1. CHAPTER 1: ELASTICITY Elasticity deals with elastic stresses and strains, their relationship and the external forces that cause them. An elastic strain is defined as strain that disappears instantaneously once the forces that cause it are removed. It is very essential to understand the micro and macromechanical problems. Examples for micromechanical problems: stress fields around dislocations, incompatibilities of stresses at the interface between grains, and dislocation interactions in work hardening. Examples for macromechanical problems: Stresses developed in drawing and rolling wire and the analysis of specimen-machine interactions in tensile for tensile strength.

2. CHAPTER 1: ELASTICITY This chapter is structured in such a way as to satisfy the needs of both the undergraduate and the graduate students. A graphical method for the solution of two-dimensional stress problems (the Mohr circle) is described. On the other hand, the stress and strain systems in tridimensional bodies can be better treated as tensors, with the indicial notation. Once this tensor approach is understood, the student will have acquired a very helpful visualization of stresses and strains a s tridimensional entities

3. CHAPTER 1: ELASTICITY 1)Specimen 2)Crosshead of the machine 3)Load Cell 4)Strain Gauge, extensometer, Assume that a certain amount of force is applied, there will be tendency to stretch the sample, breaking the internal bonds. The breaking tendency is opposed by internal reactions, stresses. The resistance is uniformly distributed over the normal section and represented by three modest arrows at A. The normal stress σ is defined as this resistance per unit area. Applying the equilibrium of forces equation from the mechanics of materials we will have: This is the internal resisting stress to the externally applied load and avoiding breaking of the specimen

4. CHAPTER 1: ELASTICITY Stress and Strain Stress, σ, is defined as the intensity of force at a point: σ=∂F/∂A as ∂A→0 If the state of stress is the same everywhere in a body, σ=F/A A normal stress (compressive or tensile) is one in which the force is normal to the area on which it acts. With a shear stress, the force is parallel to the area on which is acts. Two subscripts are required to define a stress. The first subscript denotes the normal to the plane on which the force acts and the second subscript identifies the direction of the force. For example, a tensile stress in the x- direction is denoted by σ xx, indicating that the force is in the x-direction and it acts on a plane normal to x. For a shear stress, σ xy, a force in the y- direction acts on a plane normal to x.

5. CHAPTER 1: ELASTICITY Because stresses involve both forces and areas, they are not vector quantities. Nine components of stress are needed to describe a state of stress at a point, as shown in below. The stress component σ yy =F y /A y describes the tensile stress in the y-direction. The stress component σ zy =F y /A z is the shear stress caused by a shear force in the y-direction acting on a plane normal to z. Repeated subscripts denote normal stresses (e.g:σ xx,σ yy …) whereas mixed subscripts denote shear stresses (e.g:σ xy,σ zx …). Tensor Notation of state of stress. The nine components of stress acting on an infinitesimal element. The normal stress components are σ xx, σ yy, and σ zz. The shear stress components are σ yz, σ zx, σ xy, σ zy, σ xz, and σ yx.

6. CHAPTER 1: ELASTICITY Except tensor notation is required, it is often simpler to use a single subscript for a normal stress and to denote a shear stress by τ; for example, σ x =σ xx and τ xy =σ xy Stress component expressed along one set of axes may be expressed along another set of axes. The body is subjected to a stress σ yy =F y /A y. It is possible to calculate the stress acting on a plane whose normal, y’, is at an angle θ to y. The normal force acting on the plane is F y’ =F y cosθ and the area normal to y’ is A y /cosθ, so σ y =σ y’y’ =F y’ /A y’ =(F y cosθ)/(A y /cosθ)=σ y cos 2 θ Similarly, the shear stress acting on the x’ direction τ y’x’ (=σ y’x’ ), is given by: τ y’x’ =σ y’x’ =F x’ /A y’ =(F y Sinθ)/(A y /cosθ)=σ y cosθSinθ Stresses acting on an area A’

7. CHAPTER 1: ELASTICITY In the stress convention, generally tensile stresses are positive and compressive forces are negative. σ ij is positive if either i and j are positive or both negative. On the other hand, the stress component is negative for a combination of i and j in which one is positive and the other is negative. As the applied force increases, so does the length of the specimen. For an increase dF the length l increases by dl. The nomalized increase in length is equal to: Or upon integration Where l 0 is the original length. This parameter is known as the longitudinal true strain. In many application, a simpler form of strain, commonly called engineering or nominal strain, is used. This type of strain is defined as: In materials that exhibit large amounts of elastic deformation (rubbers, soft biological tissues, etc.) it is customary to express the deformation by a parameter called ‘stretch’ or ‘stretch ratio’. It is usually expressed as λ Hence deformation starts at λ=1

8. CHAPTER 1: ELASTICITY When the strains are reasonably small the engineering and the true strains are approximately the same. Subscript t will be used for true strain and subscript e will be used for engineering strain. The engineering strain and true strain can be related as: In a likewise fashion the engineering stress can be expressed as: The engineering stress and the true stress can be related as following: During elastic deformation the change in cross-sectional area is less than 1% for most metals and ceramics, thus σ e =σ t. However, during plastic deformation the difference between true and engineering values become progressively larger.

9. CHAPTER 1: ELASTICITY The sign convention for strains is the same as that for stresses. Positive for tensile strains and negative for compressive strains. Stress-Strain curves in elastic regimen (a) for metals and ceramics (b) for rubber The solid lines describe the loading trajectory and the dashed lines describe the unloading. For perfectly elastic solid, the two kinds of lines should coincide if thermal effects are neglected. The curve of (a) is characteristics of metals and ceramics and the elastic regimen can be described by a straight line. The curve of (b) is characteristics of rubber and stress and strain are not proportional. However, the strain returns to zero once the load is removed. First the resistance to stretching increases slightly with extension. After considerable deformation the rubber band stiffens up and further deformation will eventually lead to rupture. The whole process is elastic except the failure.

10. CHAPTER 1: ELASTICITY A conceptual error often made is to assume that elastic behavior is always linear; the rubber example shows very clearly that there are notable exceptions. However, for metals, the stress and strain can be assumed to be proportional in the elastic regimen; these materials are known as Hookean solids. For polymers, viscoelastic effects are very important. Viscoelasticity results in different trajectories for loading and unloading, with formation of hysteresis loop. The area of the hysteresis loop is the energy lost per unit volume in the entire deformation cycle. Metals also exhibit some viscoelasticity but it is most often neglected. Viscoelasticity is attributed to time-dependent microscopic processes accompanying deformation.

11. CHAPTER 1: ELASTICITY As the tension goes, so does the stretch: Robert Hooke presented… E represents the Young’s modulus and is high for metals and ceramics. E mainly depends on composition, crystallographic structure and nature of the bonding of the elements. Heat and mechanic treatment have little effect on E as long as they do not affect the former parameters. Hence, annealed and cold rolled steel have the same Young’s modulus, there are, of course, small differences due to formation of cold rolled texture. E decreases slightly with increases in temperature. In monocrystals (single crystals) E shows different values for different crystallographic orientations. In polycrystalline aggregates that do not exhibit any texture, E is isotropic. It has the same value in all directions.

12. CHAPTER 1: ELASTICITY The values of E shown in tables are usually obtained by dynamic methods involving the propagation of elastic waves, not from stress- strain tests. Elastic wave is passed through a sample, the velocity of the longitudinal and shear waves, V l and V s, respectively are related to the elastic constants, by means of the mathematical expressions: ρ is the density, E is the Young’s Modulus and G is the shear modulus.

13. CHAPTER 1: ELASTICITY

15. Strain Energy (Deformation Energy) Density When work is done on a body, its dimensions change. The work done (W) is converted into a heat (Q) and an increase in internal energy (U) of the body. We can write as per the first law of thermodynamics. For most solids, the elastic work produces an insignificant amount of heat. Hence the work done on a body during deformation is converted into internal energy, which is stored in the deformed material and we call it strain energy or strain energy density when referring to the stored strain energy per unit volume. In elastic springs the energy is stored, while in a damping element the energy is dissipated as heat.

16. CHAPTER 1: ELASTICITY Consider an elemental cube under uniaxial tension σ 11. The work done is given by the product of force and the change in length. Convert the stress into force and strain into displacement. The work done is the area under the force vs. displacement curve. Where σ 11 is the tensile stress component in direction 1 and ε 11 is the corresponding tensile strain, σx 1, σx 2, and σx 3, are the lengths of the sides of the cube. The work done per unit volume is: Similar expressions for the work done can be obtained by other stress components. σ 31 and ε 31 are the shear stress and shear strain, respectively.

17. CHAPTER 1: ELASTICITY Using the principle of superposition, i.e., Combining the results for two or more stresses (or strains), we can write for the total work done per unit volume or the strain energy density as: In more, indicial compact form we can write: The units of strain energy are J/m 3 or N.m/m 3 or N/ m 2. The last one is the same as the units of stress. It should not cause any confusion if the reader will recall that the strain is a dimensionless quantity. Note that the strain energy is a scalar quantity, hence no indexes. For a linearly elastic solid under a uniaxial stress we can use the Hooke’s law to obtain an alternate expression for the strain energy density: One can extend the concept of elastic strain energy density to region of inelastic behavior by defining the strain energy density as the area under the stress-strain curve of a material. Sometimes, we take this area under the stress-strain curve as a measure of the toughness of a material

18. CHAPTER 1: ELASTICITY

19. Shear Stress and Strain The specimen is placed between a punch and a base having a cyclindirical orifice; the punch compresses the specimen. The internal resistance to the external forces now has the nature of a shear. The small cube in (b) was removed from the region being sheared between punch and base. It is distorted in such a way that the perpendicularity of the faces is lost. The shear stresses and strains are defined as: The area of the surface that undergoes shear is: A mechanical test commonly used to find the shear stresses and strains is torsion test. The relation between the torque and the shear stress is:

20. CHAPTER 1: ELASTICITY c is the radius of the cylinder and J=πc 4 /2 is the polar moment of inertia. For a hollow cylinder with b and c as a inner and outer radii we subtract the hollow part to obtain: For metals and ceramics and certain polymers (the Hookean solids), the proportionality between shear stress and shear strain is observed in the elastic regimen. In analogy with Young’s modulus, a transverse elasticity, called the rigidity or shear modulus is defined:

21. CHAPTER 1: ELASTICITY

25. Unit cube in a body subjected to tridimensional stress; only stresses on the three exposed faces if the cube are shown. Unit cube being extended in direction Ox 3 A body, upon being pulled in tension, tends to contract laterally. The stresses are defined in a tridimensional body and they have two indices. The first indicates the plane (or the normal to the plane) on which they are acting; the second indicates the direction in which they are pointing. These stresses are schematically shown acting on three faces of a unit cube in Figure a. The normal stresses have two identical subscripts: σ 11,σ 22, σ 33. The shear stresses have two different subscripts: σ 12,σ 13, σ 23.These subscripts refer to the reference system Ox 1,x 2,x 3. If this notation is used, both normal and shear stresses are designated by the same letter, lower case sigma. On the other hand, in more simplified cases where we are dealing with only one normal one shear stress component, σ and τ will be used, respectively; this notation will be maintained throughout the text. In Figure the stress σ 33 generates strains ε 11,ε 22, ε 33. Since the initial dimensions of the cube are equal to 1, the changes in length are equal to the strains. Poisson’s ratio is defined as the ratio between the lateral and the longitudinal strains. Both ε 11, ε 22, are negative (signifying a decrease in length), and ε 33 is positive. In order for Poisson’s ratio to be positive, the negative sign is used.

26. CHAPTER 1: ELASTICITY Hence, In an isotropic material, ε 11 is equal to ε 22. We can calculate value of ν for two extreme cases: (1) when the volume remains constant and (2) when there is no lateral contraction. When the volume is constant, the initial and final volumes, V 0 and V. respectively, are equal to Neglecting the cross products of the strains, because they are orders of magnitude smaller than the strains themselves, we have Since V=V 0 For the isotropic case, the two lateral contractions are the same (ε 11 = ε 22 ). Hence, Substituting equation into the previous equation, we arrive at: v=0.5

27. CHAPTER 1: ELASTICITY For the case in which there is no lateral contraction, v is equal to zero. Poisson’s ratio for metals is usually around 0.3. The values given in the table apply to the elastic regimen; in the plastic regimen, v increases to 0.5, since the volume remains constant during plastic deformation.

28. CHAPTER 1: ELASTICITY It is possible to establish the maximum and minimum for Poisson’s ratio. We know that G and E are positive. This is a consequence of the positiveness and definiteness of the strain energy function (a subject that we will not treat here-in simple words, the unloaded state of the body is the lowest energy state. In the equation below: We set, Thus: This leads to: The lower bound for Poisson’s ratio is obtained by deforming a body and assuming that its volume remains constant, as was done earlier in this section. Thus:

29. CHAPTER 1: ELASTICITY MORE COMPLEX STATE OF STRESS The generalized Hooke’s law (as the set of equations relating tridimensional stresses and strains is called) is derived next, for an isotropic solid. It is assumed that shear stresses can generate only shear strains. Thus, the longitudinal strains are produced exclusively by the normal stresses. σ 11 generates the following strain: v= -ε 22 / ε 11 = -ε 33 / ε 11 for stress σ 11, we also have: For σ 22 : For σ 33 : In this treatment, the shear stresses generates only shear strains

30. CHAPTER 1: ELASTICITY The second simplifying assumption is called the principle of superposition. The total strain in one direction is considered to be equal to the sum of the strains generated by the various stresses along that direction. Hence, the total ε 11 is the sum of ε 11 produced by σ 11, σ 22,σ 33. We obtain the generalized Hooke’s law: Applying these equations to a hydrostatic stress situation We can see that there are no distortions in the cube

31. CHAPTER 1: ELASTICITY The triaxial state of stress is difficult to treat in elasticity. Therefore, we try to assume a more simplified state of stress that resembles the tridimensional stress. This is often justified by the geometry of the body and by the loading configuration. In sheets and plates (where one dimension can be neglected with respect to the other two), the state of stress can be assumed to be bidimensional. This state of stress is also known as plane stress, because normal stresses (normal to the surface) are zero at the surface, as are shear stresses (parallel to the surface) at the surface. The opposite case, in which one of the dimensions is infinite with respect to the other two, is treated under the assumption of plane strain. If one dimension is infinite, strain in it is constrained; hence, one has two dimensions left. This state is called bidimensional or, more commonly plane strain. It also occurs when strain is constrained in one direction by some other means. A long dam is constrained. Yet another state of stress is pure shear, when there are no normal stresses.

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33. 7 - 33 The most general state of stress at a point may be represented by 6 components, Same state of stress is represented by a different set of components if axes are rotated. The first part of the chapter is concerned with how the components of stress are transformed under a rotation of the coordinate axes. The second part of the chapter is devoted to a similar analysis of the transformation of the components of strain. MOHR’S CIRCLE PRESENTATION

34. 7 - 34 Plane Stress - state of stress in which two faces of the cubic element are free of stress. For the illustrated example, the state of stress is defined by State of plane stress occurs in a thin plate subjected to forces acting in the midplane of the plate. State of plane stress also occurs on the free surface of a structural element or machine component, i.e., at any point of the surface not subjected to an external force.

35. 7 - 35 Transformation of Plane Stress Consider the conditions for equilibrium of a prismatic element with faces perpendicular to the x, y, and x’ axes. The equations may be rewritten to yield

36. 7 - 36 Principal Stresses The previous equations are combined to yield parametric equations for a circle, Principal stresses occur on the principal planes of stress with zero shearing stresses.

37. 7 - 37 Maximum Shearing Stress Maximum shearing stress occurs for

38. 7 - 38 Mohr’s Circle for Plane Stress With the physical significance of Mohr’s circle for plane stress established, it may be applied with simple geometric considerations. Critical values are estimated graphically or calculated. For a known state of plane stress plot the points X and Y and construct the circle centered at C. The principal stresses are obtained at A and B. The direction of rotation of Ox to Oa is the same as CX to CA.

39. 7 - 39 Mohr’s Circle for Plane Stress With Mohr’s circle uniquely defined, the state of stress at other axes orientations may be depicted. For the state of stress at an angle  with respect to the xy axes, construct a new diameter X’Y’ at an angle 2  with respect to XY. Normal and shear stresses are obtained from the coordinates X’Y’.

40. 7 - 40 Mohr’s Circle for Plane Stress Mohr’s circle for centric axial loading: Mohr’s circle for torsional loading:

41. 7 - 41 Example 7.02 For the state of plane stress shown, (a) construct Mohr’s circle, determine (b) the principal planes, (c) the principal stresses, (d) the maximum shearing stress and the corresponding normal stress. SOLUTION: Construction of Mohr’s circle

42. 7 - 42 Example 7.02 Principal planes and stresses

43. 7 - 43 Example 7.02 Maximum shear stress

44. 7 - 44 Sample Problem 7.2 For the state of stress shown, determine (a) the principal planes and the principal stresses, (b) the stress components exerted on the element obtained by rotating the given element counterclockwise through 30 degrees. SOLUTION: Construct Mohr’s circle

45. 7 - 45 Sample Problem 7.2 Principal planes and stresses

46. 7 - 46 Sample Problem 7.2 Stress components after rotation by 30 o Points X’ and Y’ on Mohr’s circle that correspond to stress components on the rotated element are obtained by rotating XY counterclockwise through

47. CHAPTER 1: ELASTICITY Pure Shear: Relationship between G and E There is a special case of bidimensional stress in which σ 22 =-σ 11. This state of stress is represented in the figure. It can be seen that σ 12 =0, implying that σ 11 and σ 22 are principal stresses and write σ 2 =-σ 1. In Mohr’s circle of the second figure the center coincides with the origin of the axes.We can see that a rotation of 90° (on the circle) leads to a state of stress in which the normal stresses are zero. This rotation is equivalent to a 45° rotation in the body (real space). The magnitude of the shear stress at this orientation is equal to the radius of the circle. Hence, the square shown in the last figure is deformed to a lozenge under the combined effect of the shear stresses. Such a state of stress is called pure shear.

48. CHAPTER 1: ELASTICITY It is possible, from this particular case, to obtain a relationship between G and E; furthermore, the relationship has a general nature. The strain ε 11 is, for this case, We have, for the shear stresses (using the normal, and not the Mohr, sign convention), But we also have, Eqn. 1 Eqn. 2 Eqn. 3 Substituting equations 2 and 3 into equation 1 yields It is possible, by means of geometrical consideration on the triangle ABC in Figure to show that Hence,

49. CHAPTER 1: ELASTICITY Anisotropic Effects Figure shows that a general stress system acting on a unit cube has nine components and is a symmetrical tensor. (The off-diagonal components are equal) We can therefore write: When the unit cube in the figure is rotated, the stress state at that point doesnot change; however, the components of stress change. The same applies to strains. A general state of strain is descirbed by:

50. CHAPTER 1: ELASTICITY We can also use a matrix notation for stresses and strains, replacing the indices by the following We now have stress and strain in general form as: It should be noted that: But: