Stress AnalysisPLASTICITY Dr. Maaz akhtar The theory of linear elasticity is useful for modelling materials which undergo small deformations and which return to their original configuration upon removal of load. Almost all real materials will undergo some permanent deformation, which remains after removal of load. With metals, significant permanent deformations will usually occur when the stress reaches some critical value, called the yield stress, a material property. Elastic deformations are termed reversible; the energy expended in deformation is stored as elastic strain energy and is completely recovered upon load removal. Permanent deformations involve the dissipation of energy; such processes are termed irreversible, in the sense that the original state can be achieved only by the expenditure of more energy. Classical Theory of Plasticity The classical theory of plasticity grew out of the study of metals in the late nineteenth century. It is concerned with materials which initially deform elastically, but which deform plastically upon reaching a yield stress. In metals and other crystalline materials the occurrence of plastic deformations at the micro-scale level is due to the motion of dislocations and the migration of grain boundaries on the micro-level. The deformation of microvoids and the development of micro-cracks is also an important cause of plastic deformations. Plasticity theory began with Tresca in 1864, when he undertook an experimental program into the extrusion of metals and published his famous yield criterion discussed later on. Further advances with yield criteria and plastic flow rules were made in the years which followed by Saint-Venant, Levy, Von Mises, Hencky and Prandtl etc. Classification of Plasticity Problems There are two broad groups of metal plasticity problem which are of interest to the engineer and analyst. 1. Small Plastic strains The first involves relatively small plastic strains, often of the same order as the elastic strains which occur. Analysis of problems involving small plastic strains allows one to design structures optimally, so that they will not fail when in service, but at the same time are not stronger than they really need to be. In this sense, plasticity is seen as a material failure. 2. Large Plastic Strains The second type of problem involves very large strains and deformations, so large that the elastic strains can be disregarded. These problems occur in the analysis of metals manufacturing and forming processes, which can involve extrusion, drawing, forging, rolling and so on. In these latter-type problems, a simplified model known as perfect plasticity is usually employed. The word “perfect” means that the material in this model does not strain-harden, that is the yield strength is used which is independent of the amount of plastic strain. The Bauschinger Effect If one takes a new sample and loads it in tension into the plastic range, and then unloads it and continues on into compression, one finds that the yield stress in compression is not the same as the yield strength in tension, as it would have been if the specimen had not first been loaded in tension. In fact the yield point in this case will be significantly less than the corresponding yield stress in tension. This reduction in yield stress is known as the Bauschinger effect. The effect is illustrated in Figure-1. The solid line depicts the response of a real material. The dotted lines are two extreme cases which are used in plasticity models; the first is the isotropic hardening model, in which the yield stress in tension and compression are maintained equal, the second being kinematic hardening, in which the total elastic range is maintained constant throughout the deformation. The presence of the Bauschinger effect complicates any plasticity theory. However, it is not an issue provided there are no reversals of stress in the problem under study. After large plastic deformation however. For example. It is widely used in analysing metal forming processes. the material will have become anisotropic: there will be distinct material directions and asymmetries. commonly used in theoretical analyses. assumptions can be made on the type of hardening and on whether elastic deformations are significant. rolling or drawing. In such cases the elastic strains can be neglected altogether as in the two models (c) and (d). depending on the material and circumstances. For example. In (b) work-hardening is neglected and the yield stress is constant after initial yield. in the design of steel structures. consider the hierarchy of models illustrated in Figure below. e. the other three may or may not be. In (a) both the elastic and plastic curves are assumed linear.g. The rigid/perfectly-plastic model (d) is the crudest of all – and hence in many ways the most useful. in metal working processes such as extrusion. Figure-1: Baushinger Effect Assumptions of Plasticity Theory For basic plasticity theories following assumptions are usually made: (1) Response is independent of rate effects (2) Material is incompressible in the plastic range (3) There is no Bauschinger effect (4) The material is isotropic (5) Yield stress is independent of hydrostatic pressure The first two of these will usually be very good approximations. Figure-2: Stress-strain curves for different processes . most metals can be regarded as isotropic. where up to 50% reduction ratios are common. In many areas of applications the strains involved are large. Such perfectly- plastic models are particularly appropriate for studying processes where the metal is worked at a high temperature – such as hot rolling – where work hardening is small. Together with these. for example in rolling. in many situations the plastic strain is small. the plastic strain is so large that elastic strain is negligible. The material is said to work harden (i. This model is called elastic. the material is rigid. Thus. Even though a metal is capable of arbitrarily large deformation. When plastic strain and elastic strain are comparable. At any instant of strain.e. For now we wish to include elasticity. related to the longitudinal plastic strain as In applications like metal forming. and identify the net strain entirely with plastic strain in rigid. . also known as strain hardening or cold working. perfectly plastic model. the plastic deformation of the metal can be constrained by elastic surroundings. For a metal under uniaxial stress. the two transverse plastic strains are equal. we may neglect elastic strain. For example. on the order of elastic strain. This strengthening occurs because of dislocation movements and dislocation generation within the crystal structure of the material. Figure-4: Stress-strain curve for rolling operation The stress-strain curve (Flow curve) in the region of uniform plastic deformation does not increase proportionally with strain. Empirical . is the strengthening of a metal by plastic deformation. E being the Young’s modulus.. Figure-3: Stress-strain curve for tangent modulus (K) Elastic-Perfectly Plastic deformation Behavior Plastic deformation does not change volume.Stress and strain are related through in the elastic region. Many non-brittle metals with a reasonably high melting point as well as several polymers can be strengthened in this fashion. When the stress is within the yield strength. the increment in stress is related to the increment in strain through . strain harden). we need to include both in the model. perfectly plastic model. The net strain is the sum of elastic strain and plastic strain: Strain Hardening and Flow Rule Work hardening. Dislocations on intersecting slip planes permit both elastic interactions and dislocation reactions to contribute to work hardening. The tangent modulus K is the slope of the stress-strain curve in the plastic region and in general change during a deformation. and the strain does not change. is Ramberg Osgood strain hardening coefficient. but the initial yield stress is constant.1-0. Although the viscoelastic materials can suffer irrecoverable deformation. is yield stress. The complete stress-strain behavior is strain rate dependent. Figure-6: Stress-strain behavior is strain rate dependent with constant initial yield . It has general form or where. is offset yield stress. E is Young’s modulus and is Viscoelastic and Viscoplastic Behavior of Solids Materials having rate dependent deformation are commonly known as viscoelastic material. and are different strength coefficients and is the strain hardening exponent (n=0 for perfectly plastic solids. Figure-5: Stress-strain curve showing increase in yield strength with higher strain rate Polymers and certain metallurgical alloys show softening behavior immediately after reaching the yield point. is Ramberg Osgood strength coefficient. In many forming processes the deformation rates are small enough to consider the material behavior to be independent of strain rate and to use an elastoplastic material model. is stress. At larger strains the softening is followed by hardening. In a tensile test the yield stress is seen to increase with higher strain rates. which is the characteristic property of plastic behaviour. Another flow equation known as Ramberg-Osgood law. is strain. they do not have any critical yield or threshold stress. works upto ultimate tensile strength is given by: where.5 for most metals). n=1 for perfectly elastic solids & n=0.relationships attributed to Ludik and Holloman can be used to describe the shape of plastic stress- strain curve. For high strain rates this assumption leads to faulty results. (1) Failure Theories for Ductile Material Many theories are present which gives failure criteria for ductile materials are discussed below: (a) Maximum Shear Stress Theory It is also known as Tresca criteria states that yielding begins when maximum shear stress at a point reaches the maximum shear stress at yield under uniaxial tension or compression. above equations become Above results when plotted on two dimensional principal plane. Permanent deformation in ductile material is observed when load reaches the yield point. if then Consider an element from uniaxial tension subjected to yield. Hence. using Mohr’s circle equation principal stresses will be equal to:. Maximum shear stress is largest.Failure Theories (Yield Criteria) All engineering materials are classified as ductile or brittle. hence. Generally a ductile material is one in which gross plastic deformation is greater than 5%. In brittle materials yield point is not clear (determined by offset method) and fracture usually occurs near to ultimate tensile strength (UTS). Both materials behave in different manner. . Therefore. Any further load will strain hardened the material. One single theory cannot be used to predict failure in both the materials. Maximum shear stress gives. for ductile material failure criteria is based on Yield Strength while for brittle material failure criteria is based on ultimate tensile strength. Ductile materials show large plastic deformation and observed necking prior to fracture but brittle materials fails suddenly without proper indication. while brittle material having plastic deformation less than 5%. For multiaxial state of stress shear stress is obtained by: . it gives a hexagon as shown below: . . hence they have different failure criteria. yielding under multiaxial state of stress occurs for any one of the following conditions: For two dimensional system (let . If the principal stresses are unordered. . hence strain energy for hydrostatic stat of stress become: Hence strain energy for deviatoric state of stress can be determine by . A Figure-7: Tresca Yield Locus At point ‘A’ which makes -45ᴼ with x-axis the shear stress is found to be . then total strain energy is given by: From Hooke’s law. (b) Maximum Distortion Energy Theory This theory is also known as Von-Mises theory or Octahedral shear stress theory. . It states that yielding occurs when the distortional strain energy at a point equals to the distortional strain energy at yield under uniaxial tension or compression. Consider are principal stresses and are principal strains. Total stress is the sum of hydrostatic and deviatoric states of stress as shown in figure-8. Figure-8: Total strain energy as su of hydrostatic and deviatoric state of stress Energy stated with hydrostatic state of stress gives: Also. stress-strain relationships are given by: . Comparing deviatoric relations. hence . hence. Comparison of Tresca and Von-Mises failure criteria tells that Tresca criteria is more conservative and gave 15% less yield surface area (figure-10). we get For plane stress condition above relation simplified to Above relation is an equation of ellipse. . A Figure-9: Von-Mises Yield Locus At point ‘A’ and . which gives . .Consider an element from uniaxial tension subjected to yield. which gives for plane stress condition an elliptical shape yield surface as shown in figure-9. Assume a special case where Poisson’s ratio is zero above equation represents the circle equation and yield surface will be a circular region. Failure criteria for brittle materials are discussed below. Total strain energy is given by: For a uniaxial state of stress at yield principal stresses can be given by: . According to this theory failure occurs when For plane stress condition Rankine theory can be written as or Maximum principal stress criteria for plane stress condition gives failure surface that represents a square. By comparing both expressions for . (a) Maximum Principal Stress Criteria This criteria is proposed by Rankine states that yielding occurs at a point when the maximum principal stress reaches the value equals to maximum principal stress at yield in uniaxial tension or compression. Figure-10: Comparison of Von-Mises and Tresca Yield criteria (c) Strain Energy Density or Total Strain Energy Criteria The strain energy density criteria proposed by Beltrami states that yielding occurs when the strain energy density at a point equals the strain energy density at yield in uniaxial tension or compression. Figure-11: Yield Surface for Rankine criteria . we get the following relationship: For plane stress condition above equation simplified and gives following relation: Shape of yield surface for strain energy density criteria is an ellipse in principal stress space that depends on the Poisson’s ratio. (2) Failure Theories for Brittle Material Failure of brittle materials is characterized by ultimate tensile strength. . Likewise. Also. . the right circle is for uniaxial tension at the limiting tension stress. Venant’s criteria (c) Mohr’s Failure Criteria The Mohr Theory of Failure. Equation for Mohr’s-Coulomb failure criteria for plane stress condition is given by. This envelope is shown in the figure below. Figure-13: Yield Surface (shaded) for Mohr’s criteria The left circle is for uniaxial compression at the limiting compression stress of the material. The middle Mohr's Circle on the figure-13 (dash line) represents the maximum allowable stress for an intermediate stress state. also known as the Coulomb-Mohr criterion or internal-friction theory. For uniaxial tension case at yield gives the following relation: For plane stress condition above equations for becomes.(b) Maximum Principal Strain Criteria It is also known as St. Mohr's theory is often used in predicting the failure of brittle materials. Figure-12: Yield Surface for St. Mohr's theory suggests that failure occurs when Mohr's Circle at a point in the body exceeds the envelope created by the two Mohr's circles for uniaxial tensile strength and uniaxial compression strength. is based on the famous Mohr's Circle. Venant’s criteria states that yielding occurs when maximum principal strain equals to the maximum principal strain at yield under uniaxial tension or compression. Principal strains in terms of principal stresses are given by: . Mohr's theory requires that the two principal stresses lie within the shaded zone depicted in Figure-14. a) Assuming material follows Tresca yield criteria. c) Determine which criteria is more conservative Solution Principal stresses are determined using equation of two-dimensional Mohr’s circle. b) Assuming material follows Von-Mises yield criteria. The load stress-strain are linear so that the factor of safety can be applied to either the loads or stress components. Problem-1 When the loads that act on the hub of a flywheel reach their working values. The flywheel material has a yield stress equals .Graphically. determine factor of safety against yield. used to develop yield criteria. Also shown on the figure is the maximum stress criterion (dashed line). Hill’s criteria etc. determine factor of safety against yield. Figure-14: Yield Surface (shaded) for Mohr’s criteria There are some more yield criteria such as Drucker-Pager yield criteria. the nonzero stress components at the critical point in the hub where yield is initiated are . that follows similar method of determination of yield surface with minor changes in their equations. This theory is less conservative than Mohr's theory since it lies outside Mohr's boundary. Answer . a) Tresca Criteria: b) Von-Mises Criteria: c) Tresca criteria is more conservative as it predicts yielding at smaller loads. and . Maximum compressive stress in the part = 20 ksi So from compressive stress point of view. Solution: The gray cost iron is brittle material.e. For solution we apply maximum principal stress theory. If the tensile yield strength is 700 MPa. Maximum tensile stress in the part = 10 ksi So from compressive stress point of view. what minimum thickness must be specified to prevent yielding? Consider failure based on Tresca criteria.Problem-2 A foundation of a machine is made by gray cast iron. Thus the factor of safety of the part will be smaller value i. Solution: Hoop stresses. Longitudinal stresses. such that . The mean radius of the tube is 30 cm. . Answer Problem-3 A thin-wall tube with closed ends is subjected to a maximum internal pressure of 35 MPa in service. Answer . Yielding occurs when . Find the factor of safety in the foundation. The most critical stress condition at a point in the part is shown below. Putting all the values into the above relation. Hence. we can easily derive the following expression . Use Tresca criteria.m).m) and torque (T=24kN. Calculate the required shaft diameter (d) in order to achieve a factor of safety of 2. we get Note: If Von-Mises failure criteria is used we have following relationship. Solution: Since & (Bending and Torsion equations) From Mohr’s circle equation.Problem-4 A circular shaft of tensile strength 350 MPa is subjected to a combined state of loading defined by bending moment (M=8 kN.