one has to ησε= (7.3.3) The strain is seen to increas e linearly and without bound so long as the stress is applied, Fig. 7.3.3. Note that there is no movement of the dash-pot at the onset of load; it takes time for the strain to build up. When the load is removed, there is no stress to move the piston back through the fluid, so that any stra in built up is permanent. The slope of the creep-line is ησ/o. Figure 7.3.3: Creep-Recovery Response of the Dash-pot The linear relationship between the stress and strain during th e creep-test may be expressed in the form ησεttJtJ to = = )( ),( )( (7.3.4) J here is called the creep (compliance) function ( E J /1= for the elastic spring). 7.3.2 The Maxwell Model Consider next a spring and dash-pot in series, Fig. 7.3.4. This is the Maxwell model . One can divide the total strain into one for the spring (1ε) and one for the dash-pot (2ε). Equilibrium requires that the stress be the same in both elements. One thus has the following three equations in four unknowns: ση σ t t σ 0σ stress applied stress removedε Section 7.3 Solid Mechanics Part I Kelly 2442 1 2 1 ,1,1εεεσηεσε += = = & E (7.3.5) To eliminate 1ε and 2ε, differentiate the first and thir d equations, and put the first and second into the third: εησησ &&=+E Maxwell Model (7.3.6) This constitutive equation has b een put in what is known as standard form – stress on left, strain on right, increasing order of derivatives from left to right, and coefficient of σ is 1. Figure 7.3.4: the Maxwell Model Creep-Recovery Response Consider now a creep test. Physically, when the Maxwell model is subjected to a stress 0σ, the spring will stretch immediat ely and the dash-pot will take time to react. Thus the initial strain is Eo/ )0(σε= . Using this as the initial c ondition, an integration of 7.3.6 (with a zero stress-rate2) leads to 2 there is a jump in stress from zero to oσ when the load is applied, implying an infinite stress-rate σ&. One is not really interested in this jump here because the corresponding jump in strain can be predicted from the physical response of the spring. One is more interested in what happens just "after" the load is applied. In that sense, when one speaks of initia l strains and stress-rates, one means their values at +0, just after 0=t ; the stress-rate is zero from +0 on. To be more precise, one ca n deal with the sudden jump in stress by integrating the constitutive equation across the point 0=t as follows: [] [ ] ) ( ) ( ) ( ) ( )( )/()( )( )( )/( τετετστσσηε σ ση τ ττ ττ ττ τ Δ−−Δ+=Δ−−Δ++ ∫→∫= ∫+ ∫ Δ+ Δ−Δ+ Δ−Δ+ Δ−Δ+ Δ− E dtt Edtt E dtt dtt E & & In the limit as 0→Δτ , the integral tends to zero ( σ is finite), the values of stress and strain at −0, i . e . i n the limit as 0→Δτ from the left, are zero. All that rema ins are the values to the right, giving )(0 )0(+ +=ε σ E , as expected. One can deal with this sudden behaviour more easily using integral formulations or with the La place Transform (see §7.4, §7.5) σE σ 1εη 2ε Section 7.3 Solid Mechanics Part I Kelly 245⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+=→+=→= Et tCt t oo o 1 1)()( ησεησεησε& (7.3.7) The creep-response can again be expressed in terms of a creep compliance function: EttJ tJ to1)( where)( )( += =ησε (7.3.8) When the load is removed, the spring again reacts immediately, but the dash-pot has no tendency to recover. Hence there is an immediate elastic recovery Eo/σ , with the creep strain due to the dash-pot remaining. The full creep and recovery response is shown in Fig. 7.3.5. The Maxwell model predicts creep, but not of the ever-decreasing strain-rate type, there is no anelastic recovery, but there is the elastic response and permanent strain. Figure 7.3.5: Creep-Recovery Response of the Maxwell Model Stress Relaxation In the stress relaxation test, the materi al is subjected to a constant strain 0ε at 0=t . The Maxwell model then leads to { ▲Problem 1} Et EetE tE tRtt oRηεσ = = =−, )( where)( )(/ (7.3.9) Analogous to the creep function J for the creep test, )(tE is called the relaxation modulus function. The parameter Rt is called the relaxation time of the material and is a measure of the time taken for the stress to relax; the shorte r the relaxation time, the more rapid the stress relaxation. ε t stress applied stress removedE0σE0σ Section 7.3 Solid Mechanics Part I Kelly 2467.3.3 The Kelvin (Voigt) Model Consider next the other two-element model, the Kelvin (or Voigt ) model , which consists of a spring and dash-pot in parallel, Fig. 7.3. 6. It is assumed there is no bending in this type of parallel arrangement, so that the st rain experienced by the spring is the same as that experienced by the dash-pot. This time, 2 1 2 1 ,1,1σσσσηεσε += = = & E (7.3.10) where 1σ is the stress in the spring and 2σ is the dash-pot stress. Eliminating 2 1,σσ leaves the constitutive law εηεσ &+=E Kelvin (Voigt) Model (7.3.11) Figure 7.3.6: the Kelv in (Voigt) Model Creep-Recovery Response If a load 0σ is applied suddenly to the Kelvin mode l, the spring will want to stretch, but is held back by the dash-pot, which cannot re act immediately. Si nce the spring does not change length, the stress is initially taken up by the dash-pot. The creep curve thus starts with an initial slope ησ/o. Some strain then occurs and so some of the stress is transferred from the dash-pot to the spring. The slope of the creep curve is now ησ/2, where 2σ is the stress in the dash- pot, with 2σ ever-decreasing. In the limit when 02=σ , the spring takes all the stress and thus the maximum strain is Eo/σ . Solving the first order non-homogeneous diffe rential equation 7.3.11 with the initial condition 0 )0(=ε gives ()t E oeEt)/(1 )(η σε−−= (7.3.12) σE σ η 1σ 2σ Section 7.3 Solid Mechanics Part I Kelly 247which agrees with the above physical reas oning; the creep comp liance function is now ()Et eEtJRttRη= −=−, 11)(/ (7.3.13) The parameter Rt, in contrast to the relaxation time of the Maxwell model, is here called the retardation time of the material and is a measure of the time taken for the creep strain to accumulate; the shorter the retarda tion time, the more rapi d the creep straining. When the Kelvin model is unloaded, the spring will want to contract but again the dash pot will hold it back. The spring will however eventually pull the dash-pot back to its original zero position given tim e and full recovery occurs. Suppose the material is unloaded at time τ=t . The constitutive law, with zero stress, reduces to εηε&+=E0 . Solving leads to t ECet)/()(ηε−= (7.3.14) where C is a constant of integration. The t here is measured from the point where "zero load" begins. If one wants to m easure time from the onset of load, t must be replaced with τ−t . The strain at τ=t is ()τηστε)/(1)/()(E o e E−− = . Using this as the initial condition, one finds that ()τσετη η>− =−t e eEtE t E o,1 )()/( )/( (7.3.15) The creep and recovery response is shown in Fig. 7.3.7. Th ere is a transient-type creep and anelastic recovery, but no inst antaneous or permanent strain. Figure 7.3.7: Creep-Recovery Response of the Kelvin (Voigt) Model Stress Relaxation Consider next a stress-rela xation test. Setting the st rain to be a constant 0ε, the constitutive law 7.3.11 reduces to 0εσE= . Thus the stress is taken up by the spring and is constant, so there is in fact no stress re laxation over time. Actually, in order that the ε tτ stress applied stress removed Section 7.3 Solid Mechanics Part I Kelly 248Kelvin model undergoes an instantaneous strain of 0ε, an infinite stress needs to be applied, since the dash-pot will not respond instantaneously to a finite stress3. 7.3.4 Three – Element Models The Maxwell and Kelvin models are the simple st, two-element, models. More realistic material responses can be modelled using more elements. The four possible three- element models are shown in Fig. 7.3.8 below. The models of Fig. 7.3.8a-b are referred to as “solids” since they reac t instantaneously as elastic materials and recover completely upon unloading. The models of Figs. 7.3.8c-d are referred to as “fluids” since they involve dashpots at the in itial loading phase and do not recover upon unloading. Figure 7.3.8: Three-element Models: (a) Standard Solid I, (b) Standard Solid II, (c) Standard Fluid I, (d) Standard Fluid II 3 the stress required is )( )0(0tδηεσ= , where )(tδ is the Dirac delta functio n (this can be determined using the integral representations of §7.4, §7.5 )2(E )1(E σ σ η(a) )2(E)1(E σ σ η(b) Eσ σ 2η 1η (c) E 1η σ σ 2η(d) Section 7.3 Solid Mechanics Part I Kelly 249The differential constitutive relations for the Maxwell and Kelvin models were not difficult to derive. However, even with three elements, deriving them can be a difficult task. This is because one needs to eliminat e variables from a set of equations, one or more of which is a differential equation (for ex ample, see 7.3.5). The task is more easily accomplished using integral formulations and the Laplace transform, which are discussed in §7.4-§7.5. Only results are given here: th e constitutive relations for th e four models shown in Fig. 7.3.8 are () () εηηεησηησεηηεηησησεηεσησεηε σησ &&&&&&& && && & E EE EEE EEEE EE E EEE E E 21 12 121 2 1222 1 1 22 11 2 121 2 1 (d)(c)(b)(a) +=++++=+++=++++=++ (7.3.16) The response of these models can be determin ed by specifying stress (strain) and solving the differential equations 7.3.16 for strain (stress). 7.3.5 The Creep Compliance and the Relaxation Modulus The creep compliance function and the relaxa tion modulus have been mentioned in the context of the two-element models discussed ab ove. More generally, they are defined as follows: the creep compliance is the strain due to unit stress: 1 when )( )( ),( )( = = =o o tJt tJ t σ ε σε Creep Compliance (7.3.17) The relaxation modulus is the stress due to unit strain: 1 when )( )( ),( )( = = =o o tEt tE t ε σ εσ Relaxation Modulus (7.3.18) Whereas the creep function describes the res ponse of a material to a creep test, the relaxation modulus describes the re sponse to a stress-relaxation test. 7.3.6 Generalized Models More complex models can be constructed by using more and more elements. A complex viscoelastic rheological model will usually be of the form of the generalized Maxwell model or the generalized Kelvin chain , shown in Fig. 7.3.9. The generalized Maxwell model consists of N different Maxwell units in parallel, each unit with different parameter values. The absence of the isolated spring would ensure fluid-type behaviour, whereas the absence of the isolated dash-pot would ensure an instantaneous response. Section 7.3 Solid Mechanics Part I Kelly 250The generalised Kelvin chain consists of a ch ain of Kelvin units and again the isolated spring may be omitted if a fluid-type response is required. In general, the more elements one has, the more accurate a model will be in describing the response of real materials. That said, the more complex the model, the more material parameters there are which need to be eval uated by experiment – th e determination of a large number of material parameters might be a difficult, if not an impossible, task. It is evident that, in general, a linear viscoela stic constitutive equati on will be of the form L &&&&& L &&&&&& +++++=+++++)( 4 3 2 1)( 4 3 2 1IV oIV o q q q q q p p p p p εεεεε σσσσσ (7.3.19) The more elements (springs/dashpots) one us es, the higher the orde r of the differential equation. Eqn. 7.3.19 is sometimes written in the short-hand notation εσQ P= (7.3.20) where P and Q are the linear differential operators ii n ii ii n iitq tp ∂∂= ∂∂= ∑ ∑ = = 0 0, Q P (7.3.21) A viscoelastic model can be created by si mply entering values for the coefficients ip, iq, in 7.3.19, without referring to any particular rh eological spring – das hpot arrangement. In that sense, springs and dashpots are not n ecessary for a model, all one needs is a differential equation of the form 7.3.19. Ho wever, the use of springs and dashpots is helpful as it gives one a physical feel for th e way a material might respond, rather than simply using an abstract mathema tical expression such as 7.3.19. Section 7.3 Solid Mechanics Part I Kelly 251 Figure 7.3.9: Generalised Viscoelastic Models 7.3.7 Retardation and Relaxation Spectra Generalised models can contain many parame ters and will exhibit a whole array of relaxation and retardation times. For example, consider two Kelvin units in series, as in the generalised Kelvin chain; the first unit has properties 1 1,ηE and the second unit has properties 2 2,ηE . Using the methods discussed in §7.4-§7.5, it can be shown that he constitutive equation is εηηεηηε σηησ && & & 2 121 2 112 21 2 121 2 12 1 E E E EE E E EEE E E ++++++=+++ (7.3.22) Consider the case of specified st ress, so that this is a second order differential equation in )(tε. The homogeneous solution is { ▲Problem 3} 2 1/ /)(R R tt tt h Be Aet− −+=ε (7.3.23) where 2 22 1 11/ ,/ E tE tR R η η = = are the eigenvalues of 7.3. 22. For a constant load 0σ, the full solution is { ▲Problem 3} σ E σ η 1 E NE Nη 1 η Generalized Kelvin Chain ση σ E1 E 1η NE NηGeneralized Maxwell Model Section 7.3 Solid Mechanics Part I Kelly 252()()⎥⎦⎤ ⎢⎣⎡−+− =− −2 1/ 2/ 10 1111)(R R tt tteEeEtσε (7.3.24) Thus, whereas the single Kelvin unit has a single retardation time, Eqn. 7.3.13, this model has two retardation times, which are the eigenvalues of the differential constitutive equation. The term inside the square brack ets is evidently the creep compliance of the model. Note that, for constant strain, the model predic ts a static response with no stress relaxation (as in the single Kelvin model). In a similar way, for N units, it can be shown that the response of the generalised Kelvin chain to a constant load 0σ is, neglecting the effect of the free spring/dashpot, of the form () ii i RN itt i Et eEti Rησε = − =∑ =−, 11)( 1/ 0 (7.3.25) where i iEη, are the spring stiffness and das hpot viscosity of Kelvin element i, N iK1= , Fig. 7.3.9. The response of real ma terials can be modelled by allowing for a number of different retardation times of different orders of magnitude, e.g. { }K K ,10,10,1,10,2 1 1−=i Rt . If one considers many elements with large s tiffnesses, Eqn. 7.3.25 can be expressed as ()()()i R ii RN itt i R t t e t ti R ηφ φσε1, 1 )( 1/ 0 =Δ −Δ=∑ =− (7.3.26) In the limit as ∞→N , letting ()R Rdtdtd d /φφ= one has ()()Rtt R dt e t tR∫∞ −− = 0/ 0 1 )(ϕσε (7.3.27) where ()R R dtd t /φϕ= . The representation 7.3.27 allows for a continuous retardation time, in contrast to the discrete ti mes of the model 7.3.25. The function ()Rtϕ is called the retardation spectrum of the model. Different responses can be modelled by simply choosing different forms for the retardation spectrum. A similar analysis can be carried out for the Generalise Maxwell model. For two Maxwell elements in parallel, the constitutive equation can be shown to be () εηη εηησηησηησ && & && &21 212 1 2 1 2121 2112 21 EEE E EE EEE E +++=+++ (7.3.28) Consider the case of specified st rain, so that this is a second order differential equation in )(tσ . The homogeneous solution is, analogous to 7.3.23, { ▲Problem 4} Section 7.3 Solid Mechanics Part I Kelly 253 2 1/ /)(R R tt tt h Be Aet− −+=σ (7.3.29) where again 2 22 1 11/ ,/ E tE tR R η η = = , and are the eigenvalues of 7.3.28. For a constant strain 0ε, the full solution is { ▲Problem 4} [ ]2 1/ 2/ 1 0 )(R R tt tteE eE t− −+ =εσ (7.3.30) Thus, whereas the single Maxwell unit has a sing le relaxation time, Eqn. 7.3.9, this model has two relaxation times, which are the eigenvalues of the differential constitutive equation. The term inside the square bracket s is evidently the rela xation modulus of the model. By considering a model with an indefinite number of Maxwell units in parallel, each with vanishingly small elastic moduli iEΔ, one has the expression analogous to 7.3.27, ()Rtt R dt et tR∫∞ −= 0/ 0 )(ϑεσ (7.3.31) and ()Rtϑ is called the relaxation spectrum of the model. To complete this section, note that, for the tw o Maxwell units in parallel, a constant stress 0σ leads to the creep strain { ▲Problem 5} () 212 1 2 121 2 1/ 2 1/ 2 10 , 11)(EEE EtteteE EtRtt R ttR R+ +=⎥⎦⎤ ⎢⎣⎡ ++−+−+=− − ηηηη ηη ηησε (7.3.32) 7.3.8 Problems 1. Derive the Relaxation Modulus )(tE for the Maxwell material. 2. What are the values of the coefficients i iqp, in the general differential equation 7.3.19 for (a) the Maxwell model and the Kelvin model? (b) The three-element models 3. Consider two Kelvin units in series, as in the generalised Kelvin chain; the first unit has properties 1 1,ηE and the second unit has properties 2 2,ηE . The constitutive equation is given by Eqn. 7.3.22. (a) The homogeneous equation is of the form 0=++εεε C B A&&& . By considering the characteristic equation 02=++ C B Aλλ , show that the eigenvalues are 2 22 1 11/ ,/ E tE tR R η η = = and hence that the homogeneous solution is 7.3.23. Section 7.3 Solid Mechanics Part I Kelly 254(b) Consider now a constant load 0σ. Show that the particular solution is ()21 2 1 0 / )( EE E E t +=σε . (c) One initial condition of the problem is that 0 )0(=ε . The second condition results from the fact that only the dashpots react at time 0=t (equivalently, one can integrate the constitutive equation across 0=t as in the footnote in §7.3.2). Show that this condition leads to ()21 2 1 0 / )0( ηηηησε +=& . (d) Use the initial conditions to show that the constants in 7.3.23 are given by 2 0 1 0 / ,/ E BE A σ σ −= −= and hence that the comple te solution is given by 7.3.24. (e) Consider again the constitu tive equation 7.3.22. What values do the constants 2 2,ηE take so that it reduces to the si ngle Kelvin model, Eqn. 7.3.11. 4. Consider two Maxwell units in parallel, as in the generalised Maxwell model; the first unit has properties 1 1,ηE and the second unit has properties 2 2,ηE . The constitutive equation is given by Eqn. 7.3.28. (a) The homogeneous equation is of the form 0=++σσσ C B A&&& . By considering the characteristic equation 02=++ C B Aλλ , show that the eigenvalues are 2 22 1 11/ ,/ E tE tR R η η = = and hence that the homogeneous solution is 7.3.29. (b) Consider now a constant load 0ε. Show that the particular solution is zero. (c) One initial condition results from the fact that only the springs react at time 0=t , which leads to the condition ()2 1 0 )0( E E+=εσ . A second condition can be obtained by integrating the constitutive equation across 0=t as in the footnote in §7.3.2. Show that this leads to the condition ( )0 22 2 12 1 / / )0( εηη σ E E+−=+& . (d) Use the initial conditions to show that the constants in 7.3.29 are given by 02 01,ε ε EB EA = = and hence that the complete solution is given by 7.3.30. (e) Consider again the constitu tive equation 7.3.28. What values do the constants 2 2,ηE take so that it reduces to the si ngle Maxwell model, Eqn. 7.3.6. 5. Consider the two Maxwell units in parallel, as in Problem 4. (a) From the constitutive equation 7.3.28, the differential equation to be solved is of the form 0σεε=+&&&B A . By considering the characteristic equation 02=+λλ B A , show that the eigenvalues are () ()2 1 2121 2 1 2 1 ,0E EEE ++−==ηηηηλλ and hence that the homogeneous solution is RtteC Ct/ 2 1 )(−+=ε where 2/1λ−=Rt . (b) Consider now a constant stress 0σ. By using the conditi on that only the springs react at time 0=t , show that the particular solution is ()2 1 0/ηησ+ t . (c) One initial condition results from the fact that only the springs react at time 0=t , which leads to the condition ()2 1 0/ )0( E E+=σε . A second condition can be obtained by integrating the constitutive equation across 0=t as in the footnote in §7.3.2. Show that this leads to the condition ()[ ]2 1 0/ )0( E EtR+ −=+σε& . (d) Use the initial conditions to show that the complete solution is given by 7.3.32 Section 7.4 Solid Mechanics Part I Kelly 2557.4 The Hereditary Integral In the previous section, it was shown that th e constitutive relation for a linear viscoelastic material can be expressed in the form of a linear differential equation, Eqn. 7.3.19. Here it is shown that the stress-strai n relation can also be expressed in the form of an integral, called the hereditary integral . 7.4.1 An Example: the Maxwell Model Consider the differential equati on for the Maxwell model, Eqn. 7.3.6, dtdEE dtd εσησ=+ (7.4.1) The first order differential equation can be solved using the standard integrating factor method. This converts 7.4.1 into an integral equation. Thre e similar integral equations will be derived in what follows1. Hereditary Integral over [] t,∞− It is sometimes convenient to regard 7.4.1 as a differential equation over the time interval [] t,∞− , even though the time interv al of interest is really []t,0. This can make it easier to deal with sudden “jumps” in stress or strain at time 0=t . The initial condition on 7.4.1 is then ()0=∞−σ . (7.4.2) Using the integrating factor η/Ete , re-write 7.4.1 in the form ()dttdEe t edtdEt Et )()(/ / εση η= (7.4.3) Integrating both sides over [] tˆ,∞− gives () () dtdttdEe e et Et Et tEt∫ ∞−∞−= −ˆ / / ˆ/ )(εσ ση η η (7.4.4) or ()dtdttdEe tt ttE∫ ∞−−−=ˆ /ˆ )()ˆ(εση (7.4.5) 1 note that Eqn. 7.4.1 predicts that sudden changes in the strain-rate, ε&, will lead to sudden changes in the stress-rate, σ&, but the stress σ will remain continuous. The strain ε does not appear explicitly in 7.4.1; sudden changes in strain can be dealt with by (i) integr ating across the point where the jump occurs, or (ii) using step functions and the integral formulation (see later) Section 7.4 Solid Mechanics Part I Kelly 256 Changing the notation, () τττετ σ dddtE tt ∫ ∞−−=)()( (7.4.6) where )(tE, the relaxation modulus fo r the Maxwell model, is η/)(EtEetE−= (7.4.7) This is known as a hereditary integral; given the strain history over [] t,∞− , one can evaluate the stress at the current time. It is the same constitutive equation as Eqn. 7.4.1, only in a different form. Hereditary Integral over []t0, The hereditary integral can also be e xpressed in terms of an integral over []t,0. Let there be a sudden non-zero strain )0(ε at 0=t , with the strain po ssibly varying, but continuously, thereafter. The strain, which in Eqn. 7.4.6 is to be regarded as a single function over [] t,∞− with a jump at 0=t , is sketched in Fig. 7.4.1. Figure 7.4.1: Strain with a sudden jump to a non-zero strain at 0=t There are two ways to proceed. First, write the integral over three separate intervals: () () () ⎭⎬⎫ ⎩⎨⎧−+ −+ − =∫ ∫ ∫ ++ −− ∞−→τττετ τττετ τττετ σ ϑϑ ϑϑ ϑdddtE dddtE dddtE tt)( )( )(lim)( 0 (7.4.8) With 0)(=tε over []ϑ−∞−,, the first integral is zero. With a jump in strain only at 0=t , the integrand in the third integral remains finite. The second integral can be evaluated by considering the function ) (tf illustrated in Fig. 7.4.2, a straight line with slope ϑε 2/)0(. A s 0→ϑ , it approaches the actu al strain function ) (tε, which jumps to )0(ε at 0=t . Then ()ηϑ ηϑη ϑϑ ϑϑηϑετττετ/ / / 0 0 2)0(lim)() ( limE E Ete e e dddtE− + − →+ −→− = −∫ (7.4.9) )(tε 0t)0(ε Section 7.4 Solid Mechanics Part I Kelly 257 Figure 7.4.2: A function used to approxi mate the strain for a sudden jump Using the approximation x ex+≈1 for small x, the value of this integral is ηε/)0(EtEe−. Thus Eqn. 7.4.6 can be expressed as () τττετ ε σ dddtE tEtt ∫−+ = 0)()0()( )( (7.4.10) By “0” here in the lower limit of the integral, one means +0, just after any possible non- zero initial strain. In that sense, the strain ()tε in Eqn. 7.4.10 is to be regarded as a continuous function, i.e. with no jumps, over []t,0. Jumps in strain after 0=t can be dealt with in a similar manner. A second and more elegant way to arrive at Eqn. 7.4.10 is to re-express the above analysis in terms of the Heaviside step function ) (tH and the Dirac delta function ) (t δ (see the Appendix to this section fo r a discussion of these functions). The function sketched in Fig. 7.4.1 can be expressed as ) ()( ttHε where now ) (tε is to be regarded as a con tinuous function over []t,0 – the jump is now contained within the step function )(tH . Eqn. 7.4.6 now becomes () ( ) ()()ττττετ τττεττ σ dddHtE dddH tE tt t ∫ ∫ ∞− ∞−−+ −= )()()( (7.4.11) The first integral becomes the integral in 7.4.10. From the brief discussion in the Appendix to this section, th e second integral becomes ()()() ( ) ( ) )0( )( )( εττδτετ ττττετ tE d tE dddHtEt t = −= − ∫ ∫ ∞− ∞− (7.4.12) A Third Hereditary Integral Finally, the integral can also be expressed as a function of ) (tε, rather than its derivative. To achieve this, one can integrate 7.4.10 by parts: ()∫−−+ =t dtdtdEt Et 0) () ()()0( )( ττεττε σ (7.4.13) This can be expressed as )(tf ϑ−ϑ+)0(ε Section 7.4 Solid Mechanics Part I Kelly 258 ()∫−− =t d tR t Et 0) ( )()0( )( ττετ ε σ (7.4.14) where dttdE tR /)( )(−= . Note that integration by parts is only possible when there ar e no “jumps” in the functions under the integral sign and this is assumed for th e integrand in 7.4.10. If there are jumps, the integral can either be split into separate in tegrals as in 7.4.9, or the functions can be represented in terms of st ep functions, which automatically account for jumps. The formulae 7.4.6, 7.4.10 and 7.4.14 give the stress as functions of the strain. Similar formulae can be derived for the strain in te rms of the stress (see the Appendix to this Section). Relaxation Test To illustrate the use of the hereditary integral formulae, consider a relaxation test, where the strain history is given by ⎩⎨⎧ <=otherwise,0 ,0)(0εεtt (7.4.15) Expressing the strain history as ) ( )(0tH tεε= , Eqn. 7.4.6 gives () )( )( )(0 0 tE d tE tt εττδτεσ = −=∫ ∞− (7.4.16) From 7.4.10, with the derivative in the integrand zero, one has ) ( )0()( )(0tE tEt εε σ = = . Finally, from 7.4.14, with ()ηη/ 2/ )(Ete E tR−+= , one again has )( )(0/ 0 0/) (2 0 0 tE Ee d eEEtEtt tEε ετηεεση ητ= = −=− −−∫ (7.4.17) 7.4.2 Hereditary Integrals: General Formulation Although derived for the Maxwell mode, these fo rmulae Eqns. 7.4.6, 7.4.10, 7.4.14, are in fact quite general, for example they can be derived from the differential equation for the Kelvin model (see Appendix to this section). The hereditary integral were derived dire ctly from the Maxwell model differential equation so as to emphasis that they are one and the same constitutive equation. Here they are derived more generally from first principles. Section 7.4 Solid Mechanics Part I Kelly 259The strain due to a constant step load ) 0(σ applied at time 0=t is by definition )()0( )( tJ tσε= , where ) (tJ is the creep compliance function. The strain due to a second load, σΔ say, applied at some later time τ, is ) ( )( τσε −Δ= tJ t . The total strain due to both loads is2, Fig. 7.4.3, ) ( )()0( )( τσ σε −Δ+ = tJ tJ t (7.4.18) Figure 7.4.3: Superposition of loads Generalising to an indefinite number of applied loads of infinitesimal magnitude, idσ, one has ∑∞ =− + = 1) ( )()0( )( ii itJd tJ t τσ σε (7.4.19) In the limit, the summation becomes the integral στd tJ∫−) (, o r3 (see Fig. 7.4.4) ∫−+ =t dddtJ tJ t 0)() ( )()0( )( τττστ σε Hereditary Integral (for Strain) (7.4.20) 2 this is again an application of th e linear superposition princi ple, mentioned in §4.2.3; because the material is linear, the "effect" of a sum of "causes" is equal to the sum of the individual "effects" of each "cause" 3 this integral equation allows for a sudden non-zero stress at 0=t . Other jumps in stress at later times can be allowed for in a similar manner – one would split the integral into separate integrals at the point where the jump occurs ε t tττ ) (τσ−Δ tJσΔ )0(σ )()0( tJσ Section 7.4 Solid Mechanics Part I Kelly 260 Figure 7.4.4: Formation of th e hereditary integral One can also derive a corresponding heredita ry integral in terms of the relaxation modulus { ▲Problem 1}: ()∫−+ =t dddtE tE t 0) ( )()0( )( τττετ εσ Hereditary Integral (for Stress) (7.4.21) This is Eqn. 7.4.10, which was derive d specifically from the Maxwell model. The hereditary integrals only require a knowledge of the creep function (or relaxation function). One does not need to construct a rheological model (with springs/dashpots), to determine a creep function. For example, the creep function for a material may be determined from test-data from a creep test. The hereditary integral formulation is thus not restricted to particular combinations of springs and dash-pots. Example Consider the Maxwell model and the two load histories shown in Fig. 7.4.5. The maximum stress is the same in both, σˆ, but load (1) is applied more gradually. Figure 7.4.5: two stress histories Examine load (1) first. The stress history is )(tσ τττd+τdidσ t (1) t (2) σˆ Tσˆ T Section 7.4 Solid Mechanics Part I Kelly 261⎪⎪ ⎩⎪⎪ ⎨⎧ >< = TtTt tT t ,ˆ,ˆ )( σσ σ In the hereditary integral 7.4.20, the creep compliance function ) (tJ is given by 7.3.8, E ttJ /1/)(+=η , and the stress is zero at time zero, so 0 )0(=σ . The strain is then Tt<<0 : T dtd /ˆ /σσ= ⎥⎦⎤ ⎢⎣⎡+=⎥⎦⎤ ⎢⎣⎡+−= −+ = ∫ ∫Et t TdEt TdddtJ tJ tt t ηστητστττστ σε2ˆ 1 ˆ )() ( )()0( )(2 0 0 tT<: 0 /=dt dσ ⎥⎦⎤ ⎢⎣⎡+−=⎥⎦⎤ ⎢⎣⎡+−= −=−+ −+ = ∫ ∫∫ ∫ ETtdEt TdddtJdddtJ dddtJ tJ t Tt TT 12/ˆ1 ˆ )() ()() ()() ( )()0( )( 0 00 ηστητστττσττττστ τττστ σε τ For load history (2), ⎪⎩⎪⎨⎧ >< = TtTt t ,ˆ,0 )( σσ The strain is then 0 )(=tε for Tt<. The hereditary integral 7.4.20 allows for a jump at 0=t . For a jump from zero stress to a non-zero stress at Tt= it can be modified to ⎥⎦⎤ ⎢⎣⎡+−=− =ETtTtJT t1ˆ) ()( )(ησ σε which is less than the strain due to load (1 ). (Alternatively, one could use the Heaviside step function and let ) (ˆ)( TtH t−=σσ in 7.4.20, leading to the same result, () ) (ˆ ) ( ˆ)()0( )( 0TtJ dT tJ tJ tt−=−−+ =∫σττδτσ σε .) This example illustrates two points: (1) the material has a "memory". It remember s the previous loadi ng history, responding differently to different loading histories (2) the rate of loading is important in viscoelas tic materials. This result agrees with an observed phenomenon: the strain in viscoelast ic materials is larger for stresses which grow gradually to their fi nal value, rather than when applied more quickly4. 4 for the Maxwell model, if one applied the second load at time 2/Tt= , so that the total stress applied in (1) and (2) was the same, one would have obtained the same response after time T, but this is not the case in general Section 7.4 Solid Mechanics Part I Kelly 262 7.4.3 Non-linear Hereditary Integrals The linear viscoelastic models can be extende d into the non-linear range in a number of ways. For example, generalising ex pressions of the form 7.4.14, ()() ( ) ()∫−+=t dt f tR tf t 02 1 ) ( )( τετ εσ (7.4.22) where 2 1,ff are non-linear functions of the strain history. The relaxation function can also be assumed to be a functi on of strain as well as time: ()() ( ) ()∫−+=t dt f tK tf t 02 1 ), ( )( τεετ εσ (7.4.23) 7.4.4 Problems 1. Derive the hereditary integral 7.4.21, ()∫−+ =t dddtE tE t 0) ( )()0( )( τττετ εσ 2. Use the hereditary integral form of the cons titutive equation for a linear viscoelastic material, Eqn. 7.4.20, to evaluate the re sponse of a material with creep compliance function )1 ln()(+= t tJ to a load ) 1( )( Bt to+=σσ . Sketch )(tJ, which of course gives the strain response due to a unit load 1 )(=tσ . Sketch also the load ) (tσ and the calculated strain ) (tε. [note: [] t b b tb tb dx xbt−++++−+−−=+−∫)1 ln()1()1 ln()1 ( 1) (ln 0] 3. A creep test was carried out on a certain linear viscoelastic material and the data was fitted approximately by the function ()te t21ˆ)(−−=σε , where σˆ was the constant applied load. (i) Sketch this strain response over 3 0≤≤t (very roughly, with 1ˆ=σ ). (ii) Which of the following three rheological models could be used to model the material: (a) the full generalized Kelvin chain of Fig. 7.3.9 (b) the Kelvin chain minus the free spring (c) the generalized Maxwell model minu s the free spring and free dash-pot Give reasons for your choice (and reas ons for discounting the other two). (iii) For the rheological model you chose in part (ii), roughly sketch the response to a standard creep-recovery test (the resp onse during the loading phase has already been done in part (i)). Section 7.4 Solid Mechanics Part I Kelly 263(iv) Find the material’s response to a load 1 )(2+=ttσ . 4. Determine the strain response of the Kelv in model to a stress history which is triangular in time: ⎪⎪ ⎩⎪⎪ ⎨⎧ ∞<< =<< −=<< =< = = tt tt tttt ttt tt tt t t c cc 11 1 11 1 2 ,0)(2 ,)/( 2)(0 ,)/()(0 ,0)( )( σσσσσσσ σ 7.4.5 Appendix to §7.4 1. The Heaviside Step Function and the Dirac Delta Functions The Heaviside step function )(tH is defined through ⎪⎩⎪⎨⎧ >=< =− atatat atH ,1,2/1,0 ) ( (7.4.24) and is illustrated in Fig. 7.4.6a. The derivative of the Heaviside step function, dt dH/, can be evaluated by considering ) (atH− to be the limit of the function ) (tf shown in Fig. 7.4.6b as 0→ϑ . This derivative dtdf/ is shown in Fig. 7.4.6c and in the limit is ) ( lim) ( 0atdtdf dtatdH−==− →δ ϑ (7.4.25) where δ is the Dirac delta function defined through (the integr al here states that the “area” is unity, as illustrated in Fig. 7.4.6c) ⎩⎨⎧=∞=−otherwise0,) (atatδ , () 1=−∫+∞ ∞−dtatδ (7.4.26) Integrals involving delta func tions are evaluated as follows: consider the integral σc t1 2t1 Section 7.4 Solid Mechanics Part I Kelly 264() dtbttg∫+∞ ∞−−δ)( (7.4.27) The delta function here is zero, and hence th e integrand is zero, everywhere except at bt=. Thus the integral is () () () )( )( )( )( bg dtbt bg dtbtbg dtbttg =− =− =− ∫ ∫ ∫+∞ ∞−+∞ ∞−+∞ ∞−δ δ δ (7.4.28) Figure 7.4.6: The Heaviside Step Function and evaluation of its derivative 2. The Maxwell Model: Functions of the Stress In §7.4.1, the hereditary integrals for the Maxwell model were deri ved for the stress in terms of integrals of the strain. Here, they ar e derived for the strain in terms of integrals of the stress. Consider again the differential equation for the Maxwell model, Eqn. 7.3.6, σησε 1 1+=dtd E dtd (7.4.29) Direct integration gives ()()∫ ∞−+=t d tEt ττσησε1 1)( (7.4.30) ) (atH− −∞=←t 0=t1 (a) (b))(tf (c))(tf′ ϑ211 area= ϑ−a ϑ+a• at= →+∞=t ϑ−a ϑ+a 0=t 0=t1 Section 7.4 Solid Mechanics Part I Kelly 265 Integrating by parts leads to ()()∫ ∞−−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+=t dddtt Et τττσ ητσηε1)( (7.4.31) Bringing the first term inside the integral, ()∫ ∞−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−+=t ddd t Et τττσ ητε1)( (7.4.32) or ()()∫ ∞−−=t dddtJ t τττστ ε)( (7.4.33) where the creep function is η/ /1)( tE tJ += . If there is a jump in stress at 0=t , 7.4.33 can be expressed as an integral over []t,0 by evaluating the contribution of the ju mp to the integral in 7.4.33: () () () )0(11220lim2/ 1 20lim20 1lim 02 0 0 σηηϑϑσ ητττϑστϑσ ητ ϑϑ ϑϑϑ ϑϑ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+=⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+ =⎥⎦⎤ ⎢⎣⎡−+ =⎥⎦⎤ ⎢⎣⎡−+ →+ −→+ −→∫ t Et Et Edt E (7.4.34) leading to ()()∫−+ =t dddtJ tJt 0)0()( )( τττστ σ ε (7.4.35) Alternatively, one could also have simply let ) ()( )( ttH tσ σ→ in 7.4.33, again leading to the term ( )()() ()() 0σττδτστ tJ d tJt= −∫∞−. Finally, integrating by parts, one also has () ( ) ()∫−− =t d tS t Jt 0)0( )( ττστ σ ε (7.4.36) where dttdJ tS /)( )(−= . Section 7.4 Solid Mechanics Part I Kelly 2663. The Kelvin Model: Functions of the Stress Consider the differential equation fo r the Kelvin model, Eqn. 7.3.11, σηεηε 1=+E dtd (7.4.37) Using the integrating factor η/Ete , one has () )(1)(/ /t e t edtdEt Etσηεη η= (7.4.38) Integrating both sides over [] tˆ,∞− gives () () dtt e e et Et Et tEt∫ ∞−∞−= −ˆ / / ˆ/)(1σηε εη η η (7.4.39) or ()dtt e tt ttE∫ ∞−−−=ˆ /ˆ)(1)ˆ( σηεη (7.4.40) Changing the notation, ()()ττσηεητd e tt tE∫ ∞−−−=/ 1)( (7.4.41) An integration by parts leads to ()()()τττσσεητdddeEtEtt tE∫ ∞−−−−=/ 1 1)( (7.4.42) Finally, taking the free term inside the integral: ()()τττστ ε dddtJ tt ∫ ∞−−=)( (7.4.43) where ()E e tJEt/ 1)(/η−−= is the creep compliance function for the Kelvin model. The other versions of the hereditary integral,. Eqn. 7.4.10, 7.4.14 can be derived from this as before. Section 7.5 Solid Mechanics Part I Kelly 2677.5 Linear Viscoelasticity and the Laplace Transform The Laplace transform is very useful in c onstructing and analysing linear viscoelastic models. 7.5.1 The Laplace Transform The formula for the Laplace transform of the derivative of a function is 1: etc. ),0( )0( )()0( )( 2f sffs fLffs fL & &&& −−=−= (7.5.1) where s is the transform variable, the overbar denotes the Laplace transform of the function, and )0(f is the value of the function at time 0=t . The Laplace transform is defined in such a way that )0(f refers to −=0t , that is, just before time zero. Some other important Laplace transforms are summarised in Table 7.5.1, in which α is a constant. )(tf )(sf α s/α )(tH s/1 ) (τδ−t seτ− )(tδ& s teα− ) /(1 s+α ()αα/ 1te−− ) (/1 s s+α ()2/ 1 / α ααte t−−− ) (/12s s+α nt L,1,0 , /!1=+n snn Table 7.5.1: Laplace Transforms Another useful formula is the time-shifting formula: [ ] )( ) () ( sfe tH tfLsτττ−=−− (7.5.2) 7.5.2 Mechanical models revisited The Maxwell Model The Maxwell model is governed by the set of three equations 7.3.5: 1 this rule actually only works for functions whose derivatives are continuous, although the derivative of the function being transformed may be pi ecewise continuous. Disc ontinuities in the functio n or its derivatives introduce additional terms Section 7.5 Solid Mechanics Part I Kelly 268 2 1 2 1 ,1,1εεεσηεσε += = = & E (7.5.3) Taking Laplace transforms gives 2 1 2 1 ,1,1εεεσηεσε += = = sE (7.5.4) and it has been assumed that the strain 2ε is zero at −=0t . The three differential equations have been reduced to a set of three algebraic equati ons, which may now be solved to get εησησ s sE=+ (7.5.5) Transforming back then gives Eqn. 7.3.6: εησησ &&=+E (7.5.6) Now examine the response to a sudden load. When using the Laplace transform, the load is written as ) ( )( tH toσσ= , where ) (tH is the Heaviside step function (see the Appendix to the previous s ection). Then 7.5.6 reads εηδσησ &= + )( )( tEtHo o (7.5.7) Using the Laplace transform gives t tHEts sEsE so o o o oo ησ σεησσεεησησ+ =→ +=→=+ )( )(1 1 2 (7.5.8) which is the same result as before, Eqn. 7.3.7-8. Subsequent unloading, at time τ=t say, can be dealt with most convenientl y by superimposing another load ) ( )( τσσ −−= tH to onto the first. Putting this into the cons titutive equation and using the Laplace transform gives s o s oesEesτ τσ ησε− −− −=1 1 2 (7.5.9) Transforming back, again using the time-shifting rule, gives ) ( ) () ( )( τσττησε −−−−−= tHEtH t to o (7.5.10) Section 7.5 Solid Mechanics Part I Kelly 269Adding this to the strain due to the fi rst load then gives the expected result ⎪⎪ ⎩⎪⎪ ⎨⎧ <<< + = tt tEt oo o ττηστησσ ε ,0, )( (7.5.11) The Kelvin Model Taking Laplace transforms of the three equations for the Kelvin model, Eqns. 7.3.10, gives εηεσ s E+= , which yields 7.3.11, εηεσ &+=E . The response to a load )( )( tH toσσ= is ()()t E o o o eEts EsEtH)/(1 )(/1)(η σεηησεεηε σ−−=→+=→+=& (7.5.12) The response to another load of magnitude ) ( )( τσσ −−= tH to is () () ) ( 1 )(/) ( ) )(/(τσεηησεεηετσ τητ − −−=→+−=→+=−− −−− tH eEts EseE tH t E os o o& (7.5.13) The response to both loads now gives the complete creep and recovery response: () () ⎪ ⎩⎪ ⎨⎧ > −<< − = −− τστσ ε τη ηη t e eEt eEt E t E ot E o ,10 , 1 )( )/( )/()/( (7.5.14) To analyse the response to a sudde nly applied strain, substitute ) ( )( tH toεε= into the constitutive equation εηεσ &+=E to get ) ( )(0 0 t tHE δηεεσ + = , which shows that the relaxation modulus of the Kelvin model is )( )( t EtE ηδ+= (7.5.15) The Standard Linear Model Consider next the standard linear model, which consists of a spring in series with a Kelvin unit, Fig. 7.5.1 (see Fig. 7.3.8a). Upon load ing one expects the left-hand spring to stretch immediately. The dash pot then takes up the stress, transferring the load to the second spring as it slowly opens over time. Upon unl oading one expects the left-hand spring to contract immediately and for the right-hand sp ring to slowly contract, being held back by the dash-pot. The equations for this model are, from the figure, Section 7.5 Solid Mechanics Part I Kelly 270 2 222 1112 12 1 εησεσεσεεεσσσ &===+=+= EE (7.5.16) Figure 7.5.1: the sta ndard linear model One can eliminate the four unknowns from these five equations using the Laplace transform, giving () εηεσησ sE EE s E E1 21 2 1 +=++ (7.5.17) which transforms back to (in standard form) εηε σησ & & 2 11 2 121 2 1 E EE E EEE E E +++=++ (7.5.18) which is the same as Eqn. 7.3.16a. The response to a load ) ( )( tH t oσσ= is () ()()() ()() )( )(/1 /1)( )( 2 12 1 2 11 21 2 1 tJ ts Es EE E s E EE EEt tH E E oo oo o σεη ησ ησεεηε δησ σ =→++++=→+= + + & (7.5.19) and the creep compliance is 2E σ σ 1E 1ε2εσ η1σ 2σ ε Section 7.5 Solid Mechanics Part I Kelly 271() ()()t E t EeEEE EeEtJη η / 212 1 / 12 211)(− −−++ = (7.5.20) Note that 1/ )0( Eoσε= as expected. For recovery one can superimpose an opposite load onto the first, at time τ say: () () ()() ⎭⎬⎫ ⎩⎨⎧−⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛++ −−=→+⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛+−+−=→+=−−− +− − − − −− − ) (/ 212 1 ) (/ 12 12 1 2 11 21 2 1 2 211) ( )() /(1 ) /(1) ( ) ( τη τητ τ τσεη ησησεεηετδηστσ t E t E os os oo o eEEE EeEtH tes Es EE Ees EEE EE t tH E E & (7.5.21) The response after time τ is then () ()() 1 )(/ / 22 2− =− τη ησεE t E oe eEt (7.5.22) This is, as expected, simply the recovery response of the Kelvin unit The full response is as shown in Fig. 7.5.2. This seems to be fairly close now to the response of a real material as discussed in §7.1, although it does not allow for a permanent strain. Figure 7.5.2: Creep-recovery response of the standard linear model Non-constant Loading The response to a complex loading history can be evaluated by so lving the differential constitutive differential equation (or the corresponding hereditary integral). The differential equation can be most easily solved using Lapl ace transforms. Example Consider the example treated earlier using heredi tary integrals, at the end of §7.4.2. Load (1) of Fig. 7.4.5 can be thought of as consisting of the two loads (1a) tT)/ˆ( σσ= and (1b) ) () )(/ˆ( TtHTtT −−−=σσ applied at time Tt=. Load (2) consists of a constant load applied at time Tt=. For load (1a), σ ε t t τ Section 7.5 Solid Mechanics Part I Kelly 2722 2 321 1 ˆ)( 1ˆ1 1ˆ ˆ ˆtTtETt sTE sT TEtT η σε σ ησεεησησ+=→ +=→=+ & which gives the response for Tt<. For load (1b) one has [note: ()() () 0=−−τδτt tL ] [] ⎭⎬⎫ ⎩⎨⎧−−−−−=→− −=→− −=→−+−−−−−−= − −− − ) (1) (21) (ˆ)(1 ˆ11 ˆ1 ˆ1 1 ˆ) ( ) () (ˆ) () (ˆ 22 32 TtETtTTtHtseE seseTE seTsTtH Tt TtETtHTtT Ts TsTs Ts τ η σετσ ητσεσ ησεδτση σεη& The response after time T is then given by adding the two results: ⎟⎠⎞⎜⎝⎛−+=211 ˆ)( TtEt ησε 7.5.3 Relationship between Creep and Relaxation Taking the Laplace transform of the general constitutive equation 7.3.19, εσQ P= , leads to ( )( )ε σ L L +++++=+++++4 43 32 2 14 43 32 2 1 sqsq sqsq q sp sp spsp po o (7.5.23) which can also be written in the contracted form εσ )( )( sQ sP= (7.5.24) where P and Q are the polynomials in iiin ii sq sQ sp sP ∑ ∑ = == = 0 0)( , )( (7.5.25) The Laplace transforms of the creep compliance ( ))( )( sJ tJ→ and relaxation modulus () )( )( sE tE→ can be written in terms of these polynomials as follows. First, the strain due to a unit load ) (tH=σ is ) (tJ. Since s/1=σ , substitution into the above equation gives )()()(ssQsPsJ= (7.5.26) Section 7.5 Solid Mechanics Part I Kelly 273 Similarly, the stress due to a unit strain ) (tH=ε is ) (tE and so )()()(ssPsQsE= (7.5.27) It follows that 21)()(ssEsJ= (7.5.28) Thus, for a linear viscoelastic material, there is a unique and simple relationship between the creep and relaxation behaviour. 7.5.4 Problems 1. Check that the relation 7.5.28, 2/1)()( s sEsJ= , holds for the Kelvin model 2. (a) Derive the constitutive relation (in standard form) for the three-element model shown below using the Laplace transform (thi s is the Standard Fluid II of Fig. 7.3.8d and the constitutive relation is given by Eqn. 7.3.16d) (b) Derive the creep compliance ) (tJ by considering a suddenly applied load. E σ σ 2η 1η Section 7.6 Solid Mechanics Part I Kelly 2747.6 Oscillatory Stress, Dynamic Loading and Vibrations Creep and relaxation experiments do not provi de complete information concerning the mechanical behaviour of viscoelastic material s. These experiments usually provide test data in the time-range from 10 seconds to 10 ye ars. It is often of interest to know the response of materials to loads of very short duration. For example, duration of the impact of a steel ball on a viscoelastic block may be of the order of sec 105− 1. In order to be able to determine the response for such conditions , it is necessary to know the behaviour of a material at high rates of load ing (or short duration loading). The techniques and apparatus for investigati ng the response of a mate rial to very short term loading are different to those involved in longer-term testing. For very short time loading it is more convenient to use oscillatory than static lo ading, and in order to predict the behaviour of a viscoelastic material subj ected to an oscillatory load, one needs to formulate the theory based on osci llatory stresses and strains. 7.6.1 Oscillatory Stress Consider a dynamic load of the form ) cos( )( t toωσσ= (7.6.1) where oσ is the stress amplitude and ω is the angular frequency2. Assume that the resulting strain is of the form3 ) cos( )( δωεε − = t to (7.6.2) so that the strain is an oscillation at the sa me frequency as the stre ss but lags behind by a phase angle δ, Fig. 7.6.1. This angle is referred to as the loss angle of the material, for reasons which will become clear later. Expanding the strain trigonometric terms, t t t o o ωδεωδεε sin sin cos cos )( + = (7.6.3) The first term here is completely in phase with the input; the second term is completely out of phase with the inpu t. If the phase angle δ is zero, then the stress and strain are in phase (as happens with an ideal elastic material), whereas if 2/πδ= , the stress and strain are completely out of phase. 1 dynamic experiments usually provide data from about sec. 108− to about sec. 103 so there is a somewhat overlapping region where data can be obtained from both types of experiment 2 when an oscillatory force is first applied, transient vi brations result at the natural frequency of the material – these soon die out leaving the vibrations at the source frequency 3 if one substitutes 7.6.1 into the general constitutive equation 7.3.17, one sees that the strain and its derivatives contain sine and cosine terms, so that the strain must be of the general form ) sin( ) cos( t Bt A ω ω+ , where A and B are constants. For convenie nce, this can be written as ) cos( Dt C−ω where C and D are new constants Section 7.6 Solid Mechanics Part I Kelly 275 Figure 7.6.1: Oscillatory stress and strain The Complex Compliance Define δσεδσεsin , cos2 1 oo ooJ J = = (7.6.4) so that ( )t Jt J to ω ωσε sin cos )(2 1+ = (7.6.5) The quantities 1J and 2J are a measure of how in, or out of, phase the stress is with the strain. The former, 1J, is called the storage compliance and the latter, 2J, is called the loss compliance . They are usually written as the components of a complex compliance , *J: 2 1*iJJ J−= (7.6.6) If one has a stress input in th e form of a sine function, then () t Jt Jt tt tt t oo ooo ω ωσωδεωδεδωεεωσσ cos sincos sin sin cos) sin( )() sin( )( 2 1− =− =− == (7.6.7) and again the storage compliance is a measure of th e amount "in phase" and the loss compliance is a measure of the amount "out of phase". The Complex Modulus ()tωσcos0 ()δωε−t cos00σ 0ε ωδ Section 7.6 Solid Mechanics Part I Kelly 276One can also regard of the strain as the input and the stress as the outpu t. In that case one can write ( δ is again the phase angle by which the strain lags behind the stress) t tt tt t o ooo ωδσωδσδωσσωεε sin sin cos cos) cos( )() cos( )( − =+ == (7.6.8) This is in effect the same stress-strai n relationship as that used above, only the stress/strain are shifted along the t-axis. Define next the two new quantities δεσδεσsin , cos2 1 oo ooE E = = (7.6.9) so that ( )t Et E to ω ωεσ sin cos )(2 1− = (7.6.10) Again, these quantities are a measure of how much the response is in phase with the input. The former, 1E, is called the storage modulus and the latter, 2E, is called the loss modulus . As with the compliances, they are usually written as the components of a complex modulus4, *E: 2 1*iE E E+= (7.6.11) Again, if one has a sinusoidal strain as input, one can write () t Et Et tt tt t oo ooo ω ωεωδσωδσδωσσωεε cos sincos sin sin cos) sin( )() sin( )( 2 1+ =+ =+ == (7.6.12) It is apparent from the above that 1**=EJ (7.6.13) which is a much simpler relationship than th at between the creep compliance function and the relaxation modulus (which involve d Laplace transforms, Eqn. 7.5.28). 4 typical values for the storage and loss moduli for a polymer would be around MPa 101=E , MPa 1.02=E . The ratio of the amplitudes is called the dynamic modulus , o oE εσ/*= . Section 7.6 Solid Mechanics Part I Kelly 277Complex Formulation The above equations can be succinctly written using a complex formulation, using Euler's formula θθθsin cos i ei±=± (7.6.14) Thus, for stress input, () [] ti oti oti oti oti o eJeiJJe ie te t ωωωδωω σσδδεεεσσ *2 1) ( sin cos)()( =−=− ===− (7.6.15) The creep compliance function ) (tJ is the strain response to a unit load. In the same way, from 7.6.15, the complex compliance *J can be interpreted as the strain response to a sinusoidal stress input of unit magnitude. Similarly, for a strain input, one has () ti oti oti oti o eEe ie te t ωωδωω εδδσσσεε *) ( sin cos)()( =+ === + (7.6.16) and the term in brackets is, by definition, the complex modulus *E. The relationship between the complex compliance/modulus and the differential constitutive equation Putting ti oe tωσσ=)( and the resulting strain ()t i oe tδωεε−=)( into the general differential operator form of the constitutive equation 7.3.17, one has [ ] []ti o oti o o eJ iq iq iq qe ip ip ip p ωω σω ωωσω ωω * 3 32 2 13 32 2 1 )( )( )()( )( )( LL ++ ++=+ + ++ (7.6.17) This equation thus gives the relationshi p between the complex compliance and the constants i iqp, . A similar relationship can be easily found for the complex modulus: LL + + ++++ ++=3 32 2 13 32 2 1 * )( )( )()( )( )( ω ωωω ωω ip ip ip piq iq iq qE oo (7.6.18) Section 7.6 Solid Mechanics Part I Kelly 278Again one sees that 1**=EJ . From 7.6.17-18, the complex compliance a nd complex modulus are functions of the frequency ω, and thus, from the definitions 7.6.4, 7.6.6, 7.6.9, 7.6.11, so is the phase angle δ. Thus ω is the primary variable influencing the viscoelastic properties (whereas time t was used for this purpose in th e analysis of static loading). The relationship between the complex compliance/modulus and the creep compliance/ relaxation modulus It can be shown5 that the complex compliance ) (*ωJ and the complex modulus ) (*ωE are related to the creep compliance ) (tJ and relaxation modulus ) (tE through [] []ωω ωωωω isis tELi EtJLi J == == )( )()()()()( ** , (7.6.19) Here, the Laplace transform is firs t taken and then evaluates at ωis= 6. A Note on Frequency Frequencies below 0.1 Hz are asso ciated with seismic waves. Vibrations of structures and solid objects occur from about 0.1 Hz to 10 kH z depending on the size of the structure. Stress waves from 20 Hz to 20 kHz are pe rceived as sound - above 20 kHz is the ultrasonic range. Fr equencies above 1012 Hz correspond to molecular vibration and represent an upper limit for st ress waves in real solids. 7.6.2 Example: The Maxwell Model The constitutive equation for the Maxw ell model is given by Eqn. 7.3.6, εησησ &&=+E (7.6.20) Consider an oscillatory stress ) cos( toωσσ= . We thus have7 ⎭⎬⎫ ⎩⎨⎧+ =→⎭⎬⎫ ⎩⎨⎧+ −=→= − ∫∫∫ ) sin(1) cos(1)() cos(1) sin( ) sin( ) cos( t tEtdtt dttEd tEt oo o o ωωηω σεωηωωσεεηωωσηωσ & (7.6.21) 5 using Fourier transform theory for example 6 1J and 2J are also related to each other (as are 1E and 2E) by an even more complicated rule known as the Kramers-Kronig relation 7 the constant of integration is zero (assuming that the initial strain is that in the spring, Eo/σ ). Section 7.6 Solid Mechanics Part I Kelly 279Thus the complex compliance is ωη1 1 2 1*iEiJJ J −=−= (7.6.22) This result can be obtained more easily us ing the relationship between the complex compliance and the constitutive equation: the constitutive equation can be rewritten as ηηεεσσ ==== +=+1 0 1 1 0 1 ,0 , ,1 where, q qEp p q q p po o& & (7.6.23) From Eqn. 7.6.17, ωη ωηωη ωωωω 1 1 )())(/(1 )( )()( )( 2 2 12 2 1 *iE iiE iq iq qip ip pJ oo−=+=+ +++ ++=LL (7.6.24) Also, the complex modulus is related to the complex compliance through 7.6.13, * */1J E= , so that 2 22 2 22 * )( )()( EEiEEE+++=ωηωη ωηωη (7.6.25) For very low frequencies, 0→ω , t t→ωω/) sin( , and the response, as expected, reduces to that for a static load, ()η σε / /1 )( tE to+= . For very high frequencies, 0 /1→ω , and the response is () ) cos(/ )( t E toω σε= . Thus the strain is completely in-phase with the load, but the dash-pot is not moving – it has no time to respond at such high frequencies - the spring/ dash-pot model is reacting like an isolated spring, that is, like a solid , with no fluid behaviour. 7.6.3 Energy Dissipation Because the equations 7.6.12 ) sin( )( ), sin( )( δωσσωεε + = = t t t to o (7.6.26) are the parametric equations for an ellipse, that is, they trace out an ellipse for values of t, the stress-strain curve for an oscillatory stress is an elliptic hysteres is loop, Fig. 7.6.2. The work done in stressing a materi al (per unit volume) is given by ∫=εσd W (7.6.27) The energy lost WΔ through internal friction and heat is given by the area of the ellipse. Thus Section 7.6 Solid Mechanics Part I Kelly 280 dtdtdd WTt tTt tεσεσ∫∫+ + ==Δ1 11 1 (7.6.28) where 1t is some starting time and T is the period of oscillation, ωπ/2=T . Substituting in Eqns. 7.6.26 for strain and stress then gives [] Tt tooTt tooTt too ttdt tdtt t W +++ ⎥⎦⎤ ⎢⎣⎡++− =++ =+ =Δ ∫∫ 1 11 11 1 sin2) 2cos( 21sin) 2sin(21) cos() sin( δωδωεωσδδωεωσωδωεωσ (7.6.29) Figure7.6.2: Elliptic Stress-St rain Hysteresis Loop Taking 01=t then gives8 δεπσ sinoo W=Δ Energy Loss (7.6.30) When 0=δ , the energy dissipated is zero, as in an elastic material. It can also be seen that 22 0 22 0 J E W πσπε==Δ (7.6.31) and hence the names loss modulus and loss compliance. 8 the same result is obtained for ) sin( ), sin( δωεεωσσ − = = t to o or when the stress and strain are cosine functions εσ K,/2,0ωπ=t () K,/ωδπ−=t K,/3,/ωπωπ=t Section 7.6 Solid Mechanics Part I Kelly 281Damping Energy The energy stored after one complete cycle is zero since the material has returned to its original configuration. The maximum en ergy stored during any one cycle can be computed by integrating the increment of work εσd from zero up to a maximum stress, that is over one quarter the period T of one cycle. Thus, integrating from ωδ/1−=t (where 0=σ ) to ωπ2/1 2+=tt , Fig. 7.6.39 ⎥⎦⎤ ⎢⎣⎡+ = δπδεσ sin4 2cos oo W (7.6.32) The second term is 4 / sinδεπσoo , which is one quarter of the energy dissipated per cycle, and so can be considered to represent the dissipated energy. The remaining, first, term represents the area of the shaded triangle in Fig. 7.6.3 and can be considered to be the energy stored, 2 / cosδεσoo sW= (it reduces to the elastic solution 2 /oo Wεσ= when 0=δ ). The damping energy of a viscoelastic ma terial is defined as SWW/Δ , where SW is the maximum energy the system can store in a give n stress/strain amplitude. Thus (dividing WΔ by 4 so it is consistent with the integra tion over a quarter-cycle to obtain the stored energy) δπtan2=Δ sWW Damping Energy (7.6.33) Thus the damping ability of a linearly visc oelastic material is only dependent on the phase/loss angle δ. Figure7.6.3: Elliptic Stress-St rain Hysteresis Loop 9 or one could integrate from zero to maximum strain, over []ωπ2/,0 , giving the same result ε σ ω δ / − =t0=tω δ π / ) 2/(−=t ω δ π / ) (−=tω π 2 / = t oσoε δσsinoδεcoso δεsinoδσcoso Section 7.6 Solid Mechanics Part I Kelly 282The quantity δtan is known as the mechanical loss , or the loss tangent . It can be considered to be the fundamental measure of damping in a linear material (other measures, for example δπδ tan2, , etc., are often used)10. Typical values for a range of materials at various temperatures and frequencies are shown in Table 7.6.1. Material Temperature Frequency ( v) Loss Tangent ( δtan ) Sapphire 4.2 K 30 kHz 10105.2−× Sapphire rt 30 kHz 9105−× Silicon rt 20 kHz 8103−× Quartz rt 1 MHz 710−≈ Aluminium rt 20 kHz 510−< Cu-31%Zn rt 6 kHz 5109−× Steel rt 1 Hz 0005.0 Aluminium rt 1 Hz 001.0 Fe-0.6%V 33oC 0.95 Hz 0016.0 Basalt rt 0.001-0.5 Hz 0017.0 Granite rt 0.001-0.5 Hz 0031.0 Glass rt 1 Hz 0043.0 Wood rt 1.5-8 Hz 0083.0 Wood rt ≈ 1 Hz 02.0 Bone 37oC 1-100 Hz 01.0 Lead rt 1-15 Hz 029.0 PMMA rt 1 Hz 1.0 PMMA 135oC 0.5 Hz 7.1 Table 7.6.1: Loss Tangents of Common Materials 7.6.4 Impact Consider the impact of a viscoe lastic ball dropped from a height dh onto a rigid floor. During the impact, a proportion of the initial potential energy dmgh , which is now kinetic energy2 21mv, where v is the velocity at impact, is lo st and only some is stored. The stored energy is converted back to kinetic energy which drives th e ball up on the rebound, reaching a height d rh h<, with final potential energy rmgh . The ratio of the two heights is11 d ss dr dr W WW mghmgh hhf+==≡ (7.6.34) where sW is the energy stored and dW is the energy dissipated during the impact. 10 some investigators recommend that one uses the maximum storable energy when 0=δ , in which case the stored energy is 2/ooεσ and the damping measure would be 2/ sin / δπ=ΔSWW 11 the coefficient of restitution e is defined as the ratio of the velocities before and after impact, d rvve /= , so 2ef= . Section 7.6 Solid Mechanics Part I Kelly 283 The impact event can be approximated by a ha lf-cycle of the osci llatory stress-strain curve, Fig. 7.6.4. Integrating over []ωπ/,0 or [ ]ωδπωδ /) (,/− − , one has12 [] δπδεσ sin cos21+ =oo W (7.6.35) and so the “height lost” is given by δπtan 1 1 1 −=−≈+−= sd d sd WW W WWf (7.6.36) Figure7.6.4: Impact approximated as a half-c ycle of oscillatory stress and strain Note some other approximations made: (i) energy losses due to air resistance, fr iction and radiation of sound energy during impact have been neglected (ii) in a real impact, the stress and strain are both initially zero. In the current analysis, when one of these quantities is zero, the other is finite, and this will inevitably introduce some error13. 7.6.5 Damping of Vibrations The inertial force in many applications can be neglected. However, when dealing with vibrations, the product of acceleration times ma ss can be appreciable when compared to the other forces present. 12 although it might be more accurate to integrate over []ωδπ /) (,0− 13 as mentioned, there is a transient term involved in the oscillation which has been ignored, and which dies out over time, leaving the strain to lag behind the stress at a constant phase angle εσ oscillatory (half-cycle)impact recovery after impact Section 7.6 Solid Mechanics Part I Kelly 284Vibrational damping can be examined by looking at a simple oscillator with one degree of freedom, Fig. 7.6.5. A mass m is connected to a wall by a viscoelastic bar of length L and cross sectional area A. The motion of the system is described by the equations Dynamic equation: 0=+Fxm&& Kinematic relation: Lx/=ε Constitutive relation: (depends on model) Assuming an osci llatory motion, ti oexxω= , and using the first two of these, ti oti o ti o eAmLeAmL LxA emxω ω ω ωεωσ σ ω ⎥⎦⎤ ⎢⎣⎡= =→=+ −2 2 20 (7.6.37) The quantity in brackets is the complex modulus *E (see Eqn. 7.6.16). As an example, for the Maxwell model (see Eqn. 7.6.24) 1 * 1− ⎥⎦⎤ ⎢⎣⎡−=ηωi EE (7.6.38) and so LmA i E=−ωηω21, (7.6.39) which can be solved to get ⎪⎭⎪⎬⎫ ⎪⎩⎪⎨⎧ −±=22 4 2 η ηωE LmAEiE (7.6.40) Figure7.6.5: Vibration If m is small or E is large (and η/E is not too large) the root has a real part, v say, so that v Ei± = )2/(η ω (7.6.41) L x viscoelastic bar Section 7.6 Solid Mechanics Part I Kelly 285and one has the damped vibration ( ) () () ) sin( ) cos()2/(2 1)2/(2 12 1 vt B vt A exec ec exec ecxx t E oivt ivt t E oti ti o + =+ =+ = −− − ηηω ω (7.6.42) If, on the other hand, the mass is large or th e spring compliant, one gets a pure imaginary root, iv Ei± = )2/(η ω , so that ωi is real (and less than zero) and one has the aperiodic damping ( )tv E tv E o ec ecxx) 2/( 2) 2/( 1− ++ =η η (7.6.43) 7.6.6 Problems 1. Use the differential form of the constitutive equation for a linearly viscoelastic material to derive the complex compliance , the complex modulus , and the loss tangent for a Kelvin material. (put the first two in the form βαi+). Use your expression for the complex compliance to derive the strain response to a stress ) cos( toωσ , in terms of ηωσ ,,,, Eto , in the form ( )t Bt A to ωωσε sin cos )( + = What happens at very low frequencies? Section 7.7 Solid Mechanics Part I Kelly 2867.7 Temperature-dependent Viscoelastic Materials Many materials, for example polymeric mate rials, have a response which is strongly temperature-dependent. Temperature effect s can be incorporated into the theory discussed thus far in a simple way by allowi ng for the coefficients of the differential constitutive equations to be functions of temperature. Thus, Eqn. 7.3.19 can be expressed more generally as ()( ) () ()()( ) L&&& L&& & +++=+++ εθεθεθ σθσθσθ2 1 2 1 q q q p p po o (7.7.1) where θ denotes temperature. Equivalently, one can allow for the creep and relaxation functions to be functions of temperature in the hereditary integral formulation. Thus Eqns. 7.4.20-21 read ∫∫ −+ =−+ = tt dddtE tE tdddtJ tJ t 00 )(), ( ),()0( ),()(), ( ),()0( ),( τττεθτ θεθστττσθτ θσθε (7.7.2) 7.7.1 Example: The Maxwell Model Consider a Maxwell material whose dash-pot viscosity η is a function of temperature θ. The differential constitutive equation is then ()()dtd dtd Eεθησθησ = + (7.7.3) where E is the temperature-independ ent spring stiffness. This equation is a function of both temperature and time. With temperature a function of time, ()tθθ= , it is a linear differential equation with non-c onstant coefficients. For co nstant temperature, it has constant coefficients. Consider first the case of constant temper ature. The relaxation modulus and creep compliance functions can be evaluated by appl ying unit strain and unit stress. From the previous work, one has ()()() ()()θηθθηθ θθ t EtJEt eE tERttR +== =− 1,, ),(/ (7.7.4) Thus any given material has temperature- dependent relaxation and creep functions. Consider now the change of variable Section 7.7 Solid Mechanics Part I Kelly 287()θηξtA= (7.7.5) where A is any constant (which can be chosen arbitrarily for convenience – see later). This transforms Eqn. 7.7.3 into ()ξε ξσξσddAdd EA=+ (7.7.6) This is now an equation with dependence on only one variable, ξ. From this equation, one obtains relaxation and creep functions ()AEJEAt eE ERtR ξξξξ +== =− 1, )(/ (7.7.7) These equations generate master curves from which the different temperature-dependent curves 7.7.4 can be obtained. Example Data For example, consider a viscosity wh ich varies linearly over the range C 100 C 100o o<<− θ according to the relation () ⎥ ⎦⎤ ⎢⎣⎡ ⎟⎟ ⎠⎞ ⎜⎜ ⎝⎛−−= 1 1 00θθηθηηA (7.7.8) where 0η is a constant viscosity, 2.0=ηA and C20o 0=θ (a reference temperature at which ()0ηθη=. This function is plotted in Fig. 7.7.1 below. Figure 7.7.1: linear dependence of viscosity on temperature 00.511.52 -100 -50 50 100θ0/)(ηθη Section 7.7 Solid Mechanics Part I Kelly 288Also, let mE=/0η . The resulting relaxation and cr eep functions of Eqn. 7.7.4 are plotted in Fig. 7.7.2 below (for 5=m ). Figure 7.7.2: temperature-dependent function s; (a) relaxation modulus, (b) creep compliance Note the following, referring to Fig. 7.7.2: (i) for temperatures greater than the reference temperature o 020==θθ (see Eqn. 7.7.8), the viscosity is ()0ηθη<. This implies that, for 0θθ>, the relaxation times are shorter than for 0θθ= (see Eqn. 7.7.4a), Fig. 7.7.2a, and the slope of the creep curves is greater than for 0θθ= (see Eqn. 7.7.4b)., Fig. 7.7.2b. (ii) for temperatures smaller than the reference temperature, ()0ηθη>. Thus, for 0θθ<, the relaxation times are longer than for 0θθ= and the slope of the creep curves is smaller than for 0θθ=. 01020304050 10 20 30 40 5000.20.40.60.81 10 20 30 40 50),(1θtEE t100−=θ 60−=θ 20−=θ 20=θ 60=θ 100=θ t),(θtJE 100−=θ100=θ 60=θ 20=θ(b)(a) Section 7.7 Solid Mechanics Part I Kelly 289Now choose the constant A in Eqn. 7.7.5 to be equal to 0η. This ensures that t=ξ at the reference temperature 0θ (see 7.7.8). In other words, the master curves of Eqn. 7.7.7 and the functions 7.7.4 corresponding to 0θ coincide (with the t axis and ξ axis coincident). The master relaxation and creep curves of Eqn. 7.7.7 are now meE E//)(ξξ−= and () m JE / 1ξξ+= . These are plotted in Fig. 7.7.3 below (for 5=m ). Figure 7.7.3: master curves; (a) relaxat ion modulus, (b) creep compliance All the curves of Fig. 7.7. 2 collapse onto the master curv e of Fig. 7.7.3 as follows: (i) the curves corresponding to the reference temperature, o 020==θθ , in Figs. 7.7.2 lie on the master curves (with the t axis and ξ axis coincident) (ii) for a curve with 0θθ>, if the time axis of Fig. 7.7.2a,b is “stretched” (according to 7.7.5), the curve will come to lie along the 0θθ= curve (and hence on the master curve); for a curve with 0θθ<, if the time axis of Fig. 7.7.2a is “shrunk” (according 0246810 10 20 30 40 5000.20.40.60.81 10 20 30 40 50)(1ξEE ξ )(ξJE(b)(a) ξ Section 7.7 Solid Mechanics Part I Kelly 290to 7.7.5), the curve will come to lie along the 0θθ= curve (and hence on the master curve) 7.7.2 Thermorheologically Simple Materials The fact that the relaxation and creep curves of Fig. 7.7.2 collapsed onto the master curves of Fig. 7.7.3 relied on the change of va riable, Eqn. 7.7.5, reducing the time and temperature dependent constitutive relation 7.7.3 to an equation in one variable, ξ, only, Eqn. 7.7.6. This in turn depended critically on the form of the differ ential equation 7.7.3. For example, if the spring stiffness E in the Maxwell model is temperature-dependent, the collapsing of curves is not possible. Temperature-dependent viscoelastic material s for which this collapsing of curves is possible are called thermorheologically simple materials. In this context, the parameter ξ is called the reduced time . More generally, the transfor mation 7.7.5 is expressed in the form ()θξ θat= (7.7.9) and the function ) (θθa is called the shift factor function. The shift factor is chosen so that the relaxation and creep curves corresponding to the ch osen reference temperature 0θ coincide (as in the Maxwell model example above), i.e. so that 1 )(0=θθa . The relaxation and creep functions now transform as ),( ),( ),,( ),( 0 0 θξ θ θξ θ J tJ E tE → → (7.7.10) For temperatures below the reference temperature, 0θθ<, )(0θθa will be greater than 1, and the corresponding relaxation/creep curv es collapse onto the master curve by “shrinking” the time axis t, which looks like a “shifting” of the curve “to the left” onto the 0θθ= curve. On the other hand, for 0θθ>, 1 )(0<θθa , and the corresponding curves collapse by a “stretching” of the time axis, whic h looks like a “shifting” of the curves “to the right” onto the master curve. This is summarised in Fig. 7.7.4 below. The result of this is that materials at hi gh temperatures and high strain rates behave similarly to materials at low temp eratures and low strain rates. The method discussed can also be used when the temperature is time-dependent, for then the transformation can be expressed as ()()()∫=t adt 0τθτξ θ (7.7.11) so that Section 7.7 Solid Mechanics Part I Kelly 291 ()()t a dtd θξ θ1= (7.7.12) leading to the same reduced differential equation. Figure 7.7.4: Relaxation modulus, as a functi on of (a) time, (b) reduced time The above discussion has related to the diffe rential constitutive equation 7.7.1. The analysis can also be expressed in terms of he reditary integrals of the form 7.7.2. For example, the equivalent hereditary integral in terms of reduced time, corresponding to the reduced differential equation (see Eqn. 7. 7.6 for the Maxwell model equation) is () τττετξξσξ dddE∫ ∞−−=)()( (7.7.13) where ()ξE is as before (see Eqn. 7.7.7 for the Maxwell model expression). stretch ),(θtE t0θθ=0θθ< 0θθ> ),(θtE ()θξ θat=()1>θθa ()1=θθa ()1<θθashrink Section 7.7 Solid Mechanics Part I Kelly 292 A1Answers to Selected Problems: Chapter 2 2.1 2.2 1. N 6. 20910 N,2. 3759 = =AB BC F F 2. N100 3 N6.563/10≈ g 2.3 2. 1000+ 3. N600=F , F R RyC xC = = N, 1000 4. Nm25.6−=M A1Answers to Selected Problems: Chapter 3 3.1 1. (a) N8.0=F , (b) Nm 3021.0&=M , (c) Nm3005.0&=M 2. Nm 3005.0 ,0 &== M F 3. 3/375=AR , 3/345=BR 4. 2kPa 3.2 1. 31,31==c c y x 3.3 2. 0 ,/= =s N lSσ σ 4. zyσ is negative 5. yzσ (positive) 6. bottom left: xzσ (negative), top: zxσ (negative), bottom right: yzσ (positive) 7. bottom left: xxσ (positive), top: zzσ (positive), bottom right: yyσ (positive) 3.4 2. 933 .0 ,116.2 ,884.012 22 −≈′ ≈′ ≈′ σ σ σxx 3.5 1. )( 22)2( 22wσσ≠ 2. 0 )() ( )0,() (0 )( )0,( ,0 )0,( = +− +−= − −= − ∫ ∫∫ ∫ ∫ + −+ −+ −+ −+ − dxxpbxtdxx bxtdxxptdxx t dxx t a ab byya ab byyb bxy σσ σ 5. (a) 0 ,2 1 ==σσσxx , (b) 2 / ,2/xx xy xx yy xx σσσσσ −=′ =′=′ 6. (b) o32−≈θ , (c) 85.2 ,85.32 1 −= +=σ σ 7. 0 ,22 1 ==σασ , () ασ=max12 , the original planes are the planes of maximum shearing stress. 3.6 3. Engineering strain: 3 0008.0 ,002.0 , 00225.0 &−= = =xy yy xx ε ε ε Actual strain: 4 310 3166823.8 ,10 001386.2 , 00225.0− −× −= × = =xy yy xx ε ε ε A2 Errors: % 200.0 %,069.0 %,0 −→ −→ →xy yy xx ε ε ε 3.7 3.8 1. 0 ,02.02 1 = =ε ε , 45 degrees A1Answers to Selected Problems: Chapter 4 4.1 4.2 3. 073.10,927.16,0 4. (a) normal strains: E Eo o /) 1( ,/) 1( ,02ννσνσ − −− , normal stresses: 0 , ,0 0σνσ−− , (b) 2 /oσ 5. normal strains: 0 ,) 1/() 1)(21( ,0 Eo νννσ −+−− , normal stresses: ) 1/( , ), 1/(0 0 0 ννσσννσ −−−−− 10. 6101.3−× rads 4.3 4.4 4.5 1. 0.25 MPa 3. (a) 15MPa, 30MPa, (b) 15MPa, (c) 750N 4. 0.121, 0.029, 0.064 ( 310−× ), 0.015mm 4.6 1. kN 75.092x V−= , m kN 25.0 93x x M−= 2. m kN72 24 2;2 10 2;10 10 2;22 2 2 2−+−−+−−+−−= x x x x x x x M 3. mm3.152=y 4. MPa 107 5. MPa 43.0 MPa;78.4 − 6. ()()EI wL 384/ 54 max−=δ 7. 6 / 2/ )]; 3()[6/(3 2 2Pa xPa EIvxax P EIv +−=− −= 8. 12 / ;2/2wL M wLR −= = A1Answers to Selected Problems: Chapter 6 6.1 6.2 2. 5 1 3 2 10 463.8 MPa, 5360.0 MPa, 6969.0−×−= −= −= ε σ σ 3. () () ftt f ftf tp pννννσννννσ−+−=−+−=11,11 3 2 4. (a) 82 .4 ,75.36 ,25.486 2 1 −= = = σ σ σ (b) 5 6 2 1 10 4486.3 , 1623.0 , 0317.0−×−= = = ε ε ε (c) kPa9.69=G , 0904.0 Pa,20021 2 = =ν E (d) principal stresses ( 0=θ ): 0,35,50 principal strains 02.20=θ ): 0317.0 , 1623.02 1 = =p p ε ε (e) No. 5. o2.48≈θ . A1Answers to Selected Problems: Chapter 7 7.4 2.     ( ) ln( 1) ( 1)ln( 1)o tt B t t t     3. (ii) b (iii) no instantaneous elastic rec overy; has permanent deformation 293Index A Ageing 239 Anisotropic 100, 222 Atmospheric Pressure 58 B Beams 136 Boundary Conditions 55 Buckling 215 Bulk Modulus 109 C Cantilever 22, 80 Castigliano’s First Theorem 195 Castigliano’s Second Theorem 187 Centroid 31, 38, 119 Columns 215 Compatibility 115 Complementary Energy 194 Contact 30 Continuum 93 Couple 17 Creep 236 Creep Compliance 249 Crotti-Engesser Theorem 195 D Damping 281 Degrees of Freedom 117 Deflection Curve 136, 160 Dissipation 279 Dissipative 172 Dynamics 190 E Eigenfunction 217 Eigenvalue 217 Elastica 217 Elastomer 241 Energy Methods 178 Equilibrium 12, 21 Equilibrium Equations 13, 21 F Failure 99, 163 Failure Locu 164 First Moment of Area 38, 119 294Flexural Rigidity 155 Force Body 37 Equivalent force 18, 31 Line of action of 17 Form Factor 187 Free Surface 57 Free Vibration 176 G Generalised Viscoelastic Models 249 H Heaviside Step Function 263 Hereditary Integral 255 Homogeneous 100 Hysteresis 99 I Impact 191, 282 Incompressible 110 Internal Friction 177 Isotropic 100, 229 J K Kelvin Model 246 L Laplace Transform 267 Loss Modulus 276 M Maxwell Model 243 Mode 217 Mohr’s Circle 68 Moment 17 Moment of Inertia 148 N Neutral Surface 136 Newton’s Laws 9 Newton’s First Law 9 Newton’s Second Law 11 Newton’s Third Law 11 O Orthotropic 223 295 P Plane Strain 82, 107 Plane Stress 55, 61, 106 Plasticity 99, 235 Poisson’s Ratio 101 Polymer 240 Polar Moment of Area 123 Polygon Law 10 Potential Energy 171 Pressure Vessel 127 Principal Stress 41, 64 Principal Planes 65 Principle of Conservation of Mechanical Energy 175 Principle of Minimum Potential Energy 208 Principle of Work and Kinetic Energy 173 Principle of Virtual Work 201 Q Quasi-Static 177 R Radius of Gyration 217 Rayleigh-Ritz Method 209 Relaxation Modulus 249 Relaxation Spectrum 251 Representative Volume Element (RVE) 95 Retardation Time 247 Rotations 79 S Saint-Venant’s Principle 46 Slenderness Ratio 217 Statically Indeterminate 156 Stiffness 115, 191 Storage Modulus 276 Strain 74 Engineering Strain 76 Logarithmic Strain 98 Maximum Shear Strain 86 Sign Convention 77 Principal Strain 85 Tensorial Shear Strain 77 True Strain 98 Strain Energy 222 Strain Energy Density 184 Strain Transformation Formulae 85 Stress Bending Stress 145 296 Boundary Conditions 55 Components 43 Distribution 29, 31 Engineering Stress 97 Flexural Stress 145 Internal Stress 40 Invariants 64 Maximum Principal Stress 41 Maximum Shear Stress 41, 67 Nominal Stress 97 Shear Stress 33, 49 Tensor 51 Uniform Stress 29 Stress Relaxation 237 Stress Space 164 Stress – Strain Relations 103 Stress Transformation Equations 51 Strut 215 Supports 22 Surfaces 14 T Temperature 286 Thermorheological 290 Tensor 51 Torsion 121, 181 Traction 42 Transversely Isotropic 227 Tresca Theory of Failure 163 U V Vibrations 283 Virtual Work 200 Viscoelastic 233 Volumetric Strain 105 Von Mises Theory of Failure 163 W Warping 121 Weight 10, 37 Work 170 X Y Yield 97 Young’s Modulus 101 297 Z